text
stringlengths 3
1.74M
| label
class label 2
classes | source
stringclasses 3
values |
|---|---|---|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
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Stackexchange
|
Show that Shiny is busy (or loading) when changing tab panels. <p>(Code follows after problem description)</p>
<p>I am working on making a web app with Shiny, and some of the R commands that I am executing take minutes to complete. I found that I need to provide the user with some indication that Shiny is working, or they will continuously change the parameters I provide in the side panel, which just causes Shiny to reactively restart the calculations once the initial run is completed.</p>
<p>So, I created a conditional panel that shows a "Loading" message (referred to as a modal) with the following (thanks to Joe Cheng on the Shiny Google group for the conditional statement):</p>
<pre><code># generateButton is the name of my action button
loadPanel <- conditionalPanel("input.generateButton > 0 && $('html').hasClass('shiny-busy')"),
loadingMsg)
</code></pre>
<p>This is working as intended if the user remains on the current tab. However, the user can switch to another tab (that may contain some calculations that need to be run for some time), but the loading panel appears and disappears immediately, all while R chugs away at the calculations, and then refreshing the content only after it is done.</p>
<p>Since this may be hard to visualize, I provided some code to run below. You will notice that clicking the button to start the calculations will produce a nice loading message. However, when you switch to tab 2, R starts running some calculations, but fails to show the loading message (maybe Shiny does not register as being busy?). If you restart the calculations by pressing the button again, the loading screen will show up correctly. </p>
<p>I want the loading message to appear when switching to a tab that is loading!</p>
<p><strong>ui.R</strong></p>
<pre><code>library(shiny)
# Code to make a message that shiny is loading
# Make the loading bar
loadingBar <- tags$div(class="progress progress-striped active",
tags$div(class="bar", style="width: 100%;"))
# Code for loading message
loadingMsg <- tags$div(class="modal", tabindex="-1", role="dialog",
"aria-labelledby"="myModalLabel", "aria-hidden"="true",
tags$div(class="modal-header",
tags$h3(id="myModalHeader", "Loading...")),
tags$div(class="modal-footer",
loadingBar))
# The conditional panel to show when shiny is busy
loadingPanel <- conditionalPanel(paste("input.goButton > 0 &&",
"$('html').hasClass('shiny-busy')"),
loadingMsg)
# Now the UI code
shinyUI(pageWithSidebar(
headerPanel("Tabsets"),
sidebarPanel(
sliderInput(inputId="time", label="System sleep time (in seconds)",
value=1, min=1, max=5),
actionButton("goButton", "Let's go!")
),
mainPanel(
tabsetPanel(
tabPanel(title="Tab 1", loadingPanel, textOutput("tabText1")),
tabPanel(title="Tab 2", loadingPanel, textOutput("tabText2"))
)
)
))
</code></pre>
<p><strong>server.R</strong></p>
<pre><code>library(shiny)
# Define server logic for sleeping
shinyServer(function(input, output) {
sleep1 <- reactive({
if(input$goButton==0) return(NULL)
return(isolate({
Sys.sleep(input$time)
input$time
}))
})
sleep2 <- reactive({
if(input$goButton==0) return(NULL)
return(isolate({
Sys.sleep(input$time*2)
input$time*2
}))
})
output$tabText1 <- renderText({
if(input$goButton==0) return(NULL)
return({
print(paste("Slept for", sleep1(), "seconds."))
})
})
output$tabText2 <- renderText({
if(input$goButton==0) return(NULL)
return({
print(paste("Multiplied by 2, that is", sleep2(), "seconds."))
})
})
})
</code></pre>
| 0non-cybersec
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Stackexchange
|
3 Indiana Boys Save Fallen Elderly Neighbor in Frigid Temperatures.
| 0non-cybersec
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Reddit
|
Recovering files from old internal disk without windows authentication. <p>So I have an old hard drive which ran Windows XP. The rest of the computer was scraped. I can't recall the login password for the user on that hard drive. I want to recover files from this hard drive.</p>
<p>My question:</p>
<p>If I get an adapter and connect it to my PC running windows 10 will I be able to simply access the drive as if it was a new drive and browse the files? If not how do I retrieve the files?</p>
<p>Also if I want to completely copy the content of the disk to a new hard drive can achieve that with a tool like deamon?</p>
| 0non-cybersec
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Stackexchange
|
How to solve <form> that includes file upload and other text input, on the same page. <p>I need help with my form. There's a mix of input, textarea and file upload that i want to enter into the database..</p>
<p>What do I use in the ? Do I use the normal form attribute :</p>
<pre><code> <form action="" method="">
</code></pre>
<p>or</p>
<pre><code><form enctype="" action="" method="">
</code></pre>
<p>Please have it in mind that, I have to do this in a single page, and the picture upload must be done along with other text input.</p>
<p>Thanks for your time.</p>
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Stackexchange
|
Cthulu is actually the nicest..
| 0non-cybersec
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Reddit
|
Android Hello-World compile error: Intellij cannot find aapt. <p>I'm trying to get set up with an Android development environment using IntelliJ in Ubuntu 12.04. I create an Android Application Module, but when I try to build, I get the following error:</p>
<blockquote>
<p>android-apt-compiler: Cannot run program "/home/jon/Programs/android-sdk-linux/platform-tools/aapt": java.io.IOException: error=2, No such file or directory</p>
</blockquote>
<p>Several hours of scouring the internet hasn't helped.</p>
<p>By the way, I ran <code>locate aapt</code> in the terminal and found that <code>aapt</code> is located at <code>/home/jon/Programs/android-sdk-linux/build-tools/17.0.0/aapt</code></p>
| 0non-cybersec
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Stackexchange
|
Calculating traversal position of a node in a full binary tree, given its path. <p>Given a path down a <em>full</em> binary tree to a node (for example, a sequence of $1$s and $0$s, $0$ representing "go left" and $1$ representing "go right"), how would one find the position of the node in the preorder traversal. In other words the $i$th node in the preorder traversal will end up with at this node.</p>
<p>Obviously something better than brute force would be nice.</p>
| 0non-cybersec
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Stackexchange
|
Sloth POV.
| 0non-cybersec
|
Reddit
|
How can I debug carthage (owned) dependencies in my iOS application?. <p>I have an iOS application project with couple of private(owned) <code>carthage</code> dependencies, sometimes I'm in the need of debugging bugs or to know what is the code executed by certain methods. </p>
<p>I'd like to know if there is an easy way to setup the framework source code in <code>Carthage/Checkouts</code> using a run script, so that if I do a carthage update and run my app a can debug the frameworks source code without the need of adding sub projects.</p>
<p>Can this be done by including the debugging symbols in the framework build if so what would I need to do, to make it work?</p>
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|
Stackexchange
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
|
Stackexchange
|
Getting duplicate primary key for auto increment column. <p>I have a MySQL Innodb table with 'id' as auto increment primary key column. </p>
<p>My Scripts are doing following operations:</p>
<ol>
<li>insert ignore into tablename set x='a', y='b'</li>
<li>update ignore tablename set x='a', y='b' where id = 'smthing'</li>
<li>delete from tablename where id = 'smthing'</li>
</ol>
<p>Now after running the script when I insert something manually (insert into tablename set x='ax', y='bx') I get the following error:</p>
<pre><code>Duplicate Entry for Auto Increment Value.
</code></pre>
<p>On check I see that Auto Increment is trying to put the highest possible value of int that MySql will support as new value even and it already exist in db.</p>
<p>The table has two index: id+x, id+y though idk if this is related to this issue.</p>
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Stackexchange
|
Notepad++: How to automatically set Language as Xml when load files. <p>Whenever I edit a .config file in Notepad++ (which is an XML file) I want the syntax highlighting to automatically color it like XML. How do I configure Notepad++ to do this so that I don't have to manually select it every time I open a .config file?</p>
<p>Thanks.</p>
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|
Stackexchange
|
Fatal Exception: java.lang.NoClassDefFoundError: android.support.v7.app.AppCompatDelegateImplV23. <p>On my latest release a few of my users (less than 0.2%) are getting a crash because <code>android.support.v7.app.AppCompatDelegateImplV23</code> can not be found. I can not reproduce the issue, I'm only getting reports over crashlytics. I have used apktool to extract the files out of my apk and I can see <code>AppCompatDelegateImplV23.smali</code> in there. Any idea what could be happening?</p>
<p>This is my compile line on gradle:</p>
<pre><code>compile 'com.android.support:appcompat-v7:23+'
</code></pre>
<p>This is the exception:</p>
<pre><code>Fatal Exception: java.lang.NoClassDefFoundError: android.support.v7.app.AppCompatDelegateImplV23
at android.support.v7.app.AppCompatDelegate.create(AppCompatDelegate.java:133)
at android.support.v7.app.AppCompatDelegate.create(AppCompatDelegate.java:117)
at android.support.v7.app.AppCompatActivity.getDelegate(AppCompatActivity.java:456)
at android.support.v7.app.AppCompatActivity.onCreate(AppCompatActivity.java:59)
at MyActivityClass.onCreate(MyActivityClass.java:353)
at android.app.Activity.performCreate(Activity.java:6248)
at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1125)
at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2437)
at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2544)
at android.app.ActivityThread.access$900(ActivityThread.java:150)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1394)
at android.os.Handler.dispatchMessage(Handler.java:102)
at android.os.Looper.loop(Looper.java:168)
at android.app.ActivityThread.main(ActivityThread.java:5845)
at java.lang.reflect.Method.invoke(Method.java)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:797)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:687)
at de.robv.android.xposed.XposedBridge.main(XposedBridge.java:117)
</code></pre>
<p>EDIT: This is my main gradle file:</p>
<pre><code>buildscript {
repositories {
maven { url 'https://maven.fabric.io/public' }
jcenter() // version plugin support
}
dependencies {
classpath 'io.fabric.tools:gradle:1.+'
classpath 'com.android.tools.build:gradle:1.3.1'
}
allprojects {
repositories {
jcenter()
}
}
}
apply plugin: 'com.android.application'
apply plugin: 'io.fabric'
apply plugin: 'com.google.gms.google-services'
repositories {
maven { url 'https://maven.fabric.io/public' }
}
android {
signingConfigs {
config {
......
}
}
compileSdkVersion 23
buildToolsVersion "23.0.1"
useLibrary 'org.apache.http.legacy'
productFlavors {
// Define separate dev and prod product flavors.
dev {
// dev utilizes minSDKVersion = 21 to allow the Android gradle plugin
// to pre-dex each module and produce an APK that can be tested on
// Android Lollipop without time consuming dex merging processes.
minSdkVersion 21
}
prod {
// The actual minSdkVersion for the application.
minSdkVersion 14
}
}
defaultConfig {
applicationId "mypackage"
minSdkVersion 14
targetSdkVersion 23
versionCode some number
versionName "some version number"
multiDexEnabled true
signingConfig signingConfigs.config
}
buildTypes {
release {
minifyEnabled true
proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
signingConfig signingConfigs.config
applicationVariants.all { variant ->
variant.outputs.each { output ->
output.outputFile = new File(output.outputFile.parent,"myapk.apk")
}
}
}
debug {
signingConfig signingConfigs.config
}
}
repositories {
mavenCentral()
mavenLocal()
}
compileOptions {
sourceCompatibility JavaVersion.VERSION_1_7
targetCompatibility JavaVersion.VERSION_1_7
}
packagingOptions {
exclude 'AndroidManifest.xml'
exclude 'META-INF/beans.xml'
exclude 'META-INF/LICENSE.txt'
exclude 'META-INF/NOTICE.txt'
exclude 'META-INF/LICENSE'
exclude 'META-INF/NOTICE'
exclude 'META-INF/DEPENDENCIES'
}
productFlavors {
}
dexOptions {
// preDexLibraries = false
jumboMode = true
javaMaxHeapSize "2g"
//doesn't seem to be supported with multidex
// incremental true
}
lintOptions{
disable 'MissingTranslation'
disable 'ExtraTranslation'
}
}
dependencies {
compile fileTree(dir: 'libs', include: ['*.jar'])
//multidex
compile 'com.android.support:multidex:1.0.0'
//support libs
compile 'com.android.support:support-v4:23+'
compile 'com.android.support:appcompat-v7:23+'
compile 'com.android.support:cardview-v7:23+'
compile 'com.android.support:design:23+'
//app invites
compile 'com.google.android.gms:play-services-appinvite:8+'
compile 'com.google.android.gms:play-services-analytics:8+'
//chromecast
compile 'com.google.android.gms:play-services-cast:8+'
compile 'com.android.support:mediarouter-v7:23+'
//plus button
compile 'com.google.android.gms:play-services-plus:8+'
compile('com.crashlytics.sdk.android:crashlytics:2.5.5@aar') {
transitive = true;
}
compile 'com.github.navasmdc:MaterialDesign:1.+@aar'
//material lib needs it
compile 'com.nineoldandroids:library:2.4.+'
compile 'com.facebook.stetho:stetho:1+'
compile 'com.facebook.stetho:stetho-okhttp:1+'
//debugCompile 'com.squareup.leakcanary:leakcanary-android:1+'
debugCompile 'com.squareup.leakcanary:leakcanary-android-no-op:1+'
releaseCompile 'com.squareup.leakcanary:leakcanary-android-no-op:1+'
compile 'com.makeramen:dragsortadapter:1.3+'
compile 'com.github.amlcurran.showcaseview:library:5.0.0'
}
</code></pre>
<p>EDIT: just had a new report but for a different version of the same class</p>
<pre><code>Fatal Exception: java.lang.NoClassDefFoundError: android.support.v7.app.AppCompatDelegateImplV14
at android.support.v7.app.AppCompatDelegate.create(AppCompatDelegate.java:135)
at android.support.v7.app.AppCompatDelegate.create(AppCompatDelegate.java:117)
at android.support.v7.app.AppCompatActivity.getDelegate(AppCompatActivity.java:456)
at android.support.v7.app.AppCompatActivity.onCreate(AppCompatActivity.java:59)
at MyActivity.onCreate(MyActivity.java:353)
at android.app.Activity.performCreate(Activity.java:5008)
at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1079)
at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2044)
at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2105)
at android.app.ActivityThread.access$600(ActivityThread.java:133)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1211)
at android.os.Handler.dispatchMessage(Handler.java:99)
at android.os.Looper.loop(Looper.java:137)
at android.app.ActivityThread.main(ActivityThread.java:4795)
at java.lang.reflect.Method.invokeNative(Method.java)
at java.lang.reflect.Method.invoke(Method.java:511)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:789)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:556)
at dalvik.system.NativeStart.main(NativeStart.java)
</code></pre>
<p>EDIT: I've just gone over a lot of the reports, some are for <code>AppCompatDelegateImplV23</code> and some are for <code>AppCompatDelegateImplV14</code>, some involved XPosed but most don't. The one thing that is common to all of them is that they are all for rooted phones. </p>
| 0non-cybersec
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Stackexchange
|
Home Exchange.
| 0non-cybersec
|
Reddit
|
Stop Dashlane auto-fill on specific input fields?. <p>Does anybody know how to hint to the Dashlane browser plug-in to leave an <code><input></code> text field alone?</p>
<p>For most browser password managers, setting the <code>auto-complete</code> property to <code>off</code> works, Eg: <code><input type='text' autocomplete='off' /></code>.</p>
<p>For LastPass, this additional attribute is recommended: <code>data-lpignore="true"</code>.</p>
<p>I have no idea what this might be for Dashlane. I have noticed that the attributes <code>data-kwimpalastatus="alive"</code> and <code>data-kwimpalaid="xxxx-x"</code> (where <code>xxxx-x</code> is an auto-generated identifier) are added to my page, so this looks like a sensible place to start, but I've not had much joy figuring out the magic incantation...</p>
<p>In case anyone asks, the only other browser extensions I'm using are DuckDuckGo and Google Docs offline, so it's a reasonably safe bet that the <code>datakwimpala*</code> attributes are down to Dashlane.</p>
<p>Does anyone have any ideas??</p>
<p>Thanks!</p>
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Stackexchange
|
Entire functions $f$ for which there exists a positive constant $M$ such that $|f(z)|\le M|\cos z|$. <p>Find all entire functions $f$ for which there exists a positive constant $M$ such that $|f(z)|\le M|\cos z|$ for all $z\in \mathbb{C}$.</p>
<p>I think to prove this, we need to use the Identity theorem, i.e. analytic functions are determined by their values in a convergent sequence. In this case, we can find real sequence converging to $\pi/2$. In that case, given any $\epsilon >0$, eventually $|f(z_n)|$ must be less than $M\epsilon$. But this does not guarantee that $f(z_n)=0$, which is what I need to prove that $f\equiv 0$. How can I sharpen this proof? I would greatly appreciate any help.</p>
| 0non-cybersec
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Stackexchange
|
Install Xcache for dedicated. <p>Hell I had try install xcache</p>
<blockquote>
<pre><code>http://xcache.lighttpd.net/wiki/InstallFromSource
</code></pre>
</blockquote>
<p>But in step </p>
<pre><code>~/src/xcache $ make
~/src/xcache $ su
~/src/xcache # make install
</code></pre>
<p>My dedicated showed an error</p>
<pre><code>make: *** No targets specified and no makefile found. Stop.
