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Malham Cove, Yorkshire Dales.[938x1250][OC]. | 0non-cybersec
| Reddit |
ohhh dubstep..... | 0non-cybersec
| Reddit |
Is there a standard name for this index 2 subgroup in an affine group over a finite field of odd char?. <p>Fix an odd prime power $q$, fix a generator of the multiplicative group ${\mathbb F}_q^\times$, let $H$ be the subgroup generated by the <strong>square</strong> of this element, and form the semi-direct product ${\mathbb F}_q \rtimes H$. This is a subgroup of the full affine group of ${\mathbb F}_q$, and in the ongoing work I'm doing with colleagues, it provides a useful example at one point.</p>
<p>More out of curiosity than anything else, I wondered if this group has a standard name in the literature, or is denoted by a standard symbol? In the current draft of our paper it's denoted, unimaginatively, by $G_q$, but I wouldn't be surprised if that clashes with other notation that's standard in finite group theory.</p>
<p>(I've seen the $q=7$ case in several books, usually as an exercise in determining the character table, but it is only described as the non-abelian group of order 21.)</p>
| 0non-cybersec
| Stackexchange |
Here you go, here you don't.. | 0non-cybersec
| Reddit |
How can I get aligned equations not to be spoiled by an item?. <p>Is it possible to use <code>itemize</code> between align? For instance one wants to list some equations with previous explanations following an <code>item</code>, but one wants the explanations not to break the align structure, so that, say, the equal signs are aligned? What I want is:</p>
<p><img src="https://i.stack.imgur.com/jLQwN.png" alt="enter image description here"></p>
<p>My attempts to interlace an item in an give an error.</p>
<p>In this MWE:</p>
<pre><code>\documentclass{standalone}
\usepackage{amsmath}
\begin{document}
\begin{itemize}
\item This is the expansion for $f$:
\begin{align}
f(X) = & \sum_n {t_n+x_n-s_n-2} X^n
\end{align}
\item This is the expansion for $g$:
\begin{align}
g(X) = & \sum_m {r_m} X^m
\end{align}
\end{itemize}
\end{document}
</code></pre>
<p>I get the equations unaligned.</p>
| 0non-cybersec
| Stackexchange |
Finite simple groups: smallest nonsplit automorphic extensions. <p>Let $S$ be a finite nonabelian simple group such that the exact sequence </p>
<p>$$1 \to S \to {\rm Aut}(S) \to {\rm Out}(S) \to 1$$</p>
<p>is nonsplit, where $S$ is identified with ${\rm Inn}(S)$. Then there is always a (not necessarily unique) minimal subgroup $A$ in ${\rm Out}(S)$ with respect to the condition that </p>
<p>$$1 \to S \to S.A \to A \to 1$$</p>
<p>is nonsplit. Where can I find a classification of all such pairs $(S,A)$? In particular, is it always true that $|A|=2$?
I checked it with the <a href="http://en.wikipedia.org/wiki/ATLAS_of_Finite_Groups" rel="nofollow">ATLAS</a> that $S$ cannot be sporadic and, if it is alternating, then $S\cong A_6$ and $S.A\cong M_{10}$. </p>
<p><strong>Update:</strong> (inspired by Derek Holt's example below)</p>
<p>Can there be nonisomorphic minimal nonsplit extensions $S.A_1$ and $S.A_2$ for a given simple group $S$ with $A_1\cong A_2$?</p>
| 0non-cybersec
| Stackexchange |
When I was a kid I read a book about possible life on other worlds. Can you help me figure out what it was? (some details inside). . The book on display in my elementary school's library in the late 80s in the manner dictionaries are often displayed. It was on it's own platform usually open. From what I remember it was about what types of life might exist on other worlds, given their own unique environmental evolutionary pressures. There were several well done illustrations. A couple that come to mind are balloon like creatures that would evolve on planets with dense gaseous atmospheres and tall creatures (think star wars AT-AT) that were mostly legs. If it helps the book was large, I would guess something like 15 inches by 12 inches, but only a couple inches thick.
Please bear in mind that this was over 20 years ago and these descriptions are just memories and to be held highly suspect. I Figure at best this is a long shot. | 0non-cybersec
| Reddit |
News to me, but the Gaki No Tsukai guys have done a few let's plays.. | 0non-cybersec
| Reddit |
7 cool green gadgets.. | 0non-cybersec
| Reddit |
Windows 10 upgrade activation failure. <p>I downloaded a windows 10 image from the volume licensing site and used that to upgrade a HP Elitebook from Windows 8.1 which had been activated. The license key is available in the BIOS and can be accessed from the command prompt using:
wmic path softwarelicensingservice get OA3Xoriginalproductkey
Unfortunately, after the upgrade, the license did not automatically activate, in fact, Microsoft could not activate it without using a digital product key. When I downloaded an image using the Media Creation Tool, it worked great and the key automatically activated.
I now have 25 computers upgraded but do not know which ones Microsoft had to assist me with and which were upgraded using the newer image. Is there any way of finding this out?
I am worried that if I have to reinstall after the free upgrade period, the windows key currrently installed will not be recognised and I will have to buy a new license or go back to Windows 8.1</p>
| 0non-cybersec
| Stackexchange |
Is there an error in my proof of existence of a solution to this integral equation?. <p>I have this proof to show that $$x(t) = \log(1+t) + \frac{1}{5} \int^1_0 e^{-t} \cos ^2(ts)(x(s))^2 ds $$ has a unique integral solution in $C[0,1]$.</p>
<p>I was looking at someone who had posted an identical question to this here: <a href="https://math.stackexchange.com/questions/2259735/show-that-the-integral-equation-has-a-solution-on-a-suitable-subset-of-c0-1">Show that the integral equation has a solution on a suitable subset of $C[0,1]$</a> where someone answered that you need to be careful about which subsets of $C[0,1]$ you choose. I didn't take any care about subsets at all in my own proof so I was wondering if someone could tell me precisely where I went wrong in the following sketch of my proof. Help is truly appreciated.</p>
<p>Sketch of proof at first is that I show that for all $t\in [0,1]:$ $$\frac{1}{5} \int^1_0 e^{-t} \cos ^2(ts)(x(s))^2 ds $$</p>
<p>is continuous with respect to $t$. (and thus locally Lipschitz given the closed bounded interval).</p>
<p>Then I define $$T_x(t) = \log(1+t) + \frac{1}{5} \int^1_0 e^{-t} \cos ^2(ts)(x(s))^2 ds $$</p>
<p>which has a fixed point at:</p>
<p>$$x(t) = \log(1+t) + \frac{1}{5} \int^1_0 e^{-t} \cos ^2(ts)(x(s))^2 ds $$ </p>
<p>Then from the lemma above that the integral is continuous, $T$ is also continuous and thus we have that $T: C[0,1] \rightarrow C[0,1]$.</p>
<p>Then I make the following argument that $T$ is a contraction map.</p>
<p>$$|T_{x_1}(t) - T_{x_2}(t)| = \frac{1}{5} \Big|\int ^1_0 e^{-t} \cos^2 (ts) (x_1(s)-x_2(s))ds\Big| \leq \frac{1}{5} \text{max}_{s\in[0,1]} \Big|x_1(s)-x_2(s)\Big|$$</p>
<p>Thus $\Big\| T_{x_1} - T_{x_2} \Big\| \leq \frac{1}{5} \Big| x_1 - x_2 \Big|$, so $T$ is a contraction map, and since $C[0,1]$ is complete then $T$ has a unique fixed point. Thus there is a unique solution.</p>
<p>If this is correct or if I am missing something fundamental? Thanks for the time.</p>
| 0non-cybersec
| Stackexchange |
/bin/sh: 0: Can't open sh. <p>I'm trying to run a simple C program.</p>
<pre><code>#include <stdio.h>
#include <unistd.h>
extern char** environ;
int main(){
// execl("/bin/sh","sh","-c","/bin/ls -l",(char *) NULL);
char* argv[] = {"/bin/sh","sh","-c","/bin/ls", (char*) NULL};
execve(argv[0], argv, environ);
return 0;
}
</code></pre>
<p>The commented out execl runs fine. But when I try to do the same with execve, then the compiler invokes the following error:</p>
<pre><code>/bin/sh: 0: Can't open sh
</code></pre>
<p>What am I doing wrong here?</p>
| 0non-cybersec
| Stackexchange |
Earth from above... woooooah. | 0non-cybersec
| Reddit |
Desktop shortcut doesn't work for script but script work fine. <pre><code>#!/bin/bash/
ls -1 *.mp3 > playlist.m3u
for (( ; ; )) do mplayer -playlist playlist.m3u; sleep 0.1; done
</code></pre>
<p>This is the script and it works !</p>
<pre><code>[Desktop Entry]
Version=1.0
Type=Application
Name=Play
Comment=Play music
Exec=bash /home/awesome/Desktop/test/play.sh
Icon=
Terminal=true
GenericName=Player
</code></pre>
| 0non-cybersec
| Stackexchange |
Pandas: plot multiple columns to same x value. <p>Followup to a <a href="https://stackoverflow.com/questions/21089154/pandas-selecting-multiple-columns-from-one-row">previous question</a> regarding data analysis with pandas. I now want to plot my data, which looks like this:</p>
<pre><code>PrEST ID Gene Sequence Ratio1 Ratio2 Ratio3
HPRR12 ATF1 TTPSAXXXXXXXXXTTTK 6.3222 4.0558 4.958
HPRR23 CREB1 KIXXXXXXXXPGVPR NaN NaN NaN
HPRR23 CREB1 ILNXXXXXXXXGVPR 0.22691 2.077 NaN
HPRR15 ELK4 IEGDCEXXXXXXXGGK 1.177 NaN 12.073
HPRR15 ELK4 SPXXXXXXXXXXXSVIK 8.66 14.755 NaN
HPRR15 ELK4 IEGDCXXXXXXXVSSSSK 15.745 7.9122 9.5966
</code></pre>
<p>... except there are a bunch more rows, and I don't actually want to plot the ratios but some other calculated values derived from them, but it doesn't matter for my plotting problem. I have a dataframe that looks more or less like that data above, and what I want is this:</p>
<ul>
<li>Each row (3 ratios) should be plotted against the row's ID, as points</li>
<li>All rows with the same ID should be plotted to the same x value / ID,
but with another colour </li>
<li>The x ticks should be the IDs, and (if
possible) the corresponding gene as well (so some genes will appear
on several x ticks, as they have multiple IDs mapping to them)</li>
</ul>
<p>Below is an image that my previous, non-pandas version of this script produces:</p>
<p><img src="https://i.stack.imgur.com/2zB1t.png" alt="enter image description here"></p>
<p>... where the red triangles indicate values outside of a cutoff value used for setting the y-axis maximum value. The IDs are blacked-out, but you should be able to see what I'm after. Copy number is essentially the ratios with a calculation on top of them, so they're just another number rather than the ones I show in the data above.</p>
<p>I have tried to find similar questions and solutions in the documentation, but found none. Most people seem to need to do this with dates, for which there seem to be ready-made plotting functions, which doesn't help me (I think). Any help greatly appreciated!</p>
| 0non-cybersec
| Stackexchange |
WCGW if I cut this biker off while talking on my phone.. | 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 |
Cenapede. | 0non-cybersec
| Reddit |
Vince Staples talking about Logic's "black reminders" in his music.. | 0non-cybersec
| Reddit |
Chebyshev's Theorem. <p>Hi, </p>
<p>I´m looking for Chebyshev´s theorem which says that the inequality $|x(k)-y|<3/k$ has infinitely many solutions, where $x(k)=x_0+k\alpha \pmod 1$, $\alpha$ is an irrational number, and $x_0,y\in S^1$. Does anybody know the exact formulation?</p>
| 0non-cybersec
| Stackexchange |
Where are Bitcoin online casinos hosted?. <p>With a few Bitcoin-powered online casinos popping up every now and then, I am wondering where they are hosted and if they operate legally in the given area?</p>
| 0non-cybersec
| Stackexchange |
TexStudio "Go to source" and \import. <p>When using <code>\import</code> to store chapters in separate files, I am losing the ability to go from PDF to source and vice versa (except that compiling does take me to the right spot in the PDF). I am on 2.12.4 due to the reasons out of scope of this question. Are there any workarounds?</p>
| 0non-cybersec
| Stackexchange |
Death Star in progress. | 0non-cybersec
| Reddit |
Route specific ip adress through VPN gateway. <p>I'm running a Linux Mint (client) machine, and I need to connect to a remote (server) IP (10.0.4.29) through a VPN Gateway (163.172.224.201).</p>
<p>The VPN (Softether) created a virtual network interface (named vpn_vpn).</p>
<p>I'm currently connected to a Wifi Network (livebox on below outputs), and the VPN connection is working correctly.</p>
<p>Here are the output of:</p>
<p>netstat -r</p>
<p><code>Kernel IP routing table
Destination Gateway Genmask Flags MSS Window irtt Iface
default livebox 0.0.0.0 UG 0 0 0 wlp3s0
link-local * 255.255.0.0 U 0 0 0 wlp3s0
192.168.1.0 * 255.255.255.0 U 0 0 0 wlp3s0
192.168.6.0 * 255.255.255.0 U 0 0 0 vpn_vpn</code></p>
<p>ip route list</p>
<p><code>default via 192.168.1.1 dev wlp3s0
169.254.0.0/16 dev wlp3s0 scope link metric 1000
192.168.1.0/24 dev wlp3s0 proto kernel scope link src 192.168.1.128 metric 600
192.168.6.0/24 dev vpn_vpn proto kernel scope link src 192.168.6.57</code></p>
<p>How can I route all traffic to 10.0.4.29 through my gateway, and let other IP go through the livebox gateaway?</p>
<p>I already tried to execute these commands:</p>
<p><code>sudo ip route add default via 163.172.224.201
RTNETLINK answers: Network is unreachable</code></p>
<p><code>sudo route delete default
sudo ip route add default via 163.172.224.201
RTNETLINK answers: Network is unreachable</code></p>
<p><code>sudo route add -host 10.0.4.29 gw 163.172.224.201
SIOCADDRT: Network is unreachable</code></p>
<p><code>sudo ip route add 10.0.4.29 dev vpn_vpn
ip route get 10.0.4.29
10.0.4.29 dev vpn_vpn src 192.168.6.57
cache</code></p>
<p>and then ping 10.0.4.29</p>
<p><code>PING 10.0.4.29 (10.0.4.29) 56(84) bytes of data.
