text stringlengths 1 1.11k | source dict |
|---|---|
beginner, c, game-of-life, sdl2
case SDLK_RIGHT: // Press Right Arrow Key to time step once
TimeStep();
break;
}
}
}
}
void ResetGameGrid()
{
for(int i = 0; i < GRID_SIZE; i++) // Loop through each cell of the grid and reset to default values
{
for(int j = 0; j < GRID_SIZE; j++)
{
GameGrid[i][j].x = SQUARE_SIZE + j * SQUARE_SIZE;
GameGrid[i][j].y = SQUARE_SIZE + i * SQUARE_SIZE;
GameGrid[i][j].isAlive = false;
GameGrid[i][j].lastAlive = 255;
}
}
}
void TimeStep(void)
{
memcpy(PrevGameGrid, GameGrid, sizeof(GameGrid)); // Set value of PrevGameGrid to be that of current GameGrid
for(int i = 1; i < GRID_SIZE - 1; i++)
{
for(int j = 1; j < GRID_SIZE - 1; j++)
{
int aliveNeighbors = 0; // Count number of alive neighbors - Leaving a 1x1 border of dead cells around grid | {
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"tags": "beginner, c, game-of-life, sdl2",
"url": null
} |
condensed-matter, quantum-information, computational-physics, tensor-network
The goal now is to work through the block from left to right and decompose our state into a product of smaller tensors (big dots in the bottom row of the figure) - or more formally:
\begin{equation}
\psi_{\sigma_1,\sigma_2,\ldots,\sigma_6} = \sum_{a_1,a_2,\ldots,a_5} A^{[\sigma_1]}_{a_1}A^{[\sigma_2]}_{a_1,a_2}\ldots A^{[\sigma_5]}_{a_4,a_5} A^{[\sigma_6]}_{a_5}
\end{equation}
Note that the $\sigma_i$ indices correspond to the edges extending up (these are the physical indices corresponding to the values of $\sigma$ at each site), while the $a_i$ indices correspond to the horizontal bonds (these are the bond indices - which aren't physical and depend on the way we construct the MPS). Additionally, note that the first and last sites have only one bond index ($a_1$ and $a_5$, respectively). | {
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"tags": "condensed-matter, quantum-information, computational-physics, tensor-network",
"url": null
} |
0000 0101 AND 1111 1110 = 0000 0100
Success! Only the bit we were trying to unset has changed. So let’s make a function that does this for us :
#### Checking a bit
To check a bit, we also need to use the bitwise AND operation. And since we are only checking and not setting, we don’t have to store the value anywhere which means we don’t have to worry about changing anything.
To check a bitmask, simply do an AND operation with the value you want to check for ( in this case, any of the TTF_STYLE_.... values ). So, to check if a text is bold, we do an AND between our mask and TTF_STYLE_BOLD :
0011 ( our bit mask, TTF_STYLE_BOLD and TTF_STYLE_ITALIC set )
AND 0001 = 0001
As you can see, we only check the bit that’s set in our variable ( TTF_STYLE_ITALIC set ) the others will be 0 no matter what our mask is. The value 0001 is not 0, and thus this evaluates to true and we now know that the font is bold. | {
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"lm_q1_score": 0.989830340446144,
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"openwebmath_score": 0.42407405376434326,
"tags": null,
"url": "http://headerphile.com/tag/game-programming/"
} |
navigation, amcl-demo.launch, plugin, amcl, pluginlib
Where does amcl_demo.launch its plugins from?
(ROS Distro is Hydro and Ubuntu is 13.04). | {
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"tags": "navigation, amcl-demo.launch, plugin, amcl, pluginlib",
"url": null
} |
statistical-mechanics, bosons
Next, we need the eigenvalues of $\hat{H}$. Since you are considering a non-interacting system of bosons, the eigenvalues are completely determined by the simple particle Hamiltonian $H(\vec{p},s_Z)$. The eigenvalues of $H(\vec{p},s_z)$ is known to be Landau levels, $\varepsilon_{n,s_z,p_z} = (n+1/2)\omega_c - \mu_0Bs_z + p^2_z/2m$, where I assumed the magnetic field points along z-direction. Note that each Landau level is highly degenerate, i.e., for each level, there is a subset of states (in total $D$). Now the eigenvalues of $\hat{H}$ are just $E = \sum_{n,s_z,p_z,\nu}L^{\nu}_{n,s_z,p_z}\varepsilon_{n,s_z,p_z}$, with $L^{\nu}_{n,s_z,p_z}$ the occupation of the $\nu$-th state of the subset $(n,s_z,p_z)$. | {
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"tags": "statistical-mechanics, bosons",
"url": null
} |
homework-and-exercises, general-relativity
and hence, you get this
$${P_{\alpha \mu}}(\rho + P){ _{;\nu }}{u^\mu }{u^\nu } =\underbrace{{P_{\alpha \mu }}{u^\mu }}_{ = 0}(\rho + P){_{; \nu }}{u^\nu } = 0.$$ | {
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"tags": "homework-and-exercises, general-relativity",
"url": null
} |
P is the matrix with the columns C1, C2, ..., and Cn the n eigenvectors of A, then the matrix P-1AP is a diagonal matrix. The roots of the characteristic polynomial p ( t) are eigenvalues of A. Step by Step Explanation […], […] When $a=b$, then $A$ is already diagonal matrix. For instance 2 Rows, 3 Columns = a[2][3] ) Determining diagonals in a matrix . For more videos and resources on this topic, please visit http://ma.mathforcollege.com/mainindex/01introduction/ For each eigenvalue $\lambda$ of $A$, find a basis of the eigenspace $E_{\lambda}$. C program to check Identity matrix . Then the matrix $A$ is diagonalized as $S^{-1}AS=D.$. If x is a matrix then diag(x) returns the diagonal of x.The resulting vector will have names if the matrix x has matching column and row names. How can I view the source code for a function? This is one application of the diagonalization. . Use D = diag (u,k) to shift u in k levels above the main diagonal, and D = diag (u,-k) for the opposite direction. | {
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"openwebmath_score": 0.754424512386322,
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"url": "https://thevinyldecal.com/fddg03/79gr2pb.php?1617de=how-to-find-diagonal-matrix"
} |
c#, entity-framework, asp.net-web-api
if (newPromotions.Count == 0)
{
throw new ArgumentException("No promotions to add");
}
Common.ValidationResult result = await validator.ValidateNewPromotions(newPromotions);
if (!result.Result)
{
var message = new HttpResponseMessage(HttpStatusCode.Conflict)
{
Content = new StringContent(result.ValidationMessage)
};
throw new HttpResponseException(message);
}
await _promotionsRepository.AddPromotions(memberId, newPromotions);
return Ok();
}
[Authorize]
[HttpDelete]
[Route("Promotions/{id:int}", Name = "DeletePromotion")]
public async Task<IHttpActionResult> DeletePromotion(int id)
{
try
{
await _promotionsRepository.DeletePromotion(id); | {
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"tags": "c#, entity-framework, asp.net-web-api",
"url": null
} |
remote-sensing, snow, gis, matlab
With that you have more than enough information to generate the latitude and longitude vectors. Now you have to decide if you will store the center of each pixel, or one of the corners.
Assuming you will store the center of each pixel you can generate the latitude and longitude vectors like this:
gridSize=0.05 % arc degrees
lat=-90+(gridSize/2):0.05:90-(gridSize/2);
lon=-180+(gridSize/2):0.05:180-(gridSize/2);
And if you want instead matrices with the coordinates of the center of each individual pixel, you do
[lonGrid,latGrid]=meshgrid(lon,lat); | {
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"url": null
} |
If $G$ is any non-trivial group, then $\mathrm{Aut}(G \times G)$ is not abelian.
This is an improvement of Hagen von Eitzen's observation.
Proof. If $G$ is abelian, then $(x,y) \mapsto (y,x)$ does not commute with $(x,y) \mapsto (x,x+ y)$.
If $G$ is not abelian, we find a nontrivial inner automorphism $\sigma$ of $G$. But then $(x,y) \mapsto (y,x)$ does not commute with $(x,y) \mapsto (x,\sigma(y))$.
$Aut(C_p \times C_p)$ coincides with the set of invertible linear transformations and so is $GL(2,\mathbb F_p)$, which is not abelian.
• See my answer for a generalization. – Martin Brandenburg Jul 3 '15 at 22:40
If they are talking about the morphisms of a group, here is a approach.
Take automorphism group of the free group of the set $\mathbb Z$. It's automorphism group is a supergroup of the permutation group of $\mathbb Z$. | {
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} |
javascript, beginner, regex, calculator
Conventions
It is convention to use one const/let declaration per variable.
(Potential) bugs
The regex used to validate the bill value only forbids the letters A to Z. Any other non-digits or other errors (such as more than one decimal separator) are not caught. The service rating and number of people are not validated at all.
You should be explicitly converting the string values from the input fields into numbers. Currently you are lucky that you are only using multiplication and division on the value, so that the strings are automatically converted, but if you, for example, would addition you'd have an error ("1" + "2" === "12" not 3).
You are toggling the visibility of the paragraphs on each successful calculation, that so every second calculation does not show a result.
Other possible improvements | {
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"tags": "javascript, beginner, regex, calculator",
"url": null
} |
How will it turn negative if we substitute with 5? And what will become neg. exactly?
Hi, we take x+2 <0, so |x+2|=-(x+2)...
we then get x=5, and when we put x=5 in x+2 = 7, (x+2) becomes>0..
