text
stringlengths 3
1.74M
| label
class label 2
classes | source
stringclasses 3
values |
|---|---|---|
Games you MUST own everything for..... What games do you require yourself to purchase all expansions for? Currently for me, I have everything for Smallworld, Smash Up, Pandemic, and if anything else were to come out for any of em Id buy it right away lol | 0non-cybersec
| Reddit |
Discrete Mathematics with Algorithms. | 0non-cybersec
| Reddit |
Every time, Chipotle. Every time... | 0non-cybersec
| Reddit |
Possible to make a switchless install from a password protected archive?. <p>I am using 7-zip to make a switchless install (method for doing so <a href="http://www.msfn.org/board/topic/39048-how-to-make-a-7-zip-switchless-installer/" rel="nofollow noreferrer">here</a> and <a href="http://coderdan.blogspot.com/2011/10/making-7-zip-switchless-installer.html" rel="nofollow noreferrer">here</a>). I would like to use a password protected archive to make said install to help prevent the average user from simply unzipping the archive (yes, I do realize that password protected archives are not terribly secure, I'm just trying to prevent the average user from unzipping). Is this possible to do? When I've tried to do this I just get an error box saying "Unsupported Method". Is there some way to specify the password for the archive while making the executable file (probably in the config file)?</p>
| 0non-cybersec
| Stackexchange |
Why is there no std::move_if algorithm?. <p>I've seen a few places on the internet where they describe using <code>std::copy_if</code> with <code>std::make_move_iterator</code>, but if the iterator were to be a forward iterator, that would result in having valid but unspecified (VBU) objects scattered around the source container.</p>
<p>Wouldn't it be better to have a <code>std::move_if</code> algorithm such that if an object is moved, then it would move the resulting VBU object to the end of the range, like that which is done in the <code>std::remove_if</code> algorithm, consolidating all of the VBU objects together so that they can be erased or reassigned?</p>
| 0non-cybersec
| Stackexchange |
Daily Simple Questions Thread - October 19, 2017. Welcome to the /r/Fitness Daily Simple Questions Thread - Our daily thread to ask about all things fitness. Post your questions here related to your diet and nutrition or your training routine and exercises. Anyone can post a question and the community as a whole is invited and encouraged to provide an answer.
# As always, be sure to [read the wiki](https://www.reddit.com/r/Fitness/wiki/) first. Like, all of it.
Also, there's a [handy-dandy search bar](https://www.reddit.com/r/Fitness/search?&restrict_sr=on) to your right, and if you didn't know, you can also use Google to search fittit by using the limiter "site:reddit.com/r/fitness" after your search topic.
Other good resources to search are [Exrx.net](http://www.exrx.net/index.html) for exercise-related topics and [Examine.com](https://examine.com/) for nutrition and supplement science.
Be aware that [the more relevant information you add](https://www.reddit.com/r/Fitness/wiki/getting_started#wiki_how_to_ask_r.2Ffitness_for_help), the more relevant the answers you receive will be. And if you are posting about your routine, please make sure you follow [the guidelines](https://www.reddit.com/r/Fitness/wiki/rules#wiki_routine_critiques).
**(Please note: This is not a place for general small talk or chit-chat. Also, the community decided long ago that we keep jokes, trolling, and memes outside of the Daily Q&A threads. Please use the downvote / report button when necessary.)**
| 0non-cybersec
| Reddit |
How to quantify error distribution between dataset 1 and dataset 3?. <p>thanks in advance for looking into this. </p>
<p>I have 3 datasets: two estimations (estimation 1 and estimation 2) and one measurement campaign (3). I have included some dummy data <a href="https://i.stack.imgur.com/QxxLN.png" rel="nofollow noreferrer">here</a>. I’m able to calculate the distribution of the error between the two estimations (1 and 2) and the distribution of the error between estimation 2 and the measurement campaign (3). The two distributions are shown <a href="https://i.stack.imgur.com/XAMNH.png" rel="nofollow noreferrer">here</a>. Both distributions are assumed to be Gaussian. In red, the error distribution between estimation 1 and 2 is shown (in this case, the mean is almost zero, but this is not always the case). In blue, the error distribution between estimation 2 and the measurement is shown (in this case, normal distribution with a lower mean and a higher standard deviation).</p>
<ul>
<li>How can I calculate the distribution of the error between estimation 1 and the measurement campaign based on the two aforementioned distributions? </li>
<li>How should I calculate the aforementioned error distributions? </li>
</ul>
<p>I have calculated the following quantities:</p>
<ul>
<li>Error distribution between estimation 2 and measurement = (measurement - estimation 2) / estimation 2 for all data points
<ul>
<li>Results in mean X and standard deviation deltaX</li>
</ul></li>
<li>Error distribution between estimation 2 and estimation 1 = (estimation 1 - estimation 2) / estimation 2 for all data points
<ul>
<li>Results in mean Y and standard deviation deltaY</li>
</ul></li>
</ul>
<p>How can I now quantify the mean Z and the standard deviation deltaZ of the error distribution between estimation 1 and the measurement? Also, any text book or article recommendations are welcome. </p>
<p>Thanks again for looking into this. </p>
| 0non-cybersec
| Stackexchange |
let it grow let it grow. | 0non-cybersec
| Reddit |
Prove that there's a multiple of 1997 which has only ones in its decimal expansion. <p>A problem from an exercise book on the Pigeonhole Principle</p>
<p>Prove that there's a multiple of 1997 which has only ones in its decimal expansion.</p>
<p>My progress</p>
<p>As there are infinite number of such numbers, we can pick up <span class="math-container">$1998$</span> of them and according to the Pigeonhole principle there will be at least two of them that are equal modulo <span class="math-container">$1997$</span>.
We can take their difference and it will be of a form</p>
<pre><code>111...1 (r "1"'s)
-
1...1 (s "1"'s)
-------
110...0 (r - s "1"'s, s "0"'s)
</code></pre>
<p>This yields a multiple of <span class="math-container">$1997$</span> which consists of <span class="math-container">$1$</span>'s and <span class="math-container">$0$</span>'s in its decimal expansion.</p>
<p>I also checked that there's such a number with a brute force search in a Python script</p>
<pre><code>s = 1
x = int("1" * s)
while x % 1997 != 0:
s += 1
x = int("1" * s)
</code></pre>
<p>which yielded the answer 998.
So it's the number <span class="math-container">$11\dots1$</span> (998 <span class="math-container">$1$</span>'s) that is a multiple of 1997.</p>
<p>This doesn't show the existence of a proof, of course.</p>
| 0non-cybersec
| Stackexchange |
Multi screen remote desktop. <p>I am looking for a solution (free or paid.) for our small company to improve the environment for working from home. (especially during this period)</p>
<p>We have 3-4 monitors at the remote site (homes) and would like to connect all of them via rdp like solution to a single session.</p>
<p>The only working solution i found is using ssh tunneling with "-Y" args and forwarding x11, but it is very slow and unresponsive.</p>
<p>On the other hand when working with rdp (for example Remmina) it works fast but limit me to only one screen (because currently the computers we try to longing to have 0-1 screens)</p>
<p>We can install a dedication container\vm\software in the target computer if it is required.</p>
| 0non-cybersec
| Stackexchange |
High Standards. | 0non-cybersec
| Reddit |
What's your "carrot?". A carrot, like something you'd tie to a stick and draw a horse with or on other words, a goal that's just out of reach that keeps you going.
My dad always tells me it's important to have a carrot and right now the big carrot is paying off my student loans and becoming a millionaire. My more realistic carrot is a new DSLR, something more on the professional side. It'll take a long while to save up but damn, I want a D750. | 0non-cybersec
| Reddit |
Any exercises to build muscle to help cover my collar bone?. Hello, I'm currently 6"1, Male, 18 years old. Ive gone from 145 lbs to 163 lbs over the course of 9 months of working out 4-5 times a week and I have even been eating a ton more. Yet they are still poking WAY out.
