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How to automatically change wallpaper from interfacelift?. <p>This is something I'd like to work on. There is this site called <a href="http://interfacelift.com/wallpaper/downloads/date/any/" rel="nofollow">Interfacelift</a> which offers really good wallpapers for any screen resolution and is updated regularly. The site which has a google+ page, has agreed to provide beta testers with a <strong>MediaRSS link</strong>. </p>
<p><em>Is there any way I can use this to allow ubuntu to change my wallpaper automatically to the newest wallpaper submitted on Interfacelift?</em> </p>
<p>They somehow use this to create .theme files for windows to create dynamic wallpaper changing. I was wondering if something similar can be done in Ubuntu.</p>
| 0non-cybersec
| Stackexchange |
YSK a sudden loss of vision affecting both eyes is a sign of a stroke. 30-50% of strokes involve vision loss, yet most people do not know that you can lose vision from a stroke and it is not part of the FAST (Face, Arm, Speech, Time) acronym for stroke. I work with stroke victims a many noticed the vision loss but tried to sleep it off, not knowing it was a stroke and delaying their treatment during a crucial time.
If you notice a loss of vision in the periphery of both eyes, call your doctor.
Edit: everyone has a problem with the “call your doctor” part. Let’s modify that to “call the appropriate medical professionals.” That can be different for different people, so use your best judgement | 0non-cybersec
| Reddit |
copy all files that have no extension. <p>We can copy some file by extensions like this:</p>
<pre><code>cp *.txt ../new/
</code></pre>
<p>but how can I copy all files that have no extension?</p>
| 0non-cybersec
| Stackexchange |
Will the E-Book Kill the Footnote? . | 0non-cybersec
| Reddit |
Extremely photogenic bungee jumper.. | 0non-cybersec
| Reddit |
If we're throwing around what we want from Gen 8. | 0non-cybersec
| Reddit |
References for "closed form" numeric solutions of $\tan x=-a x$. <p>I am looking for references that discuss solutions of the equation</p>
<p>$\tan x=-a x$</p>
<p>(for $x,a\in \mathbb{R}$). I know about the graphical approaches, and any number of numerical solution approaches, but I was wondering if there's a semi-closed form solution, which gives something like</p>
<p>$x = f(a,c)$</p>
<p>where $c$ is a (set of) numerically solved constants which are independent of $a$ (i.e. a set of constants corresponding to the set of roots).</p>
<p>I've come up with an approach (i.e. a function $f$), but I figure this has been solved already, so I'd prefer to reference that ``known'' solution.</p>
| 0non-cybersec
| Stackexchange |
What did the boy with no hands get for his birthday?. Gloves!!
just kidding....I don't know what he got....he hasn't opened it yet | 0non-cybersec
| Reddit |
[IMAGE] No Sony, thank you!. | 0non-cybersec
| Reddit |
Google Sheet Query returns first 3 rows in one row. <p>I have made a simple query that combines data from 4 separate sheets:</p>
<pre><code>=query({'1.MA'!A13:O200;'2.PF'!A13:O200;'3.BM'!A13:O200;'4.C'!A13:O200},"Select * where Col2 is not null ")
</code></pre>
<p>The objective is to have all data in the sheets where Column B is filled in. This works for the majority, except for the first 3 rows in the first sheet. The query returns them combined in one row.</p>
<p>All sheets are exactly the same, the error does not reproduce n the first 3 rows of the other sheets.</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 |
AT&T CEO: Net neutrality is ‘bad for the industry’. | 0non-cybersec
| Reddit |
Root user inside Composer container. <p>I have a basic question when running running a <a href="https://hub.docker.com/r/composer/composer/" rel="noreferrer">Composer</a> in a Docker container.</p>
<p>Is it OK to run the <code>composer</code> as user <code>root</code> inside the container?
I am confusing that the owner of the created files (for example when using <code>composer require</code>) is <code>root</code>.</p>
<p>Is run as <code>root</code> inside the container OK best practice?</p>
| 0non-cybersec
| Stackexchange |
How to make popover appear where my mouse enters the hover target?. <p>This is an example code to show the popover window display below my button:</p>
<pre><code>$.fn.popover.defaults = $.extend({} , $.fn.tooltip.defaults, {
placement: 'bottom', content: ''
</code></pre>
<p>Now I want the popover window appear on the place where my cursor moves on(not only top/bottom/left/right, but a specific location which depends on where user put their cursor on). </p>
<p>How to get it? </p>
| 0non-cybersec
| Stackexchange |
Using custom hadoop input format in PySpark. <p>I am trying to use a custom defined Hadoop input format for reading some data on HDFS. This is the command that I am using on pyspark,
<code>rdds=sc.newAPIHadoopFile("inputPATH","com.jet1.custom.spark.CustomInputFormat","org.apache.hadoop.io.LongWritable","com.jet1.spark.val",{"fs.defaultFS":"hdfs://localhost:8020"});</code></p>
<p>I have already passed the corresponding jars while running pyspark.
But I am getting the error</p>
<p><code>py4j.Py4JException: Method newAPIHadoopFile([class org.apache.spark.api.java.JavaSparkContext, class java.lang.String, class java.lang.String, class java.lang.String, class java.lang.String, class java.util.HashMap, null, class java.util.HashMap, class java.lang.Integer]) does not exist</code></p>
<p>I think this is because of some wrong configuration settings and not specifically because of something incorrect in the command.
Also, JAVA_HOME is set in the terminal environment. Any idea on what might be wrong.</p>
| 0non-cybersec
| Stackexchange |
LPT Request: What's the best way to get out of bed in the morning and avoid hitting the snooze button?. Is there something I should be doing the night before to prepare? Is there something I should do first thing after waking up? I get 7-9 hours of sleep per night, on a regular schedule, yet I just can't help but feel exhausted in the mornings. I routinely set 5+ alarms, and even then I still have a hard time not sleeping in.
I've tried the "hop out of bed and do 10 pushups" and "drink a glass of water immediately" and a few other methods, but nothing short of a cold shower seems to work (and I'd like to avoid taking a mandatory cold shower every morning, if possible).
Before getting coffee or an energy drink after I leave the house, what can I do to quickly get up and stay awake? | 0non-cybersec
| Reddit |
Updating distribution using Bayes' theorem?. <p><strong>I got this question as home work but i am not able to understand how I will update distribution using Bayes' rule. I want to a direction to proceed?</strong></p>
<p>Let X be the score of a
randomly selected student in this class on the quiz. Let Y denote the mean score. If we
knew Y , suppose that a reasonable model for X would be that X is <strong>normal</strong> with mean Y
and variance 10.</p>
<p><a href="https://i.stack.imgur.com/CYOzy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CYOzy.png" alt="enter image description here"></a></p>
<p>When I write the quiz, I do not have a very good idea of what the mean Y will be. I think
the mean is equally likely to be anywhere between 60 and 90, i.e.,</p>
<p><a href="https://i.stack.imgur.com/adlSN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/adlSN.png" alt="enter image description here"></a></p>
<p>Suppose that I randomly select a quiz, grade it, and it turns out to receive a 100. Using
Bayes rule, how should I update my distribution for Y ? That is, what is fY |X(y|X = 100)?
(Simplify as much as possible, but you may leave your answer in terms of the standard
normal cumulative distribution function Φ if you wish.)</p>
| 0non-cybersec
| Stackexchange |
Failed to start Snappy Daemon. <p>Tried restarting my computer and while it's booting Ubuntu 18.04, it gets hung up after:</p>
<pre><code>[FAILED] Failed to start Snappy daemon,
See 'systemct1 status snapd.service' for details,
</code></pre>
| 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 |
Do dominant males who enjoy having sex with married women do anything to control the women while they are at home with their husbands ?. I just read a comment on another web site about infidelity written by a female unfaithful spouse who said her affair partner used to control how she cut/colored her hair and the clothes she bought and wore. What is the point of this ? Control of the woman at her home ? Domination of the husband ? | 0non-cybersec
| Reddit |
Will 8 GB of RAM be enough for the next 5 years?. OK this is such a weird question that I have been hesitating before posting. I am lost and need your help.
I recently changed job. Previously I worked in a traditional environment I.e. MS Office and email via Outlook. Now, my company uses nearly in-browser tools: Google Suite (Gmail, Docs, Spreadsheets, Drive), BitBucket and JIRA. I spend all the time in Chrome.
I have a 2013 basic MacBook Air (4GB RAM, Intel Core i5 1,3 GHz with Turbo Boost to 2,6 GHz) which was more than enough for previous stuff. Apple has really worked out the efficiency for Office and desktop apps. And I was using Safari for web browsing.
Unfortunately, the Google Suite does not work well in Safari. It freezes all the time, especially when I browse the large company Drive directory with several documents open in other tabs. Chrome works much better but still far from satisfaction.
I need a new Mac but am not sure which one.
I am choosing between the MacBook and the non-Touchbar MacBook Pro with 16 GB of RAM.
The MacBook is cool but it has max 8 GB of RAM.
The Pro is heavier but can have more RAM.
As I am on 4 GB of RAM now I am not sure: is 8 GB going to be enough or should I go for the 16?
Thanks | 0non-cybersec
| Reddit |
Why are my CSS properties being overridden/ignored?. <p>I'm having some issues with the CSS "hierarchy" (not sure if it's proper to call it a hierarchy). I'm trying to style the below bit of HTML.</p>
<pre><code><body>
<section id="content">
<article>
<ul class="posts-list">
<li class="post-item">
<h2>[post title]</h2>
<p class="item-description">...</p>
<p class="item-meta">...</p>
</li>
...
</ul>
</article>
</section>
</body>
</code></pre>
<p>Since section#content changes on every page I have, I wanted to maintain consistent styles across all of them, so I wrote some "global" CSS rules.</p>
<pre><code>#content {
color: #000;
margin-left: 300px;
max-width: 620px;
padding: 0px 10px;
position: relative;
}
#content p,
#content li {
color: #111;
font: 16px / 24px serif;
}
</code></pre>
<p>I wanted to style HTML within a <code>ul.posts-list</code> differently, so I wrote these rules.</p>
<pre><code>li.post-item > * {
margin: 0px;
}
.item-description {
color: #FFF;
}
.item-meta {
color: #666;
}
</code></pre>
<p>However, I ran into some issues. Here is how Chrome is rendering the CSS:</p>
<p><img src="https://i.stack.imgur.com/NSQzG.png" alt="Screenshot of how Chrome is rendering my CSS"></p>
<p>For some reason, the rules <code>#content p, #content li</code> are overriding my rules for <code>.item-description</code> and <code>.item-meta</code>. My impression was that class/id names are considered specific and thus higher priority. However, it seems that I have a misunderstanding of how CSS works. What am I doing wrong here?</p>
<p>Edit: Also, where can I read up more about how this hierarchy works?</p>
| 0non-cybersec
| Stackexchange |
restrict team/sharepoint/group creation. Hi,
​
I want to prohibit the function that users can create groups/teams/sharepoint sites etc. I found this guide;
​
[https://docs.microsoft.com/En-GB/office365/admin/create-groups/manage-creation-of-groups?view=o365-worldwide](https://docs.microsoft.com/En-GB/office365/admin/create-groups/manage-creation-of-groups?view=o365-worldwide)
​
But here they create a security group to delegate the creation of sites/groups etc. (requires azuread prem lic). The tenant global admin should be the only person that can make groups/teams/sites.
​
Is that something I can achieve easily?
​
Thanks in advance. | 0non-cybersec
| Reddit |
Raspberry Pi running a Twitter adventure game.. I don't know if there are any other examples of this kind of thing out there, but I'd like to let you know about my little project.
It's a (very small) text adventure game that you play using Twitter. I wrote it using Python scripts, Twython and MySQL. The system checks for updates every 2 minutes, parses the messages and sends out responses. To start playing just send the word "start" (no quotes) to @soulsdarkest
I've just finished making it so apologies if it breaks on you or can't handle very complicated instructions. Twitter doesn't like duplicate tweets either so to get around this it helps to add a small random number at the end of any tweet you send to the system, for example:
@soulsDarkest examine fire 2459
I'll try to improve the parser over the next couple of days, and hopefully it can handle several people playing!
Cheers! | 0non-cybersec
| Reddit |
That's one way to not get hair on your car seats. | 0non-cybersec
| Reddit |
Whats blue and doesn't fit anymore. A dead epileptic | 0non-cybersec
| Reddit |
Where's the Food Going?. Before I started CICO/DASH, I would eat TONS of carbs all day, and binge every night--I was probably consuming around 3000-5000 calories per day.
I've been CICO for a month, I've lost 18 pounds, and I feel great.
And I'm obviously going to the bathroom a LOT less (BMs).
I used to have one "epic" BM every single day, but now I'm down to one or two BMs per week, and they're not anywhere near as big.
​
WHERE'S THE FOOD GOING?
**Am I burning (my new, high quality food) in almost REAL TIME? to FUEL my body?**
**Is that for real?**
I'll eat a TON of broccoli, or quorn or yogurt one day, and expect to have a huge BM the next day that never comes.
WHAT'S HAPPENING TO MY BODY?!
