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|---|---|---|
You Love to Fail - The Magnetic Fields. | 0non-cybersec
| Reddit |
ITAP of the Osaka castle in the evening. | 0non-cybersec
| Reddit |
Pack Chrome extension on server with only command-line interface. <p>Is it possible to pack chrome extension with key (*.pem) on server only with CLI (Ubuntu-server)? </p>
| 0non-cybersec
| Stackexchange |
TIFU and almost drowned. Today I fucked up by almost drowning
This happened yesterday, July 23rd.
This was easily the closest I’ve ever been to death.
My friends and I were on our way back from the local bike park, about an hours drive from where we all live. We decided, since it was a warmer day, that it would be a good idea to go cliff jumping at a nearby lake on the way home. Especially since it’s literally on the drive home, it seemed like the best way to cool off, rinse the dust off, and end the awesome day of biking we had just had.
When we get to the cliff, the winds are HIGH. I’ve never seen this lake this crazy. Waves were whitecapping, and the waves looked pretty sizeable. This lake is usually as calm as glass.
What all of us didn’t take into account is the height of the cliff. The cliff is probably ~35 feet above the lake. So what we interpreted as 4-5ft waves were closer to 8-10 feet in height.
Now, we’ve all jumped here before. Many times. This is usually *the* place to cliff jump. It’s easy to get up the side, by climbing up a rock face and there’s a bay just around the corner, maybe a 5 minute swim (if that) in calm weather. Our plan yesterday was to jump, and very quickly swim back to the bay, which we had already noticed, was very calm and easy to swim in. We are all good swimmers, and we thought that the waves being “4-5 feet tall” that this shouldn’t be an issue at all.
After about 10 minutes of waiting for the waters to calm down a little at the top of the cliff, we all jumped. I was last into the water out of 3 guys. My friends were already hollering at me “YEAH! The water’s great!”.
I must not have braced myself correctly, because I instantly got easily half a lung full of water when I jumped. Immediately not off to a good start. I coughed it out pretty quickly, but already this had taken some energy out of me.
After yelling out at my friends how nice the water was, about 10 seconds later I realized that we were in some serious shit. The waves were FAR bigger than we had each anticipated, and I immediately started swimming towards the bay. Maybe another thirty seconds later, I realized that I was hardly moving, and that I was essentially not moving anywhere.
I looked back at my friends, me being the closest to the bay at this point, by about 10 feet. I will never forget the sheer look of terror on my nearest friend’s face. Immediately I knew we were in some serious shit.
I turned around and immediately thought to myself: “Okay, stay calm, focus on not taking on water and timing your swimming with the waves”
This was a good idea, I just didn’t execute it well. I took on a few (I honestly can’t remember) more big gulps of water, barely able to clear my lungs before them filling again.
At this point I yelled straight up so my friends could hear me “ABORT THE BAY, CLIMB THE ROCKS” and I could hear both of them yell back in agreement.
As you could imagine, the rock face of the cliff was being absolutely thrashed by these waves. My plan was to time my swimming with the waves, latch onto one of the rocks, and get a firm handhold, stay close to the rock, and let one wave hit me, and then keep climbing. I had done something very similar when I was ocean kayaking a few years prior.
So, while fighting to keep water out of my lungs, I latch onto one of these rocks, straddle it with both of my thighs, press my chest up against it, and hack up some lake water as I feel the water receding down my legs and back out.
At this point, a giant twelve foot wave hits me, the impact itself wasn’t the problem, but it pulled me back out with it with a force I have never felt before. I was off the rock in a heartbeat and instantly down to the lake floor, roughly ten feet down.
I put both of my hands above my head, slightly spread out, to try to feel for any objects above me, and push off from the bottom, away from the cliff.
I’m now back out to the lake, once again trying to time my swimming with these giant waves. This is where I start thinking to myself: “Is this how I die? Am I going to drown today?”
Now I’m exhausted, gasping for air, barely able to stay afloat. I can’t swim in these waves much longer. I look to my right and see one of my friends, about fifteen feet over, up on one of the rocks, just as a wave hits the rock, showering him in water as he disappears from my view.
**Focus** Look at the rock, time it, touch the rock, and climb as fast as you can.
I swim forward with everything I have and latch onto the same spot I had before, only this time I waste no time, I’m 6 feet up this rock in what feels like half a second. I feel the next wave lick my feet and lower calves.
My second friend is now less than two feet to my right, he pats me on the back and asks “You good?” In between my hacking up water I nod and he continues climbing.
After about 15 seconds I climb up another 10 feet to a flat ledge, already trying to peer over the rock face to see our third friend. He’s up, about 10 feet out of the water, bent over, coughing up water. I Yell “Hey! You good?” He doesn’t lift his head, just raises his left hand and gives me a thumbs up.
We climb another 15 feet or so, and we’re at the top. All of us flop down on the warm rocks, soaking wet, shaking, and bleeding as the sensation of pain comes back into focus. All of us have scraped up shins, I must’ve stepped on a sharp rock, because my foot is red, starting to bruise, and starting to swell. I’ve never been happier to be injured in my life. It’s way better than drowning in that water. Both of the other guys have minor scrapes and bruises as well.
I fully realize that this was an incredibly stupid decision, and I’m extremely thankful to be alive. Today has been weird. I feel incredibly lucky that I survived and I cannot stop replaying the situation in my head.
TL;DR Misjudged how high the waves were. Jumped off cliff into raging water, almost didn’t make it out.
| 0non-cybersec
| Reddit |
I really like boosts but I’m not getting mugged off like this.... | 0non-cybersec
| Reddit |
How to disable grammatical correctness check on input type text fields. <p>Sometime certain words typed into an input type text box, is shown with a red underline on the browser indicating that the word could be grammatically incorrect.</p>
<p>How do I prevent this from showing up?</p>
| 0non-cybersec
| Stackexchange |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
How do I host .cur image cursor files?. <p>I'm using a <code>.cur</code> file as an image cursor. I uploaded it to the only image host that I could find that accepts <code>.cur</code> files, but lately it doesn't properly load the cursor anymore.</p>
<p>My webhost now doesn't accept <code>.cur</code> files. I converted it to <code>.png</code> but those won't show in IE</p>
<p>Basically I'm looking for a way to host/upload a <code>.cur</code> file.</p>
| 0non-cybersec
| Stackexchange |
Look at this cake. | 0non-cybersec
| Reddit |
sscanf and %i / radix detection of large unsigned integers. <p>I'm working with files containing large numbers (disk offsets) and have run into a problem using <code>scanf</code> for <code>%lli</code> - the numbers are of the format <code>0x</code>... but <code>scanf</code> doesn't read all the bits.</p>
<p>Here's example code (I tested on <a href="https://www.onlinegdb.com/online_c_compiler" rel="nofollow noreferrer">https://www.onlinegdb.com/online_c_compiler</a>).</p>
<pre><code>#include <stdio.h>
#include <stdlib.h>
int main()
{
char* str = malloc(30);
unsigned long long l;
sprintf(str, "0x%llx\n", 0xffffffffffffffffULL);
printf("%s\n", str);
sscanf(str, "%lli", &l);
printf("%llx\n", l);
return 0;
}
</code></pre>
<p>It outputs:</p>
<pre><code>0xffffffffffffffff
7fffffffffffffff
</code></pre>
<p>whereas I expect both numbers to be equal.</p>
<p>If I use the <code>%llx</code> specifier, it works properly, but I don't want to limit myself to hex-formatted numbers.</p>
| 0non-cybersec
| Stackexchange |
Is there a GUI scale mod for 1.13.2 that works on servers?. My computer is only letting me go up to 2x gui instead of the 3x I want so I'm looking for a mod to do it. | 0non-cybersec
| Reddit |
T.V. Series First World Problem. | 0non-cybersec
| Reddit |
Decrypting Encryption. | 1cybersec
| Reddit |
US government shutdown now official, "Agencies should now execute plans for an orderly shutdown due to the absence of appropriations.". | 0non-cybersec
| Reddit |
*Might* an unsigned char be equal to EOF?. <p>When using <code>fgetc</code> to read the next character of a stream, you usually check that the end-of-file was not attained by</p>
<pre><code>if ((c = fgetc (stream)) != EOF)
</code></pre>
<p>where <code>c</code> is of <code>int</code> type. Then, either the end-of-file has been attained and the condition will fail, or <code>c</code> shall be an <code>unsigned</code> char converted to <code>int</code>, which is expected to be different from <code>EOF</code> —for <code>EOF</code> is ensured to be negative. Fine... apparently.</p>
<p>But there is a small problem... Usually the <code>char</code> type has no more than 8 bits, while <code>int</code> must have at least 16 bits, so every <code>unsigned char</code> will be representable as an <code>int</code>. Nevertheless, in the case <code>char</code> would have 16 or 32 bits (I know, this is never the case in practice...), there is no reason why one could not have <code>sizeof(int) == 1</code>, so that it would be (theoretically!) possible that <code>fgetc (stream)</code> returns <code>EOF</code> (or another negative value) but that end-of-file has not been attained...</p>
<p>Am I mistaken? Is it something in the C standard that prevents <code>fgetc</code> to return <code>EOF</code> if end-of-file has not been attained? (If yes, I could not find it!). Or is the <code>if ((c = fgetc (stream)) != EOF)</code> syntax not fully portable?...</p>
<p><strong><em>EDIT: Indeed, this was a duplicate of Question #3860943. I did not find that question at first search. Thank for your help! :-)</em></strong></p>
| 0non-cybersec
| Stackexchange |
Sometimes, in surgery, the anesthesia doesn't work correctly so that the patient is paralyzed but still can feel everything and thus is in massive pain. Can doctors measure this by a sudden increasal in heart rate or something like that?. | 0non-cybersec
| Reddit |
Whooaaa. | 0non-cybersec
| Reddit |
Convert animated GIF to animated SVG and to get the code of SVG?. <p>I have animated gif.However I would need to export to SVG format.Is it possible to convert an animated GIF to an animated SVG-image?
