text stringlengths 1 1.11k | source dict |
|---|---|
newtonian-mechanics, forces, free-body-diagram, string, centrifugal-force
Title: Tension at extreme of a rotating rope of mass $m$
A uniform rope of mass per unit length $\lambda$, length $\ell$ is attached to shaft that is rotating at constant
angular velocity $\omega$. Find the tension in the rope as a function of distance from the shaft.
You may ignore the effect of gravitation. | {
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"tags": "newtonian-mechanics, forces, free-body-diagram, string, centrifugal-force",
"url": null
} |
species-identification, botany
As for distribution, Hylocereus are quite prevalent throughout Hawaii, and can also be found scattered throughout the globe.
Hawaii is the leftmost blue dot. image source
More information about this plant can be found here. | {
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"tags": "species-identification, botany",
"url": null
} |
c++, template
union { struct { T x, y; }; struct { T u, v; }; };
enum { size = 2 };
Vec2() :x(), y() {}
explicit Vec2(ctor) {}
explicit Vec2(T s) :x(s), y(s) {}
Vec2(T a, T b) :x(a), y(b) {}
// -- Conversion Constructions --
template <typename A, typename B>
Vec2(const A& a, const B& b)
: x(static_cast<T>(a)), y(static_cast<T>(b)) {}
template <typename U>
Vec2(const Vec2<U>& v)
: x(static_cast<T>(v.x)), y(static_cast<T>(v.y)) {}
template <typename U>
explicit Vec2(const Vec3<U>& v)
: x(static_cast<T>(v.x)), y(static_cast<T>(v.y)) {}
template <typename U>
explicit Vec2(const Vec4<U>& v)
: x(static_cast<T>(v.x)), y(static_cast<T>(v.y)) {}
value_type & operator[](unsigned int index) { return (&x)[index]; }
const value_type & operator[](unsigned int index) const { return (&x)[index]; } | {
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python, opencv
# return the thresholded image
return image
def grab_frame(cam):
#cv2.namedWindow("test")
#img_counter = 0
while True:
ret, color1 = cam.read()
#r = 100.0 / color1.shape[1]
r = 640.0 / color1.shape[1]
#r = 0.25
dim = (100, int(color1.shape[0] * r))
dim = (640,480)
# perform the actual resizing of the image and show it
color = cv2.resize(color1, dim, interpolation = cv2.INTER_AREA)
#color = color1.copy()
b = color.copy()
# set green and red channels to 0
b[:, :, 1] = 0
b[:, :, 2] = 0
g = color.copy()
# set blue and red channels to 0
g[:, :, 0] = 0
g[:, :, 2] = 0
r = color.copy()
# set blue and green channels to 0
r[:, :, 0] = 0
r[:, :, 1] = 0 | {
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quantum-mechanics, electric-circuits, electrons, electric-current, electrical-resistance
$$
\Lambda
=
\frac{h}{\sqrt{2\pi mkT}}
\simeq 10^{-11}\mathrm{ m}
$$
where the numerical application is for an electron at room temperature. The quantum effects are therefore already negligible at the nanometric scale. | {
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"url": null
} |
You can classify trees into (A) coverable, (B) coverable except the root and (C) inherently uncoverable.
Suppose you've computed the status of trees $$T_1,\dots,T_n$$ and would like to compute the status of the tree consisting of a vertex connected to $$T_1,\dots,T_n$$. If any of $$T_i$$ is (C), then the result is (C). If all $$T_i$$ are (A), then the result is coverable except for the root (B). If there is one (B) and everything else is (A), then you can match that vertex with the root and the result is (A). Finally, if the tree is connected to more than one (B), it can't be covered (C). This is a recursive traversal of the tree, which can be done in linear time.
• I think this works, and I can even write some c++-esque pseudocode to go along with it. Have added it as an edit to the opening question. Nov 28, 2021 at 18:44 | {
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"url": "https://cs.stackexchange.com/questions/146184/how-to-determine-if-a-tree-t-v-e-has-a-perfect-matching-in-ov-e"
} |
homework-and-exercises, electric-circuits, electrical-resistance, batteries
In the given diagram $r$ is the resistance of the piano and $R$ is the unknown resistance connected in series with the battery.
Since this is a simple series circuit the current $I$ is constant throughout the circuit and given by:
$$-Ir-IR+12=0$$
$$12=I(r+R)$$
$$I=\frac{12}{r+R}$$
The potential difference across the piano should be $9\,\rm{V}$ and so
$$V_{\rm{piano}}=Ir$$
$$V_{\rm{piano}}=\frac{12}{r+R}r$$
$$V_{\rm{piano}}r+V_{\rm{piano}}R=12r$$
$$9r+9R=12r$$
$$9R=3r$$
$$R=\frac{r}{3}$$
So according to my calculations, a resistance of $r/3$ should be connected for a voltage of $9\,\rm{V}$ across the piano. Is this right or did I just do everything wrong? | {
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"tags": "homework-and-exercises, electric-circuits, electrical-resistance, batteries",
"url": null
} |
coordination-compounds
Title: Can a high-spin complex be diamagnetic? There are two nickel complexes: one is octahedral and the other square planar. One complex is diamagnetic, the other is paramagnetic. Both are high-spin.
Now, $\ce{Ni^2+}$ that is octahedral should be paramagnetic, as should $\ce{Ni^2+}$ that is square planar. Actually any high-spin complex should be paramagnetic. Is it possible that the octahedral $\ce{Ni^2+}$ is paramagnetic, and square planar $\ce{Ni^2+}$ diamagnetic (achieved by not filling the last orbital even tho that goes against high-spin concept)? A high-spin complex cannot be diamagnetic. By definition, a high-spin system has multiple unpaired electrons. The names "high-spin" and "low-spin" refer to the total electron spin. In the following example ($d^6$ octahedral), the high spin state $S = 2$ and the low spin state $S = 0$. | {
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• Thank you very much, very elegant, but I am struggling to understand it. The solution for the first coefficient is indeed c/r^2, but the one for the second coefficient is c1/r^2+c2+c3*r, so I don't understand how you force the c2 and c3 to become zeros. Presumably it is what the code "sols=..." does, which is quite complicated. Anyway, I accept your solution and will do my best to understand it now, but if you have any other clarifying/simplifying remarks, please make them :) – Tigran Aivazian Apr 5 '18 at 16:02
• I understand what "coef=" code does, but what does "sols=" code do? Could you please explain it in a little bit more detail? Thank you. – Tigran Aivazian Apr 5 '18 at 16:11 | {
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} |
c++, object-oriented, memory-management, vectors, c++03
}
int vector::get(int i)
{
return p[i];
}
void vector::pushback(int num)
{
/* if size of vector is equal to element count; resize vector */
if (size == elements)
{
size *= 2;
int *temp = new int[size];
for (int i = 0; i < size/2; i++)
temp[i] = p[i];
delete [] p;
p = temp;
p[elements++] =num;
} else if (largest_ndx >= elements) /* if number was set to furthest index, push back after that */
{
p[largest_ndx + 1] = num;
elements++;
} else
{
p[elements++] = num;
}
}
void vector::pushfront(int num)
{
int i = size - 1;
while (i != -1)
{
p[i+1] = p[i];
--i;
}
p[0] = num;
elements++;
}
vector::~vector()
{
delete [] p;
}
vector& vector::operator= (const vector& other)
{
if (this == &other)
return *this; /* returns copy of current object */ | {
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} |
java, random
Title: Input on where and how I can make improvements on my randomizer As my title states I'm just looking for a little input on how I can improve one of my side projects when I'm not doing my CS assignments. All my program is, is a primitive randomizer for Battlefield 4 classes and weapons (because I can never choose what to play). But since I am still new to java (second semester in) I was wondering if someone more experienced than I could shoot a few concepts my way on how I can make this better, if not more re-useable for other things. Here is my driver:
package BattlefieldClassRandomizer;
import java.util.Scanner;
public class BF4ClassRandomizerDriver { | {
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Exercise: use these counts to verify that N_n = λ, so one might say that
in general there are λ "nuclei".
Examples: in P^2, "conics" = zeros of "nonsingular quadratic forms" =
zeros of homogeneous quadratic polynomials P(x,y,z) over k that are
irreducible even over the algebraic closure. [That's a necessary proviso:
consider x^2 + xy + y^2 mod 2, or x^2 + y^2 mod 3.] Note/recall that
the set of zeros of a homogeneous P is well-defined even though its value
at a point of P^2 in general is not. A conic is clearly an arc because
restricting to a line yields a homogeneous quadratic polynomial in two
variables that is not identically zero (else P had a linear factor).