</code></pre>
<p>Please help me </p>
| 0non-cybersec
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Stackexchange
|
Solution Clusters and Monotone-2SAT. <p>It is known that generic k-SAT formulas may exhibit the presence of exponentially many solution clusters.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>Is it true also for Monotone-2SAT formulas?</p>
</blockquote>
<p>For the definition of cluster, see Kaveh's answer to the question <a href="https://cstheory.stackexchange.com/questions/1115/sat-solution-space-definition-of-cluster-of-solutions">SAT Solution Space - Definition of Cluster of Solutions</a>.</p>
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Stackexchange
|
Bicycle cop patrols the skate park.
| 0non-cybersec
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Reddit
|
There's always someone like you.
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Reddit
|
How can I protect my laptop from thieves?. <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://superuser.com/questions/213650/how-to-secure-a-laptop-against-thieves">How to secure a laptop against thieves</a> </p>
</blockquote>
<p>How can I protect my laptop from thieves? Is there any software that can trace my laptop if it is stolen? What is the software? How to use it?</p>
| 0non-cybersec
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Stackexchange
|
Unable to communicate between guest OS in VirtualBox in both. <p>I was wondering that can we connect guest os in virtual box via internal networking like physical machine networking.</p>
<p>So I created two guest OS having Kali in one vm and windows in another vm, and creates a dhcp server via <code>vboxmanage</code> command</p>
<p>the command i fired is enter code here</p>
<pre><code>vboxmanage dhcpserver add --netname testnet --ip 10.0.0.1 --netmaask 255.255.255.0 --upperip 10.0.0.20 --lowerip 10.0.0.10 --enable
</code></pre>
<p>and assigned to vm via internal network and allowed the Promiscuous Mode to <code>Allow VMs</code>. Please also explain what is Promiscuous Mode.</p>
<p>Assigned static IP to Kali Linux is 10.0.0.20 and assigned static IP to windows is 10.0.0.10.</p>
<p>I was successful at one end only means able to ping Kali with windows but failed to ping Windows via Kali Linux</p>
<p><a href="https://i.stack.imgur.com/Tx47R.png" rel="nofollow noreferrer">Result of Ping from Windows</a></p>
<p><a href="https://i.stack.imgur.com/ZDIFH.jpg" rel="nofollow noreferrer">Result of Ping from Kali Linux</a></p>
<p>I also tried same by Windows 7 and Windows 8, successful in pinging Windows 8 but failed to ping Windows 7.</p>
<p>Also if static ip's not alloted, the guest os are not getting default gateway address, but only getting IP Address.</p>
<p>Any ideas how to enable both side communication.</p>
<p>Thanks in advance :):)</p>
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Stackexchange
|
ar
X
iv
:m
at
h/
06
11
81
7v
1
[
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at
h.
G
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]
2
7
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ov
2
00
6
SIMPLY CONNECTED HOMOGENEOUS CONTINUA
ARE NOT SEPARATED BY ARCS
MYRTO KALLIPOLITI AND PANOS PAPASOGLU
Abstract. We show that locally connected, simply connected ho-
mogeneous continua are not separated by arcs. We ask several
questions about homogeneous continua which are inspired by anal-
ogous questions in geometric group theory.
1. Introduction
In this paper we prove a theorem about homogeneous continua in-
spired by a result about finitely presented groups ([13]).
Theorem 1. Let X be a locally connected, simply connected, homoge-
neous continuum. Then no arc separates X.
We recall that an arc in X is the image of a 1-1 continuous map
α : [0, 1] → X . We say that an arc α separates X if X rα has at least
two connected components. We say that X is simply connected if it
is path connected and every continuous map f : S1 = ∂D2 → X can
be extended to a continuous f̄ : D2 → X (where D2 is the 2-disc and
S1 its boundary circle). The proof of Theorem 1 relies on Alexander’s
lemma for the plane (see [12]) and our generalization of this lemma to
simply connected spaces (see sec. 2).
There is a family air between continua theory and group theory.
This became apparent after Gromov’s theory of hyperbolic groups ([6]).
Gromov defines a boundary for a hyperbolic group which is a contin-
uum on which the group acts by a ‘convergence action’. The classic
‘cyclic elements’ decomposition theory of Whyburn was extended re-
cently by Bowditch ([3]) in this context and it gave deep results in
group theory. ‘Asymptotic topology’, introduced by Gromov ([7]) and
developed further by Dranishnikov ([4]), shows that the analogy goes
beyond the realm of hyperbolic groups. The ‘philosophy’ of this is that
topological questions that make sense for continua can be translated to
Date: September 7, 2018.
1991 Mathematics Subject Classification. 54F15.
Research co-funded by European Social Fund and National Resources EPEAEK
II-Pythagoras.
1
http://arxiv.org/abs/math/0611817v1
2 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
‘asymptotic topology’ questions which make sense for groups (see [4]
for a dictionary between topology and asymptotic topology).
One wonders whether Theorem 1 holds in fact for all locally con-
nected, homogeneous continua of dimension bigger than 1:
Question 1. Let X be a locally connected, homogeneous continuum of
dimension 2. Is it true that no arc separates X?
We note that by a result of Krupski ([10]), homogeneous continua
are Cantor manifolds. It follows that no arc separates a homogeneous
continuum of dimension bigger than 2.
Krupski and Patkowska ([11]) have shown that a similar property
(the disjoint arcs property) holds for all locally connected homogeneous
continua of dimension bigger than 1 which are not 2-manifolds.
We remark that by [10] if a Cantor set separates a homogeneous con-
tinuum X then dimX = 1. So question 1 is equivalent to the following
question: Is it true that if no Cantor set separates a locally connected,
homogeneous continuum X , then no arc separates X? Restated in
this way the question makes sense also for boundaries of hyperbolic
groups. In fact a similar question can be formulated for finitely gener-
ated groups too (see [13]).
Not much is known about locally connected, simply connected ho-
mogeneous continua. One motivation to study them is the analogy
with finitely presented groups. Another reason is that one could hope
for a classification of such continua in dimension 2:
Question 2. Are the 2-sphere and the universal Menger compactum of
dimension 2 the only locally connected, simply connected, homogeneous
continua of dimension 2?
We recall that S1 and the universal Menger curve are the only locally
connected homogeneous continua of dimension 1 ([1]). Prajs ([15],
question 2) asks whether S2 is the only simply connected homogeneous
continuum of dimension 2 that embeds in R3.
A related question about locally connected, simply connected con-
tinua that makes sense also for finitely presented groups is the following:
Question 3. Let X be a locally connected, simply connected homoge-
neous continuum which is not a single point. Does X contain a disc?
We remark that in the group theoretic setting the answer is affirma-
tive for hyperbolic groups ([2]). By a result of Prajs ([14]) a positive
answer to this would imply that S2 is the only locally connected, sim-
ply connected, homogeneous continuum of dimension 2 that embeds in
R
3.
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS 3
We refer to Prajs’ list of problems ([15]) for more questions on ho-
mogeneous continua.
2. Preliminaries
Definition 1. Let X be a metric space. A path p is a continuous map
p : [0, 1] → X. A simple path or an arc α, is a continuous and 1 − 1
map α : [0, 1] → X. We will identify an arc with its image.
For a path p we denote by ∂p the set of its endpoints, i.e. ∂p =
{p(0), p(1)}.
An arc α separates X if X rα has at least two connected components.
If x, y ∈ X we say that an arc α separates x from y if α separates X
and x, y belong to distinct components of X r α.
Definition 2. Let α be an arc of X. On α we define an order <α as
follows: If x = α(x′), y = α(y′) then x <α y if and only if x
′ < y′.
We denote by [x, y]
α
the set of all t ∈ α such that x ≤ t ≤ y. Similarly
we define (x, y)
α
, [x, y)α and (x, y]α. When there is no ambiguity we
write [x, y] instead of [x, y]
α
and x < y instead of x <α y. Finally, if
t ∈ [0, 1] we denote by x+ t the point α(x′ + t) (where x = α(x′)).
We recall Alexander’s lemma from plane topology (see Theorem 9.2,
p.112 of [12]).
Alexander’s Lemma (for the plane). Let K1, K2 be closed sets on
the plane such that either K1 ∩K2 = ∅ or K1 ∩K2 is connected and at
least one of K1, K2 is bounded. Let x, y ∈ R
2
r (K1 ∪K2). If there is
a path joining x, y in R2 rK1 and a path joining x, y in R
2
rK2 then
there is a path joining x, y in R2 r (K1 ∪K2).
It is easy to see that Alexander’s lemma also holds for the closed
disc D2 in the case that K1 ∩ K2 = ∅. In fact this implies that this
lemma holds in general for every simply connected space. In particular
we have the following:
Alexander’s Lemma (for simply connected spaces). Let X be
a simply connected space, K1, K2 disjoint closed subsets of X and let
x, y ∈ Xr(K1∪K2). If there is a path joining x, y in XrK1 and a path
joining x, y in XrK2 then there is a path joining x, y in Xr(K1∪K2).
4 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
Proof. Let p1, p2 be paths joining x, y such that p1∩K1 = p2∩K2 = ∅.
We consider the closed path p1 ∪ p2 and let f : S
1 = ∂D2 → X be
a parametrization of this path. Since X is simply connected, f can
be extended to a map F : D2 → X . Then F−1(K1) and F
−1(K2) are
disjoint, closed subsets of D2. Clearly, neither F−1(K1) nor F
−1(K2)
separates x, y, therefore, using Alexander’s lemma for the closed disc,
we have that there is a path p that joins x, y without meeting F−1(K2)∪
F−1(K2). This implies that F (p) is a path from x to y that does not
meet K1 ∪K2. �
For the rest of this paper we assume that X is a simply connected,
locally connected continuum.
Lemma 1. Let O be a connected open subset of X, K be a connected
component of ∂O and let x, y ∈ O such that d(x,K) < ε and d(y,K) <
ε. Then there is a path p in O connecting x to y such that p is contained
in the ε-neighborhood of ∂O.
Proof. Let U be the union of the open balls Bε(t) with center t ∈ ∂O
and radius ε. Let V be the connected component of U containing K.
Clearly x, y ∈ V so there is a path in X joining them that does not
intersect ∂V . On the other hand x, y ∈ O so there is a path in X
joining them that does not intersect ∂O.
Since ∂O ∩ ∂V = ∅ and ∂O, ∂V are closed, applying Alexander’s
lemma for the simply connected space X , we have that p is a path
lying in X joining x, y that intersects neither ∂O nor ∂V . Clearly p is
contained in O and lies in the ε-neighborhood of ∂O. �
Lemma 2. Let α be an arc that separates X and let C be a connected
component of Xrα. Then C is simply connected and ∂C is connected.
Proof. Let f : S1 = ∂D2 → C. We will show that this map can be
extended to a map f̂ : D2 → C.
X is simply connected, so there is a map F : D2 → X such that
F |S1 = f . Furthermore, X r C is an open set, therefore ∂F
−1(X r
C) ∩ F−1(X r C) = ∅ and since F is a continuous extension of f it
follows that F (∂F−1(X rC)) ⊂ α (where by ∂F−1(X rC) we denote
the boundary of F−1(X r C) in X).
Let f ′ : ∂F−1(X rC) → α be the restriction of F in ∂F−1(X rC).
Then, applying Tietze’s extension theorem, we obtain an extension for
f ′:
F ′ : F−1(X r C) → α.
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS 5
Finally we define f̂ : D2 → C as follows:
f̂(x) =
f(x) if x ∈ ∂D2,
F (x) if x ∈ D2 r F−1(X r C),
F ′(x) if x ∈ F−1(X r C).
This shows that C is simply connected.
Suppose now that S = ∂C is not connected and let p be a path that
joins two different components of S, such that if a, b are the endpoints
of p, then (pr {a, b}) ∩ S = ∅. Let x ∈ pr {a, b} and y ∈ (a, b)
α
r S.
(Figure 1).
a b
p
x
yS(0) (1)
Figure 1.
We set K1 = [α(0), y]α ∩S and K2 = SrK1. It is clear that K1, K2
are disjoint closed subsets of X and that neither K1 nor K2 separates
x from y. Then Alexander’s lemma (for the simply connected space
X) implies that there is a path joining x, y in Xr (K1 ∪K2) = X rS,
a contradiction. �
Lemma 3. Let α be an arc that separates X, x, y ∈ α and ε > 0 with
ε < d(x, y). Then for every connected component C of Xrα such that
x, y ∈ ∂C there are points x′, y′ ∈ C with d(x, x′), d(y, y′) < ε and a
path p ∈ C that joins x′, y′ and is contained in the ε-neighborhood of
[x, y]
α
.
Proof. Let B ε
2
(x) and Bε(x) be balls of center x and radius
ε
2
and ε
respectively. We consider the connected components of αr
◦
B ε
2
(x) and
we restrict to those that are not contained in Bε(x) (here we denote by
◦
B ε
2
(x) the open ball). It is clear that there are finitely many such com-
ponents, so we denote them by I1, I2, . . . , In. Let δ1 < min{d(Ii, Ij)}
for every i, j = 1, 2, . . . , n, i 6= j. Similarly, let J1, J2, . . . , Jm be the
connected components of α r
◦
B ε
2
(y) that are not contained in Bε(y)
and let δ2 < min{d(Ji, Jj)} for every i, j = 1, 2, . . . , m, i 6= j. Let
δ′ < min{δ1, δ2,
ε
2
}.
6 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
From Lemma 2, we have that C is simply connected, therefore Lemma
1, for δ = δ
′
4
, implies that there is a path q ∈ C that joins a point of
Bε(x) with a point of Bε(y) and lies in the δ-neighborhood of α (Figure
2). We will show that there is a subpath of q that has the required
properties.
x
y
B (y)
B (x)
C
q
Figure 2.