I want to export my gif as an Animated SVG so that I can use it on my website and Mobile Apps.</p>
| 0non-cybersec
| Stackexchange |
Verify the monotony of f without the use of derivative. <p>$f:\Bbb R \rightarrow (2,+\infty)$</p>
<p>and it is known that $$f^3(x)-12f(x)=3x-15$$ I need to verify f's monotony but without the use of the derivative so </p>
<p>Let there be a second function $$H(x)=x^3-12x \qquad \forall x \epsilon (2,+\infty)$$ so we got this $$H (f (x))=3x-15$$$$\forall x_1,x_2 \epsilon \Bbb R : x_1 \lt x_2 \Rightarrow 3x_1-15 \lt 3x_2-15 \Rightarrow H (f (x_1)) \lt H (f (x_2))$$ so I need to verify the monotony of $H $ but I cant do it without derivative.</p>
| 0non-cybersec
| Stackexchange |
The roommate didn't think this was as funny as I did.. | 0non-cybersec
| Reddit |
Make a deal with the universe. | 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 |
Explicit homeomorphism between quotient square and torus. <p>Let <span class="math-container">$X = [0,1] \times [0,1]$</span>, and let consider the quotient topology <span class="math-container">$X^* := X / ((x,0) \sim (x,1), (0,y) \sim (1,y))$</span>. Given <span class="math-container">$r_0 > h > 0$</span>, we define explicitly the torus as:
<span class="math-container">$$
Y_{h,r_0} = \left\{(x,y,z) : z^2 = h^2 - \left(r_0 - \sqrt{x^2 + y^2}\right)^2\right\}
$$</span>
I want to construct an explicit homeomorphism <span class="math-container">$f : X^* \to Y_{h,r_0}$</span>.</p>
<hr>
<p>I believe I've managed to come up with the homeomorphism. We express points in <span class="math-container">$Y_{h,r_0}$</span> in terms of cylindrical coordinates, and define <span class="math-container">$f$</span> as follows:
<span class="math-container">$$
f(x,y) = (r_0 - h\cos(2\pi y), 2\pi x, h\sin(2\pi y))
$$</span>
One can check that this is well-defined up to quotient topology, and is clearly continuous as it is continuous coordinate-wise. I've also constructed the inverse map <span class="math-container">$g : Y_{h,r_0} \to X^*$</span>:
<span class="math-container">$$
g(r,\theta,z) = \left(\frac{\theta}{2\pi}, \frac{1}{2\pi}\arg(r_0 - r + iz)\right)
$$</span>
One can also check that this well defined up to quotient topology, and it is both the left and the right inverse of <span class="math-container">$f$</span>. However, I'm struggling to prove that it is continuous (I'm not entirely sure if it's even continuous). I believe the second argument is indeed continuous, but it seems complicated. I'm also not sure if it's possible to define <span class="math-container">$g$</span> differently such that no complex numbers are involved, and the function will still not be <em>too</em> complicated.</p>
<p>Any help is appreciated.</p>
| 0non-cybersec
| Stackexchange |
Amazing amateur video of the oil spill in the Gulf of Mexico. It’s worse than you think…. | 0non-cybersec
| Reddit |
Today is sports final Sunday. | 0non-cybersec
| Reddit |
What's going on with President Trump, Fox News, and Judge Jeanine Pirro?. I stumbled upon this tweet today from the President:
[https://twitter.com/realDonaldTrump/status/1107269978678611969](https://twitter.com/realDonaldTrump/status/1107269978678611969)
I'm shocked to see some sort of feud between the station and the President, but I have no clue why it happened or why Pirro has seemingly been removed from Fox News. Anyone care to clarify? | 0non-cybersec
| Reddit |
Studies show "not jokes" are coming back and are likely to be funnier than ever.. Not. | 0non-cybersec
| Reddit |
Find $\lim_\limits{x\to -2}{f(x)}$. <p>Let $f:\mathbb{R}\mapsto{\mathbb{R}}$ be an odd function such that: $$\lim_{x\to 2}{(f(x)-3\cdot x+x^2)}=5$$
Find $\lim_\limits{x\to -2}{f(x)}$, if it exists. (so also prove its existence)</p>
| 0non-cybersec
| Stackexchange |
Nodes with multiple inputs aligned at predefined side. <p>I am currently trying to create the following plot (in nice) using tikz:</p>
<p><a href="https://i.stack.imgur.com/Bcyll.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bcyll.png" alt="enter image description here"></a></p>
<p>This is my first attempt, which is not working as I want:</p>
<pre><code>\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning}
\begin{document}
\tikzstyle{block} = [draw, rectangle]
\begin{tikzpicture}[auto, node distance=2cm]
\node [block] (A) {A};
\node [block, below right=of A] (B) {B};
\node [block, right= of B] (C) {C};
\draw [draw,->] (A)-|(B);
\draw [draw,->] (B)--(C);
\draw [draw,->] (C)--(B);
\end{tikzpicture}
\end{document}
</code></pre>
<p>My first problem is, that I don't know how I can tell tikz, for example for the C->B path, to not use the shortest path, but connect C with B "from the left side". Also, the arrows seem to always connect towards the middle of the blocks, how can I specify a position above or below the middle (as for the B or C blocks)?</p>
<p>Finally, how can I connect from a line directly to a block (the third input of B) with marking the line there with a black dot?</p>
<p>I have looked at some tutorials for node placement, but the above topics were not addressed unfortunatly... </p>
| 0non-cybersec
| Stackexchange |
In Laravel 5.7 React app, getting the error: 'classProperties' isn't currently enabled. <p>I'm very new to <b>React.js</b>. I installed <b>Laravel 5.7</b> and swapped Vue.js scaffolding with React by running this command: </p>
<pre><code>php artisan preset react
</code></pre>
<p>Now the problem is, I cannot assign anything to the state inside a component. </p>
<p>For example if I do the following inside the component: </p>
<pre><code>state = { foo: false };
</code></pre>
<p>I got the error:</p>
<pre><code>ERROR in ./resources/js/components/Root.js
Module build failed (from ./node_modules/babel-loader/lib/index.js):
SyntaxError: D:\xampp\htdocs\smart-school-v0.1\resources\js\components\Root.js: Support for the experimental syntax 'classProperties' isn't currently enabled (8:11):
</code></pre>
<p>I installed: </p>
<pre><code>@babel/plugin-proposal-class-properties
</code></pre>
<p>and updated <b>.babelrc</b> (Babel configuration file) like this:</p>
<pre><code>{
"presets": [
"@babel/preset-env",
"@babel/preset-react"
],
"plugins": [
"@babel/plugin-proposal-class-properties"
]
}
</code></pre>
<p>I followed <a href="https://github.com/babel/babel/issues/8577#issuecomment-439579410" rel="noreferrer" title="@babel/plugin-proposal-class-properties does not work">this</a> but no luck.</p>
<p><b>Package.json</b></p>
<pre><code>{
"private": true,
"scripts": {
"dev": "npm run development",
"development": "cross-env NODE_ENV=development node_modules/webpack/bin/webpack.js --progress --hide-modules --config=node_modules/laravel-mix/setup/webpack.config.js",
"watch": "npm run development -- --watch",
"watch-poll": "npm run watch -- --watch-poll",
