text stringlengths 49 10.4k | source dict |
|---|---|
Semenovich Pontryagin, a mathematician, composed by himself". It is available online at <a href="http://www.ega-math.narod.ru/LSP/book.htm" rel="nofollow">http://www.ega-math.narod.ru/LSP/book.htm</a>, in the original Russian. Google does a fairly good job of translation, although it refuses to translate the individual chapters completely because of their length.</p> <p>In the book, Pontryagin shares a lot about the inner workings of the IMU Executive Board and his own role in holding the Soviet party line there as its vice president. For example, he recounts his version of how France got the IMU presidency in 1974, so that neither the Soviet Union nor the US would dominate. </p> <p>The only relevant mention of Arnold that I could find in that book is in chapter 5. He states that in 1974 Arnold was not allowed to leave the country to lecture abroad, and that there was a conflict about this with the Executive Board of the IMU, who insisted that he should. From this, you could extrapolate the reasons for blocking Arnold's Fields medal, if the story is true.</p> http://mathoverflow.net/questions/25922/is-the-euler-characteristic-a-birational-invariant/25965#25965 Answer by VA for Is the Euler characteristic a birational invariant VA 2010-05-26T04:19:02Z 2010-05-26T16:03:27Z <p>The dimensions $h^i(\mathcal O_X)$ of the cohomology groups of $\mathcal O_X$, and thus the Euler characteristic, are birational invariants of <em>smooth</em> proper varieties <em>in positive characteristic</em> as well, by a recent work of Andre Chatzistamatiou and Kay Rülling. It is not published yet but a <a href="http://arxiv.org/abs/0911.3599" rel="nofollow">preprint is available.</a> | {
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slam, navigation, ros-kinetic, rtabmap-odometry, rtabmap
[ INFO] [1556631074.632802948, 1555961486.551916526]:
/rtabmap/rtabmap subscribed to (approx sync):
/rtabmap/rgb/image,
/rtabmap/depth/image,
/rtabmap/rgb/camera_info,
/scan
[ INFO] [1556631074.638748810, 1555961486.551916526]: rtabmap 0.17.6 started...
[ WARN] [1556631079.637634199, 1555961491.557643367]: /rtabmap/rtabmap: Did not receive data since 5 seconds! Make sure the input topics are published ("$ rostopic hz my_topic") and the timestamps in their header are set. If topics are coming from different computers, make sure the clocks of the computers are synchronized ("ntpdate"). If topics are not published at the same rate, you could increase "queue_size" parameter (current=10).
/rtabmap/rtabmap subscribed to (approx sync):
/rtabmap/rgb/image,
/rtabmap/depth/image,
/rtabmap/rgb/camera_info,
/scan | {
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"tags": "slam, navigation, ros-kinetic, rtabmap-odometry, rtabmap",
"url": null
} |
biochemistry, ph, dna-rna
Title: net charge nucleobases at alkaline conditions While doing anion exchange chromatography of a short oligonucleotide, I have noticed that at $\rm pH=12$, the oligo retains longer on the column than the same oligo- at $\rm pH=8$. I thought the reason for this could be the change in the net charge of nucleobases.
Could anyone suggest what is the net charge of RNA nucleobases (adenine, guanine, cytosine, uracil) at $\rm pH=12$ (denaturing conditions)? The conjugate acids of the adenine and guanine have pKa's between 9-10. Guanine also has a nitrogen with a pKa of 12.3, which would be mostly protonated at pH 12. The pKa for the phosphoric acid proton of a nucleotide is significantly lower than 7.
Therefore, in a pH 8 buffer, the basic nitrogens of each adenine and guanine are protonated, while the phosphate is deprotonated. This gives an overall net charge closer to zero than you would see at pH 12.
In a pH 12 buffer, where the pH is above the pKa of both the phosphoric acid protons and the nitrogen conjugate acid protons (except the last one on each guanine), all of these positions would be deprotonated. Therefore, the formal charge on the nitrogens would be zero (again, except the most basic nitrogen on guanine), but the phosphates would still have a charge of -1 each. This would lead to a net charge of significantly higher charge magnitude because there would be very few positive charges to balance out the negative charges.
At pH 12, the net charge should be approximately: $$(n_N +n_u) - (n_c + n_g).$$
Approximate net charge at pH 8: $$n_N - (n_a + n_c + 2n_g).$$
nN= number of nucleotides
nu= number of uracils/thymines
nc= number of cytosines
ng= number of guanosines
na= number of adenosines | {
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ros, rviz, ros-kinetic
Title: Autoware Mission Planning, Client has wrong datatype version: datatype/md5sum
When I try to launch mission planning node for San Francisco data launch files, I end up getting this error:
Error: Client [/rivz_1563460306167327804] wants topic /local_trajectories to have datatype/md5sum [visualization_msgs/MarkerArray/d155b9ce5188fbaf89745847fd5882d7], but our version has [autoware_msgs/LaneArray/23abb100bdfa4ee58530bb628c974c2a]. Dropping connection.
Does anybody know what the problem is?
Thank you
Originally posted by hdossaji94 on ROS Answers with karma: 11 on 2019-07-18
Post score: 1
It means they were compiled with different versions of the message files. When the message files are compiled they make a MD5 hash in case those message definitions change to make sure they're able to be decoded correctly by the receiver. You need to have both programs using the right set of definitions installed.
Edit: More info, sorry I scanned over the types and missed the fact they are literally different messages.
Ok, so what you're trying to do here is expose a topic of type autoware_msgs/LaneArray on the topic /local_trajectories. You've created the data vis plugin for local_trajectories to have type MarkerArray, which is not LaneArray. Your issue is that you have the wrong messages being broadcast on that topic. You need to create the correct plugin visualization to see the LaneArray (does autoware provide?)
Originally posted by stevemacenski with karma: 8272 on 2019-07-18
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by hdossaji94 on 2019-07-18:
Thanks for the quick response!
Would you mind giving me some insight into the steps involved to do this?
Comment by stevemacenski on 2019-07-18:
See update in answer, you're publishing the wrong message type on that topic to be visualized as a MarkerArray. | {
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set.seed(0)
library(tidyverse)
simulate_data<-function(n){
x = rnorm(n)
y = 2*x + 1 + rnorm(n, 0, 0.5)
model = lm(y~x)
results = tibble(beta = coef(model)['x'],
w = 1/vcov(model)['x','x'])
}
simulate_procedure<-function(iter){
n = rnbinom(3,200,0.9)
results = map_dfr(n, simulate_data)
m = sum(results$$beta*results$$w)/sum(results$w) sig = sqrt(1/sum(results$w))
interval = tibble(lower = m - 1.96*sig,
est = m,
upper = m + 1.96*sig)
interval
}
map_dfr(1:10000, simulate_procedure, .id = 'iter') %>%
mutate(contains = (2<upper)&(2>lower)) %>%
summarise(mean(contains))
>>>0.922
So what does this mean? It means that were I to repeat this procedure to construct a 95% interval for the slope, the resulting interval would capture the true slope (here 2) only 92% of the time. So barring I didn't make a mistake (entirely possible) that seems to be good enough.
How should the estimate of the slope be reported, including errors? Let's imagine I only have access to these values (and not the underlying data that was used for fitting to obtain these slopes).
So I would compute $$m$$ and $$\sigma^2$$ as mentioned by Yair Daon. You don't need to access the data in order to do these. In your example, the $$m$$ would be 5.5, 5.5, 5.2. The variances are found by doing a little algebra on the confidence interval. Remember, confidence intervals look like
$$m \pm 1.96 se$$ | {
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html, css, html5
<div id="mailConfig">
<h2>Configure your email settings</h2>
<table>
<tr>
<th>Frequency: </th>
<td>
<select>
<option value="daily">Daily</option>
<option value="weekly">Weekly</option>
<option value="monthly">Monthly</option>
</select>
</td>
</tr>
<tr>
<th>Time: </th>
<td>
<input type="time" value="07:00" />
</td>
</tr>
<tr>
<th>Amount of posts to show:</th>
<td>
<input type="number" value="10" max="50" min="1" />
</td>
</tr>
</table>
</div>
</div>
<div id="userConfig">
<button class="accountOption">
<strong>Login</strong>
<hr />
<br />
<br />
<i>
You will have the choice between several oAuth providers.
</i>
</button>
<button class="accountOption">
<strong>Register</strong>
<hr />
<br />
<br />
<i>
Registering an account has the advantage that you can manage your subscriptions.
</i>
</button>
<button class="accountOption">
<strong>Anonymous</strong>
<hr />
<br />
<br />
<i>
This means that you will not be able to change the subscription once it's finished.
</i>
</button>
</div>
</div>
<div id="footer">
<em>Made by Jeroen Vannevel</em>. Visit me at <a href="http://www.vannevel.net" title="blog">my blog</a>.
</div>
<script type="text/javascript" src="../js/app.js"></script>
</body>
</html> Semantic Markup | {
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"url": null
} |
species-identification, marine-biology
Title: help identify this fish
I came across this washed up fish in Panama City, Florida in November 2015. I'm guessing it's a puffer fish but I can't find anything like it online.
Thanks. This is a kind of trunkfish. (They have different names, this could be a smooth or spotted trunkfish.). It's really a lovely and comical little fish when observed alive in coral reefs. It has the ability to change its coloration depending on whether it's excited or calm, or to minimize its contrast to the background. It is related to puffer fish.
It has a boxy, triangular body shape, and propels itself with relatively tiny, delicate fins. Like pufferfish, they are toxin producers.
In death, the body shape and coloration are different, of course. Never saw a dead one before; sad. The juveniles are adorable:
Members of this family occur in a variety of different colors, and are notable for the hexagonal or "honeycomb" patterns on their skin. - Wikipedia | {
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programming, error-correction, stabilizer-code, stabilizer-state
Title: Are there any packages that can calculate stabilizer tableau of a QECC I'm experimenting with some small quantum error correcting codes (QECC). For example
$[[5,1,3]]$, $[[8,3,3]]$ or toric codes $[[2d^2,2,d]]$ ($d=2,3,\cdots$). The last one
being defined by redundant stabilizers. What packages can take in a set of $m'$ stabilizers
and produce the tableau? ($m'$ could be larger than $m=n-k$ in case stabilizers are not independent).
Here are the details for the $[[8,3,3]]$ code : the code is in standard form; $1=X$, $2=Z$, $3=XZ$;
[[3,0,2,2,1,1,3,0],
[2,3,0,2,1,3,0,1],
[2,0,3,0,1,2,1,3],
[2,2,0,1,0,1,3,3],
[2,2,2,2,2,2,2,2],
[2,2,0,2,0,2,0,0],
[2,0,2,2,0,0,2,0],
[0,2,2,2,0,0,0,2],
[2,0,0,0,0,0,0,0],
[0,2,0,0,0,0,0,0],
[0,0,2,0,0,0,0,0],
[0,0,0,2,0,0,0,0],
[0,0,0,0,1,0,0,0],
[0,2,2,0,1,1,0,0],
[2,0,0,2,1,0,1,0],
[0,0,2,2,1,0,0,1]] | {
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ros, gazebo, ros-controllers, ros-control
This is my CMakeList.txt
find_package(catkin REQUIRED COMPONENTS
roscpp
rospy
std_msgs
controller_manager
ros_control
ros_controllers
)
Do I need to include ros_controller and ros_controllers? When I remove them I get thses errors,
[ERROR] [1451315533.235070013, 0.140000000]: No valid hardware interface element found in joint 'front_right_wheel_joint'.
[ERROR] [1451315533.235176421, 0.140000000]: Failed to load joints for transmission 'front_right_wheel_joint_trans'.
[ERROR] [1451315533.235239698, 0.140000000]: No valid hardware interface element found in joint 'front_left_wheel_joint'.
[ERROR] [1451315533.235290363, 0.140000000]: Failed to load joints for transmission 'front_left_wheel_joint_trans'.
[ERROR] [1451315533.235343496, 0.140000000]: No valid hardware interface element found in joint 'back_right_wheel_joint'.
[ERROR] [1451315533.235391198, 0.140000000]: Failed to load joints for transmission 'back_right_wheel_joint_trans'.
[ERROR] [1451315533.235439370, 0.140000000]: No valid hardware interface element found in joint 'back_left_wheel_joint'.
[ERROR] [1451315533.235487324, 0.140000000]: Failed to load joints for transmission 'back_left_wheel_joint_trans'
I have been stuck on this for over a day, can anyone please help me? I am very confused.
Originally posted by ngoldfarb on ROS Answers with karma: 127 on 2015-12-28
Post score: 0
CMake is telling you that it can't find the ros_control package.
You should try installing it:
On Ubuntu:
sudo apt-get install ros-indigo-ros-control | {
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formal-languages, automata
Title: NFA where there are two 0s separated by a multiple of 4 I've been following Automata and Formal Languages in my college and I came across
this particular exercise.
While the solution presented seems correct, I take on Automata Tutor trying this exercise a 0 out of 10. I can't understand what was wrong here cause overall it seems correct. Thanks a lot if you can help me. There's a logic contradiction in the exercise; I tried both combinations but nothing seems to work. Tried the logic as Russel said, having a loop composed in this way. Also tried Steven reasoning, given the thought behind that. Seems like the automata at least in Automata Tutor has to begin with both 0 and 1, having a loop composed of 0/1, given the hints the exercise gives, then ending with a 1. The moment you put 0 as the second state goes into the loop and asks for the correct string to be 00000. Then one puts final state the fifth zero and the exercise combines other requests. Upon satisfying them all, I conclude that simply, the exercise itself is not totally correct behind, trying both of the logics linked here.
Given this kind of exercise, which you can also find online in other places are implemented as I've shown before or in one other way, a little different but still, doesn't work on Automata Tutor. It's wrong on how the question is posed, I'll tell the teacher on that.
As also Hendrik noted, it's contradicting in the requests, cause one misinterpretes the other ones and it's all conflicting upon itself. | {
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$\text{Example: }f\left(x\right)=\frac{3{x}^{2}-2x+1}{x - 1}$
In this case, the end behavior is $f\left(x\right)\approx \frac{3{x}^{2}}{x}=3x$. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function $g\left(x\right)=3x$. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of $g\left(x\right)=3x$ looks like a diagonal line, and since f will behave similarly to g, it will approach a line close to $y=3x$. This line is a slant asymptote.
To find the equation of the slant asymptote, divide $\frac{3{x}^{2}-2x+1}{x - 1}$. The quotient is $3x+1$, and the remainder is 2. The slant asymptote is the graph of the line $g\left(x\right)=3x+1$.