In fact it is known -- though we will not prove this (yet another possible
paper topic) -- that:
(i) all conics are equivalent under PGL_3(k);
(ii) all conics are Type I ovals (that is, they have q+1 points);
and remarkably, one can prove in general that
(iii) if q is odd then every oval is a conic (Segre [1954]). | {
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} |
annealing, adiabatic-model
$$
\hat{H}\left(t\right)~=~\left(1-\frac{t}{\tau}\right)\hat{H}_i+\frac{t}{\tau}\hat{H}_f,
\qquad \left(0\leq t\leq \tau \right)
$$
where $\tau$ is the adiabatic time scale.
So my question is: Why is it crucial that the initial Hamiltonian does not commute with the final Hamiltonian? In adiabatic QC, you encode your problem in a Hamiltonian such that your result can be extracted from the ground state. Preparing that ground state is hard to do directly, so you instead prepare the ground state of an 'easy' Hamiltonian, and then slowly interpolate between the two. If you go slow enough, the state of your system will stay in the ground state. At the end of your process, you'll have the solution. | {
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point. Here's a neat desmos script to play with. Other Graphs: Rotating Ellipse. 3blue1brown has 174 posts and 0 shared files available to view for free on yiff. That is why you only need to enter r here to get the circumference. By using this website, you agree to our Cookie Policy. - [Instructor] Let's say that X is a function of the parameter T and it's equal to cosine of T and Y is also defined as a function of T and it's equal to sin of T and we wanna find the arc length of the curve traced out, so length of curve from T is equal to zero to T is equal to pi over two. My preferred method with parametric equations also allows you to easily manipulate it (e. Because of this, absolute-value functions have graphs which make sharp turns where the graph would otherwise have crossed the x-axis. t heta 2 = − 5. Example: Both polynomials are 2 nd degree, so the asymptote is at ; If the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal | {
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"openwebmath_score": 0.5953616499900818,
"tags": null,
"url": "http://snionline.it/kfvh/rotating-ellipse-desmos.html"
} |
ros-melodic
2020-04-04 18:04:12 remove ros-melodic-image-publisher:amd64 1.14.0-1bionic.20200320.134551 <none>
2020-04-04 18:04:13 remove ros-melodic-image-common:amd64 1.11.13-0bionic.20200320.141440 <none>
2020-04-04 18:04:13 remove ros-melodic-image-transport-plugins:amd64 1.9.5-0bionic.20200320.155634 <none>
2020-04-04 18:04:14 remove ros-melodic-compressed-depth-image-transport:amd64 1.9.5-0bionic.20200320.141515 <none>
2020-04-04 18:04:14 remove ros-melodic-compressed-image-transport:amd64 1.9.5-0bionic.20200320.141529 <none>
2020-04-04 18:04:15 remove ros-melodic-urdf-sim-tutorial:amd64 0.4.0-0bionic.20200320.154215 <none>
2020-04-04 18:04:15 remove ros-melodic-gazebo-ros-control:amd64 2.8.6-1bionic.20200320.144157 <none>
2020-04-04 18:04:16 remove ros-melodic-control-toolbox:amd64 1.17.0-0bionic.20200320.131747 <none>
2020-04-04 18:04:16 remove ros-melodic-controller-manager:amd64 0.17.0-1bionic.20200320.133159 <none> | {
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"tags": "ros-melodic",
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} |
fluid-dynamics, differential-geometry, conservation-laws, vector-fields, flow
Title: Linking definition of incompressible fluid to physical intuition So the picture I have in my head of an incompressible fluid is one which if I fill a canister to the brim with the fluid, and if I try to press down on the top of the canister, the container will bulge and break.
The math definition I am comfortable with is that the fluid flow $\vec{u}$ is a divergence free vector field, i.e.
$$
\nabla\cdot \vec{u}=0
$$
I am trying to connect the two. My idea is given any volume canister (I guess technically there should be a time dimension to my canister but it shouldn't matter i think) $\Omega$, we have by the divergence theorem
$$
0=\int\int\int_\Omega\nabla\cdot \vec{u}\mathrm dv=\int\int_{\partial \Omega}\vec{u}\cdot\vec{n}ds
$$ | {
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"tags": "fluid-dynamics, differential-geometry, conservation-laws, vector-fields, flow",
"url": null
} |
I like the book "Analytic Combinatorics" by Philippe Flajolet and Robert Sedgewick Part A. It doesn't only show a way how you translate combinatorical problems in terms functional relations of generating functions (the show that unlabelled structures relate to ordinary generating functions, and labelled structures relate to exponential generating functions), but it also provides ways (asymptotic expansion, ...) such that you can actually calculate something.
-
When trying to understand a counting situation, the best thing to have is a closed form formula, a function that is easily calculable by well known methods ('closed-form' is a bit slippery in meaning, but usually means, without any recursion, but there's room for disagreement over whether summations or products are allowed (though factorial is often allowed). From such functions, it is usually straightforward to determine approximations and asymptotics and to combine with other other functions. | {
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"openwebmath_score": 0.8223768472671509,
"tags": null,
"url": "http://math.stackexchange.com/questions/25430/why-are-generating-functions-useful"
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asteroids
Title: Why did the distribution of asteroids discovered in 2010 have a radial modulation? This answer links to one of Scott Manley's excellent asteroid videos Asteroid Discovery - 1970-2015 - 8K resolution. The animation highlights the positions of the meteors at the moment of their discovery, and by watching one can see the technology improve and notice patterns as instruments are pointed in different directions to avoid the light from the Sun and (at least sometimes) the Moon. (there is music, adjust volume accordingly)
There are often fan-shaped patterns showing the directions that more sensitive telescopes with modest fields of view are pointed.
However, I noticed that only during the year 2010, roughly between asteroid numbers 500,000 and 520,000 there are radial striations at certain distances from the Sun. I don't see this happen at any other time during the video. | {
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} |
java, recursion, file, stack
}
DirectoryIteratorTest.java
import org.junit.Assert;
import org.junit.Test;
import java.io.File;
public class DirectoryIteratorTest {
private String mTestDir = "C:\\Users\\Maksim\\Downloads\\Test" + File.separatorChar;
@Test
public void singleFileIsInit() {
File initPath = new File(mTestDir + "Android_Accelerometer.png");
Assert.assertEquals(initPath.getAbsolutePath(), DirectoryIterator.getLargestFilePath(initPath));
}
@Test
public void onlyDirs() {
File initPath = new File(mTestDir + "only_dirs");
Assert.assertEquals(null, DirectoryIterator.getLargestFilePath(initPath));
}
@Test(expected = IllegalArgumentException.class)
public void nullPath() {
Assert.assertEquals(null, DirectoryIterator.getLargestFilePath(null));
}
@Test
public void emptyPath() {
File initPath = new File("");
Assert.assertEquals(null, DirectoryIterator.getLargestFilePath(initPath));
} | {
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} |
• Thank you! Because I added explanatory material, your name got erased. But probably there is at least one more typo to fix. – André Nicolas Aug 15 '14 at 15:26
• I can certainly relate regarding typos! :-| – Namaste Aug 15 '14 at 15:29
• By periodicity, $\cos(2\pi-x)=\cos(2\pi+(-x))=\cos(-x)$. But $\cos(-x)=\cos x$ for all $x$, so the book is right. – André Nicolas Aug 15 '14 at 15:31
• How is cos(2π+x) = cox? It is given in my book that cos(2π-x)= cosx, is cos(2π+x) = cosx as well? – Always Learning Forever Aug 15 '14 at 15:32
• Yes. For $\sin$ and $\cos$, and therefore also for all the other trig fumctions, we have periodicity. In general $\cos(x+2n\pi)=\cos(x)$ for any integer $n$. Same for $\sin(x+2n\pi)$. Looking at pictures of $\cos$ and $\sin$ can be very useful. Shift the curve by $2\pi$, and the picture does not change. – André Nicolas Aug 15 '14 at 15:36 | {
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} |
ros, logging
Originally posted by Daniel Stonier on ROS Answers with karma: 3170 on 2011-03-17
Post score: 5
Original comments
Comment by GeniusGeeko on 2011-03-18:
How often would you like to "rosclean" your logs?
Just to do this..
echo "find ~/.ros/log/* -mtime +5 -exec rm -r {} \;" >> ~/.bashrc
Replace the 5 with how many days you want to remove that was created before that number. This will remove files older than 5 days old.
It will search for old files every time you open a terminal.
Originally posted by GeniusGeeko with karma: 299 on 2011-03-18
This answer was ACCEPTED on the original site
Post score: 7
Original comments
Comment by joq on 2011-03-18:
Good idea using find. I probably would not add it to my ~/.bashrc, but that would make a nice alias. | {
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# Definite integral - Please point to me my mistake
This emerged while I was investigating this question, i.e. the solution to the definite integral $$I_x = \int_0^\infty\left(5x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$
In a comment, its value is given as 0.2119539... using the Maple solver, and I have verified this value by using another on-line definite integral calculator.
