text stringlengths 1 1.11k | source dict |
|---|---|
java, role-playing-game
And here is the code for the events:
public class Events {
//MOUNTAIN EVENTS START
//Event 1: Lose all drones
public void mountEvent1() {
System.out.println("------------------------------------------------------------------------------------------------------------------");
System.out.println("Your drones pick through the crags and stones of the mountain, searching for resources for your growing hive.");
System.out.println("There is a rumble and your drones get one good look at the wave of stone");
System.out.println("headed toward them before your link to them is severed.");
System.out.println("------------------------------------------------------------------------------------------------------------------");
}
//Event 2: Combat Encounter
public void mountEvent2() {
System.out.println("Your drones make their way through the mountain when they are suddenly ambushed!");
} | {
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seasons
Title: Have the seasons always been? Has the Earth had it's wobble that causes the seasonal variation in solar energy in the northern and southern hemispheres for it's entire history? Is this variation evident in the geologic record or is it an open question? Could a meteor impact like that which killed the dinosaurs 66 million years ago have altered or created the wobble? Are there any factors dampening or increasing the seasonal wobble over time?
Has the Earth had it's wobble that causes the seasonal variation in solar energy in the northern and southern hemispheres for it's entire history? | {
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organic-chemistry, inorganic-chemistry, carbon-allotropes, aromaticity
Title: Are graphite and hexagonal boron nitride aromatic Are graphite and hexagonal boron nitride aromatic?
Graphite has a planar network of 6-membered rings with each carbon connected to three other carbons. Since the valency of carbon is not satisfied, there will be some $\pi$-bonds. Can graphite act like an aromatic system? Can I explain this using Hückel's rules?
Hexagonal boron nitride has a similar structure. So if graphite exhibits aromatic character, then so will boron nitride (like inorganic benzene and benzene). Graphite is definitely aromatic and boron nitride is at least partially aromatic. For instance, in this paper, the authors calculate the percent resonance energy (%RE) of graphite as a comparison to other known aromatic systems. The %RE is a measure of the total resonance stabilization per carbon atom relative to some reference system. | {
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"url": null
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quantum-state, density-matrix, terminology-and-notation
$$
\rho_{XY} = \sum_{xy} p_{xy} |x\rangle \langle x | \otimes |y\rangle \langle y |.
$$
Note that this is consistent with the marginal states representing the marginal random variables!
The second state you mention in the question would represented the joint random variable $XY$ where $p_{xy} = 0$ whenever $x \neq y$. | {
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php, laravel
// BlockFactory, bound to the Block facade (bad naming i know)
class BlockFactory
{
protected $blocks = [];
protected $currentBlock;
public function create($title, $content, $position, $order = 0, $parameters = [])
{
$this->currentBlock = new Block($title, $content, $position, $order, $parameters); // this part is bugging me somehow...
return $this;
}
public function add()
{
$this->blocks[] = $this->currentBlock;
}
public function render()
{
return $this->blocks;
}
}
// The actual block class
class Block
{
protected $title;
protected $content;
protected $position;
protected $order;
protected $parameters; | {
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## Determine if vector b is a linear combination of vectors a1, a2, a3.
Hi guys. I've solved an exercise but the solution sheet says what doesn't make sense to me. Could you please help with this problem?
Determine if vector b is a linear combination of vectors a1, a2, a3.
a1=[1, -2, 0], a2=[0, 1, 2], a3=[5, -6, 8], b=[2, -1, 6].
b is a linear combination when there exist scalars x1, x2, x3 such that x1*a1 + x2*a2 + x3*a3 = b. right?
I put a's in a coefficient matrix and b in the augmented column. [a1 a2 a3 | b]. Row-reduced it produces a consistent system (although I get x3 a free variable - third row all zeroes). But the solution sheet says b is not a linear combination of the a vectors. Where is the catch? Should the RREF have a unique solution?
Thank you.
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire | {
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c++, file, c++17, serialization, file-structure
enum Flag : uint8_t { header = 0, body = 0xff } flag;
Then get_header could come out something like this:
std::optional<TAPHeader> get_header() const {
const int min_header_size = 17;
if (flag == header && data.size() >= min_header_size) {
return TAPHeader{data};
}
return {};
}
At least to me, this expresses the intent a bit more clearly.
Right now, you print out the raw checksums (as read, and as computed) and leave it to the reader to verify that they match. Personally, I'd prefer to have the computer do that check (and consider omitting the message entirely if they match, as well as adding ANSI code to print it in bright red if they mismatch). | {
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Last edited: Aug 2, 2013
4. Aug 1, 2013
### symbolipoint
If you want to make better use of the compound interest formula instead of the continuous growth formula, at least pick a large enough compounding period (and rate adjusted to that period).
5. Aug 2, 2013
### Staff: Mentor
No. This is not correct. The compound interest formula should be A = P(1+i/n)^nt, where i is the annual interest rate, and n is the number of compounding periods per year. So,
$A=Pe^{ntln(1+i/n)}$. In the limit of large n, this approaches the continuous compounding formula $A=Pe^{it}$.
6. Aug 2, 2013
### Ray Vickson
The difference is due to using a different value of 'r' in the two calculations. To match them up you need to use a "continuous" rate r that satisfies $e^r = 1.065 \longrightarrow r \doteq 0.0629748.$
7. Aug 2, 2013
### fakecop
Yes, I'll fix my mistake now.
8. Aug 2, 2013
### Ray Vickson | {
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## Best Algebra 2 Lab Ever
This post shares what I think is one of the best, inclusive, data-oriented labs for a second year algebra class. This single experiment produces linear, quadratic, and exponential (and logarithmic) data from a lab my Algebra 2 students completed this past summer. In that class, I assigned frequent labs where students gathered real data, determined models to fit that data, and analyzed goodness of the models’ fit to the data. I believe in the importance of doing so much more than just writing an equation and moving on.
For kicks, I’ll derive an approximation for the coefficient of gravity at the end.
THE LAB: | {
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"url": "https://casmusings.wordpress.com/tag/cas/"
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particle-physics, neutrinos, beyond-the-standard-model, majorana-fermions
Title: What mechanism will generate Dirac neutrinos? Giving mass to neutrinos is not a problem. The problem is to explain its smallness. Once right-handed neutrinos $\nu_R$'s are included in the Standard Model (SM), the Majorana mass for $\nu_R$ must also be included unless that is forbidden by some symmetry. This automatically and unavoidably leads to the robust conclusion that the mass eigenstates (both light and heavy) are Majorana type.
So my question is what mechanism will generate Dirac neutrinos? I am not entirely sure but I think without introducing new symmetries to the SM this is not possible and still then it is not quite natural. So I'll just summarize my ideas on this and see if someone can come up with something better.
To have a Dirac mass term you need to have some right handed neutrino $\nu_R$. As you said the general approach within the seesaw type 1 is to introduce a right handed singlet $\nu_R$ which gives you both a Dirac and Majorana mass term | {
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c++, overloading
First, the outer parentheses are not necessary at all. Second, the ternary operator doesn't do anything. Everything before the ? is already a boolean value, so you can just write:
return new_n.radius < n.radius && new_n.height < n.height;
operator= should return a non-const reference
operator= only makes sense on a non-const object, and the return value should also be a non-const reference.
Use () for functions that take no parameters
(void) is only necessary in C, in C++ you can just write:
Cylinder();
Prefer initializer lists to initialize member variables
While your constructors are valid, prefer using initializer lists to initialize member variables, like so:
Cylinder::Cylinder(): radius(1), height(1) {}
Cylinder::Cylinder(double radius_, double height_): radius(radius_), height(height_) {}
Or combine them using default parameter values:
Cylinder::Cylinder(double radius_ = 1, double height_ = 1): radius(radius_), height(height_) {} | {
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• you get the sequence $0\to H^1(G_k,\mathcal{O}(X_{\overline{k}})^\times)\to \text{Pic}(X)\to \text{Pic}(X_{\overline{k}})^{G_k}$. Now, since $\text{SL}_2$ is semisimple one can show that $\mathcal{O}_{\text{SL}_2}(\text{SL}_2)=k^\times$ (valid over any $k$, including $\overline{k}$) and thus this first cohomology term vanishes by Hilbert's Theorem 90. Since $\text{Pic}(\text{SL}_{2,\overline{k}})$ vanishes, so then must $\text{Pic}(\text{SL}_2)$. Nov 7 '16 at 10:51 | {
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"tags": null,
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entropy, time, reversibility, arrow-of-time
[Edit #3] The idea that thermal inertia provides a fundamental example of the asymmetry of time seems to crumble once it's examined closely. The temperature of a thermal mass (in a gas) strictly does not give information about the recent temperature of the gas, it gives information about the average kinetic energy of the atoms that recently hit the mass. It gives exactly the same amount of information about the kinetic energy of the atoms bouncing off the object in the near future. | {
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c++, animation, timer, fltk
attach (*hour_indicator);
// redraw ();
draw ();
}
Result:
after 33 seconds:
after 1 min 24 seconds:
Questions:
Is the above code looking good enough, are there any mistakes that could be eliminated?
How to make the clock update more smoothly; without "blinking" on every second?