"hot": "cross-env NODE_ENV=development node_modules/webpack-dev-server/bin/webpack-dev-server.js --inline --hot --config=node_modules/laravel-mix/setup/webpack.config.js",
"prod": "npm run production",
"production": "cross-env NODE_ENV=production node_modules/webpack/bin/webpack.js --no-progress --hide-modules --config=node_modules/laravel-mix/setup/webpack.config.js"
},
"devDependencies": {
"@babel/plugin-proposal-class-properties": "^7.2.3",
"@babel/preset-env": "^7.2.3",
"@babel/preset-react": "^7.0.0",
"axios": "^0.18",
"babel-preset-react": "^6.23.0",
"bootstrap": "^4.0.0",
"cross-env": "^5.1",
"jquery": "^3.2",
"laravel-mix": "^4.0.13",
"lodash": "^4.17.5",
"popper.js": "^1.12",
"react": "^16.2.0",
"react-dom": "^16.2.0",
"resolve-url-loader": "^2.3.1",
"sass": "^1.15.3",
"sass-loader": "^7.1.0"
},
"dependencies": {
"react-router-dom": "^4.3.1"
}
}
</code></pre>
<p><b>.babelrc</b></p>
<pre><code> {
"presets": [
"@babel/preset-env",
"@babel/preset-react"
],
"plugins": [
"@babel/plugin-proposal-class-properties"
]
}
</code></pre>
<p><b>Root.js Component</b></p>
<pre><code>import React, { Component } from 'react';
import ReactDOM from 'react-dom';
export default class Root extends Component {
state = { foo: false };
render() {
return (
<p>Loaded</p>
);
}
}
if (document.getElementById('app')) {
ReactDOM.render(<Root />, document.getElementById('app'));
}
</code></pre>
<p>It is not working as expected and I'm keep getting this error:</p>
<pre><code>Syntax Error: SyntaxError: D:\xampp\htdocs\smart-school-v0.1\resources\js\components\Root.js: Support for the experimental syntax 'classProperties' isn't currently enabled (19:11):
17 |
18 |
> 19 | state = { foo: false };
| ^
20 |
21 | render() {
22 |
Add @babel/plugin-proposal-class-properties (https://git.io/vb4SL) to the 'plugins' section of your Babel config to enable transformation.
@ ./resources/js/app.js 14:0-28
@ multi ./resources/js/app.js ./resources/sass/app.scss
Asset Size Chunks
Chunk Names
/css/app.css 0 bytes /js/app [emitted] /js/app
/js/app.js 593 KiB /js/app [emitted] /js/app
ERROR in ./resources/js/components/Root.js
Module build failed (from ./node_modules/babel-loader/lib/index.js):
SyntaxError: D:\xampp\htdocs\smart-school-v0.1\resources\js\components\Root.js: Support for the experimental syntax 'classProperties' isn't currently enabled (19:11):
</code></pre>
<p>Please help me out. </p>
<p>Thanks in advance.</p>
| 0non-cybersec
| Stackexchange |
Farmavizyon Pharmacy Fair. | 0non-cybersec
| Reddit |
Ivanka Trump has an army of White House employees scrambling to address her every need. | 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 |
Powershell - add variables inside a json string. <p>I have the following json code in my powershell script.
I set the $variable to 1111111111</p>
<pre><code>$jsonfile = '{"Version": "2012-10-17","Statement": {"Effect": "Allow","Action": "sts:AssumeRole","Resource": "arn:aws:iam::$variable:role/xxxxxx"}}'
</code></pre>
<p>The output gives ....arn:aws:iam::<strong>$variable</strong>:role/xxxxxx..... instead of ....arn:aws:iam::<strong>1111111111</strong>:role/xxxxxx</p>
<p>The problem is that I must use the single quote for the json string otherwise I will get an error. If I use single quote I wont be able to put the variables inside the string. How do I workaround this problem? </p>
| 0non-cybersec
| Stackexchange |
Android 6 bluetooth. <p>I upgraded to Android 6 and my applications who use Bluetooth doesn't work with this new API version. It's the same problem with application on Play Store: Bluetooth spp tools pro (good application to view if bluetooth works) which doesn't discovery of devices.</p>
<p>The problem seems to be in Bluetooth discovery:</p>
<pre><code>BluetoothAdapter mBluetoothAdapter = BluetoothAdapter.getDefaultAdapter();
mBluetoothAdapter.startDiscovery()
Log.i("BLUETOOTH", String.valueOf(mBluetoothAdapter.isDiscovering())); // Return false
</code></pre>
<p>My applications work well with Android 4/5 and I followed : <a href="http://developer.android.com/guide/topics/connectivity/bluetooth.html">http://developer.android.com/guide/topics/connectivity/bluetooth.html</a></p>
| 0non-cybersec
| Stackexchange |
Regulated function-higher dimensions. <blockquote>
<p>Let <span class="math-container">$\mathbb{X}$</span> be a Banach space and let the function <span class="math-container">$f:[0,1]^2 \rightarrow \mathbb{X}$</span> be continuous.</p>
<p>(a) Show that <span class="math-container">$\forall\ \epsilon>0\ \exists\ \delta>0$</span> s.t. <span class="math-container">$\|f(x)-f(y)\|<\epsilon$</span> for every <span class="math-container">$x,y \in [0,1]^2 $</span> with <span class="math-container">$\|x-y\|_{\infty}<\delta$</span>.</p>
</blockquote>
<p>Suppose <span class="math-container">$f$</span> is continuous but not uniformly continuous. Then <span class="math-container">$\exists\ \epsilon>0$</span> s.t. <span class="math-container">$\forall\ \delta>0$</span>, <span class="math-container">$\exists\ x,y \in [0,1]^2$</span> s.t. <span class="math-container">$\|x-y\|_{\infty}<\delta$</span> but <span class="math-container">$\|f(x)-f(y)\| \geq \epsilon$</span>. Take <span class="math-container">$(f_n)=\frac{1}{n} \to 0$</span> as <span class="math-container">$n \to \infty$</span>. Then take two sequences and use Bolzano theorem? Is this right? I will reach some sort of contradiction.</p>
<blockquote>
<p>(b) Show that <span class="math-container">$\forall\ \epsilon>0\ \exists\ g:[0,1]^2 \rightarrow \mathbb{X}$</span>
s.t. <span class="math-container">$\|f-g\|_{\infty} \leq \epsilon$</span> and <span class="math-container">$g$</span> is a 2-dimensional step function. (For some reals, the function <span class="math-container">$g$</span> is constant on each open rectangle)</p>
</blockquote>
<p>I have no idea about this bit.</p>
<blockquote>
<p>(c) Use part (b) to show that the order of integration does not matter.</p>
</blockquote>
| 0non-cybersec
| Stackexchange |
How to remove Getstream.fun. | 1cybersec
| Reddit |
How do I use string.tr to replace double quotes with single ones in Ruby?. <p>How do I use string.tr to replace double quotes with single ones in Ruby?</p>
| 0non-cybersec
| Stackexchange |
Making a np.einsum faster when inputs are many identical arrays? (Or any other faster method). <p>I have a piece of code of type:</p>
<pre><code>nnt = np.real(np.einsum('xa,xb,yc,yd,abcde->exy',evec,evec,evec,evec,quartic))
</code></pre>
<p>where <code>evec</code> is (say) an L x L <code>np.float32</code> array, and <code>quartic</code> is a L x L x L x L x T <code>np.complex64</code> array.</p>
<p>I found that this routine is rather slow.</p>
<p>I thought that since all the <code>evec</code>'s are identical, there might be a faster way of doing this?</p>