But manipulating the integral leads me to a different solution. Specifically,
define $y=h(x) = x^5 +x$, and we have $dy = h'dx$, with $h' = 5x^4+1$. Also, $h(x)$ admits an inverse, so $x = h^{-1}(y)$. Then
$$\int_0^\infty\left(5x^5+x\right)\operatorname{erfc}\left(x^5+x\right)\,dx = \int_0^\infty x\left(5x^4+1\right)\operatorname{erfc}\left(x^5+x\right)\,dx$$ $$= \int_0^\infty xh'\operatorname{erfc}\left(y\right)\,dx = \int_0^\infty h^{-1}(y)\operatorname{erfc}\left(y\right)\,dy$$ | {
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c++, qt
inline const QList<QImage*>& getImages() const { return m_images; }
inline const QImage& getThumbnail() const { return m_thumbnail; }
protected:
/* Protected methods */
//adds another component to this vehicle
inline void addComponent(Component* c)
{
m_components << c;
}
private:
/* Private data members */
QList<Component*> m_components;
int m_id;
int m_year;
int m_mileage;
int m_weight;
int m_length;
int m_width;
int m_height;
int m_wheelbase;
QString m_date_received;
QString m_type;
QString m_maker;
QString m_model;
QString m_color;
QString m_plate;
QList<QImage*> m_images;
QImage m_thumbnail;
};
class Car : public Vehicle
{
public:
/* Public interface */
//ctor
Car();
//copy/move
Car(const Car&) = delete;
Car(Car&&) = delete;
Car& operator=(const Car&) = delete;
Car& operator=(Car&&) = delete;
//dtor
virtual ~Car(); | {
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java, algorithm
for (int i = 0; i < n; i++) {
numbers[i] = in.nextInt();
}
System.out.println(getMinRefill(numbers,400));
}
}
Tested input:
0
0 200 375 550 750 950
0 = A and 950 = B Your algorithm looks fine, and efficient. Strictly speaking, since the stations are sorted, you could use binary search to find the next station to refill, by looking for the insertion point of lastRefill + 400. This could speed things up if gas stations are extremely dense. But that's unrealistic anyway, so I think no need to bother, your simple implementation is fine.
The implementation is a bit hard to read because of the names. I recommend these renames:
numbers -> stations
refillAt -> capacity
numOfRefills -> refills | {
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$$m\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \leq (-1)^{n+1}{R_n}\left( x \right) \leq M\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$
In your case you have $0.7 < x < 1.3$ so you can use the theorem and get using $M$ and $m$ as $f'''(0.7)$ and $f'''(1.3)$ since the derivative is monotonically decreasing.
$$0.11\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}} < {R_2}\left( x \right) < 0.63\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}}$$
So using this you can calculate what the error is for example for $x=1.3$
$$0.000495 < {R_2}\left( {1.3} \right) < 0.002835$$
You can see the plots of the cubics here (red and blue), which give upper and lower bounds of the error (green) for $x$ near $1$ (and obviously give $0$ when $x=1$).
- | {
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c++, image, template, classes, variadic
Make everything work for more than 5 dimensions
I can see why you stopped at 5 dimensions if you still have to manually write the code for each number of dimensions separately. Just spend a little bit of time to figure out how to write this in a generic way, then you'll actually have to do less typing and make the code work for any number of dimensions.
Whenever you have to loop over a parameter pack, you can always do this:
auto function = [&](auto index) {
…
};
(function(indexInput), …); | {
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coffeescript
Build your parameters from an object; don't hand-code them. Use jQuery's $.param or something similar to build the query string. It makes everything more explicit and less error-prone.
Here's how it'd work with jQuery
url = "http://...?" + $.param
{ name: "current", value: someVar }
{ name: "text", value: someVar }
... | {
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noise, eeg
Title: Howto calculate SNR for EEG data? This question is about SNR in the context of EEG. (a related question about SNR)
I am interested in calculating The SNR of an ERP. My motivation is: To calculate the "signal-to-noise ratio" and from this to extract the information (how?) on how many trials are needed to construct a stable ERP.
my Main question is: how do I calculate and SNR for such scenario ?
some more explation about EEG and ERP are enclosed:
EEG:
The on-going electrical activity of the brain measured from scalp electrodes is called the electroencephalogram or EEG. within the EEG is a signal which is more revealing about information processing in the brain. This signal can be obtained by time-locking the recording of the EEG to the onset of events such as a person reading a word on a computer screen, listening to a musical note played on an instrument, or viewing a picture in a magazine. The resulting activity is called an "event-related potential" (ERP)
SNR: | {
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number € n of ways to:. Since order does not matter, use combinations to determine the number of possible subsets. ” (a) acdbens (b) baaaben (c) aaabbba 10. Thus a 17 letter password, offering 80 bit security, could be represented by a 10 word sentence and a 7 word sentence, or a 9 and an 8 word sentence. Permutations should not be confused with combinations (for which the order has no influence) or with arrangements also called partial permutations (k-permutations of some elements). We must be satisfied with your capacity to pay the up front fee for providing Proof of Funds (known as the Arrangement Fee). If we VASTLY underestimate there are 1 billion stars in the universe and overcount the number of planets we’ve found at 500, the percentage is a mere. You can use the Fundamental Counting Principle to find the number of permutations. Find the expected number of offspring per year. Queen of Hearts) and a number (e. Credit: Getty/Barcroft Media. In your Word document, click the | {
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"openwebmath_score": 0.31705012917518616,
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"url": "http://xrayworks.de/find-the-number-of-distinguishable-arrangements-of-the-letters-of-the-word-billion.html"
} |
graph-theory, co.combinatorics, planar-graphs
or whether they have the same Euler characteristic (or both) — see http://www.maths.ed.ac.uk/~aar/surgery/zeeman.pdf for some reasonable lecture notes on this subject, that include the proof you're asking for. There are no other 2-manifolds in this classification that have the same Euler characteristic as the sphere, so if you calculate the Euler characteristic and find that it matches the formula for a sphere, you know your embedding must be onto a sphere. | {
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python, sorting, mergesort
So, what you should be doing instead is iterating through the left and right lists non-destructively, using two variables i and j to keep track of how far you've gone. | {
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c++, c++11, file, boost, checksum
std::vector<char> buf(file_size);
is.read(&buf[0], file_size);
boost::crc_32_type result;
result.process_bytes(&buf[0], file_size);
return result.checksum();
}
The problem with this is that some files are huge. You don't want to read the whole thing into memory like that. So you could define a reasonable size buffer and read chunks at a time.
std::vector<char> buffer(4096);
while(is.read(&buffer[0], buffer.size()))
{
std::size_t count = is.gcount();
result.process_bytes(&buffer[0], count);
}
Or you could use a memory mapped file:
Its been a while since I did this. Here are some instructions.
https://www.safaribooksonline.com/library/view/linux-system-programming/0596009585/ch04s03.html | {
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c++, logging, c++17
Log::~Log() {
addLog(Level::Info, "Stopped logging system.");
m_logfile.close();
}
}
Is my code thread safe? If not, how do I make it thread safe?
No, of course it's not thread-safe. You don't do anything to make it thread-safe.
A more nuanced answer would be: It's thread-safe as long as you don't use it in an unsafe way. For example, calling Logger::log(FATAL, "hello world") from two different threads concurrently would of course be unsafe. But if your program has only one thread... :)
If you want to allow calling Logger::log from two threads concurrently, you'll have to do something to eliminate the data race on m_logfile which is caused by the two threads' both calling m_logfile << levels[static_cast<int>(s)] at the same time. For example, you could throw a mutex lock around addLog.
Is there a way to make the interface a bit cleaner for use? Right now #define feels a bit hacky. | {
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complexity-theory, np-hard, randomized-algorithms, colorings, parameterized-complexity
Title: k disjoint triangles with graph splitting to two distinct groups Please note that this question is different than this question.