If I use the commented functions redraw () within increment_second () / minute () / hour () the clock shows, within the specified window, only once after the loops within run_clock () are over. On the other hand, in the current implementation, i.e. using functions draw () the clock updates every second by not in the specified window and it interferes with the rest of the open windows. | {
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rosjava, android
Originally posted by tfoote with karma: 58457 on 2011-11-30
This answer was ACCEPTED on the original site
Post score: 3
Original comments
Comment by seanarm on 2011-12-05:
... to provide some limited ROS compatibility with remote nodes, on a machine with a full ROS installation.
Comment by seanarm on 2011-12-05:
I'm looking for "drop-in" ROS capability for mobile development. That is, I don't want to have to perform the entire ROS installation process (for various convenience reasons). Specifically, in the ideal case, I want a JAR that I can include in an Eclipse Android application... | {
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javascript, jquery
// Check favorite set in setFavorite()
function checkFavorite() {
if (localStorage.getItem("favorite") !== null) {
var name = localStorage.getItem("favorite");
var favstatus = localStorage.getItem("favorite-status");
var favstatusSplit = favstatus.substr(2);
var favstatusLower = favstatusSplit.toLowerCase();
var string = $(".row-l").text().toLowerCase();
var re = new RegExp(name.toLowerCase(), 'g');
var test = string.match(re);
$(".row-l:contains(" + name + ")").parent().parent().find(".star-inside").addClass("favorite");
if (test !== null) {
$(".fav_school_inside").text(name + " - " + favstatusLower + "!");
$(".clear span").text("\"" + name + "\"");
setTimeout(function () {
$(".fav_school").addClass("top");
}, 1000);
setTimeout(function () {
$(".fav_school").removeClass("top");
}, 6000);
}
} | {
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electromagnetism, lagrangian-formalism, gauge-theory, noethers-theorem, gauge-invariance
Then, after imposing the quantum mechanical constraints (2.1), one fixes all gauge redundancies. This is the coulomb gauge in QED. It is straightforwad to verify that the passage towards canonical quantization in the coulomb gauge is achieved by replacing the Dirac bracket (1.8) by the commutation relation $$\left[A_{i}(\mathbf{x}),\Pi_{j}(\mathbf{y})\right]=-i\hbar\left(\delta_{ij}-\frac{\partial_{i}\partial_{j}}{\Delta}\right)\delta(\mathbf{x}-\mathbf{y}), \tag{2.2}$$
which is compatible with all the quantum mechanical constraints in (2.1). It is straightforward to check that in the reduced phase space where the canonical commutation relation (2.2) is defined, all gauge redundancies are removed. i.e
\begin{align}
[\Pi_{i}(\mathbf{x}),\mathcal{G}[\lambda]]=0,\quad &\mathrm{or}\quad U[\lambda]\Pi_{i}(\mathbf{x})U[\lambda]^{\dagger}=\Pi_{i}(\mathbf{x}), \\ | {
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coordinate, observable-universe
Title: Where is the North of the Universe Does it exist North of the Universe? (Which is probably used for determining coordinates of far galaxies.)
If so, then where it is?
On what is it aligned? (Is it orientated on Earth, Sun, Milky Way, Local group or on something more bigger?) Sort of... There is a system called the International Celestial Reference System (ICRS) which has center at the Solar System Barycenter (normally inside the Sun but not the same as the Sun's center) and which has the x and y axis in the plane of the Earth's equator and the z-axis ("North" if you will) pointing towards the Celestial Pole. The x-axis points towards the equinox where the plane of the Earth's orbit and the plane of the Solar system (the ecliptic) cross (on average; this gets complicated quickly as everything related to the Earth is wobbling about on different timescales). | {
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## Input Arguments
collapse all
Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Specify a differential equation by using the == operator. If eqn is a symbolic expression (without the right side), the solver assumes that the right side is 0, and solves the equation eqn == 0.
In the equation, represent differentiation by using diff. For example, diff(y,x) differentiates the symbolic function y(x) with respect to x. Create the symbolic function y(x) by using syms and solve the equation d2y(x)/dx2 = x*y(x) using dsolve.
syms y(x)
S = dsolve(diff(y,x,2) == x*y) | {
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fft, discrete-signals, fourier-transform, power-spectral-density
%% Plot results to compare both methods
figure(3);
clf;
box;
hold on;
% Plot theoretical value at top
plot(ww, ones(1,length(ww))*T/4, '--', 'Color', 'g', 'LineWidth', 3.5);
% Plot "direct" calculation
plot(ww, Sx, '.', 'Color', 'b', 'LineWidth', 3.5);
% Plot "FFT" calculation
plot(w_fft, Sx_fft, '-', 'Color', 'r', 'LineWidth', 3.5);
set(gca,'FontSize',18,'FontName', 'CMU Sans Serif');
xlabel('\omega', 'fontsize', 24, 'FontName', 'CMU Sans Serif');
ylabel('$S_x$', 'fontsize', 24, 'Interpreter', 'latex', 'FontName', 'CMU Sans Serif');
title('DTFT');
xlim([w_start, w_end]);
grid on;
end Your vector
t = linspace(0, T, num_samples);
is the problem. It should be
t = (0:num_samples-1)/fs;
Note that, even though you claim it in a comment in the code, in the current version, dt = t(2) - t(1); is not equal to 1/f_s. | {
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python, parsing
# usage:
# $ python thisfile.py text.txt searchstring insert.txt
text_file, searchstring, insert_file = sys.argv[1:]
# Directory and name of text_file.
dirname, basename = os.path.split(text_file)
# Create temporary file in same directory as text_file.
with tempfile.NamedTemporaryFile('w', dir=dirname, prefix=basename,
delete=False) as temp:
with open(text_file) as f1:
inserted = False # Have we inserted insert_file yet?
for f1_line in f1:
if not inserted and f1_line.startswith(searchstring):
with open(insert_file) as f2:
for f2_line in f2:
temp.write(f2_line)
inserted = True
temp.write(f1_line)
os.rename(temp.name, text_file)
Writing reliable code that works on different platforms is not easy! | {
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c#, design-patterns, entity-framework, repository
So, is anything blatantly wrong with this approach? Here I've injected the UoW/context directly where it's needed - in a bigger project I would have created a "service" and constructor-injected the UoW there instead. I'd second TopinFrassi's answer, that this is a more or less sensible approach, and the implementation is excellent. However, there are some potential issues, which I'll expand from craftworkgames' comments and your response. Because I can't really find anything that needs criticism in how it's written, this will be entirely design focussed.
There are two central issue in passing an interface that claims it can provide
IDbSet<TEntity> Repository<TEntity>() where TEntity : class;
You're promising an awful lot
You're exposing this outside of directly data-access related code (even if it's in service classes, these classes are unlikely to all be directly DA related except perhaps in a very small application)
And the problems this can cause are: | {
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life
The search for life on Mars has concentrated on finding evidence of autotrophic microorganisms – organisms that convert simple chemicals in the environment into complex organic compounds. Most autotrophs use water as the reducing agent, but some can use other chemicals like hydrogen sulphide.
The energy source for these autotrophs can either be sunlight, in which case we call them phototrophs (which includes algae and green plants here on Earth), or obtained through the oxidation of electron donors in their environment, in which case they're called chemotrophs.
There's a wide variety of chemotrophs here on Earth, typically living in completely dark and often hostile environments such as next to deep-sea volcanic vents. For example, iron-oxidizing bacteria even colonise new lava beds in the deep ocean, and in effect use the oxidation of available ferrous iron as their respiratory process. | {
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "life",
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} |
homework-and-exercises, electrostatics
Title: Can We Take the Potential at the Surface as the Potential at the Surface along with Potential at Infinity?
Here $V$ is the potential at surface and it is given to be zero, but I want to ask why does it hold that
$$
V_{surface} = V_{surface} + V_\infty
$$
Here V is the potential at surface.
Is nor correct.
$V$ is the potential of the surface $V_{\rm surface}$ relative to the potential at infinity $V_{\infty}$ ie the potential difference between the surface and infinity, so $V = V_{\rm surface} -V_{\infty} $ | {
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This sort of "problem" occurs quite naturally and can look this way without actually indicating a problem. (There might be some problem, but a pattern similar to this doesn't necessarily indicate one.)
It's a consequence of regression to the mean and arises directly out of fitting the conditional mean (i.e. it's exactly what you expect to see with regression).
One thing that might throw off some answerers is that you have your plot "backward" to what most of us are used to -- with the random variable on the x axis rather than the y-axis.
