text stringlengths 1 1.11k | source dict |
|---|---|
evolution, natural-selection, human-evolution, adaptation, intelligence
Title: Intelligence without natural selection? Natural selection is not the only driving force of evolution. There are other mechanisms such as genetic drift, mutations, gene flow, etc... To what extent can these different mechanisms (which don't influence adaptation to new environments as much as natural selection) evolve a more 'complex' species than when there is no natural selection?
Using a very utopian example... imagine that the first living beings that appeared in the world had access to an infinite amount of resources and space. Thus, the mechanism of natural selection would not make sense because it wouldn't be necessary. However, there could be other mechanisms that allow evolution such as genetic drift. | {
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"tags": "evolution, natural-selection, human-evolution, adaptation, intelligence",
"url": null
} |
machine-learning, deep-learning, generative-model, diffusion-models
Could you explain, please? I will suppose that you already have understood how diffusion models work. Some good resources are this blog and the DDPM paper.
If we look at Figure 3 of the paper, we see that in linear schedule the image are almost purely noise in the last quarter.
During sampling, we usually perform the same number of steps the model was trained on, but in the linear schedule the reverse process will just turn some random noise into some other random noise for 1/4 of the time. | {
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"tags": "machine-learning, deep-learning, generative-model, diffusion-models",
"url": null
} |
operators, hilbert-space, second-quantization, unitarity
Title: What unitary operator achieves Bogoliubov transformation? Let $a_1, \ldots, a_N$ be boson operators, $[a_i, a_j^\dagger ] = \delta_{ij}$. One often considers Bogoliubov transformation $a_i \to \sum_j (A_{ij} a_j + B_{ij} a_j^\dagger)$, where the matrices $A=(A_{ij}), B=(B_{ij})$ satisfy
$$AA^\dagger - BB^\dagger = I, \quad AB^T - BA^T=0,$$
to preserve the commutation relation.
Question: Can we find a unitary operator $U$ (acting on the usual bosonic Hilbert space) that satisfies $U a_i U^\dagger = \sum_j(A_{ij} a_j + B_{ij} a_j^\dagger)$?
I think finding $U$ is an entirely different issue than checking the invariance of the commutation relation. There is a general argument for the existence of the unitary operator. From the Stone-von Neumann theorem, if you know where $a$ is mapped, then there is only one unitary operator that can represent the transformation (up to a global phase). | {
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"url": null
} |
The plane is called \P^2(k) or k\P^2 (the latter notation is more
common when k is the field of real or complex numbers; please
don't use the textbook's PG(2,q)). Fix a 3-dimensional vector space
V over k. The "points" of \P^2(k) are the 1-dimensional subspaces of V
("lines through the origin", a.k.a. "nonzero vectors modulo scaling"),
and the "lines" are 2-dimensional subspaces of ("planes through the origin"),
each consisting of the "points" that are its 1-dimensional subspaces.
The design properties are then consequences of linear algebra in V:
any two distinct 1-dim. spaces are contained in a unique 2-dim. space
(their span), and any two distinct 2-dim. spaces W,W' intersect in
a 1-dim. space because
dim(W) + dim(W') = dim(W+W') + dim(W\cap W')
and here the LHS is 2+2=4 and the first term on the RHS is 3
(since W is distinct from W', the vector space sum W+W' is all of V). | {
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"url": "http://www.math.harvard.edu/~elkies/M155.15/notes.html"
} |
javascript, beginner, html, css, dom
.b147,
.b147 a {
background-color: #00C5CD
}
.r148,
r148 a {
color: #00CED1
}
.b148,
.b148 a {
background-color: #00CED1
}
.r149,
r149 a {
color: #48D1CC
}
.b149,
.b149 a {
background-color: #48D1CC
}
.r150,
r150 a {
color: #63B8FF
}
.b150,
.b150 a {
background-color: #63B8FF
}
.r151,
r151 a {
color: #00B2EE
}
.b151,
.b151 a {
background-color: #00B2EE
}
.r152,
r152 a {
color: #1E90FF
}
.b152,
.b152 a {
background-color: #1E90FF
}
.r153,
r153 a {
color: #1C86EE
}
.b153,
.b153 a {
background-color: #1C86EE
}
.r154,
r154 a {
color: #1C86EE
}
.b154,
.b154 a {
background-color: #1C86EE
}
.r155,
r155 a {
color: #1874CD
} | {
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"tags": "javascript, beginner, html, css, dom",
"url": null
} |
c#, performance, algorithm, recursion, combinatorics
Looking at Knuth's downloadable programs starting with DLX I see there are some new variants since I last read the draft of fascicle 4C of the Art of Computer Programming. I strongly recommend you download that and dedicate a good 10 hours to reading and understanding it: it's a very good technique for many kinds of puzzle. | {
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"tags": "c#, performance, algorithm, recursion, combinatorics",
"url": null
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thermodynamics, differentiation
One final remark is that people are lazy with regards to introducing new letters for new functions (especially in the context of chain rule), so that they'll use a letter like $V$ to mean both an independent variable and also to mean a function, or the same letter $U$ can mean two different functions. This double usage can be very confusing if you're unaware that this is being done... the only way around this is to practice writing statements in absolutely crystal clear unambiguous notation and then see how this relates to the more common statements. | {
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"tags": "thermodynamics, differentiation",
"url": null
} |
java, javafx
<Button layoutX="495.0" layoutY="256.0" mnemonicParsing="false" onMouseClicked="#loginClick" text="Login" />
<Label layoutX="45.0" layoutY="113.0" styleClass="content" text="Welcome to TeleMart's Enterprise Resource Planning System. Please enter your login credentials on the right so that we could verify your identity. Also, please do not share your login credentials with anyone or use another person's login credentials." textAlignment="CENTER" wrapText="true" />
<Label layoutX="220.0" layoutY="309.0" styleClass="content" text="© 2015, Hassan Althaf." />
</children>
</AnchorPane> | {
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awka. DERIVATION OF THE LOCAL STIFFNESS MATRIX CONSIDERING TIMOSHENKO BEAM THEORY (TBT) 1. The variable cross-section members have been widely used in engineering practice for many years,thus it is necessary to investigate their element stiffness matrixes. In matrix Gla protein–deficient mice, a model of human vascular calcification, mice lacking HDAC9 had a 40% reduction in aortic calcification and improved survival. In this paper, the special attention is dedicated to 3D frame analysis and enhancements of the basic beam element used for modelling. Element and System Coordinates for a Beam Element The DOFs corresponding to the element x’ (axial) and y’ (shear) axes are transformed into components in the system coordinates X and Y in a similar manner as for truss elements. The element stiffness matrix for a beam element is given by. The two quantities are related by a stiffness matrix,. the stiffness should be added to rows and columns 1 and 7. eliminate the strain term and develop the | {
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"tags": null,
"url": "http://xxqn.from.bz/stiffness-matrix-for-beam.html"
} |
beginner, c
int main(void) {
Table table;
load_data(file, &table);
// and later
for (int i = 0; i < table.rows; i++) {
free(table.arr[i]);
}
free(table.arr);
}
void load_data(char* fname, Table *table) {
// read through the file once to get the number and length of strings, then
int **arr = malloc(sizeof(*arr) * table->rows);
for (int i = 0; i < table->rows; i++) {
arr[i] = calloc(table->cols, sizeof(int));
}
table->arr = arr;
// read data
} | {
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1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0
2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)
Prove that |f '(x
)|<=M/2
Use the Taylor series with remainder $\displaystyle f(x+t) = f(x) + tf'(x) + \tfrac12t^2f''(\xi)$, where $\displaystyle \xi$ lies between x and x+t. This implies that $\displaystyle |tf'(x) +f(x) - f(x+t)| = |\tfrac12t^2f''(\xi)|\leqslant\tfrac12Mt^2$. There are now two separate cases to consider.
If f(x) and f'(x) have the same sign, take $\displaystyle t = 1-x$. Then $\displaystyle |tf'(x) +f(x)| = |(1-x)f'(x) +f(x)| \geqslant |(1-x)f'(x)|$. Also, $\displaystyle f(x+t) = f(1) = 0$. So the inequality in the previous paragraph tells us that $\displaystyle |(1-x)f'(x)| \leqslant \tfrac12M(1-x)^2$. Hence $\displaystyle |f'(x)|\leqslant \tfrac12M(1-x)\leqslant \tfrac12M$. | {
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"url": "http://mathhelpforum.com/calculus/132034-taylor.html"
} |
java, design-patterns, jpa, jsf
MyQueue:
package dur.facelets;
import dur.database.Clients;
import dur.database.ClientsQueryBean;
import java.io.Serializable;
import java.util.logging.Logger;
import javax.ejb.EJB;
import javax.inject.Named;
import javax.ejb.Singleton;
import javax.enterprise.context.ApplicationScoped;
//import javax.inject.Singleton;
@Named
@ApplicationScoped
@Singleton
public class MyQueue implements Serializable {
@EJB
private ClientsQueryBean clientsQueryBean;
private static final long serialVersionUID = 1L;
private final Logger log = Logger.getLogger(MyQueue.class.getName());
private int next = 1001;
public MyQueue() {
}
public int getNext() {
log.info("next\t" + next);
Clients c = clientsQueryBean.getClientById(next);
log.info(c.getName());
return next++;
}
}
ClientsQueryBean:
package dur.database; | {
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"tags": "java, design-patterns, jpa, jsf",
"url": null
} |
physical-chemistry
Title: Why water expands when freezes? I'm sure this is for most of you a basic question, but it really puzzles me:
How it is that, even though all materials expand as they get warmer, and contract (maybe these are not the correct terms) when get colder, water exapands when freezes.