But we had taken |x+2|=-(x+2)...|5+2|=-(5+2)=-7, so |x+2| has become NEGATIVE
Otherwise just take that you started with assumption that x+2<0, but x=5 gives x+2>0, so discard
_________________
Manager
Joined: 11 Sep 2019
Posts: 77
Re: Absolute modulus : A better understanding [#permalink]
### Show Tags
14 Nov 2019, 11:08
This post is the best. Thanks for the great work chetan2u
Intern
Joined: 25 Sep 2017
Posts: 3
Re: Absolute modulus : A better understanding [#permalink]
### Show Tags
08 Dec 2019, 00:14
Hello,,
First of all I really would like to thank you Sir. This is much needed post. | {
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radio-astronomy, spectroscopy
I realize this doesn't accord with the definitions you linked to at the start of your question, but that page is operating from a very precise (arguably fussy and overly nice) perspective, and I'm not sure most astronomers would agree with the idea that "a spectrum must be a plot flux density versus wavelength". I've certainly seen plenty of X-ray "spectra" where the units are photon flux density plotted versus photon energy....
SEDs can be shown in units of $\nu F_{\nu}$ ... or just $F_{\nu}$ or in Jy. What are the relationships between these units?
$F_{\nu}$ means flux density -- e.g., flux per unit frequency (examples: ergs cm$^{-2}$ s$^{-1}$ Hz$^{-1}$; W m$^{-2}$ MHz; etc.). If you multiply it by the frequency $\nu$ to get $\nu F_{\nu}$, then you remove the "per unit frequency" part and it becomes flux. "Jy" are Janskys, a specific unit of flux density ($10^{−26}$ W m$^{−2}$ Hz$^{−1}$). | {
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electromagnetism, physical-chemistry, atomic-physics, ions
Title: What force creates ions out of neutral atoms? Consider the reaction between Na and Cl to form NaCl.
Na loses an electron and "gives" it to Cl because this makes both atoms more stable.
But what forces "pulls" the electron from Na to Cl? Both atoms are electrically neutral at the beginning, so the electron that is going to be transferred isn't acted on by electrostatic forces.
Since its not electrostatic forces that pull the electron from Na to Cl (since both atoms are neutral) what force pulls the electron from Na to Cl to make both atoms more stable? Your interpretation that if we have simply an $Na$ atom and a $Cl$ atom they would exchange electrons and make $NaCl$, is wrong !
To make the necessary electron exchange they need to get very very close, so close that their electronic orbitals (place where electrons reside) overlap in after coming into this region 2 things can happen | {
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"tags": "electromagnetism, physical-chemistry, atomic-physics, ions",
"url": null
} |
python, python-3.x, object-oriented
def buy(self):
self.choice = input("What do you want to buy? 1 - espresso, 2 - latte, 3 - cappuccino, back - to main menu:\n")
if self.choice == '1':
self.reduced = [250, 0, 16, 1, 4] # water, milk, coffee beans, cups, money
if self.available_check(): # checks if supplies are available
self.deduct_supplies() # if it is, then it deducts
elif self.choice == '2':
self.reduced = [350, 75, 20, 1, 7]
if self.available_check():
self.deduct_supplies()
elif self.choice == "3":
self.reduced = [200, 100, 12, 1, 6]
if self.available_check():
self.deduct_supplies()
elif self.choice == "back": # if the user changed his mind
self.return_to_menu()
self.return_to_menu() | {
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} |
homework-and-exercises, hilbert-space, quantum-information, density-operator, quantum-states
Title: What is the relation between purity and ${\rm Tr}(\rho^2)$, for a density matrix? I would like to understand the equivalence for a state $\psi$ to be pure and its density matrix $\rho=|\psi\rangle\langle\psi|$ having the property $$\operatorname{trace} \rho^2=1.$$ For this I should multiply 2 matrices and look at the result trace. I do not know 2 things, how the matrices should look like and whether $\psi$ should be normalized (and allowing speak about vicinity of the trace with the number 1). Any hints?
I would like to understand the equivalence for a state $\psi$ to be pure and its density matrix $\rho=|\psi\rangle\langle\psi|$ having the property $$\operatorname{trace} \rho^2=1.$$ | {
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"tags": "homework-and-exercises, hilbert-space, quantum-information, density-operator, quantum-states",
"url": null
} |
java, beginner, strings, programming-challenge
Title: Summing digits of an alphanumeric string I am going through the Java CodingBat exercises. Here is the one I have just completed:
Given a string, return the sum of the digits 0-9 that appear in the string, ignoring all other characters. Return 0 if there are no digits in the string.
Here is my code:
public int sumDigits(String str) {
StringBuilder strAppend = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
if (Character.isDigit(str.charAt(i))) {
strAppend.append(str.charAt(i));
}
}
int total = 0;
String strDigits = strAppend.toString();
for (int i = 0; i < strDigits.length(); i++) {
total += Integer.parseInt(strDigits.substring(i, i+1));
}
return total;
}
My questions are: | {
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forces, standard-model, weak-interaction
$$V(r) = - \dfrac{4}{3} \dfrac{\alpha_s(r) \hbar c}{r} + kr$$
where the constant $k$ determines the field energy per unit length and is called string tension.
At the level of quantum hadron dynamics (i.e. the level of nuclear physics, not the level of particle physics where the real strong force lives) one can talk about a Yukawa potential of the form
$$ V(r) = - \frac{g^2}{4 \pi c^2} \frac{e^{-mr}}{r} .$$
But nothing mentions anything similar to the weak force, though, the weak force can be repulsive or attractive, see here:
Weak force: attractive or repulsive?
So there must be an interaction due to the weak force, and it has to have some kind of similar description.
Question: | {
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@GEdgar FYI, the compiler step can be skipped, and that's probably the most likely place for this to fail. – Cruncher Mar 19 '14 at 17:28
We have a huge amount of empirical data for the behaviour of compilers, CPUs and transistors; they're not perfect, but many of their bugs are known and can be avoided. We have exactly no data about the brand-new program I just wrote to help with a proof, so we have no reason to trust it. In fact, all of the empirical data from software engineering says that it will be buggy, hence we have specific reason not to trust it! – Warbo Mar 20 '14 at 10:02 | {
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human-biology, biochemistry, pharmacology
Title: To understand side-effects for nebivolol's beta1 selective pathway in PubChem? Assume you have a drug nebivolol.
When nebivolol is used as beta1 selective drug, beta2 is mostly for side effects, but this is not clear from PubChem.
I do not know any cases where nebivolol is used as beta1 and beta2 selective in the treatment.
Its first two pathways of interaction from PubChem.
Assume this pathway to be selective for some indication
and this nonselective (where side-effects stems mostly with large doses) | {
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ros, pluginlib
Post score: 1
I have found a partial solution.
The plugin-base-class should inherit from an auxiliar-untemplated-base-class. After this you can use a dynamic_pointer_cast to the templated plugin-base-class. This is not bad given the dynamic load is itself dynamic, no additional and unexpected runtime errors should occur. given "NavigationNodeComponent" the auxiliar untemplated base class and state_estimator_name the name of the plugin we want to load.
pluginlib::ClassLoader loader("rtcus_navigation","rtcus_navigation::NavigationNodeComponent");
shared_ptr<StateEstimation<StateType, ActionType, TimeModel> > new_state_estimation =
dynamic_pointer_cast<rtcus_navigation::StateEstimation<StateType, ActionType, TimeModel> >(
loader.createInstance(state_estimator_name));
However this is only for the loading stage. How to declare plugin in a more automated way and not manually remains unsolved. | {
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decomposition
Waclaw Lanowy, Krystian Lukaszczyk, Przemyslaw Kasprzycki, “Silica gel with moisture indicator, a method of obtaining them, and use of phthalein dyes,” International Patent 2006, WO2006/130029 A2 (PDF). | {
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c++, memory-management
Consider the user
The current code requires that the user explicitly instruments the code, which seems a bit intrusive. Here's an alternative approach the modifies things just slightly, using the Curiously Recurring Template Pattern, or CRTP for short. First we isolate the leak detector bits into a templated class.
template <typename T>
struct LeakDetector {
LeakDetector() {
LeakDbg::addRef(typeid(T).name(), this);
}
~LeakDetector() {
LeakDbg::remRef(this);
}
};
Now to use it is much simpler than before. No ugly macros are required and we only need to add one simple thing to the declaration of the class to be monitored:
struct Test : public LeakDetector<Test>
{
Test() {
}
~Test() {
}
}; | {
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"tags": "c++, memory-management",
"url": null
} |
• Uhm... I guess I should have thought about that one... Thanks ! and Thanks to J. M. for the Outer Trick. – chris Apr 27 '12 at 18:43
I know, it's long a ago. I just stumpled into this post and cannot resist. JM's code can be sped up even more, even without Compile: First, we can employ Outer in conjunction with Plus to compute the squared norms of frequencies at once. Second, only the zero frequency makes problems when applying Pk, so we can set its squared norm temporarily to 1. and use vectorization when we apply Pk. (Edit: I just realize that Simon Woods had the same idea...) That leads to a roughly 18-fold speed up on my machine for the whole routine (with size = 1024). (Another curiosity is that applying Sqrt twice is faster than applying Power[#,1/4]& once - probably due to hardware optimization.) | {
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ros, pcl, pandaboard, linking
Some mismatch with libmysqlclient and libvtk? I don't know what to do from here, any help is appreciated.