We assume that none of the Ii, Jj, i = 1, 2, . . . , n, j = 1, 2, . . . , m
contain [x, y]
α
, since otherwise we are done. Thus, without loss of
generality, let I1 be the connected component of α r
◦
B ε
2
(x) that is
contained in [x, y]
α
. We denote by Nδ([x, y]α) the open δ-neighborhood
of [x, y]
α
. Suppose that there is a connected component I = [a, b]q of
q r Nδ([x, y]α) with a ∈ Bδ(x) and b /∈ Bδ(x). Then there is an r > 0
such that (b− r, b)
q
/∈ Nδ(α). Indeed, if not then for every r > 0 there
is an Ii 6= I1 such that (b− r, b)q ∈ Nδ(Ii). Thus d(Ii, b) ≤ δ. But
d(I1, Ii) ≤ d(Ii, b) + d(b, I1) ≤ δ+ δ = 2δ =
δ′
2
< δ′, a contradiction. So
there is an r > 0 such that for every i 6= 1 we have (b− r, b)
q
/∈ Nδ(Ii),
therefore (b− r, b)
q
/∈ Nδ(α), which is not possible. This contradiction
proves the lemma. �
3. Proof of Theorem 1
We will prove the theorem by contradiction.
Remark: Since X is locally connected and compact, it follows that
every open connected subset of X is path connected (see Theorem 3.15,
p.116 of [9]). In particular the closure of every component of X r α is
path connected.
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS 7
Definition 3. Let α1, α2 be arcs that separate X. We say that α1
crosses α2 at x ∈ (α1 r ∂α1) ∩ (α2 r ∂α2) if for any neighborhood of x
in α2, (x− ε, x+ ε)α2, there are a, b ∈ (x− ε, x+ ε)α2 separated by α1.
More generally, if [x1, x2] is a connected component of α1 ∩ α2, which
is contained in (α1 r ∂α1) ∩ (α2 r ∂α2), we say that α1 crosses α2 at
[x1, x2] if for any neighborhood of [x1, x2] in α2, (x1−ε, x2+ε)α2, there
are a, b ∈ (x1 − ε, x2+ ε)α2 separated by α1. In this case, the endpoints
x1, x2 are also called cross points of α1, α2.
If I1 ⊂ α1, I2 ⊂ α2 are intervals of α1, α2 containing x in their
interior, we say that I1, I2 cross at x. Similarly we define what it means
for two intervals to cross at a common subarc. We call x (respectively
[x1, x2]) a cross-point (respectively cross-interval) of α1, α2. We say
that I1, I2 cross if they cross at some point x or at some interval [x1, x2].
1
2
xy
Figure 3.
For example in Figure 3, x is a cross point of α1, α2, while y is an
intersection point of α1, α2 which is not a cross point.
Lemma 4. There is an arc that separates X in exactly two components.
Proof. Suppose that this is not the case, so let α be an arc that sep-
arates X in more than two components. Since X has no cut points
there are two connected components of X r α, say C1, C2, such that
β = ∂C1 ∩ ∂C2 6= ∅ is a subarc (which is not a point) of α. Clearly β
separates X . To simplify notation we denote by C1, C2 the components
of X r β that satisfy ∂C1 = ∂C2 = β. Let C3 be another component
of X r β. By Lemma 2 we have that ∂C3 is connected, so ∂C3 = γ is
a subarc of β, which separates X .
Lemma 4.1. γ cannot be crossed by any other separating arc of X.
8 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
Proof. Suppose that there is an arc γ′ that separates X and crosses γ
at t. Then there are x, y ∈ γ, x < t < y that are separated by γ′.
Let Y = C1 ∪ C2. Since C3 is path connected, it follows that γ
′ 6⊂ Y .
We denote by Cx the connected component of Y r γ
′ that contains x.
As in Lemma 2, we may show that Y is simply connected. We show
then that Alexander’s lemma for Y implies that ∂Cx has a connected
component that separates x, y in Y .
This can be achieved as follows: Let Cy be the connected component
of Y rCx that contains y. It is clear that ∂Cy ⊆ ∂Cx. Then no proper
closed subset of ∂Cy separates x from y. Indeed, suppose that there is
a closed K ⊂ ∂Cy that separates x from y and let z ∈ ∂Cy rK. Let
U be an open neighborhood of z such that U ∩K = ∅. It is obvious
that U intersects every component of Y r ∂Cy , therefore, there are
paths q1, q2 ∈ Y r∂Cy that join x, y with points x
′, y′ ∈ U respectively.
However U is path connected and since U ∩K = ∅, it follows that there
is also a path q ∈ U that joins x′ with y′. Thus p1 ∪ q ∪ p2 is a path
joining x, y without meeting K, a contradiction. Therefore, I = ∂Cy is
connected and separates x from y.
We note now that I does not cross γ. Indeed, suppose that there is
an a ∈ I r ∂I in which γ′ crosses γ. Let V ⊂ X be sufficiently small
neighborhood of a such that (γ′ r I) ∩ V = ∅. We denote by J the
connected component of γ ∩V that contains the point a. Then we can
pick points x′, y′ ∈ J with x′ < a < y′ in γ that are separated by γ′.
Let Nx′, Ny′ be connected neighborhoods of x
′ and y′ respectively, such
that Nx′, Ny′ ⊂ V . Applying now Lemma 3 for the component C3 and
for ε < min{diam(Nx′ ∩C3), diam(Ny′ ∩C3)}, we have that every point
of Nx′ ∩C3 can be joined with every point of Ny′ ∩C3 by a path in C3
which lies in the ε-neighborhood of [x′, y′]
γ
.
Let t ∈ C3 ∩Nx′, s ∈ C3 ∩Ny′ and let q be a path that joins t and s
as above. We note now that Nx′ and Ny′ are path connected, so there
are paths q1 ∈ Nx′ and q2 ∈ Ny′ joining the endpoints of q with the
points x′ and y′ respectively. Clearly then the path p = q1∪q∪q2 joins
x′ with y′ without meeting γ′. This is however impossible, since x′ and
y′ are separated by γ′. Therefore, I does not cross γ. Thus γ r I is
contained in a single component of X r I, a contradiction. �
We return now to the proof of Lemma 4: Let G be the group of
homeomorphisms of X . For every f ∈ G we have that f(γ) separates
X and from the previous lemma it follows that f(γ) does not cross γ.
Let S = G · γ. Clearly S is uncountable. Let Q be a countable dense
set of X . We define a map R : S → Q× Q× Q as follows: Let p ∈ S
and U1, U2, U3 be three connected components of X r p. For every Ui
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS 9
we pick an ri ∈ Q and we associate p ∈ S the triple (r1, r2, r3). We
remark that R is 1 − 1 map, which is a contradiction. This completes
the proof of Lemma 4. �
Let γ be an arc that separates X in exactly two components C1, C2
with ∂C1 = ∂C2 = γ. We denote by G the group of homeomorphisms
of X . Let S = G · γ. It is clear that S is uncountable and that every
arc α ∈ S also separates X in exactly two components U1, U2 such that
∂U1 = ∂U2 = α. For an arc α ∈ S we will denote these two components
by α+ and α− (Figure 4).
Henceforth we will consider only arcs in S.
+
-
Figure 4.
Lemma 5. Let α1, α2 ∈ S such that α1 crosses α2 at x (or at [x1, x2]).
Then α2 crosses α1 at x (or at [x1, x2]).
Proof. Suppose that there are α1, α2 ∈ S such that α1 crosses α2 at x
but α2 does not cross α1 at x. Then there is an interval I ⊂ α1 con-
taining x at its interior that lies in the closure of one of the components
of X r α2, say α
+
2
. Clearly then we have that I ∩ α−
2
= ∅.
Let V ⊂ X be sufficiently small neighborhood of x such that (α1 r
I) ∩ V = ∅. We denote by J the connected component of α2 ∩ V that
contains the point x. We pick two points a, b ∈ J with a < x < b in α2
which are separated by α1 and let Na, Nb be connected neighborhoods
of a and b respectively such that Na, Nb ⊂ V . As in proof of Lemma
4.1, for ε < min{diam(Na ∩ α
−
2
), diam(Nb ∩ α
−
2
)}, we can find a path
p that joins a with b without meeting α1, a contradiction. We argue
similarly if α1 crosses α2 at an interval [x1, x2].
�
We recall now a version of Effros’ Theorem ([5], [8] p. 561):
10 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
Theorem 2. For every ε > 0 and x ∈ X the set W (x, ε) of y ∈ X
such that there is a homeomorphism h : X → X with h(x) = y and
d(h(t), t) < ε for all t ∈ X, is open.
Lemma 6. There are arcs α = [a1, a2] and β = [b1, b2] in S, such
that b1 ∈ (a1, a2)α and if A is the connected component of α ∩ β that
contains b1, then a1, a2 6∈ A.
Proof. We will need the following:
Lemma 6.1. Let α ∈ S. Then there is an arc β ∈ S that crosses α.
Proof. Let α, γ ∈ S, c ∈ ∂γ, a ∈ α r ∂α and g ∈ G such that gc = a.
By the definition of S it is not possible that gγ ⊂ α, since α separates
X in exactly two connected components. Assume now that α does not
cross gγ. We denote by A the connected component of α ∩ gγ that
contains a and let ∂α = {a1, a2}.
We distinguish two cases: Suppose that a1, a2 6∈ A. Let z ∈ gγ such
that (z, gc)gγ lies in the closure of one of the components of X r α,
say α+. Let z′ ∈ (z, gc)gγ r α and ε > 0 such that Bε(z
′) ⊂ α+.
By Theorem 2 there is a δ > 0 such that Bδ(a) ⊂ W (a, ε). Let y ∈
Bδ(a) ∩ α
− (Figure 5).
a=gc
g
+
-
z’
z yA a2a1
Figure 5.
Then there is a homeomorphism, h ∈ G, with h(a) = y such that
d(t, h(t)) < ε for every t ∈ X . We consider the arc β = h(gγ). Then
clearly β crosses α, since h(z′) ∈ α+ and h(a) ∈ α−.
Suppose now that a2 ∈ A. We consider the homeomorphism h ∈ G
of the previous case. If a2 6∈ h(A), then clearly Lemma 6.1 is proved.
So let a2 ∈ h(A) and ε
′ < min{ε, 1
2
d(α, h(gc))}. As before, by Theorem
2, there is a δ′ > 0 such that Bδ′(a2) ⊂ W (a2, ε
′). Let y′ ∈ Bδ′(a2)∩α
+
(Figure 6). Then there is an h′ ∈ G with h′(a2) = y
′ and d(t, h′(t)) < ε′
for every t ∈ X . It is obvious now that α crosses h′(β).
�
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS11
=h(g )
+
-
y’
A a2a1
h(gc)
Figure 6.
Let α = [a1, a2] ∈ S. By Lemma 6.1 there is an arc β = [b1, b2]
that crosses α at x ∈ α ∩ β. Without loss of generality, suppose that
x is the endpoint of a cross interval I of α, β. Let γ ∈ S, c ∈ ∂γ and
g ∈ G such that gc = x. We denote by A the connected component of
gγ∩α that contains c and similarly by B the component of gγ∩β that
contains c. Clearly if a1, a2 /∈ A, then Lemma 6 is proved. Otherwise,
we note that if A contains one of the endpoints of α, then b1, b2 /∈ B,
since I is a cross interval of α, β. So in this case, the required arcs are
gγ and β. �
We return to the proof of Theorem 1.
Let α = [a1, a2], β = [b1, b2] ∈ S be paths as in Lemma 6, that
is b1 ∈ (a1, a2)α and if A is the connected component of α ∩ β that
contains b1, then a1, a2 /∈ A. Let t1, t2 ∈ α r β such that b1 ∈ (t1, t2)α
and let p1, p2 be paths joining t1, t2 in α+ and α− respectively (the
points ti exist since a1, a2 /∈ A). We pick pi such that pi∩α has exactly
two connected components neither of which intersects β (this can be
achieved using Lemma 3 for ε < 1
2
min{d(t1, ∂A1), d(t2, ∂A2)}, where
Ai is the connected component of αr β that contains ti and ∂Ai is its
boundary in α).
Let ε > 0 with ε < 1
2
d(A, p1 ∪ p2). As in proof of Lemma 6.1, using
Theorem 2, we can find a homeomorphism h ∈ G such that h(β) crosses
α at x ∈ α∩h(β), with d(A, x) < ε. Then we remark that x ∈ (t1, t2)α
and that the subarc of hβ with endpoints x and hb1 does not intersect
p1 ∪ p2.
We pick now points s ∈ (t1, x)αr (p1∪p2) and t ∈ (x, t2)αr (p1∪p2)
which are separated by hβ so that they satisfy the following: If y is
a cross point of α and hβ, lying in [s, t]
α
, then the subarc [y, hb1]hβ
12 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
does not intersect the paths p1, p2. Such points exist by definition of x
(Figure 7).
h
hb2
hb1
t1 t2x
+
-
p1
p2
s t a2a1
Figure 7.
We consider now the closed paths p1 ∪ [t1, t2]α and p2 ∪ [t1, t2]α. Let
D1, D2 be discs and let f1 : D1 → α+, f2 : D2 → α− be maps so that
f1(∂D1) = p1 ∪ [t1, t2]α, f2(∂D2) = p2 ∪ [t1, t2]α (such maps exist, since
α+ and α− are simply connected by Lemma 2).
We ‘glue’ D1, D2 along [t1, t2]α and we obtain a disc D and a map
f : D → X with f(∂D) = p1 ∪ p2. More precisely, we consider the
disc D = D1 ⊔ D2/ ∼, where ∼ is defined as follows: x1 ∼ x2 if and
only if x1 ∈ ∂D1, x2 ∈ ∂D2 and f1(x1) = f2(x2). Finally, we define
f : D → X as:
f(t) =
{
f1(t), if t ∈ D1,
f2(t), if t ∈ D2.
By abuse of notation we identify points of [t1, t2]α in D with their
image under f . We note that the interior, say U , ofD is homeomorphic
to R2 and since t, s are separated by hβ in X , it follows by Alexan-
der’s lemma that t, s are separated in U by a connected component of
f−1(hβ) ∩ U . We call this component K (Figure 8).
Clearly f(K) is a subarc of hβ that contains cross points or cross
intervals of hβ with [s, t]α. Let c be such a cross point. Then we
can write f(K) as f(K) = I1 ∪ I2, where Ii, i = 1, 2, are (connected)
subarcs of hβ, such that I1 ∩ I2 = c. Furthermore, at least one of I1, I2
does not intersect p1 ∪ p2 (this is by our choice of h and c). It follows
that at least one of f−1(I1) ∩ U, f
−1(I2) ∩ U is compact.
We set I ′
1
= I1 r c, I
′
2
= I2 r c. We will define two sets K1, K2
such that the following are satisfied: K1, K2 are closed subsets of U
that contain f−1(I ′
1
) and f−1(I ′
2
) respectively, K1 ∩ K2 is connected
contained in f−1(c) and K1 ∪K2 = K.
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS13
t1 t2
D1
D2
t s
K
c
Figure 8.
We consider the connected components of f−1(c) ∩ K. We remark
that there is exactly one component of f−1(c)∩K, say C, that intersects
both D1 and D2.