Is this how normal (thin/healthy) bodies are supposed to work? | 0non-cybersec
| Reddit |
Insta models!. | 0non-cybersec
| Reddit |
Northern Ireland court to consider whether Brexit requires parliament vote. | 0non-cybersec
| Reddit |
Web 2.0 Pivot Attacks - Attacking Websites through their Third Party Providers. | 1cybersec
| Reddit |
Handle Angular 401 responses. <p>I have simple api and a authorization point</p>
<p>when i request to api i get a 401 if the token is invalid (token loses validity past five minutes).</p>
<p>i know i can intercept 401 for example with </p>
<pre><code>app.factory("HttpErrorInterceptorModule", ["$q", "$rootScope", "$location",
function($q, $rootScope, $location) {
var success = function(response) {
// pass through
return response;
},
error = function(response) {
if(response.status === 401) {
// dostuff
}
return $q.reject(response);
};
return function(httpPromise) {
return httpPromise.then(success, error);
};
}
]).config(["$httpProvider",
function($httpProvider) {
$httpProvider.responseInterceptors.push("HttpErrorInterceptorModule");
}
]);
</code></pre>
<p>but i want capture and queue the request and show a login form if is success then change the token (it's a header) and execute request again</p>
| 0non-cybersec
| Stackexchange |
Google Chrome and Microsoft Edge Chrome Both Crash Instantly Upon Launching. <p>I have a single Windows server (Windows Server 2016 / 1607) which has peculiar issues with web browsers.</p>
<p>Both Google Chrome and Microsoft Edge Chrome crash instantly when they are launched and display the <code>Aw Snap</code> / <code>This Page is having a problem</code> error respectively, even when I launch the about:chrome page. </p>
<p>IE11 on the other hand works perfectly.</p>
<p>First I thought it was the Chrome / Symantec bug from a couple of months ago, but I've already implemented policies enterprise-wide to fix that particular issue. </p>
<p>I've tried the following:</p>
<ul>
<li>Reinstalling latest version of Chrome </li>
<li>Installing latest version of Edge Chrome </li>
<li>Disabling Symantec Endpoint Protection (SEP) </li>
<li>Updating SEP to 14.2.xxxx</li>
</ul>
| 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 |
Travelling at light speed away from the sun doesn't really seem that fast. | 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
| Stackexchange |
Saas vs. "Cloud Based Service". <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://webapps.stackexchange.com/questions/301/what-is-cloud-vs-saas-vs-asp">What is Cloud vs. Saas vs. ASP?</a> </p>
</blockquote>
<p>What is the distinguishing factor(s) between these two?</p>
<p>Cloud Based Service - You can come use our Cloud to store and manage your data
SaaS - You can come use our Cloud application to store and manage your data</p>
<p>When do you call it Saas vs a Service provided to a client who has a specific processing need in mind and that processing need uses a SW application (several out there to choose from but the Cloud Provider has picked one specifically and the user can't change that) </p>
<p>In this case the SW app is not a cloud app - it does not make the cloud run, it does not link users to the systems, it is task specific. So is it always Saas if it is an application that "runs on the cloud". Or could it be considered a Computing service since customer can't choose the application used to perform the specific task</p>
| 0non-cybersec
| Stackexchange |
Finnaly completed it!. | 0non-cybersec
| Reddit |
YSK not to touch "chocolate hair" (Especially if you are "curious/just trying to be nice").. | 0non-cybersec
| Reddit |
Custom shortcut for special character - Word 2016 Mac. <p>Good evening!</p>
<p>I am trying to find a way to assign a custom shortcut (for example control + s) for special characters found in the Latin section of Emoji & Symbols tab on Word 2016 Mac (for example ț).</p>
<p>On Windows its a rather trivial task - you can simply find it in the Insert - Symbols tab. Not so easy on Mac... </p>
<p>Any help would be greatly appreciated. Thanks.</p>
| 0non-cybersec
| Stackexchange |
Vid of Kipchoge's 2 hour marathon from my perspective. Blistering pace. | 0non-cybersec
| Reddit |
I tried being friends with my ex who dumped me. Here is my experience to those who might want to try the same.. She dumped me four months ago out of the blue. It hurt me very very much and i was so thirsty of her company that i tried to have her in my life at least in some way. But every meeting with her for coffe or something hurt because “that’s not it”. It is painful to have to kiss her on the cheek, not to talk the way we used to and to endure her cold and distant eyes. There was no warmth and love in her eyes any more. I feel so degraded and with smaller value when we hangout. When i make a suggestion for something more between us she rejects it away like its some filth. Everytime when she says that we’re just friends hurts so much because i am now just a dude like any other. There is nothing special anymore. Every meeting i was hoping i will get some spark which will put us in the old days but i get nothing. I return every time sadder and depressed and sometimes even crying. I don’t think i can endure it anymore and i want to end this”friendship”. The idea of not having her at all in my life scares me too. I feel so stretched. I wish i haven’t even met her so i don’t have to go through all of this. | 0non-cybersec
| Reddit |
lldb breakpoint commands not working in XCode 8. <p><strong><em>EDIT: It now appears that it's not limited to <code>frame info</code> but in fact, NONE of the <code>breakpoint command add</code> commands work. They will only work if typed in manually at an (lldb) command prompt</em></strong></p>
<p>I set some breakpoints in LLDB via the XCode console, to examine a class and how it's used in code. </p>
<pre><code>(lldb) breakpoint set --func-regex "DVLayer"
</code></pre>
<p>Breakpoint 7: 73 locations.</p>
<pre><code>(lldb) breakpoint command add 7
</code></pre>
<p>Enter your debugger command(s). Type 'DONE' to end.</p>
<pre><code> frame info
continue
DONE
</code></pre>
<p>This has worked fine in all previous versions of XCode going back to XCode 4. Now however, all I get are these statements in the console, and no frame info.
I have tried the breakpoint commands on 5 different classes, and it worked on none of them... always this output.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Command #2 'continue' continued the target.</p>
<p>Also note that <strong>THIS ONLY FAILS TO WORK INSIDE A <code>breakpoint command add</code> statement</strong></p>
<p>If I am simply stopped at a breakpoint in LLDB at the (lldb) command prompt, I can type <code>frame info</code> and it works as expected, however, as indicated above, adding <code>frame info</code> as a breakpoint command completely fails with the output above.</p>
| 0non-cybersec
| Stackexchange |
Child Domain Migration to Parent Domain. <p>Our company is planning on moving from a single forest with multiple domains to a single forest with a single domain.</p>
<p>This is my infrastructure:</p>
<p><strong>Forest Root</strong></p>
<pre><code>ROOT-DC.techtunes.lan
ROOT-ADC.techtunes.lan
ROOT-ADC2.techtunes.lan
</code></pre>
<hr>
<p><strong>SITES:</strong></p>
<p><strong>Cambridge</strong> </p>
<pre><code>dc1cam.cambrdige.techtunes.lan
adc1cam.cambrdige.techtunes.lan
</code></pre>
<hr>
<p><strong>Oxford</strong> </p>
<pre><code>dc1oxf.oxford.techtunes.lan
adc1oxf.oxford.techtunes.lan
</code></pre>
<hr>
<p><strong>Karachi</strong></p>
<pre><code>dc1khi.karachi.techtunes.lan
adc1khi.karachi.techtunes.lan
</code></pre>
<hr>
<p>Now, we are planning to move our child domains on root-dc.techtunes.lan then we will create separate OU for each site and also place Global Catalogs on each site for backup purposes.</p>
<p>I have read that the Active Directory Migration Tool v3.2 is an important tool to migrate our users, but I have some questions about it.</p>
<p>What happens with users whose names are duplicated in each domain? For example: one user named <code>abc</code> is in Cambridge and also in Karachi. What happens when we move both <code>acb</code> accounts to a single domain?</p>
<p>What about other services like DHCP and DNS? We want to run DHCP locally at each site.</p>
<p>I'd appreciate answers on how to proceed.</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 |
Returning abstract type in base class. <p>In a design of a class hierarchy, I'm using an abstract base class that declares various methods that the derived classes would implement. In a sense, the base class is as close to an interface as you can get in C++. However, there is an specific issue. Consider the code below which declares our interface class:</p>
<pre><code>class Interface {
public:
virtual Interface method() = 0;
};
class Implementation : public Interface {
public:
virtual Implementation method() { /* ... */ }
};
</code></pre>
<p>Of course, this wouldn't compile, because you cannot return an abstract class in C++. To get around this problem I'm using the following solution:</p>
<pre><code>template <class T>
class Interface {
public:
virtual T method() = 0;
};
class Implementation : public Interface<Implementation> {
public:
virtual Implementation method() { /* ... */ }
};
</code></pre>
<p>This solution works and is all fine and dandy, however, to me, it doesn't look very elegant, because of the redundant bit of text which would be the parameter for interface. I'd be happy if you guys could point our any other technical issues with this design, but that is my only concern at this point.</p>
<p>Is there any way to get rid of that redundant template parameter? Possibly using macros?</p>
<p><strong>Note:</strong> The method in question has to return an instance. I'm aware that if <code>method()</code> returned a pointer or a reference, there would be no issue.</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 |
OS X 10.8.4 update available now.. | 0non-cybersec
| Reddit |
The European response to PRISM must be protecting privacy. | 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
| Stackexchange |
Huck's Honeymoon. | 0non-cybersec
| Reddit |
So peaceful when not disembowling. | 0non-cybersec
| Reddit |
Can wifi speed on a laptop be limited by i7 core processor?. <p>Can someone dumb this down for me or help with a solution? I have an i7 Core Processor in my Inspiron HP laptop and am not getting full wifi speeds from our Comcast/Xfinity X6 modem/router. Sitting at my desk, my laptop only gets around 100MBPS. But sitting at my desk next to my laptop, I get over 400MBPS download speed on my iPhone from that same router/modem. Also get the same higher speed from an iPad. Could it be that my laptop processor won't handle that speed? Thx for any help on this.</p>
| 0non-cybersec
| Stackexchange |
So I gave a bus tour to the Eagles Rookies about a month ago.... | 0non-cybersec
| Reddit |
Google Sheets conditional formatting with AND for value range. <p>I'm trying to set a colour range of conditional formatting based on another cells value. So if A1 is larger than A2 it will be Blue, if A1 is between 0 and A2 it will be Yellow and if A1 is less than 0 it will be red. </p>
<p>I'm struggling with the second function - if A1 is between 0 and A2 it will be Yellow. Please can someone advise how to do this?</p>
| 0non-cybersec
| Stackexchange |
Finally the true next-generation console!. | 0non-cybersec
| Reddit |
Crash of Garbage Collection Work Queue if dylib is loaded. <p>We are porting an app from 10.6 to 10.8. I am looking at dylib we load in app. I am facing very unusual crash in <strong>Garbage Collection Work Queue</strong> with following message.</p>
<pre><code>malloc: Thread::suspend(): unable to suspend a thread: err = 268435459, Thread 0x111000000: _pthread = 0x108129000, _thread = 0x8b07, _stack_base = 0x108129000, enlivening on, 0 local blocks
</code></pre>
<p>For application <code>GCC_ENABLE_OBJC_GC = required</code> is set. If I have <code>GCC_ENABLE_OBJC_GC = required</code> in dylib it will still crash. I cannot turn off garbage collector in application. I have to manage it crash from my dylib.</p>
<p>Reason for crash turns out to be that garbage collector is not able to suspend the thread. (as it says in log). This thread is created using <a href="http://web.mit.edu/darwin/src/modules/xnu/osfmk/man/thread_create.html" rel="noreferrer">thread_create()</a>. If I put a indefinite while loop (with sleep) in constructor of dylib, I dont get crash. I get crash when constructor has finished its execution.</p>
<p>Is their a way to tell garbage collector not to try and suspend the thread? Or to increase reference count of thread? or anything I can do to stop garbage collector not to interfere with my dylib code. </p>
| 0non-cybersec
| Stackexchange |
Determining if Server/Client's TLS configuration is Vulnerable to SLOTH. <p>In the context of SSL/TLS protocol and <a href="http://www.mitls.org/pages/attacks/SLOTH" rel="nofollow">SLOTH</a> vulnerability. I want to find out if certain websites(webservers) or applications that use TLS are vulnerable. I need some guidance for this. </p>
<p>My idea is to develop/find a tool(maybe openssl) that connects to the servers and in the client hello asks for RSA-MD5 SignatureAndHashAlgorithm to test if the server sends the corresponding certificate under this configuration and conclude that the server is vulnerable.</p>
<p>Likewise for the client certificates, I want to ask for client certificates signed by RSA-MD5 and if the client sends its certificates under this setup I'll conclude that the client is vulnerable to SLOTH.</p>
<p>First, is my approach right? or am I off the road?</p>
<p>Second, what types of tools do I need to use? Is openssl capable of producing such requests and responses?</p>
| 0non-cybersec
| Stackexchange |
How to find the solution to this system of integer variables?. <p>Let $x_i$ be nonnegative integer variables in $\mathbb{N}$. The system is to find $x_i$ in</p>
<p>$$\sum_{i=1}^nx_i=3n,\\\quad \quad \quad \quad \;\;\,x_i\leqslant3,\,\forall\, i\in\{1,\ldots,n\}.$$</p>
<p>I find the solution to be $x_i=3$ for all $i$ but I cannot prove that it is the unique solution.</p>
| 0non-cybersec
| Stackexchange |
So I wanted to build a pc since 5 years, and its happening now! I need help in fine-tuning the parts tho. Any suggestions? 1300 around is my budget.. So I live overseas (India, actually). I have an uncle in the US, who is willing to gather the parts and ship it here.
I will mainly be using it for gaming, streaming and a slight video editing.