Thanks a lot. The expansion upon freezing comes from the fact that water crystallizes into an open hexagonal form. This hexagonal lattice contains more space than the liquid state. | {
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homework-and-exercises, astrophysics, stars, stellar-physics, jupiter
Title: Let's make Jupiter a star It is known that Jupiter is mostly made of hydrogen, but that it is not massive enough to start nuclear fusion. In other words, Jupiter is not a star, but could be a star if someone added hydrogen to the planet.
How can the critical mass, where there is sufficient thermal energy at the core to start nuclear fusion, be calculated? Yes, your logic is correct, if we kept adding hydrogen to jupiter, it could eventually become a star.
See: http://www-star.st-and.ac.uk/~kw25/teaching/stars/STRUC5.pdf
So long as you understand the basic thermodynamics in there then you will be able to follow through to the end, if not, and you are not interested, then the critical mass for collapse into a star is approximately 0.08 * (Mass of Sun). | {
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atoms, identical-particles
Are atoms unique?
No.
Do atoms have any uniquely identifying characteristic besides their history?
Their history of a particle does not affect it*. No particles are unique. Atoms may have isotopes or spin to identify one from another, but these are not unique from another particle with the same properties.
would it contain information with which we could positively identify that they two are the same?
Yes only because we could positively identify that this carbon atom is the same as almost every other carbon atom in existence.
*Unless it does, in which case it may be considered a different particle with different properties. | {
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} |
java, android
Use the enum Values
My heaviest recommendation is to use the predefined enum values that you're assigning so that you can prevent the switch structure and clean this code up even more:
Color[] cardColors = new Color[] { Color.yellow, Color.green, Color.red };
List<Card> cards = new ArrayList<Card>();
cards.add(new Card(0, 0, "Card 1", "London", Color.black));
cards.add(new Card(0, 0, "Card 2", "Paris", Color.black));
cards.add(new Card(0, 0, "Card 3", "Dubai", Color.black));
for (int i = 0; i < cards.size(); i++) {
Card c = cards.get(i);
c.backgroundColor = cardColors[i];
applyForeColor(c, cardColors[i]);
System.out.print(c.displayName + ": " +
c.backgroundColor + ", " +
c.foreColor + "\n");
} | {
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quantum-field-theory, quantum-information, fermions, bosons, grassmann-numbers
where for a spinor or vector field there are additional factors in there that don't play a role here. The crucial question is whether the objects $a$ and $a^\dagger$ (and likewise $b$ and $b^\dagger$) commute or anticommute, since they are the creation and annihilation operators for the particle (resp. antiparticle in the case of $b$) associated to the field.
If they commute, the corresponding particle is a boson - you can pile up arbitrarily many particles into the same state just by applying the creation operator many times. Then, the operators are bosonic modes. If they anticommute, the corresponding particle is a fermion - applying the creation operator twice just gives zero, so you can't ever have more than one particle in the same state. Here, the operators are fermionic modes. They are "modes" because the Fourier transform splits up a classical oscillation into its pure-frequency modes, so by analogy, we also call the objects it produces when applied to a quanutm field "modes". | {
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mangoes = (5+1) = 6 ways. What do we do? We put different subscripts next to the different occurrences: $B_1 UB_2B_3LE$. Find the expected number of complaints per day. Bell Labs alone produced the transistor (there are several billion of them in your pocket), the cell phone, the solar cell, the laser, Telstar I, digital signal processing, and the infrastructure of most contemporary systems programming. An ordinal number is a number that indicates position or order in relation to other numbers: first, second, third, and so on. Achieveressays. (E) Nonclassicality for combinations of initial-final photon arrangements captured by the suppression laws associated with Σ 1, Σ 2, and Σ 3. 6 billion digits, but this one has 10 quadrillion digits! To put that in perspective, the digits of the previous number would be able to fit in about 358 dictionaries (assuming each page can fit 10,000 digits and each dictionary has 1000 pages), but to store the digits of this number we'll need a billion | {
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} |
special-relativity, inertial-frames, time-dilation, observers
Does it mean if the rocket were forever in the uniform motion going away from you, you will be always older than your friend, and friend will be always older than you depending on which frame of reference you take? In relativity time is no longer a universal concept, it is a quantity specific to a frame of reference. It isn't meaningful to compare the "age" of two objects in two different frames of reference using a single "frame time." "Frame time" denotes time as measured in a specific frame of reference. This frame of reference could be that of either object for example or perhaps even a third party observer. There is one way to compare ages of objects that's meaningful though but it requires a different concept of time to be measured. Proper time is the time measured in a frame of reference where the observer measures him or herself to be at rest. This definition is rather bizarre since it doesn't allow arbitrary points in space to have a proper time. Only observers with a defined | {
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} |
rotational-dynamics, equilibrium, statics
As you can see, I have considered the perpendicular components (with respect to the see-saw) of $F_1$ and $F_2$, and used them to find the torques while being tilted at an angle $\theta$. Since the tilted system is considered to be at equilibrium, the two torques (one clockwise and the other anticlockwise) must be equal. But, the $cos(\theta)$ cancels out and we arrive at a contradiction. Where is the problem?
I have found a similar (but not same) version of my problem at: Angular equilibrium on a see-saw
However, the setup of that system is a bit different. There, the pivot (center of gravity of the see-saw) is suspended and can rotate about a fixed point. The answer to that post seems reasonable to me, but my situation here is not exactly the same as that.
Any solutions or insights into my problem will be much appreciated. Thanks!
Suppose that the system is not initially balanced at the horizontal
position. We want to find out the angle at which this system rotates | {
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general-relativity, special-relativity, differential-geometry, tensor-calculus, curvature
$R^a{}_{bcd}=\Lambda^{a}{}_{\mu}\Lambda^{\nu}{}_{b}\Lambda^{\sigma}{}_{c}\Lambda^{\zeta}{}_{d}R^\mu{}_{\nu\sigma\zeta}$.
Is this the right way to do it? A Lorentz transformation is a special kind of coordinate transformation. Any tensor $T$ is invariant under coordinate transformations. What this means is that, once you decompose $T$ in terms of some basis of your (co)tangent space, say
$$
T = \sum_{I}T^I{\bf e}_I,
$$
the components $T^I$ change under a (linear) coordinate transformation ${\bf e}_I \mapsto {\bf e}_{\hat{I}} = \sum_{I}\Lambda_{\hat{I}}{}^{I} {\bf e}_I$ in order to keep $T$ as a whole invariant. I have used hats here to indicate the transformed frame. In the particular case of the Riemann tensor ${\rm Riem}$ at a given point $p$, we have
$$
{\rm Riem} = R^\rho{}_{\sigma\mu\nu} {\bf e}_\rho \otimes {\bf e}^\sigma \otimes {\bf e}^\mu \otimes {\bf e}^\nu,
$$ | {
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} |
homework-and-exercises, electrostatics, electric-fields, symmetry, gauss-law
Note that this argument only works because the slab is infinite and symmetrical in the $XZ$ plane; hence the phrase "by symmetry". | {
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} |
electromagnetism, special-relativity
$$F' = -\frac{2k_E \lambda_1 \lambda_2}{\gamma^2 r} $$
Then I get the correct result. However, I don't see why that would be the case, as I believe my charge density transformation to be correct. Can anyone help me find where I have gone wrong? Thanks in advance! A possible error in your calculations is that you wrote $\lambda_k^{'} = \gamma \lambda_k$, whereas you had better write $\lambda_k^{'} = \alpha \lambda_k$, where $\alpha=1/\gamma$. If $\lambda_k$ is the charge density of the moving charges/wire, the distances between the charges are Lorentz contracted and the density is great. Moreover, from the viewpoint of an observer who is at rest WRT the charges, the length between the charges is the proper length, which is greater than the Lorentz contracted one, and thus the density reduces. | {
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java, android, error-handling, guava
This doesn't do what you write, as the first time retry=0.
I take care to not retry after InterruptedException & UnknownHostException.
The latter may be retryable as it may be a temporary network or DNS problem. But it's your choice.
final String regId = autoRetry(new DoWork<String>() {...
I'm too lazy to copy it in my Eclipse and without it, it's unreadable as I have no idea where the expressions ends. What about a private final DoWork<String> regIdDoWork = ... or alike?
Optional.<Predicate<Boolean>>absent()).orNull(); //No need of predicate hence send absent
How nicer would the world be if there was no Optional? :D Just null.
Note that even if you like Optional, you should probably use it for as return types only. This is the place where it may prevent bugs and where it doesn't add so much verbosity. | {
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gas-laws, food-chemistry
Title: Which gas makes Orange juice bottles swell? I forgot a bottle of 1L Orange Juice (100% orange juice with pulp) in the fridge for about a week, the bottle was previously open so air got inside as well.
After a week the bottle got inflated and when I opened it I could hear the gas coming out. No odours, orange juice still the same colour (not sure if the same taste, I threw it away).
