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Disney Cars Party One of the favorite parties for boys is a Cars themes party. It is also one of the most popular boy's Disney party ideas. Lightning McQueen features prominently in this partyware set so it he is your child's favorite character, you are in luck. The plates and table wear are fun because they feature not only Lightning MqQueen but also Tow Mater, Finn McMissile and Francesco. Kids can choose a plate that features their favorite characters. We just love the way all the pieces coordinate so the entire theme comes together for a perfect finish. Centerpieces, balloons, buttons, toy cars and games with Cars themes are all things that you can add to make things more festive. With so many choices, you can pick and choose items based on your venue and your budget. Some ideas that you can incorporate into your party include having the kids dress in Cars costumes, playing a video of the Cars movie during the party and making a 3-D Cars cake. You can find ideas like this in our Party Ideas section. A Cars Birthday Party is a great idea for boys that love the movie. Give him the best birthday ever with the Cars 2 party supplies here. | Mid | [
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Members Chat This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More. I will no longer be hosting XLNation as of March 1, 2017. Skullz613 will be taking the reigns of the site in early 2017 and XLNation will be moving to a new web hosting service. During that transition there will be an interruption in service of up to two weeks. Skullz613 is now a full administrator of XLNation. Be kind to him. Offices Virus Corporation 1.0 New building model imported from City Life. I was reading this thread - City Life Buildings in Cities XL and did a bit of searching and stumbled across a bunch of downloadable buildings from City Life Game. Anyways I downloaded them and tried this one. copy from nicko's swat threadi added some textures,lights,cars etc. Recent Reviews In menu location there are 4 options for the office, T1 means small building.. T2 is bigger T3 is the bigget,T4 is for custom building. I hope i am not wrong since i dont play the game for a long time :) | Low | [
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[Cite as State v. Wilson, 2017-Ohio-8152.] STATE OF OHIO ) IN THE COURT OF APPEALS )ss: NINTH JUDICIAL DISTRICT COUNTY OF SUMMIT ) STATE OF OHIO C.A. No. 28187 Appellee v. APPEAL FROM JUDGMENT ENTERED IN THE TRAMELL RAYSHAWN WILSON COURT OF COMMON PLEAS COUNTY OF SUMMIT, OHIO Appellant CASE No. CR 2012 02 0332 DECISION AND JOURNAL ENTRY Dated: October 11, 2017 HENSAL, Judge. {¶1} Trammell Wilson appeals his sentence for multiple offenses in the Summit County Court of Common Pleas. For the following reasons, this Court affirms. I. {¶2} The substantive facts and procedural history of this case were set forth by this Court in State v. Wilson, 9th Dist. Summit No. 27361, 2015-Ohio-2023, (Wilson II) as follows: In June 2011, Wilson instigated a confrontation with two men outside of a nightclub in Akron. During the course of the confrontation, Wilson pulled out a gun, shot at one of the men, and actually shot the second man multiple times. A grand jury indicted Wilson on two counts of felonious assault and one count of having a weapon while under disability. Each felonious assault count also contained a repeat violent offender specification, pursuant to R.C. 2941.149, and a firearm specification, pursuant to R.C. 2941.145. A jury found Wilson guilty on both of his felonious assault counts, the firearm specifications linked to those counts, and his weapon under disability count. Thereafter, Wilson stipulated that he had a prior felonious assault conviction and, due to his prior conviction, qualified as a repeat violent offender. The court merged Wilson’s repeat violent offender specifications for purposes of sentencing, but sentenced him on all of his remaining counts. Specifically, the 2 court sentenced Wilson to (1) three years on his repeat violent offender specification; (2) six years on each of his felonious assault counts; (3) three years on each of his firearm specifications; and (4) 36 months on his weapon under disability count. The court ordered each of Wilson’s prison terms to run consecutively with the exception of the weapon under disability count. Thus, the court sentenced Wilson to a total of 21 years in prison. On appeal from his convictions, Wilson argued that the trial court erred by failing to merge his convictions for felonious assault and having a weapon under disability. Because there was no evidence that the trial court had analyzed the merger issue under State v. Johnson, 128 Ohio St.3d 153, 2010-Ohio-6314, we determined that the trial court had to apply Johnson in the first instance. Consequently, we remanded the matter to the trial court for it to apply Johnson and to determine whether the felonious assault and having weapons under disability offenses should merge. On remand, the trial court conducted a new sentencing hearing and ultimately concluded that Wilson’s counts for felonious assault and having a weapon under disability should not merge. In keeping with its original decision, the court once again merged Wilson’s repeat violent offender specifications and sentenced him to a total of 21 years in prison. Nevertheless, the court changed the individual prison terms that it had originally ordered Wilson to serve. The court sentenced Wilson to (1) ten years on his repeat violent offender specification; (2) eight years on each of his felonious assault counts; (3) three years on each of his firearm specifications; and (4) 36 months on his weapon under disability count. The court then ordered the ten-year repeat violent offender specification, one of the eight- year felonious assault counts, and one of the three-year firearm specifications to be served consecutively to reach the 21–year total. The court ordered the remaining counts to run concurrently. Id. at ¶ 2-5. (Internal quotations and citations omitted.) {¶3} Mr. Wilson appealed the trial court’s second sentencing entry, arguing in part that the court erred by increasing the length of his prison term on his repeat violent offender specifications and his felonious assault counts. In Wilson II, this Court concluded that the trial court had exceeded the scope of the remand and lacked the authority to conduct a de novo sentencing hearing and to increase Mr. Wilson’s sentence on his repeat violent offender specifications. Id. at ¶ 13. It determined, however, that since Mr. Wilson’s sentences for felonious assault were reversed on appeal in State v. Wilson, 9th Dist. Summit No. 26683, 2014- 3 Ohio-376 (Wilson I), those sentences were properly before the trial court and the trial court had the authority to resentence him on those counts. Id. at ¶ 15. Accordingly, this Court remanded the matter to the trial court for further proceedings consistent with Wilson II. {¶4} On remand, the trial court held a third sentencing hearing and re-imposed Mr. Wilson’s original sentence in a subsequent journal entry. Specifically, the trial court again merged Mr. Wilson’s repeat violent offender specifications for purposes of sentencing, and then sentenced Wilson to: (1) three years on the merged repeat violent offender specification; (2) six years on each of the felonious assault counts; (3) three years on each of the firearm specifications; and (4) 36 months on the weapon under disability count. The trial court ordered each of the prison sentences to run consecutively with the exception of the weapon under disability count, which was to run concurrently. In total, the trial court sentenced Mr. Wilson to 21 years. Mr. Wilson has appealed, assigning as error that the trial court failed to comply with this Court’s remand instructions in Wilson II. II. ASSIGNMENT OF ERROR THE TRIAL COURT ERRED AND FAILED TO ABIDE BY THIS COURT’S REMAND AND WENT BEYOND ITS AUTHORITY WHEN IT MODIFIED [MR. WILSON’S] SENTENCE RELATIVE TO THE REPEAT VIOLENT OFFENDER SPECIFICATION. {¶5} Mr. Wilson argues that the trial court incorrectly conducted a de novo sentencing hearing on remand from Wilson II. Specifically, he argues that only the sentence for the repeat violent offender specification was subject to review on remand since it was the only offense affected by the appealed error in Wilson II. In reviewing a felony sentence, “[t]he appellate court’s standard for review is not whether the sentencing court abused its discretion.” R.C. 2953.08(G)(2). “[A]n appellate court may vacate or modify a felony sentence on appeal only if 4 it determines by clear and convincing evidence” that: (1) “the record does not support the trial court’s findings under relevant statutes[,]” or (2) “the sentence is otherwise contrary to law.” State v. Marcum, 146 Ohio St.3d 516, 2016-Ohio-1002, ¶ 1. Clear and convincing evidence is that “which will produce in the mind of the trier of facts a firm belief or conviction as to the facts sought to be established.” Cross v. Ledford, 161 Ohio St. 469 (1954), paragraph three of the syllabus. {¶6} In general, although a remand for a new sentencing hearing anticipates a de novo hearing, “only the sentences for the offenses that were affected by the appealed error are reviewed de novo; the sentences for any offenses that were not affected by the appealed error are not vacated and are not subject to review.” State v. Wilson, 129 Ohio St.3d 214, 2011-Ohio- 2669, ¶ 15, citing State v. Saxon, 109 Ohio St.3d 176, 2006-Ohio-1245, paragraph three of the syllabus; R.C. 2929.19(A). Additionally, “the law of the case doctrine ‘provides that the decision of a reviewing court in a case remains the law of that case on the legal questions involved for all subsequent proceedings in the case at both the trial and reviewing levels.’” Neiswinter v. Nationwide Mut. Fire Ins. Co., 9th Dist. Summit No. 23648, 2008-Ohio-37, ¶ 10, quoting Nolan v. Nolan, 11 Ohio St.3d 1, 3 (1984). Accordingly, “[a]bsent extraordinary circumstances, * * * an inferior court has no discretion to disregard the mandate of a superior court in a prior appeal in the same case.” Nolan at syllabus. Nonetheless, the doctrine of law of the case “is considered to be a rule of practice rather than a binding rule of substantive law and will not be applied so as to achieve unjust results.” Id. at 3. {¶7} Since Wilson II, this Court has updated its understanding of the scope of a remand for an allied-offense analysis. In State v. Ross, 9th Dist. Lorain No. 09CA009742, 2012-Ohio- 536 (Ross I), the trial court sentenced Michael Ross to a total of nine and a half years 5 imprisonment. We noted that the trial court had not had the opportunity to determine whether his offenses were allied under State v. Johnson, 128 Ohio St.3d 153, 2010-Ohio-6314, so we remanded “the matter to the trial court for it to apply Johnson in the first instance.” Id. at ¶ 69. On remand, the trial court found that only some of the offenses were allied. Nevertheless, it altered the sentences for the other offenses as well. State v. Ross, 9th Dist. Lorain Nos. 14CA010601, 14CA010602, 2015-Ohio-3399, ¶ 5 (Ross II). In Ross II, we held that the trial court erred when it modified the sentences for the offenses that were not allied, explaining that the trial court “did not have the authority to conduct a de novo resentencing on those offenses.” Id. at ¶ 6. Specifically, “there was no allied offense sentencing error for it to correct on those sentences.” Id. at ¶ 8. We concluded that this result was consistent with the Ohio Supreme Court’s decision in State v. Wilson, 129 Ohio St.3d 214, 2011-Ohio-2669, and we, therefore, remanded the matter for the trial court to re-impose its original sentence for the offenses that were not allied. Id. at ¶ 9. {¶8} This Court has repeatedly followed Ross II. In State v. Powell, 9th Dist. Summit No. 27830, 2016-Ohio-2820, this Court noted that it had previously remanded the case for the trial court to make an initial determination whether Mr. Powell’s offenses were allied. Id. at ¶ 19. On remand, the trial court determined that the offenses were not allied, so it re-imposed Mr. Powell’s original sentence. Id. On appeal following the re-imposition of his original sentence, Mr. Powell argued that he should have been entitled to a de novo resentencing. This Court rejected his argument, explaining that, “when a trial court determines on remand that offenses are not allied, it does not have authority to conduct a de novo resentencing on those offenses.” Id. at ¶ 20, citing Ross II at ¶ 6. We also followed Ross II in State v. Copeland, 9th Dist. Summit No. 27905, 2016-Ohio-1613, in which we affirmed that, when a case is “simply remanded * * * so 6 that the trial court c[an] engage in a merger analysis in the first instance[,]” “the lack of an allied offenses error preclude[s] the trial court from disturbing the length of [the defendant’s] sentences or the imposition of consecutive sentences[.]” Id. at ¶ 13; see also State v. McIntyre, 9th Dist. Summit No. 27670, 2016-Ohio-93, ¶ 12 (“Nor did the trial court have authority [on remand] to alter the sentence on [count five], absent a conclusion that it was an allied offense that merged for purposes of sentencing.”). {¶9} In Wilson I, this Court, similar to Ross I, remanded “the matter * * * for the trial court to apply State v. Johnson in the first instance.” Wilson I, 9th Dist. Summit No. 26683, 2014-Ohio-376, at ¶ 61. On remand, the trial court determined that none of Mr. Wilson’s offenses were allied. Mr. Wilson, therefore, “was not entitled to a de novo resentencing hearing[.]” Powell at ¶ 21. In Wilson II, we should have directed the trial court to do exactly what it did do in its third sentencing entry, which was to re-impose Mr. Wilson’s original sentence. Ross II at ¶ 9. After careful review, we conclude that extraordinary circumstances exist and that the doctrine of law of the case should not be applied to the trial court’s judgment on remand from Wilson II. {¶10} At the third sentencing hearing, the trial court reasoned that “[t]o now say that the Court must either impose a higher sentence or a lower sentence, based upon what was essentially a harmless failure to consider the merger issue back in 2012, is illogical.” Instead, it decided that “at this point I am going to go back and do what the Court of Appeals said was my authority upon the first remand, and that was to consider whether the * * * counts should merge. I did that. I determined that they should not merge and therefore just leave the sentence where it was in 2012.” The trial court’s assessment is consistent with this Court’s holdings in Ross II, Powell, Copeland, and McIntyre. Accordingly, we conclude that Mr. Wilson has not demonstrated by 7 clear and convincing evidence that the trial court acted contrary to law when it re-imposed his original sentence. Mr. Wilson’s assignment of error is overruled. III. {¶11} Mr. Wilson’s assignment of error is overruled. The judgment of the Summit County Court of Common Pleas is affirmed. Judgment affirmed. There were reasonable grounds for this appeal. We order that a special mandate issue out of this Court, directing the Court of Common Pleas, County of Summit, State of Ohio, to carry this judgment into execution. A certified copy of this journal entry shall constitute the mandate, pursuant to App.R. 27. Immediately upon the filing hereof, this document shall constitute the journal entry of judgment, and it shall be file stamped by the Clerk of the Court of Appeals at which time the period for review shall begin to run. App.R. 22(C). The Clerk of the Court of Appeals is instructed to mail a notice of entry of this judgment to the parties and to make a notation of the mailing in the docket, pursuant to App.R. 30. Costs taxed to Appellant. JENNIFER HENSAL FOR THE COURT CARR, P. J. CONCURS IN JUDGMENT ONLY. 8 SCHAFER, J. DISSENTING. {¶12} As the majority in Wilson II determined that the trial court lacked authority to resentence Wilson on all of his offenses and erred by increasing Wilson’s sentence on his repeat violent offender specification, the only action the trial court was authorized to take on remand from Wilson II was to reinstate the three year prison sentence on Wilson’s repeat violent offender specification. The Supreme Court of Ohio has stated that the doctrine of the law of the case “is necessary to ensure the consistency of results in a case, to avoid endless litigation by settling the issues, and to preserve the structure of superior and inferior courts as designed by the Ohio Constitution. In pursuit of these goals, the doctrine functions to compel trial courts to follow the mandates of reviewing courts.” (Internal citation omitted.) Nolan v. Nolan, 11 Ohio St.3d 1, *3 (1984). {¶13} Although I am troubled by the potential consequences, I do not believe the circumstances of this case are so extraordinary that this Court should ignore the law of this case or the principles of finality and judicial economy. First, it is axiomatic that on remand, “the sentences for any offenses that were not affected by the appealed error are not vacated and are not subject to review.” State v. Wilson, 129 Ohio St.3d 214, 2011-Ohio-2669, ¶ 15. Second, the reduction of a sentence based upon a trial court error is not an inherently unjust result nor is it untenable in this case. Third, the Supreme Court of Ohio did not issue an intervening decision. Fourth, while this Court acknowledges that three months after our decision in Wilson II, this Court limited the scope of the remand for an allied offense analysis, see State v. Ross, 9th Dist. Lorain Nos. 14CA010601, 14CA010602, 2015-Ohio-3399, ¶ 6, the allied offenses issue was not an issue on appeal in Wilson II and was, therefore, not within the scope of the remand. Moreover, even if the allied offenses issue was within the scope of the remand, the law of the 9 case doctrine would still have precluded the trial court from applying the holding in Ross because “the decision of a reviewing court in a case remains the law of that case on the legal questions involved for all subsequent proceedings in the case at both the trial and reviewing levels.” Nolan at 3. Finally, this Court’s subsequent change of course as to the scope of the remand for an allied offenses error in Ross is not such an uncommon or extraordinary circumstance that it demands this Court disregard the law of this case. Indeed, such a determination would set a disturbing precedent and undermine the finality of countless cases. {¶14} Therefore, I would conclude that the trial court acted contrary to law when it re- imposed Wilson’s original sentence. Moreover, had the majority found that the trial court had acted contrary to law, it would have had the authority to “increase, reduce, or otherwise modify” Wilson’s sentence without the need to remand the matter to the trial court. See R.C. 2953.08(G)(2). Such an action would have satisfied the principle of judicial economy. {¶15} For the reasons stated above, I respectfully dissent. APPEARANCES: LAWRENCE J. WHITNEY, Attorney at Law, for Appellant. SHERRI BEVAN WALSH, Prosecuting Attorney, and RICHARD S. KASAY, Assistant Prosecuting Attorney, for Appellee. | Mid | [
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Teachers and support staff at Dubbo's Catholic schools will vote ‘yes’ to a new enterprise agreement this week, following an 18-month industrial campaign. The new agreement includes a 2.5 percent pay rise for 2017, 2018 and 2019 in NSW with an additional pay rise to come in line with public schools. The ‘yes’ vote will bring to an end a sometimes acrimonious dispute between the Independent Education Union of Australia (IEUA NSW/ACT) Branch and the 11 Catholic Dioceses of NSW and the ACT. Acting Secretary IEUA NSW/ACT Branch Gloria Taylor said the new enterprise agreement also enshrines access to the Fair Work Commission for arbitration. "The employer did challenge our member's access to arbitration," she said. "The new work practice agreement will allow teachers and support staff to spend more time with students and less time on data collection and input, meetings and emails." IEUA Central West Organiser Jackie Groom said St Pius, St John's, St Laurence's and St Mary's were on board with the union's argument. "I categorically would say that those schools were overwhelmingly supportive of the union's campaign," she said. “In most dioceses, there is now a commitment to limit unnecessary data processing and programming that is not directly related to teaching. Staff will not be required to respond to emails outside of normal school hours. The new work practices agreements also allow for more mentoring for new teachers and support staff will be paid an allowance for attending overnight camps.” Ms Groom said reaching an agreement was due to the determination and unity of union members. "They stuck to their guns for 18 months, despite attempts by the employers to bribe them with a pay rise, because they knew how fundamental the right to arbitration was,” she said. “They also understood that provisions addressing work intensification would benefit not only them but their students." Catholic Commission for Employment Relations (CCER) Executive Director Tony Farley said he was pleased to have finalised negotiations with the IEUA. "The main change to the proposed Dispute Resolution Procedure is that it allows for either employers or employees to refer disputes to the Fair Work Commission for arbitration without the consent of the other party, " he said. Catholic teachers vote ‘yes’ to a new enterprise agreement to end acrimonious dispute Teachers and support staff at Dubbo's Catholic schools will vote ‘yes’ to a new enterprise agreement this week, following an 18-month industrial campaign. The new agreement includes a 2.5 percent pay rise for 2017, 2018 and 2019 in NSW with an additional pay rise to come in line with public schools. The ‘yes’ vote will bring to an end a sometimes acrimonious dispute between the Independent Education Union of Australia (IEUA NSW/ACT) Branch and the 11 Catholic Dioceses of NSW and the ACT. Acting Secretary IEUA NSW/ACT Branch Gloria Taylor said the new enterprise agreement also enshrines access to the Fair Work Commission for arbitration. "The employer did challenge our member's access to arbitration," she said. "The new work practice agreement will allow teachers and support staff to spend more time with students and less time on data collection and input, meetings and emails." IEUA Central West Organiser Jackie Groom said St Pius, St John's, St Laurence's and St Mary's were on board with the union's argument. "I categorically would say that those schools were overwhelmingly supportive of the union's campaign," she said. “In most dioceses, there is now a commitment to limit unnecessary data processing and programming that is not directly related to teaching. Staff will not be required to respond to emails outside of normal school hours. The new work practices agreements also allow for more mentoring for new teachers and support staff will be paid an allowance for attending overnight camps.” Ms Groom said reaching an agreement was due to the determination and unity of union members. "They stuck to their guns for 18 months, despite attempts by the employers to bribe them with a pay rise, because they knew how fundamental the right to arbitration was,” she said. “They also understood that provisions addressing work intensification would benefit not only them but their students." Catholic Commission for Employment Relations (CCER) Executive Director Tony Farley said he was pleased to have finalised negotiations with the IEUA. "The main change to the proposed Dispute Resolution Procedure is that it allows for either employers or employees to refer disputes to the Fair Work Commission for arbitration without the consent of the other party, " he said. | High | [
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Saturday, January 31, 2009 Afghanistan - Another Iraq? The New York Times has a piece today about the difficulty Obama will face with Afghanistan. Everyone seems to agree that winning a war there is a next-to-impossible task. As President Bush placed most of the emphasis on Iraq, the Taliban grew in strength in Afghanistan, controlling huge areas of territory outside of the major urban areas. Enter Mr. Obama. During the campaign he promised to send two additional brigades — 7,000 troops — to Afghanistan. During the transition, military planners started talking about adding as many as 30,000 troops. And within days of taking office, Mr. Obama announced the appointment of Richard Holbrooke, architect of the Balkan peace accords, to execute a new Afghanistan policy. But even as Mr. Obama’s military planners prepare for the first wave of the new Afghanistan “surge,” there is growing debate, including among those who agree with the plan to send more troops, about whether — or how — the troops can accomplish their mission, and just what the mission is. I don't know about anyone else, but that sounds ominously familiar to me; an ill-defined plan, inadequate resources for the immense task at hand. Even before the election, I wondered what was going on here. Does Barack Obama really need to perpetuate the supposed man-hunt for the phantom bin Laden? Is that what it's all about? Or is Obama beholden to the military industrial complex? Perhaps this was part of the deal. Is it too cynical to suppose that deals like this are made in Washington? On Reuters there's a wonderful article by Bernd Debusmann which explores the possibility of a solution to this dilemma. Since the real problem is the illegal opium production, controlled by the Taliban, why don't we buy the entire crop? It would cost far less than the war, and would afford other opportunities concerning the world-wide heroin problem. Richard Holbrooke, the man President Barack Obama has just picked as special envoy for Afghanistan, said: “Breaking the narco-state in Afghanistan is essential or all else will fail.” The problem is it may be easier said than done. Which makes me wonder what these guys are up to. Do they really want to do what they say? Defense Secretary Robert Gates, addressing the Senate Armed Services Committee this week, described Afghanistan as “our greatest military challenge right now” but said there could be no purely military solution — not even with the additional 30,000 troops Obama plans to dispatch over the next 18 months. James Nathan, a political science professor at Auburn University in Alabama and former State Department official, outlines the radical solution. Purchasing the whole crop would take it away from the traffickers without cutting more than half the economy of Afghanistan,” Nathan said in an interview. “Such a purchase would directly confront Afghanistan’s most corrosive corruption. It would end the Taliban’s money stream.” And the cost? By Nathan’s reckoning, between $2 billion and $2.5 billion a year, no pocket change but not a large sum compared with the around $200 billion the U.S. taxpayer has already paid for the war in Afghanistan. The idea may sound startling but its logic is not far from the farm subsidies paid to U.S. and European farmers. On a more modest scale than Nathan’s buy-it-all idea, a European think tank, the International Council on Security and Development (ICOS), is lobbying for an alternative to traditional counter-narcotics policies dubbed Poppy for Medicine. What's your opinion? Is that a reasonable solution? What's wrong with it? Isn't it better than spending the next five or ten years stuck in another war? Is it too much of a stretch to suspect secret deals behind the scenes in Washington? Could the new administration be just a corrupt as the old one as far as this stuff goes? Does that make me a conspiracy theorist? I admit, I never thought Oswald was the lone gunman. going into the 2000's, it was obvious the place was a failed state, a chaotic hellhole similar to to several places in sub-Saharan Africa we could mention; a country badly in need of an intervention and some leadership. after 9/11, it was equally ovbious that that anarchy was supporting and sheltering terrorism, on a level which provided a pretty damn good argument for staging some sort of intervention. but what actually happened was an unplanned, underfunded, half-hearted mismeasure. actually fixing the real and worrisome problems of Afghanistan would have taken much more effort and cost than it was given; instead, ridiculous effort and cost went into Iraq, for absolutely no goddamn reason at all. if that money had instead been spent occupying, and then nationbuilding in, Afghanistan, we might have made some real progress against the Taliban by now. what's actually happened is that chaos and anarchy have won again. the hell with osama bin forgotten; Afghanistan is a country of over thirty million human beings and no effective government. we had the perfect excuse to do some good there, by eliminating a theocratic warlord state, ending the same's support for terrorist actions abroad, and building something vaguely like a modern country on the borders of Pakistan and Iran. (think we maybe could have benefited from such an entity, eh?) yet we pissed it away for nothing. what actually will happen is we'll declare victory and leave (or else be pushed out gradually like the Soviets were, with far more egg on our faces), the Taliban will take back over, and we'll be back at square one --- minus however many Afghanis have been killed in the interim. heckuva job, bushy. heckuva job, America. It looks to me like we're headed into another quagmire from which it'll be hard to escape. The Bush supporters should naturally transfer their allegiance to the new president, shouldn't they? The patriotism crowd should show the same blind obedience to this Commander-in-Chief as they did to the last one, shouldn't they? | Mid | [
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Grading the rotating cast in the bottom defensive pairing is an interesting task. For the purposes of this grade, the bottom pairing players are going to be Taylor Fedun, Jamie Oleksiak, Ben Lovejoy, Joel Hanley, Dillon Heatherington, and Gavin Bayreuther. It’s especially hard to grade Hanley, Heatherington, and Bayreuther with the handful of games they played, but we’ll do our best. Expected goals-for, a stat tracked by Emmanuel Perry at corsica.hockey, looks at the number of goals expected for or against while a player is on the ice. In assessing a defenseman’s usefulness, looking at the expected goals-for and against is a pretty useful measure. As you might imagine, the bottom pairing had varying results. Fedun came to the Dallas Stars in November on a trade with the Buffalo Sabres for a conditional 2020 seventh round pick. The conditions were that Fedun had to play 25 games and since that was definitely achieved, the Sabres will have that pick next year. After coming onboard in November, Fedun was the most steady member of the pair on the third line, only missing a few games down the stretch this spring. Dallas won’t be missing that seventh round pick though, because Fedun was exactly what they needed on that third pairing. In 54 games played, he made every single one of his teammates better when he was on the ice. This graph from hockeyviz.com just shows shots for and against (helpfully labeled with good/bad/dull/fun) but he also had a positive expected goals-for, which is also helpful. If this grade were for Fedun alone, it would be an A, as the man did everything asked of him by his team. But this is a group project, so on to the next player. If you need a refresher, Jamie Oleksiak was traded to the Pittsburgh Penguins in December 2017 for a 2019 fourth round pick. In January 2019, Pittsburgh and Dallas busted out a deal eraser, bringing Oleksiak back to the Stars for that same conditional fourth round pick, as if the whole thing were some weird fever dream. This means that Oleksiak’s stats for the season are padded by his time on the Penguins, with their defensive systems and their puck-hogging forwards. If you isolate just his time with Dallas, Oleksiak’s numbers aren’t quite so rosy, but they definitely aren’t all bad. He ended the season with negative possession, but a positive differential in expected goals-for. He didn’t play enough minutes with his teammates to have the same with or without the chart that Fedun has above, but his shots heat map is rather positive. The map without Oleksiak has a lot more red right in front of the goal. Oleksiak still couldn’t crack the full-time roster though and sat out quite a few games in the spring in favor of the next defenseman. Ben Lovejoy Lovejoy was an acquisition at the trade deadline for Connor Carrick. He played 51 games for the New Jersey Devils before playing 20 games for the Stars. At acquisition, Lovejoy was sold to the Stars as a penalty kill specialist, but that isn’t what he ended up being for the team. In 49 minutes of penalty kill time, he never really managed to help suppress shots. Just for comparison’s sake, this is the team penalty kill without him. His 5-on-5 shot map is worse. For a stay-at-home defenseman, he wasn’t very good at suppressing shots in front of his own goal. He had a negative impact on possession for the team when on the ice, and had a negative expected goals-for differential. In every way possible, the team was worse when Lovejoy was on the ice. Twenty games is an incredibly small sample size, but Lovejoy is an unrestricted free agent this summer and it isn’t likely that the Stars will see more of him in victory green. Bayreuther’s sample size is smaller than his cousin Lovejoy’s by a single game, but he’s also a decade younger. Bayreuther signed as a free agent in March 2017 during his senior year of college and came up to Dallas from the Texas Stars when it was announced that Marc Methot would be missing most of the season. Bayreuther played most of his games on a pair with Roman Polak, and wasn’t in the rotation at all after the Stars brought Oleksiak back to Dallas. Still, there were things to like about his game while he was here. He had positive possession and expected goals-for numbers. During those 19 games, the Stars took more shots with him on the ice than off. Still, small sample size is small, and it just gets smaller from here. The Other Guys Joel Hanley and Dillon Heatherington played 21 games between them (16 and five respectively), and not a lot can really be said with that small a sample size. Hanley is a good AHL defensemen that has proven he can hang with the Dallas Stars when necessary. Heatherington was on a bridge contract this year and is a restricted free agent this summer, but he does have one more NHL point this season than Hanley does. Bottom Defensive Pair, End-of-Season Grade: B- The game is rarely won or lost on the third pairing, and for the most part, this group did what was asked of them. Sure, some of them did it better than others, and bigger sample sizes are needed to determine the usefulness of some of these skaters. With the albatross of Ben Lovejoy weighing them down, the group as a whole gets a B-. | Low | [
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Ted Sarandos: Netflix Wants to Launch 20 Scripted Series a Year MIAMI — Netflix chief content officer Ted Sarandos said Wednesday that the netcaster aims to launch “around 20” original scripted series per year in its effort to appeal to a variety of subscribers. The goal is to serve “really diverse tastes all around the world,” Sarandos said during a Q&A with “Arrested Development” creator Mitch Hurwitz and “Breaking Bad” creator Vince Gilligan, held as part of the NATPE confab. Netflix is focused on having a broad menu of original shows that target different niches. “We really are not trying to define the brand by any one show,” he said. Sarandos also offered some insight into Chelsea Handler’s upcoming latenight talkshow, which won’t be much like her past effort for E!, “Chelsea Lately.” The new show will “reposition Chelsea away from what she was at E,” he said. “The show became something she was not happy with.” He said the focus would be on interview, filmed segments and showcasing Handler’s comedy. | Mid | [
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CDC: Greens Cause Most Food Illnesses By Associated Press |  Posted: Tue 11:08 AM, Jan 29, 2013 NEW YORK -- A government study has fingered leafy green vegetables as the leading source of food poisoning illnesses. However, the most food-related deaths were from contaminated chicken and other poultry. The Centers for Disease Control and Prevention released the study Tuesday. It's based on an analysis of food poisoning cases from 1998 through 2008. It's the agency's most comprehensive attempt to identify which foods most often carry germs that make us sick. The CDC estimates roughly 1 in 6 Americans -- or 48 million people-- gets sick from food poisoning each year. That includes 128,000 hospitalization and 3,000 deaths. Online Public Information File Viewers with disabilities can get assistance accessing this station's FCC Public Inspection File by contacting the station with the information listed below. Questions or concerns relating to the accessibility of the FCC's online public file system should be directed to the FCC at 888-225-5322, 888-835-5322 (TTY), or [email protected]. | High | [
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It is often necessary, or at least desirable, to concentrate a liquid mixture by removing a portion of the solvent, generally water, from the liquid mixture. The resulting product, therefore, is in a more concentrated form. It has been common to concentrate radioactive waste (rad waste), cooling tower blow down waste water, fruit and vegetable juices such as orange juice, grapefruit juice, grape juice, and tomato juice by evaporation to remove water. In addition, seawater and brackish water have been concentrated by evaporation, although the condensed vapor has been recovered as usable potable water rather than discarded as in concentrating fruit and vegetable juices. Nevertheless, each is a concentrating process. In the case of juice, the concentrate is the desirable product whereas in obtaining potable water from seawater or brackish water the concentrate is discarded. Evaporative concentration as described, as well as evaporation of chemical solutions or liquid dispersions, requires substantial energy since it relies on the latent heat of vaporization. Scaling of equipment and enhanced corrosion are often inherent at the temperatures involved in evaporative concentration. Loss of flavor and aroma also result during evaporative concentration of food products. Because of the shortcomings involved in evaporative concentration, it has been found advantageous to freeze concentrate many products, particularly those having water as the liquid carrier. In such a process, water is removed by first producing ice crystals which are then separated from the concentrate. Next, the ice crystals are washed to remove the remaining concentrate from them. The ice crystals can then be discarded or melted if potable water is desired. Engdahl U.S. Pat. No. 4,314,455 discloses a freeze concentration apparatus and process. In FIG. 2 of that patent a two stage freeze concentration system is disclosed. The aqueous mixture which is separated from the ice formed in the process is recirculated to the freeze exchanger. The mixture of ice and water exiting the freeze exchanger is very dilute in ice concentration so that it is very desirable to increase the solids concentration before it is sent to an ice washer. Apparatus and methods suitable for such concentration do not appear available in the art even though a need exists for them not only to concentrate ice in water but to concentrate any solids from a liquid having a relatively different density than the solids i.e., the solids can be more dense or less dense than the liquid. | Mid | [
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The transformative nature of President Trump's election in 2016 -- by way of a unique coalition of support -- had given Republicans a degree of optimism heading into a challenging midterm election season. Though the president has historically low approval ratings nationally, he is still overwhelmingly popular among the party's base, which includes voters Trump lured away from the Democrats. But as Democrats discovered over the past few elections with Barack Obama as president, star power doesn't necessarily trickle down. If anything, it fuels and unites the opposition. Now, after losing a congressional district Tuesday that Trump had won by 20 points, the GOP faces a similar reckoning. Democrat Conor Lamb's apparent defeat of Rick Saccone in a Western Pennsylvania district that combines steel country with the suburbs, comes on the heels of Democrats winning a U.S. Senate seat in Alabama for the first time in a quarter-century. Trump endorsed the Republican candidate in both cases. He visited Pennsylvania twice to tout his agenda and tie it to Saccone, who ran as an ally of the president. Trump's son Don Jr. told voters during a campaign stop earlier this week that "everything [my father] stands for and represents for the future" would be on the ticket. Republicans say Trump's visit was helpful, and argue that Saccone was a uniquely weak candidate and Lamb as a uniquely strong one. But the results raise questions about the president's utility for his party in the midterms and the transferability of his coalition of support. “This is a sign that the congressional map is more likely to expand than narrow over the coming months," said GOP strategist Ken Spain, who served as communications director for the National Republican Congressional Committee during the 2010 cycle -- when his party took control of the House. “In midterm elections, voters are voting ‘yes’ or ‘no’ to the party in charge," Spain said. "While the president will likely be an asset for Republicans in rural and exurban districts, it does not appear – at least at this point in the election cycle – that he is a net positive asset on the 2018 campaign trail.” Former NRCC Chairman Tom Davis said Republicans shouldn't be counting on Trump to turn out voters anyway. "Obama couldn't turn out Democrats in midterms. Why would it be different for Trump? That's the history of midterm elections. They don't come out in off years," Davis said. "It's a strong, strong base, but the problem is Trump-centric voters aren't coming out for other Republicans." As midterm campaigns heat up, Republicans will be analyzing where the president can be useful and where he will be a liability. There is some consideration that resistance to Trump among Democrats, though palpable, may already be baked in, so Republican candidates might as well have him visit to drum up support among his voters. But there is some nuance in primaries versus general elections. “If you're in a primary election, you better be sticking close to Trump wherever you are. But in the general in many of these districts, being attached to Trump doesn't help,” Davis said. “It depends where you are.” Republican operatives say Trump and Republicans have to carefully strategize about where and how to deploy the president, especially since control of Congress would determine the rest of their shared agenda and serve as a firewall against impeachment. While Republicans debate Trump's usefulness in turning out voters, some strategists say the party needs to more closely align with Trump in order for the transfer factor to pay dividends. "What he needed to do is get a lot of people to say, 'I'm not necessarily voting Republican, I'm voting for the Trump agenda,’" said GOP strategist John Brabender, who has served as an adviser in many Pennsylvania races. "The real challenge is how to take those conservative Democrats who like the president's agenda and are dissatisfied with both parties and help them understand the president is going to need reinforcement." Lamb's campaign "was smart enough to figure out that a Democrat running as supportive of the Trump agenda has a better chance of winning in this district. And they got a twofer: a Democratic base fired up, but also those swing Democratic voters choosing the Trump agenda," said Brabender. Though Lamb didn't run as a Trump ally, he didn't run against him either. Unlike many in his party, Lamb didn't even mention the president much on the campaign trail and instead ran as a conservative Democrat, appealing to union workers and also to suburban voters more inclined to his demeanor than to Trump's. He also embraced Trump's plan to impose tariffs on steel and aluminum imported to the United States, ran a pro-gun message, and said he would not back Democratic Leader Nancy Pelosi. But he also ran against the GOP tax bill and as a supporter of Obamacare. Lamb's union support was also integral, and demonstrated that a coalition inclined toward Trump in a presidential election isn't easily swayed by him in other races. "It shows what a candidate who is willing to stand together with workers and embrace our agenda, what kind of activism that can motivate," said Josh Goldstein, spokesman for the AFL-CIO. "There are a lot of races coming up in these midterms where there will be candidates who are willing to or able to inspire working people." While Republicans argue Lamb's candidacy was singular enough to win a district like PA-18, Democrats point to ratings that show over 100 congressional districts as more competitive than PA-18 was on paper. And they note districts like the 12th in southern Illinois, where veteran and prosecutor Brendan Kelly is challenging Republican Mike Bost in an area Trump won by 15 points. Democrats have also touted a flood of military veteran recruitments. And they point to the Pennsylvania race as a test case for whether the GOP message on tax cuts and the economy is working. This notion also raises the question of whether Trump can be an effective messenger on the agenda items Republicans wish he would talk about or concentrate on. "Democrats want 2018 to be a race about the president and his personality. Republicans want to run on the issues like the tax bill and the improving economy. But the president won in 2016 in large part because he ran on personality," said Spain. "As a result you have diverging messaging efforts playing out on the GOP side." | Mid | [
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Salt Lake City 911 dispatch Director Scott Freitag — who is also a Layton City Council member — was arrested Wednesday afternoon while allegedly driving drunk in a government vehicle. He was subsequently fired, Salt Lake City announced Thursday. Matthew Rojas, spokesman for the mayor’s office, said a review of Freitag’s schedule indicates that he was on the clock at the time of the arrest — 1:25 p.m. Wednesday — though he added that Freitag’s schedule is somewhat fluid, due to occasional long days. “There is no acceptable reason for anyone to put innocent lives in danger by getting behind the wheel of a car while intoxicated, especially an individual leading a critical public safety agency,” Mayor Jackie Biskupski said in a news release announcing the firing. “While I am angry and disappointed in Scott’s behavior, I do hope he gets help to address his problems, and that he finds the support of loved ones that he needs at this time.” A Centerville Police Department sergeant saw the Salt Lake City Corporation vehicle entering the southbound Interstate 15 Kaysville on-ramp and noticed that the vehicle was driven erratically, according to a Centerville news release. Freitag was pulled over and failed field sobriety tests. He registered a 0.214 percent blood alcohol content during a breathalyzer test, the release said, and the officer found an open mixed drink in the center console of the vehicle. Utah’s legal blood-alcohol-content limit for driving is 0.08 percent. It is scheduled to change to 0.05 percent on Dec. 30. Freitag was arrested on suspicion of driving under the influence of alcohol, having an open container of alcohol in a moving vehicle, and possession of a firearm by an intoxicated person. He was cited and released. Court records show that he has not been officially charged, and that he has no scheduled court date. Layton Mayor Bob Stevenson said Thursday that he is still working to figure everything out, and to talk to the rest of the council about the situation. A news release from the city said Layton declined to comment on the matter. “The City has very limited information and is not in a position to either comment on the situation or present any position,” the release states. Freitag has been a member of the Layton City Council since 2007. He has been elected to the seat three times, and his current term ends in 2019. Freitag is the council’s liaison to the police and fire departments and to the Legislature. Freitag was initially appointed director of SLC911 in May 2013, by Mayor Ralph Becker. He was reappointed in July 2016, Rojas said. Prior to that, he worked for the Salt Lake City Fire Department for 16 years, serving as communications director, among other posts. In 2003, he received the Fire Department's Medal of Gallantry for apprehending an arson suspect at the scene of a fire. | Low | [
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Related Content Local "Biggest Loser" Patrick House was on mile 23, and nearly finished with the marathon, when the explosions rang out. "We kept running. We made it to about mile 24 1/2 before they finally, the National Guard stepped in and said they were stopping the race," House said. Fifty-nine Mississippi residents were registered to compete in Monday's Boston Marathon, said Mississippi Department of Public Safety spokesman Warren Strain. None of the Mississippi runners had reported injuries, Strain said. Madison runner Martha Davis said she crossed the finish line about four minutes ahead of the explosion. "There was a loud boom and a huge plume of white smoke followed by an explosion," Davis told 16 WAPT's Erin Kelly over the phone. "It was, first of all, a loud boom. It sounded like a cannon. It sounded like a cannon that goes off at the beginning of a festive event. And I thought, 'That seems odd to have a cannon,' and then it was followed shortly by a second loud sound." Davis said there was the constant sound of sirens following the explosions. "I'm real anxious," she said. "I have a friend that I still haven't heard from out there. There's pretty much a constant barrage of phone calls from my friends." Donna Bruce, of Madison, crossed the finish line 20 minutes before the blasts. "I had already gotten across the finish line, met my husband, and was starting to walk back toward the hotel, and the emergency vehicles just started swarming in," Bruce said. Many people 16 WAPT News spoke to in Boston were waiting anxiously for word from other runners as cell phone service came to a standstill. "No one's phones are working. I can't believe you got through to this phone because we're not able to call out," said runner Jim George of Brandon. House was able to contact his family Monday afternoon to let them know he was alright. "You assume that you're safe at all times and that you're almost invincible. And when you see things like this, you realize how fragile life really is," House said. | Mid | [
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INTRODUCTION {#S1} ============ Estrogens, acting via estrogen receptor α (ERα), stimulate cell proliferation and tumor growth.^[@R1]--[@R3]^ The importance of estrogens and ERα in breast cancer is illustrated by the central role of endocrine therapy targeting estrogens and ERα in treatment of ERα^+^ breast cancer.^[@R1]--[@R5]^ To help fold and sort the increased protein required for estrogen-ERα induced cell proliferation, cells must increase chaperone levels. The endoplasmic reticulum (EnR) stress sensor, the unfolded protein response (UPR) monitors and maintains protein-folding homeostasis.^[@R6],\ [@R7]^ The UPR responds to misfolded proteins, or other forms of stress, by activating three signal transduction pathways, which reduce protein production and increase EnR protein-folding capacity. Protein production is regulated by autophosphorylation of the stress-activated transmembrane kinase, PERK.^[@R6],\ [@R7]^ P-PERK phosphorylates eukaryotic initiation factor 2α (eIF2α), resulting in transient inhibition of protein synthesis. The other UPR arms initiate with proteolytic activation of the transcription factor ATF6α, leading to increased chaperone production and activation of the EnR splicing factor IRE1α, which alternatively splices the transcription factor XBP1, leading to production of active spliced-XBP1, increased protein folding capacity and altered mRNA decay and translation.^[@R6],\ [@R7]^ The UPR is usually inactive in normal cells, but is overexpressed in several cancers.^[@R8]^ Chronic UPR activation leads to increased expression of EnR chaperones, such as BiP (GRP78/HSAP5), p58^IPK^ and calreticulin that facilitate protein folding and promote survival, proliferation, angiogenesis, and resistance to chemotherapy and endocrine therapy.^[@R9]--[@R12]^ In the widely studied "reactive mode", the UPR in tumor cells is activated in response to accumulation of stress from rapid cell division, hypoxia and therapy. A few studies in immune cells describe a different type of UPR activation; in this "anticipatory mode", the UPR is activated in the absence of EnR stress and prior to the accumulation of unfolded proteins.^[@R13],\ [@R14]^ We explored whether estrogen induces anticipatory activation of the UPR in the absence of EnR stress, increasing protein folding capacity prior to the increased protein production and protein folding load that accompanies activation of the genomic estrogen-ERα cell proliferation program. Previous studies of the UPR and of estrogen-ERα action focused on the estrogen-inducible UPR gene, XBP1. XBP1 binds to and activates ERα; XBP1 expression is associated with tamoxifen resistance in ERα^+^ breast cancer.^[@R15]--[@R18]^ The plasma membrane enzyme phospholipase C γ (PLCγ) hydrolyzes PIP~2~ to diacyglycerol (DAG) and inositol 1,4,5-triphosphate (IP~3~). We show that the mitogen estrogen, 17β-estradiol (E~2~), acting through a rapid extranuclear action of ERα, elicits a PLCγ-mediated opening of EnR IP~3~R calcium channels, increasing cytosol calcium and triggering anticipatory activation of each arm of the UPR. Opening the IP~3~R calcium channel and activating the ATF6α arm of the UPR, resulting in BiP induction, are important for subsequent E~2~-ERα induced cell proliferation. Consistent with an important role in cancer for anticipatory activation of the UPR, analysis of data from \~1,000 ERα^+^ breast cancer patients demonstrates that elevated expression of a UPR gene signature is tightly correlated with subsequent resistance to tamoxifen therapy, time to tumor recurrence and poor survival. RESULTS {#S2} ======= Estrogen Activates all 3 Arms of the UPR {#S3} ---------------------------------------- To evaluate the ability of E~2~-ERα to activate the UPR, we focused on production of spliced and modified proteins that result from activating the three arms of the UPR ([Supplementary Figure 1](#SD1){ref-type="supplementary-material"}). E~2~ rapidly activated the IRE1α arm of the UPR, as shown by increases in spliced-XBP1 (sp-XBP1) mRNA in T47D and MCF-7 breast and PEO4 ovarian cancer cells ([Figure 1a and b](#F1){ref-type="fig"}), and by induction of downstream sp-XBP1 targets, SERP1 and ERDJ ([Supplementary Figure 2a](#SD1){ref-type="supplementary-material"}).^[@R19]^ The antiestrogens ICI 182,780/Faslodex/fulvestrant (ICI) and 4-hydroxytamoxifen, (4-OHT), which compete with E~2~ for binding to ERα, blocked the E~2~-mediated increase in sp-XBP1 ([Figure 1a](#F1){ref-type="fig"}). Consistent with E~2~-ERα activating the IRE1α arm of the UPR, RNAi knockdown of ERα blocked E~2~-induction of sp-XBP1 mRNA ([Figure 1c](#F1){ref-type="fig"}), and induction of GREB1 by nuclear E~2~-ERα ([Supplementary Figure 2b](#SD1){ref-type="supplementary-material"}). We next assessed whether estrogen activates the ATF6α arm of the UPR. ATF6α is a 90 kDa protein (p90-ATF6α) that translocates from the EnR to the Golgi in response to stress, where it undergoes proteolytic cleavage to its active 50 kDa form (p50-ATF6α) ([Supplementary Figure 1b](#SD1){ref-type="supplementary-material"}).^[@R6],\ [@R7],\ [@R20]^ Increased ATF6α proteolysis in T47D cells and PEO4 cells demonstrates that E~2~-ERα transiently activates the ATF6α arm of the UPR ([Figure 1d](#F1){ref-type="fig"}; [Supplementary Figure 2c](#SD1){ref-type="supplementary-material"}). Since pretreatment with ICI, abolished the E~2~-mediated increase in p50-ATF6α, this effect is mediated through ERα ([Figure 1d](#F1){ref-type="fig"}). Active cleaved ATF6α regulates induction of BiP and other EnR chaperones.^[@R20],\ [@R21]^ Consistent with this, ATF6α knockdown in T47D cells blocked BiP induction ([Figure 1e](#F1){ref-type="fig"}). BiP increases EnR protein folding capacity, contributing to resolution of the stress, and helps reverse UPR activation; likely preventing the cytotoxicity that would result if UPR activation was sustained. Consistent with its antiapoptotic role, in several cancers, elevated levels of BiP are associated with a poor prognosis.^[@R9]^ Estrogen rapidly induced BiP mRNA in breast and ovarian cancer cells ([Figure 1f](#F1){ref-type="fig"}), leading to a 2.3-fold increase in BiP protein ([Figure 1g](#F1){ref-type="fig"}). RNAi knockdown of ERα prevented E~2~-induction of BiP mRNA ([Figure 1h](#F1){ref-type="fig"}). PERK activation leads to inhibition of protein synthesis ([Supplementary Figure 1c](#SD1){ref-type="supplementary-material"}). Surprisingly, E~2~ induces a rapid and transient increase in PERK phosphorylation ([Figure 2a](#F2){ref-type="fig"}), resulting in increased phosphorylation of eIF2α ([Figure 2b](#F2){ref-type="fig"}) and a modest transient decline in overall protein synthesis ([Figure 2c](#F2){ref-type="fig"}). Consistent with p-PERK catalyzing formation of p-eIF2α, PERK knockdown inhibited formation of p-eIF2α ([Figure 2d](#F2){ref-type="fig"}). Consistent with E~2~ acting through ERα, ICI inhibited E~2~-stimulated phosphorylation of PERK and eIF2α and largely reversed the E~2~-mediated inhibition of protein synthesis ([Figure 2a, b, and c](#F2){ref-type="fig"}). PERK activation leads to ATF4 expression, and we observed a transient increase in ATF4 expression ([Figure 2e](#F2){ref-type="fig"}). However, the proapoptotic protein CHOP was not induced because mild and transient activation of PERK does not induce CHOP ([Figure 2f](#F2){ref-type="fig"}; [Supplementary Figure 2d](#SD1){ref-type="supplementary-material"}).^[@R22]^ Together, this data demonstrates that E~2~, acting through ERα, activates all three UPR arms. E~2~-ERα Rapidly Increases Cytosol Ca^2+^ by a PLCγ-mediated Opening of the EnR IP~3~R Ca^2+^ Channel, Activating the UPR {#S4} ------------------------------------------------------------------------------------------------------------------------- Rapid UPR activation by E~2~-ERα suggested accumulation of unfolded protein was not triggering UPR activation. Some UPR activators, such as thapsigargin, rapidly activate the UPR by depleting Ca^2+^ stores in the lumen of the EnR, increasing intracellular Ca^2+^. To test whether E~2~ rapidly alters cytosol Ca^2+^, we monitored cytosol calcium using the sensor dye Fluo-4 AM. In the presence or absence of extracellular Ca^2+^, estrogen produced a rapid and transient increase in fluorescence in T47D breast cancer cells ([Figure 3a and b](#F3){ref-type="fig"}). Since E~2~ increases cytosol Ca^2+^ when there is no extracellular Ca^2+^, and the EnR lumen is the major Ca^2+^ store available to increase cytosol Ca^2+^, E~2~ is acting by depleting the EnR Ca^2+^ store. Estrogen also increased cytosol calcium in PEO4 ovarian cancer cells ([Supplementary Figure 3](#SD1){ref-type="supplementary-material"}). Inhibition of the IP~3~R channel with 2-APB, which locks the IP~3~R Ca^2+^ channels closed, and RNAi knockdown of the three isoforms of the IP~3~R channels ([Figure 3c](#F3){ref-type="fig"}), abolished the rapid E~2~-ERα-mediated increase in cytosol Ca^2+^ ([Figure 3a, b, and d](#F3){ref-type="fig"}). In contrast, high concentration ryanodine (Ry), which closes the ryanodine receptor (RyR) Ca^2+^ channels, did not block the increase in cytosol Ca^2+^ ([Figure 3a and b](#F3){ref-type="fig"}). We next assessed whether Ca^2+^ release was necessary for UPR activation using 2-APB and ryanodine individually, or in combination. 2-APB, but not ryanodine, inhibited E~2~-ERα activation of the PERK arm of the UPR, as shown by inhibition of formation of p-eIF2α ([Supplementary Figure 4a](#SD1){ref-type="supplementary-material"}). RNAi knockdown of IP~3~R ([Figure 3c](#F3){ref-type="fig"}) blocked E~2~-induced Ca^2+^ release ([Figure 3d](#F3){ref-type="fig"}), activation of the IRE1α arm of the UPR ([Supplementary Figure 4b](#SD1){ref-type="supplementary-material"}), and blocked E~2~-induction of BiP ([Figure 3c](#F3){ref-type="fig"}), which is a commonly used surrogate readout for UPR activation. We next tested the possibility that activation of PLCγ, which hydrolyzes PIP~2~ to DAG and IP~3~, plays a role in E~2~-mediated opening of the IP~3~R Ca^2+^ channels. Treating T47D cells with the PLCγ inhibitor, U73122, or siRNA knockdown of PLCγ, abolished the rapid E~2~-ERα-mediated increase in cytosol Ca^2+^ ([Figure 3e and f](#F3){ref-type="fig"}; [Supplementary Figure 5](#SD1){ref-type="supplementary-material"}). Since PLCγ mediates E~2~-dependent opening of the IP~3~R Ca^2+^ channels and calcium release ([Figure 3f](#F3){ref-type="fig"}), we examined the effect of siRNA knockdown of PLCγ on E~2~-ERα-dependent activation of the UPR. siRNA knockdown of PLCγ blocked E~2~-ERα activation of the ATF6α arm of the UPR, as shown by a reduction in p50-ATF6α, and inhibition of BiP induction ([Figure 3e](#F3){ref-type="fig"}). To evaluate the role of ERα in the E~2~-mediated increase in cytosol calcium, we performed siRNA knockdown. In T47D cells, RNAi knockdown of ERα, in the absence of extracellular Ca^2+^, prevented E~2~-stimulated calcium release ([Figure 3g and h](#F3){ref-type="fig"}; [Supplementary Movie 1 and 2](#SD1){ref-type="supplementary-material"}). PLCγ is on the inner leaflet of the plasma membrane and the E~2~-ERα-mediated increase in cytosol Ca^2+^ occurs in \<2 min. Thus, the E~2~-ERα-mediated increase in intracellular Ca^2+^ that leads to UPR activation is a rapid, extranuclear action of ERα at the plasma membrane. The UPR and E~2~-ERα Action in E~2~-ERα Stimulated Cell Proliferation {#S5} --------------------------------------------------------------------- We explored the role of Ca^2+^ release from the EnR in promoting E~2~-ERα induced gene expression, UPR activation, and subsequent cell proliferation. Consistent with a possible role for intracellular Ca^2+^ in E~2~-ERα action,^[@R23]^ chelating intracellular Ca^2+^ with BAPTA-AM blocked E~2~-stimulated cell proliferation ([Supplementary Figure 6a](#SD1){ref-type="supplementary-material"}). In T47D cells, PLCγ or IP~3~R knockdown, or locking IP~3~R with 2-APB, strongly inhibited the increase in cytosol Ca^2+^ ([Figure 3a, b, d, and f](#F3){ref-type="fig"}), UPR activation ([Figure 3c and e](#F3){ref-type="fig"}; [Supplementary Figure 4](#SD1){ref-type="supplementary-material"}), and E~2~-ERα-stimulated cell proliferation ([Figure 4a and b](#F4){ref-type="fig"}). However, IP~3~R knockdown did not inhibit E~2~-dependent down-regulation of ERα or E~2~-induction of GREB1 or pS2 mRNA ([Figure 4c](#F4){ref-type="fig"}; [Supplementary Figure 6b](#SD1){ref-type="supplementary-material"}).^[@R24],\ [@R25]^ Similarly, 2-APB did not abolish E~2~-ERα induced expression of stably transfected ERE-luciferase in T47D cells, while 2-APB and Ry together, strongly inhibited reporter gene expression ([Figure 4d](#F4){ref-type="fig"}). This suggests there are different intracellular Ca^2+^ requirements for E~2~-ERα-mediated UPR activation and E~2~-ERα-mediated gene expression. Importantly, the IP~3~R knockdown data uncouples UPR activation from E~2~-ERα-mediated gene expression, and demonstrates that blocking UPR activation is sufficient to inhibit estrogen-stimulated cell proliferation. We next evaluated the role of E~2~-induction of EnR chaperones in E~2~-ERα-stimulated cell proliferation. Knockdown of PLCγ or IP~3~R strongly inhibited E~2~-induction of BiP and E~2~-ERα-stimulated cell proliferation ([Figures 3c](#F3){ref-type="fig"}, [3e](#F3){ref-type="fig"}, and [4a](#F4){ref-type="fig"}). Knockdown of the primary UPR regulator of EnR chaperones, ATF6α, also strongly inhibited E~2~-induction of BiP and E~2~-ERα-stimulated cell proliferation ([Figure 1e](#F1){ref-type="fig"} and [4a](#F4){ref-type="fig"}). Thus, UPR activation and subsequent induction of EnR chaperones plays an important role in E~2~-ERα-stimulated cell proliferation. We further evaluated the effects of PLCγ, IP~3~R, ATF6α, XBP1, and PERK knockdown on E~2~-stimulated proliferation of MCF-7 cells ([Supplementary Figure 7](#SD1){ref-type="supplementary-material"}). Knockdown of the ATF6α and XBP1 arms of the UPR produced 40% declines in E~2~-stimulated in cell proliferation, while PERK knockdown produced a 20% decline ([Figure 4e](#F4){ref-type="fig"}). IP~3~R knockdown produced a 50% decline in E~2~-ERα-stimulated MCF-7 cell proliferation ([Figure 4e](#F4){ref-type="fig"}). This is consistent with the 40% decline in proliferation following 2-APB treatment ([Supplementary Figure 6c](#SD1){ref-type="supplementary-material"}), which did not fully abolish E~2~-induction of pS2 and GREB1 mRNA ([Figure 4f](#F4){ref-type="fig"}; [Supplementary Figure 6d](#SD1){ref-type="supplementary-material"}). Targeting IP~3~R in MCF-7 cells produced less dramatic inhibition of E~2~-ERα-stimulated cell proliferation compared to T47D cells or BG-1 ovarian cancer cells ([Figure 4a, b, and e](#F4){ref-type="fig"}; [Supplementary Figure 6c and e](#SD1){ref-type="supplementary-material"}). Knockdown of PLCγ in MCF-7 cells nearly abolished E~2~-ERα-stimulated cell proliferation ([Figure 4e](#F4){ref-type="fig"}). Together, this data demonstrates that weak anticipatory activation of the UPR, resulting in induction of chaperones, plays an important role in E~2~-ERα-stimulated cell proliferation. This novel E~2~-ERα pathway leading to cancer cell proliferation is shown ([Figure 4g](#F4){ref-type="fig"}). E~2~-ERα Action Increases Levels of UPR Sensors and Downstream Targets {#S6} ---------------------------------------------------------------------- We investigated whether E~2~-ERα facilitates UPR activation by inducing the sensors that trigger activation of the three UPR arms. E~2~ rapidly induced mRNAs encoding sensors for all 3 UPR arms and the chaperones BiP and GRP94 ([Figure 5a](#F5){ref-type="fig"}). These were early responses, usually visible within 2 hours. Although some responses declined at later times, estrogen produced sustained increases in resident chaperones and some UPR components, such as eIF2α ([Figure 5a](#F5){ref-type="fig"}). E~2~-ERα-regulated Gene Expression and UPR Activation are Correlated *In Vivo* {#S7} ------------------------------------------------------------------------------ To assess *in vivo* relevance, we used growing MCF-7 tumors receiving estrogen and regressing MCF-7 tumors receiving only cholesterol vehicle ([Figure 5b](#F5){ref-type="fig"}) and compared expression of classical measures of E~2~-ERα activity to markers of UPR activation.