Here's my newegg list:
https://drive.google.com/file/d/1RziKg7aEGNlsAZQ4cFMn99oU8rTwoR1X/view?usp=drivesdk
For the monitor, I was done with my 1600x900 60hz one, so went for the most popular, cheapest one
ASUS TUF Gaming VG24VQ 24" https://www.newegg.com/product/N82E16824281027 | 0non-cybersec
| Reddit |
Paul Peschisolido on Neil Warnock taking the Sheffield Utd players bowling: "He convinced us all to put £10 in a pot, winner takes all. We agreed for some fun, then he pulled out his own bowling shoes and custom ball, shot 250 and took all our money".. | 0non-cybersec
| Reddit |
If $A^2=I$ and $\lambda \not =-1$, then $A=I$. <blockquote>
<p>Given an <span class="math-container">$n\times n$</span> matrix <span class="math-container">$A$</span> with <span class="math-container">$A^2=I$</span>, assume that <span class="math-container">$-1$</span> is not an eigenvalue of <span class="math-container">$A$</span>. Prove <span class="math-container">$A=I$</span>.</p>
</blockquote>
<p><strong>Proof attempt:</strong> Since <span class="math-container">$A^2=I,$</span> we have <span class="math-container">$A^{-1}=A$</span>. Using that fact that <span class="math-container">$\det(A)=\prod_{i=1}^n\lambda_i$</span>, we see that each of the eigenvalues is nonzero. For any eigenvector <span class="math-container">$v$</span> of <span class="math-container">$A$</span>, we see <span class="math-container">$Av=\lambda v$</span>, so <span class="math-container">$\frac{1}{\lambda} v = A^{-1}v=Av$</span>. This implies <span class="math-container">$\lambda^2=1$</span>. By assumption, we conclude that each eigenvalue is <span class="math-container">$1$</span>. Hence, <span class="math-container">$\det A=1$</span>.</p>
<p>(1) How do we know that <span class="math-container">$A$</span> actually has any eigenvectors? E.g., the map which rotates a vector in the plane by <span class="math-container">$\pi/2$</span> clockwise is invertible but has no eigenvectors. (Perhaps there should be some given information about the underlying field.)</p>
<p>(2) How do I use <span class="math-container">$\lambda=1$</span> to conclude <span class="math-container">$A=I$</span>? </p>
<p>I tried using <span class="math-container">$A=A^{-1} = \frac{C^T}{\det A}=C^T$</span>, where <span class="math-container">$C$</span> is the matrix of cofactors of <span class="math-container">$A$</span>. From this, I conclude that each entry of <span class="math-container">$A$</span> equals its corresponding co-factor -- i.e., <span class="math-container">$A_{i,j} = \det(\overline{A_{i,j}})$</span>, where <span class="math-container">$\overline{A_{i,j}}$</span> is the sub-matrix of <span class="math-container">$A$</span> obtained by deleting row <span class="math-container">$i$</span> and column <span class="math-container">$j$</span>. Using the co-factor expansion of the determinant along row <span class="math-container">$i$</span>, we also conclude <span class="math-container">$$1=\det A=\sum_{j=1}^n A_{i,j}(-1)^{i+j}\det(\overline{A_{i,j}})=\sum_{j=1}^n(A_{i,j})^2(-1)^{i+j}.$$</span></p>
| 0non-cybersec
| Stackexchange |
Iceland.. | 0non-cybersec
| Reddit |
Find the ratio between the radius and the height. <blockquote>
<p>A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10 cm<span class="math-container">$^3$</span>/min. When the depth of water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to the radius?</p>
</blockquote>
<p>So,</p>
<p><span class="math-container">${dV \over dt} = -10$</span></p>
<p><span class="math-container">$h=8$</span></p>
<p><span class="math-container">${dh \over dt}= -2$</span></p>
<p>Now,</p>
<p><span class="math-container">$${V}= {1 \over 3} \pi r^2h$$</span></p>
<p>However, it seems like I am not given enough information.</p>
<p>What do I do?</p>
<hr>
<p>I tried:</p>
<p><span class="math-container">$${dV \over dh} = {2\over 3} \pi r {dr \over dh}$$</span></p>
<p>However, it feels like I am not given enough information since I don't have <span class="math-container">${dr \over dh}$</span></p>
| 0non-cybersec
| Stackexchange |
Sets finding biggest value and minimum value using inclusion exclusion principle. <p>Consider the set A1, A2, A3 of | A1 | = 15, | A2 | = 12, | A3 | = 6, <span class="math-container">$| A1 \cap A2 \cap A3 | =3$</span> What is the minimum and maximum value of <span class="math-container">$|A1\cup A2 \cup A3|$</span>?</p>
<p>My attempt: the minimum value is 15
Because A1 is biggest set can contain 9 from A2 + 3 from A3 and 3 from <span class="math-container">$| A1 \cap A2 \cap A3 | =3$</span> so all satisfy</p>
<p>But for biggest possible value</p>
<p><span class="math-container">$| A1 | + |A2 | + |A3 | - (|A1 \cup A2 | + |A2 \cup A3| + |A3 \cup A1) +3 $</span> </p>
<p>Suppose a is <span class="math-container">$|A1 \cup A2 |$</span> and b= <span class="math-container">$|A2 \cup A3| $</span> and c= <span class="math-container">$|A3 \cup A1|$</span> </p>
<p><span class="math-container">$15+12+6-(a+b+c)+3=$</span></p>
<p><span class="math-container">$a +b+c =36$</span></p>
<p>But we know also that </p>
<p><span class="math-container">$a+b \leq 12$</span></p>
<p><span class="math-container">$a+c \leq 9$</span></p>
<p><span class="math-container">$b+c \leq 3$</span></p>
<p>Sum above <span class="math-container">$ 2a+2b+2c=24$</span></p>
<p>So <span class="math-container">$a+b+c=12$</span></p>
<p>Then how can i conclude what is the biggest value that is possible?</p>
<p>Consider the sets as people and there are 3 circle for every sets in venn diagram</p>
| 0non-cybersec
| Stackexchange |
I see your Pulse Bomb Volleyball and present to you Hammond Uppercut Volleyball!. | 0non-cybersec
| Reddit |
So that’s what Apple did with Workflow. Anyone notice how Siri shortcuts looks oddly familiar? | 0non-cybersec
| Reddit |
django admin save_model not saving model changes. <p>This is my admin.py:</p>
<pre><code>class TransactionInfoAdmin(admin.ModelAdmin):
search_fields=['transactionId','subscriptionInfo__customerNumber']
save_on_top = False
def has_add_permission(self,request):
return False
def has_delete_permission(self, request, obj=None):
return False
def get_readonly_fields(self, request, obj=None):
if obj:
return self.readonly_fields + ('subscriptionInfo','transactionId','authorizationId','responseCode','operation','batchNumber','merchantId')
return self.readonly_fields
def save_model(self, request, obj, form, change):
#SUCCESS ACTIONS
obj.subscriptionInfo.paid = 1
obj.save()
pass
</code></pre>
<p>When I am saving the <code>obj.subscriptionInfo.paid = 1</code> its not updating the data row, any ideas why?</p>
<p><strong>NOTE: subscriptionInfo is foreign key in the above model TransactionInfo.</strong> </p>
<p>-------------------------- UPDATE --------------------------</p>
<p>My model.py</p>
<pre><code>class SubscriptionInfo(models.Model):
subscriptionId = models.CharField(max_length=50)
customerNumber = models.IntegerField()
subscriptionType = models.CharField(max_length=50, default="basic")
currency = models.CharField(max_length=50, default="NOK")
paid = models.IntegerField(default=0)
class Meta:
verbose_name = "Subscription Information"
verbose_name_plural = "Subscription Information"
def __unicode__(self):
return u'%s' % self.customerNumber
class TransactionInfo(models.Model):
subscriptionInfo = models.ForeignKey(SubscriptionInfo, db_column='f')
authorizationId = models.IntegerField()
transactionId = models.TextField()
batchNumber = models.IntegerField()
merchantId = models.IntegerField()
operation = models.CharField(max_length=50)
responseCode = models.CharField(max_length=50)
paidDate = models.DateField(_("Date"), default=datetime.date.today)
class Meta:
verbose_name = "Transaction Information"
verbose_name_plural = "Transaction Information"
def __unicode__(self):
return u'Transaction : %s' % self.pk
</code></pre>
<p>Anyone?</p>
| 0non-cybersec
| Stackexchange |
Getting function Content and function name in C with regular expression in python. <p>I am trying to get function content (body) if the function's name matches a defined pattern</p>
<p>what I tried so far:</p>
<blockquote>
<p>(Step1) get with a recursion all function bodies in a define C file
{(?:[^{}]+|(?R))*+}</p>
<p>(Step2) find all matches of wanted function' s name</p>
<p>(Step3) Combine both steps. This where I am struggling</p>
</blockquote>
<h3>Input</h3>
<pre><code>TASK(arg1)
{
if (cond)
{
/* Comment */
function_call();
if(condIsTrue)
{
DoSomethingelse();
}
}
if (cond1)
{
/* Comment */
function_call1();
}
}
void FunctionIDoNotWant(void)
{
if (cond)
{
/* Comment */
function_call();
}
if (cond1)
{
/* Comment */
function_call1();
}
}
</code></pre>
<p>I am looking for the function TASK. When I add the regex to match TASK in front of "{(?:[^{}]+|(?R))*+}", nothing works.</p>
<pre><code>(TASK\s*\(.*?\)\s)({((?>[^{}]+|(?R))*)})
</code></pre>
<h3>Desired Output</h3>
<pre><code>Group1:
TASK(arg1)
Group2:
if (cond)
{
/* Comment */
function_call();
if(condIsTrue)
{
DoSomethingelse();
}
}
if (cond1)
{
/* Comment */
function_call1();
}
</code></pre>
| 0non-cybersec
| Stackexchange |
Poodles love to help with instructions.. | 0non-cybersec
| Reddit |
Applebee's mirror twitter comments about them on their home page - they often aren't very complimentary at the moment.. | 0non-cybersec
| Reddit |
iPhone max users with small hands. How is using the max compared with the X? | 0non-cybersec
| Reddit |
Yellowfin Tuna Burrito - My Ceviche, Miami. | 0non-cybersec
| Reddit |
Use a password to protect a text file. <p>Is there any easy/simple way to protect a .txt file with a password? So whenever open the file, password is required, just like opening .pdf with a password protected. Though I still want to be able to edit the text file in general text editors (notepad).</p>
| 0non-cybersec
| Stackexchange |
How to access dataset in current scope generated by a call to a stored procedure in TSQL?. <h2>Problem Background</h2>
<p>Generating and accessing data of a fixed column layout is easy. You can create local temp tables up-front, and populate them by calling stored procedures.</p>
<p>On the other hand, if you want to generate data with a dynamic column layout, you must generally build an SQL statement dynamically and execute it with "exec sp_executesql". Since the data layout is unknown at run-time, you cannot create a temp-table up-front, and once inside the "exec sp_executesql" statement, any temporary tables created there are bound to that scope and vanish when the call returns, so it's much more difficult to access the data (i.e. your options are more limited).</p>
<h2>My Specific Situation</h2>
<p>I have a query that needs to access data in a dynamically generated table.</p>
<p>The table is generated by a stored procedure, which dynamically builds a query, stores it in a variable "@sql nvarchar(max)", and runs it by calling "exec sp_executesql @statement = @sql".</p>
<p>The @sql statement was something like "select * into #temptable from...", but #temptable was destroyed by the time "exec sp_executesql" returned. A quick fix for this was to just use "##temptable" instead (i.e. a global temp table), because it survives when the stored procedure returns AND I can easily access it in the calling scope (because it has a known/static name).</p>
<p>I don't like this solution because global temp tables aren't thread-safe (name collistion-wise), and I don't want to have to mess with dynamically-generated unique names, because I'll just end up having to use more dynamic SQL to access them... which puts me right back at square one, leaving the data inaccessible outside the SP.</p>
<p>I don't think returning table variables (through output parameters) is an option (new to SQL Server 2008 too), unless it can be done without having to define a static table type. The tables my stored procedure generates are dynamic, and depend on the input parameter(s) passed.</p>
<p>Inline table-valued functions are not an option, because I'm running code loops to build the @sql query and calling "exec sp_executesql".</p>
<p>Multi-statement table-valued functions (instead of the stored procedure), is also not an option, because such a function must have a well-defined table format, whereas I'm running dyanmic SQL to return a table with a variable number of columns and column names depending on the input parameter values.</p>
<p>All I really want to do is select the result set of the dynamic query into a new table, but I'm finding it difficult, as none of the above works; particularly irritating is how local temporary tables aren't local to the session, but local to the stored procedure so that they vanish upon returning. The only <a href="https://stackoverflow.com/questions/653714/how-to-select-into-temp-table-from-stored-procedure">solution</a> I've seen insists that using OPENROWSET is the only way, but I don't want to mess with connection strings inside my stored procedure, for the same reason I don't want to include unique-name-management code... it's just way more complicated than it ought to be.</p>
<p>In summary, I just want to execute dynamic SQL that generates a dataset of an unknown format, and be able to easily access it from the calling scope.</p>
| 0non-cybersec
| Stackexchange |
Is java.lang.OutOfMemoryError: Metaspace normal in sbt REPL?. <p>I am new to Scala, sbt and its <a href="https://en.wikipedia.org/wiki/Read%E2%80%93eval%E2%80%93print_loop" rel="noreferrer">REPL</a>.
One thing I really love is the <code>~</code> option to loop a run or compile when editing files. So I end up running <code>~run</code> quite often.</p>
<p>But it leaks memory. As after 20-30 runs the whole sbt interactive shell crashes with:</p>
<pre><code>[info] Compiling 1 Scala source to /home/[redacted]
sbt appears to be exiting abnormally.
The log file for this session is at /tmp/sbt853875123365456892.log
java.lang.OutOfMemoryError: Metaspace
Error during sbt execution: java.lang.OutOfMemoryError: Metaspace
</code></pre>
<p>Honestly it's not a big deal, but while I understand what an OOME is, the question I am trying to figure out - is it really what it's supposed to be? The answer is "no", but...</p>
<ul>
<li>is it a sbt bug? (should I try to report it?)</li>
<li>is something wrong with my program? (memory leak?)</li>
</ul>
| 0non-cybersec
| Stackexchange |
J=?. | 0non-cybersec
| Reddit |
Proof that number of definable sets is countable. <p>I did the following two related exercises in Just/Weese on page 114:</p>
<p>18: Show that if $z$ is a finite set and $y \subset z$ then there exists a formula $\varphi(x)$ of $L_S$ such that $y = \{x \in z: \varphi (x) \}$.</p>
<p>19: Let $Y$ be the collection of all $y \subset \omega$ for which there exists a formula $\varphi (x)$ without parameters such that $y = \{x \in \omega : \varphi (x) \}$. Show that $Y$ is countable. </p>
<p>One problem I have is that the first one is rated "difficult" and the second one is rated "extra difficult" but both of my answers are not only very simple but it also only took me a minute to come up with each so both are probably wrong and I am missing something. Here they are:</p>
<hr>
<p>18: Let $y$ be the set $y = \{y_1 , \dots , y_n \} \subset z$. Let $\varphi(x) = (x = y_1) \lor \dots (x=y_n)$. Then $y = \{ x \in \omega : \varphi (x) \}$.</p>
<hr>
<p>19: Let $F_n$ be the set of all formulas $\varphi$ in $L_S$ such that $\varphi$ consists of $n$ characters. Since there are only finitely many symbols in $L_S$, logical and non-logical combined, $F_n$ is finite. Then $F = \bigcup_{n \in \mathbb N} F_n$, the set in which each element corresponds to a definable $y \subset \omega$, is a countable union of finite sets and hence countable. </p>
<p>Could you point out what I am missing? Thanks for your help!</p>
| 0non-cybersec
| Stackexchange |
Can't manually install a driver (.inf). <p>I have an AOC monitor with Freesync and I'm trying to install the driver the driver that they provide (<a href="http://aoc-europe.com/fr/products/g2260vwq6#product-detail-accessoires" rel="nofollow noreferrer">http://aoc-europe.com/fr/products/g2260vwq6#product-detail-accessoires</a>)</p>
<p>It's a .inf file, I read many tutorials on how to install them (for example <a href="https://superuser.com/questions/316206/installing-a-drivers-inf-file">Installing a driver's INF file?</a>) </p>
<p>The problem I have is that I can not see the file (though I can see the directory that contains it) and when I enter the path manually it gives me an error saying "The specified path does not exist or can not be accessed" </p>
<p>Here's a gif demonstrating this : <a href="https://i.imgur.com/OYiGmiM.gifv" rel="nofollow noreferrer">http://i.imgur.com/OYiGmiM.gifv</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 |
I hear people from Illinois get mad when you pronounce the S,. It really ill-annoys them. | 0non-cybersec
| Reddit |
Completely intentional in Dirt 3. | 0non-cybersec
| Reddit |
Samba Domain Member Slow Authentication. <p>I have a Samba 4 domain controller set up with about 100 member computers connected to it on a private LAN. </p>
<p>As I was rolling this out, I found a couple clients that were improperly configured with the wrong DNS address after a reboot and it took about 10 minutes to log in as a local user to fix it. Is there any kind of timeout that you can configure to make that 30s or less? I need all of these machines to allow local users to log on relatively quickly. </p>
| 0non-cybersec
| Stackexchange |
Weight Watchers Super Bowl ad narrated by Aaron Paul. | 0non-cybersec
| Reddit |
YOUNG MAN.... | 0non-cybersec
| Reddit |
Methodologies for performance-testing a WAN link. <p>We have a pair of new diversely-routed 1Gbps Ethernet links between locations about 200 miles apart. The 'client' is a new reasonably-powerful machine (HP DL380 G6, dual E56xx Xeons, 48GB DDR3, R1 pair of 300GB 10krpm SAS disks, W2K8R2-x64) and the 'server' is a decent enough machine too (HP BL460c G6, dual E55xx Xeons, 72GB, R1 pair of 146GB 10krpm SAS disks, dual-port Emulex 4Gbps FC HBA linked to dual Cisco MDS9509s then onto dedicated HP EVA 8400 with 128 x 450GB 15krpm FC disks, RHEL 5.3-x64).</p>
<p>Using SFTP from the client we're only seeing about 40Kbps of throughput using large (>2GB) files. We've performed server to 'other local server' tests and see around 500Mbps through the local switches (Cat 6509s), we're going to do the same on the client side but that's a day or so away.</p>
<p>What other testing methods would you use to prove to the link providers that the problem is theirs?</p>
| 0non-cybersec
| Stackexchange |
No fat shaming please. | 0non-cybersec
| Reddit |
1
Multi-Entity and Multi-Enrollment Key Agreement
with Correlated Noise
Onur Günlü, Member, IEEE
Abstract—A basic model for key agreement with a remote
(or hidden) source is extended to a multi-user model with joint
secrecy and privacy constraints over all entities that do not
trust each other. Multiple entities using different measurements
of the same source through broadcast channels (BCs) to agree
on mutually-independent local secret keys are considered. Our
model is the proper multi-user extension of the basic model
since the encoder and decoder pairs are not assumed to trust
other pairs, unlike assumed in the literature. Strong secrecy
constraints imposed on all secret keys jointly, which is more
stringent than separate secrecy leakage constraints for each secret
key considered in the literature, are satisfied. Inner bounds for
maximum key rate, and minimum privacy-leakage and database-
storage rates are proposed for any finite number of entities.
Inner and outer bounds for degraded and less-noisy BCs are
given to illustrate cases with strong privacy. A multi-enrollment
model that is used for common physical unclonable functions is
considered to establish inner and outer bounds for key-leakage-
storage regions that differ only in the Markov chains imposed.