Which gas was produced and how did it happen? The air that got in seems to have played a part on it (these bottles stay in the supermarket closed for a long time without getting inflated). I can't imagine the gas was anything except $\ce{CO2}$ formed by the action of microorganisms contaminating the orange juice. They degrade the organic components of the juice to power their metabolisms, and produce carbon dioxide as a byproduct. | {
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ros
Thank you
Comment by Abbas on 2016-10-30:
Hi ahendrix,
I installed the driver from https://github.com/ros-drivers/driver_common (for indigo). Then, catkin_make was ran without an error.
Thank you,
Abbas | {
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space, galaxies, popular-science
You can propose that their whole town/city, easily consisting of 1,000,000 people, each take a single car, and start driving. These are of course special cars; they do not stop, they do not need to refuel; they just go and drive on for 24 hours a day, 7 days a week. Now, the combined distance all 1,000,000 cars travel in 8.3 years, is one light year.
There is still room for improvement, as most people have never counted to 1,000,000 and therefore have no intuition for how much a million truly is. You can break that down again and come to your final story outline: | {
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complexity-theory, logic, propositional-logic
My question is: Is the length and size of a formula the same? There are two possible ways of defining what the length or size of the formula is:
According to the size of a reasonable (and fixed!) binary encoding.
According to the size of a reasonable (and fixed!) encoding over an infinite alphabet which includes symbols for all possible variables. | {
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compressed-sensing
Title: Proof of non-existence of the universal archiver? Does anybody knows a proof that no algorithm $A$ exists that can reversibly transform every possible finite sequence $S$ to the sequence $C$ of smaller size?
Here I assume $S$ and $C$ to be a finite bit sequences (or more generally some finite sequences of elements from certain finite set), algorithm should be executed in the finite time for each sequence S and use finite memory. The same constraints applies for the reverse algorithm $A^{-1}$ - it should consume finite memory and "unpack" certan sequence in the finite time.
I guess such a proof would be trivial one, but I forgot how the formal proof is done. Assume there is a program that maps every sequence of $n$ bits to a sequence of $n-1$ bits. There are $2^n$ sequences with $n$ bits, but only $2^{n-1}$ sequences with $n-1$ bits. Hence there are two sequences $S,S'$ that get mapped to the same sequence $C$. Therefore there can be no algorithm that reverses the transformation. | {
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javascript, asp.net, url
Avoid throwing
You are throwing an error at entry yet ignore the very same as you recurse. Why is the error not handled before you call the function. Having the error there means you need to wrap the call in a try catch. Are you unsure of the object you are passing, Where is it coming from and why would it not be as expected? Surely if its not as expected you would expect an empty array returned, rather than having to deal with an error.
If you must throw dont throw strings. See points below.
Be efficient
It cost money to run code so always be efficient
It is more efficient to close over a recursive function with objects that are shared. You create an new array urlPairs each time you recure. Then on returning you create yet another array when you concat the result. You can eliminate that overhead using closure. | {
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organic-chemistry, photochemistry, radicals, electromagnetic-radiation
Title: Why can UV light initiate a reaction between hydrogen and chlorine gas? Can someone explain me how does UV light help combine chloride gas and hydrogen to produce hydrochloric acid?
$$\ce{Cl2(g) + H2(g) -> 2HCl(g)}$$ Before going into the mechanism of this reaction, I suggest you look up free radical mechanism, as this reaction takes place through that.
$\ce{Cl-Cl}$ bond in $\ce{Cl2}$ is weak enough to be broken by mere UV rays (present in sunlight), and hence they undergo homolytic cleavage (the resultant products are $\ce{Cl}$ atoms, not ions) to form two $\ce{Cl}$ free radicals. Now these $\ce{Cl}$ free radicals are extremely reactive (due to one free electron that it can share in a covalent bond) and hence attack hydrogen gas. | {
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quantum-chemistry, computational-chemistry, intermolecular-forces, software
As you can see, using SAPT energies provides a much better fit than naively using the total interaction or binding energies. I'm not sure why the total SAPT energy works so well, but it may be due to the absolute magnitudes at each point being so small.
The importance of only fitting exchange and dispersion is obvious for $\ce{Na^+---Cl^-}$ (using def2-SVP for the basis).
This is an unfair example, since the dispersion interaction is dwarfed by exchange, and it is an ionically-bound molecule. As a final example, consider $\ce{V(III)---Cl^-}$ (using def2-SV(P) for the basis, vanadium as a quintet). | {
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object-oriented, game, c++17, cmake
// Board editing utility
void AddPiece(Coordinate coordinate, Color color);
void InvertCell(Coordinate coordinate);
// Getters & commonly used board state
Color GetCellColor(Coordinate coordinate) const;
int GetBoardSize() const;
bool IsGameOver() const;
bool CanMove(Player* player) const;
bool IsValidMove(Move move) const;
bool TurnValid() const;
std::vector<Coordinate> GetTilesFlipped(Move move) const;
// Active Player Methods
Color GetActiveColor() const;
std::string GetActiveName() const;
void ChangeTurn();
private:
Player* black_;
Player* white_;
std::vector<std::vector<Cell>> board_;
int board_size_;
Player* active_player_;
bool Inbounds(int coordinate) const;
};
}
#endif //OTHELLO_MODEL_H
Model.cpp
#include "Model.h"
#include <utility>
Othello::Model::Model() : board_(0)
{} | {
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zoology, physiology, brain, ethology, behaviour
Rogers, L. J. & Andrew, R. J. (Eds) (2002). Comparative Vertebrate Lateralization. Cambridge: Cambridge University Press.
Roth, E. D. (2003). ‘Handedness’ in snakes? Lateralization of coiling behaviour in a cottonmouth, Agkistrodon piscivorus leucostoma, population. Animal behaviour, 66(2), 337-341.
Shine, R., Olsson, M. M., LeMaster, M. P., Moore, I. T., & Mason, R. T. (2000). Are snakes right-handed? Asymmetry in hemipenis size and usage in gartersnakes (Thamnophis sirtalis). Behavioral Ecology, 11(4), 411-415.
Sovrano, V. A., Rainoldi, C., Bisazza, A. & Vallortigara, G. (1999). Roots of brain specializations: preferential left-eye use during mirror-image inspection in six species of teleost fish. Behavioural Brain Research, 106, 175–180.
Sovrano, V. A., Bisazza, A. & Vallortigara, G. (2001). Lateralization of response to social stimuli in fishes: a comparison between different methods and species. Physiology & Behavior, 74, 237– 244. | {
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ros, ros2, release
Title: Can nodes from different ROS 2 distributions communicate compatibly?
Do all nodes passing messages need to be from the same ROS 2 release to reliably communicate? Can a node compiled for and running say the Bouncy release on one machine safely send messages to a node compiled and running say the Dashing release on another machine? Do releases reserve the right to break compatibility so it would be unwise to mix releases running in one system? Or do the releases remain compatible with each other as long as they are communicating over DDS? | {
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How and why vector spaces are defined?
Is it true that vector spaces are defined to check if system of linear equations is solvable or not?
Explanation: Goal is to solve system of linear equations.
In matrix form: $Ax = b$. As $A = [C_1 \; C_2 \; ... \; C_n]$, where $C_n$ is a column and $x = [x_1 \; x_2 \; ... \; x_n]$.
Therefore, $C_1x_1 + C_2x_2 + ....+ C_nx_n = b$. Linear combination of column vectors produce vector $b$.
Because of above statement (linear combination) we choose a set of vectors that have closure under addition and scalar multiplication (closure under linear combination) and call that set of vectors a vector space. Now, if vector $b$ lies in that set of vectors (vector space) then only system of linear equations is solvable. | {
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When $n \gt 2,$ there is at most one point $p \in P$ such that $P \subseteq S(p).$ Therefore, we can sum the individual probabilities for each point $p \in P$ to obtain the probability that there exists any $p \in P$ such that $P \subseteq S(p).$ This sum is $n(1/2)^{n-1}.$ Therefore, the probability that all $n \gt 2$ points lie on the same semicircle is $n(1/2)^{n-1}.$
Combining cases
Note that $n(1/2)^{n-1} = 1$ when $n$ is either $1$ or $2.$ That is the same answer obtained in Case 1. Therefore, the expression from Case 2 gives the correct result for any positive integer $n,$ and the probability that all $n$ points lie on the same semicircle is $n(1/2)^{n-1}.$
### Solution, Question 2
We make an assumption that $\phi\lt\pi.$
For a point $X\in(C),$ let $\Phi(X)$ be the arc subtending angle $\phi$ that extends counterclockwise from $X,$ inclusive of $X.$ The probability of a point uniformly distributed on $(C)$ to fall into $\Phi(X)$ equals $\displaystyle \frac{\phi}{2\pi}.$ | {
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deep-learning, cnn, overfitting, convolutional-neural-network
This thread is a great resource on the difference between the two questions.
Let me know if it works.
EDIT: Further explanation
Although we normally talk about “binary classification”, the way the outcome is usually modeled is as a Bernoulli random variable, conditioned on the input data. So:
$$P(y = 1|\mathbf{x}) = p, \ 0\leq p\leq1$$
A Bernoulli random variable takes on values between 0 and 1. So that’s what our network should produce. One idea might be to just clip all values of $wth+b$ outside that interval. But if we do this, the gradient in these regions will be 0: The network cannot learn.