<p>Thanks in advance.</p>
| 0non-cybersec
| Stackexchange |
Indiana Shut Down Its Rural Planned Parenthood Clinics And Got An HIV Outbreak. | 0non-cybersec
| Reddit |
Monotone Convergence theorem for decreasing sequence. <p>Suppose $f_n: X\to [0, \infty]$ is measurable for $n = 1, 2, 3, \dots$, $f_1 \geqslant f_2 \geqslant f_3 \geqslant \dots \geqslant 0,$ $f_n(x) \to f(x)$ as $n\to \infty$, for every $x\in X$, and $f_1 \in L^1(\mu)$. Prove that then
$$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu= \int \limits_{X}fd\mu$$
and show that this conclusion <em>does not</em> follow if the condition "$f_1 \in L_1 (\mu)$" is omitted.</p>
<p><strong>Proof:</strong> $$f_1 \geqslant f_2 \geqslant f_3 \geqslant \dots \geqslant 0 \implies -f_1 \leqslant -f_2 \leqslant -f_3 \leqslant \dots \leqslant 0 \implies 0\leqslant f_1-f_2\leqslant f_1-f_3\leqslant \dots\leqslant f_1.$$ In other words, sequence $g_n=f_1-f_n$ is increasing & measurable and $g_n(x)\to f_1(x)-f(x)$ for each $x\in X$ and we can use Monotone convergence theorem: $$\lim \limits_{n\to \infty}\int \limits_{X}g_nd\mu=\lim \limits_{n\to \infty}\int \limits_{X}(f_1-f_n)d\mu=\int \limits_{X}(f_1-f)d\mu.$$ If $f_1\in L^1(\mu)$ then $f_n\in L^1(\mu)$ for each $n\in \mathbb{N}$ and $f\in L^1(\mu)$ then: $$\int \limits_{X}f_1d\mu-\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}f_1d\mu-\int \limits_{X}fd\mu$$ since $\int \limits_{X}f_1d\mu$ is finite we can subtract it and we get what we need!</p>
<p>$\color{red}{Wrong \quad Counterexample:}$ Condition $f_1\in L^1(\mu)$ is crucial! Suppose the we omit this condition. Let $X=\mathbb{N}, \mathfrak{M}=2^{\mathbb{N}}$ and $\mu=|\cdot|$ is counting measure on $\mathfrak{M}$. Suppose $f_n(x)=\dfrac{1_{A_n}(x)}{n}$ where $A_n=\{1,2,\dots, n\}$. It's easy to check that $f_n(x)\to 0$ as $n\to \infty$ for $x\in X$. But $\int \limits_{X}f_nd\mu=\frac{1}{n}\mu(A_n)=1.$ So $$1=\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu\neq \int \limits_{X}fd\mu=0$$</p>
<p>Is my proof and its counterexample correct?
Would be very grateful for any suggestions & comments.</p>
<p><strong>EDIT:</strong> Let's consider triple $(X,\mathfrak{M},\mu):=(\mathbb{N},2^{\mathbb{N}},|\cdot|)$, where $|\cdot |$ - counting measure on $2^\mathbb{N}$. Let $A_n=\{n, n+1,\dots\}$.
Suppose that $f_n(x)=\dfrac{1_{A_n}}{n}$. It's easy to see that $f_1\geqslant f_2\geqslant \dots \geqslant f_n\geqslant \dots \geqslant 0$. </p>
<p>Also note that $f_n(x)\to f(x)=0$ as $n\to \infty$ for $x\in X$. Also $\int \limits_{X}f_nd\mu=\dfrac{1}{n}\mu(A_n)=\infty.$ Hence $$\infty=\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu\neq\int \limits_{X}fd\mu=0.$$</p>
<p>Is it true?</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 Supreme Court is Quietly Changing The Status of Religion in American Life. | 0non-cybersec
| Reddit |
Method (alts?) organised a HFC pug, blatantly ninja loot. Even the best of us fall?. EDIT4: Removed the screenshot of an angry message PM i received, as there is no proof that it was from this guy.
Doing a HFC heroic pug with leader who advertised himself as World 6. He and some of others in raid are from Method on Tarren Mill (i armoried, legit from the guild).
The run was advertised as nothing reserved - but after Gorefiend kill, one of the people (his friend, seen on screenshot) started spamming in chat. Suspicious, i checked chat logs - turns out i was right. Smack in the middle of the spam, the leader gave the tier piece to his other friend (also seen on screenshot).
Now, stuff like this (unfortunately) happens often, and im not even mad because it wasnt tier piece for my class. But im just feeling so dissapointed, that even top tier guilds have people who behave like literal assholes. There are plenty of runs on LFG who reserve an item they need, but this was just so blatantly low, it made me sad.
Is this what the WoW community is coming down to?
http://imgur.com/N32vlAM
Edit: I should clarify that it wasnt just one person, there were 4-5 method people in the run, who were all obviously okay with this. Screenshot only shows the spammer and the loot-receiver.
Edit2: So far i have been unsuccessful in reaching out to the leadership. The only person i talked with said he was busy & will look into it later. If anyone from leadership wants the names, let me know & i will PM you the names & full details.
Edit3: Sco has responded in this thread, and i let him know the details.
https://www.reddit.com/r/wow/comments/3zpnki/method_alts_organised_a_hfc_pug_blatantly_ninja/cyo7otp | 0non-cybersec
| Reddit |
Log Canonical Divisors. <p>I know the definition of a log canonical divisor and the conditions that a divisor needs to satisfy to be one. But I would like to know why this concept is important. Where is this being applied and what role do these divisors play?</p>
<p>I have just begun reading Miles Reid Undergraduate Geometry and a couple of other basic reading material. So kindly explain in layman terms with some suggestions to additional reading material if possible.</p>
<p>Thank you!!!</p>
| 0non-cybersec
| Stackexchange |
Dirty energy's quiet war on solar panels. | 0non-cybersec
| Reddit |
Don't know if this is trashy, but it definitely looks stupid.. | 0non-cybersec
| Reddit |
Apples response to Google Glass should be called iSight. | 0non-cybersec
| Reddit |
No icons on main screen on VirtualBox. <p>I've tried everything, I've uninstalled and reinstalled, I've tried adding guest additions, nothing is working. </p>
<p>EDIT: I should also add I've been using this guide to getting VB set up: <a href="http://alexmk.hubpages.com/hub/How-To-Install-Ubuntu-Steam-TF2-under-VirtualBox-on-Windows-to-get-Linux-Tux-Promo" rel="nofollow">http://alexmk.hubpages.com/hub/How-To-Install-Ubuntu-Steam-TF2-under-VirtualBox-on-Windows-to-get-Linux-Tux-Promo</a></p>
| 0non-cybersec
| Stackexchange |
ELI5: Why do we get the wierd colourful tingling images when we rub our closed eyes?. | 0non-cybersec
| Reddit |
The relation between measure and outer measure.. <p>I have small question, I have a given in a question that my set <span class="math-container">$E$</span> has a finite measure and because the Lebesgue measure and the outer measure coincide on measurable sets, I can see that my set <span class="math-container">$E$</span> also has a finite outer measure, am I correct?</p>
<p>Thanks! </p>
| 0non-cybersec
| Stackexchange |
C Minishell - Need Help Killing Zombies. Hey guys,
So I'm building a Unix Minishell in C, and have just implemented pipelines, so now I can do things like this:
ps aux | grep dh | grep -v grep | cut -c1-5
However I'm having a Zombie problem, sometimes control returns the the shell when there are still several children out there, some of them zombies, some of them actively running.