- My question is, could I do something cover them up and maybe protect them? Seems like they are very exposed and could get broken easily. | 0non-cybersec
| Reddit |
Sherman-Morrison formula and a sum of outer products. <p>A specific form of the general Sherman-Morrison formula reads</p>
<p>$(1+u v^T)^{-1}$ = $1 - \frac{u v^T}{1+v^T u}$</p>
<p>where $1$ is the identity matrix, $u,v$ are vectors (say with length n) and T denotes the transpose.</p>
<p>Is there a generalization of the formula to sums of outer products, i.e. for $(1+\sum\limits_{i=1}^{n} u_i v_i^T )^{-1} $ ?</p>
| 0non-cybersec
| Stackexchange |
Parsing content-disposition header's filename in multipart/from-data. <p>According to RFC, in multipart/form-data content-disposition header
filename field receives as parameter HTTP quoted string - string between quites where
character '\' can escape any other ascii character.</p>
<p>The problem is, web browsers don't do it.</p>
<p>IE6 sends:</p>
<pre><code>Content-Disposition: form-data; name="file"; filename="z:\tmp\test.txt"
</code></pre>
<p>Instead of expected</p>
<pre><code>Content-Disposition: form-data; name="file"; filename="z:\\tmp\\test.txt"
</code></pre>
<p>Which should be parsed as <code>z:tmptest.txt</code> according to rules instead of <code>z:\tmp\test.txt</code>.</p>
<p>Firefox, Konqueror and Chrome don't escape " characters for example:</p>
<pre><code>Content-Disposition: form-data; name="file"; filename=""test".txt"
</code></pre>
<p>Instead of expected</p>
<pre><code>Content-Disposition: form-data; name="file"; filename="\"test\".txt"
</code></pre>
<p>So... how would you suggest to deal with this issue?</p>
<p><em>Does Anybody have an idea?</em></p>
| 0non-cybersec
| Stackexchange |
Interquartile range. <p>I have grouped data ranges that I need to find the interquartile range for, I think i have found the groups that the upper and lower quartiles are in, but as they are ranged, where do I go from here? Do I find the mid point of each range and then subtract them like you normally would or am I completely off track?</p>
<p>Also could someone possible explain the difference between the interquartile range and the semi interquartile range please?</p>
<p>Any help anyone could give me would be greatly appreciated.</p>
<p>Many thanks :)</p>
| 0non-cybersec
| Stackexchange |
Raspberry Pi C++11 std::thread: pure virtual method called. <p>Ubuntu 14.04. I've installed gcc-arm-linux-gnueabihf, g++-arm-linux-gnueabihf (tried 4.8 and 4.9 from utopic).</p>
<p>Code that uses std::thread:</p>
<pre><code>#include <iostream>
#include <chrono>
#include <future>
void secondList()
{
const std::chrono::seconds twoSeconds(2);
for (size_t i = 0; i != 300; ++i)
{
std::this_thread::sleep_for(twoSeconds);
std::cout << "2s\n";
}
}
int main(int, const char *[])
{
auto secondThr = std::async(std::launch::async, secondList);
return 0;
}
</code></pre>
<p>Compiling with:</p>
<pre><code>arm-linux-gnueabihf-g++ --std=c++11 main.cpp -lpthread -o main
</code></pre>
<p>On RPI it fails:</p>
<pre><code>pi@raspberrypi ~ $ ./main
pure virtual method called
terminate called without an active exception
Aborted
</code></pre>
<p>Compiling on RPI works:</p>
<pre><code>pi@raspberrypi ~ $ g++ --std=c++0x main.cpp -lpthread -o main
</code></pre>
<p>Pi image 2015-02-16-raspbian-wheezy, g++ on Pi (Debian 4.6.3-14+rpi1) 4.6.3.</p>
<p>I've tried compiler options <code>-mcpu=cortex-a7</code>, <code>-mcpu=cortex-a8</code> and <code>-D__GCC_HAVE_SYNC_COMPARE_AND_SWAP_{1,2,4,8}</code> that were mentioned in similar questions.</p>
<p>Also tried g++ from ppa: <a href="http://ppa.launchpad.net/linaro-maintainers/toolchain/ubuntu" rel="nofollow">http://ppa.launchpad.net/linaro-maintainers/toolchain/ubuntu</a> precise</p>
<p>Why it is happening and how to get a cross-compiler that works?</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 |
Why do initial algebras correspond to data and final coalgebras to codata?. <p>If I understand correctly, we can model inductive data types as initial F-algebras and co-inductive data types as final F-coalgebras (for an appropriate endofunctor <code>F</code>) [<a href="https://en.wikipedia.org/wiki/Initial_algebra#Use_in_computer_science" rel="noreferrer">1</a>]. I understand that according to Lambek's lemma the initial algebras (and final coalgebras) are fixed point solutions of the isomorphism <code>T ≅ F T</code>, but I don't see why the initial algebra is the <em>least</em> fixed point, while the final coalgebra is the <em>greatest</em> fixed point. (Is it obvious that the isomorphism <code>T ≅ F T</code> has a solution?)</p>
<p>Also I'm not really clear on how are inductive and co-inductive data types defined in type theory. Are there any recommended resources on this topic and maybe their relationship to category theory?</p>
<p>Thank you!</p>
| 0non-cybersec
| Stackexchange |
Steven Universe Fan Art: Malachite Speed Drawing. | 0non-cybersec
| Reddit |
Hotels in chania greece, chania hotels crete
. | 0non-cybersec
| Reddit |
Best practice in use of static variables in C. <p>Global variables are generally considered to be a poor programming practice</p>
<p>In C, are static variables (i.e with module (file) scope) considered OK?</p>
<p>My thought was that member variables in an object oriented language cannot be much less dangerous than static variables in C and member variables seem to be considered to be a good thing.</p>
<p>I'm tiring of passing parameters through multiple functions and can see the attraction of static variables for this, especially if they are <code>const</code>.</p>
<p>But I'm keen to know if this is frowned upon - and also whether there is really any difference in level of programming naughtiness between a big object with a member variable used in several of its methods and a C file containing a few functions that utilize a static variable?</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 |
The Things They Carried. Can we take a second to talk about how fucking haunting this book is? I have to read it for a class and decided I would finish the bulk of it tonight. Boy was that a bad idea. It feels like Apocalypse Now condensed into writing, complete with stories about remote outposts where no one seems to be in charge and everyone including Green Berets losing their minds. The one vignette where he talks about the kid he shot has to be the most powerful depiction of PTSD I've ever seen. Anyway it's scaring the ever living shit out of me (while also being extremely enlightening), feels vaguely like an acid trip. | 0non-cybersec
| Reddit |
Homemade slider buns. | 0non-cybersec
| Reddit |
Run JavaScript in Windows. <p>I thought for some simple tests that just run a few commands i would try using some JavaScript and run it from the command line in Windows XP.</p>
<p>So for a quick test I created a script </p>
<pre><code>alert('Hello, World!');
</code></pre>
<p>Then tried to run it</p>
<pre><code>D:\>Cscript.exe hello.js
Microsoft (R) Windows Script Host Version 5.7
Copyright (C) Microsoft Corporation. All rights reserved.