Here I have generated some data according to a regression model (with a normally distributed predictor and conditionally normal response) and fitted a model of the same form as the one that generated the data. Here's the corresponding plot to yours drawn the other way around: | {
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python, algorithm, python-3.x, graph
next_vertices hold integer vertex values. Instead of:
next_vertices = [[0] * len(vertices) for i in range(len(vertices))] | {
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terminology
There may be some other niche term for this, but if you relate it to basic stats there's a high probability everyone in the room will understand what you're talking about. | {
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java, strings, cryptography
/**
* this method will decypher and encrypted string message
* @param message
* @return
*/
private static String DecypherEncryption(String message) {
String[] code_and_case = message.split("¤¤¤¤");
String codeMessage;
String caseMessage;
if(code_and_case.length == 2) {
codeMessage = code_and_case[0];
caseMessage = code_and_case[1];
}
else {
codeMessage = message;
caseMessage = null;
} | {
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the number of different helmet designs drivers can use for each season. The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) can be found by the forumla: Comments Poor, please add more definition. It is based on using parabolas at the top instead of straight lines. K 4 represents the maximum value of the absolute value of the 4th derivative of f over. Rule 3: The DPD of the last course is numerically equal to the latitude of that course but with the opposite sign. A modified midpoint rule for the approximate calculation of weighted integrals /* p(x)f(x)dx, where p(x) ^ 0 is the weight function, has been recently proposed by Jagermann [1]. a + ih fori = O, 1. Mathematics HL and further mathematics formula booklet 1. Jul 29, 2019 · Finding the midpoint of a line segment is easy as long as you know the coordinates of the two endpoints. Customers sometimes use these equations to get a sense of whether something is overpriced, although an accurate picture requires information | {
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"url": "http://smsm.computer-service-hilden.de/midpoint-rule-formula.html"
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quantum-mechanics, mathematical-physics, operators, hilbert-space, discrete
$$||A\psi_\epsilon - \lambda \psi_\epsilon|| < \epsilon$$
In other words there is a class of approximated eigenvectors, though no proper eigenvector exists for $\lambda \in \sigma_c(A)$.
Under suitable further hypotheses on the Hilbert space (rigged Hilbert space) the set of $\psi_\epsilon$ admits a limit outside the Hilbert space. The distributions $\delta(x-x_0)$ are typical examples of this situation when referring to the position operator $X$ in $L^2(\mathbb R)$, since $\sigma(X)=\sigma_c(X)= \mathbb R$. | {
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javascript, jquery
tab += 1;
}
};
tab = Math.min(tabs.length, Math.max(tab, 0)); //stay in range
tabs.eq(tab).focus()
}); One thing you could consider is making this a little more generic and thus reusable. Currently you seem to assume, that there is only one .tab element in your document. If you had multiple (independent) .tab elements, each with their own set of select elements and you did't want to allow "tabbing" between then, then $(".tab select") will select all selects and not only the ones in the "current" .tab. So instead of
let tabs = $(".tab select"); | {
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algorithms, history
Title: Is there an algorithm which was originally invented to solve a contrived problem but later found practical use? This is probably a computer science history question.
Is there an algorithm which was originally invented to solve a contrived problem but later found practical use? The Ternary Golay Code has parameters $[n,k,d]_3=[11,6,5]_3$ and thus is a linear code over $\mathbb{Z}_3=\{0,1,2\}$ with $3^6=729$ codewords. It is perfect so that all the Hamming spheres of radius $2$ around codewords fill the whole space $\{0,1,2\}^{11}$ with no gaps left over. This means that the code has covering radius $2,$ any ternary string is at most Hamming distance 2 from a codeword.
This code was discovered by a Finnish football (soccer) enthusiast, which gives one a scheme of playing 729 guesses of 11 football matches (1 denotes home win, 0 denotes a draw and 2 denotes an away win) and being guaranteed to guess at least $11-2=9$ matches correctly. | {
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$\theta=ae+b\pi$
With $a\in\Bbb{Z}$ and $b\in\Bbb{Z}$?
I want to write:
$\theta\in e\Bbb{Z}+\pi\Bbb{Z}$
But I know that that is wrong because $a$ and $b$ could be different integers.
• The same concept in functional programming languages is called 'lifting': The first $+$ operator and scalar multiplication are ordinary addition on complex numbers $\mathbb C$. The second $+$ operator and scalar multiplication are 'lifted' versions of those ordinary operations, to subsets of $\mathbb C$, or in other words, lifted to $P(\mathbb C)$. See, e.g., wiki.haskell.org/Lifting and stackoverflow.com/q/2395697/223837. – MarnixKlooster ReinstateMonica Feb 8 '18 at 0:20
• The answers below do a good job, but I might note that "coset" a common name for $\beta + 2\pi \mathbb Z$. That is, it is the translation of the subgroup $2\pi\mathbb Z$ by $\beta$. (There is also "sumset", which is what you have in the second example, but it'd be weird to see in this context) – Milo Brandt Feb 8 '18 at 4:14 | {
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machine-learning, protein-protein-interaction
Title: Machine learning using protein-sequences I'm participating in a bioinformatics machine-learning seminar at my university. The main task is predicting binary classification of protein-protein interactions using sequence data as input.
One of the subtasks is familiarization with the dataset and presenting the dataset. Now I'm wondering which additional information I can get out of the sequence data.
I want to start out with binary classification for the protein interaction with the labels “Interact” and “Non-Interact. The dataset that I got provided consists of two fasta files. One containing the sequence header with the species and the corresponding amino-acid sequences ~300 entries. The other containing the exact same species header and the label “Interact” or "Non-interact”. Species are completely mixed. | {
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times, and all these trials might have different outcomes. Practicing probability and statistics? A coin toss is a tried-and-true way for your fifth grader to understand odds. Coin Toss Probability Date: 05/26/2007 at 09:16:18 From: David Subject: approximate probability Suppose that you toss a balanced coin 100 times. When the game is played using patterns of length 3, no matter what sequence Player A chooses, Player B can always make a winning selection. Binomial probability is a way of calculating the probability of an event happening in a binomial trial (i. I then tried tossing it a hundred times, and ended up getting a probability of 0. You will conduct a probability experiment by flipping a coin 10 times. Tosses are independent so the probability of heads then tails is (2/3) * (1/3) = 2/9. A common topic in introductory probability is solving problems involving coin flips. For example, when coins 1 and 2 are chosen, we can get an outcome which is (white, red), which was not | {
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black-holes, gravitational-waves, estimation, galaxies, structure-formation
The ambient background seismic activity on Earth is given by the NLNM (new low noise model). At 0.434 mHz, this gives $1.63\cdot 10^{-17} (\mathrm{m}/\mathrm{s}^2)^2/\mathrm{Hz}$. Consequently, the signal would come in just above this noise floor, meaning it might be just measurable by sensitive seismic monitoring stations at quiet locations.
Some caveats: | {
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sensors, failure, motor
Following error
Another tool to look for failing motors is the maximum following error over a move. This is the maximum difference between where the motor was supposed to be and where it actually was at any point during the move. It is a single value for the whole move and is something that many motion controllers keep track of for their own use. In fact, many have soft limits on maximum following error and error if you exceed them.
Again, tripping a max follow error can just be an indication that you just need to retune your PID loop, but the frequency with which you need to retune your PID and the difficulty you have in achieving your desired performance can be valuable tools in determining when a motor is getting to the end of it's life. | {
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html, css, html5
I find 200+ errors on one page to be UNACCEPTABLE.
And therefore I find jQuery to be Unacceptable.
It's not just this site, it's every site using jQuery that has these errors.
The number of errors keep on piling up I now have many hundreds of errors on just this one page we are on now. I had to remove many of the errors as this post allows a max of 30,000 characters.
Responsive Design 101:
All, whenever possible, horizontal CSS widths should be em instead of px or a percentage. All font-size are specified in em. Height should not be specified, so when the width is reduced the height will auto increase.
The basic structure of the page:
<body><div id="page">
</div></body>
The page width should not be 100%. Even in a mobile design.
Where the "experts" (e.g. Google PageSpeed Insights) say a mobile design should always have this meta tag:
<meta name="viewport" content="width=device-width, initial-scale=1.0" /> | {
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# Expected number of items looked at
Currently failing at basic probability…
Given a sequence of items, linear search means looking at each in turn and seeing if it's the one we're looking for. (I.e. in the worst case, it's the last item in the sequence or not in the sequence at all and we'll have to look at all items to find it/determine it's not there.)
How many elements of the input sequence need to be checked on average, assuming that the element being searched for is equally likely to be any element in the array?
Ok, we know two things:
• the element we're looking for is in the array
• P[the i-th item is the one we're looking for] = $\frac{1}{n}$ for all $i\in\{1,..,n\}$
And I think I'm supposed to get $\frac{n+1}{2}$ or $\frac{n}{2}$ or something thereabout.