Let now C1 be a connected component of f
−1(c)∩K different from C
and suppose that C1 ⊂ D1. We consider the closure ofK, K, inD1∪D2.
K is connected thus the closure of the component of K∩D1 containing
C1 intersects ∂D1. Indeed, we consider the set K ∩ (D1 − [t1, t2]α)
as an open subset of the continuum K. Let K ′ be the component of
K ∩ (D1 − [t1, t2]α) that contains C1. We recall that if U is an open
subset of a continuum and C is a component of U then the frontier of
U contains a limit point of C (Theorem 2.16, p.47 of [9]). It follows
that the closure of K ′ intersects [t1, t2]α.
Therefore, we have that f(K ′) ⊂ α+ so f(K ′) ⊂ I1 or f(K
′) ⊂ I2.
We remark that if f(K ′) ⊂ I1 then a non trivial interval of I1 containing
c lies in α+.
We have a similar conclusion if f(K ′) ⊂ I2. Therefore if a connected
component of f−1(c) ∩ K different from C lying in D1 intersects the
closure of both f−1(I ′
1
), f−1(I ′
2
) we have that an open interval of I1
around c lies in α+. This is impossible since c is a cross point. We
argue similarly for connected components of f−1(c) ∩ K contained in
D2.
We conclude that the union of the components of f−1(c) ∩K which
lie in D1, intersect exactly one of f
−1(I ′
1
), f−1(I ′
2
). Clearly the same is
true for the union of the components of f−1(c) ∩ K contained in D2.
In particular exactly one of the following two holds:
14 MYRTO KALLIPOLITI AND PANOS PAPASOGLU
1. If C1 is a connected component of f
−1(c) ∩ K different from C
lying in D1 then the component of D1 ∩ K containing C1 intersects
f−1(I ′
1
), while if C1 lies in D2 the component of D2 ∩K containing C1
intersects f−1(I ′
2
).
2. If C1 is a connected component of f
−1(c) ∩ K different from C
lying in D1 then the component of D1 ∩ K containing C1 intersects
f−1(I ′
2
), while if C1 lies in D2 the component of D2 ∩K containing C1
intersects f−1(I ′
1
).
Assume that we are in the first case. Then we define K1 to be the
union of the components of f−1(c) ∩K intersecting D1 together with
f−1(I ′
1
). We define K2 to be the union of the components of f
−1(c)∩K
intersecting D2 together with f
−1(I ′
2
). It is clear that K1, K2 are closed
and that K1 ∩K2 = C, K1 ∪K2 = K. Since K is connected, K1, K2
are connected too. We define K1, K2 similarly in the second case.
We note now that at least one of K1, K2 is compact subset of U , thus
bounded in U . Since K separates s, t and K1, K2 are closed subsets of
D, applying Alexander’s lemma for the plane we have that at least one
of K1, K2 separates s, t in U .
It follows that either f−1(I1) or f
−1(I2) separates s, t. We remark
that the same argument holds in the case c is replaced by a cross interval
J : We have then that I = I1 ∪ I2 with I1 ∩ I2 = J and as before either
f−1(I1) or f
−1(I2) separate s, t in U . Now we can continue subdividing
intervals along cross points (cross intervals) that lie in [s, t]α as follows:
Let’s say that f−1(I1) separates s, t. We have that there is a connected
component of f−1(I1), sayM , that separates them. We note that f(M)
is a subinterval of I1 and if there is a cross point (or cross interval) of
[t, s]α, hβ contained in its interior, we repeat the previous procedure
replacing K by M . If not we have a contradiction. Therefore, either
s, t are separated in U by the inverse image of an interval f(K) of hβ
which does not contain in its interior any cross point of hβ, α lying
in [s, t]α, or by iterating this procedure we conclude that the inverse
images under f of intervals of hβ of arbitrarily small diameter separate
s from t in U . It is clear that both are impossible, so the theorem is
proven.
References
[1] R.D. Anderson, One-dimensional continuous curves and a homogeneity theo-
rem Ann. of Math. (2) 68, p.1-16 (1958).
[2] M. Bonk, B. Kleiner, Quasi-hyperbolic planes in hyperbolic groups Proc. Amer.
Math. Soc. 133, no. 9, p.2491-2494 (2005).
SIMPLY CONNECTED HOMOGENEOUS CONTINUA ARE NOT SEPARATED BY ARCS15
[3] B.H. Bowditch, Cut points and canonical splittings of hyperbolic groups, Acta
Math. 180, No.2, p.145-186 (1998).
[4] A. Dranishnikov, Asymptotic topology, Russian Math.Surveys 55, No 6, p.71-
116 (2000).
[5] E.G. Effros, Transformation groups and C∗-algebras Ann. of Math. (2) 81, p.
38-55 (1965).
[6] M. Gromov, Hyperbolic groups, Essays in group theory (S. M. Gersten, ed.),
MSRI Publ. 8, Springer-Verlag, p. 75-263 (1987).
[7] M. Gromov, Asymptotic invariants of infinite groups, ’Geometric group the-
ory’, (G.Niblo, M.Roller, Eds.), LMS Lecture Notes, vol. 182, Cambridge Univ.
Press (1993).
[8] C.L. Hagopian, No homogeneous tree-like continuum contains an arc Proc.
Amer. Math. Soc. 88, no. 3, p.560-564 (1983).
[9] J.G. Hocking, G.S. Young, Topology, Dover (1961).
[10] P. Krupski, Homogeneity and Cantor manifolds Proc. Amer. Math. Soc. 109,
no. 4, p.1135-1142 (1990).
[11] P. Krupski, H. Patkowska, Menger curves in Peano continua Colloq. Math.
70, no. 1, p.79-86 (1996).
[12] M.H.A. Newman, Elements of the topology of plane sets of points, Cambridge
University Press (1951).
[13] P. Papasoglu, Quasi-isometry invariance of group splittings, Annals of Math.
161, pp. 759-830 (2005).
[14] J.R. Prajs, Homogeneous continua in Euclidean (n+1)-space which contain an
n-cube are n-manifolds Trans. Amer. Math. Soc. 318, no. 1, p.143-148 (1990).
[15] J.R. Prajs, Thirty open problems in the theory of homogeneous contiunua, in
Problems from Topology Proceedings, Ed. by E. Pearl vi+216 pages, Topology
Atlas (2003).
E-mail address, Myrto Kallipoliti: [email protected]
E-mail address, Panos Papasoglu: [email protected]
(Myrto Kallipoliti) Mathematics Department, University of Athens,
Athens 157 84, Greece
(Panos Papasoglu) Mathematics Department, University of Athens,
Athens 157 84, Greece
1. Introduction
2. Preliminaries
3. Proof of Theorem ??
References
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arXiv
|
SparkR bottleneck in createDataFrame?. <p>I'm new to Spark, SparkR and generally all HDFS-related technologies. I've installed recently Spark 1.5.0 and run some simple code with SparkR:</p>
<pre><code>Sys.setenv(SPARK_HOME="/private/tmp/spark-1.5.0-bin-hadoop2.6")
.libPaths("/private/tmp/spark-1.5.0-bin-hadoop2.6/R/lib")
require('SparkR')
require('data.table')
sc <- sparkR.init(master="local")
sqlContext <- sparkRSQL.init(sc)
hiveContext <- sparkRHive.init(sc)
n = 1000
x = data.table(id = 1:n, val = rnorm(n))
Sys.time()
xs <- createDataFrame(sqlContext, x)
Sys.time()
</code></pre>
<p>The code executes immediately. However when I change it to <code>n = 1000000</code> it takes about 4 minutes (time between two <code>Sys.time()</code> calls). When I check these jobs in console on port :4040, job for <code>n = 1000</code> has duration 0.2s, and job for <code>n = 1000000</code> 0.3s. Am I doing something wrong?</p>
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Stackexchange
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Can a har file be programmatically generated from headless chrome using Puppeteer?. <p>I would like to control a headless chrome instance using puppeteer, taking snapshots and clicking on various page elements, while capturing a har file. Is this possible? I have looked at the <a href="https://github.com/GoogleChrome/puppeteer/blob/master/docs/api.md" rel="noreferrer">API</a> but haven't found anything useful. </p>
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Stackexchange
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How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
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Stackexchange
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Can I use different log files for FortiGate rules?. <p>Fortigate firewalls use different log files per type and device. <a href="https://help.fortinet.com/fos50hlp/54/Content/FortiOS/fortigate-logging-reporting-54/logs.htm" rel="nofollow noreferrer">Here is the log file name format</a>:</p>
<pre><code> <logtype> - <logdevice> - <date> T <time> . <id>.log
</code></pre>
<p>For example: <code>AntiVirusLog-disk-2012-09-13T11_07_57.922495.log</code></p>
<p>Is there a way to specify a log file per rule or rules group? Ideally, I would like a separate log file for a group of rules with a given minimal priority level.</p>
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Stackexchange
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Dear Reddit: What is your profession? Where do you work and what do you do there?. Me? I own a printing company based in Baltimore MD, and I also work part-time for the USPS. How about you?
| 0non-cybersec
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Reddit
|
Tom Vassal talks about why unboxing videos are stupid...then proceeds to unbox his own game..
| 0non-cybersec
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Reddit
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Flatten a list in Prolog. <p>I've only been working with Prolog for a couple days. I understand some things but this is really confusing me.</p>
<p>I'm suppose to write a function that takes a list and flattens it.</p>
<pre><code>?- flatten([a,[b,c],[[d],[],[e]]],Xs).
Xs = [a,b,c,d,e]. % expected result
</code></pre>
<p>The function takes out the inner structures of the list.</p>
<p>This is what I have so far:</p>
<pre><code>flatten2([],[]).
flatten2([Atom|ListTail],[Atom|RetList]) :-
atom(Atom), flatten2(ListTail,RetList).
flatten2([List|ListTail],RetList) :-
flatten2(List,RetList).
</code></pre>
<p>Now, this works when I call:</p>
<pre><code>?- flatten2([a,[b,c],[[d],[],[e]]], R).
R = [a,b,c,d,e]. % works as expected!
</code></pre>
<p>But when I call to see if a list that I input is already flattened, is returns <code>false</code> instead of <code>true</code>:</p>
<pre><code>?- flatten2([a,[b,c],[[d],[],[e]]], [a,b,c,d,e]).
false. % BAD result!
</code></pre>
<p>Why does it work on one hand, but not the other? I feel like I'm missing something very simple.</p>
| 0non-cybersec
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Stackexchange
|
How to use the IF ALL statement in Python. <p>I have a function named checker(nums) that has an argument that will later receive a list. What i want to do with that list is to check if each other element is greater or equal to the previous one.
Example:
I have a list <code>[1, 1, 2, 2, 3]</code> and i have to check if it fulfills the condition.
Since it does, the function should return <code>True</code></p>
<p>My code:</p>
<pre><code>def checker(nums):
for x in range(len(nums)):
if x+1<len(nums):
if nums[x] <= nums[x+1] and nums[-1] >= nums[-2]:
return True
</code></pre>
<p>This will only run once and return True if the first condition is true.
I've seen a statement if all and am unsure of how to use it.</p>
| 0non-cybersec
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Stackexchange
|
Interface Builder was unable to determine the type of "Main.storyboard". This may be due to a missing SDK. <p>I am getting a strange error whenever I try to view a particular storyboard of a project in Xcode. When I click the storyboard in the file directory I get an alert that reads "Interface Builder was unable to determine the type of "Main.storyboard". This may be due to a missing SDK."</p>
<p>This only happens with this particular storyboard, I tried creating another storyboard and I was able to view it just fine. I tried reverting back to a previous version of the project when I was able to open the storyboard, but oddly enough this doesn't fix the error and I get the same alert. I also tried opening the project in an earlier version of Xcode, but to no avail.</p>
<p>If it provides more context, I get an error for the storyboard I get when I try to build the project that reads "The operation couldn’t be completed. (com.apple.InterfaceBuilder error 2001.)"</p>
<p>Any ideas on what causes this or how to fix it?</p>
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Stackexchange
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Forbes Highest Paid Athletes from 1990-Now.
| 0non-cybersec
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Reddit
|
CakePHP: Queries in Controller or Models?. <p>If I really go through the MVC approach, then the queries should be in the Model, in CakePHP case the Table Classes, but when I go through the tutorials and documentation that Cake Provides they simply state that queries should be in the Controller.</p>
<p>As you can see in the example over here on Cake's websites: <a href="https://book.cakephp.org/3.0/en/tutorials-and-examples/blog/part-two.html" rel="noreferrer">https://book.cakephp.org/3.0/en/tutorials-and-examples/blog/part-two.html</a></p>
<p>But if I go through this link or many other I have come across, the query part should be in the Models: <a href="https://www.toptal.com/cakephp/most-common-cakephp-mistakes#common-mistake-3-keeping-business-logic-in-controllers-instead-of-models" rel="noreferrer">https://www.toptal.com/cakephp/most-common-cakephp-mistakes#common-mistake-3-keeping-business-logic-in-controllers-instead-of-models</a></p>
<p>It is not just about what Cake displays in the examples or some developers opinion, but what should really be the genuine way to code in Cake when dealing with database queries. I have found almost 90% people doing query related tasks in Controllers only for Cake, as they quote "Cake mentions the same in their examples". But what about the MVC way, we create Table Classes just to mention the associations then? If Cake's own website does that, then somehow it means they have done it intentionally.</p>
| 0non-cybersec
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Stackexchange
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Dropdown in Django Model. <p>I want to create a field in Django <code>models.py</code> which will render as a dropdown and user can select the options from there. </p>
<p>If I have 5 choices:</p>
<ul>
<li>GREEN</li>
<li>BLUE</li>
<li>RED</li>
<li>ORANGE</li>
<li>BLACK</li>
</ul>
<p>How should I write my code in <code>models.py</code> and <code>Forms.py</code> so that the template renders it like a dropdown element?</p>
| 0non-cybersec
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Stackexchange
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How to duplicate a section in Mailchimp?. <p>I'm creating a template on mailchimp and I would like to add sections to it.
For now it is divided in "subheader/header/body/..."
I would like to add another section, so I will be able to change the background color just on this section.</p>
<p>How can I do that?</p>
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Stackexchange
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This is what I woke up to at ACen 2011, Day 1.
| 0non-cybersec
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Reddit
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Unique.
| 0non-cybersec
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Reddit
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Controlling the screenshot in the iOS 7 multitasking switcher. <p>I've been trying to find some information regarding the new multitasking switcher in iOS 7 and especially the screenshot that the OS takes when the app is going into hibernation.</p>
<p><img src="https://i.stack.imgur.com/tUNHz.jpg" alt="enter image description here"></p>
<p>Is there any way to completely turn off this feature or screenshot? Or can I hide the app altogether from the switcher? The app needs to run in the background, but we do not want to show any screenshot from the app.</p>
<p>The screenshot is potentially a security-risk, think along the lines for banking-apps where your card number or account summary will be available to anyone that double-click on the home button on the device.</p>
<p>Anyone with any insight into this? Thanks.</p>
| 0non-cybersec
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Stackexchange
|
Remote workstation management. <p>Some of my family members asked me to install Ubuntu on their laptops (because I just refuse to do anything windows related anymore), but helping them over the phone can be a pain.</p>
<p>Is there some kind of spacewalk/landscape alternative just for workstation management?</p>
<p>Things like Foreman, Salt, Ansible, ... all seem overkill, and don't seem to support dynamic ip's anyway.</p>
| 0non-cybersec
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Stackexchange
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The Art of Debate: Never Lose An Argument Again .
| 0non-cybersec
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Reddit
|
Anyone using VMware NSX T load balancer and VMware WAF (Avi Networks) in their environment. Hey folks as the title mentions does anyone use the new VMware WAF along with it\`s load balancing functionality.