At first I thought it was merely zombies, so I made a SIGCHLD handler that increments a global volatile_atomic_t variable called "zombie_count" each time it is called. Then when control returns to the shell, I merely wait zombie_count times, and then prompt again for the next line(or collect the next line if we're executing a shell script). However, I have discovered that this doesn't always work, because what if we get back to the shell, and there are children that are not yet zombies, then zombie count will be less than the number of total children. If this happens, my shell prints a prompt, "% ", and *then* prints the output of the pipe. This only happens with certain commands(usually commands that take a long time).
So, I was wondering what the best way to deal with this is, I'm tired of using global variables, as I'm already using all these:
extern volatile sig_atomic_t sig_int;
extern volatile sig_atomic_t zombie_count;
extern int last_exit_status;
extern int m_argc;
extern int m_shifted_argc;
extern char **m_argv;
extern char **m_shifted_argv;
extern pid_t cpid;
However, I can't think of any other way to do it, besides maybe using a static variable that increments each time I fork, and then waiting that many times, but then that would make my entire SIGCHLD handler and related components entirely useless, and I liked the elegance of the whole thing.
Anyways, I was hoping somebody here knew a nice elegant solution to this problem as it must be something everybody building a shell comes across at some point.
Thanks :)
PS: If you guys want me to post my program, just ask, although it's pretty big. | 0non-cybersec
| Reddit |
How to apply multiple transforms in CSS?. <p>Using CSS, how can I apply more than one <code>transform</code>?</p>
<p>Example: In the following, only the translation is applied, not the rotation.</p>
<pre><code>li:nth-child(2) {
transform: rotate(15deg);
transform: translate(-20px,0px);
}
</code></pre>
| 0non-cybersec
| Stackexchange |
Additive functors and zero homomorphisms. <p>The question:</p>
<p>Let $F:\mbox{Ab} \to \mbox{Ab}$ be an additive functor; if $f$ is a zero homomorphism, then so is $F(f)$; if $A$ is the zero group, then so is $F(A)$.</p>
<p>This boils down to showing that, for an additive functor $F(0)=0$. Obviously this is true, but my category theory is very basic, and I can't quite yet get a handle on what is allowed.</p>
<p>Is it simply just the case that</p>
<p>$$F(f+g)=F(f)+f(g) \implies F(0) = F(f+(-f))=F(f)+F(-f)=F(f)-F(f)=0?$$</p>
| 0non-cybersec
| Stackexchange |
How to do jquery code AFTER page loading?. <blockquote>
<p>If you want an event to work on your
page, you should call it inside the
$(document).ready() function.
Everything inside it will load as soon
as the DOM is loaded and <strong>before</strong>
the page contents are loaded.</p>
</blockquote>
<p>I want to do javascript code only <strong>after</strong> the page contents are loaded how can I do that?</p>
| 0non-cybersec
| Stackexchange |
Apple's "cheap" iPhone 5C will cost $735 in China (x-post from /r/china). | 0non-cybersec
| Reddit |
WPF DataGrid: CommandBinding to a double click instead of using Events. <p>I know how to use the MouseDoubleClick event with my DataGrid to grab the selectedvalue, but how would one go about using command bindings instead? That way my ViewModel can handle the logic.</p>
<p>So far I have the following:</p>
<pre><code><DataGrid Name="TestGrid" Grid.Row="2" Grid.ColumnSpan="2" AutoGenerateColumns="True" MouseDoubleClick="TestGrid_MouseDoubleClick"
ItemsSource="{Binding Registrations}" SelectedValue="{Binding CurrentRegistration}" IsReadOnly="True" AlternationCount="2" GridLinesVisibility="None">
</code></pre>
<p>I want to get rid of MouseDoubleClick and replace it appropriately.</p>
| 0non-cybersec
| Stackexchange |
Disable "Semi-maximize left" in Ubuntu 13.10. <p>As the title says,</p>
<p>How can I disable or re-assign the "semi-maximize left" keyboard shortcut in Ubuntu 13.10?
By default it is bound to <kbd>Ctrl</kbd> + <kbd>Super</kbd> + <kbd>←</kbd></p>
<p>I have tried to change the bindings for the "put left" action with <code>compizconfig-settings-manager</code> as suggested by the previous question <a href="https://askubuntu.com/questions/147105/how-can-i-reassign-the-semi-maximize-keyboard-shortcut">for Ubuntu 12.04</a>. This however no longer works in Ubuntu 13.10.</p>
<p>This question is motivated by the keyboard shortcut <a href="https://askubuntu.com/questions/412046/unable-to-use-intellij-idea-keyboard-shortcuts-on-ubuntu">conflicts</a> between the default <strong>Unity</strong> desktop and <strong>IntelliJ IDEA</strong>.</p>
| 0non-cybersec
| Stackexchange |
Gets what she deserves.. | 0non-cybersec
| Reddit |
MISSMALINI CELEBRITY SITE AWARDS ADMEDIA GATE & ANGLER EXPLOIT KIT DURING THE OSCARS. | 1cybersec
| Reddit |
Punch-Drunk Love: Adam Sandler delivers a surprisingly good performance in this romantic dramedy. Written/Directed by Paul Thomas Anderson.. | 0non-cybersec
| Reddit |
How to call VS Code Editor from terminal / command line. <p>The question says it all.</p>
<p>How can I open VS Code editor from</p>
<ul>
<li>windows cmd</li>
<li>linux and mac terminal</li>
</ul>
<p>e.g. for notepad++ I write</p>
<p><code>> start notepad++ test.txt</code></p>
<p>By the way, the editor is awesome (cross-platform)! Thank you Nadella!</p>
<p>You can download it from <a href="https://code.visualstudio.com/" rel="noreferrer">microsoft</a></p>
| 0non-cybersec
| Stackexchange |
It is known, they cant handle it. | 0non-cybersec
| Reddit |
Red's $15,000 'Red Raven' Camera Kit Available Exclusively at Apple Stores. | 0non-cybersec
| Reddit |
Computation of characteristic polynomial. <p>So we had an assignment in which the following question was asked :</p>
<blockquote>
<p>Let $A ∈ Z^{nxn}$ such that each absolute value of an entry is bounded by $U$. Show that the coefficients of the characteristic polynomial of $A$ have binary encoding length polynomial in $log(U)$ and $n$.</p>
</blockquote>
<p>and here's the solution :</p>
<blockquote>
<p>Without loss of generality we suppose $U ≥ 1$. We denote the coefficient of largest absolute value
of the characteristic polynomial of $A$ by $c(U, n)$. Applying Laplace rule to $|A − λI|$ we obtain the
recurrence $c(U, n) ≤ U · n · c(U, n − 1)$, which suggests the bound $c(U, n) ≤ n!U^n$
. The latter bound
is trivial to prove by induction and it implies that the binary encoding length of $c(U, n)$ is at most
$n log(Un)$.</p>
</blockquote>
<p>The thing is, I really don't see how different the answer would be if the question was to find an upper bound for the determinant's encoding length. Shouldn't it be different for the characteristic polynomial's coefficients ? </p>
<p>Thank you for your answers. </p>
<p>Ps : for the bravest of you, the following question was about finding a way to compute the characteristic polynomial of a $(n·n)$ matrix by using $n+1$ determinants and the Vandermonde matrix. What would be the complexity ? Could I use Hadamar Bound for a matrix of the form $A-t·I_n$ ? If so How ?