^[@R26]^ In the +E~2~ tumors, the markers for E~2~-ERα activity, pS2 and GREB1 mRNAs,^[@R24],\ [@R25]^ were induced 12-fold and 17-fold and all three UPR arms were moderately activated ([Figure 5c and d](#F5){ref-type="fig"}). Consistent with activation of the IRE1α arm of the UPR, sp-XBP1 increased 3-fold, while total XBP1 declined ([Figure 5d](#F5){ref-type="fig"}). Consistent with E~2~-activation of the ATF6α arm of the UPR, +E~2~ tumors displayed 2.0 and 1.8-fold increases in BiP and GRP94 mRNAs, respectively ([Figure 5d](#F5){ref-type="fig"}). Levels of CHOP and GADD34 mRNA were 2.1-fold and 1.4-fold higher in the +E~2~ group, respectively, indicating weak activation of the PERK arm ([Figure 5d](#F5){ref-type="fig"}). While levels of primary UPR sensors IRE1α and PERK were reduced in these tamoxifen-sensitive tumors, their immediate targets eIF2α and sp-XBP1 were increased ([Figure 5d](#F5){ref-type="fig"}). To assess UPR activity early in ERα^+^ breast cancer development, we compared E~2~-ERα activity and UPR pathway activity in samples of histologically normal breast epithelium and invasive ductal carcinoma (IDC). Compared to normal epithelium from IDC patients, IDC samples displayed elevated levels of ERα mRNA and E~2~-ERα induced pS2 and GREB1 mRNAs, and reduced levels of E~2~-ERα downregulated IL1-R1 mRNA ([Figure 5e](#F5){ref-type="fig"}). IDC samples displayed elevated SERP1 mRNA, a marker for IRE1α activation;^[@R19]^ CHOP and GADD34, which are markers of PERK activation; and BiP and GRP94 chaperones, which are markers of ATF6α activation ([Figure 5f](#F5){ref-type="fig"}). These data suggest UPR activation occurs very early in tumor development. Using data from an independent cohort of 278 ERα^+^ breast cancers we explored whether expression of ERα mRNA and protein, or E~2~-ERα-regulated genes, correlates with expression of UPR genes. Expression of several UPR genes displayed highly significant correlation with expression of ERα and ERα-target genes ([Supplementary Table 1](#SD1){ref-type="supplementary-material"}). Prior Estrogen Activation of the UPR Protect Cells from Subsequent Exposure to Cell Stress {#S8} ------------------------------------------------------------------------------------------ Weakly activating, non-toxic, concentrations of the UPR activator, tunicamcyin (TUN), elicit an adaptive stress response that increases EnR chaperones, and renders cells resistant to subsequent exposure to an otherwise lethal concentration of tunicamycin.^[@R27],\ [@R22]^ Consistent with weak E~2~ activation of the UPR, E~2~ induces a 2.3-fold increase in BiP protein compared to a 5.5-fold increase in BiP following maximal UPR activation by a lethal concentration of tunicamycin ([Figure 1g](#F1){ref-type="fig"} and [Supplementary Figure 8](#SD1){ref-type="supplementary-material"}). We tested whether prior exposure of T47D cells to E~2~, or a low concentration of tunicamycin, altered the concentration of tunicamycin required to subsequently induce substantial cell death. Pre-treating cells with estrogen or TUN had nearly identical effects; each elicited an \~10 fold increase in the concentration of tunicamycin required to induce apoptosis ([Figure 6a](#F6){ref-type="fig"}). Thus, the E~2~-induced weak anticipatory activation of the UPR both facilitates tumor cell proliferation and is a potential mechanism by which estrogen might protect ERα^+^ breast tumors against subsequent apoptosis due to hypoxia, nutritional deprivation and therapy. A UPR Gene Signature Predicts Clinical Outcome in ERαPositive Breast Cancer {#S9} --------------------------------------------------------------------------- To explore UPR activation as a potential prognostic marker in ERα^+^ breast cancer, we developed a UPR gene signature consisting of genes encoding components of the UPR pathway and downstream targets of UPR activation ([Supplementary Table 2](#SD1){ref-type="supplementary-material"}). Using data from 261 ERα^+^ breast cancer patients, each assigned to a high- or low-genomic UPR grade, we observed reduced time to relapse for patients overexpressing the UPR signature (hazard ratio (HR) = 5.5, 95% CI: 3.1--9.8) ([Supplementary Figures 9a and b](#SD1){ref-type="supplementary-material"}). To evaluate the UPR signature in patients undergoing tamoxifen therapy, samples collected from 474 ERα^+^ breast cancer patients, prior to starting 5-years of tamoxifen therapy, were assigned to low, medium, or high UPR risk groups. Increased prior expression of the UPR gene signature was tightly correlated with subsequent reduced time to recurrence ([Figure 6b and d](#F6){ref-type="fig"}; [Supplementary Figure 9c](#SD1){ref-type="supplementary-material"}). Hazard ratios increased from 2.2 to 3.7 for the medium and high-risk groups, respectively, suggesting that recurrence risk is sensitive to levels of the UPR gene signature ([Figure 6b](#F6){ref-type="fig"}). The UPR index provides prognostic information beyond current clinical covariates. In a cohort of 236 ERα^+^ breast cancer patients, UPR overexpression was strongly predictive of reduced survival (HR 2.69, 95% CI: 1.3--5.6), over and above clinical covariates alone (tumor grade, node involvement, tumor size and ERα status) ([Figures 6c and d](#F6){ref-type="fig"}; [Supplementary Figure 9d](#SD1){ref-type="supplementary-material"}). Thus, the UPR index is a powerful prognostic gene signature in ERα^+^ breast cancer with predictive power to stratify patients into high and low risk groups. DISCUSSION {#S10} ========== In contrast to the well-studied "reactive mode" of UPR activation that occurs in response to endoplasmic reticulum stress, there are few studies of UPR activation that anticipates the future need for increased capacity to fold and sort proteins, and occurs in the absence of endoplasmic reticulum stress.^[@R7]^ Anticipatory UPR activation is observed in B-cell differentiation where UPR activation in plasma cells precedes the massive production and secretion of immunoglobulins.^[@R13],\ [@R14]^ Because the signals responsible for anticipatory activation of the UPR are largely unknown, it is poorly understood. In the absence of cell stress or misfolded proteins, the mitogen, estrogen, acting via ERα, triggers anticipatory activation of the UPR in breast and ovarian cancer cells. In less than 2 minutes, E~2~-ERα triggers PLCγ-mediated opening of EnR IP~3~R calcium channels and release of Ca^2+^ into the cytosol. This increase in cytosol Ca^2+^ stimulates activation of all three arms of the UPR and is required for E~2~-ERα-stimulated cell proliferation. Anticipatory activation of the UPR by E~2~-ERα enhances EnR protein folding capacity, and thereby primes cells to meet the higher protein folding and sorting demands that characterize the later growth phases of the cell cycle. The major EnR chaperone BiP, plays a central role in EnR homeostasis, protein processing, and UPR signaling. Since BiP knockdown stimulates UPR activation and promotes EnR stress-induced apoptosis,^[@R10],\ [@R28]^ and cells undergoing E~2~-mediated apoptosis have lower levels of chaperones,^[@R29]^ we assessed the consequences of abrogating the expansion of EnR protein-folding capacity by blocking anticipatory activation of the UPR. PLCγ, IP~3~R or ATF6α knockdown blocked E~2~-induction of BiP and inhibited E~2~-ERα stimulated proliferation of T47D cells. While IP~3~R knockdown nearly abolished E~2~-ERα-stimulated Ca^2+^ release from the EnR, and this blocked UPR activation, it did not inhibit E~2~-ERα-mediated gene expression. Thus, inhibition of E~2~-ERα-stimulated UPR activation and chaperone induction is sufficient to inhibit E~2~-ERα-stimulated cell proliferation. Using 2-APB and ryanodine together, or chelating intracellular calcium with BAPTA, completely abrogated the increase in intracellular calcium, and blocked E~2~-ERα-regulated gene expression. Based on the inhibitor and knockdown data, we hypothesize that very small increases in intracellular calcium are sufficient to enable E~2~-ERα-regulated gene expression and that somewhat larger increases in intracellular calcium are likely required for E~2~-ERα activation of the UPR. E~2~-ERα induces a substantial increase in intracellular calcium, which may promote coordination between the nucleus and endoplasmic reticulum, and couple activation of the E~2~-ERα genomic program with UPR activation and expansion of the EnR protein-folding capacity. We further validated the importance of this novel extranuclear pathway of E~2~-ERα action using MCF-7 cells to assess how knockdown of each pathway component affects E~2~-ERα-stimulated cell proliferation. PERK knockdown produced a 20% decline in E~2~-ERα-stimulated cell proliferation. Although seemingly detrimental to promoting cell proliferation, PERK activation may be required to fully activate the ATF6α arm of the UPR.^[@R30]^ Knockdown of XBP1 or ATF6α produced a 40% decline in E~2~-ERα-stimulated cell proliferation. IP~3~R knockdown produced an even larger reduction in E~2~-ERα stimulated cell proliferation, while PLCγ knockdown had the largest effect. Thus, anticipatory activation of the UPR plays an important role in E~2~-ERα dependent proliferation of cancer cells. As expected,^[@R1],\ [@R3]^ IDC tumor samples exhibited increased ERα expression and activation compared to normal breast epithelial tissue. Consistent with a role for the UPR in this proliferative phase of early tumor development, increased UPR expression and activation was observed in IDC tumor samples. This suggests that increased UPR expression occurs early in tumor development, long before detection, diagnosis, and the initiation of treatment. Activation of the UPR by E~2~-ERα exerts a long-term impact on the pathology of ERα positive breast cancer. Weak activation of the UPR by estrogen, or by tunicamcyin, elicits an adaptive response that protects cells from subsequent exposure to higher levels of cell stress. We explored whether the effects of E~2~-ERα on the UPR correlated with clinical resistance to tamoxifen therapy. Increased UPR activation and elevated expression of UPR components were predictive of a poor response to tamoxifen-therapy, shorter time to recurrence, and decreased overall survival. If UPR expression promotes resistance to tamoxifen therapy, some UPR genes should exhibit differential regulation in our tamoxifen-sensitive MCF-7 tumors,^[@R26]^ compared to their expression in the tamoxifen-resistance gene signature. Supporting this view, several genes encoding UPR components were E~2~-downregulated in tamoxifen-sensitive MCF-7 tumors, but elevated in the human tumors expressing the tamoxifen-resistance gene signature (PERK, p58^IPK^). For ERα^+^ breast cancers resistant to endocrine therapies, an important objective is development of more specific biomarkers that predict therapeutic response and identification of new therapeutic targets. The UPR is a new biomarker and therapeutic target in ERα^+^ breast cancer; validated through mechanistic studies in culture, a mouse xenograft, and bioinformatics analysis of patient tumor samples. Anticipatory estrogen activation of the UPR is a novel extranuclear action of ERα, a previously undescribed early component of the estrogen-ERα cell proliferation program and a new paradigm by which estrogens may influence tumor development and resistance to therapy. MATERIALS AND METHODS {#S11} ===================== Cell Culture and Reagents {#S12} ------------------------- Cell culture medium and conditions were previously described.^[@R31]--[@R33]^ MCF-7, T47D, and T47D-kBluc cells were obtained from the ATCC. Drs. S. Kaufmann and K. Korach provided PEO4 cells and BG-1 cells, respectively. E~2~, 4-OHT, U73122, 2-APB, and tunicamycin were from Sigma Aldrich. ICI 182,780 was from Tocris Biosciences and ryanodine was from Santa Cruz Biotechnology. Phospho-eIF2α (\#3398), eIF2α (\#5324), Phospho-PERK (\#3179), PERK (\#5683), and BiP (\#3177) antibodies were from Cell Signaling. Pan-IP~3~R (sc-28613), XBP1 (sc-7160), and ERα (sc-56836) antibodies were from Santa Cruz Biotechnology. Other antibodies used were ATF6α (Imgenex) and β-Actin (Sigma). Cell Proliferation Assays {#S13} ------------------------- Cells proliferation assays were carried out as described.^[@R31]--[@R33]^ Protein Synthesis {#S14} ----------------- Protein synthesis was evaluated by measuring incorporation of ^35^S-Methionine into newly synthesized protein. Cells were incubated in 96 well plates for 20 minutes with 3 μCi of ^35^S-methionine per well (PerkinElmer), lysed, and clarified by centrifugation. The appropriate volume, normalized to total protein, was spotted onto Whatman 540 filter paper discs and immersed in cold 10% TCA and washed in 5% TCA. Trapped protein was solubilized and filters counted. Calcium Imaging {#S15} --------------- Cytoplasmic Ca^2+^ concentrations were measured using the calcium-sensitive dye, Fluo-4 AM.^[@R34],\ [@R35]^ Cells were grown on 35 mm-fluorodish plates (World Precision Instruments) for two days prior to experiments. Cells were loaded with 5 μM Fluo-4 AM (Life Technologies) in buffer (140 mM NaCl, 4.7 mM KCl, 1.13 mM MgCl~2~, 10 mM HEPES, 10 mM Glucose, pH = 7.4) for 30 minutes at 37° C. The cells were washed three times with buffer and incubated with either 2 mM or 0 mM CaCl~2~ for 10 minutes. Images were captured for one minute to determine basal fluorescence intensity, and then the appropriate treatment was added. Measurements used a Zeiss LSM 700 confocal microscope with a Plan-Four 20X objective (N.A. = 0.8) and 488-nM laser excitation (7% power). Images were obtained through monitoring fluorescence emission at 525 nM, and analyzed with AxioVision and Zen software (Zeiss). Luciferase Assays, qRT-PCR, and siRNA Transfections {#S16} --------------------------------------------------- Reporter gene assays and qRT-PCR were previously described.^[@R31],\ [@R32]^ siRNA knockdowns were performed using DharmaFECT1 Transfection Reagent and 100 nM ON-TARGETplus non-targeting pool or SMARTpools for ERα (ESR1), PLCγ (PLCG1), PERK (EIF2AK3), ATF6α (ATF6), XBP1, or pan-IP3R (Dharmacon). The pan-IP~3~R SmartPool consisted of three individual SmartPools, each at 33 nM, directed against each isoform of the IP~3~R (ITPR1, ITPR2, and ITPR3). MCF-7 Xenograft {#S17} --------------- Experiment were approved by the Institutional Animal Care Committee (IACUC) of the University of Illinois at Urbana-Champaign. The MCF-7 cell mouse xenograft model has been described previously.^[@R26]^ Estrogen pellets (1 mg:19 mg estrogen:cholesterol) were implanted into 30 athymic female OVX mice at 7 weeks of age. Three days later, 1 million MCF-7 human breast cancer cells suspended in matrigel were subcutaneously injected into two sites on each flank, for a total of 4 tumors per mouse. When average tumor size reached 17.6 mm^2^, E~2~ pellets were removed and a lower dose of E~2~ in sealed silastic tubing (1:31 estrogen:cholesterol, 3 mg total weight) was implanted. When average tumor size reached 23.5 mm^2^, 15 mice retained E~2~ silastic tubes (+E~2~ group) and 15 mice received silastic tubes containing only cholesterol (−E~2~ group). Tumors were measured every 4 days with a caliper. Tumor cross sectional area was calculated as (a/2)\*(b/2)\*3.14, where a and b were the measured diameters of each tumor. Upon termination of the experiments, mice were euthanized and tumors were excised. Tumor Microarray Data Analysis {#S18} ------------------------------ Analysis was performed using publically available tumors cohorts. ERα and UPR gene expression profiles of histologically normal breast epithelium (GSE20437)^[@R36]^ were compared to IDC tumors from ERα^+^ breast cancer patients (GSE20194). ERα and UPR correlation analysis was performed on 278 invasive ductal carcinoma samples (GSE20194).^[@R37]^ A "UPR Gene Signature" was constructed to carry out risk prediction analysis. The UPR gene signature was evaluated for its ability to predict: (i) tumor relapse in 261 early-stage ERα^+^ breast cancers (GSE6532),^[@R37]^ (ii) tumor relapse in 474 ERα^+^ patients receiving solely tamoxifen therapy for 5 years (GSE6532, GSE17705),^[@R38],\ [@R39]^ and (iii) overall survival in a mixed-cohort of 236 breast cancer patients (GSE3494).^[@R40]^ Microarray data analysis was performed using BRB ArrayTools (version 4.2.1) and R software version 2.13.2. Gene expression values from CEL files were normalized by use of the standard quantile normalization method.^[@R41]^ Pearson correlation tests and Spearman log rank tests were used to determine gene expression correlation coefficients. Wald tests were used to test whether UPR genes were predictive of tumor recurrence and overall survival. Univariate and multivariate hazard ratios were estimated using Cox regression analysis. Covariates statistically significant in univariate analysis were further assessed in multivariate analysis. A patient was excluded from multivariate analysis, if data for one or more variables was missing. Risk prediction using the UPR gene signature was carried out using the supervised principle components method,^[@R42]^ and visualized using Kaplan-Meier plots and compared using log-rank tests. Statistical Analysis {#S19} -------------------- Calcium measurements are reported as mean ± SE. All other data is reported as mean ± S.E.M. Two-tailed student's t-test used for comparisons between groups. One-way ANOVA followed by Fisher's LSD or Tukey's post hoc test used for multiple comparisons. P\< 0.05 was considered significant. Supplementary Material {#S20} ====================== We thank Mr. J. Hartman for assistance with xenografts, and Drs. S. Kaufmann and K. Korach for cell lines. Supported by NIH RO1DK 071909 (to D.S.) and Westcott and Carter predoctoral fellowships (to N.A.). Analyses were performed using BRB-ArrayTools, developed by Dr. Richard Simon and BRB-ArrayTools Development Team at the National Cancer Institute. **CONFLICTS OF INTEREST** The authors declare no conflicts of interest. [Supplementary Information](#SD1){ref-type="supplementary-material"} accompanies the paper on the *Oncogene* website (<http://www.nature.com/onc>). ![E~2~-ERα activates the IRE1α and ATF6α arms of the UPR in breast and ovarian cancer cells, resulting in the induction of the major EnR chaperone, BiP. (**a**) qRT-PCR comparing the effect of estrogen (E2), ICI 182,780 (ICI) and 4-hydroxytamoxifen (4-OHT) on E~2~-ERα induction of spliced-XBP1 (sp-XBP1) in ERα^+^T47D breast cancer cells (n = 3; −E~2~ set to 1). Different letters indicate a significant difference among groups (p \< 0.05) using one-way ANOVA followed by Tukey's post hoc test. (**b**) qRT-PCR showing the effect of E~2~-ERα on sp-XBP1 mRNA in ERα^+^MCF-7 breast and PEO4 ovarian cancer cells (n = 3; −E~2~ set to 1). P-values testing for significance between indicated group and -E2 group. (**c**) RNAi knockdown of ERα abolishes E~2~-induction of sp-XBP1 in MCF-7 cells (n = 3). Cells treated with 100 nM non-coding control (NC) or ERα siRNA SmartPool, followed by treatment with E~2~ for the indicated times (**d**) Western blot analysis showing full-length 90 kDa ATF6α (p90-ATF6α) and proteolytically cleaved 50 kDa ATF6α (p50-ATF6α) in E~2~-treatedT47D breast cancer cells. (**e**) RNAi knockdown of ATF6α blocks E~2~-induction of BiP in T47D cells. Cells treated with 100 nM non-coding control (NC) or ATF6α siRNA SmartPool, followed by treatment with E~2~ for 4 hours. (**f**) qRT-PCR showing the effect of E~2~ on BiP mRNA in MCF-7 cells and in PEO4 ovarian cancer cells (n = 3; −E~2~ set to 1). (**g**) Western blot analysis of BiP protein levels in MCF-7 cells treated with E~2~. The fold-change in BiP protein levels is shown below each lane and was determined by quantifying BiP and β-Actin signals, and calculating the ratio of BiP/β-Actin (t=0, \[−E~2~\], set to 1). (**h**) RNAi knockdown of ERα abolishes E~2~-induction of BiP in MCF-7 cells (n = 3). Cells treated with 100 nM non-coding control (NC) or ERα siRNA SmartPool, followed by treatment with E~2~ for the indicated times. Concentrations: E~2~, 1 nM (a, d), 10 nM (b, c, e--h); ICI, 1 μM (a, d); 4-OHT, 1 μM (a). Data is mean ± S.E.M. \* p \< 0.05; \*\* p \< 0.01; \*\*\* p\< 0.001.](nihms618571f1){#F1} {#F2} {#F3} {#F4} {#F5} {#F6} | Mid | [
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Increased total and mite-specific immunoglobulin E in patients with aspirin-induced urticaria and angioedema. An increased prevalence of atopy has been observed in patients with intolerance of aspirin and nonsteroidal anti-inflammatory drugs (NSAIDs). To investigate total and mite-specific immunoglobulin (Ig) E in serum from patients with hypersensitivity to NSAIDs and healthy controls. Patients who reacted to 2 or more chemically unrelated NSAIDs with urticaria and angioedema, confirmed by a double-blinded provocation test with aspirin, were skin tested with inhalant allergens. Total and specific IgE to Dermatophagoides pteronyssinus (Dp) and Blomia tropicalis (Bt) in the serum was quantified by enzyme-linked immunosorbent assay (ELISA) in patients and a control group of healthy blood donors. One-hundred-and-fourteen patients and 74 controls were studied. Skin tests were positive in 95 patients (83.3%). Total mean IgE levels were 107.1 (91.3) IU/mL in controls and 161.0 (150.8) IU/mL in patients (P = .006). Mean (SD) levels of IgE to Dp were 0.210 (0.17) optical density (OD) units in controls and 0.473 (0.65) OD units in patients (P = .001). Levels of specific IgE to Bt were 0.230 (0.20) OD units in controls and 0.522 (0.8) OD units in patients (P =.0001). Positive ELISA results for IgE to Dp were found for 29.6% of controls and 70.4% of patients (P =.0001); the corresponding percentages for Bt were 32.4% of controls and 67.6 % of patients (P = .0001). Cross-reactive patients with NSAID-induced urticaria and angioedema exhibit an increased prevalence of sensitization to Dp and Bt and increased total serum IgE. Further research is necessary to determine the reasons for this association. | Mid | [
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Algal endosymbiosis in brown hydra: host/symbiont specificity. Host/symbiont specificity has been investigated in non-symbiotic and aposymbiotic brown and green hydra infected with various free-living and symbiotic species and strains of Chlorella and Chlorococcum. Morphology and ultrastructure of the symbioses obtained have been compared. Aposymbiotic Swiss Hydra viridis and Japanese H. magnipapillata served as controls. In two strains of H. attenuata stable hereditary symbioses were obtained with Chlorococcum isolated from H. magnipapillata. In one strain of H. vulgaris, in H. oligactis and in aposymbiotic H. viridis chlorococci persisted for more than a week. Eight species of free-living Chlorococcum, 10 symbiotic and 10 free-living strains of Chlorella disappeared from the brown hydra within 1-2 days. In H. magnipapillata there was a graded distribution of chlorococci along the polyps. In hypostomal cells there were greater than 30 algae/cell while in endodermal cells of the mid-section or peduncle less than 10 algae/cell were found. In H. attenuata the algal distribution was irregular, there were up to five chlorocci/cell, and up to 20 cells/hydra hosted algae. In the dark most cells of Chlorococcum disappeared from H. magnipapillata and aposymbiotic hydra were obtained. Chlorococcum is thus an obligate phototroph, and host-dependent heterotrophy is not required for the preservation of a symbiosis. The few chlorococci that survived in the dark seem to belong to a less-demanding physiological strain. In variance with known Chlorella/H. viridis endosymbioses the chlorococci in H. magnipapillata and H. attenuata were tightly enveloped in the vacuolar membrane of the hosting cells with no visible perialgal space. Chlorococcum reproduced in these vacuoles and up to eight daughter cells were found within the same vacuole. We suggest that the graded or scant distribution of chlorococci in the various brown hydra, their inability to live in H. viridis and the inability of the various chlorellae to live in brown hydra are the result of differences in nutrients available to the algae in the respective hosting cells. We conclude that host/symbiont specificity and the various forms of interrelations we show in green and brown hydra with chlorococci and chlorellae are based on nutritional-ecological factors. These interrelations demonstrate successive stages in the evolution of stable obligatoric symbioses from chance encounters of preadapted potential cosymbionts. | High | [
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Exception in rendering! Message: window is not defined ReferenceError: window is not defined at new c (/tmp/execjs20161208-80463-3dp2f9js:136:3912) at m.mountComponent (/tmp/execjs20161208-80463-3dp2f9js:47:15602) at /tmp/execjs20161208-80463-3dp2f9js:49:31860 at a.r.perform (/tmp/execjs20161208-80463-3dp2f9js:47:12503) at Object.a [as renderToString] (/tmp/execjs20161208-80463-3dp2f9js:49:31821) at r (/tmp/execjs20161208-80463-3dp2f9js:50:21164) at Object.S.ReactOnRails.serverRenderReactComponent (/tmp/execjs20161208-80463-3dp2f9js:32:6073) at eval (eval at <anonymous> (/tmp/execjs20161208-80463-3dp2f9js:173:8), <anonymous>:10:23) at eval (eval at <anonymous> (/tmp/execjs20161208-80463-3dp2f9js:173:8), <anonymous>:17:3) at /tmp/execjs20161208-80463-3dp2f9js:173:8 Q: Car makes grinding noise - 1996 Chevrolet Corsica asked by Simon H on September 19, 2016 I've noticed some sight noise in my car lately. It is coming from the bottom of the vehicle, where my feet are. This is an intermittent grinding and thumping noise. There is no noise when I first put it in drive upon starting the car. However, as I drive, the noise starts occurring up to the point when I reach around 45 mph. The noise then disappears again. When I put the car in reverse, it doesn't make any noise. What could make this noise? Are there any parts I need to replace? Get an instant quote for your car What others are asking Q: Car shakes when put in park You should start by checking for a failed head gasket. With the overheating and coolant loss you describe, followed by very rough running suggests one or more cylinders have lost compression, causing the engine to shake. It's also worth having... Q: sluggish, stalling, check engine light came on With the car not being able to upshift and the sluggishness issue that you have, there are a few potential causes. The most common would be a shift solenoid, torque converter, MAF sensor or a fuel pressure issue. You will... | Low | [
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On 22 November 2016, the American Herald Tribune web site published an article (which was uncritically aggregated by other sites such as MintPress News under the misleading headline “Media Blackout as Millions of Muslims March Against ISIS in Iraq”). In reality, millions of Muslims did not march against the Islamic State, nor did a “media blackout” result in news agencies’ ignoring such an event. Although the original article was published under a misleading clickbait title, it was at least based on a real event: millions of Shia Muslims did embark on a pilgrimage in November 2016 to visit the shrine of Imam Hussein in Karbala, Iraq. This pilgrimate is an annual undertaking for Arbaeen, which marks the end of forty days of mourning of the martyrdom of Hussein: Millions of Shia Muslims have taken part in one of the biggest marches in the world, as they travel through Iraq in celebration of a famous Muslim martyr. The marchers made their way to the city of Karbala, 62 miles south west of Baghdad, on Sunday and Monday for the holy day of Arbaeen, which marks the end of a 40-day mourning period following Ashura, the religious ritual that commemorates the death of the Prophet Mohammad’s grandson Imam Hussein in 680 AD. Large crowds visit the shrines of Imam Hussein and his half-brother Abbas in Karbala, where they were killed in a revolt against the Umayyad ruler Yazeed in the 7th century AD when they refused to pledge allegiance to Yazeed’s Umayyad caliphate. The pilgrimage was reported (as it is every year) by multiple mainstream news outlets. While it is a religious event and not specifically a political one, some news outlets reported that the annual pilgrimage has taken on a defiant tone as attacks from the Islamic State have plagued the area: Large-scale security operations, involving 24,000 soldiers and police, were put in place during this year’s march due to fears Isis forces in and around its last major stronghold of Mosul may seek to strike Baghdad or Karbala during Arbaeen. Karbala, around 50 miles, south west of Baghdad, faces the desert of Anbar, a vast province that was until recently an Isis bastion and where jihadists still carry out frequent attacks. Last week, a suicide bomber killed six people near Karbala in an attack claimed by the so-called Islamic State. The pilgrimage has taken on additional meaning in response to this threat, becomimg a protest in recent years against Isis terrorists. Millions of Muslims did embark on a religious pilgrimage for Arbaeen in November 2016, but that phenomenon is an annual event which is duly covered by the news media every year, not an unprecedented gathering involving huge crowds of protesters whom reporters all over the world collectively decided to ignore. | Mid | [
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Does diabetes affect outcome after arthroscopic repair of the rotator cuff? We compared the outcome of arthroscopic repair of the rotator cuff in 32 diabetic patients with the outcome in 32 non-diabetic patients matched for age, gender, size of tear and comorbidities. The Constant-Murley score improved from a mean of 49.2 (24 to 80) pre-operatively to 60.8 (34 to 95) post-operatively (p = 0.0006) in the diabetic patients, and from 46.4 (23 to 90) pre-operatively to 65.2 (25 to 100) post-operatively (p = 0.0003) in the non-diabetic patients at six months. This was significantly greater (p = 0.0002) in non-diabetic patients (18.8) than in diabetics (11.6). There was no significant change in the mean mental component of the Short-Form 12, but the mean physical component increased from 35 to 41 in non-diabetics (p = 0.0001), and from 37 to 39 (p = 0.15) in diabetics. These trends were observed at one year. Patients with diabetes showed improvement of pain and function following arthroscopic rotator cuff repair in the short term, but less than their non-diabetic counterparts. | Mid | [
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Q: Read XLS and convert it to CSV using Python I need to convert XLS files to CSV in order to the data contained in a PostgreSQL database, I used the following code to do the conversion : import xlrd import unicodecsv def xls2csv (xls_filename, csv_filename): # Converts an Excel file to a CSV file. # If the excel file has multiple worksheets, only the first worksheet is converted. # Uses unicodecsv, so it will handle Unicode characters. # Uses a recent version of xlrd, so it should handle old .xls and new .xlsx equally well. wb = xlrd.open_workbook(xls_filename) sh = wb.sheet_by_index(0) fh = open(csv_filename,"wb") csv_out = unicodecsv.writer(fh, encoding='utf-8') for row_number in xrange (sh.nrows): csv_out.writerow(sh.row_values(row_number)) fh.close() The XLS files I'm using contains 212 columns and at least 100 rows, when I test the code with just 4 rows it works fine, but when nrows>5 the interpreter raises the following errors : xls2csv ('e:/t.xls', 'e:/wh.csv') WARNING *** file size (353829) not 512 + multiple of sector size (512) WARNING *** OLE2 inconsistency: SSCS size is 0 but SSAT size is non-zero *** No CODEPAGE record, no encoding_override: will use 'ascii' *** No CODEPAGE record, no encoding_override: will use 'ascii' Traceback (most recent call last): File "<ipython-input-14-ccae93f2d633>", line 1, in <module> xls2csv ('e:/t.xls', 'e:/wh.csv') File "C:/Users/hey/.spyder/temp.py", line 10, in xls2csv wb = xlrd.open_workbook(xls_filename) File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\__init__.py", line 441, in open_workbook ragged_rows=ragged_rows, File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\book.py", line 119, in open_workbook_xls bk.get_sheets() File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\book.py", line 678, in get_sheets self.get_sheet(sheetno) File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\book.py", line 669, in get_sheet sh.read(self) File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\sheet.py", line 804, in read strg = unpack_string(data, 6, bk.encoding or bk.derive_encoding(), lenlen=2) File "C:\Users\hey\Anaconda2\lib\site-packages\xlrd\biffh.py", line 269, in unpack_string return unicode(data[pos:pos+nchars], encoding) UnicodeDecodeError: 'ascii' codec can't decode byte 0xb2 in position 2: ordinal not in range(128) A: There is decoding issue when you open the xls file, I suspect the 5th line of xls file has special character, based on the xlrd documentation, you can use encoding_override="cp1251" to translate to Unicode: wb = xlrd.open_workbook(xls_filename, encoding_override="cp1251") | Mid | [
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tag:blogger.com,1999:blog-14114712.post216789509622738085..comments2016-12-09T07:26:42.786-08:00Comments on ADD / XOR / ROL: Improving Binary Comparison (and it's implication for malware classification)halvar.flakehttp://www.blogger.com/profile/[email protected]:blogger.com,1999:blog-14114712.post-70389454089842354522008-10-31T02:17:00.000-07:002008-10-31T02:17:00.000-07:00@forever: This should be in BinDiff v3.0, due out ...@forever: This should be in BinDiff v3.0, due out in q1 2009. <BR/><BR/>@nate: False positives are usually identifiable -- they tend to have low "confidence" scores in the matching algorithm. Generally, with larger graphs the false positive rate decreases drastically. <BR/><BR/>As far as I can see, they are backporting patches to 0.9.5a, but I only had a cursory look.<BR/><BR/>Concerning gnutls: I would be _extremely_ surprised if gnutls vs openssl matches in any way -- but I will try, thanks for the hint ! :)<BR/><BR/>@ryan: No, not yet -- the confidence of the neighboring matches does not influence the confidence of the current match yet. In debug builts of the code we construct a "reasoning tree" though -- e.g. "this node was matched because of this other node here". This would allow us to base the current confidence on the confidence of the things that "led" the algorithm to decide the way it did. <BR/><BR/>Unfortunately, keeping this "reasoning tree" in memory is not easily feasible for very large diffs, so we don't build it unless we're trying to diagnose an error...halvar.flakehttp://www.blogger.com/profile/[email protected]:blogger.com,1999:blog-14114712.post-61358321196153611492008-10-06T16:46:00.000-07:002008-10-06T16:46:00.000-07:00Asking out of ignorance, so apologies in advance.Y...Asking out of ignorance, so apologies in advance.<BR/><BR/>You are already taking advantage of callpath info to bring up neighbor functions? I.e. A calls B. B by itself in both binaries is 0.5 confidence. But A is 0.95. If both A's call both B's in the same place, do you bump B up to 0.725?Ryan Russellhttp://www.blogger.com/profile/[email protected]:blogger.com,1999:blog-14114712.post-58808055763825277752008-10-04T10:39:00.000-07:002008-10-04T10:39:00.000-07:00Hey Halvar, your posts do have titles now! :)Hey Halvar, your posts do have titles now! :)Thierry Zollerhttp://www.blogger.com/profile/[email protected]:blogger.com,1999:blog-14114712.post-36400510122799502512008-10-01T10:19:00.000-07:002008-10-01T10:19:00.000-07:00Halvar, amazing stuff. I like your thesis too, al...Halvar, amazing stuff. I like your thesis too, although I've only read a few pages so far.<BR/><BR/>While you've demoed some libraries that likely match the one used in the target, what about the false positive ratio? It would be interesting to see your comparison of multiple versions of the OpenSSL library against PIX to see which one it is most similar to. Are they really backporting patches or is it stock 0.9.5a?<BR/><BR/>Also, perhaps compare gnutls to the PIX OpenSSL implementation and see how much various internal functions (say, derive master secret) match even though the libraries are different implementations.Natehttp://www.blogger.com/profile/[email protected]:blogger.com,1999:blog-14114712.post-57433736688550422812008-10-01T07:01:00.000-07:002008-10-01T07:01:00.000-07:00I'm going to be the annoying one and ask when we c...I'm going to be the annoying one and ask when we can expect to see this behavior in BinDiff or some other publicly available product?Foreverhttp://www.blogger.com/profile/[email protected] | Mid | [
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Sunlight mediated synthesis of silver nanoparticles by a novel actinobacterium (Sinomonas mesophila MPKL 26) and its antimicrobial activity against multi drug resistant Staphylococcus aureus. Synthesis of silver nanoparticles using microorganism are many, but there are only scanty reports using actinobacteria. In the present study, the actinobacterium of the genus Sinomonas was reported to synthesis silver nanoparticles for the first time. A photo-irradiation based method was developed for the synthesis of silver nanoparticles, which includes two day old cultural supernatant of novel species Sinomonas mesophila MPKL 26 and silver nitrate solution, exposed to sunlight. The preliminary synthesis of silver nanoparticles was noted by the color change of the solution from colorless to brown; the synthesis was further confirmed using UV-visible spectroscopy which shows a peak between 400 and 450nm. Spherical shape silver nanoparticles of size range 4-50nm were synthesized, which were characterized using transmission electron microscopy. The Fourier transform infrared spectroscopy result indicates that, the metabolite produced by the novel species S. mesophila MPKL 26 was the probable reducing/capping agent involved in the synthesis of silver nanoparticles. The synthesized silver nanoparticles maintained consistent shape with respect to different time periods. The synthesized silver nanoparticles were evaluated for the antimicrobial activity against multi drug resistant Staphylococcus aureus which show good antimicrobial activity. The method developed for synthesis is easy, requires less time (20min) and produces spherical shape nanoparticles of size as small as 4nm, having good antimicrobial activity. Hence, our study enlarges the scope of actinobacteria for the rapid biosynthesis of silver nanoparticles and can be used in formulating remedies for multi drug resistant S. aureus. | High | [
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Trump to visit victims of unprecedented floods in Texas and Louisiana – Business Insider Local residents affected by tropical storm Harvey received donations at a parking lot in East Houston, Texas, U.S.Thomson Reuters By Emily Flitter and Daniel Trotta HOUSTON (Reuters) – U.s. President Donald Trump travels to Houston and Lake Charles, Louisiana on Saturday to meet victims of catastrophic storm Harvey, one of the worst natural disasters in U.S. history that is presenting a test of his administration. While Trump visits, attention will also be focused on Minute Maid Park, where baseball’s Houston Astros play their first home games since Harvey devastated the fourth-most populous U.S. city. The Saturday doubleheader with the New York Mets is expected to be wrought with emotion and punctuated with moments to honor the dozens who died as a result of Harvey. The storm, one of the costliest to hit the United States, has displaced more than 1 million people, with 50 feared dead from flooding that paralyzed Houston, swelled river levels to record highs and knocked out the drinking water supply in Beaumont, Texas, a city of 120,000 people. Hurricane Harvey came ashore last Friday as the strongest storm to hit Texas in more than 50 years. Much of the damage took place in the Houston metropolitan area, which has an economy about the same size as Argentina’s. Seventy percent of Harris County, which encompasses Houston, at one point was covered with 18 inches (45 cm) or more of water, county officials said. For graphic on Harvey’s energy impact, click http://tmsnrt.rs/2xzso1S For graphic on hurricane costs, click http://tmsnrt.rs/2vGkbHS For graphic on storms in the North Atlantic, click http://tmsnrt.rs/2gcckz5 Trump first visited the Gulf region on Tuesday, but stayed clear of the disaster zone, saying he did not want to hamper rescue efforts. Instead, he met with state and local leaders, and first responders. He was criticized, however, for not meeting with victims of the worst storm to hit Texas in 50 years, and for largely focusing on the logistics of the government response rather than the suffering of residents. The White House said Trump will first travel to Houston to meet with flood survivors and volunteers who assisted in relief efforts and then move on to Lake Charles, another area hammered by the storm. The Trump administration in a letter to Congress asked for a $7.85 billion appropriation for response and initial recovery efforts. White House homeland security adviser Tom Bossert has said aid funding requests would come in stages as more became known about the impact of the storm. Texas Governor Greg Abbott has said that his state may need more than $125 billion. The storm, which lingered around the Gulf of Mexico Coast for days, dumped record amounts of rain and left devastation across more than 300 miles (480 km) of the state’s coast. As water receded, many returned to survey the damage and left hundreds of thousands wondering how they can recover. In Orange, Texas, about 125 miles (200 kms) east of Houston, Sam Dougharty, 36, returned on Friday where waist-high water remained in his backyard and barn. His family’s house smelled like raw sewage and was still flooded to the ankles. A calf and a heifer from their herd of 15 were dead. The chickens were sagging on the top two roosts of their coop. “We never had water here. This is family land. My aunt’s owned it for 40 years and never had water here,” he said. FROM THE SHELTER TO THE STADIUM Harvey came on the 12th anniversary of Hurricane Katrina, which killed about 1,800 around New Orleans. Then-U.S. President George W. Bush’s administration was roundly criticized for its botched early response to the storm. Some of the tens of thousands of people forced into shelters by Harvey will attend the Astros game where Houston Mayor Sylvester Turner will throw out the first pitch and a moment of silence in planned for those who perished. Sports have helped other cities rebound from catastrophe, such as when the New York Mets played the first baseball game in their damaged city 10 days after the attacks of Sept. 11, 2001, or when the New Orleans Saints returned to the Superdome in 2006 for football a year after Hurricane Katrina. In the Harris County town of Clear Creek, the nearly 50 inches (127 cm) of rain that fell there equated to a once in a 40,000 year event, Jeff Lindor, meteorologist with the Harris County Flood Control District, said. Some 440,000 Texans have already applied for federal financial disaster assistance, and some $79 million has been approved so far, Abbott said. The storm shut about a fourth of U.S. refinery capacity, much of which is clustered along the Gulf Coast, and caused gasoline prices to spike to a two-year high ahead of the long Labor Day holiday weekend. The national average for a gallon of regular gasoline has risen 17 cents since the storm struck, hitting $2.519 as of Friday morning, according to motorists group AAA. Meanwhile a new storm, Irma, had strengthened on Friday into a Category 3 hurricane on the five-step Saffir-Simpson scale. It remained hundreds of miles from land but was forecast to possibly hit Puerto Rico, the Dominican Republic and Haiti by the middle of next week. For graphic on Harvey’s energy impact, click http://tmsnrt.rs/2xzso1S For graphic on hurricane costs, click http://tmsnrt.rs/2vGkbHS For graphic on storms in the North Atlantic, click http://tmsnrt.rs/2gcckz5 | Low | [
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Q: Show that $(\mathbf{x}, \boldsymbol\Theta\mathbf{x}+\boldsymbol\eta)$ is jointly normal This is from Theodoridis' Machine Learning, exercise 3.16. Suppose $\mathbf{x}$ is a vector of jointly normal random variables with covariance matrix $\boldsymbol\Sigma_x$. Let $$\mathbf{y} = \boldsymbol\Theta\mathbf{x}+\boldsymbol\eta$$ where $\boldsymbol\Theta$ is a $k \times l$ matrix of [I assume, known or fixed] parameters and $\boldsymbol\eta$ is normal with mean $\mathbf{0}$ and covariance matrix $\boldsymbol\Sigma_\eta$, independent of $\mathbf{x}$. Show that $\mathbf{y}$ and $\mathbf{x}$ are jointly Gaussian with covariance matrix $$\boldsymbol\Sigma = \begin{bmatrix} \boldsymbol\Theta\boldsymbol\Sigma_x\boldsymbol\Theta^{T}+\boldsymbol\Sigma_\eta & \boldsymbol\Theta\boldsymbol\Sigma_x \\ \boldsymbol\Sigma_x\boldsymbol\Theta^{T} & \boldsymbol\Sigma_x \end{bmatrix}\text{.}$$ It is clear that $\mathbf{y} \sim \mathcal{N}_k(\boldsymbol\Theta\boldsymbol\mu_x,\boldsymbol\Theta\boldsymbol\Sigma_x\boldsymbol\Theta^{T}+\boldsymbol\Sigma_\eta)$. This explains the top-left diagonal of $\boldsymbol\Sigma$. The $\boldsymbol\Theta\boldsymbol\Sigma_x$, as well as the $\boldsymbol\Sigma_x\boldsymbol\Theta^{T}$, are easy to explain as well. However, I'm not sure how to show that $\mathbf{y}$ and $\mathbf{x}$ are jointly Gaussian. In particular, I considered looking at the distribution of $\mathbf{y} \mid \mathbf{x}$, but using this assumes bivariate normality of $(\mathbf{x}, \mathbf{y})$ to start. A: It's simply an application of linear algebra to start working on with $x$ and $y$ jointly (no need to study the conditional distribution). Since we are interested in looking for the joint distribution of $[y', x']'$, it is natural by condition to express it as \begin{align} \begin{bmatrix} y \\ x \end{bmatrix} = \begin{bmatrix} \Theta x + \eta \\ x \end{bmatrix} = \begin{bmatrix} \Theta & I \\ I & 0 \end{bmatrix} \begin{bmatrix} x \\ \eta \end{bmatrix}. \tag{1} \end{align} Since $x$ and $\eta$ are independent normal vectors respectively, it follows that \begin{align} \begin{bmatrix} x \\ \eta \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ 0 \end{bmatrix}, \begin{bmatrix} \Sigma_x & 0 \\ 0 & \Sigma_\eta \end{bmatrix}\right). \end{align} Using the well-known affine transformation property of normal random vector, in view of $(1)$, $[y', x']'$ has multivariate normal distribution with the mean vector $$\begin{bmatrix} \Theta & I \\ I & 0 \end{bmatrix} \begin{bmatrix} \mu_x \\ 0 \end{bmatrix} = \begin{bmatrix} \Theta \mu_x \\ \mu_x \end{bmatrix}$$ and the covariance matrix $$\begin{bmatrix} \Theta & I \\ I & 0 \end{bmatrix} \begin{bmatrix} \Sigma_x & 0 \\ 0 & \Sigma_\eta \end{bmatrix} \begin{bmatrix} \Theta & I \\ I & 0 \end{bmatrix}' = \begin{bmatrix} \Theta\Sigma_x & \Sigma_\eta \\ \Sigma_x & 0 \end{bmatrix} \begin{bmatrix} \Theta' & I \\ I & 0 \end{bmatrix} = \begin{bmatrix} \Theta\Sigma_x\Theta' + \Sigma_\eta & \Theta\Sigma_x \\ \Sigma_x\Theta' & \Sigma_x \end{bmatrix}.$$ This completes the proof. | Mid | [
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Effect of superficial harrowing on surface properties of sand with rubber and waxed-sand with fibre riding arena surfaces: a preliminary study. A recent epidemiological study identified various aspects of arena surfaces and arena surface maintenance that were related to risk of injury in horses and that arena maintenance is important in reducing injury risk. However, there has been little research into how properties of arena surfaces change with harrowing. This study aimed to compare the properties of different arena surface types pre- and post-harrowing. The Orono Biomechanical Surface Tester fitted with accelerometers and a single- and a three-axis load cell was used to test 11 arenas with two different surfaces types, sand with rubber (SR) and waxed-sand with fibre (WSF). Three drop tests were carried out at 10 standardised locations on each arena. Mixed models were created to assess the effect of surface type, pre- or post-harrowing, and drop number on the properties of the surface, including maximum horizontal deceleration, maximum vertical deceleration, maximum vertical load and maximum horizontal load. Post-harrowing, none of the parameters were altered significantly on SR. On WSF, maximum vertical deceleration and maximum vertical load significantly decreased post-harrowing. The differences in the effects of superficial harrowing on SR and WSF could be attributed to the different compositions and sizes of the surface material. The results suggest that different maintenance techniques may be more suitable for different surface types and that the effects of superficial harrowing are short-lived due to the rapid re-compaction of the surface with repeated drops on WSF. Further work is required to determine the effects of other maintenance techniques, and on other surface types. | Mid | [
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470 EUROPEAN CHAMPIONSHIP HEADLINES IN MONACO 07 May 2017 The week ahead in Monaco sees the sailing focus on the 2017 470 European Championship hosted at Yacht Club de Monaco. The 2017 470 Europeans marks the first time the Yacht Club de Monaco (YCM) has ever hosted an Olympic Class Championship, and there is no doubt the organizers are determined to put on an impressive experience for all. An international fleet has descended at this first 470 Championship of the season, with 178 sailors representing 25 nations in the line-up. Monaco is a glamorous venue for the Championship, combining the stunning superyacht marina backdrop to the 470 boat park, with racing unfolding in front of the world famous Monte Carlo landscape. Italy’s Sveva Carraro summed up the ambience perfectly, saying, “These are very big yachts and our 470 is probably the smallest here in the harbour, so it’s quite strange for us. But we have felt at home since we arrived, it is something really nice here.” Saturday night’s Opening Ceremony, hosted by Bernard d’Alessandri, CEO of YCM, together with board members Ernst Klaus and Alain Ucari, reflected the spirit and motto of the YCM, “One Spirit, One Team, One Club”. After the presentation of nations, oaths were made on behalf of the Sailors by Cheryl Teo (SIN) and the Race Officials by Patrice Clerc (MON), before enjoying canapes and cocktails. Who to Watch Favourites in the 470 Men have to be Australia’s Mat Belcher/Will Ryan who swept the board at the recently held SWC Series Round 2 – Hyères. Fresh from a post-Olympic six month break, the pair will be looking to defend their 470 Open European title. Earlier this month, Belcher and Ryan’s long held world #1 ranking transferred to the hands of Sweden’s Carl Fredrik Fock/Marcus Dackhammar, who have shown a massive escalation in performance since launching their full-time campaign for Tokyo 2020. “It felt good,” smiled Fock on achieving their career best world #1 ranking. “It’s a step forwards for us this year and it’s nice that we are there. We are really happy. There is a little bit of pressure, but I think we can handle it. “We did a lot of training in Rio with the other Swedish team and I think it was really good for us,” he added on their progress in the fleet. “We improved a lot last year and the hard work has really paid off this year. Commenting on their strategy at the Europeans, Fock explained, “We want to keep it simple in the beginning and really focus on the small details.” The world #4 ranked pair, Greece’s Panagiotis Mantis/Pavlos Kagialis are always a force to be reckoned with, as are the #2 partnership of Jordi Xammar/Nicolas Rodrigues. Last year’s European Champions were Sofian Bouvet/Jeremie Mion, with half of the team returning to defend their title. Rio 2016 Olympian Mion has partnered up with Guillaume Piroulle, and they have every ability to mount a credible bid for the Championship. Amongst the newcomers to the 470 fleet are Portugal’s Costa brothers, Diogo and Pedro, who won the 2016 420 World Championships, and now make their debut appearance at a 470 Championship. Since transitioning to the 470 Class, their intensive training schedule has paid dividends, finishing 12th in a fleet packed full of Olympians at this year’s Trofeo Princesa Sofia. They have certainly put the 470 fleet on notice of their future potential. In the 470 Women, the world #1 partnership of Afrodite Zegers/Anneloes van Veen (NED) have proved unbeatable since returning to post-Rio 2016 competition in January 2017. They have claimed five back to back wins, securing victories including the North Americans, SWC Series Round 1 – Miami, Trofeo Princesa Sofía and SWC Series Round 2 – Hyères. The closest they have previously come to victory at the European Championships was a 5th place last year, and the podium looks certain to feature their names. The 470 Women’s fleet features six other Rio 2016 teams, so plenty of experience in the ranks. With a new Olympic quadrennial, the next generation of talent moves on up. Italy’s Elena Berta finished 19th in Rio, and returns to the campaign trail with Sveva Carraro. The pair have raced two elite events so far this year, securing 5th place finishes in each. “We have been very pleased with our recent results,” reflected Berta. “We hope to achieve the medal race here, but we can’t yet measure ourselves as there are many new teams. After the second day of racing we will have a clearer idea.” Hoping to become Malaysia’s first ever female sailors to qualify to the Olympic Games in the 470 are Nuraisyah Jamil/Ashikin Sayed. Since pairing up in the 470, successes include a gold medal at the 2015 South-East Asian Games. They missed out on qualification to Rio 2016 and resume their mission for Tokyo 2020. Never too Late Proving you are never too old to sail a 470 are Greece’s Christos Mamtoumidis/ Giorgios Doulas. Sixty-one year old Mamtoumidis is a lifelong 470 sailor, whose past form rewarded him with three Greek National Championship titles. Racing with Giorgios Doulas, the pair came to Monaco straight from Bourgas Sailing Week in Bulgaria, where they finished 5th out of an 8 boat 470 fleet. Reflecting on his time in the 470, Mamtoumidis said, “The first time the 470 Worlds were held in Thessaloniki in 1974 I raced there. And in 1981, 1982 and 1985 I won the Greek Nationals and then raced at 470 events around the world after that. Last year I competed at the 470 Masters Cup and we won the gold medal in the Grand-grand masters. And now we are here!” Expectations are modest, as Mamtoumidis laughed, “I expect we will come last!” A diverse fleet, newcomers and Olympians, young and those in more senior years, will make for an interesting leaderboard mix. The forecast indicates a predominantly lighter breeze regatta, and the renowned unpredictability of the conditions here on the Bay of Monaco are sure to heighten the race track game ABOUT THE 2017 470 OPEN EUROPEAN CHAMPIONSHIP The 2017 470 Open European Championship is organized by Yacht Club de Monaco in co-operation with the International 470 Class Association. Racing starts on Monday 8 May, with an 11 race series scheduled for the 470 Men and 470 Women fleets, before the top 10 teams advance to the final Medal Race on Saturday 13 May 2017. Titles will be awarded in the 470 Men and Women for the overall Open European Championship leaderboard and medals will be awarded in the European Championship for European nations. Video – ICARUS Sports is the official TV production and distribution company for the Championship. They will produce daily highlights videos for social media and websites, and a selection of raw footage. Sailor interviews for TV and Olympic teams may be available on request, subject to priority and time commitments. Please send requests to [email protected] and [email protected] Videos will be available to download from the following server options: | Mid | [
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<?php /** * @author Adam Charron <[email protected]> * @copyright 2009-2019 Vanilla Forums Inc. * @license GPL-2.0-only */ namespace VanillaTests\Forum\EmbeddedContent\Factories; use Garden\Web\RequestInterface; use Vanilla\Contracts\Site\SiteSectionInterface; use Vanilla\Forum\EmbeddedContent\Factories\DiscussionEmbedFactory; use Vanilla\Site\DefaultSiteSection; use Vanilla\Site\SiteSectionModel; use VanillaTests\APIv2\AbstractAPIv2Test; use VanillaTests\Fixtures\MockConfig; use VanillaTests\Fixtures\MockSiteSection; use VanillaTests\Fixtures\MockSiteSectionProvider; /** * Tests for the discussion/quote embed. */ class DiscussionEmbedFactoryTest extends AbstractAPIv2Test { /** * Test that all domain types are supported. * * @param string $urlToTest * @param bool $isSupported * @param string $customRoot * @param SiteSectionInterface[] $siteSections * @dataProvider supportedDomainsProvider */ public function testSupportedDomains(string $urlToTest, bool $isSupported, string $customRoot = '', array $siteSections = []) { $discussionApi = $this->createMock(\DiscussionsApiController::class); /** @var RequestInterface $request */ $request = self::container()->get(RequestInterface::class); $request->setAssetRoot($customRoot); $sectionProvider = new MockSiteSectionProvider(new DefaultSiteSection(new MockConfig(), new \Gdn_Router())); $sectionProvider->addSiteSections($siteSections); $sectionModel = new SiteSectionModel(new MockConfig(), new \Gdn_Router()); $sectionModel->addProvider($sectionProvider); $factory = new DiscussionEmbedFactory($request, $sectionModel, $discussionApi); $this->assertEquals($isSupported, $factory->canHandleUrl($urlToTest)); } /** * @return array */ public function supportedDomainsProvider(): array { $bootstrapBase = 'http://vanilla.test'; return [ // Allowed 'Correct' => [ $bootstrapBase . '/discussion/41342', true ], // Not allowed 'Correct webroot' => [ $bootstrapBase . '/actual-root/discussion/41342', true, '/actual-root' ], 'Correct section' => [ $bootstrapBase . '/actual-root/actual-section/discussion/41342', true, '/actual-root', [new MockSiteSection('test', 'en', '/actual-section', 'test1', 'test1')] ], 'Correct section w/ regex character in it' => [ $bootstrapBase . '/actual-root/regex!^$-()-section/discussion/41342', true, '/actual-root', [new MockSiteSection('test', 'en', '/regex!^$-()-section', 'test1', 'test1')] ], // Not allowed 'Wrong webroot' => [ $bootstrapBase . '/wrong-root/discussion/41342', false, '/actual-root' ], 'Wrong section' => [ $bootstrapBase . '/actual-root/actual-section/discussion/41342', false, '/actual-root', ], 'wrong host' => [ 'https://otherdomain.com/discussion/41342', false ], 'wrong url (typo)' => [ $bootstrapBase . '/discussions/41342', false ], 'Wrong url (is a comment)' => [ $bootstrapBase . '/discussion/comment/41342', false ], 'bad ID' => [ $bootstrapBase . '/discussion/asdfads', false ], ]; } } | Low | [
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// RUN: %clang_cc1 -std=c++98 -emit-llvm %s -o - -triple=x86_64-apple-darwin10 | FileCheck %s // RUN: %clang_cc1 -std=c++11 -emit-llvm %s -o - -triple=x86_64-apple-darwin10 | FileCheck %s struct S { virtual ~S() { } }; // PR5706 // Make sure this doesn't crash; the mangling doesn't matter because the name // doesn't have linkage. static struct : S { } obj8; void f() { // Make sure this doesn't crash; the mangling doesn't matter because the // generated vtable/etc. aren't modifiable (although it would be nice for // codesize to make it consistent inside inline functions). static struct : S { } obj8; } inline int f2() { // FIXME: We don't mangle the names of a or x correctly! static struct { int a() { static int x; return ++x; } } obj; return obj.a(); } int f3() { return f2(); } struct A { typedef struct { int x; } *ptr; ptr m; int a() { static struct x { // FIXME: We don't mangle the names of a or x correctly! int a(ptr A::*memp) { static int x; return ++x; } } a; return a.a(&A::m); } }; int f4() { return A().a(); } int f5() { static union { int a; }; // CHECK: _ZZ2f5vE1a return a; } #if __cplusplus <= 199711L int f6() { static union { union { int : 1; }; int b; }; // CXX98: _ZZ2f6vE1b return b; } #endif int f7() { static union { union { int b; } a; }; // CHECK: _ZZ2f7vE1a return a.b; } // This used to cause an assert because the typedef-for-anonymous-tag // code was trying to claim the enum for the template. enum { T8 }; template <class T> struct Test8 { typedef T type; Test8(type t) {} // tested later }; template <class T> void make_test8(T value) { Test8<T> t(value); } void test8() { make_test8(T8); } // CHECK-LABEL: define internal void @"_ZNV3$_85test9Ev"( typedef volatile struct { void test9() volatile {} } Test9; void test9() { Test9 a; a.test9(); } // CHECK-LABEL: define internal void @"_ZN5Test8I3$_7EC1ES0_"( | Low | [
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dnl Intel P6 mpn_mod_34lsub1 -- remainder modulo 2^24-1. dnl Copyright 2000, 2001, 2002, 2004 Free Software Foundation, Inc. dnl dnl This file is part of the GNU MP Library. dnl dnl The GNU MP Library is free software; you can redistribute it and/or dnl modify it under the terms of the GNU Lesser General Public License as dnl published by the Free Software Foundation; either version 2.1 of the dnl License, or (at your option) any later version. dnl dnl The GNU MP Library is distributed in the hope that it will be useful, dnl but WITHOUT ANY WARRANTY; without even the implied warranty of dnl MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU dnl Lesser General Public License for more details. dnl dnl You should have received a copy of the GNU Lesser General Public dnl License along with the GNU MP Library; see the file COPYING.LIB. If dnl not, write to the Free Software Foundation, Inc., 51 Franklin Street, dnl Fifth Floor, Boston, MA 02110-1301, USA. include(`../config.m4') C P6: 2.0 cycles/limb C mp_limb_t mpn_mod_34lsub1 (mp_srcptr src, mp_size_t size) C C Groups of three limbs are handled, with carry bits from 0mod3 into 1mod3 C into 2mod3, but at that point going into a separate carries total so we C don't keep the carry flag live across the loop control. Avoiding decl C lets us get to 2.0 c/l, as compared to the generic x86 code at 3.66. C defframe(PARAM_SIZE, 8) defframe(PARAM_SRC, 4) dnl re-use parameter space define(SAVE_EBX, `PARAM_SIZE') define(SAVE_ESI, `PARAM_SRC') TEXT ALIGN(16) PROLOGUE(mpn_mod_34lsub1) deflit(`FRAME',0) movl PARAM_SIZE, %ecx movl PARAM_SRC, %edx subl $2, %ecx C size-2 movl (%edx), %eax C src[0] ja L(three_or_more) jb L(one) C size==2 movl 4(%edx), %ecx C src[1] movl %eax, %edx C src[0] shrl $24, %eax C src[0] high andl $0xFFFFFF, %edx C src[0] low addl %edx, %eax movl %ecx, %edx C src[1] shrl $16, %ecx C src[1] high andl $0xFFFF, %edx addl %ecx, %eax shll $8, %edx C src[1] low addl %edx, %eax L(one): ret L(three_or_more): C eax src[0], initial acc 0mod3 C ebx C ecx size-2 C edx src C esi C edi C ebp movl %ebx, SAVE_EBX movl 4(%edx), %ebx C src[1], initial 1mod3 subl $3, %ecx C size-5 movl %esi, SAVE_ESI movl 8(%edx), %esi C src[2], initial 2mod3 pushl %edi FRAME_pushl() movl $0, %edi C initial carries 0mod3 jng L(done) C if size < 6 L(top): C eax acc 0mod3 C ebx acc 1mod3 C ecx counter, limbs C edx src C esi acc 2mod3 C edi carrys into 0mod3 C ebp addl 12(%edx), %eax adcl 16(%edx), %ebx adcl 20(%edx), %esi leal 12(%edx), %edx adcl $0, %edi subl $3, %ecx jg L(top) C at least 3 more to process L(done): C ecx is -2, -1 or 0 representing 0, 1 or 2 more limbs respectively cmpl $-1, %ecx jl L(done_0) C if -2, meaning 0 more limbs C 1 or 2 more limbs movl $0, %ecx je L(done_1) C if -1, meaning 1 more limb only movl 16(%edx), %ecx L(done_1): addl 12(%edx), %eax C 0mod3 adcl %ecx, %ebx C 1mod3 adcl $0, %esi C 2mod3 adcl $0, %edi C carries 0mod3 L(done_0): C eax acc 0mod3 C ebx acc 1mod3 C ecx C edx C esi acc 2mod3 C edi carries 0mod3 C ebp movl %eax, %ecx C 0mod3 shrl $24, %eax C 0mod3 high initial total andl $0xFFFFFF, %ecx C 0mod3 low movl %edi, %edx C carries shrl $24, %edi C carries high addl %ecx, %eax C add 0mod3 low andl $0xFFFFFF, %edx C carries 0mod3 low movl %ebx, %ecx C 1mod3 shrl $16, %ebx C 1mod3 high addl %edi, %eax C add carries high addl %edx, %eax C add carries 0mod3 low andl $0xFFFF, %ecx C 1mod3 low mask addl %ebx, %eax C add 1mod3 high movl SAVE_EBX, %ebx shll $8, %ecx C 1mod3 low movl %esi, %edx C 2mod3 popl %edi FRAME_popl() shrl $8, %esi C 2mod3 high andl $0xFF, %edx C 2mod3 low mask addl %ecx, %eax C add 1mod3 low shll $16, %edx C 2mod3 low addl %esi, %eax C add 2mod3 high movl SAVE_ESI, %esi addl %edx, %eax C add 2mod3 low ret EPILOGUE() | Low | [