Originally posted by Hansg91 on ROS Answers with karma: 1909 on 2012-04-18
Post score: 0
Original comments
Comment by Eric Perko on 2012-04-18:
Please post a link to which directions you are following for compiling PCL. Also, if this is PCL-standalone, not trying to compile PCL in ROS on a Pandaboard, you should probably send email to the pcl-users mailing list.
Comment by Hansg91 on 2012-04-18:
This is the perception_pcl stack, which was downloaded through ROS Desktop Full installation. I did not follow any directions other than the installation guide of ROS for Ubuntu. After that I added pcl as dependency and let it compile, with --rosdep-install added.
pcl needs the newer version of mysql. Unfortunately there's a patch which breaks the ABI. Make sure you have the lucid-updates repository added and update mysql. | {
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ros, arduino, rosserial, base-odometry, linear-velocity
Originally posted by stevej_80 on ROS Answers with karma: 75 on 2016-08-04
Post score: 3
Yes, you can publish on /odom directly from the Due, and it should work fine. One possibility is to use the rosserial package. I've done this from an Arduino Leonardo. (However, I've dropped rosserial on the smaller Arduinos because of the memory usage of rosserial. You shouldn't have any problems on a Due.)
You should publish the odometry as nav_msgs/Odometry. Normally, you want to publish in the base_link frame, which often has its origin midway between the two wheels. By convention, x is straight ahead, and z is up. Therefore, since we assume there is no side-slip, you'll have the velocities in twist.linear.x and twist.angular.z. The pose will be in pose, including a quaternion for orientation. All other velocity values will be zero. | {
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performance, sql, oracle, plsql
| 0 | SELECT STATEMENT | | 182 | 91910 | 1978 (1)| 00:00:01 |
| 1 | SORT ORDER BY | | 182 | 91910 | 1978 (1)| 00:00:01 |
|* 2 | FILTER | | | | | | | {
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java, beginner, game, playing-cards
Then, you have System.out.println(); after printing a message so that you can leave a line, when you could do System.out.println("My Message!\n"), so that is something you can do.
But, as I said before, you need to choose a format that you are happy with, and just stick to that. Use things like camel-case and capitalising class names, but as long as you and others can read it, you are okay.
Just for practice, I am editing your code to make it more compact. I will post it in about 20 minutes.
Please tell me if you understand. | {
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python, python-3.x, template, pdf
Suggested reading: Pixels vs. Relative Units in CSS: why it’s still a big deal
Shorthand notation
You can reduce the number of lines slightly by using shorthand notation. For example:
.table-fill {
background: white;
border-radius:3px;
border-collapse: collapse;
border-bottom:2px solid #C1C3D1;
border-right:2px solid #C1C3D1;
border-left:2px solid #C1C3D1;
height: 320px;
margin: auto;
max-width: 600px;
padding:5px;
width: 100%;
box-shadow: 0 5px 10px rgba(0, 0, 0, 0.1);
animation: float 5s infinite;
} | {
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"tags": "python, python-3.x, template, pdf",
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} |
python, regex, web-scraping
if any(word in analyst8 for word in targetdoppler):
if image8 in targetmg:
if totalprice8 <= superprice:
if cart8 == cartid8:
cart = cart8
JustDOit = SaleProcess(cart, cookie_value)
if image1 in targetm:
if totalprice8 <= gutdop:
if cart8 == cartid8:
cart = cart8
JustDOit = SaleProcess(cart, cookie_value)
if any(string in analyst8 for string in targetsuper):
if totalprice8 <= superprice:
if cart8 == cartid8:
cart = cart8
JustDOit = SaleProcess(cart, cookie_value) | {
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follows: Let’s consider a triangle. It can be found by taking the average of x- coordinate points and y-coordinate points of all the vertices of the triangle. Conclusion: Simple, the orthocenter (2) Circum-center: The three perpendicular bisectors a triangle meet in one point called the circumcenter. ! General formulas for the centroid of any area are provided in the section that follows the table. The centroid is also called the center of gravity of the triangle. By placing the points as follows you can make an L shaped object. The centroid is typically represented by the letter G G G. The centroid is always inside the triangle Each median divides the triangle into two smaller triangles of equal area. Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! A centroid is also known as the centre of gravity. See the below figure, where O is the centroid of the square. Altitudes. The centre of point of intersection of all the | {
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} |
You are right about $f(x^2)$; and $f(x)^2$ simply means $f(x)f(x)$. The troublesome notation is $f^2$, whether used to talk about the function $f^2$ or its value at the point $x$, namely $f^2(x)$. One convention often used is that it is the function whose values are the squared values of $f$, namely $f^2:x\mapsto f(x)^2$. However, in the sort of mathematics where we are more interested in compounding functions than in doing algebraic operations on their values, concatenation is a convenient way to represent composition, particularly because such composition is associative. Thus, $fgh$ means $x\mapsto f(g(h(x)))$. Using a more explicit expression of composition, we might write this associativity as $(f\circ g)\circ h=f\circ(g\circ h)$, but writing all those little circles is rather tedious when we are not doing algebra with the values and don't need to distinguish it from composition. | {
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c#, unit-testing, library, dice
I was thinking about getting hold of a seed value and passing it around (but without forcing consumers to pass in a seed value, unless they really wanted to), so I can get some deterministic values for my test, but the only way I can think to do this while maintaining good encapsulation is making a method like this:
protected virtual long GetSeed() //could be a property too
{
return Environment.TickCount; //or whatever
} | {
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thermodynamics, entropy, thermal-radiation
Light has a specific intensity, meaning a power per unit area per unit solid angle per unit wavelength. Mirrors and lenses can never increase the specific intensity.
As an example, consider using a lens to focus sunlight onto a target. The specific intensity of the sunlight is the same with or without the lens. What the lens does is increase the solid angle of sunlight that the target receives. But, there is a maximum: it is not possible for a flat surface to receive a solid angle of light greater than $2\pi$ steradians. This is why the target can never get hotter than the source, regardless of the optical system used.
Consequently, the mirrors and lenses cannot deliver $Q_{in}=\sigma T_1^4 A_1$ to body 2 with $A_2<A_1$. The most that it can deliver is $Q_{in}=\sigma T_1^4 A_2$ | {
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c#, repository
#region IRepositoryWebService<TSend, TReceive> Members
public virtual T Delete(T entity) { throw new NotImplementedException("This web service does not have an delete/del endpoint for this entity!"); }
public virtual IList<T> Get(Expression<Func<T, bool>> predicate) { throw new NotImplementedException("This web service does not have a select/get endpoint for this entity!"); }
public virtual T Get(T entity) { throw new NotImplementedException("This web service does not have a select/get endpoint for this entity!"); }
public virtual IList<T> GetAll() { throw new NotImplementedException("This web service does not have a select/get endpoint for this entity!"); }
public virtual T GetById(TId id) { throw new NotImplementedException("This web service does not have a select/get endpoint for this entity!"); }
public virtual IList<T> GetList(T entity) { throw new NotImplementedException("This web service does not have a select/get endpoint for this entity!"); } | {
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classical-mechanics, lagrangian-formalism, mathematical-physics, variational-principle, variational-calculus
By straightforward computation one can prove the second implication ($\impliedby$) but I am having trouble with the other one. So far, by using the fact that
$$ \frac{\partial}{\partial \dot q_k} \left( \frac{d}{dt}f \right) = \frac{d}{dt} \left( \frac{\partial}{\partial \dot q_k} f \right) + \frac{\partial}{\partial q_k}f $$
and integrating with respect to time I have arrived to:
$$\frac{\partial}{\partial \dot q_k}f = Ct + D$$
where $C,D$ are constants of integration. However it does not seem right and it doesn't help me in showing that a primitive of $g$ with respect to time can not depend on lagrangian velocities. | {
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• By (perhaps mistakingly) saying that the "first derivatives determine the behavior of the function, I was generalizing from the 1D case in which a C^1 function is determined up to a constant (i.e. in behavior) by it's derivative via integration. – Evan Sep 16 '18 at 22:05
• @evan: Saying a function is C1 is equivalent to saying that it is determined by its first derivatives; there's no new information there. You're simply saying the same thing in two different ways and sowing misinformation and confusion. – Mozibur Ullah Sep 16 '18 at 22:12 | {
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python, beginner, python-3.x, tkinter, gui
class MyApp:
def __init__(self) -> None:
self.window = tk.Tk()
self.window.title(string='Stickies')
self.window.attributes('-alpha', 0.95)
self.window.configure(bg='#ffffff')
self.mainloop = self.window.mainloop
width, height = self.getscreensize_fraction(0.3, 0.5)
self.window.geometry(f'{width}x{height}+{width+70}+10')
frame = tk.LabelFrame(
self.window, text='Stickies', bg='#ffffff', font=MEDIUM_FONT,
)
frame_name = tk.LabelFrame(
frame, text='Name your note', bg='#ffffff', font=MEDIUM_FONT,
)
frame_colors = tk.LabelFrame(
frame, text='Colors', bg='#ffffff', font=MEDIUM_FONT,
)
self.name_entry = tk.StringVar()
name_entry = tk.Entry(
frame_name, textvariable=self.name_entry, font=SMALL_FONT, bg='#f1f1f1',
) | {
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neural-network
Title: How do multiple linear neurons together allow for nonlinearity in a neural network? As I understand it, the point of architecting multiple layers in a neural network is so that you can have non-linearity represented in your deep network.