A better way is to squish the complete incoming interval into the range (0,1), using the sigmoid function:
$$\sigma(x) = \frac{1}{1 + e^{(-x)}}$$ | {
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## Does Pointwise Convergence to a Continuous Function Imply Uniform Convergence?
Recently in my advanced calculus class we discussed how the uniform limit of a sequence of continuous functions is itself continuous. One of my students turned the question around: If the limit of a pointwise sequence of continuous functions is continuous, does that mean that the convergence is uniform?
The answer is “No,” although I couldn’t think of a counterexample off the top of my head. I found one (not surprisingly) on Math.SE. Although there are some good discussion and multiple good answers to this question given there, I’m going to discuss Ilmari Karonen’s answer.
Let $\displaystyle f_n: [0,1] \mapsto \mathbb{R}$ be given by
$\displaystyle f_n(x) = \begin{cases} nx, & 0 \leq x \leq 1/n; \\ 2 - nx, & 1/n < x \leq 2/n; \\ 0, & \text{ otherwise}. \end{cases}$ | {
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Staff - 1 year, 8 months ago
Thus the payoffs are same, so i can choose either of the payoff be it payoff A or Payoff B
- 1 year, 8 months ago
So, what you're saying is that between
Payoff C: Get $0 Payoff D: 50% chance to get$1,000, 50% chance to lose $1,000 Because the expected payoff for both is$0, you are equally happy to choose either?
Staff - 1 year, 8 months ago
we need to look at different aspects of payoff like standard deviation etc but I haven't thought about it yet..
I think I need to look at the variance of values in payoff 1 and payoff 2. Then if the expected value is same then we will go for the one which is less riskier.
- 1 year, 8 months ago
Great. Proceed from there, and then think about what matters to you.
Staff - 1 year, 8 months ago | {
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water-resources
Title: Is it possible to create a sub-surface pond? Has anyone designed a pond that extends beyond its surface perimeter?
Example cross section:
brown - existing surface (or brought to grade)
black - designed pond
blue - capacity I would describe this negative incline portion of the pond as a "shelf". They are common and need to be self-supporting with a material that has compression and tensile strength; concrete and rebar, rock, fiberglass, etc.
There are lots of examples of this:
covered irrigation canals
dam water turbine intakes
water diversion intakes
zoo habitats
natural shelf under waterfalls
naturally in water caves | {
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fluid-dynamics, pressure, flow, bernoulli-equation
At a more exacting level, the particles are colliding a great many times every second. If there is a substantial pressure difference, such as when one suddenly opens a valve at the bottom of a tank, the fluid particles at the leading edge of this suddenly find more collisions from behind than in front, and that imparts a net movement. This information is "relayed" back to the molecules further into the tank in the form of the average of a bunch of these collisions.
What we end up seeing is that there is a relationship between pressure and velocity, based on continuity It isn't inherently causal in one direction or the other. The causality stems from parts of the problem which provide a limiting factor. For example, in a long enough tube, drag limits velocity, so velocity drives pressure. In an air cannon, the amount of pressurized gas provides a major limitation, so we often talk of the pressure driving the velocity. | {
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mathematical-physics, integration
Title: When is Lebesgue integration useful over Riemann integration in physics? Riemann integration is fine for physics in general because the functions dealt with tend to be differentiable and well behaved. Despite this, it's possible that Lebesgue integration can be more powerfully used even in physical situations that can be solved by Riemann integration. So my questions is the following:
In solving physics problems, when is Lebesgue integration useful over Riemann integration? An important example in quantum mechanics is e.g. the Hilbert space
$$H~=~L^2(\mathbb{R}^3)$$
of Lebesgue square integrable wave functions $\psi$ in the position space $\mathbb{R}^3$. The Lebesgue square integrable functions (as opposed to just the Riemann square integrable functions) are needed to complete the Hilbert space with respect to the square norm
$$||\psi||_2~:=~\sqrt{\int d^3x ~ |\psi(x)|^2}.$$
Concerning completeness, see also this Phys.SE post. | {
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VR1 is the voltage across resistor R1.
Thus, using Kirchhoff's Voltage Law, we can write
V1 = VR1 + VLED1 + VLED2
#### WBahn
Joined Mar 31, 2012
28,182
My question is whether in this circuit and when using V = IR, is Vtotal = Vsupply - (LED1fv + LED2fv) or is V just the sum of LED1fv + LED2fv? I'm pretty sure its Vsupply - (LED1fv + LED2fv) but I am getting hung up on the difference between the two. Also is Power calculation using the voltage source (Vsupply - Vforward voltages = LED1fv + LED2fv).
When in doubt, rely on the fundamentals.
For Ohm's Law, you need the voltage across THAT resistor, the current through THAT resistor, and the resistance of THAT resistor.
Kirchhoff's Voltage Law says that the some of the voltage drops around the circuit is zero, so start there:
Vs - Vr - Vf1 - Vf2 = 0
So the voltage across the resistor is
Vr = Vs - Vf1 - Vf2 = Vs - (Vf1 + Vf2)
Then plug this into Ohm's Law:
Vr = Ir · R
Vs - (Vf1 + Vf2) = Ir · R | {
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general-relativity, gravity, metric-tensor
Because it seems to be going straight against the fundamental motives and principles of GR.
The motive for GR was for large masses and energies. When masses and energies are low Newtonian mechanics is sufficient to fit the data.
Or, is it because this view allows us to introduce gravitons and helps a bit in the unification of forces?
Unification of forces happens in the microcosm of particles, which uses dimensionsan immense order of magnitude away from general relativity requirements. It will be enough if one can show that the QFT model of gravity melds with the general relativity requirements at large masses and energies.But this has not much to do with the concept of force=dp/dt and its universal applicability classically and in QFT. | {
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c#, performance, multithreading, compression
decompressor.LinkTo(outputWriter, new DataflowLinkOptions { PropagateCompletion = true });
...
decompressor.Post(new InputBlock(compressedData, uncompressedBlockSize));
Post data to the decompressor, it will process it in parallel and pass it on to the outputWriter in the same order as it was posted.
This needs a synchronous Decompress:
public static byte[] Decompress(byte[] inputBytes, long decompressedSize)
{
var compressed = new MemoryStream(inputBytes);
var decoder = new Decoder();
var properties2 = new byte[5];
if (compressed.Read(properties2, 0, 5) != 5)
{
throw (new Exception("input .lzma is too short"));
}
decoder.SetDecoderProperties(properties2);
var compressedSize = compressed.Length - compressed.Position;
var decompressed = new MemoryStream();
decoder.Code(compressed, decompressed, compressedSize, decompressedSize, null);
if (decompressed.Length != decompressedSize)
throw new Exception("Decompression Error"); | {
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"openwebmath_score": null,
"tags": "c#, performance, multithreading, compression",
"url": null
} |
performance, c, security, cryptography
Usability
I'm not a security expert, so I'm probably not qualified to comment much on that. But one thing I do notice is that you aren't guaranteeing the user that the password won't have repeating numbers or letters. It's common for password verifiers on websites to disallow any more than 2 consecutive instances of the same character. It might be worth checking to make sure you don't have 3 or more in a row if that's a concern for the types of websites you use. | {
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sampling, nyquist, reconstruction
Title: Random (Over) Sampling signal and perfect Reconstruction in Nyquist form? Imagine we have band limited signal with bandwidth of $B$, so the required Nyquist rate would be $f_{nyq}>2B$ that is oversampled with rates $f_s$ where $f_s = M*f_{nyq}$ and $M$ is random and $M>=1$. By that, I mean the sampling rate is at least at Nyquist rate but most of the time its way more than Nyquist and it is random. I have heard and read about iterative My questions are:
Is there a method to construct the signal as if it was taken in Nyquist more, without degradation of signal quality? After all, we have more samples than needed, so I do not expect any degradation, right?
Is there are some loss, how much is it, what factors it depends on?
Is there a toolbox or something that I can feed my data in and have some experiments of my own? ,
4.Is the technique used in MATLAB lossy? | {
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"tags": "sampling, nyquist, reconstruction",
"url": null
} |
java, javafx
public class Fifty extends Application {
private String filePath = null;
private TextArea text;
public void init() throws Exception {
super.init();
Parameters parameters = getParameters();
List<String> unnamedParams = parameters.getUnnamed();
if (unnamedParams.size() > 0) {
filePath = unnamedParams.get(0);
}
}
public void start(Stage stage) throws Exception {
SplitPane splitPane = new SplitPane();
text = new TextArea();
final WebView web = new WebView();
try {
text.setText(readFile(filePath, Charset.defaultCharset()));
web.getEngine().loadContent(text.getText());
} catch (Exception e) {
alertException(e);
}
text.setStyle("-fx-font-family: monospace");
text.textProperty().addListener(new ChangeListener<String>() {
@Override
public void changed(final ObservableValue<? extends String> observable, | {
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angular-momentum, mass, rotation
Title: Rate of angular momentum at the center of mass I'm currently trying to calculate the Zero Moment Point for a game I develop, however im terrible at physics and therefore have trouble calculating the “rate of angular momentum at the center of mass”.
As I understood the angular momentum is defined as:
$\qquad \qquad L =$ distance from the center of rotation $\times$ mass $\times$ velocity
The mass of my object is $10$ kg.