I'm not even sure I understood the question that's why I did not post this separately. </p>
| 0non-cybersec
| Stackexchange |
1 slice of apple pie will cost you $2.45 in Jamaica. A slice of apple pie costs $3.75 in Trinidad and the same slice costs $4.45 in Barbados.. And those are the Pie-Rates of the Caribbean. | 0non-cybersec
| Reddit |
Generalized version of L’Hospital’s rule?. <p>I was wondering if the following inequalities are true:$\liminf_{x \to c} \frac{f'(x)}{g'(x)} \leq \liminf_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f'(x)}{g'(x)}$ under some conditions on $f$ and $g$.</p>
<p>Under assumptions for L'Hospital's rule, if $\lim_{x \to c} \frac{f'(x)}{g'(x)}$ exists, then the above inequalities imply the conclusion of L'Hospital's rule. I think the above inequalities are the generalized version of L'Hospital's rule if they are right. Is it true?</p>
<p>Would you give me any comment about it? Thanks in advance.</p>
| 0non-cybersec
| Stackexchange |
Best Use Of Pallet Wood I've Seen. | 0non-cybersec
| Reddit |
Broke my bowl in microwave today. | 0non-cybersec
| Reddit |
Find necessary and sufficent conditions for A to be bijective function. <p>I'm studying functional analysis. I have a problem in this exercise.</p>
<p>In Hilbert space <span class="math-container">$(X,<,>)$</span>, <span class="math-container">$\{e_n\}$</span> is an orthonormal basis, <span class="math-container">$\{\lambda_n\}$</span> is a bouded sequence in <span class="math-container">$\mathbb K$</span></p>
<p><span class="math-container">$A(x)=\displaystyle \sum_{k=1}^{\infty}\lambda_k.<x,e_k>.e_k$</span></p>
<p><span class="math-container">$A_n(x)=\displaystyle \sum_{k=1}^{n}\lambda_k.<x,e_k>.e_k$</span></p>
<p>I proved A is linear and continuous and <span class="math-container">$||A||=\sup_{k \in \mathbb{N}}|\lambda_k|$</span>.</p>
<p>But the problem is:</p>
<blockquote>
<p>Find necessary and sufficent condtions for A to be a bijective function.</p>
</blockquote>
<p>Any help is appreciated.</p>
| 0non-cybersec
| Stackexchange |
While the Nation Protested Gun Violence, Trump Went Golfing and Pence Tweeted About a Movie. | 0non-cybersec
| Reddit |
ADFS stop working. <p>So I tried to secure our domain controllers with a narrowed list of Ciphers and now the Office 365 ADFS is broken for Chrome and Firefox. I need some help trying to figure out how to put it back. I used a GPO to narrow the list and I have unlinked that GPO from the OU and still the problem persists. </p>
<p>Does anyone have any ideas?</p>
<p>Here is the list of Ciphers I narrowed it down to...</p>
<blockquote>
<p>TLS_RSA_WITH_AES_128_CBC_SHA256,TLS_RSA_WITH_AES_256_CBC_SHA256,TLS_ECDHE_RSA_WITH_AES_128_CBC_SHA256_P256,TLS_ECDHE_RSA_WITH_AES_128_CBC_SHA256_P384,TLS_ECDHE_RSA_WITH_AES_128_CBC_SHA256_P521,TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA384_P256,TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA384_P384,TLS_ECDHE_RSA_WITH_AES_256_CBC_SHA384_P521,TLS_ECDHE_ECDSA_WITH_AES_128_CBC_SHA256_P256,TLS_ECDHE_ECDSA_WITH_AES_128_CBC_SHA256_P384,TLS_ECDHE_ECDSA_WITH_AES_128_CBC_SHA256_P521,TLS_ECDHE_ECDSA_WITH_AES_256_CBC_SHA384_P384,TLS_ECDHE_ECDSA_WITH_AES_256_CBC_SHA384_P521,TLS_DHE_DSS_WITH_AES_128_CBC_SHA256,TLS_DHE_DSS_WITH_AES_256_CBC_SHA256,TLS_RSA_WITH_NULL_SHA256,TLS_ECDHE_ECDSA_WITH_AES_128_GCM_SHA256_P256,TLS_ECDHE_ECDSA_WITH_AES_128_GCM_SHA256_P384,TLS_ECDHE_ECDSA_WITH_AES_128_GCM_SHA256_P521,TLS_ECDHE_ECDSA_WITH_AES_256_GCM_SHA384_P384,TLS_ECDHE_ECDSA_WITH_AES_256_GCM_SHA384_P521</p>
</blockquote>
| 0non-cybersec
| Stackexchange |
For a DokuWiki site, how to force new users to have an email address from a specific domain?. <p>I am trying to manage a DokuWiki site designed for school members.
The problem is that I have to check whether the new users are actually members of our school.</p>
<p>Since DokuWiki sends temporary password to the email written by a new user, I think it is the best to use our school's domain (e.g., <code>@example.edu.com</code>) to confirm the identity. </p>
<p>Is there a way to 'force' all the new ids to have a 'fixed' email domain? I wish the "sign up" page to show something like this:</p>
<pre><code>User ID : [_______]
Name : [_______]
Email : [_______]@example.edu.com
[Register Button]
</code></pre>
<p>Or are there other ways to solve my problem?</p>
| 0non-cybersec
| Stackexchange |
[Spoiler] All in all you're just another.... | 0non-cybersec
| Reddit |
Advice on dating for a girl with a job at unsociable hours. So I've been working at a bar/music venue/night club for a year and a half now. I love my job and it's brought me some incredible and unique opportunities (as well as really teaching me a sense of self - which it seems bar work does).
Anyway, I have a pretty cool and specific job due to where I work, a big friend group and interests that keep me busy enough; but since I switched to night time I've felt really lonely. I dated a couple of people and get lots of compliments but I spend so much time around drunk dudes I just feel like I'll never meet anyone.
I'm cool with my own company (and not lacking things to do). I genuinely don't mind being single in the sense that I have good friends, a wicked job and interests. I'm just beginning to feel like I'd like to be in a relationship.
Any guys gotten into relationships with female bartenders or got any advice? Dunno if I really worded this well but whatever.
TLDR; lonely female bartender plz help etc | 0non-cybersec
| Reddit |
Using only movie titles and quotes, where do you live?. | 0non-cybersec
| Reddit |
Eigenvalue distribution of a band matrix. <p>Let <span class="math-container">$\mathbf M_i$</span> be rectangular matrices of dimensions <span class="math-container">$N_{i-1}\times N_i$</span>. We assume that their entries are random, with zero mean and variance <span class="math-container">$\sigma_i^2$</span>.</p>
<p>For some positive integer <span class="math-container">$k$</span>, I define the matrix <span class="math-container">$\mathbf{A}_k$</span>:</p>
<p><span class="math-container">$$\mathbf{A}_k = \left(\begin{array}{cccccc}
0 & \mathbf{M}_1 & 0 & \cdots & 0 & 0\\
\mathbf{M}_1^{\top} & 0 & \mathbf{M}_2 & \cdots & 0 & 0\\
0 & \mathbf{M}_2^{\top} & 0 & \ddots & 0 & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 0 & \mathbf{M}_n\\
0 & 0 & 0 & \cdots & \mathbf{M}_k^{\top} & 0
\end{array}\right)$$</span></p>
<p>Notice that <span class="math-container">$\mathbf{A}$</span> is symmetric. But unfortunately it is not from a rotationally invariant random ensemble, which has complicated things for me.</p>
<p>I have two questions.</p>
<ol>
<li>What is the eigenvalue density of <span class="math-container">$\mathbf{A}_k$</span>?</li>
<li>What happens to the eigenvalue density when I set <span class="math-container">$N_i=c_iN$</span>, with positive constants <span class="math-container">$c_i$</span> and we take the limit <span class="math-container">$N\rightarrow\infty$</span>? I presume one must perform an appropriate re-scaling here to get a meaningful result.</li>
</ol>
<p>Even if the answer to 1. is not tractable, maybe the limiting form 2. can still be obtained.</p>
<p><strong>Update:</strong> Say I have access to the singular value decomposition of <span class="math-container">$\mathbf{M}_i$</span>. Can we say anything about the eigenvalues/eigenvectors of <span class="math-container">$\mathbf{A}_k$</span>?</p>
| 0non-cybersec
| Stackexchange |
df reports several gb delta between total and used, but 0 available space. <p>Filesystem Size Used Avail Use% Mounted on</p>
<p>/dev/sdb 246G 234G 0 100% /backup_data</p>
<p>on the above fs:
as root i can create non-empty files in the above tree, /backup_data.
as the owner of a dir in the above tree, with perms 775 on the dir which i
am attempting to write to,
i can create only empty files. any attempt to write data to a new or existing
file, fails.
it seems as if root has the capability to utilize space which regular users (even with appropriate perms) cannot utilize.</p>
<p>any explanation for this?</p>
<p>tnx
ams</p>
| 0non-cybersec
| Stackexchange |
What is your System Update policy?. The company I work for has had a long history of doing "install and leave it" systems, where the system gets installed, racked, and the only software updates it gets might be security related for the next 6-8 years (because it's working, it's stable, and we can't risk any downtime by breaking it).