Let $X$ be the number of items the linear search algorithm is looking at. | {
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whereas the first is obvious since $$\sum_{n=1}^{\infty}\frac{3}{n^{2}+n} = 3\lim_{k\to\infty }\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+1}\right) =3\lim_{k\to\infty }(1-\frac{1}{k+1}) = 3~~$$
and
• I see now. This is based off the telescoping series, and the fact that the first and second term are of this form. Thanks so much! – numericalorange Jan 18 '18 at 19:03
• @numericalorange you are welcome don't forget to vote up it is useful for future users – Guy Fsone Jan 18 '18 at 19:07
• I upvoted you and pressed the green check mark to show this is the best answer. Is that what you mean by upvoting it? Or do I upvote my own question? – numericalorange Jan 18 '18 at 19:13
• @numericalorange that is all thanks – Guy Fsone Jan 18 '18 at 19:19
Hint:
The given expression can be written as $$\sum_{n=1}^\infty \frac {3}{n(n+1)} - \frac {2}{(2n+3)(2n+5)}$$
On partial decomposition it becomes | {
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c#, .net
Console.WriteLine("Worker: work completed");
if(Completed != null)
{
Completed(this, EventArgs.Empty);
}
}
}
public class Boss
{
public async void WorkCompleted(object sender, EventArgs e)
{
await Task.Delay(TimeSpan.FromSeconds(3));
Console.WriteLine("Better...");
Grader.Grade(5);
}
}
public static class Grader
{
public static void Grade(int grade)
{
Console.WriteLine("Worker grade = {0}", grade);
}
}
public class Universe
{
public void WorkerStartedWork(object sender, EventArgs e)
{
Console.WriteLine("Universe notices worker starting work");
}
public async void WorkerCompletedWork(object sender, EventArgs e)
{
await Task.Delay(TimeSpan.FromSeconds(10));
Console.WriteLine("Universe pleased with worker's work");
Grader.Grade(7);
} | {
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neural-networks, machine-learning, unsupervised-learning, time-series, anomaly-detection
7289999 09:15:36.820 50.04882813 1.66015625
7289999 09:15:37.904 333.2519531 1.85546875
7289999 09:15:38.924 377.1972656 1.953125
7289999 09:15:39.994 377.1972656 1.7578125
7289999 09:15:41.94 388.671875 1.85546875
7289999 09:15:42.136 388.671875 1.85546875
7290025 09:18:00.429 381.5917969 1.85546875
7290025 09:18:01.448 381.5917969 1.85546875
7290025 09:18:02.488 381.5917969 1.953125
7290025 09:18:03.549 381.5917969 14.453125
7290025 09:18:04.589 381.5917969 46.77734375 | {
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```sind(asind([2 3])) ```
```ans = 2.0000 3.0000 ```
Graph of Inverse Sine Function
Plot the inverse sine function over the domain .
```x = -1:.01:1; plot(x,asind(x)) grid on ```
Input Arguments
collapse all
`X` — Sine of anglescalar value | vector | matrix | N-D array
Sine of angle, specified as a real-valued or complex-valued scalar, vector, matrix, or N-D array. The `asind` operation is element-wise when `X` is nonscalar.
Data Types: `single` | `double`
Complex Number Support: Yes
Output Arguments
collapse all
`Y` — Angle in degreesscalar value | vector | matrix | N-D array
Angle in degrees, returned as a real-valued or complex-valued scalar, vector, matrix, or N-D array of the same size as `X`. | {
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rosparam
<rosparam param="map_frame" subst_value="True">$(arg tf_pre)/map </rosparam> # prefix the tf name for the map
<rosparam>
odom_frame: odom
base_frame: base_link
map_frame: map
...
</rosparam>
</node>
</launch> You haven't provided a fully reproducible example to show what is actually going wrong. But based on your description it looks like you're setting that parameter twice and the one that you're trying to parameterize is lower in precedence.
The last instance of the parameter is the one used. http://wiki.ros.org/roslaunch/XML#Evaluation_order | {
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From "Today’s Cartoon by Randy Glasbergen", displayed with special permission. For many more cartoons, please visit Randy's site @ www.glasbergen.com
Ok, Dan is a really creative guy, but he didn't actually create that one. Randy, the guy who did, however, was a nice guy and let me post a copy here. Thanks, Randy. He has a lot of funny stuff at his web site, so please go there to view them.
## Tuesday, 16 September 2008
### A Graphic Method of Calculating the Standard Deviation | {
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quantum-mechanics, wavefunction, schroedinger-equation
Title: What is the relationship between the wave functions on both sides of the one-dimensional local non-zero potential energy? In a one-dimensional space, the potential $V(x)\neq 0$ if $0<x<a$ and $V=0$ if $x <0$ or $x>a$.
If we set the wave function in $x<0$ as $\phi_1 = Ae^{ikx}+Be^{-ikx}$, and the wave function in $x>a$ as $\phi_2 = Ce^{ikx}+De^{-ikx}$.
Now I want to know if there is a general relation between $\phi_1$ and $\phi_2$, or a relation between $A,B,C,D$.
Here is my solution:
Since $\phi_1$ satisfies the stationary Schroedinger's equation,
$$
\frac{d^2\phi_i}{dx^2} + \frac{2m}{\hbar^2}E\phi_i =0, i=1,2 (x<0).
$$
Therefore,
$$
(\frac{2m}{\hbar^2}E-k^2) (Ae^{ikx}+ Be^{-ikx}) = 0,
$$
and
$$
(\frac{2m}{\hbar^2}E-k^2) (Ce^{ikx} + De^{-ikx}) = 0.
$$
Since the energy at both regions should be the same, I just obtain
$$
E=\frac{\hbar^2k^2}{2m}
$$
for both $x<0$ and $x>a$. But this expression does not include the constants $A,B,C$ or $D$. | {
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graphs
/** Implementation of dijkstra's algorithm using a binary heap. */
private void dijkstra(final NavigableSet<Vertex> q) {
Vertex u, v;
while (!q.isEmpty()) {
u = q.pollFirst(); // vertex with shortest distance (first iteration will return source)
if (u.dist == Integer.MAX_VALUE) break; // we can ignore u (and any other remaining vertices) since they are unreachable
//look at distances to each neighbour
for (Map.Entry<Vertex, Integer> a : u.neighbours.entrySet()) {
v = a.getKey(); //the neighbour in this iteration
final int alternateDist = u.dist + a.getValue();
if (alternateDist < v.dist ) { // shorter path to neighbour found
q.remove(v);
v.dist = alternateDist;
v.previous = u;
q.add(v);
}
}
}
} | {
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java, algorithm
}
Question
How can this code be improved?
One of the main problems with this implementation is that it has a pretty bad time-complexity. How can this be improved?
I would appreciate any suggestions.
[1] Wikipedia contributors. (2018, March 6). Sardinas–Patterson algorithm. In Wikipedia, The Free Encyclopedia. Retrieved April 30, 2020, from https://en.wikipedia.org/w/index.php?title=Sardinas%E2%80%93Patterson_algorithm&oldid=829124957 The Sardinas-Patterson algorithm described in wikipedia as you already said is based on the use of sets for codewords and there is an explanation for every set operation. Consequently I will use the java Set class for every operation involved. I start from the first defined operator:
In general, for two sets of strings D and N, the (left) quotient is defined as the residual words obtained from D by removing some prefix in N. | {
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quantum-mechanics, wavefunction, schroedinger-equation, calculus, dirac-delta-distributions
The limit as $\epsilon\to 0$ quantifies the discontinuity of $f(x)$ at zero. That is the reason why $$\lim_{\epsilon\to 0}\int_{-\epsilon}^\epsilon -\frac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}dx=-\frac{\hbar^2}{2m}\bigg(\dfrac{d\psi}{dx}\bigg|_{x=0^+}-\dfrac{d\psi}{dx}\bigg|_{x=0^-}\bigg).$$
On the other hand, $$\int_{-\epsilon}^\epsilon \delta(x)f(x)dx=f(0)$$
and there is no dependence on $\epsilon$ whatsoever, so the limit is trivial. The thing is that the integral of $\delta(x)f(x)$ over any interval $[a,b]$ containing $0$ is $f(0)$, with no further dependence on $a$ and $b$. So that is the reason why $$\lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}-\delta(x)\alpha \psi(x)=-\alpha \psi(0).$$ | {
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time-dilation
Asking about the spatial distance along the straight line is possible but requires you to chose a frame of reference in which to ask it as there is no unique answer. | {
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forces, space, vacuum
Drag equation
As innisfree said in the comments, the usual formula to quickly get estimates is the drag equation:
$$F_D=0.5 \rho v^2C_DA$$
Let's fill in the gaps, what do we have, what do we want?
Cross section
Loading the above picture in an image editor, Superman's head diameter is ~200px; a lasso select yields a cross sectional area perpendicular to the velocity of ~124900px². Assuming Superman has a head 1.5 times bigger than the average male (just because), this is 725px/m, which means the area is really 0.238m².
Coefficient of drag
This is an experimental parameter linked to the airflow around the geometry, but we'll take it as 0.82 for that of the side of a cylinder.
Drag force / Thrust force equilibrium | {
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Let us continue with the range of the function. The range is the set of all the values that $f \left(x\right)$ can take for any $x$ in the domain. In your case, we can find the range by somewhat sloppy reasoning as follows:
For very small $x$ (as $x$ tends to $- \infty$), $f \left(x\right)$ tends to $0$, because we divide $6$ by something very large. As $x$ becomes larger and approaching $4$ from below, we divide $6$ by something which is negative and almost $0$. As a result $f \left(x\right)$ approaches $- \infty$.
Since $f \left(x\right)$ is continuous everywhere except at $x = 4$, we know that $f \left(x\right)$ assumes all values between $- \infty$ and $0$ for $x \in \left(- \infty , 4\right)$. (This is a consequence of the Intermediate value theorem ).