Curious to see how it performs.
Lastly a little background. I work for a bank we got Fortinet/Palo alto that traffic would pass through towards our DMZ but i\`m not sure if to get the WAF functions of VMware for our Jack Henry core banking and our other digital banking products such as imaging etc, it seems IPS is close to a WAF so i\`m wondering if to add it.
Thanks
| 1cybersec
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Reddit
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Bending Wood. **Background**: (feel free to skip this)
Hey guys, this is my first post in here. I am looking to create a bed frame because I think it'll be a fun little project this spring. I'm very hopeful about how it'll look when I'm all done, but since I have never had much success in getting things *exactly* the length/angle I wanted in the past, I'm realistic that the outcome could be less than ideal. I have a feeling I'll be posting questions around here through the whole process.
**Question**:
Here is a [sketch](http://swigg.net/files/foot.png) of what I am trying to make. The red piece was just to figure out angles, in here it comes out to a point, I'd want it to be one continuous piece, just curved.
So for the "feet" of the bed I'm planning 6 or 8 curved pieces. My main question is how I can get this kind of bend. I saw you can steam wood to get it flexible, or use several thin pieces and glue them together in the curve you want.
* Is one of these methods stronger than the other?
* These will be supporting a lot of weight, the mattress/box frame, two people, and the rest of the bed which is going to be all wood, is there a way to figure out how thick these would have to be to support? I'm not opposed to having more than 8 though.
Edti:
Rough [model](http://imgur.com/a/IoqHc) of what i'm trying to do.
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Reddit
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My terrible Morty cosplay meets deadpool at a local comic con.
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Reddit
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Running command multiple times and trigger report on failure. <p>I am trying to run the "make" command on my project 100 times. But, obviously I wouldn't be monitoring the results but I would like to know when the build fails.</p>
<p>How would I go about making it trigger a notification or a mail or a report when the build failed ?</p>
<p>I am doing this to run "make" 100 times</p>
<pre><code> for run in {1..100}
do
make
done
</code></pre>
| 0non-cybersec
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Stackexchange
|
Yesterday I shit myself worse than I've ever shit myself.. As soon as I left the house I felt it stirring, but I thought I had plenty of time. Went to the store and on the way back realized the urgency. Everything was OK, right up to seeing my front door. Suddenly, I knew I couldn't hold it. It poured out into my pants, but it was contained. Until I got to the bathroom, pulled down my pants, and got shit everywhere. It was not a log. More like a pudding.
Threw out the underwear. Shorts were moist in the back, no chunks. Threw them in the laundry. Paper towels and cleaning spray took care of the bathroom.
There's a part of me that thinks this is hilarious. But there's a part of me that is super embarrassed. I haven't told anyone else about this. I will, some day, but I need a little time before I can admit this offense to my friends.
| 0non-cybersec
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Reddit
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Set theory construction. <p>What is N(U(P(2)-2)?</p>
<p>where N is the intersection of, U is the union of, and P(A) is the power set of A,and the minus sign represents relative complement or set difference</p>
<p>I get 0, Exercises in Set Theory by L.E. Sigler (Ex. 1.23) gives 1 - I reason as follows:</p>
<p>P(2) = {0,1,2, {1}}</p>
<p>P(2) - 2 = {0,1,2, {1}} - {0,1} = {2, {1}}</p>
<p>U((P(2)-2) = U({2, {1}} = { {0,1}, {1}} = {0,1} = 2</p>
<p>N2 = N{0,1} = 0</p>
<p>Am I missing something?</p>
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Stackexchange
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Since yall wanted to see more of the small venator.
| 0non-cybersec
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Reddit
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WikiLeaks Reveals Sinister Relations between Erdogan’s Family and ISIS.
| 0non-cybersec
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Reddit
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Does everyone get sleepy three times per day?. As someone who usually has an extremely erratic sleeping schedule, I've noticed something oddly constant in the middle of the chaos: no matter if I'm well-rested or have been up for 36 hours, I become sleepy and slow *exactly* three times per day, at *exactly* the same times:
* Around midnight.
* 7 AM.
* 2 PM.
Between these hours I'm fully alert -- no matter if I've been awake for just a few hours or for two days.
I remember hearing someplace that if you isolate humans entirely from the usual solar cycle, then they start living according to a 28-hour "day". Since there are roughly 7 hours of alertness between my sleepy episodes, then a 28-hour "day" would make a lot of sense actually.
So, basically, am I weird or is there research to back up my experience?
| 0non-cybersec
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Reddit
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It's the Apocalypse and your glasses just broke. What do you do?.
| 0non-cybersec
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Reddit
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Unordered layers in tikz-3dplot (3d effect). <h2>Short formulation of the question</h2>
<p>Is there a way to order layers in tikz-3dplot according to their distance to the observer's position? The picture should look realistic in order to give some intuition to the reader. But I prefer a solution without rearranging all items of the graphic.</p>
<h2>Example</h2>
<p>The following working example does not show the correct 3D view, i.e. the balls are wrong ordered in the second picture:</p>
<pre><code>\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\begin{document}
\begin{tikzpicture}
\node[fill=red,circle] (c1) at (0,0,0) {};
\node[fill=gray,circle] (c2) at (1,0,0) {};
\node[fill=blue,circle] (c3) at (1,0,1) {};
\node[fill=green,circle] (c4) at (0,0,1) {};
\node[fill=black,circle] (c5) at (0,1,0) {};
\node[fill=orange,circle] (c6) at (1,1,0) {};
\node[fill=yellow,circle] (c7) at (1,1,1) {};
\node[fill=magenta,circle] (c8) at (0,1,1) {};
\draw[line width=2pt] (c1) -- (c2) -- (c3) -- (c4) -- (c1);
\draw[line width=2pt] (c5) -- (c6) -- (c7) -- (c8) -- (c5);
\draw[line width=2pt] (c1) -- (c5);
\draw[line width=2pt] (c2) -- (c6);
\draw[line width=2pt] (c3) -- (c7);
\draw[line width=2pt] (c4) -- (c8);
\end{tikzpicture}
\tdplotsetmaincoords{45}{0}
\begin{tikzpicture}[tdplot_main_coords]
\node[fill=red,circle] (c1) at (0,0,0) {};
\node[fill=gray,circle] (c2) at (1,0,0) {};
\node[fill=blue,circle] (c3) at (1,0,1) {};
\node[fill=green,circle] (c4) at (0,0,1) {};
\node[fill=black,circle] (c5) at (0,1,0) {};
\node[fill=orange,circle] (c6) at (1,1,0) {};
\node[fill=yellow,circle] (c7) at (1,1,1) {};
\node[fill=magenta,circle] (c8) at (0,1,1)
\draw[line width=2pt] (c1) -- (c2) -- (c3) -- (c4) -- (c1);
\draw[line width=2pt] (c5) -- (c6) -- (c7) -- (c8) -- (c5);
\draw[line width=2pt] (c1) -- (c5);
\draw[line width=2pt] (c2) -- (c6);
\draw[line width=2pt] (c3) -- (c7);
\draw[line width=2pt] (c4) -- (c8);
\end{tikzpicture}
\tdplotsetmaincoords{54.7}{45}
\begin{tikzpicture}[tdplot_main_coords]
\node[fill=red,circle] (c1) at (0,0,0) {};
\node[fill=gray,circle] (c2) at (1,0,0) {};
\node[fill=blue,circle] (c3) at (1,0,1) {};
\node[fill=green,circle] (c4) at (0,0,1) {};
\node[fill=black,circle] (c5) at (0,1,0) {};
\node[fill=orange,circle] (c6) at (1,1,0) {};
\node[fill=yellow,circle] (c7) at (1,1,1) {};
\node[fill=magenta,circle] (c8) at (0,1,1) {};
\draw[line width=2pt] (c1) -- (c2) -- (c3) -- (c4) -- (c1);
\draw[line width=2pt] (c5) -- (c6) -- (c7) -- (c8) -- (c5);
\draw[line width=2pt] (c1) -- (c5);
\draw[line width=2pt] (c2) -- (c6);
\draw[line width=2pt] (c3) -- (c7);
\draw[line width=2pt] (c4) -- (c8);
\end{tikzpicture}
\end{document}
</code></pre>
<h2>Further outlook</h2>
<p>Maybe it is crucial to note that I want to use different view angles (feature of <code>[tikz-3dplot][1]</code>) without ordering the entire crystal with respect to the appearing layers.</p>
| 0non-cybersec
|
Stackexchange
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
|
Stackexchange
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
|
Stackexchange
|
i3bar spawns many more processes than expected. <p>I'm using i3blocks as my i3 status command and I've noticed abnormally high idle CPU usage (5-10% compared to 2-3% idle in Xfce) and decided to look at my task manager, and there seems to be <a href="https://i.imgur.com/U1QJn7d.pn" rel="nofollow noreferrer">many processes</a> being spawned and killed all the time.</p>
<p>I see why this is happening (CPU usage update, RAM usage update, clock update etc) however I am wondering if there is a way (or even different status command) that it can be avoided.</p>
<p>Thanks in advance.</p>
| 0non-cybersec
|
Stackexchange
|
Sydney.
| 0non-cybersec
|
Reddit
|
Control point border thickness in ggplot. <p>When using ggplot, I can set <code>shape</code> to 21-25 to get shapes that have independent setting for the internal (<code>fill</code>) and border (<code>col</code>) colors, like so:</p>
<pre><code>df <- data.frame(id=runif(12), x=1:12, y=runif(12))
ggplot(df, aes(x=x, y=y)) +
geom_point(aes(fill=id, size=id), colour="black", shape=21)
</code></pre>
<p><img src="https://i.stack.imgur.com/9eVZa.png" alt="enter image description here"></p>
<p>However, I can't figure out how to control the thickness of the shape borders, either setting them absolutely or as an aesthetic mapping. I note that if I set an <code>lwd</code> value, it overrides the <code>size</code> aesthetic:</p>
<pre><code>ggplot(df, aes(x=x, y=y)) +
geom_point(aes(fill=id, size=id), colour="black", shape=21, lwd=2)
</code></pre>
<p><img src="https://i.stack.imgur.com/ogboS.png" alt="enter image description here"></p>
<p>How do I control the border thickness?</p>
| 0non-cybersec
|
Stackexchange
|
Is there a way to phrase "there does not exist a universal set" in structural language?. <p>Both ZFC (which is a good example of a material set theory) and ETCS (which is a good example of a structural set theory) prove the sentence "there is no set having maximum cardinality" as an easy corollary of Cantor's theorem. However, ZFC also proves the sentence "there is no universal set," or more precisely: $$\neg \exists V \forall x(x \in V).$$</p>
<p>The question then arises of whether we can phrase this proposition for a structural set-theory like ETCS.</p>
<p><strong>Question.</strong> Is there a way to express "there does not exist a universal set" in structural language?</p>
| 0non-cybersec
|
Stackexchange
|
get dictionary key by value. <p>How do I get a Dictionary key by value in C#?</p>
<pre><code>Dictionary<string, string> types = new Dictionary<string, string>()
{
{"1", "one"},
{"2", "two"},
{"3", "three"}
};
</code></pre>
<p>I want something like this:</p>
<pre><code>getByValueKey(string value);
</code></pre>
<p><code>getByValueKey("one")</code> must be return <code>"1"</code>.</p>
<p>What is the best way do this? Maybe HashTable, SortedLists?</p>
| 0non-cybersec
|
Stackexchange
|
arXiv:quant-ph/9505018v1 24 May 1995
ar
X
iv
:q
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95
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99
5
Universality in Quantum Computation
David Deutsch, Adriano Barenco, and Artur Ekert
Clarendon Laboratory, Department of Physics
University of Oxford, Oxford, OX1 3PU, UK
February 1, 2008
Abstract
We show that in quantum computation almost every gate that operates on two
or more bits is a universal gate. We discuss various physical considerations
bearing on the proper definition of universality for computational components
such as logic gates.
To be published in Proc.R.Soc.London A, June 1995.
Introduction
It has been known for several years that the theory of quantum computers — i.e.
machines that rely on characteristically quantum phenomena to perform computa-
tions [1] — is substantially different from the classical theory of computation, which
is essentially the theory of the universal Turing machine. We may identify three im-
portant differences. Firstly, the properties of quantum computers are not postulated
in abstracto but are deduced entirely from the laws of physics. Logically, this was
already true of the classical theory, as Landauer [2] has pointed out, but the intuitive
nature of the classically-available computational operations, and the millennia-long
history of their study, allowed pioneers such as Turing, Church, Post and Gödel to
capture the correct classical theory by intuition alone, and falsely to assume that its
foundations were self-evident or at least purely abstract. (There is an analogy here
with geometry, another branch of physics that was formerly regarded as belonging to
mathematics.) Secondly, quantum computers can perform certain classical tasks, such
as factorisation [3], using quantum-mechanical algorithms [4] which have no classical
analogues and can be overwhelmingly more efficient than any known classical algo-
rithm. Thirdly, quantum computers can perform new computational tasks, such as
http://arXiv.org/abs/quant-ph/9505018v1
quantum cryptography [5], which are beyond the repertoire of any classical computer.
(The class of quantum computable functions is precisely the set of classical recursive
functions. However, unlike in the classical case, not all quantum computations can
be re-interpreted as function evaluations.)
Computation may be defined as the systematic creation of symbols (the “output”)
which, under a given method of interpretation, have abstract properties that were
specified in other symbols (the “input”). “Symbols” here are physical objects. A
universal set of components is one that is adequate for the building of computers
to perform any physically possible computation. A universal computer is a single
machine that can perform any physically possible computation. The concept of uni-
versality for computers on the one hand and for components on the other, and the
concept of computation itself, are all closely linked. For if the solution to a problem
could be created by a certain physically possible computer, but there were no system-
atic method of building that computer, then the solution would not necessarily be
“computable” in any useful sense. But if there were a systematic method of building
a computer to solve each such problem, then the factory that manufactured such com-
puters to order would in effect be a universal computer. And if there is a universal
computer, there must be a universal set of components, namely the components that
are needed to build it. Similarly, if there is a finite universal set of components then a
factory for manufacturing and assembling them into given configurations would also
be a universal computer.
Both the classical and the quantum theory of computation admit universal computers.
But the ability of the respective universal computers to perform any computation that
any other machine could perform under the respective laws of physics, could in the
classical case only be conjectured (the Church-Turing conjecture). In the quantum
theory it can be proved [1], at least for quantum systems of finite volume. This is
one of the many ways in which the quantum theory of computation has turned out
to be inherently simpler than its classical predecessor.
In this paper we concentrate on universality for components, and in particular for
quantum logical gates. These are the active components of quantum computational
networks [6], which are computers in which quantum two-state systems (i.e. quantum
bits or qubits) are carried inertly from one gate to another. In an n-bit quantum gate,
n qubits undergo a coherent interaction. Barenco [7] has shown that any two-bit gate
A(φ, α, θ) that effects a unitary transformation of the form
A(φ, α, θ) =
1 0 0 0
0 1 0 0
0 0 eiα cos θ −iei(α−φ) sin θ
0 0 −iei(α+φ) sin θ eiα cos θ
, (1)
on the state of two qubits is universal, where the representation (1) is in terms of the
2
computation basis {| 00〉 , | 01〉 , | 10〉 , | 11〉}, and α, φ, and θ are irrational multiples
of π and of each other.