D:\hello.js(1, 1) Microsoft JScript runtime error: Object expected
</code></pre>
<p>Google has not helped and I am sure I am missing something silly, can any of you guys shed any light on why this simple script doesn't run?</p>
| 0non-cybersec
| Stackexchange |
Does App transfer between developers necessitate a re-purchase by end users?. <p>If an app gets transferred from one developer to another, does it necessitate a re-purchase?</p>
<p>From the App Store Connect, <a href="https://help.apple.com/app-store-connect/#/deved688524f" rel="nofollow noreferrer">App transfer overview</a> page:</p>
<blockquote>
<p>You transfer an app when you’ve sold the app to another developer or you want to move it to another App Store Connect account or organization.</p>
<p>You can transfer the ownership of an app to another developer without removing the app from the App Store. The app retains its reviews and ratings during and after the transfer, and users continue to have access to future updates. Additionally, when an app is transferred, it maintains its Bundle ID — it's not possible to update the Bundle ID after a build has been uploaded for the app.</p>
</blockquote>
<p>I have purchased a free/paid app from the App Store but it is not currently installed on the device. In the meanwhile, the original developer of the app decides to sell away the ownership to someone else.</p>
<p>As an end user who previously bought the app, (and any associated in-app purchases,)will this necessitate a re-purchase? i.e. will the app remain linked to my Apple ID or not?</p>
<p>Emphasis on the fact that the app is not <strong>currently installed</strong> on device. The linked documentation does mention that the users will continue to have access to future updates, but I am looking for an authoritative documentation which answers the query. (Or someone who had personal experience with this scenario)</p>
| 0non-cybersec
| Stackexchange |
A man balancing on a piece of wood on the roof of a skyscraper 1939...[1175x1280]. | 0non-cybersec
| Reddit |
Probability of getting a sufficiently long piece from multiple stick breaking. <p><a href="https://math.stackexchange.com/q/262388/357390">This question</a> asks:</p>
<blockquote>
<p>Start with a stick of length <span class="math-container">$1$</span>. Repeatedly remove some fraction <span class="math-container">$U$</span> of the remaining stick, where <span class="math-container">$U$</span> is uniform in <span class="math-container">$[0,1]$</span>. What is the probability that at least one of the removed pieces has length at least <span class="math-container">$\frac12$</span>?</p>
</blockquote>
<p>Eventually I managed to solve it, but the result was not very enlightening. Thus I present here a generalisation of the question where <span class="math-container">$\frac12$</span> is replaced with an arbitrary <span class="math-container">$0\le x\le1$</span>. Let <span class="math-container">$f(x)$</span> be the probability that we eventually break off a stick of length at least <span class="math-container">$x$</span>; it has been shown that for <span class="math-container">$\frac12\le x\le1$</span>, <span class="math-container">$f(x)=-\ln x$</span>.</p>
<blockquote>
<p>What, then, is <span class="math-container">$f(x)$</span> for <span class="math-container">$0\le x\le\frac12$</span>?</p>
</blockquote>
<p>This generalisation is more difficult because now we can cut off <em>more than one</em> stick longer than <span class="math-container">$x$</span>. Let's use <a href="https://math.stackexchange.com/a/262421/357390">joriki's approach</a> to the source question: when <span class="math-container">$x\le\frac12$</span> either we cut off a piece longer than <span class="math-container">$x$</span> at the start, with probability <span class="math-container">$1-x$</span>, or we leave a stick of length <span class="math-container">$t$</span> and replace <span class="math-container">$x$</span> with <span class="math-container">$x/t$</span>:
<span class="math-container">$$f(x)=1-x+\int_{1-x}^1f(x/t)\,dt$$</span>
This is the same formula as in joriki's answer except that the lower bound <span class="math-container">$x$</span> has become <span class="math-container">$1-x$</span>. We can make the same manipulations that follow there to get
<span class="math-container">$$f(x)=1-x+x\int_x^{x/(1-x)}\frac{f(u)}{u^2}\,du$$</span>
<span class="math-container">$$1-f(x)=x\left(1-\int_x^{x/(1-x)}\frac{f(u)}{u^2}\,du\right)$$</span>
and eventually the functional differential equation
<span class="math-container">$$x(1-x)^2f''(x)=f'\left(\frac x{1-x}\right)-f'(x)(1-x)^2$$</span>
but how would I proceed from here?</p>
<p>I've determined the following approximate values for <span class="math-container">$f(x)$</span> with one million trials for each <span class="math-container">$x$</span>.</p>
<p><img src="https://i.stack.imgur.com/Cp8LO.png" alt=""></p>
<pre><code>.49 .712676
.48 .732251
.47 .751341
.46 .769896
.45 .787595
.44 .805675
.43 .822355
.42 .838964
.41 .854620
.40 .869991
.39 .883803
.38 .897808
.37 .910511
.36 .923427
.35 .934614
.34 .945107
.33 .954406
.32 .963026
.31 .969796
.30 .976313
.29 .981963
.28 .986094
.27 .989759
.26 .992816
.25 .995101
.24 .996743
.23 .997929
.22 .998779
.21 .999327
.20 .999609
.19 .999815
.18 .999924
.17 .999958
.16 .999990
.15 .999995
<=.14 1.000000
</code></pre>
| 0non-cybersec
| Stackexchange |
Is this Function Continuous at the origin?. <p>Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $$f(x)=-2\sum_{n=1}^\infty xe^{-n^2x^2}$$</p>
<p>Is $f$ continuous at the origin? I think that $f$ is not continuous at the origin. Any help is appreciated.</p>
| 0non-cybersec
| Stackexchange |
Moonlighter - Between Dimensions DLC Announcement Trailer - Nintendo Switch. | 0non-cybersec
| Reddit |
Restoring xfce4-power-manager to default settings. <p>After unchecking "Handle display power management" in Xfce Power Manager, the system locked up. (I was troubleshooting an issue with the laptop where it would lock up upon the display turning off.) </p>
<p>Upon rebooting, the system behaves normally until a password is entered in the greeter. At this point the desktop will fail to load and the system will become unresponsive. </p>
<p>Is there any way via TTY that I can re-enable this option? </p>
<p>I've attempted restarting the power manager instance and then starting an xsession but the result is the same. </p>
| 0non-cybersec
| Stackexchange |
Simply, running an Ubuntu Server through Virtualbox?. <p>When I go through official documentation (which isn't so great) I cannot figure out how to run an Ubuntu Sever through Virtual box. I cannot find an option, and my iso is never a proper iso to boot from in the storage options... any help would be great or maybe pointing me to a good tutorial would be awesome. I just want to mess around and see what up. Thanks.</p>
| 0non-cybersec
| Stackexchange |
This is my "go to" meal. What's yours? Post your bread and butter! Well not really bread and butter, you know what I mean.. http://imgur.com/Cw5xnNQ
After a long day this takes me about 7 minutes to whip up. I never get tired of it
Whats your favorite go to meal? Recipes and techniques are greatly apprecaited. | 0non-cybersec
| Reddit |
C - Incrementation not updating variable value. <p>I am working on a simple C program, but ran in to some confusion. Below is the code: </p>
<pre><code>int main(void) {
int i, j, k;
i = 3;
j = 4;
k = 5;
printf("%d ", i < j || ++j < k);
printf("\n"); // LINE 1
printf("%d %d %d", i, j, k); // LINE 2
return 0;
}
</code></pre>
<p>In the above program, the variable j starts off being 4. Then in the printf statement of line 1 we increment the value of j by 1(++j = 5). </p>
<p>So theoretically, I would assume that when j is printed in printf(line 2) it prints as 5, since we did an incrementation in line 1 for j. However, every time I run the code, line 2 prints the original value of j which was 4, and NOT 5.</p>
<p>Is there something I am missing?</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 |
Sorry if this is long, but I need to talk about the awful experience I had yesterday at the Apple Store. Still fuming about it.. **Edit: Thanks for the feedback. I revised the below message and forwarded it to Exec. Relations and Tim Cook. Will update when I hear back.**
**TLDR: Awful battery on new iPhone 6 that multiple restores wont fix. AppleCare tells me to go to Store to swap out for new phone, due to hardware issue, in what should have been a 5 minute appointment. After 2.5 hours in store, after being called a liar and shuffled around to three different people, I finally got a new phone.**
I don’t know where exactly it went wrong, but yesterday was the worst experience I have ever had with any retailer and has put a really bad taste in my mouth, as far as my experiences with Apple go. Ive been a lifelong apple customer, and even worked for apple and serviced apple products at points in my life and am still in disbelief in my experience with both Applecare and the Cherry Creek apple store. What used to be an aim to please, and help the customer, seemed to turn into finger pointing and blame shifting at the customers expense, all to avoid what should have been a simple phone-swap.
I bought my iPhone on launch day after waiting for 7 hours in line. After getting my iPhone 6, I was quickly disappointed in the poor battery life. After a day or two of usage, I found it wouldn’t even last a full day (when my 5S never had a problem). Calling back on my past experience, I DFU restored it (as a new phone) and started using it for another few days. Still, the battery life was terrible and apps were crashing (Both system and 3rd party) left and right. Perplexed, I contacted AppleCare via chat and asked them to run a diagnostic. They confirmed what I suspected: tons of apps in the background were crashing and causing poor battery life. I did another DFU restore (as a new phone) and used it for the rest of the week. Still, my battery life was terrible and after having my phone die halfway through the day on Friday, I decided it was time to get rid of my 6. I figured, maybe this is just how the 6 battery life is and decided to head to the Verizon store, where I purchased it, to see about returning it and maybe getting a 6+ when they became available. I didn’t really like the size, but the bigger battery seemed to be a necessity. Since friday was the last day of the 14 day return policy, I had to make a decision. After talking to someone at Verizon, I fell back on my earlier suspicion that I just got a defective iPhone 6. I could return my iPhone to Verizon, no question, but they were out of 6’s and 6+’s for the near future. So I decided to run home (there was still a few hours til verizon closed) and contact Apple one more time (no Apple Store near me).
This is where it took a turn for the worse. I got on with Apple Chat support and explained that today was the last day I could return my phone, I needed help making a decision whether to return it to verizon or to see if Apple could help me. He ran a diagnostic and confirmed (again) that I was having battery issues. The tech specifically said that I needed to get serviced swapped and told me that the Apple Store should see this in their system. So I made an appointment for the closest Apple Store near me. Knowing that I would be sacrificing most of my day, and driving a ways to get there, I wanted to verify they even had service phones to swap. So I called the store and after talking to a specialist on the phone who had no idea what I was talking about, or how to use the phone (she hung up on me once), I got frustrated and asked for a manager. The manager was understanding and assured me there were service units available, but then said something odd: he said they probably wouldn’t need to replace it anyway, so not to worry. He apparently didn’t see anything from Applecare in the system.