For this special case, the encoder and decoder measurement
channels have the same channel transition matrix and secrecy
leakage is measured for each secret key separately. We illustrate
cases for which it is useful to have multiple enrollments as
compared to a single enrollment and vice versa.
Index Terms—Information theoretic privacy, multiple enroll-
ments, multiple entities, physical unclonable functions.
I. INTRODUCTION
A natural source of randomness is biometric identifiers such
as fingerprints that are generally transformed into a frequency
domain and quantized to obtain bit sequences that are unique
to an individual [1]. Similarly, physical identifiers such as fine
variations of ring oscillator (RO) outputs or random start-
up values of static random access memories (SRAMs) that
are caused by uncontrollable manufacturing variations, are
safer and cheaper alternatives to key storage in a non-volatile
memory [2]. Physical identifiers for digital devices, such as
Internet-of-Things (IoT) devices, can be implemented using
physical unclonable functions (PUFs) [2]. One can use PUFs
in various coding schemes as a source of local randomness
[3, Chapter 1], e.g., in the randomized encoder of the wiretap
channel [4] and of the strong coordination problem [5], [6].
We use the basic source model for key agreement from
[7], [8] to find achievable rate regions for key agreement
with PUFs and biometric identifiers. In this classic model, an
encoder observes a source output to generate a secret key and
Manuscript received April 30, 2020. O. Günlü is supported by the German
Federal Ministry of Education and Research (BMBF) within the national
initiative for “Post Shannon Communication (NewCom)” under the Grant
16KIS1004.
O. Günlü is with the Information Theory and Applications Chair, Tech-
nische Universität Berlin, 10623 Berlin, Germany (e-mail: guenlue@tu-
berlin.de).
sends public side information, i.e., helper data, to a decoder,
so the decoder can reliably reconstruct the same secret key by
observing another source output and the helper data. The main
constraints are that the information leaked about the secret
key, i.e., secrecy leakage, is negligible and the information
leaked about the identifier output, i.e., privacy leakage, is small
[9], [10]. Furthermore, the amount of storage should also be
minimized to limit the hardware cost [11].
Suppose the encoder generates a key from a noisy measure-
ment of a hidden (or remote) source output, and a decoder
has access to another noisy measurement of the same source
and the helper data to reconstruct the same key. We call this
model the generated-secret (GS) model with a hidden source.
This model is introduced in [12] as an extension of the visible
(noiseless) source outputs observed by the encoder, considered
in [9], [10]. Similarly, for the chosen-secret (CS) model, an
embedded (or chosen) key and noisy identifier measurements
are combined by the encoder to generate the public helper data.
We consider both models to address different applications.
A. Related Work and Motivation
The same identifier is used by multiple encoder and decoder
pairs in [13], where the identifier outputs observed by different
encoders are the same because the encoder measurements are
assumed to be noiseless. Therefore, the multiple use of the
same noiseless source output allows all encoders to know the
secret key of the other encoders. This model does not fit well to
the practical key agreement with identifiers scenarios because
there is noise in every identifier measurement.
Multiple enrollments of a hidden source using noisy mea-
surements are considered in [14], where weakly secure secret
keys are generated without privacy leakage and storage con-
straints. Furthermore, there is a causality assumption in [14] on
the availability of the helper data, i.e., any decoder has access
to all previously-generated helper data. This assumption is not
necessarily realistic as a decoder of, e.g., an IoT device that
embodies a PUF should be low complexity and the amount
of data to process increases linearly with the number of
enrollments. In addition, any manipulation in any of the helper
data can cause the complete multi-enrollment system to fail.
A classic method used for key agreement, i.e., the fuzzy
commitment scheme (FCS) [15], is used in [16] in combina-
tion with an SRAM PUF to enroll the noisy outputs of the
same SRAM multiple times. The symmetry condition in [16,
Eq. (16)] conditioned on a fixed SRAM cell state is entirely
similar to the symmetry satisfied by binary-input symmetric
output (BISO) channels; see e.g., [17, p. 613], [12, Eq. (14)].
For SRAM outputs that satisfy this symmetry, the normalized
ar
X
iv
:2
00
5.
08
21
0v
2
[
cs
.I
T
]
2
3
Ju
l
20
20
2
(weak) secrecy leakage about each separate secret key is
shown to be zero. It is discussed in [18, Section 3.4] that
any uniformly-distributed hidden identifier output with BISO
measurement channels satisfies the results in [16]. In [18,
Theorem 1] the secret-key capacity of the two-enrollment key
agreement problem is established for measurement channels
with the same channel transition matrix. However, these multi-
enrollment models do not consider the privacy leakage and
storage constraints, there is no constraint on the independence
of the secret keys of different enrollments, and the secrecy
leakage constraint is weak and is not applied jointly on all se-
cret keys. Furthermore, optimal random linear code construc-
tions that achieve the boundaries of the key-leakage-storage
regions are given in [19], where the classic code constructions
FCS and code-offset fuzzy extractors [20] are shown to be
strictly suboptimal. Therefore, the multi-enrollment models
and constructions in the literature are strictly suboptimal and
not necessarily realistic. We therefore list stronger secrecy
constraints jointly on all entities, which approximates the re-
ality better in combination with storage rate and joint privacy-
leakage rate constraints. These constraints define the multi-
entity key agreement problem, where the entities that use the
same identifier do not have to trust other entities. Therefore,
the multi-entity key agreement problem is a proper multi-user
extension of single-enrollment models.
Every measurement of an identifier is considered to be
noisy due to, e.g., local temperature and voltage changes
in the hardware of the PUF circuit or a cut on the finger.
Noise components at the encoder and decoder measurements
of a hidden source can be also correlated due to, e.g., the
surrounding logic in the hardware [21] or constant fingertip
moisture. This correlation between the noise sequences is
modeled in [22] as a broadcast channel (BC) [23] with an
input that is the hidden source output and with outputs that
are the noisy encoder and decoder measurements. We use this
model for multi-entity key agreement with identifiers, where
each entity (i.e., each encoder and decoder pair) observes
noisy identifier outputs of the same hidden source through
different BCs. We allow the BCs to be different as trusted
entities generally use different hardware implementations of
the encoder and decoder pairs, which results in different
correlations between noise components.
We also consider physically-degraded (PD) and less-noisy
(LN) BCs to give finer inner and outer bounds to the key-
leakage-storage regions for the GS and CS models of the
multi-entity key agreement problem. For the considered PD
and LN BCs, we prove that strong privacy can be achieved.
In [9], [10], [24], an extra common randomness that is
available to the encoder and decoder and that is hidden from
the eavesdropper is required to obtain strong privacy. This
assumption is not realistic since such a common randomness
requires hardware protection against invasive attacks, and if
such a protection is feasible, then it is not necessary to use an
identifier for key agreement.
B. Models for Identifier Outputs
We study physical and biometric identifier outputs that are
independent and identically distributed (i.i.d.) according to a
given probability distribution. These models are reasonable if
one uses transform-coding algorithms from [25] that occupy a
small hardware area to extract almost i.i.d. bits from PUFs
under varying environmental conditions. Similar transform-
coding based algorithms have been applied to biometric
identifiers to obtain independent output symbols [26]. These
transform-coding algorithms provide almost i.i.d. identifier
outputs and noise sequences; however, the correlation between
the noise components on the encoder and decoder components
are not removed using these methods. Furthermore, PUFs are
used for on-demand key reconstruction and physical attacks
on PUFs permanently change the identifier outputs [27], so
we assume that the eavesdropper cannot obtain information
correlated with the PUF outputs, unlike biometric identifiers.
C. Summary of Contributions
We extend the key-leakage-storage rate tuple analysis of
the single-enrollment model for hidden identifier outputs mea-
sured through general BCs in [22] to consider multi-entity
key agreement with a set of stringent secrecy constraints. A
summary of the main contributions is as follows.
• We derive achievable key-leakage-storage rate tuples for
the GS model with strong secrecy for any finite number of
entities using the same identifier’s measurements through
different BCs for key agreement. Separate identifier mea-
surements considered in [12], [28] correspond to a PD BC
and the visible source model in [9], [10] corresponds to
a semi-deterministic BC.
• For a set of PD and LN BCs, the privacy-leakage rates
for the two-entity key agreement problem are calculated.
These PD and LN BCs are shown to provide strong
privacy without the need of a common randomness. An
outer bound is given for the considered PD and LN BCs.
• We consider a special case of the multi-enrollment key
agreement problem, where all measurement channels are
separate (i.e., PD BCs) and they have the same transition
matrix. This is a common model used for SRAM PUFs
[16]. Using a less stringent secrecy leakage constraint that
bounds the information leakage for each secret key sep-
arately and without the mutual independence constraint
on the secret keys, we establish inner and outer bounds
for the strong-secrecy key-leakage-storage region for this
two-enrollment key agreement problem. The bounds dif-
fer only in the Markov chains imposed. This result is
a significant improvement to the two-enrollment secret-
key rate region (without storage and privacy-leakage rate
constraints) established in [18] for weak secrecy, which
is recovered by eliminating auxiliary random variables in
the proposed rate regions.
• All inner and outer bounds for the GS model are extended
to the CS model, which comprises secret-key binding
methods that embed a chosen secret key to the encoder.
• We give two scenarios to compare single-enrollment and
two-enrollment models and illustrate that for different
assumptions on measurement channels, either of the two
models can perform better in terms of the privacy-leakage
vs. secret-key rate boundary tuples.
3
D. Organization
This paper is organized as follows. In Section II, we describe
the multi-entity key agreement problem with BC measure-
ments. We give achievable key-leakage-storage regions for the
GS and CS models with strong secrecy and BC measurements
for any finite number of entities in Section III in addition
to inner and outer bounds for PD and LN BCs that satisfy
strong privacy. The proposed inner bounds for the two-entity
key agreement problem in Section IV are shown to differ
from the outer bounds only in the Markov chains imposed
for a special case with less stringent secrecy constraints. In
Sections V and VI, proofs of the given rate regions for the
general multi-entity key agremeent problem and for the two-
enrollment key agreement problem, respectively, are given.
Section VII concludes the paper.
E. Notation
Upper case letters represent random variables and lower
case letters their realizations. A superscript denotes a string of
variables, e.g., Xn=X1, X2, . . . , Xi, . . . , Xn, and a subscript
i denotes the position of a variable in a string. A random
variable X has probability distribution PX . Calligraphic letters
such as X denote sets, set sizes are written as |X | and their
complements as X c. [1 : J ] denotes the set {1, 2, . . . , J}
for an integer J ≥ 1 and [1 : J ] \ {j} denotes the set
{1, 2, , . . . , j − 1, j + 1, . . . , J} for any j ∈ [1 : J ]. Hb(x) =
−x log x− (1− x) log(1− x) is the binary entropy function,
where we take logarithms to the base 2, and H−1b (·) denotes its
inverse with range [0, 0.5]. X ∼ Bern(α) is a binary random
variable with Pr[X = 1] = α. A binary symmetric channel
(BSC) with crossover probability p is denoted by BSC(p). Q(·)
is the Q-function that gives the tail probability for the standard
normal distribution.
II. PROBLEM DEFINITIONS
Consider hidden identifier outputs Xn that are i.i.d. ac-
cording to a probability distribution PX . If there are a finite
number J of trusted entities that use the same identifer, the j-th
encoder and decoder pair observes noisy source measurements
that are outputs of a BC P
X̃jYj |X
, with abuse of notation, for
j ∈ [1 : J ], where X̃j , Yj , and X are finite sets.
For the GS model shown in Fig. 1(a) for J = 2 trusted
entities, the j-th encoder fGS,j(·) generates helper data Wj
and a secret key Sj from X̃nj . All secret keys are stored in
a secure database, whereas helper data are stored in a public
database so that an eavesdropper has access only to the helper
data. Using the helper data Wj and the noisy measurements
Y nj , the j-th decoder gj(·, ·) generates the key estimate Ŝj .
Similar steps are applied for the CS model in Fig. 1(b) also for
J = 2 trusted entities, except that each Sj should be embedded
into the j-th encoder fCS,j(·, ·).
Denote a set of secret keys as
SK = {Sj : j ∈ K} (1)
and a set of helper data as
WK = {Wj : j ∈ K} (2)
PX
(S1,W1)
(a)
= fGS,1(X̃
n
1 )
W1
(b)
= fCS,1(X̃
n
1 , S1)
Key
Database
Public
Database = ?
P
X̃1Y1|X
Ŝ1 = g1 (Y
n
1 ,W1)
Xn
Y n1
S1
S1
W1 Ŝ1
X̃n1
(b)(a)
(a) (b)
(S2,W2)
(a)
= fGS,2(X̃
n
2 )
W2
(b)
= fCS,2(X̃
n
2 , S2)
Key
Database
Public
Database
= ?
P
X̃2Y2|X
Ŝ2 = g2 (Y
n
2 ,W2)
Y n2
S2
S2
W2 Ŝ2
X̃n2
(b)(a)
(a) (b)
Fig. 1. Illustration of the multi-entity key agremeent problem for J = 2
entities with encoder and decoder measurements through BCs for (a) the GS
model and (b) the CS model.
for any K ⊆ [1 : J ]. A (secret-key, privacy-leakage, storage),
or key-leakage-storage, rate tuple is denoted as (Rs, R`, Rw).
Similarly, we denote a set of secret-key rates, for any K ⊆
[1 : J ], as
Rs,K = {Rs,j : j ∈ K} (3)
and a set of storage rates as
Rw,K = {Rw,j : j ∈ K}. (4)
We next define the key-leakage-storage regions for the GS
and CS models.
Definition 1. A key-leakage-storage rate tuple
(Rs,[1:J], R`,Rw,[1:J]) is achievable for the GS and CS
models with j-th encoder and decoder measurements through
a BC P
X̃jYj |X
if, given any δ>0, there is some n≥1, and J
encoder and decoder pairs for which Rs,j =
log |Sj |
n
for all
j ∈ [1 : J ] and
Pr
⋃
j∈[1:J]
{Sj 6= Ŝj}
≤ δ (reliability) (5)
4
1
n
H(Sj) ≥ Rs,j − δ, ∀j∈ [1 : J ] (key uniformity) (6)
I (SK;SKc) ≤ δ, ∀K⊆ [1 : J ] (key ind.) (7)
1
n
I(Xn;W[1:J])≤R`+δ (privacy) (8)
I
(
S[1:J];W[1:J]
)
≤ δ (strong secrecy) (9)
1
n
log |Wj | ≤ Rw,j+δ, ∀j∈ [1 : J ] (storage). (10)
The key-leakage-storage regions Cgs for the GS model and
Ccs for the CS model are the closures of the set of all
achievable rate tuples (Rs,[1:J], R`,Rw,[1:J]).
Both secret-key uniformity (6) and storage rate (10) con-
straints correspond to J separate constraints. However, relia-
bility (5), strong and mutual key independence (7), privacy-
leakage rate (8), and secrecy leakage (9) constraints are joint
constraints for all J trusted entities.
The mutual key independence constraint in (7) is not im-
posed in the multi-enrollment key agreement problem consid-
ered in [16]. Furthermore, a normalized (weak) version of this
constraint is imposed in the multi-enrollment key agreement
problem considered in [14], where the j-th decoder gj(·, ·) is
assumed to have access to the set of helper data W[1:j] for
all j ∈ [1 : J ]. The lack of the mutual key independence
constraint and the assumption of availability of all previous
helper data require that different encoder and decoder pairs
should trust each other. This can be the case, e.g., if all
enrollments are made by the same entity. Therefore, the multi-
entity key agreement problem imposes strictly more stringent
constraints than the multi-enrollment key agreement problem.