The fact that the laws of physics support computational universality is a profound
property of Nature. Since any computational task that is repeatable or checkable
may be regarded as the simulation of one physical process by another, all computer
programs may be regarded as symbolic representations of some of the laws of physics,
specialised to apply to specific processes. Therefore the limits of computability coin-
cide with the limits of science itself. If the laws of physics did not support computa-
tional universality, they would be decreeing their own un-knowability. Since they do
support it, it would have been strangely anthropocentric if universality had turned
out to be a property of a very narrowly-defined class of interactions (such as (1)).
For then the physical processes in which such interactions occurred, presumably in-
cluding certain human artefacts, would have required a fundamentally more general
mathematical description than most other physical processes in the universe. But it
turns out that the opposite is the case. Almost every class of physical processes must
instantiate the same, standard set of mathematical relationships, namely those that
are quantum computable. For we shall prove that universality is not confined to the
special class (1) of gates, but that almost all two-bit quantum gates are universal.
This confirms, and betters, the conjecture of Deutsch [6] that almost all three-bit
quantum gates are universal.
Proof that almost all 2-bit gates are universal
Consider a two-bit quantum gate U that effects a unitary transformation U of U(4).
We shall prove that a generic U is universal, i.e. that the set of transformations
in U(4) corresponding to non-universal gates is of lower dimensionality than U(4)
itself, where U(4) is considered as a 16-dimensional manifold with the natural metric
√
1 − 1
4
Re(Tr(P †Q)). Define the repertoire of U as the set of gates whose effects on
their input qubits can be approximated with arbitrary accuracy by networks contain-
ing only U-gates.
U has the form
U = eiĤ1 , (2)
where Ĥ1 is a Hermitian operator. U and its generator Ĥ1 are diagonal in the same
basis, so n successive applications of the gate on the same pair of qubits effects the
unitary operation Un given in the diagonal basis by
einφ1 0 0 0
0 einφ2 0 0
0 0 einφ3 0
0 0 0 einφ4
, (3)
3
where the φj are the eigenvalues of Ĥ1. If, as will be the case for generic U, the
φj are irrational multiples of π and of each other, n can be chosen so that the four
expressions
φ̃j = nφj mod 2π j = 1 . . . 4 (4)
approximate arbitrarily well any set of values φ̃j in the interval [0, 2π[. In particular,
for any real λ and ǫ, there exists an integer n such that φ̃j = (λφj mod 2π) + O(ǫ)
for j = 1 . . . 4. In other words, any operation of the form
Uλ = eiλĤ1 (5)
is in our repertoire.
A second gate Ũ defined by
Ũ = TUT = eiĤ2 (6)
is obtained directly from U by exchanging the two input qubits just before they
enter the gate and the two output qubits just after they emerge. T (“twist”) is the
unitary operation corresponding to each of these exchanges. In the computation basis
{|00〉, |01〉, |10〉, |11〉} it has the representation
T =
1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
(7)
The generator Ĥ2 = TĤ1T of Ũ is linearly independent of Ĥ1 provided that Ĥ1 does
not commute with T , which again, generically, it does not.
Now note that if all operations generated by a pair of Hermitian operators P̂ and Q̂
are in a given repertoire, and α and β are real, then every operation generated by
αP̂ + βQ̂ is also in the repertoire [12]. This is because
ei(αP̂+βQ̂) = lim
n→∞
(
eiαP̂ /neiβQ̂/n
)n
. (8)
Likewise, every operation generated by the commutator i[P̂ , Q̂] is in the repertoire
because
e[P̂ ,Q̂] = lim
n→∞
(
e−iP̂ /
√
neiQ̂/
√
neiP̂ /
√
ne−iQ̂/
√
n
)n
. (9)
Thus we can use the generator
Ĥ3 = i[Ĥ1, Ĥ2] (10)
to generate a third class of operations in our repertoire. All the operations generated
by arbitrary linear combinations of Ĥ1, Ĥ2 and Ĥ3 are in the repertoire too. Physically
4
these are all obtained by acting on a single pair of qubits with a long chain of U-gates,
some connected directly and others by a twisted pair of wires.
This procedure can be repeated and new generators obtained by commuting ones that
have already been derived. If at any stage sixteen linearly independent generators Ĥj
have been constructed in this way from U, the universality of U is established.
Consider the scheme
Ĥ2 = TĤ1T
Ĥj = i[Ĥ1, Ĥj−1] j = 3 . . . 14
Ĥ15 = i[Ĥ2, Ĥ3]
Ĥ16 = i[Ĥ2, Ĥ5].
(11)
Linear independence of the sixteen generators Ĥj is equivalent to the non-vanishing
of the determinant of a 16 × 16 matrix consisting of the coefficients of the decom-
position of the Ĥj in an orthonormal basis. Suppose that, for a particular gate U,
this determinant does vanish. We must show that such gates lie in a lower- (less
than 16-) dimensional sub-manifold of U(4). To this end, consider a universal gate A
of the form (1). Straightforward but tedious calculation verifies that the generators
ĤA1 . . . Ĥ
A
16 formed according to the scheme (11) are linearly independent. The sim-
pler scheme defined by the first two lines of Eq.(11) with j = 3 . . . 16 does not have
this property. Then consider the one-parameter family of generators Ĥj(k) formed
according to the same scheme (11), but with
Ĥ1(k) = Ĥ1 + k(Ĥ
A
1 − Ĥ1). (12)
Let ∆(k) be the corresponding determinant. This is a polynomial of degree 100 in k
which, in the unfavourable case we are considering, vanishes at k=0 where Ĥ1(0) =
Ĥ1. But this polynomial cannot be identically zero, since we have checked that
∆(1) 6= 0. Therefore it can have at most 99 zeros in addition to the one at k = 0. For
every other value of k, the gate generated by the corresponding Ĥ1(k) is universal,
and in particular there is an interval around k = 0 in which every generator Ĥ1(k)
other than Ĥ1 generates universal gates.
A similar argument applies to an entire 16-dimensional neighbourhood of the genera-
tor Ĥ1. The generators in a sufficiently small neighbourhood can be parametrised by
sixteen coordinates, which can be chosen in the manner of (12) so that a generator of
A lies at a finite point in the coordinate space. The determinant formed according
to the scheme (11) from each of these generators is a polynomial in each of these
sixteen coordinates — so it is an analytic function which, even if it vanishes at Ĥ1, is
not identically zero. Hence it can at worst vanish on a 15-dimensional sub-manifold
(more precisely on a 15-dimensional variety) of the neighbourhood of Ĥ1, and our
result is proved.
5
It also follows that every generator of a universal gate is surrounded by a neighbour-
hood containing only such generators.
Analogous results hold for n-bit gates for all n > 2.
(Elements of the above proof have been derived independently by Lloyd [13].)
Which gates are not universal?
Clearly there can be no 1-bit universal gate because a 1-bit gate, and indeed any
number of 1-bit gates, cannot place two initially un-entangled qubits into an entangled
state. Likewise no classical gate can be universal because, by definition, a classical
gate evolves computation-basis states to other computation-basis states and never to
superpositions of them. Similarly, no gate that had that property with respect to any
fixed bases in the state spaces of the qubits that it acted upon, could be universal.
Such gates are properly called “classical” also. The question arises whether there are
any other non-universal gates.
Our proof leaves that question open because at various stages we have imposed generic
conditions on parameters to prove the universality of certain classes of gates. In most
cases, however, failure to meet those conditions is no guarantee of non-universality.
For example if, for a particular gate, the scheme (11) does not yield sixteen linearly
independent generators, there may be other schemes that do. Even if there are none,
that does not rule out constructions involving more general networks than the simple
chains we have considered, in which instances of the gate could be composed to
make a universal computer. Likewise, if the parameters φj violate the irrationality
conditions, the corresponding gate may nevertheless be universal. For instance the
gate A(π, π/2, θ) is known [8] to be universal even though the second parameter is a
rational fraction of π.
We conjecture that the non-universal gates are precisely
• the 1-bit gates and collections of 1-bit gates; and
• the classical gates.
If true, this would reveal an interesting connection between the existence of a “classical
level” in physics (i.e. a regime in which classical physics is a good approximation to
quantum physics) and the existence of classical computation as a closed and stable
regime within quantum computation.
6
How to define universality for components
We believe that the quantum computational network model, with its active “gates”,
passive “wires” and moving “qubits”, is a robust and reliable idealisation for analysing
a wide class of possible technologies for quantum computation. It might well cover all
possible technologies, but we wish to stress that this is neither proven nor self-evident.
The proper definition of “universality”, like everything else in the quantum theory
of computation, depends on what the laws of physics are. Therefore we shall now
point out some of the assumptions we have made about the physics and technology
of computation, which will have to be justified or perhaps amended by future, deeper
analyses.
One of our most fundamental assumptions is that the most general possible compu-
tations can indeed be performed by machines — i.e. by well-defined physical systems
which can be constructed to order and which maintain their identity during com-
putations. This assumption is presumably secure, given the necessity for unitary
operations in quantum computation. Unitarity can only be maintained in systems
from which the rest of the universe is isolated.
It is not quite so clear that these systems must in turn be composed of well-defined,
albeit interacting, subsystems, i.e. computational components such as gates. All we
can say is that it is hard to conceive of a technology to manufacture complex comput-
ing machines other than from simpler subsystems which are themselves computing
machines. That the computational state should be carried by two-state quantum sys-
tems rather than three- or higher-state ones is obviously not a necessary assumption,
but all our conclusions still hold when straightforwardly generalised to computers and
components that use more complex information carriers, so long as their state spaces
are of finite dimension.
In the theory of quantum computational networks, a gate is considered to be universal
if instances of it are the only computational components required to build a universal
computer. That does not mean that those gates would be the only physical compo-
nents of such a computer. At the very least there must also be “wires”, or some other
means of presenting the qubits to the gates at the right times and in the right com-
binations. And there must be input and output devices and presumably many other
components that form the environment in which the gates and bits interact in the nec-
essary ways. Yet however indispensable such components are, they perform no strictly
computational function in that they do not change the state of the computation. For
instance, a wire may be regarded as a gate, but it is the trivial “identity gate” which
does not affect the quantum state of the qubits that pass through it. Only compo-
nents that affect the computational state count as gates in discussions of universality.
However, the distinction between “computational” and “non-computational” opera-
tions can only be made relative to a given physical and technological implementation.
7
Different laws of physics, or different technologies, would lead us to draw the line
differently. Consider, for example, the proposed quantum-dot-based technology [9] in
which qubits are stored as the states of individual electrons trapped at fixed locations
(“dots”) and interacting only with their nearest neighbours in an array. The form
of the interaction (i.e. the type of gate) is determined by externally applied electric
fields and radiation. There are no physical “wires” to move qubits into adjacent po-
sitions so that they can undergo a gate-type interaction. Instead this is achieved by
successively swapping the states of adjacent dots, each swap involving three elemen-
tary “controlled-not” gate operations [9]. Thus the computational state of each qubit
is transferred unchanged from one physical dot to another, so in that sense such an
operation, taken as a whole, is computationally trivial. Nevertheless the only way
of realising it is as the net effect of several non-trivial computations. Therefore in
quantum dot technology, the operations that move qubits around are computational
operations.
So we see that in principle a gate may be universal or not according to the physics
and technology with which it is realised, and in particular, universality for quantum
dot technology has to be defined slightly differently from universality for quantum
networks. We may define a universal 2-bit operation U on a quantum dot array as
follows: U is universal if for every integer n, any unitary transformation of n qubits
can be effected with arbitrary precision by successive applications of U to suitable
pairs of adjacent quantum dots. The proof we have given above must be adapted to
show that in quantum-dot and related cellular-automaton-like technologies, almost
all 2-bit operations are universal: The key difference is that we cannot assume ab
initio that the “twist” operation T is in the repertoire. We can only assume that
the given operation A can be applied to an arbitrary pair of adjacent dots. But that
means in particular that it can be applied in two senses to a given pair, regarding
them in either order as the “first” and “second” input bits of the gate. So TAT is
automatically in the repertoire even though T itself is not (initially). It then follows
from the proof we have given that every 2-bit gate, and in particular the controlled-
not gate, is in the repertoire. From this, as we have said, both T and the “wire” or
qubit-moving operation can be constructed, and our result follows.
Another way of expressing the distinction between computational and non-computational
operations is through the concept of composition. Compositions are operations which,
though necessary in the construction and operation of computing machines, do not
affect the computational state (even transiently). We compose components when we
join one to another to form a more complex machine. We compose computational
operations when we use the output of one sub-computation as the input of another.
As we have seen, presenting the appropriate qubits as the inputs of a gate is mere
composition in a quantum network technology, but not in a quantum dot technology.
Strictly speaking, definitions of universality refer to computations of arbitrary length
8
and complexity, using arbitrarily large amounts of storage. This means that we must
contemplate mechanisms for providing a supply of qubits in a standard state (the
equivalent of Turing’s unlimited “blank tape”), and for maintaining the computer in
operation for an arbitrarily long period. Before we can certify that a given component
or operation is universal, we must satisfy ourselves that these mechanisms do not
themselves affect the state of the computation.
In addition to “blank tape”, some computations require “sources” of qubits in fixed
states. The constructions in our proof do not require such sources; moreover any
source can be constructed from the “blank tape” by suitable operations from our
universal repertoire. However, this too need not be true in every technology. It
is possible that there are technologies in which a given component or operation is
universal if sources of qubits in given states are available, but not otherwise.
If any quantum computation, required to produce its output state with a given accu-
racy, is intractable when it is performed entirely by composing instances of a universal
operation U, but is tractable using a larger, finite set of elementary operations, then
the universality of U may well be unphysical. In this paper we have so far paid no
attention to issues of complexity, and the specific constructions in our proofs are very
inefficient. But we should expect efficient constructions to exist. For suppose that we
want to effect a given n-bit unitary operation X, with probability no less than 1− ǫ,
using a universal n-bit unitary operation U. From U and suitable compositions we
can, in O(n) steps, create a second operation V that does not commute with U (for
instance TUT when n = 2). The total number of k-step sequences of U’s and V’s
is 2k, and generically, exponentially many of them are different. Now, the manifold
U(2n) of n-bit unitary operations, with metric
√
1 − 2−nRe(Tr(P †Q)), is compact.
Therefore the average error probability ǫ when the best of the 2k sequences is used to
approximate X must fall exponentially with k, and conversely the number of steps (or
gates) required to perform X with probability 1−ǫ is polynomial in log ǫ. This heuris-
tic argument about average efficiencies does not cover the worst case, which strictly
speaking is the relevant one. Nevertheless it gives us excellent reason to believe that
polynomial efficiency is possible even in the worst case and in general technologies
(see also Yao [10]).
The “error probability” we have just been discussing is that due to the approximation
of a continuous group U(2n) by finite sets of elements. Practical computing machines
are of course also subject to physical sources of error, such as thermal noise, unwanted
interactions and imperfect machining of components. Error-correction is itself a form
of computation, and error-correction strategies for coherent quantum computations
must themselves involve quantum computation [11]. In general, unlike in classical
computation, the best error correction strategy depends on the specific computation
that is being corrected. Therefore in principle it is possible that a gate or operation
might be universal only under the idealisation that errors are absent.