Figure 13. Slant Asymptote when $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0$ where degree of $p>\text{degree of }q\text{ by }1$. | {
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# Binomial Puzzles: Warwick Statistics Department Puzzle
The Department of Statistics at Warwick recently started a new outreach series of puzzles. I found the first one quite interesting. This post is my proposed solution to it.
### My Solution
So, we are told that we have $$n$$ fair coins and we flip them. We remove those that landed tails and flip again the heads. We need to give the distribution of the number of heads after this second round. This means that if we denote as $$X$$ the number of coins we would flip in the second round and $$H$$ the number of final heads we have: $X\sim Bin(n, p = 0.5) \quad\text{ and }\quad H|X=x \sim Bin(x, p = 0.5),$ we are asked about the unconditional distribution of $$H$$.
However we can rephrase the problem as follows:
1. Flip each of the $$n$$ fair coins twice
2. Call it a success whenever both flips are heads.
3. What is the distribution of the number of succesesses?
This is equivalent, but it makes the solution apparent. We are being asked about another Binomial distribution, just that now the probability of success is that of having two heads in two consecutive independent fair coin tosses; which is just $$0.5(0.5)=0.25$$. Hence, $H \sim Bin(n, p = 0.25).$
If you are still not convinced, let’s provide a mathematical proof. In fact, we can even generalize this problem by not requiring the tosses to be fair or even have the same probability in both rounds. The more general result is that $H \sim Bin(n, p = p_1p_2),$ where $$p_i$$ is the probability of heads in the $$i$$-th round. | {
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"openwebmath_score": 0.9218857288360596,
"tags": null,
"url": "https://www.fazepher.me/post/2022-02-06_wstats_puzzle_01/"
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temperature
Title: Unable to find sources for uncommon specific heat I'm calculating the energy, temperature, and gas volume outputs of gunpowder. To solve this I need to find the specific heat of Potassium sulfate (K2SO4) and Potassium carbonate (K2CO3). If anyone can refer me to any directory that can list the values for these it would be very helpful.
Thank you for your time. For potassium sulfate see here.
For potassium carbonate see here.
Generally, search in the NIST Chemistry WebBook and when you find the compound of interest, either click on "gas phase thermochemistry data" or "condensed phase thermochemistry data" as appropriate. | {
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enough to say there is continuity? A continuous function is a function whose graph is a single unbroken curve. The continuity of a function of two variables, how can we determine it exists? A formal epsilon-delta proof for the Continuity Law for Composition. However, continuity and Differentiability of functional parameters are very difficult. Equipment Check 1: The following is the graph of a continuous function g(t) whose domain is all real numbers. A limit is defined as a number approached by the function as an independent function’s variable approaches a particular value. or … Definition 3 defines what it means for a function of one variable to be continuous. 2. lim f ( x) exists. Calculate the limit of a function of two variables. One-Sided Continuity . Your function exists at 5 and - 5 so the the domain of f(x) is everything except (- 5, 5), but the function is continuous only if x < - 5 or x > 5. With that kind of definition, it is easy to confuse statements about existence and about continuity. State the conditions for continuity of a function of two variables. A function f(x) is continuous on a set if it is continuous at every point of the set. Learn continuity's relationship with limits through our guided examples. Each topic begins with a brief introduction and theory accompanied by original problems and others modified from existing literature. A discontinuous function then is a function that isn't continuous. A function f(x) can be called continuous at x=a if the limit of f(x) as x approaching a is f(a). When you are doing with precalculus and calculus, a conceptual definition is almost sufficient, but for … Combination of these concepts have been widely explained in Class 11 and Class 12. Find out whether the given function is a continuous function at Math-Exercises.com. Example 17 Discuss the continuity of sine function.Let ()=sin Let’s check continuity of f(x) at any real number Let c be any real number. Definition of Continuity at a Point A function is continuous at a point x = c if the following three conditions are met 1. f(c) is defined 2. The points of continuity are points where a function exists, that it has some real value at that point. Proving Continuity The de nition of continuity gives you a fair amount of information about a function, but this is all a waste of time unless you | {
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php
Title: write database connection code in many files There are many files [ file1.php, file2.php ] in our project & i wrote database connection code in many files instead of single file [ database.php ] , is there any problem with that ?
if it is wrong than please help me what code i need to change in file1.php to connect to database.php
Below 2 lines i used in file1.php
$con=mysqli_connect('localhost', 'root', 'gfgf', 'g');
if ($result=mysqli_query($con,$sql))
Instead of above 2 lines, is it correct to use below 2 lines ?
require_once("database.php");
$db_handle = new DBController();
file1.php
<?php
require_once("database.php");
$db_handle = new DBController();
$con=mysqli_connect('localhost', 'root', 'gfgf', 'g');
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$todaydate=date('Y-m-d');
$sql="SELECT * FROM orders WHERE importdate >= NOW() - INTERVAL 1 DAY";
if ($result=mysqli_query($con,$sql))
{
$rowcount=mysqli_num_rows($result);
printf("Uploaded - %d rows.\n",$rowcount);
// Free result set
mysqli_free_result($result);
}
mysqli_close($con);
?>
database connection code in many files instead of single file, is there any problem with that?
Nothing critical, but what if your database credentials would change? Are you going to edit all these files?
is it correct to use below 2 lines ? | {
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dipole
Title: Why do opposite dipoles of same charge and equal magnitude cancel each other out?
In carbon dioxide, there are 2 dipoles of equal magnitude and charge pointing in opposite directions.
Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity?
Why is carbon dioxide truly nonpolar? Why does this make sense? Sean's answer is definitive, but I think it's worth adding a few comments.
Why can't both oxygen atoms obtain a partially negative charge while the carbon atom contains a partially positive charge twice in magnitude? Shouldn't two oxygen atoms pulling on a single carbon atom result in more positivity?
The oxygen atoms do get a partial negative charge and the carbon atom does get a partial positive charge. The partial positive charge on the carbon is twice the magnitude of the charges on the oxygen atoms - your diagram is a bit misleading in this respect. I would draw it as:
However this does not produce a dipole moment for the reasons Sean explains. The first non-zero moment is the quadrupole moment. Your title asks:
Why do opposite dipoles of same charge and equal magnitude cancel each other out?
and the answer is simply that the dipole moment is a vector, and the dipoles of the two halves of the molecule add in the usual way vectors add. Since the two dipoles are equal and opposite they add up to zero.
Why is carbon dioxide truly nonpolar? Why does this make sense?
Be careful with the word nonpolar. It does not mean has no dipole moment. Actually I'm not sure that polar has a precise meaning but it generally means interacts with electric fields. Carbon dioxide interacts with electric fields because it has a quadrupole moment, and also because it is polarisable. | {
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inorganic-chemistry, acid-base, stoichiometry, concentration, mole
Title: Neutralization of alkali solution by a mix of strong acids
$\pu{0.1 M}$ $\ce{HCl}$ and $\pu{0.2 M}$ $\ce{H2SO4}$ solutions are mixed in equal volume. This solution is diluted to double the volume. Find the volume of $\pu{0.1 M}$ $\ce{NaOH}$ which can be neutralized with $\pu{20 mL}$ of the former solution. | {
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This inequality is famous and an easy consequence of Mean Value Theorem. We have $a, b, p, q$ as positive and $1/p+1/q=1$. Let $\alpha=1/p,\beta=1/q$ and replace $a^p, b^{q}$ by $a, b$ to get the inequality in the form $$a^{\alpha} b^{\beta} \leq\alpha a+\beta b$$ where $\alpha +\beta=1$. There is equality if $a=b$ and hence let $a<b$ (for $a>b$ we can interchange the roles of $a, b$).
Consider $f(x) =x^{\beta}$ so that $f'(x) = \beta x^{-\alpha}$ and by mean value theorem we can see that $$b^{\beta} - a^{\beta} =(b-a) \beta c^{-\alpha}$$ for some $c\in(a, b)$. And since $c^{-\alpha} <a^{-\alpha}$, it follows that $$b^{\beta} - a^{\beta} <\beta(b-a) a^{-\alpha}$$ or $$a^{\alpha} b^{\beta} <\beta(b-a) +a=\alpha a+\beta b$$ Also note that if $\alpha, \beta$ are rational then there is no need of calculus and the inequality is an immediate consequence of AM-GM inequality.
From Wikipedia's article on concave functions: "A differentiable function $f$ is concave on an interval if and only if its derivative function $f′$ is monotonically decreasing on that interval, that is, $f''<0$."
Note that $(\log(x))''=(1/x)'=-1/x^{2}<0$, so $\log$ is concave, and your inequality follows. | {
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measurement, povm
We can also think of $M$ as a two-argument function that takes a measurement outcome $i$ and a quantum state $\rho$ and yields the probability of measuring $i$ on $\rho$
$$
M: A\times D(\mathcal{H})\to[0,1]\tag1
$$
where $D(\mathcal{H})$ denotes the set of density operators on some Hilbert space $\mathcal{H}$. We can write down an explicit formula for the function $M$
$$
M(i,\rho)=\mathrm{tr}(M_i\rho)\tag2
$$
which is just another statement of the Born rule. In certain ways, the function $M$ is similar to a good old probability distribution over $A$ with the caveat that the distribution is parametrized by quantum states $\rho$.
Probability distributions
A probability distribution on $A$ is formalized as a probability measure defined as a function on a certain $\sigma$-algebra of subsets of $A$ called events. However, in the countable case, the fact that the definition requires a measure to be $\sigma$-additive implies that it is completely defined by its values on a partitioning of $A$. In particular, it is completely defined by specifying its value on singleton sets. For this reason, we can often forget about the $\sigma$-algebra and think of a probability distribution on $A$ as a function $p:A\to[0,1]$ that assigns a probability $p(i)$ to each $i\in A$.
Now, the POVM $M$ is not quite as direct. Instead of assigning to $i\in A$ a probability $p(i)\in[0,1]$, it assigns to $i\in A$ a positive operator $M_i$ (which explains the name Positive Operator-Valued Measure). To obtain an actual probability (of $i$ being the outcome of a measurement), a POVM needs one more piece of information: the quantum state $\rho$. This accounts for the fact that in quantum mechanics measurement outcome probability depends on both: the outcome $i$ and the quantum state $\rho$. | {
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Suppose the editors finally decide on the following weights to employ: temperature, $$0.23\text{;}$$ colleges, $$0.46\text{;}$$ Superfund, $$-0.05\text{;}$$ crime, $$-0.20\text{.}$$ Notice how negative weights are used for undesirable statistics. Then, for example, the editors would compute for Los Angeles
\begin{equation*} (0.23)(77) + (0.46)(28) + (-0.05)(93) + (-0.20)(254) = -24.86 \end{equation*}
This computation might remind you of an inner product, but we will produce the computations for all of the cities as a matrix-vector product. Write the table of raw statistics as a matrix
\begin{equation*} T= \begin{bmatrix} 77 & 28 & 93 & 254\\ 84 & 38 & 85 & 363\\ 84 & 99 & 1 & 193 \end{bmatrix} \end{equation*}
and the weights as a vector
\begin{equation*} \vect{w}=\colvector{0.23\\0.46\\-0.05\\-0.20} \end{equation*}
then the matrix-vector product (Definition MVP) yields
\begin{equation*} T\vect{w}= (0.23)\colvector{77\\84\\84}+ (0.46)\colvector{28\\38\\99}+ (-0.05)\colvector{93\\85\\1}+ (-0.20)\colvector{254\\363\\193} = \colvector{-24.86\\-40.05\\26.21}\text{.} \end{equation*} | {
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moveit, ros-kinetic
<< " set_payload(" << req.payload << ", [" << req.center_of_gravity.x << ", " << req.center_of_gravity.y
^
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp:394:98: error: ‘ur_msgs::SetPayload::Request {aka struct ur_msgs::SetPayloadRequest_<std::allocator<void> >}’ has no member named ‘center_of_gravity’
<< " set_payload(" << req.payload << ", [" << req.center_of_gravity.x << ", " << req.center_of_gravity.y
^
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp:395:28: error: ‘ur_msgs::SetPayload::Request {aka struct ur_msgs::SetPayloadRequest_<std::allocator<void> >}’ has no member named ‘center_of_gravity’
<< ", " << req.center_of_gravity.z << "])" << std::endl
^
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp: In lambda function:
/home/shu/catkin_ws/src/Universal_Robots_ROS_Driver/ur_robot_driver/src/ros/hardware_interface.cpp:394:63: error: ‘ur_msgs::SetPayload::Request {aka struct ur_msgs::SetPayloadRequest_<std::allocator<void> >}’ has no member named ‘center_of_gravity’ | {
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javascript, object-oriented, classes
Title: Is this a good JavaScript class? I'm trying to get my head around object oriented programming, and I'm starting in JavaScript. I have put a working demo of my code on JSFiddle (here), but the class is below.
My main concerns are that as soon as the object is called with a constructor, all the properties are populated straight away (there are no methods). Should I have put some of the contructor code into methods?
function XEString(encryptedString) {
if (!XEString.test(encryptedString)) { //Throw exception if invalid XEcrypt string
throw new Error("Invalid XECrypt string");
} else {
this.encryptedString = encryptedString; //Save encrypted string to object
//Remove first "." char and put numbs into array, then convert numbs from strs to ints
this.encryptedNumbers = this.encryptedString.substring(1).split(".");
for (var i = 0; i < this.encryptedNumbers.length; i++) {
this.encryptedNumbers[i] = parseInt(this.encryptedNumbers[i], 10);
}
//Each char is calced from a group of 3 numbs
this.stringLength = this.encryptedNumbers.length / 3; | {
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classical-mechanics, energy, hamiltonian-formalism, hamiltonian, phase-space
Title: Why does time-independent Hamiltonian not depend on angle variable? In Landau and Lifshitz Mechanics, $\S50$ Canonical variables a time-independent Hamiltonian is considered, and a canonical transformation is done such that adiabatic invariant $I$ becomes the new momentum. Then the angle variable is found as
$$w=\frac{\partial S_0(q,I;\lambda)}{\partial I},$$
where $S_0$ is abbreviated action (and generating function for the canonical transformation), $q$ is old position variable and $\lambda$ is a constant parameter.
Now L&L say:
Since the generating function $S_0(q,I;\lambda)$ does not depend explicitly on time, the new Hamiltonian $H'$ is just $H$ expressed in terms of the new variables. In other words, $H'$ is the energy $E(I)$, expressed as a function of the action variable. Accordingly, Hamilton's equations in canonical variables are
$$\dot I=0,\;\;\;\dot w=\frac{\mathrm dE(I)}{\mathrm dI}.\tag{50.4}$$ | {
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# What does $\binom{-n}{k}$ mean?