From 192.168.6.57 icmp_seq=1 Destination Host Unreachable</code></p>
<p>I know this is a really simple case, and I know I'm missing something, I don't know what. Can someone please help me?</p>
<p>Regards,</p>
<p>Julien</p>
| 0non-cybersec
| Stackexchange |
Comment on Cowboy Bebop Session 10, “Ganymede Elegy”. | 0non-cybersec
| Reddit |
Discrete sets inside compact sets. <p>If $K$ is an infinite, compact, Hausdorff topological space, must $K$ contain a discrete, closed, countably infinite set?</p>
| 0non-cybersec
| Stackexchange |
Can Ghidra show me the p-code generated for an instruction?. <p>I'm trying to verify the correctness of a custom processor module in Ghidra. The instruction that I am currently analyzing does some fancy stuff with its <a href="https://ghidra.re/courses/languages/html/pcoderef.html" rel="nofollow noreferrer">p-code</a>. Specifically, it uses a bunch of <a href="https://ghidra.re/courses/languages/html/sleigh_constructors.html#sleigh_tables" rel="nofollow noreferrer">tables</a> that end up forming a <a href="https://ghidra.re/courses/languages/html/sleigh_constructors.html#idm140310874886224" rel="nofollow noreferrer">p-code tree</a> that is several levels deep when it disassembles that instruction. Since p-code trees are walked in a depth-first order, that means the p-code of the child nodes will be emitted first before their parents. This could result in the p-code of the instruction not working in the order that you might expect.</p>
<p>I would like to be able to have Ghidra (or SLEIGH, or whatever affiliated tool) spit out the full generated p-code after it has walked the tree so I can verify that things are happening in the order that I think they are.</p>
| 0non-cybersec
| Stackexchange |
ar
X
iv
:1
00
5.
22
80
v1
[
cs
.C
R
]
1
3
M
ay
2
01
0
Modelling Nonlinear Sequence Generators in
terms of Linear Cellular Automata
Amparo Fúster-Sabater(1) and Dolores de la Gúıa-Mart́ınez(2)
(1) Instituto de F́ısica Aplicada, C.S.I.C.
Serrano 144, 28006 Madrid, Spain
[email protected]
(2) Centro Técnico de Informática, C.S.I.C.
Pinar 19, 28006 Madrid, Spain
[email protected]
Abstract
In this work, a wide family of LFSR-based sequence generators, the
so-called Clock-Controlled Shrinking Generators (CCSGs), has been an-
alyzed and identified with a subset of linear Cellular Automata (CA). In
fact, a pair of linear models describing the behavior of the CCSGs can
be derived. The algorithm that converts a given CCSG into a CA-based
linear model is very simple and can be applied to CCSGs in a range of
practical interest. The linearity of these cellular models can be advanta-
geously used in two different ways: (a) for the analysis and/or cryptanal-
ysis of the CCSGs and (b) for the reconstruction of the output sequence
obtained from this kind of generators.
Keywords: Cellular automata, Clock-controlled generators, Pseudo-
random sequence, Linear modelling
1 Introduction
Cellular Automata (CA) are discrete dynamic systems characterized by a simple
structure but a complex behavior, see [9], [13], [15], [19] and [21]. They are built
up by individual elements, called cells, related among them in many varied
ways. CA have been used in application areas so different as physical system
simulation, biological process, species evolution, socio-economical models or test
pattern generation. Their simple, modular, and cascable structure makes them
very attractive for VLSI implementations. CA can be characterized by several
parameters which determine their behavior e.g. the number of states per cell,
0Research supported by Ministerio de Educación y Ciencia (Spain) under grant SEG2004-
02418 and SEG2004-04352-C04-03.
Applied Mathematical Modelling. Volume 31, Issue 2, pp. 226-235. February 2007.
DOI:10.1016/j.apm.2005.08.013
1
http://arxiv.org/abs/1005.2280v1
the function Φ (the so-called rule) under which the cellular automaton evolves to
the next state, the number of neighbor cells which are included in Φ, the number
of preceding states included in Φ, the geometric structure and dimension of the
automaton (the cells can be arranged on a line or in a square or cubic lattice in
two, three or more dimensions), ... etc.
On the other hand, Linear Feedback Shift Registers (LFSRs) [10] are elec-
tronic devices currently used in the generation of pseudorandom sequences. The
inherent simplicity of LFSRs, their ease of implementation, and the good sta-
tistical properties of their output sequences turn them into natural building
blocks for the design of pseudorandom sequence generators with applications
in spread-spectrum communications, circuit testing, error-correcting codes, nu-
merical simulations or cryptography.
CA and LFSRs are special forms of a more general mathematical structure:
finite state machines [18]. In recent years, one-dimensional CA have been pro-
posed as an alternative to LFSRs ([1], [3], [15] and [20]) in the sense that every
sequence generated by a LFSR can be obtained from one-dimensional CA too.
Pseudorandom sequence generators currently involve several LFSRs combined
by means of nonlinear functions or irregular clocking techniques (see [14], [17]).
Then, the question that arises in a natural way is: are there one-dimensional
CA able to produce the sequence obtained from any LFSR-based generator?
The answer is yes and, in fact, this paper considers the problem of given a par-
ticular LFSR-based generator how to find one-dimensional CA that reproduce
its output sequence. More precisely, in this work it is shown that a wide class
of LFSR-based nonlinear generators, the so-called Clock-Controlled Shrinking
Generators (CCSGs) [12], can be described in terms of one-dimensional CA
configurations. The automata here presented unify in a simple structure the
above mentioned class of sequence generators. Moreover, CCSGs that is gen-
erators conceived and designed as nonlinear models are converted into linear
one-dimensional CA. Once the generators have been linearized, all the theoret-
ical background on linear CA found in the literature can be applied to their
analysis and/or cryptanalysis. The conversion procedure is very simple and can
be realized in a range of practical interest.
The paper is organized as follows: in section 2, basic concepts e.g. one-
dimensional CA, CCSGs or the Cattel and Muzio cellular synthesis method are
introduced. A simple algorithm to determine the pair of CA corresponding to a
particular shrinking generator and its generalization to Clock-Controlled Shrink-
ing Generators are given in sections 3 and 4, respectively. A simple approach to
the reconstruction of the generated sequence that exploits the linearity of the
CA-based model is presented in section 5. Finally, conclusions in section 6 end
the paper.
2 Basic Structures
In the following subsections, we introduce the general characteristics of the basic
structures we are dealing with: one-dimensional cellular automata, the shrinking
2
generator and the class of clock-controlled shrinking generators. Throughout
the work, only binary CA and LFSRs will be considered. In addition, all the
LFSRs we are dealing with are maximal-length LFSRs whose output sequences
are PN-sequences [10].
2.1 One-Dimensional Cellular Automata
One-dimensional cellular automata can be described as n-cell registers [9], whose
cell contents are updated at the same time according to a particular rule; that is
to say a k -variable function denoted by Φ. If the function Φ is a linear function,
so is the cellular automaton. When k input binary variables are considered, then
there is a total of 2k different neighbor configurations. Therefore, for cellular
automata with binary contents there can be up to 22
k
different mappings to the
next state. Moreover, if k = 2r+1, then the next state xt+1i of the cell x
t
i depends
on the current state of k neighbor cells xt+1i = Φ(x
t
i−r , . . . , x
t
i, . . . , x
t
i+r) (i =
1, ..., n).
CA are called uniform whether all cells evolve under the same rule while
CA are called hybrid whether different cells evolve under different rules. At the
ends of the array, two different boundary conditions are possible: null automata
when cells with permanent null contents are supposed adjacent to the extreme
cells or periodic automata when extreme cells are supposed adjacent.
In this paper, all the automata considered will be one-dimensional null hybrid
CA with k = 3 and linear rules 90 and 150. Such rules are described as follows:
Rule 90 Rule 150
xt+1i = x
t
i−1 ⊕ x
t
i+1 x
t+1
i = x
t
i−1 ⊕ x
t
i ⊕ x
t
i+1
For an one-dimensional null hybrid cellular automaton of length n = 10
cells, configuration rules ( 90, 150, 150, 150, 90, 90, 150, 150, 150, 90 ) and initial
state (0, 0, 0, 1, 1, 1, 0, 1, 1, 0), Table 1 illustrates the formation of its output
sequences (binary sequences read vertically) and the succession of states (binary
configurations of 10 bits read horizontally). For the above mentioned rules, the
different states of the automaton are grouped in closed cycles. The number of
different output sequences for a particular cycle is ≤ n as the same sequence
(although shifted) may appear simultaneously in different cells. At the same
time, all the sequences in a cycle will have the same period and linear complexity
[13] as well as any output sequence of the automaton can be produced at any
cell provided that we get the right state cycle.
2.2 The Shrinking Generator
The shrinking generator is a binary sequence generator [7] composed by two
LFSRs : a control register, called R1, that decimates the sequence produced by
the other register, called R2. We denote by Lj (j = 1, 2) their corresponding
lengths and by Pj(x) ∈ GF (2)[x] (j = 1, 2) their corresponding characteristic
polynomials [10].
3
Table 1: An one-dimensional null hybrid linear cellular automaton of 10 cells
with rule 90 and rule 150 starting at a given initial state
90 150 150 150 90 90 150 150 150 90
0 0 0 1 1 1 0 1 1 0
0 0 1 0 0 1 0 0 0 1
0 1 1 1 1 0 1 0 1 0
1 0 1 1 1 0 1 0 1 1
0 0 0 1 1 0 1 0 0 1
0 0 1 0 1 0 1 1 1 0
0 1 1 0 0 0 0 1 0 1
1 0 0 1 0 0 1 1 0 0
0 1 1 1 1 1 0 0 1 0
1 0 1 1 0 1 1 1 1 1
...
...
...
...
...
...
...
...
...
...
The sequence produced by the LFSR R1, that is {ai}, controls the bits of
the sequence produced by R2, that is {bi}, which are included in the output
sequence {cj} (the shrunken sequence), according to the following rule P :
1. If ai = 1 =⇒ cj = bi
2. If ai = 0 =⇒ bi is discarded.
A simple example illustrates the behavior of this structure.
Example 1: Let us consider the following LFSRs:
1. Shift register R1 of length L1 = 3, characteristic polynomial P1(x) =
1 + x2 + x3 and initial state IS1 = (1, 0, 0). The sequence generated by
R1 is {ai} = {1, 0, 0, 1, 1, 1, 0} with period T1 = 2
L1 − 1 = 7.
2. Shift register R2 of length L2 = 4, characteristic polynomial P2(x) =
1+x+x4 and initial state IS2 = (1, 0, 0, 0). The sequence generated by R2
is {bi} = {1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1}with period T2 = 2
L2−1 = 15.
The output sequence {cj} is given by:
• {ai} → 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 .....
• {bi} → 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 .....
• {cj} → 1 0 1 0 1 1 0 1 1 0 0 1 0 .....
The underlined bits 0 or 1 in {bi} are discarded. In brief, the sequence
produced by the shrinking generator is an irregular decimation of {bi} from the
bits of {ai}.
4
According to [7], the period of the shrunken sequence is
T = (2L2 − 1)2(L1−1) (1)
and its linear complexity [17], notated LC, satisfies the following inequality
L2 2
(L1−2) < LC ≤ L2 2
(L1−1). (2)
In addition, it can be proved [7] that the output sequence has some nice dis-
tributional statistics too. Therefore, this scheme is suitable for practical imple-
mentation of stream cipher cryptosystems and pattern generators.
2.3 The Clock-Controlled Shrinking Generators
The Clock-Controlled Shrinking Generators constitute a wide class of clock-
controlled sequence generators [12] with applications in cryptography, error cor-
recting codes and digital signature. An CCSG is a sequence generator composed
of two LFSRs notated R1 and R2. The parameters of both registers are defined
as those of subsection 2.2. At any time t,the control register R1 is clocked nor-
mally while the second register R2 is clocked a number of times given by an
integer decimation function notated Xt. In fact, if A0(t), A1(t), . . . , AL1−1(t)
are the binary cell contents of R1 at time t, then Xt is defined as
Xt = 1 + 2
0Ai0(t) + 2
1Ai1(t) + . . .+ 2
w−1Aiw−1(t) (3)
where i0, i1, . . . , iw−1 ∈ {0, 1, . . . , L1 − 1} and 0 < w ≤ L1 − 1.
In this way, the output sequence of an CCSG is obtained from a double
decimation. First, {bi} the output sequence of R2 is decimated by means of Xt
giving rise to the sequence {b′i}. Then, the same decimation rule P , defined in
subsection 2.2, is applied to the sequence {b′i}. Remark that if Xt ≡ 1 (no cells
are selected in R1), then the proposed generator is just the shrinking generator.
Let us see a simple example of CCSG.
Example 2: For the same LFSRs defined in the previous example and the
function Xt = 1+2
0A0(t) with w = 1, the decimated sequence {b
′
i} is given by:
• {bi} → 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 .....