It's velocity is $(3, 4, -1)$.
However, I don't get how to calculate the distance from the center of rotation, since my object has no fixed point its rotating around.
I'm referencing the following article, where I want to calculate HG:
Zero moment point.
Thanks for any help in advance! A particle of mass $m$ located at $\vec{r} = \pmatrix{x & y & z}$ from the origin, and having velocity $\vec{v} = \pmatrix{ vx & vy & vz} $ has the following properties | {
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python, configuration
So don't think how you will make the code work with configuration, but how will you make configuration work with code if that makes sense. The way you can go about reading configuration is basically endless and the structure of the configuration could also be independent of your app.
Edit:
Here is the example:
def read_toml_config(path):
def read(path):
with open(self.file, 'r') as f:
return toml.load(f) | {
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ros, velodyne-pointcloud, velodyne-driver, velodyne, transform
Here's my question: Is there any advantage to transforming point clouds from individual VelodynePackets (in Case 2), as opposed to the full point cloud (Case 1)? When would I use one over the other?
Thanks!
Originally posted by Venkat Ganesh on ROS Answers with karma: 57 on 2016-05-31
Post score: 1
The main use of transform_node is when the device is moving fast enough for the transforms to change significantly during the 0.1 second required for a full revolution (assuming 600 RPM). This is especially important if the heading changes.
Originally posted by joq with karma: 25443 on 2016-06-01
This answer was ACCEPTED on the original site
Post score: 1 | {
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"tags": "ros, velodyne-pointcloud, velodyne-driver, velodyne, transform",
"url": null
} |
c++, c++11, multithreading, concurrency, queue
void IncrementPositionAfterRead()
{
if(isEmpty[m_read_position_] == false)
{
isEmpty[m_read_position_] = true;
if(m_read_position_ == m_queue_capacity_ - 1)
{
m_read_position_ = 0;
}
else
{
m_read_position_++;
}
}
}
CircularPtrQueue& operator=(const CircularPtrQueue&) = delete;
CircularPtrQueue(const CircularPtrQueue&) = delete;
~CircularPtrQueue()
{
for(size_t i = 0; i < m_queue_capacity_; i++)
{
delete m_queue_buffer_[i];
}
}
};
#endif
And here is the test file:
#include "circularptrqueue.h"
#include <tuple>
#include <iostream>
#include <thread>
#include <chrono>
#define QUEUE_SIZE 100000
#define NUM_CONSUMERS 1
struct data
{
long int timestamp;
std::tuple<long int, int, unsigned short int> glue;
}; | {
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} |
Distance, Rate, and Time. Problems: In order to receive credit for this assignment you MUST show your work. Also, the answer key and explanations are given for the same. Time is usually denoted by t in equations. Sign in now Join now More. Difficult Time, Speed & Distance Question - 5 Q5. Distance-Time Graphs Worksheet Speed Problems. 00: 00: 00: hr min sec; Challenge Stage 1 of 4 Get 3 correct in a row . How much time does he take to walk a distance of 20 km? Solution. Lesson Summary Distance is the amount of space between two things or people, measured as a continuous line. Example 1. It travels for 8 seconds. Let d, s and t denote distance, speed and time respectively. Calculate Time from Distance and Speed. Practice Questions on Time, Speed & Distance. Draw a distance vs. A 11 minute read. "Time required to cover the distance" = "Speed"/"Distance" = 1000/5 = 200 sec = 3 min 20 sec City A is 620 km apart from city B. One entity usually starts after the other entity and travels at a | {
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"lm_q1_score": 0.9845754506337407,
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"openwebmath_score": 0.5783324241638184,
"tags": null,
"url": "http://carlaandme.dk/atgando/wt2jtp2.php?gakoombkk=speed-distance-time-problems"
} |
javascript, object-oriented, d3.js
{country: 'DE', year: '2000', rate: '0.1'},
{country: 'DE', year: '1999', rate: '0'},
{country: 'DE', year: '1998', rate: ''},
{country: 'DE', year: '1997', rate: '813'},
{country: 'DE', year: '1996', rate: '222'},
{country: 'DE', year: '1995', rate: '181'},
{country: 'DE', year: '1994', rate: '11'},
{country: 'DE', year: '1993', rate: '2.3'},
{country: 'DE', year: '1992', rate: '1'},
{country: 'DE', year: '1991', rate: '0'},
{country: 'DE', year: '1990', rate: 'NaN'}*/
]; | {
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"tags": "javascript, object-oriented, d3.js",
"url": null
} |
zoology, entomology, ant, lifespan, collective-behaviour
Within eusocial colonies, there is often a protective environment (Buffenstein & Jarvis 2002): there is usually a physical structure (nest or burrow); symbiotic bacteria and/or fungi creating a more hygienic microflora; and queens are also protected by other castes. This gives queens a lower incidence of death due to accident or attack which (from the reasoning in the previous paragraph) supports the selection of a longer reproductive life. This is likely to lead to longer lived workers since they share the same genes. | {
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"tags": "zoology, entomology, ant, lifespan, collective-behaviour",
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to be. If A and B are dependent events, then the probability of. Thus, Equations 1. You can only calculate the probability if you know that the occurence of one event is independent from the occurence of the other event. Let A be the event “the sum of the points is 7”, B the event “die #1. We see that two events A and B are either both dependent or independent one from the other. gl/9WZjCW If A, B are two independent events, show that bar A and bar B are also independent. Event A: Spinning an odd number on the first spinner. The probability of one event does not change the probability of the other event. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. We offer free resources including Writing and Teaching Writing, Research, Grammar and. Suppose you select a number at random from the set {90, 91, 92,. Definition: Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. The outcome of the first roll does not | {
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"url": "http://incommunity.it/xyhy/a-and-b-are-two-independent-events.html"
} |
classical-mechanics, lagrangian-formalism, conservation-laws, symmetry, noethers-theorem
Now, bonus question. Is there any profound reason why these invariants are all proportional to mass? And proportional to powers of velocity? Within the Newtonian framework of mechanics conservation laws are tricky to develop and are not obvious at first glance. Lagrangian mechanics generalises the concept of conservation laws by exploiting "symmetries". The connection between symmetries and conservation laws is made by Noether's theorem.
An object has a symmetry if it is invariant under a transformation. The transformation could be discrete or continuous, local or global and the object could be the action, Lagrangian, equations of motion or even the coordinates themselves.
The relationship between symmetries and conservation laws in Noether's theorem holds only for continuous symmetries, however it encompasses both global and local transformations through the first and second Noether theorems. | {
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navigation, ekf, odometry, gps, ros-kinetic
twist:
twist:
linear:
x: 3.34638040916
y: -0.000663673175027
z: 0.0
angular:
x: 0.0
y: 0.0
z: 0.0427353974701
covariance: [0.029937610018694723, -1.748788884986028e-06, 0.0, 0.0, 0.0, -1.9526198758763966e-06, -1.7487888849861702e-06, 0.031781089816912814, 0.0, 0.0, 0.0, -0.00029328346609445733, 0.0, 0.0, 4.996875475492092e-07, -4.146782564862857e-23, 1.0922949666341516e-23, 0.0, 0.0, 0.0, -4.146782564862858e-23, 4.987548512708783e-07, 2.3359823844380062e-27, 0.0, 0.0, 0.0, 1.0922949666341511e-23, 2.335982408268371e-27, 4.987548512708783e-07, 0.0, -1.9526198758802387e-06, -0.0002932834660944581, 0.0, 0.0, 0.0, 0.035568126302575054] | {
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"tags": "navigation, ekf, odometry, gps, ros-kinetic",
"url": null
} |
By means of the $\ds{Stoltz\!-\!Ces\grave{a}ro\ Theorem}$:
\begin{align} S & \equiv \lim_{n \to \infty}\sum_{k = 1}^{n}{\ln\pars{k} \over nk} = \lim_{n \to \infty}{\sum_{k = 1}^{n + 1}\ln\pars{k}/k - \sum_{k = 1}^{n}\ln\pars{k}/k \over \pars{n + 1} - n} = \lim_{n \to \infty}{\ln\pars{n + 1} \over n + 1} \\[5mm] & = \lim_{n \to \infty}{\ln\pars{\bracks{n + 1} + 1} - \ln\pars{n + 1} \over \bracks{\pars{n + 1} + 1} - \pars{n + 1}}= \lim_{n \to \infty}\ln\pars{n + 2 \over n + 1} = \lim_{n \to \infty}\ln\pars{1 + {1 \over n + 1}} = \bbx{\ds{0}} \end{align} | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1976716/evaluate-lim-n-to-infty-sum-k-1n-frac-logknk"
} |
ros-melodic
Originally posted by mgruhler with karma: 12390 on 2021-06-28
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by akash12124234 on 2021-06-28:
the suggested techman works perfect | {
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ros, catkin-make, catkin, overlay
Title: Catkin fails to chain workspaces
Hi guys, I have a problem and cannot find a permanent solution to it:
I am trying to chain two workspaces A and B but catkin fails to recognize workspace A.
So what i did was create worspace A in:
/usr/share/ros/workspaceA
This works fine and i can build my packages.
Then i created another workspace B to work in, this should overlay workspace A, in:
~/ros/workspaceB
My packages in workspaceB depend on packages in workspace A.