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Zirconium—cerium-based mixed oxides as promoters or catalyst supports are generally known to perform better than zirconia or ceria alone. A number of processes have been proposed for the preparation of such zirconium—cerium-based oxides: for example, a sol process which comprises mixing zirconium sol and cerium sol, adding alkali to the mixed sols to form precipitates, and calcining the precipitates [Japan Kokai Tokkyo Koho Hei 6-279,027 (1994)]; a process which comprises heating the particles of zirconium hydroxide with cerium sol in the presence of nitric acid to effect dissolution and reprecipitation, adding alkali, allowing the mixture to react further, and calcining the product followed by pulverising [Japan Kokai Tokkyo Koho Hei 10-194,742 (1998)]; and a process which comprises adding oxalic acid to an aqueous solution of acidic salts of zirconium and cerium to precipitate zirconium—cerium oxalate, thermally decomposing the resulting oxalate in a non-oxidising atmosphere and then heating the decomposed product in an oxidising atmosphere [Japan Kokai Tokkyo Koho Hei 11-165,067 (1999)]. The sol process has an advantage of yielding zirconium—cerium-based mixed oxides with a relatively large specific surface area, but it faces problems such as the necessity for advance preparation of sol which is a disadvantage in respect to cost and the dried and calcined product being hard and easy to aggregate. The process which involves heating in the presence of nitric acid is advantageous in that the use of the particles of zirconium hydroxide keeps the product from aggregating and solidifying. However, a single crystal phase becomes difficult to obtain and the range of composition which can be prepared becomes narrow when the amount of CeO2 is increased relative to that of ZrO2. The process involving thermal decomposition of the oxalate has an advantage of readily yielding zirconium—cerium-based mixed oxides of single crystal phase because of the formation of the oxalate by coprecipitation. This process, however, requires a heat treatment at elevated temperatures in order to conduct the thermal decomposition sufficiently which causes problems such as a decrease in specific surface area and an increase in calcining cost. Japan Kokai Tokkyo Koho Hei 10-212,122 (1998) proposes fine zirconia-ceria particles and a process for preparing the same: the ZrO2 particles by themselves are primary particles with a BET specific surface area of 40-200 m2/g, an average particle diameter determined by electron microscopy of 0.1 μm or less and a ratio of the average particle diameter determined by electron microscopy to the average particle diameter determined from BET specific surface area of 0.9 or more and the molar ratio CeO2/ZrO2 is 5/95-60/40; the fine zirconia-ceria particles show high oxygen supply efficiency in exhaust gas, adsorb or release oxygen well at low exhaust gas temperatures and can be mixed homogeneously with three-way catalysts for purifying automotive exhaust gas. The process of preparation described in the specification of the aforementioned patent, however, requires a long period of boiling, occasionally extending over several hundred hours, to effect the hydrolysis of an aqueous solution of zirconium salt to hydrated zirconia sol with an average particle diameter of 0.1 μm or less. Moreover, the hydrated zirconia sol thus obtained is too fine to permit the application of usual industrial procedures for solid-liquid separation such as filtration under reduced pressure, filtration under pressure and centrifugal separation. In consequence, a troublesome procedure such as sedimentation and separation of supernatant liquid would be required in the steps for filtration and water washing of the hydrated zirconia sol. An operation such as aggregation, if carried out during these steps, causes a sharp decrease in specific surface area. Furthermore, the specific surface area tends to diminish rapidly when calcination is effected at an elevated temperature close to the working temperature. A process for efficiently preparing thermally durable zirconium—cerium-based mixed oxides is proposed in Japan Kokai Tokkyo Koho Hei 11-292,539 (1999): the process comprises dispersing basic zirconium sulphate in water, mixing the dispersion with a solution containing cerium ions such as a solution of cerium nitrate, adding alkali to the mixture to yield hydroxides, effecting solid-liquid separation of the hydroxides and calcining the hydroxides. This process utilises basic zirconium sulphate with an average particle diameter of 0.5˜20 μm and hence has an advantage of yielding zirconium—cerium-based mixed oxides with a relatively large specific surface area at elevated temperatures, for example, 100 m2/g or more at 400° C. and 30 m2/g or more at 1,000° C. However, basic zirconium sulphate with an average particle diameter of 0.5˜20 μm must be prepared in advance, which requires an extra manufacturing step with a concomitant rise in cost. Besides, zirconium—cerium-based mixed oxides of single crystal phase become difficult to obtain as the addition of ceria and a third component oxide increases. Hence, the desired performance becomes difficult to obtain and, as a result, the performance of the mixed oxides as a catalyst/promoter deteriorates. A further process of coprecipitation by means of ammonia or ammonium carbonate or the like, starting from a mixed solution of zirconium nitrate and cerium nitrate is also known (Japan Kokai Tokkyo Koho Hei 9-278444). However, the precipitate obtained by this process is a bulky mixed hydroxide in the form of a gel with a high water content; therefore productivity is poor and can hardly be regarded as suited to industrial scale production. In addition it states that it is necessary to have the cerium salt in the tetravalent state, which is difficult to control and is not necessary in the present invention. Thus, a filtration process is essential in order to remove impurities from the gel precipitate, and the bulkiness of the precipitate means that unit treatment speed is also invariably slow. Moreover, the high water content increases the energy needed in order to convert it to the oxides. The use of sulphate as a precipitation modifier has been used in Japan Kokai Tokkyo Koho Hei 8-34613 and 8-34614 (1996). These, together with Japan Kokai Tokkyo Koho Hei 8-34612 (1996), describe the production of yttria-doped zirconias. In 8-34612 and 8-34613 hydrogen peroxide is added to the zirconium salt as a masking agent in order to bring the pH's of precipitation of the zirconium and yttrium salts closer together to allow homogeneous precipitation. Sulphate is added to modify the precipitation in 8-34613 and 8-34614. In the former urea is used as the precipitant and in the latter ammonia, but in three of these cases the use of alkali metal hydroxides is proscribed. Furthermore, the use of ammonia is to be deprecated because of its adverse environmental effects. Accordingly, an object of this invention is to provide a process, easily practicable on a commercial scale, for preparing zirconium—cerium-based mixed oxides which not only possess good thermal stability at elevated temperatures but are also highly homogeneous in their crystal phase. | Mid | [
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Jean-Léonard Touadi Jean-Léonard Touadi (born 25 January 1959 in Brazzaville, Republic of the Congo) is a Congolese Italian journalist, author and politician. Touadi was raised in France and immigrated to Italy in 1979. There, he rose to prominence as a television journalist and as Rome's deputy mayor in charge of security. In the 2008 general election he was elected to the Italian Parliament with Italy of Values (IdV), becoming Italy's second black Member of Parliament (MP) and the first MP from sub-Saharan Africa. On 11 July 2008, few months after his election, he left the IdV and joined the Democratic Party (PD). He is married to Cristina Bacillieri and they have two daughters: Sophie-Claire (2005) and Sandrine (2007). Notes References Profile at Italian Chamber of Deputies http://www.italymag.co.uk/italy/politics/italy-gets-third-black-mp http://www.adnkronos.com/AKI/English/Politics/?id=1.0.2074789726 Category:1959 births Category:Living people Category:People from Brazzaville Category:Italian people of Republic of the Congo descent Category:Democratic Party (Italy) politicians Category:21st-century Italian politicians Category:Italy of Values politicians | Mid | [
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Q: macOS documentation for structs in Security.h I'm trying to use the Security.h macOS framework via Java and JNA. That means I need to reconstruct certain structs as Java classes. The problem is, when I look at the docs for a struct (this one, for example), all I see is a brief description of the struct without any mention of its fields. Where can I get a full description of a struct in Apple's documentation? A: For a quick look, you can find the headers on Apple's open source site, but it's difficult to navigate, especially as the headers are under different locations depending on the version of the OS you want to check. In all cases I've found it's defined in SecBase.h. For example here is the one for latest macOS. And there you get this: typedef struct CF_BRIDGED_TYPE(id) SECTYPE(SecKeychainItem) *SecKeychainItemRef; So you'll probably need other headers to track down the exact fields of the struct. A better way to do that would be to install XCode with frameworks for the OS you want, and you'll get the headers on your local system. For example: $ ls /Applications/Xcode.app/Contents/Developer/Platforms/*.platform/Developer/SDKs/*.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/AppleTVOS.platform/Developer/SDKs/AppleTVOS.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/AppleTVOS.platform/Developer/SDKs/AppleTVOS9.2.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/AppleTVSimulator.platform/Developer/SDKs/AppleTVSimulator.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/AppleTVSimulator.platform/Developer/SDKs/AppleTVSimulator9.2.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.11.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/WatchOS.platform/Developer/SDKs/WatchOS.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/WatchOS.platform/Developer/SDKs/WatchOS2.2.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/WatchSimulator.platform/Developer/SDKs/WatchSimulator.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/WatchSimulator.platform/Developer/SDKs/WatchSimulator2.2.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS9.3.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs/iPhoneSimulator.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h /Applications/Xcode.app/Contents/Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs/iPhoneSimulator9.3.sdk/System/Library/Frameworks/Security.framework/Headers/SecBase.h I don't have a good solution with the online docs, though. Another way, from memory it was more helpful for the task of recreating the struct in Java for JNA, would be to build a minimal C program (But I'm not sure how to do that on macOS, linking with Security framework, perhaps you do), and give it to gdb to print the structure layout using ptype: (gdb) whatis v type = struct complex (gdb) ptype v type = struct complex { double real; double imag; } But as noted in comments, if we try this here, we get this: (gdb) ptype SecKeychainItemRef type = struct OpaqueSecKeychainItemRef { <incomplete type> } I'm afraid this symbol is voluntarily made opaque... Confirmed by Brendan in comments: every macOS type I can think of that ends in Ref is an opaque type (really a pointer), only meant to be passed to functions Here is a debug session with Xcode: | Mid | [
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Q: Tomcat CentOS log4j2 На удалённом сервере CentOS по какой-то причине не создаётся папка с логами, и не пишутся сами логи. На локальной машине Win10, всё работает. На удалёнке перед томкатом стоит nginx, есть ли какие-то идеи? Вот конфигруационный файл log4j2: <?xml version="1.0" encoding="UTF-8"?> <Configuration status="WARN" monitorInterval="30"> <Properties> <Property name="log_dir">LOG</Property> <Property name="name_file">${date:yyyy_MM_dd}</Property> </Properties> <Appenders> <Console name="Console" target="SYSTEM_OUT"> <PatternLayout pattern="[%d{MM-dd HH:mm:ss,SSS}] %-5p %-40.40c %x : %m%n"/> </Console> <RollingFile name="RollingFile" fileName="${log_dir}/${name_file}.log" filePattern="${log_dir}/server-%d{yyyy-MM-dd}-%i.log"> <PatternLayout> <pattern>[%d{MM-dd HH:mm:ss,SSS}, %r] %c %m%n</pattern> </PatternLayout> <Policies> <SizeBasedTriggeringPolicy size="50MB"/> </Policies> <DefaultRolloverStrategy max="20"/> </RollingFile> <Async name="AsyncFile"> <AppenderRef ref="RollingFile"/> </Async> <Async name="AsyncConsole"> <AppenderRef ref="Console"/> </Async> </Appenders> <Loggers> <Root level="warn"> <AppenderRef ref="AsyncFile"/> <AppenderRef ref="AsyncConsole"/> </Root> </Loggers> </Configuration> A: Cоздайте папку <Property name="log_dir">LOG</Property> руками и сделайте ее доступной для записи пользователю, под которым бежит Tomcat. Или что-то вроде этого из под рута: chgrp -R tomcat7 /etc/tomcat7 chmod -R g+w /etc/tomcat7 /etc/tomcat7 - это должен путь, куда у вас томкэт установлен. Можно еще ${catalina.base} использовать как префикс пути в <Property name="log_dir">, что бы не гадать, куда это все будет писаться. | Mid | [
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THE Seattle City Council approved regulations that cover the Seattle Police Department’s use of unmanned aircraft systems. But the department has to start all over again, under the new rules. That much has apparently not changed. Last week the SPD secured a last-minute revision of pending City Council legislation that laid out the operating conditions for the use of surveillance technology, including drones. The unanimously approved council bill allows the police to use drones under three sets of conditions: when they have a warrant to do so; under certain “exigent” emergency circumstances; and in the course of a criminal investigation when the courts would not require a warrant for specified kinds of surveillance in public spaces. Seattle Police Chief John Diaz asked the council for the exemptions. He noted a council requirement to always secure a warrant could create an impediment to investigations because the courts are not inclined to issue warrants when they are not needed. The sought-after language was included in the council bill, but its last-minute inclusion offered the public virtually no opportunity to comment during a brief public hearing. Acquisition and use of drones by the police department will start over. Last month, Mayor Mike McGinn directed the department to return two drones to the vendor. That order still stands. The department also will have to start over in pursuit of grants and acquisition. Councilmember Tim Burgess said the council is also interested in learning about and screening where the money comes from. Civilian use of drones in the coming years is expected to flourish across the country, and that is raising privacy issues. The Federal Aviation Administration is looking at creating six drone test sites around the country. The FAA has said all applicants must follow federal and state privacy laws. The Washington State Legislature contemplated privacy legislation but dropped it over objections by Boeing, which wants to build the unmanned systems. Drones were the subject of a U.S. Senate Judiciary Committee hearing last week. The concerns that echo from Seattle City Hall were repeated in Congress by the committee chairman, Sen. Patrick Leahy, D-Vt.: “It is not hard to imagine the serious privacy problems that this type of technology could cause.” Seattle police need to start from scratch with a rationale for drone use that fits the need and the precautions. | Mid | [
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The deadline “for fullest consideration” for the position of next president of Seton Hall University was Nov. 22, the first day of Thanksgiving Break, according to the school’s presidential search webpage. However, the search committee has not yet released an update on the status of the search process. Dan Kalmanson, associate vice president for public relations and marketing, relayed a message from a representative of the... The Voice “I was a stranger and you welcomed me.” These are the first words the Seton Hall community saw on Tuesday when it opened a statement from University President Dr. A. Gabriel Esteban. It is a quote from the Bible, but those words could have just as easily come from Esteban himself, a native of the Philippines. Disregard context and those words apply to each and every one of us. Whether it be as a new student at Seton... | Mid | [
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Q: Update google latitude I have built a GPS tracker that updates a homepage with its positions (and webcam images). How can I update a the current location of a google latitude user? A simple bash script invoking curl or a c-program would be nice! Update: I also need to know how to do the authentication. A: When you say you want to "update" a Google Latitude user, you want to "update their current location", right? For some Google services, Google will set up a Google API project on Google Code. In this case, you're in luck, because there is a Google Latitude API that describes the different actions you can do using a REST interface (REST is always compatible with curl). Here's their example code for updating a user's location: POST https://www.googleapis.com/latitude/v1/currentLocation?key=INSERT-YOUR-KEY /* Authorization header here */ Content-Type: application/json { "data": { "kind":"latitude#location", "latitude":37.420352, "longitude":-122.083389, "accuracy":130, "altitude":35 } } The Google Latitude API webside describes the full details. You'll need to get an API key before you can start writing code, and you'll need to do an OATH 2.0 authentication handshake before you can actually update the user's location. Update If you don't want to write the authentication code yourself, Google provides several pre-packaged client libraries, in .NET, GWT, Java, PHP, Python, and Ruby. They each support the full API, including authentication. Google has a full example that uses their Java API to do authentication. Follow the instructions at http://samples.google-api-java-client.googlecode.com/hg/latitude-json-oauth-sample/instructions.html?r=default and try it out. A: Is this what you're looking for? In the API Console, be sure to request access to Latitude under the Services tab. This script will prompt for API Key, Client ID, Client Secret and then launch a browser for login (you may need to tweak that line for your system, see below). Once you're logged in and grant access to your application, you'll get a code you'll paste in when prompted by the script. Then you'll enter your lat/long/elev which will be posted to the service. #!/bin/sh LoginUrl="https://accounts.google.com/o/oauth2/auth" TokenUrl="https://accounts.google.com/o/oauth2/token" RedirectUri="urn:ietf:wg:oauth:2.0:oob" Scope="https://www.googleapis.com/auth/latitude.all.best https://www.googleapis.com/auth/latitude.all.city https://www.googleapis.com/auth/latitude.current.best https://www.googleapis.com/auth/latitude.current.city" CurlocUrl="https://www.googleapis.com/latitude/v1/currentLocation" read -s -p "Enter your API Key: " APIKey echo "" read -s -p "Enter your Client ID: " ClientId echo "" read -s -p "Enter your Client Secret: " ClientSecret echo "" # this next line may need to be tweaked in order to launch the browser open "${LoginUrl}?response_type=code&client_id=${ClientId}&redirect_uri=${RedirectUri}&scope=${Scope}" read -s -p "Log in, grant permission, enter the code: " Code echo "" resp=`curl -is "${TokenUrl}" -d "code=${Code}&client_id=${ClientId}&client_secret=${ClientSecret}&redirect_uri=${RedirectUri}&grant_type=authorization_code"` AccessToken=`echo "${resp}" | sed -e '/access_token/ !d; s/ *"access_token" *: *"\(.*\)",*/\1/'` TokenType=`echo "${resp}" | sed -e '/token_type/ !d; s/ *"token_type" *: *"\(.*\)",*/\1/'` ExpiresIn=`echo "${resp}" | sed -e '/expires_in/ !d; s/ *"expires_in" *: *"\(.*\)",*/\1/'` RefreshToken=`echo "${resp}" | sed -e '/refresh_token/ !d; s/ *"refresh_token" *: *"\(.*\)",*/\1/'` echo "Enter the location details." read -p "Latitude in degrees (nn.nnnn): " latitude read -p "Longitude in degrees (nn.nnnn): " longitude read -p "Elevation in feed (nnnn): " altitude curl -is "${CurlocUrl}" -H "Content-Type: application/json" -H "Authorization: OAuth ${AccessToken}" -d "{ 'data': { 'kind': 'latitude#location', 'latitude': '${latitude}', 'longitude': '${longitude}', 'accuracy': 0, 'altitude': ${altitude} } }" | Mid | [
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Q: How to alternate row style without javascript for DisplayFor/EditorFor bound to a list of model How can I pass the index of the current element to the View for DisplayFor/EditorFor so that I can decide if the row should show the alternate style or not? My main view looks like this: <table> @Html.EditorFor(model => model.MyListOfItems) </table> The view used for the EditorFor looks like: @Html.HiddenFor(model => model.IdOfTheItem) <tr class="shouldBeChangedDependingOnRowEvenOrNot"> <td>@Html.CheckBoxFor(model => model.MarkForBatchEdit)</td> <td>@Html.DisplayFor(model => model.NameOfThisItem)</td> <td>@Html.DisplayFor(model => model.StateOfThisItem)</td> </tr> Well I am aware of this similar question and the suggested solution: alternating row color MVC But I can't apply them to this case. Any suggestions? A: The EditorFor is overloaded to take an additionalViewData parameter, so you could pass the index in the ViewData, which is a collection of key/value pairs. @Html.EditorFor( model => model.MyListOfItems , new { CurrentIndex = SomeNumber } ) In your view you would get the value using ViewData["CurrentIndex"]. Also, instead of passing the element index, why not do the calculation in your controller and pass whether you have an even or odd row in your ViewData. bool isEvenRow = ((CurrentElementIndex % 2) == 0); ViewData["isEvenRow"] = isEvenRow; Then you will just toggle your CSS in the view based on whether the value is true or false. | Mid | [
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66 F.3d 8 150 L.R.R.M. (BNA) 2346, 130 Lab.Cas. P 11,414 Thomas QUESNEL, Plaintiff-Appellant,v.PRUDENTIAL INSURANCE COMPANY, Defendant-Appellee. No. 95-1178. United States Court of Appeals,First Circuit. Heard Aug. 2, 1995.Decided Sept. 25, 1995. John F. Moriarty, Jr., with whom Moriarty & Neves was on brief, for appellant. Burton J. Fishman, with whom Tucker, Flyer & Lewis and Kevin Patrick Reilly were on brief, for appellee. Before TORRUELLA, Chief Judge, LYNCH, Circuit Judge, and CASELLAS,* District Judge. TORRUELLA, Chief Judge. 1 Plaintiff-appellant Thomas Quesnel challenges the district court's dismissal of his wrongful termination action, originally brought in state court, against his former employer, Prudential Insurance Company ("Prudential"). The district court found that Quesnel's claim necessitated analysis of the collective bargaining agreement binding the parties, and accordingly held the claim to be preempted by federal labor law. For the following reasons, we affirm. BACKGROUND 2 Quesnel began his employment at Prudential as a district agent, and became a sales manager in 1991. He later returned to the level of district agent in September of that year. 3 Throughout Quesnel's period of employment, Prudential had a collective bargaining agreement (the "CBA") with the Union of Food and Commercial Workers (the "Union") which covered all "district agents" working for Prudential, regardless of union membership.1 This CBA contained the terms by which Prudential's agents were employed and compensated. The CBA also set forth, inter alia, grievance procedures, which provided for the arbitration of grievances for wrongful termination. Quesnel and Prudential were also parties to a standard Agent's Agreement, which set forth the scope of the agency relationship. 4 Quesnel was terminated in March 1992. He filed this action in Massachusetts state court in May 1994, claiming that Prudential had terminated him for the purpose of denying him his earned commissions, which, under Massachusetts law, is considered a wrongful termination. See Fortune National Cash Register Co., 373 Mass. 96, 104-05, 364 N.E.2d 1251 (1977). Prudential removed the case to the United States District Court for the District of Massachusetts, and moved to dismiss Quesnel's claim on the grounds that it was preempted by federal labor law and that Quesnel had failed to exhaust his administrative remedies available to him under the terms of the CBA. Quesnel responded that he was not a member of the Union and therefore not a party to the CBA. Instead, he argued, his employment relationship with Prudential was controlled by the Agent's Agreement, which, he asserted, was independent of the CBA, and thus his claims were not preempted. Accordingly, Quesnel moved for remand to state court. 5 On February 1, 1995, the district court granted Prudential's motion to dismiss and denied Quesnel's request for remand. The district court ruled that because "no court could begin to address [Quesnel's] claims here without immersing itself in the CBA," Quesnel's state law claims were preempted under principles of federal labor law. The court then dismissed Quesnel's claim because it was governed by the NLRA and because Quesnel was time barred from any recovery.2 DISCUSSION A. Standard of Review 6 Appellate review of a district court's dismissal under Fed.R.Civ.P. 12(b)(6) is plenary. We therefore apply the same standard as did the district court, that " 'a complaint should not be dismissed for failure to state a claim unless it appears beyond doubt that the plaintiff can prove no set of facts in support of his claim which would entitle him to relief.' " Miranda v. Ponce Fed'l Bank, 948 F.2d 41, 44 (1st Cir.1991) (quoting Conley v. Gibson, 355 U.S. 41, 45-46, 78 S.Ct. 99, 102, 2 L.Ed.2d 80 (1957)). B. Preemption of Quesnel's Claim 7 The sole issue before us is whether Quesnel's state law claims are preempted as a matter of law under Sec. 301(a) of the Labor-Management Relations Act, 29 U.S.C. Sec. 185(a).3 It is well-established that Sec. 301 completely preempts a state law claim if the resolution of the claim necessitates analysis of, or substantially depends on the meaning of, a collective bargaining agreement. Lingle v. Norge Division of Magic Chef, Inc., 486 U.S. 399, 405-06, 108 S.Ct. 1877, 1881-82, 100 L.Ed.2d 410 (1988); Allis-Chalmers Corp. v. Lueck, 471 U.S. 202, 220, 105 S.Ct. 1904, 1915-16, 85 L.Ed.2d 206 (1985); Magerer v. John Sexton & Co., 912 F.2d 525, 528 (1st Cir.1990). 8 1. Does Quesnel's claim require interpretation of the CBA? 9 Assuming Quesnel is subject to the CBA, we must determine whether resolution of his claims in the instant case necessitates analysis of, or substantially depends upon the meaning of, the CBA. If so, then his claims must be dismissed as preempted in light of the foregoing principles. Having carefully examined the CBA, we think that the district court correctly found that the CBA is directly implicated in any resolution of Quesnel's claims. The CBA sets forth the terms and scope of the employment relationship of all district agents, encompassing rates of pay, wages, and conditions of employment. Significantly, the CBA sets forth grievance procedures for alleged wrongful termination. Determination of whether Quesnel was indeed wrongfully terminated, and whether his failure to follow grievance procedures set forth in the CBA nonetheless precludes his claim would require a court, as the district court found, to immerse itself in the CBA's terms. Interpretation of the CBA is therefore crucial to any resolution of Quesnel's claim. 10 2. Is Quesnel subject to the CBA? 11 Because we find that resolution of Quesnel's claims require interpretation of the CBA, his claims are preempted if Quesnel is indeed subject to the CBA's terms. Quesnel wisely does not dispute this; rather, he claims that his employment relationship with Prudential was governed not by the CBA, but solely by the Agent's Agreement, and therefore his claims should be adjudicated in state court. At the very least, Quesnel argues, there exists a genuine issue of material fact as to whether he is subject to the CBA.4 12 Whether Quesnel is subject to the CBA is in this case, however, a question of law, not of fact. See Coll v. PB Diagnostic Systems, Inc., 50 F.3d 1115, 1122 (1st Cir.1995) (interpretation of contract is a question of law); Whitney Bros. v. Sprafkin, 3 F.3d 530, 534 (1st Cir.1993) (same). After examining the CBA and the Agent's Agreement, we conclude that it is clear that Quesnel is indeed subject to the CBA's terms. First, the CBA was effective on the date Quesnel became a district agent, and by its terms encompasses "all District Agents employed or hereafter to be employed" by Prudential, including those agents employed in the company's Massachusetts offices. 13 Second, regardless of the fact that Quesnel was not a Union member, he is a member of the bargaining unit for whose benefit the CBA was created. The Union was and is obligated under Sec. 9(a) of the National Labor Relations Act, 29 U.S.C. Sec. 159(a), to represent the interests of all employees in collective bargaining, including nonmembers. See Vaca v. Sipes, 386 U.S. 171, 87 S.Ct. 903, 17 L.Ed.2d 842 (1967) (unions must fairly represent all employees in a unit for which it is exclusive bargaining representative). Therefore, the fact that Quesnel is not a Union member does not remove him from the bargaining unit for whose benefit the CBA was created. See Saunders v. Amoco Pipeline Co., 927 F.2d 1154, 1156 (10th Cir.1991) (individual employee is bound by terms of collective bargaining agreement even if not a union member). Indeed, in his brief Quesnel essentially concedes that he was a member of the bargaining unit of the CBA, and maintains that he could have invoked the CBA's grievance and arbitration procedures, but properly chose not to. Quesnel cannot pick and choose among his avenues of remedy, however; having been a member of the bargaining unit and received the benefits of the CBA while employed, Quesnel cannot now disclaim it. The grievance procedures set forth in the CBA is exclusive of other dispute resolution mechanisms. 14 Finally, we do not think that the Agent's Agreement displaces or in any way substitutes for the CBA. The Agent's Agreement does not deal with terms and conditions of employment or with grievance procedures, as does the CBA. Rather, the Agent's Agreement merely delineates Prudential's business policies applicable to district agents. Moreover, Articles XXVI and XXVII of the CBA specifically reference and amend the Agent's Agreement, a strong indication that the Agent's Agreement is not intended to supplant, but merely to supplement, the CBA. We therefore find that Quesnel is subject to the terms of the CBA, and that his claims are, accordingly, preempted. CONCLUSION 15 For the foregoing reasons, we affirm. * Of the District of Puerto Rico, sitting by designation 1 Specifically, Article I of the CBA states: The Employer agrees to and hereby does recognize, to the extent required by the National Labor Relations Act, as amended, The Union as the exclusive representative for the purposes of collective bargaining in respect to rates of pay, wages, hours of employment, or other conditions of employment, of all District Agents employed or hereafter to be employed by the Employer.... (Emphasis added). 2 The CBA sets forth grievance and arbitration procedures for wrongful termination. Quesnel did not pursue these remedies, and the time limit for seeking relief under the CBA has lapsed. Finding that Quesnel was subject to the CBA, the district court accordingly dismissed his claims for failure to exhaust these administrative remedies 3 Section 301(a) provides: Suits for violation of contracts between an employer and a labor organization representing employees in an industry affecting commerce as defined in this chapter, or between any such labor organizations, may be brought in any district court of the United States having jurisdiction of the parties, without respect to the amount in controversy or without respect to the citizenship of the parties. 4 Quesnel contends that because the district court went beyond the pleadings by considering the CBA, the Agent's Agreement, and an affidavit of Quesnel, it was actually treating Prudential's motion to dismiss as one for summary judgment, and that we must therefore apply the standard of review applicable to summary judgment decisions. As we explain, however, because we conclude as a matter of law that Quesnel's claim is preempted, his arguments regarding the appropriate standard of review are irrelevant | Low | [
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Q: Performance concerns when storing multiple bitmaps I have created a custom View object which overrides the onDraw method to paint a fairly involved UI. I am adding 5 of these custom views onto a LinearLayout, but only one View is visible at any one time. Depending on the users' actions inside my application, I am toggling the View.Visibility property on each so that only one is visible. Just for clarification, the method I'm using works for me and it seems to be fairly responsive. I'm just a little concerned how this method would affect lower-end or lower-spec devices. Here is a sample of my current code: Custom View public class MyDrawingView extends View { private Bitmap mViewBitmap; private int mWidth = 1024; // The width of the device screen private int mHeight = 600; // Example value, this is dynamic @Override protected void onDraw(Canvas canvas) { // Copy the in-memory bitmap to the canvas. if(mViewBitmap != null) canvas.drawBitmap(mViewBitmap, 0, 0, mCanvasPaint); } private void drawMe() { if(mViewBitmap == null) mViewBitmap = Bitmap.createBitmap(mWidth, mHeight, Bitmap.Config.ARGB_8888); Canvas c = new Canvas(mViewBitmap); c.drawBitmap(...); c.drawText(...); // Multiple different methods here drawing onto the canvas c.save(); } } Layout XML <LinearLayout> <com.company.project.ui.MyDrawingView android:id="@+id/myCustomView1" android:layout_width="wrap_content" android:layout_height="wrap_content" /> <com.company.project.ui.MyDrawingView android:id="@+id/myCustomView2" android:layout_width="wrap_content" android:layout_height="wrap_content" /> <com.company.project.ui.MyDrawingView android:id="@+id/myCustomView3" android:layout_width="wrap_content" android:layout_height="wrap_content" /> <com.company.project.ui.MyDrawingView android:id="@+id/myCustomView4" android:layout_width="wrap_content" android:layout_height="wrap_content" /> <com.company.project.ui.MyDrawingView android:id="@+id/myCustomView5" android:layout_width="wrap_content" android:layout_height="wrap_content" /> </LinearLayout> Questions Should I be keeping these 5 separate instances of my View, with Bitmaps of size 1024x600, in memory all the time? Should I merge the functionality, so that I need only add one View to my layout XML then re-generate a Bitmap each time a View needs to be updated? Which option is better for performance, bearing in mind that redrawing my Bitmap may take some time due to its complexity? Documentation I have already read the Android documentation on Managing Bitmap Memory, however I feel I have implemented the points outlined already in my custom View and I don't think it quite covers my scenario. A: OK, so I took on-board Geobits' comments on heap size and came up with a solution to the issue: Go with one View, which will dynamically draw data onto a single internal Bitmap as required Change the bitmap type to RGB_565 in order to make the bitmap slightly smaller Delete/recycle the bitmap where possible After the above, I managed to reduce the required RAM to 25Mb for the entire application which I'm more than happy with. | Mid | [
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// DO NOT EDIT THIS FILE - it is machine generated -*- c++ -*- #ifndef __org_omg_IOP_ServiceContextHolder__ #define __org_omg_IOP_ServiceContextHolder__ #pragma interface #include <java/lang/Object.h> extern "Java" { namespace org { namespace omg { namespace CORBA { class TypeCode; namespace portable { class InputStream; class OutputStream; } } namespace IOP { class ServiceContext; class ServiceContextHolder; } } } } class org::omg::IOP::ServiceContextHolder : public ::java::lang::Object { public: ServiceContextHolder(); ServiceContextHolder(::org::omg::IOP::ServiceContext *); void _read(::org::omg::CORBA::portable::InputStream *); void _write(::org::omg::CORBA::portable::OutputStream *); ::org::omg::CORBA::TypeCode * _type(); ::org::omg::IOP::ServiceContext * __attribute__((aligned(__alignof__( ::java::lang::Object)))) value; static ::java::lang::Class class$; }; #endif // __org_omg_IOP_ServiceContextHolder__ | Low | [
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second chances 04.02.13 Sanford Wins SC's GOP Primary Sanford with supporters in Charleston, S.C. on Tuesday. (Bruce Smith/AP) Former South Carolina Governor Mark Sanford snagged the GOP nomination for the state's House seat (the same one he held in the nineties) on Tuesday, beating opponent Curtis Bostic. He'll now face Elizabeth Colbert, sister of the famed Stephen, in a special election this May. Sanford, whose reputation was marred after he confessed to an extramarital affair, says voters are overlooking those indiscretions. "I think tonight's verdict will say a lot as to where people are or are not on that," Sanford said. "I suppose at some level, I will never completely move beyond that." Sanford made his first public appearance with his paramour, now fiancée, Maria Belen Chapur, Tuesday night. | Low | [
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Race differences in the pattern of familial aggregation for dehydroepiandrosterone sulfate and its responsiveness to training in the HERITAGE Family Study. Using a familial correlation model to assess familial influences, baseline dehydroepiandrosterone sulfate (DHEAS) and its change (post-training minus baseline) in response to a 20-week endurance exercise training program were analyzed in 85 black families who participated in the HERITAGE Family Study (HERITAGE). Baseline levels were adjusted for a polynomial in age, and the training response was adjusted for a polynomial in age, as well as the baseline values, within 4 sex-by-generation groups before genetic analysis. We found that the maximal heritability for baseline DHEAS reached 66% (with no sex and generation differences) in black families, which is slightly (but not significantly) higher than the estimate (58%) reported previously in 99 white families in HERITAGE. Whereas weak, but significant, familial effects (26%) for the training response were previously reported for whites in HERITAGE, they were undetectable in the present study. Furthermore, we found heterogeneity in the pattern of familial aggregation (primarily due to different spouse and parent-offspring correlations) for both the baseline and its training response between blacks and whites. In conclusion, baseline DHEAS levels in blacks were also determined by substantial familial factors (just as for whites), independent of the effects of age and sex. Genetic and nongenetic familial components influencing baseline DHEAS levels in both races may be different. | High | [
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The AAA(+) (ATPases associated with a variety of cellular activities) superfamily protein ClpC is a key regulator of cell development in Bacillus subtilis. As part of a large oligomeric complex, ClpC controls an array of cellular processes by recogni ... The AAA(+) (ATPases associated with a variety of cellular activities) superfamily protein ClpC is a key regulator of cell development in Bacillus subtilis. As part of a large oligomeric complex, ClpC controls an array of cellular processes by recognizing, unfolding, and providing misfolded and aggregated proteins as substrates for the ClpP peptidase. ClpC is unique compared to other HSP100/Clp proteins, as it requires an adaptor protein for all fundamental activities. The NMR solution structure of the N-terminal repeat domain of ClpC (N-ClpCR) comprises two structural repeats of a four-helix motif. NMR experiments used to map the MecA adaptor protein interaction surface of N-ClpCR reveal that regions involved in the interaction possess conformational flexibility and conformational exchange on the microsecond-to-millisecond timescale. The electrostatic surface of N-ClpCR differs substantially from the N-domain of Escherichia coli ClpA and ClpB, suggesting that the electrostatic surface characteristics of HSP100/Clp N-domains may play a role in adaptor protein and substrate interaction specificity, and perhaps contribute to the unique adaptor protein requirement of ClpC. Organizational Affiliation:  Department of Molecular Genetics, College of Medicine, University of Cincinnati, Cincinnati, OH 45267, USA. | High | [
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BEIJING (Reuters) - China announced a series of measures on Thursday to revive slumping car sales, but failed to meet market expectations as it included no plans to relax controls over the issuance of new licenses for traditional-fuel cars in major cities. FILE PHOTO: A carrier trailer transports newly manufactured cars at a port in Dalian, Liaoning province, China May 21, 2019. REUTERS/Stringer/File Photo Beijing has been trying to boost consumption of goods ranging from eco-friendly appliances to big-ticket items such as cars to fire up growth, as the world’s second-largest economy is expected to slow further in 2019 amid a bruising trade spat with the United States. Auto industry executives have expected China’s car market, the world’s biggest, would return to growth this year thanks to government support after sales contracted for the first time last year since the 1990s. The National Development and Reform Commission (NDRC), China’s state planner, said in a statement it is stopping local governments from imposing new restrictions on car purchases and cancelling existing ones that apply to new energy vehicles. The measures, which apply to 2019-2020, include support to encourage car purchases in rural areas. The NDRC statement also called for more local governments to allow pickup trucks to enter their cities. Financial magazine Caixin, citing an NDRC document, reported in April that China is planning to increase the number of newly issued car licenses in major cities including Beijing, Shanghai and Guangzhou by 50 percent this year from 2018 levels, and double that next year. The NDRC, however, did not mention such steps on Thursday. “The (April) draft was quite strong so I am quite disappointed about the policies,” said an industry association official who declined to be named as she was not permitted to speak to the media. Yale Zhang, head of Shanghai-based consultancy Automotive Foresight, said the document would likely support some new energy vehicle sales, but would not have a big impact on combustion engine cars. “Now the question comes to whether local governments are willing to support auto sales.” BREAKING MARKET BARRIERS Vehicle sales in China fell 14.6% in April from the same month a year earlier, marking the 10th consecutive month of decline. Automakers have been lowering prices after the government introduced tax cuts to spur consumer spending, but customers are holding off purchases in the hope of more favorable policies. “This document has positive impact on the industry in the long term, but will not boost sales very quickly,” said the industry association official. Yet, the measures, which were unveiled after mainland markets shut, helped boost shares of Hong Kong-listed automakers. Shares in Geely Automobile Holdings rose by more than 3% after the announcement, while BYD Co Ltd climbed as much as 5.2%. Some local governments had already started to offer supportive policies in the run-up to NDRC’s announcement. Authorities in the big southern Chinese cities of Guangzhou and Shenzhen said that they will increase quotas for new car registrations from this month till the end of next year which will allow at least 180,000 more car sales. The NDRC said the new moves “strive to break the market barriers that restrict consumption and protect consumers’ legitimate interests.” | Low | [
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books.google.com - In the wake of both the semiotic and the psychoanalytic revolutions, how is it possible to describe the object of religious worship in realist terms? Semioticians argue that each object is known only insofar as it gives birth to a series of signs and interpretants (new signs). From the psychoanalytic...https://books.google.com/books/about/Nature_s_Religion.html?id=MCy0N7tzSdYC&q=structures&utm_source=gb-gplus-shareNature's Religion | Mid | [
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From Child on Street to Nobel Laureate - charzom http://www.washingtonpost.com/wp-dyn/content/article/2007/10/08/AR2007100800252.html?hpid=sec-health ====== axiom What an incredible life! makes me feel spoiled for being born in Soviet Russia. | Low | [
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Rebuild the devastated islands as you take control of the Rescue Team in this fantastic Time Management game! fter a terrifying tornado leaves storms through the islands, itâ€s up to the Rescue Team to save the day and save lives. Use strategy to figure out the quickest way to complete your goals, and then race against the clock to remove debris, repair bridges, put out fires, and protect the civilians in Rescue Team 2! Try full 2-D cartoon-style games. Exaggerated style, cute characters and bright colors impress you. Although the games are simple, it is very rich in levels and a lot of available upgrades will made it even more beautiful. Rescue Team 2 music is easy with the beautiful countryside scenery, you will relax the spirit. Delightful picks in music will help you to sound more convenient for the game. Now download and start play the game. Here you can download a demo version of Rescue Team 2 game for free with 1 hour expiry date of a trial period. It means that after 1 hour playing you must pay us $6,99 or uninstall our game. It is conformed to Mac OS. The version, presented at our site, is proved from any viruses and spies, so you can download it without any worries. Besides this will take you only few minutes, if you have Cable/ADSL connection. Take our Rescue Team 2 game here and enjoy it every free minute of your time! | Mid | [
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promises come cheap, dear reader Menu Growing up watching Star Wars marathons late in to the night with my brother Kevin, I never stopped to think whether the movies were good or bad. In fact, I never stopped to recognize that they were separate movies at all. Only later in life when I would discuss the movies with friends did I realize that people ranked the films separately. To me I couldn’t see the point in watching just one of the three, so the idea was just crazy. Our marathons happened several times each year and we never started earlier than 10pm. This was to ensure that we were properly sleep-deprived by the time we saw dancing Ewoks. We watched full-screen VHS copies of the movies, that were taped from HBO, on a 4:3 CRT television. The quality was poor and got worse with each screening due to the magnetic tape stretching more and more. Keeping the VCR remote handy for adjusting tracking was absolutely necessary. About halfway through, we would take a break to make a pot of Kraft Mac&Cheese with the dayglow powder cheese, as none other would suffice. The pot we cooked it in was so old that the teflon coating would often scrape off and we would find what we dubbed “teflon surprises” and sometimes competed to see who had the most. A full 12 pack of Mtn Dew was also required, though it did little to keep us awake due to that being our normal intake of caffeine on any other night. Kevin always made it to the end, while I developed the ability to occasionally wake up enough to react to one of his jokes so that he wouldn’t realize I was sleeping already (he always knew). This was my Star Wars experience. Then the prequels came out and while I now have my own opinions of those movies, I was one of the people who left the theater after Episode I with a great feeling. I actually enjoyed the movie because I wasn’t comparing it at all to the originals. Sure, I hated Jar Jar, but it was the first Star Wars movie I got to see in the theater when it was released. Episode I showed me a familiar world that I knew from childhood and so long as Jedi were hitting things with lightsabers and spaceships were flying around, I was overjoyed. It was at this time that I really started to realize that people had very strong opinions of the OT and the new prequels. Suddenly the innocence that I experienced with the OT was brought in to sharp focus. I came to understand that these movies did not exist only in late night sleep-deprived marathon sessions filled with Kraft Mac&Cheese and teflon surprises. Why do I mention any of this now? Nostalgia, for one. This new film brought back so many of the feelings I had surrounding the OT and in turn took me back to that innocent time. But, also because I have been reading a lot of reactions on r/StarWars that were making me feel upset until I looked at this from this new perspective. Everyone’s experience with Star Wars is totally different and unique and valid. Make up your own mind about this new episode and don’t let trolls ruin it for you. Enjoy your two hours in a galaxy far, far away and say hi to some old friends one more time. Most drivers in Los Angeles have experienced the horrors of the 110 West through Downtown LA. Most specifically the point where the left lanes (110W) and the right lanes (off-ramps and on-ramps from the 101) come together. Smart drivers know that if you miss your chance to merge left, there is a dedicated off-ramp for the 110W you can use instead. As most of us know, there’s always that one asshole who waits until the very last moment to move left in order to get on the 110W. They usually do this in some dramatic display that reenforces how important it is that they make this lane change, usually including them driving over the rough median. On my way to work, I was a few cars behind a CHP cruiser as it was passing this hectic area of the freeway. The cruiser was in the right lane of the 110W with a line of cars to his right that were all presumably heading for the off-ramps ahead. At the very last possible moment, a black SUV had their expected moment of panic and suddenly changed lanes (across the solid white lines) and found himself DIRECTLY in front of the CHP cruiser. The lights on the cruiser immediately lit up, as though the officer knew this would happen… because he did. My laughter at this point was completely uncontrollable, as I have waited years to see this exact scenario play out and it finally did. As I passed the douchebag, I of course had to get a look at them to see what kind of idiot would pull such a stupid move. What did I find but a white guy driving while ACTIVELY texting on his phone. This really made my day. I can only hope that the officer pointed out the off-ramp that would have saved him not only time, but now money. So, Brian isn’t dead after all. In case you missed it, Family Guy had a plan all along and the internet fell for it. While watching “Life of Brian” a few weeks ago, I was not moved by the death scene. Maybe it was the whole time machine plot point that caused me not to be affected by the scene, as I thought Stewie would just rebuild the time machine and bring Brian back. Well, I wasn’t too far off, it seems. Though, it looks like Stewie will be using the magical power of Santa to bring back his best friend! Something to keep in mind with these shows is that they are written and created months in advance. Which means the internet outrage that was sparked by the death of Brian must have caused the Family Guy writers to sit back in their expensive lounge chairs, light up a doobie and laugh. I found the whole online petition to bring back Brian laughable myself. If you actually signed the petition, you should probably go outside and play in traffic. Seriously. You signed a petition (online, so it’s meaningless anyway) to resurrect a FICTIONAL CHARACTER that you did nothing to help create or develop. Think about that and take a moment to reflect on your priorities in life. Please be sad that Brian is dead(ish). That’s great! Someone else’s art made you feel that way. It probably reminded you of your dead dog, or something else you have lost. Isn’t that amazing? That someone else’s art can have that powerful an effect on you? Yeah, well… remember the important part is that it’s SOMEONE ELSE’S. And ultimately, what did all that complaining and petitioning do? Nothing, because they were going to bring Brian back anyway. But, I for one have enjoyed the break. Seth McFarlane killed his dog. Well really, he killed our dog, as Brian had become the epitome of the anthropomorphized canine we’ve always wanted. While this plot-twist in the Family Guy universe is sad for anyone who agrees that Brian was the truest character on the show, it’s also about time. The problem with Brian on the show is that he is essentially just Seth in dog form. Seth didn’t choose a different voice for this character, so it allowed him to make Brian the singer and the performer. But, when you’re over-exposed to Seth’s voice on TV and movies that aren’t Family Guy, it gets a little tiring. TED was a great example of this, as Seth just used a thicker accent than normal, but was essentially re-creating Brian for the big screen. TED was a drunk, womanizing, anthropomorphized teddy bear. Not a huge leap from Brian in the end. Through your tears, and beating your fists against the TV, remember that Brian was a dog. Dogs live about 10 years, on average, and Brian lasted 11. We will miss him dearly, but knowing what we know about animated shows, this likely isn’t the last we’ve seen of our animated best friend. Vinne may be annoying, but is his accent really any worse than Lois’s? Newly married and newly employed, I am taking back the web with a post to a dead WordPress site! Hooray! This will lead me down the rabbit hole of WP again and hopefully lead to even better things! Nice short post this time, but you can expect more from thePlayland in the very near future! It’s Droid X Day!! I was so excited that I drove down to my local Verizon Wireless store this morning to join the crowds and get my early upgrade. When I arrived, a customer service agent let me know that since my free upgrade wasn’t until Jan 2011, I wasn’t going to be eligible for an upgrade! Oh no! Dejected, I headed outside and called 611 on my cell. I was connected to a Verizon agent, working on his day off, who let me know that I DID qualify for a rebate-based early-upgrade discount! That’s all I really wanted, anyway, so I was elated! He said I could order it from him, so I asked if the same deal would be available in-store (since I am impatient and didn’t want to wait for shipping). He said yes, but that he would just waive the shipping and deliver overnight if I bought from him right then. To which I replied, “yes, please!” At this point, I was already a happy customer. Then, he waived my early upgrade fee, too! Now I was really happy, but a little worried because I didn’t really want to pay the whole cost today. Tomorrow, yes. When it came time for the payment, he said I could pay now, or defer it to my monthly bill! Meaning I didn’t have to drop a dime today to get my new Droid X shipped overnight! When I am asked why I don’t have an iPhone, my answer is Verizon. Not only do they provide me with coverage in my apartment (something AT&T cannot do), they just seem to give a crap about whether I am a happy customer. That’s something I NEVER experienced with Cingular/AT&T. So, while I enjoy using my iPod Touch and can see why people like iPhones so much, I can’t get behind Apple with their decision to limit themselves to AT&T. | Mid | [
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Q: Blinking a LED through a Web application I'm doing some experiments with some Linux-based embedded systems (Beaglebone, FriendlyARM mini6410, Embest Devkit 8000). I would like to write a web application using some kind of web technology PHP, Javascript,...whatever it is, whose purpose is just to blink a LED. I have to do this in order to see if I'm able to control some hardware resources through a Web application. I know that for the Beaglebone I can use node.js and bonescript but I would like to devise a solution that I can easily bring on other targets (with no or limited modifications to the application) since I have to replicate this on different embedded targets. I know that I have to go through sysfs in order to be able to control hardware resources from a user space application. I could do this in PHP (through sysfs I can drive a led just by opening and reading/writing files) or I could write a C CGI application wich does the same...But my question is, what are the other options? I would like to identify different possibilities for designing the web application in order to chose the proper one. A: write a small cgi. It can be even done with shell, where you can send commands to sysfs You could make something like #!/bin/sh #this will parse the sent parameters eval $(echo "$QUERY_STRING"|awk -F'&' '{for(i=1;i<=NF;i++){print $i}}') # this has to be set to whatever you want GPIO=22 echo $GPIO > /sys/class/gpio/export echo "out" /sys/class/gpio/gpio$GPIO/direction echo 1 /sys/class/gpio/gpio$GPIO/value cat << EOF Content-Type: text/html <!DOCTYPE html> <html> <body> <h1>pin on</h1> </body> </html> EOF finally you need to be sure that you have the rights to write on sysfs and that the webserver is configured to use the shell as the interpreter of the cgi A: Well, if I was doing it I would probably choose socket communication. Perhaps serial for really piddly pieces of hardware (that may not have an ethernet/wifi port). I would possibly enable support for both if the device is interchangable by abstracting the comms layer so the server could use sockets or serial connection strategies. Most all languages have support for both one way or another. | Mid | [
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name = 'ESMF' version = '6.1.1' homepage = 'http://sourceforge.net/projects/esmf' description = """The Earth System Modeling Framework (ESMF) is software for building and coupling weather, climate, and related models.""" toolchain = {'name': 'goolf', 'version': '1.4.10'} toolchainopts = {'usempi': True} source_urls = [SOURCEFORGE_SOURCE] sources = ['%s_%s_src.tar.gz' % (name.lower(), '_'.join(version.split('.')))] dependencies = [ ('netCDF', '4.2.1.1'), ('netCDF-Fortran', '4.2'), ('netCDF-C++', '4.2'), ] moduleclass = 'geo' | Mid | [
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Log In Google+ spams users after disk space shortfall Some Gmail accounts hit. Google has explained why its social network started spamming users over the weekend - the service ran out of disk space. Some Google+ users may have noticed repeated notifications filling their inboxes on Saturday. One user said he received the same message 12 times, but those who had turned off notifications or not received any during the time will not have been affected. Google's senior vice-president for social Vic Gundotra apologised for the fault. "For about 80 minutes we ran out of disk space on the service that keeps track of notifications," he said in a post via his Google+ account. "Hence our system continued to try sending notifications. Over, and over again. Yikes." "We didn't expect to hit these high thresholds so quickly, but we should have," he said, thanking users while reminding them the system was still a "field trial". Google also had to stress the system was not a final release, warning business users last week not to use it yet. All rights reserved. This material may not be published, broadcast, rewritten or redistributed in any form without prior authorisation.Your use of this website constitutes acceptance of nextmedia's Privacy Policy and Terms & Conditions. | Low | [
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For a new sysadmin, storage can be one of the more confusing aspects of infrastructure. This confusion can be caused by lack of exposure to new or different technologies, often because storage needs may be managed by another team. Without a specific interest in storage, an admin might find one’s self with a number of misconceptions, questions, or concerns about how or why to implement different solutions. When discussing enterprise storage, two concepts are at the core of most conversations: storage area networks (SAN) and network-attached storage (NAS). Both options provide storage to clients across a network, which offers the huge benefit of removing individual servers as single points of failure. Using one of these options also reduces the cost of individual clients, as there is no longer a need to have large amounts of local storage. Storing all of this important data in a specially designed system provides a centralized place to manage—and back up your data, build access controls, provide security contexts to secure your data, and more in one single place, rather than having to build all of these processes into a fleet of machines. Scaling storage up to meet future needs is also much easier when storage is centrally located. You no longer need to spend as much energy tracking individual server disk usage. Instead, you manage the larger central pool and increase capacity by adding disks or shelves of disks as needed. These expansions can even be tiered, using drives with different performance capabilities to offer a more tailored experience to the different clients using this storage. Regardless of the performance required, both a SAN and a NAS use the same basic building blocks for their underlying storage: drives. These drives can be anything from inexpensive consumer-grade 3.5-inch platter drives to 10K RPM SAS, and all the way up through solid-state and NVM Express (NVMe) devices. Speed, scale, and budget requirements determine the right design, but this is all commonly available hardware and nothing too exotic is required. When looking at these two big-picture ideas from afar, they seem interchangeable, but they have many differences worth considering. A NAS, from an architectural standpoint, is usually a single server. It can be built as a virtual machine on a hypervisor, but is more often a physical machine itself, for scaling and performance reasons. The NAS machine runs one or more file-sharing protocols that are exposed to an internal network. Subsequently, those shares are presented by protocols like NFS or SMB (CIFS) to allow clients to attach to the NAS for reading/writing files as if each client had a large local filesystem. A network-available filesystem like this is a pretty common need in a business environment, so a NAS is an easy entry point into the world of shared storage. A SAN, on the other hand, is rarely a single machine. The SAN philosophy is to build a storage system from a handful of independent parts. Even with the most inexpensive options, you usually have a single physical chassis containing a pair of controllers that can failover to one another for maintenance (upgrades, etc.), or in case of failure. SAN storage is based on the idea of providing block-level access for hosts that need control over their own storage details (filesystems, etc.), rather than a simple file share like NFS provides. A machine would normally use an internal disk as a block device, and upon that it would create filesystems. The SAN abstracts this issue away and provides that block device across a network. This access is almost always provided with either iSCSI or Fibre Channel (including Fibre Channel over Ethernet, or FCoE) as the communication protocol between the clients and the SAN. The client consuming that block device can then partition it and create filesystems on it as necessary, without having to worry about another team managing those details. A good SAN use case is a VMware hypervisor using SAN storage to hold virtual machine data, rather than on its own local drives. VMware's native filesystem (VMFS) requires block-level access to its storage, which means it cannot use a file share (like NFS) to store this data; though an NFS datastore can be created instead if required. While a NAS is most often a single machine, the components of a SAN can include dedicated switches (or VLANs on a shared network), controller nodes, disk shelves, tape backup units, or gateway devices. The added complexity provides better scalability, redundancy, and tiering for individual services running on the SAN. Because of this cluster-like approach to the hardware, it is usually easier to add additional resources to a SAN than it is to a NAS, in the form of new switches or disk shelves. Adding resources to a NAS requires the space, connectivity, and power to be free in a machine since the NAS model usually depends on a single chassis (or single virtual machine). There are certainly ways to scale a NAS out to large sizes, but the SAN model is much more suited for growth and scale. It can even make sense for an administrator to build out a large SAN for many different groups to use, including backing storage for a virtual NAS if there is limited space to deploy a large physical one. This approach is possible because a SAN and a NAS effectively live at different layers of abstraction; the SAN providing the block storage that a NAS inherently needs, and the NAS managing the filesystem and network shares on top of that block storage. These technologies are not mutually exclusive, nor is one inherently better than the other. They both provide valuable storage capabilities for different needs. Many organizations end up running one or more of each type for different workloads, levels of redundancy, and availability. If you need a place for a group of users to store and share their files, a NAS is probably the right answer. Trying to use a NAS to provide shared storage for workloads that require lower-level access to storage (like block), or that have a less common filesystem (like vmfs) can introduce a lot more complexity, and consequently, a SAN would be a much better fit. * Image credits: "Storage 5x10s" by Meathead Movers is licensed under CC BY-SA 2.0 and "Storage Units" by JeepersMedia is licensed under CC BY 2.0. | Mid | [
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Rohan Silva In an occasional series on the state of the capital - as the race for London’s mayoralty begins - an ex-Number 10 policy adviser celebrates its enterprise and momentum, but argues that it could give more people a fairer deal | Mid | [