By similar reasoning, we can consider all $x$ between $4$ and $+ \infty$ to find that $f \left(x\right)$ assumes all values between $0$ and $+ \infty$. | {
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} |
optimization, sql, postgresql
| 4 | DEPENDENT SUBQUERY | attributes_products | ref | ap_product,ap_attribute | ap_attribute | 4 | const | 1 | Using where |
| 3 | DEPENDENT SUBQUERY | attributes_products | ref | ap_product,ap_attribute | ap_attribute | 4 | const | 1 | Using where |
| 2 | DEPENDENT SUBQUERY | attributes_products | ref | ap_product,ap_attribute | ap_attribute | 4 | const | 1 | Using where |
+----+--------------------+---------------------+-------+-------------------------+--------------+---------+-------+------+--------------------------+ | {
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ros, roslaunch, static-transform-publisher
It is probably advisable to use the tf2_ros version. Can someone explain, why we don't need to specify a frame rate for the new version? Are there any other benefits of the second version?
Originally posted by ticotico on ROS Answers with karma: 212 on 2021-11-03
Post score: 1
For a discussion about static transforms I would like to refer to #q346664, which goes into quite some detail (and links to the relevant wiki pages/documentation) and #q324820.
As to the difference in command line arguments: as explained in #q346664, static transforms are not broadcast periodically.
If there is no periodic broadcast, the period argument is no longer needed.
It is probably advisable to use the tf2_ros version
Yes, and for multiple reasons:
it significantly reduces the traffic (ie: messages/sec) on the /tf topic for typical usage scenarios (where many frames are actually static), and
tf is deprecated (and has been for a long time) | {
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java, beginner, ascii-art
enum Line {
FULL (" ------- "),
EMPTY(" "),
LEFT ("| "),
RIGHT(" |"),
BOTH ("| |");
private final String text;
Line(String txt) {
this.text = txt;
}
public String toString() {
return text;
}
} | {
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ros-kinetic, turtlebot3
PI= 3.1415926535897
repeat = 0
theta = 0.0
x = 0.0
y = 0.0
msg = 0.0
index_same=0.0
target = 90
kp=0.5
class MovetoGoals:
def __init__(self):
# Creates a node with name 'final_projekt_node' and make sure it is a
# unique node (using anonymous=True).
rospy.init_node('final_projekt_node', anonymous=True)
# Publisher which will publish to the topic 'Twist'.
self.pub = rospy.Publisher('cmd_vel',Twist, queue_size=10) | {
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Then we must select $k$ black balls and $n-k$ white balls in whatever way we do.(for $0\le k\le n$)
For a fixed $k\in N,0\le k\le n$ we can do this in $\binom{R}{k}\binom{M}{(n-k)}$ ways.
so to get the total no. of ways we must add the above for all $k:0\le k\le n$
So we have the total no. of ways $=\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}$.
But if we think about it in a different way we can say that we have to select $n$ balls from a collection of $R+M$ balls and this can be done in $\displaystyle \binom{R+M}{n}$ ways.
So ,
$$\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}=\displaystyle \binom{R+M}{n}$$
(Reproduced from there.)
Since ${R\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^R$ and ${M\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^M$, the sum $S(R,M,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^R\cdot(1+x)^M=(1+x)^{R+M}$. | {
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"url": "https://math.stackexchange.com/questions/337923/how-to-prove-vandermondes-identity-sum-k-0n-binomrk-binommn-k"
} |
two-wheeled, stability
Is it possible to both propel the robot forward and keep it stable using a feedback controller on the wheels, or is a gyroscope necessary? You could use other ways of measuring orientation, such as an accelerometer, optical tracking of markers, or a depth sensor pointed at the floor. | {
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pcr, plasmids
So why doesn't the whole plasmid get copied when using PCR? Let's assume the primers are designed to bind 1000 bases apart from each other. The first round of PCR will probably produce longer pieces of DNA, because the oligos can only bind to the plasmid and the Polymerase can keep going. But the polymerase will fall off at some point, either randomly or when the temperature is pushed up for the next round of PCR. But in round 2, there should be equal numbers of circular plasmid and linear copy. The proportion of linear copy will increase every round. As the linear copies increase, the oligos are more and more likely to bind to linear copies than to circular plasmid. | {
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At least one of the triangles in an optimal configuration must cover two of the corners. Observe that the proposed configuration does this.
Now, suppose we have a configuration with a triangle covering the bottom two corners, and one more for each top corner. We are free to move them around, subject to the constraints that they must continue to cover those corners. We concentrate on the triangle covering two corners. Our goal is to use as much of the area of that triangle as possible to cover the square, which is achieved when it's side is flush with the square (100% usage). We can shrink the the triangle all the way down to $s=1$ and still have it covering both corners, but as the linked problem's solution showed, $s=1$ is not sufficient, so $s>1$. | {
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} |
time, everyday-life, popular-science, time-travel
I decide to check my future, and (god-permitting) I see that I have passed with flying colors. Now, I "come back in time" and enjoy life, and don't study. So, it is just Impossible that I succeed. So, where is the flaw?? | {
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botany, plant-physiology, respiration
In the mid-1990s, a later phase of technology was introduced at the Grassland, Soil, and Water Research Laboratory (USDA-Agricultural Research Service) that involves an outdoor tunnel system, whereby plants are grown across a continuum of modern to low [CO2] (Mayeux et al., 1993; Fig. 3b,c). During the day, air of known [CO2] is pumped into one end of the tunnel where plants experience relatively high [CO2]. Plants near the end of the tunnel, on the other hand, experience low [CO2] (similar to glacial values) as a result of photosynthetic removal of CO2 from air as it moves progressively through the tunnel. At night, airflow is reversed while plants are solely respiring, and this serves to equalize [CO2] throughout the tunnel.
Despite those studies not focusing explicitly on what's the minimum possible concentration, we can still draw some conclusions based on them. For instance, in Fig.4 (ibidem): | {
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python, tkinter
Element(root,'Hydrogen').pack(padx=5,pady=20)
Key(root,e_type='Halogen').pack()
Demo(root,'Hydrogen').pack()
root.mainloop()
Here I can clearly see that widgets.py is "executable" (interpretable?) and that this file is intended to start a tkinter loop. I'm guessing that this just an artifact from when you first learnt to use tkinter, but this is basically what we should have done in your main.py file, so let's do that.
main.py
The point behind having if __name__ == "__main__": is that it means that the code wrapped in that condition only will be called upon if the current file is executed directly, like this:
$ python3 main.py # We want this to run our program
But not be loaded if the module is imported, like this:
from periodic_table import main # We do not want this to run our program | {
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machine-learning, data-mining, clustering, unsupervised-learning, k-means
Title: How to deal with with rows with zero in every feature while clustering? I am working on a clustering problem which has 13000 observations and 15 features. Around 3000 observations in the dataset has zero in every features ( i.e all values zero in 3000 rows). I am trying to do clustering on top of it. What is a better way to deal with it ? I have few things in my mind but would like to get clarity on :
Check for number of rows with all zero and remove them ?
Include the rows with zero value in every feature and let the clustering
algorithm handle the same? | {
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} |
python
Title: How to sort legends alphabetically in plotly dash with category_orders function I am using Plotly dash for creating a scatter plot. The legends I am getting in the final figure are randomly placed. I want to sort labels alphabetically from A to Z with category_orders.
fig = px.box(
df,
x=selected_x,
y=selected_y,
points='all',
hover_data=hover_data,
color=colour_by,
width=800,
height=600,
labels={
selected_y: "{} {}".format(selected_y_gene, selected_y),
selected_x: "{} {}".format(selected_x_gene, selected_x),
}, Before you construct the figure could you try sorting the values in the dataframe in the order you require? Thus set the True/False flag according to what you want.
orderedDF = df.sort_values(
by=col_label,
ascending=True)
then add the following into the plot denoting the labels:
sort=False | {
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bash
have_command() {
type -p "$1" >/dev/null
}
in_terminal() {
[ -t 0 ]
}
in_terminal && start_in_terminal=1
info_user() {
local msg=$1
local windowicon="info" # 'error', 'info', 'question' or 'warning' or path to icon
[ "$2" ] && windowicon="$2"
if in_terminal; then
echo "$msg"
else
zenity --info --text="$msg" --icon-name="${windowicon}" \
--no-wrap --title="Virtualhost" 2>/dev/null
fi
}
notify_user () {
echo "$1" >&2
in_terminal && return
local windowicon="info" # 'error', 'info', 'question' or 'warning' or path to icon
local msgprefix=""
local prefix="Virtualhost: "
[ "$2" ] && windowicon="$2" && msgprefix="$2: "
if have_command zenity; then
zenity --notification --text="${prefix}$1" --window-icon="$windowicon"
return
fi
if have_command notify-send; then
notify-send "${prefix}${msgprefix}$1"
else
xmessage -buttons Ok:0 -nearmouse "${prefix}${msgprefix}$1" -timeout 10
fi
} | {
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# A fourth proof
Let me now give a fourth proof, using exterior algebra. The proof is probably not new (the method is definitely not), but I find it instructive.
This proof would become a lot shorter if I didn't care for the signs and would only prove the weaker claim that $$\det\left( A_J^I \right) = \pm \det A\cdot \det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$$ for some value of $$\pm$$. But this weaker claim is not as useful as Theorem 1 in its full version (in particular, it would not suffice to fill the gap in Macdonald's book that has motivated this question).