• Xt → 2 1 1 2 2 2 1 2 1 1 2 2 2 1 2 1 1 2 2 .....
• {b′i} → 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 1 0 1 1 .....
According to the decimation function Xt, the underlined bits 0 or 1 in {bi}
are discarded in order to produce the sequence {b′i}. Then the output sequence
{cj} of the CCSG output sequence is given by:
• {ai} → 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 0 1 .....
• {b′i} → 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 1 0 1 1 .....
5
• {cj} → 1 1 0 1 0 1 0 1 1 0 1 1 .....
The underlined bits 0 or 1 in {b′i} are discarded.
In brief, the sequence produced by an CCSG is an irregular double decima-
tion of the sequence generated by R2 from the function Xt and the bits of R1.
This construction allows one to generate a large family of different sequences by
using the same LFSR initial states and characteristic polynomials but modifying
the decimation function. Period, linear complexity and statistical properties of
the generated sequences by CCSGs have been established in [12].
2.4 Cattel and Muzio Synthesis Algorithm
The Cattell and Muzio synthesis algorithm [4] presents a method of obtaining
two CA (based on rules 90 and 150) corresponding to a given polynomial. Such
an algorithm takes as input an irreducible polynomial Q(x) ∈ GF (2)[x] defined
over a finite field and computes two reversal linear CA whose output sequences
have Q(x) as characteristic polynomial. Such CA are written as binary strings
with the following codification: 0 = rule 90 and 1 = rule 150. The theoretical
foundations of the algorithm can be found in [5]. The total number of operations
required for this algorithm is listed in [4](Table II, page 334). It is shown that
the number of operations grows linearly with the degree of the polynomial, so
the method does not suffer from any sort of exponential blow-up. The method
is efficient for all practical applications (e.g. in 1996 finding a pair of length
300 CA took 16 CPU seconds on a SPARC 10 workstation). For cryptographic
applications, the degree of the irreducible (primitive) polynomial is L2 ≈ 64, so
that the consuming time is negligible.
Finally, a list of One-Dimensional Linear Hybrid Cellular Automata of De-
gree Through 500 can be found in [6].
3 CA-Based Linear Models for the Shrinking
Generator
In this section, an algorithm to determine the pair of one-dimensional linear CA
corresponding to a given shrinking generator is presented. Such an algorithm is
based on the following results:
Lemma 3.1 The characteristic polynomial of the shrunken sequence is of the
form P (x)N , where P (x) ∈ GF (2)[x] is a L2-degree polynomial and N is an
integer satisfying the inequality 2(L1−2) < N ≤ 2(L1−1).
Sketch of proof. The idea of the proof consists in demonstrating the uniqueness
of the polynomial P (x) that defines the linear recurrence relation satisfied by
{cj} for both the upper and lower bounds on the linear complexity. The values
of such bounds are given in equation (2). ✷
6
Lemma 3.2 Let P2(x) ∈ GF (2)[x] be the characteristic polynomial of R2 and
let α be a root of P2(x) in the extension field GF (2
L2). Then, P (x) ∈ GF (2)[x]
is the characteristic polynomial of cyclotomic coset 2L1 − 1, that is
P (x) = (x+ αE)(x+ α2E) . . . (x+ α2
L1−1E) (4)
being E an integer given by
E = 20 + 21 + . . .+ 2L1−1 . (5)
Sketch of proof. The shrunken sequence can be written as an interleaved se-
quence [11] made out of an unique PN -sequence repeated 2(L1−1) times where
2(L1−1) is the number of 1′s in a full period of {ai}. Such a PN -sequence is
obtained from {bi} taking digits separated a distance 2
L1 − 1. That is the
PN -sequence is the characteristic sequence associated with the cyclotomic coset
2L1 − 1 whose characteristic polynomial is P (x). ✷
Remark that P (x) depends exclusively on the characteristic polynomial of
the register R2 and on the length L1 of the register R1. In addition, the poly-
nomial P (x) will be the input to the Cattell and Muzio synthesis algorithm [4].
Based on such an algorithm, the following result is derived:
Proposition 3.3 Let Q(x) ∈ GF (2)[x] be a polynomial defined over a finite
field and let s1 and s2 two binary strings codifying the two linear CA obtained
from the Cattell and Muzio algorithm. Then, the two binary strings correspond-
ing to the polynomial Q(x) ·Q(x) are:
S′i = Si ∗ S
∗
i i = 1, 2
where Si is the binary string si whose least significant bit has been complemented,
S∗i is the mirror image of Si and the symbol ∗ denotes concatenation.
Sketch of proof. The result is just a generalization of the Cattell and Muzio
synthesis algorithm, see [4] and [5]. The concatenation is due to the fact that
rule 90 (150) at the end of the array in null automata is equivalent to two
consecutive rules 150 (90) with identical sequences. ✷
According to the previous results, the following linearization algorithm is
introduced:
Input: A shrinking generator characterized by two LFSRs, R1 and R2, with
their corresponding lengths, L1 and L2, and the characteristic polynomial P2(x)
of the register R2.
Step 1: From L1 and P2(x), compute the polynomial P (x) as
P (x) = (x+ αE)(x+ α2E) . . . (x + α2
L1−1E) (6)
with E = 20 + 21 + . . .+ 2L1−1.
7
Step 2: From P (x), apply the Cattell and Muzio synthesis algorithm [4] to
determine the two linear CA (with rules 90 and 150), notated si, whose
characteristic polynomial is P (x).
Step 3: For each si separately, we proceed:
3.1 Complement its least significant bit. The resulting binary string is
notated Si.
3.2 Compute the mirror image of Si, notated S
∗
i , and concatenate both
strings
S′i = Si ∗ S
∗
i .
3.3 Apply steps 3.1 and 3.2 to each S′i recursively L1 − 1 times.
Output: Two binary strings of length n = L2 2
L1−1 codifying the linear
CA corresponding to the given shrinking generator.
Remark 3.4 The characteristic polynomial of the register R1 is not needed.
Thus all the shrinking generators with the same R2 but different registers R1
(all of them with the same length L1) can be modelled by the same pair of one-
dimensional linear CA.
Remark 3.5 It can be noticed that the computational requirements of the lin-
earization algorithm are minimum. In fact, it just consists in the application
of the Cattell and Muzio synthesis algorithm whose consuming time is negligible
plus (L1 − 1) concatenations of binary strings. Both procedures can be carried
out on a simple PC.
In any case, thanks to this simple algorithm a linear model producing the
output sequence of the shrinking generator is obtained. In order to clarify the
previous steps a simple numerical example is presented.
Input: A shrinking generator characterized by two LFSRs R1 of length
L1 = 3 and R2 of length L2 = 5 and characteristic polynomial P2(x) = 1 + x+
x2 + x4 + x5. Now E = 23 − 1
Step 1: P (x) is the characteristic polynomial of the cyclotomic coset 7. Thus,
P (x) = 1 + x2 + x5 .
Step 2: From P (x) and applying the Cattell and Muzio synthesis algorithm,
two reversal linear CA whose characteristic polynomial is P (x) can be
determined. Such CA are written in binary format as:
0 1 1 1 1
1 1 1 1 0
8
Step 3: Computation of the required pair of CA.
For the first automaton:
0 1 1 1 1
0 1 1 1 0 0 1 1 1 0
0 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 (final automaton)
For the second automaton:
1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 (final automaton)
For each automaton, the procedure in Step 3 has been carried out twice
as L1 − 1 = 2.
Output: Two binary strings of length n = 20 codifying the required
CA.
In this way, we have obtained a pair of linear CA among whose output se-
quences we can obtain the shrunken sequence corresponding to the given shrink-
ing generator. Remark that the model based on CA is a linear one. In addition,
for each one of the previous automata there are state cycles where the shrunken
sequence is generated at any one of the cells.
4 CA-Based Linear Models for the Clock-Controlled
Shrinking Generators
In this section, an algorithm to determine the pair of one-dimensional linear CA
corresponding to a given CCSG is presented. Such an algorithm is based on the
following results:
Lemma 4.1 The characteristic polynomial of the output sequence of a CCSG
is of the form P ′(x)N , where P ′(x) ∈ GF (2)[x] is a L2-degree polynomial and
N is an integer satisfying the inequality 2(L1−2) < N ≤ 2(L1−1).
Sketch of proof. The idea of the proof is analogous to that one developed in
Lemma 3.1. ✷
Remark that, according to the structure of the CCSGs, the polynomial P ′(x)
depends on the characteristic polynomial of the register R2, the length L1 of the
register R1 and the decimation function Xt. Before, P (x) was the characteristic
polynomial of the cyclotomic coset E, where E = 20 + 21 + . . . + 2L1−1 was
a fixed separation distance between the digits drawn from the sequence {bi}.
Now, this distance D is variable and is a function of Xt. The computation of
D gives rise to the following result:
9
Lemma 4.2 Let P2(x) ∈ GF (2)[x] be the characteristic polynomial of R2 and
let α be a root of P2(x) in the extension field GF (2
L2). Then, P ′(x) ∈ GF (2)[x]
is the characteristic polynomial of cyclotomic coset D, where D is given by
D = 2L1−w (
2w∑
i=1
i) − 1 = (1 + 2w) 2L1−1 − 1. (7)
Sketch of proof. The idea of the proof is analogous to that one developed in
Lemma 3.2. In fact, the distance D can be computed taking into account that
the function Xt takes values in the interval [1, 2, . . . , 2
w] and the number of
times that each one of these values appears in a period of the output sequence
is given by 2L1−w. A simple computation, based on the sum of the terms of an
arithmetic progression, completes the sketch. ✷
From the previous results, it can be noticed that the algorithm to determine
the CA corresponding to a given CCSG is analogous to that one developed in
section 3; just the expression of E in equation (4) must be here replaced by the
expression of D in equation (7). A simple numerical example is presented.
Input: A CCSG characterized by: Two LFSRs R1 of length L1 = 3 and R2
of length L2 = 5 and characteristic polynomial P2(x) = 1 + x + x
2 + x4 + x5
plus the decimation function Xt = 1+ 2
0A0(t) + 2
1A1(t) + 2
2A2(t) with w = 3.
Step 1: P ′(x) is the characteristic polynomial of the cyclotomic coset D. Now
D ≡ 4 mod 31, that is we are dealing with the cyclotomic coset 1. Thus,
the corresponding characteristic polynomial is:
P ′(x) = 1 + x+ x2 + x4 + x5 .
Step 2: From P ′(x) and applying the Cattell and Muzio synthesis algorithm,
two reversal linear CA whose characteristic polynomial is P ′(x) can be
determined. Such CA are written in binary format as:
1 0 0 0 0
0 0 0 0 1
Step 3: Computation of the required pair of CA.
For the first automaton:
1 0 0 0 0
1 0 0 0 1 1 0 0 0 1
1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 (final automaton)
For the second automaton:
0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 (final automaton)
10
For each automaton, the procedure in Step 3 has been carried out twice
as L1 − 1 = 2.
Output: Two binary strings of length n = 20 codifying the required
CA.
Remark 4.3 From a point of view of the CA-based linear models, the shrinking
generator or any one of the CCGS are entirely analogous. Thus, the fact of
introduce an additional decimation function does neither increase the complexity
of the generator nor improve its resistance against cryptanalytic attacks since
both kinds of generators can be linearized by the same class of CA-based models.
5 A Simple Approach to the Output Sequence
Reconstruction for this Class of Sequence Gen-
erators
Since CA-based linear models describing the behavior of CCSGs have been
derived, a cryptanalytic attack that exploits the weaknesses of these models has
been also developed. It consists in reconstructing the CCSG output sequence
from an amount of such a sequence (the intercepted subsequence). The key idea
of this attack is based on the study of the repeated sequences in the automata
under consideration and the relative shifts among such sequences. In fact, the
sequence at a extreme cell of the automaton is repeated on average once out
of L2 cells. In order to determine these shifts, the algorithm of Bardell [2] to
phase-shift analysis of CA is applied. The approach is composed by several
steps:
• Step 1: The portion of M intercepted bits of the output sequence is placed
at the most right (left) cell of one of the automata. This provides shifted
portions of the same output sequence produced at different cells. The
lengths of these subsequences are (on average) (M −L2), (M −2L2), (M−
3L2), . . . , (M − pL2) where p = ⌊M/L2⌋.
• Step 2: The locations of the different cells that generate the same output
sequence as well as the relative shifts among these sequences are detected
via Bardell’s algorithm.
• Step 3: Repeat Steps 1 and Step 2 for every one of the subsequences
obtained above.
Summing up the contributions of the bits provided by each automaton, we
obtain that the total number of bits reconstructed is
NT ≈ Mp
2 = M(M/L2)
2 (8)
We know not only this number of bits but also the precise location of such bits
along the sequence. Notice that we have two different CA plus an additional
11
pair of CA corresponding to the reverse version of the output sequence (the
pair associated to the reciprocal polynomial of P2(x)). In addition, for each
automaton the intercepted M-bit sequence can be placed either at the most
right cell or the most left cell producing different locations of the same sequence.
Thus, each one of the different automata will contribute to the reconstruction of
the output sequence with a number of bits given by the equation (8). Moreover,
remark that the output sequence for these generators is an interleaved sequence
[11] made out of a fixed PN-sequence. Hence, the portions of the reconstructed
subsequence allow us to fix the starting point of many of these PN-sequences.
The rest of the bits of each PN-sequence can be easily derived.
Once the previous steps are accomplished, the original output sequence can
be reconstructed by concatenating all different reconstructed subsequences.
Finally, let us see a simple example of application of Bardell’s algorithm.