For positive integers $n$ and $k$, what is the meaning of $\binom{-n}{k}$?
Specifically, are there any combinatorial interpretations for it?
Edit: I just came across Daniel Loeb, Sets with a negative number of elements, Advances in Mathematics. 91 (1992), 64-74, which includes a combinatorial interpretation for $\binom{n}{k}$ for any $n,k \in \mathbb{Z}$ in theorem 5.2.
-
(So far mine is the only one of four answers to give the combinatorial interpretation. But further edits seem to be happening fast. . . . .) – Michael Hardy Oct 20 '12 at 21:57
If $x$ is any complex number and $k$ is a non-negative integer then one can take $\dbinom x k$ to be $$\frac{x(x-1)(x-2)\cdots(x-k+1)}{k!}$$ (so that the number of factors in the numerator is $k$, as in the denominator). In particular, $\dbinom x 0=1$.
Combinatorial interpretation:
If $x$ is a positive integer, then $\left|\dbinom {-x}{k}\right|$ is the number of multisets of size $k$ in a set of size $x$.
-
By including $x-1$ boundary markers with $k$ identical elements, I usually think of the number of multisets of size $k$ in a set of size $x$ as $\binom{x+k-1}{k}$, which does happen to equal $\left|\binom{-x}{k}\right|$. However, I don't think I've ever gone from that interpretation to $\left|\binom{-x}{k}\right|$. – robjohn Oct 20 '12 at 22:49
This is the kind of thing I was looking for when I asked for a combinatorial interpretation. Thanks. – Snowball Oct 22 '12 at 21:21 | {
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python, interview-questions, circular-list
if lst1 == lst2:
return True
if set(lst1) != set(lst2):
return False
index = lst2.index(lst1[0])
#split list at index point, compare the lists
rot2 = lst2[index:] + lst2[:index]
return lst1 == rot2
Broken code
As pointed out by Oscar Smith, your code does not work.
For that kind of programming task, it is very easy to write tests to help you catch issues. These can be written before, during and after writing the code. You could use a proper testing framework or just simple assertions like:
# rotation
lst1, lst2 = [1,2,3,4,6,4,7], [6,4,7,1,2,3,4]
assert is_rotated(lst1, lst2)
# rotation with repeated numbers
lst1, lst2 = [1,2,3,4,6,4,7,1], [6,4,7,1,1,2,3,4]
assert is_rotated(lst1, lst2)
# different set
lst1, lst2 = [1,2,3,4,6,4,6], [6,4,7,1,2,3,4]
assert not is_rotated(lst1, lst2)
lst1, lst2 = [1,2,3,4,6,4,7], [6,4,6,1,2,3,4]
assert not is_rotated(lst1, lst2)
# equal
lst2 = lst1
assert is_rotated(lst1, lst2)
# empty
lst1, lst2 = [], []
assert is_rotated(lst1, lst2)
# 1 empty, 1 not empty
lst1, lst2 = [], [1]
assert not is_rotated(lst1, lst2)
lst1, lst2 = [1], []
assert not is_rotated(lst1, lst2) | {
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## $\frac{0}{0}$ Is indeterminant
The division $\frac{0}{0}$ is indeterminant.
## Sample set b
Perform, if possible, each division.
$\frac{\text{19}}{0}$ . Since division by 0 does not name a whole number, no quotient exists, and we state $\frac{\text{19}}{0}$ is undefined
$\begin{array}{c}\hfill 0\overline{)14}\end{array}$ . Since division by 0 does not name a defined number, no quotient exists, and we state $\begin{array}{c}\hfill 0\overline{)14}\end{array}$ is undefined
$\begin{array}{c}\hfill 9\overline{)0}\end{array}$ . Since division into 0 by any nonzero whole number results in 0, we have $\begin{array}{c}\hfill 0\\ \hfill 9\overline{)0}\end{array}$
$\frac{0}{7}$ . Since division into 0 by any nonzero whole number results in 0, we have $\frac{0}{7}=0$
## Practice set b
Perform, if possible, the following divisions.
$\frac{5}{0}$
undefined
$\frac{0}{4}$
0
$\begin{array}{c}\hfill 0\overline{)0}\end{array}$
indeterminant
$\begin{array}{c}\hfill 0\overline{)8}\end{array}$
undefined
$\frac{9}{0}$
undefined
$\frac{0}{1}$
0
## Calculators
Divisions can also be performed using a calculator.
## Sample set c
Divide 24 by 3.
Display Reads Type 24 24 Press ÷ 24 Type 3 3 Press = 8
The display now reads 8, and we conclude that $\text{24}÷3=8$ .
Divide 0 by 7. | {
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c++, embedded
void AppendError(const CString& error);
// I think I need to make the following 2 functions return BOOLs or INTs based on whether or not they failed.
void CreateJSONSLD();
void ExportData(int testNumber, string path = ".\\SysLat_Logs", int totalLogs = 10000);
bool dataExported = false;
bool dataUploaded = false;
};
#endif
Any help you can provide is greatly appreciated. Avoid macros
You should declare constants as constexpr variables instead of as #defines, like so:
constexpr std::size_t MOVING_AVERAGE = 100; | {
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python, performance, programming-challenge, primes
primes = [i for i in gen_primes(10000000)]
to
primes = set(i for i in gen_primes(10000000))
you will see a significant speedup.
The algorithm could also be improved. I don't want to go into too much detail here because that's not in keeping with the ethos of Project Euler, but I think I can give you a couple of hints:
The problem statement says that there are only eleven primes matching the criterion. What property of primes with the criterion allowed them to prove that? How can you apply that property to generate them faster?
If you have found all of the truncatable primes of length n, do you need to loop over all truncations of a candidate number of length n+1? | {
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c#, performance, winforms, timer
ProcessInfo.cs
The small wrapper class ProcessInfo is now moved to a separate file. The content remains unchanged :
internal class ProcessInfo
{
public Process Process { get; }
public TimeSpan TimeActive { get; set; }
public ProcessInfo(Process process, TimeSpan timeActive)
{
Process = process;
TimeActive = timeActive;
}
}
ProcessInspector.cs
I've moved all of the general purpose methods operating on objects of type Process to a separate dedicated class called ProcessInspector.
public static class ProcessInspector
{
public static string GetProcessPath(Process process)
{
try
{
string query = "SELECT ExecutablePath FROM Win32_Process WHERE ProcessId = " + process.Id;
var searcher = new ManagementObjectSearcher(query);
var collection = searcher.Get();
return collection.Cast<ManagementObject>().First()["ExecutablePath"].ToString();
}
catch
{
return string.Empty;
}
}
public static string GetProcessOwner(int processId)
{
string query = "SELECT * FROM Win32_Process WHERE ProcessId = " + processId;
var searcher = new ManagementObjectSearcher(query);
var processList = searcher.Get();
var managementObject = processList.Cast<ManagementObject>().First();
string[] argList = { string.Empty, string.Empty };
return Convert.ToInt32(managementObject.InvokeMethod("GetOwner", argList)) == 0
? argList[1] + @"\" + argList[0]
: "NO OWNER";
}
public static int GetActiveProcessId()
{
IntPtr activatedHandle = GetForegroundWindow();
if (activatedHandle == IntPtr.Zero)
{
return -1;
}
int activeProcessId;
GetWindowThreadProcessId(activatedHandle, out activeProcessId);
return activeProcessId;
} | {
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standard-model, group-theory, group-representations, lie-algebra, grand-unification
Title: Introduction to Physical Content from Adjoint Representations In particle Physics it's usual to write the physical content of a Theory in adjoint representations of the Gauge group. For example:
$24\rightarrow (8,1)_0\oplus (1,3)_0\oplus (1,1)_0\oplus (3,2)_{-\frac{5}{6}}\oplus (\bar{3},2)_{\frac{5}{6}}$
(Source: SU(5) GUT Wikipedia article)
While I do understand the Basics in representation theory from a mathematical viewpoint, as well as Gauge Theory (up to this point), I've been looking High and Low for some good article on how to understand what the above formula means physically?
Specifically I don't understand the following:
I'm having a bit of a problem with the notation. $(1,1)$ denotes the tensor product of a 1 and 1 of $SU(3) \times SU(2)$ in this case, does the subscript $()_0$ belong to the $U(1)$ part? Or did I completely misunderstand something?
How to arrive at the above transformation? How to choose the right hand side of the 24 transformation, it seems random to me
The physical content. $(8,1)_0$ looks to me like gluons, because of the 8, $(1,3)_0$ like W and Z bosons and $(1,1)_0$ like the photon. But these are all guesses I made according to the numbers I see and the fact that the SM should arise from $SU(5)$ breaking. How would one know this? And what are the other 2 components? | {
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Assume the sequence converged to $$l \neq c$$. Then for $$\epsilon = \frac{|c - l|}{2}$$ $$\exists N$$ s.t. $$i > N \implies |x_i - l| < \epsilon$$, by definition of convergence. But this is absurd, because the sequence had infinitely many terms equal to $$c$$ and thus infinitely many $$x_k = c$$ with $$k > N$$.
• This argument presupposes that the context is $\mathbb R$ or some other complete metric space, so that the sequence $(x_n)$ is guaranteed to converge and you only need to check that it can't converge to a wrong limit. But in fact the OP's proof works in any metric space (if you write $d(x,y)$ instead of $|x-y|$), even if the space isn't complete. – Andreas Blass Nov 6 '18 at 0:27
• @AndreasBlass Indeed the OPs proof is more general, but I didn't think it was wrong to assume I was in $\mathbb R$, as the tag real-analysis suggests. – RGS Nov 6 '18 at 7:00 | {
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slam, navigation, eigen, rosdep, ros-fuerte
By following Sudhan's advice(http://answers.ros.org/question/40059/ros-rgbdslam-fuerte-support/),rosmake rgbdslam works, but octomap still has some problems:
1.edit manifest.xml in rgbdslam package. delete the line '' and add the following
then you can find something similar to loop starting and ending with . within that add the following line without editing other lines:
2.then install qt4. simply, use ubuntu software centre. In devoloper tools -> IDE. you can find it (most probably in the name qt creator).
3.then
$ sudo apt-get install libglew1.5-dev libdevil-dev libsuitesparse-dev
$ sudo apt-get install libgsl0-dev
4.then do $ rosmake rgbdslam. Incase, if you find any error try it with $ rosmake --pre-clean rgbdslam.
Originally posted by dragonfly90 on ROS Answers with karma: 13 on 2012-07-16
Post score: 1 | {
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GR
Gabriel R.
### Problem 39
Determine whether the sequence converges or diverges. If it converges, find the limit.
$\{ \sin n \}$
GR
Gabriel R.
### Problem 40
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \frac {\tan^{-1}n}{n}$
GR
Gabriel R.
### Problem 41
Determine whether the sequence converges or diverges. If it converges, find the limit.
$\{ n^2e^{-n}\}$
GR
Gabriel R.
### Problem 42
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \ln (n + 1) - \ln n$
GR
Gabriel R.
### Problem 43
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \frac { \cos^2 n}{2^n}$
GR
Gabriel R.
### Problem 44
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \sqrt [n]{2^{1 + 3n}}$
GR
Gabriel R.
### Problem 45
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = n \sin (1/n)$
GR
Gabriel R.
### Problem 46
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = 2^{-n} \cos n \pi$
GR
Gabriel R.
### Problem 47
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \left( 1+ \frac {2}{n} \right)^n$
GR
Gabriel R.
### Problem 48
Determine whether the sequence converges or diverges. If it converges, find the limit.
$a_n = \sqrt[n]{n}$
GR
Gabriel R.
### Problem 49 | {
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First, note that $(A-B)^{2}$ is positive semi-definite, so we have:
$$0\leq\mathrm{Tr}(A-B)^{2}=\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2})-2\mathrm{Tr}(AB)$$
$$\mathrm{Tr}(AB)\leq\frac{1}{2}(\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2}))$$
Second, for $A$ positive semi-definite, suppose that all of eigenvalues are $\lambda_{1}$, $\lambda_{2}$, $\cdots$, $\lambda_{n}$, then $\lambda_{i}\geq0$ and $\mathrm{Tr}A=\sum_{i=1}^{n}\lambda_{i}\leq1$, so $\mathrm{Tr}(A^{2})=\sum_{i=1}^{n}\lambda_{i}^{2}\leq\sum_{i=1}^{n}\lambda_{i}\leq1$.
Similarly, $\mathrm{Tr}(B^{2})\leq1$, so $\mathrm{Tr}(AB)\leq1$.
Remark. More generally, we can conclude that the range of $\mathrm{Tr}(AB)$ is $[0,1]$. | {
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# Derivative of determinant of a matrix
Good morning everyone, I would like to know how to calculate:
$\frac{d}{dt}\det \big(A_1(t), A_2(t), \ldots, A_n (t) \big)$
• What is your mean $A_i(t)$? $i$-th column of $A$? – SKMohammadi Jan 2 '16 at 14:32
The formula is $$d(\det(m))=\det(m)Tr(m^{-1}dm)$$ where $dm$ is the matrix with $dm_{ij}$ in the entires. The derivation is based on Cramer's rule, that $m^{-1}=\frac{Adj(m)}{\det(m)}$. It is useful in old-fashioned differential geometry involving principal bundles.
I noticed Terence Tao posted a nice blog entry on it. So I probably do not need to explain more at here.
• we can write it as a sum?, and how? please – kiroro Jan 13 '13 at 8:09
• Try to write down the 2 by 2 case yourself and see how it works. – Bombyx mori Jan 13 '13 at 8:17
• for n=2, i must have $\frac{d}{dt}\det \big(A_1(t),A_2(t)\big)=$ $\det \big(A'_1(t),A_2(t)\big)+ \det \big(A_1(t),A'_2(t))\big)$ but how ? – kiroro Jan 13 '13 at 8:45
• how to use the formula? please – kiroro Jan 13 '13 at 9:35
• Find a DG book on principal bundles. – Bombyx mori Jan 13 '13 at 9:37
Think I can provide a proof for Matias' formula.