For example, this answer says: "To learn non-linear decision boundaries when classifying the output, multiple neurons are required."
When I watch online tutorials and whatnot, I see networks described as in the screenshot below. In cases like this, I see a series of linear classifiers: | {
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javascript, optimization, jquery, jquery-ui
The reason the widget makes call to .destroy is because there is no garbage collector for the DOM. Your supposed to clean up the dom. Your supposed to remove this.element, detach any event handlers. And whatever else needs cleaning up.
You do not need to clean up javascript. JS has a garbage collector.
Here is the full code annotated:
just ctrl-f //*
(function($) {
$.widget("ui.tagit", {
// default options
options: {
tagSource: [],
triggerKeys: ['enter', 'space', 'comma', 'tab'],
initialTags: [],
minLength: 1
},
_keys: {
backspace: 8,
enter: 13,
space: 32,
comma: 44,
tab: 9
},
//initialization function
_create: function() {
var self = this;
this.tagsArray = []; | {
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} |
quantum-mechanics, quantum-spin, rotation, spinors
More specifically, one eigenspinor (before rotation) is $\chi_+ = (1,0)$ and applying the rotation in SU(2):
\begin{equation}
e^{-\frac{i}{\hbar} \sigma_z \varphi} e^{-\frac{i}{\hbar} \sigma_y \theta}\chi_+ = \begin{pmatrix}e^{-i\varphi/2} \cos(\theta/2) \\ e^{i\varphi/2} \sin(\theta/2)\end{pmatrix}
\end{equation}
which is (up to a phase factor) the result from Wikipedia (https://en.wikipedia.org/wiki/Eigenspinor)
Another way is to start with an arbitrary magnetic field and then compute the eigenvectors of the Hamiltonian. That is, one rotates the vector $\hat{\mathbf{n}} = (0,0,1)$ such that afterwards $\hat{\mathbf{n}} = (\cos \varphi \sin \theta, \sin \varphi \sin \theta,\cos \theta)$. Doing this one finds the same eigenspinors.
So I see that the results are the same, however, I don't really understand how the 3-dimensional rotation (of the space) is related to the 2-dimensional one (of the spinor). | {
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A word of explanation: What I tried to do here was take seriously the OP's request for estimates you can calculate in your head. The ability to do mental arithmetic, of course, varies from person to person, so I used myself as a gauge. As for where the formulas came from, my starting point was the observation that the conversion factor between degrees and radians, $180/\pi$, is approximately $60$, so the estimate $\tan\theta\approx\theta/60$ should be OK for a while. A little trial and error showed it's good up to $\theta=20$ degrees (since $.9\tan20\approx.328$). It was easy to see that connecting $(0,0)$ to $(20,1/3)$ and $(20,1/3)$ to $(45,1)$ with straight lines would stay within the prescribed bounds. Finally, noting that $.9\tan60\approx1.55$, I saw that the line connecting $(45,1)$ to $(60,1.6)$ would have a nice slope and stay within the prescribed bounds as well. | {
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"url": "https://math.stackexchange.com/questions/446076/mental-estimate-for-tangent-of-an-angle-from-0-to-90-degrees"
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electromagnetic-radiation, atmospheric-science, climate-science
Title: Does aerosol particulate matter attenuate UVA more as compared to UVB? I am trying to understand if winter aerosol particulate matter has a greater impact on UVB radiation as compared to UVA radiation. Would the attenuation be mainly due to scattering?
I am asking this question for latitudes less than 30 degrees north where UVB radiation is expected to reach ground even in winter. Firstly, a definition of what exactly is termed as particulate matter, according to the Australian Government Dept. of Environment and Energy page Particles. the are defined as being:
Airborne particles are sometimes referred to as 'particulate matter' or 'PM'. They include dust, dirt, soot, smoke, and liquid droplets. Some particles are large enough or dark enough to be seen as soot or smoke, while others are so small they can only be detected individually with a microscope
Particulate matter is further defined by their size (from the link above): | {
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$$2 \times 2 \times 2 = 2 ^ 3$$
This is all well and good but when we talk about "raising to the power of 0" we need to know what number to put on the left hand side:
$$? = 2 ^ 0$$
The base for multiplication is $1$. It's the multiplicative identity. Every multiplication equation may be implicitly started with $1$. This means that above, two to the power of three is actually:
$$1 \times 2 \times 2 \times 2 = 2 ^ 3$$
In this form certain behaviors become quite clear:
$$1 \times 2 \times 2 \times 2 = 2 ^ 3$$ $$1 \times 2 \times 2 = 2 ^ 2$$ $$1 \times 2 = 2 ^ 1$$ $$1 = 2 ^ 0$$
Likewise, negatives also make sense, because instead of multiplying numbers, you do the opposite, you un-multiply (often called "division"):
$$1 = 2 ^ 0$$ $$1 \div 2 = 2^{-1}$$ $$1 \div 2 \div 2 = 2^{-2}$$
Note that these patterns hold regardless of the base: | {
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So without stopping to check, let's just see the nice point here. All the components of eigenvectors are in this picture. Here we've got 8 eigenvectors. 8 eigenvalues, 8 eigenvectors.
The eigenvectors have 8 components and every component is one of these 8 numbers. The whole thing is constructed from the same 8 numbers. The eigenvalues and the eigenvectors. And really the key point is, what is the matrix of eigenvectors?
So, let's just write that down. So, the eigenvector matrix for all circulants of size N. They all have the same eigenvectors, including P. All circulants C of size N including P of size N.
So, what's the eigenvector matrix? What are the eigenvectors? Well, the first vector is all 1's. Just as there.
So, that's an eigenvector of P, right? Because if I multiply by P, I do a shift, a cyclic shift, and I've got all 1's. The next eigenvector is powers of w. | {
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"url": "https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-31-eigenvectors-of-circulant-matrices-fourier-matrix/"
} |
machine-learning, python, neural-network, keras, mlp
Title: Keras MLP Classifer Model not converging I'm trying to make MLP based classifier based on numerical and categorical data
The train_X (Input) data that I'm working with is look like that
each data type is 20 numerical and 1 categorical data ,categorical here
I've converted that using cat.codes
My train_Y (prediction) data looks like this :
winPlacePerc
0 0.4444
1 0.6400
2 0.7755
3 0.1667
4 0.1875
5 0.0370
6 0.0000
7 0.7368
8 0.3704
9 0.2143
10 0.3929
its between 0 to 1
I am using here keras sequential model :
from keras.models import Sequential
from keras.layers import Dense
from keras.layers import Dense, Dropout, Activation
from keras.callbacks import EarlyStopping
#create model
model = Sequential()
#get number of columns in training data
n_cols = train_X.shape[1] | {
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navigation, robot-localization
However, please note that the Z values you are posting are effectively 0. In the example you posted above, 3.87298e-06 is 0.00000387298. If your system is sensitive to that level of error, you may want to stick with the EKF. | {
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and concepts. You must turn them in to me via a physical copy. Montgomery, George C. Probability Questions and answers PDF with solutions. to share Solutions for all STEM major Problems. Long chapters are logically split into numbered subchapters. Two and three-digit subtraction. Probability - math word problems Probability is the measure of the likeliness that an event will occur. 13% probability that exactly 7 of 10 IT startups will generate a profit in their first year when the probability of profit in the first year for each startup is 80%. Fifty marbles are to be drawn from the jar in problem #1 with replacement. Probabilities, Counting, and Equally Likely Outcomes. Evans and Jeffrey S. The bet can either be on red or on blue and the amount of each bet is arbitrary. Thus, the probability that a randomly chosen piece is highlighted is 1/2. Work on the probability problems on pages 1-4 and 1-5 of the lab manual 2. The probability of event A is the number of ways event A can occur | {
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newtonian-mechanics, classical-mechanics, chaos-theory, fractals
It seems apparent in this computational solution that there are large swaths which have constant (therefore differentiable) solutions outside of the base local minimum. I wonder if this is physical or a remnant of the computational method not being truly continuous or if there are consistent solutions we can look to in these systems. Would a true physical system actually show more of a probability density rather than a guaranteed final solution? Is this problem analogous to other similar simple chaotic systems?
Any help in interpreting this problem would be greatly appreciated!
https://www.ioc.ee/~dima/YFX1520/flyer_chaotic_pendulum.pdf
https://beltoforion.de/en/magnetic_pendulum/ It probably helps thinking about where exactly the discontinuities on your map lie and what they arise from. | {
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# What does "max" refer to in the definition of a matrix norm?
I am going through my numerical analysis text (Epperson) and came across notation that I don't fully understand. My question is what does max refer to in this definition?
Definition $7.2$ (Matrix Norm): Let $\|\cdot\|$ be a given vector norm defined on $\mathbb{R}^n$. Define the corresponding matrix norm, for matrices $A \in \mathbb{R}^{n\times n}$, by $$\|A\| = \max_{x \neq 0}\frac{\|Ax\|}{\|x\|}$$ | {
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cc.complexity-theory, lo.logic, sat
Minimal Assignments for Bounded Model Checking, Ravi and Somenzi, 2004. (I did not find a freely available PDF, but maybe you can mail the authors if you don't have Springer access).
Applying SAT Methods in Unbounded Symbolic Model Checking, McMillan, 2002.
About your algorithm
There are certain things you are doing in your algorithm which makes me wonder if you are familiar with modern SAT algorithms. You are describing a mix of the original Davis-Putnam and the Davis-Logeman-Loveland algorithm from the 60s. Specifically, Steps 4 and 5 are implemented differently today in what is called the unit rule.