The unnormalized secrecy leakage constraint (9) provides
strong secrecy, which is a stronger notion than the weak
secrecy considered in [9], [10], [12], [14], [16], [28]. Further-
more, (9) is more stringent than the set of individual secrecy
leakage constraints I(Sj ;W[1:J]) imposed for all j ∈ [1 : J ],
considered in [16] for symmetric SRAM PUF outputs in
combination with the suboptimal FCS.
The unnormalized privacy leakage I(Xn;W[1:J]) cannot be
bounded by a finite number in general. We illustrate special
strong privacy cases in the next section.
III. INNER BOUNDS
We are interested in characterizing the optimal trade-off
among the secret-key, privacy-leakage, and storage rates with
strong secrecy for BC measurements at the encoders and
decoders of any finite number J of entities that use the same
hidden identifier outputs. We give achievable rate regions for
the GS and CS models in Theorem 1. The proofs are given in
Section V.
Denote
UK = {Uj : j ∈ K} (11)
and define a function max{·, ·} that gives the maximum of the
input values as its output.
Theorem 1 (Inner Bounds for Multi-entity GS and CS Mod-
els). An achievable rate region Rgs for the GS model with J
entities is the union over all P
Uj |X̃j
for all j ∈ [1 : J ] of the
rate tuples such that Rs,j ≥ 0 for all j ∈ [1 : J ] and
Rs,j ≤ I(Uj ;Yj)− I(Uj ;U[1:J]\{j}), ∀j ∈ [1 : J ] (12)
R` ≥
J∑
j=1
max{0, I(Uj ;X)−I(Uj ;Yj)}, (13)
Rw,j≥I(Uj ; X̃j)− I(Uj ;Yj), ∀j ∈ [1 : J ] (14)
Rs,j +Rw,j ≤ H(Uj | U[1:J]\{j}), ∀j ∈ [1 : J ]. (15)
An achievable rate region Rcs for the CS model with J
entities is the union over all P
Uj |X̃j
for all j ∈ [1 : J ] of the
rate tuples such that Rs,j ≥ 0 for all j ∈ [1 : J ], (12), (13),
and
Rw,j≥I(Uj ; X̃j)− I(Uj ;U[1:J]\{j}), ∀j ∈ [1 : J ] (16)
Rw,j ≤ H(Uj | U[1:J]\{j}), ∀j ∈ [1 : J ]. (17)
For the achievable rate regions Rgs and Rcs, we have
PU[1:J]X̃[1:J]XY[1:J]
= PX
J∏
j=1
P
Uj |X̃j
P
X̃jYj |X
. (18)
Corollary 1. Suppose for all j ∈ [1 : J ] that
• X̃j−Yj−X form a Markov chain, i.e., X is a PD version
of Yj with respect to X̃j , or
• P
XYj |X̃j
is a LN BC with I(Uj ;Yj) ≥ I(Uj ;X) for all
P
Uj |X̃j
.
For these cases, strong privacy, i.e.,
R` ≥ 0 (19)
can be achieved for the GS and CS models in combination
with the other corresponding bounds given in Theorem 1.
The proof of Corollary 1 follows from Theorem 1 because
I(Uj ;X) − I(Uj ;Yj) ≤ 0 for all j ∈ [1 : J ] for BCs
considered in Corollary 1.
Corollary 1 illustrates that it is possible to obtain strong
privacy, i.e., negligible unnormalized privacy leakage, without
the requirement of a common randommness that is hidden
from an eavesdropper assumed in [9], [10], [24]. This is the
case because the observation Y nj of each decoder is “better”
than the observation X̃nj of the corresponding encoder with
respect to the hidden source Xn for all entities.
Remark 1. The rate regions for our problem depend on the
joint conditional probability distributions P
XYj |X̃j
rather than
only the marginal conditional distributions. Thus, the key-
leakage-storage regions for the stochastically-degraded BCs
are not necessarily equal to the regions for the corresponding
PD BCs, unlike in the classic BC problem. Furthermore, since
PX̃[1:J]XY[1:J]
is fixed, the distinction between the LN BCs and
essentially-less noisy BCs [29], is not necessary.
We next give simple outer bounds for the key-leakage-
storage regions Cgs for the GS model and Ccs for the CS model
when the BCs P
XYj |X̃j
for all j ∈ [1 : J ] are PD BCs or LN
BCs as defined in Corollary 1. These simple outer bounds give
insights into the reason for different bounds on the secret-key
rates. Based on these insights, we show a special case in the
5
next section with a less stringent secrecy constraint, for which
the inner and outer bounds differ only in the Markov chains
imposed and we illustrate that they match for simpler models.
Lemma 1. Suppose one of the cases given in Corollary 1 is
satisfied by the BCs P
XYj |X̃j
for all j ∈ [1 : J ]. An outer
bound on the key-leakage-storage region Cgs is the union over
all P
Uj |X̃j
, where Uj − X̃j − (X,Yj) form a Markov chain,
for all j ∈ [1 : J ] of the rate tuples such that Rs,j ≥ 0 for all
j ∈ [1 : J ], (14), (19), and
Rs,j ≤ I(Uj ;Yj), ∀j ∈ [1 : J ]. (20)
An outer bound to the key-leakage-storage region Ccs for
the same BCs P
XYj |X̃j
is the union over all P
Uj |X̃j
, where
Uj − X̃j − (X,Yj) form a Markov chain, for all j ∈ [1 : J ]
of the rate tuples such that Rs,j ≥ 0 for all j ∈ [1 : J ], (19),
(20), and
Rw,j ≥ I(Uj ; X̃j), ∀j ∈ [1 : J ]. (21)
The proof of Lemma 1 follows straightforwadly by fol-
lowing the steps in [12, Section VI], defining the auxiliary
random variables Uj,i = (Sj ,Wj , Y
i−1
j ) for all j ∈ [1 : J ] and
i ∈ [1 : n], and by bounding I(Xn;W[1:J]) ≥ 0; therefore,
we omit the proof.
The outer bounds do not include the inequalities in (15) and
(17). Furthermore, the secret-key rate achieved by the inner
bound in (12) is smaller than the outer bound given in (20),
where the difference is the term −I(Uj ;U[1:J]\{j}). This term
is a result of the constraint in (44) that is imposed to satisfy the
strong and mutual key independence constraint given in (7).
Therefore, we next consider a model without the constraint in
(7) and use a secrecy-leakage constraint that is less stringent
than the one in (9), i.e., replace (9) by
I(Sj ;W[1:J]) ≤ δ, ∀j ∈ [1 : J ] (22)
which is also a strong secrecy metric. Due to the lack
of a mutual key independence constraint, the model in the
next section is not a multi-entity model but rather a multi-
enrollment model. For a special case of this multi-enrollment
key agreement problem, we establish inner and outer bounds
for the key-leakage-storage regions that comprise the same
bounds but for different Markov chains.
IV. BOUNDS FOR A MULTI-ENROLLMENT MODEL
Consider J = 2 entities that measure the noisy outputs of
the same hidden source Xn through separate channels that
have the same channel transition matrices, i.e., for all j ∈ [1 :
2], x̃j ∈ X̃ , and yj ∈ X̃ we have
P
X̃jYj |X
(x̃j , yj |x) = PX̃|X(x̃j |x)PX̃|X(yj |x). (23)
This model is commonly used for, e.g., SRAM PUFs [30].
Using (23), we define a new multi-enrollment model.
Definition 2. A key-leakage-storage rate tuple
( sRs,1, sRs,2, sR`, sRw,1, sRw,2) is achievable for the GS and
CS models with measurements through a BC P
X̃Y |X(x̃, y|x)
as in (23) if, given any δ > 0, there is some n ≥ 1, and
two encoder and decoder pairs for which sRs,1 =
log |S1|
n
,
sRs,2 =
log |S2|
n
, sRw,1 =
H(W1)
n
, sRw,2 =
H(W2)
n
, and
Pr
[
{S1 6= Ŝ1}
⋃
{S2 6= Ŝ2}
]
≤ δ (reliability) (24)
1
n
H(Sj) = sRs,j − δ, j = 1, 2 (key uniformity) (25)
1
n
I(Xn;W1,W2)= sR`+δ (privacy) (26)
I (Sj ;W1,W2) ≤ δ, j = 1, 2 (strong secrecy) (27)
1
n
log |Wj | = sRw,j+δ, j = 1, 2 (storage) (28)
I(W1;W2) ≤ δ (storage ind.). (29)
The key-leakage-storage regions sCgs,J=2 for the GS model
and sCcs,J=2 for the CS model are the closures of the set of all
achievable rate tuples ( sRs,1, sRs,2, sR`, sRw,1, sRw,2).
We characterize in Theorem 2 inner and outer bounds for
key-leakage-storage regions sCgs,J=2 and sCcs,J=2. The proofs
of Theorem 2 are given in Section VI, where the reason for
the necessity of the secrecy-leakage constraint in (27) that is
less stringent than the joint secrecy-leakage constraint in (9)
is given in Remark 2. Similarly, the reason for the necessity
of the strong helper data (storage) independence constraint in
(29) is discussed in Remark 4.
Denote
j′ = 3− j, j = 1, 2. (30)
Theorem 2. (Inner Bounds): An achievable key-leakage-
storage region ĎRgs,J=2 is the union over all PU1|X̃1 and
P
U2|X̃2
of the rate tuples such that sRs,j ≥ 0 for j = 1, 2
and
sRs,j ≤ I(Uj ;Yj), j = 1, 2 (31)
sR` ≥
2∑
j=1
(
I(Uj ;X)−I(Uj ;Yj)
)
, (32)
sR` ≤
2∑
j=1
(
I(Uj ;X)−I(Uj ; X̃j)+ sRw,j
)
, (33)
sRw,j≥I(Uj ; X̃j)− I(Uj ;Yj), j = 1, 2 (34)
sRs,j + sRw,j ≤ H(Uj), j = 1, 2 (35)
sRs,j + sRw,j + sRw,j′ ≤ H(Uj , Uj′), j = 1, 2. (36)
An achievable key-leakage-storage region ĎRcs,J=2 is the
union over all P
U1|X̃1
and P
U2|X̃2
of the rate tuples such that
sRs,j ≥ 0 for j = 1, 2, (31)-(33), and
sRw,j≥I(Uj ; X̃j), j = 1, 2 (37)
sRw,j ≤ H(Uj), j = 1, 2 (38)
sRw,j+ sRw,j′≤H(Uj , Uj′)+ sRs,j′ , j = 1, 2. (39)
For both achievable rate regions ĎRgs,J=2 and ĎRcs,J=2, we
6
have
P
U1U2X̃1X̃2XY1Y2
(u1, u2, x̃1, x̃2, x, y1, y2)
= P
U1|X̃1
(u1|x̃1)PU2|X̃2(u2|x̃2)PX̃|X(x̃1|x)PX̃|X(x̃2|x)
× P
X̃|X(y1|x)PX̃|X(y2|x)PX(x). (40)
(Outer Bounds) An outer bound for the key-leakage-storage
region sCgs,J=2 is the union over all PU1|X̃1 and PU2|X̃2 of the
rate tuples such that sRs,j ≥ 0, (31) - (36), and Uj − X̃j −
X − Yj form a Markov chain for j = 1, 2.
An outer bound for the key-leakage-storage region sCcs,J=2
is the union over all P
U1|X̃1
and P
U2|X̃2
of the rate tuples such
that sRs,j ≥ 0, (31) - (33), (37) - (39), and Uj − X̃j −X −Yj
form a Markov chain for j = 1, 2.
The inner and outer bounds differ because the outer bounds
define rate regions for the Markov chains U1 − X̃1 −X − Y1
and U2− X̃2−X −Y2, which are larger than the rate regions
defined by the inner bounds that satisfy (40). For instance, in
the achievability proof of Theorem 2, we apply the properties
of the Markov chain U2 − X̃2 − U1 in (87)(b), which does
not form a Markov chain for the choice of U1 and U2 in the
outer bounds. Therefore, inner and outer bounds do not match
in general.
Corollary 2. Choosing U1 = X̃1 and U2 = X̃2, it is straight-
forward to show that inner and outer bounds in Theorem 2
match if we do not impose any storage or privacy constraints,
i.e., impose only (24), (25), and (27). This result improves on
the secret-key capacity region given in [18, Theorem 1] for a
weak secrecy constraint.
Example 1. Consider the RO PUF model from [25, Section
4.1] where a transform-coding method is applied to conser-
vatively model the measurement channels PY |X = PX̃|X
as independent BSCs with the same crossover probability
of pA and where the hidden source output is Bern(
1
2
). We
therefore can apply the achievability results from Theorem 2
to this RO PUF model. Using [12, Theorem 3] to evaluate the
boundary tuples of ĎRgs,J=2, it suffices to consider probability
distributions P
Uj |X̃j
for j = 1, 2 such that P
X̃j |Uj
are BSCs
with crossover probabilities
x̃j =
H−1b (H(X|Uj))− pA
1− 2pA
. (41)
Consider the projection of the boundary tuples of ĎRgs,J=2 onto
key-leakage plane, i.e., (31) and (32). We plot in Fig. 2 single-
enrollment results where the privacy-leakage rate is measured
with respect to single helper data and two-enrollment results
for the sum rate of the two keys, both for pA = 0.06 [25].
To achieve a total secret-key rate of I(X̃1;Y1) = I(X̃2;Y2),
the privacy-leakage rate for the two-enrollment model is
approximately 13.5% less than the privacy-leakage rate for
the single-enrollment model for RO PUFs. The reason for this
gain is the information bottleneck problem that arises from
(31) and (32) to find the boundary tuples.
Example 2. Consider uniform binary antipodal measurements
over an additive white Gaussian noise (AWGN) channel.
0 0.05 0.1 0.15 0.2
0
0.1
0.2
0.3
0.4
0.5
0.6
Privacy-leakage Rate sR` (bits/symbol)
S
ec
re
t-
ke
y
R
at
e
s R
s
(b
it
s/
sy
m
bo
l)
Single Enrollment
Two-enrollment Sum Rate
Fig. 2. Privacy-leakage vs. secret-key rate projection of the boundary tuples
of the single- and two-enrollment RO PUF models with BSCs(pA = 0.06).
Define the signal power as PS and the noise power as PN,
so we have a signal-to-noise ratio (SNR) of SNR =
PS
PN
. If
a matched filter, which maximizes the SNR at the sampling
instant for the AWGN channel, is applied at the encoder and
decoder, the bit error probability Pb is given by [31, pp. 96]
Pb = Q
(√
SNR
)
. (42)
The channel between the input binary symbols and the outputs
of the matched filter is a BISO channel. Using [12, Theorem
3], we have that P
X̃j |Uj
for j = 1, 2 that are BSCs with
crossover probabilities given in (41) by replacing pA with Pb,
suffice to obtain the boundary tuples of ĎRgs,J=2. We remark
that pA = 0.06 used in Example 1 corresponds to an SNR of
approximately 3.83dB.
In Fig. 3, the privacy-leakage rate vs. secret-key rate bound-
ary tuples are depicted for two cases. First, a two-enrollment
model at SNR = 3.83dB with a sum rate for secret keys is
depicted. For comparison, we plot a single-enrollment model
where the signal power Ps is doubled, i.e., we have an SNR
of approximately 6.84dB.