The $k$-disjoint triangles problem is as follows:
Input: A graph $G=(V,E)$ and an integer $k\in \mathbb{N}$
Output: Are there $k$ vertex-disjoint triangles in $G$? | {
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javascript, node.js, functional-programming, postgresql
// MODIFIED FROM https://medium.com/@garychambers108/better-logging-in-node-js-b3cc6fd0dafd
["log", "warn", "error"].forEach(method => {
const oldMethod = console[method].bind(console) ;
console[method] = logging.createConsoleMethodOverride({oldMethod, console}) ;
}) ;
const pool = new pg.Pool({
user
, database
, host
, log: pglogging.createPgLogger()
}) ;
pool.query('SELECT 1 ;').then(result => console.log(result)).then(()=>pool.end()) ;
Functional approach
const {user, database, host} = require('./mydatabaseconnection') ;
const pg = require('pg') ;
const logging = require('./logging') ;
const pglogging = require('./pglogging') ;
// MODIFIED FROM https://medium.com/@garychambers108/better-logging-in-node-js-b3cc6fd0dafd
["log", "warn", "error"].forEach(method => {
const oldMethod = console[method].bind(console) ;
console[method] = logging.createConsoleMethodOverride({oldMethod, console}) ;
}) ; | {
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dft
Title: Goertzel algorithm produces incorrect phase I've implemented Goertzel algorithm according to the Wikipedia page (https://en.wikipedia.org/wiki/Goertzel_algorithm) and another page (http://www.mstarlabs.com/dsp/goertzel/goertzel.html), which are consistent with each other. I am then testing it using a naive implementation for a single DFT frequency (https://en.wikipedia.org/wiki/Discrete_Fourier_transform). I am getting correct amplitude, but incorrect phase. Please help me find the mistake in my code, I've spent a couple of days on it now, and can't see the problem...
/**
* Complex imaginary number i.
*/
template<typename T>
constexpr auto COMPLEX_i = std::complex<T> (
static_cast<T> (0.0), static_cast<T> (1.0));
/**
* Convert frequency in Hertz to Omega (radian per sample).
*/
template<typename T>
T
hertz_to_omega (T hertz, T sample_rate)
{
return hertz / sample_rate * static_cast<T> (2.0) * M_PI;
} | {
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immunology, pathology
Source
Quite simply, at the beginning of an infection, the viruses / bacteria will be destroyed by macrophages.
The dendritic cell will "pick up" some of the leftover pieces.
When it finds a T-Cell that has the same configuration, it orders lots of identical T-Cells to be made. These T cells have multiple jobs, one of which is activating the B cells. It is the B cells that release the antibodies.
So Ebola.
If the dendritic cells are "invaded", then the only line of defence is the macrophages. This is often not enough to stop an attack, and it multiplies out of control.
The reason the Dendritic cells can be invaded is because they are just as vulnerable as other cells. They haven't activated the T cells, and they haven't activated the B cells, so there is nothing protecting them.
Images Source (YouTube video) | {
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entanglement, textbook-and-exercises, quantum-operation, locc-operation, majorization
Why can we start from simultaneously diagonal states $\rho_{d}, \sigma_{d}$? Hint: the partial trace.
In several quantum resource theories the state transformation reduces to ``classical majorization'', i.e., majorization of vectors (as opposed to say matrix majorization), for example, resource theory of coherence, non-uniformity, etc. | {
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astrophysics, terminology, conventions, stars, stellar-physics
The reason for the 8 solar mass division (it is uncertain by about 1 solar mass and also depends to a certain extent on rotation and the initial metallicity of the star, so is not a sharp threshold) is that this is where the carbon/oxygen core (during He shell burning)$^{1}$ becomes hot enough to ignite further fusion. Core burning continues through to iron-peak elements, then there is a core mass collapse, a violent supernova and large quantities of processed material (O, Mg, Ne, Si, r-process elements) are ejected at high speeds. A neutron star or black hole remnant is formed. In lower mass stars, the core becomes degenerate, supported by electron degeneracy pressure, and core nucleosynthesis halts. The star ends its life by expelling the majority of its envelope (mostly H and He, with some enrichment with C, N and s-process elements) at low speeds through stellar winds. The degenerate core becomes a white dwarf. | {
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ros, ros-canopen
and
vcan0 000 [2] 82 01
vcan0 000 [2] 02 01
and
[ INFO] [1510308088.698080279]: Initializing XXX
[ INFO] [1510308088.698592687]: Current state: 1 device error: system:0 internal_error: 0 (OK)
[ INFO] [1510308088.701974798]: Current state: 2 device error: system:0 internal_error: 0 (OK)
[ INFO] [1510308098.703534765]: Current state: 2 device error: system:125 internal_error: 0 (OK)
[ INFO] [1510308098.703622135]: Current state: 0 device error: system:125 internal_error: 0 (OK)
[ INFO] [1510308098.703703969]: Current state: 0 device error: system:0 internal_error: 0 (OK)
[ INFO] [1510308098.703794303]: Current state: 0 device error: system:0 internal_error: 0 (OK)
Is it because I am using a virtual can?
also as this is a bit from the original question, should I open this up as a new question rather than continuing here?
Originally posted by rhas on ROS Answers with karma: 37 on 2017-11-08
Post score: 0 | {
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c#, .net
Title: From production to staging The story here is about a console application that replays to a test system string clauses already imported in the production system.
Looking forward to your comments.
namespace MyCompany.Department.ProjectXYZ;
{
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.ServiceModel;
using MyCompany.XYZ; | {
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thermodynamics, materials, heat, temperature
Title: Generally available fluid with good heat transport properties I have designed a container which is heated up and has to heat up some tube-coils of water inside of it. This container is filled with a fluid that can help store and transfer the heat received from the container to the tube-coils.
My plan was to find some material that I can easily buy and which is not very expensive to fill the $636cm^3$ container.
It has to have a high coefficient of heat transfer, high heat capacity and an operating temperature of at least the range of 10-400 degrees Celsius. I will also need to calculate how much the volume changes when it goes from around -35 to 500 degrees Celsius.
I also think a high density and if the material is non-toxic might be a plus. | {
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c#, .net, linq
Title: Take N elements from List of Lists I've come with a solution super tricky for a simple requirement. I think I could solve the problem using LinQ, but I'm not seeing it so clearly at all.
What's sure, I'm not comfortable with my code.
The requirement is the following:
Given a list of lists of string (IEnumerable>), take the first N elements from each but one by one.
So given this list of list and N=10:
{aaa, bb, ccc, ddd}
{eee, fff}
{ggg, hhhhh, iii, jjj, kkk}
{lll}
{1111, 22, 333, 444, 55555, 66666}
This is the output:
{aaa, eee, ggg, lll, 111, bb, fff, hhhhh, 22, ccc}
And here is the code:
private IEnumerable<string> Extract(
IEnumerable<IEnumerable<string>> listOfList,
int N
)
{
var result = new List<string>();
for (int i = 0; i < N; i++)
{
foreach (IEnumerable<string> list in listOfList)
{
if (list.Count() > i)
{
result.Add(list.ElementAt(i)); | {
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### CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.) | {
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binary-star, gravitational-waves
In slightly more friendly units
$$ \frac{dr}{dt} = 7.8\times10^{-19} \frac{(M_1 M_2)(M_1+M_2)}{r^3}\ au/yr,$$
where $r$ is in astronomical units and the masses are in solar masses, and
$$ t = 3.2\times10^{17} \frac{R^4}{(M_1 M_2)(M_1+M_2)}\ yr. $$
Thus you can see that the effect gets much stronger (and the merger times shorter) if you have two high mass objects (it depends on the product of the masses) with a small orbital separations.
For "normal" stars you can never get them closer than their stellar radii and generally means that the effect is too small to be important. However, compact objects (white dwarfs and neutron stars) can be brought much closer together and this is why the effect has been seen there. The Hulse-Taylor binary components | {
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python, comparative-review
type hints help with documentation
def get_freq(self):
return self.instr.query("freq?")
def set_freq(self, freq):
self.instr.write(f"freq:{freq}Hz")
Consider adding optional type hints,
which mitigate the need for a """docstring""":
def get_freq(self) -> float:
return self.instr.query("freq?")
def set_freq(self, freq: float) -> None:
self.instr.write(f"freq:{freq}Hz")
Or, IDK, maybe those are actually integer quantities?
As someone calling into your library,
I would wish to resolve such ambiguities.
Maybe the type depends on what kind of resource address the ctor received?
Do you possibly wish to insist on serving just certain types of instruments?
So many questions, so few docstrings.
missing import
rm = pyvisa.ResourceManager(sim_fpath)
I don't know what sim_fpath means here.
make method call explicit
f2 = inst2.freq
time.sleep(0.5)
f3 = inst2.freq | {
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electricity, electrons, charge, electronics, carrier-particles
Title: Can we have electronics with charge carriers OTHER than electrons? Ions perhaps? Or other elementary particles with charge? Depending on your view, there is electronics with other charge carriers. It is commonplace to have semiconductor devices where the relevant carriers are holes!
Furthermore, batteries and electrolysis relies heavily on ions as charge carriers (but hardly count as electronics). I guess genuine electronics with ions will be difficult as charge carrier mobility will be much lower in solid state systems (and high carrier mobility is highly relevant for effective devices).
Furthermore, we are beginning to build electronics, where not the charge, but the spin carries the information ("spintronics"). Some spintronics devices are already in production, namely read-heads for hard disk drives based on the giant magnetoresistance. | {
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newtonian-mechanics, oscillators, perpetual-motion
hammering something against friction and saying "gee, it moved - that proves overunity"
rolling a cart against friction and saying "gosh it moved farther"
making a gadget with a pendulum, standing there pushing it, something heavy at the other end moves a little bit, and saying "gosh - free energy"
I do have to admit, those toys were well made.