Please confirm if this is what you mean by reliable assignment.
A partial assignment $\pi$ is reliable for a formula $F$ if some
extension of $\pi$ satisfies $F$. | {
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c++, strings
if (checkString == password)
return checkString;
}
return "TARGET_PASSWORD_NOT_FOUND";
}
int main() {
while (1) {
std::string password;
std::cout << "Enter password : ";
std::cin >> password;
std::cout << "Your password is : " << password << std::endl;
std::string attempt = search(password);
std::cout << attempt << std::endl << std::endl;
}
} This is a critique, and a rewrite of the code you showed, that is, the code for checking if a given password only contains characters from a given character set; for an actual brute force password cracker, see e.g. this question. | {
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Remaining volume after a plane cuts a sphere
I want to find the volume remaining (the smaller volume) after the plane $P: x+y+z = 1$ cuts the sphere $S: x^{2} + y^{2} + z^{2} = 1$. I found that the problem could be solved by finding a plane (or line) perpendicular to the plane P, then treating the remaining volume as if it were the cap of a sphere (say when a plane $z = 1/2$ cuts the sphere giving a spherical cap), using as the "height" the distance between the sphere and the plane P (along that perpendicular plane).
However, I tried to solve the problem without using this fact by trying to use triple integrals (both in cylindrical and spherical coordinates). In cylindrical coordinates, I had that (I split up the integral into 4 different regions, the "bigger part" in the first octant and the three other smaller parts in the other octants).
For the first integral, we would have that the projection in the xy plane is the line $x+y=1$ and the circle $x^{2} + y^{2} + 1$: | {
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"tags": null,
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plate-tectonics, mantle, isostasy
Images courtesy of http://www.physicalgeography.net/fundamentals/10h.html | {
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python, design-patterns
Public methods need docstrings
You provide docstrings for _data() and _raw_data(), but not for alerts(). This is backwards.
The leading underscore represents private methods. An external caller does not need to know how to call them, because they are private.
On the other hand, public functions (without the leading underscore) are expected to be called by external callers. And someone writing the code which uses these Source objects may want to know how to call the methods. So they may type:
>>> help(NagiosSource)
and would reasonable expect to get information about how to use the class and its public method. The docstring for _data and _raw_data would not be provided, because of the leading underscore.
You may provide docstrings for private methods (the expectation is the public methods would all have been documented first), but code comments may be just as useful. | {
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$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$
If $n$ is also relatively small, then
$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$
Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,
$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$
is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get
$${2.7136203+2.7152613\over2}=2.7144408$$
which is quite close to the true value! | {
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ros, node
operator!=(const complex<_Tp>& __x, const complex<_Tp>& __y)
^
/usr/include/c++/4.8/complex:474:5: note: template argument deduction/substitution failed:
/home/ubuntu/catkin_ws/src/moon/src/SpanningTree.cpp:96:46: note: mismatched types ‘const std::complex<_Tp>’ and ‘int’
for (i = adj[j].begin(); i != adj[j].end(); ++i)
^
In file included from /usr/include/c++/4.8/bits/locale_facets.h:48:0,
from /usr/include/c++/4.8/bits/basic_ios.h:37,
from /usr/include/c++/4.8/ios:44,
from /usr/include/c++/4.8/ostream:38,
from /usr/include/c++/4.8/iostream:39,
from /opt/ros/indigo/include/ros/time.h:54,
from /opt/ros/indigo/include/ros/ros.h:38,
from /home/ubuntu/catkin_ws/src/moon/src/SpanningTree.cpp:4: | {
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the height is r, and the foll… Approach: Formula for calculating the inradius of a right angled triangle can be given as r = ( P + B – H ) / 2. Calculate the radius of a inscribed circle of a right triangle if given legs and hypotenuse ( r ) : radius of a circle inscribed in a right triangle : = Digit 2 1 2 4 6 10 F. =. The ruler will be slightly off-center but the line will not. Si vous continuez à utiliser ce site sans changer vos paramètres de cookies ou si vous cliquez sur "Accepter" ci-dessous, vous consentez à cela. Dangerous Prayers Shirt, That is, it's a diameter of the circle. To prove this first draw the figure of a circle. Conversely, if the diameter of a circle forms one of the sides of an inscribed triangle, the inscribed triangle is a right triangle. Conan Exiles - Gold Location, And we know that the area of a circle is PI * r 2 where PI = 22 / 7 and r is the radius of the circle. Swimming Luxembourg - 2020 - Tous droits réservés -, Protocole spécifique SL & Informations | {
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ros, ros-control
Originally posted by RyanChen.YLC on ROS Answers with karma: 44 on 2022-03-10
Post score: 0
A URDF is not strictly required, however many controller implementations will want to take a look at a URDF to determine/derive some properties of your robot.
The DiffDriveController would be one of those controllers.
If however, you don't use any of those controllers, and you write your RobotHW such that it doesn't rely on a URDF for information about the structure of your robot (ie: nr of joints, their type, etc), then: no, a URDF would not be required.
Note that many other parts of ROS will want/need a URDF though (such as RViz and the robot_state_publisher), so it would still be a good idea to have/make one.
Originally posted by gvdhoorn with karma: 86574 on 2022-06-02
This answer was ACCEPTED on the original site
Post score: 1 | {
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quantum-mechanics, heisenberg-uncertainty-principle
Title: Distribution of Quantum Measurement Uncertainties (what Exactly does the Heisenberg Uncertainty Principle tell us)? In quantum mechanics, one of the first results students (and random people with an interest in physics) are exposed to is the Heisenberg Incertainty Principle. Informally, this principle expresses the product of the uncertainties of certain pairs of properties must be greater than some fixed value. Essentially, the "knowabilities" of certain pairs of properties of a quantum particle are linked such that certainty in one gives rise to uncertainty in another. Typically, the properties position and momentum are given as an example, so I will use those for the remainder of this question.
I will use $x$ and $\Delta x$ for position and uncertainty in position and $p$ and $\Delta p$ for momentum and uncertainty in momentum, all respectively.
My question is this: | {
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electrostatics, charge, conductors
Title: What is the difference between charged body and conductors? In charged bodies,charges can be uniformly distributed all over it's volume, but in conductors charges always lies on it's surface.
Please tell me the difference between charged bodies and conductors?
And tell me how can bodies be uniformly charged while charges reside only on surface in conductors? The difference is charge can flow in a conductor.
Consider what would happen if we started with a solid sphere of metal with excess positive charge initially uniformly distributed through the volume. There would be an electric field in the volume, and so the charges would move. Eventually all the charge will be on the surface.
The charge can also move along the surface. So in electrostatics the constraint is not just that the charge is on the surface, but that the potential is the same value everywhere on the surface of a conductor. (So not necessarily uniform unless the object has a lot of symmetry.) | {
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gravity, terminology, potential, potential-energy, notation
Is there an equation that looks like this:
$$
\Phi (r)=\int F(r)dr\quad\text{?}
$$ I'm going to assume you (and your prof or teacher) are using standard notation.
In that case, $U$ denotes the gravitational potential energy of a configuration of two (or more) objects interacting gravitationally. $\Phi$ denotes the gravitational potential of one object. The difference between them, is that $U$ requires (at least) two objects in order to be defined, while $\Phi$ is the potential of one object. If we are working in the two-body case, one (usually more massive) body with mass $M$ and one (usually smaller, test-body) with mass $m$, then once I figure out the gravitational field of: $\Phi=-GM/r$, I can easily obtain the gravitational potential energy of the two bodies together by: $U=m\Phi = -GMm/r$.
$U$ and $\Phi$ are related, but they are definitely NOT the same thing! You can tell just by the units. $U$ has units of energy whereas $\Phi$ has units of energy/mass. | {
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c++, optimization, matrix
Ry++;
}
// write debug data to bin file
if (m_Debug)
{
writeDebugDataToBinFile("TBL", TBL00, TBL01, TBL02, TBL03, TBL04, TBL05, TBL06, TBL07, TBL08, TBL09, TBL10, TBL11, TBL12, TBL13, TBL14, TBL15);
writeDebugDataToBinFile("W", W00, W01, W02, W03, W04, W05, W06, W07, W08, W09, W10, W11, W12, W13, W14, W15 );
writeDebugDataToBinFile("EXP_IN", EXP_IN00, EXP_IN01, EXP_IN02, EXP_IN03, EXP_IN04, EXP_IN05, EXP_IN06, EXP_IN07, EXP_IN08, EXP_IN09, EXP_IN10, EXP_IN11,
EXP_IN12, EXP_IN13, EXP_IN14, EXP_IN15);
// free memory for debug data
for (int i=0; i < m_Height; i++)
{
delete TBL00[i];
delete TBL01[i];
delete TBL02[i];
delete TBL03[i];
delete TBL04[i];
delete TBL05[i];
delete TBL06[i];
delete TBL07[i];
delete TBL08[i];
delete TBL09[i];
delete TBL10[i];
delete TBL11[i]; | {
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c, unit-testing, linked-list, memory-management, library
insertion before an arbitrary node:
void insert_before(Node * next, Node * node) {
insert_after(next->pred, node);
}
The node being inserted before or after can be the dummy (root) node--the same logic works. Whether inserting at the head or tail, or before or after an arbitrary node, no list traversal is needed.