Fig. 3 shows for the two cases with the same total signal
power of 2Ps that the single enrollment boundary tuple can
result in a gain of approximately 228.55% at the corner point
in terms of the secret-key rates achieved for a given privacy-
leakage rate. For such an AWGN channel with a fixed total
signal power; therefore, the single-enrollment model can result
in significant gains in terms of achieved secret-key rates as
compared to the two-enrollment model for small sR` values.
V. PROOF OF THEOREM 1
We provide a proof that follows from the output statistics of
random binning (OSRB) method, proposed in [32] and further
extended in [33], by applying the steps in [34, Section 1.6].
7
0 0.02 0.04 0.06 0.08 0.1
0
0.2
0.4
0.6
0.8
1
Privacy-leakage Rate sR` (bits/symbol)
S
ec
re
t-
ke
y
R
at
e
s R
s
(b
it
s/
sy
m
bo
l)
Single Enrollment @SNR=6.84dB.
Two-enrollment Sum Rate @SNR=3.83dB
Fig. 3. Privacy-leakage vs. secret-key rate projection of the boundary tuples
of the single- and two-enrollment RO PUF models with different SNRs.
A. Proof for the GS Model
Proof Sketch: Fix P
U1|X̃1
, P
U2|X̃2
, . . . , P
UJ |X̃J
. Let
(Un
[1:J]
, X̃n
[1:J]
, Xn,Yn
[1:J]
) be i.i.d. according to (18). Assign
three random bin indices (Sj ,Wj , Cj) to each realization
unj for all j ∈ [1 : J ], where Sj represents the secret
key, Wj the helper data, and Cj a public index. Assume
Sj ∈ [1 : 2nRs,j ], Wj ∈ [1 : 2nRw,j ], and Cj ∈ [1 : 2nRc,j ]
such that Rs,j , Rw,j , Rc,j ≥ 0 for all j ∈ [1 : J ].
Apply the union bound to the reliability constraint in (5) to
obtain the sum of J error probabilities. This sum vanishes for
any finite number J when n → ∞ by using a Slepian-Wolf
(SW) [35] decoder to estimate Unj from (Cj ,Wj , Y
n
j ) if [32,
Lemma 1]
Rc,j +Rw,j > H(Uj |Yj), ∀j ∈ [1 : J ]. (43)
The key uniformity (6), mutual key independence (7), and
strong secrecy (9) constraints are satisfied if [32, Theorem 1]
Rs,j+Rw,j+Rc,j < H(Uj | U[1:J]\{j}), ∀j ∈ [1 : J ] (44)
since (44) ensures that the three random indices
(Sj ,Wj , Cj) are almost mutually independent and
uniformly distributed, and they are almost independent
of U[1:J]\{j}. Therefore, (Sj ,Wj , Cj) are almost independent
of
(
S[1:J]\{j},W[1:J]\{j}, C[1:J]\{j}
)
because Unk determines
(Sk,Wk, Ck) for all k ∈ [1 : J ].
Similarly, the public randomness Cj is almost independent
of X̃nj , so it is almost independent of (X̃
n
[1:J]
, Xn,Yn
[1:J]
), if
we have [32, Theorem 1]
Rc,j < H(Uj |X̃j), ∀j ∈ [1 : J ]. (45)
Thus, the public indices C[1:J] can be fixed by generating them
uniformly at random. The j-th encoder can generate Unj ac-
cording to P
Un
j
|X̃n
j
Cj
obtained from the binning scheme above
to compute the bins (Sj ,Wj) from Unj for all j ∈ [1 : J ]. This
procedure induces a joint probability distribution that is almost
equal to PU[1:J]X̃[1:J]XY[1:J] fixed in (18) [34, Section 1.6].
Applying the Fourier Motzkin elimination [36] using the
software available in [37] to (43)-(45) for each j ∈ [1 : J ]
separately, we obtain the inequalities
Rw,j > I(Uj ; X̃j)− I(Uj ;Yj) (46)
Rs,j < I(Uj ;Yj)− I(Uj ;U[1:J]\{j}) (47)
Rw,j +Rs,j < H(Uj |U[1:J]\{j}) (48)
for all j ∈ [1 : J ].
To satisfy the constraints (46)-(48), we can fix the rates to
Rs,j = I(Uj ;Yj)−I(Uj ;U[1:J]\{j})−2�, ∀j ∈ [1 : J ] (49)
Rw,j = I(Uj ; X̃j)− I(Uj ;Yj) + 2�, ∀j ∈ [1 : J ] (50)
Rc,j = H(Uj |X̃j)− �, ∀j ∈ [1 : J ] (51)
for some � > 0 such that �→ 0 when n→∞.
Consider the privacy leakage. Since C[1:J] are public, we
can bound the privacy leakage as follows.
I(Xn;W[1:J], C[1:J])
(a)
≤ I(Xn;W[1:J]|C[1:J]) + J�n
≤ H(W[1:J])−H(W[1:J], C[1:J]|Xn) +H(C[1:J]) + J�n
(b)
= H(W[1:J])−
J∑
j=1
H(Wj , Cj |Xn) +H(C[1:J]) + J�n
≤
J∑
j=1
(
H(Wj)+H(Cj)−H(Wj , Cj |Xn)
)
+J�n (52)
where
(a) follows from the fact that Cj is almost independent of
(X̃n
[1:J]
, Xn,Yn
[1:J]
) for all j ∈ [1 : J ] due to (45). Therefore,
Cj is almost independent of (Xn, C[1:J]\{j}) because Unk
determines Ck and Unk − X̃
n
k − X̃
n
j − U
n
j form a Markov
chain for all k 6= j and j, k ∈ [1 : J ], so we have
I(Xn; C[1:J])
= I(Xn;C1) + I(X
n;C2|C1) + . . .+ I(Xn;CJ |C[1:J−1])
≤ J.�n (53)
for some �n > 0 such that �n → 0 when n→∞;
(b) follows because (Wj , Cj)−Xn−(W[1:j−1], C[1:j−1]) form
a Markov chain for all j ∈ [2 : J ].
Consider two cases for the privacy leakage analysis.
Case 1: Suppose for any j ∈ [1 : J ] that we have
Rc,j +Rw,j < H(Uj |X) (54)
i.e., H(Uj |X) > H(Uj |Yj), so (Wj , Cj , Xn) are almost
mutually independent [32, Theorem 1]. Therefore, we have
H(Wj)+H(Cj)−H(Wj , Cj |Xn)
≤ H(Wj)+H(Cj)−(H(Wj) +H(Cj)− �′n) = �
′
n (55)
for some �′n > 0 such that �
′
n → 0 when n→∞. Combining
(52) and (55) proves strong privacy.
Case 2: Suppose for any j ∈ [1 : J ] that we have
Rc,j +Rw,j ≥ H(Uj |X) (56)
8
i.e., H(Uj |X) ≤ H(Uj |Yj), so (Wj , Cj , Xn) can reliably
estimate Unj [32, Lemma 1]. Therefore, we have
H(Wj)+H(Cj)−H(Wj , Cj |Xn)
(a)
≤ H(Wj)+H(Cj)−nH(Uj |X) + n�′′n
(b)
≤ n(I(Uj ;X)− I(Uj ;Yj) + �+ �′′n) (57)
where (a) follows because Unj determines (Wj , Cj),
(Wj , Cj , X
n) can realiably estimate Un for some �′′n > 0
such that �′′n → 0 when n→∞, and (Unj , X
n) are i.i.d., and
(b) follows by (50) and (51).
Combining (52) and (57), we obtain
I(Xn;W[1:J], C[1:J])− J�n
≤
J∑
j=1
j:
H(Uj |X)≤H(Uj |Yj)
n(I(Uj ;X)−I(Uj ;Yj)+�+�′′n). (58)
Using the selection lemma [38, Lemma 2.2], these prove
the achievability of the rate region Rgs.
B. Proof for the CS Model
We use the achievability proof for the GS model. Suppose
the key S′j , generated as in the GS model together with the
helper data W ′j and public index C
′
j , have the same cardinality
as the corresponding embedded secret key Sj , i.e., |S ′j | =
|Sj | for all j ∈ [1 : J ]. The chosen key Sj is uniformly
distributed and independent of (Xn, X̃n
[1:J]
,Yn
[1:J]
,S[1:J]\{j})
for all j ∈ [1 : J ]. Consider the j-th encoder fcs,j(·, ·) with
inputs (X̃nj , Sj) and output Wj = (S
′
j +Sj ,W
′
j), and the j-th
decoder gj(·, ·) with inputs (Y nj ,Wj) and output Ŝj = S
′
j +
Sj − Ŝ′j . All addition and subtraction operations are modulo-
|Sj | for all j ∈ [1 : J ]. The j-th decoder of the GS model is
used to obtain Ŝ′j for all j ∈ [1 : J ].
We have the error probability
Pr
⋃
j∈[1:J]
{Sj 6= Ŝj}
= Pr
⋃
j∈[1:J]
{S′j 6= Ŝ
′
j}
(59)
which is small due to the proof of achievability for the GS
model.
Using (49) and (50), and from the one-time padding oper-
ation applied above, we can achieve a storage rate of
Rw,j ≥ I(Uj ; X̃j)− I(Uj ;U[1:J]\{j}), ∀j ∈ [1 : J ] (60)
for the CS model.
We have the secrecy leakage of
I(S[1:J];W[1:J], C′[1:J])
(a)
= I(S[1:J];W[1:J]|C′[1:J])
=I(S[1:J];W ′[1:J]|C
′
[1:J])+I(S[1:J]; (S
′+S)[1:J]|W
′
[1:J], C
′
[1:J])
(b)
= H((S ′+S)[1:J]|W
′
[1:J], C
′
[1:J])−H(S
′
[1:J]|W
′
[1:J], C
′
[1:J])
(c)
≤ n
( J∑
j=1
Rs,j
)
−H(S ′[1:J]|C
′
[1:J]) + I(S
′
[1:J];W
′
[1:J]|C
′
[1:J])
(d)
≤ n
( J∑
j=1
Rs,j
)
−
(
n
( J∑
j=1
Rs,j
)
− �′′′n
)
+ I(S ′[1:J];W
′
[1:J]|C
′
[1:J])
(e)
≤ �′′′n + �
(4)
n (61)
where (a) follows since S[1:J] are chosen independently of
the public indices C[1:J], (b) follows because S[1:J] are chosen
independently of (W ′
[1:J]
, C′
[1:J]
,S ′
[1:J]
), (c) follows because
|S ′j | = |Sj | for all j ∈ [1 : J ], (d) follows because S
′
[1:J]
and
C′
[1:J]
are almost mutually independent and each S′j is almost
uniformly distributed due to (44) for some �′′′n > 0 such that
�′′′n → 0 when n→∞, and (e) follows because the GS model
satisfies the strong secrecy constraint (9) due to (44) for some
�
(4)
n > 0 such that �
(4)
n → 0 when n→∞.
Consider the privacy leakage:
I(Xn;W[1:J], C′[1:J])
(a)
≤ I(Xn;W[1:J]|C′[1:J]) + J�n
≤ I(Xn;W ′[1:J]|C
′
[1:J]) +H((S + S
′)[1:J]|W ′[1:J], C
′
[1:J])
−H((S + S ′)[1:J]|Xn,W ′[1:J], C
′
[1:J],S
′
[1:J]) + J�n
(b)
≤ I(Xn;W ′[1:J]|C
′
[1:J])+
( J∑
j=1
log(|Sj |)
)
−H(S[1:J])+J�n
(c)
= I(Xn;W ′[1:J]|C
′
[1:J]) + J�n (62)
where (a) follows by (53), (b) follows because S[1:J] are
chosen independently of (Xn,W ′
[1:J]
,S ′
[1:J]
, C′
[1:J]
) and |S ′j | =
|Sj | for all j ∈ [1 : J ], and (c) follows from the uniformity
and mutual independence of S[1:J].
Using the selection lemma, these prove the achievability of
the rate region Rcs.
VI. PROOF OF THOREM 2
We use the OSRB method steps in [34, Section 1.6].
A. Achievability Proof for the GS Model
Fix
P
U1|X̃1
= P
U2|X̃2
= P
U |X̃ . (63)
Let (Un1 , U
n
2 , X̃
n
1 , X̃
n
2 , X
n, Y n1 , Y
n
2 ) be i.i.d. according to
(40). Assign three random bin indices (Sj ,Wj , Cj) to each
realization unj for all j = 1, 2. Assume Sj ∈ [1 : 2
nĎRs,j ],
Wj ∈ [1 : 2n
ĎRw,j ], and Cj ∈ [1 : 2n
ĎRc,j ] such that
sRs,j , sRw,j , sRc,j ≥ 0 for j = 1, 2.
Apply the union bound to the reliability constraint in (24),
which vanishes when n → ∞ by using an SW decoder to
estimate Unj from (Cj ,Wj , Y
n
j ) if [32, Lemma 1]
sRc,j + sRw,j > H(Uj |Yj), j = 1, 2. (64)
The key uniformity (25) constraint is satisfied if [32, The-
orem 1]
sRs,j+ sRw,j+ sRc,j < H(Uj), j = 1, 2 (65)
since (65) ensures that the three random indices (Sj ,Wj , Cj)
are almost mutually independent and uniformly distributed.
9
Suppose a virtual joint encoder assigns six indices
(S1,W1, C1, S2,W2, C2) to each realization pair (un1 , u
n
2 ).
This virtual encoder is an operational dual of the virtual
decoder used in the proof of [18, Theorem 1]. Using the virtual
joint encoder, the strong secrecy constraint in (27) and the
strong helper data independence constraint in (29) are satisfied
if [32, Theorem 1]
sRs,1+ sRw,1+ sRc,1+ sRw,2+ sRc,2 < H(U1, U2) (66)
and
sRs,2+ sRw,2+ sRc,2+ sRw,1+ sRc,1 < H(U1, U2) (67)
because (66) ensures that (S1,W1, C1,W2, C2) are al-
most mutually independent; whereas, (67) ensures that
(S2,W2, C2,W1, C1) are almost mutually independent.
Remark 2. The set of equations considered in (65)-(67)
cannot be imposed for the joint secrecy-leakage constraint in
(9) for general probability distributions P
X̃1X̃2XY1Y2
, since to
impose (9) one would replace (66) and (67) with
sRs,1+ sRw,1+ sRc,1+ sRs,2 + sRw,2+ sRc,2 < H(U1, U2) (68)
which would also imply the mutual independence of secret
keys in (7). However, the inequalities in (65) and (68) cannot
be satisfied simultaneously in general as H(U1) + H(U2) ≥
H(U1, U2). This problem is avoided in the proof of Theorem 1
by imposing the inequality in (44) rather than (65).
The public randomness Cj is almost independent of X̃nj ,
so it is almost independent of (X̃n1 , X̃
n
2 , X
n, Y n1 , Y
n
2 ), if we
have [32, Theorem 1]
sRc,j < H(Uj |X̃j), j = 1, 2. (69)
Thus, the public indices (C1, C2) can be fixed by generating
them uniformly at random. Unj can be generated according
to P
Un
j
|X̃n
j
Cj
for j = 1, 2 obtained from the binning scheme
above to compute the bins (Sj ,Wj) from Unj for j = 1, 2.
This procedure induces a joint probability distribution that is
almost equal to P
U1U2X̃1X̃2XY1Y2
that is fixed in (40) [34,
Section 1.6].