Example 3: Let us consider a cellular automaton with the following char-
acteristics:
• Number of cells n = 10
• Automaton under study in binary format: 0011001100
• Characteristic polynomial (1 + x+ x3 + x4 + x5)2 .
Let S be the shift operator defined on Xi (i = 1, . . . , 10), the state of the
i-th cell , such as follows:
SXi(t) = Xi(t+ 1) .
Thus, the corresponding difference equation system for the previous automaton
can be written as follows:
SX1 = X2 SX2 = X1 +X3 . . . SX10 = X9 .
Next expressing eachXi as a function ofX10, we obtain the following system:
X1 = (S
9 + S4 + S3 + S2 + S + 1)X10
X2 = (S
8 + S6 + S5 + S4 + S3 + S + 1)X10
...
X9 = (S)X10 .
Analogous results can be obtained expressing each Xi as a function of X1.
Now taking logarithms in both sides of the equalities,
log(X1) = log(S
9 + S4 + S3 + S2 + S + 1) + log(X10)
log(X2) = log(S
8 + S6 + S5 + S4 + S3 + S + 1) + log(X10)
...
log(X9) = log(S) + log(X10) .
12
The base of the logarithm is R(S) and the values of the logarithms are
integers over a finite domain. According to the Bardell’s algorithm, we deter-
mine the integers m (if there exist) such that Sm modR(S) equal the different
polynomials in S included in the above system. For instance,
S26 modR(S) = S2 + 1 .
Or simply, S26 = S2 + 1 and 26 log(S) = log(S2 + 1) with log(S) ≡ 1. Now
substituting in the previous system, the following equations can be derived:
log(X9)− log(X10) = 1
log(X8)− log(X10) = 26
log(X4)− log(X10) = 6
log(X2)− log(X1) = 1
log(X3)− log(X1) = 26
log(X7)− log(X1) = 6 .
The phase-shifts of the outputs 9, 8 and 4 relative to cell 10 are 1, 26 and
6 respectively. Similar values are obtained in the other group of cells, that is
cells 2, 3 and 7 relative to cell 1. The other cells generate different sequences.
Further contributions to phase-shift analysis of CA based on 90/150 rules can
be found in [16] and [8].
6 Conclusions
Awide family of LFSR-based sequence generators, the so-called Clock-Controlled
Shrinking Generators, has been analyzed and identified with a subset of linear
cellular automata. In this way, sequence generators conceived and designed
as complex nonlinear models can be written in terms of simple linear models.
An easy algorithm to compute the pair of one-dimensional linear hybrid cellu-
lar automata that generate the CCSG output sequences has been derived. A
cryptanalytic approach based on the phase-shift of cellular automata output
sequences is proposed. From the obtained results, we can create linear cellu-
lar automata-based models to analyse/cryptanalyse the class of clock-controlled
generators.
References
[1] Bao, F.: Crytanalysis of a New Cellular Automata Cryptosystem. In: 8th
Australasian Conference on Information Security and Privacy–ACISP 2003.
Lecture Notes in Computer Science, Vol. 2727. Springer Verlag, Berlin Hei-
delberg New York (2003) 416-427
13
[2] Bardell, P.H.: Analysis of Cellular Automata Used as Pseudorandom Pat-
tern Generators. Proceedings of the IEEE International Test Conference.
Paper 34.1 (1990) 762-768
[3] Blackburn, S., Merphy, S., Paterson, K.: Comments on ’Theory and Ap-
plications of Cellular Automata in Cryptography’. IEEE Transactions on
Computers. 46 (1997) 637-638
[4] Cattell K., Muzio, J.: Synthesis of One-Dimensional Linear Hybrid Cellular
Automata. IEEE Transactions on Computer-Aided Design. 15 (1996) 325-
335
[5] Cattell K., Muzio, J.: A Linear Cellular Automata Algorithm: Theory. Dept.
of Computer Science. University of Victoria, Canada, Tech. Rep. DCS-161-
IR, 1991.
[6] Cattell K., Shujian, Z.: Minimal Cost One-Dimensional Linear Hybrid Cel-
lular Automata of Degree Through 500. J. of Electronic Testing: Theory
and Applications. 6 (1995) 255-258
[7] Coppersmith, D., Krawczyk, H., Mansour, Y.:The Shrinking Generator. In:
Advances in Cryptology–CRYPTO’93. Lecture Notes in Computer Science,
Springer Verlag, Berlin Heidelberg New York. 773 (1994) 22-39
[8] Cho, S., et al.: Computing Phase Shifts of 90/150 Cellular Automata Se-
quences. In: ACRI 2004. Lecture Notes in Computer Science, Vol. 3305.
Springer Verlag,(2004) 31-39
[9] Das, A.K., Ganguly, A., Dasgupta, A., Bhawmik, S., Chaudhuri,P.P.: Ef-
ficient Characterisation of Cellular Automata. IEE Proc., Part E. 1 (1990)
81-87
[10] Golomb, S.: Shift-Register Sequences (revised edition). Aegean press (1982)
[11] Gong, G.: Theory and Applications of q-ary interleaved sequences. IEEE
Transactions on Information heory. 41 (1995) 400-411
[12] Kanso, A.: Clock-Controlled Shrinking Generator of Feedback Shift Regis-
ters. In: 8th Australasian Conference on Information Security and Privacy–
ACISP 2003. Lecture Notes in Computer Science, 2727 (2003) 443-451
[13] Martin, O., Odlyzko, A.M., Wolfram, S.: Algebraic Properties of Cellular
Automata. Commun. Math. Phys. 93 (1984) 219-258
[14] A.J. Menezes et al. Handbook of Applied Cryptography, CRC Press, New
York, 1997
[15] Nandi, S., Kar, B.K., Chaudhuri, P.P.: Theory and Applications of Cellular
Automata in Cryptography. IEEE Transactions on Computers. 43 (1994)
1346-1357
14
[16] Nandi, S., Chaudhuri, P.P.: Additive Cellular Automata as an on-chip test
pattern generator. Test Symposium 1993. Proc. of the Second Asian. (1993)
166-171
[17] Rueppel, R.A.: Stream Ciphers, in Gustavus J. Simmons, Editor, Contem-
porary Cryptology, The Science of Information. IEEE Press (1992) 65-134
[18] Stone, H.S.: Discrete mathematical Structures and Their Applications.
Chicago, IL. Science Research (1973)
[19] Wolfram, S.: Cellular Automata as Models of Complexity. Nature. 311
(1984) 419
[20] Wolfram, S.: Cryptography with Cellular Automata. In: Advances in
Cryptology–CRYPTO’85. Lecture Notes in Computer Science, 218 (1994)
22-39
[21] Wolfram, S.: Random Sequence generation by Cellular Automata. Ad-
vances in Applied Mathematics. 7 123 (1986)
15
1 Introduction
2 Basic Structures
2.1 One-Dimensional Cellular Automata
2.2 The Shrinking Generator
2.3 The Clock-Controlled Shrinking Generators
2.4 Cattel and Muzio Synthesis Algorithm
3 CA-Based Linear Models for the Shrinking Generator
4 CA-Based Linear Models for the Clock-Controlled Shrinking Generators
5 A Simple Approach to the Output Sequence Reconstruction for this Class of Sequence Generators
6 Conclusions
| 1cybersec
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| Reddit |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
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George Lucas sends letter to 'Lost'. | 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 |
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 |
From House Elves to Hunger Games: Labor Issues in Sci-Fi and Fantasy. | 0non-cybersec
| Reddit |
Disable Asset Pipeline/Sprockets Rails 4.1. <p>I cannot seem to find a way to disable the Asset Pipeline in Rails 4.1. I see a lot of information for Rails 3.2. </p>
<p>I want to use Grunt/GulpJs and bower to handle all of my assets and I cannot seem to find something for this. Rather I find a decent amount but most of it doesn't apply to me or is broken.</p>
<p>There is the <a href="https://rubygems.org/gems/half-pipe"><code>half-pipe</code> gem</a>. However half-pipe relies on Rails 4.0 and I'm on Rails 4.1 and I can't find where to force a Gem to install in different version of Rails, if that's even possible.</p>
<p>This question expanded into a few more questions than I intended. Basically I just need to know how to disable the Asset Pipeline, barring that how to integrate GruntJS.</p>
| 0non-cybersec
| Stackexchange |
With macOS High Sierra, you can finally pull an app window down to the very bottom of the screen. The damned gap between the app window and the bottom of the screen is finally gone with High Sierra! | 0non-cybersec
| Reddit |
Error 'Printer Settings could not be saved' after driver upload (Samba). <p>When I upload driver to Samba, I often get the message "Printer Properties" - "Printer settings could not be saved. Operation could not be completed (error 0x0000007a)".</p>
<p>Since driver seems to be correctly uploaded to my Samba 3.6.27 server, I do not understand what is the point of this error message. I also manage to change default configuration of the printer by using the contextual properties dialog (which change on driver upload).</p>
<p>Could you explain what is the actual consequence of the error? What i can do to avoid this error?</p>
| 0non-cybersec
| Stackexchange |
How can I disable _moz_resizing?. <p>I am using nicEdit editor and I have added my own custom image resizing script to it. But I want to disable the default _moz_resizing that appears in Firefox. </p>
<p>I wanted to have finer control over the image being resized. ( Eg: Allow only the image to resize and inherit the width of the parent container. ) </p>
<p>So I wrote a custom script. But since Firefox has its own image resizing control (_moz_resizing) how do I disable it? If there is no way to do so, I have a very simple workaround where I detect if Firefox and turn off my custom script and use the _moz_resizing instead.</p>
<p>But I won't have fine-grained control and I will have to rely on there being browser bugs in Firefox. :(</p>
| 0non-cybersec
| Stackexchange |
When they put the names in a different order than the actors.. | 0non-cybersec
| Reddit |
Surjective maps of Sheaves. <p>My question is mainly along the lines of another post:</p>
<p><a href="https://math.stackexchange.com/questions/3361084">Surjectivity on stalks implies surjectivity on sheaves</a></p>
<p>I’m trying to prove that given a map of sheaves <span class="math-container">$\psi : \mathcal{F} \rightarrow \mathcal{G}$</span> the followings true:</p>
<p><span class="math-container">$(\mathcal{Im}\psi)_p = \mathcal{G}_p$</span> for every p <span class="math-container">$\Rightarrow \mathcal{Im}(U) = \mathcal{G}(U)$</span> for any open subset U of X</p>
<p>I found a solution </p>
<p><a href="https://www.math.arizona.edu/~cais/CourseNotes/AlgGeom04/Hartshorne_Solutions.pdf" rel="nofollow noreferrer">https://www.math.arizona.edu/~cais/CourseNotes/AlgGeom04/Hartshorne_Solutions.pdf</a></p>
<p>Which states that <span class="math-container">$\psi$</span> is surjective
iff <span class="math-container">$\mathcal{Im}\psi = \mathcal{G}$</span>
iff <span class="math-container">$(\mathcal{Im}\psi)_p = \mathcal{G}_p$</span> For every p
iff <span class="math-container">$\mathcal{Im}\psi_p = \mathcal{G}_p$</span> For every p
iff <span class="math-container">$\psi_p$</span> is surjective</p>
<p>Most of this is fine, but why does step 2 hold? I.e why is it true that </p>
<p><span class="math-container">$\mathcal{Im}\psi = \mathcal{G}$</span> Iff <span class="math-container">$(\mathcal{Im}\psi)_p = \mathcal{G}_p$</span> For every p</p>
<p>I can see why the forward implication holds but why the backward? </p>
<p>The answer and comments in the post above didn’t seem to involve the fact any specific properties of the image sheaf (as was pointed out in the comments) and so the answer seemed weird (given that the statement shouldn’t hold for arbitrary sheafs F and G)</p>
<p>My questions are then </p>
<p>1) what is the proof for the reverse implication of the above mentioned step 2?</p>
<p>2) where (if at all) is there a mistake in the answer/comments in the above mentioned post? And if there are no mistakes, then where in both inclusions in the proof for the reverse implication are the hypotheses (that we are working in the image sheaf) used? </p>
<p>3) is there a more elegant approach to proving the initial surjectivity statement? </p>
| 0non-cybersec
| Stackexchange |
What does "closed" mean in relation to vectors?. <p>I am reading a book on Matrix Algebra and saw this:
<a href="https://i.stack.imgur.com/CTeZr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CTeZr.png" alt="enter image description here" /></a></p>
<p>What does "closed" mean? The author has not defined that term.</p>
| 0non-cybersec
| Stackexchange |
Sexiest pancakes ever made - Raspberry, banana, chocolate bits, syrup [500x700]. | 0non-cybersec
| Reddit |
Birthday paradox for large numbers. <p>I'm attempting to test the claim:
"Every card deck shuffled is unique. A shuffled deck of cards will exist once and never again."</p>
<p>Assumption: A perfect world where a deck of cards is perfectly random all of the time. 52 cards are used in the deck.</p>
<p>The birthday paradox and it's easy enough to calculate for small numbers. 23 people and 365 birthdays as used in the 50% examples. But how do you approach (or approximate) the birthday paradox for values like 52!?</p>
<p>I understand 52! is a large (~226 bit) number but I would like to get a feel of the order of magnitude of the claim. Is it 1% or 0.00001%? </p>
<p>To calculate the probability of a shuffle collision would be:
(52!)!/(52!^n*(52!-n)!)</p>
<p>I understand the formula. But 52!! is incomputable so where do I go from here? How to approach a problem like this?</p>
<p>This is just for my own curiosity and not homework. If it can be done for a deck of cards I'd want to give it a try on collisions in crypto keys. (RSA, and AES256 etc.)</p>
| 0non-cybersec
| Stackexchange |
How do you talk to your SO about financial stuff?. Greetings Relationships. I have come to ask you for advice about a very sensitive subject.