## The permutation $$w\left( K\right)$$
Let us first introduce some more notations: | {
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"url": "https://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices"
} |
machine-learning, deep-learning, nlp, word-embeddings, bert
Furthermore, I realize that using the WordPiece tokenizer is a replacement for lemmatization so the standard NLP pre-processing is supposed to be simpler. However, since we are already only using the first N tokens, and if we are not getting rid of stop words then useless stop words will be in the first N tokens. As far as I have seen, in the examples for Hugging Face, no one really does more preprocessing before the tokenization.
[See example below of the tokenized (from Hugging Face), first 64 tokens of a document]
Therefore, I am asking a few questions here (feel free to answer only one or provide references to papers or resources that I can read): | {
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"url": null
} |
ros, navigation, move-base, clear-costmap
Originally posted by fergs with karma: 13902 on 2014-06-07
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by Yuichi Chu on 2014-06-08:
Thank you very much for your answer,Michael.The problem I came across is described here.
http://answers.ros.org/question/172915/why-the-costmaps-dont-clear-obstacles-in-time-when-using-move_base-package/
Comment by Yuichi Chu on 2014-06-08:
And I thought maybe calling clear costmaps periodically can solve the problem.Then the error occured like I have described above. Is there another way to solve the previous problem? | {
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spectroscopy, analytical-chemistry
Title: Wavelength extention in AAS Theoreticly, the width of the spectral line in AAS (atomic absorbtion spectroscopy) is 10^-5, but in fact there happens an extention of it and it becomes 0.002-0.005. There are some factors which lead to this extention. Most common types of extentions are :
Doppler extention
Pressure extention
Self-absorbance extention.
Does anyone know how to explain in a simple way why do these 3 kinds of extentions take place?
Do these deviations happen all the time in these measurments or are they some deviations that can happen in case the concentration of atoms is to high? From this source broadenings occur due to | {
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statistical-mechanics, entropy, quantum-information, information
and has the advantage of simplicity and completeness, compared to the ignorance interpretation, which needs the additional qualitative concept of ignorance and with it all sorts of questions that are too imprecise or too difficult to answer. I wouldn't say the ignorance interpretation is a relic of the early days of statistical mechanics. It was first proposed by Edwin Jaynes in 1957 (see http://bayes.wustl.edu/etj/node1.html, papers 9 and 10, and also number 36 for a more detailed version of the argument) and proved controversial up until fairly recently. (Jaynes argued that the ignorance interpretation was implicit in the work of Gibbs, but Gibbs himself never spelt it out.) Until recently, most authors preferred an interpretation in which (for a classical system at least) the probabilities in statistical mechanics represented the fraction of time the system spends in each state, rather than the probability of it being in a particular state at the present time. This old | {
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formal-languages, automata, parsers
Bison/yacc-style precedence declarations do not introduce any extra power into the CFG formalism, which is evident from the way they are implemented. Precedence declarations are used only to resolve parsing conflicts; the parser generator produces a pushdown automaton for the grammar as written, and then uses the declarations to remove all but one transition from any state which has multiple transitions on the same terminal. (More accurately, the parser generator tries to remove conflicting transitions. Since precedence declarations are only used to resolve shift-reduce conflicts, not reduce-reduce conflicts, not all grammars can be resolved in this way.) The result is still a pushdown automaton, with the same states but fewer transitions, and is now deterministic. Thus, the language it accepts could be accepted by a deterministic context-free grammar; DPDAs and DCFGs accept the same set of languages. | {
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java, algorithm, sorting, reinventing-the-wheel
StdOut.println("Length of array to be generated");
int length = StdIn.readInt();
StdOut.println("Range random values");
int range = StdIn.readInt();
int[] masterArray = new int[length];
for (int i = 0; i < length; i++) {
masterArray[i] = StdRandom.uniform(range);
}
StdOut.println();
StdOut.println();
//Will display the array if it's reasonably short
if(length <= 20) {
StdOut.println("Master Array");
for(int i=0; i < length; i++) {
StdOut.print(masterArray[i] + " ");
}
StdOut.println();
StdOut.println();
}
//Selection Sort
StdOut.println("Selection Sort");
Integer[] selectionArray = new Integer[length];
arrayCopy(masterArray, selectionArray);
Stopwatch t1 = new Stopwatch();
selectionSort(selectionArray); | {
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quantum-mechanics, quantum-field-theory
$$\tag{A} \Delta_F(t)~=~ \int \frac{\mathrm{d}\nu}{2\pi}~ e^{i\nu t} \frac{i}{\nu^2-\omega^2}~=~ \frac{1}{2\omega} \sum_{\pm}\theta(\pm t)e^{\mp i\omega t}~=~\Delta_F(-t), $$
where $\theta$ denotes the Heaviside step function, and $\omega>0$ is the characteristic angular frequency of the harmonic oscillator. The time derivative is
$$\tag{B} \dot{\Delta}_F(t)~=~ \frac{1}{2i} \sum_{\pm}\pm\theta(\pm t)e^{\mp i\omega t}, $$
where the two contributions in eq. (B) proportional to the delta function $\delta(t)$ have cancelled out. Hence
$$\tag{C} \Delta^3_F(t)~=~ \frac{1}{8\omega^3} \sum_{\pm}\theta(\pm t)e^{\mp 3i\omega t}, $$
and
$$\tag{D} \dot{\Delta}^2_F(t)\Delta_F(t)~=~ \frac{-1}{8\omega} \sum_{\pm}\theta(\pm t)e^{\mp 3i\omega t}. $$
It should be stressed that the Feynman $i\epsilon$ prescription is implicit understood in the above eqs. (A-D). Here it means that we should substitute
$$\tag{E} \omega \to \omega-i\epsilon $$ | {
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nomenclature, stereochemistry, esters
An Evans aldol reaction, the enolate of which is also shown in figure 1, can be analysed similarly. Oxygen has a higher priority than nitrogen, so the same basic rules apply.
A Masamune aldol reaction, however, has two oxygens attached to the enolate carbon, formally making it an O,O-ketene acetal. The CIP rules would dictate to follow the connectivities atom by atom. The Masamune auxilliary, like any ester, has a carbon attached to the oxygen. The enolate anion may be strongly attached to a certain atom (e.g. when using a boron- or titanium(IV)-mediated aldol reaction) but also may not be, or the attachment may be ambiguous (e.g. a potassium salt used for enolating in the presence of $\ce{LiCl}$). | {
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teleportation, key-distribution
Is there any product that really uses quantum teleportation?
Are there any other methods/principles commercially used other than Heisenberg's principle? 1. Existing Products
I've not yet come across commercial systems using entanglement based schemes. The commercial products I've seen, like ID Quantique and MagiQ seem to be using BB84/B92 type protocols. It may be relevant to the question to note that I'm only aware of these particular companies because their commercial QKD systems got hacked.
The infrastructure for long distance entanglement on a commercial scale doesn't exist yet. There's a lot of money going into it, but the no cloning theorem makes quantum repeater design highly non-trivial, and a robust solution has not yet emerged. | {
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The function $\arctan\colon \mathbb{R}\to (-\frac{\pi}{2},\frac{\pi}{2})$ is the inverse of $\tan$. (for the right domain of definition). As $\tan \frac{\pi}{4} = 1$, this means that $\arctan 1=\frac{\pi}{4}$.
Regarding your question about angles: angles are (in mathematics) measured in radians (in $[0,2\pi)$ or $[-\pi,\pi)$), not in degrees: you should expect a value or order $\pi$ or so, not ranging between $0$ and $360$.
• You can see this readily if you consider that $\theta = \dfrac \pi 4$ satisfies $\sin \theta = \cos \theta$ – GFauxPas Mar 29 '15 at 17:30
• the terms used for the inverse function of $tan,$ i know of, are either $\tan^{-1}$ or $arctan.$ the reciprocal is reserved for $\frac{1}{\tan}$ – abel Mar 29 '15 at 17:51
• @abel : my bad, translation issues. Will edit my post. – Clement C. Mar 29 '15 at 18:01
• clement, no problem. native english speaking students stumble over this all the time. – abel Mar 29 '15 at 18:39 | {
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astronomy, experimental-physics, measurements, signal-processing
where
$$ \mathrm{DM} = \int_0^D n_\mathrm{e} \, \mathrm{d}x $$
is known as the dispersion measure along the line of sight. The spread in arrival times is then seen to obey
$$ \Delta t \approx -\frac{4\pi e^2\mathrm{DM}}{cm_\mathrm{e}} \omega^{-3} \Delta\omega = -\frac{e^2\mathrm{DM}}{\pi cm_\mathrm{e}} \nu^{-3} \Delta\nu. \tag{1} $$
Here the negative sign indicates higher frequencies will arrive earlier.
Plugging in Numbers
A typical line of sight terminating in our own galaxy will have a dispersion measure of something like $100\ \mathrm{pc}/\mathrm{cm}^3$. If you are dealing with sources in other galaxies, there will be a similar contribution from the other galaxies' ISM, as well as a contribution from the intergalactic medium. This latter value can exceed $1000\ \mathrm{pc}/\mathrm{cm}^3$ for very distant galaxies.3
With this in mind, we can rewrite (1) to be | {
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newtonian-mechanics, classical-mechanics, rotational-dynamics
For example if the rod had $\alpha=0$ each side would only need to apply a force of $F_g/2$. But it is also possible to apply $F_g/2$ in the vertical while pushing arbitrarily hard in the horizontal so long as the other side applies the same force.