9
The problem of errors due to imperfect machining is somewhat ameliorated by the
results of this paper. For they show that the strategy of building quantum computers
need not be, as in the classical case, to design a set of abstract operations and then
to try to realise them as accurately as possible. Instead we can find almost any 2-
qubit operation that can be conveniently and accurately performed and composed,
and then be confident that this operation can be used as the elementary operation of
a universal quantum computer.
Conclusion
The results of this paper are good evidence that universality in a strong, robust and
practical sense is generic in arbitrary quantum computer technologies. But a full
proof must await a deeper theoretical integration of physics and computation than is
yet available.
Acknowledgements
The authors wish to thank Richard Jozsa and David DiVincenzo for intersting discus-
sions and comments on the subject of this paper. This work was partially supported
by the Royal Society. A.B. acknowledges the financial support of the Berrow’s fund
at Lincoln College, Oxford.
References
[1] D. Deutsch, Proc. R. Soc. London A 400, 97 (1985).
[2] R. Landauer, IBM J. Res. Develop. 5, 183 (1961).
[3] P.W. Shor, Proc. 35th Ann. Symp. on the Foundations of Computer Science
(IEEE Computer Society, Los Alamitos, 1994), p.124.
[4] D. Deutsch and R. Jozsa, Proc. R. Soc. Lond. A 439, 553 (1992); E. Bernstein
and U. Vazirani, Proc. 25th ACM Symp. on Theory of Computation, p.11 (1993).
[5] S. Wiesner, SIGACT News, 15, 78 (1983); C. H. Bennett and G. Brassard,
in Proc. IEEE Int. Conference on Computers, Systems and Signal Process-
ing, (IEEE, New York, 1984).; A. K. Ekert, Phys. Rev. Lett. 67, 661 (1991);
C. H. Bennett, Phys. Rev. Lett. 68, 3121 (1992).
[6] D. Deutsch, Proc. R. Soc. Lond. A 425, 73 (1989).
10
[7] A. Barenco, University of Oxford preprint (1994).
[8] T. Sleator and H. Weinfurter, preprint (1994).
[9] A. Barenco, D. Deutsch, A. Ekert, and R. Jozsa, Logic gates for quantum com-
putation, Phys. Rev. Lett. (in print).
[10] A. Yao, Proc. 34th IEEE Symp. on Foundation of Computer Science (IEEE
Computer Society Press, Los Alamitos, 1993), pp.352-360.
[11] A. Berthiaume, D. Deutsch and R. Jozsa, The stabilization of quantum com-
putations , in Proceedings of the Workshop on the Physics and Computation—
PhysComp ’94 (IEEE Computer Society Press, Dallas, Texas).
[12] D. DiVincenzo, Phys. Rev. A 50, 1015 (1995).
[13] S.Lloyd, preprint (1994).
11
| 0non-cybersec
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arXiv
|
How can a dhcp client in linux can send a custom option?. <p>I need a DHCP client to send a string to the server as part of the DHCP request. I have tried setting DHCP options in the client, but they are not being sent during the initial DHCP handshake. </p>
<p>The dhcp-options manpage shows how new options can be defined. But these seem to be sent only from the DHCP server to the DHCP client. How can I send options from the DHCP client to the server instead?</p>
| 0non-cybersec
|
Stackexchange
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
|
Stackexchange
|
[No Spoilers] 'Game of Thrones' Star Hafthor Bjornsson Breaks Deadlift World Record [Video].
| 0non-cybersec
|
Reddit
|
Stop downloading Lion from the App store. <p>So, I bought a new Mac book Air recently, opened up the AppStore and found "Lion" in the featured updates.</p>
<p>I thought that was shown because I was not running lion and clicked on it to upgrade. Later mucked around the system to figure that I was running Lion and my system had been updated to the latest version of Lion.</p>
<p>But because I clicked the Install on the Lion in the appstore,</p>
<ul>
<li>I am charged with $30, which I don't really have to pay.</li>
<li>My launchpad now displays the Lion as downloading.</li>
</ul>
<p>I can't seem to stop the launchpad icon from downloading and can't remove it from there.</p>
<p>Now my queries:</p>
<ul>
<li>Is it possible to "un-install" from the appstore. How?</li>
<li>Once the Lion completes download, will my system be updated again to the same version of the OS? What happens to the installed apps</li>
<li>Is it possible that somehow I can get back my $30, which is a wrong charge.</li>
<li>How do I stop the download from happening, prevent re-install and remove that downloading icon from Launchpad?</li>
</ul>
<p>I probably acted a little stupid by clicking to buy that in the first place without even knowing my system runs Lion. But in my defense, I didn't think an app that I don't need to install would download and charge me; not by Apple.</p>
<p>Thanks in advance.</p>
| 0non-cybersec
|
Stackexchange
|
Support of pushforward of a sheaf. <p>Let <span class="math-container">$X$</span> be a topological space with <span class="math-container">$j : U\hookrightarrow X$</span> an open immersion, and let <span class="math-container">$\mathcal{F}$</span> a sheaf of abelian groups on <span class="math-container">$U$</span>.
It is easy to see that <span class="math-container">$\text{supp}(j_{!}\mathcal{F}) = \overline{\text{supp}(\mathcal{F})}$</span>, but what can we say about <span class="math-container">$\text{supp}(j_{\ast}\mathcal{F})$</span>? Being <span class="math-container">$j_{!}\mathcal{F}\hookrightarrow j_{\ast}\mathcal{F}$</span>, we have the inclusion <span class="math-container">$\text{supp}(j_{!}\mathcal{F})\subset\text{supp}(j_{\ast}\mathcal{F})$</span>. Can we say more than this in the full generality, or in some special case (excluding the obvious one in which <span class="math-container">$j$</span> is proper)? For example, is there any situation in which we would have that <span class="math-container">$\text{supp}(j_{\ast}\mathcal{F}) = \text{supp}(j_{!}\mathcal{F})\cup K$</span>, with <span class="math-container">$K$</span> compact?</p>
| 0non-cybersec
|
Stackexchange
|
Tikz crashes with vertical lines in plot - Error: Dimension too large. <p>I had to shorten the code because it would not fit here. I am trying to make 4 subplots in Tikz. I used Matlab and Matlab2Tikz to get the figure in Latex. I cannot compile and keep getting an error message saying dimensions are too large. In general I cannot compile when I have vertical lines in a plot. It seems to work if I do not plot vertical lines around the edges. But even after doing this I cannot get this to run. I tried using logaxis but it does not work. I can see that the numerical precision might be an issue. Is there any way I can upload the whole code (41,000 characters).</p>
<pre><code>\documentclass[12pt, a4paper]{article}
\usepackage[dvipsnames]{xcolor}
\usepackage{tikz,pgfplots}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{listings}
\usepackage[a4paper]{geometry}
\usepackage{caption}
\newlength\figH
\newlength\figW
\setlength{\figH}{14cm}
\setlength{\figW}{14cm}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage{color}
\begin{document}
\definecolor{mycolor1}{rgb}{0.00000,0.44700,0.74100}%
\definecolor{mycolor2}{rgb}{0.85000,0.32500,0.09800}%
%
\begin{tikzpicture}
\begin{axis}[%
width=0.411\figW,
height=0.419\figH,
at={(0\figW,0.581\figH)},
scale only axis,
separate axis lines,
every outer x axis line/.append style={white!15!black},
every x tick label/.append style={font=\color{white!15!black}},
xmin=1.995,
xmax=2.002,
xtick={1.995,1.996,1.997,1.998,1.999,2,2.001,2.002},
every outer y axis line/.append style={white!15!black},
every y tick label/.append style={font=\color{white!15!black}},
ymin=0.4972,
ymax=0.4988,
ytick={0.4968,0.497085714285714,0.497371428571429,0.497657142857143,0.497942857142857,0.498228571428571,0.498514285714286,0.4988},
yticklabels={{0.4974},{0.4977},{0.4979},{0.4982},{0.4985},{0.4988}},
axis background/.style={fill=white},
title style={font=\bfseries},
title={Leverage - $\omega$ menu - 1st case},
axis on top
]
\addplot [color=mycolor1,solid,line width=3.0pt,forget plot]
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};
\addplot [color=black,dotted,line width=0.5pt,forget plot]
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\addplot [color=black,dashed,line width=0.5pt,forget plot]
table[row sep=crcr]{%
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\addplot [color=mycolor2,dashdotted,line width=0.5pt,forget plot]
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5.65 0.497841641027983\\
5.7 0.497841641027983\\
5.75 0.497841641027983\\
5.8 0.497841641027983\\
5.85 0.497841641027983\\
5.9 0.497841641027983\\
5.95 0.497841641027983\\
6 0.497841641027983\\
6.05 0.497841641027983\\
6.1 0.497841641027983\\
6.15 0.497841641027983\\
6.2 0.497841641027983\\
6.25 0.497841641027983\\
6.3 0.497841641027983\\
6.35 0.497841641027983\\
6.4 0.497841641027983\\
6.45 0.497841641027983\\
6.5 0.497841641027983\\
6.55 0.497841641027983\\
6.6 0.497841641027983\\
6.65 0.497841641027983\\
6.7 0.497841641027983\\
6.75 0.497841641027983\\
6.8 0.497841641027983\\
6.85 0.497841641027983\\
6.9 0.497841641027983\\
6.95 0.497841641027983\\
7 0.497841641027983\\
7.05 0.497841641027983\\
7.1 0.497841641027983\\
7.15 0.497841641027983\\
7.2 0.497841641027983\\
7.25 0.497841641027983\\
7.3 0.497841641027983\\
7.35 0.497841641027983\\
7.4 0.497841641027983\\
7.45 0.497841641027983\\
7.5 0.497841641027983\\
7.55 0.497841641027983\\
7.6 0.497841641027983\\
7.65 0.497841641027983\\
7.7 0.497841641027983\\
7.75 0.497841641027983\\
7.8 0.497841641027983\\
7.85 0.497841641027983\\
7.9 0.497841641027983\\
7.95 0.497841641027983\\
8 0.497841641027983\\
};
\end{axis}
\end{tikzpicture}%
\end{document}
</code></pre>
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Stackexchange
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TIL that Jim Carrey used to send funny letters to Tupac when he was in prison.
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Reddit
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Knock knock.
| 0non-cybersec
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Reddit
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Fix Approved Pulled Pork - SOOO YUM :).
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Reddit
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Does Tensorflow cache results between session runs?. <p>I am using the Python API of Tensorflow, and I am profiling my computational graph. Say the end node of the graph is a variable called <strong>tf_v</strong>, I profiled the wall clock run time of <strong>sess.run(tf_v)</strong> 10 times. </p>
<p>I consistently noticed that the first run takes much longer than subsequent runs. Is tensorflow caching results between session runs? (i.e if the input variables values in the graph didn't change, tensorflow returns a cached result from previous run? ) </p>
| 0non-cybersec
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Stackexchange
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
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Stackexchange
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[SV] I'm so happy right now I could cry!. I stepped on the scale this morning, and I'm finally under 200lbs. I haven't been under 200lbs in *years*. And it feels so good. I've come so far from where I was at the beginning of the year. I cut out soda and alcohol completely. I quit eating foods that left me feeling bloated, and uncomfortable. My relationship with food has changed so much. Food isn't the enemy any more, and it's not my greatest comfort.
I'm just really proud of myself and I needed to share it with someone, even if its a lot of someones I don't know. I've tried and failed countless times. And this time I've stuck with it, I made smarter choices, I figured my shit out, and I got stuff done. I've lurked on this sub a lot and it's always given me inspiration. I hope this post can give someone out there like me some inspiration too.
ETA: Thank you everyone for your kind words! It means a lot to me <3!
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Reddit
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How can I get a Linux *partition* image to boot in a XenServer virtual machine?. <p>I am trying to migrate a Linux PC to a virtual machine running in XenServer. Normally I would just create an image of the boot disk (/dev/sda) with dd and use that to overwrite an empty virtual disk image. Unfortunately the boot disk is rather large and therefore I only created an image of the boot partition (/dev/sda1) which is only about 250 GB. I then created a new virtual disk of 500 GB, attached it to an existing Linux virtual machine (/dev/xvdc), partitioned it to create a primary partition (/dev/xvdc1) and dd'd the partition image to that.</p>
<p>I could mount that partition. The files were where I expected them, but of course the grub boot code was missing.</p>
<p>So I used the method described in <a href="http://karuppuswamy.com/wordpress/2010/06/02/how-to-chroot-to-ubuntu-using-live-cd-to-fix-grub-rescue-prompt/" rel="nofollow noreferrer" title="How to chroot Ubuntu using Live CD to fix GRUB rescue prompt">How to chroot Ubuntu using Live CD to fix GRUB rescue prompt</a> to install grub on /dev/xvdc.</p>
<pre><code>$ sudo su
$ mount /dev/sdax /mnt/newroot
$ mount ‐‐bind /dev /mnt/newroot/dev
$ mount ‐‐bind /proc /mnt/newroot/proc
$ mount ‐‐bind /sys //mnt/newroot/sys
$ chroot /mnt/newroot
$ grub-install /dev/xvdc
</code></pre>
<p>It seemed to work fine, no error messages were displayed.</p>
<p>Now I attached the new virtual disk to a new virtual machine as the the only drive and tried to boot it. No such luck. I do not get any error message, no grub prompt, it simply starts and then stops.</p>
<p>EDIT: There was an error message, I just overlooked it:</p>
<pre><code>"The Bootloader for this VM returned an error -- did the VM installation
succeed? INVALID_SOURCE
cdrom repo
/dev/sm/backend/[some long number, probably a GUID]
</code></pre>
<p>What am I missing here?</p>
<p>EDIT: Attaching the virtual drive to an existing Linux virtual machine solved the problem. I got it to boot from that drive, made a full copy of that VM and finally reverted the original back to its original boot drive.</p>
<p>That still leaves the question: Why did XenServer insist on trying to boot from a non existing DVD?</p>
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Stackexchange
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Russian opposition crushed by wave of repression.
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Reddit
|
Xen DomU ethernet driver seems not loading. <p><strong>Problem</strong></p>
<p>The Ethernet virtual device of my DomU <strong>doesn't seems to load</strong> and I can't explain why.</p>
<p><strong>Context</strong></p>
<p><strong><em>Dom0 Info</em></strong></p>
<p>Dom0 : Debian Wheezy with Xen</p>
<pre><code>uname -r : 3.2.0-4-amd64
</code></pre>
<p>Dom0 dmesg</p>
<pre><code>[11:55]root@Dopamine2:~[0]# dmesg | grep Network
[ 11.550947] Intel(R) Gigabit Ethernet Network Driver - version 3.2.10-k
[ 11.627242] igb 0000:03:00.0: Intel(R) Gigabit Ethernet Network Connection
[ 11.705513] igb 0000:03:00.1: Intel(R) Gigabit Ethernet Network Connection
</code></pre>
<p>Dom0 lspci</p>
<pre><code>03:00.0 Ethernet controller: Intel Corporation I350 Gigabit Network Connection (rev 01)
03:00.1 Ethernet controller: Intel Corporation I350 Gigabit Network Connection (rev 01)
</code></pre>
<p><strong><em>One Working DomU</em></strong></p>
<p>I have 1 x fresh DomU (wheezy), installed with "xen-create-image", working perfectly.