Applying the Fourier Motzkin elimination to (64)-(67) and
(69), we obtain the inequalities
sRw,1 > H(U1|Y1)−H(U1|X̃1) (70)
sRw,2 > H(U2|Y2)−H(U2|X̃2) (71)
sRs,1 < I(U1;Y1) (72)
sRs,2 < I(U2;Y2) (73)
sRs,1 < −H(U1|Y1)−H(U2|Y2) +H(U1, U2) (74)
sRs,2 < −H(U1|Y1)−H(U2|Y2) +H(U1, U2) (75)
sRs,1 + sRw,2 < −H(U1|Y1) +H(U1, U2) (76)
sRs,1 + sRw,1 < H(U1) (77)
sRs,1 + sRw,1 < −H(U2|Y2) +H(U1, U2) (78)
sRs,1 + sRw,1 + sRw,2 < H(U1, U2) (79)
sRs,2 + sRw,2 < −H(U1|Y1) +H(U1, U2) (80)
sRs,2 + sRw,2 < H(U2) (81)
sRs,2 + sRw,1 < −H(U2|Y2) +H(U1, U2) (82)
sRs,2 + sRw,2 + sRw,1 < H(U1, U2). (83)
Observe that we have
H(U1|X̃2) = H(U1|Y1) = H(U2|X̃1) = H(U2|Y2) (84)
H(U1|X̃1) = H(U2|X̃2) (85)
H(U1) = H(U2) (86)
due to (23) and (63). We therefore obtain
H(U1, U2)−H(U1|Y1)
(a)
= H(U2)+H(U1|U2)−H(U1|X̃2)
(b)
≥ H(U2) (87)
where (a) follows by (84) and (b) follows from the Markov
chain U2−X̃2−U1. A similar result can be shown by swaping
the indices. Therefore, the constraints in (78) and (80) are
inactive due to the constraints, respectively, in (77) and (81) for
the two-enrollment model given in Definition 2. Similarly, the
constraints in (74) and (75) are inactive due to the constraints,
respectively, in (72) and (73).
Replace the inequalities in (76) and (82), respectively, with
2 sRs,1 + sRw,1 + sRw,2 < I(U1;Y1) +H(U1, U2) (88)
2 sRs,2 + sRw,2 + sRw,1 < I(U2;Y2) +H(U1, U2). (89)
Then, (88) is inactive because (72) and (79) imply (88), and
(89) is inactive because (73) and (83) imply (89). We remark
that rate region represented by (70)-(83) is the same as the
region represented by replacing (76) and (82) with (88) and
(89) because the corner points of the two rate regions are the
same. Therefore, the inequalities in (76) and (82) are inactive.
To satisfy the constraints (70)-(83), we can fix the rates to
sRs,j = I(Uj ;Yj)−5�, j = 1, 2 (90)
sRw,j = I(Uj ; X̃j)− I(Uj ;Yj) + 2�, j = 1, 2 (91)
sRc,j = H(Uj |X̃j)− �, j = 1, 2 (92)
for some � > 0 such that �→ 0 when n→∞ due to (84)-(87).
Since C1 and C2 are public, we can bound the privacy
leakage as follows.
I(Xn;W1,W2, C1, C2)
(a)
≤ I(Xn;W1,W2|C1, C2) + 2�n
(b)
≤ H(W1,W2)−H(W1, C1|Xn)−H(W2, C2|Xn)
+H(C1, C2) + 2�n
(c)
≤ H(W1) +H(W2)−H(Un1 |X
n)−H(Un2 |X
n) + 2n�′′n
+H(C1) +H(C2) + 2�n (93)
(d)
≤ n(I(U1;X)− I(U1;Y1) + I(U2;X)− I(U2;Y2))
+ 2n�′′n + 2n�+ 2�n (94)
where (a) follows by (53) for J = 2, (b) follows because
(W1, C1)−Xn− (W2, C2) form a Markov chain, (c) follows
for some �′′n > 0 such that �
′′
n → 0 when n→∞ because for
the two-enrollment model considered, (56) is satisfied due to
the Markov chain Uj −X − Yj for j = 1, 2, and (d) follows
by (91) and (92), and because (Un1 , U
n
2 , X
n) are i.i.d.
10
Using (93) for general rate tuples that satisfy the constraints
(70)-(83), i.e., not only (90)-(92), we can bound the privacy
leakage as
I(Xn;W1,W2, C1, C2)
(a)
≤ n sRw,1 + n sRw,2 + nI(U1;X)− nI(U1; X̃1)
+ nI(U2;X)− nI(U2; X̃2) + 2n�′′n + 2�n (95)
where (a) follows by (92) and because (Un1 , U
n
2 , X
n) are i.i.d.
Using the selection lemma, these prove the achievability of
the key-leakage-storage region ĎRgs,J=2.
B. Achievability Proof for the CS Model
The achievability proof for the CS model follows by apply-
ing the one-time padding step used in Section V-B.
C. Outer Bound Proofs for the GS and CS Models
Suppose for some δn>0 and n, there is a pair of encoders
and decoders such that (24)-(29) are satisfied by some key-
leakage-storage tuple ( sRs,1, sRs,2, sR`, sRw,1, sRw,2). Using (24)
and Fano’s inequality, we obtain
H(Sj |Wj , Y nj )
(a)
≤ H(Sj |Ŝj)≤n�n, j = 1, 2 (96)
where (a) permits randomized decoding, �n =
δn max{ sRs,1, sRs,2}+Hb(δn)/n such that �n→0 if δn→0.
Let Uj,i , (Sj ,Wj , Xi−1), which satisfies the Markov
chain Uj,i − X̃j,i −Xi − Yj,i for all i ∈ [1 : n] and j = 1, 2.
Remark 3. For the choice of Uj,i = (Sj ,Wj , Xi−1) (and
similarly for Uj,i = (Sj ,Wj , Y
i−1
j )) for j=1, 2, U1,i−X̃1,i−
U2,i do not form a Markov chain for all i ∈ [1 : n] although
for the inner bound we use this Markov chain. This is the
reason why inner and outer bounds do not match in general.
Proof for (31): We obtain for the GS and CS models for
j = 1, 2 that
n( sRs,j − δn)
(a)
≤ H(Sj)−H(Sj |Wj , Y nj ) + n�n
(b)
≤ I(Sj ;Y nj |Wj) + n�n + δn
≤
n∑
i=1
[
I(Sj ,Wj , Y
i−1
j ;Yj,i) + �n +
δn
n
]
(c)
≤
n∑
i=1
[
I(Sj ,Wj , X
i−1;Yj,i) + �n +
δn
n
]
(d)
=
n∑
i=1
[
I(Uj,i;Yj,i) + �n +
δn
n
]
(97)
(a) follows by (25) and (96), (b) follows by (27), (c) follows
by applying the data-processing inequality to the Markov chain
Y i−1j − (Wj , Sj , X
i−1)− Yj,i, j = 1, 2, ∀i ∈ [1 :n] (98)
and (d) follows from the definition of Uj,i.
Proof for (32): We have for the GS and CS models that
n( sR` + δn)
(a)
= H(W1,W2)−H(W1|Xn)−H(W2|Xn)
(b)
= H(W1|Y n1 )−H(W1|X
n) +H(W2|Y n2 )−H(W2|X
n)
+ I(W1; X̃
n
2 ) + I(W2;Y
n
2 )− I(W1;W2)
(c)
≥
2∑
j=1
[
H(Wj |Y nj )−H(Wj |X
n)
]
≥
2∑
j=1
[
H(Sj ,Wj , Y
n
j )−H(Sj |Wj , Y
n
j )−H(Y
n
j )
−H(Sj ,Wj |Xn)
]
(d)
≥
2∑
j=1
[
I(Sj ,Wj ;X
n)− I(Sj ,Wj ;Y nj )− n�n
]
(e)
≥
2∑
j=1
n∑
i=1
[
I(Sj ,Wj , X
i−1;Xi)−I(Sj ,Wj , Xi−1;Yj,i)−�n
]
(f)
=
2∑
j=1
n∑
i=1
[
I(Uj,i;Xi)− I(Uj,i;Yj,i)− �n
]
(99)
where (a) follows by (26) and from the Markov chain W1 −
Xn −W2, (b) follows because I(W1;Y n1 ) = I(W1; X̃n2 ) due
to (23), (c) follows from the Markov chain W1−X̃n2 −W2, (d)
follows by (96), (e) follows because the channel and source
are memoryless and from the Markov chain in (98), and (f)
follows from the definition of Uj,i.
Proof for (33): We have for the GS and CS models that
n( sR` + δn)
(a)
≤ H(W1) +H(W2)−H(W1|Xn)−H(W2|Xn)
(b)
≤
2∑
j=1
[
n sRw,j +H(Sj ,Wj |X̃nj )−H(Sj ,Wj |X
n) + n�n
]
(c)
=
2∑
j=1
[
n sRw,j +
n∑
i=1
I(Sj ,Wj , X
i−1;Xi)
−
n∑
i=1
I(Sj ,Wj , X̃
i−1
j ; X̃j,i) + n�n
]
(d)
≤
2∑
j=1
[
n sRw,j +
n∑
i=1
I(Sj ,Wj , X
i−1;Xi)
−
n∑
i=1
I(Sj ,Wj , X
i−1; X̃j,i) + n�n
]
(e)
≤
2∑
j=1
[
n sRw,j +
n∑
i=1
(I(Uj,i;Xi)− I(Uj,i; X̃j,i))
+ n�n
]
(100)
where (a) follows by (26) and from the Markov chain W1 −
Xn−W2, (b) follows by (96) and from the Markov chain Sj−
(Wj , X
n)− Y n for j = 1, 2, (c) follows because the channel
and source are memoryless, (d) follows from the Markov chain
Xi−1−(Wj , Sj , X̃i−1j )−X̃j,i, j = 1, 2, ∀i ∈ [1 :n] (101)
11
and (e) follows from the definition of Uj,i.
Proof for (34): Observe for the GS model for j = 1, 2 that
n( sRw,j + δn)
(a)
≥ H(Wj |Y nj ) + I(Wj ;Y
n
j )
(b)
≥ H(Sj ,Wj , Y nj )−H(Y
n
j )−H(Sj |Wj , Y
n
j )
−H(Sj ,Wj |X̃nj ) + I(Wj ;Y
n
j )
(c)
≥ I(Sj ,Wj ; X̃nj )− I(Sj ,Wj ;Y
n
j )− n�n
(d)
=
n∑
i=1
[I(Sj ,Wj , X̃
i−1
j ; X̃j,i)−I(Sj ,Wj , Y
i−1
j ;Yj,i)−n�n]
(e)
≥
n∑
i=1
[I(Sj ,Wj , X
i−1; X̃j,i)−I(Sj ,Wj , Xi−1;Yj,i)−n�n]
(f)
=
n∑
i=1
[I(Uj,i; X̃j,i)−I(Uj,i;Yj,i)−n�n] (102)
where (a) follows by (28), (b) follows from the encoding
steps, (c) follows by (96), (d) follows because the source and
channel are memoryless, (e) follows from the data-processing
inequality applied to the Markov chains in (98) and (101), and
(f) follows from the definition of Uj,i.
Proof for (37): Observe for the CS model for j = 1, 2 that
n( sRw,j+δn)
(a)
≥ I(Sj ,Wj ; X̃nj )−H(Sj |Wj)+H(Sj ,Wj |X̃
n
j )
(b)
≥ I(Sj ,Wj ; X̃nj ) + I(Sj ;Wj)
(c)
≥
n∑
i=1
I(Sj ,Wj , X̃
i−1
j ; X̃j,i)
(d)
≥
n∑
i=1
I(Sj ,Wj , X
i−1; X̃j,i)
(e)
=
n∑
i=1
I(Uj,i; X̃j,i) (103)
where (a) follows by (28), (b) follows because X̃n is indepen-
dent of Sj and from the encoding step, (c) follows because the
source and channel are memoryless, (d) follows by applying
the data-processing inequality to the Markov chain in (101),
and (e) follows from the definition of Uj,i.
Proof for (35): We have for the GS model for j = 1, 2 that
n( sRs,j + sRw,j)
(a)
= H(Sj ,Wj) + I(Sj ;Wj) + nδn
(b)
≤
n∑
i=1
[
H(Sj ,Wj , X
i−1) +
δn
n
+ δn
]
(c)
=
n∑
i=1
[
H(Uj,i) +
δn
n
+ δn
]
(104)
where (a) follows by (25), (b) follows by (27), and (c) follows
from the definition of Uj,i.
Proof for (38): Similarly, we have for the CS model for
j = 1, 2 that
n sRw,j ≤
n∑
i=1
H(Sj ,Wj , X
i−1)
(a)
=
n∑
i=1
H(Uj,i) (105)
where (a) follows from the definition of Uj,i.
Proof for (36): We obtain for the GS model for j = 1, 2
and j′ as defined in (30) that
n( sRs,j + sRw,j + sRw,j′)
(a)
= H(Sj ,Wj ,Wj′) + I(Sj ;Wj ,Wj′) + I(Wj ;Wj′) + nδn
(b)
≤
n∑
i=1
[
H(Sj ,Wj ,Wj′ , Sj′ , X
i−1) +
2δn
n
+ δn
]
(106)
(c)
=
n∑
i=1
[
H(Uj,i, Uj′,i) +
2δn
n
+ δn
]
(107)
where (a) follows by (25), (b) follows by (27) and (29), and
(c) follows from the definitions of Uj,i and Uj′,i.
Proof for (39): We have for the CS model for j = 1, 2 and
j′ as defined in (30) that
n( sRw,j+ sRw,j′)
≤
n∑
i=1
H(Wj ,Wj′ , Sj , Sj′ , X
i−1) + I(Wj ;Wj′) + n sRs,j′
(a)
≤
n∑
i=1
[
H(Wj ,Wj′ , Sj , Sj′ , X
i−1) +
δn
n
+ sRs,j′
]
(108)
(b)
=
n∑
i=1
[
H(Uj,i, Uj′,i) +
δn
n
+ sRs,j′
]
(109)
where (a) follows by (29) and (b) follows from the definitions
of Uj,i and Uj′,i.
Remark 4. (106) and (108) are the only places we use the
strong helper data independence constraint in (29) and it does
not seem straightforward to obtain the inequalities in (106)
and (108) without (29).
Introduce a uniformly distributed time-sharing random vari-
able Q ∼ Unif[1 : n] independent of other random variables.
Define X=XQ, X̃j=X̃j,Q, Y =Yj,Q, and Uj=(Uj,Q,Q) so
that Uj−X̃j−X−Yj form a Markov chain for j = 1, 2. The
outer bound for the GS model follows by using the introduced
random variables in (97), (99), (100), (102), (104), and (107),
and letting δn → 0. Similarly, the outer bound for the CS
model follows by using the introduced random variables in
(97), (99), (100), (103), (105), and (109), and letting δn → 0.
VII. CONCLUSION
We derived inner bounds for the multi-entity key-leakage-
storage regions for GS and CS models with strong secrecy,
a hidden identifier source, and correlated noise components
at the encoder and decoder measurements that are modeled
as BCs. The inner bounds are valid for any finite number
of entities that use the same hidden source to agree on a
secret key. We argued that the mutual key independence
constraint we impose makes the proposed multi-entity key
agreement problem a proper multi-user extension of the classic
single-enrollment key agreement problem, unlike the multi-
enrollment key agreement problem considered in the literature.
A set of degraded and less-noisy BCs was shown to provide
strong privacy without a need for a common randomness.