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) = -2*h**2. Calculate m(x(u)). -450*u**4 - 60*u**2 - 2 Let q(c) = -c**2. Let t(x) = -x + 2. Let l(j) = 2*j - 3. Let v(k) = 2*l(k) + 3*t(k). Determine v(q(m)). -m**2 Let f(k) = -2*k**2. Let t(y) = -5*y**2 - 4. Let w(p) = 4*p**2 + 3. Suppose -c + 5*c = 12. Let g(n) = c*t(n) + 4*w(n). What is g(f(a))? 4*a**4 Let v(t) = t + 5*t - 3*t - 6*t. Let r(i) = i. What is v(r(b))? -3*b Let c(w) = -23*w. Let m(r) = 19*r. Calculate c(m(h)). -437*h Let i(u) = 2*u - 2*u + 4*u + 3*u. Let l(p) = p**2. Give l(i(n)). 49*n**2 Let d = -4 - -6. Suppose -12 = -5*c + d*c. Let t(h) = 0*h**2 + h**2 - c*h**2. Let g(p) = -2*p**2. Calculate t(g(q)). -12*q**4 Let n(z) = 2*z. Let j(s) = -18*s. What is j(n(q))? -36*q Let p(d) = -6*d**2 + d**2 + 2*d**2 + 0*d**2. Let v(f) = -f + 0*f - f. Determine v(p(x)). 6*x**2 Let v(a) = -10*a**2. Let l(j) = 36*j. Calculate v(l(p)). -12960*p**2 Let x(m) = -19*m**2. Let s(t) = -53*t**2. Give x(s(i)). -53371*i**4 Let v(n) = -n. Let w(z) be the third derivative of -13*z**5/60 - 14*z**2. Calculate v(w(i)). 13*i**2 Let v(c) = -4*c**2. Let f(s) = 5*s + 42. Give v(f(d)). -100*d**2 - 1680*d - 7056 Let k(t) be the second derivative of -t**4/6 + 2*t. Let p(w) be the first derivative of 0*w - 1 + 1/2*w**2. Give k(p(f)). -2*f**2 Let c(v) = 1903*v**2. Let x(j) = j**2. Calculate x(c(o)). 3621409*o**4 Let u(p) = -6*p. Let w(t) = -14*t. Calculate u(w(d)). 84*d Let i(o) be the third derivative of -7*o**5/60 + 2*o**2. Let t(w) = 2*w. Give t(i(y)). -14*y**2 Let d(b) = 255*b**2. Let r(w) = 5*w**2. Give d(r(m)). 6375*m**4 Let w(d) = 7*d**2 - 18*d**2 - 9*d**2. Let m(l) = -2*l. Calculate m(w(a)). 40*a**2 Let m(v) = v. Suppose 2 - 6 = -4*s. Suppose 2*j + i = 3 + s, j = -3*i + 2. Let h(n) = -n**2 + 0*n**2 + 4*n**j. What is h(m(g))? 3*g**2 Let g(a) = 6*a - 19 + 11 + 8. Let x(n) = -2*n**2 + 3*n**2 + 0*n**2. What is x(g(r))? 36*r**2 Let c(d) be the second derivative of -7*d**4/12 + 4*d. Let r(j) = -2*j**2. Calculate r(c(b)). -98*b**4 Suppose 20 + 5 = -5*m. Let y(n) = 3*n + 5. Let u(h) = -4*h - 7. Let x(q) = m*u(q) - 7*y(q). Let p(t) = 4*t**2. What is x(p(d))? -4*d**2 Let c(h) = 133*h**2 - 114*h. Let t(x) = x**2 - x. Let i(z) = -c(z) + 114*t(z). Let l(q) = 2*q**2. What is i(l(w))? -76*w**4 Let k(p) = p**2. Let q(i) = 0*i + 1 + 0*i + 6*i - 15*i. Give k(q(w)). 81*w**2 - 18*w + 1 Let i(t) = 2*t. Let s(k) = 1702*k**2. Calculate s(i(m)). 6808*m**2 Let d(k) be the first derivative of k**3 + 1. Let u(y) = -5*y**2 - 2*y + 4. Let g(a) = -a + 2. Let c(m) = -2*g(m) + u(m). What is c(d(b))? -45*b**4 Let o(m) be the second derivative of -m**3/3 - 20*m. Let n(v) = -8*v. Let u(a) = 5*a + 3*a - a. Let d(i) = -5*n(i) - 6*u(i). Determine d(o(b)). 4*b Let u(d) = 31*d. Let y(w) = w**2. Calculate y(u(r)). 961*r**2 Let b(d) = -2*d. Let q(g) = -51*g - 1. Calculate q(b(i)). 102*i - 1 Let r(h) = 2*h - 3*h - 2*h. Let a(f) = 6 - 6 + 2*f**2. Give a(r(b)). 18*b**2 Let z(v) = -5*v - 4. Let s(f) = -f - 1. Let t(d) = -4*s(d) + z(d). Let w(x) = 5 - 5 - 2*x - 2*x. What is t(w(m))? 4*m Let k(o) = -4*o**2. Let u(z) = 48*z - 12. Calculate u(k(d)). -192*d**2 - 12 Let z(m) = 2*m + 2. Suppose 0*j - 10 = 5*j. Let f = -5 - -10. Let k(s) = s**2 + 5*s + 5. Let o(r) = f*z(r) + j*k(r). Let t(i) = 2*i. Determine o(t(l)). -8*l**2 Let x(i) = -23*i. Let q(f) = -8*f. Calculate x(q(l)). 184*l Let r(m) = 633*m**2. Let v(h) = h. Give r(v(g)). 633*g**2 Let a(c) be the second derivative of -c**4/12 + c**2 - 4*c. Let m(u) be the first derivative of a(u). Let v(s) = -2*s. Determine m(v(n)). 4*n Let w(d) = 2*d - 7. Let p(v) = 2*v. Let t be p(3). Let x(z) = z - 3. Let j(s) = t*w(s) - 14*x(s). Let n(o) = -14*o**2 + 6*o**2 + o**2. What is n(j(f))? -28*f**2 Let y(h) = 33*h + 33*h - 97*h + 33*h. Let z(o) = -4*o + 7. Let b(g) = -3*g + 5. Let p(v) = 7*b(v) - 5*z(v). Determine y(p(w)). -2*w Let s(o) = 2*o**2. Let p(b) be the third derivative of 5*b**4/12 - b**3/2 - 13*b**2 - 2. Determine p(s(y)). 20*y**2 - 3 Let m(g) = -g**2 + 3*g**2 + 3*g**2 - 3*g**2. Let q(y) = 4*y. Give q(m(u)). 8*u**2 Let a(x) = 3*x. Let b = -5 - -7. Let o(k) = -k + 4 - b - 1. Let q(g) = -7*g + 6. Let n(s) = -6*o(s) + q(s). Give n(a(z)). -3*z Let f(y) = -6*y - 4*y + 8*y. Let j(u) = u. What is j(f(z))? -2*z Let x(s) = -5*s - 3*s + 5*s - 3*s. Let p(u) = 2*u. Determine p(x(g)). -12*g Let w(l) = 4*l**2. Let m(k) = 13*k - 11. Let s(o) = 7*o - 6. Let y(f) = 6*m(f) - 11*s(f). Give w(y(r)). 4*r**2 Let b(a) = a. Let l = -10 + 15. Let y(t) = 16*t**2 + 5*t. Let g(u) = 15*u**2 + 4*u. Let x(j) = l*g(j) - 4*y(j). Give b(x(w)). 11*w**2 Let r(q) = 4*q**2. Let y be (0/7)/(-1*2). Let i(h) be the first derivative of 3 + y*h - 1/3*h**3 + 0*h**2. What is i(r(s))? -16*s**4 Let a(c) = -25*c. Let q(o) = -22*o**2. Determine a(q(v)). 550*v**2 Let b(v) = v + 7. Let g be b(-7). Let u(p) = -3*p**2 - 2*p**2 + g*p**2. Let q(i) = -2*i. Calculate q(u(z)). 10*z**2 Let y(h) = -355*h**2. Let g(z) = z**2. Give g(y(u)). 126025*u**4 Let t(q) = 158 - 158 + 2*q. Let x(f) = 2*f. Calculate x(t(p)). 4*p Let o(a) be the third derivative of a**5/20 - 39*a**2. Let f(i) be the second derivative of -i**3/3 + i. Calculate f(o(n)). -6*n**2 Let v(i) = 7*i - 5. Let q(g) = -4*g + 3. Let b(k) = -10*q(k) - 6*v(k). Let o(z) = 5*z**2. Calculate b(o(s)). -10*s**2 Let h(c) = -c. Suppose 4 + 0 = s. Let y(w) = w**2 + w + 1. Let x = 17 + -19. Let v(p) = p**2 + 2*p + 2. Let a(f) = s*y(f) + x*v(f). Calculate h(a(j)). -2*j**2 Let o(g) = -g. Let b(h) = 21*h**2 + 4*h - 4. Let u(w) = -1450*w**2 - 275*w + 275. Let t(l) = 275*b(l) + 4*u(l). Determine o(t(r)). 25*r**2 Let y(o) = -1627*o**2. Let a(r) = -2*r. Give y(a(b)). -6508*b**2 Let f(h) = 4*h. Let u be 8/12 - (-26)/6. Let k(y) = y + 2*y - u*y. What is f(k(o))? -8*o Let y(r) = 5*r**2. Let v(c) = -2*c. Let p(d) = d + 0*d + 10*d. Let z(o) = -6*p(o) - 34*v(o). Give z(y(j)). 10*j**2 Let v(f) = -137*f. Let h(s) = -2*s. Determine v(h(r)). 274*r Let x(q) = -3*q**2. Let h = 13 - 11. Let g(f) = -2 - 5 + 7 - 2*f**h. Calculate x(g(d)). -12*d**4 Let t(m) = -6*m. Let u(v) = 7*v. Let z be 10/(-2)*(-1 + 0). Let h(x) = z*u(x) + 6*t(x). Let n(y) = -y**2 - 3*y**2 + 3*y**2. What is h(n(s))? s**2 Let f(j) = -29 + 11 - 12*j**2 + 18. Let c(h) = -h. Give f(c(n)). -12*n**2 Let r(m) = 26*m. Let d(z) = -8*z. Calculate r(d(w)). -208*w Let f(c) = -c**2 + 612*c. Let b(y) = -4*y. Determine f(b(x)). -16*x**2 - 2448*x Let n(l) = 12*l**2. Let g(i) = -3*i - 5. Let b(z) = -4*z - 7. Let s(y) = -5*b(y) + 7*g(y). Calculate n(s(k)). 12*k**2 Let g(r) be the second derivative of 101 + r**3 - 2*r**3 - 101 - 2*r. Let p(y) = -y**2. Determine p(g(w)). -36*w**2 Let j(t) be the third derivative of -t**4/6 + 3*t**2. Let v(o) = -2*o. Calculate j(v(h)). 8*h Let n(v) = 2*v. Let q(r) = -43*r. Calculate q(n(p)). -86*p Let b(q) = 95*q**2 + 35. Let g(i) = -8*i**2 - 3. Let l(d) = 3*b(d) + 35*g(d). Let o(h) = -3*h**2. What is l(o(p))? 45*p**4 Let d(o) = -o**2 - 13797*o - 1. Let u(l) = l**2. Determine d(u(i)). -i**4 - 13797*i**2 - 1 Let j(a) = a**2. Let v(p) = 114*p + 1. What is j(v(m))? 12996*m**2 + 228*m + 1 Suppose -4 = -5*r + 6. Let k(w) = -2 + w**r - 1 + 3. Let q(t) = 2*t**2. Calculate q(k(z)). 2*z**4 Let d(h) be the second derivative of -h**2 + 1/24*h**4 + 0*h**3 - h + 0. Let q(w) be the first derivative of d(w). Let c(r) = 3*r**2. Calculate q(c(b)). 3*b**2 Let o(x) = 19*x. Let a(h) = -2*h**2 + 5. Let l(n) = -n**2 + 3. Let t(m) = -3*a(m) + 5*l(m). Give t(o(y)). 361*y**2 Let a(z) be the first derivative of z**2/2 + 1. Let w(h) = 1. Let g(r) = -5*r - 2. Let u(j) = g(j) + 2*w(j). What is u(a(k))? -5*k Let y(r) = -7*r**2 - 4*r + 4. Let j(c) = 8*c**2 + 5*c - 5. Let v(n) = -4*j(n) - 5*y(n). Let f(p) = 3*p. What is v(f(t))? 27*t**2 Let v(q) = 3*q. Let n(o) = -o**2. Let w(j) = 2*j**2. Let c(z) = 7*n(z) + 5*w(z). Determine c(v(u)). 27*u**2 Let y be 10/4 + 2/(-4). Let h = 35 - 33. Let i(x) = -y*x + 5*x - h*x. Let z(w) = -w**2. Calculate i(z(d)). -d**2 Let x be 3/(0 - 1/(-3)). Let w(l) = 0 - 3*l + 0*l + 2. Let q(f) = 12*f - 9. Let i(n) = x*w(n) + 2*q(n). Let k(m) = -2*m**2. Give i(k(r)). 6*r**2 Let y be 1/(2/12*3). Suppose 2*s + 45 = 3*m - y*s, 4*m + s = 41. Let l(u) = -m - u + 11. Let x(v) = 2*v. Determine l(x(g)). -2*g Let y(o) = -23*o. Let w(v) = 2*v**2 + 5. Let c(q) = -3*q**2 - 7. Let m(l) = 5*c(l) + 7*w(l). Determine y(m(p)). 23*p**2 L | Low | [
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<?php class TestHTML2Md extends \PHPUnit\Framework\TestCase { protected $mdTestCases = array(); protected $restore_txt = false; protected $backupGlobalsBlacklist = ['user_info']; /** * Prepare what is necessary to use in these tests. * * setUp() is run automatically by the testing framework before each test method. */ public function setUp() { global $txt; if (!isset($txt['link'])) { $txt['link'] = 'Link'; $this->restore_txt = true; } $this->mdTestCases = array( array( 'Test bold', '<strong class="bbc_strong">bold</strong>', '**bold**', ), array( 'Named links', '<a href="http://www.elkarte.net/" class="bbc_link" target="_blank">ElkArte</a>', '[ElkArte](http://www.elkarte.net/)', ), array( 'URL link', '<a href="http://www.elkarte.net/" class="bbc_link" target="_blank">http://www.elkarte.net/</a>', '[Link](http://www.elkarte.net/)', ), array( 'Lists', '<ul class="bbc_list"><li>item</li><li><ul class="bbc_list"><li>sub item</li></ul></li><li>item</li></ul>', "* item\n* * sub item\n* item", ), array( 'Table', '<h3>Simple table</h3><br /><table><thead><tr><th>Header 1</th><th>Header 2</th></tr></thead><tbody><tr><td>Cell 1</td><td>Cell 2</td></tr><tr><td>Cell 3</td><td>Cell 4</td></tr></tbody></table>', "### Simple table\n\n| Header 1 | Header 2 |\n| -------- | -------- | \n| Cell 1 | Cell 2 |\n| Cell 3 | Cell 4 |", ), array( 'Quotes', '<blockquote><br /><p>Example:</p><pre><code>sub status {<br /> print "working";<br />}<br /></code></pre><p>Or:</p><pre><code>sub status {<br /> return "working";<br />}<br /></code></pre></blockquote>', "> \n> \n> Example:\n> \n> sub status {\n> print \"working\";\n> }\n> \n> \n> Or:\n> \n> sub status {\n> return \"working\";\n> }" ), array( 'Combo', '<ol><li><strong><em>test test</em></strong></li></ol>', '1. **_test test_**' ), array( 'PlainTxtLink', 'Thank you for registering at Awesome Forum. Your username is SomeUser. If you forget your password, you can reset it by visiting https://www.awesomeforum.com/?action=reminder Before you can login, you first need to activate your account. To do so, please follow this link: <a href="https://www.awesomeforum.com/?action=register;sa=activate;u=12345;code=S5cv#4Xh">Reg Link</a> Should you have any problems with activation, please visit https://www.awesomeforum.com/?action=register;sa=activate;u=12345 use the code "S5cv#4Xh". Gracias', 'Thank you for registering at Awesome Forum. Your username is SomeUser. If you forget your password, you can reset it by visiting [Link](https://www.awesomeforum.com/?action=reminder) Before you can login, you first need to activate your account. To do so, please follow this link: [Reg Link](https://www.awesomeforum.com/?action=register;sa=activate;u=12345;code=S5cv#4Xh) Should you have any problems with activation, please visit [Link](https://www.awesomeforum.com/?action=register;sa=activate;u=12345) use the code "S5cv#4Xh". Gracias' ), ); } public function tearDown() { global $txt; if ($this->restore_txt) unset($txt['link']); } /** * testToMarkdown, parse html to MD text and checks that the results are what we expect */ public function testToMarkdown() { foreach ($this->mdTestCases as $testcase) { $name = $testcase[0]; $test = $testcase[1]; $expected = $testcase[2]; $parser = new \ElkArte\Html2Md($test); // Convert the html to bbc $result = $parser->get_markdown(); // See if its the result we expect $this->assertEquals($expected, $result); } } } | Low | [
0.514435695538057,
24.5,
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Q: How many hours would a 38000 mAh powerbank will last? i bought a 38000 mAh powerbank and to my surprise when I could only use it for a total of 6 hours and then I have to recharge it again. My battery's capacitu is 1 700 mAh. By the way I'm taking up education major in biosci, so I'm not really expert with this. PLS. HELP. THanks. A: As Plasma says in a comment this site is overflowing with calculations about batteries and warnings about power banks (some of which my own posts), I'm in a procrastinatory mood, because I have an algorithmic task that needs to back-of-mind bubble, so I shall rehash the important stuff. That would depend on many factors: Is it a main-brand 38000mAh, or a cheap eBay/Amazon third party at a fistful of dollars? What device you connect How they rated the power (linked to 1 in many ways). Fist, 1: If it's eBay/Amazon-Third-Party there's no guarantee it's actually storing 38000mAh in any respect. Might be half, might be a quarter, might be even less. Might also be more, by the way, happened to me. Luck of the draw I guess. Second, the device you connect draws a certain amount of power. This is not directly to do with your internal battery. Some devices have a 2000mAh battery inside, but will happily pull 2A continuously on the cable as well, when in active use. Third and most imporantly, they often use 38000mAh on the internal battery. If they use normal Lithium Ion, that's a median voltage of 3.7V. mAh for a storage device with a constant voltage output with the intend of delivering a given amount of power is a stupid indication. It's why many laptop batteries quote Wh instead, because it is chemistry independent. So assuming it's LiIon, it will be about 38Ah * 3.7V =~ 140Wh. But that's its internal battery The outward voltage is regulated with electronics, which in the case of an A-brand may reach upto 95% efficiency for 5V output, or upto 90% efficiency for higher voltages. B-brand (eBay and/or Amazon-third-party sales) may be as low as 80% efficient in either case or even lower. Let's assume 80%. That leaves only 0.8*140Wh =~ 112Wh. Then if you take this out at 5V that is about: 112Wh / 5V =~ 22Ah. So if that is drained in 6 hours, the device should be taking an approximate: 22Ah / 6h =~ 3.75A. Which is very high for a 5V device, but not absolutely unheard of. But still, more likely that you actually did buy a B-brand and it's only half of what is advertised, in which case I can only say: You get what you pay for, or in this case, not what you didn't pay for. If you are using a laptop-type device at 19V or similar on the bank, then it will only be draining: 112Wh / 19V =~ 5.9Ah --> 5.9Ah / 6h =~ 0.99A. Which in turn is very low for a 19V-ish device such as a laptop or similar, in which case I'd say you probably got what you ordered. ((Note: All intermediary numbers are rounded in the post, but not on my calculator, so if you get different results from just typing the last equation in a series: Use your Memory Store key between steps.)) | Mid | [
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Google aiming to build $82 million aircraft facility at SJC, probably launch more Project Glass demos from it It's no secret that Google has an interest in the automotive industry, but over the years the popular search engine has also managed to amass quite the collection of aircrafts. So much in fact, the company is in the process of inking an $82 million construction deal that would bring its fleet to Mineta San Jose International Airport. Pending city council approval, the privately funded facility would generate an annual $2.6 million rent lease, around $400,000 in fuel revenues and create 236 jobs. If agreed upon, the 29-acre Googleport will take up to two years to build and will include an executive terminal along with hangers to house the company's private Boeing 737 and 747 jets. Google currently parks its jets at Moffett Federal Airfield, where the company has offered torenovateNASA Ames' Hanger One in exchange for two-thirds of its facility space to house its planes. There's no word if either deal will affect the other, but as it stands San Jose's city council is expected to vote on its proposal sometime in April. | Mid | [
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The invention relates to an automatic testing apparatus for testing the circumferential spacing of gears, in which for the testing operation, the gear is rotationally driven by its own power source, via a slip coupling as needed, in one rotational direction and can be indexed from one measuring position to another, and in which a primary slide for the individual testing operations is displaceable on the frame of the apparatus by a drive mechanism substantially radially toward the gear and back away from it between stops which may be adjustable; provision is made for positioning the gear for the measuring operation and for measuring the spacing deviation of one tooth edge or flank (right or left) approachable in the vicinity of the pitch circle by means of a feeler disposed on the primary slide. Also, provision is made for the automatic insertion, continuing from one tooth gap to another, of the feeler into the measuring position and for retracting the feeler back out of this position, as well as for controlling the pick-up, emission and processing of the thereby coordinated measurement value. Furthermore, the invention relates to a method for testing the circular spacing, and, in a further development, to a method for measuring deviation and gear concentricity, tooth thickness deviation and tooth gap deviation on gears which can be indexed from one measuring position to another. The testing apparatuses addressed above have varying designs and modes of operation in terms of the details thereof. In one case, the gear is rotationally connected with an incremental rotational drive means, by means of which the gear can be further divided in increments from one tooth gap to another by the spacing dimension ascertained without error, that is, by computer, which is the desired spacing dimension for the ideal case. A measuring feeler, preferably cooperating with an inductive transducer is pivotably supported on the primary slide and is retracted from the gear with the aid of the primary slide between the individual spacing steps brought about by the inductive transducer and, after the spacing step has been performed, is reinserted into the gear. For one testing revolution, the measuring feeler is in contact with one tooth flank, for example the left one, in the vicinity of the pitch circle for each measuring operation, and the deviation of this tooth flank from a zero-balance of the measuring feeler effective during a first measuring operation is ascertained and after processing is expressed by way of example by the electronic portion of the measuring apparatus. Once such a testing revolution has ended, the measuring feeler is shifted to the other tooth flank, that is, in this case, to the right tooth flank of the gear and again balanced to zero, the testing operation takes the same course as described above for one entire gear circumference. In another measuring apparatus, the gear is driven via a slip coupling in one rotational direction, and two feelers are disposed on the primary slide, for example, one fixed feeler and one pivotable feeler cooperating with an inductive transducer. In the status where they are inserted into the gear, the feelers come into contact with the same tooth flanks (right or left) in adjacent tooth gaps, the flank at which the contact takes place being dependent on the rotational direction of the gear. Now in a first measuring operation the feelers are adjusted to the same circle in the vicinity of the pitch circle of the gear, which as a rule is effected by making them just touch, and the pivotable measuring feeler is then balanced to zero. Then with the aid of the primary slide, the feelers are retracted from the gear and the gear rotates under the effect of a drive mechanism until the feelers, being shifted by one tooth gap or space, are reinserted into the gear. Here the fixed feeler then holds the gear firmly counter to the action of the slip coupling, and the pick-up of the measurement value is effected by means of the pivotable feeler, being accordingly accomplished for the next spacing. Once the gear wheel has been tested in this manner over one revolution on one edge of the teeth, the feelers are then shifted to the other tooth flank, and the rotational direction of the gear is also reversed. The adjustment of the feelers and the course of the testing are then the same as described above. In a third apparatus, the test object, again rotationally driven via a slip coupling, is firmly held by a detent device for the individual measuring operations, the detent device conventially being a ball head which is inserted into one tooth gap until it is in contact without play. On the other side of the primary slide, a transverse slide is disposed which can be displaced at a tangent to the gear counter to spring force. One fixed and one pivotable feeler, the latter cooperating with an inductive transducer, are again seated on the slide, in this case the transverse slide. Again in a first measuring operation, and with the gear held in place by means of the detent device, the two feelers are adjusted to the same circle in the vicinity of the pitch circle of the gear against one flank of the teeth, and the pivotable measuring feeler is again balanced to zero. The transverse slide in this case is shifted somewhat counter to the spring force acting upon it. For the next measuring operation, the detent device and the feelers are now retracted from the gear, and the gear rotates further by the amount of one spacing, driven by its drive mechanism, until the detent device dips into the next tooth gap and thus firmly holds the gear counter to the action of the slip coupling. The feelers are then driven into the gear teeth, and the spacing deviation is picked up via the pivotable measuring feeler and further processed in the measuring apparatus until it is transmitted. One complete testing revolution has ended, the feelers are newly adjusted against the opposite tooth flanks of two adjacent tooth gaps and balanced to zero, and the testing of the spacing deviation takes place during a new revolution of the gear. Further details of these known apparatuses for testing spacing of gears will be explained later in detail in reference to the prior art drawings. The known testing apparatuses have the disadvantage that two complete gear revolutions are required in order to ascertain errors in circumferential spacing, and furthermore the measuring feelers must be readjusted between the two revolutions from one tooth flank to the other tooth flank. This means high cost in terms of both time and money. In order to measure deviation in gear concentricity, tooth thickness and tooth gaps, which are also of interest in assessing gear quality, a different testing apparatus is required. With it, in order to measure deviation in tooth gaps and gear concentricity, a ball-like measuring feeler is inserted into each tooth gap of the test object and the depth to which it is inserted at a given time is measured, providing information as to deviation from one tooth gap to another and finally as to the deviation in concentricity. With respect to the deviations in tooth thickness, a fork-like measuring feeler is placed on each tooth of the gear wheel one after the other, and here again the depth of insertion or the deviation of the depth of insertion from an initially established zero balance is ascertained. Thus, in order to measure the three last-named values, at least further revolution of the gear and one further testing apparatus are required, still further increasing costs in time and money for testing gears beyond what was described above. | Mid | [
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Three Killed In Fresh Attack On Benue Community Yusuf 2 weeks ago Three persons have been confirmed killed while many others are still missing following Thursday night attack on Tsokwa village in Anyii, Logo Local Government Area by some armed men. The Chairman of Logo local government, Nyajo Richard, told Channels Television on Friday morning that the attackers suspected to be herdsmen stormed the village around 8:00 pm in large number shooting at different directions and burning down houses in the village, in the process. One of the villagers died at the scene of the attack while two victims who were rescued died in the process of reviving them. Three other survivors are currently receiving treatment at the local NKST hospital in Anyii, Logo Local Government Area of the state. The Local Government Chairman explained further that the Divisional Police Officer has mobilised policemen to the community for recovery operations and to search for those who are still missing since the attack. As at the time of filing this report, the Commissioner of Police is yet to respond to the fresh attack, as calls put through to his mobile phone could not go through. Many villagers were killed on January 1, 2018, where some armed men attacked Logo Local Government Area. The survivors have been displaced from their homes as they remain largely at various IDP camps in the local government because their villages are still not safe for their return. | Low | [
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Sulfur is an essential element to many organisms. Sulfur is a component of various compounds essential to life including methionine, cysteine, the B-vitamins, thiamine, and biotin, amongst others. Sources of dietary sulfur can include feed stocks with significant amounts of sulfur. As an example, some byproducts from ethanol production, such as distillers dried grains, may sometimes include significant amounts of sulfur. Sources of dietary sulfur can also include water that has significant sulfur content. In some geographic regions, groundwater can be contaminated with high amounts of naturally occurring sulfur. In many cases, sulfur is ingested in the form of sulfate compounds. In the rumen of an animal, the sulfate compounds can then be converted by microbes into hydrogen sulfide. Unfortunately, an excess of dietary sulfur (or dietary sulfur overload) can lead to various negative effects. Excess sulfur can result in restlessness, diarrhea, muscular twitching, dyspnea, and even death. Sulfur in the form of sulfide is a particularly potent neurotoxin that can rapidly produce unconsciousness and death. In the agricultural context, elevated concentrations of sulfur have been shown to result in reduced feed intake and diminished growth of animals. Accordingly, a need exists for methods and compositions for mitigating dietary sulfur. | High | [
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Having unsuccessfully tried everything listed on the forums to get an Origo wifi card working on my 860 with the latest Cacko ROM I want to know what wifi cards are you all using that will work with Cacko? I'm after a card that can be plugged in and will work with a limited amount of setting up as my knowledge of fiddling with Linux is rather limited! | Mid | [
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Voters in Washington state are making their choice for the Republican party presidential nominee. The caucus in that western state Saturday comes ahead of the so-called "Super Tuesday" voting in 10 states that could determine who will face President Barack Obama, a Democrat, in November's election. In the latest national poll (Gallup), former Massachusetts governor Mitt Romney has 35 percent support, while former U.S. senator Rick Santorum has 24 percent. The other candidates, former House speaker Newt Gingrich and Texas Representative Ron Paul, trail with 15 percent and 11 percent, respectively. At stake in Washington are 40 delegates to the Republican national convention. | Mid | [
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Plz Don't use Hotmail to Register. You might not receive Activation mail. Use Other free mail provider like Gmail or Yahoo. Topics - bandara this is my first story EVER that i've ever wrote out it's been 3 months since i came up with the plot but i've just started working on it today and my grammar isnt very good but i'll try anyways this story is called THE LAST REQUEST Keichi a novel writer with a normal family. A sister , father and mother living happily together with a normal life. Keichi talks to his sister Naomi“Hey!! Sis do you think you can read this story and give me feed back ?”“sure why not”“Children dinner is ready”“We’re coming mom”After dinner Keichi came back to his room and slept. He had a dream of running through darkness and at the end of the tunnel he saw light as he quickly ran towards the light he notices that he isn’t getting tired when he got through the light he saw blood covering his body and the room filled with the corpses of his family. His sister, father and mother was mutilated on the floor. Looking down at his hand, he was holding a blood covered axe as he rethink of the situation he realized that he doesn’t know what happened as he tries to think harder his senses was starting to fade and again he saw darkness. When he woke up he was hand cuffed and in a police car, Keichi began to panic he asked the police about what happened.“Kid you killed your family, why did you do it?”“What!!! What are you talking about ?”“Kid don’t act crazy on me you killed your family in cold blood”“But!!!...”Keichi realized that he can’t argue to the police officer . Keichi sat quietly through the whole trip. When he reached the police station he was locked up and was told that he will be prosecuted tomorrow. He slept quietly through the night.On the day of the prosecution he was brought before the judge and was instantly deemed guilty and sentenced to death. Keichi was shocked. In his cell his began thinking through the events and coming up with nothing and suddenly he hears a voice of his sister Naomi he turns around and he saw her. Keichi began to speak“but….but I thought you were dead”“I am dead brother…..i was killed by you!!!!”“I didn’t kill anyone at all”“You did!!! You killed our family and now I have come as a shinigami to look at your last moments”“shinigami what are you talking about ?”“After you killed me, a shinigami came to grant me my last request which can only be granted to dying people. I wished for the highest punishment for you brother after that I signed up to become a shinigami”“…..”“before your execution brother I want you to see my work”“what ?”In an instant Keichi was transported to a unfamiliar neighborhood with Naomi.“what are we doing here Naomi?”“I’am a shinigami remember I am here to grant a last request to a women that has only 7 days to live”Then both of them continued walking “we’re here”Naomi knocked the door and the door opens after half an hour waiting “Um...Hello sorry you had to wait but I’m very sick and I can’t really move that well”Naomi catching her breath “Hello miss Yukari I’m a shinigami that was sent to grant you last request”“last….request ?”“a last request a wish that will be granted by a shinigami to a dying person that doesn’t have a long time to live”“ohhh”“you don’t seem surprise by it miss Yukari”“well I have a terminal heart disease and I already knew that I was going to die soon”Keichi was standing there silently“So what is your wish miss Yukari”“I wish for a healthy body”“I can grant you that wish but it will only be temporary because us shinigami can’tchange the fate of a person”“ok please grant me a healthy body I was sick for my whole life at least I want to know how it feels to be healthy”Naomi stated to chant a spell and a bright light was flashing in the room“ok it’s done”Yukari started to stand up and walk and run around her house like a healthy person totally different from when they first saw her. this isnt the end of the chapter yet. more will come soon. Feed back is really appreciated a young man last sight was of his sister holding a big knife splashed with the blood of their family and then a shinigami(reaper) came to take him but as a rule among the shinigami they must grant one last wish to the person who is going to die or was murdered so without hesitation he wishes that his sister received the highest punishment possible BUT did his sister Really killed their family call me Ban i'm an amature in drawing (took a little bit of art in school) i like to draw manga when i'm bored 01. Male02. I found this website on google03. i'm just a super amature at drawing04. i'm 1605. i'm from chicago06. favorite Manga/Anime Shakugan No Shana07. i would like to post my drawings in the forum | Mid | [
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We are pleased to see the Government of Ontario reaffirm its support of the Climate Change Action Plan and outline strategies to accelerate the transition to an innovation-based, low-carbon economy through investments in affordable housing, retrofit activities in social housing units, renewable energy technologies, cycling infrastructure and electric vehicle (EV) charging. Contributing to the acceleration of the low-carbon economy is a focus for Evergreen. This year, we are embarking on the redevelopment of the historic kiln building at Evergreen Brick Works to create a year round hub that will ignite, accelerate and celebrate new innovations, practices and policies to shape our cities for the better. This retrofit is one of the first in Canada to set and strive for a carbon neutral target. The project began in early 2017 and continues through the fall. The Government’s plan to leverage provincial land assets to build more affordable housing units across the province is exciting. Through our research as part of the GTA Housing Action Lab, Evergreen has been advocating for programs and policies that support the affordability of housing to ensure residents of all incomes have the best chance to live in a suitable home and have a choice in their housing. Piloting their program at multiple sites in the Greater Toronto and Hamilton Area to develop up to 2,000 new housing units, in addition to the changes outlined in the Fair Housing Plan signal Ontario’s commitment to remain a leader in building and supporting flourishing cities. We look forward to learning more. More needs to be done to ensure that the environment remains a priority. We hope to see more clarity on matching federal investments in transit and social housing to support the transition to a low carbon economy, creating flourishing cities across the province. | Mid | [
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§§ §§ cf 612 No. 27953 w~ iv ¢"~J INtTHE INTERMEDIATE coURT oF APPEALs §§ ,y §§ “§ oF THE sTATE oF HAwAI%c §§ le *~f CITY BANK, Plaintiff/Counterclaim-Defendant/Appellee, V. ARTEMIO MARCOS ABAD, INDIVIDUALLY AND AS TRUSTEE OF THE ARTEMIO M. ABAD REVOCABLE TRUST; GLORIA PASCUA ABAD, INDIVIDUALLY AND AS TRUSTEE OF THE GLORIA P. ABAD REVOCABLE TRUST; JENNIFER ABAD, Defendants/Cross-Claim Defendants/Appellants, and_ FINANCE FACTORS, LIMITED, Defendant/CrOSS- claimant/Counterclaimant/Appellee; DIRECTOR OF BUDGET AND FISCAL SERVICES, CITY AND COUNTY OF HONOLULU, Defendants/Cross-Claim Defendants/Appellees; and FELIX PASCUA; JOHN DOES l-50; JANE DOES l-50; DOE PARTNERSHIPS l-50; DOE CORPORATIONS, DOE "NON-PROFIT" CORPORATIONS 1-lO, and DOE GOVERNMENTAL UNITS l-50, Defendants; and FINANCE FACTORS, LIMITED, Third-Party Plaintiff- Appellee, v. AMERICAN GENERAL FINANCE, INC., Third-Party Defendant-Appellee, and ERLINDA GONZALES and WILSON U. PASCUA, individually and doing business as HONLAND REALTY, Third-Party Defendants APPEAL FRoM THE cIRcU1T coURT oF THE FIRsT cIRcUIT (civ. No. 01-1-0995) ORDER DENYING THE MOTION FOR RECONSIDERATION (By: Foley, Presiding Judge, Fujise and Leonard, JJ.) Upon consideration of "Appellants' Motion for Reconsideration of Order Granting Request for Attorney's Fees and in the Adjusted Amount for Attorney's Costs, Filed June 29, 20lO" (Motion for Reconsideration) submitted by Gary victor Dubin, on behalf of Defendant-Appellants Artemio Marcos Abad, individually and as trustee of the Artemio M. Abad Revocable Trust, Gloria Pascua Abad, individually and as Trustee of the Glora P. Abad Revocable Trust, and Jennifer Abad, HRS §§ 607-14 and 607-9, Hawafi Rules of Appellate Procedure Rule 39, the papers in support, and the records and files herein, IT IS HEREBY ORDERED that the Motion for Reconsideration is denied. DATED: Honolulu, HawaiUq July 2l, 20l0. Presiding Ju:;:§§L2 dev Associate Judge | Low | [
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Invert the colors of a picture in Windows 7 One of the easiest ways to add interest and visual impact to a stock photo, digital picture or clipart image is to simply invert its colors; you can do that in Windows 7 without having to download any additional software, as the MS Paint graphic editor (which ships as part of the operating system) includes this very functionality. In this tutorial, you will learn how to invert the colors of entire pictures, and how to selectively invert the colors of portions of an image. Invert colors of part of an image of the entire picture Follow these steps to invert colors inside Paint on Windows 7: First, open the picture whose colors you want to invert; right-click on the image file, and choose "Open With"; then, choose "Paint" from the submenu. Once the photo is opened inside Paint, choose the "Select" tool under the Home tab. You can now right-click anywhere inside the image, and choose "Invert Colors" at the very bottom of the context menu, as shown below: MS Paint will apply the inverted colors effect and Windows 7 will display an updated image: (note that unlike previous versions of Windows, you cannot use the Ctrl+I keyboard shortcut to invert colors) To undo the invert-color command, simply press Ctrl+Z on your keyboard (or click on the "Undo" button) - press Ctrl+Y to redo and toggle back and forth. Once you are satisfied with the new colors of your image, you can save it. We will now explain how to invert the colors of a select portion of the picture, which makes for impactful results: With the Select tool, drag around the area of the image where you want to invert colors; then, right-click inside the selected area, and choose "Invert Colors" from the context menu: As you can see, you can combine the inverted colors effect and layer it as many times as you like. Once you are done inverting colors, click on the "Save" button. This is all it takes to invert the colors of an entire image, or a portion of it. Since you can copy and paste pictures between MS Paint and other programs, like Microsoft Word, you can add some visual effects to graphic files inside Paint, and then re-use them in other applications. | High | [
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Conjoined double internal mammary artery grafting. Double internal mammary artery (IMA) grafting to the coronary arteries was performed on 82 patients. In ten of them one IMA was used as a free graft, and was proximally connected to the other ("conjoined" double IMA), the indications being insufficient supply of adequate veins, diseased aortic wall, availability of only a short right IMA segment for free grafting, occluded left subclavian artery, and when distal and scattered lesions of relatively important vessels had to be bypassed. Additional sequential IMA anastomoses were performed in four patients and an additional sequential vein graft in a fifth. All patients became angina-free postoperatively and have remained so during observation up to 16 months. Recatheterization studies were performed in six patients and in all of them the IMA-to-IMA (n = 6) and distal anastomoses (n = 26) were patent. Conjoined double IMA grafting is an important option available to the cardiac surgeon for management of selected patients with coronary artery disease. | Mid | [
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Friday, October 9, 2015 I received these markers for review from Chalkola. I definitely give them 5 out of 5 stars. The colors are bright and vibrant. The markers are easy to use and clean up nicely off chalkboard, glass, and mirrors with just a wet cloth. Even my children were excited about using them as you can see. They are non toxic and do not have a strong smell. They are also acid free. I did a technique comparison and the markers worked for all of them but I thought the blending was the most fun. I highly recommend these. You can purchase them here: Amazon USA link - http://www.amazon.com/dp/B00PM473TK Amazon UK Link - http://www.amazon.co.uk/dp/B00VVOWCSC You can read more about the markers on their website http://www.chalkola.com/ Tuesday, October 14, 2014 These layouts were created using the Scaredy Cat paper pack, plus the matching the complements and assortment pack. I had to add the purple glitter paper and some slate and black cardstock in order to make all 5 layouts. What Else Do I Offer... Want Freebies? Simply, host a home gathering and get a free stamp set, PLUS free product based on sales! Also, ask me how to get a free stamp each month! You can also get lots of free product with our constant campaigns that change each month. Need an Idea Book? Filled with over 600 card ideas and scrapbook layouts, the Idea Book helps inspire creativity and is available for only $3.95 US plus $2 shipping. It is an amazing book full of beautiful card and scrapbooking ideas. I am happy to mail you a copy. What I can offer scrapbookers? Do you want to leave a lasting legacy for your family? Close To My Heart has everything you need to preserve your family’s legacy and celebrate lasting relationships. Many people have told me they just don’t know where to start; designing pages that look nice is overwhelming. Created to be simple and easy, the pre-printed My Reflections® Scrapbook Kits have something for everyone. Not only are the My Reflections® Kits perfect for beginners, they also save valuable time and money for experienced scrapbookers. Scrapbooking and stamping go hand in hand. Stamps allow you the freedom to create pages around your pictures instead of arranging your pictures around a page. Every stamp can be used in a variety of different themes and looks, and they’re an investment you can use over and over again. You’ll be amazed and pleased with Close To My Heart’s EXCLUSIVE colors available in cardstock, background & texture paper, stamp pads and markers. All the colors are divided into five different seasons that coordinate perfectly—you’ll never have to guess which colors go together. Why should you make cards? Did you know that most of us spend 17 minutes shopping for a card, only to leave with something we think is less than perfect? Think of how rarely you’ve found a card with a message and artwork perfect for the person in mind. To get a unique card that delivers just the right message, why not make it yourself? There’s something truly special about receiving a handmade card—it says volumes about the giver and the receiver. Store-bought cards can be costly, and finding one you like can take too much time. With the right supplies and guidance you can quickly create a gorgeous card that conveys exactly the message you want for a fraction of the cost! Contact Me Today! I would love to show you and your friends how to create beautiful artwork to honor the special relationships in your lives. Please contact me to see the latest exclusive products from Close To My Heart, or to schedule a party today! | Mid | [
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Why is there often a disconnect between the religions with which people claim affiliation and the behaviour of those people in everyday life? This seems to be more common in direct proportion to the fervour of the believer. I've personally known people who will festoon every available surface with pictures of HHDL and yet embody none of the virtues of their idol. Similarly I have met people who claim staunch adherence to the Church (Christianity) who will offer harsh reaction to a hungry beggar, and yet think nothing of turning up to the house of their loving Christ in obscenely expensive vehicles. So, from where does this disparity arise? “Not till your thoughts cease all their branching here and there, not till you abandon all thoughts of seeking for something, not till your mind is motionless as wood or stone, will you be on the right road to the Gate.” Once I used to ask people: so, what do you like about (say) HHDL? And after the stock answer (it really is predictable) concerning world peace and being kind and so on, I would ask how they bring this teaching into their daily lives. This would be met by a lot of hostility. I wasn't really trying to be provocative rather simply interested in meeting people who at least attempted to enjoin with the practice of their chosen mentors. I guess I am unskillful in asking questions along with much else besides. These days I just observe and don't ask. Actions speak louder and all that. “Not till your thoughts cease all their branching here and there, not till you abandon all thoughts of seeking for something, not till your mind is motionless as wood or stone, will you be on the right road to the Gate.” Once I used to ask people: so, what do you like about (say) HHDL? And after the stock answer (it really is predictable) concerning world peace and being kind and so on, I would ask how they bring this teaching into their daily lives. This would be met by a lot of hostility. I wasn't really trying to be provocative rather simply interested in meeting people who at least attempted to enjoin with the practice of their chosen mentors. I guess I am unskillful in asking questions along with much else besides. These days I just observe and don't ask. Actions speak louder and all that. Seems very skilful to me - if you provoke such a reaction then this highlights something about the respondent. I think in your OP the word "fervour" is particularly striking, and suggests to me an identification which renders any actual practice useless. we cannot get rid of God because we still believe in grammar - Nietzsche Because humans are flawed, until they are absolutely 100% enlightened and perfect. And statistically speaking, we've probably seen a lot more unenlightened than enlightened human beings. And once we're enlightened, there's no need to practice or aspire to be like those idols and gurus anymore. Most of us are unwilling to find the nature independent of externals. So for example we go for obvious forms of behaviour, kindliness, beauty etc.As we begin to realise our own superficial assessment of other as reflective of our ignorance, we start to look beyond the limitations of form. In essence we practice without the need for these qualities to be acknowledged, to have a crass form but to be intangible and changeable . . . Once I used to ask people: so, what do you like about (say) HHDL? And after the stock answer (it really is predictable) concerning world peace and being kind and so on, I would ask how they bring this teaching into their daily lives. This would be met by a lot of hostility. I wasn't really trying to be provocative rather simply interested in meeting people who at least attempted to enjoin with the practice of their chosen mentors. I guess I am unskillful in asking questions along with much else besides. These days I just observe and don't ask. Actions speak louder and all that. Seems very skilful to me - if you provoke such a reaction then this highlights something about the respondent. the reaction also highlights something about the questioner. This isn't specific to Qing Tian, but in all interactions. Once I used to ask people: so, what do you like about (say) HHDL? And after the stock answer (it really is predictable) concerning world peace and being kind and so on, I would ask how they bring this teaching into their daily lives. This would be met by a lot of hostility. I wasn't really trying to be provocative rather simply interested in meeting people who at least attempted to enjoin with the practice of their chosen mentors. I guess I am unskillful in asking questions along with much else besides. These days I just observe and don't ask. Actions speak louder and all that. Seems very skilful to me - if you provoke such a reaction then this highlights something about the respondent. the reaction also highlights something about the questioner. This isn't specific to Qing Tian, but in all interactions. Of course, but if the question asked is simply how they plan on applying the teaching in their everyday life, then a hostile reaction would seem to indicate exactly the lack of applied practice that was enquired about in the first place. It also depends how you ask, I guess. we cannot get rid of God because we still believe in grammar - Nietzsche Why is there often a disconnect between the religions with which people claim affiliation and the behaviour of those people in everyday life? This seems to be more common in direct proportion to the fervour of the believer. I've personally known people who will festoon every available surface with pictures of HHDL and yet embody none of the virtues of their idol. Similarly I have met people who claim staunch adherence to the Church (Christianity) who will offer harsh reaction to a hungry beggar, and yet think nothing of turning up to the house of their loving Christ in obscenely expensive vehicles. So, from where does this disparity arise? Spiritual materialism, dissatisfaction with oneself, and an attempt to create a better "spiritual version" of that self. Trying to mold the world into something it's not. The rest of it is just Samsara I think, it's really hard to check our own motivations behind things, sometimes exhausting, and I think it's easy to just give up in favor of the fantasy version. I've seen some of the most sanctimonious garbage come from "religious" people, on here, other forums, in person...I do it myself commonly too i'm sure. Sorry to be so blunt but yeah, with a few really inspiring exceptions, most religious folks don't impress me with their ethical discipline, especially as regards the simplest day to day interactions...not that the non-religious impress too much either, mind you. The kindest people i've known (this is just subjective experiences so don't jump on me) have been people who have kind of been through the ringer and came out stronger, with different priorities, heavy mental illness, addiction, abuse, facing death etc. I have know a few really inspiring, genuinely good and gentle people that fit this description, i'd say they were "religious" in the most important meaning of the word as they certainly had some kind of revelation about what is important, and what isn't that I think most of us have hard time seeing in daily life. For the most part, failure to practice what we preach is the norm I think, maybe that's all the more reason to work towards consistency. "it must be coming from the mouthy mastermind of raunchy rapper, Johnny Dangerous” Johnny Dangerous wrote:The kindest people i've known (this is just subjective experiences so don't jump on me) have been people who have kind of been through the ringer and came out stronger, with different priorities, heavy mental illness, addiction, abuse, facing death etc. I have know a few really inspiring, genuinely good and gentle people that fit this description, i'd say they were "religious" in the most important meaning of the word as they certainly had some kind of revelation about what is important, and what isn't that I think most of us have hard time seeing in daily life. For the most part, failure to practice what we preach is the norm I think, maybe that's all the more reason to work towards consistency. I think in large part people lack a high degree of empathy, or empathy is not as strong in human beings as we like to believe. While I think empathy is innate, how it's applied (and who it is applied towards) is learned. We learn to direct our empathy towards certain people or situations, etc. A very broad example, in the US, is how much people empathized with the victims of the Boston bombings. Yet many of these same people don't give the victims of similar bombings in other countries (and carried out by our own government!) a second thought. Unless we are taught at an early age, it seems we lack the imagination to put ourselves into the role of the other until we experience their situation for ourselves. So until we are divorced, robbed, lose our jobs, are betrayed, have major health issues, etc. we don't feel the suffering other people go through. On a larger scale, we identify with our tribe, school, town, state, country. We can go to war and kill people against "outsiders" because they are not one of us. We don't even see them as human beings. Then there is cognitive dissonance. We know we "should" feel a certain way, but we don't (see above), so to resolve the dissonance we justify how we behave with what we say we believe. Basically it's human nature. It's probably results from life being so full of suffering, from the very beginnings for homo sapiens, if we were to really feel it, we'd be unable to function. That would not be a very good survival trait for a species. You don't see a lot of angst amongst the herds on the Serengeti after one of their own becomes dinner for a pack of lions - in some ways we're just a another herd clearbluesky, "And statistically speaking, we've probably seen a lot more unenlightened than enlightened human beings" Statistically speaking ...let me remind you how rare enlightenment is; perhaps in the crush of 7 billion people there maybe a handful of fully enlightened beings and most of us wouldn't know or recognise a fully awakened person if we tripped over one! Uan, We all need sympathy. Real sympathy is largly lacking amongst the religious and secular as far as I can see. Unfortunately consume is the buzzword, money is the passport to fickle status, a better life, a happy life and consequently for many people there's not much energy left over for deeper and more meaningful matters. greentara wrote:clearbluesky, "And statistically speaking, we've probably seen a lot more unenlightened than enlightened human beings" Statistically speaking ...let me remind you how rare enlightenment is; perhaps in the crush of 7 billion people there maybe a handful of fully enlightened beings and most of us wouldn't know or recognise a fully awakened person if we tripped over one! You're reminding me by agreeing you're saying? I assume that's what you mean, but yep, a handful in 7 billion, and that's just one realm of Samsara! So rare in fact, that even those lucky enough to have a picture of the Dalai Lama, disparity or not, could be said to be somewhat fortunate in this life. Once I used to ask people: so, what do you like about (say) HHDL? And after the stock answer (it really is predictable) concerning world peace and being kind and so on, I would ask how they bring this teaching into their daily lives. This would be met by a lot of hostility. I wasn't really trying to be provocative rather simply interested in meeting people who at least attempted to enjoin with the practice of their chosen mentors. ... Alot of what people know about the Dalai Lama is a marketing image. For example, recently in the Tibetan exile community in India, one five year old girl was raped and the perpetrators went free: Sources further said that the father of the minor initially wanted to report to the police but later declined to do so after discussions with the camp administration which suggested that the matter be internally resolved as "it would bring bad name to the Tibetan leader the Dalai Lama." ... So two Tibetan rapist pedophiles will go free to salvage the image of Tibetan society as the Shangri-la created by Western Romantics in the 19th Century and carried into this day by religious Western Neo-Romantics. Essentially this is the negotiating position of the Tibetan government in exile: this false false Shangri-la depiction of Tibetan society with Dalai Lama as the spiritual saint who can do no wrong, and no wrong can be done in his name. He even threatens Tibetans that he will die earlier, from time to time. They know what impact this will have, as their entire negotiating strategy with China as an exile community depends on his personality cult. So rapists go free, and dirty laundry, stays dirty, to uphold an image that lacks substance. Thrasymachus, The Dalai Lama is caught between a rock and a hard place. Trying to keep the small Tibetan enclave afloat in India which is a hotbed of corruption and 'negotiating' with the hostile Chinese is a supreme juggling act. All say he is the spiritual leader but he is both a spiritual and political icon, very like Mahatma Gandhi. The Tibetans and their Western supporters have a childish Saint-Bodhisattva image they created out of the Dalai Lama. And the Chinese on the other side have a childish demonization. The truth of the actual man himself, is forever murdered between the two extremes. From the documentary "10 Questions for the Dalai Lama," you can see that the Tibetan authorities carefully vet who gets a chance for an interview or to produce a documentary or book involving the Dalai Lama™. Needless to say if anyone asked inconvenient questions, they would be blacklisted, under the guise of censorship for the Tibetans & Western buddhists own good via "right speech". Two commenters on the news article I previously published, sent this in: Bodgirl(UK), stherchin(Dharamasala) wrote:Bodgirl, Tibetan community:Just to paint glossy community look to the world, all the criminals are always made to go scot free whether it is rape, theft or any kind of abuses. Just add Dalai Lama's name and victims are emotionally blackmailed to follow the village head orders or else they will be shunned by the community. If these two men are not given harsher punishment, there will be more cases like these in the future. Covering the crime is also a crime and using H.H name in covering up crime is completely wrong. No wonder we lost our country. stherchin, Should be handed over to policeDo we really want to encourage such heinous crime at the cost of fearing negative publicity? Settlement office, or for that matter even CTA does not have mechanism and right to decide on such crimes….We are what we are and shouldn't shy away from projecting the truth. The camp administration is being hypocritical with comments like “it would bring bad name to the Tibetan leader the Dalai Lama." If they really cared about His Holiness image then they would have taken her to nearby hospital and reported to the police station. Because that’s what a law abiding resident does. Would the camp administration give the same comment if this had happened to their daughter or sisters? I don’t think so. This isn’t the first time that rape cases have happened in our community. It’s just that we don’t have practice and guts to go to police. If today we decide for internal resolve fearing over exaggerating by Indian media then tomorrow there will be more of such crimes, perhaps even murders. In fact we already have murder cases and lots of stabbing cases right from north Dharamsala to south Mundgod. If we live in India then we must follow Indian law. We can’t be like animals ‘ramalook’. Remember, withholding information or not reporting such heinous crime is also a crime. Let this incident be a medium to put fear on wrongdoers in our community. My support and sympathy to the victim. Those monsters should be punished as per the law of the land It seems more Tibetans know more about this type of nonsense and are fatigued that it is still going on, compared to the Western religious romantics with their more outdated, obsolete image upheld largely intact from their 19th Century Romanticist antecedents. This probably due to proximity: the Tibetans in exile are experiencing this type of hypocrisy and image construction at their expense, while the distant Western have the luxury to uphold the hypocritical image that will not impact their life one bit, other than giving them an ideal they would like to believe in. Mental afflictions aren't active in those people when they express admiration for wholesome qualities. When they're dealing with possessions and beggars the mental afflictions are actively distorting their reasoning. Unfortunately you can't kill mental afflictions with just admiration of the wholesome. If there is a radical inconsistency between your statements and the position you claim to hold,you are a sock puppet.Make as many accounts as you want; people can identify your deception with this test. | Mid | [
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Bile evacuation induced by hypothalamic stimulation in dogs. To investigate the role of the central nervous system in the cephalic phase of gallbladder emptying, the effects of electrical stimulation of the hypothalamus on the gallbladder and sphincter of Oddi and on bile flow rate were studied in anesthetized dogs. Stimulation of the anterior hypothalamic area induced gallbladder contraction and simultaneous relaxation of the sphincter without a significant increase in plasma cholecystokinin concentration. The bile flow rate was facilitated by stimulation of the same area. These responses were obtained even in dogs with cervical cord transection and were abolished by cervical vagotomy. The contraction response in the gallbladder was prevented by atropine, but a slight relaxation response in the sphincter remained. On the other hand, stimulation of the posterior hypothalamic area induced gallbladder relaxation and contraction or relaxation in the sphincter. The bile flow rate was decreased by stimulation of the same area. These responses were still obtained after vagotomy. These results indicate that the anterior hypothalamic area and the vagus nerves play an important role in gallbladder emptying in the cephalic phase. | Mid | [
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Arturo Boiocchi Arturo Boiocchi (; 7 October 1888 – 1964) was an Italian professional footballer who played as a striker. At club level he played for U.S. Milanese for 15 years. At international level he played in the first ever game of the Italy national football team on 15 May, 1910, against France. External links Category:1888 births Category:1964 deaths Category:Italian footballers Category:Italy international footballers Category:Association football forwards | Mid | [
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Q: Form Resizing Responsive issue textarea I've got an issue with my responsive textarea. It works fine (across different sizes) until the user tries to expand the form (I've currently got it locked in size with a height/max-height/width/max-width combo). Is there a solution to this? What am I doing wrong - haven't seen this issue before. Album - http://imgur.com/a/9ICnF#0 Website - http://pxlprfct.co.uk/ Html <textarea name="message" id="message" placeholder="Message"></textarea> Css form{ border: none; outline: none; text-align: center; text-decoration: none; font-family: 'Open Sans', sans-serif; font-weight: 700; color: white; text-transform: uppercase; background: black; /*Android Hack*/ margin-top: -1px; padding-bottom: 2em; } #message { border: none; outline: none; color: black; display: block; background: white; padding: 1em; margin: auto; font-size: 1em; margin-bottom: 1em; font-family: 'Open Sans', sans-serif; height: 12em; max-height: 12em; width: 70%; max-width: 70%; } Sorry about the messy code - it's all been hacked together quickly! :) Thanks in advance! http://jsfiddle.net/qfYQN/1/ A: I can't think of a solid resolution to this issue. The way I see it, you have two options: Disable the resize css property or use some javascript to reset the margin to auto when the user resizes the box. css: resize:none; jquery: Unfortunately there is not an event listener for a text box resizing so this would be the simplest method: $('#message').mouseup(function(){ $(this).css('margin', 'auto'); }); http://jsfiddle.net/WLBde/ Otherwise you could set an interval that checks the width of the text box every X seconds and if it is different then reset the margins like above. Hope this helps! | Low | [
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Q: Non-GUI Emacs with cscope So, i'm running emacs over a crappy ssh connection and I have it set up to use cscope. I can not use X because of this...hence I'm running emacs inside putty. However, when I search for something with cscope and it opens up the other buffer, I can not follow the links where cscope tells me which file and line number the item is on. When I go t a line number and hit enter, emacs tells me 'buffer is read-only' (it is trying to actually put in a new line instead of following the link). anyone know how I can follow those links? A: I don't know about cscope for sure - but you should be able to find out the appropriate key binding by doing a "Ctrl-h m" in the buffer with all the links. This should open another buffer showing you help/key bindings on all the active modes. E.g. if you do the same thing in a grep result buffer it indicates the key binding "C-c C-c compile-goto-error" which is used to open file at the grep line number (so it may be the same keys for cscope). | Mid | [
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Santa should be worried. A recent trip to Grafton Street brought us past the Disney store, outside which a huge crowd gathered. We approached, and the younger daughter cried out, "Happy snowman!" Ah. Frozen. My two girls joined the crowd and gazed at the windows. Olaf, the aforementioned happy snowman - it's a song; if you don't know, you don't need to - was present in several guises, including as a moving figure. It was, I thought, like Switzers' Christmas windows, which provided the magic of my mid-1970s childhood. My girls were wide-eyed at the snowy spectacle of Frozen brought to life, and it was pretty cute to see them run inside to pet the nose of Sven, Frozen's Rudolph. Inside the shop, Let It Go was on a loop and grown adults sang along mindessly, but word-perfect. Read More The shop in which Frozen merchandise far outnumbered anything else was pretty exciting to me, too, as I reach the end of a year-long immersion in all things Anna and Elsa. And, albeit silently, I sang along with Elsa's 'free-at-last' anthem with the best of them. This may become what Christmas is about, or Christmas 2014, at least. Frozen toys are at the top of most-wanted lists; Disney has re-released the film into cinemas; Poundland in the UK sold out of Frozen merchandise within 10 minutes, and the elevated price of Frozen dress-up outfits was not deterring Irish parents. I wonder if somewhere, someone is rubbing their hands together with Disney-villain glee at how their plan worked out: Frozen has achieved world domination; part of me wonders if Frozen was created by an algorithm, such is the extent to which it has everything that not only a little girl wants, but, to a great extent, what parents want for their little girls, too. Read More In case you've been living in a cave, or have absolutely no contact with the under-10s, this is the deal. Anna and Elsa are sisters. Princesses, obviously. Elsa has the power to freeze things, and when she hurts her little sister Anna, she locks herself away for all of their childhood and then, later, banishes herself to an icy kingdom where she can never hurt anyone again. Cue Let It Go, a song about leaving the past behind and being immune to hurt and pain, but overlooking the fact that this cuts you off from love, too. The millions of little girls who know this song by heart might not fully grasp the subtleties, but their amazing mimicry of Anna's high-emotion arm-waving and head-shaking suggests that her defiance speaks to them on some instinctive level. Which works wonders on parents who worry about where their little girls will get their self-worth, and how they can help to boost same, before they're unleashed on Instagram. Then, basically, and because she simply won't let her go, Anna goes in search of Elsa and sisterly love conquers all. Yes, there are romantic subplots, but they're not what matters. What matters is the girls, and this is what works for children and parents alike. Kids aren't into Disney movies for the adult notions of my-handsome-prince love. They care about kids' relationships with each other, and they love Anna and Elsa taking charge of their lives without interfering adults. Read More For the parents, it's craftier again. So Anna and Elsa have impossible, improbable bodies - huge eyes and tiny noses, super-skinny limbs and tiny waists - but their self-worth lies in sisterly love, not any pesky prince, and they tell little girls to be brave and strong and assertive. Honestly, if it wasn't so cute and catchy, you'd have to hate the unabashed manipulation of an entire generation, who will be around this Christmas in their Frozen PJs, playing with their Anna and Elsa dolls, and singing along to the soundtrack. The red and white of Santa and Christmas - admittedly foisted upon us by Coca Cola - is being usurped by the pale blue of Frozen, and, humming Let it Go like our mantra, we're blithely going along with it. Happy snowman is right. Sunday Independent | Mid | [