When I do a dmesg on this DomU, we can see inside : "Initialising Xen virtual ethernet driver"</p>
<p><strong><em>Failing DomU (the problem)</em></strong></p>
<p>But another existing DomU (that I want to migrate from an old Dom0 to the new one), also a Debian wheezy, seems to be not working. Network is not loaded :</p>
<pre><code>/etc/init.d/networking restart
[warn] Running /etc/init.d/networking restart is deprecated because it may not re-enable some interfaces ... (warning).
[....] Reconfiguring network interfaces...Cannot find device "eth0"
Failed to bring up eth0.
</code></pre>
<p>When I do a dmesg on this failing box, the "Initialising Xen virtual ethernet driver" is NOT inside.</p>
<p>(I don't know what other information I can provide ?)</p>
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Stackexchange
|
Move a hunk of a commit to another commit in a single rebase. <p>Using GIT, I'd like to move a hunk of a commit to a different one, while automatically retaining commit metadata, using a single interactive rebase.</p>
<p>For example, having these 2 commits:</p>
<p>first commit X:</p>
<pre><code> context
+ foo
context
</code></pre>
<p>second commit Y:</p>
<pre><code> context
+ bar
context
+ baz
context
</code></pre>
<p>I would like to move the <code>bar</code> hunk from commit Y into the previous commit X, without having to manually reset author, committer, log, etc.</p>
<p>The simplest way I can think of requires 2 interactive rebases:</p>
<ol>
<li>First rebase: split Y commit into Z (<code>bar</code>) and Y' (<code>baz</code>), the last one using <code>commit -c Y</code> to retain the metadata.</li>
<li>Second rebase: squash X and Z together into X', and leave Y' as is.</li>
</ol>
<p>Which would leave this result:
first commit X' (same metadata as X):</p>
<pre><code> context
+ foo
context
+ bar
context
</code></pre>
<p>second commit Y' (same metadata as Y):</p>
<pre><code> context
+ baz
context
</code></pre>
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Stackexchange
|
Embedded movies, animations in beamer and article document class. <p>I would like to use embedded animations and videos in my presentation which has the <code>beamer</code> document class. As far as I am concerned there are two packages for doing this <strong>media9</strong> and <strong>movie15</strong>; however, I could not manage to have the PDF reader on Ubuntu 14.04.3 LTS 64-bit play the embedded videos or animations. Although I could see that the video is attached and there is an interactive pin symbol on the page and it can be clicked upon, when that button was clicked it launched the video on a separate window and application which is not what I want. So far I have tried Okular, Adobe Reader, Evince and Qpdfviewer, but none of them were able to play the embedded videos without launching a separate window. The minimal working example is as follows:</p>
<pre><code>\documentclass[12pt,landscape]{article}
\usepackage{geometry}
\geometry{verbose,letterpaper}
\usepackage{media9}
\usepackage{hyperref}
\begin{document}
Whispering-gallery mode in a quarter circle:
\includemedia[
width=0.4\linewidth,
height=0.3\linewidth,
activate=pageopen,
addresource=Circle-m-increase3.mp4,
flashvars={source=Circle-m-increase3.mp4}
]{}{StrobeMediaPlayback.swf}
\end{document}
</code></pre>
| 0non-cybersec
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Stackexchange
|
Latitude and Longtitude Distance Calculation. <p>I'm using MySQL and I am trying to calculate the distance in meters between two different longitude, latitude coordinates. I wrote a stored function in MySQL but it calculates just 1 for each row in select query. There is my stored function in the following segment:</p>
<pre><code>DELIMITER $$
CREATE DEFINER=`root`@`localhost` FUNCTION `CALCULATE_DISTANCEE`(Lati varchar(20),Longi varchar(20),MyLat decimal, MyLon decimal) RETURNS int(11)
BEGIN
declare Lat decimal(12,12);
declare Lon decimal(12,12);
declare RadiusOfWorld decimal(12,12);
declare DistanceLat decimal(12,12);
declare DistanceLon decimal(12,12);
declare A decimal(12,12);
declare R decimal(12,12);
declare D decimal(12,12);
declare DistanceAsMeter decimal(12,12);
select CAST(Lati as Decimal(12,4)) into Lat;
select CAST(Longi as Decimal(12,4)) into Lon;
select 6378.137 into RadiusOfWorld;
set DistanceLat = (MyLat - Lat) * PI() / 180;
set DistanceLon = (MyLon - Lon) * PI() / 180;
select SIN(DistanceLat/2) * SIN(DistanceLat/2) + COS(Lat*PI()/180) * COS(MyLat*PI() / 180) * SIN(DistanceLat / 2) * SIN(DistanceLat / 2) into A;
Select (2*ATAN2(SQRT(A),SQRT(1 - A))) into R;
select (RadiusOfWorld * R) into D;
select D*'1000' into DistanceAsMeter;
RETURN DistanceAsMeter;
END
</code></pre>
<p>And also I am calling the function like this:</p>
<pre><code>select *,`brain_db`.`CALCULATE_DISTANCEE`(stop.Stop_lat,stop.Stop_lon,'-157.818079','21.260340') AS Distance from stop;
</code></pre>
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Stackexchange
|
What is this foreign language registry key?. <p>Does anyone recognize this?</p>
<p><img src="https://i.stack.imgur.com/OeLZc.jpg" alt="Foreign registry key"></p>
<p>It's under HKEY_CURRENT_USER\Software (and HKEY_USERS\S-1-5-....\Sodtware) on an English WinXP machine.<br/>
I don't remember ever having installed something called 'PC Sync', definitely not the <a href="http://www.laplink.com/index.php/individuals/file-transfer-remote-management/pcsync" rel="nofollow noreferrer">LapLink product with the same name</a> (and no synching software anyway).<br/>
There are no other keys named 'PC Sync' in the registry.</p>
<p>Added:</p>
<p>It's Korean, Franks comment is correct - Google translate now converts it to:<br/>
Local AppWizard-generated application.</p>
<p>Googling for <em>Local AppWizard</em>, I find these <a href="http://www.codefighter.com/registry.html" rel="nofollow noreferrer">two</a> <a href="http://www.tech-archive.net/Archive/VC/microsoft.public.vc.mfc/2007-07/msg00138.html" rel="nofollow noreferrer">pages</a> which seem to suggest that they may be leftovers from an installation program (written in C?) that failed to call 'SetRegistryKey("appname"); in InitInstance.' ?</p>
| 0non-cybersec
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Stackexchange
|
I've too much body hair. I don't know what to do and it's destroying my confidence. Please, help.. I'm 18 and I don't think I'm normal. My hormones must be messed up or something. I've too much body hair and I don't know how to deal with it.
The worst part is my face and my neck. I've lots of black hairs on my upper lip, on my cheeks and under my chin, down to my neck. I use to wax my moustache and bleach the hairs on my neck, but I don't like the result. The hairs on my neck are quite long (about 0.5 in), and even if they're bleached, they're still very noticeable especially against the light. It's not like I can wax my whole neck and chin. So I was thinking of some permanent hair removal method. I've heard it exists...? I really don't know who to ask. Do you have any info? Any advice?
I also hate my legs. I've dark, thick hairs there, too (yay çAç). I use an epilator, and the result is pretty mediocre. I've a lot, I mean, *a lot* of ingrown hairs and because of them my legs are always coarse (paradoxically, they're smoother when the hairs are long). So, I don't use any lotion and I've never used a loofah pad. But I bought one! So, I wanted to ask you─ what is the right procedure to use it? How many times a week? Should I use some lotion, too?
I'm sorry to ask you so many things, but my parents aren't so supportive and I've no friends (yay çAç). I've succeeded in losing some weight and now I can say I look very nice... if it wasn't for this body hair! Please, please help me.
| 0non-cybersec
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Reddit
|
I've entered into the mainframe and disabled the firewall!.
| 0non-cybersec
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Reddit
|
iOS 11 UINavigationBar Position Loses Status Bar Height During Transition. <p>This bug has cropped up in iOS 11. There is a UINavigationController embedded as a child UIViewController of another UIViewController. There's a UIPanGestureRecogniser setup that affects the y axis of the child UIViewController's view's frame allowing you to drag it down to reveal the UIViewController below </p>
<p>The problem that's cropped up is that now as soon as the UIViewController's frame gets adjusted to move it downwards iOS seems to remove the inset on the UINavigationBar for the UIStatusBar, which results in an unpleasant jerk upwards. </p>
<p><a href="https://i.stack.imgur.com/wtyfZ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wtyfZ.gif" alt="Status Bar Animation Bar"></a></p>
<p>How do I prevent this from happening? </p>
| 0non-cybersec
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Stackexchange
|
Hand position for ring support. It is not clear to me what is the optimal wrist alignment for support work on rings. In many videos-- including Antranik's in the Recommended Routine-- the wrist seems to be extended rather than directly in line/underneath the forearm. I a few videos that I have seen the wrist is slightly flexed.
Extended or flexed-- it seems to me-- pulls the forearms away from the support of the straps (a good thing), and also increases the surface area of the hand on the rings.
What is considered "proper" form here?
| 0non-cybersec
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Reddit
|
Need help identifying what is making this noise in my pc..
| 0non-cybersec
|
Reddit
|
I can now say that i was able to catch them all!.
| 0non-cybersec
|
Reddit
|
Recovering 3 months worth of code. <p>I tried today to install <code>xubuntu</code> on my PC but I had some problems in partitioning, while I was trying to fix it, I accidentally deleted everything from the partitions .</p>
<p>So is it possible to recover from ntfs with Xubuntu .</p>
<p>The data that was on my hard drive exceed 400GB of songs, family photos, my work !! </p>
| 0non-cybersec
|
Stackexchange
|
Is there a simple way in PL/pgSQL to check if a query returned no result?. <p>I'm currently experimenting a bit with PL/pgSQL and want to know if there is a more elegant way to do something like this:</p>
<pre><code>select c.data into data from doc c where c.doc_id = id and c.group_cur > group_cur order by c.id desc limit 1;
EXCEPTION
WHEN NO_DATA_FOUND THEN
select c.data into data from doc c where c.doc_id = id and c.global_cur > global_cur order by c.id desc limit 1;
EXCEPTION
WHEN NO_DATA_FOUND THEN
RETURN NULL;
</code></pre>
| 0non-cybersec
|
Stackexchange
|
Sea For Yourself (2015) - Exploring the fishing industry in Northern Ireland... and giving 10 teenagers first-hand experience.
| 0non-cybersec
|
Reddit
|
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
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Stackexchange
|
Playing hide and seek with my daughter.
| 0non-cybersec
|
Reddit
|
What is the sed comand to fix this file so last value on each line is double quoted. <p>I have a file containing two million lines of the form:</p>
<pre><code>"00005cea-668e-4475-9e19-92a25c8b74fb",129.24728",D#
</code></pre>
<p>the last value should actually be:</p>
<pre><code>"00005cea-668e-4475-9e19-92a25c8b74fb",129.24728,"D#"
</code></pre>
<p>Please , how do I use sed command to fix this file so the spurious " is removed and the last value is double quoted</p>
| 0non-cybersec
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Stackexchange
|
What to use instead of Smartmatch?. <p>I just upgraded to Perl 5.26.1, now I get</p>
<pre><code>Smartmatch is experimental at check_tr.pl line 67.
</code></pre>
<p>where the code in question is</p>
<pre><code>my @patterns = (qr/summary:\s+(.+?) in\s+(.+?) - average of\s+(.+?)$/,
qr/summary:\s+(.+?) in\s+(.+) (.+? .+?)$/);
my $r = "";
opendir(DIR, $dir) or die $!;
while(my $file = readdir(DIR)) {
next if ($file =~ m/^\./);
open(F, '<', "$dir/$file") or die $!;
if (<F> ~~ @patterns) {
$r .= <F>;
...
</code></pre>
<p><strong>Question</strong></p>
<p>Ideally with as few changes to the code as possible. What should I do instead of smart matching?</p>
| 0non-cybersec
|
Stackexchange
|
HTTP Request Smuggling + IDOR.
| 1cybersec
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Reddit
|
Node.JS - any side effects of setting max-old-space-size too large?. <p>Our server runs Node.JS on a cloud PaaS (specifically Bluemix).
We can configure the amount of allocated memory through the PaaS dashboard, but I noticed that for values above ~1.4GB you also need to execute node with the <code>--max-old-space-size</code> option (explanation <a href="http://blog.caustik.com/2012/04/11/escape-the-1-4gb-v8-heap-limit-in-node-js/" rel="noreferrer">here</a>).</p>
<p>This means that whenever I want to change the size of allocated memory, I have to change it in two places in the settings.</p>
<p>What would happen if I call <code>node --max-old-space-size 99999999999 app.js</code>? Will Node.JS try to allocate 99999999999MB, or will it consider the actual memory-limit of the VM\Container in which it runs? Does it affect the behavior of the GC (i.e. if it looks like there is a lot of free space, the GC will run less times)? Is there a <code>--max-old-space-size use-machine-limits</code> option?</p>
<p>Thanks</p>
| 0non-cybersec
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Stackexchange
|
Can someone explain me the flow of this JavaScript function? (Closure concept). <p>I'm reading "<a href="http://eloquentjavascript.net/" rel="nofollow">Eloquent JavaScript</a>". Chapter 3 introduces "<a href="http://eloquentjavascript.net/03_functions.html#h_hOd+yVxaku" rel="nofollow">Closure</a>" concept and gives you a couple of examples. One of these is next one:</p>
<pre><code>function multiplier(factor) {
return function(number) {
return number * factor;
};
}
var twice = multiplier(2);
console.log(twice(5));
// → 10
</code></pre>
<p>I think I understood the concept. If first I execute <code>console.log(twice)</code>, since variable <code>number</code> is undefined, what I get is <code>[Function]</code>. What I don't understand is how <code>twice(5)</code> works. Why local variable <code>number</code> is initialized with value <code>5</code>? </p>
<p>Also, why if I execute <code>console.log(multiplier(2,5))</code> I don't get 10 as a result?</p>
<p>Thanks. </p>
| 0non-cybersec
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Stackexchange
|
Ubuntu hangs on ryzen 5 1600 processor. <p>I bought a computer of the following assembly:</p>
<ul>
<li>AMD Ryzen 5 1600</li>
<li>16 GB DDR4-2666</li>
<li>NVIDIA GeForce GT 710</li>
</ul>
<p>when installing versions of ubuntu 18.4.3, 18.4.4, 20.4, hangs are observed, a way out of which is possible only with the computer turned off.</p>
<p>Hangs can occur when the system boots up or a few minutes after the start.</p>
<p>On windows 10 there is no such problem.</p>
<p>Currently installed ubuntu 18.4.4</p>
<p>Tell me, plz, how to get rid of freezes?</p>
| 0non-cybersec
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Stackexchange
|
#buttstuff2016.
| 0non-cybersec
|
Reddit
|
How can I network a Windows 7 PC with a MacBook Pro running Lion over FireWire?. <p>I always thought this was possible but after searching around a little bit all the articles that even come close are using OSes pre Windows 7 and OSX 10.7 (Lion). I tried just connecting them with a FireWire cable and Windows just reported the MacBook as an unknown device.</p>
<p>Has anyone got this working?</p>
| 0non-cybersec
|
Stackexchange
|
Completely Remove isearch.claro-search.com Redirect Virus from Your PC.
| 1cybersec
|
Reddit
|
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