I am mid-30s and my SO is late 30s. We have been dating 1 and 1/2 years and currently live together. Marriage has come up and the only thing holding me back at this point is financial worries.
Currently we are both employed. I make enough to pay my bills and put money in the bank. I have a very good credit rating and my only debts are student loans and 4 more payments on one credit card. My SO is kind of a different story. I have never point blank asked him about his financial background. What I get from him is that he has a poor credit rating, no substantial savings, a good amount of debt, and is not the best at handling money. However, as far as I can tell he pays his bills every month and, after we had our first mini discussion, is working on a savings account.
I have been advised by numerous friends and family that I should ask him to fully disclose his financial background as well as ask him about the amounts of debt he owes. I am really torn because although we have discussed marriage, we have also said we don’t want to rush the relationship. I honestly don’t know if I should force this discussion before we really sit down and decide on a marriage timeline. He has already said that he wants to marry me and I am a bit nervous that he might try to propose before I have this discussion with him. If he does, I will say no, because I want this resolved before we even get engaged.
To be clear, I am not very concerned about his debt. I also have some debt. I am concerned because he has debt and a low credit rating. This sets off a red flag that he might not be handling that debt properly. Since we have been together, we have worked through several major issues. I feel that we can work together on this as well. I am just at a loss as to how to approach the discussion.
**TL;DR I have a high credit rating and good financial sense, my SO does not. Marriage has been discussed. How do I approach this discussion?**
| 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 |
Logic behind an eBay phishing attempt. Wonder if someone can explain the logic behind a phishing attempt against me on eBay. I'm writing a book, not about phishing/hacking, but one of the characters in the book suffers from a cyberattack so I'd like to understand some of the whys.
Scenario: I placed a $2000 camera for sale on eBay. Within an hour, it's sold. No questions, nothing. The buyer sends me a message with their email address and asks me to complete the sale using our emails. I refuse and insist to be paid via PayPal. Couple of emails back and forth and eBay tells me the buyer's account has been compromised and that I should relist the camera.
I googled the buyer's address and it's a real person, a retired 60-something lady living in a small, quiet country town.
So,
1) What the buyer was trying to do by making me to deal with them via email and not via eBay messaging system?
2) What's the con? I'm the seller, I'm supposed to be paid. What did they want? My camera? I wouldn't send anything without first being paid.
3) How did they get hold of the lady's eBay account? | 1cybersec
| Reddit |
Idea for endorsement. So endorsement levels are nice - but they don't necessarily affect too much.
I had a few thoughts.
A) change private profile settings to allow only people of x endorsement level to view it.
B) for each 5 endorsement levels (don't even know if there's a cap somaybe adjust this) you get 1 additional avoid teammate slot
Just a thought. | 0non-cybersec
| Reddit |
Well she now she dead. | 0non-cybersec
| Reddit |
infinite loop, csname and forloop function. <p>I want to define some commands that finish with "a", "b", "c"... and totalize the number of them, how can I do this??</p>
<pre><code>\documentclass{article} \usepackage[utf8]{inputenc}
\usepackage{ifthen}
\usepackage{pgffor}
\usepackage{alphalph}
\usepackage{etoolbox}
\usepackage{forloop}
\newcounter{totpecas}
\newcommand \peca {This is the first.}
\newcommand \pecab {This is the
second.}
\newcommand \pecac {This is the third.}
\newcommand \npecatot{1}
\begin{document}
\forloop{totpecas}{2}{{\ifcsdef{peca\alphalph{\value{totpecas}}}{1}{0}}
= 1}{\renewcommand \npecatot {\value{totpecas}}}
The number of defined commands is {\npecatot}.\\
\end{document}
</code></pre>
| 0non-cybersec
| Stackexchange |
[Spoilers][Rewatch] Clannad: After Story Ep. 18: "The Ends of the Earth". #Clannad: After Story Ep. 18: "The Ends of the Earth"
####[**Table of contents**](https://www.reddit.com/r/anime/comments/4pi3hh/we_start_the_clannad_rewatch_tomorrow_at_2pm_est/)
<-------------------------------Previous Episode|Next Episode----------------------------------->
:--|--:
[Clannad: After Story Ep. 17: "Summertime"](https://www.reddit.com/r/anime/comments/4w5us1/spoilersrewatch_clannad_after_story_ep_17/)|[Clannad: After Story Ep. 19: "The Road Home"](https://www.reddit.com/r/anime/comments/4wgyao/spoilersrewatch_clannad_after_story_ep_19_the/)
***
[MAL](http://myanimelist.net/anime/2167/Clannad)
[MAL:AS](http://myanimelist.net/anime/4181/Clannad__After_Story)
[Free legal streaming @ Hulu: Clannad](http://www.hulu.com/clannad)
[Free legal streaming @ Hulu: Clannad: After Story](http://www.hulu.com/clannad-after-story)
/r/Clannad
***
With regards to spoilers: Please be aware we have both first time watchers and re-watchers, so please tag any spoilers as such. Also, please try to avoid limiting yourself to just spoiler discussions, doing so will make first time watcher's experience much more enjoyable. | 0non-cybersec
| Reddit |
How can I break referential integrity briefly, within a transaction, without disabling the foreign key constraint?. <p>I have a table with 3 columns:</p>
<pre><code>ID, PARENT_ID, NAME
</code></pre>
<p><code>PARENT_ID</code> has a foreign key relationship with <code>ID</code> in the same table. This table is modeling a hierarchy.</p>
<p>Sometimes the <code>ID</code> of a record will change. I want to be able to update a record's <code>ID</code>, then update the dependent records' <code>PARENT_ID</code> to point to the new <code>ID</code>.</p>
<p>The problem is, when I attempt to update the <code>ID</code> of a record it breaks the integrity and fails immediately.</p>
<p>I realize I could insert a new record with the new <code>ID</code>, then update the children, then delete the old record, but we have a lot of triggers in place that would get screwed up if I did that.</p>
<p>Is there any way to temporarily update the parent with the promise of updating the children (obviously it would fail on commit) without disabling the foreign key briefly?</p>
| 0non-cybersec
| Stackexchange |
Passionfruit Souffle. | 0non-cybersec
| Reddit |
Proof of an infinite sum of probabilities. <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/74186/alternative-expected-value-proof">Alternative Expected Value Proof</a> </p>
</blockquote>
<p>If $X$ is a random variable that takes values in the range $\left \{ 1,2,3,4,5,6,\ldots \right \}$ how can I prove the following statement?</p>
<p>$$\mathbb{E}[X]=\sum_{i=1}^\infty \mathbb{P}[X \ge i]$$</p>
<p>I honestly have no idea where to even begin. Even just something to kick off the process would be great.</p>
| 0non-cybersec
| Stackexchange |
"The City Tempest, Jeremy Mann, Oil on Panel, 2013". | 0non-cybersec
| Reddit |
Ohio boy, 5, saves overdosed parents after walking two blocks, alone and barefoot, for help. | 0non-cybersec
| Reddit |
Can one make Code Analysis understand Code Contracts?. <p>When using Code Analysis and Code Contracts in combination, I get a lot of warnings like</p>
<p><a href="http://msdn.microsoft.com/en-us/library/ms182182.aspx" rel="noreferrer">CA1062</a>: Microsoft.Design : In externally visible method 'Foo.Bar(Log)', validate parameter 'log' before using it.</p>
<p>In Foo.Bar, I have a contract that validates <code>log</code>.</p>
<pre><code>public Bar(Log log)
{
Contract.Requires(log != null);
log.Lines.Add(...);
// ...
}
</code></pre>
<p>Is there a way to make FxCop understand code contracts?</p>
| 0non-cybersec
| Stackexchange |
Access OneDrive personal vault through API. <p>Microsoft introduced a special folder in OneDrive (I only see it in a personal OneDrive account, not in OneDrive for Business) called <a href="https://www.engadget.com/2019/10/01/microsoft-onedrive-personal-vault/" rel="noreferrer">"Personal Vault"</a>. I searched the <a href="https://docs.microsoft.com/en-us/graph/overview" rel="noreferrer">documentation of MS Graph API</a> but could not find this mentioned.</p>
<p>So my question is: is there any way to access this personal vault as a third-party app?</p>
| 0non-cybersec
| Stackexchange |
A set theory questions regarding functions.. <p>Q: Let $A=\{1,2,3\}$ and $B=\mathcal{P}(A)$. Let $F: B \rightarrow B$ be the function defind by the formula $F(X)=A\setminus X$. What is $F(\{1,3\})$ ?</p>
<p>A:
I have no idea where to start. I assume $X$ is an arbitrary $X$. I originally tried to define the function as $F=\{(x,y) \in B \times B | y=x+2\}$ however, the function is already defined. Im also sure my original idea is something completely different than what the topic is. </p>
<p>I also figured to draw a "bubble" picture for sets $B \rightarrow B$ but this would be just an arrow point from 1 to 3. </p>
<p>Any info is greatly appreciated. </p>
| 0non-cybersec
| Stackexchange |
Approximation of $u \in C^{1,2}([0,T]\times \mathbb{R}^d)$ with $u(0,x)= 0$ by smooth functions vanishing close to $t=0$. <p>My question is the following:
Let <span class="math-container">$T>0$</span>, <span class="math-container">$u: [0,T]\times \mathbb{R}^d \to \mathbb{R}$</span> be continuously differentiable once w.r.t. <span class="math-container">$t \in (0,T)$</span> and twice w.r.t. <span class="math-container">$x \in \mathbb{R}^d$</span> such that all partial derivatives can be extended to <span class="math-container">$[0,T]\times \mathbb{R}^d$</span> continuously. Further, suppose <span class="math-container">$u$</span> along with all its partial derivatives is bounded.</p>
<p>Does there exist a sequence <span class="math-container">$(u_n)_n$</span> with <span class="math-container">$u_n \in C^{\infty}_c((0,T]\times \mathbb{R}^d)$</span> which approximates <span class="math-container">$u$</span> pointwise up to all <span class="math-container">$(1,2)$</span>-derivatives such that each sequence of derivatives is bounded in <span class="math-container">$n$</span>?</p>
<p>The answer is affirmative, if we do not require <span class="math-container">$u_n$</span> to vanish close to <span class="math-container">$0$</span> (note that there is no assumption on vanishing close to <span class="math-container">$T$</span>), because clearly <span class="math-container">$u$</span> as above can be approximated by smooth functions in the desired sense. However, to me it seems possibly troublesome to prove what I'm looking for, essentially because if we require <span class="math-container">$\partial_tu_n$</span>to be bounded uniformly in <span class="math-container">$n$</span>, we cannot allow each <span class="math-container">$u_n$</span> to be constantly <span class="math-container">$0$</span> on some <span class="math-container">$(\epsilon_n,T]\times \mathbb{R}^d$</span>, because this might require us to approximate <span class="math-container">$u$</span> by functions <span class="math-container">$u_n$</span> with increasing (not bounded) derivative <span class="math-container">$\partial_tu_n$</span> close to <span class="math-container">$t=\epsilon_n$</span>.</p>
<p>Is it still possible to obtain what I'm looking for? My hope is that due to <span class="math-container">$u(0,x)=0 \forall x$</span> the situation is actually not too bad. Any help is appreciated!</p>
| 0non-cybersec
| Stackexchange |
Android on-screen keyboard auto popping up. <p>One of my apps has an "opening screen" (basically a menu) that has an <code>EditText</code> followed by several <code>Button</code>s. The problem is that several of my users are reporting that when they open the app it's automatically popping up the on-screen keyboard without them even touching the <code>EditText</code>. As far as I can tell, all of these users are using the <a href="http://www.htc.com/www/product/hero/overview.html" rel="noreferrer">HTC Hero</a>. </p>
<p>Is this a bug in 1.5? Is there anything I can do about it?</p>
| 0non-cybersec
| Stackexchange |
refreshing the page results in 404 error- Angular 6. <p>I am building an application with the help of <strong>Angular6</strong> and facing problems in routing. All the <strong>routes</strong> are working when I click on a particular tab but whenever I <strong>refresh</strong> the current page, it is throwing <strong>404</strong> error. I have seen many posts regarding this issue on Stack overflow but failed to overcome from this problem. </p>
<p>Below is my app.module.ts</p>
<pre><code>import {BrowserModule} from '@angular/platform-browser';
import {NgModule} from '@angular/core';
import {RouterModule, Routes} from '@angular/router';
import {AppComponent} from './app.component';
import {FetchApiComponent} from './fetch-api/fetch-api.component';
import {FormsModule} from '@angular/forms';
import {HttpClientModule} from '@angular/common/http';
import {UserServiceLatest} from './fetch-latest/app.service';
import {UserServiceTop} from './fetch-top/app.service';
import {YoutubePlayerModule} from 'ngx-youtube-player';
import {SidebarComponent} from './sidebar/sidebar.component';
import {FetchLatestComponent} from './fetch-latest/fetch-latest.component';
import { FetchTopComponent } from './fetch-top/fetch-top.component'
import {UserService} from './fetch-api/app.service';
import { ServiceWorkerModule } from '@angular/service-worker';
import { environment } from '../environments/environment';
import { AngularFireModule } from 'angularfire2';
import * as firebase from 'firebase';
import { firebaseConfig } from './../environments/firebase.config';
import { AngularFireDatabaseModule } from 'angularfire2/database';
import {PushService} from './push.service';
const appRoutes: Routes = [
{
path: '',
component: FetchApiComponent
},
{
path: '/latest',
component: FetchLatestComponent
},
{
path: '/top',
component: FetchTopComponent
},
{
path :'*',
component: FetchApiComponent
}
];
firebase.initializeApp(firebaseConfig);
@NgModule({
declarations: [
AppComponent,
FetchApiComponent,SidebarComponent, FetchLatestComponent, FetchTopComponent
],
imports: [
RouterModule.forRoot(appRoutes),
BrowserModule, YoutubePlayerModule,
FormsModule,
AngularFireModule.initializeApp(firebaseConfig),
AngularFireDatabaseModule,environment.production ?ServiceWorkerModule.register('firebase-messaging-sw.js'):[],ServiceWorkerModule.register('/firebase-messaging-sw.js', { enabled: environment.production }),
HttpClientModule,environment.production ? ServiceWorkerModule.register('ngsw-worker.js') : [], ServiceWorkerModule.register('/ngsw-worker.js', { enabled: environment.production })
],
providers: [UserService,UserServiceTop,UserServiceLatest,PushService],
bootstrap: [AppComponent]
})
export class AppModule {}
</code></pre>
<p>Can you point me in right direction?</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 |
Geometry Conjecture, Interpretation.. <p>I've been participating in a message group on geometry and the following conjecture has been presented. I am only trying to understand it fully. First the conjecture is the following:</p>
<p><strong>"ABC is a triangle with circumcenter O. A1B1C1 is the cevian triangle of O with respect to ABC. A2B2C2 is cevian triangle of O with respect to A1B1C1.