All of our new deploys run the current version of our chosen Linux distribution and there have been pushes in recent months to replace older hardware and that has also included (as both an objective and a bi-product) updating to the latest versions of software. Still, I have been coming across systems in our rather large organization that I just feel are unnecessarily outdated, like a couple of Fedora Core 4 systems, or a Debian Etch machine.
That leads back to my question: What is your policy on keeping systems up to date, not just in terms of patches but running current, supported, major OS versions? Do you have tools or systems in place to track deployed major OS versions, or when they last had any form of patching performed? | 0non-cybersec
| Reddit |
Proving the Cauchy-Schwarz integral inequality in a different way. <p>Suppose $\alpha$ is monotonically increasing on $[a,b]$. Also suppose $f,g: [a,b] \rightarrow \mathbf{R}$, and both functions are Riemann-Stieltjes integrable with respect to $\alpha$. That is, $\forall \epsilon > 0$ there is a partition $P_\epsilon$ such that for all upper and lower sums:
\begin{equation*}
U(P_\epsilon,f,\alpha) - L(P_\epsilon,f,\alpha) < \epsilon
\end{equation*}</p>
<p>Use the Cauchy-Schwarz inequality
<a href="https://i.stack.imgur.com/7KGph.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7KGph.png" alt="enter image description here"></a></p>
<p>To prove the following:
\begin{equation*}
\left|\int_a^b f(x)g(x)d\alpha (x)\right|^2 \leq \left(\int_a^b [f(x)]^2 d\alpha(x)\right) \left(\int_a^b [g(x)]^2 d\alpha(x)\right)
\end{equation*}</p>
<p>I've seen this proof using done by looking at $0 \leq \int_a^b (\lambda f(x) + g(x))^2 dx$, and then looking at the resulting quadratic equation. This problem, however, seems to be a more general case. How could I approach this?</p>
| 0non-cybersec
| Stackexchange |
Log Log Integrals. <p>Evaluate the integrals
\begin{align}
I_{1} &= \int_{0}^{1} \ln^{2n}(x) \ \ln\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
and
\begin{align}
I_{2} = \int_{0}^{1} \ln^{2n}(x) \ \ln^{2}\left(\ln\left(\frac{1}{x}\right)\right) dx.
\end{align}
From the resulting values of the integrals above is it possible to evaluate the integral
\begin{align}
I_{3} = \int_{0}^{1} \left( x^{a} + \frac{1}{x^{a}} \right) \ \ln^{p}\left(\ln\left(\frac{1}{x}\right)\right) dx
\end{align}
where $p=1,2$ in a compact form? Please show all work in in the solutions. </p>
| 0non-cybersec
| Stackexchange |
Proxypass for certain folder. <p>Using <code>Apache (Apache/2.4.18 (Ubuntu)</code>, <code>ProxyPass (mod_proxy)</code>, how can I serve content from one host to a folder on another host?</p>
<h3>Example:</h3>
<pre><code>pears.com/folder
oranges.com/folder
</code></pre>
<p>I would like requests made on <strong>oranges.com/folder</strong> to be served the <em>content</em> that is in <strong>pears.com/folder</strong>. oranges.com/folder does not exist.</p>
<p>I tried some of the examples from, <a href="https://httpd.apache.org/docs/2.4/urlmapping.html" rel="nofollow noreferrer">apache.org/docs</a>. <br>
In <strong>orange's</strong> virtualhost config:<br></p>
<pre><code>ProxyHTMLStripComments on
ProxyRequests off
SetOutputFilter proxy-html
ProxyHTMLDoctype XHTML
SSLProxyEngine on
SetEnv no-gzip 1
<Location /folder>
ProxyPass pears.com/folder
ProxyPassReverse pears.com/folder
Order allow,deny
Allow from all
</Location>
</code></pre>
<p>The error I receive is, </p>
<blockquote>
<p>The webpage at <a href="https://pears.com/folder/" rel="nofollow noreferrer">https://pears.com/folder/</a> might be temporarily down or
it may have moved permanently to a new web address.<br>
ERR_CONTENT_DECODING_FAILED</p>
</blockquote>
<hr>
<p><strong>Edit</strong> - The solution was to disable compression on the pears server</p>
| 0non-cybersec
| Stackexchange |
[Social] Guys and girls, let me see your desktops/home screens!. Post a screenshot of your Laptop/PC/Tablet/smartphone/whatever. Please not just your wallpaper. | 0non-cybersec
| Reddit |
Having pdflatex clean up after itself in WinEdt?. <p>How do I get <code>pdflatex</code> (via WinEdt) to remove all the extra output files (e.g., <code>.lof</code>, <code>.blg</code>, etc.) after compiling my <code>.tex</code>/<code>.bib</code>/<code>.etc</code> to a <code>.pdf</code>?</p>
<p>If there is a solution, I suspect it will come via "<em>Options</em>" -> "<em>Execution Mode</em>". I'm just not finding anything though :(</p>
<p>I was told by someone that <code>pdflatex</code> won't do this and that I would need to create a Makefile. Is that true?</p>
| 0non-cybersec
| Stackexchange |
Baseball bat vs water balloon. | 0non-cybersec
| Reddit |
Show context menu from code behind. <p>This might be a simple question, but I've been looking around and can't find the answer. Is there any code to show the context menu on Android from a code, instead of pressing the menu button? E.g. when I touch the screen then it'll call the context menu?</p>
| 0non-cybersec
| Stackexchange |
Configure Windows for an internet connection that's only metered during the day. <p>Can the Windows 8/10 metered connection setting be scheduled to automatically turn on and off depending on the time of day in order to take advantage of an overnight unlimited traffic window from an otherwise metered ISP? Having to turn it on and off manually is tedious and error prone; is there any way to automate doing so?</p>
| 0non-cybersec
| Stackexchange |
Console application not torn down by an unhandled task exception. <p>I'm trying to figure out why my console app is not torn down by an unhandled task exception. All I do is create a task where I immediately throw an exception. Finally I force GC. In the first example I have a handler for the <code>TaskScheduler.UnobservedTaskException</code> event and I can see the exception get handled.</p>
<pre><code> static async Task ThrowsException()
{
Console.WriteLine("Throwing!");
throw new Exception("Test exception");
}
static void Main(string[] args)
{
TaskScheduler.UnobservedTaskException += TaskScheduler_UnobservedTaskException;
ThrowsException();
Console.WriteLine("Collecting garbage.");
GC.Collect();
GC.WaitForPendingFinalizers();
Console.WriteLine("Done collecting garbage.");
Console.ReadKey();
}
static void TaskScheduler_UnobservedTaskException(object sender,
UnobservedTaskExceptionEventArgs e)
{
Console.WriteLine("Unobserved task exception occured in finalizer.");
Console.WriteLine(e.Exception.InnerException.Message);
}
</code></pre>
<p>Output:
</p>
<pre><code>Throwing!
Collecting garbage.
Unobserved task exception occured in finalizer.
Test exception
Done collecting garbage.