But a perpetuum mobile has to run "forever" with no input, indeed with net output.
EDIT: The apparent energy output is just that - apparent. Not one of the gadgets could run by itself. | {
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probability-theory, information-theory
$$ = \{\vec x \in \mathcal{X}^n: |\frac{-1}{n}\sum_{i=1}^n\log P^i_X(x_i) - H(\tilde P_X)| \leq \epsilon \}$$
I have no idea how to prove/disprove this. It seems like I can not use Weak law of large numbers. Suppose that $X_n = n$ with probability $1$. Then $H(\tilde{P}_X) = \log n$, but the only vector with non-zero probability has $\frac{1}{n} \log \frac{1}{P(\vec{x})} = 0$. | {
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and used to study variability. ( also known as sigma or σ ) determines the spread around this mean to the sigma symbol the! First we should start with what a sigma is it is a data-driven approach to problem-solving model. As sigma or σ ) determines the spread around this mean than.... And most common formula related to the sigma symbol is the population standard deviation whereas Sx is the for! Proportion, which is approximated by p-hat, the sample standard deviation whereas Sx is the population proportion which! Less common formula is the summation notation using the upper case sigma of... Sample estimator mean/central tendency a fitted model, i.e., sigma^ approach is a which. In Ancient Greece, the sigma symbol is the summation notation what does sigma mean in stats the upper case sigma chart, on,! Of past data to make judgements about how the data is spread and used to study variability. Data set a fitted model, i.e., sigma^ of data to make judgements about how the data spread! ) | {
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php, object-oriented, interview-questions
"SQL_Helper are functions which help with... SQL... :)"
This is like going to an interview with your shirt untucked and instead of properly shaking hands you greet them with 'yo bro'. Simply unprofessional. Won't get you hired.
Anyways... onward...
Documentation
Your documentation is messy. Most great developers care about how their code and documentation
looks. They treat writing them as craft. You want the look/layout to be pleasing to the eye and the contents to be pleasing to the brain.
This means you need to keep things consistent looking and to the point. | {
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calibration, stereo
Title: Large black areas on rectified stereo images
I'm using two RGB cameras. They are rotated, one clock-wise and the other counter clock-wise. I'm calibrating them with Ros Stereo Calibration node which is based on OpenCv. The output I get the camera matrix, projection matrix and R and T from cv::stereoCalibrate. The stereo_image_proc node outputs the rectified images that you can see below:
image description http://i68.tinypic.com/292p3wi.png
They seem rectified, they are both "facing" the right way but there is a HUGE area of only black pixels......why is that? Does it have to do with some parameter in cv::stereoRectify that I should change?
Does it matter if I hold the calibration patter horizontally or vertically when calibrating? Could that be the issue?
Thanks for the help.
ps: This is a continuation of this question (when I was having completely black rectified images because I was doing the calibration wrong). | {
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black-holes
So, it looks as if eiher of the Pioneer probes could have spotted a black hole up to 5,825 au away.
The gravitational anomaly goes down as the square of the distance, so they’d spot a hundred times the anomaly at 580 au.
There is no way they missed a hundred times this anomaly. Or to go for a round number, there is also surely no way they’d miss a black hole as close as 1000 au either, which would be 34 times larger. The only possibility would be a black hole in a direction at right angles to both. But though I am not sure how to calculate how much of an anomaly the Voyagers or New Horizon would spot, I’d be surprised if they couldn’t notice a black hole within a few hundred au.
So, I think we can be reasonably confident there isn’t an old five solar mass black hole currently within a few hundred au of Earth and unlikely to be one within a thousand au.
NOTICING IT GRAVITATIONALLY WITH OBSERVATIONS OF SEDNA AND OTHER TRANS NEPTUNIAN OBJECTS | {
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power series. Explanation:. Intervals of Convergence of Power Series. As a study aid, I have listed below the major. If the terms of a sequence being summed are power functions, then we have a power series, defined by Note that most textbooks start with n = 0 instead of starting at 1, because it makes the exponents and n the same (if we started at 1, then the exponents would be n - 1). Abel's theorem is typically applied in conjunction with the alternating series theorem which is used to show the conditional convergence at one or both endpoints. Remember that integrating or differentiating the terms will not change the radius of convergence of the series. Briefly, we have our function and on each element or interval from xk minus 1 to xk, we replace the integral by the area of the rectangle which has the height of f (xk minus 1) at the left endpoint. Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0). | {
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php, html
The defaut case is the fallback case. To me, logic dictates that the default/fallback procedure is the last resort, and it should therefore be written as the last case label. Maybe that's just me, and my (sometimes over eager) tendency to use case fallthroughs:
switch ($foo)
{
case 1:
case 2:
bothDoTheSame();
break;
case 3:
threeSpecific();
case 4:
threeAndFour();
break;
default: defaultCase();//no break required
} | {
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java, chess
}
for (int i = 1, max = ChessBoard.SIZE; i < max; i++) {
int x = pos.getX() + i;
int y = pos.getY() - i;
if (x <= ChessBoard.SIZE && y > 0) {
Position posMove = new Position(x, y);
if (board.getPiece(posMove) == null) {
result.add(posMove);
} else if (board.getPiece(posMove).isWhite() ^ !isWhite) {
max = i;
} else {
result.add(posMove);
max = i;
}
} else {
max = i;
}
}
for (int i = 1, max = ChessBoard.SIZE; i < max; i++) {
int x = pos.getX() - i;
int y = pos.getY() + i;
if (x > 0 && y <= ChessBoard.SIZE) {
Position posMove = new Position(x, y);
if (board.getPiece(posMove) == null) {
result.add(posMove); | {
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digital-communications, amplitude
From the link the OP gave, the intention here is to derive an amplitude and phase imbalance specification and for that the relationship between SNR and imbalance is that the SNR will be the normalized error vector. For example, if the error vector normalized to ideal was 0.1 radians, the SNR would be $20Log_{10}(0.1) = 20$ dB. For small angle the amplitude error and phase error would have the same effect, so if the error vector was 0.1 due to amplitude imbalance, we would also get a 20 dB SNR. (Universally, it is the magnitude of the error vector). An error vector of 0.1 dB is an amplitude imbalance of $20Log_{10}(1\pm 0.1) = +0.8$ or $-0.9$ dB. | {
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javascript, node.js, google-cloud-platform
Unused arguments at the end of the list can be omitted safely in JS, so function( file, index ) can be simply function( file), or even file => if you use the new ES syntax (which you are) | {
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newtonian-mechanics, forces, rotational-dynamics, rotation, rotational-kinematics
Title: Confusion in rotating frames my question is,if we very slightly put a ball on smooth(frictionless) rotating disc then what would happen to ball. It will fall or not.
because with respect to rotating disc the ball will be acted upon by two forces centrifugal and coriolis force because of which we conclude that it will fall ,specifically by following a curved path. | {
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turing-machines, computability, reductions, semi-decidability
If A $\leq_m$ B and A is not Turing recognizable then B is not Turing-recognizable.
A $\leq_m$ B if and only if $A^c \leq_m B^c$.
For language $L_{2}$,
If I can show that $A_{TM}$ is mapping reducible to $L_{\not=2}$, according to property 2 we could get that $\overline{A_{TM}}$(the complement of $A_{TM}$) is mapping reducible to $L_{2}$, and for the fact that $\overline{A_{TM}}$ is not Turing recognizable, we could get that $L_{=2}$ is not Turing recognizable.
For language $L_{\not=2}$,
If I can show that $\overline{A_{TM}}$ is also mapping reducible to $L_{\not=2}$, and for the fact that $\overline{A_{TM}}$ is not Turing recognizable, we could get that $L_{\not=2}$ is not Turing recognizable.
However, I don't know how to prove the mapping reducible relationships between them. Or my direction could be totally wrong.
Could somebody give me the hints or possible solutions?
If I can show that $A_{TM}$ is mapping reducible to $L_{=2}$, ... | {
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stereo
Originally posted by Vijeet on ROS Answers with karma: 11 on 2016-08-17
Post score: 1
There are many open source algorithms using visual SLAM. Use this link to check for a few implementation. I am not sure what you mean by applying it for cars, theoretically they should work if you are using stereo vision to get data, be it robots or cars.