You should reword the question to ask, "what is the minimum force needed to keep the object up." We can do that by setting $F_{2x}=0$.
Edit
I would like to add that $F_{2x}$ could be negative (a pull) which could result in $F_{1x}=0$. We probably need a better constraint than what I gave. One may be "have both lifters perform the same work and assume the rob only moves in the X direction." In this case the X components of the force must be equal in magnitude. | {
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machine-learning, reinforcement-learning, game-ai, javascript
In your case, the function that takes in an observation of the environment and spits out a vector of decision choices is your SVM.
The question linked above contains a description of the training algorithm. It boils down to playing the game multiple times while storing the input vector, output vector and output decision for each step until you reach a termination condition (ie hitting an asteroid) and receiving a score (in this case a negative score since you want to avoid the asteroids). Then, going backwards through the the output vectors, assigning a slowly reducing amount of the output score to the particular cell in each output vector corresponding to the decision that was made at that step of the game. | {
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For cube roots of non-real numbers that appear in this table, one has to take the principal value, that is the cube root with the largest real part; this largest real part is always positive. Big Idea After many days of investigation, students will finally apply their previous knowledge to this new problem--and take the first steps to extend right triangle trigonometry to all points on the unit circle. Free Online FERRIS WHEEL TRIGONOMETRY PROBLEM 1 Practice & Preparation Tests. Our mission is to provide a free, world-class education to anyone, anywhere. Before beginning this activity, students should have been introduced to sine and cosine. 10 Mathematics in School, March 1993. From Trigonometry For Dummies, 2nd Edition. Learning trigonometry will help you understand visualize and graph these relationships and cycles. FREE (4) sjhenners The Great Ice Cream Game. This trig cheat sheet uses Θ, A and B to represent angles. When we include negative values, the x and y axes divide the | {
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quantum-mechanics, density-operator
A quick example
A good example to keep in mind is the delayed choice quantum eraser experiment. Recall that this consists of
photons pass through a dual slit experiment
those slits are outfitted with plates that rotate their polarizations 45°, so that the polarization difference is 90° between the two slits and we see a sum of two overlapping bell-shaped curves for the intensity on the screen, and
we generate the photons as entangled pairs, the initial polarization is entangled with the polarization of another photon which spins round and round in a long coil of fiber-optic cable. | {
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python, python-3.x, sqlite
P2_Score = P2_Score - int(P2_Hit) #player 2 scoring function
if P2_Score == 0:
print("Points of {0} after his turn: {1}\n".format(Player_2, P2_Score))
print("\x1b[1;33;40m===============***WIN! WIN! WIN!***===============")
print("\x1b[1;33;40m===============***{0} IS A GOD!!!***===============\n".format(Player_2.upper()))
P2_Game_Wins += 1
unix = int(time.time())
date = str(datetime.datetime.fromtimestamp(unix).strftime('%Y-%m-%d %H:%M:%S'))
cursor.execute("INSERT INTO game_overview(game_id, player1, player2, winner, gamemode, checkout) VALUES (?, ?, ?, ?, ?, ?)",
(date, Player_1, Player_2, Player_2, points_default, P2_Hit))
darts_db.commit()
break
elif P2_Score >= 2:
print("Points of {0} after his turn: {1}".format(Player_2, P2_Score))
elif P2_Score <= 1: | {
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organic-chemistry, nomenclature
isopropyl-
methyl-
I have come across two compounds: 1-methyl-3-methoxy-4-isopropylbenzene and 1-isopropyl-4-methylcyclohexane
So, is “iso” taken care of in case of alphabetisation? And, are there any special criteria for this operation? (Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)
Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows: | {
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electromagnetism, momentum, conservation-laws
Title: How can there be net linear momentum in a static electromagnetic field (not propagating)? I understand from basic conservation of energy and momentum considerations, it is clear in classical electrodynamics that the fields should be able to have energy and momentum. This leads to the usual Poynting vector and energy density relations for electromagnetic fields.
However, I do not know how to interpret situations where there is a net linear momentum in a static electromagnetic field. The fields aren't propagating. It doesn't make sense to me that momentum can be divorced from motion.
As a concrete example to discuss: | {
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The Kaplansky conjecture [for Banach algebras] fails. There exists a compact Hausdorff space $X$ such that there is a discontinuous homomorphism from $C(X)$ into another Banach algebra.
It was shown that the conjecture fails when one assumes the continuum hypothesis; but it is consistent that it holds, and the continuum hypothesis fails. (See Wikipedia for a bit more, and references.)
• It's worth noting that Farah's result is using OCA, which follows from PFA, which is another nice example of what the OP was looking for (think of it in terms of the Baire category theorem if you insist on having examples outside of set theory). – Haim Jan 23 '14 at 11:00
The statement $"\mathfrak{x}_0= \mathfrak{x}_1 "$ for many pairs of cardinal invariants of the continuum $\mathfrak{x}_0,\mathfrak{x}_1$ such that $"\mathfrak{x}_0=\mathfrak{x}_1"$ is not provable in $ZFC$ (for example, $\mathfrak a$ and $\mathfrak d$). | {
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lagrangian-formalism, hamiltonian-formalism
$$
T^{*}X=\coprod_{x\in X}T^{*}_{x}X
$$
where $T^{*}_{x}X$ is the dual space of $T_{x}X$. We have
$$
T^{*}X\ni(x^{i},p_{i})
$$
where the $p_{i}$'s are the components of the element in $T_{x}^{*}X$. The Hamiltonian $\mathscr{H}$ is a function on $T^{*}X$. To go from Lagrangian to Hamiltonian dynamics, we define the so called Legendre transform associated to $\mathscr{L}$. This is the function
$$
\Bbb{F}\mathscr{L}:TX\to T^{*}X
$$
defined by
$$
\Bbb{F}\mathscr{L}(q^{i},\dot{q}^{i})=\left(q^{i},\frac{\partial\mathscr{L}}{\partial \dot{q}^{i}}(q^{i},\dot{q}^{i})\right)
$$
Let's suppose that $\Bbb{F}\mathscr{L}$ is invertible. Then we have a function $\Bbb{F}\mathscr{L}^{-1}:T^{*}X\to TX$ and we can define the Hamiltonian $\mathscr{H}$ as
$$
\mathscr{H}:T^{*}X\to \Bbb{R},\qquad\quad \mathscr{H}(q^{i},p_{i})=\min_{\dot{q}}\left\{p_{i} \dot{q}^{i}-\mathscr{L}\left(q^{i},\dot{q}^{i}\right)\right\}
$$
It's easily seen that
$$ | {
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c#, object-oriented, interview-questions
public List<Order> GetOrders() {
return _orders;
}
public Dictionary<int, Schedule> GetSchedules()
{
return _schedules;
}
/// <summary>
/// Generate Flight itineraries and return what hasbeen generate.
/// This function return only orders that are loaded. In other words,
/// Orders that have scedule. associated to them.
/// </summary>
/// <returns></returns>
public List<Flight> GenerateFlights() {
var loadedSchedules = GetLoadedSchedules();
List<Schedule> sortedSchedule = new List<Schedule>();
sortedSchedule = loadedSchedules.Values.ToList();
sortedSchedule.Sort((emp1, emp2) => emp1.Day.CompareTo(emp2.Day));
int flightNumber = 1;
List<Order> ordersByPriority = _orders.OrderBy(o => o.Priority).ToList(); | {
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electromagnetism, water
Submarines communicate with frequencies as low as 100 Hz.To test for a "good conductor" we compare conductivity $\sigma$ with $\omega \epsilon_0 \epsilon_r$,
where $\omega$ is the angular frequency, and $\epsilon_r$ is the
relative permittivity (about 80 for water). This comparison suggests
both freshwater and seawater are good conductors for waves at this
frequency. (Indeed you would not try mixing AC electricity with your bathwater, right?)
Given that, there is an easy expression for the characteristic length,
the "skin depth", that an EM wave will travel before being attenuated
by $1/e$: i.e. $ l = \sqrt{2/\mu_0 \sigma \omega}$. At these
frequencies, the skin depth in freshwater is 355m, whilst for seawater
it is 25m.
OK, but I guess you knew all that, and your question is why does
conductivity affect the skin depth in this way when you might
intuitively have thought that better conductors would somehow allow EM
waves to pass more freely? | {
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homework-and-exercises, photons, photoelectric-effect
Title: Finding the maximum kinetic energy of any photoelectrons? An incident photon, $f=5.5\times 10^{14}\ Hz$, hits a metal with a work function of $2.8\ eV$.
How do I find the maximum kinetic energy of any photo-electrons?
I'm confused exactly how to do this, because I keep getting a negative kinetic energy. I have gotten 2.3 eV for the photon's energy, and I know the equation is $E_{\ in}$ - Work Function = $KE_{\ max}$, but 2.3-2.8 is a negative energy. What am I doing wrong? You're not doing this wrong. As you know energy of each photon is $E = hf = 2.27eV$ so they can't produce any photoelectrons on a metal with work function greater than that. | {
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According to no free lunch theorm, implicit methods must have some additional costs (half joking. that’s another theorm). The cost for an implicit method is solving a nonlinear system. In general we will have $$f(y, t)$$ on the right-hand side of the ODE, not simply $$-ay$$.