When trying
source /usr/share/ros/workspaceA/devel/setup.bash
source ~/ros/workspaceB/devel/setup.bash
cd ~/ros/workspaceB/
catkin_make
I get
...
-- This workspace overlays: ~/ros/workspaceB/devel;/opt/ros/indigo
... | {
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"tags": "ros, catkin-make, catkin, overlay",
"url": null
} |
solid-state-physics, semiconductor-physics, electronic-band-theory, metals
Title: Why are there electron and hole currents in semiconductor, but only electron current in metal? Why are there electron and hole current in semiconductor, but only electron current in metal ?
I know that in metal there is an overlap between valence and conduction band,
while for semiconductor, there is no overlap, but that does not help me to understand my point of the question.
I have read the thread :
Electron holes in metals
and
Why doesn’t a conductor have hole current?
but the explanations were not so much convincing to me. | {
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"url": null
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Mathematics behind this card trick
1. Suppose I have $21$ playing cards. I distribute them in $3$ columns and tell you to choose mentally a card. Then just indicate in which column the card is.
2. I pick up one of the columns which doesn't contain your card, then the column which contains your card then the remaining column.
3. Now I deal the cards in $3$ columns again, starting from the left to the right, and repeating the process until there are no cards left in my hand. I ask you to indicate to me in which column your card is.
4. I repeat step 2 then 3.
5. I repeat step 2.
Now by counting either way in the deck of $21$ cards, your card will be the 11th card.
I've tried using modulo to understand the problem but I'm stuck doing integer divisions. So does anyone have a simpler way to explain the trick. | {
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"url": "https://math.stackexchange.com/questions/2055398/mathematics-behind-this-card-trick"
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c++, design-patterns, game, classes
////////////////////////////////////////////////////////////////
// Main function
////////////////////////////////////////////////////////////////
int main() {
manager m;
m.get_instance("Window").print("Open up");
m.get_instance("Asset").print("Load textures");
return 0;
}
Here is a working example of my code. What do you think about the class structure and architecture? Is there an easier or more expressive way? I'd like to get feedback and improve the code structure before implementing other managers the same way. About the architecture and thinking exclusively on the given example application (logging), I believe manager does not need to track all manager::instance instances. Maybe in other applications this might be needed, but not in this one. Thus, I would just turn get_instance into a convenience method (make_instance) to create new manager::instances:
class manager {
public:
class instance; | {
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"tags": "c++, design-patterns, game, classes",
"url": null
} |
programming, qiskit, simulation, qaoa
``` It turns it out it sure doesn't. I was using the QAOA qiskit method and I think this function takes out the solution bitstring that gives the lowest objective value. As I was trying only for small $ N $, there would be a decent probability attached to the solution even without the algorithm ! | {
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$$\color{green}{P_0} \color{blue}{\Rightarrow} P_1 \color{blue}{\Rightarrow} \dots \color{blue}{\Rightarrow} P_n \color{blue}{\Rightarrow} \dots .$$
At no point are you directly proving that $P_1$, $P_2$, $\dots$ are true, instead they follow from base case and induction step. | {
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"url": "https://math.stackexchange.com/questions/1645967/what-do-we-actually-prove-using-induction-theorem/1646299"
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Free Online Scientific Notation Calculator. Roots as Rational Exponents. A rational exponent is an exponent that is a fraction. Step 2: : Change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. VOCABULARY nth root of a For an integer n greater than 1, if bn 5 a, then b is an nth root of a. A rational equation is any equation that involves at least one rational expression. In this lesson we will not be adding and subtracting frac-tions, just building them up to a common denominator. Free Exponents Calculator - Simplify exponential expressions using algebraic rules step-by-step. Using this technique a number is represented in bits by three parts: sign, exponent, and fraction. (vÃî)10 10/6 S 32 Rewrite the ex ression using radical nota ion. TI-SmartView™ emulator for MathPrint™ calculators This simple software complements the TI-30XS MultiView™ scientific calculator, letting the | {
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"lm_q2_score": 0.8267118026095991,
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"openwebmath_score": 0.7000554203987122,
"tags": null,
"url": "http://miguelsal.it/rational-exponents-calculator.html"
} |
experimental-chemistry, home-experiment, polymers, terminology, materials
Wahab, M. A.; Kim, I.; Ha, C.-S. Silica- and Silsesquioxane-Containing Polymer Nanohybrids. In Macromolecules Containing Metal and Metal-Like Elements; Abd-El-Aziz, A. S., Carraher, C. E., Pittman, C. U., Zeldin, M., Eds.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 2005; pp 133–160. DOI: 10.1002/0471712566.ch6.
Perov, B. V.; Khoroshilova, I. P. Hybrid Composite Materials. In Polymer Matrix Composites; Shalin, R. E., Ed.; Soviet Advanced Composites Technology Series; Springer Netherlands: Dordrecht, 1995; pp 269–304. DOI: 10.1007/978-94-011-0515-6_6.
Encyclopedic Dictionary of Polymers; Gooch, J. W., Ed.; Springer New York: New York, NY, 2011. DOI: 10.1007/978-1-4419-6247-8. | {
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"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "experimental-chemistry, home-experiment, polymers, terminology, materials",
"url": null
} |
declination, ascension
Title: Calculating Right Ascension and Declination from Latitude / Longitude / Time I have a latitude / longitude / time of a location on Earth, I need to calculate the Right Ascension / Declination that's overhead at the time provided.
So far I've found out that Right Ascension is the same as the sidereal time for the location but I haven't found any information about declination.
I'm trying to set my overhead viewing location in the web portal of World Wide Telescope. It defaults to "now" but I want to show 10PM and there's no way (that I've found) to set time through the javascript interface. I can however set Right Ascension and Declination, hence my question. The declination of the point overhead (zenith) is the same as the observer's latitude. The RA of the point transiting zenith at any given time is the equivalent of the local sidereal time.
(Alternatively, the local sidereal time is RA of the observer's meridian.) | {
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"tags": "declination, ascension",
"url": null
} |
star, galaxy, rotation, constellations
Title: How are constellations intact if the stars are rotating around galactic nuclei? From what I understood, the Milky Way (or stars in the Milky Way) doesn't rotate like a collection of points in a disc due to the presence of some invisible matter. In theory, the angular velocities of all the stars should be the same while linear velocities should decrease as the radial distance increases. But in reality, linear velocities stay almost the same as we move farther from the galactic center. But that will decrease the angular velocity yes, right? | {
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object-oriented, game, c++17, cmake
Which is an interface that is both simple, and extremely hard to misuse.
Okay, on to the Player class.
Currently, you’re just using the player class to hold information about the player—their name and piece colour. You are not using the player class to be the player.
The player class should have a view (to view the state of the model/game) and a controller (to control the state of the model/game). Anything else is gravy.
In particular, there’s no harm in storing the player’s name in the player class, because if the player changes their own name… meh. Doesn’t really affect the game. But storing any game state information in the player class is probably not wise. For example, storing the piece colour there is dodgy. What happens if the player changes their own piece colour mid-game? They shouldn’t have that control. It would probably be better to store any game state info connected to a player in the model. The game needs to know what each player’s colour is, after all. | {
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"url": null
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ros, ros-control, wiki
Original comments
Comment by mug on 2022-01-06:
When I set it to 0.01, ros controllers will have this warning message: Controller'/th6/arm_controller' failed with error GOAL_TOLERANCE_VIOLATED. But if it is set to 0, ros controllers will prompt SUCCESS. This result seems to be inconsistent with its definition.
Comment by Akhil Kurup on 2022-01-06:
I don't have much experience with this. Maybe answer #272328 will help.
Comment by mug on 2022-01-06:
@Akhil Kurup Thanks.
Comment by Mike Scheutzow on 2022-01-06:
@mug on a real robot arm, the joint rotation velocity is being measured by a sensor. The joint motor has to fight gravity, so the value is noisy and usually non-zero, even if your eyes tell you the joint is "not moving". This parameter sets the max. velocity for moveit to consider the joint to be "not moving". If you set the value too high, moveit will give up too early as the arm movement slows down as it approaches the specified goal.
Comment by Mike Scheutzow on 2022-01-06: | {
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javascript, html
})();
// GLOBAL APP CONTROLLER
var controller = ((UICtrl) => {
var setupEvenetListeners = () => {
// Get dom strings from UI Controller
var DOM = UICtrl.getDOMStrings();
// On change
document.querySelector(DOM.room_options).addEventListener('change', () => {
UICtrl.hideDivElements();
});
}
return {
init: () => {
console.log("Application has started");
UICtrl.hideDivElements(); // Hide elements at start of program
setupEvenetListeners();
}
}
})(UIController);
controller.init();
</script> | {
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"url": null
} |
ros, catkin-make, build
Title: Beginner Q re: catkin_make of turtlesim (hydro)
Hi,
I'm new to ROS, and have been going through the tutorials. I decided I wanted to modify the behaviour of the Turtlesim demo application (having gotten the source from GitHub). I wanted to try 'catkin_make' using a target that has a bit more meat than the original example in
ROS / Tutorials / BuildingPackages. I increased the precision of 'Pi' and made some changes to the 'draw_square' publisher.
The files I got from GIT are located in a folder inside my home directory. But 'catkin_make' gives an error when I try to make the project.