So I got back on with Applecare again and went through the entire process again (3rd time that day) and made sure that the guy put it in my file. To quote him: “This case number, xxxxxxxxx, shows that a hardware repair is needed, meaning repair/swapped out.” Knowing I would have problems the next day, I took [screen shots](http://imgur.com/hTlCuTk.png) of all this.
Saturday rolls around and I go to the genius bar. The wait wasn’t very long, and soon I was put with a FRS. As soon as he started to help me, I knew it was going to be a long day. He started from scratch, running diagnostics, and told me how this and that was crashing and how when Usage & Standby is the exact same, something is wrong (no shit), and lectured me on turning off location services, etc, and that I just needed to restore it. I told him all about whats happened already and that, as per applecare, it needed to be swapped. He refused, so I got taken to a Genius. Rinse and repeat. Lead genius refused to do so, and said he had no record of AppleCare saying anything about this. I showed him screenshots from my interaction with AppleCare, directly saying a case number and the need for replacement, and was essentially called a liar. He didn’t see it in his system and refused to move forward at all, and then further lectured me on battery management techniques. When I asked him if we could get AppleCare on the line to verify it, he told me that they didn’t really have anything to do with AppleCare and if thats what AppleCare said, I should go home and talk to them. I asked again to talk to them at the store, at which point he walked away and told me I could speak to a manger instead. The manager was rather indignant, and not helpful at all. He too refused to move forward without any notes from AppleCare. I asked *again* to speak with them, and he told me to grab a computer on the other side of the store and deal with it. It had been more then an hour, at this point, at what should have been a 5 minute appointment.
So I did. I got AppleCare on Chat and had them go through my previous conversations and verify everything that was said, then got the manger over. He still, apparently, didn’t see anything in the system and told them to update it. Then refused to talk to them any further via chat, and instead told AppleCare that he would talk to them on the phone (this, by itself I thought was strange. My only thought on this was that he knew he could speak to them via phone, away from me, in his office. Sketchy?). The chat person said she would try to call him. 20 minutes go by, with nothing. Finally, she says she’s having trouble with her phone (weird?) and will transfer me to her Senior Advisor. The SA gets on and has no idea whats going on, so I explain (AGAIN) what the deal is. Then the manager at the store tells him to call, three times, before he even acknowledges the request. All the AC says is “just a moment” then 20 minutes goes by with nothing. So I tell him that I have been waiting a very long time, and that the manager refuses to move forward until he calls. Another wait. Finally, he tells me he can’t make calls. By this time, its been about 2 hours in the store and Im at my wits end and tell him as much. Finally, he just agrees to put an exception in the system, in order to facilitate a swap out. Two and a half hours later, I walked out with a new phone.
Im sorry this is so long, but the entire ordeal was extremely long, and Ive tried to condense it. Between sub-par Apple Care support and downright awful, unhelpful, and condescending in-store support, I am just baffled by what happened. I would love to send a message to someone in Corporate so this can be prevented in the future. Does anyone know what address would be best to send it to?
Oh. And my new phone is great. My old phone, by this time would be down to 80%. My new phone is currently at 95%.
| 0non-cybersec
| Reddit |
Altopias: Speculative Art Explores Both Dark & Light Futures. | 0non-cybersec
| Reddit |
MacBook restarted now no HD to work from. <p>I was using my 2012 era MacBook Pro when it suddenly restarted, to a folder with a ? On it. When I try any of the internet recovery methods, it just says disk 0 and base recovery system. So I can’t reinstall any OS X. And if I try to reinstall it no disk pops up. So is my laptop toast? </p>
| 0non-cybersec
| Stackexchange |
[WP] The death is coming to "collect" a young girl - but he's there too soon and he finds her standing on top of a high building. Write about their talk, how it's going and how it will end.. | 0non-cybersec
| Reddit |
Free online SSL/TLS server test for compliance with PCI requirements and NIST guidelines. | 1cybersec
| Reddit |
I steal peoples/friends' passwords to just about everything. AMA. I do it for the rush, and for finding out what type of passwords each of my friends use. It tells a lot about a person. I usually only steal passwords of people I know (facebook/other social networking, email, bank accounts on rare occasions, etc). Sometimes I'll get bored and set up shop in a coffee shop or something for easy prey. I generally never log in using it, I just enjoy knowing people's passwords. AMA - techniques, reasons, ethical questions, etc.
**Edit** Bed for now, will answer more in the morning ( 6pm GMT ).
**Edit #2** Off to work for a few hours. Will be back to answer more later. | 0non-cybersec
| Reddit |
admob cpc is very low in 2017? how to solve?. <p>my admob cpc is very low only 0.02.my page view is 150000.page impression 70000 click 986.page rpm 17.</p>
<p>then.</p>
<p>*How to increase my cpc?</p>
<p>*If I BLOCK LOW CPC ADSENSE ADVERTISERS URL then face any problem with admob?</p>
<p>please help.
advance thanks </p>
| 0non-cybersec
| Stackexchange |
If A and B are diagonalizable then so is AB. <p>When we have to n×n matrices that can be made diagonal (maybe not in the same basis), is it true that the same works for their product?</p>
| 0non-cybersec
| Stackexchange |
Microsoft LogParser: How to use parameters in a file. <p>I have started to use Microsoft LogParser in order to analyse IIS logs.</p>
<p>LogParser allows using SQL query from a file and supply parameters to the query directly in a command line, for example:</p>
<pre><code>LogParser file:query.sql?date=2010-12-29 -i:IISW3C
</code></pre>
<p><em>query.sql</em> is a file name with SQL query</p>
<p><em>date=2010-12-29</em> is a parameter that is supplied to the SQL query</p>
<p>Here is content of query.sql:</p>
<pre><code>select cs-uri-stem
,count(*)
from logs.log
where date = <date>
group by cs-uri-stem
</code></pre>
<p>The problem is that I don't know how to call parameter in the file. Does anybody know this?</p>
| 0non-cybersec
| Stackexchange |
properness of stack. <p>Hi,</p>
<p>assume we have an algebraic stack $A$ over $Sch(\mathbb{Z})$ which is quasi-compact and with separated diagonal. Assume that I have a stack $B$ which is obtained by rigidifying $A$ by a subgroup of the inertia group. Assume moreover that $A\rightarrow B$ is faithfully-flat and smooth over $\mathbb{Z}[\frac{1}{d}]$ and that the diagonal of $B$ is finite.
Question : is $B$ proper over $\mathbb{Z}$, or at least over $\mathbb{Z}[\frac{1}{d}]$ ?? If so is it trivial to prove it?</p>
| 0non-cybersec
| Stackexchange |
Combos just got inexpensive!. | 0non-cybersec
| Reddit |
"open -t" vs. "open -ef". <p>the Command:</p>
<pre><code>syslog | tail | open -ef
</code></pre>
<p>works perfectly on my osx 10.8.5 and opens the required lines from syslog in textEdit, however when I run</p>
<pre><code>syslog | tail | open -t
</code></pre>
<p>which should open the same lines in my default text editing program I get nothing and just the lines --No lines in buffer--</p>
<p>I am using MacVim as my default text editor.</p>
| 0non-cybersec
| Stackexchange |
[Spoiler][Clip] FUNimation is clearly having fun dubbing Highschool DxD HERO.. | 0non-cybersec
| Reddit |
When you shine light at a mirror, does the amount of light in the room change?. This seems like a dumb question. But what if you looked at the light and it's reflection in the same visual frame? Then would the amount of light be double? | 0non-cybersec
| Reddit |
The Most Detailed Map Ever Made of the Milky Way in Radio Waves. | 0non-cybersec
| Reddit |
Escape Plan Official Trailer #1 (2013) - Sylvester Stallone/Schwarzenegger Movie HD - YouTube. | 0non-cybersec
| Reddit |
Do you want to see a magic trick?. | 0non-cybersec
| Reddit |
Penetration testing projects. Hi,
I am finishing my uni and I am at point of choosing my bachelor thesis, which has to be hard to do and invloves coding. I want to do something involving penetration testing, since I am working in this field. However, I lack ideas what topic could I write about. I was thinking about something like MitM tool, but there are already some of them on the field. I can't find a niche in this field. Do you have any ideas?