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Berluti Men's Cotton Jacket This designer fashion was made by French fashion designer Berluti. It was first offered for sale by online retailer Neiman Marcus on March 2, 2018. Men's cotton and denim material describe this item. Fashion type has been filed under jacket, coat and clothing. Jacket price from online retailer Neiman Marcus is now 1490.00 USD. Neiman Marcus currently offers a FREE SHIPPING promotion good until January 1, 2020. They also have a COUPON CODE: OCTGC good until October 20, 2018.Gift Card Event! Receive Up to a $500 Gift Card with Your $200 or More Purchase at Neiman Marcus.Com! Beauty and Fragrance Included. Use Code OCTGC. Offer Valid 10/17-10/19. Information on this item, current pricing, and online store promotions were last checked by our editor October 18, 2018. Browse the profile page of Berluti for designer label information and other Berluti brand fashion online. | Low | [
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Q: In Python, all literal values result in the creation of objects What does this mean "In Python, all literal values result in the creation of objects"? I am learning Python, and got this sentence from here: http://python4java.necaiseweb.org/Fundamentals/FunctionsAndMethods But I does not really understand its meaning. First, what are "literal values"? Second, please explain this sentence. If you could make some examples, that would be helpful! A: Python actually has a few definitions for "literal" in different contexts, but if you just want the basic idea: Numbers like 123 and strings like "abc" are literals; expressions like 123 + 456 are not. In Java, when you write 123, that's not a reference to a Java object, it's a "native" integer. In Python, it is a reference to a Python object. Because everything is an object in Python, an integer has methods, can be stuck in collections, etc. There's no "boxing" and "unboxing" anywhere. If I want to stick 123 into a list, I just do it: >>> my_list = list() >>> my_list.append(123) >>> my_list [123] And if I want to use the value from the list as an integer, I just do it: >>> my_list[0] - 120 3 For that matter, I can just write a list display, using literals just like other objects: >>> my_other_list = [my_list, 2] (Just don't ask whether a list display is also a literal, because that's when the "different definitions for different contexts" actually matters…) It's worth pointing out that it's not really true that "all literal values result in the creation of objects". A literal may be a new object, but it may also be a reference to existing object with the same value that. For example: >>> a = 3 >>> b = 3 >>> a is b True >>> id(a) == id(b) True (This isn't guaranteed to be true by the language, but it usually will be on most Python implementations.) So, b = 3 did not result in the creation of an object, just another reference to the same object in a = 3. (And in fact, that 3 was most likely already pre-built and pre-cached by the interpreter before even looking at your code.) But you don't need to care about this, because 3 is immutable. It doesn't matter if you get the same 3 object or a different one because, short of is and id, there is no way to tell the difference. And the same is true for strings, floats, etc. | High | [
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# See docs/tracing.txt for syntax documentation. # linux-user/signal.c user_setup_frame(void *env, uint64_t frame_addr) "env=%p frame_addr=%"PRIx64 user_setup_rt_frame(void *env, uint64_t frame_addr) "env=%p frame_addr=%"PRIx64 user_do_rt_sigreturn(void *env, uint64_t frame_addr) "env=%p frame_addr=%"PRIx64 user_do_sigreturn(void *env, uint64_t frame_addr) "env=%p frame_addr=%"PRIx64 user_force_sig(void *env, int target_sig, int host_sig) "env=%p signal %d (host %d)" user_handle_signal(void *env, int target_sig) "env=%p signal %d" user_host_signal(void *env, int host_sig, int target_sig) "env=%p signal %d (target %d(" user_queue_signal(void *env, int target_sig) "env=%p signal %d" user_s390x_restore_sigregs(void *env, uint64_t sc_psw_addr, uint64_t env_psw_addr) "env=%p frame psw.addr %"PRIx64 " current psw.addr %"PRIx64 | Mid | [
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酒店信息 Green Gallery Hua Hin Toward the end of Soi 51 in the renowned resort town of Hua Hin is a green house which has been well maintained over several decades. Once, this house was a vacation home of Princess Raohinavadi Dissakul ,a daughter of Prince Damrong Rajanubhab and Saeng Satarat. The house was built in traditional Thai style, blending with colonial architecture. Daily breakfast will serve you all. Your pet is welcome for Courtyard zone and Beach side zone (Limited weight 5Kg.) ***No pet allowed in Gallery Zone Green Gallery Hua Hin Sea-Thru pool view Sea-Thru pool view This room is for any pink color lovers. The room uses pink curtains, pillows, and even floral painting on the bare brick wall above the bed. The highlight is the glass-walled bathroom seen through from the bed. The room’s deck has a chair for your pleasure.Room size 28 sq m. 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Dogs welcome for under weight 5Kg. rate 500 Baht/Night/Dog 1特大床 禁烟 客房服务 DVD 播放器 迷你冰箱 吹风机 iPod Dock 淋浴 — 单独用 提供亚麻织物和毛巾 茶/咖啡制作 电视 无线网 空调 室内影院 有线/卫星电视 免费水果篮 日常客房服务 阳台 迷你酒吧 CD 播放器 Green Gallery Hua Hin Erotico sea side view Erotico sea side view Assuming that the length warred design concept reflects the symbolic gestures of love and music notes. Next to this room is just a few meters from the sea you can see the sea from the bedroom window or take a few step to get a 180 degree view of the sea from private deck. Room size 25 sqm. Dogs welcome for under weight 5Kg. rate 500Baht/Night/Dog 1特大床 禁烟 服务台 DVD 播放器 茶/咖啡制作 户外设置 日常客房服务 客房服务 提供亚麻织物和毛巾 有线/卫星电视 电视 无线网 空调 迷你酒吧 CD 播放器 免费水果篮 淋浴 — 单独用 阳台 迷你冰箱 吹风机 iPod Dock Green Gallery Hua Hin Monalisa sea side view Monalisa sea side view Inspired from the world most popular art with the smiling knockout of this woman Mona Lisa. Inside you will experience a unique getaway in chic d?cor. You can see the smile of the Mona Lisa from almost every angle with the panoramic sea view on the private Room Size 25 sq.m. Dogs welcome for under weight 5Kg. rate 500 Baht/Night/Dog 1大床 禁烟 客房服务 提供亚麻织物和毛巾 CD 播放器 免费水果篮 室内影院 空调 迷你酒吧 吹风机 日常客房服务 有线/卫星电视 阳台 迷你冰箱 电视 无线网 淋浴 — 单独用 DVD 播放器 茶/咖啡制作 Green Gallery Hua Hin Blue and White beachfront suit Blue and White beachfront suit Fill your holiday with blue and white color matching with the ocean and clouds in a blue vintage room style. It is a view of the adjacent marine species that stepping away from the balcony is exposed to sand. All the rooms have only two colors like the two of us at once into the room I don’t want to go out, size 40 sq.m. Your lovely dogs under weight 5Kg. can stay 500THB/Night/Dog. Green Gallery Hua Hin Silhouette standard room Silhouette standard room This most popular room uses sweet vintage decors, mixing with silhouette painting on the bare brick wall which gives this room its name. The room also features pool access and swinging chair good for your reading pleasure that that you may not want to step outside.Room size 25 sq m. 1大床 禁烟 客房服务 提供亚麻织物和毛巾 迷你冰箱 电视 有线/卫星电视 空调 室内影院 CD 播放器 日常客房服务 免费水果篮 DVD 播放器 迷你酒吧 吹风机 无线网 淋浴 — 单独用 Green Gallery Hua Hin Chandelier standard room Chandelier standard room This room has simple look and feel but with dark-color curtains and white furnishing. Our designer wants the curtains to look like the glittering sea which is the view seen from the window of this room. Room size 30 sq m. 1特大床 1单人床 禁烟 客房服务 休息区 有线/卫星电视 免费水果篮 服务台 空调 室内影院 CD 播放器 日常客房服务 吹风机 DVD 播放器 迷你酒吧 电视 无线网 淋浴 — 单独用 提供亚麻织物和毛巾 迷你冰箱 Green Gallery Hua Hin Green Suit family room 4 paxs Green Suit family room 4 paxs This classic-modern style room furnishes in dark-brown tone, distinct by green wall. Areas within the room are well designated with two bedrooms and one bathroom making it like staying in your own home. Room size 40 sq. 2大床 禁烟 客房服务 提供亚麻织物和毛巾 CD 播放器 无线网 电视 空调 迷你酒吧 吹风机 室内影院 淋浴 — 单独用 DVD 播放器 迷你冰箱 免费水果篮 有线/卫星电视 日常客房服务 Green Gallery Hua Hin Lighthouse Suites family room 4 paxs Lighthouse Suites family room 4 paxs This room is as colorful as the sea view seen through the room’s window. The highlight of this room is the whale and lighthouse wall painting. Also in this room is a small connecting room suitable for group or family stay.Room size 30 sq m. 1大床 1特大床/双床 禁烟 客房服务 室内影院 CD 播放器 免费水果篮 迷你冰箱 空调 迷你酒吧 吹风机 日常客房服务 有线/卫星电视 DVD 播放器 冰箱 — 酒冰柜 电视 无线网 淋浴 — 单独用 休息区 冰箱 - 全尺寸 规则和条款 1. Our room rates are inclusive of tax, service charge and breakfast for 2 persons and 4 persons for Green and Lighthouse suit 2. Check-in time is 14:00 hours and check-out time is 12:00 hours. Guests staying after 15:00 hours will be charged the full rate. This charge does not oblige us to provide guests an additional room night, however. Should you wish to extend your stay, please make arrangement in advance with us and we would do our best to accommodate your need. 3. Each room can accommodate up to the maximum number of people specified for that room. Each additional person will be charged THB900/night. Up to 2 children under 10 may stay in the same room with their parents for no charge. 4. Your small pets are welcome with conditions. Please notify us and get our approval should you wish to bring pets to our premise. As a gratuity to our housemaids, the extra charge for pets is THB500/night (bigger size than 5Kg. charge THB900) Pets must be under your command at all times. 5. Room and deposit payment may be made by cash, money transfer to specified bank account, money transfer via an approved payment service, or with your credit cards. 6. For deposit and payment via money transfer, please complete the transfer within 3 days once we notify you of room availability. In the meantime, we will reserve the room for you in the order we have received. However, we reserve the right to change your reservation request until we have received the deposit in full. Please also keep the transfer/pay-in slip as evidence and prepare to present it upon arrival. 7. Your room reservation is considered complete only when we have received the deposit payment in full. We will contact you via telephone or email to confirm your reservation. 8. For deposit and payment with credit cards, if we are able to reserve a room for you, we will charge the deposit amount to your credit cards, and contact you via email to confirm the reservation and the approval. Also, we will charge your credit cards for the remaining sum of room charges after checking-in unless you notify us that you wish to pay the remaining sum with an alternate agreed method. 9. Reservation via an agency is considered complete only when we have received the payment from your agency in full. Deposit Payment 1. Staying during Low Season (June - October except Friday, Saturday, holiday or long weekend) - For 1 or 2 nights stay, please deposit the amount equal to full room charges at least 15 days prior to the check-in date - For more than 2 nights stay, please deposit 50% of room charges at least 15 days prior to the check-in date 2. Staying during High Season (November - May including Friday, Saturday, hoiliday and long weekend) - For 1 or 2 nights stay, please deposit the amount equal to full room charges at least 30 days prior to the check-in date - For more than 2 nights stay, please deposit 50% of room charges at least 30 days prior to the check-in date No Show Policy 1. We cannot make a refund if you fail to show up at the time and on the day specified in the reservation without notifying us in advance and receving our confirmation of cancellation. The cancellation charges are calculated by reference to the schedule below. 2. If you fail to show up before 18:00 hours on the day specified in the reservation, we reserve the right to change your reservation and your depoist will be non-refundable. Cancellation and Charges You may cancel or change your reservation under the following payment conditions: 1. Staying during Low Season (June - October except Friday, Saturday, holiday or long weekend) - More than 45 days prior to the check-in date, no room charge - More than 30 days prior to the check-in date, 50% of room charges - More than 15 days prior to the check-in date, one night room charges + 10% - 0-15 days prior to the check-in date, full room chargess 2. Staying during High Season (November - May including Friday, Saturday, hoiliday and long weekend) - More than 60 days prior to the check-in date, no room charge - More than 45 days prior to the check-in date, 50% of room charges - More than 30 days prior to the check-in date, one night room charges + 10% - 0-30 days prior to the check-in date, full room charges Changes to the Reservation 1. Additional stay or room request is subject to availability. 2. Additioanl stay or room request is not guaranteed until you have received our confirmation via email. Refund Policy 1. Any refundable amount may be subject to financial charges and fees but not exceeding 5% of the refund amount. 2. All deposit payments are non-refundable. ***No pet allowed in Gallery Zone Dog's regulation - Please allow your dog for our Ticks & Fleas Control Spray before check - in - Please keep your dog droppings every time. Fine start from 1,000THB for any damage caused or your dogs dropped in room or hotel area. Cancellation and Charges You may cancel or change your reservation under the following payment conditions: 1. Staying during Low Season (June - October except Friday, Saturday, holiday or long weekend) - More than 45 days prior to the check-in date, no room charge - More than 30 days prior to the check-in date, 50% of room charges - More than 15 days prior to the check-in date, one night room charges + 10% - 0-15 days prior to the check-in date, full room chargess 2. Staying during High Season (November - May including Friday, Saturday, hoiliday and long weekend) - More than 60 days prior to the check-in date, no room charge - More than 45 days prior to the check-in date, 50% of room charges - More than 30 days prior to the check-in date, one night room charges + 10% - 0-30 days prior to the check-in date, full room charges Changes to the Reservation 1. Additional stay or room request is subject to availability. 2. Additioanl stay or room request is not guaranteed until you have received our confirmation via email. Refund Policy 1. Any refundable amount may be subject to financial charges and fees but not exceeding 5% of the refund amount. 2. All deposit payments are non-refundable. | Mid | [
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USDA-Approved Agent Orange: It’s Coming to a Farm Near You Pin 443 1K Shares If you're new here, you may want to subscribe to my RSS feed. Thanks for visiting! (Psst: The FTC wants me to remind you that this website contains affiliate links. That means if you make a purchase from a link you click on, I might receive a small commission. This does not increase the price you'll pay for that item nor does it decrease the awesomeness of the item. ~ Daisy) By Daisy Luther Ah, the sweet smell of Agent Orange in the morning. Our benevolent agricultural guardians at the USDA have announced that they are allowing the introduction of new corn and soybean seeds that have been designed specifically to withstand a dousing with 2,4-D, a key ingredient in the infamously deadly Agent Orange. Now we get to be the unwilling guinea pigs while USDA-approved test fields are planted. It seems that our government did not learn its lesson from the millions of people gruesomely affected by the ingredients of Agent Orange the first time around. Agent Orange, you may recall, was brought to wartime Vietnam by the evil masterminds at Dow and Monsanto. American forces sprayed it all over the countryside of Vietnam from 1961 to 1971. Its purpose was to defoliate trees and shrubs that were providing cover to enemy forces, and to kill food crops that were providing sustenance. This caused damage to the ecosystem of Vietnam that is still present today. More than 5 million acres of forests were destroyed, and half a million acres of farmland were tainted. It will take centuries of nurturing for the land to recover. The environment was not the only thing affected. Exposure to Agent Orange resulted in five horrible illness in those exposed: soft-tissue sarcoma, non-Hodgkin’s lymphoma, chronic lymphocytic leukemia (including hairy-cell leukemia), Hodgkin’s disease, and chloracne. (source) What’s even worse is that the damage may not be limited to those directly exposed – it can affect offspring even up to 3rd and 4th generations. Over a million US veterans were also exposed: In 2010, the U.S. Department of Veterans Affairs provided $16.2 billion in compensation to 1,095,473 Vietnam-era veterans.[i] The agency does not relate these service-connected benefit figures directly to Agent Orange/dioxin exposure or to any other possible cause of illness, nor does it provide data on total compensation for the years since the war ended. Thousands of U.S. veterans returning from Vietnam reported health problems almost immediately and rapidly associated them with Agent Orange/dioxin exposure. Controversy over these assertions began just as fast, and continues now. Many questions remain: Whether (and how to test whether) the illnesses of veterans and their offspring are related to Agent Orange and other herbicide exposure; Levels of dioxin present in the chemicals; The accuracy of data about veterans’ exposure; Levels of corporate, military and government awareness of dioxin’s presence; Fixing of responsibility for the contaminant’s presence and liability for its damages; Details of research protocols, accuracy of findings and reliability of interpretations; and Decisions on who should pay what to whom for which possible courses of remedial action. This “blame game” has blocked action in both the U.S. and Vietnam, needlessly prolonging the suffering of millions of U.S. veterans and Vietnamese. – (source) And now, the USDA, in all of their infinite wisdom, intends to expose Americans to one of the deadly ingredients via our food supply. Corn and soybeans are present in some form in up to 90% of the processed foods available today. So not only will we be exposed to the effects environmentally, anyone who eats processed food will be directly consuming it. Mmmm…Corn with Agent Orange Sauce…Yummy. Some scientist argue that 2,4-D is not responsible for the horrible human toll extracted by Agent Orange, while others claim the weed-killer is deadly. According to the Associated Press, scientists don’t believe 2,4-D to be responsible for health complications caused by Agent Orange, and have instead pinpointed the ingredient 2,4,5-T – banned by the EPA in 1985 – as the culprit. Previous findings by the EPA have also declared the weed killer safe to use, but other groups aren’t as confident. As RT reported in the past, the Natural Resources Defense Council has linked 2,4-D to cases of cancer, genetic mutations and more. In addition the impact on humans, the Save Our Crops Coalition believes it will be extremely difficult to contain the application of the herbicide to a particular area. “These herbicides have been known to drift and volatilize to cause damage to plants over ten miles away from the point of application,” the coalition claimed. (source) Proven in the island petri dish of Molokai, the danger of GMO crops is not limited to the consumption of those foods. The farming methods themselves cause an epidemic of deadly health problems to those near the fields, including cancer, respiratory illness, and horrible skin disorders. The EPA review of these experimental new seeds will occur over the next few months, and if approved (and we all know it will be since the EPA is as much of a sell-out as the USDA) farmers across the country will then be able to plant the new seeds douse the fields with 2,4-D throughout the growing season. When the very air you depend on to survive is poisoned, what can you do? How can you prep for this? Agent Orange. It’s coming to a farm near you. | Mid | [
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Previous Chapter Next Chapter The truck stopped at the gate, producing the occasional sputter and knocking sound as it sat there. The driver extended a hand out the window, waving for the camera, and the gate opened by way of remotely operated pulley. It was another minute to the top of the hill, where the truck rolled to a stop in the parking lot, an expanse of gravel without any defined parking spots. The three people within remained where they were, warily observing the restaurant from a distance. It was a log cabin writ large, the cedars stripped of their bark and stained with something that made them almost glossy, a warm yellow under the sun. The third floor was half the size of the other two, allowing the other half of what would have been the third floor to be a rooftop patio instead. A series of tables was scattered around the building, some close together and others a considerable distance away, as if they were trying to escape into the woods. Beyond the building was a cliff, and a vast expanse of forests, hills, mountains, and a small lake. “Nice view,” Moose said, from the backseat. Linc was settled in the passenger seat, reclining a bit with his seat angled back and his legs folded under him. With Moose in the back, he’d had to slide his seat as far forward as it could go, and it didn’t leave him much legroom. “Just think, past that view there isn’t nothing at all. If you headed straight ahead and kept going, you might not find any habitation until you ended up on the backside of this settlement here.” “If you headed straight ahead,” the driver said, pausing to take a swig of the bottle of water she’d wedged into her cup holder, making a face at how warm it was, “You’d put yourself into that lake down there. Or you’d end up in the ocean. You’d drown either way.” Linc smirked. “People would call you an idiot,” she said. “Why would you go straight ahead like that? Are you proving a point?” “I don’t think that’s what Linc was getting at,” Moose said. “Harper knows what I’m getting at,” Linc said, turning around in his seat to look at Moose as he said it. Moose was a big guy, with tousled blond hair. He’d undone some of the straps of his mask and had the mask laying over one muscular shoulder. The mask was metal, crude, and Moose wore something cloth under it for padding, which he had on now. He wore a sleeveless undershirt, jeans, boots, and had two gauntlets sitting next to him on the car seat. Even with the truck being large and Moose lying down across the length of the seat, he barely fit. He didn’t seem to mind much. Behind Moose and the truck was the gravel road that led up the hill, the gate checkpoint, and a ways below that, the simple settlement where most visitors would be made to feel unwelcome. One to two thousand people would be living there at most. “They built this place and situated it on the very edge of civilization,” Linc said, to round off his earlier thought. “You two always seemed like the kind of edge of civilization people to me,” Moose asked. “We do okay,” Linc said. “Put us in the middle of a city, we do fine, eh babe?” “Mm,” Harper made a vaguely affirmative sound. “This a trap, y’think?” Linc turned his attention to the building at the top of the hill. “Nah. Why would you build a nice place like this and use it for a nefarious purpose.” “Well, y’know, it’s gonna be nefarious. That’s why we’re here,” she said. “It’s a question of if it’s a murderous sort of nefarious.” “That’s a good question, I admit,” Linc said. “I knew a guy,” Moose said. “He had a mansion. Inherited or somethin’. Super nice.” “The guy or the house?” Linc asked. “Hm?” “The guy was nice or the house was nice?” “The house. That’s what I’m gettin’ at. The guy was as nefarious as they get. He renovated the insides. He wanted to make a whole business of holdin’ people that needed holdin’. For ransom. Said he’d deal with ’em and clean up the mess if ransoms weren’t paid. Wanted to be a contractor for disposin’ of people in horrible ways.” “You’re supposed to just drop them off at the nice, conspicuous mansion, hand over cash?” Harper asked. “That was it, I think. He’d make sure they died slow and horrible for you, clean them up, make sure they weren’t found.” “Definitely not a nice guy then,” Linc said. “I dare say he wasn’t,” Moose said. “That’s a terrible idea for a business,” Harper said. “It kind of is,” Linc said. “Might’ve been,” Moose said. “He didn’t seem in it for the money, gotta say. I highly suspect he was more focused on the part where he would do horrible things to people. Guy has a nice place, he wants to do creatively bad things to people, and he wanted a bit of pocket money. Draw lines between each of those things and you end up with something shaped like his game plan there.” “A triangle?” Linc asked, looking back at Moose. At Moose’s shrug, he elaborated, “If you draw straight lines between three things, you get a triangle.” “Maybe the lines weren’t straight,” Moose said. “But if you’re wondering if this is a trap, I don’t think it being fancy is ruling anythin’ out.” “It’s a log building, Moose. Nothing that fancy.” “Fancy to me.” Harper leaned forward against the steering wheel, to get closer to the windshield, squinting against the sun. “What do you think, babe?” Linc asked. “Is it a murderous nefarious or a prosperous nefarious.” “It’s something,” she said. “The people on the roof are in costume and some of them are looking at us. I think we better get ourselves inside or they’re going to start laughing at us.” “They’re going to laugh whatever happens,” Linc said. “Your truck has seen better days-” “Don’t go talking about my truck, Linc.” “And we’ve got Moose with us, no offense Moose.” “Some offense taken, thank you very much,” Moose said, indignant. “You call yourself Moose. People are going to laugh. That’s when you show your merits and make them stop laughing, is the way it works.” “People shouldn’t laugh in the first place,” Moose said. “The Moose is a terrifying and noble creature. If you wouldn’t fuck with a rhino, you shouldn’t fuck with a moose. It’s one of the only proper prehistoric, giant animal species to have the grit to last to today.” Harper turned off the truck. The truck sputtered, coughed, and died abruptly, in a way that suggested it wouldn’t revive again. “I know, bud,” Linc said, taking his eye off the truck’s much-abused, dust-caked dash. “I know that much, I’ve seen one up close. I’ve seen one run through snow that a normal person couldn’t walk in and hit a car hard enough to roll it. I have a healthy respect.” “Damn right,” Moose said. Harper gave Linc a look, pulling her full mask on and flipping up her hood. “But they don’t all know it,” Linc said. “You gotta work with that. You picked a jokey name, you gotta put up with the jokes.” “Hope was they’d be laughing with, not laughing at,” Moose said. “At least I’d hope you weren’t the ones poking fun at me. It’s unkind, Linc.” Harper climbed out of the truck. “I’m not laughing at, bud. I’m just saying they might be. That’s all,” Linc said. He pulled his mask on, fixed his hair and beard with few sweeps of his hand, and climbed out, then hit the lever to flip his seat forward and give Moose room to squeeze out. Moose kept the cloth mask on over his upper face, leaving the metal mask on his shoulder. He stretched, his joints popping audibly, and pulled his fur-lined gauntlets on. “You’re going to have to take those off again if we end up eating,” Linc said. “It’s about presentation,” Moose said. “Besides, the name doesn’t make sense if I don’t got ’em.” Harper was in costume, though the costume part was mostly a hooded, sleeveless top in her namesake velvet color, lopsided in how it trailed down over one leg in a robe-like aesthetic. She wore skintight shorts underneath. A black mask covered her upper face, and had truncated, forking horns that poked out through the top of the hood and kept the hood from falling back. Linc wore a mask like Velvet’s, but his traced the area around his eye sockets and eyebrows, with the edges tracing back and into his hair, forking as they did. He wore a bodysuit for the upper body and pants. His costume had always been meant to be layered, but the heat had forced him to strip down to the base layer, with the pants only because he felt like a clown if he wore only the skintight stuff. People leaning against and over the railing on the roof watched them as they approached the door. “This place is a hell of a lot better than the last couple we visited,” Prancer remarked. “More expensive too,” Velvet said. She was looking at the blackboard posted by the door, with prices. “Twenty dollars for a chicken sandwich?” “Come on,” Prancer said, pushing the door open. The inside was expansive, with the kitchen as an island in the middle, counters and surrounding it, booths around the edges of the room, and tables in the space in between. There were only eleven non-staff people within, but the ground floor could have seated a hundred or more. Prancer approached the kitchen island. He spoke to a black, twenty-something woman in a tan polo shirt and apron, “Who do I talk to for the rules?” She jerked a thumb over her shoulder, indicating an elderly black man who was wiping out a glass. The man was watching, squinting with one eye, as he carried out the routine motion of cleaning the glass. “He’s in charge?” Prancer asked. The employee gave Prancer a single nod. “What can I do for you?” the man asked, as the three approached. “We’ve been around the block a couple of times, I’m just looking for a primer on customs, and any special rules.” “Payment up front for what you’re ordering, have the money ready when you order if it’s busy. Don’t cause trouble, don’t draw weapons, don’t be loud, give us a heads up and use the side door or the patio if your power is going to bother anyone. Upstairs is the bar, you don’t go upstairs unless you’re invited or you already know you qualify to go upstairs.” “What kind of qualification?” Velvet asked. “If you have to ask, you don’t have it,” the old man said. He put the glass down and picked up another. “Roof is for more private meetings than you’d have on the second floor. Don’t go taking yourself up there if you wouldn’t be allowed on the second floor.” “Noted,” Prancer said. “Anyone to avoid, watch out for, anything like that?” “That’s more for you to watch out for than for me to bother with. If they’re causing a problem or being a bother to others, they’ll get kicked out. If you help with the kicking out, I’ll give you something on the house.” “Right,” Prancer said. “Got it.” “Do you serve drinks down here?” Moose asked. “We do. Anything fancier than beer or wine, we’ll have to send someone upstairs to fetch it.” “Could I grab a mightyman?” Moose asked. He pulled off a gauntlet and retrieved a wallet from his pocket. He held out a twenty. “Long, hot drive.” “Name?” the old man asked, gesturing at an employee. The employee set to getting the beer. The old man pulled a pad and pen out of his apron. “Name? Uh, M.K.,” Moose said. “No initials,” the man said. “Just Moose then,” Moose said, sliding the twenty across the bar. “We can order food at the tables?” “You can,” the old man said. He picked up the money, then pulled out a fiver from the pocket of his apron and passed it back to Moose. He looked at the others. “Names?” “Prancer. She’s Velvet.” “Do I need to worry about you?” “Nah. We’re pretty tame. We’re here to make contacts and get our names out there for the small stuff.” “If you do any business, be discreet enough I and my staff don’t see it. If you use powers, don’t bother the person next to you.” “Got it. Can I grab a beer? What my buddy Moose is having,” Prancer handed over the bills. “Me too,” Velvet said. Prancer withdrew a larger bill from his wallet, and set it on the counter, sliding it toward the old man. “Gratuity?” “No need,” the old man said. “Service fees and peace of mind are worked into the food prices. Order something, if you want to thank me.” Beers in hand, they briefly considered sitting at the counter before Moose took a seat at one of the tables. “Where you sit is important,” Prancer said. “Booth, you’re minding your own business, you’ve got walls around you. Sitting at the counter, you’re open to people approaching you and joining you, I think. Not entirely sure how it works here in particular.” “I hear you,” Moose said, “But I was sittin’ funny the entire drive here, and if I sit on one of those stools then I’m going to have my back spasming the entire way back. I need a sturdy chair, here.” “Sure, doesn’t matter that much,” Prancer said. He twisted around in his seat, one hand on his beer, taking a look at the others who were present. “Pretty laid back here.” “Could be a quieter time of the day,” Velvet said. “Out of the way place, too,” Prancer said. “You heard what he said about using powers? How many places have we been to, and how many allowed use of powers at all?” “Ten. Ten places,” Velvet said, hunkering down over her beer. “This is the only one, I think. Might have been a rule in The Well, but that was more the kind of place where you don’t know the rules until someone’s punching your face in for violating them.” Prancer watched as a faint speck of dust traveled across his vision, pink-tinted. He smiled. Four teenagers in the corner booth. They wore dark clothes with symbols and designs spray painted on and bleached into the fabric. One with a bandanna on his head looked their way, and Prancer flashed the guy a smile. Three in another booth, against a wall. Tinkers. There was a cloth strewn out over the table, and parts were laid out. They ranged from twists of metal to a glass tube housing something that looked like a large, chewed wad of gum. The wad was throwing itself against the sides of its glass cage. He wondered how that worked with the ‘no business’ rule. Were they only talking shop? Where was the line drawn? Sitting alone in one booth was a woman with a mask covering her lower face, long black hair, and a long red dress with a slit down one side, exposing a tantalizing bit of leg. She wore an intricate framework of metal at her arms and hands, a series of bands at the elbow, wrist, knuckles, and rings at the finger, with thin rods of steel extending between each, along the back of each finger, and stopping at each finger and thumbtip. Each tip was enveloped by an ornate claw. Her heels were much the same, Prancer noted. Heels were unusual for someone in costume, and hers were more unusual still. She wore something similar to her gloves, with the same bands at her leg, ankle, and foot, with the thin metal bars extending between each. Her toes were covered with the same metal claws, there was a strap of metal below the balls of her feet, and at her heel, one large claw-point served as the ‘heel’ of her heel, stabbing straight down. When she moved one leg to fold one knee over the other, the claw tips moved on their own, twitching, recalibrating, the heel shifting back to stay pointed at the ground, flick back and away, then flick down. She undid one side of her mask so it swung toward Prancer, still blocking his view of her mouth, helped by the draping of long hair, and she leaned down, taking a bite of her wrap. She put one hand to the loose end of her mask while she chewed, and fastened the end as she swallowed. She saw him looking, turning her head his way. He smiled at her. She only stared. “Someone’s coming,” Moose said. One of the spray painted kids. The guy Prancer had smiled at. “You’re new.” “Prancer, Velvet, M.K.,” Prancer introduced the group. “Where are you from?” “Alaska, believe it or not.” “Long way,” the teenager said. “Especially when you’re driving it,” Prancer said. “Who are you guys?” “The group’s Ripcord. I’m Gorgos. We raid stores and resell, mostly. We’re nobodies. It’s the B-listers and small fry down here. The people with name recognition go upstairs.” “Meaning the people we want to do business with are upstairs,” Velvet said, still leaning heavily over her beer. “It’s fine,” Prancer said. “We’ll work it out.” “What do you guys do?” “We wheel and deal,” Prancer said. “Prancer likes to be clever, but he doesn’t get that sometimes you have to explain why it’s clever, otherwise you only confuse people,” Velvet said. “It’s why I have you, babe.” “The wheel part is getaway driving and transporting,” she explained. The kid leaned forward. The decoration on his outfit looked like the sort done with a stencil and a spray bottle filled with bleach, strategically bleaching fabric. Snakes and a woman’s face as a recurring motif. He had a bandanna over the top of his head and one over his nose and mouth. “What do you deal?” “Grass, mostly,” Prancer said. “You actually have some?” “Not here, but we have it. Brand new and in high demand, given the times,” Prancer said. “Are you looking for resellers?” “For the right price. Mostly we’re looking for new friends, and we’re trying to get the lay of the land before we do anything too enterprising.” “Can I get back to you?” the guy asked. “You’re welcome to,” Prancer said. “We wouldn’t mind company either, if you guys wanted to join us.” “I’d have to get back to you on that too,” the guy said. “We’re trying to find our way these days. We agreed in the beginning we wouldn’t have one leader, and that was great then, but right now we’ve got two different leadership styles butting heads.” Prancer looked over at the table, where those seated were having a very fierce, hushed discussion. “If you want to just sit and trade stories, we’d be happy to have you,” Prancer said. “Get away from all that, maybe come away with some fresh perspective.” “I might take you up on that. For now I’d better get back and make sure nobody reaches across the table to strangle someone.” “Question before you go,” Moose said. “Is it always this quiet?” “It’s about to get noisier,” Gorgos said. “Keep an eye on the guy at the end of the kitchen there. He communicates with people in town. He was talking to the boss about something and the boss put another cook on the stove. Wait ten minutes and I bet he’ll hit the button to open the gate. If he holds it down it’s a lot of people. My guess is the ferry from NYC hit the shore near the G-N portal twenty minutes ago.” “Good to know, thanks,” Prancer said. Gorgos jogged back to his team. “You’re dwellin’ a lot on going upstairs,” Moose observed. “Reminds me of being a kid and being told I had to stay downstairs with my cousins and their friends during the holidays. My cousins were assholes,” Velvet said. “One good thing about Gold Morning is it took them out of the picture.” Moose whistled. “She’s wearing the purple cloak, that’s a sign of royalty, don’t you know?” Prancer plucked at Velvet’s hood. Velvet batted his hand away. “And royalty doesn’t not go upstairs. Royalty doesn’t show mercy.” “Y’know I went to prison because of you, Prance,” she said, quiet. “Well, yeah. I will point out we survived Gold Morning because we weren’t home when Alaska got hit.” “I went to prison for you,” she said, again. “That counts for a lot.” “‘Course. How does that connect, though?” “Just sayin’,” she said, her accent thicker as her voice became softer. “You said things would be different.” “They will,” he said. He put a hand around her shoulders and pulled her closer, then kissed the top of her head. “We’ve got a decent crop, a lot of demand. We’ll do okay. We’ll make inroads.” “I’m optimistic,” Moose said. “I’m not unoptimistic,” Velvet said. “You’re not enthused either, doesn’t sound like,” Moose said. “Just sorta hoping for more, sooner,” she said. “Yeah,” he said. There wasn’t much more to say. Velvet reached out, and the menu flew from the tabletop to her hand. It was tinted red and dusty, but much of their table and glasses were, now. Prancer took stock of the other three capes in the room before the newest batch of arrivals reached the front door. There was one, who might have been a bouncer, who had stepped out the side door momentarily and was now taking a seat by that same door. He wore a mask of metal bars that looked like they’d been welded to one another, all vertical, but he also wore a black apron. That left only the couple at the bar. Matching costumes, white armor with jet black iconography, multiple circles and crescents in various patterns, with the armor sprayed black around each icon, so it looked like the darkness glowed. The man wore full armor, the woman wore only scattered pieces of armor, with white chainmail to cover the rest of her. They drank white wine, in the middle of the afternoon. Capes were strange people, Prancer mused. “I want to be the kind of person who earns her way upstairs,” Velvet said. Her head still rested against his shoulder, where he’d pulled her close. “That’s really stuck in you, huh?” “It’s stuck,” she said. Without moving her head, she raised the beer to her lips and took a careful sip. “You might have to lose the beater of a truck, babe, if you want to dress the part.” “Don’t go talkin’ about my truck, Prance.” “Every time you turn it off, it sounds like it’s off for good. I say a little prayer to myself that it will be, even knowing it’s a long, long walk back to home. Then I can take the money I’ve got saved up and buy you something nice. All the bells and whistles.” “When I got out of prison, I only had two things, babe. That truck, and you. I wasn’t feeling especially fond of you at the time, either. It’s the only thing I’ve had for myself since I was old enough to have anything, that I’ve been able to keep.” “Counts for somethin’, that,” Moose said. “It does,” Velvet said, frowning down at her beer. Prancer frowned at Moose, who only shrugged. Guy wasn’t making it any easier. “What if we overhauled the outside, got someone to give the engine a real solid lookin’-at?” Prancer asked. “So long as it stays my truck. I don’t want you ship-of-Theseusing it.” Prancer resisted swearing under his breath. So that tactic wouldn’t work. There was more of the pink dust in the air, now. He gave Velvet a kiss on top of the head, then shrugged slightly. She moved her head off of his shoulder, sitting upright. “Things will be better,” he said. She reached for his hand and squeezed it. “I’m going to go find the ladies’ room. Order food before things get hairy. I’ll have the chicken caesar sandwich and grab a few bottled waters while you’re at it. For the drive back.” “You know the markup on those will be insane,” Prancer said. “Just get me my water,” she said. She walked away. Prancer watched her walk away, feeling wistful. He signaled the waitress. He made sure to give Velvet’s order while he remembered it, and then gave his own. Moose put in enough of an order for two people. When they were alone again, Moose commented, “Sorry, for interjectin’.” “Interjecting?” “When you were talking about the truck, and about prison.” “Ah. Yep. Apology accepted.” “Hard to be the third wheel sometimes. Especially when things get complicated, relationship-wise.” “Can’t speak about the third-wheeling. That’s for you to figure out. But for the relationship part, it’s the simplest thing in the world, Moose,” Prancer said. “She’s my girlfriend, I’m her boyfriend. Sometimes you and she enjoy each other’s company, sometimes I enjoy someone else’s company, but that doesn’t change that it’s me and it’s her as the boyfriend and girlfriend.” His voice had become progressively more stern as he’d talked. He paused, meeting Moose’s eyes. “Makes sense,” Moose said. Prancer smiled. “Doesn’t seem like you’ve had anyone but her keeping you company, gotta say,” Moose said. Prancer looked at the woman with the mask on her lower face and the claw-heels. “Trying to be better.” “Good for you,” Moose said, before taking a drink of his beer. “I’m going to marry that Velvet sometime soon,” Prancer said. “I’ve just got to make amends for old wrongs first. Can’t ask her to marry me when the last momentous event in our lives was me being a screwup.” “The prison thing?” “Everything before, too. Trying to be better.” “I don’t want to step on any toes or get into anythin’ too sensitive here,” Moose said. “But can I ask? Would be easier to not step on toes if I can ask.” “It’s the whole thing. Get powers as a kid, sixteen years old, make friends with the right people, start dealing. It’s an elevation in status, y’know? I was the guy who the cool kids in high school went to for product. Had money, had girls throwing themselves at me, I was invited to all the parties, and I meet Velvet there. One of many girls in one of many cities. But she gets powers and comes back to me, wants in, wants out of her house, especially. I oblige, and she doesn’t make me regret it.” Moose nodded. “Years pass, we find our fit. She’s got more financial sense, I’ve got the salesmanship. Most capes, there’s going to be conflict. She’s got her thing, you know how her power works. She hangs around somewhere, and this dust collects, and she can telekinetically control stuff, more dust there is on it. It’s how she gets that fucking truck going again, when it refuses to move. She makes us sit there for five minutes and then gives it another try, and it works, and she’ll fiddle with it later and get it tuned up just enough it starts going.” “She must care an awful lot about it,” Moose observed. “She does. But that’s her whole psychology. She wants to settle in, wants to have a place she can call hers, whether it’s that truck cab or, I don’t know, going upstairs. I get restless. The mover thing. That causes friction. But we work despite it. We’re as soulmate as you can get when you’ve got… whatever these things are giving us our powers. Parasites. You had the visions when we were on the battlefield, that day.” “Sure,” Moose said. “As yin-yang soulmate as you can get with these things screwing up the fit,” Prancer said. “But we got comfortable. I graduate school, barely, she graduates a year after me, we keep up the routine. Some wheeling, mostly dealing. The parties every weekend, tooth and nail fights because we’re both the type to flirt with others, before we realized we were fine just not worrying about it. Couple more years pass, I’m twenty-one, she’s twenty, still in the routine.” “A rut?” Moose asked. “Just the way things were. Somewhere along the line, you know, I’m twenty-seven, she’s twenty-six, and I’m still boning boys and girls from high school. Still partying.” Moose’s eyes had widened. “Legal, mind you,” Prancer said. “But… sketchy, in retrospect.” “More than a little, no offense,” Moose said. “None taken. I deserve it. I didn’t realize until they came after us. Capes, police. You get into a groove and you don’t think about things and somehow a decade gets away from you. You’re not the cool guy people are excited to get to know. You’re the guy they’re into because they have to be if they want a discount, or if they want someone accessible that’s older. Sad. Pathetic. Slapped me in the face while people were talking to and about me in court. Forced me to take a long, hard look at who I was and who I wanted to be.” “That’s good,” Moose said. A young woman entered the restaurant. Prancer almost thought it was the first of the influx, but she was alone. She was an older teenager or twenty-something, with long white hair, wearing a black dress and black makeup, and she took a seat alone at the table. She rummaged in a bag to find a book. The waitress approached her, kettle already in hand. The money was passed across the table, and the tea was poured. A regular. Her mask was so simple it might as well have not been there. Curious, too, that she’d come this far to read a book. Maybe someone would be joining her. Prancer watched the new arrival, but he kept talking, “She told me, over and over again, I needed to be better. That she wanted better. That we needed to be careful. I didn’t listen. We got out of prison, she took me back, and I owe her for that.” “If your critical flaw was not listening, might be you’ve gotta listen when she’s saying she loves that vehicle out there.” Prancer nodded slowly. Then he let his head loll back, and he groaned. “I’ve put up with that thing for so many damn years.” Velvet’s glass of beer slid across the table, and Prancer caught it just before it could reach the edge of the table and tip into his lap. “You’re talking about my truck?” Velvet asked, making her way back from the restroom. “Moose is telling me to let it go,” Prancer said. “I’m trying to come to terms with the idea.” “You’re a good boy, Moose,” Velvet said, taking her seat. The glass pulled out of Prancer’s hands, sliding across the table to slap into Velvet’s hand. “Appreciate that, Velvet.” “Did you order or did you forget?” she asked. “Remembered,” Prancer said. “It’s coming soon.” The front door opened. A large collection of capes entered and immediately headed off to find their booths and tables. One of the new arrivals stepped inside and loitered by the door. She was a woman with a slender figure and a bag over her head, for lack of a better description. The bag was cloth, with a pink animal pattern on it. The rest of her form-fitting outfit matched, including the shawl she wore. Prancer leaned in the direction of the door, putting his mouth near Velvet’s ear. “I see Nursery. I wonder if Blindside is around.” “I hope the kid’s okay,” Velvet said. She looked at Moose. “Were you there when we talked to ’em?” “I was.” “They were up to something.” “I remember.” A man in armor was one of the last to arrive. The armor was white, and looked like it was fashioned of strips, woven and wound around him, the ends left frayed and sticking out to the sides and behind him. There was no face to it. Only a Y-shaped set of ridges. He stood between Nursery, a man in a black outfit and heavy hood, and a heavyset man with long hair, a dense beard, and a mechanical arm that extended to the ground. At his arrival, people across the room started applauding, from Ripcord to the people at the counter, to the white haired girl and the woman with the mask. Even the kitchen staff. The man in armor laughed, the sound mingling with the general applause. Moose joined in, and Velvet and Prancer offered their own light, confused applause. “Thank you. Thank you. Is Marquis here?” the man in armor asked. The old man at the kitchen pointed skyward. “Roof.” The man in armor saluted, then ducked back through the door. Velvet raised her hand to get Nursery’s attention. The woman’s group was already splitting up. The man in black joined the people in white armor. The bearded man with the mechanical arm walked over to the woman with the claws, sitting in her booth. Nursery approached the table. “Good to see you,” Prancer said. “I didn’t think I would see you three all the way out here,” Nursery said. “We’re trying to see who’s out there. The other places have been a little seedy.” “They are. Seedy can be fun, though,” Nursery said. “Reminds me of the old days.” “You keep updating your costume,” Velvet said. “Silly thing, isn’t it? It’s easier to make a new one than to wash the blood and slime out. I feel ridiculous.” “What was happening with the applause?” Prancer asked. “Mission success,” Nursery said. “In a roundabout, unexpected way, but that’s often how these things go.” “Congratulations,” Moose said. “Thank you, Moose. It was a thing. We took a week to figure out what we were doing, we had to check with a few people, a number of thinkers, make sure we weren’t stepping on toes. The peace being what it is, we didn’t want to cause too much trouble.” “Was it a big mission?” Prancer asked. “Big,” Nursery said. “Plenty of room for things to go very wrong, with some bad repercussions that could be felt by everyone.” Prancer’s eyebrows went up. “But we were careful, we had the right people-” “You included among those people,” Velvet said. “Yes,” Nursery said, clasping her hands together. “What was the job?” Prancer asked. “To kidnap someone, and have her disappear for long enough that people would get upset about it.” “Huh,” Prancer said. “They’re anxious out there. They feel powerless. The idea was to pick someone controversial, and take them out of the picture. Make them the topic of debate. Is vigilante justice right or wrong? In this case, where the wrong isn’t so terribly wrong? Well, that was their idea. I do think she did something horrible. It’s why I agreed to the job.” “What was that?” “Hurt a woman and made her miscarry. They say it was a mistake.” “I can see where that hits close to home.” “Sorry to hear,” Moose said. “Thank you. You’re kind. The plan was to provoke the debate and raise the issue before things reached a more critical point. Venting off the steam before things exploded. The debate seems to be trending that way.” “Sounds like it needed a fine hand,” Velvet said. “That’s some good work.” “I didn’t do it alone,” Nursery said. “Your first time working with the others?” “It was. Lord of Loss is sweet, good at what he does.” “He went straight to the roof, I’m guessing that means he isn’t the type to work with B-listers like us.” “No, I suppose not. He doesn’t like being indoors. You’re recruiting? That’s what you’re asking after?” “Or looking for a spot of work,” Prancer said. “Snag, sitting over there, is looking to hire people for a project down the road. He wants to do test runs first, make sure he succeeds on the first try. Those two hired the same information broker we worked with for that job.” “You had an information broker?” “She was ops too. Talked to us on the earpieces. A little shaky on some things, surprisingly quick on others. But I think you run into that with any thinker.” Prancer nodded. “Snag is a few months new, a rookie, with a rookie’s mindset, but he has good instincts. If I can say this in confidence…” “Of course,” Velvet said. “…I wouldn’t want to be on a team with him long-term,” Nursery said. “He’s mean. Unkind, impatient. Emotional. You get that with a lot of the new ones. Too close to whatever set them off.” Prancer nodded slowly. “Old ones have their own problems. Ruts and routines.” “They do. Um, I should hurry. Blindside has a mouth but I do like them. They do a decent job, if you can work around the limitations. They’re outside now, sitting on the patio by the side door. Can’t come inside without turning a few heads.” Prancer smiled at the bad joke. “Kingdom Come likes his bible verses, I earned some considerable brownie points by knowing the names and numbers to go with most of them. Benefit of bible school until I was eighteen. He’s a consummate professional. Very gentle, very efficient.” “Expensive?” Velvet asked. “Not too bad, I don’t think. I don’t know what he was paid, but if it’s close to my own wage, it shouldn’t be horrendous. He’s very selective about the jobs he’ll accept.” “What about you?” Prancer asked. “Me? I’m boring. I’m not even a parahuman, not really.” “Wait, what?” Moose asked. “I’m not,” Nursery said. She had a light tone of voice, like she was smiling from the other side of the cloth mask. “It’s why I feel so out of place in costume.” Prancer watched as others came through the door. He recognized Biter but failed to get Biter’s attention with a wave. “How does that work?” Moose asked. “Show him the bump,” Velvet said, smiling. “The bump?” Moose asked. “Oh.” Prancer glanced over at Nursery, who was holding her cloth costume tighter against her stomach, showing her slightly protruding belly. “They’re the parahuman,” Nursery said. “I’m the ride.” “Oh,” Moose said. “Oh wow. Sorry.” “No need to be sorry. It’s a bond unlike any other,” Nursery said. She gave Moose a pat on the cheek. “It’s hard sometimes, but I owe it to them. Making up for mistakes I’ve made.” “Yeah,” Prancer said, staring at his beer. He looked from his glass to Nursery. “Don’t be too hard on yourself, hm?” “I’ll try,” she said. “I should go. Take care and wish me luck.” “Good luck,” Moose said. He still looked shell-shocked. “What’s next?” Velvet asked, smiling. “What we did yesterday is only one instance. They’ll have to do it again when the pressure builds. Sooner or later, however many thinkers they work with, however good the people they hire, there will be a mistake. Something will happen, it could be too much, too little, and then everything goes to hell in a handbasket.” “Heavy,” Prancer said. “But I’ve stayed too long. My baby and I earned ourselves an invite upstairs, because they might hire me again, and because we showed our stuff, I don’t want the offer to expire,” Nursery said, excited. “We can’t drink at the bar, but it’s still a chance to meet some of the people running the corner worlds, the major players. A huge opportunity.” “That’s amazing,” Moose said, looking down at the bump. “Congratulations. Both of you.” “You’re so sweet. I should go, excuse me,” Nursery said, leaving. “We’ll talk again,” Prancer said. Velvet raised a hand, her smile frozen on her face. Prancer reached over to squeeze her thigh. “I think I hate her now,” Velvet said. He gave her leg another squeeze. His thoughts turned over as he watched the people enter. Some headed upstairs. Ones with nice costumes, scary ones. He recognized quite a few. There were also the others. The B-listers, the dregs, the people who weren’t yet established, filling up the ground floor, ordering their food and drinks. “Hey Moose,” he said. Moose stared off into space, in the direction of the stairs. “Moose,” Velvet said. “I’m pretty sure she’s a loon. I wouldn’t worry too much about it.” “Moose,” Prancer tried again. Moose frowned, glancing back at the stairs. “Yeah?” “Look at the room. Tell me, who do you know here?” “Some of the big guys. Biter, you and I had drinks with him. Etna, Crested, Beast of Burden, Bitter Pill, Nailbiter, Hookline, Kitchen Sink.” “Do me a favor?” “Sure.” “Round ’em up. Anyone you get along with, who you think wouldn’t cause a fuss.” “What are you doing?” “Still figuring it out,” Prancer said. “You recognize anyone?” “Few people. You want me to gather ’em?” “Please. We might have to take it outside. Actually, let’s definitely take it outside. By the side door. So the owner doesn’t complain.” “You’ve piqued my interest,” Velvet said. Prancer nodded, still lost in thought. He watched as she walked away, pausing to feel a moment of fondness for her, and then resumed his thinking. He made his way to the side door. “Hey,” Blindside said. “Hey. Gathering some people. Thought we’d come to you, invite you to hear me out.” “Thanks, Prancer. What’s this about?” “Give me a second to think. I’m a salesman, and I’ve got to figure out exactly what I’m pitching.” “Sure.” Prancer stuck his fist out, stopped where Blindside’s power made it stop. He felt Blindside tap a fist against the side of his hand. The others assembled. The people who had been invited, then the people who hadn’t, who were curious. “I want to organize,” he started. “I’m not the person to lead it, don’t get me wrong, this isn’t a power play. I’m not a power player. But I think, right now, while we’re still at peace, while there aren’t so many people who have beefs with one another, or the beefs have had two years to go quiet, this would be the chance to do it.” “You wouldn’t be the first person to think about doing this,” Biter said. “No?” “No, some other groups, some small, some large. They’re banding together, a mutual peace. Forming a set of rules and expectations that aren’t unwritten, that we actually discuss and work out.” “With the little guys?” Prancer asked. “B-listers? Those of us who aren’t being focused on while the big guys are laying out tracts of territory and settlements?” “Some of them. Those groups are smaller than you’d be talking, if you’re talking about everyone here.” Prancer nodded. He glanced at Velvet, and he saw the way she was looking at him, and he felt like a proper man for the first time. She reached for his hand and squeezed it, hard. Then, more alive and excited than he’d seen her in a long, long time, Velvet spoke, “You think they’d be open to talking?” “I think they might,” Biter said. Previous Chapter Next Chapter | Low | [
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Q: How do I create a specific process using fork()? I have multiple processes; a controller and a person. The controller needs to spawn off multiple person processes based on user inputted value. How using fork() do I specify to create a person process? pid_t child = fork(); // How do I make this a `person`? From what I understand, fork() just makes a copy of the current process. What am I missing? A: Fork indeed makes a copy of the current process. Once this is done, you can use execve() in order to execute your new process. | Mid | [
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Photo yellowing of human hair. In general, human hair is claimed to turn yellower after sun exposure. This is particularly affirmed for white hair. However, quantitative data relating yellowness to hair type and to the radiation wavelength are missing. This work shows results of the effect of full or UVB-filtered radiation of a mercury vapor or a xenon-arc lamp on the yellowness of virgin white, dark-brown, blond and red hair. All hair types showed a substantial change in yellowness after irradiation, which is dependent on the hair type and radiation wavelength. Surprisingly, white hair turns less yellow after both full and UVB-filtered radiation exposure. This effect is more pronounced when UVB is filtered from the radiation system. The only radiation that shows a photo-yellowing effect on white hair is infrared. As the yellowness of white hair is commonly related to tryptophan degradation, fluorescence experiments with hair solutions were performed to identify the natural degradation of tryptophan which occurs in hair after light irradiation. Pigmented hairs were also studied, as well as hair treated with a bleaching solution. Although we observe a decrease in tryptophan content of hair after lamp radiation, a direct correlation with hair yellowness was not achieved. Results are discussed in terms of hair type, composition and melanin content. | Mid | [
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Evidence-based public health: a fundamental concept for public health practice. Despite the many accomplishments of public health, a greater attention to evidence-based approaches is warranted. This article reviews the concepts of evidence-based public health (EBPH), on which formal discourse originated about a decade ago. Key components of EBPH include making decisions on the basis of the best available scientific evidence, using data and information systems systematically, applying program-planning frameworks, engaging the community in decision making, conducting sound evaluation, and disseminating what is learned. Three types of evidence have been presented on the causes of diseases and the magnitude of risk factors, the relative impact of specific interventions, and how and under which contextual conditions interventions were implemented. Analytic tools (e.g., systematic reviews, economic evaluation) can be useful in accelerating the uptake of EBPH. Challenges and opportunities (e.g., political issues, training needs) for disseminating EBPH are reviewed. The concepts of EBPH outlined in this article hold promise to better bridge evidence and practice. | High | [
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[Sensitivity to the differential dose of erythromycin and oleandomycin of various holarctic strains of tularemia microbe isolated in the Kazakh SSR]. The presence of strains of holarctic race of tuleremia microbe, both sensitive and resistant to the differential doses of erythromycine and oleandomycine was revealed at the territoryn of Kazakh SSR. The existence of such strains does not permit to differentiate them into geographical races by the sensitivity to erythromycine and oleandomycine. | Mid | [
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The present invention relates generally to high power magnetron oscillators and more particularly, to improved magnetron oscillators that are tuned using plasmas, and frequency tuning methods for use with magnetron oscillators. High-power magnetrons are efficient generators of microwave power. They convert the kinetic energy from an electron beam into microwave- or millimeter-wave energy within a resonant cavity. The oscillation frequency is determined by the cavity, electron-beam voltage and current, and the externally applied magnetic field. The high efficiency of magnetrons make them an attractive RF source for use in many applications; CW magnetrons have demonstrated efficiencies in excess of 80%. As an oscillator, however, the magnetron is inherently a narrowband device. While mechanically-tuned magnetrons are available, they suffer from several drawbacks. The maximum tuning rate is limited by the inertia of the tuning mechanism, whose moving parts penetrate the vacuum envelope of the magnetron, thus reducing the reliability of the magnetron. The oscillation frequency of a tunable magnetron is varied by changing the resonant frequency of its resonant structure. In a mechanically-tuned magnetron this is achieved by mechanically altering the geometry of the resonant structure. This involves mechanically changing the dimensions of the cavity, with corresponding changes to beam voltage and magnetic field as needed, which changes the oscillation frequency of the magnetron. Mechanical tuning is relatively slow and requires a movable vacuum element so that the high-vacuum integrity of the tube can be maintained. Accordingly, it is an objective of the present invention to provide for improved magnetron oscillators that may be rapidly tuned. It is another objective of the present invention to provide for improved magnetron oscillators that are tuned using plasmas. It is a further objective of the present invention to provide for frequency turning methods for use with magnetron oscillators. | Mid | [