I would suggest starting with ORB-SLAM, Open RatSLAM and OKVIS since they seem to provide better results in most cases. You can of course try other algorithms for comparing the performance. Some of the algorithms may be for monocular camera but later might release a version for stereo camera, so look out for that. Other algorithms that may be interesting based on your question would be LSD SLAM, LibViso, PTAM to list a few. You also mentioned finding some papers on the related topic, it is quite likely that they already have some kind of implementation in place to take a look at. | {
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&=\sum_{j=0}^{\left\lfloor\frac{n-1}7\right\rfloor}\binom{n-6j-1}{j}\left(\frac76\right)^{n-7j-1}\left(-\frac16\right)^j\\ \end{align} For $$n=30$$, this gives \begin{align} &\textstyle\binom{29}{0}\left(\frac76\right)^{29}-\binom{23}{1}\left(\frac76\right)^{22}\left(\frac16\right)^1+\binom{17}{2}\left(\frac76\right)^{15}\left(\frac16\right)^2-\binom{11}{3}\left(\frac76\right)^8\left(\frac16\right)^3+\binom{5}{4}\left(\frac76\right)^1\left(\frac16\right)^4\\[6pt] &=\textstyle\frac1{6^{29}}\left(\binom{29}{0}7^{29}-\binom{23}{1}7^{22}6^6+\binom{17}{2}7^{15}6^{12}-\binom{11}{3}7^86^{18}+\binom{5}{4}7^16^{24}\right)\\[3pt] &=\frac{333366007330230566034343}{36845653286788892983296}\\[9pt] &\doteq9.0476345943840229021 \end{align} | {
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electric-circuits, electric-current, electrical-resistance, voltage, conductors
Title: How will the current flow in the figure drawn below?
Consider the green material in the figure to be conducting. So, I was wondering how the current will flow in the rod, as the battery is not connected at the ends, but on the surface of the rod. Thus, will there be no current at the ends of the rod and what about voltage? An approximate numerically calculated figure is attached.
The figure is a two-dimensional result.
Although 3D calculations is possible, the figure is easier to see in 2D.
$\phi$ is potential in volts.
$\vec{J}$ is the electric current density.
(Edit #1)
I add a figure highlighting the elements in the two left-most columns. | {
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python, python-3.x, programming-challenge, recursion, dynamic-programming
debuggability, and flexibility in the face of evolving requirements. But that's
just a guideline, and there can be valid exceptions. But notice that if we
decide to make Hydra eager during initialization, we could also decide to
design the class to discourage mutation by users of the class. In other words,
if the class solves the whole problem upon initialization, there's much less
need for a bunch of setters that carefully align internal state. Eager can
often be paired with a policy of treating the objects as non-modifiable, at
least in spirit.
The Hydra class does not readily answer the primary question. Your stated
goal is to minimize the number of hits necessary to kill the hydra. But the
class does not appear to have a simple way to answer that question: how many
hits were required? It also does not appear to have an easy way to see how many
hits of each kind were required. Maybe there is a way and I overlooked it. In | {
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general form of a parabola is: y = ax² + bx + x. The parabola equation is y=9 -x^2. Based on the quadratic function shown in the table below, which of the following is the range of this function? (2) (4) y £11 For Problems 3 — 5, use tables on your calculator to help you investigate these functions. If you like the worksheet you can print it straight from your browser. The maximum value is 1 since that is the hightes point, or the greatest y-value, of the graph X Advertisement. Therefore, a quadratic function may have one, two, or zero roots. Best mix for a given depth. For a quadratic Bezier, this is actually quite simple. This Parabola equation solver calculator helps you to solve your academic equations and engineering algebraic problems with ease. Report a problem. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4). In algebra, a quadratic equation is any polynomial equation of the second degree with the | {
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game, sql, postgresql
Ah, much better! It can even be extended to find the number of wins and losses in the same query without too much effort. Can also easily join with the player table to find out the player id for a specific player name.
Game Actions
turn INT NOT NULL,
If you would consider "End Turn" itself an action, then this "turn" integer isn't really needed.
action_attack INT NULL,
Umm... I'm sure there are other possible actions except attack, right? And what is the exact purpose of this column? How much attack damage that was dealt? It has a very confusing name and I'm not sure you know yourself what it should be used for.
I don't see an "action_id" or "action_name" anywhere here. If one card at any point have multiple possible actions, then an action_id column is required.
Action Targets
target_player INT NULL,
FOREIGN KEY (action_player) REFERENCES player(id),
target_card INT NULL,
FOREIGN KEY (action_player) REFERENCES player(id) | {
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quantum-mechanics, superposition, chaos-theory
@CuriousOne: This is absolutely wrong. Quantum mechanics is not only applicable to emsembles but also to single quantum systems. | {
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- Scenario ONE: https://i.imgur.com/atiQ9yT.png
- Scenario TWO: https://i.imgur.com/1AmZgHZ.png
Explanations:
- Scenario ONE: this one is comparatively complex, but it still does make sense. If we choose x as the side adjacent to theta, then we will end up with -arccos(x / sqrt(u)) + c. Although this appears different, if you look at this graph (drag the slider at the top-left at your own leisure) ==> https://www.desmos.com/calculator/3gfueg1fmv <== then you will see that -arccos(x / sqrt(u)) + pi/2 is actually equal to arcsin(x / sqrt(u)). Since after we integrate we are left with a constant of integration (I have used c to denote said constant), this "absorbs" the extra pi / 2 that we would need to add for the graphs to match exactly. Regardless, the curves for -arccos(x / sqrt(u)) and arcsin(x / sqrt(u)) are identical except that one is vertical translation of the other, and that is all we are trying to prove with integration, anyway. | {
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natural-language-processing, terminology, books, test-datasets, validation-datasets
This paragraph clearly says that development set is used for hyperparameter tuning.
#3: held-out corpora for hyperparameter tuning
How are these $\lambda$ values set? Both the simple interpolation and
conditional interpolation $\lambda'$s are learned from a held-out
corpus. A held-out corpus is an additional training corpus that we
use to set hyperparameters like these $\lambda$ values, by choosing
the $\lambda$ values that maximize the likelihood of the held-out
corpus. | {
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quantum-mechanics, thermodynamics, statistical-mechanics, bose-einstein-condensate, quantum-statistics
Title: Pressure of Bose-Einstein-Condensate in decreasing volume Consider a Bose-Gas where $T_0$ is the critical temperature at which the temperature dependant chemical potential is $\mu(T_0) = 0$. Looking at the interval $ 0 \leq T < T_0 $ one can show that the pressure of the system is:
$$ P = \dfrac{2}{3}\dfrac{E}{V} = \zeta \bigg(\dfrac{5}{2}\bigg) \dfrac{k_BT_0}{\lambda_{T_0}^3}\bigg(\dfrac{T}{T_0}\bigg)^{\dfrac{5}{2}}$$
where $\zeta$ is the Riemann zeta function and $\lambda_T$ the termal de-Broglie-wavelength of the wavepackages. As one can see
$$P \propto T^{\dfrac{5}{2}} $$
But the interesting fact is that the pressure of an ideal bose gas in the interval $ 0 \leq T < T_0 $ is independant of the particle density N/V, where N is the number of particles and V the volume. One can deduce that there is no resistence of the system when decreasing the volume in which the gas is placed. | {
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dynamic-programming, knapsack-problems
But we can split them in other ways, keeping the condition that we can take a subset that sums whatever sum we want up to the maximum. For example you can split $10$ like $(1, 2, 3, 4)$, and this partition is useful since you can get any sum up to $10$. There is a canonical way to split a number in $log_2(n)$ numbers that has this property. Split the number in power of 2 starting from $1, 2, ...$ until you can't split it anymore, and add the remainder to the partition too.
For example:
5 -> (1, 2, 2)
10 -> (1, 2, 4, 3)
15 -> (1, 2, 4, 8)
1000 -> (1, 2, 4, 8, 16, 32, 64, 128, 256, 489) | {
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java, linked-list, quick-sort
Node[] n = list.convertA(list);
quickSort(n, 0, (n.length-1));
System.out.println("Sorted list is:");
for(int q =0;q<n.length;q++){
System.out.println(n[q] + " ");
}
}
public static int partition(Node arr[], int left, int right) {
int i = left, j = right;
Node tmp;
Node pivot = arr[(left + right) / 2];
while (i <= j) {
while (arr[i].getStudent().getAge() < pivot.getStudent().getAge()) {
i++;
}
while (arr[j].getStudent().getAge() > pivot.getStudent().getAge()) {
j--;
}
if (i <= j) {
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
i++;
j--;
}
}
return i;
}
public static void quickSort(Node arr[], int left, int right) {
int index = partition(arr, left, right-1); | {
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organic-chemistry, computational-chemistry
This does not say that it is not a ortho-para director. It's a ortho-para director but it deactivates the ortho and para places of the benzene cycle. | {
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sic inverse compositional algorithm for dense image align-ment. We can find inverse of a matrix in following way. The identity matrix for the 2 x 2 matrix is given by. Steps involved in the Example. The value at cell [r][c] of the result matrix is the product of the values in row r of the first matrix and the values in column c of the second matrix. This is a fun way to find the Inverse of a Matrix: Play around with the rows (adding, multiplying or swapping) until we make Matrix A into the Identity Matrix I. In the first case the algorithm gives the ordinary inverse of A. This is true even if you are solving $Mx=b$ repeatedly, unless the number of solves is close to the dimension of $M$. The A chosen in the much praised explanation does not do that. A popular method for computing the matrix logarithm is the inverse scaling and squaring method, which essentially carries out the steps of the scaling and squaring method for the matrix exponential. And by ALSO doing the changes to an | {
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thermodynamics, energy, pressure, volume
Title: Deriving a thermodynamic relation using the Helmholtz free energy I'm supposed to derive the following thermodynamic identity:
$$
- \left( \frac{\partial U}{\partial V} \right)_{T,N} + T \left( \frac{\partial P}{\partial T} \right)_{V,N} = P
$$
...by starting with the Helmholtz free energy $F = U - TS$.