I've tried copying the files over to the /opt/ros/hydro/share/turtlesim folder and using catkin_make there, but I get the same error as before.
The error I'm getting is:
The specified base path "/opt/ros/hydro/share/turtlesim" contains a CMakeLists.txt but "catkin_make" must be invoked in the root of workspace | {
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c++, strings, c++11
// Destroy
~CStringOwner()
{
free(data);
}
void swap(CStringOwner& other) noexcept
{
using std::swap;
swap(data, other.data);
swap(size, other.size);
}
// Other shit goes here.
}; | {
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heat-engine, carnot-cycle
The pressure in the evaporator is controlled by a pressure controller on the evaporator’s exit-vapor line, which is also the compressor suction line. This device controls a valve which “modulates” the pressure of the evaporator, but cannot be fully closed because the compressor should never be isolated from the evaporator. In the case where the refrigeration system runs low on refrigerant, the pressure in the evaporator exit-vapor line will drop very low, and the pressure switch shown in the drawing will disconnect power to the compressor. | {
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} |
automata-theory, fl.formal-languages, grammars
Title: Are variables in RHS of Chomsky Normal Form productions distinct? I'm wondering if it is permitted for production rules in a context-free grammar (CFG) in Chomsky Normal Form (CNF) to have multiple occurences of the same variable in the right-hand side of the productions.
e.g. is this permitted:
$A \to BB$
or must one do something like
$A \to B_1 B_2$
$B_1 \to B$
$B_2 \to B$ | {
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is correctly prepared. For a given set of distinct points and numbers. % Sample calls % [C] = newpoly(X,Y) % [C,D] = lnewpoly(X,Y) % Inputs % X vector of abscissas % Y vector of ordinates % Return % C coefficient list for the Newton polynomial % D divided difference table % % NUMERICAL METHODS: MATLAB Programs, (c) John H. 4 Interpolation and Approximation Dr. Runge, Über empirische Funktionen und die Interpolation zwischen äquidistanten Ordinaten, Z. ynew = interp1 (x, y, xnew, method). See Examples 1 and 2. It is also called Waring-Lagrange interpolation, since Waring actually published it 16 years before Lagrange [312, p. So the solution exists and is unique $$\blacksquare$$. For example, consider the function y = x 4 with x ⩾ 0 tabulated at the four points where y = 0, 1, 16 and 81. | {
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"tags": null,
"url": "http://cphq.tergolandia.it/newton-interpolation-example.html"
} |
algorithms, search-algorithms
$$\begin{align*}v_1 &=[150, 0, 255] \\ v_2 &= [255, 150, 0] \\ v_3 &= [0, 255, 150] \\ t &= [62, 63, 184]\end{align*}$$
The obvious thing to try was a memoization / search approach where we keep a data-structure containing each vector seen so far (initially containing only $v_1, v_2, v_3$) and recursively apply the operator on each pair of vectors stored in my data-structure until the target is found. However this appears to be much to time consuming as the number of possibilities grow super quickly. Is there any better approach that one might try? It doesn't help that there appears to be little structure - For example, starting with our three vectors, there appears to be many combinations that we cannot even form. | {
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} |
black-hole, supermassive-black-hole
Obviously jets of material travelling away from black holes have high speeds relative to the black holes. And they will be slowing down as they get farther and farther away from the black hole. Either they are travelling slower than the escape velocity of the black hole, and will eventually fall back toward the black hole, or they wiil be traveling faster than the escape velocity of the black hole and will continue to move farther away forever.
The great galaxy M8 has a giant black hole at its center with a mass of about 6.5 billion solar masses. The black hole has a jet of ejected material extending at least 5,000 light years from the black hole. According to this online calaculator, the escape velocity from an object with a mass of about 6,500,000,000 suns at a distance of about 5,000 light years is about 191.009 kilometers per second, which is only 17.075 times the escape velocity of Earth.
http://www.calctool.org/CALC/phys/astronomy/escape_velocity | {
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} |
RandomNiederreiterComplexes[pairsList__, number_] :=Map[
Function[randomsList,
pairsList[[All,1]]+Complex@@@Times[
ReIm[pairsList[[All,2]]-pairsList[[All,1]]],
randomsList
]
],
BlockRandom[
SeedRandom[Method->{"MKL",Method->{"Niederreiter", "Dimension" -> 2Length[pairsList]}}];
SeedRandom[];
RandomReal[{0, 1}, {number,Length[pairsList],2}]
]
]
RandomNiederreiterComplexes[{zmin_?NumericQ,zmax_?NumericQ},number_]:=RandomNiederreiterComplexes[{{zmin,zmax}},number][[All,1]]
To keep things on a more even footing, and to allow for a deterministic option, here is a corresponding deterministic uniform grid with the same syntax:
DeterministicComplexGrid::usage="DeterministicComplexGrid[{zmin, zmax}, n] generates a grid of about n equally spaced complex numbers in the rectangle with corners zmin and zmax. | {
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"url": "https://mathematica.stackexchange.com/questions/36898/using-nsolve-in-the-complex-plane"
} |
java, linked-list
public void addNodeToEnd(int data) {
Node toAdd = new Node(data, null);
addNodeToEnd(toAdd);
}
public void addNodeToFront(Node node) {
head.setPrevious(node);
node.setNext(head);
this.head = node;
numNodes++;
}
public void addNodeToFront(int data) {
Node toAdd = new Node(data, null);
addNodeToFront(toAdd);
}
public boolean addNodeAfterData(Node toAdd, int data) {
Node n = head;
while(n.getNext() != null) {
if(n.getData() == data) {
toAdd.setPrevious(n);
toAdd.setNext(n.getNext());
n.setNext(toAdd);
numNodes++;
return true;
}
n = n.getNext();
}
return false;
}
public boolean addNodeAfterData(int data, int dataSearch) {
Node toAdd = new Node(data, null);
return addNodeAfterData(toAdd, dataSearch);
} | {
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} |
collision, antimatter, coulombs-law
Title: Need for acceleration in particle-antiparticle annihilation? If an electron and positron are accelerated towards each other, at distances quite far away, there wouldn't be any significant electrostatic attraction, hence they need to be accelerated. But when they do come close, the Coulomb force is significant. So why do we accelerate the particle antiparticle pair, when they get attracted by electromagnetic forces? What is the need for getting them to collide at high speeds? Doesn't annihilation occur when an electron comes in the field of the positron? Pairs of charged particles and/or objects attract via the $Q_1Q_2/R^2$ Coulomb's law. This is a classical approximation that quantifies how their velocities are changing when the objects are large or distances are much longer than the Compton wavelength etc. | {
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"tags": "collision, antimatter, coulombs-law",
"url": null
} |
general-relativity, differential-geometry, metric-tensor, gauge-theory, kaluza-klein
\begin{align}
F_{\mu\nu} F^{\mu\nu} \longrightarrow &\, \lambda^2 F_{\mu\nu} F^{\mu\nu} \nonumber\\
\nabla_\mu \sigma \longrightarrow& \, \nabla_\mu \sigma
\end{align}
The latter relation follows from $\sigma \longrightarrow \sigma - \ln \lambda$.
How does the curvature scale under this? It turns out that $\tilde R$ remains unchanged under this scaling, and so does $R$ as well.
This is not too hard to see if we investigate the scaling of the metric components. The metric is
\begin{align}
\tilde G_{\mu\nu} = &\,G_{\mu\nu} +e^{2\sigma} A_\mu A_\nu \nonumber\\
\tilde G_{\mu d} = &\, e^{2\sigma} A_\mu \nonumber\\
\tilde G_{dd} = &\,e^{2\sigma}
\end{align}
The first think we notice is that
the determinant of the $D$ dimensional metric is very simply related to the determinant of the $d$ dimensional metric. One easily works out that
$$ \tilde G = e^{2\sigma} \det G $$
The second observation is that the inverse of $\tilde G_{MN}$ is very simple
\begin{align} | {
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organic-chemistry, resonance
The following is based on NBO analyses from DF-BP86/def2-SVP.
\begin{array}{crr}\hline
\text{Structure} & \text{% Lewis} & \text{% non-Lewis}\\\hline
\mathbf{1} & 96.4 & 3.6 \\
\mathbf{2} & 96.8 & 3.2 \\
\mathbf{3} & 96.2 & 3.8 \\
\mathbf{4} & 95.8 & 4.2 \\
\mathbf{5} & 96.3 & 3.7 \\
\mathbf{6} & 95.2 & 4.7 \\\hline
\end{array}
The first statement that should be made is that this molecule is quite well described by Lewis resonance structures. However, delocalised bonding would describe the molecule much better. All structures are reasonably close to each other, as one would expect for a very delocalised system.
The ionic structure has the smallest impact, which is feasible because nitrogen-hydrogen bonds are better described as covalent. However, it proves that there can (in the right medium) be a proton shift from nitrogen to oxygen, and is therefore worth consideration. | {
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rust, concurrency, queue, lock-free
false
}
/// Retrieves the first item in the queue.