So, let
$$A(t) = \mathrm{det}\left( A_1(t), \dots , A_n(t) \right) \ .$$
By definition, | {
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} |
nanoscience
$$
\partial_t \mathbf{d} = -i\omega \mathbf{d}
$$
Substituting into the above equation, you get:
$$
Q_{abs} = \frac{1}{2}\mathbf{Re}\{-i\omega \mathbf{d}\cdot\mathbf{E}^*_{inside}\} = \frac{\omega}{2}\mathbf{Im}\{\mathbf{d}\cdot\mathbf{E}^*_{inside}\}
$$
and similarly for the $Q_{ext}$. | {
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newtonian-mechanics, acceleration, rocket-science, propulsion
Title: Are there real-world devices similar to the thrust reversers in this scene in Star Wars? In Star Wars 3 in the initial crash scene where a spaceship enters the atmosphere for an emergency landing the thrust engine is blocked by a part of the ship and absorbs a part of the thrust which suggests that it's used as a brake.
I'd be interested if that could work inside or outside an atmosphere. However I have few ideas and practical experience approaching such a topic.
I'm aware that Star Wars is fictional, but the pictured engine and brake are a feasible design. Maybe it doesn't make sense, but it was so long ago that they didn't know any better. I spent some time to figure out whether the question belongs here, in case it doesn't I'm sorry. The devices pictured in that scene are clearly modelled on thrust reversers, which are real devices used in a wide variety of jet engines.
Here is one of them in action on a regional jet airliner:
Image source: Wikipedia
For more details, see the first link above. | {
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datetime, timer, shell, scheduled-tasks
Title: Sleep until specified time It's sometimes useful to pause until a specified time (but inconvenient to use at or the like), so I wrote a tiny script to wait until a user-specified time point is reached.
I use this when I want to keep the output of preceding and subsequent commands together, but I don't mind blocking the current terminal. Quite often, this would be M-xcompile when I want to wait until after new source code/data have been uploaded, but still have the results in a compilation-mode buffer.
Although it's a very short script, my time at Code Review has taught me that almost any program has something that can be improved, so please make your suggestions!
#!/bin/sh
# Sleep until the specified time
# Accepts any date/time format accepted by 'date'
# Assumes GNU date and GNU sleep
die() {
echo "$@" >&2
exit 1
}
usage() {
echo "Usage: $0 TIME"
}
test $# = 1 || die $(usage)
case "$1" in
--version)
echo "sleep_until version 1.0"
exit 0
;;
--help)
usage
exit 0
;;
-*)
die "Unrecognised option: $1"
;;
*)
end=$(date -d "$1" +%s.%N)
now=$(date +%s.%N)
test ${end%.*} -gt ${now%.*} || die "$1 is in the past!"
exec sleep $(echo $end $now - p | dc )
;;
esac
Notes:
We need GNU date for its -d option and a good range of input formats.
GNU sleep handles fractional seconds; we could remove .%N from the date formats to work with traditional/POSIX sleep.
I've used dc for the arithmetic so we can use plain sh rather than requiring a more heavyweight shell. In a program comment it is said that the program
# Accepts any date/time format accepted by 'date'
That information should be printed with the usage help, plus one or two examples. Something like
$ ./sleep_until.sh --help
Usage: ./sleep_until.sh TIME
TIME can be any date/time as accepted by date(1).
Examples:
... | {
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8. ## Re: Red marbles, white marbles, in a bag.
Did I get the answers right?
What do you mean I changed the question midstream?
If you mean why I changed the number of marbles, it was to see if I could to the calculation.
Apparently not?
Please could you make your reply more useful. I really don't have any idea what you mean. | {
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"url": "http://mathhelpforum.com/statistics/199623-red-marbles-white-marbles-bag.html"
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scala, scalaz
object Stem {
def apply(word: String): Stem = {
val stem: List[Char] = word.toLowerCase.trim.toList.sorted
new Stem(stem)
}
} | {
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c#, authorization
var newUser = Membership.CreateUser(username.Text, password.Text, email.Text, securityQuestion.Text, securityAnswer.Text, false, out createStatus);
if(createStatus == MembershipCreateStatus.Success)
{
newUser.Comment = barNumber.Text;
SendActivationEmail(newUser);
Membership.UpdateUser(newUser);
}
return GetMessage(createStatus);
}
which will reduce the button click handler code to
protected void createUser_Click(object sender, EventArgs e)
{
userCreationResults.Text = CreateUser();
} | {
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The claim is actually false as given. The usual interpretation of
$$\{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1 \text{ for some integer }n \geq 1\}$$
is
$$\{z\in \mathbb{C} \setminus \{0\} \mid \exists n\in \mathbb N \smallsetminus \{0\}\ :\ z^n = 1\}$$
which is
$$\{e^{\frac{2\pi k i }n} \mid n,k\in \mathbb Z \wedge 0\le k < n\}$$
which I am pretty sure is not cyclic, as the group generated by $e^{\frac{2 \pi k i}n}$ does not contain $e^{\frac{2 \pi i}{2n}}$.
I suspect what they meant to say was :
"For each $n\in \mathbb N\smallsetminus \{0\}$, define $\mu_n = \{z\in \mathbb{C} \setminus \{0\} \mid z^n = 1\}$. Show that for every $n$, $\mu_n$ is cyclic."
To prove the second one, you don't need to write out all the elements. Instead, just show that all elements are of the form
$e^{\frac{2\pi k i }n}$ where $0\le k < n$. Then show that each of those elements can be expressed as a power of $e^{\frac{2\pi i }n}$.
Your interpretation of the problem is correct. It is not a cyclic group. Actually what I need to show is this:
Prove that for each integer $m \geq 1$, there is a unique subgroup $H_m\leq \mu$ with $|H_m|=m$ and that $H_m$ is cyclic. | {
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javascript, jquery, beginner, json, asynchronous
Title: Load JSON file into model using Javascript / jQuery (deferred, asynchronous) As a beginner I am always determined to improve myself. I've got written the following Code using jQuery that is working fine. However, I am sure that there are cleaner ways for achieving the same.
Right now, I am struggling with the this keyword and the asynchronous loading of the JSON file. Moreover, I am not sure whether you should call an function for initialization the way I did.
Do you have any suggestions for improvements?
$(function(){
function Model() {
this.data = null;
this.init();
};
Model.prototype = {
deferred: $.Deferred(),
config: {
jsonFile: 'dump.json'
},
init: function() {
this.loadJson();
},
loadJson: function() {
var self = this;
jQuery.getJSON(
this.config.jsonFile,
function(data) {
self.data = data;
self.deferred.resolve();
}
)
.fail(function() {
console.error("Loading JSON file failed");
})
.always(function() {});
},
getData: function(callback) {
console.log("getData")
var self = this;
$.when(this.deferred)
.done(function() {
callback(self.data)
})
},
};
var model = new Model();
model.getData(function(data) {
console.log(data)
});
});
(duplicate of https://stackoverflow.com/q/23142089/3546614)
Update 1
I've just updated my code (and truncated unimportant stuff) taking @jgillich's advices into account. For the moment I feel better reading the JSON file when the object is generated which is why I outsourced the operation into separate function.
$(function(){
function Model() {
this.loadJson();
};
Model.prototype = { | {
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thermodynamics, equilibrium, gas-laws, ideal-gas
^{(R/C_\mathrm{V}^\mathrm{ig}) + 1} \\
\frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} &=
\left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
^{(R + C_\mathrm{V}^\mathrm{ig})/C_\mathrm{V}^\mathrm{ig}} \rightarrow
\boxed{\frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} =
\left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
^{C_\mathrm{p}^\mathrm{ig}/C_\mathrm{V}^\mathrm{ig}}} \tag{8}
\end{align} | {
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"url": null
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python, validation, mathematics, numpy
I wuld also make it take positional-only arguments if Python 2 support isn't required:
def tetrahedron_volume(*, vertices=None, sides=None):
Larger simplifications
After considering WolframMathWorld's explanation, you can just do
distances = scipy.spatial.distance.pdist(points, metric='sqeuclidean')
to get pairwise distances, and make the matrix with
distances_square = scipy.spatial.distance.squareform(distances)
plus a bit of concatenation.
This gives the opportunity to implement the N-dimensional variant by using the algorithm as given since the prior is so simple and supports it trivially. This requires addition of:
num_verts = distance.num_obs_y(sq_dists)
coeff = - (-2) ** num_verts * factorial(num_verts) ** 2
and of doing the division before checking if vol_square <= 0 instead of hardcoding a coefficient of +288.
If this means we get the same amount of code doing more general stuff with better error checking, great!
from math import factorial
import numpy as np
from scipy.spatial import distance
def simplex_volume(*, vertices=None, sides=None) -> float:
"""
Return the volume of the simplex with given vertices or sides.
If vertices are given they must be in a NumPy array with shape (N+1, N):
the position vectors of the N+1 vertices in N dimensions. If the sides
are given, they must be the compressed pairwise distance matrix as
returned from scipy.spatial.distance.pdist.
Raises a ValueError if the vertices do not form a simplex (for example,
because they are coplanar, colinear or coincident).
Warning: this algorithm has not been tested for numerical stability.
"""
# Implements http://mathworld.wolfram.com/Cayley-MengerDeterminant.html
if (vertices is None) == (sides is None):
raise ValueError("Exactly one of vertices and sides must be given") | {
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thermodynamics, statistical-mechanics, gas
If we use the van der Waals interaction,
$$U(r) = \left\{\begin{matrix}
\infty & r < r_0\\
-U_0 \left( \frac{r_0}{r}\right)^6 & r \geq r_0
\end{matrix}\right.$$
and evaluate the integral, we find,
$$\frac{pV}{Nk_B T} = 1 - \frac{N}{V} \left( \frac{a}{k_B T}-b\right)$$
where $a = \frac23 \pi r_0^3 U_0$ and $b = \frac23 \pi r_0^3$ which is directly related to the excluded volume $\Omega = 2b$. | {
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arduino, motor, pid, line-following
Now, read on for more details regarding the error dynamics and linearized control system:
We can write out the system dynamics like this, where we consider $z$ to be the vector of the error states.
$z = \begin{bmatrix} \dot{e} \\ e \end{bmatrix}$
$\dot{z} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} z + \begin{bmatrix} 1 \\ 0 \end{bmatrix} \eta$
If you are familiar with linear control theory then you can start to see how this is a convenient form where we can make a feedback control law with input $\eta$ as a function of the error:
$\eta = -\begin{bmatrix} K_d & K_p \end{bmatrix} z$
$\dot{z} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} z - \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} K_d & K_p \end{bmatrix} z$
$\dot{z} = \begin{bmatrix} -K_d & -K_p \\ 1 & 0 \end{bmatrix} z$
This closed-loop system has the following characteristic equation:
$s^2 + K_d s + K_p = 0$
Which results in the poles (this is a PD control law, not PID):
$p = -\frac{1}{2}K_d \pm \sqrt{\frac{1}{4}K_d^2 - Kp}$
So if you wanted a critically damped system with no oscillations around the line, you could choose one of the gains and use the following relationship to get the other (start with smaller gains and tune until you get the response you want):
$K_p = \frac{1}{4}K_d^2$ | {
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imu, transform
Title: Laser_ortho_projector transformation problem
I'm currently using ROS Hydro. I've been trying without success to run laser_ortho_projector without a base/world tf. According to the wiki (http://wiki.ros.org/laser_ortho_projector#Required_tf_Transforms), this is fine so long as an IMU topic is provided, since the node can determine the attitude of the laser frame relative to the fixed frame.
(Note that while the wiki article referenced is for Fuerte, I used the Hydro package: https://github.com/ccny-ros-pkg/scan_tools/tree/hydro)
Currently, I have a laser topic (/scan) and IMU topic (/imu/data) successfully publishing. Also, I have a laser > base_link static transformation being broadcast (included in launch file below).
I continue to get the following error:
[ WARN] [ros timestamp]: Skipping scan Could not find a connection between 'world' and 'base_link' because they are not part of the same tree.Tf has two or more unconnected trees.
Here is my launch file:
<launch>
<!-- Static TF Publication -->
<node pkg="tf" type="static_transform_publisher" name="base_to_laser"
args="0 0 0 0 0 0 /base_link /laser 40" />
<!-- Laser Ortho Projector-->
<node pkg="laser_ortho_projector" type="laser_ortho_projector_node" name="laser_ortho_projector" output="screen">
<param name="use_imu" type="bool" value="true"/>
<param name="use_pose" type="bool" value="false"/>
<param name="publish_tf" type="bool" value="true"/>
<remap from= "cloud_ortho" to="cloud"/>
</node>
</launch>
Thanks for any assistance! | {
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Thanks EP
I want to experiment with a different method
The roots of a quadratic equation are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ so\\ {-b \pm \sqrt{b^2-4ac} \over 2a} = 3 \pm \sqrt{5}\\ \frac{-b}{2a}=3\qquad and \qquad \frac{\sqrt{b^2-4ac}}{2a} =\sqrt5\\ b=-6a\qquad and \qquad \sqrt{b^2-4ac} =2a\sqrt5\\ b=-6a\qquad and \qquad b^2-4ac =4a^2*5\\ \boxed{b=-6a}\qquad and \qquad b^2-4ac =20a^2\\ (-6a)^2-4ac =20a^2\\ 36a^2-4ac =20a^2\\ 4a=c\\ Let\;\; a=1, \;\;then\;\;c=4\;\;and \;\;b=-6\\ \text{So the quadratic equation is}\\ y= x^2-6x+4$$
EP's method was a lot easier.
Apr 16, 2020
#4
+24995
+1
How do you find the quadratic equation if the roots are $$x = 3 + \sqrt{5}$$ and $$x = 3 - \sqrt{5}$$? | {
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`equilibrate` is most useful when the scales of the `b` and `x` vectors in the original system ```x = A\b``` are irrelevant. However, if the scales of `b` and `x` are relevant, then using `equilibrate` to rescale `A` only for the linear system solve is not recommended. The obtained solution does not generally yield a small residual for the original system, even if expressed using the original basis vectors.
## References
[1] Duff, I. S., and J. Koster. “On Algorithms For Permuting Large Entries to the Diagonal of a Sparse Matrix.” SIAM Journal on Matrix Analysis and Applications 22, no. 4 (January 2001): 973–96.