AnBnCn is cevian triangle of O with respect to A(n-1)B(n-1)C(n-1).
I conjecture that when n tends to infinity then sequence centroid of triangles AnBnCn tends to O. Is it true?"</strong></p>
<p>It is the last part regarding "sequence centroid" that I am failing to understand. If the centroid is the concurrency point of the medians, then what is meant by "sequence centroid"? Does it mean to ask whether the successive centroid points of the triangles AnBnCn go towards the circumcenter O, and become synonymous with it? That would be an interesting thing to prove, perhaps, if that is in fact the meaning. As n approaches infinity the triangles become infinitely small it would seem. I suppose the question is whether such a succession of triangles condenses to the point O. Is this the correct way to interpret this question? </p>
| 0non-cybersec
| Stackexchange |
How can I get the terminal in full resolution when running Ubuntu CLI in VirtualBox?. <p>I've installed Ubuntu (latest version) with CLI only, inside VirtualBox running in Windows 7 x64. Switching to fullscreen (<code>CTRL + F</code> in my setup) works, but of course the terminal windows has a low resolution (it's just a box inside a black background).</p>
<p>I think that guest additions are missing. How to enjoy the full 1080p terminal in Windows?</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 |
If r/trees gets to 1,000,000 subs today, I will smoke an entire oz tomorrow.. | 0non-cybersec
| Reddit |
Disks Application Formatting. <p>In the Formatting Option in the Disks application for Ubuntu 14.04, it gives the option to not overwrite existing data or overwrite existing data with 0's. I've tried both, and both clear the disk and all its data. So what exactly is the difference? What is the point of using the second one, as it takes SO much longer? </p>
| 0non-cybersec
| Stackexchange |
Insane person on an Anon post.. | 0non-cybersec
| Reddit |
Cops of Reddit, what is the most stupid criminal you have ever met?. | 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 |
When you live with a monster, you need your own toothpaste. | 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 |
Set Mouse Cursor to Left Center of Currently Active Display With Command/Script. <p>I am trying to figure out how to make a script that operates in the following way:</p>
<ol>
<li>finds dimensions of the display which the mouse pointer is currently on</li>
<li>moves the mouse pointer to the center of the left edge on that display</li>
<li>I'd like to be able to add command arguments when running the script that would add or subtract to the center left edge position and move the mouse accordingly. For example if I ran:</li>
</ol>
<blockquote>
<p>python3 /path/to/script.py</p>
</blockquote>
<p>then the script would place the mouse pointer on left edge of display and in the center of the vertical axis. If I ran this:</p>
<blockquote>
<p>python3 /path/to/script.py 10 -10</p>
</blockquote>
<p>then the script would place the mouse pointer 10px --> from left edge of screen, and -10px (down) from center of vertical axis.</p>
<p>I'm pretty sure xdotool and python can accomplish this, but I am a complete noob when it comes to programming anything, and even though I found <a href="https://askubuntu.com/questions/792911/how-to-set-cursor-position-with-command-script">this link</a>, I'm still not sure how to proceed. If anyone feels so kind as to help me out with this, it would be greatly appreciated, but if not then no worries. </p>
<p>If anyone is curious as to why I want to do this: I'm setting up a tablet to use i3wm. I'm creating touch button shortcuts for i3 window management using jgmenu. Jgmenu has an option to launch at the location of the mouse pointer. I would like to use a touchegg gesture to first move the mouse pointer with this script, then launch jgmenu. </p>
| 0non-cybersec
| Stackexchange |
[D&D 5E] Some drawings of my players' characters. I've been DM'ing a homebrew D&D 5E campaign since last summer. As DM who's also an illustrator, I draw some of the NPCs my players face to give them a unique visual, but I've also drawn their own characters for fun and inspiration. I thought there might be interest in seeing some of those drawings here along with a bit of their background.
**[Aetius Scipio](http://pasiphilo.deviantart.com/art/Aetius-Scipio-Human-Wizard-586023516)** is a high-born human wizard who had to study spellcasting secretly because arcane magic is outlawed in the tyranny he grew up in. His player portrayed him as very CN - almost machiavellian. The expression on his face is meant to evoke both his privileged position and his potential for ruthlessness (though never against the party). The magic tome he holds is supposed to be one of fey origin he discovered during their first quest.
**[Ellora Sera](http://pasiphilo.deviantart.com/art/Ellora-Sera-Human-Cleric-546059177)** is a human cleric (Life domain) whose deity is focused on healing and restoration. She comes from the same tyranny that Aetius does and even served in its ruthless legions as a War Priest until a higher and more benevolent destiny called to her and she retired to answer that call. She has sworn off the empire she once served and now works to undo it. The glyphs on her arms glow when she calls upon her deity's power.
**[Valmar Astorio](http://pasiphilo.deviantart.com/art/Valmar-Astorio-Human-Fighter-545302699)** is a human fighter who, like Ellora, was a legionary in the tyrannical empire he was born into, continuing the proud familiy tradition of serving in the legions but he, too, became disillusioned with the empire. His time with the legions ended when he struck a superior officer. Only his father's clout saved him from certain death in a gulag. He now, like Ellora, seeks to undermine the empire from the outside. The smouldering menace in his eyes reflects his anger at the injustice he once helped enable.
**[Lemeria Leafringer](http://pasiphilo.deviantart.com/art/Lemeria-Leafringer-Halfling-Rogue-544346474)** is a halfling rogue who lived most of her life as the captive servant of a wealthy merchant family in the same tyranny the others come from. As an infant, she was kidnapped from her home in the fey lands where her people live and raised among humans. The moment of her escape served as the catalyst for the start of our campaign, and her flight lead her into contact with each of the other heroes, all of whom banded together to secure her freedom, thereby solidifying their loyalty to each other. Liberating Lemeria became their first act against the empire.
Aetius and Lemeria have since left the story (because their players had to go back to university) and two more player's have filled the void, so I now have a couple more characters to illustrate eventually. :) But it's been an incredidly fun and rewarding campaign to run and (according to my players) to play as well.
Anyway, feedback welcome. | 0non-cybersec
| Reddit |
Math request!. Hey everyone!
I’m getting back into programming after graduating in computer science 10 years ago from a university.
I forgotten a lot of things!
What can you recommend I look up regarding the math side of things? I’ve decided and started to learn Swift, however I would like to improve my math.
Can anyone recommend any good websites/courses/videos for this with regards to programming/algorithms please?
Thanks | 0non-cybersec
| Reddit |
Friendly Strangers. The other day, I was talking to an online friend of mine over Skype.
We're both pretty big fans of the paranormal. we were CONSTANTLY swapping stories and testing fates.
That day, my friend seemed rather freaked out-like a happy freaked out.
"Dude, I have some one here I need you talk to. You're going to love this!!!"
She typed to me.
I was a bit puzzled.
"??" I replied.
She started explaining.
"Some weird things have been happening to my cousin lately. I have her over here right now and I'm gonna have her tell you all about them! But I'm telling her not to tell you whats odd about it until the end."
I replied with a quick "LOL k, bring it!" and waited a few moments for my friend and her cous to switch places.
Soon, the screen blipped.
"Hey. You're <screenname>, <friends name> friend? She talks a lot about you. I'm gonna be honest, I never really believed in this whole ghost or paranormal things up until now so I'm a bit nervous and I think I might just be over reacting but the doctors haven't found anything psychologically wrong with me so???If you're just gonna think I'm crazy then tell me now cuz I'll just leave."
I could tell from her reluctance that this must have been really freaking her out.
"Don't worry, I believe pretty much anything anyone tells me. I doubt you're crazy."
I assured her.
"Ok. here it goes then."
She said and then typed up her story.
"It began a few weeks ago after I read some ridiculous chain letter that I didn't forward. I didn't think anything of it because, like I said, I didn't believe in that stuff and the chain letter was something about creatures wearing human skin that blend in with humans I don't know, I don't really think it's related but it started after I deleted it.
"It first started at the market. While i was in line, I noticed that the bagger was staring at me.
He was smiling a bit too uncomfortably at me. Like, he wasn't using his muscles to smile, it was more like some one took two fingers to his cheeks and pulled upwards to draw the corner of his lips into a smile??
I know I know that sounds way stupid, doesn't it?"
"Anyways, I smiled politely at him and took my groceries from the end of the belt when he finished bagging them.I freed up a hand and I thanked him for his service.
And that's when it happened.
His lips moved and he said in a deep voice "You're welcome, miss."
I stood there in complete shock for a moment. His smile grew wider before he then ignored me completely and went back to bagging groceries."
I figured I must have been exhausted because there's no way that just happened."
"The next day, I was walking down the street. The air was chill and cold, and I was enjoying all the smells and sensations of the upcoming fall.
That's when it happened again.
I saw a woman staring at me . She didn't take her eyes off me. She had that same eerie smile.
It was uncomfortable. As she got closer to me, she parted her lips and spoke to me:
"Nice weather today, isn't it miss?"
And then nodded and kept walking.
I stopped in my tracks once more. It was dizzying because of its impossibility.
And it wasn't just those few times either.
Several other people spoke to me since then.
It was always innocuous things like "Need any help today, ma'am?" or "You're looking lovely today", nothing vicious, the strangers were all very friendly albeit all with that same creepy grin, but the mere fact it was HAPPENING was terrifying and yet also strangely hope bringing? Does that sound right?
I don't know, but soon, everywhere I went, I would hear voices of certain people I was near exchanging pleasantries with me.
I even told my doctor this and he assured me that no,it isn't possible that this could happen.And yet I know it is. It IS happening to me!"
After that wall of text, my friends cousin stopped typing.
Frankly, I was confused as shit.
"Look that's all great and all but uhm, people talking to you and saying nice things isn't at all a paranormal experience, even if they smile weird.."
It was the next thing she said that changed my mind.
"And THAT's the response most people would have. But here's the thing:
Somehow I am hearing these people talk when I can't hear anything else; I don't hear birds, I don't hear cars driving down the street,
I don't hear the tapping of the keys on my keypad. I don't hear anything.
I don't hear anything because I've been fully deaf since I was 4."
I stared at my screen for a bit before she typed again.
"What do you think is going on?"
| 0non-cybersec
| Reddit |
Validation error messages without the attribute. <p>I'm trying to show custom error messages without the attribute name in front. I used to do this with the custom_error_message gem, however it doesn't work with Rails 3.1</p>
<p>What I'm trying now in create.js.erb:</p>
<pre><code>alert("<%= @post.errors[:title] %>")
</code></pre>
<p>Which returns </p>
<pre><code>[&quot;Here goes my custom message?&quot;]
</code></pre>
<p>My question is -- how can I remove the brackets and the &quot so only the message is left. I'll insert it into the page using jquery.</p>
| 0non-cybersec
| Stackexchange |
What is the behavior of these functions linear or logarithmic or neither?. <p>Please consider the following functions $F$ and $G$.
\begin{align*}
F(K) = \log_2 \left(\frac{( 2\sqrt{K-1}+K-2)^2}{(\sqrt{K-1}-1)^2}\right)+(K-1) \log_2(2)
\end{align*}
and the function
\begin{align*}
G(K)=
\log \left( \bigg( \frac{K^2-2}{(K-1)^2}\bigg) \ \frac{(2\sqrt{K-1}+K-2)^2}{(\sqrt{K-1}-1)^2} \right)
+(K-2)\log_{2}\left(\frac{(K^2-2)}{2(K-1)^2} \frac{(\sqrt{K-1}+1)^2}{(\sqrt{K-1}-1)^2}\right)
+(K-2)\log(2)
\end{align*}</p>
<p>I should note that $K$ is a positive integer greater than 2.</p>
<p>My question is I don't really understand how these functions are behaving a a function of $K$? Are these functions linear or logarithmic as function of $K$ or something else...</p>
<p>Any help would be much appreciated.</p>
| 0non-cybersec
| Stackexchange |
ELI5: PVP in WoD. So I've decided to pursue PVP as my endgame in WoD, having said that there's a lot of different avenues Ashran, Rated and Unrated BG's, Arena's, Honor Points, Conquest points.