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There has been a continued implied sense in the use of the term "academic studies" (and related terms) throughout this thread to mean the study of history, the search for so-called historical facts. I would like to point out that this is only one possible line of academic approach, and definitely not the only one. While such searches for historical facts, including the search for the so-called "historical Buddha" were more popular in the latter 19th and early 20th centuries, they are not in vogue these days at all. Post modern approaches to such topics have shown so many problems that few scholars will touch these things, and those that do usually end up with a huge number of holes. (Having spent over a semester studying and teaching Nakamura's Gotama Buddha, it really is a methodological mess at times. Bechert et al's efforts to date the Buddha show a huge range of problems, too.) What was scholarship 50 years ago is still in vogue among a more popular level, and a kind of pseudo scholarship, however. A look at Part 2 of Hans Penner's recent Rediscovering the Buddha, for example, should be enough to point out many of the problems, and also shows that much of modern Buddhist studies academia is not working along the lines of history or searches for historical facts. Huseng, my friend, I must assure you that I am not anti-intellectual in the slightest. And I have definitely come across the type of anti-intellectual person with "faith" and beliefs they can't handle being questioned. But I am not that sort of person. I think part of the fact is that I wasn't putting myself and my views clearly in context earlier. So lemme try to clarify. I am strongly pro-intellectual in general--in fact I am a nursing student and up to my eyeballs, gladly, in evidence-based practice... and I'm specifically pro-intellectual in the case of most western people first exploring Buddhism and seeing what the teachings actually consist of (instead of some vaguely feel-good notion of what the Buddhist path is) and taking some time to test it out to see if it holds up to analysis and to their own observations, experiences, and conclusions. I am especially for this in the case of people considering and exploring Vajrayana. I think it's an excellent idea to study the foundational Buddhist texts, particularly the distillations of the main points re: aggregates, elements, and sense fields and the major mental afflictions and their derivatives, etc; and especially the distillations of the Prajnaparamita sutras such as the Madhyamakavatara and Mulamadhyamakarikas, and the main points of the Tathagatagarbha sutras like the Uttaratantra shastra; likewise for Buddhist logic, and so on. These are just some of the main things I personally think are important for most every beginning western Buddhist interested in the Tibetan Vajrayana tradition. Then, in terms of inner tantra ala Nyingma, you can't go wrong with shastras like Do Drupchen's Key to the Precious Treasury or Mipham's Essence of Clear Light, both on the Guyhagarbha tantra. And I fully support people studying Davidson and the like if they're moved to. When I seemed to poo poo academic study of tantra, I didn't make it clear I was speaking from the POV of someone who has already done tons of research and studying and asked many, many, many questions of scholars and contemplated things and came to my own conclusions. I WAS that "doggedly curious and doubtful person" I mentioned in an earlier post and which you took to be a pejorative label. It wasn't--some of us are just that way and we need extra help coming to terms authentically with Vajrayana practice and squaring it with our intellects and their notions. That's not a judgmentally qualitative statement, only a matter of fact one. But now I've been practicing inner tantric creation stage for 12 yrs--not a long time by most accounts, and there are no impressive claims I can make about my practice, but I have practiced enough to have gotten into my own groove with practice. And I eventually came to realize that--within the realm of creation stage in Nyingma, which is intertwined with the view and practice of Dzogchen--a whole bunch of intellectual focus is beside the point in creation stage because the aim of practice isn't to build something up with the intellect, but to gradually rely less and less on the intellect and ease more into one's real condition, which is perfected and endowed with all qualities from the beginning. So it's not that this is anti-intellectual in the conventional foolish, threatened, self-conscious way we all too often see in the world today. Rather it's deeper than the intellect--it's about tapping into the unstained condition the intellect is a distortion of. So, to restate briefly, I think intellectual study is excellent in the beginning and until one comes to grasp that the practice is more profound than intellectual data. At that latter point, if one wants to do analysis, the supreme analysis is along the lines of "who is visualizing this seed syllable and mantra mala?" and the like. That way one can go straight to the point. Yudron wrote:... But, as a practitioner, the feeling state of being in a fortified palace--for those deities that have one--is simply evocative of a certain style of enlightenment. I feel the qualities of the deities mandala.. this is enlivening... For Nyingmpas, the deity is rigpa, and we are resting in that. IMHO analyzing the historical under-pinnings of the imagery and so forth just leads to a splitting off in one's practice, contributing to the idea that these practices are far away Asian things. To me, the setting of these practices is my house, and my body, and there is no time. Yudron, you've said in far fewer words what I was unable to clearly convey in several paragraphs! I'm thankful for that (as I'm sure everyone else is, too! ) intellectualizing, analysing seems giving debates so often while those need no rejection at all! There is taught; to remain intellectualizing when terma is 'given', is not exactly helpful. Since terma reveals nondual nature, keeping intellectualizing/conceptualizing is still believing natures' truth is to reveal by thoughts and so there is grasping to them. Awakened nature is always "talk" from nature's clarity wisdom. Without awakened nature/ Guru, mind remains keeping track of coming/going of thoughts, conditioned intellect. But merely rejecting the navigation intellectual system is stupid. All paths are to open our nature, there is no distinction at all in nature. Deep bowing for ALL. Huifeng wrote:There has been a continued implied sense in the use of the term "academic studies" (and related terms) throughout this thread to mean the study of history, the search for so-called historical facts. I would like to point out that this is only one possible line of academic approach, and definitely not the only one. While such searches for historical facts, including the search for the so-called "historical Buddha" were more popular in the latter 19th and early 20th centuries, they are not in vogue these days at all. Post modern approaches to such topics have shown so many problems that few scholars will touch these things, and those that do usually end up with a huge number of holes. (Having spent over a semester studying and teaching Nakamura's Gotama Buddha, it really is a methodological mess at times. Bechert et al's efforts to date the Buddha show a huge range of problems, too.) What was scholarship 50 years ago is still in vogue among a more popular level, and a kind of pseudo scholarship, however. A look at Part 2 of Hans Penner's recent Rediscovering the Buddha, for example, should be enough to point out many of the problems, and also shows that much of modern Buddhist studies academia is not working along the lines of history or searches for historical facts. | Mid | [
0.612068965517241,
35.5,
22.5
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--- abstract: 'We consider an oriented surface $S$ and a cellular complex $X$ of curves on $S$, defined by Hatcher and Thurston in 1980. We prove by elementary means, without Cerf theory, that the complex $X$ is connected and simply connected. From this we derive an explicit simple presentation of the mapping class group of $S$, following the ideas of Hatcher–Thurston and Harer.' address: | Department of Mathematics\ Technion, 32000 Haifa, Israel author: - Bronislaw Wajnryb title: | An elementary approach to the mapping class\ group of a surface --- Introduction ============ Let $S$ be a compact oriented surface of genus $g\ge 0$ with $n\ge 0$ boundary components and $k$ distinguished points. The mapping class group $\cM_{g,n,k}$ of $S$ is the group of the isotopy classes of orientation preserving homeomorphisms of $S$ which keep the boundary of $S$ and all the distinguished points pointwise fixed. In this paper we study the problem of finding a finite presentation for $\cM_{g,n}=\cM_{g,n,0}$. We restrict our attention to the case $n=1$. The case $n=0$ is easily obtained from the case $n=1$. In principle the case of $n>1$ (and the case of $k>0$) can be obtained from the case $n=1$ via standard exact sequences, but this method does not produce a global formula for the case of several boundary components and the presentation (in contrast to the ones we shall describe for the case $n=0$ and $n=1$) becomes rather ugly. On the other hand Gervais in [@Gervais] succeeded recently to produce a finite presentation of $\cM_{g,n}$ starting from the results in [@Wajnryb] and using a new approach. A presentation for $g=1$ has been known for a long time. A quite simple presentation for $g=2$ was established in 1973 in [@Birman-Hilden], but the method did not generalize to higher genus. In 1975 McCool proved in [@McCool], by purely algebraic methods, that $\cM_{g,1}$ is finitely presented for any genus $g$. It seems that extracting an explicit finite presentation from his proof is very difficult. In 1980 appeared the groundbreaking paper of Hatcher and Thurston [@Hatcher-Thurston] in which they gave an algorithm for constructing a finite presentation for the group $\cM_{g,1}$ for an arbitrary $g$. In 1981 Harer applied their algorithm in [@Harer] to obtain a finite (but very unwieldy) explicit presentation of $\cM_{g,1}$. His presentation was simplified by Wajnryb in 1983 in [@Wajnryb]. A subsequent Errata [@Errata] corrected small errors in the latter. The importance of the full circle of ideas in these papers can be jugded from a small sample of subsequent work which relied on the presentation in [@Wajnryb], eg [@Kohno], [@Lu], [@Matsumoto], [@Matveev-Polyak]. The proof of Hatcher and Thurston was deeply original, and solved an outstanding open problem using novel techniques. These included arguments based upon Morse and Cerf Theory, as presented by Cerf in [@Cerf]. In this paper we shall give, in one place, a complete hands-on proof of a simple presentation for the groups $\cM_{g,0}$ and $\cM_{g,1}$. Our approach will follow the lines set in [@Hatcher-Thurston], but we will be able to use elementary methods in the proof of the connectivity and simple connectivity of the cut system complex. In particular, our work does not rely on Cerf theory. At the same time we will gather all of the computational details in one place, making the result accesible for independent checks. Our work yields a slightly different set of generators and relations from the ones used in [@Wajnryb] and in [@Errata]. The new presentation makes the computations in section 4 a little simpler. We shall give both presentations and prove that they are equivalent. A consequence of this paper is that, using Lu [@Lu] or Matveev and Polyak [@Matveev-Polyak], the fundamental theorem of the Kirby calculus [@Kirby] now has a completely elementary proof (ie, one which makes no appeal to Cerf theory or high dimensional arguments). This paper is organised as follows. In section 2 we give a new proof of the main theorem of Hatcher and Thurston. In section 3 we derive a presentation of the mapping class group following (and explaining) the procedure described in [@Hatcher-Thurston] and in [@Harer]. In section 4 we reduce the presentation to the simple form of Theorem \[simple presentation\] repeating the argument from [@Wajnryb] with changes required by a slightly different setup. In section 5 we deduce the case of a closed surface and in section 6 we translate the presentation into the form given in \[20\], see Remark 1 below. We start with a definition of a basic element of the mapping class group. \[Dehn twist\] [A (positive) Dehn twist with respect to a simple closed curve $\alpha$ on an oriented surface $S$ is an isotopy class of a homeomorphism $h$ of $S$, supported in a regular neighbourhood $N$ of $\alpha$ (an annulus), obtained as follows: we cut $S$ open along $\alpha$, we rotate one side of the cut by 360 degrees to the right (this makes sense on an oriented surface) and then glue the surface back together, damping out the rotation to the identity at the boundary of $N$. The Dehn twist (or simply twist) with respect to $\alpha$ will be denoted by $T_\alpha$. If curves $\alpha$ and $\beta$ intersect only at one point and are transverse then $T_\alpha(\beta)$, up to an isotopy, is obtained from the union $\alpha\cup\beta$ by splitting the union at the intersection point.]{} 1.05 We shall say that two elements $a,b$ of a group are braided (or satisfy the braid relation) if $aba=bab$. \[simple presentation\] Let $S_{g,1}$ be a compact, orientable surface of genus $g\geq 3$ with one boundary component. Let $a_i,b_i,e_i$ denote Dehn twists about the curves $\alpha_i,\beta_i,\epsilon_i$ on $S_{g,1}$ depicted on Figure \[general surface\]. The mapping class group $\cM_{g,1}$ of $S_{g,1}$ is generated by elements $b_2,b_1,a_1,e_1,a_2,e_2, \dots,a_{g-1},e_{g-1},a_g$ and has defining relations: [(M1)]{}Elements $b_2$ and $a_2$ are braided and $b_2$ commutes with $b_1$. Every other pair of consecutive elements (in the above order) is braided and every other pair of non-consecutive elements commute. [(M2)]{} $(b_1a_1e_1a_2)^5=b_2a_2e_1a_1b_1^2a_1e_1a_2b_2$ [(M3)]{} $d_3a_1a_2a_3=d_{1,2}d_{1,3}d_{2,3}$, where $d_{1,2}=(a_2e_1a_1b_1)^{-1}b_2(a_2e_1a_1b_1)$, $d_{1,3}=t_2d_{1,2}t_2^{-1}$, $d_{2,3}=t_1d_{1,3}t_1^{-1}$, $\strut t_1=e_1a_1a_2e_1$, $t_2=e_2a_2a_3e_2$, $d_3 =b_2a_2e_1b_1^{-1}d_{1,3}b_1e_1^{-1}a_2^{-1}b_2^{-1}$. \[M2 presentation\] The mapping class group $\cM_{2,1}$ of an orientable surface $S_{2,1}$ of genus $g=2$ with one boundary component is generated by elements $b_2,b_1,a_1,$ $e_1,a_2$ and has defining relations [(M1)]{} and [(M2)]{}. \[presentation closed\] The mapping class group $\cM_{g,0}$ of a compact, closed, orientable surface of genus $g>1$ can be obtained from the above presentation of $\cM_{g,1}$ by adding one more relation: [(M4)]{} $[b_1a_1e_1a_2\dots a_{g-1}e_{g-1}a_ga_ge_{g-1}a_{g-1} \dots e_1a_1b_1,d_g]=1$, where $t_i=e_ia_ia_{i+1}e_i$, for $i=1,2,\dots,g-1$, $d_2=d_{1,2}$, $\strut d_i=(b_2a_2e_1b_1^{-1}t_2t_3\dots t_{i-1}) d_{i-1}(b_2a_2e_1b_1^{-1}t_2t_3\dots t_{i-1})^{-1}$, for $i=3,4,\dots,g$ The presentations in Theorems 1 and 3 are equivalent to but not the same as those in [@Wajnryb] and [@Errata]. We now give alternative presentations of Theorems 1 and 3, with the goal of correlating the work in this paper with that in [@Wajnryb] and [@Errata]. See Remark 1, below, for a dictionary which allows one to move between Theorems 1$^\prime$ and 3$^\prime$ and the results in [@Wajnryb] and [@Errata]. See Section 6 of this paper for a proof that the presentations in Theorems 1 and $1^\prime$, and also Theorems 3 and $3^\prime$, are equivalent. [**Theorem 1$^\prime$**]{} [**Theorem 3$^\prime$**]{} We now explain how to move back and forth between the results in this paper and those in [@Wajnryb] and [@Errata]. 1. The surface and the curves in [@Wajnryb] look different from the surface and the curves on Figure \[general surface\]. However if we compare the Dehn twist generators in Theorem $1^\prime$ with those in Theorem 1 of [@Wajnryb] and [@Errata] we see that corresponding curves have the same intersection pattern. Thus there exists a homeomorphism of one surface onto the other which takes the curves of one family onto the corresponding curves of the other family. The precise correspondence is given by: $(b_2,b_1,a_1,e_1,a_2,e_2, \dots,a_{g-1},e_{g-1},a_g)$ $\longleftrightarrow\ \ (d,a_1,b_1,a_2,b_2,a_3,\dots,b_{n-1},a_n,b_n)$, where the top sequence refers to Dehn twists about the curves in Figure 1 of this paper and the bottom sequence refers to Dehn twists about the curves in Figure 1 on page 158 of [@Wajnryb]. 2. Composition of homeomorphisms in [@Wajnryb] was performed from left to right, while in the present paper we use the standard composition from right to left. 3. The element $d_g$ in this paper represents a Dehn twist about the curve $\delta_g$ in Figure \[general surface\] of this paper. The element ${\tilde d}_g$ in relation (D) of Theorem $3^\prime$ represents a Dehn twist about the curve $\beta_g$ in Figure 1. We wrote $d_g$ as a particular product of the generators in $\cM_{g,1}$. It follows from the argument in the last section that any other such product representing $d_g$ will also do. 4. In the case of genus $g=2$ we should omit relation (C) in Theorems $1^\prime$ and $3^\prime$. Here is the plan of the proof of the theorems. Following Hatcher and Thurston we define a 2–dimensional cell complex $X$ on which the mapping class group $\cM_{g,1}$ acts by cellular transformations and the action is transitive on the vertices of $X$. We give a new elementary proof of the fact that $X$ is connected and simply connected. We then describe the stabilizer $H$ of one vertex of $X$ under the action of $\cM_{g,1}$ and we determine an explicit presentation of $H$. Following the algorithm of Hatcher and Thurston we get from it a presentation of $\cM_{g,1}$. Finally we reduce the presentation to the form in Theorem \[simple presentation\] and as a corollary get Theorem \[presentation closed\]. I wish to thank very much the referee, who studied the paper very carefully and made important suggestions to improve it. This research was partially supported by the Fund for the Promotion of Research at the Technion and the Dent Charitable Trust — a non military research fund. Cut-system complex {#cut-system complex} ================== We denote by $S$ a compact, connected oriented surface of genus $g> 0$ with $n\ge 0$ boundary components. We denote by $\bar S$ a closed surface obtained from $S$ by capping each boundary component with a disk. By a [*curve*]{} we shall mean a simple closed curve on $S$. We are mainly interested in the isotopy classes of curves on $S$. The main goal in the proofs will be to decrease the number of intersection points between different curves. If the Euler characteristic of $S$ is negative we can put a hyperbolic metric on $S$ for which the boundary curves are geodesics. Then the isotopy class of any non-separating curve on $S$ contains a unique simple closed geodesic, which is the shortest curve in its isotopy class. If we replace each non-separating curve by the unique geodesic isotopic to it we shall minimize the number of intersection points between every two non-isotopic curves, by Corollary \[hass-scott\]. So we can think of curves as geodesics. In the proof we may construct new curves, which are not geodesics and which have small intersection number with some other curves. When we replace the new curve by the corresponding geodesic we further decrease the intersection number. If $S$ is a closed torus we can choose a flat metric on $S$. Now geodesics are not unique but still any choice of geodesics will minimize the intersection number of any pair of non-isotopic curves. If two curves are isotopic on $S$ then they correspond to the same geodesic, but we shall call them disjoint because we can isotop one off the other. Geodesics are never tangent. If $\alpha$ and $\beta$ are curves then $|\alpha\cap\beta|$ denotes their geometric intersection number, ie, the number of intersection points of $\alpha$ and $\beta$. If $\alpha$ is a curve we denote by $[\alpha]$ the homology class represented by $\alpha$ in $H_1(\bar S,\ints)$, up to a sign. We denote by $i(\alpha,\beta)$ the absolute value of the algebraic intersection number of $\alpha$ and $\beta$. It depends only on the classes $[\alpha]$ and $[\beta]$. We shall describe now the cut-system complex $X$ of $S$. To construct $X$ we consider collections of $g$ disjoint curves $\gamma_1,\gamma_2,\dots,\gamma_g$ in $S$ such that when we cut $S$ open along these curves we get a connected surface (a sphere with $2g+n$ holes). An isotopy class of such a collection we call a cut system $\lan \gamma_1,\dots,\gamma_g\ran$. We can say that a cut system is a collection of geodesics. A curve is contained in a cut-system if it is one of the curves of the collection. If $\gamma_i^\prime$ is a curve in $S$, which meets $\gamma_i$ at one point and is disjoint from other curves $\gamma_k$ of the cut system $\lan \gamma_1,\dots,\gamma_g\ran$, then $\lan \gamma_1,\dots,\gamma_{i-1},\gamma_i^\prime, \gamma_{i+1},\dots,\gamma_g\ran$ forms another cut system. In such a situation the replacement $\lan \gamma_1,\dots, \gamma_i,\dots,\gamma_g\ran\to \lan \gamma_1,\dots,\gamma_i^\prime,\dots,\gamma_g\ran$ is called a simple move. For brewity we shall often drop the symbols for unchanging circles and shall write $\lan \gamma_i\ran\to\lan \gamma_i^\prime\ran$. The cut systems on $S$ form the 0–skeleton (the vertices) of the complex $X$. We join two vertices by an edge if and only if the corresponding cut systems are related by a simple move. We get the 1–skeleton $X^1$. By a [*path*]{} we mean an edge-path in $X^1$. It consists of a sequence of vertices ${\bf p}=( v_1, v_2,\dots,v_k)$ where two consecutive vertices are related by a simple move. A path is closed if $ v_1=v_k$. We distinguish three types of closed paths:\ If three vertices (cut-systems) have $g-1$ curves $\gamma_1,\dots,\gamma_{g-1}$ in common and if the remaining three curves $\gamma_g,\gamma_g^\prime,\gamma_g^{\pp}$ intersect each other once, as on Figure \[faces\], C3, then the vertices form a closed triangular path: (C3)a triangle$\lan \gamma_g\ran\to\lan\gamma_g^\prime\ran\to\lan \gamma_g^{\pp}\ran\to\lan \gamma_g\ran$.\ If four vertices have $g-2$ curves $\gamma_1,\dots,\gamma_{g-2}$ in common and the remaining pairs of curves consist of $(\gamma_{g-1},\gamma_g)$, $(\gamma_{g-1}^\prime,\gamma_g)$, $(\gamma_{g-1}^\prime,\gamma_g^\prime)$, $(\gamma_{g-1}, \gamma_g^\prime)$ where the curves intersect as on Figure \[faces\], C4, then the vertices form a closed square path: (C4)a square$\lan \gamma_{g-1},\gamma_g\ran\to\lan\gamma_{g-1}^\prime, \gamma_g\ran \to\lan\gamma_{g-1}^\prime,\gamma_g^\prime\ran \to\lan \gamma_{g-1},\gamma_g^\prime\ran\to\lan \gamma_{g-1},\gamma_g\ran$.\ If five vertices have $g-2$ curves $\gamma_1,\dots,\gamma_{g-2}$ in common and the remaining pairs of curves consist of $(\gamma_{g-1},\gamma_g)$, $(\gamma_{g-1},\gamma_g^\prime)$, $(\gamma_{g-1}^\prime,\gamma_g^\prime)$, $(\gamma_{g-1}^\prime, \gamma_g^{\pp})$ and $(\gamma_g,\gamma_g^{\pp})$ where the curves intersect as on Figure \[faces\], C5, then the vertices form a closed pentagon path: (C5)a pentagon$\lan \gamma_{g-1},\gamma_g\ran\to\lan \gamma_{g-1}, \gamma_g^\prime\ran\to\lan\gamma_{g-1}^\prime,\gamma_g^\prime\ran \to\lan\gamma_{g-1}^{\prime},\gamma_g^{\pp}\ran\to$$\strut\lan\gamma_g^{\pp}, \gamma_g\ran \to\lan \gamma_{g-1},\gamma_g\ran$.\ $X$ is a 2–dimensional cell complex obtained from $X^1$ by attaching a 2–cell to every closed edge-path of type (C3), (C4) or (C5). The mapping class group of $S$ acts on $S$ by homeomorphisms so it takes cut systems to cut systems. Since the edges and the faces of $X$ are determined by the intersections of pairs of curves, which are clearly preserved by homeomorphisms, the action on $X^0$ extends to a cellular action on $X$. In this section we shall prove the main result of [@Hatcher-Thurston]: \[Hatcher-Thurston\] $X$ is connected and simply connected. We want to prove that every closed path $\bf p$ is null-homotopic. If $\bf p$ is null-homotopic we shall write ${\bf p\sim o}$. We start with a closed path ${\bf p}=(v_1,\dots,v_k)$ and try to simplify it. If $\bf q$ is a short-cut, an edge path connecting a vertex $v_i$ of $\bf p$ with $v_j$, $j>i$, we can split $\bf p$ into two closed edge-paths:${\bf p}_1=(v_j,v_{j+1},\dots, v_k,v_2,\dots,v_{i-1},{\bf q})$ and ${\bf p}_2=({\bf q}^{-1},v_{i+1},v_{i+2}, \dots,v_j)$.If both paths are null-homotopic in $X$ then $\bf p\sim o$. We want to prove Theorem \[Hatcher-Thurston\] by splitting path $\bf p$ into simpler paths according to a notion of complexity which is described in the next definition. \[radius\] [Let ${\bf p}=(v_1,v_2,\dots,v_k)$ be a path in $X$. Let $\alpha$ be a fixed curve of some fixed $v_j$. We define [*distance*]{} from $\alpha$ to a vertex $v_i$ to be $d(\alpha,v_i)=min\{|\alpha\cap\beta|:\beta\in v_i\}$. The [*radius of ${\bf p}$ around $\alpha$*]{} is equal to the maximum distance from $\alpha$ to the vertices of ${\bf p}$. The path $\bf p$ is called a [*segment*]{} if every vertex of $\bf p$ contains a fixed curve $\alpha$. We shall write $\alpha$–segment if we want to stress the fact that $\alpha$ is the common curve of the segment. If the segment has several fixed curves we can write $(\alpha,\beta,\dots,\gamma)$–segment.]{} Theorem \[Hatcher-Thurston\] will be proven by induction on the genus of $S$, for the given genus it will be proven by induction on the radius of a path $\bf p$ and for a given radius $m$ around a curve $\alpha$ we shall induct on the number of segments of $\bf p$ which have a common curve $\gamma$ with $|\gamma\cap\alpha|=m$. The main tool in the proof of Theorem \[Hatcher-Thurston\] is a reduction of the number of intersection points of curves. \[def 2-gon\] [Curves $\alpha$ and $\beta$ on $S$ have an [*excess intersection*]{} if there exist curves $\alpha^\prime$ and $\beta^\prime$, isotopic to $\alpha$ and $\beta$ respectively, and such that $|\alpha\cap\beta|>|\alpha^\prime\cap\beta^\prime|$. Curves $\alpha$ and $\beta$ form a 2–[*gon*]{} if there are arcs $a$ of $\alpha$ and $b$ of $\beta$, which meet only at their common end points and do not meet other points of $\alpha$ or $\beta$ and such that $a\cup b$ bound a disk, possibly with holes, on $S$. The disk is called a 2–gon. We can [*cut off*]{} the 2–gon from $\alpha$ by replacing the arc $a$ of $\alpha$ by the arc $b$ of $\beta$. We get a new curve $\alpha^\prime$ (see Figure \[2-gon\]).]{} \[hass\] If $\alpha,\beta$ are two curves on a surface $S$ having an excess intersection then they form a 2–gon (without holes) on $S$. \[hass-scott\] Two simple closed geodesics on $S$ have no excess intersection. In particular if we replace two curves by geodesics in their isotopy class then the number of intersection points between the curves does not increase. If there is a 2–gon we can shorten one geodesic in its homotopy class, by first cutting off the 2–gon and then smoothing corners. \[cut off 2-gon\] Consider a finite collection of simple closed geodesics on $S$. Suppose that curves $\alpha$ and $\beta$ of the collection form a minimal 2–gon (which does not contain another 2–gon). Let $\alpha^\prime$ be the curve obtained by cutting off the minimal 2–gon from $\alpha$ and passing to the isotopic geodesic. Then $|\alpha\cap\alpha^\prime|=0$, $|\beta\cap\alpha^\prime|<|\beta\cap\alpha|$ and $|\gamma\cap\alpha^\prime| \leq |\gamma\cap\alpha|$ for any other curve $\gamma$ of the collection. In particular if $ |\gamma\cap\alpha|=1$ then $ |\gamma\cap\alpha^\prime|=1$. Also $[\alpha]=[\alpha^\prime]$. Since the 2–gon formed by arcs $a$ and $b$ of $\alpha$ and $\beta$ is minimal every other curve $\gamma$ of the collection intersects the 2–gon along arcs which meet $a$ and $b$ once. Thus cutting off the 2–gon will not change $|\alpha\cap\gamma|$ and it may only decrease after passing to the isotopic geodesic. Clearly $\alpha$ and $\alpha^\prime$ are disjoint and homologous on $\bar S$ and by passing to $\alpha^\prime$ we remove at least two intersection points of $\alpha$ with $\beta$. The case of genus 1 ------------------- In this section we shall assume that $S$ is a surface of genus one, possibly with boundary. By $\bar S$ we shall denote the closed torus obtained by glueing a disk to each boundary component of $S$. We want to prove: \[genus1\] If $S$ has genus one then the cut system complex $X$ of $S$ is connected and simply connected. On a closed torus $\bar S$ the homology class $[\alpha]$ of a curve $\alpha$ is defined by a pair of relatively prime integers, up to a sign, after a fixed choice of a basis of $H_1(\bar S,\ints)$. If $[\alpha]=(a_1,a_2)$ and $[\beta]=(b_1,b_2)$ then the absolute value of their algebraic intersection number is equal $i(\alpha,\beta)=|a_1b_2-a_2b_1|$. If $\alpha$ and $\beta$ are geodesics on $\bar S$ then $|\alpha\cap\beta|= i(\alpha,\beta)$, therefore it is also true for curves on $S$ which form no 2–gons, because then they have no excess intersection on $\bar S$. Let $\alpha$, $\beta$ be nonseparating curves on $S$ and suppose that $|\alpha\cap\beta|=k\neq 1$. Then there exists a nonseparating curve $\delta$ such that if $k=0$ then $|\delta\cap\alpha|=|\delta\cap \beta|=1$ and if $k>1$ then $|\delta\cap\alpha|<k$ and $|\delta\cap\beta|<k$. If $|\alpha\cap\beta|=0$ then the curves are isotopic on $\bar S$ and they split $S$ into two connected components $S_1$ and $S_2$. We can choose points $P$ and $Q$ on $\alpha$ and $\beta$ respectively and connect them by simple arcs in $S_1$ and in $S_2$. The union of the arcs forms the required curve $\delta$. If $k>1$ and if the curves have an excess intersection on $\bar S$ then they form a 2–gon on $S$. We can cut off the 2–gon decreasing the intersection, by Lemma \[cut off 2-gon\]. If there are no 2–gons then the algebraic and the geometric intersection numbers are equal. In particular all intersections have the same sign. Consider two intersection points consecutive along $\beta$. Choose $\delta$ as on Figure \[equal signs\]. Then $|\delta\cap\alpha|=1$ so it is nonseparating and $|\delta\cap\beta|<k$. .7in If surface $S$ has genus 1 then the cut system complex of $S$ is connected. A cut system on $S$ is an isotopy class of a single curve. If two curves intersect once they are connected by an edge. It follows from the last lemma by induction that any two curves can be connected by an edge-path in $X$.We now pass to a proof that closed paths are null-homotopic. \[square\] A closed path ${\bf p}=(\delta_1,\delta_2,\delta_3,\delta_4,\delta_1)$ such that $|\delta_2\cap \delta_4|=0$ is null-homotopic in $X$. Let $\beta=T_{\delta_2}^{\pm 1}(\delta_3)$ be the image of $\delta_3$ by the Dehn Twist along $\delta_2$. Recall that as a set $\beta=\delta_2\cup \delta_3$ outside a small neighbourhood of $\delta_2\cap\delta_3$. From this we get that $|\beta\cap \delta_2|=1$, $|\beta\cap \delta_3|=1$ and $|\beta\cap \delta_4|=1$. Thus ${\bf p^\prime}=(\delta_1,\delta_2,\beta,\delta_4,\delta_1)$ is a closed path which is homotopic to ${\bf p}$ because ${\bf p-p}^\prime$ splits into a sum of two triangles ${\bf t_1}=(\beta,\delta_2,\delta_3,\beta)$ and ${\bf t_2}=(\beta,\delta_3,\delta_4,\beta)$. We also have $i(\beta,\delta_1)=|i(\delta_3,\delta_1)\pm i(\delta_2,\delta_1)|$, so for a suitable choice of the sign of the twist we have $i(\delta_3,\delta_1)>i(\beta,\delta_1)$ unless $i(\delta_3,\delta_1)=0$. We may assume by induction that $i(\delta_3,\delta_1)=0$. If $|\delta_1\cap\delta_3|>0$ then $\delta_1$ and $\delta_3$ have an excess intersection on $\bar S$ and form a 2–gon on $S$. We can cut off the 2–gon from $\delta_3$ getting a new curve $\beta$ such that $[\beta]=[\delta_3]$, $|\beta\cap \delta_3|=0$, $|\beta\cap \delta_1|<|\delta_3\cap \delta_1|$ and $|\beta\cap\delta_i|=|\delta_3\cap\delta_i|$ for $i\neq 1$, by Lemma \[cut off 2-gon\]. We get a new closed path ${\bf p^\prime}=(\delta_1,\delta_2,\beta,\delta_4,\delta_1)$ and the difference between it and the old path is a closed path ${\bf q}=(\beta,\delta_2,\delta_3,\delta_4,\beta)$ with $|\beta\cap \delta_3|=0$ and $|\delta_2\cap \delta_4|=0$. So by induction it suffices to assume that $|\delta_1\cap \delta_3|=0$. If we now let $\beta=T_{\delta_2}(\delta_3)$ then $\beta$ intersects each of the four curves once so our path is a sum of four triangles and thus is null-homotopic in $X$. \[no 2-gons\] If ${\bf p}=(\alpha_1,\dots,\alpha_k)$ is a closed path then there exists a closed path ${\bf p^\prime}=(\alpha_1^\prime,\dots,\alpha_k^\prime)$ such that ${\bf p}^\prime$ is homotopic to $\bf p$ in $X$, $[\alpha_i]=[\alpha_i^\prime]$ for all $i$ and the collection of curves $\alpha_1^\prime,\dots,\alpha_{k-1}^\prime$ forms no 2–gons. Suppose that there exists a 2–gon bounded by arcs of two curves in ${\bf p}$. Then there also exists a minimal 2–gon bounded by arcs of curves $\alpha_i$ and $\alpha_j$. If we cut off the 2–gon from $\alpha_i$ we get a curve $\alpha_i^\prime$ such that $[\alpha_i]=[\alpha_i^\prime]$, $\alpha_i^\prime$ is disjoint from $\alpha_i$, $|\alpha_j\cap\alpha_i|>|\alpha_j\cap\alpha_i^\prime|$ and $|\alpha_m\cap\alpha_i^\prime|\leq|\alpha_m\cap\alpha_i|$ for any other curve $\alpha_m$, by Lemma \[cut off 2-gon\]. It follows that if we replace $\alpha_i$ by $\alpha_i^\prime$ we get a new closed path ${\bf p^\prime}$ with a smaller number of intersection points between its curves. The difference of the two paths is a closed path ${\bf q}=(\alpha_{i-1},\alpha_i,\alpha_{i+1}, \alpha_i^\prime,\alpha_{i-1})$ which is null-homotopic by Lemma \[square\]. Thus ${\bf p^\prime}$ is homotopic to ${\bf p}$ in $X$. Lemma \[no 2-gons\] follows by induction on the total number of intersection points between pairs of curves of ${\bf p}$.[**Proof of Proposition \[genus1\]**]{} Let ${\bf p}=(\alpha_1,\dots,\alpha_k,\alpha_1)$ be a closed path in $X^1$. We may assume that the path has no 2–gons. By Lemma \[hass\] there is no excess intersection on the closed torus $\bar S$. It means that the geometric intersection number of two curves of ${\bf p}$ is equal to their algebraic intersection number. Let $m=max\{i(\alpha_1,\alpha_j)|j=1,\dots,k\}$ be the [*radius*]{} of ${\bf p}$ around $\alpha_1$. Suppose first that $m=1$. Two disjoint curves on $\bar S$ have the same homology class, and two curves representing the same class have algebraic intersection equal to 0. It follows that two consecutive curves in a path cannot be both disjoint from $\alpha_1$. If $k>4$ then either $|\alpha_1\cap\alpha_3|=1$ or $|\alpha_1\cap\alpha_4|=1$. We get an edge which splits ${\bf p}$ into two shorter closed paths with radius $1$. If $k=4$ and $|\alpha_1\cap\alpha_3|=0$ then ${\bf p\sim o}$ by Lemma \[square\]. If $k=3$ then ${\bf p}$ is a triangle. Suppose now that $m>1$. We may assume by induction that every path of radius less than $m$ is null-homotopic and every path of radius $m$ which has less curves $\alpha_j$ with $|\alpha_1\cap\alpha_j|=m$ is also null-homotopic. Choose the smallest $i$ such that $i(\alpha_1,\alpha_i)=m$. Then $i(\alpha_1,\alpha_{i-1})< m$ and $i(\alpha_1,\alpha_{i+1})\leq m$. Choose a basis of the homology group $H_1(\bar S)$ which contains the curve $\alpha_1$. A homology class of a curve is then represented by a pair of integers $(a,b)$. We consider the homology classes and their intersection numbers up to a sign. We have $[\alpha_1]=(1,0)$, $[\alpha_{i-1}]=(a,b)$, $[\alpha_i]=(p,m)$ and $[\alpha_{i+1}]=(c,d)$. The intersection form is defined by $i((a,b),(c,d))=|ad-bc|$. Thus $am-bp=\pm 1,cm-dp=\pm 1, |b|< m, |d|\leq m$. We get $m(ad-bc)=(\pm d\pm b)$. Since $2m>|b|+|d|$ we must have $|ad-bc|=1$ or $ad-bc=0$. In the first case $i(\alpha_{i-1},\alpha_{i+1})=|\alpha_{i-1}\cap \alpha_{i+1}|=1$. We can cut off" the triangle ${\bf q}=(\alpha_{i-1},\alpha_i,\alpha_{i+1},\alpha_{i-1})$ getting a path which is null-homotopic by the induction hypothesis. If $ad-bc=0$ then $i(\alpha_{i-1},\alpha_{i+1})=|\alpha_{i-1}\cap \alpha_{i+1}|=0$. Let $\beta=T_{\alpha_{i-1}}^{\pm 1}(\alpha_i)$. Then $|\beta\cap \alpha_{i-1}|=1$ and $|\beta\cap \alpha_{i+1}|=1$. We can replace $\alpha_i$ by $\beta$ getting a new closed path. Their difference is the closed path ${\bf q}=(\alpha_{i-1},\alpha_i,\alpha_{i+1},\beta,\alpha_{i-1})$ which is null homotopic by lemma \[square\]. Thus the new path is homotopic to the old path. For a suitable choice of the sign of the Dehn twist we have $i(\beta,\alpha_1)<m$. It may happen that $|\beta\cap\alpha_1|\geq m$. We can get rid of 2–gons by Lemma \[no 2-gons\] and thus get rid of the excess intersection. We get a homotopic path which is null-homotopic by the induction hypothesis. This concludes the proof of Proposition \[genus1\] Paths of radius $0$ ------------------- From now until the end of section \[cut-system complex\], we assume that $S$ is a surface of genus $g>1$ with a finite number of boundary components. We denote by $X$ the cut-system complex of $S$. We assume: [**Induction Hypothesis 1**]{}The cut-system complex of a surface of genus less than $g$ is connected and simply-connected. We want to prove that every closed path in $X$ is null-homotopic. We shall start with paths of radius zero. The simplest paths of radius zero are closed segments. A closed segment is null-homotopic in $X$. When we cut $S$ open along the common curve $\alpha$ the remaining curves of each vertex form a vertex of a closed path in the cut-system complex of a surface of a smaller genus. By Induction Hypothesis 1 it is a sum of paths of type (C3), (C4) and (C5) there. When we adjoin $\alpha$ to every vertex we get a splitting of the original paths into null-homotopic paths.In a similar way we prove: \[common curve\] If two vertices of $X$ have one or more curves in common we can connect them by a path all of whose vertices contain the common curves. If we cut $S$ open along the common curves the remaining collection of curves form two vertices of the cut-system complex on the new surface of smaller genus. They can be connected by a path. If we adjoin all the common curves to each vertex of this path we get a path in $X$ with the required properties. \[non-separating pair\] [ If $\alpha$ and $\beta$ are two disjoint non-separating curves on $S$ then $\alpha\cup\beta$ does not separate $S$ if and only if $[\alpha]\neq[\beta]$. In this case the pair $\alpha$, $\beta$ can be completed to a cut-system on $S$.]{} We shall now construct two simple types of null-homotopic paths in $X$. \[hexagon relation\] Let $\alpha_1$, $\alpha_2$, $\alpha_3$ be disjoint curves such that the union of any two of them does not separate $S$ but the union of all three separates $S$. Then there exist disjoint curves $\beta_1$, $\beta_2$, $\beta_3$ and a closed path\ [(C6)]{}$\lan\alpha_1,\alpha_2\ran\to\lan\alpha_1,\beta_2\ran\to\lan\alpha_1, \alpha_3\ran\to\lan\beta_1,\alpha_3\ran\to\lan\alpha_2,\alpha_3 \ran$$\to\lan\alpha_2,\beta_3\ran\to\lan\alpha_2,\alpha_1\ran$,\ which is null-homotopic in $X$. Let $\gamma_3$,…, $\gamma_g$ be a cut system on a surface $S-(\alpha_1\cup\alpha_2\cup\alpha_3)$ (not connected), ie, a collection of curves which does not separate the surface any further. Then $\lan\alpha_1,\alpha_2,\gamma_3,\dots,\gamma_g\ran$ is a cut system on $S$. Let $S_1$ and $S_2$ be the components of a surface obtained by cutting $S$ open along all $\alpha_i$’s and $\gamma_j$’s. An arc connecting different components of the boundary does not separate the surface so we can find (consecutively) disjoint arcs $b_1$ connecting $\alpha_1$ with $\alpha_2$, $b_2$ connecting $\alpha_2$ with $\alpha_3$ and $b_3$ connecting $\alpha_3$ with $\alpha_1$ in $S_1$, and similar arcs $b_1^\prime$, $b_2^\prime$ and $b_3^\prime$ in $S_2$ with the corresponding ends coinciding in $S$. The pairs of corresponding arcs form the required curves $\beta_1$, $\beta_2$ and $\beta_3$ and we get a closed path ${\bf p}$ described in (C6). Moreover the curves $\beta_2$ and $\beta_3$ are disjoint and $[\beta_2]\neq[\beta_3]$. To prove that the path is null-homotopic in $X$ we choose a curve $\delta=T_{\alpha_2}(\beta_1)$. Then $|\delta\cap\alpha_1|=|\delta\cap\alpha_2|=|\delta\cap\beta_1|= |\delta\cap\beta_2|=1$ and $|\delta\cap\alpha_3|=|\delta\cap\beta_3|=0$. Figure \[hexagon\] shows how ${\bf p}$ splits into a sum of triangles (C3), squares (C4) and pentagons (C5) and therefore is null-homotopic in $X$. 0.9pt If $\bf p$ is a path of radius $0$ around a curve $\alpha$ then $\bf p\sim o$. Let $v_0$ be a vertex of ${\bf p}$ containing $\alpha$. We shall prove the proposition by induction on the number of segments of $\bf p$ having a fixed curve disjoint from $\alpha$. Consider the maximal $\alpha$–segment of ${\bf p}$ which contains the vertex $v_0$. We shall call it the first segment. Let $v_1$ be the last vertex of the first segment. The next vertex contains a curve $\beta$ disjoint from $\alpha$ such that $\beta$ is the common curve of the next segment of ${\bf p}$. Since $|\alpha\cap\beta|=0$ the simple move from $v_1$ to the next vertex does not involve $\beta$ hence $v_1$ also contains $\beta$. Let $v_2$ be the last vertex of the second segment. If there are only two segments then $v_2$ also contains both $\alpha$ and $\beta$. By Lemma \[common curve\] there is an $(\alpha,\beta)$–segment connecting $v_1$ and $v_2$. Then ${\bf p}$ is a sum of a closed $\alpha$–segment and a closed $\beta$–segment. So we may assume that there is a third segment. The vertex $\tilde v$ of $\bf p$ following $v_2$ contains a curve $\gamma$ disjoint from $\alpha$ and $\gamma$ is the common curve of the third segment. Let $v_3$ be the last vertex of the third segment. We shall reduce the number of segments. There are three cases. [**Case 1**]{}Vertex $v_2$ does not contain $\gamma$.Since $\tilde v$ contains $\gamma$ and does not contain $\beta$ we have $|\beta\cap\gamma|=1$. Let $S_1$ be a surface of genus $g-1$ obtained by cutting $S$ open along $\beta\cup\gamma$. Vertices $v_2$ and $\tilde v$ have $g-1$ curves in common and the common curves form a cut system $u$ on $S_1$. The union $\beta\cup\gamma$ cannot separate $S-\alpha$ hence $\alpha$ does not separate $S_1$ and it belongs to a cut-system $u^\prime$ on $S_1$. Vertices $u$ and $u^\prime$ can be connected by a path $\bf q$ in the cut-system complex of $S_1$. If we adjoin $\beta$ (respectively $\gamma$) to each vertex of $\bf q$ we get a path $\bf q_2$ (respectively $\bf q_1$). Path $\bf q_2$ connects $v_2$ to a vertex $u_2$ containing $\alpha$ and $\beta$ and path $\bf q_1$ connects $\tilde v$ to a vertex $u_1$ containing $\alpha$ and $\gamma$. The corresponding vertices of $\bf q_1$ and $\bf q_2$ are connected by an edge so the middle rectangle on Figure \[radius 0, case 1\] splits into a sum of squares of type (C4) and is null-homotopic. We can connect $v_1$ to $u_2$ by an $(\alpha,\beta)$–segment so the triangle on Figure \[radius 0, case 1\] is a closed $\beta$–segment and is also null-homotopic. The part of $\bf p$ between $v_1$ and $\tilde v$ can be replaced by the lower path on Figure \[radius 0, case 1\]. We get a new path ${\bf p}^\prime$, which has a smaller number of segments (no $\beta$–segment) and is homotopic to $\bf p$ in $X$. \[ht!\]0.9pt [**Case 2**]{}Vertex $v_2$ contains $\gamma$ and $\alpha\cup\gamma$ does not separate $S$.If there exists a vertex $v$ which contains $\alpha$ and $\beta$ and $\gamma$ we can connect it to $v_1$ and $v_2$ using Lemma \[common curve\]. We get a closed segment and the remaining path has one segment less (Figure \[radius 0, case 2\]). Otherwise $\alpha\cup\beta\cup\gamma$ separate $S$ and we can apply Lemma \[hexagon relation\]. There exist vertices $w_1$ containing $\alpha$ and $\beta$, $w_2$ containing $\beta$ and $\gamma$ and $w_3$ containing $\alpha$ and $\gamma$ and a $\beta$–segment from $w_1$ to $w_2$, a $\gamma$–segment from $w_2$ to $w_3$ and an $\alpha$–segment from $w_3$ to $w_1$. The sum of the segments is null-homotopic. We now connect $v_1$ to $w_1$ by an $(\alpha,\beta)$–segment and $v_2$ to $w_2$ by a $(\beta,\gamma)$–segment. Thus the second segment of $\bf p$ can be replaced by a sum of an $\alpha$–segment and a $\gamma$–segment, and the difference is a closed $\beta$–segment plus a null-homotopic hexagonal path of Lemma \[hexagon relation\] (see the right side of Figure \[radius 0, case 2\]).\ \[ht!\]0.9pt [**Case 3**]{}Vertex $v_2$ contains $\gamma$ and $\alpha\cup\gamma$ separates $S$ into two surfaces $S_1$ and $S_2$.If there were only three segments then, as at the vertex $v_1$, the first vertex of the first segment would contain both $\alpha$ and $\gamma$ and their union would not separate $S$. This contradicts our assumptions. It follows in particular that every closed path of radius zero with at most three segments (where the common curve of each segment is disjoint from a fixed curve of the first segment) is null-homotopic. We may assume that the path $\bf p$ has a fourth segment with a fixed curve $\delta$ disjoint from $\alpha$. Since $[\gamma]=[\alpha]$ we cannot have $|\gamma\cap\delta|=1$. Therefore $\delta$ is not involved in the simple move from $v_3$ to the next vertex and $v_3$ contains $\delta$. In particular $[\gamma]\neq[\delta]$ and $[\alpha]\neq[\delta]$. We may assume that $\beta$ lies in $S_1$. If $\delta$ lies in $S_2$ then there is a vertex $w$ which contains $\alpha$ and $\beta$ and $\delta$. We can connect $w$ to $v_1$ by an $(\alpha,\beta)$–segment, to $v_2$ by a $\beta$–segment and to $v_3$ by a $\delta$–segment. We get a new path, homotopic to $\bf p$, which does not contain $\beta$–segment nor $\gamma$–segment (see Figure \[radius 0, case 3\], left part.) Suppose now that $\delta$ lies in $S_1$. Consider the cut-system complex $X_1$ of $S_1$ and choose a vertex $s$ of $X_1$ which contains $\delta$ and a vertex $s^\prime$ of $X_1$ which contains $\beta$. Let $\bf q$ be a path in $X_1$ which connects $s$ to $s^\prime$. Let $t$ be a fixed vertex of the cut-system complex $X_2$ of $S_2$ (if $X_2$ is not empty.) We add $\alpha$ and all curves of $t$ to each vertex of the path $\bf q$ and get an $\alpha$–segment in $X$ connecting a vertex $w_2$, containing $\delta$, to a vertex $w_2^\prime$, containing $\beta$. Then we add $\gamma$ and all curves of $t$ to each vertex of the path $\bf q$ and get a $\gamma$–segment in $X$ connecting a vertex $w_3$, containing $\delta$, to a vertex $w_3^\prime$, containing $\beta$. We now connect $v_1$ to $w_2$ by an $\alpha$–segment, $v_2$ to $w_2^\prime$ by a $\beta$–segment, $v_3$ to $w_3^\prime$ by a $\gamma$–segment and $v_4$ to $w_3$ by a $\delta$–segment (see Figure \[radius 0, case 3\], the right side.) Corresponding vertices of the two vertical segments on Figure \[radius 0, case 3\], the right side, have a common curve $\delta_i$, a curve of a vertex of the path $\bf q$ disjoint from $\alpha$ and from $\gamma$, and can be connected by a $\delta_i$–segment. We get a laddersuch that each small rectangle in this ladder has radius zero around $\gamma$ and consists of only three segments. Therefore it is null-homotopic. Every other closed path on Figure \[radius 0, case 3\], the right side, has a similar property. We get a new path, homotopic to $\bf p$, which does not contain $\beta$–segment nor $\gamma$–segment. 0.9pt The general case ---------------- We now pass to the general case and we want to prove it by induction on the radius of a closed path. [**Induction hypothesis 2**]{}A closed path of radius less than $m$ is null-homotopic. We want to prove that a closed path $\bf p$ of radius $m$ around a curve $\alpha$ is null-homotopic. The general idea is to construct a [*short-cut*]{}, an edge-path which splits $\bf p$ and is close to $\alpha$ and to a fixed $\beta$–segment. The first step is to construct one intermediate curve. \[curve with n big\] Let $\gamma_1$ and $\gamma_2$ be non-separating curves on $S$ such that $|\gamma_1\cap\gamma_2|=n>1$. Then there exists a non-separating curve $\delta$ such that $|\gamma_1\cap\delta|<n$ and $|\gamma_2\cap\delta|<n$. Suppose that we are also given non-separating curves $\alpha$, $\beta$ and an integer $m>0$ such that $|\alpha\cap\beta|\leq m$, $|\gamma_1\cap\alpha|<m$, $|\gamma_2\cap\alpha|\leq m$, and $|\beta\cap\gamma_1|=|\beta\cap\gamma_2|=0$. Then we can find a curve $\delta$ as above which also satisfies $|\delta\cap\alpha|<m$ and $|\delta\cap\beta|=0$. We orient the curves $\gamma_1$ and $\gamma_2$ and split the union $\gamma_1\cup\gamma_2$ into a different union of oriented simple closed curves as follows. We start near an intersection point, say $P_1$, on the side of $\gamma_2$ after $\gamma_1$ crosses it and on the side of $\gamma_1$ before $\gamma_2$ crosses it. Now we move parallel to $\gamma_1$ to the next intersection point with $\gamma_2$, say $P_2$. We do not cross $\gamma_2$ at $P_2$ and move parallel to $\gamma_2$, in the positive direction, back to $P_1$. We get a curve $\delta_1$. Now we start near $P_2$ and move parallel to $\gamma_1$ until we meet an intersection point, say $P_3$, which is either equal to $P_1$ or was not met before. We do not cross $\gamma_2$ at $P_3$ and move parallel to $\gamma_2$, in the positive direction, back to $P_2$. We get a curve $\delta_2$. And so on. Curve $\delta_i$ meets $\gamma_1$ near some points of $\gamma_1\cap\gamma_2$, but not near $P_i$ and it meets $\gamma_2$ near some points of $\gamma_1\cap\gamma_2$, but not near $P_{i+1}$. So $\delta_i$ meets both curves less than $n$ times. Let $\bar\gamma$ denote the (oriented) homology class represented by an oriented curve $\gamma$ in $H_1(\bar S,\ints)$. We have $\bar\gamma_1+\bar\gamma_2=\bar\delta_1+\dots+\bar\delta_k$. Now we repeat a similar construction for the opposite orientation of $\gamma_2$ starting near the same point $P_1$. We get new curves $\epsilon_1$,…,$\epsilon_r$ and $\bar\gamma_1-\bar\gamma_2=\bar\epsilon_1+\dots+\bar\epsilon_r$. Also $\bar\delta_1-\bar\epsilon_1=\bar\gamma_2$. Combining these equalities in $H_1(\bar S,\ints)$ we get $\bar\epsilon_1+\sum_{i\neq 1}\bar\delta_i=\bar\gamma_1$, $\bar\delta_1+\sum_{i\neq 1}\bar\epsilon_i=\bar\gamma_1$, $\sum_{i\neq 1}\bar\delta_i-\sum_{i\neq 1}\bar\epsilon_i=\bar\gamma_2$. A simple closed curve separates $S$ if and only if it represents $0$ in $H_1(\bar S,\ints)$. Since $\gamma_1$ and $\gamma_2$ are non-separating it follows that either $\delta_1$ and some $\delta_i$, $i\neq 1$, are not separating or $\epsilon_1$ and some $\epsilon_i$, $i\neq 1$, are not separating. And each of them meets $\gamma_1$ and $\gamma_2$ less than $n$ times, so it can be chosen for $\delta$. If we are also given curves $\alpha$ and $\beta$ and integer $m>0$ which satisfy the assumptions of the Lemma then $|\gamma_1\cap\alpha|+|\gamma_2\cap\alpha|=\Sigma|\delta_i\cap\alpha|= \Sigma|\epsilon_i\cap\alpha|\leq 2m-1$ therefore one of the constructed nonseparating curves intersects $\alpha$ less than $m$ times and is disjoint from $\beta$. \[curve disjoint\] Let $\gamma_1$ and $\gamma_2$ be disjoint non-separating curves on $S$ such that $\gamma_1\cup\gamma_2$ separates $S$. Then there exists a non-separating curve $\delta$ such that $|\gamma_1\cap\delta|=1$ and $|\gamma_2\cap\delta|=1$. Suppose that we are also given non-separating curves $\alpha$, $\beta$ and an integer $m>0$ such that $|\alpha\cap\beta|\leq m$, $|\alpha\cap\beta|=1$ if $m=1$, $|\gamma_1\cap\alpha|<m$, $|\gamma_2\cap\alpha|\leq m$, and $|\beta\cap\gamma_1|=|\beta\cap\gamma_2|=0$. Then we can find a curve $\delta$ as above which also satisfies $|\delta\cap\alpha|<m$ and $|\delta\cap\beta|<m$. By our assumptions $\gamma_1\cup\gamma_2$ separates $S$ into two components $S_1$ and $S_2$. We can choose a simple arc $d_1$ in $S_1$ which connects $\gamma_1$ to $\gamma_2$ and a simple arc $d_2$ in $S_2$ which connects $\gamma_1$ to $\gamma_2$. Then we can slide the end-points of $d_1$ and $d_2$ along $\gamma_1$ and $\gamma_2$ to make the end-points meet. We get a nonseparating curve $\delta$ which intersects $\gamma_1$ and $\gamma_2$ once. Suppose that we are also given curves $\alpha$ and $\beta$ and an integer $m>0$. We need to alter the arcs $d_1$ and $d_2$, if necessary, in order to decrease the intersection of $\delta$ with $\alpha$ and $\beta$. We may assume that $\beta$ lies in $S_1$. We have to consider several cases. [**Case 1**]{}Curve $\alpha$ lies in $S_1$.Then $d_2$ is disjoint from $\alpha$. If $m=1$ then $|\alpha\cap\beta|=1$ so the union $\alpha\cup\beta$ does not separate its regular neighbourhood and does not separate $S_1$. We can choose $d_1$ disjoint from $\alpha$ and $\beta$ and then $\delta$ is also disjoint from $\alpha$ and $\beta$.\ Suppose that $m>1$. If $\alpha$ separates $S_1$ (but does not separate $S$) then it separates $\gamma_1$ from $\gamma_2$ in $S_1$. There exists an arc $d$ in $S_1$ which connects $\gamma_1$ with $\gamma_2$ and is disjoint from $\alpha$, if $\alpha$ does not separate $S_1$, or meets $\alpha$ once, if $\alpha$ separates $S_1$. We choose such an arc $d$ which has minimal number of intersections with $\beta$. If $|\beta\cap d|>m$ then there exist two points $P$ and $Q$ of $\beta\cap d$, consecutive along $\beta$, and not separated by a point of $\beta\cap\alpha$. We can move along $d$ to $P$ then along $\beta$, without crossing $\beta$, to $Q$, and then continue along $d$ to its end. This produces an arc which meets $\alpha$ at most once and has smaller number of intersections with $\beta$. So we may assume that $|\beta\cap d|= m$ and that every pair of points of $\beta\cap d$ consecutive along $\beta$ is separated by a point of $\beta\cap\alpha$. We now alter $d$ as follows. Consider the intersection $d\cap (\alpha\cup\beta)$. If the first or the last point along $d$ of this intersection belongs to $\alpha$ we start from this end of $d$. Otherwise we start from any end. We move along $d$ to the first point, say $P$, of intersection with $\alpha\cup\beta$. If $P\in \alpha$ we continue along $\alpha$, without crossing it, to the next point of $\alpha\cap\beta$. Then along $\beta$, without crossing it, to the last point, say $Q$, of $\beta\cap d$ on $d$, and then along $d$ to its end. The new arc crosses $\beta $ at most once, near $Q$, and crosses $\alpha$ less than $m$ times. If $P\in \beta$ we continue along $\beta$, without crossing $\beta$, to $Q$ and then along $d$ to its end, which produces a similar result. We can choose such an arc for $d_1$ and then the curve $\delta$ satisfies the Lemma. [**Case 2**]{}Curve $\alpha$ meets $\gamma_1$ or $\gamma_2$.Then $m>1$, because if $m=1$ and $|\gamma_1\cap\alpha|=0$ and $\alpha$ crosses $\gamma_2$ into $S_1$ then it must cross it again in order to exit $S_1$, and this contradicts $|\gamma_2\cap\alpha|\leq m$. The arcs of $\alpha$ split $S_1$ (and $S_2$) into connected components. One of the components must meet both $\gamma_1$ and $\gamma_2$ (Otherwise the union of all components meeting $\gamma_1$ has $\alpha$ for a boundary component and then $\alpha$ is disjoint from $\gamma_1$ and $\gamma_2$.) Choosing $d_1$ (respectively $d_2$) in such a component we can make them disjoint from $\alpha$. Now we want to modify $d_1$ in such a way that $|d_1\cap\alpha|=0$ and $| d_1\cap\beta|<m$. There are three subcases. [**Case 2a**]{}There exists an arc $a_1$ of $\alpha$ in $S_1$ which connects $\gamma_1$ and $\gamma_2$.Choose $d_1$ parallel to this arc. It may happen that $d_1$ meets $\beta$ $m$ times. Then $a_1$ is the only arc of $\alpha$ which meets $\beta$. We then modify $d_1$ as follows. We move from $\gamma_1$ along $d_1$ until it meets $\beta$. Then we turn along $\beta$, away from $a_1$, to the next point of $a_1$. We turn before crossing $a_1$ and move parallel to $a_1$ to $\gamma_2$. The new arc does not meet $\alpha$ and meets $\beta$ less than $m$ times. [**Case 2b**]{}There exists an arc of $\alpha$ in $S_1$ which connects $\gamma_1$ and $\beta$ and there exists an arc of $\alpha$ which connects $\gamma_2$ and $\beta$.Then there exist points $P$ and $Q$ of $\alpha\cap\beta$, consecutive along $\beta$, and arcs $a_1$ and $a_2$ of $\alpha$ such that $a_1$ connects $\gamma_1$ to $P$ and $a_2$ connects $Q$ to $\gamma_2$. We move along $a_1$ to $P$ then along $\beta$, without crossing $\beta$, to $Q$, and then along $a_2$ to $\gamma_2$. The new arc does not meet $\alpha$ and meets $\beta$ less than $m$ times. [**Case 2c**]{}If an arc of $\alpha$ in $S_1$ meets $\beta$ then it meets only $\gamma_1$.(The case of $\gamma_2$ is similar.) We consider an arc $d$ in $S_1$ which is disjoint from $\alpha$ and connects $\gamma_1$ and $\gamma_2$. We start at $\gamma_2$ and move along $d$ to the first point of intersection with $\beta$. Then we move along $\beta$, without crossing it, to the first point of intersection with $\alpha$. Then we move along $\alpha$, away from $\beta$, to $\gamma_1$. The new arc does not meet $\alpha$ and meets $\beta$ less than $m$ times. If $\beta$ is disjoint from $\alpha$ then $\beta$ is either disjoint from a component of $S_1-\alpha$ which connects $\gamma_1$ to $\gamma_2$ or is contained in it. We can find an arc in the component (disjoint from $\alpha$) which connects $\gamma_1$ with $\gamma_2$ and meets $\beta$ at most once. So in each case we have an arc $d_1$ which is disjoint from $\alpha$ and meets $\beta$ less than $m$ times. We now slide the end-points of $d_1$ along $\gamma_1$ and $\gamma_2$ to meet the end-points of $d_2$. Each slide can be done along one of two arcs of $\gamma_i$. Choosing suitably we may assume that $d_1$ meets at most $m/2$ points of $\alpha$ sliding along $\gamma_2$ and at most $(m-1)/2$ points of $\alpha$ sliding along $\gamma_1$. The curve $\delta$ obtained from $d_1$ and $d_2$ meets $\alpha$ and $\beta$ less than $m$ times. [**Case 3**]{}The curve $\alpha$ lies in $S_2$.Then $|\alpha\cap\beta|=0$ so we must have $m>1$. We can choose $d_2$ which is disjoint from $\beta$ and meets $\alpha$ at most once and we can choose $d_1$ which is disjoint from $\alpha$ and meets $\beta$ at most once. The curve $\delta$ obtained from $d_1$ and $d_2$ meets $\alpha$ and $\beta$ less than $m$ times. \[connecting path\] Let $\delta_1$ and $\delta_2$ be non-separating curves on $S$ and let $w_1$ be a vertex of $X$ containing $\delta_1$ and let $w_2$ be a vertex of $X$ containing $\delta_2$. Then there exists an edge-path ${\bf q}=(w_1=z_1,z_2,\dots,z_k=w_2)$ connecting $w_1$ and $w_2$. Suppose that we are also given non-separating curves $\alpha$, $\beta$ and an integer $m>0$ such that $|\alpha\cap\beta|\leq m$, $|\alpha\cap\beta|=1$ if $m=1$, $|\delta_1\cap\alpha|<m$, $|\delta_2\cap\alpha|\leq m$, and $|\beta\cap\delta_1|=|\beta\cap\delta_2|=0$. Then there exists a path $\bf q$ as above and an integer $j$, $1\leq j < k$, such that $d(z_i,\beta)<m$ for all $i$, $d(z_i,\alpha)<m$ for $1\leq i\leq j<k$ and $z_i$ contains $\delta_2$ for $j<i\leq k$. We shall prove the lemma by induction on $|\delta_1\cap\delta_2|=n$.\ If $\delta_1=\delta_2$ we can connect $w_1$ and $w_2$ by a $\delta_1$–segment, by Lemma \[common curve\].\ If $n=1$ there exist vertices $u_1$, $u_2$ in $X$ which are connected by an edge and such that $\delta_1\in u_1$, $\delta_2\in u_2$. Now we can connect $u_1$ to $w_1$ and $w_2$ to $u_2$ as in the previous case.\ If $n=0$ and $\delta_2\cup\delta_1$ does not separate $S$ then there exists a vertex $v$ containing both curves $\delta_2$ and $\delta_1$. We can connect $v$ to $w_1$ and $w_2$ as in the first case.\ Suppose now that $n=0$ and that $\delta_2\cup\delta_1$ separates $S$. Then, by Lemma \[curve disjoint\], there exists a curve $\delta$ such that $|\delta_2\cap\delta|=|\delta_1\cap\delta|=1$. We can find a vertex $v$ containing $\delta$ and we can connected $v$ to $w_1$ and $w_2$ as in the second case. If we are also given curves $\alpha$, and $\beta$ and an integer $m$ we can choose $\delta$ which also satisfies $|\alpha\cap\delta|<m$ and $|\beta\cap\delta|<m$. Then the path obtained by connecting $v$ to $w_1$ and $w_2$ have all vertices in a distance less than $m$ from $\beta$ and in a distance less than $m$ from $\alpha$, except for the final $\delta_2$–segment which ends at $w_2$ (Curve $\delta_2$ may have distance $m$ from $\alpha$.)\ If $n>1$ then by Lemma \[curve with n big\] there exists a curve $\delta$ such that $|\delta_1\cap\delta|<n$ and $|\delta_2\cap\delta|<n$. We choose a vertex $v$ containing $\delta$. By induction on $n$ we can connect $v$ to $w_1$ and $w_2$. If we are also given curves $\alpha$ and $\beta$ and an integer $m$ then we can find $\delta$ which also satisfies $|\delta\cap\alpha|<m$ and $|\delta\cap\beta|=0$. By induction on $n$ we can connect $w_1$ to $v$ and $v$ to $ w_2$ by a path the vertices of which are closer to $\beta$ than $m$, and closer to $\alpha$ than $m$ except for a final $\delta_2$–segment which ends at $w_2$. As an immediate corollary we get: \[X is connected\] Complex $X$ is connected. We need one more lemma before we prove that every closed path is null-homotopic in $X$. \[better curve\] Let $\alpha$, $\beta$, $\gamma$ be non-separating curves on $S$ such that $|\alpha\cap\beta|=m$, $|\alpha\cap\gamma|\leq m$, $|\beta\cap\gamma|=1$. There exists a non-separating curve $\delta$ such that $|\delta\cap\alpha|<m$, $|\delta\cap\beta|=0$ and $|\delta\cap\gamma|\leq 1$. If $m=1$ then $|\delta\cap\gamma|=0$ and $[\delta]$ is different from $[\alpha]$, $[\beta]$ and $[\gamma]$. When we split $S$ along $\gamma\cup\beta$ we get a surface $S_1$ with a rectangular" boundary component $\partial$ consisting of two $\beta$–edges (vertical) and two $\gamma$–edges (horizontal on pictures of Figure \[constructing better curve\]). We can think of $S_1$ as a rectangle with holes and with some handles attached to it. Curve $\alpha$ intersects $S_1$ along some arcs $a_i$ with end-points $P_i$ and $Q_i$ on $\partial$. If, for some $i$, points $P_i$ and $Q_i$ lie on the same $\beta$–edge then $m>1$ and we can construct a curve $\delta$ consisting of an arc parallel to $a_i$ and an arc parallel to the arc of $\beta$ which connects $P_i$ and $Q_i$ passing through the point $\gamma\cap\beta$. Then $|\delta\cap\beta|=0$, $|\delta\cap\alpha|<m$ and $|\delta\cap\gamma|=1$. Recall that if two curves intersect exactly at one point then they are both non-separating on $S$. Therefore $\delta$ satisfies the conditions of the Lemma. If for some $i$ points $P_i$ and $Q_i$ lie on different $\gamma$–edges of $\partial$ then we can modify the arc $a_i$ sliding its end-point $P_i$ along the $\gamma$ edge to the point corresponding to $Q_i$. We get a closed curve $\delta$ satisfying the conditions of the Lemma. So we may assume that there are no arcs $a_i$ of the above types. Suppose that for every pair $i,j$ the pairs of end-points $P_i,Q_i$ and $P_j,Q_j$ do not separate each other on $\partial$. Then we can connect the corresponding end points by nonintersecting intervals inside a rectangle. In the other words a regular neighbourhood of $\alpha\cup\partial$ in $S_1$ is a planar surface homeomorphic to a rectangle with holes. Since $S_1$ has positive genus there exists a subsurface of $S_1$ of a positive genus attached to one hole or a subsurface of $S_1$ which connects two holes of the rectangle. Such a subsurface contains a curve $\delta$ which is non-separating on $S$ and is disjoint from $\alpha$, $\beta$ and $\gamma$ and the homology class $[\delta]$ is different from $[\alpha]$, $[\beta]$ and $[\gamma]$. This happens in particular when $m=1$ because then there is at most one point on every edge and the pairs of end points of arcs do not separate each other. So we may assume that $m>1$ and that there exists a pair of arcs, say $a_1$ and $a_2$, such that the pair $P_1,Q_1$ separates the pair $P_2,Q_2$ in $\partial$. Since an arc $a_i$ does not connect different $\gamma$–edges we must have two points, say $P_1$ and $P_2$, on the same edge. Suppose that they lie on a $\beta$–edge, say the left edge. Choosing an intermediate point, if there is one, we may assume that $P_1$ and $P_2$ are consecutive points of $\alpha$ along $\beta$. We have different possible configurations of pairs of points. For each of them we construct curves $\delta_i$, as on Figure \[constructing better curve\]. Each $\delta_i$ is disjoint from $\beta$ and intersects $\gamma$ at most once, and if it is disjoint from $\gamma$ it intersects some other curve once. So $\delta_i$ is not-separating. We shall prove that we can always choose a suitable $\delta_i$ with $|\delta_i\cap\alpha|<m$. Observe that $\delta_i$ may meet $\alpha$ only along the boundary $\partial$, and not along the arc connecting $P_1$ to $P_2$.\ [**Case 1**]{}Points $Q_1$ and $Q_2$ lie on the same $\gamma$–edge, say lower edge.If there is no point of $\alpha$ on $\gamma$ to the left of $Q_1$ then $|\delta_1\cap\alpha|<m$. If there is a point of $\alpha$ on $\gamma$ to the left of $Q_1$ then $|\delta_2\cap\alpha|<m$.\ [**Case 2**]{}Points $Q_1$ and $Q_2$ lie on different $\gamma$–edges.Then $|\delta_3\cap\alpha|<m$.\ [**Case 3**]{}Points $Q_1$ and $Q_2$ lie on the right edge.Then $|\delta_4\cap\alpha|<m$.\ [**Case 4**]{}One of the points $Q_i$, say $Q_1$, lies on a $\gamma$–edge and the other lies on a $\beta$–edge.Let $u_i$, $i=1,\dots,6$ denote the number of intersection points of $\alpha$ with the corresponding piece of $\partial$ on Figure \[constructing better curve\]. Then $u_3+u_4=u_5+u_6=|\alpha\cap\beta|= m$ and $u_1+u_2\leq m$. Also $|\delta_1\cap\alpha|=u_1+u_4$, $|\delta_5\cap\alpha|=u_1+u_3$, $|\delta_6\cap\alpha|=u_2+u_5$ and $|\delta_7\cap\alpha|=u_2+u_6$. Moreover, since $P_2$ and $Q_2$ are connected by an arc of $\alpha$, they represent different points on $S$ (otherwise it would be the only arc of $\alpha$) and $u_4\neq u_6$. It follows that $|\delta_i\cap\alpha|<m$ for $i=1,5,6$ or $7$. We may assume now that for every pair of arcs $(i,j)$ whose end-points separate each other no two end-points lie on the same $\beta$–edge. If $P_i$ and $Q_i$ lie on a $\gamma$–edge and $P_j$ lie in between then $a_i$ together with the interval of $\gamma$ between $P_i$ and $Q_i$ form a nonseparating curve which meets $\alpha$ less than $m$ times. So we may assume that $P_1$ and $P_2$ lie on different $\beta$–edges, say $P_1$ on the left edge and $P_2$ on the right edge, and $Q_1$ and $Q_2$ lie on the lower edge. Replacing $Q_1$ or $Q_2$ by an intermediate point, if necessary, we may also assume that for every point $Q_i$ between $Q_1$ and $Q_2$ the corresponding point $P_i$ also lies between $Q_1$ and $Q_2$. Now if there is no point of $\alpha$ on the left edge below $P_1$ then $|\delta_1\cap\alpha|<m$. If there is such point of $\alpha$ consider the one closest to $P_1$ and call it $P_3$. Then, by our assumptions, point $Q_3$ lies on the lower edge to the left of $Q_2$ and $|\delta_8\cap\alpha|<m$.