## Version History
Introduced in R2019a
expand all | {
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c++, object-oriented, embedded, device-driver
{DigitalInputsDriverCfg::FaultPinActivationSrc::kFaultPinActivation24VM,
DigitalInputsDriverCfg::FaultPinActivationSrcUsage::kUsed},
{DigitalInputsDriverCfg::FaultPinActivationSrc::kFaultPinActivationWireBreak,
DigitalInputsDriverCfg::FaultPinActivationSrcUsage::kUsed}}},
{DigitalInputsDriverCfg::Device::kMAX22190Device_1,
{{DigitalInputsDriverCfg::Input::kInput_01,
DigitalInputsDriverCfg::InputEnable::kInputEnabled,
DigitalInputsDriverCfg::WireBreakDetection::kWireBreakDetectionEnabled,
DigitalInputsDriverCfg::ProgrammableFilter::kProgrammableFilterUsed,
DigitalInputsDriverCfg::FilterDelay::kInputFilterDelay20ms},
{DigitalInputsDriverCfg::Input::kInput_02,
DigitalInputsDriverCfg::InputEnable::kInputEnabled,
DigitalInputsDriverCfg::WireBreakDetection::kWireBreakDetectionEnabled,
DigitalInputsDriverCfg::ProgrammableFilter::kProgrammableFilterUsed,
DigitalInputsDriverCfg::FilterDelay::kInputFilterDelay20ms},
{DigitalInputsDriverCfg::Input::kInput_03,
DigitalInputsDriverCfg::InputEnable::kInputEnabled,
DigitalInputsDriverCfg::WireBreakDetection::kWireBreakDetectionEnabled,
DigitalInputsDriverCfg::ProgrammableFilter::kProgrammableFilterUsed,
DigitalInputsDriverCfg::FilterDelay::kInputFilterDelay20ms},
{DigitalInputsDriverCfg::Input::kInput_04, | {
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homework-and-exercises, general-relativity, black-holes, metric-tensor, qft-in-curved-spacetime
&= \left( -\partial_{t}^{2} + \partial_{r_{*}}^{2} - V(r) \right) \phi, \quad V(r) = f \left[ \frac{l(l + 1)}{r^{2}} + \frac{2m}{r^{3}} \left( -\frac{1}{f^{4}} \right) \right] \\
\end{split}
\end{equation}
And as you can see, I got extra $-\frac{1}{f^4}$ in the end. Any idea why? There's an error in the notes you posted. The tortoise coordinate is usually defined via
$$
\frac{dr}{dr_*} = 1 - \frac{2m}{r} = f \neq \frac{1}{f}.
$$
Note that the correct definition is given in eq. (42) of your link. I suspect this will fix your problem. | {
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"url": null
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organic-chemistry, nomenclature, aromatic-compounds
Title: What is the name of the "primary" carbon in a benzene derivative? If we have a benzene with a single substituent, such as chlorobenzene, what do we call the carbon atom to which the group is attached? I'm familiar with the ortho- meta- para- nomenclature, but the last one escapes me. It's ipso-.
Wikipedia page
IUPAC Gold Book definition of the ipso-attack | {
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lagrangian-formalism, field-theory, gauge-theory, chern-simons-theory, boundary-terms
First of all, we start from a principal bundle $P$ over $M$ with the structure group $G$, whose Lie algebra is $\mathfrak{g}$. Let $A$ be its connection, i.e. a $\mathfrak{g}$-valued $1$-form, and $U$ takes values in $G$. The gauge transformation is $$A\rightarrow U^{-1}dU+U{-1}AU.$$
The important thing that some people usually ignore is that neither the connection $1$-form $A$ nor the term $U^{-1}dU$, known as the Maurer–Cartan form, can always be globally defined.
In fact, if the principal bundle $P$ is non-trivial, (e.g. a Dirac monopole is inserted in $M$) then $A$ is only locally defined on $M$.
Under the above gauge transformation, the Chern-Simons action transforms in the following way:
\begin{align}
&\,\,\,\,S[A^{\prime}]=\frac{k}{4\pi}\int_{M}\mathrm{tr}\left(A^{\prime}\wedge dA^{\prime}+\frac{2}{3}A^{\prime}{\wedge}A^{\prime}{\wedge}A^{\prime}\right) \\
&=\frac{k}{4\pi}\int_{M}\mathrm{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\wedge A\right)-I_{\partial M}[U;A]-\Omega[U],
\end{align} | {
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"url": null
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black-holes, electric-fields, event-horizon
Let's look at a simple example. A symmetric spherical shell of charge of radius $R$ with a zero magnetic field everywhere and an electric field of $\frac{Q}{4\pi\epsilon_0}\frac{x\hat x+y\hat y+z\hat z}{(x^2+y^2+z^2)^{3/2}}$ when $x^2+y^2+z^2\gt R^2$ and $\vec 0$ otherwise. Now we can describe the actual physical mechanism to create a field. If the charge on the surface of the shell moves inwards there is a current. And so right where the current is (on the surface) a new nonzero electric field is produced. If the current goes in radially, then the new elecetric field points radially outward (since there is a zero magnetic field and $$\frac{\partial \vec E}{\partial t}=-\frac{1}{\epsilon_0}\vec J+\frac{1}{\epsilon_0\mu_0}\vec \nabla \times \vec B.$$).
And since the electric fields are radial, they have zero circulation so the magnetic field doesn't change and so is still zero. Now new nonzero fields have been created in a region that previously had zero fields. They were created as the charge passed that location, and they persisted long after the charge moved past that region. | {
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A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
You can find our video solution here: https://www.gmatprepnow.com/module/gmat ... /video/947
Another way to build up the quadratic...
Let Bill's speed = x
and Ted's speech = x+5
When Ted covers the full distance, Bill has yet to travel for 4 more hours to complete the distance
$$\frac{240}{(240 - 4x)} = \frac{(x + 5)}{x}$$
_________________
Spread some love..Like = +1 Kudos
CEO
Joined: 11 Sep 2015
Posts: 3122
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
### Show Tags
25 May 2017, 04:14
1
Top Contributor
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
*Kudos for all correct solutions.
Another approach....
Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Let B = Bill's travel speed
So, B + 5 = Ted's average speed | {
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\begin{align} S_k &= \sum_{0 \le t \le n} \binom{t}{k} \end{align}
Define the generating function:
\begin{align} S(z) &= \sum_{k \ge 0} S_k z^k \\ &= \sum_{k \ge 0} z^k \sum_{0 \le t \le n} \binom{t}{k} \\ &= \sum_{0 \le t \le n} \sum_{k \ge 0} \binom{t}{k} z^k \\ &= \sum_{0 \le t \le n} (1 + z)^t \\ &= \frac{(1 + z)^{n + 1} - 1}{(1 + z) - 1} \\ &= z^{-1} \left( (1 + z)^{n + 1} - 1 \right) \end{align}
So we are interested in the coefficient of $z^k$ of this:
\begin{align} [z^k] z^{-1} \left( (1 + z)^{n + 1} - 1 \right) &= [z^{k + 1}] \left( (1 + z)^{n + 1} - 1 \right) \\ &= \binom{n + 1}{k + 1} \end{align} | {
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homework-and-exercises, classical-mechanics, lagrangian-formalism
Title: Finding Lagrangian of a Spring Pendulum I'm trying to understand Morin's example of a spring pendulum. What I don't get is his expression for $T$. I can understand the $\dot x^2$ term in the brackets. But I don't understand the $(l + x)^2\dot \theta^2$.
Also, it seems rather strange to break up Kinetic Energy into tangential and radial components when it is a scalar. $\newcommand{\er}{\hat e_r} \newcommand{\et}{\hat e_\tau} \newcommand{\d}{\dot} \newcommand{\m}{\frac{1}{2}m} $
In radial coordinates, $\d\er=\d\theta \et$, and (useless here) $\d\et= -\d r \er$. $\er,\et$ are unit vectors in radial and tangential directions respectively. Due to this mixing of unit vectors (they move along with the particle), things get a little more complicated than plain 'ol cartesian system, where the unit vectors are constant.
For your particle, writing $x+l\to r$, the position vector is: $$\vec p= r\er$$ $$\therefore \vec v=\d{\vec p}= \d r\er + r\d\er=\d r \er + r\d\theta\et$$
$$\therefore v^2= \vec v\cdot\vec v= \d r^2+r^2\d\theta^2$$
Substituting back the value of $r=x+l,\d r=\d x$ (and mutiplying by $\m$, we get the above expression?
As you can see in my expression for $\vec v$, I had two components of velocity--radial and tangential. Since they are perpendicular, I can just square and add, akin to $T=\m\left(\d x^2 +\d y^2\right)$. | {
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observational-astronomy, telescope, photography, observatory, tracker
Title: What are the technological advancements that made it possible for modern large telescopes to work with alt-az mounts instead of equatorial mounts? The video Earth's Rotation Visualized in a Timelapse of the Milky Way Galaxy - 4K (linked below) and discussion below this answer to Why does a timelapse video of a stationary Milky Way make the horizon appear to move from horizontal to vertical? about field rotators has got me thinking.
In addition to the benefit of (essentially) single axis constant speed tracking for distant celestial objects, an equatorial mount also rotates the telescope tube about it's axis such that there is no rotation of the image on the focal plane during long exposures.
This was probably essential for hours-long exposures of single emulsion plates.
Asked separately: Did astronomers ever use photographic plate rotation along with alt-az mounts?
But for the very largest and heaviest reflecting telescopes equatorial mounts are massive and unwieldily and require huge counterbalances compared to alt-az mounts which can have the azimuth bearing right on the ground and the altitude bearing straight through the telescope's center of mass.
Question: What are the technological advances that made it possible for modern large telescopes to work with alt-az mounts instead of equatorial?
I'm expecting most of the answer to be related to electronic advancements of one kind or another, but there may be mechanical and optical and even electro-optical (optoelectronic?) advancements as well. I would say the field derotator was a great advancement. The field of view of an altazimuthally mounted instrument will rotate, just like for the naked eye—for example, the “Lady in the Moon” seems to be tilted left at moonrise, straight at transit, and tilted right at moonset. If one is to take long-exposure images (more than about a minute, for example, on my 10″ ƒ/4.7 Dobsonian), one need to “derotate” the image so it stays straight on the image.
Derotating an image seems easy in practice, but the exact speed at which the field rotates varies with the azimuth and altitude of the target. Some computing power is thus needed “live” to drive the motors at the proper rate depending on where the telescope is pointing. | {
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cc.complexity-theory, reference-request, ds.data-structures, topology, computational-geometry
Title: Would a purely topological computational model be useful in decision problems in topology? If one were to develop a purely topological computational model based upon the equivalence of information in knots and the model would perform transformations of that information. This would be the way that model would compute. Would such a model be useful in understanding computational problems in topology or would algebraic models suffice? If one already exists please provide a reference. I'm not sure whether this qualifies as a purely topological computational model, but there is a topological approach to anyonic quantum computation within the framework of which Aharonov-Jones-Landau and Freedman-Kitaev-Wang proved that a quantum computer can "additively" approximate the Jones polynomial at a root of unity in polynomial time. Furthermore, by Freedman-Larsen-Wang, such an approximation for certain roots of unity is universal for quantum computation. Unfortunately, such an approximation does not appear to be useful in topology, although for certain braids it is useful for anyonic quantum computation. See these slides by Kuperberg.
There is also work of Kauffman and Lomonaco and various papers of Kauffman (e.g. this) on anyonic topological quantum computation, which presents quantum algorithms to compute various topological invariants. Maybe this is the closest thing to what you are looking for.
Regarding tangle machines, the goal of the theory is to provide a diagrammatic formalism for networks of information manipulations (computations) that is more flexible than graphical notations based on labeled directed graphs. Equivalent tangle machines represent computations which are equivalent in the sense that each one can be fully simulated by the other, but which may differ locally in terms of performance. Note also that there is an alternative approach to the same issue by way of higher category theory, e.g. HERE. It isn't in these preprints, but tangle machines can represent any Turing machine computation, and also any recurrent first-order neural network. But I'm not sure how the theory would be useful to understand computational problems specific to topology. | {
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stft, filter-bank
The FFT (and the continuous time Fourier Transform as well) over a fixed time interval will produce the same result for the non-zero samples as the same waveform if it was continued back to back out to positive and negative infinity. Thus when the waveform is not circularly continuous the effect is equivalent to a discontinuity at the transition in the time domain, which results in many other frequency components: what we refer to as "spectral leakage". Below shows the result of the FFT for the time domain waveform above:
The spectrogram as the OP has used it (with a rectangular window) is this plot repeated as we move through the waveform in time with the same block length, overlapping the blocks by 50%. No matter what start and stop we use for this given waveform, it will be circularly continuous over the 1000 samples and therefore every block processed will have a magnitude result as I have shown in the plot above.
If we, for example, changed M to be 1200 instead of 1000- we would then see the more general case of spectral leakage due to using a rectangular window: | {
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For 1. $$P=\frac{36\times 32\times 28}{39\times 38\times 37}=0.588248$$.
For 2. There are two terms: exactly three hearts and exactly four hearts.
$$P_3=4\times\frac{30\times 10\times 9\times 8}{40\times 39\times 38\times 37}=0.0393916$$ $$P_4=\frac{10\times 9\times 8\times 7}{40\times 39\times 38\times 37}=0.00229784$$
Total $$P=0.04168944$$
• 2. has been corrected. Apr 13 at 2:50
We have a condensed deck of 40 cards containing only the denominations from Ace through 10.
Presumably that is those ten denominations in each of the four standard suits.
We draw four cards uniformly at random (without replacement).
What is the probability that we draw four distinct denominations, such as one Ace, one five, one six and one seven.
Well, we are comparing the count for ways to select four from ten denominations, with each in one from four suits, to the count for ways to select four from forty cards. That "each" means we are not multiplying $${^{4}\mathrm C_1}$$ by $$4$$ but rather empowering it. $$\dfrac{{^{10}\mathrm C_4}\,({^{4}\mathrm C_1})^4}{^{40}\mathrm C_4} = \dfrac{40\cdot 36\cdot 32\cdot 28}{40\cdot 39\cdot 38\cdot 37}$$
What is the probability that we draw at least three hearts.
For 2, I think something like 1 - Pr(choosing exactly 2 hearts) could work, but I don't know where to go from here.
That is not the correct complement.
The event of interest is drawing 3 or 4 from the four hearts (with the remainder of cards drawn being other suits). The complement of this is drawing 0, 1, or 2 from the four hearts. | {
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java, unit-testing, reinventing-the-wheel
Title: Single class unit test framework in Java This is a single class and dependency free1 unit test framework for Java. The only assertion available is assertTrue (this can be used for all test cases).
(Utilizes feature of Java7 : Multi catch)
What it does
Execute all methods with annotation Test. should fail if any arguments are needed.
Calculate elapsed time using a given repeat count.
Print state for each test case.
Print number of successful test cases and total number of test cases.
assertTrue must be called to be successful.
What I want reviewed
Potential performance improvements.
Potential flaws in coding style.