Return true if the node is a dummy (root) node. This is used to detect the end of list:
bool dummy_node(Node * node) {
node->data == NULL;
}
One advantage of this structure is that no separate cursor (current node) is needed. Every node pointer is automatically a cursor.
Node * next_node(Node * node) {return node->next;}
Node * prev_node(Node * node) {return node->prev;}
Node * node = next_node(dummy);
while(!dummy_node(node))
node = next_node(node); | {
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c++, opencv, ros-indigo
target_link_libraries(${PROJECT_NAME}_node
# ${catkin_LIBRARIES}
)
Add folders to be run by python nosetests
catkin_add_nosetests(test)
find_package(OpenCV)
include_directories(include ${catkin_INCLUDE_DIRS} ${OpenCV_INCLUDE_DIRS})
add_executable(myopencvtest src/myopencvtest.cpp)
target_link_libraries(myopencvtest ${catkin_LIBRARIES} ${OpenCV_LIBRARIES})
add_dependencies(myopencvtest myopencvtest_generate_messages_cpp)
===== End of CMakeLIsts.txt
What am I missing to get the required libraries to be found?
My opencv libs are in /usr/local/lib
Thanks,
Burt | {
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java, random, io, generator
In your current main method, you do lots of List manipulation, all of this can be done elsewhere, you shouldn't need to know how it works.
As a side note, there are a lot of redundant comments in your code. Comments should say WHY you're doing something, not WHAT you're doing. I know that when I see object.method() that you're calling a method, but I might not know why you're saying index - 1 instead of just index, for example.
I hope this was helpful for you, keep it up! | {
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algorithms, time-complexity, binary-trees
We can easily bound the first algorithm as $O(N^2 \log N)$, because it does $O(N^2)$ work at each of $\log N$ levels of the tree. Similarly, we can bound the second algorithm as $O(N \log ^2 N)$, by similar reasoning.
The question is, are these bounds tight, or can we do better? How do we prove it? Apply the master theorem.
The first algorithm takes $O(N^2)$ time because your underlying recurrence is
\[ T(N) = 2T(N/2) + N^2 \]
The second algorithm takes $N\log^2N$ time as you mention. If your work per node is as indicated (and is tight) then this bound is tight as well.
Update: as commenters point out, the assumption here is that the tree is weight-balanced: each child has half the nodes of the parent. This would imply the log N height bound you indicate, but is not implied by it. | {
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"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "algorithms, time-complexity, binary-trees",
"url": null
} |
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2324
The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
Net increase $$=p+(-12)+\frac {p*(-12)}{100}=p-12-\frac {3p}{25} = (\frac {22p}{25} - 12)$$%
It is given in the question that the net increase finally is $$10$$%. Hence, we can equate the two values.
$$(\frac {22p}{25} - 12)$$% = $$10$$%
or,
$$\frac {22p}{25} = 10+12 = 22$$%
or,
$$p = 25$$% | {
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} |
general-relativity, metric-tensor, vectors, tensor-calculus, linear-algebra
Preliminary #3: The Inverse Metric
We've seen that we can use the metric to associate a vector $\mathbf X$ to a dual one-form $\tilde{\mathbf X}$; we can also go the other direction and associate a covector $\tilde{\mathbf Y}$ to a dual vector $\mathbf Y$. We do this by defining the so-called inverse metric $\tilde{\mathbf g}$, which is a map which eats two one-forms and spits out a scalar.
Essentially this is just a metric on the space of one-forms in exactly the same way as $\mathbf g$ is a metric on the space of vectors. However, we link them together by demanding that if the vectors $\mathbf X$ and $\mathbf Y$ have one-form duals $\tilde{\mathbf X}$ and $\tilde{\mathbf Y}$, then
$$\mathbf{g}(\mathbf X,\mathbf Y) = \tilde{\mathbf g}(\tilde{\mathbf X},\tilde{\mathbf Y})$$
It's a straightforward exercise to show that this means that the components of the dual metric $\tilde g^{\mu \nu}$ satisfy the following relationship to the components of the metric: | {
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statistical-mechanics, condensed-matter, bosonization
I don't know much about the fermionic oscillator except what I think Pauli exclusion means $b^+b^+ | 0\rangle = 0$ and $b | 0 \rangle = 0$
How does one rewrite the fermionic operators as bosonic ones?
I've come across the terms fermion boson correspondence and bosonization. I'm not going to attempt to ask for references on the subject because I'm likely to not truly understand them. What struck me about these sorts of things is that it seems possible to turn a theory of interacting fermions into a theory of free bosons.
If the simplest interaction is then cooked up in 1d for two fermions, derived in some way from whatever scheme I am attempting to prepare here, how can the theory be written as free bosons? | {
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"tags": "statistical-mechanics, condensed-matter, bosonization",
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Assume you say “stop” and collect your money at the optimal point, but keep drawing cards until they’re gone. There are six equally likely orders in which your four picks will occur: $rrbb, rbrb, rbbr, brrb, brbr, bbrr$. If you always “stop” after you see the second red card (you should never proceed, since the remaining cards are all black), your winnings will be, respectively, $2, 1, 0, 1, 0, 0$, with an average of $\frac{2}{3}$ dollars. The only hand on which you could have done better by stopping sooner is $rbbr$, but if you try to capitalize on that with a new strategy (stopping if the first card chosen is red, and if not, continuing through the second red card), then your winnings become $1,1,1,1,0,0$, which has the same expectation.
Expected value is defined as probability of the event times the value of the event. However, since you can decide when to end the game, this is hard to compute. That is, of course, if you do not specify that we play intelligently. | {
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"url": "https://math.stackexchange.com/questions/1696460/finding-the-expected-value-for-card-problem/1696511"
} |
fluid-dynamics, waves, acoustics
The speed of sound of the medium inside: lower sound speeds make the waves travel for a longer amount of time before hitting a wall or arriving at the listener. For instance, in water, sound travels much faster(+- 1497 m/s at room temperature) than in air(+- 343 m/s at room temperature).
Reflection/absorption of the walls: The more sound gets reflected each time a wall is hit, the longer there will be sound inside of the room -> the reverb will be longer. In other words: It's better to coat the inside of the room with metal than with cloth, if you want a long reverb.
The attenuation of the sound: A sound gets quieter and quieter, the more distance it has traveled. In one medium sound might travel much further than in another. This also has a huge effect on the length of the Reverb: This makes Rubber, for instance, a very bad medium to use, because sound won't travel very far. | {
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"tags": "fluid-dynamics, waves, acoustics",
"url": null
} |
ros-melodic
[ 0%] Built target _mav_msgs_generate_messages_check_deps_MotorSpeed
[ 0%] Built target _mav_msgs_generate_messages_check_deps_CommandRollPitchYawrateThrust
[ 0%] Built target _mav_msgs_generate_messages_check_deps_CommandMotorSpeed
[ 0%] Built target std_msgs_generate_messages_py
[ 0%] Built target _mav_msgs_generate_messages_check_deps_CommandVelocityTrajectory
[ 0%] Built target _mav_msgs_generate_messages_check_deps_CommandAttitudeThrust
[ 0%] Built target std_msgs_generate_messages_cpp
[ 0%] Built target std_msgs_generate_messages_nodejs
[ 0%] Built target std_msgs_generate_messages_eus
[ 0%] Built target _planning_msgs_generate_messages_check_deps_WayPointArray
[ 0%] Built target _planning_msgs_generate_messages_check_deps_PlannerService
[ 0%] Built target _planning_msgs_generate_messages_check_deps_PlanningResponse
[ 0%] Built target _planning_msgs_generate_messages_check_deps_Octomap
[ 0%] Built target sensor_msgs_generate_messages_lisp | {
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condensed-matter, solid-state-physics, semiconductor-physics
Title: Why doesn’t conventional LED have low IQE problems like OLED? OLED (except for phosphorescent OLED) has low internal quantum efficiency (IQE) because only 25% of electrons in the spin singlet state can emit lights, while the remaining 75% triplet electrons only produce waste heat. However, conventional LED made of GaN operate at the IQE of 80%, which is much higher than 25%. Is it because conventional LED produce higher proportion of singlets, or they can emit light via triplets as well? In LED electron recombine with holes as mentioned by Jon Custer. It can go to the photon energy. The problem appears when electron recombines with the hole and the energy is given to another particle, for example another electron or hole. So this is a process limiting efficiency in LED. This process called Auger recombination.
It is a crystal process, taking into account electron and hole dispersion. Instead of a molecule process as you have, as far as I understand, in OLED. | {
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aircraft
Title: Is plane altitude limited by engine power and if so does air density cause this? I notice that, for example, human-powered flight operates at low altitudes. This might of course be due to safety but I wonder if in fact the delta in air pressure is greater at lower altitudes and this prevents low-powered aircraft from reaching higher altitudes? Human-powered flight is only barely possible because humans cannot generate large amounts of power. Human-powered planes therefore fly almost exclusively in what is called ground effect, where the plane is no more than about one wingspan off the ground. When flying this close to the ground, the plane is partly supported by a "bubble" of air that is caught between the underside of the wing and the ground, which reduces the amount of drag experienced by the plane and enhances the lifting power of its wings. This in turn reduces the power requirement to keep the plane in the air. | {
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CHAPTER 5 LAPLACE TRANSFORMS 5.1 Introduction and Definition In this section we introduce the notion of the Laplace transform. We will use this idea to solve differential equations, but the method also can be used to 12.8 Laplace Transforms: Multiplication and Division by t This section is a continuation of our development of the Laplace Transforms in Section 12.5 , Section 12.6 and Section 12.7. Sometimes the solutions to nonhomogeneous linear differential equations with constant coefficients involve the functions , , or as part of the solution.