We established inner and outer bounds for the key-lekage-
storage regions for a two-enrollment model with measurement
12
channels that are valid for SRAM and RO PUFs. Inner and
outer bounds were shown to differ only in the Markov chains
imposed and they match if the storage and privacy-leakage
rate constraints are removed. Two examples illustrated that
depending on the constraints of the practical scenario, single
or multiple enrollment might perform better in terms of the
secret-key vs. privacy-leakage rate ratio. In future work, we
will find a set of probability distributions for which the strong
helper data independence constraint in the two-enrollment
model can be eliminated.
ACKNOWLEDGMENT
O. Günlü thanks Rafael F. Schaefer for fruitful discussions.
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[20] Y. Dodis, R. Ostrovsky, L. Reyzin, and A. Smith, “Fuzzy extractors:
How to generate strong keys from biometrics and other noisy data,”
SIAM J. Comput., vol. 38, no. 1, pp. 97–139, Jan. 2008.
[21] D. Merli, F. Stumpf, and C. Eckert, “Improving the quality of ring
oscillator PUFs on FPGAs,” in ACM Workshop Embedded Sys. Security,
New York, NY, Oct. 2010, pp. 9:1–9:9.
[22] O. Günlü, R. F. Schaefer, and G. Kramer, “Private authentication with
physical identifiers through broadcast channel measurements,” in IEEE
Inf. Theory Workshop, Visby, Sweden, Aug. 2019.
[23] T. M. Cover and J. A. Thomas, Elements of Information Theory, 2nd ed.
Hoboken, NJ: John Wiley & Sons, 2012.
[24] R. A. Chou, M. R. Bloch, and E. Abbe, “Polar coding for secret-key
generation,” IEEE Trans. Inf. Theory, vol. 61, no. 11, pp. 6213–6237,
Nov. 2015.
[25] O. Günlü, T. Kernetzky, O. İşcan, V. Sidorenko, G. Kramer, and R. F.
Schaefer, “Secure and reliable key agreement with physical unclonable
functions,” Entropy, vol. 20, no. 5, May 2018.
[26] J. Wayman, A. Jain, D. Maltoni, and D. M. (Eds), Biometric Systems:
Technology, Design and Performance Evaluation. London, U.K.:
Springer-Verlag, 2005.
[27] R. Pappu, “Physical one-way functions,” Ph.D. dissertation, M.I.T.,
Cambridge, MA, Oct. 2001.
[28] O. Günlü, K. Kittichokechai, R. F. Schaefer, and G. Caire, “Controllable
identifier measurements for private authentication with secret keys,”
IEEE Trans. Inf. Forensics Security, vol. 13, no. 8, pp. 1945–1959, Aug.
2018.
[29] C. Nair, “Capacity regions of two new classes of two-receiver broadcast
channels,” IEEE Trans. Inf. Theory, vol. 56, no. 9, pp. 4207–4214, Sep.
2010.
[30] R. Maes, P. Tuyls, and I. Verbauwhede, “A soft decision helper data
algorithm for SRAM PUFs,” in IEEE Int. Symp. Inf. Theory, Seoul,
South Korea, June 2009, pp. 2101–2105.
[31] J. Hagenauer, “Lecture Notes in Digital Communications 1,” G. Kramer
and O. Günlü, Eds. Singapore: TU Munich Asia, Feb. 2019.
[32] M. H. Yassaee, M. R. Aref, and A. Gohari, “Achievability proof via
output statistics of random binning,” IEEE Trans. Inf. Theory, vol. 60,
no. 11, pp. 6760–6786, Nov. 2014.
[33] M. Nafea and A. Yener, “A new wiretap channel model and its strong
secrecy capacity,” IEEE Trans. Inf. Theory, vol. 64, no. 3, pp. 2077–
2092, Mar. 2018.
[34] M. Bloch, Lecture Notes in Information-Theoretic Security. Atlanta,
GA: Georgia Inst. Technol., July 2018.
[35] D. Slepian and J. Wolf, “Noiseless coding of correlated information
sources,” IEEE Trans. Inf. Theory, vol. 19, no. 4, pp. 471–480, July
1973.
[36] A. Schrijver, Theory of Linear and Integer Programming. Chichester,
England: John Wiley & Sons, June 1998.
[37] I. B. Gattegno, Z. Goldfeld, and H. H. Permuter, “Fourier-Motzkin
elimination software for information theoretic inequalities,” 2016.
[38] M. Bloch and J. Barros, Physical-layer Security. Cambridge, U.K.:
Cambridge Uni. Press, 2011.
I Introduction
I-A Related Work and Motivation
I-B Models for Identifier Outputs
I-C Summary of Contributions
I-D Organization
I-E Notation
II Problem Definitions
III Inner Bounds
IV Bounds for a Multi-Enrollment Model
V Proof of Theorem 1
V-A Proof for the GS Model
V-B Proof for the CS Model
VI Proof of Thorem 2
VI-A Achievability Proof for the GS Model
VI-B Achievability Proof for the CS Model
VI-C Outer Bound Proofs for the GS and CS Models
VII Conclusion
References
| 1cybersec
| arXiv |
Nintendo Switch will release March 3 with an MSRP of $299.99 USD. | 0non-cybersec
| Reddit |
ST_MakeEnvelope with Google Maps Coordinates problem. <p>In postgress I have the following function that uses <code>ST_MakeEnvelope</code> based on the Google Maps viewport coordinates to returns all projects within that viewport.</p>
<pre><code>CREATE OR REPLACE FUNCTION public.search_projects_area(
lngw double precision,
lats double precision,
lnge double precision,
latn double precision
) RETURNS SETOF project_locations LANGUAGE sql STABLE AS $ function $
SELECT
A.*
FROM
project_locations A
WHERE
st_intersects(
A.location :: geography,
ST_MakeEnvelope(lngw, lats, lnge, latn, 4326) :: geography
) $ function $
</code></pre>
<p>Below is how I get the coordinates to create the envelope:</p>
<pre><code>const bounds = this.map.getBounds();
const sw = this.map.getBounds().getSouthWest();
const ne = this.map.getBounds().getNorthEast();
const lngw = sw.lng();
const lats = sw.lat();
const lnge = ne.lng();
const latn = ne.lat();
</code></pre>
<p>The problem is that projects in specific areas are not fetched sometimes depending on where I have the map zoomed or positioned. The projects are defenitely inside the viewport though in all cases.</p>
<p>Its as if the <code>ST_MakeEnvelope</code> method doesnt create the correct rectangle based on the map viewport.</p>
<p>Is my code above correct?</p>
| 0non-cybersec
| Stackexchange |
history manipulation in tcsh, substitute every match. <p>I know everybody will tell me that I should switch from using <code>tcsh</code> to <code>bash</code> or something else, but hysteresis is a powerful force, so I'm not going to. Most of the time I find <code>tcsh</code> perfectly adequate but there's one thing I can't figure out how to do in it, but is easy to do in <code>bash</code>: globally substitute one variable for another in a history entry. For example: suppose history says</p>
<pre><code>1 mv dog.tex cat.tex
</code></pre>
<p>and I want to execute the command</p>
<pre><code>mv dog.pdf cat.pdf
</code></pre>
<p>In <code>bash</code> this is easy:</p>
<pre><code>!1:gs/tex/pdf/
</code></pre>
<p>But the global operator doesn't work in <code>tcsh</code>. Is there something else that does? Thanks</p>
| 0non-cybersec
| Stackexchange |
How do I setup “name based” virtual hosts using Ubuntu 12.04?. <p>How do I setup “name based” virtual hosts using Ubuntu 12.04?</p>
<p>I have followed the instructions given at <a href="https://help.ubuntu.com/12.04/serverguide/httpd.html#http-configuration" rel="noreferrer">https://help.ubuntu.com/12.04/serverguide/httpd.html#http-configuration</a> .</p>
<p>I've done the following...</p>
<ol>
<li><code>cp default newsite</code></li>
<li>replace “/var/www” with “/var/www/newsite” in newsite</li>
<li>add “ServerName newsite.example.com” to newsite</li>
</ol>
<p>After some research, I found a blog entry that stated I needed to disable the default site using <code>a2dissite default</code>. After I did that, it worked. Is that correct? This is never mentioned in the Ubuntu server guide. The guide also includes this line... </p>
<blockquote>
<p>“The default virtual host has no ServerName directive specified, so it
will respond to all requests that do not match a ServerName directive
in another virtual host.”</p>
</blockquote>
<p>Which seems to imply that both the default site and others can co-exist.</p>
<p>I'm running a fresh install of 12.04 Server and have reloaded the apache config each time I made an adjustment.</p>
<p><strong>In summary... after adding a new file under /etc/apache2/sites-available (an altered copy of the “default” file with the ServerName directive added) and the corresponding symlink under /etc/apache2/sites-enabled, is it necessary to disable or rename the default site symlink in order for the new site to function? The documentation and one answer given below seems to infer that it is not necessary to do so, but if so, what have I done wrong? Using the config below, when trying to access newsite.example.com I get the default site.</strong></p>
<p>/etc/apache2/sites-available$ cat default</p>
<pre><code><VirtualHost *:80>
ServerAdmin webmaster@localhost
DocumentRoot /var/www
<Directory />
Options FollowSymLinks
AllowOverride None
</Directory>
<Directory /var/www/>
Options Indexes FollowSymLinks MultiViews
AllowOverride None
Order allow,deny
allow from all
</Directory>
ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/
<Directory "/usr/lib/cgi-bin">
AllowOverride None
Options +ExecCGI -MultiViews +SymLinksIfOwnerMatch
Order allow,deny
Allow from all
</Directory>
ErrorLog ${APACHE_LOG_DIR}/error.log
# Possible values include: debug, info, notice, warn, error, crit,
# alert, emerg.
LogLevel warn
CustomLog ${APACHE_LOG_DIR}/access.log combined
Alias /doc/ "/usr/share/doc/"
<Directory "/usr/share/doc/">
Options Indexes MultiViews FollowSymLinks
AllowOverride None
Order deny,allow
Deny from all
Allow from 127.0.0.0/255.0.0.0 ::1/128
</Directory>
</VirtualHost>
</code></pre>
<p>/etc/apache2/sites-available$ cat newsite</p>
<pre><code><VirtualHost *:80>
DocumentRoot /var/www/newsite
ServerName newsite.example.com
<Directory />
Options FollowSymLinks
AllowOverride None
</Directory>
<Directory /var/www/newsite/>
Options Indexes FollowSymLinks MultiViews
AllowOverride None
Order allow,deny
allow from all
</Directory>
ErrorLog ${APACHE_LOG_DIR}/error.log
# Possible values include: debug, info, notice, warn, error, crit,
# alert, emerg.
LogLevel warn
CustomLog ${APACHE_LOG_DIR}/access.log combined
</VirtualHost>
</code></pre>
<p>/etc/apache2/sites-enabled# ls -l</p>
<pre><code>root@sandbox:/etc/apache2/sites-enabled# ls -l
total 0
lrwxrwxrwx 1 root root 26 Mar 18 09:56 000-default -> ../sites-available/default
lrwxrwxrwx 1 root root 26 Mar 7 13:36 newsite -> ../sites-available/newsite
</code></pre>
| 0non-cybersec
| Stackexchange |
Number of different positions of rooks on chessboard. <p>I know that this topic as been mentioned before, but no accurate answer has been provided.</p>
<p>Suppose we have to place $n$ rooks on $n \times n$ chessboard so that no one attacks another. How to count the number of different ways to place them up to rotations of the chessboard?</p>
<p>I have been trying the Burnside's lemma but did not obtain satisfactory results...</p>
| 0non-cybersec
| Stackexchange |
Boston fire fire museum can’t figure out where this helmet is from. Can we figure it out?. | 0non-cybersec
| Reddit |
How to disable the button next to Activities in GNOME?. <p>I'm using Ubuntu 18.04 version, there is a GNOME extension that allows you to disable the 'close button' next to "Activities" corner on the left side of top bar. I really dislike to see a button which covers a lot of space and just has a 'close' effect, nothing else.</p>
<p>Do you guys know the extension name? I just can't find it.</p>
| 0non-cybersec
| Stackexchange |
Let R be the intersection of two lines, how do I prove that it lies on the radical axis?. <p>Given two circles $C_1$ and $C_2$ with centre $O_1$ and $O_2$ respectively. Choose two arbitrary points, one on $C_1$ and one on $C_2$. We name them $P_1$ and $P_2$. Draw the line $P_1P_2$ and mark the midpoint, $M$ of the line. Draw two lines $O_1M$ and $O_2M$. Then, drop a perpendicular line from $P_1$ to $O_1M$ and a perpendicular line from $P_2$ to $O_2M$.
How can I show that the intersection of the two perpendicular lines lies on the radical axis of two circles?</p>
| 0non-cybersec
| Stackexchange |
How to resolve app uninstall without adhoc. <p>When I installed the app with help of this <a href="http://www.securitylearn.net/2012/12/26/build-ipa-file-using-xcode-without-provisioning-profile/" rel="nofollow noreferrer">link</a> onto my device but this procedure has problems under iOS 7 which i have reported to apple but i'm wondering if someone knows of a fix in the mean time?</p>
<p>When you click the link to install, the app installing icons appears and after a short time a second icon appears with the proper app icon but the install one does not disappear (see image below) even if i delete the app the installing icon stays. Any ideas on how to fix this would be much appreciated.</p>
<p><img src="https://i.stack.imgur.com/xAcv5.jpg" alt="enter image description here"></p>
| 0non-cybersec
| Stackexchange |
[No Spoilers] Game of Thrones Wedding Cake. | 0non-cybersec
| Reddit |
This one time when someone accidentally bought a Linux computer from Dell, couldn't install Microsoft Office, so called the news to tell them about it.. | 0non-cybersec
| Reddit |
speech -100. | 0non-cybersec
| Reddit |
Converting false to pointer type void*?. <p>Can someone explain me what is going on here...?</p>
<p>I had this code:</p>
<pre><code>#include <fstream>
#include <string>
#include <iostream>
int main(){
std::ifstream file("test.txt");
std::string x;
while (true) {
if (!(file >> x)) return 0;
std::cout << x << "\n";
}
}
</code></pre>
<p>...compiles fine, does what it is supposed to do, no problem so far. Sometimes I dont like the <code>!</code> so much, because it can be overlooked easily, so I replaced the <code>if</code> with</p>
<pre><code>if ((file >> x)==false) return 0;
</code></pre>
<p>..and suddenly my compiler (gcc 4.8.5) complains with a warning:</p>
<pre><code> warning: converting ‘false’ to pointer type ‘void*’ [-Wconversion-null]
if ((file >> x)==false) return 0;
</code></pre>
<p>and this is where I am starting to be puzzled. Where is the <code>void*</code> coming from? Doesnt <code>>></code> return a reference that should be casted to a <code>bool</code>? Why is <code>false</code> converted to <code>void*</code>? Why isnt the same warning triggered when I dont explicitly write <code>false</code>?</p>
<p>Out of curiosity I also tried this:</p>
<pre><code>if ((file>>x)==true) return 0;
</code></pre>
<p>which causes a storm of errors starting with </p>
<pre><code>error: no match for ‘operator==’ (operand types are ‘std::basic_istream<char>’ and ‘bool’)
if ((file>>x)==true) return 0;
^
</code></pre>
<p>and now I am completely lost. How is <code>false</code> a differnt <code>bool</code> than <code>true</code>? Different values of course, but I always thought <code>true</code> and <code>false</code> are of same type.</p>
| 0non-cybersec
| Stackexchange |
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