$\ $
How do I begin this problem? To me it seems like the only way I would be able to relate $U, T, S$ to the variables $P,V$ would be through the use of some kind of equation of state, but I'm not explicitly given one. | {
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performance, memory-management, vb.net
for a couple of reasons. First you are creating a string array out of this ArrayList (however this is created) and then store it in an Array so casting it to an object to cast it again to a String on the next lines.
But then you use the valLength String's to compare with Length which is an Integer.
The valName variable is never used, so you can just remove it along with the commented code. Code which isn't used or is commented is just dead code which should be removed to increase readability.
Another thing bothering me is the use of abbreviations for variables names. You won't gain any performancy increase by doing so, but you loose a lot of readability.
Dim validationValues As String() = validationsArray(i).ToString.ToUpper.Split("|")
Dim valueDataType As String = validationValues(1).Trim()
Dim valueLength As String = validationValues(1).Trim()
Dim currentValue As String = values(i).ToString()
Select Case valDataType
Case "VARCHAR2" | {
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nuclear-physics
Similarly, $m\left(^A_{Z+1}Y\right)-(Z+1)m_e$ is really just a long way of writing $m_{\rm nuc,Y}$.
If you already had the masses of the nuclei directly, you would only need to include the mass of the single electron in the final state. | {
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newtonian-mechanics, reference-frames, centrifugal-force, angular-velocity, machs-principle
How, then, can one know how what the centrifugal force is? Is it defined in relation to all the other mass in the universe, in such a way that it's negligible in classical mechanics problems? The answer is that not everything is relative. Indeed, you've just shown that it is possible to detect rotation in an absolute sense.
More generally, the principle of relativity says that all inertial frames are equivalent. In other words, it is impossible to detect (or even to define) absolute motion at a uniform velocity; it doesn't make sense to say that you're moving in a straight line at constant speed if you don't tell me with respect to what. | {
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zoology
Title: What kind of animal can do this?
When I was moving to another house, I found this plastic bag on the top of the cabinet of my old house. It looks like someone cut it all with a scissor, but no one have access to my bedroom, and it was not me.
I don't thing it could be rats. That bedroom was very tiny and closed, there's no way for a rat to enter.
The house is very near a university with a little "forest", there are many types of mosquito and little flying animals, but I can't imagine which could do this.
I'm just curious. (and not sure if "zoology" is the correct tag here, but I had to choose one) Although your plastic bag tears are most likely due to light damage as @Chris mentioned above, there is a type of ant that chews its way through plastic, rubber and other manufactured polymers: Monomorium pharaonis, the Pharaoh ant. | {
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object-oriented, tic-tac-toe, c++17, classes, inheritance
return true;
}
else if (board.at(7) == player->GetName() && board.at(8) == player->GetName() && board.at(9) == player->GetName()) {
return true;
}
else if (board.at(1) == player->GetName() && board.at(4) == player->GetName() && board.at(7) == player->GetName()) {
return true;
}
else if (board.at(2) == player->GetName() && board.at(5) == player->GetName() && board.at(8) == player->GetName()) {
return true;
}
else if (board.at(3) == player->GetName() && board.at(6) == player->GetName() && board.at(9) == player->GetName()) {
return true;
}
else if (board.at(1) == player->GetName() && board.at(5) == player->GetName() && board.at(9) == player->GetName()) {
return true;
}
else if (board.at(7) == player->GetName() && board.at(5) == player->GetName() && board.at(3) == player->GetName()) {
return true;
} | {
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c#, wpf, async-await
private void ShowDialogCore(object viewModel, ...)
{
...
}
}
The upside with implementing it like this is that the caller will not need to remember to call it on the UI thread. It is the job of the windowmanager to schedule if needed. Hence fixing many potential bugs in one place.
Regarding the pattern Using a flag like this is fine and will work well if you refactor to relaycommand. A try-finally block makes it clear in my opinion. | {
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Example 3: The power law $$f(x) = \tfrac{1}{2\sqrt{x}}$$ is an example of a density function which is not in $$L^2$$. In this case we can evaluate $$\mathbb{P}[|X-Y| < \varepsilon]$$ exactly (well, if you believe in inverse hyperbolic trig functions). The asymptotic behavior is $$\mathbb{P}[|X-Y| < \varepsilon] = \tfrac{1}{2} (-\log \varepsilon) \varepsilon + (\tfrac{1}{2} + \log 2) \varepsilon + O(\varepsilon^2).$$ So the limit in question diverges logarithmically. | {
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ros, installation
Originally posted by gvdhoorn with karma: 86574 on 2017-02-05
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by suforeman on 2017-02-05:
As a companion comment,when running ROS + Ubuntu on a Raspberry Pi the SD card is not different than a USB memory stick. As long as the solid state memory storage is fast, it works well. Running with RAM based temp storage makes a significant improvement. | {
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### Show Tags
29 Jun 2015, 04:40
3
(1) The tens digit of N is 5.
We know that all of the digits should be factors of N, hence this statement effectively tells us that last digit is also 5 (any number divisible by 5 should end with 5 or 0, as 0 can't be a factor of any number, we are left with 5).
Now, let's plug numbers! First digit should be >1 (question stem) and a can't be even (as our number ends with 5, hence it's not divisible by 2):
355- sum of digits = 13, not divisible by 3.
555 - works
755 - not divisible by 7 (just use 7*100 = 700 + 7*7 = 49 = 749, then 7*8=756).
955 - sum = 19, not divisible by 9.
Sufficient.
(2) The units digit of N is 5.
First, notice that we already have this info from A, this gives us that xy can't be even.
Let's plug numbers:
xy5 - where x and y are digits, x>1.
315, sum of digits 9, hence divisible by 3 and obviously by 1 and 5. Works.
515, divisible by 5's and 1's. Works.
We already have 2 solutions. Not sufficient.
_________________ | {
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and how do you justify that it is differentiable?
I do not. I just write it and then ask students what will the derivative of this function be?
All that stuff you wrote before it useless to make a first time student who sees the natural exponential to understand what makes it special to other exponential functions. He will get lost in the middle and not understand what you are doing it. My approach does not suffer from that. Now if you tell a student the natural exponential is its own derivative while other exponentials are almost their own derivative they see what makes the natural exponent the nicer base to use.
#### Deveno
##### Well-known member
MHB Math Scholar
Let me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea? | {
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python, reinventing-the-wheel, cryptography
Title: Encryption/Decryption algorithm #2 This is a follow-up question to this one.
I have tried to implement all the recommended things in the answers (except commenting, and not being OS specific). Again, if you see anything that needs improvement, or would make it faster, please tell me!
Here's the code. Only 222 lines this time :).
#!/usr/bin/env python
# not sure if I did this right
import base64
import random
import os
def add_padding(plain_text, block_size=128):
plain_text = plain_text.encode()
padding = -(len(plain_text) + 1) % block_size # Amount of padding needed to fill block
padded_text = plain_text + b'=' * padding + bytes([padding + 1])
return decimal_to_binary(padded_text)
def xor_string(key, secret):
xored_secret = ''
for i in range(len(secret) // len(key)):
if i > 0:
key = get_round_key(key) | {
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waves, vacuum, matter
Title: Matter waves require a medium for propagation I have read:
Matter waves require a medium to for propagation.
Matter waves do not leave the moving particle, i.e. are not emitted.
But when a particle is moving with some velocity in vacuum, what happens to the matter wave? Matter wave can neither leave the particle nor can they move in vacuum. Where is my interpretation of matter waves wrong? Perhaps the simplest answer would be to say that #1 is wrong. Where did you hear that matter waves required a medium to propagate? Since I agree that what you mean by "matter wave" sounds like a wavefunction, those don't require a medium. One might even say this was the biggest surprise in the history of physics-- we find a form of wavelike behavior without a medium. | {
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# How to linearize membership in a finite set
Given finite set $$S$$ and variable $$x$$, how do I linearize the set membership constraint $$x\in S$$?