Thank you in advance. | 1cybersec
| Reddit |
Spark Streaming Kafka backpressure. <p>We have a Spark Streaming application, it reads data from a Kafka queue in receiver and does some transformation and output to HDFS. The batch interval is 1min, we have already tuned the backpressure and <code>spark.streaming.receiver.maxRate</code> parameters, so it works fine most of the time.</p>
<p>But we still have one problem. When HDFS is totally down, the batch job will hang for a long time (let us say the HDFS is not working for 4 hours, and the job will hang for 4 hours), but the receiver does not know that the job is not finished, so it is still receiving data for the next 4 hours. This causes OOM exception, and the whole application is down, we lost a lot of data.</p>
<p>So, my question is: is it possible to let the receiver know the job is not finishing so it will receive less (or even no) data, and when the job finished, it will start receiving more data to catch up. In the above condition, when HDFS is down, the receiver will read less data from Kafka and block generated in the next 4 hours is really small, the receiver and the whole application is not down, after the HDFS is ok, the receiver will read more data and start catching up.</p>
| 0non-cybersec
| Stackexchange |
There are NO Shortcuts. Boom! Eagle smack down by awesome designer James Victore. He also does motivational, philosophical youtube videos, mostly made for other designers but applicable to all kinds of work and creativity. [Check them out here.](http://www.youtube.com/user/JamesVictore?feature=watch) | 0non-cybersec
| Reddit |
How to answer matrices problems involving "meet" and "join"?. <p><strong>Problem:</strong> Given the matrix below find the <strong>meet</strong> and <strong>join</strong> of A and B. </p>
<p>\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}</p>
<p>\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 0 & 1 \\
\end{bmatrix}</p>
<p>How to answer this kind of questions? I have learned the basics to advance math and know how to answer matrices involving problems, but I am new to this "meet" and "join" type of questions. Any help would be appreciated.</p>
| 0non-cybersec
| Stackexchange |
Rick and Morty now on 4chan [xpost from /r/4chan]. | 0non-cybersec
| Reddit |
Quasicompact over affine scheme. <p>Let $X$ be a scheme and $f : X \rightarrow \mathrm{Spec}\, A$ a quasicompact morphism. Are there any easy conditions on $A$ under which we can say that $X$ is quasicompact?</p>
<p>Quasicompact morphism means only that there is an affine cover $\cup_{i \in I} \mathrm{Spec}\, A_i$ where $f^{-1}(\mathrm{Spec}\, A_i)$ is quasicompact. It doesn't seem to be enough.</p>
| 0non-cybersec
| Stackexchange |
Multi-Monitor with multiple computers. <p>I need to know if there's an free application that does the same as the MaxiVista "Extended Screen" functionality.</p>
<p>Basically is to use a 2nd computer screen as an extra monitor of the main machine.
<a href="http://www.maxivista.com/multi_monitor.htm" rel="nofollow noreferrer">http://www.maxivista.com/multi_monitor.htm</a></p>
<p>This is different from Synergy or Input Director that "only" able to manage multiple computer with a single mouse/keyboard.</p>
<p>Thanks!
Alex</p>
| 0non-cybersec
| Stackexchange |
Any idea why icons would arbitrarily move around on a Windows desktop?. <p>Virus-scan checks out; if someone's hacking in, they're just moving icons. No apparent pattern, except icons being in different places.</p>
<p>Thoughts?</p>
| 0non-cybersec
| Stackexchange |
Mate on fb...... | 0non-cybersec
| Reddit |
The rarest eyes you ever did see. Not clear from the photo, but they're almost blue.. | 0non-cybersec
| Reddit |
Dog swims to the bottom of a pool for 2 frisbees at once... great success!. | 0non-cybersec
| Reddit |
The lilac-breasted roller bird. | 0non-cybersec
| Reddit |
What's the connection between persistent homology and tensor networks?. <p><a href="https://tensornetwork.org/" rel="nofollow noreferrer">Tensor networks</a> are mathematical representations of quantum many-body systems. </p>
<p><a href="https://math.stackexchange.com/questions/1932207/applications-of-persistent-homology">Persistent homology</a> is a method for computing topological features. </p>
<p>Are these two related? </p>
<p>It has at least two implications that I could think of in the context of <a href="https://en.wikipedia.org/wiki/Renormalization_group" rel="nofollow noreferrer">renormalization groups</a> and quantum computing & quantum complexity theory.</p>
| 0non-cybersec
| Stackexchange |
How to modify the message string being logged by syslog?. <p>I have a few linux systems with the keyboard and console shared via a KVM switch. When I switch between them, I get a lot of unwanted syslog messages regarding the keyboard connecting and disconnecting. This is a particular issue for me because I like to keep the console of each system permanently monitoring <em>"live"</em> syslog messages with <code>tail -f /var/log/syslog</code>.</p>
<p>Here's one, for example:</p>
<pre><code>Nov 7 01:03:37 PIHOSTNAME kernel: [1648435.194330] usb 1-1.3.2: new low-speed USB device number 55 using dwc_otg
</code></pre>
<p>I'm able to remove all the keyboard messages with a bunch of rules in <code>/etc/rsyslog.conf</code>. For the above example I use:</p>
<pre><code>:msg, contains, "new low-speed USB device " STOP
</code></pre>
<p>That works fine, but I'd like to keep ONE message to confirm when the keyboard is connected <strong>- and modify it a bit for user-friendliness</strong>. For example:</p>
<pre><code>:msg, contains, "new low-speed USB device " :msg+" - Keyboard Successfully Connected!"
</code></pre>
<p><em>(The above rule doesn't work, of course)</em></p>
<p>I had hoped it would output something like this:</p>
<pre><code>Nov 7 01:03:37 PIHOSTNAME kernel: [1648435.194330] usb 1-1.3.2: new low-speed USB device number 55 using dwc_otg - Keyboard Successfully Connected!
</code></pre>
<ol>
<li>How can I modify my rule to add the extra text?</li>
<li>Also, how can I modify my rule to replace certain characters with the new text?</li>
</ol>
<p>Thanks.</p>
| 0non-cybersec
| Stackexchange |
Check if computer is activated through Wake On Lan. <p>I'm working on a solution where machines are activated through Wake On Lan after which System Center pushes updates to the client pc's (running Windows 7).</p>
<p>Now I'm working at a script (PowerShell/C#), that checks if the machine should be shutdown after the updates finishes. </p>
<p>If the machine is activated through Wake On Lan and no user has logged on to the machine since activation, the machine can be safely closed. Otherwise, the machine should stay on.</p>
<p>Is there some way to check how the computer got activated? </p>
| 0non-cybersec
| Stackexchange |
My own little slice of heaven, complete with cats and coffee 😌. | 0non-cybersec
| Reddit |
Youve seen my toe beans, now please cover me back up.. | 0non-cybersec
| Reddit |
XSS - Quote Breaking. <p>First of all i don't know if this question already mention here, believe me i search already.</p>
<p>so i was trying to break string inside single and double quote of attr below;</p>
<pre><code>onclick="window.location='{$url4stat}en'"
</code></pre>
<p>now lets say the webapps urls like <a href="http://example.com/index/" rel="nofollow">http://example.com/index/</a></p>
<p>i was able to inject random param after the /index/ path which is /index/?rand=123 and was included in respond boby such; <code>onclick="window.location='/index/?rand=123&stat=en'"</code></p>
<p>so i though it might work somehow putting a double quote <code>"</code> and see if the html respond body breaks. turns out it gets encoded <code>%22</code> <code>onclick="window.location='/index/?rand=%22&stat=en</code> so putting lazy <code>"><script>alert(1)</script></code> payload will useless ( i've tried )</p>
<p>But, using burpsuite i was able to edit the encoded request before being send to webserver, when open the respond body in browser the alert just pop! is this means that it's prone to XSS ? if so, my question is it possible to bypass this? i already try double encode, using hex, %23x, and many other payload but nothing's work. turns out double encode payload will stay, if i put <code>%2522</code> the respond body becomes <code>onclick="window.location='/index/?rand=%2522&stat=en</code> nothing change, same with other payload. it looks like only a few char such <code>", ', >, <,</code> will be encoded. if i ran into IE Browser, the payloads work. only Chrome,FF, have not yet tried on Opera and Safari.</p>
<p>Will appreciate any suggestion,technique,payloads.</p>
| 1cybersec
| Stackexchange |
Long wait. | 0non-cybersec
| Reddit |
How can I make my app send out notifications when it's loaded but not running in the foreground?. <p>I have an app that I use sometimes. I must have left it there in the background before I slept. When I woke up I saw this notification on my screen. </p>
<p><a href="https://i.stack.imgur.com/MIIVh.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/MIIVh.jpg" alt="enter image description here"></a></p>
<p>Does anyone have any suggestions on how I can make a notification like this appear with my <strong>XF</strong> application? </p>
<p>Also, do these notifications appear on <strong>Android</strong> also? I've never seen them on my <strong>Android</strong> phone but that could be because I use it much less.</p>
| 0non-cybersec
| Stackexchange |
SICStus Prolog JIT compiler. <p>SICStus Prolog 4.3 added a JIT compiler for x86-64 processors.</p>
<p>I have two questions regarding the JIT compiler.</p>
<p>First, where can I find some documentation (papers, notes, or memos) on the capabilities (and limitations, design, and techniques) of the JIT compiler?</p>
<p>And second, can SICStus be directed to emit the x86-64 assembly code produced by the JIT compiler?</p>
| 0non-cybersec
| Stackexchange |
Distribution of certain variable - can't find mistake.. <p>I'm trying to solve one problem, I arrive with a very good looking answer, but it's precisely a half of what it is supposed to be. Here is a problem:</p>
<blockquote>
<p>We take a stick of length 4 and we break it in any random place on it.