I want to get started in PVP, and have no idea how to get from 0 to "Top of the pvp food chain" in terms of gear and skill, I know PVP is a constantly evolving and changing beast, and there's always more to learn from so being bad at first isn't an issue if it matters I'm a rogue, so where do people also duel all day to get better? Where am I on that proverbial food chain and how do I get to the top? I appreciate all responses. | 0non-cybersec
| Reddit |
When I can reverse the logical operators?. <p>I heard say that is logically equivalent to say it:
$$\neg (p \vee q) = p \land q$$</p>
<p>So every time you have a negation operator in front can make a "distributive" altering the operator from within? And if $\neg(p \vee \neg q)$ then result is $p \land \neg q$? Can anyone give some examples?</p>
| 0non-cybersec
| Stackexchange |
Creating a SOAP call using PHP with an XML body. <p>I'm trying to call a SOAP method using PHP.</p>
<p>Here's the code I've got:</p>
<pre><code>$data = array('Acquirer' =>
array(
'Id' => 'MyId',
'UserId' => 'MyUserId',
'Password' => 'MyPassword'
));
$method = 'Echo';
$client = new SoapClient(NULL,
array('location' => 'https://example.com/ExampleWebServiceDL/services/ExampleHandler',
'uri' => 'http://example.com/wsdl', 'trace' => 1));
$result = $client->$method($data);
</code></pre>
<p>Here's the request it creates:</p>
<pre><code> <?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://example.com/wsdl" xmlns:ns2="http://xml.apache.org/xml-soap" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<SOAP-ENV:Body>
<ns1:Echo>
<param0 xsi:type="ns2:Map">
<item>
<key xsi:type="xsd:string">Acquirer</key>
<value xsi:type="ns2:Map">
<item>
<key xsi:type="xsd:string">Id</key>
<value xsi:type="xsd:string">mcp</value>
</item>
<item>
<key xsi:type="xsd:string">UserId</key>
<value xsi:type="xsd:string">tst001</value>
</item>
<item>
<key xsi:type="xsd:string">Password</key>
<value xsi:type="xsd:string">test</value>
</item>
</value>
</item>
</param0>
</ns1:Echo>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>
</code></pre>
<p>And here's what I <em>want</em> the request to look like:</p>
<pre><code> <?xml version="1.0" encoding="UTF-8"?>
<SOAP-ENV:Envelope SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns1="http://example.com/wsdl" xmlns:ns2="http://xml.apache.org/xml-soap" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<SOAP-ENV:Body>
<Echo>
<Acquirer>
<Id>MyId</Id>
<UserId>MyUserId</UserId>
<Password>MyPassword</Password>
</Acquirer>
</Echo>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>
</code></pre>
| 0non-cybersec
| Stackexchange |
Stardew Valley's Creator on Self-Publishing, the "Everything" Update, and His Future as a Solo Developer. | 0non-cybersec
| Reddit |
My Grand Pa tells the worst jokes but this is his best.. Grand Pa: What does Mr. Potato Head and Cuba have in common?
Me:*expecting the worst*What's that.....
Grand Pa: They both have a dictator. | 0non-cybersec
| Reddit |
how to handle firebaseauth token expiration?. <p>I use firebase 2 years so far and was never clear to me how to handle firebasse auth token expiration. </p>
<p>I'm using android and once i do <code>FirebaseAuth.getInstance().signInWithCustomToken(token);</code> interanally firebase receives a message like that:</p>
<pre><code>{
"t":"d",
"d":{
"r":5,
"b":{
"s":"ok",
"d":{
"auth":{
"uid":"test",
"token":{
"exp":1592230969,
"user_id":"test",
"iat":1592227369,
"sub":"test",
"aud":"test",
"auth_time":1592227369,
"iss":"https://securetoken.google.com/igibo-b0b27",
"firebase":{
"identities":{
},
"sign_in_provider":"custom"
}
},
"provider":"custom",
"user_id":"test"
},
"expires":1592230969
}
}
}
}
</code></pre>
<p>after this request any future calls to <code>FirebaseAuth.getInstance().getCurrentUser()</code> will return info about this authenticated user.<br>
It is clean in this json that this auth token expires sometime, but it isn't clear to me what will be firebase behaviors for that...</p>
<p>will <code>FirebaseAuth.getInstance().getCurrentUser()</code> return null after token expiration?<br>
will firebase automatially renew the token so it never expires?<br>
if i need to monitore and revalidate token manually HOW I DO THAT? </p>
<p>if im not debugging i cant even find the information of expiration anywhere.</p>
| 0non-cybersec
| Stackexchange |
Windows Server 2012 R2 network Failover?. <p>i have two windows server 2012 R2 using to running software (sms gateway) , i want to configure a network failover , if the first server goes down the second server goes up.</p>
<p>note: the both server need internet connection to connect with the mobile operators , i need a way to enable the internet for second server if the first server goes down , or any thing like that.</p>
<p>can i make this by using windows server 2012, or not.</p>
<p>Thanks</p>
| 0non-cybersec
| Stackexchange |
maximum order of an element in symmetric group. <p>While doing my homework i find out that the maximum order of an element in $S_3$ is 3 (the element $(123)$) and the maximum order of an element in $S_4$ is 4 (the element $(1234)$)</p>
<p>Can i generalize that and say that the maximum order of an element in $S_n$ is n?</p>
| 0non-cybersec
| Stackexchange |
I accidentally deleted /usr/bin/python. How do I restore it?. <p>Trying to set the default Python version to 2.7, I ran this command:</p>
<pre><code>sudo rm /usr/bin/python*
</code></pre>
<p>Now when I type <code>python</code> or <code>python2.7</code>, I see these error messages: </p>
<pre><code>bash: /usr/bin/python: No such file or directory
bash: /usr/bin/python2.7: No such file or directory
</code></pre>
<p>What happened? Should I have run the <code>rm</code> command? How can I undo it?</p>
| 0non-cybersec
| Stackexchange |
Cursed_covid. | 0non-cybersec
| Reddit |
Kirby's Epic Yarn on the Wii and Yoshi's Wooly World on the Wii U. What's next for the Feel-Good developers on the Nintendo NX? [POLL]. | 0non-cybersec
| Reddit |
While we drop the constant in big O notation, does it matter in real-life situations?. <p>While I understand that <a href="https://en.wikipedia.org/wiki/Big_O_notation" rel="noreferrer">big O</a> notation simply describes the growth rate of an algorithm, I'm uncertain if there is any difference in efficiency in real life between the following O(n) algorithms.</p>
<p>To print the value of a node in a linked list k places from the end of the list.</p>
<p>Given a node:</p>
<pre><code>/* Link list node */
struct node
{
int data;
struct node* next;
};
</code></pre>
<p><strong>Solution 1 O(n)</strong></p>
<p>This solution iterates over the list twice, once to find the length of the list, and the second time to get to the <strong>end of the list - N</strong>.</p>
<pre><code>void printNthFromLast(struct node* head, int n)
{
int len = 0, i;
struct node *temp = head;
// 1) Count the number of nodes in Linked List
while (temp != NULL)
{
temp = temp->next;
len++;
}
// Check if value of n is not more than length of the linked list
if (len < n)
return;
temp = head;
// 2) Get the (n-len+1)th node from the begining
for (i = 1; i < len-n+1; i++)
{
temp = temp->next;
}
printf ("%d", temp->data);
return;
}
</code></pre>
<p><strong>Solution 2 O(n)</strong></p>
<p>This solution only iterates over the list once. The ref_ptr pointer leads, and a second pointer (main_ptr) follows it k places behind. When ref_ptr reaches the end of the list, main_ptr should be pointing at the correct node (k places from the end of the list).</p>
<pre><code>void printNthFromLast(struct node *head, int n)
{
struct node *main_ptr = head;
struct node *ref_ptr = head;
int count = 0;
if(head != NULL)
{
while( count < n )
{
if(ref_ptr == NULL)
{
return;
}
ref_ptr = ref_ptr->next;
count++;
}
while(ref_ptr != NULL)
{
main_ptr = main_ptr->next;
ref_ptr = ref_ptr->next;
}
}
}
</code></pre>
<p>The question is: Even though both solutions are O(n) while leaving big O notation aside, is the second solution more efficient that the first for a very long list as it only iterates over the list once?</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 map $\phi : \ell^{1}(\mathbb{N}) \to L^{1}(\mathbb{R})$ where $\left\{ a_n \right\} \mapsto \sum_n a_n 1_{[n,n+1]}$ is norm preserving?. <p>I'm actually just trying to show $\phi$ is continuous by showing that it is Lipschitz. However, I believe the map $\phi : \ell^{1} \to L^{1}(\mathbb{R})$ where $\left\{ a_n \right\} \mapsto \sum_n a_n 1_{[n,n+1]}$ may also be norm-preserving. Here is what I have so far:</p>
<p>Take $x = \left\{ a_n \right\} \in \ell^{1}$ and $f = \sum_n a_n 1_{[n,n+1]} \in L^{1}(\mathbb{R})$ so that</p>
<p>\begin{align*}
\|\ \phi(x) \|_{L^{1}(\mathbb{R})} &= \|\ \sum_{n} a_n 1_{[n,n+1]} \|_{L^{1}(\mathbb{R})} \\
&= \|\ f \|_{L^{1}(\mathbb{R})} \\
&= \int_{\mathbb{R}} |f(x)| dx \\
&= \sum_{n=1}^{\infty} \left( \int_{-n}^{-n+1} |f(x)|dx + \int_{n-1}^{n} |f(x)|dx \right)
\end{align*}</p>
| 0non-cybersec
| Stackexchange |
So, I murdered my ceramic stove.... It got in a fight with my mortar. The mortar won. Now I'm stuck without a stove for a while. I have an oven that I almost never use. Expand my horizons! What's some good food that requires oven only? | 0non-cybersec
| Reddit |
Icedtea in Chrome won't load java applets. <p>Icedtea in Chrome won't load java applets such as minecraft and runescape. It shows a gray screen instead. I've tried and failed to install the java plugin for chrome and it's left me wondering why icedtea wont's work!?</p>
| 0non-cybersec
| Stackexchange |
Match Thread: Mexico vs Uruguay - Copa America Centenario Group C. ~~ok wait im new to this give me some time to set it up...~~ Ok shoutout to /u/deception42 [](#icon-alien-big) because I basically stole his whole layout for this thread. Thanks man!
#[](#sprite6-p112) 3-1 [](#sprite6-p185)
**[](#icon-net-big) Venue:** University of Phoenix Stadium, Glendale, Arizona
**[](#icon-info-big) TV:** [Find your channel here](http://www.livesoccertv.com/match/1895921/mexico-vs-uruguay/)
**[](#icon-tv-big) Stream:** [Find a stream here](http://www.reddit.com/r/soccerstreams)
**[](#icon-whistle-big) Referee:** E. Cáceres [](#sprite6-p24)
----
Starting XIs
[](#sprite6-p112) **Mexico:** (5-2-3) Talavera, Layún, Araujo, Reyes, Moreno, Marquez, Hector Herrera, Guardado, Jesus "Tecatito" Corona [](#icon-down-big), Aquino [](#icon-down-big), and Javier "Chicharito" Hernandez
[](#sprite6-p185) **Uruguay:** (4-2-3-1) Muslera, Giménez, Godín, Sánchez, A. Pereira, Lodeiro [](#icon-down-big), Vecino, M. Pereira, Arévalo, Cavani and Rolan [](#icon-down-big).
***
Substitutes
[](#sprite6-p112) **Mexico:** Jesus Corona, G. Ochoa, Paul Aguilar, Jorge Torres, Y. Corona, J. Molina, H. Lozano [](#icon-up-big), C. Pena, J. Duenas [](#icon-up-big), O. Peralta , R. Jiménez
[](#sprite6-p185) **Uruguay:** M. Silva, M. Campaña, G. Silva, M. Victorino, J. Fucile, M. Corujo, Álvaro González [](#icon-up-big), D. Laxalt, G. Ramírez, A. Hernández [](#icon-up-big), C. Stuani
***
[](#icon-star) **MATCH UPDATES** [](#icon-star) ^^I'm ^^basically ^^copying [^^Goal.com's ^^commentary.](http://www.goal.com/en-us/match/mexico-vs-uruguay/2178766/live-commentary?ICID=MP_LC_1)
--
1' [](#icon-clock-big) **KICKOFF!** WE'RE UNDERWAY IN GLENDALE! Mexico in their white shirts, attack left to right, with Uruguay sporting their light blue.
3' A bit of a tactical move from Juan Carlos Osorio, who gave every indication of a 4-3-3, but as the match kicks-off, Marquez drops into the centre of defence, making it more of a 5-2-3.
4' [](#sprite6-p112) ***GOOOOOOOAAAAALLLLLLLL FOR MEXICO!!!*** [](#sprite6-p112) Own Goal Álvaro Daniel Pereira Barragán
DREAM START FOR EL TRI! Mere seconds after the clock strikes three minutes, Guardado lifts a sensational cross in from the left and with Herrera running on to meet it, Alvaro Pereira heads into his own goal!
[](#icon-ball-big) [GIF 1-0](https://streamable.com/xnwe), credit to /u/Omar_Til_Death [](#icon-alien-big)
6' Mexico have a real spring in their step, undoubtedly buoyed by the wealth of supporters in attendance. Uruguay have hardly had a touch of the ball, which has been firmly contained in the Celeste end, and they're already down a goal. This isn't at all going according to plan for the 2011 champions.
8' [](#icon-notes-big) ***STAT***: In the last meeting in 2011 Uruguay fired 22 total shots (only six for Mexico); more than Uruguay have managed in any other game in the last two editions of the Copa America.
10' Corona races 30 yards to chase down an errant pass, keeping it in play and allowing Mexico to retain possession. That one bit of effort epitomises Mexico's work ethic in these opening 10 minutes - they came to play.