• Based on the timestamps, this question appears to be made as a more generalized response to this question. Very clever and very sly. – SecretAgentMan Jul 9 at 17:02
• Yes, there were several recent questions related to this. I didn't find any existing question/answer to reference, so I created one. – RobPratt Jul 9 at 17:04
• Brilliant. Very useful for future visitors. – SecretAgentMan Jul 9 at 17:59
First, some special cases:
• If $$S=\{c\}$$, fix $$x=c$$.
• If $$S=\{0,1\}$$, declare $$x$$ to be binary.
• If $$S=\{a,b\}\not=\{0,1\}$$, introduce binary variable $$y$$ and impose linear constraint $$x=a(1-y)+by$$.
• If $$S=\{a,a+c,a+2c,\dots,a+kc\}$$, introduce integer variable $$y\in[0,k]$$ and impose linear constraint $$x=a+cy$$. (The previous bullet is the special case $$c=b-a$$ and $$k=1$$.)
Otherwise proceed as follows. | {
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soft-question, education
The reason this skim-and-discover method works so effectively, is that you can rediscover all the important insights for yourself, but do it without taking centuries, because you already have the starting point, but can still develop the experience of having discovered them via the "think about how you could've discovered the starting point yourself" stuff. It's also important to study mathematics alongside physics -- there are often extremely strong connections -- almost equivalences -- between certain areas of math and certain areas of physics (perhaps it's because the math was developed for the physics). E.g. special relativity and linear algebra (the basic geometric stuff), quantum mechanics and linear algebra (the more abstract, general stuff), general relativity and differential (at least Riemannian) geometry, etc. Learning them side-by-side helps you get the full set of insights, it's like learning two similar languages (or languages that share a similar script) side-by-side. | {
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Challenges between $D'_b$ and some $D'_x$ can be checked similarly, as can challenges between $D'_x$ and $D'_y$ where $x$ and $y$ designate dice other than $a$ and $b$.
• Cool. Is there a way to do this so no two dice have the same side (just wondering)? Jan 24, 2016 at 15:53
• Oh wait, I see. Instead of $1$ and $4$, add $1+\epsilon$ and $4-\epsilon$ (which will maintain all necessary ties), and then multiply by a common denominator at the end. Jan 24, 2016 at 15:56
• That seems like it would work. You don't have to keep the numbers proportional, you just have to preserve their relations. So you could just iterate through all the faces from lowest to highest and replace them with integers. Jan 25, 2016 at 7:41 | {
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organic-chemistry, aromatic-compounds
I've built a Dreiding model of the bis-allene product derived from naphthalene and it does not appear to be particularly strained. I suspect the reason we do not see this process occurring is because the starting polycyclic aromatic is highly (resonance) stabilized, whereas the bis-allene product, while not strained, has no resonance stabilization due to the lack of pi-overlap in the cyclic structure due to the orthogonality imposed by the allene units. In other words, the activation energy would be too high and the reaction would not proceed. However, if we were to pass naphthalene through a tube furnace at very high temperature (this has been done), it would be interesting to see if any of the resultant products could be explained on the basis of the bis-allene intermediate.
EDIT Here's the bicyco[4.4.0] skeleton. Using these numbers tell me where you want the 4 double and 1 triple bond.
Ah, so this (the structure on the right) is what you mean! There are 2 issues with it | {
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java, linked-list, time-limit-exceeded, insertion-sort
Please give me some help, I thought this question was easy and then I spent two days testing it. And it is quite disappointing when I got "Time limit exceed" error. The culprit is i = jn; in the last else branch.
This is not needed because you know that the entire list from head to i should be sorted and you need to continue from the old i.next.
Take the following step:
a->b->d->f-> ... y->z->c->h
^ ^
j i
this is where you do the insert which results in
a->b->c->d->f->... y->z->h
^ ^
j i
This causes you to need to repeat a full scan for every time you need to advance i. This can easily cause the time complexity to rise to O(n^3) for a reverse sorted list.
You should have.
a->b->c->d->f->... y->z->h
^ ^
j i | {
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machine-learning, neural-network, deep-learning, tensorflow, cnn
doing so, if you use the current regime, you may have lots of trainable parameters between dense layers and convolutional layers. Consequently, try to employ more convolutional layers and some pooling layers among them. Also, try to add more dense layers with more neurons in each. | {
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linked-list, kotlin
Payload (data)
Iteration/Traversal
Insertion and Removal
Your interface is too low-level. | {
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The Normal Distribution (aka "Gaussian Distribution") has a firm mathematical foundation. The Central Limit Theorem says that if you have a finite set of n independent and identically distributed random variables having a specific mean and variance, and you take the average of those random variables, the distribution of the result will converge to a Gaussian Distribution as n goes to infinity. There is no guesswork here, since the mathematical derivation leads to this specific distribution function and no other.
To put this into more tangible terms, consider a single random variable, such as flipping a fair coin (2 equally possible outcomes). The odds of getting a particular outcome is 1/2 for heads and 1/2 for tails. | {
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"url": "https://stats.stackexchange.com/questions/227034/how-did-scientists-figure-out-the-shape-of-the-normal-distribution-probability-d"
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java, parsing, exception
(Note, there's something important in there, I use double and not Double. Using primitives in Java when you can is almost always a better idea for performance than using the full Object version, like Double).
So, the above system begs forgiveness when the data is broken.... what about prevalidating the number...? The code is basically copied from the Javadoc:
private static final double parsePermission(String value) {
if (DOUBLE_RE.matcher(value).matches()) {
return Double.parseDouble(value);
}
return 0.0;
} | {
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c++, game
}
}
cout<<endl;
for(int i = 0; i < l; i++)//Loops through all Lines Based on the length of the card
{
coln = 0; pcn = 0;
string beg = (i == 0 || i == (l- 1)) ? " " : "|";//starts with " " if its the first or last line else starts with "|"
cout<<spacerb<<SetTextColor(ac[0], 4)<<beg;//spaces each line based on spacerb
for(int j = 0; j < ac.size(); j++)//Loops through the Cards
{
if(i == 0 || i == (l - 1))//if its the first or last line
cout<<SetTextColor(ac[j], (i == 0 ? 1 : 2))<<f.fillwith("", w);//fills with '-' based on w
else if(i == 1 || i == (l - 2))//if its second or second last line
{
cout<<SetTextColor(ac[j], 5)<<f2.fwcent(getColor(ac[coln]), w);//prints and centers the card color based on w | {
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electromagnetism, electrons, magnetic-moment
After an initial kick, the ball keeps in a circular motion, touching the outer bearing race, that provides the centripetal force.
After turn on a perpendicular and uniform magnetic field, the Lorentz force adds an additional centripetal force. But the inner race doesn't allow a decrease of the radius. So the ball keeps moving with the same velocity, but now touching the inner race. | {
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thermodynamics, kinetic-theory-of-gases
Since describing the motions of all atoms in a molecule also completely describes the motion of the molecule as a whole, no additional coordinates are needed. Hence the number of degrees of freedom (DOF's) in a molecule is equal to 3N, independent of the molecule's structure.
Structure does matter when it comes to vibrational and rotational DOF's, however. A linear molecule needs two angles to describe its orientation in space, while a non-linear molecule needs three. Hence these have 2 and 3 rotational DOF's, respectivey.
From this, and from the fact that a molecule as a whole has 3N DOF's, we can calculate the number of vibrational DOF's by simple subtraction:
Linear molecule:
Total DOF's = 3 N
Translational DOF's = 3
Rotational DOF's = 2
Vibrational DOF's = 3N – 5
Non-linear molecule:
Total DOF's = 3 N
Translational DOF's = 3
Rotational DOF's = 3
Vibrational DOF's = 3N – 6 | {
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, then the function can be interpolated by a polynomial of. Out of range values are returned as NaNs. KEY BENEFITS. Both options are set for a smooth term that is set with s(). Defining spline curves • Generally f(t) is a piecewise polynomial - for this lecture, the discontinuities are at the integers - e. Pre-Requisites for Spline Method of Interpolation Objectives of Spline Method of Interpolation TEXTBOOK CHAPTER : A Textbook Chapter on Spline Method of Interpolation? DIGITAL AUDIOVISUAL LECTURES : Linear Spline Interpolation: Theory [YOUTUBE 7:56]. Spline이란 것은 여러 점을 잇는 것을 뜻한다. An introduction into the theory and application of cubic splines with accompanying Matlab m-file cspline. yy = spline(x,y,xx) uses cubic spline interpolation to find yy, the values of the underlying function y at the points in the vector xx. • The requirement that it is to be a cubic spline gives us 3(n −1) equations. B = spline3eval(A,XX) // This performs the interpolation, and store the values in B. We need | {
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