The place of breaking is a random variable with uniform distribution
U[0,4]. We then take the shorter of these two parts, and set it as one
side of a rectangle; the rest of the stick is broken in two more
parts, to create other sides of rectangle. Calculate the distribution
of area of that rectangle.</p>
</blockquote>
<p>Let $X$ be the variable representing the breaking point. If so, then variable $Y$ - area looks like this:
$$Y=
\begin{cases}
X(2-X), &X\leq 2\\
(4-X)(X-2), &X>2\\
\end{cases}
$$ </p>
<p>I calculate the distribution in a tradiltional way. We clearly see that $Y\in [0,1]$, so for any $t\in[0,1]$
\begin{align*}
P(Y\leq t)&=P(X(2-X)\leq t)P(X\leq 2)+P((4-X)(X-2)\leq t)P(X> 2)=\\
&=\frac{1}{2}\left[\int_{0}^{1-\sqrt{1-t}}\frac{1}{4}dt+\int_{1+\sqrt{1-t}}^{2}\frac{1}{4}dt+\int_{2}^{3-\sqrt{1-t}}\frac{1}{4}dt+\int_{3+\sqrt{1-t}}^{4}\frac{1}{4}dt\right]=\\
&=\frac{1}{2}(1-\sqrt{1-t})
\end{align*}</p>
<p>And the answer clearly should be $(1-\sqrt{1-t})$. It looks like the $\frac{1}{2}$ factor is not needed, but it comes from the total probability of event $P(Y\leq t)$ after partitioning it into two events. </p>
| 0non-cybersec
| Stackexchange |
Tomcat Server Error - Port 8080 already in use. <p>I received the following error while attempting to execute a Servlet program in Eclipse Mars EE. </p>
<blockquote>
<p>'Starting Tomcat v8.0 Sever at localhost' has encountered a problem.</p>
<p>Port 8080 required by Tomcat v8.0 Server at localhost is already in
use. There may already be running in another process, or a system
process may be using the port. To start this server you will need to
stop the other process or change the port number(s).</p>
</blockquote>
<p>What should I do to stop the process? I'm assuming that Tomcat 7 server must be stopped. How shall I do it if my operating system is Windows 8?</p>
<p>Error Screenshot:</p>
<p><a href="https://i.stack.imgur.com/u5Z87.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/u5Z87.jpg" alt="error screen shot"></a></p>
| 0non-cybersec
| Stackexchange |
I woke up last night and checked the baby monitor. He was just staring at me like this.. | 0non-cybersec
| Reddit |
proof for $\frac{1}{i} = -i$?. <p>My physical chemistry textbook seems to be making the implicit assumption that $\cfrac{1}{i} = -i$.</p>
<p>I'm not sure how this is valid. Here is the snippet of relevant steps:</p>
<p>$\cfrac{i\hbar}{f(t)}$$\cfrac{df(t)}{dt}$$ = E$ ---I can see multiplying by $f(t)$ here and dividing by $i\hbar$</p>
<p>$\cfrac{df(t)}{dt} = $$ -\cfrac{i}{\hbar}$$Ef(t)$ </p>
<p>[If you're wondering,this is part of the time-dependent Schrodinger equation.]</p>
| 0non-cybersec
| Stackexchange |
Prob. 19, Chap. 5 in Baby Rudin: If $f$ is defined in $(-1, 1)$, $f^\prime(0)$ exists, $-1<\alpha_n<\beta_n<1$, and . . .. <p>Here is Prob. 19, Chap. 5 in the book <em>Principles of Mathematical Analysis</em> by Walter Rudin, 3rd edition: </p>
<blockquote>
<p>Suppose $f$ is defined in $(-1, 1)$ and $f^\prime(0)$ exists. Suppose $-1 < \alpha_n < \beta_n < 1$, $\alpha_n \to 0$, and $\beta \to 0$ as $n \to \infty$. Define the difference quotients $$ D_n = \frac{f\left(\beta_n\right)-f\left(\alpha_n\right)}{\beta_n-\alpha_n}.$$
Prove the following statements: </p>
<p>(a) If $\alpha_n < 0 < \beta_n$, then $\lim D_n = f^\prime(0)$. </p>
<p>(b) If $0 < \alpha_n < \beta_n$ and $\left\{ \beta_n / (\beta_n-\alpha_n) \right\}$ is bounded, then $\lim D_n = f^\prime(0)$. </p>
<p>(c) If $f^\prime$ is continuous in $(-1, 1)$, then $\lim D_n = f^\prime(0)$. </p>
<p>Give an example in which $f$ is differentiable in $(-1, 1)$ (but $f^\prime$ is not continuous at $0$) and in which $\alpha_n$, $\beta_n$ tend to $0$ in such a way that $\lim D_n$ exists but is different from $f^\prime(0)$. </p>
</blockquote>
<p>My Attempt: </p>
<blockquote>
<p>Part (a): Since $$f^\prime(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0}$$ exists, so, given any real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that
$$ \left| \frac{f(x) - f(0)}{x-0} - f^\prime(0) \right| < \frac{\varepsilon}{4} \tag{1} $$
for all $x \in (-1, 1)$ for which $0 < |x| < \delta$. </p>
<p>Now as $\alpha_n \to 0$ and $\beta_n \to 0$ as $n \to \infty$, so we can find natural numbers $M$ and $N$ such that
$$ \left| \alpha_n \right| < \delta \ \mbox{ for all } n > M,$$
and
$$ \left| \beta_n \right| < \delta \ \mbox{ for all } n > N,$$
So
$$ \left| \alpha_n \right| < \delta \ \mbox{ and } \left| \beta_n \right| < \delta \ \mbox{ for all } n > \max \left\{ M, N \right\}. \tag{2} $$ </p>
<p>As $\alpha_n < 0 < \beta_n$, so we have $0 < - \alpha_n < \beta_n - \alpha_n$ and $ 0 < \beta_n < \beta_n - \alpha_n$, which imply
$$ 0 < \frac{\beta_n }{ \beta_n - \alpha_n } < 1 \ \mbox{ and } \ 0 < \frac{ - \alpha_n }{ \beta_n - \alpha_n } < 1. \tag{3} $$
Now<br>
$$
\begin{align}
D_n &= \frac{ f\left( \beta_n \right) - f \left( \alpha_n \right) }{ \beta_n - \alpha_n } \\
&= \frac{ f\left( \beta_n \right) - f(0) + f(0) - f \left( \alpha_n \right) }{ \beta_n - \alpha_n } \\
&= \frac{f\left( \beta_n \right) - f(0) }{ \beta_n - \alpha_n} + \frac{ f(0) - f \left( \alpha_n \right)}{\beta_n - \alpha_n} \\
&= \frac{\beta_n }{ \beta_n - \alpha_n } \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } + \frac{ - \alpha_n }{ \beta_n - \alpha_n} \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n }.
\end{align}
$$
So, for all $n > \max \left\{ M, N \right\}$ [Refer to (2) above.], we obtain<br>
$$
\begin{align}
& \left| D_n - f^\prime(0) \right| \\
&= \left| \frac{ f\left( \beta_n \right) - f \left( \alpha_n \right) }{ \beta_n - \alpha_n } - f^\prime(0) \right| \\
&= \left| \frac{\beta_n }{ \beta_n - \alpha_n } \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } + \frac{ - \alpha_n }{ \beta_n - \alpha_n} \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right| \\
&= \left| \frac{\beta_n }{ \beta_n - \alpha_n } \left[ \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } - f^\prime(0) \right] + \frac{ - \alpha_n }{ \beta_n - \alpha_n} \left[ \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right] \right| \\
&\leq \frac{\beta_n }{ \beta_n - \alpha_n } \left| \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } - f^\prime(0) \right| + \frac{ - \alpha_n }{ \beta_n - \alpha_n} \left| \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right| \qquad \mbox{ [ using (3) ] } \\
&\leq \left| \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } - f^\prime(0) \right| + \left| \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right| \qquad \mbox{ [ using (3) again ] } \\
&= \frac{\varepsilon}{4} + \frac{\varepsilon}{4} \qquad \mbox{ [ using (1) and (2) ] } \\
&< \varepsilon.