12' Slowly Uruguay are starting to find their footing, looking to recover from a horrid start to proceedings. They're getting a bit more time on the ball, though have yet to penetrate the attacking third. Cavani is shockingly isolated up top, but Rolan is starting to move higher up the pitch to provide a bit of support.
14' The tempo has slowed a bit after a rampant start, but Mexico are still beaming with confidence - a 19-game unbeaten run will do that to a team. Their interplay in midfield is excellent, as is their movement off the ball.
16' Aquino wins a corner, and the right-hand side is turning into a real area of concern for Uruguay. Alvaro Pereira is struggling to find the pace he needs to contain Aquino and Araujo on the overlap. Herrera and Guardado are spraying the ball about brilliantly.
18' [](#icon-notes-big) ***STAT***: Uruguay have not conceded more than one goal in their last 10 Copa America games, winning all four matches in which they kept a clean sheet in (D4, L2).
20' Mexico aren't on the ascendency anymore, they're flat out dominating. Uruguay look completely overwhelmed by the waves of Mexican pressure, and their loose passing in midfield is keeping them hemmed in at the back.
22' Aquino drifts over to the left now, where he's making waves yet again! The UANL winger picks up possession and cuts in onto his right, but his well-struck effort is straight at Muslera! Decent effort, that, from Aquino who has been excellent in this first half.
24' For what it's worth, Alvaro Pereira's own goal was the fastest one of it's kind in Copa America history. El Tri continue to run riot in the attacking half, with Uruguay struggling to close their gaps and take space away from these Mexican midfielders.
25' [](#icon-yellow-big) for Andrés Guardado [](#sprite6-p112)
Guardado picks up the first booking of the match, catching Sanchez dangerously from behind with a slightly belated challenge.
27' [](#icon-yellow-big) for Matías Vecino Falero [](#sprite6-p185)
Bit of revenge for Uruguay on Guardado as Vecino trips him up at the halfway line, rather cynically, with his trailing foot.
30' [](#icon-net-big) GREAT SAVE from Talavera!
HE HAD TO FINISH THAT! Cavani times his diagonal run brillaintly to get played through behind the defence, but he's thwarted at the near post by Talavera! The PSG striker had only the keeper to beat, but couldn't ripple the back of the net.
33' After a decent spell in attack for Uruguay, Mexico are camped out again in the attacking half. Aquino is sprung through down the right and sends in an inviting cross, but Gimenez manages to sniff it out.
35' CLOSE TO A SECOND! Aquino can't be stopped! Another sensational run and delivery into the area is poised to be met by Chicharito, but Herrera arrives first and deflects a weak header out the other side of the area and away from the striker - he would have had a sitter!
37' The longer Uruguay are made to sit back and defend, the more frustration starts to build in their side. The foul count is rising at an alarming rate, and now Aquino is down after a cynical challenge right in front of the Mexican bench.
41' Into the last five minutes now and Uruguay desperately need to get this game into the interval and regroup. They have plenty of options off the bench to draw upon and change up the formation a bit, with Cavani in desperate need of a strike partner.
43' HE CAN'T TAKE IT SOON ENOUGH! Aquino, because who else, makes a terrific run and picks out a superb diagonal ball for Chicharito, who's clean through on goal. The Leverkusen striker hesitates momentarily, however, which allows Jimenez to make an exquisite recovery challenge!
44' [](#icon-red-big) for Matías Vecino Falero after getting his 2nd yellow. [](#sprite6-p185)
URUGUAY DOWN TO 10 MEN! It's an incredibly harsh call on Vecino! It looked a lot worse originally than it did in the replay. The midfielder misses the ball and kicks out at a Mexican player, but hardly makes contact! He goes down in a heap, and the referee has a second to think before sending him packing!
***
[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big) [](#icon-whistle-big) **HALFTIME** [](#icon-whistle-big) [](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)
[](#sprite6-p112) MEXICO 1-0 URUGUAY [](#sprite6-p185) (OG Alvaro Pereira 4')
***
[](#icon-notes-) *STAT*: Uruguay (36) is the country with more players expelled throughout the history of the Copa America ahead of Peru (26) and Argentina (23). [^^Source: ^^MisterChip.](https://twitter.com/2010MisterChip/status/739622435188834305)
--
[](#icon-star) **SECOND HALF** [](#icon-star)
45' [](#icon-sub-big) Álvaro Rafael González Luengo [](#icon-up) for Marcelo Nicolás Lodeiro Benítez [](#icon-down)
45' KICKOFF!! [](#icon-whistle)
48' Uruguay start off encouragingly, actually working to get in possession and keep it - and it's working. They win a pair of free-kicks on the flanks, both of which are defended admirably by Mexico, however.
50' BUNDLED WIDE! A real cluster in the Mexico penalty area following a corner, and it looks like Cavani gets the final touch as the ball trickles inches wide of the near post! Mexico break quickly, and Guardado is brought down with a really harsh challenge that forces the training staff to come on and take a look.
54' After a nervy start to the second half, Mexico are back in possession and dictating the tempo in the attacking half. Guardado is back on the pitch as well, which is a great sight for the Mexican supporters.
54' [](#icon-sub-big) Hirving "Chuky" Lozano [](#icon-up) for Javier Aquino [](#icon-down)
55' Uruguay, despite playing with just 10 men, are looking an entirely different side since the restart. They're playing much more direct in possession, and threatening to create. Mexico need to get back to what they were doing so well in the first half, dominating the midfield third.
57' Attempt blocked. Giménez (Uruguay) right footed shot from the centre of the box is blocked.
57' [](#icon-flag-big) Offside, Uruguay. Giménez tries a through ball, but Diego Rolan is caught offside.
58' SHOCKING MISS! I mean that's almost unbelievable! Uruguay break in numbers and have three against one in the final third. Rolan is played clean through by Cavani, but he rolls his finish wide of the near post!
59' [](#icon-yellow-big) for José María Giménez de Vargas [](#sprite6-p185)
60' [](#icon-sub-big) Abel Hernández [](#icon-up) for Diego Rolan [](#icon-down)
60' [](#icon-sub-big) Jesús Dueñas [](#icon-up) for Jesús Manuel Corona [](#icon-down)
61' Attempt missed. Héctor Herrera (Mexico) left footed shot from outside the box misses to the right following a set piece situation
63' [](#icon-flag-big) Offside, Mexico. Jesús Dueñas tries a through ball, but Chicharito is caught offside.
65' It hardly looks like Uruguay are playing down a man. It's been very even out on the pitch, and the tempo is quickly picking up as we approach the final 25 minutes. Shots on target are proving hard to come by, but it looks like we're in for an exciting finish.
68' [](#icon-yellow-big) for Maxi Pereira [](#sprite6-p185)
Maxi Pereira slides in with a late challenge in an attempt to win the ball back, and he goes into the referee's book.
70' TERRIFIC SAVE! Lozano sprints down the left on a probing run and picks out Chicharito with an equally impressive cross. The striker slides in to meet it, but his re-direction is kept out by a diving Muslera! [](#icon-net-big)
72' ONTO THE ROOF OF THE NET! Talavera ends up in no man's land as he rushes out to collect a free-kick, but fails. It's kept alive at the far post and headed back into the danger area where Cavani meets it, but lifts his effort over the bar!
73' [](#icon-red-big) for Andres Guardado after getting his 2nd yellow. [](#sprite6-p112)
GUARDADO SENT PACKING! Oh my word what a shocking turn of events this is! Something happens behind the play after a Mexico foul, and the referee turns around and hands Guardado a second yellow card! This is absolutely unbelievable! It happens minutes after a member of the Mexico staff was sent to the stands.
--
74' [](#sprite6-p185) ***GOOOOOOOAAAAALLLLLLLL FOR URUGUAY!!!*** [](#sprite6-p185) What a remarkable turn of events! Godin latches onto the end of a free-kick and powers a header beyond Talavera! This is absolutely unbelievable!
[](#icon-ball-big) [GIF 1-1](https://streamable.com/cuhp), credit to /u/Omar_Til_Death [](#icon-alien-big)
78' It's 10 against 10 out on the pitch as we enter the final 10 minutes, and it's all square. We're headed for a terrific finish in Arizona, in what has been the best game of the 2016 Copa America to date
80' Mexico test Muslera again from range, this time through Herrera, but it's Uruguay who are on top at the moment. Mexico are readying Raul Jimenez, as both teams appear prepared to have an honest go at it over the final 20 minutes.
83' [](#icon-sub-big) Raul Jimenez [](#icon-up) for Javier "Chicharito" Hernandez [](#icon-down)
84' [](#icon-sub-big) Gastón Ramírez [](#icon-up) for Carlos Sánchez [](#icon-down)
84' [](#icon-yellow-big) for Diego Godín [](#sprite6-p185)
THE REFEREE HAS TAKEN THIS GAME OVER! Emotion is running high and Godin lashes out at a Mexican player, and the referee rushes across to show him yellow with real emphasis. He's showboating now, it seems.
--
85' [](#sprite6-p112) ***GOOOOOOOAAAAALLLLLLLL FOR MEXICO!!!*** [](#sprite6-p112) THE SKIPPER COMES THROUGH AT THE DEATH! Absolute scenes in this Group C match! The ball comes to Marquez following a corner, and he doesn't have much room to put the finish from a tight angle, but he punishes a sensational strike into the top corner on the near side
[](#icon-ball-big) [GIF 2-1](https://streamable.com/h7d7), credit to /u/Omar_Til_Death [](#icon-alien-big)
--
86' Muted celebrations from Marquez, the savvy vet, who acts like he's been there before - and he absolutely has. Uruguay protest furiously as they claim there was an offside in the build-up, but they hardly have much of a case.
88' The noise level in the stadium is off the charts. The Mexican supporters are screaming their hearts out in celebration, urging their side to see this absolutely crazy game through to the final whistle. It's all hands on deck now for Uruguay.
90' There will be four minutes added on at the end of this match... plenty of time for an equaliser... or a third for Mexico.
--
92' [](#sprite6-p112) ***GOOOOOOOAAAAALLLLLLLL FOR MEXICO!!!*** [](#sprite6-p112) The stadium sent to euphoria as El Tri hit back on the counter and put the game to bed! Lozano does so well down the left, picking out Jimenez at the near post, and he flicks it on for Herrera who slots a header into the empty net!
[](#icon-ball-big) [GIF 3-1](https://streamable.com/xatf), credit to /u/Omar_Til_Death [](#icon-alien-big)
--
93' [](#icon-yellow-big) for Raúl Jiménez [](#sprite6-p112)
Jimenez is booked, but he hardly cares at this point. Mexico on course for three points!
***
[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big) [](#icon-whistle-big) **FULLTIME** [](#icon-whistle-big) [](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)[](#icon-beer-big)
--
[](#sprite6-p112) MEXICO 3-1 URUGUAY [](#sprite6-p185)
--
Wild scenes in Glendale as late goals from Rafa Marquez and Hector Herrera send the 10-man El Tri off to a flying start against the 10-man Celeste. Mexico went ahead early, and looked poised to add to their lead when Vecino was issued a second yellow, but Guardado was shown red in the 73rd minute and Uruguay promptly equalised through Diego Godin.
That all set the scene for a memorable finish, as El Tri score two late in front of a predominantly Mexican crowd to move to the top of Group C after one round of matches. That's all for our live coverage, thanks for following along. Goodbye!
| 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 |
How to get images in Bootstrap's card to be the same height/width?. <p>So here is my code, it displays 6 cards, three across and two rows. I would like for the images to all be the same size without having to manually resize the images. The responsiveness does work, I use "img-fluid" as a class and when I go to a mobile or smaller browser, they all have the same width, but the height is still off. </p>
<pre><code><h1 class="display-4 text-xs-center m-y-3 text-muted" id="speakers">Ice Cream</h1>
<div class="row">
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/brownie.jpg" />
<div class="card-block">
<h4 class="card-title">Brownie Delight</h4>
<p class="card-text">Our customer favorite chocolate ice cream jam packed with pieces of brownies and fudge</p>
</div>
</div>
</div>
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/butterPecan.jpg" />
<div class="card-block">
<h4 class="card-title">Butter Pecan</h4>
<p class="card-text">Roasted pecans, butter and vanilla come together to make this wonderful ice cream</p>
</div>
</div>
</div>
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/bCherry.jpg" />
<div class="card-block">
<h4 class="card-title">Black Cherry</h4>
<p class="card-text">Our classic vanilla loaded with plump black cherries to give flavor and color</p>
</div>
</div>
</div>
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/mintChip.jpg" />
<div class="card-block">
<h4 class="card-title">Mint Chip</h4>
<p class="card-text">Our signiture mint ice cream jam packed with mint chocolate chips</p>
</div>
</div>
</div>
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/pistachio.jpg" />
<div class="card-block">
<h4 class="card-title">Pistachio</h4>
<p class="card-text">Our classic pistachio is loaded with nuts to give it that great flavor, and of course comes in your favorite color</p>
</div>
</div>
</div>
<div class="col-md-6 col-lg-4">
<div class="card"><img alt="Card image cap" class="card-img-top img-fluid" src="img/IceCream.jpg" />
<div class="card-block">
<h4 class="card-title">More Flavors</h4>
<p class="card-text">We couldn not fit all of our wonderful flavors on one page, click here to see more!</p>
</div>
</div>
</div>
</div>
</code></pre>
<p>Here is an <a href="https://imgur.com/cxEkvmc" rel="noreferrer">image</a> of what I am getting and this is what I want to <a href="https://imgur.com/pT8Uctr" rel="noreferrer">get</a> where they are all the same size.</p>
| 0non-cybersec
| Stackexchange |
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