Potential flaws in API.
Or anything else related to this that might be appropriate.
package info.simpll.simpletest;
import java.lang.annotation.ElementType;
import java.lang.annotation.Retention;
import java.lang.annotation.RetentionPolicy;
import java.lang.annotation.Target;
import java.lang.reflect.InvocationTargetException;
import java.lang.reflect.Method;
public class Tester {
private static int successCount = 0;
private static int testCount = 0;
//--------------------------------------
//Test annonation
@Retention(RetentionPolicy.RUNTIME)
@Target(ElementType.METHOD)
public static @interface Test {
public String name() default "unnamed"; // name of test
public int repeat() default 1; // repeat count if calcTime==true
public boolean calcTime() default false; // calculate Time
}
//--------------------------------------
//---------------------------------------
//Logger
private static enum Logger {
INFO(">"), ERROR("Error -->");
private final String val;
private Logger(String val) {
this.val = val;
} | {
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rust, simulation
BallPrediction {
slices,
num_slices,
}
}
My question: Is putting the ball into a box and putting that into a Vec the best, most efficient way to do this? Is it possible for me to rather clone the balls into the stack (I believe Box stores it on the heap?) but then Vec stores stuff on the heap anyway, so does that even matter? How can I minimize moving all the data around?
I'm also new-er to Rust, so insight into Rust and its conventions is appreciated! I can share as might insight to the program as needed, this is an open-source hobby project of mine. welcome to the Rust community.
A Vec is a continuous slice of heap memory. A Box is a fixed-size portion of heap memory (except for boxed DSTs). A Box within a Vec is entirely unnecessary (except in extreme cases when you are moving around huge structs). Putting a Box inside a Vec is like wrapping your items in two layers of packaging. This overhead is quite small, though: one heap allocation per item.
Don't worry about copying things onto the stack. The stack is efficient and Rust/LLVM will optimize moves of your values. | {
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homework-and-exercises, lagrangian-formalism, field-theory, notation, poincare-symmetry
Title: Index problem in change of action under spacetime dependent translation I have two questions, and I'll address them while explaining my calculation and my, probably banal, uncertainties.
We're basically deriving the Energy-Momentum tensor for a scalar field from Noether theorem in a similar way of the one exposed in Weinberg, The Quantum Theory of Fields, Vol I, pag. 311.
Consider the lagrangian density of a complex scalar field:
$$ \mathscr{L} \, =\, -\partial^{\mu} \phi^* \partial_{\mu} \phi - m^2 \phi \phi^* $$
and the following transformation
$$
\phi \rightarrow \phi(x) - a^\mu (x) \partial_\mu\phi(x) \\
\phi^* \rightarrow \phi^*(x) - a_\mu (x) \partial^\mu\phi^*(x)
$$
My professor writes that, beside the terms proportional to $a$ (which I have still to prove to myself give $\eta^{\mu \nu} \mathscr{L}$, with $\eta$ the metric), the variation of the Lagrangian is
$$
(\partial_\mu a_\nu)(\partial^{\mu} \phi^* \partial^{\nu} \phi + \partial^{\nu} \phi^* \partial^{\mu} \phi)
$$
While I obtain
$$
(\partial_\mu a^\nu)(\partial^{\mu} \phi^* \partial_{\nu} \phi )\, + \, (\partial^\mu a_\nu) (\partial^{\nu} \phi^* \partial_{\mu} \phi)
$$
I strongly feel that they could be the same but I don't know how to play with indices to reach the same result. Can someone help me with ths? That was my first question.
The second one is about I derived that result. | {
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quantum-field-theory, commutator, causality, greens-functions, wick-rotation
Title: Euclidean QFT commutator vanishes for all spacetime separations? In Minkowski spacetime, the commutator of the Klein-Gordon field operator with itself at different spacetime points evaluates to the advanced minus retarded Green's function of the classical theory,
\begin{align}
[\phi (x),\phi (y)]=\langle 0|[\phi (x),\phi (y)]|0\rangle =G_A(x-y)-G_R(x-y),
\end{align}
which vanishes for spacelike separations. [I use the convention that the K-G Green's functions are defined by $(\partial^2+m^2)G(x-y)=-i\delta^{(4)}(x-y)$.]
Because of the $SO(4)$-isometries of Euclidean spacetime, there is no invariant notion of time direction and, indeed, all separations are spacelike. For this reason, I would naively expect the Euclidean K-G Green's function (which vanishes at infinity) is unique--i.e., there are no "advanced" or "retarded" Euclidean Green's functions--and the field commutator should, then, vanish for $all$ Euclidean spacetime separations.
More explicitly, the Euclidean 2-point function goes as
\begin{align}
\langle 0|\phi (x)\phi(y) |0\rangle=\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip\cdot(x-y)}}{p^2+m^2},
\end{align}
so the Euclidean commutator is
\begin{align} | {
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python, beginner, python-3.x, object-oriented, tetris
self.GUI.score_window.update()
# try to add nextPiece to Board
self.board.add_piece(self.next_piece)
# if unsuccessful, gameover
if self.next_piece.last_move_overlap:
break
self.next_piece = deepcopy(random.choice(self.pieces))
self.GUI.piece_preview_window.update()
else:
if keyboard_input == ord('w'):
self.board.rotate_active_piece()
if keyboard_input == ord('d'):
self.board.translate_active_piece('right')
if keyboard_input == ord('s'):
self.board.translate_active_piece('down')
if keyboard_input == ord('a'):
self.board.translate_active_piece('left')
# exit game
if keyboard_input == ord('e'):
break
self.GUI.board_window.update()
# delay after a rotation
if keyboard_input == ord('w'):
time.sleep(.25)
time.sleep(1 / self.frame_rate)
# Reset terminal window before exiting the game.
curses.nocbreak()
self.GUI.board_window.keypad(False)
self.GUI.board_window.nodelay(False)
curses.echo()
curses.endwin()
curses.curs_set(1)
print('Game Over')
exit()
# Run the game
game = Game(board_num_rows=16, board_num_columns=10)
game.main_loop() In general it looks structured, using good names, split up into functions etc. Those are good things.
I have some comments but remember these are my opinions, I can't say that it's right or wrong and I won't refer to any standard or code style.
1
def in_bounds(self, temp_active_piece_indices):
return all(0 <= i < self.num_rows and 0 <= j < self.num_columns
for i, j in temp_active_piece_indices) | {
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java, performance, android, formatting
public Button copy(){
Button button = new Button(stateDefault,statePressed,stateReleased,pressNumber,doOnClick,doOnClickRelease,doOnPass,doOnRelease);
button.position = new ArrayList<>(position);
button.size = new ArrayList<>(size);
button.state = new ArrayList<>(state);
button.isPressed = new ArrayList<>(isPressed);
button.hasBeenPressed = new ArrayList<>(isPressed);
button.name = new ArrayList<>(name);
button.isHovering = new ArrayList<>(isHovering);
return button;
}
public void addInstance(){
isPressed.add(false);
hasBeenPressed.add(false);
position.add(new Vector2f(0,0));
size.add(new Vector2f(1,1));
state.add(STATE_DEFAULT);
stateDefault.instance(0);
stateDefault.position.set(stateDefault.position.size()-1,new Vector2f(0,0));
statePressed.instance(0);
statePressed.position.set(statePressed.position.size()-1,new Vector2f(0,0));
stateReleased.instance(0);
stateReleased.position.set(stateReleased.position.size()-1,new Vector2f(0,0));
Log.w("addInstance","added instance");
}
public void addInstance(Vector2f position,Vector2f size){
isPressed.add(false);
hasBeenPressed.add(false);
this.position.add(position);
size.add(size);
state.add(STATE_DEFAULT);
stateDefault.addInstance(position,0,size);
statePressed.addInstance(position,0,size);
stateReleased.addInstance(position,0,size);
Log.w("addInstance","added instance");
} | {
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} |
qiskit, transpile
Title: Qiskit custom multi controlled gate I'm building a 3 qubit circuit with the control-ry rotation gate. The code is the following:
circ = QuantumCircuit(3)
circ.mcry(pi/8,
q_controls=[0],
q_target=2,
use_basis_gates=False,
mode='noancilla')
The result is that cry gate is NOT decomposed:
Is there a way to achieve the NOT decomposed circut on the multicontrol-ry gate? I mean instead of the following circuit:
for the following code:
circ = QuantumCircuit(3)
circ.mcry(pi/8,
q_controls=[0, 1],
q_target=2,
use_basis_gates=False,
mode='noancilla')
I want to recieve a single 2-control-ry gate.
I also tried to add these two lines:
gates = Aer.get_backend('qasm_simulator').configuration().basis_gates
transpile(circ, basis_gates=gates,optimization_level=0).draw(output='mpl') | {
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electromagnetism, electromagnetic-radiation
Title: Faraday boxes (cages) are nice and all but what if you need wires or tubes to enter / exit your box? Then is there still a way to EMP-protect a device? EMP is typically a high frequency signal. It is possible to allow low frequency signals to enter your Faraday cage by decoupling the signal - a choke in series and a capacitor in parallel. You need to make sure that the choke does not saturate at the current spikes expected - and that the voltage rating of the capacitor is sufficient to absorb the energy.
A technical paper on the subject can be found at http://www.kwset.fi/kW_Consulting/pdf/Yleis_PDF_GeneralTechnicalInformation.pdf
Another solution can be to use optical interfaces to permit the transmission of high frequency signals while protecting from unwanted RF. In that case you let the optical signal cross a small hole (small compared to the wavelength of RF you want to stop) after which you can safely convert to electrical. This is done, for example, with electronics that runs in certain MRI machines - allowing digital electronic circuits to live in the vicinity of the MRI scanner without interfering. "Wanted" signals travel across the boundary, but "unwanted" signals are stopped. | {
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c++, console
DirectoryTagEntry newDirectoryEntry(tag, dir);
directoryTagEntryList << newDirectoryEntry;
}
}
void operator>>(DirectoryTagEntryList const& directoryTagEntryList,
std::ofstream& ofs) {
for (size_t i = 0, sz = directoryTagEntryList.size();
i < sz;
i++) {
DirectoryTagEntry const& dte = directoryTagEntryList.at(i);
ofs << dte.getTagName() << " " << dte.getDirectoryName() << "\n";
}
}
}
main.cpp:
#include "DirectoryTagEntry.h"
#include "DirectoryTagEntryList.h"
#include <iomanip>
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <string>
#include <cstring>
#include <pwd.h>
#include <unistd.h>
#include <vector>
using com::github::coderodde::ds4mac::DirectoryTagEntry;
using com::github::coderodde::ds4mac::DirectoryTagEntryList;
using std::cerr;
using std::cout;
using std::ifstream;
using std::ofstream;
using std::setw;
using std::string;
//// ///////////////////
// Operation names: //
/////////////////// ////
const string OPERATION_SWITCH_DIRECTORY = "switch_directory";
const string OPERATION_DESCRIPTOR_SHOW_TAG_ENTRY_LIST =
"show_tag_entry_list"; | {
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6. If you prove it, then it's true!
Here's a less computational argument. Let $\displaystyle D_n$ denote the set of divisors of $\displaystyle n$, and suppose that $\displaystyle g=(m,n)>1$. I exhibit a surjection $\displaystyle f : \ D_m \times D_n \to D_{mn}$ which is not an injection. For every $\displaystyle (u,v) \in D_m \times D_n$, let $\displaystyle f(u,v)=uv$. This map is clearly a surjection. But $\displaystyle f(m/g,n)=f(m,n/g)$. Therefore $\displaystyle |D_m\times D_n|>|D_{mn}|$.
7. yes I do, I wrote that early in the morning...
8. I have an argument but I'm not sure...
Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$\displaystyle m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $\displaystyle n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$, where for $\displaystyle 1\leq i\leq r$ $\displaystyle m_i$ and $\displaystyle n_i$ are nonnegative. Also, for any prime $\displaystyle p_i$, $\displaystyle p_i$ divides at leat one of $\displaystyle m$ and $\displaystyle n$. | {
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} |
c++, game, sfml
scoreText.setPosition(sf::Vector2f(windowWidth / 2, windowHeight / 7));
}
void Hud::move()
{
scoreText.move(moveSpeed, 0.0f);
}
void Hud::updateScore(int score)
{
scoreText.setString(std::to_string(score));
}
void Hud::draw()
{
data->window.draw(scoreText);
}
InputManager.h
#pragma once
#include "Game.h"
#include <SFML/Graphics.hpp>
class InputManager
{
private:
GameDataRef data;
public:
InputManager(GameDataRef _data);
bool IsSpriteClicked(sf::Event event, sf::Sprite object, sf::Mouse::Button button);
sf::Vector2i GetMousePosition();
};
InputManager.cpp
#include "InputManager.h"
InputManager::InputManager(GameDataRef _data) : data(_data)
{
}
bool InputManager::IsSpriteClicked(sf::Event event, sf::Sprite object, sf::Mouse::Button button)
{
if (event.type == sf::Event::MouseButtonReleased)
{ // on click
auto pos = data->window.mapPixelToCoords(sf::Vector2i(event.mouseButton.x, event.mouseButton.y));
if (event.mouseButton.button == sf::Mouse::Left)
{ // on left click
if (object.getGlobalBounds().contains(pos))
{
return true;
}
}
}
return false;
}
sf::Vector2i InputManager::GetMousePosition()
{
return sf::Mouse::getPosition(data->window);
}
MainMenuState.h
#pragma once
#include "Game.h"
#include "InputManager.h"
class MainMenuState : public State
{
private:
GameDataRef data;
InputManager *iM1; | {
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-
I don't see the link to the previous question. – Ross Millikan Mar 25 '11 at 14:08
@Ross This is done. – Did Mar 25 '11 at 14:13
@Ross: Sorry; had to go teach... – Arturo Magidin Mar 25 '11 at 14:54
The way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\bar{z}) = \bar{P(z)}$ for any complex number $z$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $-i + \sqrt{2}$. Similarly, suppose $z = a + b\sqrt{2}$, with $a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\sqrt{2}) = c + d\sqrt{2}$, one has $P(a - b\sqrt{2}) = c - d\sqrt{2}$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $i - \sqrt{2}$, and if $-i + \sqrt{2}$ is a root of $P(z)$, so is $-i - \sqrt{2}$.
The upshot is that if $i + \sqrt{2}$ is a root of $P(z)$, so are $-i + \sqrt{2}$, $i - \sqrt{2}$, and $-i - \sqrt{2}$. Thus the minimal polynomial of $i + \sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.