\end{align}
$$
Thus, for every real number $\varepsilon > 0$, we can find a natural number $K \colon= \max\left\{ M, N \right\}$ such that $$ \left| D_n - f^\prime(0) \right| < \varepsilon $$ for all natural numbers $n > K$. Hence $$\lim_{n \to \infty} D_n = 0.$$</p>
</blockquote>
<p>Am I right?</p>
<blockquote>
<p>Part (b): If $0 < \alpha_n < \beta_n$, then $$0 < \frac{\alpha_n}{\beta_n - \alpha_n} < \frac{\beta_n}{\beta_n - \alpha_n}; \tag{4}$$ furthermore if there is a real number $r > 0$ such that $$\frac{\beta_n }{\beta_n - \alpha_n } \leq r,$$ then we also have $$ \frac{\alpha_n }{\beta_n - \alpha_n } \leq r.$$
Now as $f^\prime(0) = \lim_{x \to 0 } \frac{ f(x) - f(0) }{x-0}$ exists, so, for every real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$\left| \frac{ f(x) - f(0) }{x-0} - f^\prime(0) \right| < \frac{\varepsilon}{4r} \ \mbox{ for all real numbers } x \in (-1, 1) \ \mbox{ for which } \ 0 < | x | < \delta. \tag{5} $$
And, since $\alpha_n \to 0$ and $\beta_n \to 0$ as $n \to \infty$, therefore we can find natural numbers $M$ and $N$ such that
$$ \left| \alpha_n \right| < \delta \ \mbox{ for all natural numbers } n > M, \tag{6} $$
and
$$ \left| \beta_n \right| < \delta \ \mbox{ for all natural numbers } n > N. \tag{7} $$
So, for all natural numbers $n > \max\left\{ M, N \right\}$, we see that
$$
\begin{align}
& \left| D_n - f^\prime(0) \right| \\
&= \left| \frac{\beta_n }{ \beta_n - \alpha_n } \left[ \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } - f^\prime(0) \right] + \frac{ - \alpha_n }{ \beta_n - \alpha_n} \left[ \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right] \right| \\
&\leq \frac{\beta_n }{ \beta_n - \alpha_n } \left| \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } \right| + \frac{ \alpha_n }{ \beta_n - \alpha_n}
\left| \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right| \qquad \mbox{ [ using (4) ] } \\
&\leq r \left| \frac{ f\left( \beta_n \right) - f(0) }{ \beta_n } \right| + r \left| \frac{ f \left( \alpha_n \right) - f(0) }{ \alpha_n } - f^\prime(0) \right| \qquad \mbox{ [ using our hypothesis ] } \\
&< r \frac{\varepsilon}{4r} + r \frac{\varepsilon}{4r} \qquad \mbox{ [ using (5), (6), and (7) above ] } \\
&< \varepsilon.
\end{align}
$$
Since $\varepsilon$ was an arbitrary positive real number, therefore it follows that $$\lim_{n \to \infty} D_n = f^\prime(0),$$
as required. </p>
</blockquote>
<p>Am I right? </p>
<blockquote>
<p>Part (c): For each $n \in \mathbb{N}$, as $-1 < \alpha_n < \beta_n < 1$ and as $f^\prime$ is continuous in $(-1, 1)$, so $f$ satisfies the hypothesis of the mean value theorem on the interval $\left[ \alpha_n, \beta_n \right]$. So, for every $n \in \mathbb{N}$, we can find a real number $\gamma_n \in \left( \alpha_n, \beta_n \right)$ such that $$ D_n = \frac{ f\left( \beta_n \right) - f \left( \alpha_n \right) }{ \beta_n - \alpha_n } = f^\prime\left(\gamma_n\right). $$
Now as $\alpha_n < \gamma_n < \beta_n$ for all $n \in \mathbb{N}$ and as $\alpha_n \to 0$ and $\beta_n \to 0$ as $n \to \infty$, so by the sandwiching theorem we can conclude that $\gamma_n \to 0$ as $n \to \infty$, and since $f^\prime$ is continuous in $(-1, 1)$ and hence at $x=0$, therefore we can conclude that $f^\prime\left(\gamma_n\right) \to f^\prime(0)$ as $n \to \infty$; that is, $D_n \to f^\prime(0)$ as $n \to \infty$. </p>
</blockquote>
<p>Am I right?</p>
<blockquote>
<p>Let $f$ be defined on $(-1, 1)$ by $$ f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) \ & \ (x \neq 0), \\ 0 \ & \ (x=0). \end{cases} $$ Let $$\alpha_n = \frac{1}{2\pi (n + 1/4)}, \qquad \beta_n = \frac{1}{2 \pi n} \qquad \mbox{ for all } n \in \mathbb{N}.$$
Then $$ f^\prime(x) = \begin{cases} 2x \sin \left( \frac{1}{x} \right) + x^2 \cos \left( \frac{1}{x} \right) \ & \ (x \neq 0), \\ 0 \ & \ (x=0). \end{cases} $$
Moreover, $$0 < \alpha_n < \beta_n < 1 \ \mbox{ for all } n \in \mathbb{N},$$ and $$ \lim_{n \to \infty} \alpha_n = 0, \qquad \lim_{n \to \infty} \beta_n = 0.$$
Now
$$
\begin{align}
D_n &= \frac{ f\left( \beta_n \right) - f\left( \alpha_n \right) }{\beta_n - \alpha_n } \\
&= \frac{ - \frac{1}{4 \pi^2 (n+1/4)^2 } }{\frac{1}{2 \pi n} - \frac{1}{2\pi (n + 1/4)} } \\
&= - \frac{ 1}{2 \pi } \frac{ \frac{1}{(n+1/4)^2} }{ \frac{1}{n} - \frac{1}{n+1/4} } \\
&= - \frac{ 1}{2 \pi } \frac{ \frac{1}{n+1/4} }{ \frac{1/4}{ n } } \\
&= - \frac{ 1}{2 \pi }\frac{16 n}{4n + 1} \\
&= - \frac{ 8 }{ \pi } \frac{1}{ 4+1/n }.
\end{align}
$$
So
$$\lim_{n \to \infty} D_n = - \frac{2}{\pi} \neq f^\prime(0).$$ </p>
</blockquote>
<p>Am I right? </p>
<p>Are my proofs correct and rigorous enough for Rudin? If not, then where are the pitfalls? </p>
<p>Is my example appropriate in the situation described by Rudin? If so, is my calculation correct also? </p>
| 0non-cybersec
| Stackexchange |
Is there any identification for common encoding/encryption?. <p>I come across lots of tokens which might be simply encoded and not random, but I don't know any way to identity if they are encoded or random, how to find is there any cheatsheet for the same.</p>
<p>For example, I know about base64 encoding, I can tell just by looking if its base64 or not. Identification for base64 is it would be in pair of 3 so total count of the characters would be in multiple for 3, if for some reason say there are 19 characters then to make it multiple of 3 two "=" (without quotes) would be added at the end.</p>
<p>Like this are you aware about identification of other encoding/encryption?</p>
| 0non-cybersec
| Stackexchange |
How to query a Dynamo DB having a GSI with only hashKeys using DynamoDBMapper. <p>I am very new to <strong>Dynamo DB</strong> and may be this is very trivial question, but i went through the documents of Dynamo DB and stack overflow questions but i couldnt find a single link which tells how to query DDB for <strong>GSI</strong> which has only hash key and there are no range key specified for the same.</p>
<p>I get the exception Illegal query expression: No hash key condition is found in the query. </p>
| 0non-cybersec
| Stackexchange |
Charming.. | 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 |
Selecting 'edge' records in a sequence of data. <p>I have a table of items which run sequentially by date with both a start and end for each entry:</p>
<pre><code>ID | Start | End
----------------------------------------
1 | 2012-08-20 00:00 | 2012-08-20 23:59
2 | 2012-08-21 00:00 | 2012-08-21 23:59
3 | 2012-08-22 00:00 | 2012-08-22 23:59
4 | 2012-08-23 00:00 | 2012-08-23 23:59
5 | 2012-08-24 00:00 | 2012-08-24 23:59
6 | 2012-08-28 00:00 | 2012-08-28 23:59
7 | 2012-08-29 00:00 | 2012-08-29 23:59
8 | 2012-08-30 00:00 | 2012-08-30 23:59
</code></pre>
<p>Note that there are gaps in the data (for a weekend and a bank holiday). Is it possible to write a query which selects the records at the edges of the gaps? What I want is a general purpose way to end up with something like this:</p>
<pre><code>ID | Start | End
----------------------------------------
1 | 2012-08-20 09:00 | 2012-08-20 23:59
2 | 2012-08-21 00:00 | 2012-08-21 23:59
3 | 2012-08-22 00:00 | 2012-08-22 23:59
4 | 2012-08-23 00:00 | 2012-08-23 23:59
5 | 2012-08-24 00:00 | 2012-08-24 17:00
6 | 2012-08-28 09:00 | 2012-08-28 23:59
7 | 2012-08-29 00:00 | 2012-08-29 23:59
8 | 2012-08-30 00:00 | 2012-08-30 17:00
</code></pre>
<p>Note that the start time of 1 and 6 has been updated, as has the end time of 5 and 8.</p>
<p>In the real database I am forced to used this day by day data layout as it has to retain compatibility with a legacy client, so I can't simply insert the ranges directly. Also in the real data I could be dealing with several hundred of these records at a time with multiple gaps.</p>
<p>The target platform is SQL2005/2008, though bonus points for solutions which work in SQL2000 compatibility mode.</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 |
Run X applications as another user under Linux. <p>Visiting the old topic of <em>running X applications as another user under Linux</em>, as the solution had always been "<em>to use <code>gksu</code></em>" to me, but today, when I need it and tried it, it doesn't work. </p>
<p>Following <a href="https://www.lifewire.com/what-is-gksu-2189402" rel="nofollow noreferrer">What Is gksu And Why Would You Use It</a>, I tried, </p>
<pre><code>gksu -u otherusername xterm
</code></pre>
<p>After passing the dialog that asks for the password, I got:</p>
<pre><code>$ gksu -u otherusername xterm
xterm: Xt error: Can't open display: :2
</code></pre>
<p>I.e., it doesn't work for me. So, </p>
<p>How to run X applications as another user under Linux? Thx. </p>
<p>PS, this is Ubuntu 17.04:</p>
<pre><code>$ lsb_release -a
No LSB modules are available.