### General form¶
Using $$-ay$$ on the right-hand side allows a simple analysis, but the idea of ODE stability/instability applies to general ODEs. For example considering a system like
\begin{align} \frac{dy}{dt} &= -f(t) y \end{align}
You can use the typical magnitude of $$f(t)$$ as the “$$a$$” in the previous analysis.
Even for
\begin{align} \frac{dy}{dt} &= -f(t) y^2 \end{align}
You can consider the typical magnitude of $$yf(t)$$
## Stiff system¶
Consider an ODE system
\begin{align} \frac{dy_1}{dt} &= -a_1y_1 \\ \frac{dy_2}{dt} &= -a_2y_2 \end{align} | {
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"openwebmath_perplexity": 1129.366850515747,
"openwebmath_score": 0.9788032174110413,
"tags": null,
"url": "https://am111.readthedocs.io/en/latest/session8_stabiilty.html"
} |
image-processing, fft, filters, python, sampling
plt.imshow(image, cmap=plt.cm.hot)
plt.colorbar()
plt.show()
Fx = np.fft.fftshift(np.fft.rfft2(image), axes=0)
higher_resolution = len(image)
lower_resolution = 64
rate = higher_resolution / lower_resolution
lower_resolution_wavenumbers = np.fft.fftshift(np.fft.fftfreq(lower_resolution) * lower_resolution)
lowest_wavenumber_y, highest_wavenumber_y = lower_resolution_wavenumbers.astype(int)[[0, -1]]
higher_resolution_wavenumbers = np.fft.fftshift(np.fft.fftfreq(higher_resolution) * higher_resolution)
mid = higher_resolution // 2
assert higher_resolution_wavenumbers[mid] == 0
lowest_wavenumber_y_idx, highest_wavenumber_y_idx = mid + lowest_wavenumber_y, mid + highest_wavenumber_y
assert np.allclose(higher_resolution_wavenumbers[lowest_wavenumber_y_idx:highest_wavenumber_y_idx+1], lower_resolution_wavenumbers)
highest_wavenumber_x_idx = len(np.fft.rfftfreq(lower_resolution)) - 1 | {
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"tags": "image-processing, fft, filters, python, sampling",
"url": null
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slam, navigation, frames, hector-slam, frame-id
<!-- Advertising config -->
<param name="advertise_map_service" value="true"/>
<param name="scan_subscriber_queue_size" value="5"/>
<param name="scan_topic" value="/laser/scan"/>
<param name="tf_map_scanmatch_transform_frame_name" value="scanmatcher_frame" />
</node>
<node pkg="tf" type="static_transform_publisher" name="laser_link" args="0.031 0.0 0.251 0.0 0.0 0.0 /base_link /laser 50" />
<node pkg="tf" type="static_transform_publisher" name="imu_link" args="-0.049 0.0 0.180 0.0 0.0 0.0 /base_link /imu 50" />
When I launch my launch file I have the following errors : | {
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"tags": "slam, navigation, frames, hector-slam, frame-id",
"url": null
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ros-kinetic, gazebo-7
auto-starting new master
process[master]: started with pid [17081]
ROS_MASTER_URI=http://localhost:11311
setting /run_id to 925dced2-fa25-11e8-bbc5-d8cb8a119b1a
process[rosout-1]: started with pid [17094]
started core service [/rosout]
process[gazebo-2]: started with pid [17102]
process[spawn_turtlebot_model-3]: started with pid [17114]
process[mobile_base_nodelet_manager-4]: started with pid [17124]
process[cmd_vel_mux-5]: started with pid [17125]
process[bumper2pointcloud-6]: started with pid [17126]
process[robot_state_publisher-7]: started with pid [17127]
process[laserscan_nodelet_manager-8]: started with pid [17134]
process[depthimage_to_laserscan-9]: started with pid [17143]
[ INFO] [1544189992.342004025]: Finished loading Gazebo ROS API Plugin.
[ INFO] [1544189992.342448714]: waitForService: Service [/gazebo/set_physics_properties] has not been advertised, waiting... | {
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"tags": "ros-kinetic, gazebo-7",
"url": null
} |
least one zero row because the only square RREF matrix that has no zero rows is the identity matrix, and the latter is row equivalent only to non-singular matrices. Determinant of a Identity matrix () is 1. » O.S. The value of α for which det(P) = 0 is _____. Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. » LinkedIn Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. This matrix accounts for the entry just below the mesh value (y, z). : We are given a matrix with a determinant of $1$. Can we infer anything else? » Android This means that if one column of a matrix A | {
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"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9822877002595527,
"lm_q1q2_score": 0.8098588011385428,
"lm_q2_score": 0.8244619177503205,
"openwebmath_perplexity": 695.6402764700179,
"openwebmath_score": 0.7781877517700195,
"tags": null,
"url": "https://www.internosilfilm.com/8yq7dmc/identity-matrix-determinant-6d3455"
} |
Last edited: Nov 21, 2009
13. Nov 21, 2009
### deancodemo
Thankyou everyone for your help. :)
14. Nov 22, 2009
### deiki
I'm not sure the proof I'm giving here is relevant to the topic, but since there are tons of definition of the trigonometric functions with different proofs to find their properties ( which end up being the same at the end ), you might want to have a look at this one. This one might help if you have defined trigonometric with the help of the trigonometric circle.
http://ultraxs.com/image-63EE_4B090CFF.jpg
( click the img, S1, S2, S3 are the areas in grey, x is the angle. The squeeze theorem is applied when the limits appear )
15. Nov 22, 2009
### HallsofIvy
It is perfectly valid to define sin(x) to be
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!)}x^{2n+ 1}$$
and define cos(x) to be
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$$
and, since those are power series with infinite radius of convergence, the derivatives follow from term-by-term differentiation. | {
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"url": "https://www.physicsforums.com/threads/proof-of-lim-x-to-0-of-sinx-x-and-circular-proofs.356830/"
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Let $A:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. We know that there is a unique transformation $A^*:\mathbb{R}^n \to \mathbb{R}^m$ such that $$\langle Ax,y\rangle = \langle x,A^*y \rangle, \forall x \in \mathbb{R}^m,y\in \mathbb{R}^n.$$ And we know also that the matrix of $A^*$ is $A^t$. | {
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"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9895109099773436,
"lm_q1q2_score": 0.8043410748784988,
"lm_q2_score": 0.8128673133042217,
"openwebmath_perplexity": 141.57580585088158,
"openwebmath_score": 0.9523378014564514,
"tags": null,
"url": "https://math.stackexchange.com/questions/1419137/operatornameim-a-operatornameker-a-perp"
} |
compilers, program-optimization, continuations
The problem with CPS is tied to one of its main benefits. CPS transforming allows you to implement control operators like call/cc, but this means every non-local function call in the CPS intermediate code has to be treated as potentially performing control effects. If your language includes control operators, then this is just as it should be (though even then most functions probably aren't doing any control shenanigans). If your language doesn't include control operators, then there are global invariants on the use of continuations that are not evident locally. This means there are optimizations that are unsound to perform on general CPS code that would be sound to perform on this particularly well-behaved use of CPS. One way this manifests is in the change in precision of data and control flow analyses. (CPS transforming helps in some ways, hurts in others, though the ways it helps are mostly due to duplication rather than the CPS aspect itself.)1 You could, of course, add rules and | {
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"tags": "compilers, program-optimization, continuations",
"url": null
} |
navigation, mapping, rviz, ros-hydro, base-link
Comment by Moda on 2014-08-20:
And my robot is not seen in rviz, not in is real place. Rviz says that the robot is out of the map. The map is a real one.
And then, I can not send the robot somewhere, with 2D Nav Goal, in rviz,
However, I don't know where the issues really come from
Comment by Moda on 2014-08-22:
Sorry even without use_set_time, I have the very same error
Comment by Moda on 2014-08-22:
The message is
[ WARN] [1408693106.228430308]: Failed to transform initial pose in time (Lookup would require extrapolation into the future. Requested time 1408693106.228264741 but the latest data is at time 1408693106.204777002, when looking up transform from frame [map] to frame [base_link]) | {
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"openwebmath_score": null,
"tags": "navigation, mapping, rviz, ros-hydro, base-link",
"url": null
} |
python, docker, git
docker-server-1 | File "/usr/local/lib/python3.9/importlib/__init__.py", line 122, in import_module
docker-server-1 | raise TypeError(msg.format(name))
docker-server-1 | TypeError: the 'package' argument is required to perform a relative import for '.venv.lib.python3.9.site-packages.eventlet.wsgi'
docker-server-1 | [2023-09-22 11:09:20 +0000] [9] [INFO] Worker exiting (pid: 9)
docker-server-1 | [2023-09-22 11:09:20 +0000] [8] [ERROR] Worker (pid:9) exited with code 3
docker-server-1 | [2023-09-22 11:09:20 +0000] [8] [ERROR] Shutting down: Master
docker-server-1 | [2023-09-22 11:09:20 +0000] [8] [ERROR] Reason: Worker failed to boot.
docker-server-1 exited with code 3 | {
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "python, docker, git",
"url": null
} |
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