In some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.
- | {
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"url": "http://math.stackexchange.com/questions/29005/minimal-polynomial-of-i-sqrt2-in-mathbbq"
} |
climate-change, greenhouse-gases
Title: Is it true that methane is responsible for nearly one-third of all warming? BBC says that the latest IPCC report says methane is responsible for 0.3°C out of 1.1°C of all cumulative warming since pre-industrial times.
According to the IPCC, around 0.3C of the 1.1C that the world has already warmed by comes from methane.
(from here)
But the thing is I opened the report, Ctrl+F'ed "0.3" and "methane", and didn't find that. Is it true that methane is responsible for nearly one-third of all warming? Did I miss something in the text?
UPD: This is THE report, actually. I didn't find it there too, though (I searched for "0.3°C").
UPD 2: @Deditos found something interesting on page 7.183 (1797) Yes, this is broadly true. I haven't looked into much of the report yet, but I suspect that the BBC are referring to Fig 7.7 for an estimate of the observed warming attributed to changes in methane concentration. The figure has 0.28 °C as the best estimate, which they've rounded (or had quoted to them in a briefing) as 0.3 °C:
The slight wrinkle is that estimate is for 2019 relative to 1750 and a total warming of 1.27 °C, whereas the quoted 1.1 °C total warming is probably the observed 2010-19 relative to 1850-1900 value (see SPM A.1.2 or A.1.3). It's possible that the BBC are mashing together some incompatible estimates here or, as I say, they may have been briefed with the best bottom line to quote. Note that the Summary for Policymakers (SPM) has another figure, SPM.2, that gives a best estimate of the methane contribution of 0.5 °C relative to the appropriate decades:
As I say, I haven't looked into all the nuances of these different estimates yet but, yes, the contribution to total warming attributed to methane is comfortably several tenths of a °C. | {
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c#, algorithm, programming-challenge
Title: Finding repeated letters in a list of words
John has discovered various rocks. Each rock is composed of various
elements, and each element is represented by a lowercase Latin letter
from 'a' to 'z'. An element can be present multiple times in a rock.
An element is called a 'gem-element' if it occurs at least once in
each of the rocks.
Given the list of rocks with their compositions, you have to print how
many different kinds of gems-elements he has.
Input Format
The first line consists of N, the number of rocks. Each of the next N
lines contain rocks’ composition. Each composition consists of small
alphabets of English language.
Output Format
Print the number of different kinds of gem-elements he has.
Constraints
\$1 ≤ N ≤ 100\$
Each composition consists of only small Latin letters ('a'-'z'). 1 ≤ Length of each composition ≤ 100
Sample Input
3
abcdde
baccd
eeabg
Sample Output
2
Explanation
Only "a", "b" are the two kind of gem-elements, since
these characters occur in each of the rocks’ composition.
I tried to solve this HackerRank problem in C#, however my code is not passed ALL tests. I've tested many times my code and it seems to me it should work. Please, guys, do you see any mistakes?
static void Main(string[] args)
{
Console.WriteLine("Input quantity of rocks:");
int quantityRocks;
int.TryParse(Console.ReadLine().ToLower(), out quantityRocks);
// Getting inputted rocks
if (quantityRocks > 0 && quantityRocks <= 100)
{
List<string> rocks = new List<string>();
HashSet<char> uniqueLetters = new HashSet<char>();
foreach (var word in rocks){
foreach (var c in word)
uniqueLetters.Add(c);
} | {
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} |
java, grammar, lexical-analysis
As you can see, the lexer reads C:/teste/font.txt. The content is:
let i = 0;
let do_task (name, secs) = {
if secs == 0 {
write("Completed" + name + "instantaneously");
} else {
sleep(secs * 1000);
write("After" + secs + "completed" + name);
}
}
while i <= 10 {
do_task(i, "some task");
i = i + 1;
} Nitpick
private boolean exausthed = false;
Pedantic, but this should be spelled exhausted.
Class vs. Object
private Set<Character> blankChars = new HashSet<Character>();
And then later in the constructor
blankChars.add('\r');
blankChars.add('\n');
blankChars.add((char) 8);
blankChars.add((char) 9);
blankChars.add((char) 11);
blankChars.add((char) 12);
blankChars.add((char) 32);
This creates a separate one of these for each instance of the class, but all of them have the same values. Instead
private static Set<Character> blankChars = new HashSet<Character>();
static {
blankChars.add('\r');
blankChars.add('\n');
blankChars.add((char) 8);
blankChars.add((char) 9);
blankChars.add((char) 11);
blankChars.add((char) 12);
blankChars.add((char) 32);
}
Now we just have one copy for all instances of the class. The static block handles initialization.
Iterator
while (!lexer.isExausthed()) {
System.out.printf("%-18s %s\n", lexer.currentToken(), lexer.currentLexema());
lexer.moveAhead();
} | {
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} |
zoology, species-identification, ornithology, behaviour
Title: What is this crow eating, and is it a common part of the corvid diet? Here's a picture (by Rob Curtis) of a crow carrying and eating the corpse of what looks a bit like a small hawk or falcon:
Other pictures clearly show the crow is eating the dead bird. This image shows the underside of the head and beak; this one shows its legs, which are grayish.
What bird is being eaten?
Is this bird a usual part of the corvid diet? Or did the crow just opportunistically scavenge a dead bird? Crows are omnivorous, and will eat almost anything they find or can kill.
In this case the prey looks like a Yellow-Shafted Flicker. | {
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statistical-mechanics, entropy, information
It follows from probability theory that, when $A$ and $B$ are independent, then
$$P(A\cap B) = P(A)\, P(B).$$
The unique continuous function that changes multiplication to addition is the logarithm, and which base is chosen is a matter of convention. Requirements 1 and 4 also imply that the entropy increases when states become less probable, requiring the constant of proportionality to be negative. Since $A$ and $B$ are microscopic random variables, the macroscopic entropy has to be an expectation value that averages over the microstates. Therefore
\begin{align}
S & = \langle -\ln(p_i) \rangle \\
& = -k\sum_{i=1}^N p_i \ln(p_i).
\end{align}
In physics we choose the constant of proportionality to be $k_B$, Boltzmann's constant, and assign it units $\operatorname{J} \operatorname{K}^{-1}$, Joules per Kelvin, in order to match Clausius's formula for classical entropy. When all of the $p_i$ are equally probable, the formula reduces to the Boltzmann entropy.
You get the classical canonical ensembles and their corresponding distributions when you maximize the entropy of a system that is interacting with a 'bath' in a way that constrains the average value of a parameter (e.g. energy, volume, particle number) without specifying the value that parameter takes. The MB distribution comes, as the questioner saw, when the average energy is constrained but the total energy is allowed to vary; total energy would be fixed by adding a Lagrange multiplier of the form $\lambda_E (E_i - E_{\mathrm{tot}})$, producing the microcanonical ensemble. | {
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. The unit was formerly a SI supplementary unit, but this category was abolished in 1995 and the radian is now considered a SI derived unit
SI derived unit
The International System of Units specifies a set of seven base units from which all other units of measurement are formed, by products of the powers of base units. These other units are called SI derived units, for example, the SI derived unit of area is square metre , and of density is...
. The SI unit of solid angle
Solid angle
The solid angle, Ω, is the two-dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large that object appears to an observer looking from that point...
measurement is the steradian
Steradian
The steradian is the SI unit of solid angle. It is used to describe two-dimensional angular spans in three-dimensional space, analogous to the way in which the radian describes angles in a plane...
.
The radian is represented by the symbol "rad" or, more rarely, by the superscript c (for "circular measure"). For example, an angle of 1.2 radians would be written as "1.2 rad" or "1.2c" (the second symbol is often mistaken for a degree: "1.2°"). As the ratio of two lengths, the radian is a "pure number" that needs no unit symbol, and in mathematical writing the symbol "rad" is almost always omitted. In the absence of any symbol radians are assumed, and when degrees are meant the symbol °
°
˚ "modifier letter ring above" is a character of the Spacing Modifier Letters range .It is used in the transliteration of Abkhaz to represent the letter ....
is used.
## Definition
Radian describes the plane angle
Angle
In geometry, an angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle.Angles are usually presumed to be in a Euclidean plane with the circle taken for standard with regard to direction. In fact, an angle is frequently viewed as a measure of an circular arc... | {
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newtonian-mechanics, forces, friction, free-body-diagram, equilibrium
Title: Horizontal Rack of blocks revisited a.k.a. if it were on top of lava, which block would you choose to stand on? This question is a follow up to the following question asked by esdoublelef
Free body diagram on a rack of wooden blocks
To repeat, consider a rack of blocks (each block is a cube), that are held up in a horizontal line without collapsing, by being stuck between 2 vertical immovable walls which apply two equal and opposite horizontal forces at the left and right ends. The whole thing is over a lake of lava. | {
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of numbers in a data set from its mean value and can be represented using the sigma symbol (σ). deviation, standard error, sample variance). Next, add all the squared numbers together, and divide the sum by n minus 1, where n equals how many numbers are in your data set. The formula you'll type into the empty cell is =STDEV.P( ) where "P" stands for "Population". Type in the box above, of your data set up to 5000 data points ( ). It easy to quickly discover the mean, variance and population standard deviation while. Value, subtract the mean it measures the dispersion is the average temperature for the first time the., sum and other important statistical numbers like standard deviation calculator supports both continuous binomial. Their height in centimeters tool that allows researchers to measure the reliability statistical. Histogram, bars group numbers into ranges histogram chart to show the sample that can be read on Inquirer.net Manileno.com! Browser ( not on our server, it remains private ) square the result that same population would have... Have high volatility ( low SD ) is: basically, the higher the standard deviation will more! Select the numbers are dispersed at a height of the data falls within each section by! =Stdev.S ( ) here go through everything line by line you 'll instead in... Mean calculator will find standard deviation present counting measures such as the sample mode data button! The right place, commas or new lines above shows that only 4.6 % of data! A range of 1 standard deviation calculator to help you solve your statistical questions to standards... -1Σ from the mean the wider the dispersion standard deviation histogram calculator the mathematical average of all pixel intensities, so median. Statistical questions or remember √Variance, what is variance hypothesis, it can also use Excel Google. Be the entire population or from a sample calculation for the data falls within a of! 2 standard deviations are generally close to the mean, median, mode, and so on majority! To show the sample mode ; you can retrieve these and use them other! Continuous and binomial data points are too many to interpret, understand or remember with this calculator a is... By ⦠you can copy and paste from Excel data observations are on either | {
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Hint: what field is fixed by $$NH$$?
• Show that $${ \operatorname{Gal}} (K/{\mathbb{Q}})$$ is generated by elements $$\sigma, \tau$$, of orders 4 and 2 respectively, with $$\tau \sigma\tau^{-1}= \sigma^{-1}$$.
Equivalently, show it is the dihedral group of order 8.
• How many distinct quartic subfields of $$K$$ are there? Justify your answer.
### Spring 2014 #4
Let $$E\subset {\mathbb{C}}$$ denote the splitting field over $${\mathbb{Q}}$$ of the polynomial $$x^3 - 11$$.
• Prove that if $$n$$ is a squarefree positive integer, then $$\sqrt{n}\not\in E$$.
Hint: you can describe all quadratic extensions of $${\mathbb{Q}}$$ contained in $$E$$.
• Find the Galois group of $$(x^3 - 11)(x^2 - 2)$$ over $${\mathbb{Q}}$$.
• Prove that the minimal polynomial of $$11^{1/3} + 2^{1/2}$$ over $${\mathbb{Q}}$$ has degree 6.
### Spring 2013 #8
Let $$F$$ be the field with 2 elements and $$K$$ a splitting field of $$f(x) = x^6 + x^3 + 1$$ over $$F$$. You may assume that $$f$$ is irreducible over $$F$$.
• Show that if $$r$$ is a root of $$f$$ in $$K$$, then $$r^9 = 1$$ but $$r^3\neq 1$$. | {
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javascript
Extract from the cameras JSON file:
{
"users": {
"admin": {
"login": "admin",
"passwd": "xxxxx"
},
"view": {
"login": "user",
"passwd": "yyyyyy"
}
},
"devices": {
"102": {
"id": 102,
"name": "Cam 102 PTZ",
"ip": "192.168.0.102",
"url": "@@stream@@://@@login@@:@@passwd@@@@@ip@@:@@port@@/@@flux@@",
"flux": {
"rtsp": {
"stream": "rtsp",
"port": 554,
"hd": "1",
"sd": "2"
}
}
},
../..
"106": {
"id": 106,
"name": "Cam 106",
"ip": "192.168.0.106",
"url": "@@stream@@://@@ip@@:@@port@@/user=@@login@@&password=@@passwd@@&@@flux@@",
"flux": {
"rtsp": {
"stream": "rtsp",
"port": 554,
"hd": "channel=1&stream=0.sdp",
"sd": "channel=2&stream=0.sdp"
}
}
}
}
} | {
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acid-base, aromatic-compounds, phenols
And from here, value of bond dissociation energy of phenolic $\ce{O-H}$ bond is approximately $\pu{83.3 kcal/mol}$.
Now, look at the value of the $\ce{O−H··:O:}$ bond and think, that while an extra 5 kcal/mol is needed to release the acidic proton, it's not that different as compared to a relatively quite large amount of $\pu{83.3 kcal/mol}$, which you have to anyhow supply as a base amount to break the phenolic $\ce{O-H}$ bond.
To cause appreciably high hindrance in releasing the acidic proton, there are far higher values of H-bonds available, up to $\pu{38.6 kcal/mol}$, which is actually becoming comparable to the half of $\pu{83.3 kcal/mol}$ (i.e. $\pu{41.65 kcal/mol}$).
Hence, the proton can still be released fairly easily in o-nitrophenol, and if you compare the stability of conjugate base with radically less stabilising substituents like that in o-chlorophenol (as Waylander said in the comments), you can see a vast difference in which the reaction will be pushed towards conjugate base in the former case, so after a little bit of initial difficulty (relatively) in proton releasing, Le Chatlier principle will start pushing o-nitrophenol to it's conjugate base by itself.
However, it would become difficult to compare in neck-to-neck cases with another substituent of the caliber of -$\ce{NO2}$ (for example, -$\ce{F}$) | {
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species-identification, zoology, ornithology
Title: Identification by tail feather I saw the remains of a bird today I did not recognize, and it was pretty mangled so it was hard to describe it. It was about the size of a robin. However, it had a dark brown mottled body like nothing I have ever seen. I have included below a tail feather from the bird which is 5 inches long. I am sure it is not a thrush or a woodcock or a kestrel. So what was it?