///
/// # Example
/// ```
/// use cqi::LinkedQueue;
///
/// let lq = LinkedQueue::<usize>::new();
/// let guard = lq.guard();
/// lq.enqueue(42, &guard);
/// assert_eq!(lq.peek(&guard), Some(&42));
/// ```
pub fn peek<'g>(&self, guard: &'g Guard) -> Option<&'g T> {
// Here we don't need to update the `mod_count` field in the `tail` node since we aren't doing any mutations.
let head = self.head.load(Ordering::Acquire, guard);
if head.is_null() {
None
} else {
let item = unsafe { &head.deref().item };
Some(item)
}
} | {
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black-hole, astrophysics, terminology, theories, no-hair-conjecture
Perhaps they mean something specific, or perhaps they don't know any better... you'd have to examine it on a case-by-case basis. Perhaps a better question is: are there reasons why someone would be motivated to say theorem instead of conjecture?
Sometimes people call unproven mathematical statements "theorems" before there are publicly known proofs, typically because there are strong motivations to think it is true (often utilizing Occam's razor, leaning on empirical data, etc...). As @Peter Erwin pointed out, Fermat's theorem was called such long before it was proven rigorously mathematically. Laplace was famous for omitting proofs of his "theorems," which were found to most often be (mostly) correct. I think this kind of terminological inconsistency is not as bad as some other kinds, such as how string models are called "string theory" even though the fundamental assumption is not testable and the correspondence of the models to previously known theories is not well understood. | {
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} |
r, seurat
Content type 'unknown' length 39130 bytes (38 KB)
downloaded 38 KB
trying URL 'https://cran.rstudio.com/src/contrib/Seurat_2.3.0.tar.gz'
Content type 'unknown' length 846065 bytes (826 KB)
downloaded 826 KB
* installing *source* package ‘igraph’ ...
** package ‘igraph’ successfully unpacked and MD5 sums checked
installing to /sfs/qumulo/qhome/nv4e/R/x86_64-pc-linux-gnu-library/3.3/igraph/libs
** R
** demo
** inst
** preparing package for lazy loading
** help
*** installing help indices
** building package indices
** testing if installed package can be loaded
Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared object '/sfs/qumulo/qhome/nv4e/R/x86_64-pc-linux-gnu-library/3.3/igraph/libs/igraph.so': libicui18n.so.58: cannot open shared object file: No such file or directory | {
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} |
c++, object-oriented, template, interface
for (int i = 0; i < length; i++) {
result += this->_peek(buffer++);
}
return result;
}
std::streamsize MemoryDataStream::peek(int8_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(uint8_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(int16_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(uint16_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(int32_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(uint32_t *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(float *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(double *buffer) {
return this->_peek(buffer);
}
std::streamsize MemoryDataStream::peek(std::string *value) {
int8_t c;
std::streamsize i;
int size;
std::stringstream strm; | {
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} |
c#, sql, sql-server
sql2k5comm.Parameters.Add(new SqlParameter("@Max_Queue_Time", row["Max_Queue_Time"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Calls_Handled", row["Calls_Handled"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Avg_Speed_Answer", row["Avg_Speed_Answer"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Avg_Handle_Time", row["Avg_Handle_Time"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Max_Handle_Time", row["Max_Handle_Time"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Calls_Abandoned", row["Calls_Abandoned"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Avg_Time_Abandon", row["Avg_Time_Abandon"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Max_Time_Abandon", row["Max_Time_Abandon"]));
sql2k5comm.Parameters.Add(new SqlParameter("@Avg_Calls_Abandoned", row["Avg_Calls_Abandoned"])); | {
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quantum-field-theory, fermions
As far as the labeling of kets, the change is that it is no longer possible to simply label one state with $n$ particles $|n\rangle$. One has to keep track of the spins and momentum individually, so if one has some number of particles you would track each one as $|p_1, p_2,\dots,p_n\rangle$
Hope this helps. | {
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c++, object-oriented, classes
if (row < 6 && col > 0 && board[row + 2][col - 1] <= 0) //two down one left
pseudomoves.push_back(push(row, col, row + 2, col - 1));
if (row < 7 && col > 1 && board[row + 1][col - 2] <= 0) // one down two left
pseudomoves.push_back(push(row, col, row + 1, col - 2));
if (row > 1 && col > 0 && board[row - 2][col - 1] <= 0) // two up one left
pseudomoves.push_back(push(row, col, row - 2, col - 1));
if (row > 0 && col > 1 && board[row - 1][col - 2] <= 0) // one up two left
pseudomoves.push_back(push(row, col, row - 1, col - 2));
} | {
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"tags": "c++, object-oriented, classes",
"url": null
} |
c#, sorting, winforms
}
private get { return arrayToSort; }
}
private void InitializeSorting()
{
current = 0;
assumeSortedFrom = ArrayToSort.Length;
hasSwapped = false;
}
public void PerformSortStep()
{
OnSortStepPerformed(current, PointOfStep.BeforeStep);
if (current + 1 == assumeSortedFrom)
{
assumeSortedFrom = current;
OnSortStepPerformed(current, PointOfStep.ReAssignedAssumeSorted);
current = 0;
OnSortStepPerformed(current, PointOfStep.CurrentStep);
if (!hasSwapped)
{
OnSortFinished();
}
else
{
hasSwapped = false;
}
}
else
{
if (ArrayToSort[current] > ArrayToSort[current + 1])
{
SwapIndexes(current, current + 1);
hasSwapped = true;
} | {
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"tags": "c#, sorting, winforms",
"url": null
} |
python, python-3.x
imports = defaultdict(set)
with filename.open('r') as file:
for line in file:
if not (line := line.strip()):
continue
if (match := fullmatch(FROM_IMPORT, line)) is not None:
module, members = match.groups()
members = set(filter(None, members.split(',')))
elif (match := fullmatch(IMPORT, line)) is not None:
module = match.group(1)
members = set()
else:
continue
imports[module] |= members
return imports
def get_imports_from_files(paths: Iterable[Path]) -> dict:
"""Returns the imports from multiple files."""
imports = {}
for path in paths:
LOGGER.info('Checking: %s', path)
imports.update(get_imports(path))
return imports
def get_module_roots(imports: dict) -> Set[str]:
"""Returns a set of imported module roots."""
roots = set() | {
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c#, performance, collections, unity3d
foreach (var ship in FoundShips)
{
if (!TrackedShips.Contains(ship))
{
// Do the removal logic
}
}
TrackedShips = FoundShips; | {
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} |
metabolism, diet, carbohydrates
From a biochemical perspective we know that fatty acids and acetyl-CoA cannot be converted back into glucose or other carbohydrate intermediates. This is because of the irreversible biochemical reaction catalyzed by pyruvate dehydrogenase, which converts pyruvate to acetyl-CoA. Thus, fatty acids (lipids) can be oxidized to acetyl-CoA (for the TCA/Krebs Cycle) but cannot be further converted to glucose within the body.
In terms of protein, however, amino acids are either glucogenic, ketogenic, or both. If amino acids are ketogenic, then this means they can be converted into acetyl-CoA for the Krebs Cycle. If amino acids are glucogenic, then it means that they can be broken down into glucose. The breakdown of amino acids can be used to synthesize glucose or for anapleurotic reactions of the Krebs Cycle. | {
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"url": null
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Walmart Canada Gaming Twitter, Kenwood Parameter Customize, Prawn Noodle Soup Recipe Meatmen, Marco Rossi Instagram, How To Play Glorious Day By Casting Crowns On Guitar, Auto-stop Screw Depth Setter, | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9615338112885302,
"lm_q1q2_score": 0.8153496631742639,
"lm_q2_score": 0.8479677506936879,
"openwebmath_perplexity": 603.5606432501596,
"openwebmath_score": 0.6828079223632812,
"tags": null,
"url": "https://elingrelsson.se/cteen-choice-qqptpb/27194d-exponential-distribution-expected-value"
} |
The above can also be rewritten as \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*)) + d_1 \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*)) + d_1 & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*)) + d_2 \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2)) + d_2 & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}
Because $d_1$ and $d_2$ are constants, we can take them inside the expectation and rewrite the above inequalities as: \begin{eqnarray*} \mathbb{E}(u_1(\sigma_1^*, \sigma_2^*) + d_1) \geq \mathbb{E}(u_1(\sigma_1, \sigma_2^*) + d_1) & & \forall \sigma_1 \in \Delta (A_1) \\ \mathbb{E}(u_2(\sigma_1^*, \sigma_2^*) + d_2) \geq \mathbb{E}(u_2(\sigma_1^*, \sigma_2) + d_2) & & \forall \sigma_2 \in \Delta (A_2) \end{eqnarray*}
By definition of $v_1$ and $v_2$, | {
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"tags": null,
"url": "https://economics.stackexchange.com/questions/15464/zero-sum-game-constant-sum-game/15466"
} |
optics, waves, spherical-harmonics
At any fixed value of time, this represents a cluster of concentric spheres filling all space. Each wavefront, or surface of constant phase, is given by
$$kr = \text{constant}$$
Notice that the amplitude of any spherical wave is a function of $r$, ware the term $r^{-1}$ serves as an attenuation factor. Unlike the plane wave, a spherical wave decreases in amplitude, thereby changing its profile, as it expands and moves out from the origin. Figure 2.27 illustrates this graphically by showing a "multiple exposure" of a spherical pulse at four different times. The pulse has the same extent in space at any point along any radius $r$; that is, the width of the pulse along the $r$-axis is a constant. | {
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"url": null
} |
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