\ This concludes the proof of the Lemma. \[radius m\] A path $\bf p$ of radius $m$ around $\alpha$ is null-homotopic. Let $v_0$ be a vertex of $\bf p$ containing $\alpha$. We say that $\bf p$ begins at $v_0$. Let $v_1$ be the first vertex of $\bf p$ which has distance $m$ from $\alpha$. Let $\bf q$ be the maximal segment of $\bf p$, which starts at $v_1$ and contains some fixed curve $\beta$ satisfying $|\beta\cap\alpha|=m$ and such that no vertex of $\bf q$ contains a curve $\beta^\prime$ satisfying $|\beta^\prime\cap\alpha|<m$. Let $v_2$ be the last vertex of $\bf q$. Let $u_1$ be the vertex of $\bf p$ preceding $v_1$ and let $u_2$ be the vertex of $\bf p$ following $v_2$. Vertex $u_1$ contains a curve $\gamma_1$ such that $|\gamma_1\cap\alpha|<m$. Vertex $u_2$ is the first vertex of the second segment which has a fixed curve $\gamma_2$ such that $|\gamma_2\cap\alpha|\leq m$. If $u_1$ contains $\beta$ then $|\gamma_1\cap\beta|=0$. Otherwise, since $v_1$ does not contain $\gamma_1$, the move from $u_1$ to $v_1$ involves $\gamma_1$ and $\beta$, so $|\gamma_1\cap\beta|=1$. If $v_2$ contains $\gamma_2$ then $|\gamma_2\cap\beta|=0$. It may also happen that $|\gamma_2\cap\alpha|<m$ and that $\beta\in u_2$. Then also $|\gamma_2\cap\beta|=0$. Otherwise $|\gamma_2\cap\beta|=1$. We want to construct vertices $w_1$, $w_1^\prime$, $w_2$ and $w_2^\prime$ and edge paths connecting them, as on Figure \[reducing a path of radius m\], so that the rectangles are null homotopic. Then we can replace the part of $\bf p$ between $u_1$ and $u_2$ by the path connecting consecutively $u_1$ to $w_1^\prime$, $w_1^\prime$ to $w_1$, $w_1$ to $w_2$, $w_2$ to $w_2^\prime$ and $w_2^\prime $ to $u_2$. We denote the new path by ${\bf p}^\prime$.\ In our construction vertex $w_i$ contains a nonseparating curve $\delta_i$ disjoint from $\beta$. If $|\gamma_i\cap\beta|=0$ we let $\delta_i=\gamma_i$, $w_i=w_i^\prime=u_i$ and the corresponding rectangle degenerates to an edge. If $|\gamma_i\cap\beta|=1$ we proceed as follows. By Lemma \[better curve\] there exists a nonseparating curve $\delta_i$ such that $|\delta_i\cap\beta|=0$, $|\delta_i\cap\alpha|<m$, and $|\delta_i\cap\gamma_i|\leq 1$. If $|\delta_i\cap\gamma_i|=0$ (this is always the case if $m=1$) then $[\delta_i]\neq [\gamma_i]$ and $[\delta_i]\neq[\beta]$ because they have different intersections. There exists a vertex $w_i^\prime$ containing $\delta_i$ and $\gamma_i$ and a vertex $w_i$ containing $\delta_i$ and $\beta$. We can connect $u_i$ to $w_i^\prime$ by a $\gamma_i$–segment, $w_i^\prime$ to $w_i$ by a $\delta_i$–segment and $w_i$ to $v_i$ by a $\beta$–segment. The corresponding rectangle has radius zero around $\delta_i$, so it is null-homotopic by the Induction Hypothesis 2.\ If $|\delta_i\cap\gamma_i|=1$ there exist vertices $w_i^\prime$ and $w_i$ which are connected by an edge and contain $\gamma_i$ and $\delta_i$ respectively. We connect $w_i^\prime$ to $u_i$ by a $\gamma_i$–segment. We now apply Lemma \[connecting path\] to vertices $w_i$ and $v_i$ with $\delta_1,\delta_2,\alpha,\beta$ replaced by $\delta_i,\beta,\gamma_i,\beta$ respectively and $m>1$. There exists a path connecting $w_i$ to $v_i$ such that all vertices of the path have distance less than $m$ from $\gamma_i$ and $\beta$. The corresponding rectangle has radius less than $m$ around $\gamma_i$ so it is null-homotopic, by the Induction Hypothesis 2. We now apply Lemma \[connecting path\] to vertices $w_1$ and $w_2$. There exists a path ${\bf q}=(w_1=z_1,z_2,\dots,z_k=w_2)$ connecting $w_1$ and $w_2$ such that $d(z_i,\beta)<m$ for all $i$, $d(z_i,\alpha)<m$ for $1\leq i\leq j<k$ and $z_i$ contains $\delta_2$ for $j<i\leq k$. In particular the middle rectangle on Figure \[reducing a path of radius m\] has radius less than $m$ around $\beta$ so it is null-homotopic by the Induction Hypothesis 2. All vertices of the new part of path ${\bf p}^\prime$ have distance less than $m$ from $\alpha$ except for the final $\gamma_2$–segment from $w_2^\prime$ to $u_2$, if $|\gamma_2\cap\beta|=1$, or final $\delta_2=\gamma_2$–segment of $\bf q$, if $|\gamma_2\cap\beta|=0$ and the right rectangle degenerates. Thus ${\bf p}^\prime$ has smaller number of segments at the distance $m$ from $\alpha$, it has no $\beta$–segment, and it is null-homotopic by induction on the number of segments of $\bf p$ at the distance $m$ from some curve $\alpha$. 0.9pt This concludes the proof of Theorem \[Hatcher-Thurston\]. A presentation of $M_{g,1}$ =========================== In this section we shall consider a surface $S$ of genus $g>1$ with one boundary component $\partial$. Let ${\mathcal M}_{g,1}$ be the mapping class group of $S$. Let $X$ be the cut-system complex of $S$ described in the previous section. We shall establish a presentation of ${\mathcal M}_{g,1}$ via its action on $X$. The action of ${\mathcal M}_{g,1}$ on $X$ is defined by its action on vertices of $X$. If $v=\lan C_1,\dots,C_g\ran$ is a vertex of $X$ and $g\in{\mathcal M}_{g,1} $ then $g(v)=\lan g(C_1),\dots,g(C_g)\ran$. We start with some properties of homeomorphisms of a surface. Then we describe stabilizers of vertices and edges of the action of ${\mathcal M}_{g,1}$ on $X$. Finally we consider the orbits of faces of $X$ and determine a presentation of $X$. In order to shorten some long formulas we shall adopt the following notation for conjugation: $a*b=aba^{-1}$. As usually $[a,b]=aba^{-1}b^{-1}$. \[proofs\] [Some proofs of relations between homeomorphisms of a surface are left to the reader. The general idea of a proof is as follows. We split the surface into a union of disks by a finite number of curves (and arcs with the end-points on the boundary if the surface has a boundary). We prove that the given product of homeomorphisms takes each curve (respectively arc) onto an isotopic curve (arc), preserving some fixed orientation of the curve (arc). Then the product is isotopic to a homeomorphism pointwise fixed on each curve and arc. But a homeomorphism of a disk fixed on its boundary is isotopic to the identity homeomorphism, relative to the boundary, by Lemma of Alexander. Thus the given product of homeomorphisms is isotopic to the identity.]{} Dehn proved in [@Dehn] that every homeomorphism of $S$ is isotopic to a product of twists. We start with some properties of twists. \[conjugation\] Let $\alpha$ be a curve on $S$, let $h$ be a homeomorphism and let $\alpha^\prime = h(\alpha)$. Then $T_{\alpha^\prime}=hT_\alpha h^{-1}$. Since $h$ maps $\alpha$ to $\alpha^{\prime}$ we may assume that (up to isotopy) it also maps a neighborhood $N$ of $\alpha$ to a neighborhood $N^\prime$ of $\alpha^\prime$. The homeomorphism $h^{-1}$ takes $N^\prime$ to $N$, then $T_\alpha$ maps $N$ to $N$, twisting along $\alpha$, and $h$ takes $N$ back to $N^\prime$. Since $T_\alpha$ is supported in $N$, the composite map is supported in $N^\prime$ and is a Dehn twist about $\alpha^\prime$. \[basic relations\] Let $\gamma_1,\gamma_2,\dots,\gamma_k$ be a chain of curves, ie, the consecutive curves intersect once and non-consecutive curves are disjoint. Let $N$ denote the regular neighbourhood of the union of these curves. Let $c_i$ denote the twist along $\gamma_i$. Then the following relations hold: 1. The “commutativity relation": $c_ic_j=c_jc_i$ if $|i-j|>1$. 2. The “braid relation": $c_ic_j(\gamma_i)= \gamma_j$, and $c_ic_jc_i=c_jc_ic_j$ if $|i-j|=1$. 3. The chain relation": If $k$ is odd then $N$ has two boundary components, $\partial_1$ and $\partial_2$, and $(c_1c_2\dots c_k)^{k+1}=T_{\partial_1} T_{\partial_2}$. If $k$ is even then $N$ has one boundary component $\partial_1$ and $(c_1c_2\dots c_k)^{2k+2}=T_{\partial_1}$. 4. $(c_2c_1c_3c_2)(c_4c_3c_5c_4)(c_2c_1c_3c_2)= (c_4c_5c_3c_4)(c_2c_1c_3c_2)(c_4c_3c_5c_4)$. 5. $(c_1c_2\dots c_k)^{k+1}=(c_1c_2\dots c_{k-1})^k(c_kc_{k-1}\dots c_2c_1^2c_2\dots c_{k-1}c_k)=$ $(c_kc_{k-1}\dots c_2c_1^2c_2\dots c_{k-1}c_k)(c_1c_2\dots c_{k-1})^k$. Relation (i) is obvious. It follows immediately from the definition of Dehn twist that $c_2(\gamma_1)=c_1^{-1}(\gamma_2)$. Both statements of (ii) follow from this and from Lemma \[conjugation\]. Relation (iii) is a little more complicated. It can be proven by the method explained in Remark \[proofs\]. Relations (iv) and (v) follow from the braid relations (i) and (ii) by purely algebraic operations. We shall prove (iv). $(c_2c_1c_3c_2)(c_4c_3c_5c_4)(c_2c_1c_3c_2)= c_2c_3c_1c_2c_4c_5c_3c_4c_2c_3c_1c_2=$$\strut c_2c_3c_4c_1c_5c_2c_3c_2c_4c_3c_1c_2= c_2c_3c_4c_3c_1c_5c_2c_3c_4c_3c_1c_2=$$\strut c_4c_2c_3c_4c_1c_5c_2c_4c_3c_1c_2c_4= c_4c_2c_3c_1c_4c_5c_4c_2c_3c_1c_2c_4=$$\strut c_4c_5c_2c_3c_1c_4c_2c_3c_1c_2c_5c_4= c_4c_5c_2c_3c_1c_2c_1c_4c_3c_2c_5c_4=$$\strut c_4c_5c_3c_2c_3c_4c_1c_2c_3c_2c_5c_4= c_4c_5c_3c_2c_3c_4c_3c_1c_2c_3c_5c_4=$$\strut c_4c_5c_3c_4c_2c_3c_4c_1c_2c_3c_5c_4= (c_4c_5c_3c_4)(c_2c_1c_3c_2)(c_4c_3c_5c_4)$. We now prove (v). We prove by induction that for $s\leq k$ we have $(c_1c_2\dots c_k)^s=(c_1c_2\dots c_{k-1})^s(c_kc_{k-1}\dots c_{k-s+1})$. Using (i) and (ii) one checks easily that for $i>1$ we have $c_i(c_1c_2\dots c_k)=(c_1c_2\dots c_k)c_{i-1}$. Now $(c_1c_2\dots c_k)^{s+1}=(c_1c_2\dots c_{k-1})^s(c_kc_{k-1}\dots c_{k-s+1}) (c_1c_2\dots c_k)=$ $\strut (c_1c_2\dots c_{k-1})^sc_1c_2\dots c_kc_{k-1}\dots c_{k-s}= (c_1c_2\dots c_{k-1})^{s+1}c_kc_{k-1}\dots c_{k-s}$.\ For $s=k+1$ we get $(c_1c_2\dots c_k)^{k+1}=(c_1c_2\dots c_{k-1})^k(c_kc_{k-1}\dots c_1 c_1c_2\dots c_k)$. This proves the first equality in (v). By (i) and (ii) $(c_kc_{k-1}\dots c_2c_1^2c_2\dots c_{k-1}c_k)$ commutes with $c_i$ for $i<k$, which implies the second equality. The next lemma was observed by Dennis Johnson in [@Johnson] and was called a [*lantern relation*]{}. \[lantern\] Let $U$ be a disk with the outer boundary $\partial$ and with $3$ inner holes bounded by curves $\partial_1,\partial_2,\partial_3$ which form vertices of a triangle in the clockwise order. For $1\leq i<j\leq 3$ let $\alpha_{i,j}$ be the simple closed curve in $U$ which bounds a neighbourhood of the edge" $(\partial_i,\partial_j)$ of the triangle (see Figure \[fig lantern\]). Let $d$ be the twist along $\partial$, $d_i$ the twist along $\partial_i$ and $a_{i,j}$ the twist along $\alpha_{i,j}$.\ Then $dd_1d_2d_3=a_{1,2}a_{1,3}a_{2,3}$. We now describe a presentation of the mapping class group of a disk with holes. \[disk with holes presentation\] Let $U$ be a disk with the outer boundary $\partial$ and with $n$ inner holes bounded by curves $\partial_1,\partial_2,\dots,\partial_n$. For $1\leq i<j\leq n$ let $\alpha_{i,j}$ be the simple closed curve in $U$ shown on Figure \[disk with holes\], separating two holes $\partial_i$ and $\partial_j$ from the other holes. Let $d$ be the twist along $\partial$, $d_i$ the twist along $\partial_i$ and $a_{i,j}$ the twist along $\alpha_{i,j}$. Then the mapping class group of $U$ has a presentation with generators $d_i$ and $a_{i,j}$ and with relations\ [(Q1)]{}$[d_i,d_j]=1$ and $[d_i,a_{j,k}]=1$ for all $i,j,k$.\ [(Q2)]{}pure braid relations [(a)]{}$a_{r,s}^{-1}*a_{i,j}=a_{i,j}$ if $ r<s<i<j$ or $i<r<s<j$, [(b)]{}$a_{r,s} ^{-1}*a_{s,j}=a_{r,j}*a_{s,j}$ if $r<s<j$, [(c)]{}$a_{r,j}^{-1}*a_{r,s}=a_{s,j}*a_{r,s}$ if $r<s<j$, [(d)]{}$[a_{i,j},a_{r,j}^{-1}*a_{r,s}]=1$ if $r<i<s<j$.\ Relations (Q2) come in place of standard relations in the pure braid group on $n$ strings and we shall first prove the equivalence of (Q2) to the standard presentation of the pure braid group. The standard presentation has generators $a_{i,j}$ and relations (this is a corrected version of the relations in [@Birman]): (i)$a_{r,s}^{-1}*a_{i,j}=a_{i,j}$ if $ r<s<i<j$ or $i<r<s<j$, (ii)$a_{r,s} ^{-1}*a_{s,j}=a_{r,j}*a_{s,j}$ if $r<s<j$, (iii)$[a_{r,j},a_{s,j}]=[a_{r,s}^{-1},a_{r,j}^{-1}]$ if $r<s<j$, (iv)$a_{r,s}^{-1}*a_{i,j}=[a_{r,j},\ a_{s,j}]*a_{i,j}$ if $r<i<s<j$.\ So relations (a) and (b) are the same as (i) and (ii) respectively. We can substitute relation (iii) in (iv) and get (d), after cancellation of $a_{r,s}$. When we substitute (ii) for the first three terms of (iii) we get (c), after cancellation of $a_{r,s}^{-1}$.\ We now consider the disk $U$ with holes. When we glue a disk with a distinguished center to each curve $\partial_i$ we get a disk with $n$ distinquished points. Its mapping class group is isomorphic to the pure braid group $P_{n}$ with generators $a_{i,j}$ and with relations (Q2). In the passage from the mapping class group of $U$ to the mapping class group of the punctured disk we kill exactly the twists $d_i$, which commute with everything. One can check that the removal of the disks does not affect the relations (Q2) so the mapping class group of $U$ has a presentation with relations (Q1) and (Q2). We now consider the surface $S$. We shall fix some curves on $S$. The surface $S$ consists of a disk with $g$ handles attached to it. For $i=1,\dots,g$ and $j=1,\dots,g-1$ we fix curves $\alpha_i$, $\beta_i$, $\epsilon_j$ (see Figure \[general surface\]). Curve $\alpha_i$ is a meridian curve across the $i$-th handle, $\beta_i$ is a curve along the $i$-th handle and $\epsilon_i$ runs along $i$-th handle and $(1+i)$-th handle. Curves $\alpha_1,\alpha_2,\dots,\alpha_g$ form a cut-system. We denote by $I_0$ the set of indices $I_0=\{-g,1-g,2-g,\dots,-1,1,2, \dots,g-1,g\}$. When we cut $S$ open along the curves $\alpha_1,\dots,\alpha_g$ we get a disk $S_0$ with $2g$ holes bounded by curves $\partial_i$, $i\in I_0$, where curves $\partial_i$ and $\partial_{-i}$ correspond to the same curve $\alpha_i$ on $S$ (see Figure \[delta curves\]). The glueing back map identifies $\partial_i$ with $\partial_{-i}$ according to the reflection with respect to the $x$–axis. Curves on $S$ can be represented on $S_0$. If a curve on $S$ meets some curves $\alpha_i$ then it is represented on $S_0$ by a disjoint union of several arcs. In particular $\epsilon_i$ is represented by two arcs joining $\partial_{-i}$ to $\partial_{-i-1}$ and $\partial_i$ to $\partial_{i+1}$. We denote by $\delta_{i,j}$, $i<j\in I_0$, curves on $S$ represented on Figure \[delta curves\]. Curve $\delta_{i,j}$ separates holes $\partial_i$ and $\partial_j$ from the other holes on $S_0$. We now fix some elements of ${\mathcal M}_{g,1}$. \[generators\] We denote by $a_i,b_i,e_j$ the Dehn twists along the curves $\alpha_i,\beta_i,\epsilon_j$ respectively. We fix the following elements of ${\mathcal M}_{g,1}$: $s=b_1a_1a_1b_1$.$\strut t_i=e_ia_ia_{i+1}e_i$ for $i=1,\dots,g-1$.$\strut d_{1,2}=(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2$. For $i<j\in I_0$ we let $d_{i,j}=(t_{i-1}t_{i-2}\dots t_1t_{j-1}t_{j-2}\dots t_2)*d_{1,2}$ if $i>0$, $\strut d_{i,j}=(t_{-i-1}^{-1}t_{-i-2}^{-1}\dots t_1^{-1}s^{-1}t_{j-1}t_{j-2} \dots t_2)*d_{1,2}$ if $i<0$ and $i+j>0$, $\strut d_{i,j}=(t_{-i-1}^{-1}t_{-i-2}^{-1}\dots t_1^{-1}s^{-1}t_{j}t_{j-1} \dots t_2)*d_{1,2}$ if $i<0$, $j>0$ and $i+j<0$, $\strut d_{i,j}=(t_{-j-1}^{-1}t_{-j-2}^{-1}\dots t_1^{-1}t_{-i-1}^{-1} t_{-i-2}^{-1} \dots t_2^{-1}s^{-1}t_1^{-1}s^{-1})*d_{1,2}$ if $j<0$, $\strut d_{i,j}=(t_{j-1}^{-1}d_{j-1,j}t_{j-2}^{-1}d_{j-2,j-1}\dots t_1^{-1}d_{1,2})* (s^2a_1^4)$ if $i+j=0$. The products described in the above definition represent very simple elements of ${\mathcal M}_{g,1}$ and we shall explain their meaning now. We shall first define special homeomorphisms of $S_0$. \[half-twist\] [A [*half-twist*]{} $h_{i,j}$ along a curve $\delta_{i,j}$ is an isotopy class (on $S_0$ relative to its boundary) of a homeomorphism of $S_0$ which is fixed outside $\delta_{i,j}$ and is equal to a counterclockwise rotation" by $180$ degrees inside $\delta_{i,j}$. In particular $h_{i,j}$ switches the two holes $\partial_i$ and $\partial_j$ inside $\delta_{i,j}$ so it is not fixed on the boundary of $S_0$, but $h_{i,j}^2$ is fixed on the boundary of $S_0$ and is isotopic to the full Dehn twist along $\delta_{i,j}$.]{} \[action of tk\] The result of the action of $t_k$ (respectively $s$) on a curve $\delta_{i,j}$ is the same as the result of the action of the product of half-twists $h_{k,k+1}h_{-k-1,-k}$ (respectively the result of the action of $h_{-1,1}$) on $\delta_{i,j}$. So $t_k$ rotates $\delta_{i,j}$ around $\delta_{k,k+1}$ counterclockwise and around $\delta_{-k-1,-k}$ conterclockwise and switches the corresponding holes. If the pair $(i,j)$ is disjoint from $\{k,k+1,-k,-k-1\}$ than $t_k$ leaves $\delta_{i,j}$ fixed. It also leaves curves $\delta_{k,k+1}$ and $\delta_{-k-1,-k}$ fixed. In a similar way $s$ rotates $\delta_{i,j}$ counterclockwise around $\delta_{-1,1}$ and switches the holes. If $(i,j)$ is disjoint from $(-1,1)$ than $s$ leaves $\delta_{i,j}$ fixed. Also $\delta_{-1,1}$ is fixed by $s$. The result of the action can be checked directly. \[dij is a twist\] The element $d_{i,j}$ of ${\mathcal M}_{g,1}$ is equal to the twist along the curve $\delta_{i,j}$ for all $i,j$. We start with the curve $\beta_2$ and apply to it the product of twists $b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1}$. We get the curve $\delta_{1,2}$. Therefore, by Lemma \[conjugation\], $d_{1,2}$ is equal to the twist along $\delta_{1,2}$. For $i\neq -j$ we start with $\delta_{1,2}$ and apply consecutive factors $t_i$ and $s$, one at a time. We check that the result is $\delta_{i,j}$ so $d_{i,j}$ is the twist along $\delta_{i,j}$, by Lemma \[conjugation\]. For $d_{-1,1}$ we use (iii) of Lemma \[basic relations\]. The curve $\delta_{-1,1}$ is the boundary of a regular neighbourhood of $\alpha_1\cup\beta_1$ and $(a_1b_1)^6=s^2a_1^4$ by (i) and (ii) of Lemma \[basic relations\], therefore $d_{-1,1}=(a_1b_1)^6$ is the twist along $\delta_{-1,1}$ by (iii) of Lemma \[basic relations\]. Now we apply a suitable product of $t_i$’s and $d_{i,i+1}$’s to $\delta_{-1,1}$ and get $\delta_{-j,j}$. Therefore $d_{-j,j}$ is equal to the twist along $\delta_{-j,j}$ by Lemma \[conjugation\]. We can explain now the relations in Theorem \[simple presentation\]. The relations [(M1), (M2)]{} and [(M3)]{} from Theorem \[simple presentation\] are satisfied in ${\mathcal M}_{g,1}$. Relations (M1) follow from Lemma \[basic relations\] (i) and (ii). Curves $\beta_1,\alpha_1,\epsilon_1$ form a chain. One boundary component of a regular neighbourhood of $\alpha_1\cup\beta_1\cup\epsilon_1$ is equal to $\beta_2$. It is easy to check that $a_2e_1a_1b_1^2a_1e_1a_2(\beta_2)$ is equal to the other boundary component. By Lemma \[basic relations\] (iii) and by Lemma \[conjugation\] we have $(b_1a_1e_1)^4=b_2a_2e_1a_1b_1^2a_1e_1a_2b_2(a_2e_1a_1b_1^2a_1e_1a_2)^{-1}$. This is equivalent to relation (M2) by Lemma \[basic relations\] (v). Consider now relation (M3). Applying consecutive twists one can check that $(b_2a_2e_1b_1^{-1})(\delta_{1,3})=\delta_3$, where $\delta_3$ is the curve on Figure \[general surface\]. Thus $d_3$ represents the twist along $\delta_3$. When we cut $S$ along curves $\alpha_1$, $\alpha_2$, $\alpha_3$ and $\delta_3$ we split off a sphere with four holes from surface $S$. Since elements $d_{i,j}$ represent twists along curves $\delta_{i,j}$, relation (M3) follows from lantern relation, Lemma \[lantern\].Our first big task is to establish a presentation of a stabilizer of one vertex of $X$. Let $v_0$ be a fixed vertex of $X$ corresponding to the cut system $\lan\alpha_1,\alpha_2,\dots,\alpha_g\ran$. Let $H$ be the stabilizer of $v_0$ in ${\mathcal M}_{g,1}$. \[H presentation\] The stabilizer $H$ of vertex $v_0$ admits the following presentation: The set of generators consists of $a_1$, $a_2, \dots,a_g$, $s$, $t_1$, $t_2$, $\dots$, $t_{g-1}$ and the $d_{i,j}$’s for $i<j, \ \ i,j\in I_0$. The set of defining relations consists of: [(P1)]{}$[a_i,a_j]=1$ and $[a_i,d_{j,k}]=1$ for all $i,j,k\in I_0$.\ [(P2)]{}pure braid relations [(a)]{}$d_{r,s}^{-1}*d_{i,j}=d_{i,j}$ if $ r<s<i<j$ or $i<r<s<j$, [(b)]{}$d_{r,s} ^{-1}*d_{s,j}=d_{r,j}*d_{s,j}$ if $r<s<j$, [(c)]{}$d_{r,j}^{-1}*d_{r,s}=d_{s,j}*d_{r,s}$ if $r<s<j$, [(d)]{}$[d_{i,j},d_{r,j}^{-1}*d_{r,s}]=1$ if $r<i<s<j$.\ [(P3)]{}$t_it_{i+1}t_i=t_{i+1}t_it_{i+1}$ for $i=1,\dots,g-2$ and $[t_i,t_j]=1$ if $1\leq i<j-1<g-1$.\ [(P4)]{}$s^2=d_{-1,1}a_1^{-4}$ and $t_i^2=d_{i,i+1}d_{-i-1,-i}a_i^{-2}a_{i+1}^{-2}$ for $i=1,\dots,g-1$.\ [(P5)]{}$[t_i,s]=1$ for $i=2,\dots,g-1$.\ [(P6)]{}$st_1st_1=t_1st_1s$.\ [(P7)]{}$[s,a_i]=1$ for $1\leq i\leq g$, $t_i*a_i=a_{i+1}$ for $1\leq i\leq g-1$,$\strut[a_i,t_j]=1$ for $1\leq i \leq g$, $j\neq i,i-1$.\ [(P8)]{}$s*d_{i,j}=d_{i,j}$ if $i\neq \pm1$ and $j\neq\pm1$ or if $i=-1$ and $j=1$, $\strut s*d_{-1,j}=d_{1,j}$ for $2\leq j\leq g$, $s*d_{i,-1}=d_{i,1}$ for $-g\leq i\leq -2$, $\strut t_k*d_{i,j}=d_{i,j}$ if $1\leq k\leq g-1$ and $(\,j=i+1=k+1$, or $j=i+1=-k$ or $i,j\notin\{\pm k,\pm (k+1)\}\,)$, $\strut t_k*d_{k,j}=d_{k+1,j}$ for $1\leq k\leq g-1$ and $k+2\leq j\leq g$, $\strut t_k*d_{i,-k-1}=d_{i,-k}$ for $1\leq k\leq g-1$ and $-g\leq i\leq -k-2$, $\strut t_k*d_{-k-1,k}=d_{-k,k+1}$, $t_k*d_{-k-1,k+1}=d_{k,k+1}*d_{-k,k}$, for $1\leq k\leq g-1$ $\strut t_k*d_{-k-1,j}=d_{-k,j}$ for $1\leq k\leq g-1$ and $j> -k$, $j\neq k,k+1$, $\strut t_k*d_{i,k}=d_{i,k+1}$ for $1\leq k\leq g-1$ and $i<k$, $i\neq -k,-k-1$. An element of $H$ leaves the cut-system $v_0$ invariant but it may permute the curves $\alpha_i$ and may reverse their orientation. Clearly $a_i$ belongs to $H$. One can easily check that $t_i(\alpha_i)=\alpha_{i+1}$, $t_i(\alpha_{i+1})=\alpha_i$, $t_i(\alpha_k)=\alpha_k$ for $k\neq i,i+1$. $s(\alpha_1)$ is equal to $\alpha_1$ with the opposite orientation and $s$ is fixed on other $\alpha_i$’s. Thus $t_i$’s and $s$ belong to $H$. By Lemma \[dij is a twist\] we know that $d_{i,j}$ is a twist along the curve $\delta_{i,j}$ so it also belongs to $H$. We shall prove in the next section that the relations (P1) – (P8) follow from the relations (M1) – (M3), so they are satisfied in ${\mathcal M}_{g,1}$ and thus also in $H$. The group $H$ can be defined by two exact sequences. $$\label{sigma sequence} 1\rig \ints_2^g\rig \pm\Sigma_g\rig \Sigma_g\rig 1.$$ $$\label{H sequence} 1\rig H_0\rig H\rig \pm \Sigma_g\rig 1.$$ Before defining the objects and the homomorphisms in these sequences we shall recall the following fact from group theory. \[general presentation\] Let $1\rig A\rig B\rig C\rig 1$ be an exact sequence of groups with known presentations $A=\lan a_i|Q_j\ran$ and $C=\lan c_i| W_j\ran$. A presentation of $B$ can be obtained as follows: Let $b_i$ be a lifting of $c_i$ to $B$. Let $R_j$ be a word obtained from $W_j$ by substitution of $b_i$ for each $c_i$. Then $R_j$ represents an element $d_j$ of $A$ which we write as a product of generators $a_i$ of $A$. Finally for every $a_i$ and $b_j$ the conjugate $b_j*a_i$ represents an element $a_{i,j}$ of $A$, which we write as a product of the generators $a_i$.\ Then $B=\lan a_i,b_j|Q_j, R_j=d_j, b_j*a_i=a_{i,j}\ran$. We now describe the sequence (\[sigma sequence\]) and the group $\pm\Sigma_g$. This is the group of permutations of the set $I_0=\{-g,1-g,\dots,-1,1,2,\dots,g\}$ such that $\sigma(-i)=-\sigma(i)$. The homomorphism $\pm\Sigma_g\to\Sigma_g$ forgets the signs. A generator of the kernel changes the sign of one letter. The sequence splits, $\Sigma_g$ can be considered as the subgroup of the permutations which take positive numbers to positive numbers. Let $\tau_i=(i,i+1)$ be a transposition in $\Sigma_g$ for $i=1,2,\dots,g-1$. Then\ (S1)$ [\tau_i,\tau_j]=1$ for $|i-j|>1$,\ (S2)$\tau_i*\tau_{i+1}=\tau_{i+1}^{-1}*\tau_i$ for $i=1,\dots,g-2$,\ (S3)$\tau_i^2=1$ for $i=1,\dots,g-1$. This defines a presentation of $\Sigma_g$. Further let $\sigma_i$ for $i=1,\dots,g$ denote the change of sign of the $i$-th letter in a signed permutation. Then $\sigma_i^2=1$ and $[\sigma_i,\sigma_j]=1$ for all $i,j$. Finally the conjugation gives $\tau_i*\sigma_i=\sigma_{i+1}$, $\tau_i*\sigma_{i+1}=\sigma_i$ and $[\sigma_j,\tau_i]=1$ for $j\neq i$ and $j\neq i+1$. In fact it suffices to use one generator $\sigma=\sigma_1$ and the relations $\sigma_i=(\tau_{i-1}\tau_{i-2}\dots \tau_1)*\sigma$. We get the relations (S4)$\sigma^2=1$,\ (S5)$[\sigma,\tau_j]=1$ and $[(\tau_i\tau_{i-1}\dots \tau_1)*\sigma,\tau_j]=1$ for $1\leq i\leq g-1$ and $j\neq i$ and $j\neq i+1$,\ (S6)$[(\tau_i\tau_{i-1}\dots \tau_1)*\sigma,\sigma]=1$ and $[(\tau_i\tau_{i-1}\dots \tau_1)*\sigma, (\tau_j\tau_{j-1}\dots \tau_1)*\sigma]=1$ for $1\leq i,j\leq g-1$. Group $\pm\Sigma_g$ has a presentation with generators $\sigma,\tau_1,\dots,\tau_{g-1}$ and with defining relations (S1) – (S6). We shall describe now the sequence (\[H sequence\]). A homeomorphism in $H$ may permute the curves $\alpha_i$ and may change their orientations. We fix an orientation of each curve $\alpha_i$ and define a homomorphism $\phi_1\co H\to \pm\Sigma_g$ as follows: a homeomorphism $h$ is mapped onto a permutation $i\mapsto \pm j$ if $h(\alpha_i)=\alpha_j$ and the sign is $+$“ if the orientations of $h(\alpha_i)$ and of $\alpha_j$ agree, and $-$” otherwise. If $h$ preserves the isotopy class of $\alpha_i$ and preserves its orientation then it is isotopic to a homeomorphism fixed on $\alpha_i$. The kernel of $\phi_1$ is the subgroup $H_0$ of the elements of $H$ represented by the homeomorphisms which keep the curves $\alpha_1,\alpha_2,\dots, \alpha_g$ pointwise fixed. We want to find a presentation of $H$ from the sequence (\[H sequence\]). We start with a presentation of $H_0$. An element of $H_0$ induces a homeomorphism of $S_0$. When we glue back the corresponding pairs of boundary components of $S_0$ we get the surface $S$. This glueing map induces a homomorphism from the mapping class group of $S_0$ onto $H_0$. We shall prove that the kernel of this homomorphism is generated by products $d_id_{-i}^{-1}$, where $d_i$ is the twist along curve $\partial_i$, so both twists are identified with $a_i$ in $H_0$. It suffices to assume, by induction, that we glue only one pair of boundaries $\partial_i$ and $\partial_{-i}$ on $S_0$ and get a nonseparating curve $\alpha_i$ on $S$. Homeomorphism $d_id_{-i}^{-1}$ induces a spin map $s$ of $S$ along $\alpha_i$ (see [@Birman], Theorem 4.3 and Fig. 14). Let $\gamma$ be a curve on $S$ which meets $\alpha_i$ at one point $P$. Let $h_0$ be a homeomorphism of $S_0$ which induces a homeomorphism $h$ of $S$ isotopic to the identity and fixed on $\alpha_i$. We shall prove that for a suitable power $k$ the map $s^kh$ is fixed on $\gamma$ (after an isotopy of $S$ liftable to $S_0$). If $h(\gamma)$ forms a 2–gon with $\gamma$ we can get rid of the 2–gon by an isotopy liftable to $S_0$ (fixed on $\alpha_i$). If $h(\gamma)$ and $\gamma$ form no 2–gons then they are tangent at $P$. Let us move $h(\gamma)$ off $\gamma$, near the point $P$, to a curve $\gamma^\prime$. If $\gamma^\prime$ and $\gamma$ intersect then they form a 2–gon. But then $h(\gamma)$ and $\gamma$ form a self-touching 2–gon (see Figure \[fake 2-gon\]). Clearly the spin map $s$ or $s^{-1}$ removes the 2–gon. If $\gamma^\prime$ and $\gamma$ are disjoint then they bound an annulus. If the annulus contains the short arc of $\alpha_i$ between $\gamma^\prime$ and $\gamma$ then $h(\gamma)$ is isotopic to $\gamma$ by an isotopy liftable to $S_0$. If the annulus contains the long arc of $\alpha_i$ between $\gamma^\prime$ and $\gamma$ then the situation is similar to Figure \[fake 2-gon\], but $h(\gamma)$ does not meet $\gamma$ outside $P$. The spin map $s$ or $s^{-1}$ takes $h(\gamma)$ onto a curve isotopic to $\gamma$ by an isotopy liftable to $S_0$. So we may assume, by induction on $|h(\gamma)\cap\gamma|$, that $h(\gamma)=\gamma$. We proceed as in Remark \[proofs\]. We spilt $S$ into disks by curves $\alpha_i$, $\gamma$, and other curves disjoint from $\alpha_i$. Homeomorphism $h$ takes each curve to an isotopic curve and the 2–gons which appear do not contain $\alpha_i$. Thus all isotopies are liftable to $S_0$. It follows that the kernel is generated by $d_id_{-i}^{-1}$’s. By Lemma \[disk with holes presentation\] the mapping class group of $S_0$ has a presentation with generators $d_k$ and $d_{i,j}$ and with relations (Q1) and (Q2), where $a_{i,j}=d_{i,j}$, $i,j\in I_0$. Therefore $H_0$ has a presentation with generators $a_1,\dots, a_g$ and $d_{i,j}$ for $i<j\in I_0$ and with relations (P1), (P2). In these relations $d_{i,j}$ is represented by a Dehn twist along $\delta_{i,j}$. We come back to sequence (\[H sequence\]). We see from the action of $t_i$ and $s$ on $\alpha_j$ that we can lift $\tau_i$ to $t_i$ and $\sigma$ to $s$. Relations (S1) and (S2) lift to (P3). Relations (S3) and (S4) lift to (P4). Relation (S5) lifts to $[(t_it_{i-1}\dots t_1)*s,t_j]=1$ for $j\neq i$ and $j\neq i+1$ and it follows from (P3) and (P5). We shall deal with (S6) a little later. We now pass to the conjugation of the generators of $H_0$ by $s$ and $t_k$. Since $s^2\in H_0$ and $t_k^2\in H_0$, by (P4), it suffices to know the result of the conjugation of each generator of $H_0$ by either $s$ or by $s^{-1}$, and by either $t_k$ or by $t_k^{-1}$, the other follows. The result of the conjugation is described in relations (P7) and (P8). Finally we lift the relation (S6). We start with the case $[\tau_1*\sigma,\sigma]$. It lifts to $t_1st_1^{-1}st_1s^{-1}t_1^{-1}s^{-1}=$ (by (P6)) $t_1st_1^{-1}ss^{-1}t_1^{-1}s^{-1}t_1=t_1st_1^{-2}s^{-1}t_1^{-1}t_1^2$. We have $t_1^2\in H_0$, by (P4), and the conjugation of an element of $H_0$ by $s$ and $t_i$ is already determined by (P7) and (P8). So we know how to lift $[\tau_1*\sigma,\sigma]=1$. In the general case we have a commutator $[(t_it_{i-1}\dots t_1)*s,(t_jt_{j-1}\dots t_1)*s]$. If $i>j$ then, by (P3) and (P5), this commutator is equal to the conjugate of $[t_1*s,s]$ by $t_jt_{j-1}\dots t_1t_it_{i-1}\dots t_2$. This is a conjugation of an element of $H_0$ by $t_k$’s, so the result is determined by (P7) and (P8). This concludes the proof of Proposition \[H presentation\]. We shall prove now that ${\mathcal M}_{g,1}$ acts transitively on vertices of $X$, so there is only one vertex orbit, and that $H$ acts transitively on edges incident to $v_0$, so there is only one edge orbit. When we cut $S$ open along the curves $\alpha_1,\dots,\alpha_g$ we get a disk with $2g$ holes. It is homeomorphic to any other such disk and we can prescribe the homeomorphism on the boundary components, hence every two cut systems can be transformed into each other by a homeomorphism and ${\mathcal M}_{g,1}$ acts transitively on $X^0$. Let $w$ be a vertex connected to $v_0$ by an edge. Then $w$ contains a curve $\beta$ which intersects some curve $\alpha_i$ at one point and is disjoint from the other curves of $v_0$. We cut $S$ open along $\alpha_1,\alpha_2,\dots,\alpha_g,\beta$ and get a disk with $2g-1$ holes, including one big" hole bounded by arcs of both $\alpha_i$ and $\beta$. If we cut $S$ along curves belonging to another edge incident to $v_0$ we get a similar situation. There exists a homeomorphism of one disk with $2g-1$ holes onto the other which preserves the identification of curves of the boundary and takes the set $\alpha_1,\alpha_2,\dots,\alpha_g$ onto itself. It induces a homeomorphism of $S$ which leaves $v_0$ invariant and takes one edge onto the other. Thus $H$ acts transitively on the edges incident to $v_0$. We now fix one such edge and describe its stabilizer. If we replace curve $\alpha_1$ of cut-system $v_0$ by curve $\beta_1$ we get a new cut-system connected to $v_0$ by an edge. We denote the cut-system by $v_0^\prime$ and the edge by ${\bf e}_0$. Let $H^\prime$ be the stabilizer of ${\bf e}_0$ in $H$. \[H prime generators\] The stabilizer $H^\prime$ of the edge ${\bf e}_0$ is generated by $a_1^2s$, $t_1st_1$, $a_2$, $d_{2,3}$, $d_{-2,2}$, $d_{-1,1}d_{-1,2}d_{1,2}a_1^{-2}a_2^{-1}$, and $t_2,\dots,t_{g-1}$. An element $h$ of $H^\prime$ may permute curves $\alpha_2,\dots,\alpha_g$ and may reverse their orientations. It may also reverse simultaneously orientations of both curves $\alpha_1$ and $\beta_1$ (it preserves the orientation of $S$ so it must preserve algebraic intersection of curves). We check that $a_1^2s$ reverses orientations of $\beta_1$ and $\alpha_1$, $t_1st_1$ reverses orientation of $\alpha_2$ and leaves $\beta_1$ and $\alpha_1$ invariant. Elements $t_i$ permute the curves $\alpha_2,\dots,\alpha_g$. Modulo these homeomorphisms we may assume that $h$ is fixed on all curves $\alpha_i$ and $\beta_1$. It induces a homeomorphism of $S$ cut open along all these curves. We get a disk with holes and, by Lemma \[disk with holes presentation\], its mapping class group is generated by twists around holes and twists along suitable curves surrounding two holes at a time. Element $d_{-1,1}= (a_1^2s)^2$ represents twist around the big hole (the cut along $\alpha_1\cup\beta_1$). Conjugates of $a_2$ by $t_j$’s produce twists around other holes. The conjugate of $d_{2,3}$ by $(t_1st_1)^{-1}$ is equal to $d_{-2,3}$ and the conjugate of $d_{-2,3}$ by $(t_2t_1st_1)^{-1}$ is equal to $d_{-3,-2}$. Every $d_{k,k+1}$ can be obtained from these by conjugation by $t_i$’s, $i>1$. It is now clear that every curve $\delta_{i,j}$ with $i,j\neq\pm1$ can be obtained from $d_{2,3}$ and $d_{-2,2}$ by conjugation by the elements chosen in the Lemma. So the corresponding twists are products of the above generators. We also need twists along curves which surround the big" hole and another hole. Consider such a curve on $S_0$. It must contain inside both holes $\partial_{-1}$ and $\partial_1$, including the arc between them corresponding to $\beta_1$, and one other hole. One such curve, call it $\gamma$, contains $\partial_{-1},\partial_1,\partial_2$. By Lemma \[lantern\] the twist along $\gamma$ is equal to $d_{-1,1}d_{-1,2}d_{1,2}a_1^{-2}a_2^{-1}$. Any other curve which contains $\partial_{-1},\partial_1,\partial_i$ with $i>1$ is obtained from $\gamma$ by application of $t_i$’s. The curve which contains $\partial_{-1},\partial_1,\partial_{-2}$ is obtained from $\gamma$ by application of $(t_1st_1)^{-1}$ and a curve which contains $\partial_{-1},\partial_1,\partial_i$ with $i<-2$ is obtained from the last curve by application of $t_i^{-1}$’s. Therefore all generators of $H^\prime$ are products of the generators in the Lemma. We now distinguish one more element of ${\mathcal M}_{g,1}$, which does not belong to $H$. Let $r=a_1b_1a_1$. The element $r$ is a quarter-twist". It switches curves $\alpha_1$ and $\beta_1$, $r^2=sa_1^2=h_{-1,1}$ is a half-twist around $\delta_{-1,1}$ and $r^4=d_{-1,1}$. Also $r$ leaves the other curves $\alpha_i$ fixed, so it switches the vertices of the edge ${\bf e}_0$, $r(v_0)=v_0^\prime$ and $r(v_0^\prime)=v_0$. We now describe precisely a construction from [@Laudenbach] and [@Hatcher-Thurston] which will let us determine a presentation of ${\mathcal M}_{g,1}$. This construction was very clearly explained in [@Heusner]. Let us consider a free product $(H*\ints)$ where the group $\ints$ is generated by $r$. An $h$-[*product*]{} is an element of $(H*\ints)$ with positive powers of $r$, so it has a form $h_1rh_2r\dots h_krh_{k+1}$. We have an obvious map $\eta\co (H*\ints)\to {\mathcal M}_{g,1}$ through which the $h$–products act on $X$. We shall prove that $\eta$ is onto and we shall find the $h$ products which normally genarate the kernel of $\eta$. To every edge-path ${\bf p}=(v_0,v_1,\dots, v_k)$ which begins at $v_0$ we assign an $h$–product $g=h_1rh_2r\dots h_krh_{k+1}$ such that $h_1r\dots h_mr(v_0)=v_m$ for $m=1,\dots,k$. We construct it as follows. There exists $h_1\in H$ such that $h_1(v_0)=v_0$ and $h_1(v_0^\prime)=v_1$. Then $h_1r(v_0)=v_1$. Next we transport the second edge to $v_0$. $(h_1r)^{-1}(v_1)=v_0$ and $(h_1r)^{-1}(v_2)=v_1^\prime$. There exists $h_2\in H$ such that $h_2r(v_0)=v_1^\prime$ and $h_1rh_2r(v_0)=v_2$. And so on. Observe that the elements $h_i$ in the $h$–product corresponding to an edge-path $\bf p$ are not uniquely determined. In particular $h_{k+1}$ is an arbitrary element of $H$. The construction implies: The generators of $H$ together with the element $r$ generate the group ${\mathcal M}_{g,1}$. Let $g$ be an element of ${\mathcal M}_{g,1}$. Then $g(v_0)$ is a vertex of $X$ and can be connected to $v_0$ by an edge-path $ {\bf p}=(v_0,v_1,\dots, v_k =g(v_0))$. Let $g_1=h_1rh_2r\dots h_kr$ be an $h$–product corresponding to $\bf p$. Then $g_1(v_0)=v_k=g(v_0)$, therefore $\eta(g_1^{-1})g=h_{k+1}$ leaves $v_0$ fixed and belongs to the stabilizer $H$ of $v_0$. It follows that $g=\eta(h_1rh_2r\dots h_krh_{k+1}) $ in ${\mathcal M}_{g,1}$. By the inverse process we define an edge-path induced by the $h$–product $g=h_1rh_2r\dots h_krh_{k+1}$. The edge-path starts at $v_0$ then $v_1=h_1r(v_0)$, $v_2=h_1rh_2r(v_0)$ and so on. The last vertex of the path is $v_k=g(v_0)$. [An $h$–product $g$ represents an element in $H$ if and only if $v_k=v_0$. This happens if and only if the corresponding edge-path is closed. We can multiply such $g$ by a suitable element of $H$ on the right and get an $h$–product which represents the identity in ${\mathcal M}_{g,1}$ and induces the same edge-path as $g$.]{} We now describe a presentation of ${\mathcal M}_{g,1}$. \[G presentation\] The mapping class group ${\mathcal M}_{g,1}$ admits the following presentation: The set of generators consists of $a_1$, $a_2, \dots,a_g$, $s$, $t_1$, $t_2$, $\dots$, $t_{g-1}$, $r$ and the $d_{i,j}$’s for $i<j, \ \ i,j\in I_0$. The set of defining relations consists of relations [(P1) – (P8)]{} of Proposition \[H presentation\] and of the following relations:\ [(P9)]{}$r$ commutes with $a_1^2s$, $t_1st_1$, $a_2$, $d_{2,3}$, $d_{-2,2}$, $d_{-1,1}d_{-1,2}d_{1,2}a_1^{-2}a_2^{-1}$, and $t_2,\dots,t_{g-1}$.\ [(P10)]{}$r^2=sa_1^2$.\ [(P11)]{}$(k_ir)^3=(k_isa_1)^2$ for $i=1,2,3,4$, where$\strut k_1=a_1$, $k_2=d_{1,2}$, $k_3=a_1^{-1}a_2^{-2}d_{1,2} d_{-2,1}d_{-2,2}$, $k_4=a_1^{-1}a_2^{-1}a_3^{-1}d_{1,2} d_{1,3}d_{2,3}$,$\strut (rk_5rk_5^{-1})^2=sa_1^2k_5sa_1^2k_5^{-1}$, where $\strut k_5=a_2t_1d_{1,2}^{-1}$, $(ra_1t_1)^5=(sa_1^2t_1)^4$. Relations (P9) – (P11) say that some elements of $H*\ints$ belong to $ker(\eta)$, so Theorem \[G presentation\] claims that $(H*\ints)$ modulo relations (P9) – (P11) is isomorphic to ${\mathcal M}_{g,1}$. Relations (P9) tell us that $r$ commutes with the generators of the stabilizer $H^\prime$ of the edge ${\bf e}_0$. We shall prove this claim now. An element $h$ of $H^\prime$ leaves the edge ${\bf e}_0$ fixed but it may reverse the orientations of $\alpha_1$ and $\beta_1$. The element $r^2=sa_1^2$ does exactly this and commutes with $r$. Modulo this element we may assume that $h$ leaves $\alpha_1$ and $\beta_1$ pointwise fixed. But then we may also assume that it leaves some neighbourhood of $\alpha_1$ and $\beta_1$ pointwise fixed. On the other hand $r$ is equal to the identity outside a neighbourhood of $\alpha_1$ and $\beta_1$ so it commutes with $h$. From relations (P9) and (P10) we get information about $h$–products. [**Claim 1**]{}If two $h$–products represent the same element in ${\mathcal M}_{g,1}$ and induce the same edge-path then they are equal in $(H*\ints)/(P9)$. If two $h$–products $g_1=h_1r\dots rh_{k+1}$ and $g_2=f_1r\dots rf_{k+1}$ induce the same edge-path ${\bf p}=(v_0,v_1,\dots,v_k)$ then $h_1r(v_0)=f_1r(v_0)$. Therefore $h_1^{-1}f_1r(v_0)=r(v_0)$ and $h_1^{-1}f_1\in H^\prime$ commutes with $r$ in $(H*\ints)/(P9)$. Now $f_1rf_2=h_1h_1^{-1}f_1rf_2=h_1rf_2^\prime$ in $(H*\ints)/(P9)$. Therefore $g_2$ and a new $h$–product $h_1rf_2^\prime rf_3r\dots rf_{k+1}$ are equal in $(H*\ints)/(P9)$ and induce the same edge-path $\bf p$. If we apply $r^{-1}h_1^{-1}$ to the vertices $(v_1,v_2,\dots,v_k)$ of $\bf p$ we get a shorter edge-path which starts at $v_0$ and is induced by both shorter $h$–products $h_2r\dots rh_{k+1}$ and $f_2^\prime r\dots rf_{k+1}$. Claim 1 follows by induction on $k$. Two different edge-paths may be homotopic in the $1$–skeleton $X^1$. This means that there is a backtracking $v_i,v_{i+1},v_i$ along the edge-path. [**Claim 2**]{}If two $h$–products represent the same element in ${\mathcal M}_{g,1}$ and induce edge-paths which are equal modulo back-tracking then the $h$–products are equal in $(H*\ints)/((P9),(P10))$. Consider an $h$–product $g=g_ih_{i+1}rh_{i+2}r$, where $g_i$ is an $h$–product inducing a shorter edge-path $\bf p$ and the edge-path induced by $g$ has a back-tracking at the end: $g_i(v_0)=v_i$, $g_ih_{i+1}r(v_0)=v_{i+1}$, and $g_ih_{i+1}rh_{i+2}r(v_0)=v_i$. Clearly the $h$–product $g_ih_{i+1}rr$ induces the same edge-path. In particular $g_ih_{i+1}r^2(v_0)=v_i$ hence there exists $h^\prime\in H$ such that $\eta(g_ih_{i+1}r^2h^\prime)=\eta(g_ih_{i+1}rh_{i+2}r)$. Now by Claim 1 the $h$–products are equal in $(H*\ints)/(P9)$. But $g_ih_{i+1}r^2h^\prime$ is equal in $(H*\ints)/(P10)$ to a shorter $h$–product which induces the edge-path $\bf p$. Claim 2 follows by induction on the number of back-trackings. Relations (P11) correspond to edge-paths of type (C3), (C4) and (C5). Six relations correspond to six particular edge-paths. As in (P11) we let $k_1=a_1$, $k_2=d_{1,2}$, $k_3=d_{1,2}d_{-2,1} d_{-2,2}a_2^{-2}a_1^{-1}$, $\strut k_4=a_1^{-1}a_{2}^{-1}a_3^{-1}d_{1,2}d_{1,3}d_{2,3}=d_3$, $k_5=a_2d_{1,2}^{-1}t_1$, $k_6=a_1t_1$. We now choose six $h$–products. For $i=1,2,3,4$ let $g_i=(k_ir)^3$, $g_5=(rk_5rk_5^{-1})^2$ and $g_6=(rk_6)^5$. Product $g_i$ appears in the corresponding relation in (P11). For $i=1,\dots,6$ let $\gamma_i$ be the corresponding curve on Figure \[gamma curves\], represented on surface $S_0$ ($\gamma_5=\beta_2$). For $i=1,2,3,4$ homeomorphism $k_i$ leaves all curves $\alpha_i$ invariant. Also $k_ir(\alpha_1)=\gamma_i$, $k_ir(\gamma_i)=\beta_1$, $k_ir(\beta_1)=\alpha_1$. It follows that the $h$–product $g_i$ represents the edge path ${\bf p}_i=(\lan\alpha_1\ran\to\lan\gamma_i\ran\to\lan\beta_1 \ran\to\lan\alpha_1\ran)$.\ It is not hard to check that for $i=5,6$ the $h$–product $g_i$ represents the edge-path ${\bf p}_i$, where\ ${\bf p}_5=(\lan\alpha_1,\alpha_2\ran\to\lan\beta_1, \alpha_2 \ran\to\lan\beta_1,\gamma_5 \ran\to\lan\alpha_1,\gamma_5\ran\to \lan\alpha_1,\alpha_2\ran)$.\ ${\bf p}_6=(\lan\alpha_1,\alpha_2\ran\to\lan\beta_1,\alpha_2 \ran\to\lan\beta_1,\epsilon_1 \ran\to\lan\gamma_6,\epsilon_1\ran\to\lan\gamma_6,\alpha_1\ran\to \lan\alpha_2,\alpha_1\ran)$. Since $g_i$ represents a closed edge-path it is equal in ${\mathcal M}_{g,1}$ to some element $h_i\in H$. Then $V_i=g_ih_i^{-1}$ also represents ${\bf p}_i$ and is equal to the identity in ${\mathcal M}_{g,1}$. We shall prove in the next section that $h_i$ is equal in $H$ to the right hand side of the corresponding relation in (P11) and thus $V_i=1$ in $(H*\ints)/((P9),(P10),(P11))$. We already know, by Theorem \[Hatcher-Thurston\], that every closed path in $X$ is a sum of paths of type (C3), (C4) and (C5). Some of these paths are represented by the paths ${\bf p}_1$ – ${\bf p}_6$. We say that a closed path is [*conjugate*]{} to a path ${\bf p}_i$ if it has a form ${\bf q}_1{\bf q}_2{\bf q}_1^{-1}$, where ${\bf q}_1$ starts at $v_0$ and ${\bf q}_2$ is the image of ${\bf p}_i$ under the action of some element of ${\mathcal M}_{g,1}$. We shall prove that every closed path is a sum of paths conjugate to one of the paths ${\bf p}_1$ – ${\bf p}_6$ or their inverses. This will imply Theorem \[G presentation\]. We shall start with Harer’s reduction for paths of type (C3) (see [@Harer]). We fix curves $\alpha_2,\dots,\alpha_g$ and consider cut-systems containing one additional curve. Consider a triangular path $(\lan\alpha\ran,\lan\beta\ran,\lan\gamma\ran)$. Give orientations to curves $\alpha,\beta,\gamma$ so that the algebraic intersection numbers satisfy $(\alpha,\beta)=(\alpha,\gamma)=1$. Switching $\beta$ and $\gamma$, if necessary, we may also assume $(\beta,\gamma)=-1$. We cut $S$ open along $\alpha_2,\dots,\alpha_g$ and get a torus $S_{1,2g-1}$ with $2g-1$ holes. Each square on Figure \[squares\] represents a part of the universal cover of a closed torus, punctured above holes of $S_{1,2g-1}$. We show more than one preimage of some curves in the universal cover. A fundamental region is a square bounded by $\alpha$ and $\beta$ with $2g-1$ holes, one of them bounded by the boundary $\partial$ of $S$. We may assume that $\gamma$ crosses $\alpha$ and $\beta$ in two distinct points. Then $\gamma$ splits the square into three regions: $F_1$ to the right of $\gamma$ after $\gamma$ crosses $\alpha$, $F_2$ to the left of $\gamma$ after $\gamma$ crosses $\beta$, and $F_0$ (see Figure \[squares\] (a)). Reversing the orientations of all curves we can switch the regions $F_1$ and $F_2$. Let $l_i$ be the number of holes in $F_i$ for $i=0,1,2$. We want to prove that every triangular path is a sum of triangular paths with $l_1\leq 2$, $l_2=0$, and with the hole $\partial$ not in the region $F_1$. We shall prove it by induction on $l_1+l_2$. A thin circle on Figure \[squares\] denotes a single hole and a thick circle denotes all remaining holes of the region. Suppose first that $l_1>1$. Move curves $\alpha$, $\beta$, $\gamma$ off itself (on a closed torus) in such a way that the region bounded by $\alpha$ and $\alpha^\prime$ contains only one hole, belonging to $F_1$, the region bounded by $\beta$ and $\beta^\prime$ contains another hole of $F_1$ and the region bounded by $\gamma$ and $\gamma^\prime$ contains both of these holes. Up to an isotopy (translating holes and straightening curves) the situation looks like on Figure \[squares\] (c). Now consider Figure \[squares\] (b) — an octahedron projected onto one of its faces. All faces of the octahedron are triangles" in $X$. In order to understand regions $F_0$, $F_1$, $F_2$ corresponding to each face we translate the fundamental domain to a suitable square in the universal cover. For every face different from $(\alpha,\beta,\gamma)$ the region $F_1$ has at least one hole less than $l_1$ and the region $F_0$ has all of its original holes and possibly some more. Clearly the boundary of the face $(\alpha,\beta,\gamma)$ is a sum of conjugates of the boundaries of the other faces. So by induction we may assume that $l_1\leq 1$. By symmetry we may assume that $l_2\leq 1$ and $l_1>2$. We chose new curves $\alpha^\prime$, $\beta^\prime$ and $\gamma^\prime$ whose liftings are shown on Figure \[squares\] (d). Consider again the octahedron on Figure \[squares\] (b). Now region $F_2$ is fixed for all faces of the octahedron and region $F_0$ has all of its original holes and some additional holes for all faces different from $(\alpha,\beta,\gamma)$. So by induction we may assume that $l_1\leq 2$ and $l_2\leq 1$. Suppose $l_1=2$ and $l_2=1$ and choose new curves $\alpha^\prime$, $\beta^\prime$ and $\gamma^\prime$ as on Figure \[squares\] (e). Again consider the octahedron. Now for each face different from $(\alpha,\beta,\gamma)$ at least one of the regions $F_1$ and $F_2$ looses at least one of its holes. Suppose now that each region has exactly one hole. Consider curves $\alpha^\prime$, $\beta^\prime$ and $\gamma^\prime$ on Figure \[squares\] (f). Again for each face of the corresponding octahedron different from the face $(\alpha,\beta,\gamma)$ at least one of the regions $F_1$ and $F_2$ looses at least one of its holes. So we may assume that $l_2=0$ and $l_1\leq 2$. Finally it may happen that a hole in $F_1$ is bounded by $\partial$. We can isotop $\gamma$ (on the torus with holes) over $F_2$ to the other side of the intersection of $\alpha$ and $\beta$. Clearly $F_1$ becomes now $F_0$. We can repeat the previous reduction and the hole $\partial$ will remain in $F_0$. Thus we are left with four cases:\ $l_1=l_2=0$,\ $l_1=1$, $l_2=0$,\ $l_1=2$, $l_2=0$ and the two holes in $F_1$ correspond to the same curve $\alpha_i$,\ $l_1=2$, $l_2=0$ and the two holes in $F_1$ correspond to different curves $\alpha_i$ and $\alpha_j$. Clearly each case is uniquely determined up to homeomorphism (of one square with holes onto another, preserving $\gamma$). Paths ${\bf p}_i$, $i=1,2,3,4$ represent triangle paths of these four types with $\alpha$ corresponding to $\alpha_1$, $\beta$ corresponding to $\beta_1$ and $\gamma$ corresponding to $\gamma_i$. Given any path of type (C3) we can map it onto a path which starts at $v_0$. Applying an element of $H$ we may assume that the second vertex is $v_0^\prime$. Then the path is of the type considered in the above reduction and it is a sum of conjugates of ${\bf p}_1$, ${\bf p}_2$, ${\bf p}_3$, ${\bf p}_4$ and their inverses (we may need to switch $\beta$ and $\gamma$ in the proof). Consider now the path ${\bf p}_5$. When we cut $S$ open along $\alpha_1,\alpha_2, \dots,\alpha_g,\beta_1$ and $\gamma_5$ we get a disk with two big“ holes and $2g-4$ small” holes. Any other path $\bf q$ of type (C4) produces a similar configuration. There exists $g\in G$ which takes ${\bf p}_5$ on $\bf q$. Therefore every path of type (C4) is a conjugate of ${\bf p}_5$. Consider now ${\bf p}_6$ and another path $\bf q$ of type (C5). When we cut $S$ open along the first four changing curves of ${\bf p}_6$: $\alpha_1, \alpha_2,\beta_1,\epsilon_1$ and along all fixed curves of the cut systems we get a disk with one big“ hole and $2g-4$ small” holes. When we do the same for the curves in $\bf q$ we get a similar configuration. We may map $\bf q$ onto a path $(\lan\alpha_1,\alpha_2\ran\to\lan\beta_1,\alpha_2 \ran\to\lan\beta_1,\epsilon_1 \ran\to\lan\gamma,\epsilon_1\ran\to\lan\gamma,\alpha_1\ran\to \lan\alpha_2,\alpha_1\ran)$\ in which only the curve $\gamma$ is different from the corresponding curve $\gamma_6$ in ${\bf p}_6$ and all other curves are as in ${\bf p}_6$. Consider curve $\gamma_5=\beta_2$ on Figure \[gamma curves\]. Curve $\gamma_5$ can be homotop onto the union of $\beta_1$, $\epsilon_1$ and a part of $\alpha_1$. Since $\gamma$ intersects $\beta_1$ once and does not intersect $\alpha_1$ nor $\epsilon_1$ it must intersect $\gamma_5$ once. Therefore we may form a subgraph of $X$ as on Figure \[pentagon\]. 0.9pt Here bottom and middle square edge-paths are of type (C4). In the square edge-path on the left side (and on the right side) only one curve changes so the path is a sum of triangles by Proposition \[genus1\]. The top pentagon edge-path is equal to ${\bf p}_6$. The new edge-path coincides with the outside pentagon. It is a sum of conjugates of the other edge paths. We can now complete the proof of Theorem \[G presentation\]. Let $W\in H*\ints$ be such that $\eta(W)=1$. We want to prove that $W=1$ in $(H*\ints)/((P9),(P10),(P11))$. Modulo (P10) we can write $W$ as an $h$–product $g$ which represents a closed edge path $\bf p$. By Theorem \[Hatcher-Thurston\] the edge-path $\bf p$ is a sum of conjugates of path of type (C3), (C4) and (C5). By the above discussion it is a sum of conjugates of the special paths ${\bf p}_1$ – ${\bf p}_6$ and their inverses, modulo backtracking. So, modulo backtracking, ${\bf p}=\Pi{\bf q}_if_i({\bf p}_{j_i}^{\pm1}){\bf q}_i^{-1}$ for some $f_i\in {\mathcal M}_{g,1}$. Each path ${\bf q}_if_i({\bf p}_{j_i}^{\pm1}){\bf q}_i^{-1}$ can be represented by an $h$–product $g_i$. Since the path is closed we have $\eta(g_i)\in H$, so we can correct the $h$–product $g_i$ and assume that $\eta(g_i)=1$. The product $\Pi g_i$ represents the path $\Pi{\bf q}_if_i({\bf p}_{j_i}^{\pm1}){\bf q}_i^{-1}$, so, by Claims 1 and 2, $W$ is equal to $\Pi g_i$ in $(H*\ints)/((P9),(P10))$. It suffices to prove that $g_i=1$ in $(H*\ints)/((P9),(P10),(P11))$. Let $k=j_i$ and suppose that $g_i$ represents a path ${\bf q}_if_i({\bf p}_k){\bf q}_i^{-1}$. Let $u_i$ be an $h$–product representing ${\bf q}_i$. Then $u_i(v_0)=f_i(v_0)$ is the first vertex of $f_i({\bf p}_k)$, hence $\eta(u_i)^{-1}f_i=h_i\in H$. Recall that ${\bf p}_k$ can be represented by an $h$–product $V_k$ which is equal to the identity in $(H*\ints)/((P9),(P10),(P11))$. The $h$–product $u_ih_iV_k$ represents ${\bf q}_if_i({\bf p}_k)$ and there exists an $h$–product $u_ih_iV_kw_i$ such that $\eta(u_ih_iV_iw_i)=1$ and $u_ih_iV_kw_i$ represents the edge-path ${\bf q}_if_i({\bf p}_k){\bf q}_i^{-1}$. Clearly $u_ih_iw_i$ represents the edge-path ${\bf q}_1{\bf q}_1^{-1}$ which is null-homotopic by backtracking. By Claims 1 and 2 the $h$–product $g_i$ is equal to $u_ih_iV_kw_i$ in $(H*\ints)/((P9),(P10))$, $u_ih_iV_kw_i$ is equal to $u_ih_iw_i$ in $(H*\ints)/((P9),(P10),(P11))$ and $u_ih_iw_i=1$ in $(H*\ints)/((P9),(P10))$. The path inverse to ${\bf p}_k$ is represented by some other $h$–product $V_k^{\prime}$ but then $V_kV_k^{\prime}$ represents a path contractible by back-tracking. Thus $V_kV_k^\prime=1$ in $(H*\ints)/((P9),(P10))$ and $V_k=1$ in $(H*\ints)/((P9),(P10),(P11))$ hence also $V_k^{\prime}=1$\ in $(H*\ints)/((P9),(P10),(P11))$. This concludes the proof of Theorem \[G presentation\]. Reduction to a simple presentation ================================== Let us recall the following obvious direction of Tietze’s Theorem. \[different presentations\] Consider a presentation of $G$ with generators $g_1,\dots,g_k$ and relations $R_1,\dots,R_s$. If we add another relation $R_{s+1}$, which is valid in $G$, we get another presentation of $G$. If we express some element $g_{k+1}$ of $G$ as a product $P$ of the generators $g_1,\dots,g_k$ we get a new presentation of $G$ with generators $g_1,\dots,g_{k+1}$ and with defining relations $R_1,\dots,R_s,g_{k+1}^{-1}P$. Conversely suppose that we can express some generator, say $g_k$, as a product $P$ of $g_1,\dots,g_{k-1}$. Let us replace every appearence of $g_k$ in each relation $R_i$, $i=1,\dots,s$ by $P$, getting a new relation $P_i$. Then $G$ has a presentation with generators $g_1,\dots,g_{k-1}$ and with relations $P_1,\dots,P_s$. We start with the presentation of the mapping class group $\cM_{g,1}$ established in Theorem \[G presentation\]. The generators represent the mapping classes of corresponding homeomorphisms of the surface $S=S_{g,1}$ represented on Figure \[general surface\]. We adjoin additional generators $$\label{extra generators} b_1=a_1^{-1}ra_1^{-1},\ \ b_2=(t_1a_1b_1)*d_{1,2},\ \ e_1=(rd_{1,2}a_2^{-1})*b_2$$ $\kern 1.85cm e_{i+1}=(t_it_{i+1})*e_i \ \ \hbox{for} \ \ i=1,\dots,g-2$. These generators also represent the corresponding twists in $\cM_{g,1}$. We adjoin the relations (M1) – (M3). Now, by Theorem \[G presentation\], generators $s,t_i,d_{i,j},r$ can be expressed in $\cM_{g,1}$ by the formulas from Definition \[generators\]. We substitute for each of these generators the corresponding product of $b_2,b_1,a_1,e_1,a_2,\dots,a_{g-1},e_{g-1},a_g$ in all relations (P1) – (P11) and in (\[extra generators\]). We check easily that the relations (\[extra generators\]) become trivial modulo the relations (M1) – (M3) (the last one will be proven in (\[A12\]).) We want to prove that all relations (P1) – (P11) follow from relations (M1) – (M3). [We shall establish many auxiliary relations of increasing complexity which follow from the relations (M1) – (M3). We shall explain some standard technique which one can use (some proofs will be left to the reader). From the braid relation $aba=bab$ one can derive several other useful relations, like: $a*b=b^{-1}*a$, $a*(b^2)=b^{-1}*(a^2)$, $(ab)*a=b$. When we want to prove that $[a,b]=0$ we shall usually try to prove that $a*b=b$. The relation $aba=bab$ tells us that $a$ can jump" over $ba$ to the right becoming $b$. By consecutive jumping to the right we can prove that $a_1(e_1a_1a_2e_1e_2a_2)= (e_1a_1a_2e_1e_2a_2)e_2$. We also get $e_1a_1a_2e_1e_2a_2=e_1a_2e_2a_1e_1a_2$ and $(b_1a_1e_1a_2)*b_2=(b_2^{-1}a_2^{-1}e_1^{-1}a_1^{-1})*b_1$. We shall say that some relations follow by (J) – jumping, if they follow [*easily*]{} from (M1) by the above technique.]{} We start the list of the auxiliary relations. $$\label{A1} t_i*a_i=a_{i+1}, \ t_i*a_{i+1}=a_i, \ t_i*a_k=a_k \ \ \hbox{for} \ \ k\neq i,i+1,$$ $s*a_i=a_i \ \ \hbox{for} \ \ i=1,\dots,g \ \ \ {\rm by \ (J).}$ Let $w_0=a_ge_{g-1}a_{g-1}e_{g-2}\dots e_1a_1b_1$. $$\label{A2} w_0^{-1}*b_2=d_{1,2}, \ w_0^{-1}*b_1=a_1, \ w_0^{-1}*a_i=e_i, \ w_0^{-1}*e_i=a_{i+1}, \$$ $d_{1,2}*b_1=b_1^{-1}*d_{1,2}, \ d_{1,2}*e_2=e_2^{-1}*d_{1,2}, \ [d_{1,2},a_i]=1, \ [d_{1,2},e_j]=1, \ \hbox{for} \ j\neq 2,$ $\strut[d_{1,2},t_j]=1 \ \hbox{for} \ \ j\neq 2.$ We have $w_0^{-1}*b_2=$ (by (M1)) $(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2= d_{1,2}$. Other results of conjugation by $w_0$ follow by (J). Now $\strut d_{1,2}*b_1=(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1}b_2 a_2e_1a_1b_1)*b_1=$ (by J)$\strut(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1}b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1}) *b_2=$ (by jumping from left side to the right) $\strut(b_1^{-1}b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2 = b_1^{-1}*d_{1,2}$. Other relations follow from (M1) by conjugation by $w_0^{-1}$.$$\label{A3} a_k*d_{i,j}=d_{i,j} \ \ \hbox{for all}\ \ i,j,k, \ {\rm by \ (J)\ and \ \ (\ref{A1}) \ and \ \ (\ref{A2}).}$$ $$\label{ti,tj} t_i*t_{i+1}=t_{i+1}^{-1}*t_i \ for\ i=1,2,\dots,g-2, \$$ (by the calculations similar to the proof of Lemma \[basic relations\] (iv)),$\strut[t_i,t_k]=1\ \hbox{for} \ |i-k|>1, \ [t_i,s]=1 \ \hbox{for} \ i>1, \ {\rm by\ (M1).}$ Using relations (\[A2\]) and (\[ti,tj\]) we can write the elements $d_{i,j}$ in a different way. $$\label{new dij} d_{i,i+1}=(t_{i-1}t_it_{i-2}t_{i-1}\dots t_1t_2)*d_{1,2} \ \hbox{for} \ i>0,$$ $ d_{-i-1,-i}=(t_{i-1}^{-1}t_{i}^{-1}t_{i-2}^{-1}t_{i-1}^{-1}\dots t_1^{-1}t_2^{-1})*d_{-2,-1} \ \hbox{fo}r \ i>0,$$\strut d_{i,i+1}=(t_{i-1}t_i)*d_{i-1,i}, \ \ t_k*d_{i,i+1}=d_{i,i+1} \ \ \hbox{for}\ \ |k-i|\neq 1. $ Let $w_1=a_2e_1a_1b_1^2a_1e_1a_2$. $$\label{A5} (b_1a_1e_1)^4=b_2w_1b_2w_1^{-1}, \ w_1*b_2=w_1^{-1}*b_2, \ [w_1*b_2,b_2]=1$$ We have $b_2w_1b_2=$ (by (M2)) $(b_1a_1e_1a_2)^5=$ (as in the proof ofLemma \[basic relations\] (v)) $\strut(b_1a_1e_1)^4w_1=$ (by (J)) $w_1(b_1a_1e_1)^4$.Also $\strut b_2(b_1a_1e_1)^4=(b_1a_1e_1)^4b_2$, by (M1). Therefore $w_1b_2w_1^{-1}=b_2^{-1}(b_1a_1e_1)^4=(b_1a_1e_1)^4b_2^{-1}= w_1^{-1}b_2w_1$ commutes with $b_2$. $$\label{A6} st_1s=b_1a_1e_1a_2^2e_1a_1b_1t_1=t_1b_1a_1e_1a_2^2e_1a_1b_1 \ \ \hbox{hence}\ \ st_1st_1=t_1st_1s$$ We have a sequence of transformations by (J). $\strut st_1s=$$\strut b_1a_1a_1b_1e_1a_1a_2e_1b_1a_1a_1b_1=b_1a_1a_1e_1b_1a_1b_1a_2e_1a_1a_1b_1=$ $\strut b_1a_1a_1e_1a_1b_1a_2a_1e_1a_1a_1b_1=b_1a_1e_1a_1e_1b_1a_2e_1e_1a_1e_1b_1=$ $\strut b_1a_1e_1a_1b_1a_2e_1a_2e_1a_1b_1e_1=b_1a_1e_1a_2a_1b_1a_2e_1a_1b_1a_2e_1=$ $\strut b_1a_1e_1a_2a_2e_1a_1b_1t_1.$ The second equality follows immediately by symmetry.Let $w_2=e_2a_2e_1a_1^2e_1a_2e_2$. $$\label{A7} (a_1e_1a_2)^4= t_1^2a_1^2a_2^2=d_{1,2}d_{-2,-1}, \ \ [d_{1,2},d_{-2,-1}]=1.$$ $d_{-2,-1}=w_2*d_{1,2}=w_2^{-1}*d_{1,2}=(b_1a_1e_1a_2)*b_2 $ We have $d_{-2,-1}=(s^{-1}t_1^{-1}s^{-1})*d_{1,2}=$ (by (\[A6\])) $\strut ((b_1a_1e_1a_2a_2e_1a_1b_1)^{-1}t_1^{-1} b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2=$ (by (J)) $\strut ((b_1a_1e_1a_2a_2e_1a_1b_1)^{-1}b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2=$ (by (\[A5\])) $(b_1a_1e_1a_2)*b_2$. Conjugating (\[A5\]) by $(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})$ we get $(a_1e_1a_2)^4=d_{1,2}d_{-2,-1}$. Also, by (M1), $t_1^2a_1^2a_2^2=(a_1e_1a_2)^4$. This proves the first relation. The second relation follows from it by (\[A2\]). Conjugating (\[A5\]) by $w_0^{-1}$ we get, by (\[A2\]), $(a_1e_1a_2)^4=d_{1,2}w_2d_{1,2}w_2^{-1}$ and $w_2*d_{1,2}=w_2^{-1}*d_{1,2}$. Therefore, from the first relation, $d_{-2,-1}=w_2*d_{1,2}=w_2^{-1}*d_{1,2}=(b_1a_1e_1a_2)*b_2 $. [If $A$ is a product of the generators we denote by $A^\prime$ the element obtained from $A$ by replacing each generator by its inverse. We call $A^\prime$ the element [*symmetric to*]{} $A$. ]{} \[symmetry\][Relations (M1) and (M2) are symmetric. They remain valid when we replace each generator by its inverse. Therefore every relation between some elements of $\cM_{g,1}$ (products of generators) which follows from (M1) and (M2) remains valid if we replace each element by the element symmetric to it.]{} $$\label{A8} \hbox{For} \ i+ j\neq 0 \ \ d_{i,j} \ \ \hbox{is symmetric to} \ d_{-j,-i}^{-1}.$$ Element $d_{1,2}$ is symmetric to $d_{-2,-1}^{-1}$, by (\[A7\]). Also $t_1^\prime=t_1^{-1}$ and $s^\prime = s^{-1}$. We see immediately that $d_{i,j}^\prime=d_{-j,-i}^{-1}$ for $i>0$. If $i<0$ and $i+j>0$ then $d_{i,j}=(t_{-i-1}^{-1}\dots t_1^{-1}s^{-1}t_{j-1}\dots t_2)*d_{1,2}$and $\strut d_{-j,-i}^{-1}=(t_{j-1}^{-1}\dots t_1^{-1}s^{-1}t_{-i} \dots t_2)*d_{1,2}^{-1}$. Jumping with the positive powers of $t_k$ to the left we get $d_{-j,-i}^{-1}=(t_{-i-1}\dots t_1st_{j-1}^{-1}\dots t_2^{-1}s^{-1}t_1^{-1}s^{-1})* d_{1,2}^{-1}=$$\strut(t_{-i-1}\dots t_1st_{j-1}^{-1}\dots t_2^{-1})*d_{1,2}^\prime= d_{i,j}^\prime$. $$\label{A9} d_{i+1,i+2}=(t_i^{-1}t_{i+1}^{-1})*d_{i,i+1}= (t_it_{i+1})*d_{i,i+1}\ \ \hbox{for}\ \ i=1,\dots,g-2.$$ For $i=1$ we have $(t_2t_1^2t_2)*d_{1,2}=$ (by \[A7\]) $\strut(e_2a_2a_3e_2e_1a_1a_2e_1e_1a_1a_2e_1e_2a_2a_3 e_2e_2^{-1}a_2^{-1}e_1^{-1}a_1^{-2} e_1^{-1}a_2^{-1}e_2^{-1})*d_{-2,-1}$$=$ (by (J)) $\strut (e_2a_2a_3e_2e_1a_1a_2e_1e_1a_2e_2a_3a_1^{-1}e_1^{-1}a_2^{-1}e_2^{-1}) *d_{-2,-1}=$ (by (J)) $\strut(e_2a_2a_3e_2e_1a_1a_2e_1e_1a_1^{-1}a_2e_1^{-1}e_2a_2^{-1}a_3e_2^{-1}) *d_{-2,-1}=$ (by (J)) $\strut(e_2a_2a_3e_2e_1a_2e_1^{-1}a_1a_1e_1a_2e_1^{-1}e_2a_2^{-1}a_3e_2^{-1}) *d_{-2,-1}=$ (by (J)) $\strut(a_3^{-1}e_2a_2e_1a_1^2e_1a_2e_2a_3)*d_{-2,-1}=$ (by (\[A3\]) and (\[A7\])) $a_3^{-1}*d_{1,2}=$ (by (\[A3\])) $d_{1,2}$.\ So (\[A9\]) is true for $i=1$. We continue by induction. Conjugating relation (\[A9\]) by $t_it_{i+1}t_{i+2}$ we get, by (\[new dij\]) and (\[ti,tj\]), relation (\[A9\]) for index $i+1$. $$\label{A10} t_i^2=d_{i,i+1}d_{-i-1,-i}a_i^{-2}a_{i+1}^{-2}.$$ The relation is true for $i=1$, by (\[A7\]). We proceed by induction.\ $t_{i+1}^2=(t_i^{-1}t_{i+1}^{-1})*t_i^2=(t_i^{-1}t_{i+1}^{-1})* (d_{i,i+1}d_{-i-1,-i}a_i^{-2}a_{i+1}^{-2})$$= $ (by (\[ti,tj\]), (\[A9\]), (\[new dij\]) and (\[A1\])) $\strut d_{i+1,i+2}d_{-i-2,-i-1}a_{i+1}^{-2}a_{i+2}^{-2}$. $$\label{A11} d_{-i,i} \ \ \hbox{is symmetric to} \ \ d_{-i,i}^{-1}.$$ The relation is true for $i=1$. The general case follows from (\[A10\]), (\[A1\]), (\[A2\]), (\[A8\]) and the definitions. $$\label{b_1,d22} [b_1,d_{-2,2}]=1$$ By the definition $d_{-2,2}=(t_1^{-1}d_{1,2})*(s^2a_1^4)=$ (by (\[A1\]))$\strut a_2^4((d_{1,2}t_1^{-1})*s^2)$.Now $\strut t_1^{-1}*s=t_1^{-1}st_1ss^{-1}=$ (by (\[A6\])) $ b_1a_1e_1a_2^2e_1a_1^{-1}b_1^{-1}$. Taking squares we get $t_1^{-1}*s^2=b_1a_1e_1a_2^2e_1^2a_2^2e_1a_1^{-1} b_1^{-1}$ and $\strut d_{-2,2}=a_2^4d_{1,2}b_1a_1e_1a_2^2e_1^2a_2^2e_1a_1^{-1} b_1^{-1}d_{1,2}^{-1}$. Now $b_1$ commutes with $d_{-2,2}$, by (\[A1\]), $\strut$(\[A2\]) and (J). $$\label{A12} (t_it_{i+1})*e_i=e_{i+1}\ \ \hbox{for}\ \ i=1,\dots,g-2.$$ We have, by (J), $(t_it_{i+1})*e_i=$$\strut(e_ia_ia_{i+1}e_ie_{i+1} a_{i+1}a_{i+2}e_{i+1})*e_i=$ $(e_ia_ia_{i+1}e_{i+1}e_i a_{i+1})*e_i=e_{i+1}$. $$\label{A13} [b_1,d_{i,i+1}]=1 \ if \ i>1,$$ $[e_k,d_{i,i+1}]=1 \ \ \hbox{if}\ \ |k-i|\neq 1, i>0,\ \ d_{i,i+1}*e_k=e_k^{-1}*d_{i,i+1} \ \ \hbox{if} \ \ |k-i|=1.$ By (\[A9\]) and (J) $\strut d_{2,3}=(e_1^{-1}a_2^{-1}e_2^{-1}a_3^{-1}a_1^{-1}e_1^{-1}a_2^{-1} e_2^{-1}b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2=$ $\strut(e_1^{-1}a_2^{-1}e_2^{-1}a_3^{-1}a_1^{-1}b_1^{-1}e_1^{-1}a_1^{-1}a_2^{-1} e_1^{-1}e_2^{-1}a_2^{-1})*b_2$. Now $b_1$ commutes with $d_{2,3}$, by (J). For $i>2$ we have $b_1$ commutes with $d_{i,i+1}$ by (M1) and (\[new dij\]). Conjugating by $a_1b_1t_1$ we get, by (\[ti,tj\]) and (J), $[e_1,d_{i,i+1}]=1$, for $i>2$. We also have $d_{1,2}*b_1=b_1^{-1} *d_{1,2}$, by (\[A2\]). We conjugate this equality by $u=a_1b_1t_1t_2$ and get $d_{2,3}*e_1=e_1^{-1}*d_{2,3}$, by (\[new dij\]) and the first part of the proof. Conjugating relations (\[A2\]) and the above relations by suitable products $t_it_{i+1}$ we get all remaining relations, by (\[A12\]). $$\label{A14} [b_1,d_{1,2}sd_{1,2}]=1, \ \ \hbox{hence} \ \ [s,d_{1,2}sd_{1,2}]=1,$$ $[e_j,d_{i,i+1}t_jd_{i,i+1}]=1, \ \ \hbox{hence} \ \ [t_j,d_{i,i+1}t_jd_{i,i+1}]=1 , \ \ \hbox{if} \ \ |i-j|=1.$ By (\[A2\]) we have $(d_{1,2}sd_{1,2})*b_1= (d_{1,2}b_1a_1a_1b_1b_1^{-1})*d_{1,2}=(d_{1,2}b_1)*d_{1,2}=b_1$. The other case is similar, but we use (\[A13\]) instead of (\[A2\]). [Relation $uvuv=vuvu$ implies $(uv)*u=v^{-1}*u$ and $(u^{-1}v^{-1})*u=v*u$. Relations (\[A14\]) will be often used in this form.]{} Observe that relations (\[A1\]) – (\[A10\]) imply in particular that relations (P1) and relations (P3) – (P7) follow from (M1) and (M2). We shall prove now that relations (P8) follow from (M1) and (M2). \[proper move\] [We say that homeomorphism $t_k$ (respectively $t_k^{-1}d_{k+1,k+2}$ or $s$) moves a curve $\delta_{i,j}$ properly if it takes it to some curve $\delta_{p,q}$.]{} \[t-conjugation\] If $t_k$ (respectively $s$) moves curve $\delta_{i,j}$ to some curve $\delta_{p,q}$ then $t_k*d_{i,j}=d_{p,q}$ (respectively $s*d_{i,j}=d_{p,q}$). [Since the action of $t_k$ and $s$ on a curve $\delta_{i,j}$ is described by Lemma \[action of tk\] and is easily determined, Lemma \[t-conjugation\] helps us to understand the result of the conjugation. In fact the action corresponds exactly to relations (P8), so we have to prove that relations (P8) follow from (M1) and (M2).]{} [**Proof of Lemma \[t-conjugation\]**]{}We know that $[t_i,d_{i,i+1}]=1$, by (\[new dij\]), (\[ti,tj\]), and (\[A2\]). If $i<0$ and $i+j=1$ then $t_{j-1}^{-1}*d_{i,j}= (t_{j-1}^{-1}t_{j-2}^{-1}\dots t_1^{-1}s^{-1}t_{j-1}\dots t_2)*d_{1,2}= d_{i-1,j-1}$. For $i>0$ or $i<0, \ i+j>0$ all other cases of conjugation by $t_k$ follow from (\[A2\]), (\[ti,tj\]) and the definitions. The other cases of $i\neq -j$ follow by symmetry.\ Consider conjugation by $s$ for $i\neq -j$. Again it suffices to consider $i>0$ or $i<0, \ i+j>0$. The other cases follow by symmetry. We have $s^{-1}*d_{1,j}=d_{-1,j}$.\ If $i>1$ then $d_{i,j}=$ (by (\[ti,tj\])) $(t_{i-1}\dots t_2t_{j-1}\dots t_3) *d_{2,3}$ and $s*d_{i,j}=d_{i,j}$, by (\[ti,tj\]), (\[A3\]) and (\[A13\]). If $i<1$ then $d_{i,j}=$ (by \[ti,tj\]) $(t_{-i-1}^{-1}\dots t_2^{-1}t_{j-1}\dots t_3)*d_{-2,3}$. Also $s^{-1}*d_{-2,3}=$ $(s^{-1}t_1^{-1}s^{-1}t_2)*d_{1,2}= (s^{-1}t_1^{-1}s^{-1}t_1^{-1})*d_{2,3}=$ (by (\[A6\])) $(t_1^{-1}s^{-1}t_1^{-1}s^{-1})*d_{2,3}= d_{-2,3}$ (by (\[A3\]) and (\[A13\])). So $s^{-1}*d_{i,j}=d_{i,j}$, by (\[ti,tj\]).\ We now consider conjugation of $d_{-j,j}$. Clearly $s*d_{-1,1}=d_{-1,1}$, by (M1). Also $s*d_{-2,2}=d_{-2,2}$, by (\[A3\]) and (\[b\_1,d22\]). For $i>1$ we have $[s,t_i]=1$, by (M1), hence $s*d_{-j,j}=d_{-j,j}$ for all $j$, by the first part of the proof.\ Consider conjugation by $t_k$.\ For $k>j$ we have $t_k*d_{-j,j}=d_{-j,j}$, by (\[ti,tj\]) and the first part.\ Curves $t_j(\delta_{-j,j})$ and $t_{j-1}(\delta_{-j,j})$ are not of the form $\delta_{p,q}$.