</code></pre>
<p>But if I comment out the line <code>TaskScheduler.UnobservedTaskException += TaskScheduler_UnobservedTaskException</code> the program still runs to completion. In this case the output is:
</p>
<pre><code>Throwing!
Collecting garbage.
Done collecting garbage.
</code></pre>
<p>Why doesn't the application crash in this case?</p>
| 0non-cybersec
| Stackexchange |
How to manage threads in a framework. <p>I'm developing a framework with C++, and it contains three layers:</p>
<ul>
<li>Low level functions which do the hard work</li>
<li>An upper layer which uses the low level functions to accomplish tasks (functions at this level generally uses only one lower level function and explicitly accepts the pointer of the function to know which function to call)</li>
<li>Top level classes which utilizes these middle layer functions to accomplish complete tasks.</li>
</ul>
<p>The problem is, I'm introducing concurrency into the framework and need a way to tune this concurrency per function basis. I can design an object to transfer concurrency parameters, but I don't want to pass an object to functions which are not concurrent.</p>
<p>Concurrent functions can be at any level and their concurrency needs to be controlled independently.</p>
<p>Is there a better pattern or method to allow only concurrent functions to access this information?</p>
| 0non-cybersec
| Stackexchange |
$H$ is a normal subgroup (containing an element of infinite order) of the permutation group over positive integers , then is $H$ the whole group?. <p>Let $S(\mathbb N)$ be the set of all bijections on $\mathbb N$ ( the set of positive integers ) endowed with the natural group structure of function composition . If $H$ is a normal subgroup of $S(\mathbb N)$ such that $H$ contains an element of infinite order then is it true that $H=S(\mathbb N)$ ? </p>
| 0non-cybersec
| Stackexchange |
Full page ad in Washington Post today. | 0non-cybersec
| Reddit |
Fantasy Scouting Notebook: 41 observations to help make sure you're ready to draft. | 0non-cybersec
| Reddit |
Curved torsional springs and dampers in TikZ mechanical system drawing. <p>I am trying to draw some curved torsional springs and curved dampers in a mechanical system drawing. There are a number of questions on this site which deal with how to draw spring and damper components (see, for example, <a href="https://tex.stackexchange.com/questions/13933/drawing-mechanical-systems-in-latex">Drawing Mechanical Systems in LaTeX</a>), however, these all deal with springs and dampers which are not curved. </p>
<p>Specifically, I am trying to replicate this diagram:</p>
<p><a href="https://i.stack.imgur.com/TvwYY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TvwYY.png" alt="enter image description here"></a>
Here's what I've got so far:</p>
<p><a href="https://i.stack.imgur.com/gbcig.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gbcig.png" alt="enter image description here"></a></p>
<p><strong>Code</strong></p>
<pre><code>\documentclass[tikz,margin=1cm]{standalone}
\usetikzlibrary{decorations.pathreplacing}
\begin{document}
\begin{tikzpicture}
\tikzstyle{interface}=[postaction={draw,decorate,decoration={border,angle=-30,amplitude=0.1cm,segment length=0.6mm}}]
\tikzstyle{interfaceflipped}=[postaction={draw,decorate,decoration={border,angle=30,amplitude=0.1cm,segment length=0.6mm}}]
\newcommand\beamL{5cm}
\newcommand\beamH{0.1cm}
\newcommand\massL{0.5cm}
\newcommand\massH{0.5cm}
\draw (0,0)--++(0,-\beamH)--++(\beamL,0)--++(0,2*\beamH)--++(-\beamL,0)--cycle;
\draw (0,0)--++(0,-\massH)--++(-\massL,0)--++(0,2*\massH)--++(\massL,0)--cycle;
\draw [xshift=\beamL+\massL] (0,0)--++(0,-\massH)--++(-\massL,0)--++(0,2*\massH)--++(\massL,0)--cycle;
\coordinate (G1) at (5.5,-1); % Coordinates of ground 1
\path (G1) +(150:0.2) coordinate (G1start);
\draw [interface] (G1start)--++(-30:0.4);
\draw plot [smooth, tension=1] coordinates {(\beamL,\beamH) ([yshift = 0.5cm,xshift = -0.08cm]\beamL,\beamH) ([yshift = 0.8cm]\beamL+\massL,\beamH) ([yshift = 1.1cm,xshift = 0.5cm]G1) (G1)};
\coordinate (G2) at (6.1,-1); % Coordinates of ground 2
\path (G2) +(150:0.2) coordinate (G2start);
\draw [interface] (G2start)--++(-30:0.4);
\draw plot [smooth, tension=1] coordinates {(\beamL-0.2cm,\beamH) ([yshift = 0.8cm,xshift = -0.08cm]\beamL-0.2cm,\beamH) ([yshift = 1.3cm]\beamL+\massL,\beamH) ([yshift = 1.3cm,xshift = 0.5cm]G2) (G2)};
\begin{scope}[xshift=\beamL,xscale=-1]
\coordinate (G3) at (5.5,-1); % Coordinates of ground 3
\path (G3) +(150:0.2) coordinate (G3start);
\draw [interfaceflipped] (G3start)--++(-30:0.4);
\draw plot [smooth, tension=1] coordinates {(\beamL,\beamH) ([yshift = 0.5cm,xshift = -0.08cm]\beamL,\beamH) ([yshift = 0.8cm]\beamL+\massL,\beamH) ([yshift = 1.1cm,xshift = 0.5cm]G3) (G3)};
\coordinate (G4) at (6.1,-1); % Coordinates of ground 4
\path (G4) +(150:0.2) coordinate (G4start);
\draw [interfaceflipped] (G4start)--++(-30:0.4);
\draw plot [smooth, tension=1] coordinates {(\beamL-0.2cm,\beamH) ([yshift = 0.8cm,xshift = -0.08cm]\beamL-0.2cm,\beamH) ([yshift = 1.3cm]\beamL+\massL,\beamH) ([yshift = 1.3cm,xshift = 0.5cm]G4) (G4)};
\end{scope}
\end{tikzpicture}
\end{document}
</code></pre>
| 0non-cybersec
| Stackexchange |
Stephen Curry in college. | 0non-cybersec
| Reddit |
Comparison between mipmap on four and zero.. | 0non-cybersec
| Reddit |
Reminds me of my elementary presentations. | 0non-cybersec
| Reddit |
Crew of 17 aboard HMS Bounty abandons ship about 90 miles southeast of Hatteras, North Carolina . | 0non-cybersec
| Reddit |
Today is the day. I will submit my 2 week notice today. I am expected to be walked off, except my boss is working from home today, haha.
As I've explained before in other threads, my current job has some really great aspects to it, but overall it's not sustainable to be a solo systems administrator in an environment this large. Management is ignorant or willfully ignorant of IT and it's resource needs.
I'm super stoked to get out of here but am dreading submitting my notice. I've been reading up on how to do this sort of thing the best way but am interested to see if anyone has any advice for me here.
I'm fairly certain I'm going to get counter offers and phone calls from people/owners around the country begging me to stay. Don't worry though, it won't work.
Update: I called the boss at home and gave him my notice. It went better than expected. I kept it positive and when he pressed for a reason I told him that my position isn't a one man job. He agreed. He didn't try to keep me around, and thanked me for my service. He also wants me for the two weeks and said he may end up contracting with me down the road. It all worked out! | 0non-cybersec
| Reddit |
Please prove the given expression. <p>Where $\theta =\frac π7$,</p>
<p>$$4\cos\theta = \frac{4 \cos^2 \theta - 1}{2\cos^2 \theta - 1}$$</p>
<p>To prove RHS=LHS. (Begin from RHS not LHS)</p>
<p>It is a problem of trigonometry, and I have the solution of the problem. However, after seeing the solution, I don't quite understand how is one supposed to know how to approach this problem. My request is:</p>
<p>Don't just solve the problem, but also tell me from where do you come to know which approach would work. (Tell me from where do you come to know which trick or approach to be used to solution this problem.) And if possible please give both full subjective solution and short cut method.</p>
<p>Thanks in advance.</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 |
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