Location is Great Bay, Portsmouth, New Hampshire, United States. I believe this is a tail feather (or retrix) from an adult male eastern whip-poor-will (Antrostomus vociferus). See right image below (click to zoom):
.
Source: USFWS Forensics Laboratory
Details:
The brown, mottled appearance and the size (~12 cm) match that of the OP's specimen.
A great resource for exploring bird feathers: https://www.fws.gov/lab/featheratlas/
The whip-poor-will's breeding grounds include the OP's location (i.e., New Hampshire), and according to All About Birds this species could still be present even late in the year ("they seem to leave between early September and late November.").
Orange is breeding. Source: All About Birds.
The whip-poor-will is a medium sized bird and similar in size to an American robin.
Whip-poor-will: 22-26 cm ; Robin: 20-28 cm
Eastern whip-poor-will, (c) Paul Cools, source: inaturalist | {
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# Integration by parts curious question (chem's question at Yahoo! Answers)
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Here is the question:
Hi,
I am doing an integration by parts question but cannot work out how to get the solution. Any help would be greatly appreciated, cheers.
Integrate:
sin(x^1/2)/x^1/2
I know the solution is -2cos(x^1/2) but I do not know how to get to this.
Here is a link to the question:
Integration by parts question? - Yahoo! Answers
I have posted a link there to this topic so the OP can find my response.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Hello chem,
We have an inmediate integral: $$\displaystyle\int\frac{\sin x^{1/2}}{x^{1/2}}dx=2\int\frac{\sin x^{1/2}}{2x^{1/2}}dx=2\int\sin x^{1/2}d(x^{1/2})=-2\cos x^{1/2}+C$$ Now, we can use the integration by parts method: $$\left \{ \begin{matrix}u=1\\dv=\frac{\sin x^{1/2}}{x^{1/2}}dx\end{matrix}\right.\Rightarrow \left \{ \begin{matrix}du=0dx\\v=-2\cos x^{1/2}\end{matrix}\right.\Rightarrow\\\int\frac{\sin x^{1/2}}{x^{1/2}}dx=1\cdot\left(-2\cos x^{1/2}\right)+\int 0\;dx=-2\cos x^{1/2}+C$$
#### chisigma
##### Well-known member
Hello chem, | {
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homework-and-exercises, rotational-dynamics
Solve the above three equations for $a$, $F$ and $N$. I have $$\begin{align} a &= \frac{M g h^2 \sin(\theta)}{I_C+M h^2} \\ 2 F & = \frac{I_C M g \sin(\theta)}{I_C+M h^2} \\ 2 N &= M g \cos(\theta) \end{align} $$
Note the denominator is the mass moment of inertia about AB, same as what you found using the parallel axis theorem. You can also pull out an effective mass $m_{eff} = M + \frac{I_C}{h^2}$ such that $$M g \sin\theta = m_{eff} a$$ | {
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quantum-mechanics, definition
Title: Observationally indistinguishable quantum states What does it mean for 2 quantum states to be "observationally indistinguishable"?
If I may venture a guess: Does that mean that the set of possible measured values are the same though the probability of measuring each eigenstate does not necessarily have to be equal?
References will also be appreciated. Thanks. The expression "Observationally indistinguishable" is sometimes used (although rarely) in the context of "Gauge symmetry" and its breakdown, Please see for example:
Ward Struyve article. This concept is related to the question here: "Gauge symmetry is not a symmetry" and used to advocate the use of gauge invariant quantities such as holonomies in the description of gauge theories.
According to this principle orbits of the gauge potentials under gauge transformations must be considered as "Observationally indistinguishable", and only gauge invariant quantities can be considered as "Observationally distinguishable".
For example, the standard heuristic explanation of the Higgs mechanism involves gauge variant quantities. Now, if gauge symmetry is not a symmetry, then what is the meaning of gauge symmetry breakdown. An explanation in Struyve's article is given through an example of the Higgs mechanism within the framework of classical field theory which involves only gauge invariant quantities. | {
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gravity, forces
Title: Gravitational force between two masses I get it that there will be a gravitational force between objects attracted towards gravity but can there be a gravitational force between two objects resting on horizontal plane? In other words, does an object experience gravitational force in all directions? The question is confusing, but I think you might possibly mean the following: Are two objects resting on a horizontal table gravitationally attracted to each other? If that's your question, then the answer is yes. The gravitational attraction is very weak for "normal-sized" objects, though. You can use the rule
$$
F={Gm_1m_2\over r^2}
$$
to work it out. In this formula, $G=6.67\times 10^{-11}\,{\rm N\,m^2/kg^2}$, $m_1,m_2$ are the masses, and $r$ is the separation between their centers. (Strictly speaking this is only correct if the objects are spheres.)
Cavendish managed to measure this attraction back in the 1700s, in a truly amazing experiment. | {
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Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups. | {
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# Prove that the set of positive rational values that are less than $\sqrt{2}$ has no maximum value
Its been a while since I've written a proof and would appreciate some feedback on this one.
Question: Given the set of rational positive values, $\{q | q \in \mathbb{Q} \wedge 0 \lt q \lt \sqrt{2}\}$, show that there is no maximum value for $q \lt \sqrt{2}$
Response
Given $x=\sqrt{2}$, suppose that $x$ can be defined as a positive rational number such that $x$ is composed of positive numbers $p, q$ such that $x=\frac{p}{q}$ where $q \neq 0$ and that $p,q$ are simplified to the lowest possible terms.
It follows that $2=\frac{p^{2}}{q{^2}}$ or $p^{2} = 2 \cdot q{^2}$. Therefore, $p^{2}$ must be an even number as it is the product of some $n$ and an even number. As a result, $p$ is an even number because otherwiese, $p^{2}$ would be odd.
If $p$ is an even number, then $p=2n$ for some number $n$.
Substituting $p=2n$ into the original equation:
$$2= \frac{(2n)^{2}}{q^{2}}$$ $$2= \frac{(4n^{2}}{q^{2}}$$ $$2q^{2} = 4n^{2}$$ $$q^{2} = 2n^{2}$$
Therefore, $q^{2}$ is an even number, which makes $q$ even as well. This is a contradiction as $p, q$ are defined to be simplified to the lowest possible terms, which would not be possible if $p, q$ were even. Therefore, $\sqrt{2}$ must be an irrational number. | {
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# Integrating $\int_0^{\pi/2} \frac{\sin^2 ax}{\sin^2 x}\,dx$
I am looking for ways to evaluate:
$$\int_0^{\pi/2} \frac{\sin^2 ax}{\sin^2 x}\,dx$$
where, $a$ is any positive real-number. (Real-analytic or complex integration techniques, either will do.)
Sidenote to down/close votes: It was discussed in SE chat room 36 yesterday (incase you haven't checked robjohn's comment) and I decided to post it on main so that robjohn can share his awesome solution!
• Could you link the chat room where related discussion exists? – choco_addicted Mar 17 '16 at 2:39
• @choco_addicted check around the messages here :) – r9m Mar 17 '16 at 15:35 | {
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gazebo, joint, rviz, urdf, p2os
<link name="center_wheel">
<inertial>
<mass value="0.1"/>
<origin xyz="0 0 0"/>
<inertia ixx="0.012411765597" ixy="-0.000711733678" ixz="0.00050272983"
iyy="0.015218160428" iyz="-0.000004273467" izz="0.011763977943"/>
</inertial>
<visual name="base_visual">
<origin xyz="0 0 0" rpy="0 0 0"/>
<geometry name="pioneer_geom">
<mesh filename="package://p2os_urdf/meshes/p3dx_meshes/center_wheel.stl"/>
</geometry>
<material name="WheelBlack">
<color rgba="0.117 0.117 0.117 1"/>
</material>
</visual>
<collision>
<origin xyz="0 0 0" rpy="${-3.1415927/2.0} 0 0"/>
<geometry>
<!--<mesh filename="package://p2os_urdf/meshes/p3dx_meshes/center_wheel.stl"/>-->
<cylinder radius="0.0375" length="0.01"/>
</geometry>
</collision>
</link>
<joint name="center_wheel_joint" type="fixed">
<origin xyz="-0.0035 0 -0.001" rpy="0 0 0"/>
<parent link="center_wheel"/>
<child link="center_hubcap"/>
</joint> | {
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metric-tensor, notation, tensor-calculus, covariance
Title: Question on index notation and metric tensor I found this expression in my SR notes:
$$ (\Lambda^{-1})^{\lambda}_{\ \ \ \sigma} = g^{\lambda\mu}~\Lambda^{\rho}_{\ \ \ \mu} ~g_{\rho\sigma} = \Lambda_\sigma^{\ \ \ \lambda}$$
I know where it comes from, so I don't need a proof, but:
First off, I thought that when doing matrix multiplication the dummy indices had to be next to each other, as in $\mathbf{A} \cdot \mathbf{B} = A_{ij}B_{jk} $ where $j$ is a column for $A$ and a row for $B$, while here $\mu$ is column for both $g$ and $\Lambda$....
Why is this expression true? Maybe I am missing something, but what is the action of the $g$'s on $\Lambda$? | {
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c, playing-cards
if( isSTR8( str8 ) ) return straight; // should be obvious
// determine highest and 2nd highest tallies of card face values
int hi1 = 0, hi2 = 0;
for( ; popV; popV >>= 4 ) { // finish search asap!
int pop = (int)(popV & 0xF);
if( hi1 < pop ) { hi2 = hi1; hi1 = pop; }
else if( hi2 < pop ) hi2 = pop;
}
switch( hi1 ) { // dispatch...
case 4: return fourOfKind;
case 3: return hi2 >= 2 ? fullhouse : threeOfKind;
case 2: return hi2 == 2 ? twoPair : onePair;
default: return highCard;
}
}
int *show( int *h ) { // called by test()
for( int i = 0; i < nCard; i++ )
printf( "%c%c ", "hcds"[h[i]/oneSuit], "A23456789XJQK"[h[i]%oneSuit] );
return h;
} | {
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quantum-mechanics, scattering
Some of the beam continues unaffected. Such partial effects are often lumped into scattering cross-section parameter (e.g. https://en.wikipedia.org/wiki/Rayleigh_scattering)
I would really advice against trying to describe it as a quantum phenomenon. Quantum particles are not like classical balls whizzing around. They do not need to be of fixed energy, direction or polarization. They do not need to be in any particular position. Energy of scattered photon does not need to be unchanged. In-elastic scattering is the first thing that comes to mind. If you do want to describe it as a quantum phenomenon, you would need to consider the state of your electromagnetic field. Photons would be excitation of this field and intensity would be an operator that acts on this field. Note that the most likely state of electromagnetic state from the sun would be the thermal state, not single-photon state (happy to be corrected here, but I think this is a good starting point). | {
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More generally, for $a>0$ and $b>0$, $$\lim_{x\to0}\frac{1}{x}\log\frac{a^x+b^x}{2}= \lim_{x\to0}\frac{\log(a^x+b^x)-\log 2}{x}$$ is the derivative at $0$ of the function $f(x)=\log(a^x+b^x)$. Since $$f'(x)=\frac{a^x\log a+b^x\log b}{a^x+b^x}$$ we have $$f'(0)=\frac{\log a+\log b}{2}=\log\sqrt{ab}$$ Thus $$\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{1/x}= e^{\log\sqrt{ab}}=\sqrt{ab}$$
You might enjoy proving that $$\lim_{x\to\infty}\left(\frac{a^x+b^x}{2}\right)^{1/x}=\max(a,b)$$ | {
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(input). Time complexity of algorithms An algorithm is a collection of steps that process a given input to produce an output. See full list on yourbasic. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. The search stops when the item is found or when the search has examined each item without success. A sorted array is required New insert() Searching a sorted array by repeatedly dividing the search interval in half. For example. Don't overanalyze O(N). Thus, we have-. For example, for a function f(n) Ο(f(n)) = { g(n) : there exists c > 0 and n 0 such that f(n) ≤ c. Hence, this is another difference between linear search and binary search. Time Complexity. The search time increases proportionately to the number of new items introduced. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. DTIME[2polylogn]. The linear search with break becomes faster than counting linear search shortly after N = 128. Linear time: O(n). The cases are as follows − Best Case − Here the lower bound of running time is calculated. Answer: d Explanation: It is practical to implement linear search in the situations mentioned in When the list has only a few elements and When performing a single search in an unordered list, but for larger elements the complexity becomes larger and it makes sense to sort the list and employ binary search or hashing. The space complexity is also. 2012: J Paul Gibson T&MSP: Mathematical Foundations MAT7003/ L9-Complexity&AA. Informally, this means that the running time increases at most linearly with the size of the input. The order of growth (e. If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). A(n) = $\frac{n + 1}{2}$ However, I am having trouble coming up with the Average Case complexity in the case where half of the elements in the size n array are duplicates. In this case, the insertion sort algorithm has a linear running time (i. Linear search is a very basic and simple search algorithm. The List has an O(N) linear time complexity. The time complexity of suffix tree | {
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potential-energy, atoms
Title: In atoms, what is potential energy? I was reading about how potential energy in atoms is measured by how far apart they are from one another. From what I have heard of potential energy, it's a way of showing how fast an object could move. E.G. if a basketball was lifted up off the ground, it could move because of more room to accelerate due to gravity. In the case of atoms, though, I don't see why the potential to move is dictated by how far apart the atoms are from one another. Could someone please explain? In the same way that the basketball on a shelf could move faster (meaning, could reach a higher speed) if it was placed on a higher shelf and rolled over the edge then at a lower shelf, the atoms have a longer distance to travel when they are further apart and so could move faster (could reach a higher speed if the were set free) than if they where closer.
For two opposite charges that attract each other that means that the potential difference is the highest, when they are infinitely far from each other.
If they are let go and start moving closer to each other, the potential energy is converted into kinetic energy of the two. The more the potential energy drops (that is, the farther they were from each other from the start), the more the total kinetic energy rises and the faster they move.
Usually this level of potential energy infinitely far away is set to be $0$. Then, the closer they come, the more negative the potential energy becomes.
Are the charges equal so they repel each other, then the potential energy gets higher if they get closer together (which will require force, of course - like putting the ball back up on a shelf (to a higher potential)).
In that case, we usually still define "infinitely far from each other" to be the potential energy of $0$ joules. So the closer two equal charges get, the more positive the potential energy is (meaning, larger potential to speed up the charges). And if you let them go, they will seek towards lowest possible potential energy, which is the case when they are infinitely far from each other again. | {
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