\ Consider $k=j-2$ (the other cases follow by conjugation and by the first part of the proof). We have $\strut d_{-j,j}=(d_{j-1,j}t_{j-1}^{-1}t_{j-2}^{-1}d_{j-2,j-1})*d_{2-j,j-2}=$ (by (\[new dij\]) and (\[A9\])) $\strut (t_{j-2}^{-1}t_{j-1}^{-1}d_{j-2,j-1}t_{j-1}t_{j-2}t_{j-1} ^{-1}t_{j-2}^{-1}d_{j-2,j-1})*d_{2-j,j-2}=$ (by (\[ti,tj\]) and (\[new dij\])) $\strut (t_{j-2}^{-1}t_{j-1}^{-1}t_{j-2}^{-1}d_{j-2,j-1}t_{j-1} d_{j-2,j-1})*d_{2-j,j-2}$. Now, by (\[ti,tj\]) and (\[A14\]), $\strut t_{j-2}^{-1}*d_{-j,j}= (t_{j-2}^{-1}t_{j-1}^{-1}t_{j-2}^{-1}d_{j-2,j-1}t_{j-1} d_{j-2,j-1}t_{j-1}^{-1})*d_{2-j,j-2}=d_{-j,j}$ (by the previous case). We now pass to the biggest task of this section: the relations (P2). $$\label{A15} d_{i,j} \ \hbox{commutes with} \ d_{-1,1} \ \hbox{if} \ i,j\neq \pm1,$$ $ d_{i,j} \ \hbox{commutes with} \ d_{k,k+1}$ if all indices are distinct. We know by Lemma \[t-conjugation\] that $d_{i,j}$ commutes with $s$, hence also with $d_{-1,1}$. Consider the other cases. We assume first that $k>0$. Consider curves $\delta_{i,j}$ and $\delta_{k,k+1}$. We want to move the curves properly to some standard position by application of products of $s$ and $t_m$’s. This moves holes (see definition \[half-twist\]) and corresponds to conjugation of $d_{i,j}$ and $d_{k,k+1}$, by the previous lemma. [**Case 1**]{}$i\neq -j$Observe that for every $k$ either $t_{k+1}t_{k}$ or $t_{k+1}^{-1}t_k^{-1}$ moves $\delta_{i,j}$ properly. Both products take $\delta_{k+1,k+2}$ onto $\delta_{k,k+1}$ and conjugation of $d_{k+1,k+2}$ by either product produces $d_{k,k+1}$. If either $|i|$ or $|j|$ is bigger than $k+1$ we may assume, moving $\delta_{i,j}$ properly, and not moving $\delta _{k,k+1}$, that either $|i|$ or $|j|$ is equal to $k+2$. Then either $t_kt_{k+1}$ or $t_k^{-1}t_{k+1}^{-1}$ moves $\delta_{i,j}$ properly and moves $\delta_{k,k+1}$ to $\delta_{k+1,k+2}$ and leaves at most one index $|i|$ or $|j|$ bigger than $k+1$. Applying this procedure again, if necessary, we may assume $j<k$. If $j>0$ we can move $\delta_{i,j}$ properly, without moving $\delta_{k,k+1}$, and get a curve $\delta_{i,j}$ with $j<0$. Now applying consecutive products $t_{m+1}t_{m}$ or $t_{m+1}^{-1}t_m^{-1}$ we reach $k=1$. Further moves will produce one of the following three cases: [**Case 1a**]{}$d_{1,2}$ commutes with $d_{-2,-1}$.True, by (\[A7\]). [**Case 1b**]{}$d_{1,2}$ commutes with $d_{-3,-1}$ or $d_{-3,-2}$.We can conjugate $d_{-3,-2}$ by $t_1$ and get $d_{-3,-1}$. Now $d_{-3,-1}=t_2^{-1}*d_{-2,-1}=$ (by (\[A7\])) $(e_2^{-1}a_3^{-1} a_2^{-1}e_2^{-1}w_2)*d_{1,2}=$ (by (J) and (\[A2\])) $(e_2^{-1}a_3^{-1}e_1a_1d_{1,2}^{-1}e_2^{-1}a_2^{-1}e_1^{-1})*a_1=$ (by (J) and (\[A2\]))$\strut(e_2^{-1}d_{1,2}^{-1}a_3^{-1}e_2^{-1}e_1a_1a_2^{-1} e_1^{-1})*a_1$. The last expression commutes with $d_{1,2}$, by (J) and (\[A2\]). [**Case 1c**]{}$d_{-2,-1}$ commutes with $d_{3,4}$.The proof is rather long. We consider relation (M3): $d_3=a_3^{-1}a_2^{-1}a_1^{-1} d_{1,2}d_{1,3}d_{2,3}$, where $\strut d_3=(b_1^{-1}b_2a_2e_1e_2a_2a_3e_2b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1}) *b_2=$ $\strut(b_1^{-1}b_2a_2e_1e_2a_2a_3e_2b_2a_2e_1a_1)*b_1$. It follows, by (J), that $d_3$ commutes with $a_2e_1e_2a_2a_3e_2$, hence $d_3=(b_1^{-1}((a_2e_1e_2a_2a_3e_2)^{-1}*b_2) b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2$. We now conjugate relation (M3) by $u=(a_4e_3a_3e_2a_2e_1a_1b_1)^{-1}a_3e_2a_2e_1a_1b_1$. When we write $d_{1,2}=(b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2$ we see that $d_{1,2}$ commutes with $u$, by (J). All other factors on the RHS commute with $u$ by (J), so we may replace $d_3$ in the relation (M3) by $u*d_3=$ $\strut(a_4e_3a_3e_2a_2e_1a_1b_1)^{-1}* ((a_3e_2a_2e_1a_1((a_2e_1e_2a_2a_3e_2)^{-1}*b_2) b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2)$. We conjugate each term by $(a_4e_3a_3e_2a_2e_1a_1b_1)^{-1}$ and get$\strut u*d_3=(e_3a_3e_2a_2e_1((e_2a_2a_3e_2e_3a_3)^{-1}*d_{1,2}) a_1^{-1}e_1^{-1}a_2^{-1}e_2^{-1})*d_{1,2}$. We now conjugate each term of relation (M3) by $t_2^{-1}t_3^{-1}=$$\strut e_2^{-1}a_3^{-1}e_3^{-1}a_4^{-1} a_2^{-1}e_2^{-1}a_3^{-1}e_3^{-1}$. The RHS becomes $a_4^{-1}a_3^{-1}a_1^{-1} (t_2^{-1}*d_{1,2}) ((t_2^{-1}t_3^{-1}t_2)*d_{1,2})d_{3,4}$. The LHS becomes$\strut(e_2^{-1}a_3^{-1}e_3^{-1}a_4^{-1})* ((e_1((e_2a_2a_3e_2e_3a_3)^{-1}*d_{1,2}) (a_1^{-1}e_1^{-1}a_2^{-1}e_2^{-1}))*d_{1,2})$. We conjugate each bracket.$\strut(e_2^{-1}a_3^{-1}e_3^{-1}a_4^{-1})*e_1=e_1$.$\strut(e_2^{-1}a_3^{-1}e_3^{-1}a_4^{-1}(e_2a_2a_3e_2e_3a_3)^{-1})*d_{1,2} =$ (by (J) and (\[A2\])) $(t_3^{-1}t_2^{-1})*d_{1,2}$. $\strut(e_2^{-1}a_3^{-1}e_3^{-1}a_4^{-1}a_1^{-1}e_1^{-1}a_2^{-1}e_2^{-1}) *d_{1,2}=$ (by (J) and (\[A2\])) $(a_1^{-1}e_1^{-1}t_2^{-1})*d_{1,2}$. We shall prove that all terms of the obtained equation commute with $d_{-2,-1}$, except possibly for $d_{3,4}$. Therefore $d_{3,4}$ also commute with $d_{-2,-1}$. Clearly $a_1,e_1,a_2,a_3,a_4,t_1,t_3$ commute with $d_{-2,-1}$ by (\[A2\]) and symmetry. Also $d_{1,2}$ commutes with $d_{-2,-1}$ by Case 1a and $d_{1,3}$ and $d_{2,3}$ commute with $d_{-2,-1}$ by Case 1b. Therefore $t_2^{-1}*d_{1,2}=t_1*d_{2,3}$ commute with $d_{-2,-1}$ and $(t_2^{-1}t_3^{-1}t_2)*d_{1,2}=$ (by (\[ti,tj\]) and (\[A2\])) $(t_3t_2^{-1})*d_{1,2}$ commutes with $d_{-2,-1}$. It follows that $d_{3,4}$ commutes with $d_{-2,-1}$ [**Case 2**]{}$i=-j$ We have to prove that $d_{k,k+1}$ commutes with $d_{-j,j}$ if $j\neq k,k+1$. If $j<k$ then the result follows by Lemma \[t-conjugation\] and by Case 1. If $j>k+2$ we can properly move $\delta_{k,k+1}$ to $\delta_{j-2,j-1}$, without moving $\delta_{-j,j}$. So we may assume $j=k+2$. We have $\strut d_{-j,j}=(d_{k+1,k+2}t_{k+1}^{-1} t_k^{-1}d_{k,k+1})*d_{-k,k}=$ (by (\[A3\]) and (\[A10\]))$\strut(d_{-k-2,-k-1}^{-1}t_{k+1}t_kd_{-k-1,-k}^{-1})*$ $d_{-k,k}$. We conjugate $d_{k,k+1}$ by $(d_{-k-2,-k-1}^{-1}t_{k+1}t_kd_{-k-1,-k}^{-1})^{-1}$ and get, by Case 1 and (\[A9\]), $(d_{-1-k,-k}t_{k}^{-1}t_{k+1}^{-1}d_{-k-2,-k-1})*d_{k,k+1}= d_{-1-k,-k}*d_{k+1,k+2}= d_{k+1,k+2}$, which commutes with $d_{-k,k}$ by the first part of Case 2. Suppose now that $k<0$. Cases 1a and 1b follow by symmetry. In the Case 1c we arrive by symmetry at the situation where we have to prove that $d_{1,2}$ commutes with $d_{-4,-3}$. Conjugating by $t_2t_1t_3t_2$ we get a pair $d_{3,4}$, $d_{-2,-1}$, which commutes by Case 1c. Now Case 2 follows by symmetry. \[td-conjugation\] If $t_k^{-1}d_{k,k+1}$ takes a curve $\delta_{i,j}$ to $\delta_{p,q}$ then $t_k^{-1}d_{k,k+1}*d_{i,j}= d_{p,q}$. We shall list the relevant cases. If $i$ or $j$ is equal to $k$ it becomes $k+1$, if $i$ or $j$ is equal to $-k$ it becomes $-k-1$. In particular $t_k^{-1}d_{k,k+1}(\delta_{-k,k})=\delta_{-k-1,k+1}$. Indices $k+1$ and $-k-1$ are forbidden, they do not move properly, except for $\delta_{k,k+1}$ and $\delta_{-k-1,-k}$ which are fixed by $t_k^{-1}d_{k,k+1}$. Other indices $i$, $j$ do not change.\ We now pass to the proof of the Lemma. If $i=-j$ the result follows from the definitions and from (\[A15\]) and Lemma \[t-conjugation\]. Suppose $i\neq -j$. If $i$ and $j$ are different from $k$ then $d_{i,j}$ commutes with $d_{k,k+1}$, by (\[A15\]), and $t_k^{-1}$ moves $\delta_{i,j}$ properly, so we are done by Lemma \[t-conjugation\]. If $i$ and $j$ are different from $-k$ we can replace $t_k^{-1}d_{k,k+1}$ by $t_kd_{-k-1,-k}^{-1}a_k^2a_{k+1}^2$, using (\[A10\]), and we are done by a similar argument. Lemmas \[t-conjugation\] and \[td-conjugation\] allow us to reduce relations (P2) to relatively small number of cases. We can apply product of half-twists $h_{k,k+1}$ and $h_{-k-1,-k}$ either in the same direction, conjugating by $t_k$, or in opposite directions, conjugating by $t_k^{-1}d_{k,k+1}$, and move properly curves corresponding to elements $d_{p,q}$ in the relations (P2) into a small number of standard configurations. $$\label{A16} d_{i,j}\ \ \hbox{commutes with}\ \ d_{r,s}\ \ \hbox{if}\ \ r<s<i<j\ \ \hbox{or}\ \ i<r<s<j.$$ Moving curves $\delta_{i,j}$ and $\delta_{r,s}$ properly we can arrive at a situation $s=r+1$ or $j=i+1$ or $-r=s=1$ or $-i=j=1$. In particular if $i<r<s<j$ and $r=-s$ then conjugating by $t_2d_{2,3}^{-1}\dots t_{s-1}d_{1-s,s-1}^{-1}$ we get a pair $d_{-1,1}$, $d_{i,j}$. Then (\[A16\]) follows from (\[A15\]). $$\label{A17} d_{r,i}^{-1}*d_{i,j}=d_{r,j}*d_{i,j}\ \ \hbox{if}\ \ r<i<j.$$ Indices $(i,-i)$ move together (either remain fixed or move to $(i-1,1-i)$ or to $(i+1,-i-1)$) when we conjugate by $t_k$ or $s$ or $t_k^{-1}d_{k,k+1}$. Moving curves properly we can arrive at one of the following four cases depending on the pairs of opposite indices. [**Case 1**]{}There is no pair of opposite numbers among $r,i,j$.We may assume that $(r,i,j)=(1,2,3)$.\ $d_{1,2}^{-1}*d_{2,3}=$ (by (\[A9\])) $(d_{1,2}^{-1}t_1^{-1}t_2^{-1})*d_{1,2}=$ (by (\[A2\]) and (\[A14\])) $(t_1^{-1}t_2)*d_{1,2}$.$\strut d_{1,3}*d_{2,3}=(t_2d_{1,2}t_2^{-1}t_1t_2)*d_{1,2}=$ (by (\[A2\]) and (\[ti,tj\])) $(t_2t_1d_{1,2}t_2)*d_{1,2}=$ (by (\[A14\])) $(t_2t_1t_2^{-1})*d_{1,2} =$ (by (\[A2\]) and (\[ti,tj\])) $(t_1^{-1}t_2)*d_{1,2}$. [**Case 2**]{}$i=-r$We may assume that $(r,i,j)=(-1,1,2)$.\ $d_{-1,1}^{-1}*d_{1,2}=(s^{-2}a_1^{-4})*d_{1,2}=$ (by (\[A14\])) $(s^{-1}d_{1,2}s)*d_{1,2}=d_{-1,2}*d_{1,2}$. [**Case 3**]{}$r=-j$We may assume that $(r,i,j)=(-2,1,2)$. We have to prove that $d_{-2,1}^{-1}*d_{1,2}=d_{-2,2}*d_{1,2}$. We conjugte by $d_{1,2}^{-1}t_1$. The right hand side becomes $s^2*d_{1,2}$ and the left hand side becomes $(d_{1,2}^{-1}t_1t_1^{-1}s^{-1}d_{1,2}^{-1}st_1)*d_{1,2}=$ (by (\[A2\]) and (\[A14\])) $(sd_{1,2}^{-1}s^{-1}d_{1,2}^{-1})*d_{1,2}=$ (by (\[A14\])) $s^2*d_{1,2}$. [**Case 4**]{}$i=-j$We may assume that $(r,i,j)=(-2,-1,1)$.\ $d_{-2,-1}^{-1}*d_{-1,1}=a_1^4(d_{-2,-1}^{-1}*s^2)=$ (by (\[A14\]) and symmetry) $a_1^4((sd_{-2,-1}s^{-1})*s^2)=d_{-2,1}*d_{-1,1}$. $$\label{A18} d_{r,j}^{-1}*d_{r,i}=d_{i,j}*d_{r,i} \ \ \hbox{if}\ \ r<i<j.$$ Let us apply symmetry to relation (\[A17\]). We get$\strut d_{-i,-r}*d_{-j,-i}^{-1}=d_{-j,-r}^{-1}*d_{-j,-i}^{-1}$ if $-j<-i<-r$. This is relation (\[A18\]) after a suitable change of indices. $$\label{A19} [d_{i,j},d_{r,j}^{-1}*d_{r,s}]=1 \ \ \hbox{if}\ \ r<i<s<j.$$ Again we have to consider different cases depending on pairs of opposite indices. For each of them we move curves properly to some standard position. If we apply symmetry we get $[d_{-j,-i},d_{-j,-r}*d_{-s,-r}]=1$. Now conjugate by $d_{-j,-r}^{-1}$ and get $[d_{-s,-r},d_{-j,-r}^{-1}*d_{-j,-i}]=1$ if $-j<-s<-i<-r$. This is again relation (\[A19\]) with different pairs of opposite indices. So $i=-j$ is equivalent to $r=-s$ and $s=-j$ is equivalent to $r=-i$. We are left with the following five cases. [**Case 1**]{}There is no pair of opposite numbers among $r,i,s,j$.We may assume that $(r,i,s,j)=(1,2,3,4)$. Conjugate by $t_3^{-1}$. $\strut t_3^{-1}*d_{2,4}=d_{2,3}$.$\strut(t_3^{-1}d_{1,4}^{-1})*d_{1,3}=(t_2d_{1,2}^{-1}t_2^{-1}t_3^{-1}t_2) *d_{1,2}=$ (by (\[ti,tj\])) $(t_2d_{1,2}^{-1}t_3t_2^{-1}t_3^{-1})*d_{1,2}=$ (by (\[A2\])) $(t_2t_3d_{1,2}^{-1}t_2^{-1})*d_{1,2}=$ (by (\[A14\])) $(t_2t_3t_2)* d_{1,2}=d_{1,4}$, and it commutes with $d_{2,3}$, by (\[A16\]). [**Case 2**]{}$r=-i$We may assume that $(r,i,s,j)=(-1,1,2,3)$. We conjugate by $t_2^{-1}$.$\strut t_2^{-1}*d_{1,3}=d_{1,2}$.$\strut(t_2^{-1}d_{-1,3}^{-1})*d_{-1,2}=(t_2^{-1}s^{-1}t_2d_{1,2}^{-1}t_2^{-1}) *d_{1,2}=$ (by (\[ti,tj\]) and (\[A14\])) $(s^{-1}t_2)*d_{1,2}=d_{-1,3}$, and it commutes with $d_{1,2}$, by (\[A16\]). [**Case 3**]{}$r=-s$We may assume that $(r,i,s,j)=(-2,1,2,3)$. By (\[A18\]) we have $d_{-2,3}^{-1}*d_{-2,2}=d_{2,3}*d_{-2,2}$, so we have to prove that $d_{1,3}$ commutes with $d_{2,3}*d_{-2,2}$. We conjugate by $t_2^{-1}$ and get$\strut t_2^{-1}*d_{1,3}=d_{1,2}$.$\strut(t_2^{-1}d_{2,3})*d_{-2,2}=d_{-3,3}$, and it commutes with $d_{1,2}$, by (\[A16\]). [**Case 4**]{}$r=-j$ and $i\neq -s$ We may assume that $(r,i,s,j)=(-3,1,2,3)$. By (\[A18\]) we have $d_{-3,3}^{-1}*d_{-3,2}=d_{2,3}*d_{-3,2}$. After conjugation by $t_1^{-1}d_{2,3}^{-1}$ we have to prove that $t_1^{-1}*d_{-3,2}=d_{-3,1}$ commutes with $(t_1^{-1}d_{2,3}^{-1})*d_{1,3}=(t_1^{-1}d_{2,3}^{-1}t_1^{-1})*d_{2,3}=$ (by (\[A14\])) $d_{2,3}$. This is true by (\[A16\]). [**Case 5**]{}$i=-s$We may assume that $(r,i,s,j)=(-2,-1,1,m)$, where $m=2$ or $m=3$. By (\[A18\]) we have $d_{-2,m}^{-1}*d_{-2,1}=d_{1,m}*d_{-2,1}$. After conjugation by $s^{-1}d_{1,m}^{-1}$ we have to consider $s^{-1}*d_{-2,1}=d_{-2,-1}$ and $(s^{-1}d_{1,m}^{-1})*d_{-1,m}$. For $m=2$ the last expression is equal $(s^{-1}d_{1,2}^{-1}s^{-1})*d_{1,2}= d_{1,2}$, by (\[A14\]). For $m=3$ we get $(s^{-1}d_{1,3}^{-1}s^{-1})*d_{1,3}= (s^{-1}t_2d_{1,2}^{-1}t_2^{-1}t_2s^{-1})*d_{1,2}=$ (by (\[A14\])) $d_{1,3}$. Both elements commute with $d_{-2,-1}$, by (\[A16\]). This concludes the proof of the fact that relations (P2) follow from (M1) – (M3). We now pass to the relations (P9) – (P11).\ Consider first relations (P9). Clearly $a_1$ commutes with all the elements in (P9), by (\[A1\]) and (\[A3\]), so it suffices to prove that $b_1$ commutes with these elements. It commutes with $a_1^2s$, $t_1st_1$, $a_2$, and $t_i$, for $i>1$, by (J). Also $b_1$ commutes with $d_{-2,2}$, by (\[b\_1,d22\]), and commutes with $d_{2,3}$ by (\[A13\]). Finally it commutes with $d_{1,2}sd_{1,2}$, by (\[A14\]), hence also commutes with $d_{-1,1}d_{-1,2}d_{1,2}a_1^{-2}a_2^{-1}= a_1^4s^2s^{-1}d_{1,2}sd_{1,2}a_1^{-2}a_2^{-1}=a_1^2sd_{1,2}sd_{1,2}a_2^{-1}$. This proves relations (P9). Relation (P10) follows from the definitions and (M1). We now pass to relations (P11). For $i=1,2,3,4$ we have $g_i=(k_ir)^3$. By (\[A3\]) $k_i*a_1=a_1$ therefore it suffices to prove that $k_i*b_1=b_1^{-1}*k_i$. Then $\strut g_i=(k_ia_1b_1a_1)^3=$ (by (\[A3\])) $k_ia_1b_1a_1^2k_ib_1k_ia_1^2b_1a_1=k_ia_1sk_isa_1$ and this is exactly relation (P11) for $i=1,2,3,4$. $$k_1*b_1=b_1^{-1}*k_1.$$ Since $k_1=a_1$ the result follows from (M1). $$k_2*b_1=b_1^{-1}*k_2.$$ Since $k_2=d_{1,2}$ the result follows from (\[A2\]). $$k_3*b_1=b_1^{-1}*k_3.$$ We have $\strut k_3=a_1^{-1}a_2^{-2}d_{1,2}d_{-2,1}d_{-2,2}= a_1^{-1}a_2^{-2}d_{1,2}t_1^{-1}s^{-1}d_{1,2}st_1t_1^{-1} d_{1,2}s^2a_1^4d_{1,2}^{-1}t_1=$(by (\[A2\]) and (J)) $\strut a_1^{-1}t_1^{-1}d_{1,2}s^{-1}d_{1,2}sd_{1,2}s^2 a_1^2d_{1,2}^{-1}t_1=$ (by (\[A14\])) $\strut a_1^{-1}t_1^{-1}d_{1,2}sd_{1,2}sa_1^2t_1=$ (by (\[A2\]) and (J)) $a_1^{-1}t_1^{-1}(d_{1,2}b_1a_1)^4t_1$. Let $u=t_1b_1d_{1,2}a_1b_1$. It follows from (\[A2\]) and (J) that $u*a_1=d_{1,2}$, $u*e_1=b_1$, $u*a_2=a_1$, $u*d_{1,2}=a_2$. Conjugating (\[A7\]) by $u$ we get $(d_{1,2}b_1a_1)^4=a_2(u*d_{-2,-1})$, hence $k_3=$ (by (\[A1\])) $(t_1^{-1}u)*d_{-2,-1}=(b_1d_{1,2}a_1b_1)*d_{-2,-1}$. We want to prove that $b_1$ and $k_3$ are braided. It suffices to prove it for their common conjugates. We conjugate by $b_1^{-1}$ and get $b_1$ and $(d_{1,2}a_1b_1)*d_{-2,-1}$. Now we conjugate by $a_1^{-1}b_1^{-1}d_{1,2}^{-1}$ and get, by (\[A2\]), (\[A3\]) and symmetry, $d_{1,2}$ and $b_1*d_{-2,-1}=d_{-2,-1}^{-1}*b_1$. Now we conjugate by $d_{-2,-1}$ and get, by (\[A15\]), $d_{1,2}$ and $b_1$, which are braided. $$k_4*b_1=b_1^{-1}*k_4 \$$ By the definition and by (M3) we have $k_4=d_3=(b_2a_2e_1b_1^{-1})*d_{1,3}$. Conjugating $k_4$ and $b_1$ by $t_2^{-1}b_1e_1^{-1}a_2^{-1}b_2^{-1}$ we get $d_{1,2}$ and $b_1$, which are braided, by (\[A2\]). $$g_5=sa_1^2k_5sa_1^2k_5^{-1}$$ By the definition $g_5=(rk_5rk_5^{-1})^2$ where $k_5=a_2d_{1,2}^{-1}t_1$. We shall prove that $r$ commutes with $k_5rk_5^{-1}$. Then $g_5=r^2k_5r^2k_5^{-1}= sa_1^2k_5sa_1^2k_5^{-1}$, as required. $k_5rk_5^{-1}=a_2d_{1,2}^{-1}e_1a_1a_2e_1a_1b_1a_1e_1^{-1}a_1^{-1}a_2^{-1} e_1^{-1}d_{1,2}a_2^{-1}=$ (by (J) and (\[A2\]))$\strut a_2^2d_{1,2}^{-1}e_1a_1b_1a_1^{-1}e_1^{-1}d_{1,2}=$ (by (J))$\strut a_2^2d_{1,2}^{-1}b_1^{-1}a_1^{-1}e_1a_1b_1d_{1,2}$. The last expression commutes with $a_1$ and $b_1$, by (J) and (\[A2\])). $$g_6=(sa_1^2t_1)^4$$ By the definition $g_6=(ra_1t_1)^5$. The required relation is proved by a rather long computation, using (J). Observe first that $sa_1^2=(b_1a_1)^3$, and that $ra_1=(b_1a_1)^2$. We also have $\strut t_1(b_1a_1)^2t_1=e_1a_1a_2e_1b_1a_1b_1a_1e_1a_1a_2e_1=$$\strut e_1a_1b_1a_2e_1a_1b_1a_1e_1a_1a_2e_1=b_1e_1a_1b_1a_2e_1a_1b_1a_1e_1a_2e_1=$ $\strut b_1a_1e_1a_1b_1a_2e_1a_1b_1a_1e_1a_2=b_1a_1e_1a_1a_2e_1b_1a_1b_1a_1e_1a_2=$ $(b_1a_1)t_1(b_1a_1)^2e_1a_2$\ and $e_1a_2(b_1a_1)^2t_1=b_1e_1a_1a_2b_1a_1e_1a_1a_2e_1= b_1e_1a_1a_2e_1b_1a_1e_1a_2e_1=$$\strut b_1e_1a_1a_2e_1b_1a_2a_1e_1a_2= b_1a_1e_1a_1a_2e_1b_1a_1e_1a_2= (b_1a_1)t_1(b_1a_1)e_1a_2=$$\strut b_1a_1e_1a_1b_1a_2e_1a_1e_1a_2=b_1a_1b_1e_1a_1b_1a_2e_1a_1a_2= b_1a_1b_1a_1e_1a_1b_1a_2e_1a_1=$$\strut (b_1a_1)^2t_1(b_1a_1)$. The required result follows from the above relations. This concludes the proof of Theorem \[simple presentation\]. Mapping class group of a closed surface ======================================= We shall consider in this section the mapping class group $\cM_g$ of a closed surface $S_{g,0}$ of genus $g>1$. We shall keep the notation from the previous section. In particular ${\mathcal M}_{g,1}$ is the mapping class group of $S=S_{g,1}$ and $S_{g,0}$ is obtained from $S$ by capping the boundary $\partial$ of $S$ by a disk $D$ with a distinguished center $p$, and then forgetting $p$. We have two exact sequences $$1\to\ints\mapright{\psi} \cM_{g,1}\mapright{\phi}\cM_{g,0,1}\to 1$$ $$1\to \pi_1(S_{g,0,1},p)\mapright{\sigma} \cM_{g,0,1} \mapright{e_*} \cM_{g,0,0}\to 1$$ In the first sequence the Dehn twist $\Delta=T_\partial$ belongs to the kernel of $\phi$. We shall prove now that it generates the kernel. When we split the surface $S_{g,1}$ open along the curves $\beta_1,\alpha_1,\epsilon_1,\alpha_2, \dots,\epsilon_{g-1},\alpha_g$ we get an annulus $N$, and one boundary of $N$ is equal to $\partial$. If $h\in ker(\phi)$ than $h$ takes each curve $\gamma\in\{\beta_1,\alpha_1,\epsilon_1,\alpha_2, \dots,\epsilon_{g-1},\alpha_g\}$ onto a curve $h(\gamma)$ which is isotopic to $\gamma$ in $S_{g,0}$ by an isotopy fixed on $p$. Therefore $\gamma$ and $h(\gamma)$ form 2–gons which are disjoint from $p$ and hence from $D$. It follows that $h$ is isotopic in $S_{g,1}$ to a homeomorphism equal to the identity on all curves $\beta_1,\alpha_1,\epsilon_1,\alpha_2, \dots,\epsilon_{g-1},\alpha_g$. But then it is a homeomorphism of the annulus $N$ so it is isotopic to a power of $\Delta$. The second sequence is described in [@Birman], Theorem 4.3. The kernel of $e_*$ is generated by spin-maps $T_{\gamma^\prime}T_\gamma^{-1}$, where $\gamma$ and $\gamma^\prime$ are simple, nonseparating curves which bound an annulus on $S_{g,0,1}$ containing the distinguished point $p$. The composition $e_*\phi$ is an epimorphism from the group $\cM_{g,1}$ onto the group $\cM_g=\cM_{g,0,0}$ and its kernel is generated by $\Delta$ and by the spin maps $T_{\gamma^\prime}T_\gamma^{-1}$, where $\gamma$ and $\gamma^\prime$ are simple, nonseparating curves on $S$ which bound an annulus with a hole bounded by $\partial$. Clearly all such annuli are equivalent by a homeomorphism of $S$, hence all spin maps are conjugate in ${\mathcal M}_{g,1}$. It suffices to consider one spin map $T_{\delta_g^\prime}T_{\delta_g}^{-1}$, where $\delta_g$ and $\delta_g^\prime$ are curves on Figure \[spin maps\]. $T_{\delta_g}$ is equal to the element $d_g$ in the relation (M4). Let $w=b_1a_1e_1a_2\dots a_{g-1}e_{g-1}a_g^2e_{g-1}a_{g-1}\dots a_2e_1a_1b_1$. It is easy to check, drawing pictures, that $w(\delta_g)=\delta_g^\prime$. Therefore, by Lemma \[conjugation\], relation (M4) is equivalent, modulo relations in ${\mathcal M}_{g,1}$, to $T_{\delta_g}=T_{\delta_g^\prime}$. By the above argument $\cM_g$ has a presentation with relations (M1) – (M4) and relation $\Delta=1$. We have to prove that the last relation follows from the others. Let $\cM^\prime=({\mathcal M}_{g,1})/(M4)$. Let $d_g$, $d_g^\prime$, $d_{g+1}$, $c$, $c^\prime$ be twists along curves $\delta_g$, $\delta_g^\prime$, $\delta_{g+1}$, $\gamma$, $\gamma^\prime$ respectively, depicted on Figure \[spin maps\]. Each element of ${\mathcal M}_{g,1}$ represents an element in $\cM^\prime$ which we denote by the same symbol. From now on, till the end of this section, symbols denote elements in $\cM^\prime$. We want to prove that $\Delta=1$. All relations from Lemma \[basic relations\] are true in $\cM^\prime$. We have $d_g*b_1=b_1^{-1}*d_g$ and $d_g$ commutes with every $a_i$ and $e_i$. By Lemma \[basic relations\], (iii) and (v) we have: $\Delta=(a_ge_{g-1}a_{g-1}\dots e_1a_1b_1d_g)^{2g+2}=$$\strut(a_ge_{g-1}a_{g-1}\dots e_1a_1)^{2g}(b_1a_1\dots a_ga_g\dots a_1b_1) (d_gb_1a_1\dots a_ga_g\dots a_1b_1d_g)$,$\strut d_gd_g^\prime=(a_ge_{g-1}a_{g-1}\dots e_1a_1)^{2g}=$ $\strut(a_ge_{g-1}a_{g-1}\dots e_2a_2)^{2g-2} (e_1a_2\dots a_ga_g\dots a_2e_1) (a_1e_1a_2\dots a_ga_g\dots a_2e_1a_1)$,$\strut(a_ge_{g-1}a_{g-1}\dots e_2a_2)^{2g-2}=cc^\prime$,$\strut(d_gb_1a_1)^4=cd_{g+1}$. We also see that $c^\prime d_{g+1}^{-1}$ and $d_g^\prime d_g^{-1}$ are spin maps, hence $c^\prime=d_{g+1}$ and $d_g=d_g^\prime$. Therefore $(a_ge_{g-1}a_{g-1}\dots e_2a_2)^{2g-2}=(d_gb_1a_1)^4$,$\strut d_g^2=(a_ge_{g-1}a_{g-1}\dots e_2a_2)^{2g-2} (e_1a_2\dots a_ga_g\dots a_2e_1) (a_1e_1a_2\dots a_ga_g\dots a_2e_1a_1)$, hence $(a_1e_1a_2\dots a_ga_g\dots a_2e_1a_1)=(e_1a_2\dots a_ga_g\dots a_2e_1)^{-1} (d_gb_1a_1)^{-4}d_g^2$.\ Further $\Delta=(a_ge_{g-1}a_{g-1}\dots e_1a_1)^{2g}(b_1a_1\dots a_ga_g\dots a_1b_1) (d_gb_1a_1\dots a_ga_g\dots a_1b_1d_g)=$ $\strut d_g^2b_1(a_1e_1a_2\dots a_ga_g\dots a_2e_1a_1)b_1 (d_gb_1a_1\dots a_ga_g\dots a_1b_1d_g)=$ $\strut d_g^2b_1(e_1a_2\dots a_ga_g\dots a_2e_1)^{-1} (d_gb_1a_1)^{-4}d_g^2b_1(d_gb_1a_1\dots a_ga_g\dots a_1b_1d_g)$. Now $(d_gb_1a_1\dots a_ga_g\dots a_1b_1d_g)$ commutes with $d_g$, by (M4), and commutes with $b_1$ and $a_1$, by (J), hence $\Delta=$$\strut =d_g^2b_1a_1b_1d_g(a_1^{-1}b_1^{-1}d_g^{-1})^3d_g^2b_1=1$, by (J). This concludes the proof of Theorem \[presentation closed\]. Equivalence of presentations ============================ In this section we shall prove that the presentations of $\cM_{g,1}$ in Theorems 1 and $1^\prime$ are equivalent. The relations (M1) coincide with the relations (A). It follows from relations (A) that $b_2$ commutes with the left hand side of the relation (B). Thus (B) is equivalent to $(b_2a_2e_1a_1b_1^2a_1e_1a_2b_2) (a_2e_1a_1b_1^2a_1e_1a_2)^{-1}=(b_1a_1e_1)^4$. Multiplying by $(a_2e_1a_1b_1^2a_1e_1a_2)$ on the right we get $(b_2a_2e_1a_1b_1^2a_1e_1a_2b_2)=(b_1a_1e_1)^4(a_2e_1a_1b_1^2a_1e_1a_2)= (b_1a_1e_1a_2)^5$, as in the proof of Lemma \[basic relations\] (v), and we get a relation identical with (M2). We now pass to relation (M3). We shall transform it using relations (M1) and (M2) and then we shall conjugate it by $w=a_3e_2a_2e_1a_1b_1$ to get the relation (C). Since (M1)=(A) and (M2) is equivalent to (B) in presence of (M1), it will prove that (M3) is equivalent to (C) in presence of (M1) and (M2). It follows from (M1) and the definitions that each factor on the right hand side of (M3) commutes with $a_1a_2a_3$, therefore $d_3$ also commutes with $a_1a_2a_3$. Recall the relations (\[A2\]), (\[ti,tj\]) and (\[A14\]) from section 4, which follow from the relations (M1) and (M2). (\[A2\]) $d_{1,2}t_1=t_1d_{1,2}$, (\[ti,tj\]) $t_1t_2t_1=t_2t_1t_2$, (\[A14\]) $t_2d_{1,2}t_2d_{1,2}=d_{1,2}t_2d_{1,2}t_2$. We now have $d_{1,2}d_{1,3}d_{2,3}= d_{1,2}t_2d_{1,2}t_2^{-1}t_1t_2d_{1,2}t_2^{-1}t_1^{-1}=$ (by \[ti,tj\]) $\strut d_{1,2}t_2d_{1,2}t_1t_2t_1^{-1}d_{1,2}t_2^{-1}t_1^{-1}=$ (by \[A2\]) $d_{1,2}t_2t_1d_{1,2}t_2d_{1,2}t_1^{-1}t_2^{-1}t_1^{-1}=$ (by \[ti,tj\]) $\strut d_{1,2}t_2t_1d_{1,2}t_2d_{1,2}t_2^{-1}t_1^{-1}t_2^{-1}=$ (by \[A14\]) $d_{1,2}t_2t_1t_2^{-1}d_{1,2}t_2d_{1,2}t_1^{-1}t_2^{-1}=$ (by \[ti,tj\] and \[A2\]) $\strut t_1^{-1}d_{1,2}t_2t_1d_{1,2}t_2d_{1,2}t_1^{-1}t_2^{-1}=$ (by \[A2\]) $t_1^{-1}d_{1,2}t_2d_{1,2}t_1t_2t_1^{-1}d_{1,2}t_2^{-1}=$ (by \[ti,tj\] and \[A14\]) $\strut t_1^{-1}t_2^{-1}d_{1,2}t_2d_{1,2}t_1t_2d_{1,2}t_2^{-1}=$ (by \[A2\] and \[A14\]) $t_1^{-1}t_2^{-1}d_{1,2}t_2t_1t_2^{-1}d_{1,2}t_2d_{1,2}$. We now conjugate everything by $w$ and, using (M1), we get $w*a_1=b_1$, $w*e_1=a_1$, $w*a_2=e_1$, $w*e_2=a_2$, $w*a_3=e_2$, $\strut w*t_1=a_1b_1e_1a_1=\tilde t_1$, $w*t_2=a_2e_2e_1a_2=\tilde t_2$, $w*d_{1,2}=b_2$. Therefore after conjugation by $w$ the right hand side of (M3) becomes the right hand side of (C). We have shown in the proof of relations (\[A15\]), Case 1c, using only relations (M1), that $d_3=(b_1^{-1}((a_2e_1e_2a_2a_3e_2)^{-1}*b_2) b_1^{-1}a_1^{-1}e_1^{-1}a_2^{-1})*b_2$. When we conjugate the last expression by $w$ we get exactly the expression for $\tilde d_3$ in Theorem $1^\prime$. This proves the equivalence of the presentations in Theorems 1 and $1^\prime$. In order to compare Theorems 3 and $3^\prime$ we need another set of generators. Let us call the curves $\beta_2,\beta_1,\alpha_1, \epsilon_1,\alpha_2\dots,\epsilon_{g-1},\alpha_g$ — [*the generating curves*]{}. Let $\beta_g$ be the curve shown on Figure \[general surface\] and let $\beta_2^\prime$ be a curve which intersects $\epsilon_{g-2}$ once and intersects $\beta_g$ once and is disjoint from the other [*generating curves*]{}. Then the curves $\beta_2^\prime, \alpha_g,\epsilon_{g-1},\alpha_{g-1},\dots,\epsilon_1,\alpha_1,\beta_1$ have the same intersection pattern as the [*generating curves*]{} and the curve $\beta_g$ plays the same role with respect to these curves as the curve $\delta_g$ with respect to the [*generating curves*]{}. Let $b_g$ and $b_2^\prime$ be twists along the curves $\beta_g$ and $\beta_2^\prime$ respectively. Then, by Theorem 1, we have a new presentation of ${\mathcal M}_{g,1}$ with generators $b_2^\prime,a_g,e_{g-1},a_{g-1},\dots,a_1,b_1$ and with defining relations (M1$^\prime)$, (M2$^\prime)$, (M3$^\prime)$ corresponding to (M1), (M2), (M3). It is a presentation of the same group and therefore, when we express $b_2^\prime$ in terms of the generators from Theorem 1, it is equivalent to the presentation ((M1), (M2), (M3)) and to the presentation ((A), (B), (C)). By Theorem 3 the group ${\mathcal M}_{g,0}$ has a presentation with relations (M1$^\prime)$, (M2$^\prime)$, (M3$^\prime)$ and one more relation (M4$^\prime)$ $[a_ge_{g-1}a_{g-1}\dots e_1a_1b_1^2a_1e_1\dots a_{g-1}e_{g-1}a_g,b_g]=1.$ Here $b_g$ is some product of generators which represents the Dehn twist of $S_{g,1}$ along the curve $\beta_g$. All such products are equivalent modulo relations (M1$^\prime)$, (M2$^\prime)$, (M3$^\prime)$. Relation (D) of Theorem $3^\prime$ has the same form with $b_g$ replaced by $\tilde d_g$. Therefore in order to check that the presentations in Theorems 3 and $3^\prime$ are equivalent it suffices to prove that the expression for $\tilde d_g$ in (D) also represents the Dehn twist with respect to the curve $\beta_g$. This task (of drawing very many pictures) is left to the reader. , [**H Hilden**]{}, [*On mapping class groups of closed surfaces as covering spaces*]{}, from: “Advances in the theory of Riemann surfaces”, Ann. Math. Stud. 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Math. 51 (1985),90–120. , [**W Thurston**]{}, [*A presentation for the mapping class group of a closed orientable surface*]{}, Topology, [19]{} (1980) 221–237 , [*Eine Präsentation der Abbildungsklassengruppe einer geschlossenen, orientierbaren Fläche*]{}, Diplom-Arbeit, University of Frankfurt , [*Generators for the mapping class group*]{}, from: “Topology of Low-dimensional Manifolds”, LNM [722]{} (1979) 44–47 , [*Homeomorphisms of a surface which act trivially on homology*]{}, Proc. Amer. Math. Soc. [75]{} (1979) 119–125 , [*A calculus of framed links in $S^3$*]{}, Invent. Math. [45]{} (1978) 35–56 , [*Topological invariants for 3–manifolds using representations of mapping class groups I*]{}, Topology, [31]{} (1992) 203–230 , [*Presentation du groupe de diffeotopies d’une surface compacte orientable*]{}, Travaux de Thurston sur les surfaces, Asterisque 66–67 (1979) 267–282 , [*A simple proof of the fundamental theorem of Kirby calculus on links*]{}, Trans. Amer. Math. Soc. [331]{} (1992) 143–156 , [*A presentation of mapping class groups in terms of Artin groups and geometric monodromy of singularities*]{}, preprint , [**M Polyak**]{}, [*On a tangle presentation of the mapping class groups of surfaces*]{}, Contemporary Mathematics, [164]{} (1994) 219–229 , [*Some finitely presented subgroups of the automorphism group of a free group*]{}, J. Algebra, [35]{} (1975) 205–213 , [*A simple presentation for the mapping class group of an orientable surface*]{}, Israel J. Math. [45]{} (1983) 157–174 | Mid | [
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title: //title body: //div[@id='content'] strip: (//div[@id='content']/h2)[1] strip: //h2[contains(., 'mehr News')]/following::* strip: //h2[contains(., 'mehr News')] strip: //div[contains(@class, 'indizar')]/following::* strip: //div[contains(@class, 'indizar')] strip: //h1[contains(@class, 'single')]/preceding::* strip: //h1[contains(@class, 'single')] strip_id_or_class: plista_widget prune: no next_page_link: //a[contains(., 'Weiter')] test_url: http://www.fertigung.de/2013/04/igus-neuer-energiekettenkatalog/ test_url: http://www.fertigung.de/2013/04/dynamisch-und-hochpraezise/ | Mid | [
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1Description: The total statewide number of students ages 19-21 who were served in a local education agency neglected program.; Academic Year: 2013-14; Grade: N/A; Note: The total statewide number of students ages 19-21 who were served in a local education agency neglected program. A neglected program serves neglected children and youth and is a public or private residential facility, other than a foster home, that is operated primarily for the care of children who have been committed to the institution or voluntarily placed under applicable State law due to abandonment, neglect, or death of their parents or guardians. A dash (-) indicates that the data are missing or not available. The symbol "nSource: EDFacts/Consolidated State Performance Report, 2013-14: http://www2.ed.gov/admins/lead/account/consolidated/index.html; Data Uploaded On: 07/07/2015 | Mid | [
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Officials join pellet clean-up Secretary for the Environment Wong Kam-sing and Secretary for Security Lai Tung-kwok joined 30 volunteers from the disciplined services to help clean up the plastic pellet spill today. They inspected the clean-up operation at Shek Pai Wan on Lamma Island. Mr Wong met the media afterwards and thanked the volunteers. He said the Government will investigate who is responsible for the spill and initiate follow-up action. He said it will avoid putting the pellets in landfills. Concerning the spill’s effect on marine ecology, he said the Environment Bureau will take enforcement action, together with the Agriculture, Fisheries & Conservation Department. So far, no problems with green turtles and coral reefs have been reported. Mr Lai said the Government will work with volunteer groups. Teams from the Police, Customs & Excise, Immigration, Fire Services, and Correctional Services Departments are assisting in the clean-up. “As today is not a public holiday, we expected the volunteers from other places will not be as many as on Saturdays and Sundays. [As far as] the disciplined volunteers [are] concerned, because of their work shift, they can spare more volunteers to come to assist. I believe it is a better use of resources, to clean our beaches as soon as possible.” Mr Lai said it is difficult to forecast how long the clean-up will take, but it is the common aim of Hong Kong people that the mission finish as soon as possible. | Mid | [
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TAMPA, Fla. – A Florida family says they have unwanted house guests that they just can’t get rid of: aggressive squirrels. WFLA in Tampa reports that the rodents moved into Annie Peavey’s condo about a month ago and since then, they’ve destroyed furniture, run amok in the living room and have gotten more and more threatening. Now, Peavey says she’s worried about her 9-year-old son’s safety. The squirrels became trapped inside the condo around Dec. 20, 2019 when a hole on the outside of the property was closed. After the TV station got involved, the property manager promised to bring someone in to removed the rodents. Click here to read more on this story at WFLA.com. | Low | [
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The Dorr Township Board voted in March to raise the annual wage of the township supervisor from $19,500 to $24,800, bringing the position in line with the clerk and treasurer. Pictured Township Supervisor Tammy VanHaitsma. After enduring a summer-long sewer and water construction project last year, residents were faced with another inconvenience this summer with the replacement the 142nd Avenue bridge crossing U.S. 131. The bridge project is expected to be completed Sept. 1. Austin Pallett, 14, sands the arm of a red oak and ash chair he created in woodworking class. Pallett and several classmates were honored at the Michigan Industrial Technical Education Society regional competition in May. | High | [
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A Reformed Christian asks: What do you believe, and why? Broadcasting from Homer, Alaska [email protected] http://truthforhomer.blogspot.com [email protected] http://truthforhomer.blogspot.com Disputatio Justifying Non-Christian Objections Douglas Wilson & Farrell Till Whenever we object to something, we always assume some standard or rule that the thing violates. Similarly, when non-Christians object to the Christian faith, they assume some standard that Christianity violates. But can non-Christians justify these standards that they so readily use? In the following interchange, the editor of Credenda/Agenda, Douglas Wilson and Farrell Till, editor of The Skeptical Review , discuss the topic of justifying non-Christian standards of ethics and reason. For the past thirty years, Farrell Till has been an English instructor at Spoon River College in Canton, Illinois. Prior to this, he was a preacher and foreign missionary for the Church of Christ. He attended two Bible colleges and received his bachelor's and master's degrees from Harding University. His preaching career spanned twelve years, five of which were spent in missionary work in France. After becoming an agnostic, he quit the ministry in 1963 and began a teaching career. For the past five years, he has edited The Skeptical Review , a quarterly journal that focuses on the doctrine of biblical inerrancy. He has regularly debated inerrancy-related issues in various public forums, including radio and television. Having begun this work as an agnostic, he now considers himself an atheist. DW: Many unbelievers commonly object to the God of the Bible on the basis of ethical "problems" with the character of God as revealed in the Scriptures. Whether they use psalms of imprecation, the slaughter of the Canaanites, the eternal wrath of God on the impenitent, etc ., the central theme is usually the same "Who would want to worship a God like that !" But despite the surface plausibility of the objection, a careful examination of it shows their Achilles attacking our Hector with his bare heel. Far from being the unbeliever's strongest case against the true God, this objection actually reveals the radical futility of unbelief; without God there are no ethical objections to anything . FT: Although you didn't expressly state the "objective-morality" position of evangelical apologists, you certainly implied it when you said that "without God there are no ethical objections to anything." The fallacy of this position is its failure to recognize that morality is an intellectual abstraction. As such, it is no different from abstractions of tragedy, sorrow, or any of many other abstractions the human mind has formulated from its broad range of experience. Arguing that human intelligence cannot determine if acts are immoral without a god to tell us they are is as illogical as arguing that we cannot tell if events are tragic without a god of tragedy to tell us they are. DW: Fine, I'll bite. If there is no God, then all the things you mention are in the same meaningless category. Morality, tragedy, and sorrow are equally evanescent. They are all empty sensations created by the chemical reactions of the brain, in turn created by too much pizza the night before. If there is no God, then all abstractions are chemical epiphenomena, like swamp gas over fetid water. This means that we have no reason for assigning truth and falsity to the chemical fizz we call reasoning or right and wrong to the irrational reaction we call morality. If no God, mankind is a set of bi-pedal carbon units of mostly water. And nothing else. FT: You bit too hard. In equating all human abstractions with "swamp gas over fetid water," you overlook verifiable facts. The human mind can think; swamp gas can't. Human intelligence can evaluate situations and formulate abstractions of beauty, happiness, sorrow, fairness and morality; swamp gas can't. Are these abstractions valid? Well, what IQ level is needed to conceptualize abstractions like beautiful, sad, fair, right or wrong? Can one with an IQ of 100 do it, or must his IQ be infinite? The existence of moral An examination of Evidential Apologetics. Looking at an article by Louis D. Whitworth, the former senior editor at Probe Ministries. He is a graduate of Northeast Louisiana University (B.A., Sociology and English, and M.A., English) and Dallas Theological Seminary (Th.M., Pastoral Theology). Prior to joining Probe, Lou taught English literature and composition at the college level and served with Campus Crusade for Christ in the Military Ministry as well as the Singles Ministry. He is the author of the Probe booklet, Literature Under the Microscope: A Christian Look at Reading. This guy sounds really smart, but he gets it all wrong!! Romans 1:19ff For what can be known about God is plain to them, because God has shown it to them. For his invisible attributes, namely, his eternal power and divine nature, have been clearly perceived, ever since the creation of the world, in the things that have been made. So they are without excuse. For although they knew God, they did not honor him as God or give thanks to him, but they became futile in their thinking, and their foolish hearts were darkened. Claiming to be wise, they became fools ... My wife and I discuss her recent encounter with an Open Theist. A Reformed and confessional response... 1689 London Baptist Confession of Faith Chapter 3: Of God's Decree 1. God hath decreed in himself, from all eternity, by the most wise and holy counsel of his own will, freely and unchangeably, all things, whatsoever comes to pass; yet so as thereby is God neither the author of sin nor hath fellowship with any therein; nor is violence offered to the will of the creature, nor yet is the liberty or contingency of second causes taken away, but rather established; in which appears his wisdom in disposing all things, and power and faithfulness in accomplishing his decree. ( Isaiah 46:10; Ephesians 1:11; Hebrews 6:17; Romans 9:15, 18; James 1:13; 1 John 1:5; Acts 4:27, 28; John 19:11; Numbers 23:19; Ephesians 1:3-5 ) 2. Although God knoweth whatsoever may or can come to pass, upon all supposed conditions, yet hath he not decreed anything, because he foresaw it as future, or as that which would come to pass upon such conditions. ( Acts 15:18; Romans 9:11, 13, 16, 18 ) 3. By the decree of God, for the manifestation of his glory, some men and angels are predestinated, or foreordained to eternal life through Jesus Christ, to the praise of his glorious grace; others being left to act in their sin to their just condemnation, to the praise of his glorious justice. ( 1 Timothy 5:21; Matthew 25:34; Ephesians 1:5, 6; Romans 9:22, 23; Jude 4 ) 4. These angels and men thus predestinated and foreordained, are particularly and unchangeably designed, and their number so certain and definite, that it cannot be either increased or diminished. ( 2 Timothy 2:19; John 13:18 ) Concluding with music from ApologetiX (from their free podcast) The confusions, delusions and unintelligible elocutions of former Homer resident John Crowder. This is what I have to deal with people! Plus some clips from ApologetiX. http://truthforhomer.blogspot.com [email protected] Finishing up an analysis of John Loftus' opening statements in the D'Souza debate. A Reformed confessional defense of basic Christian doctrines. Concluding with Bahnsen and the problem of evil. (Sorry, ran out of time abruptly, the apple rainbow wheel of rage popped up and told me my computer was about to freeze - working on a very old iMac.) Email comments, questions or metaphysical concerns to: [email protected] http://truthforhomer.blogspot.com Peace out dogs CHAPTER 27 OF THE COMMUNION OF SAINTS CHAPTER 28 OF BAPTISM AND THE LORD'S SUPPER CHAPTER 29 OF BAPTISM CHAPTER 30 OF THE LORD'S SUPPER CHAPTER 31 OF THE STATE OF MAN AFTER DEATH, AND OF THE RESURRECTION OF THE DEAD CHAPTER 32 OF THE LAST JUDGMENT Of the Church 1. The catholic or universal church, which (with respect to the internal work of the Spirit and truth of grace) may be called invisible, consists of the whole number of the elect, that have been, are, or shall be gathered into one, under Christ, the head thereof; and is the spouse, the body, the fulness of him that filleth all in all. ( Hebrews 12:23; Colossians 1:18; Ephesians 1:10, 22, 23; Ephesians 5:23, 27, 32 ) 2. All persons throughout the world, professing the faith of the gospel, and obedience unto God by Christ according unto it, not destroying their own profession by any errors everting the foundation, or unholiness of conversation, are and may be called visible saints; and of such ought all particular congregations to be constituted. ( 1 Corinthians 1:2; Acts 11:26; Romans 1:7; Ephesians 1:20-22 ) 3. The purest churches under heaven are subject to mixture and error; and some have so degenerated as to become no churches of Christ, but synagogues of Satan; nevertheless Christ always hath had, and ever shall have a kingdom in this world, to the end thereof, of such as believe in him, and make profession of his name. ( 1 Corinthians 5; Revelation 2; Revelation 3; Revelation 18:2; 2 Thessalonians 2:11, 12; Matthew 16:18; Psalms 72:17; Psalm 102:28; Revelation 12:17 ) 4. The Lord Jesus Christ is the Head of the church, in whom, by the appointment of the Father, all power for the calling, institution, order or government of the church, is invested in a supreme and sovereign manner; neither can the Pope of Rome in any sense be head thereof, but is that antichrist, that man of sin, and son of perdition, that exalteth himself in the church against Christ, and all that is called God; whom the Lord shall destroy with the brightness of his coming. ( Colossians 1:18; Matthew 28:18-20; Ephesians 4:11, 12; 2 Thessalonians 2:2-9 ) 5. In the execution of this power wherewith he is so intrusted, the Lord Jesus calleth out of the world unto himself, through the ministry of his word, by his Spirit, those that are given unto him by his Father, that they may walk before him in all the ways of obedience, which he prescribeth to them in his word. Those thus called, he commandeth to walk together in particular societies, or churches, for their mutual edification, and the due performance of that public worship, which he requireth of them in the world. ( John 10:16; John 12:32; Matthew 28:20; Matthew 18:15-20 ) 6. The members of these churches are saints by calling, visibly manifesting and evidencing (in and by their profession and walking) their obedience unto that call of Christ; and do willingly consent to walk together, according to the appointment of Christ; giving up themselves to the Lord, and one to another, by the will of God, in professed subjection to the ordinances of the Gospel. ( Romans. 1:7; 1 Corinthians 1:2; Acts 2:41, 42; Acts 5:13, 14; 2 Corinthians 9:13 ) 7. To each of these churches thus gathered, according to his mind declared in his word, he hath given all that power and authority, which is in any way needful for their carrying on that order in worship and discipline, which he hath instituted for them to observe; with commands and rules for the due and right exerting, and executing of that power. ( Matthew 18:17, 18; 1 Corinthians 5:4, 5; 1 Corinthians 5:13; 2 Corinthians 2:6-8 ) 8. A particular church, gathered and completely organized according to the mind of Christ, consists of officers and members; and the officers appointed by Christ to be chosen and set apart by the church (so called and gathered), for the peculiar administration of ordinances, and execution of power or duty, which he intrusts them with, or calls them to, to be continued to the end of the world, are bishops or elders, and deacons. ( Acts 20:17, 28; Philippians 1:1 ) 9. The way appointed by Christ for the calling of any person, fitted and gifted by the Holy Spirit, unto the office of bishop or elder in a church, is, that he be CHAPTER 20 OF THE GOSPEL, AND OF THE EXTENT OF THE GRACE THEREOF CHAPTER 21 OF CHRISTIAN LIBERTY AND LIBERTY OF CONSCIENCE CHAPTER 22 OF RELIGIOUS WORSHIP AND THE SABBATH DAY CHAPTER 23 OF LAWFUL OATHS AND VOWS CHAPTER 24 OF THE CIVIL MAGISTRATE CHAPTER 25 OF MARRIAGE Of Free Will 1. God hath endued the will of man with that natural liberty and power of acting upon choice, that it is neither forced, nor by any necessity of nature determined to do good or evil. ( Matthew 17:12; James 1:14; Deuteronomy 30:19 ) 2. Man, in his state of innocency, had freedom and power to will and to do that which was good and well-pleasing to God, but yet was unstable, so that he might fall from it. ( Ecclesiastes 7:29; Genesis 3:6 ) 3. Man, by his fall into a state of sin, hath wholly lost all ability of will to any spiritual good accompanying salvation; so as a natural man, being altogether averse from that good, and dead in sin, is not able by his own strength to convert himself, or to prepare himself thereunto. ( Romans 5:6; Romans 8:7; Ephesians 2:1, 5; Titus 3:3-5; John 6:44 ) 4. When God converts a sinner, and translates him into the state of grace, he freeth him from his natural bondage under sin, and by his grace alone enables him freely to will and to do that which is spiritually good; yet so as that by reason of his remaining corruptions, he doth not perfectly, nor only will, that which is good, but doth also will that which is evil. ( Colossians 1:13; John 8:36; Philippians 2:13; Romans 7:15, 18, 19, 21, 23 ) 5. This will of man is made perfectly and immutably free to good alone in the state of glory only. ( Ephesians 4:13 ) Of Creation 1. In the beginning it pleased God the Father, Son, and Holy Spirit, for the manifestation of the glory of his eternal power, wisdom, and goodness, to create or make the world, and all things therein, whether visible or invisible, in the space of six days, and all very good. ( John 1:2, 3; Hebrews 1:2; Job 26:13; Romans 1:20; Colossians 1:16; Genesis 1:31 ) 2. After God had made all other creatures, he created man, male and female, with reasonable and immortal souls, rendering them fit unto that life to God for which they were created; being made after the image of God, in knowledge, righteousness, and true holiness; having the law of God written in their hearts, and power to fulfil it, and yet under a possibility of transgressing, being left to the liberty of their own will, which was subject to change. ( Genesis 1:27; Genesis 2:7; Ecclesiastes 7:29; Genesis 1:26; Romans 2:14, 15; Genesis 3:6 ) 3. Besides the law written in their hearts, they received a command not to eat of the tree of knowledge of good and evil, which whilst they kept, they were happy in their communion with God, and had dominion over the creatures. ( Genesis 2:17; Genesis 1:26, 28 ) Of Divine Providence 1. God the good Creator of all things, in his infinite power and wisdom doth uphold, direct, dispose, and govern all creatures and things, from the greatest even to the least, by his most wise and holy providence, to the end for the which they were created, according unto his infallible foreknowledge, and the free and immutable counsel of his own will; to the praise of the glory of his wisdom, power, justice, infinite goodness, and mercy.( Hebrews 1:3; Job 38:11; Isaiah 46:10, 11; Psalms 135:6; Matthew 10:29-31; Ephesians 1:11 ) 2. Although in relation to the foreknowledge and decree of God, the first cause, all things come to pass immutably and infallibly; so that there is not anything befalls any by chance, or without his providence; yet by the same providence he ordereth them to fall out according to the nature of second causes, either necessarily, freely, or contingently. ( Acts 2:23; Proverbs 16:33; Genesis 8:22 ) 3. God, in his ordinary providence maketh use of means, yet is free to work without, above, and against them at his pleasure. ( Acts 27:31, 44; Isaiah 55:10, 11; Hosea 1:7; Romans 4:19-21; Daniel 3:27 ) 4. The almighty power, unsearchable wisdom, and infinite goodness of God, so far manifest themselves in his providence, that his determinate counsel extendeth itself even to the first fall, and all other sinful actions both of angels and men; and that not by a bare permission, which also he most wisely and powerfully boundeth, and otherwise ordereth and governeth, in a manifold dispensation to his most holy ends; yet so, as the sinfulness of their acts proceedeth only from the creatures, and not from God, who, being most holy and righteous, neither is nor can be the author or approver of sin. ( Romans 11:32-34; 2 Samuel 24:1, 1 Chronicles 21:1; 2 Kings 19:28; Psalms 76;10; Genesis 1:20; Isaiah 10:6, 7, 12; Psalms 1:21; 1 John 2:16 ) 5. The most wise, righteous, and gracious God doth oftentimes leave for a season his own children to manifold temptations and the corruptions of their own hearts, to chastise them for their former sins, or to discover unto them the hidden strength of corruption and deceitfulness of their hearts, that they may be humbled; and to raise them to a more close and constant dependence for their support upon himself; and to make them more watchful against all future occasions of sin, and for other just and holy ends. So that whatsoever befalls any of his elect is by his appointment, for his glory, and their good. ( 2 Chronicles 32:25, 26, 31; 2 Corinthians 12:7-9; Romans 8:28 ) 6. As for those wicked and ungodly men whom God, as the righteous judge, for former sin doth blind and harden; from them he not only withholdeth his grace, whereby they might have been enlightened in their understanding, and wrought upon their hearts; but sometimes also withdraweth the gifts which they had, and exposeth them to such objects as their corruption makes occasion of sin; and withal, gives them over to their own lusts, the temptations of the world, and the power of Satan, whereby it comes to pass that they harden themselves, under those means which God useth for the softening of others.( Romans 1:24-26, 28; Romans 11:7, 8; Deuteronomy 29:4; Matthew 13:12; Deuteronomy 2:30; 2 Kings 8:12, 13; Psalms 81:11, 12; 2 Thessalonians 2:10-12; Exodus 8:15, 32; Isaiah 6:9, 10; 1 Peter 2:7, 8 ) 7. As the providence of God doth in general reach to all creatures, so after a more special manner it taketh care of his church, and disposeth of all things to the good thereof. ( 1 Timothy 4:10; Amos 9:8, 9; Isaiah 43:3-5 ) A valuable resource for basic training in genuine biblical Christianity ...and a possible cure for insomnia ;-) How would you answer the question: "What is justification?" or "How do we know the Bible to be the word of God?" Find the answers in this edition of Truth Talk! Further reading at http://truthforhomer.blogspot.com [email protected] A valuable resource for basic training in genuine biblical Christianity ...and a possible cure for insomnia ;-) How would you answer the question: "What is justification?" or "How do we know the Bible to be the word of God?" Find the answers in this edition of Truth Talk! Further reading at http://truthforhomer.blogspot.com Concluding comments on continuationism, A "language instructor" on How to Speak in Tongues, Todd Bentley, Why I believe cessationsim is correct. What do you believe? [email protected] http://truthforhomer.blogspot.com http://puritan.podomatic.com | Low | [
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Little Duckling Watering Can - Cream For something a little different, send your best wishes with this charming arrangement. The decorative watering can has a sweet baby duckling design and looks stunning filled with fresh flowers in shades of cream, white and peach. Adorable – just like their new arrival! Add new review Saving review... A bit about us Welcome to Willis Flowers, ( S.G.Willis & Sons Ltd. ) a 4th Generation family business with more than 50 years experience in providing first class floristry and flowers to the surrounding areas of Stony Stratford in Milton Keynes and Leighton Buzzard in Bedfordshire, both original market towns stepped in history. Having started selling mainly fruit and vegetables with flowers as a side line as the generations have worked their way up through the company the flowers have increased their place in our business to where we are concentrating 100% on flowers but as times are ever changing, so we are changing too. | Mid | [
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Mice with a null mutation of the TGF alpha gene have abnormal skin architecture, wavy hair, and curly whiskers and often develop corneal inflammation. Mice homozygous for a disrupted transforming growth factor alpha (TGF alpha) gene are healthy and fertile, although some older mice show evidence of corneal inflammation. In contrast with TGF alpha +/- and +/+ animals, TGF alpha -/- mice have a pronounced waviness of the coat. Histological examination of the skin from TGF alpha -/- mice reveals a dramatic derangement of hair follicles. Mice with a disrupted TGF alpha gene also have curly whiskers, first evident on the day of birth. The phenotype of TGF alpha -/- mice is remarkably similar to that of the mouse mutant waved-1 (wa-1). Offspring resulting from crosses between TGF alpha -/- and wa-1 mice display the curly whisker-coat phenotype, indicating that the basis of the wa-1 phenotype is a mutation in the TGF alpha gene. These observations suggest that TGF alpha plays a pivotal role in determining skin architecture and in regulating hair development. | Mid | [
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nearest integer? 193 What is 15065230 to the power of 1/3, to the nearest integer? 247 What is the cube root of 1291458 to the nearest integer? 109 What is 13403198 to the power of 1/4, to the nearest integer? 61 What is 19937 to the power of 1/2, to the nearest integer? 141 What is 7383780 to the power of 1/2, to the nearest integer? 2717 What is 52535 to the power of 1/8, to the nearest integer? 4 What is 9306354 to the power of 1/2, to the nearest integer? 3051 What is 398523 to the power of 1/3, to the nearest integer? 74 What is the seventh root of 1129942 to the nearest integer? 7 What is 6990551 to the power of 1/6, to the nearest integer? 14 What is the ninth root of 1280297 to the nearest integer? 5 What is the square root of 1135670 to the nearest integer? 1066 What is the eighth root of 273088 to the nearest integer? 5 What is 918645 to the power of 1/10, to the nearest integer? 4 What is the cube root of 316707 to the nearest integer? 68 What is 1702711 to the power of 1/3, to the nearest integer? 119 What is 7802314 to the power of 1/2, to the nearest integer? 2793 What is the square root of 10456220 to the nearest integer? 3234 What is the square root of 1010565 to the nearest integer? 1005 What is 1049295 to the power of 1/4, to the nearest integer? 32 What is the square root of 2591963 to the nearest integer? 1610 What is the square root of 682489 to the nearest integer? 826 What is the square root of 39063 to the nearest integer? 198 What is 2793687 to the power of 1/9, to the nearest integer? 5 What is 956179 to the power of 1/10, to the nearest integer? 4 What is 1539813 to the power of 1/9, to the nearest integer? 5 What is 2313976 to the power of 1/3, to the nearest integer? 132 What is 9065258 to the power of 1/5, to the nearest integer? 25 What is 2799636 to the power of 1/9, to the nearest integer? 5 What is the cube root of 300854 to the nearest integer? 67 What is 17985655 to the power of 1/4, to the nearest integer? 65 What is 148342 to the power of 1/6, to the nearest integer? 7 What is the square root of 925963 to the nearest integer? 962 What is 21886071 to the power of 1/3, to the nearest integer? 280 What is 350437 to the power of 1/7, to the nearest integer? 6 What is the tenth root of 1647258 to the nearest integer? 4 What is 838627 to the power of 1/3, to the nearest integer? 94 What is 253289 to the power of 1/8, to the nearest integer? 5 What is the third root of 4783239 to the nearest integer? 168 What is 93001 to the power of 1/3, to the nearest integer? 45 What is the cube root of 598777 to the nearest integer? 84 What is the square root of 5339380 to the nearest integer? 2311 What is the third root of 127510 to the nearest integer? 50 What is the eighth root of 943915 to the nearest integer? 6 What is 4858 to the power of 1/10, to the nearest integer? 2 What is the square root of 1629365 to the nearest integer? 1276 What is 1032592 to the power of 1/5, to the nearest integer? 16 What is the cube root of 374573 to the nearest integer? 72 What is the square root of 12882643 to the nearest integer? 3589 What is the eighth root of 13030852 to the nearest integer? 8 What is the third root of 826580 to the nearest integer? 94 What is 12965067 to the power of 1/8, to the nearest integer? 8 What is 162302 to the power of 1/5, to the nearest integer? 11 What is the tenth root of 5629621 to the nearest integer? 5 What is the cube root of 698470 to the nearest integer? 89 What is the third root of 1075786 to the nearest integer? 102 What is the eighth root of 1200816 to the nearest integer? 6 What is the third root of 3091905 to the nearest integer? 146 What is the square root of 1078364 to the nearest integer? 1038 What is the eighth root of 27454 to the nearest integer? 4 What is 579679 to the power of 1/7, to the nearest integer? 7 What is the eighth root of 673309 to the nearest integer? 5 What is the cube root of 3005479 to the nearest integer? 144 What is the third root of 582415 to the nearest integer? 84 What is 4001047 to the power of 1/5, to the nearest integer? 21 What is 19372433 to the power of 1/2, to the nearest integer? 4401 What is 1387515 to the power of 1/8, to the nearest integer? 6 What is 217471 to the power of 1/4, to the nearest integer? 22 What is the fifth root of 3073679 to the nearest integer? 20 What is 344120 to the power of 1/2, to the nearest integer? 587 What is the square root of 587377 to the nearest integer? 766 What is 13974778 to the power of 1/8, to the nearest integer? 8 What is 153701 to the power of 1/5, to the nearest integer? 11 What is 13174211 to the power of 1/4, to the nearest integer? 60 What is 11046720 to the power of 1/2, to the nearest integer? 3324 What is 749134 to the power of 1/7, to the nearest integer? 7 What is 8322743 to the power of 1/3, to the nearest integer? 203 What is the third root of 7698991 to the nearest integer? 197 What is 5459429 to the power of 1/6, to the nearest integer? 13 What is 311080 to the power of 1/3, to the nearest integer? 68 What is 1564662 to the power of 1/3, to the nearest integer? 116 What is the third root of 12415525 to the nearest integer? 232 What is 1600801 to the power of 1/8, to the nearest integer? 6 What is 194140 to the power of 1/8, to the nearest integer? 5 What is the eighth root of 2047064 to the nearest integer? 6 What is the sixth root of 819641 to the nearest integer? 10 What is 974224 to the power of 1/2, to the nearest integer? 987 What is the seventh root of 5597409 to the nearest integer? 9 What is the fourth root of 22237 to the nearest integer? 12 What is 20071 to the power of 1/5, to the nearest integer? 7 What is the third root of 23011040 to the nearest integer? 284 What is the third root of 543903 to the nearest integer? 82 What is the ninth root of 1662094 to the nearest integer? 5 What is 580764 to the power of 1/5, to the nearest integer? 14 What is the square root of 1334194 to the nearest integer? 1155 What is 1484135 to the power of 1/2, to the nearest integer? 1218 What is the third root of 42321 to the nearest integer? 35 What is the square root of 155581 to the nearest integer? 394 What is the square root of 164 to the nearest integer? 13 What is 111897 to the power of 1/2, to the nearest integer? 335 What is the cube root of 9753089 to the nearest integer? 214 What is 119213 to the power of 1/7, to the nearest integer? 5 What is the tenth root of 1637981 to the nearest integer? 4 What is 283432 to the power of 1/3, to the nearest integer? 66 What is the sixth root of 271584 to the nearest integer? 8 What is 30364 to the power of 1/5, to the nearest integer? 8 What is the fourth root of 2296731 to the nearest integer? 39 What is the tenth root of 1325002 to the nearest integer? 4 What is the cube root of 100266 to the nearest integer? 46 What is the cube root of 14878530 to the nearest integer? 246 What is 299253 to the power of 1/8, to the nearest integer? 5 What is 6345037 to the power of 1/4, to the nearest integer? 50 What is the third root of 12638032 to the nearest integer? 233 What is the ninth root of 488838 to the nearest integer? 4 What is the sixth root of 779529 to the nearest integer? 10 What is the sixth root of 3225432 to the nearest integer? 12 What is 11890640 to the power of 1/7, to the nearest integer? 10 What is the square root of 781171 to the nearest integer? 884 What is the sixth root of 250378 to the nearest integer? 8 What is the cube root of 137788 to the nearest integer? 52 What is the square root of 204931 to the nearest integer? 453 What is the square root of 6803666 to the nearest integer? 2608 What is 38388 to the power of 1/6, to the nearest integer? 6 What is the eighth root of 73381 to the nearest integer? 4 What is the tenth root of 316294 to the nearest integer? 4 What is the cube root of 563074 to the nearest integer? 83 What is the tenth root of 1226143 to the nearest integer? 4 What is the square root of 6748951 to the nearest integer? 2598 What is the square root of 1301909 to the nearest integer? 1141 What is the square root of 1930782 to the nearest integer? 1390 What is the cube root of 2700463 to the nearest integer? 139 What is the ninth r | Mid | [
0.5827067669172931,
38.75,
27.75
] |
Islet-neogenesis-associated protein enhances neurite outgrowth from DRG neurons. Islet-neogenesis-associated protein, INGAP, is a 175-amino-acid pancreatic acinar protein that stimulates pancreatic duct cell proliferation in vitro and islet neogenesis in vivo. To date, the mitogenic activity of INGAP has been identified only in nonneural tissues. The aim of this study was to examine the effects of a pentadecapeptide of INGAP (INGAP peptide), the biologically active portion of the native protein, in cultured dorsal root ganglia (DRG) explants from C57BL/6 mice. The present study provides evidence that INGAP peptide acts as a mitogen in the peripheral nervous system (PNS), and that it enhances neurite outgrowth from DRGs in vitro in a time- and dose-dependent manner. The neuritogenic action of INGAP peptide correlates with an increase in [(3)H]thymidine incorporation (P < 0.0001) and mitochondrial activity (P < 0.001). Results from these studies suggest that INGAP peptide promotes Schwann cell proliferation in the DRG which releases trophic factors that promote neurite outgrowth. | High | [
0.6883852691218131,
30.375,
13.75
] |
Q: Track email activity in C# code I am using sendgrid to send email. I can see sent mails activities in my sendgrid portal. Now I want track those activities in my c# code. Like what happened to mail - > delivered or bounced or blocked. Is there any solution? A: It is possible using Sendgrid Events and it turns out to be rather simple. Check out https://sendgrid.com/docs/API_Reference/Webhooks/event.html You can decide how you want the receive the events, in my case I selected JSON. In your C# web application, you can create a class like: public class EmailEvent { [JsonProperty("sg_message_id")] public string MessageId { get; set; } [JsonProperty("email")] public string Email { get; set; } [JsonProperty("timestamp")] public long Timestamp { get; set; } [JsonProperty("smtp-id")] public string smtpid { get; set; } [JsonProperty("event")] public string Event { get; set; } [JsonProperty("url")] public string Url { get; set; } [JsonProperty("category")] public string Category { get; set; } [JsonProperty("userid")] public string UserId { get; set; } } and a controller action like: [HttpPost] [AllowAnonymous] public async Task<IActionResult> IncomingNotification([FromBody] EmailEvent[] events) { // be prepared to handle an array of events as sendgrid can send batches } You are free to name the action the way you want as long at it matches how you configured the SendGrid webhook. | High | [
0.7388316151202751,
26.875,
9.5
] |
package org.spiderflow.common;
import org.spiderflow.model.JsonBean;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import com.baomidou.mybatisplus.core.conditions.query.QueryWrapper;
import com.baomidou.mybatisplus.core.mapper.BaseMapper;
import com.baomidou.mybatisplus.core.metadata.IPage;
import com.baomidou.mybatisplus.extension.plugins.pagination.Page;
import com.baomidou.mybatisplus.extension.service.impl.ServiceImpl;
public abstract class CURDController<S extends ServiceImpl<M, T>,M extends BaseMapper<T>, T> {
@Autowired
private S service;
@RequestMapping("/list")
public IPage<T> list(@RequestParam(name = "page",defaultValue = "1")Integer page, @RequestParam(name = "limit",defaultValue = "1")Integer size){
return service.page(new Page<T>(page, size), new QueryWrapper<T>().orderByDesc("create_date"));
}
@RequestMapping("get")
public JsonBean<T> get(String id) {
return new JsonBean<T>(service.getById(id));
}
@RequestMapping("delete")
public JsonBean<Boolean> delete(String id){
return new JsonBean<Boolean>(service.removeById(id));
}
@RequestMapping("save")
public JsonBean<Boolean> save(T t){
return new JsonBean<Boolean>(service.saveOrUpdate(t));
}
} | Mid | [
0.6385809312638581,
36,
20.375
] |
Iraq forces regain control of villages in Anbar Press TV- The Iraqi army has liberated seven villages in Anbar Province close to the border with Syria from Daesh control. The gains, which were announced on Friday, were achieved during a push underway to flush the terrorists out of the towns of Rawa, Aanah and al-Qaim, the last main populated areas held by Daesh in the western province, AFP reported. “Our military units liberated seven villages from Daesh control between the town of Haditha and the town of Aanah,” said Staff Major General Qassem al-Mohammadi. Staff Major General Noman Abed al-Zobai, the commander of the Army’s 7th Division, also said the troops had reached the outskirts of al-Sagra, an area southeast of Aanah. Iraqi forces have already liberated the key cities of Ramadi, the provincial capital, Fallujah, and Hit in the province. Ramadi’s liberation collided head-on with Daesh’s come-what-may terror drive against the country, which it had launched in 2014. On October 17, 2016, joint Iraqi forces launched an operation to retake the northern city of Mosul, which fell to Daesh in 2014. The advance has, however, been slowed down due to the presence of hundreds of thousands of civilians, many of whom are being prevented by the terrorists from leaving the city, the last Daesh stronghold in Iraq. Iraq’s Prime Minister Haider al-Abadi has said it could take the country three months to vanquish the terrorists. | Low | [
0.5335968379446641,
33.75,
29.5
] |
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