Distributor ID: Ubuntu
Description: Ubuntu 17.04
Release: 17.04
Codename: zesty
</code></pre>
| 0non-cybersec
| Stackexchange |
Cytus, one of the best music games out there was just updated!. | 0non-cybersec
| Reddit |
Thinking Ahead.... | 0non-cybersec
| Reddit |
Technical meaning of two alike combinatorial problems. <p>I am confused in how to interpret two alike combinatorial problems, because to me they both look the same. These are the problems:</p>
<ol>
<li>How many ways are there to put $24$ distinguishable flags on $18$ distinguishable flagpoles?</li>
</ol>
<p>The answer to this problem is $\binom{18+24-1}{24}24!$, so first is to arrange the flags as if they were indistinguishable and the arrange them in order.</p>
<ol start="2">
<li>How many ways are there to put 8 distinguishable balls into 10 distinguishable boxes?</li>
</ol>
<p>The answer to this question is $10^8$.</p>
<p>If the first problem I consider it as distinguishable balls in distinguishable boxes, both problems should be the same, what's the catch then?</p>
| 0non-cybersec
| Stackexchange |
The Grothendieck-Serre Correspondence : Obstructions to the construction of the cyclic group of order 8 as a Galois group?. <p>I am reading this note called as <a href="https://webusers.imj-prg.fr/%7Eleila.schneps/corr.pdf" rel="nofollow noreferrer">The Grothendieck-Serre Correspondence by Leila Schneps</a> where this quote occurs:</p>
<blockquote>
<p>the author still recalls Serre’s unexpected reaction of spontaneous delight upon being shown a very modest lemma on obstructions to the construction of the cyclic group of order 8 as a Galois group, simply because he had never spotted it himself.</p>
</blockquote>
<p>I am quite confused. I read the statement as:</p>
<blockquote>
<p>obstruction to the [construction of cyclic group of order 8] as a Galois group</p>
</blockquote>
<p>We can create a cyclic group of order 8: <span class="math-container">$(\mathbb Z/8\mathbb Z, +, 0)$</span>. So I'm not sure what the "galois group obstruction is". Is the above quote to be read as:</p>
<blockquote>
<p>obstruction to the [construction of cyclic group of order 8 as a Galois group] ?</p>
</blockquote>
<p>I am still confused, because I thought the field <span class="math-container">$\mathbb Q(\zeta_8)/\mathbb Q$</span> has galois group <span class="math-container">$\mathbb Z/8\mathbb Z$</span> (where <span class="math-container">$\zeta_8$</span> is the <span class="math-container">$8$</span>-th root of unity), from Kummer theory?</p>
<p>So what obstruction is the quote referring to which delighted Serre?</p>
| 0non-cybersec
| Stackexchange |
Rotation rates and tilts of planets. | 0non-cybersec
| Reddit |
Will we be able to install, update and remove snap packages in the Ubuntu Softwares?. <p>The only way I know to install, remove and update snap packages is to input commands in the terminal, will I be able to do it in Ubuntu Softwares or some other GUI way?</p>
<p>If we can do that, how do we know whether it is snap packages or not?</p>
| 0non-cybersec
| Stackexchange |
Implement intersection products. <p>I am doing a counting problem, and it comes to compute intersection products ( Chow ring ) on some varieties. Is there any computer algebra that deals with this?</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 |
The bee, thurhain_doodle, marker drawing, 2019. | 0non-cybersec
| Reddit |
How to automatically rename music files?. <p>I have around 700 audio songs in .m4a format which I downloaded from YouTube, but they are named as videoplayback.m4a. Is there any way they can be given proper titles automatically? </p>
<p>All are famous songs with millions of views on YouTube. I was thinking of some music recognition site or software which can be automated to do this task.</p>
| 0non-cybersec
| Stackexchange |
USB ports not working after restoring back. <p>After restoring this computer with a backup it appears that 2 of the 5 usb ports no longer work. I'm thinking that there is something corrupted on the backup disk causing the failure. When I plug in a usb keyboard to these ports the system will not recognize the new hardware on the 2 bad ports but does recognize new hardware on the remaining three ports. All the ports were working prior to the restoration.</p>
| 0non-cybersec
| Stackexchange |
Traffic.. | 0non-cybersec
| Reddit |
Different Redis instances for different microservices?. <p>Say that we have Kubernetes running with <em>n</em> pods, each being a separate microservice. Now, some of those services need a caching layer. When we take into consideration: </p>
<ul>
<li>the fact that Redis was tested to hundred of millions of keys and that it was performant</li>
<li>the fact that you can tag <code>keys</code> by <code>namespaces</code> (which would be microservice names in this instance)</li>
<li>the fact that the data would have a relatively short expiration time (we'd hardly hit the number of 10M+ keys at a given moment)</li>
<li>the fact that we have a single (although replicated) database from which data is fetched. PS: I know this goes against reactive principles. (Thanks to <a href="https://softwareengineering.stackexchange.com/users/177980/ewan">Ewan</a> for reminding me to add this)</li>
</ul>
<p>Would we spin up separate Redis pods, each corresponding to a single service or would a single Redis master pod with a separate pod for slave containers be a better architectural choice? Would having separate Redis pods be an antipattern of some sort?</p>
<p>A diagram to help you visualize the dilemma: </p>
<p><a href="https://i.stack.imgur.com/jl9OV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jl9OV.png" alt="enter image description here"></a></p>
| 0non-cybersec
| Stackexchange |
Someone paid for groceries today, and I couldn't stop crying.. I don't know if this the place to say it, but I need to tell someone this; because I haven't cried happy tears like that in a long time.
Things have been rough lately. Like, grocery shopping at the dollar store rough. Borrowing money from my boss to pay rent rough.
We were shopping for some little things today; burger buns, potatoes and some cheese. The last behind us had a grocery basket of food and told us to go ahead of her, because we only had a few things. After we were all rung up, she said she would pay for our items with hers.
I started to choke up, so I thanked her and we left. I don't know why it hit so hard but I sincerely was on the verge of sobbing by the time we got to my car.
But I went right back after to thank her properly. I was already in tears, and the checkout lady was still ringing her up. I asked her if I could hug her. She said yes and actually really held me. I was crying and telling her how much her gesture meant to me, and that things had been so rough.
I know there were lines of people, and I got quite a few looks. I probably looked ridiculous. Some kid hugging a stranger and sobbing her eyes out in a crowded grocery store, and thanking her. But I didn't care. I just wanted her to know how much it meant, and how badly I needed a boost like that. A reminder that there is still good, even when things feel the worst.
I promise that once I am back on my feet, I will return the favor as best as I can. I will do what she did. Hopefully, I will help someone in need. Or maybe inspire them to do the same.
Just wanted to share. This sweet old lady made my day. Made my week. Made me so happy, and emotional; when I have been so stressed, and scared.
It made me smile more than I have in what feels like a long time. Even though I was almost sobbing the whole time, haha. | 0non-cybersec
| Reddit |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.