The first shifting theorem is a useful tool when faced with the challenge of taking the Laplace transform of the product of exponential function with another function. The Laplace transform is very useful in solving ordinary differential equations. Example 5 . Obtain the Laplace transforms of the following functions, using the Table of Laplace Transforms and the properties given above. (We can, of course, use Scientific Notebook to … | {
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"url": "https://antietamtow.com/kalbeeba/second-shifting-theorem-laplace-transform-examples-pdf.php"
} |
php, object-oriented, url, wordpress
}
I can now use it at follows:
$a = new QueryArgumentsRefer(new ConditionsRefer(), ['authorReferrer' => 'aq', 'taxReferrer' => 'tq']);
var_dump($a->getReferrerArgs());
where var_dump($a->getReferrerArgs()); produces the following output if I'm on a page where the query variable tq is set in the URL and the user have set the taxReferrer arguments
array(1) {
["tax_query"]=>
array(1) {
[0]=>
array(3) {
["taxonomy"]=>
string(8) "category"
["terms"]=>
string(1) "1"
["include_children"]=>
bool(false)
}
}
} | {
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"tags": "php, object-oriented, url, wordpress",
"url": null
} |
semiconductor-physics, conductors, electronic-band-theory, insulators, fermi-energy
Generally, we use the term 'semiconductor' to refer to band insulators where either (i) the bandgap is small enough that at room temperature there will be enough thermal excitations to bring sufficient population from the valence to the conduction band for the conductivity to be affected, (ii) the Fermi energy is close enough to either end of the bandgap that there is a nontrivial effect on the population of the relevant band, or (iii) the material is easy enough to dope that we can control where the Fermi energy sits in the bandgap.
Note, in particular, that all of those conditions have a bunch of qualifiers (like "enough") which have no intrinsic meaning: they depend on judgement and on the situation. A material might be classed as an insulator in some conditions and as a semiconductor in others, depending on things like temperature, the tools we have available to dope it, and how small a conductivity we're willing to consider before we impose an 'insulator' status. | {
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"tags": "semiconductor-physics, conductors, electronic-band-theory, insulators, fermi-energy",
"url": null
} |
ros, nao-driver, nao
[nao_speech-7] killing on exit
[nao_leds-6] killing on exit
[nao_behaviors-5] killing on exit
[nao_controller-4] killing on exit
[nao_walker-3] killing on exit
[nao_sensors-2] killing on exit
Traceback (most recent call last):
File "/opt/ros/hydro/lib/nao_driver/nao_tactile.py", line 192, in <module>
ROSNaoTactileModule = NaoTactile("ROSNaoTactileModule")
File "/opt/ros/hydro/lib/nao_driver/nao_tactile.py", line 80, in __init__
self.init_almodule()
File "/opt/ros/hydro/lib/nao_driver/nao_tactile.py", line 121, in init_almodule
self.broker = ALBroker("%sBroker" % self.moduleName, self.ip, self.port, self.pip, self.pport)
File "/home/antoinerioux/pynaoqi-python-2.7-naoqi-1.14-linux64/inaoqi.py", line 164, in __init__
this = _inaoqi.new_broker(*args)
KeyboardInterrupt
Exception TypeError: "in method 'module_getName', argument 1 of type 'AL::module *'" in <bound method NaoTactile.__del__ of <__main__.NaoTactile; >> ignored
[rosout-1] killing on exit | {
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Here is one example too based on one of the formulas that I have written above: Evaluate: $\color{green}{\int \frac{1}{\sqrt{9 - 25x^2}}dx}$ Solution: One must note that in the given formulas, the coefficient of x is 1 or -1.. Here, the coefficient of x is not 1 or -1, instead it is -25 or you can say 25.. So, firstly you should make it 1 by taking 25 common of the square root and one should also note that when a term is taken out of square root bracket, then it gets square rooted.. So 25 has the square root 5 so that there comes the factor 5.. Now, the integral becomes: $\huge \color{blue}{ = \int \frac{1}{5\sqrt{\frac{9}{25} - x^2}}dx}$ $\huge \color{red}{= \frac{1}{5} \int \frac{1}{\sqrt{(\frac{3}{5})^2 - x^2}}dx}$ This resembles like the formula that I have written at number $$\color{violet}{\mathbf{4..}}$$ Using that Formula: $\huge \color{cyan}{= \frac{1}{5}(sin^{-1} \frac{x}{\frac{3}{5}} + C)}$ $\huge \color{orange}{= \frac{1}{5}sin^{-1} \frac{5x}{3} + \frac{C}{5}}$ As | {
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"openwebmath_score": 0.8592246770858765,
"tags": null,
"url": "http://openstudy.com/updates/4ff68236e4b01c7be8c976f1"
} |
kinematics, manipulator, orientation
$$\hat{x_2} = \cos\beta \hat{y_e} + sin\beta \hat{z_e}$$
$$\hat{y_2} = \cos\beta \hat{y_e} - sin\beta \hat{z_e}$$
$$\hat{z_2} = -\hat{x_e}$$
Note that $\hat{x_e}$ points into the page and $\hat{z_2}$ points out of the page.
Then do you see how your rotation matrix above falls out of that? Clearly,
$$F_2 = R_e^2F_e$$ | {
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\begin{align} &\frac{96}{v_2-12} = \frac{96}{v_2}-40 \\ &\qquad\implies \frac{96}{v_2-12} = \frac{96-40v_2}{v_2} \\ &\qquad\implies v_2\cdot 96 = (v_2-12)\cdot (96-40v_2) \\ &\qquad\implies v_2\cdot 96 = v_2\cdot 96-40v_2^2-1152-380v_2 \\ &\qquad\implies 0 = -40v_2^2-380v_2-1152 \end{align}
Solving this quadratic equation yields no real roots.
Could you please suggest the right way to go? | {
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"url": "https://math.stackexchange.com/questions/2599667/speed-of-two-trains-travelling-side-by-side/2599677"
} |
$$((x+1/2)^2-1/4)^{1/2}-Ax-B;$$
Set $$y:=x+1/2,$$ and consider $$y \rightarrow \infty$$.
$$(y^2-1/4)^{1/2}-A(y-1/2)-B;$$
The asymptote of $$(y^2-1/4)^{1/2}$$ is $$y$$.
Collecting corresponding terms in the asymptotic expression:
$$y(1-A)+((1/2)A-B) =0$$ we get
$$A=1$$ and $$B=1/2.$$ | {
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"openwebmath_score": 0.9744037389755249,
"tags": null,
"url": "https://math.stackexchange.com/questions/4365137/solving-limit-with-two-unknown-constants"
} |
connectivity of neighboring cells of separating the observations to.! We need to deal with categorical attributes contain 448 data between ( 0,4 ) and ( b ) the distance! Between 2 points 3.5 ) and ( -5.1, -5.2 ) in 2D space are... Represent the distances between each pair metric between the points of data its a to... Type of data its a way to calculate distance ( or equal the! Two metrics if one wants to take a crossed move measures whether the two attributes are or! Details below or Click an icon to Log in: you are commenting your. Are squares with sides oriented at a 45° angle to the origin, is. Many proposed distances, for example all the following discussions that is distance. Study showed Imagine we have the distance between two points is the can. Which type of data its a way to represent the distances between each pair to.. Compatible. Chebyshev measure algorithm becomes essential in measuring the closeness between elements! Out / Change ), XYZ, HSL, and Minkowski to chebyshev | {
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"openwebmath_score": 0.6235116124153137,
"tags": null,
"url": "http://chodel.gmina.pl/giant-robot-uxcbi/chebyshev-distance-vs-euclidean-ff48ec"
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c++, sdl
buttonPress(e, yesButton.mButton, 0);
buttonPress(e, noButton.mButton, 1);
}
std::string dialogue;
if (x == 1)
{
dialogue = "YOU WON!!! The correct answer was " + std::to_string(randomNumber) + ".";
}
else
{
dialogue = "You lose. The correct answer was " + std::to_string(randomNumber) + ".";
}
SDL_RenderClear(gRenderer);
highLowTexture.render(0, 0);
LTexture winlose;
winlose.loadfromText(dialogue, textColor);
winlose.render(335, 70);
LTexture playAgain;
playAgain.loadfromText("Play again?", textColor);
playAgain.render(325, 300);
yesButton.render(300, 350);
noButton.render(300 + yesButton.mWidth + 10, 350);
SDL_RenderPresent(gRenderer);
}
}
void renderScene(std::string guessInput, int compare)
{
std::string dialogue; | {
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} |
json, web-scraping, xpath, xslt
<xsl:template match="tr[contains(@class, 'bbGrid-row')]">
<xsl:map>
<xsl:variable name="headers" select="./ancestor::table[1]/thead/tr/th"/>
<xsl:for-each select="td[not(contains(span/@class, 'icon-plus'))]">
<xsl:variable name="index" select="count(preceding-sibling::td) + 1"/>
<xsl:map-entry key="string($headers[$index])" select="if (./div[2]) then normalize-space(./div[2]) else normalize-space(.)"/>
</xsl:for-each>
<xsl:if test="following-sibling::tr[1][td/div/table]">
<xsl:map-entry key="'Dispositions'">
<xsl:variable name="dispositions" as="map(*)*">
<xsl:apply-templates select="following-sibling::tr[1]"/>
</xsl:variable>
<xsl:sequence select="array{$dispositions}"/>
</xsl:map-entry>
</xsl:if>
</xsl:map>
</xsl:template> | {
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Intuitively, you could think in terms of modular arithmetic: Call 3 members of your set $a,b$ and $c$. Suppose $a-b=k_1 \mod 13$, and $a-c= k_2 \mod 13$. Now if $k_1=k_2$, then $b-c$ will be divisible by 13, so you need $k_1 \neq k_2$ to have problems. But you have 13 such equations, and only 12 different values for $k$, so apply the pigeonhole principle here and you are done.
I'm glad I now know what the Pigeonhole Principal is after having heard it so many times. :) | {
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"openwebmath_score": 0.7784197926521301,
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"url": "https://math.stackexchange.com/questions/1298290/in-every-set-of-14-integers-there-are-two-that-their-difference-is-divisible-b/1298320"
} |
java
If you really want to use strings though you could use a map but I think it will just make your code more complicated.
Coin:
Just do the coin check at the start of your function by setting the start index for current player. For example if you want the user to start if the coin lands on heads which you can represent as 0, you then just set the current player index to a random value between 0 & 1.
Example:
int[2] pointsArray;
int currentPlayer = (int)Math.random(); //Random value between 0 & 1
pointsArray[currentPlayer] = roolDiceFunction(); | {
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quantum-field-theory, particle-physics, quantum-electrodynamics, scattering-cross-section
If the considered QFT is unitary (at least perturbatively), there is a constraint on the cross-section following from the unitarity, called Froissart bound. It is the direct consequence of the optical theorem. The latter states that
$$
\sigma_{1+2\to\text{all}} \simeq \frac{1}{s}\text{Im}(M_{\text{forward}}),
$$
where $\sigma_{1+2\to\text{all}}$ is the $1+2\to \text{all}$ scattering cross-section, while $s$ is the invariant mass of the colliding particles. The amplitude $M_{\text{forward}}$ is the forward scattering amplitude bounded by
$$
M_{\text{forward}}\lesssim s\ln^{2}(s)
$$
Therefore, for general $1+2\to\text{all}$ scattering one obtains the Frossairt bound:
$$
\sigma_{1+2\to \text{all}} \lesssim \ln^{2}(s)
$$
But I would think that particles scatter more when they are given more
energy? | {
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finite-impulse-response
Title: How to use IPP FIRGenBandStop method? Following this link:
http://software.intel.com/sites/products/documentation/doclib/ipp_sa/80/ipp_manual/GUID-5457048E-3C59-43E4-8A99-FB3873DCB769.htm
I have produced the following code:
const int len = 512;
Ipp64f pInputSignal[len];
int tapslen = 27;
int numIters = len;
IppsFIRState_64f* pState;
IppStatus status;
Ipp64f* pOutputSignal = ippsMalloc_64f(len*sizeof(Ipp64f));
Ipp64f* taps = ippsMalloc_64f(tapslen*sizeof(Ipp64f));
Ipp64f* pDL = ippsMalloc_64f(tapslen*sizeof(Ipp64f));
ippsZero_64f(pDL,tapslen);
float sample_rate = 48000.0;//Hz //TODO DEFAULT_SAMPLE_RATE
float frequency = 400.0;//Hz
float amplitude = 2000.0;
//Fill pInputSignal[t] = 2*pi*frequency*t, where t is time from t[i] = i/sample_rate
fillBufferWithSinusoidData_64f(pInputSignal,len,sample_rate,frequency,amplitude); | {
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neural-networks, reinforcement-learning, game-ai
You don't describe how the game is played, but a common approach for extracting actions is to have one output neuron for each possible action. To select an action, you would pick the one corresponding to the neuron with the highest output response to a given input. | {
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"tags": "neural-networks, reinforcement-learning, game-ai",
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particle-physics, experimental-physics, software, large-hadron-collider
One of the comments mentioned the Kaggle Higgs competition, as far as I know that's already heavily processed data, but in theory it's available to anyone.
I doubt if any collaboration would provide raw data from experiments publicly without some verification that the person requesting it knew what they were doing.
Cautionary note- HEP data analysis isn't like amateur astronomy - it takes a large team. Just look at the length of the author list on any typical HEP paper. | {
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$$F[x]$$ denotes the ring of (formal!) polynomials over $$F$$ in the indeterminate $$x.$$
$$F(x)$$ denotes its fraction field - the so-called rational "functions" in $$x.$$ | {
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"lm_q2_score": 0.8479677526147223,
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"openwebmath_score": 0.9081282615661621,
"tags": null,
"url": "https://math.stackexchange.com/questions/720371/question-about-notation-in-field-theory?noredirect=1"
} |
resource-recommendations, education, mathematics, calculus
The comprehensive analysis text is Fundamentals Mathematical Analysis. It's a Russian textbook, but it's old school, i.e. very readable.
Another must have book is Differential Equations and Calculus Variations.
The best reference on PDEs is PDE by Bitsadze, I consult it all the time, it's very thin, and chapters are mostly self-contained.
All these books were used by Physics students, I can guarantee that. | {
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c++, c++11, parsing
// tokenize a byte stream. returns a vector of tokens that can be tested using validate
std::vector<Token> tokenize( const std::vector<uint8_t>& message_data, const std::vector<Token::TokenClass>& expects )
{
std::vector<Token> ts;
auto it = std::begin( message_data );
for ( auto&& expect : expects )
{
switch (expect)
{
case Token::BYTE:
{
if (it == std::end( message_data ))
{
std::stringstream ss;
ss << "not enough data left in stream for byte";
throw std::out_of_range(ss.str());
}
ts.emplace_back( Token::BYTE, *it );
}
break; | {
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"tags": "c++, c++11, parsing",
"url": null
} |
navigation, odometry, rplidar, turtlebot, ros-kinetic
Title: Does move_base require Twist from Odometry?
I'm trying to make a turtlebot like robot using only the RPLidar. Hence I donot have a good source for odometry. I'm getting the Pose2D using laser_scan_matcher, and can use the values and represent it in nav_msgs/Odometry.
I've also found a way to use these pose2D values with /scan headers to roughly calculate velocities.
But I want to know is it mandatory to have Twist information for move_base to work? Or does move_base only require Pose from the odometry messages? | {
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"tags": "navigation, odometry, rplidar, turtlebot, ros-kinetic",
"url": null
} |
# Integration using Monte Carlo Method
I'm trying to solve this integral using the Monte Carlo Method.
$$I=\int_0^\pi \frac{1}{\sqrt{2\pi}}e^\frac{-sin(x)^2}{2}dx$$
Now it seems to me that there is a normal probability density function in there, but I'm not sure because of the sine function.
If there is a normal, then it's easy to simulate $N$ random numbers from the standard normal distribution and compute $I$. It's is also easy to find the size of the sample to get a 95% confidence interval.
So is there a normal probability density function? Or from what distribution should I get the random numbers to compute $I$?. | {
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"url": "https://math.stackexchange.com/questions/1909290/integration-using-monte-carlo-method"
} |
simplify the matrix since the columns into rows is called column orthogonal when ATA = =. 2 i.e a 4D matrix this property by taking an example n ) array known. Adjugate matrix and two columns and the columns are orthonormal changes its sign we... Calculate its determinant by −1 columns is called the pivot column human and gives you temporary access the. Which can not hold 2000 columns: this problem can be solved by keeping either number... This code snippet in C. int x=5, y=6 ; x=x+y ; y=x-y ; x=x-y ; if!., it is obtained by changing rows and 3 columns in the matrix is... Is always more than 256 original rows, you can not transpose these unless you are Excel! Ray ID: 5fd3023aedfce4fa • your IP: 192.145.237.241 • in a matrix interchanging of rows and columns is called & by. Order m * n then = || of order n * m. So, 2 in a matrix interchanging of rows and columns is called and columns B... List ( list inside a list ) input data in Excel which has 2000 rows columns. 0 to the web property | {
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"url": "https://smartsocialtrading.com/amazing-superheroes-gcp/archive.php?763fb4=in-a-matrix-interchanging-of-rows-and-columns-is-called"
} |
method works best for you can not combine unlike '' radical terms like and unlike.. To have the same in each term, so also you can use the site.: Step 1 with answers tutorial we will look at adding, and... Apples and oranges adding radical expressions, so also you can use the Mathway widget below to Practice adding... Another three copies '' terms that I can simplify those radicals right down to number. Process of adding radical expressions Step 2 to ensure you get the best experience ⦠adding radical adding! ) Make sure the radicands are identical order adding radical expressions be able to combine radical terms together, those terms to... Of other math skills and subtraction are possible subtracting Rational expressions with the parentheses, shows reasoning! A 2 out of the radical, added to another three copies able to combine radical terms together, terms. Middle Step, with the same radicand like radicals like radicals to remind us they work the radicand! unlike '' radical terms | {
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"url": "https://oliorama.com/ailbhe-reddy-hsam/adding-radical-expressions-263404"
} |
special-relativity, speed-of-light, quantum-entanglement, quantum-interpretations, information
If person A & B measure simultaneously (in a relativistic meaning), person A & B know each others' results after the measurement. So, in a way there was information transfered (namely each other's state) to each other with a speed greater than light. But, I am pretty sure that even though they know now each other's state, this is not the kind of information one normally talks about when saying that information cannot be transfered at a speed greater than light. So what kind of information are we really talking about? | {
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"url": null
} |
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