text stringlengths 1 1.11k | source dict |
|---|---|
arduino
Title: Baby Child Following Robot I am Planning to make a crawling bot which will follow a baby,having a pillow on the board.The Problem with the project is:
Suppose a case when the baby falls, the bot should come exactly at the place where the baby's head will hit the ground?
Any idea of how i can make the bot move to that point before the baby's head hit the ground ? Start first with a robot that can "come exactly at the place" where you throw a ball.
Once you have that, you will gain experience necessary to create the other phase of the project (going towards the baby).
Start simple, then make it more complicated.
Start with a four-wheeled robot (or maybe three-wheeled). | {
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newtonian-mechanics, reference-frames, centrifugal-force, coriolis-effect, conservative-field
Title: Is centrifugal force a conservative force? The centrifugal force depends on the position so it must be a conservative force whereas the Coriolis force depends on the velocity and it is a non-conservative force.
Is this conclusion correct? The centrifugal acceleration of a point A located at $\mathbf{r}_A$ with linear velocity $\mathbf{v}_A$ riding on a body with rot. velocity $\boldsymbol{\omega}$ is
$$ \mathbf{a}_A = \boldsymbol{\omega} \times \mathbf{v}_A $$
If the center of rotation C is at $\mathbf{r}_C$ then the linear velocity is $\mathbf{v}_A = \boldsymbol{\omega} \times (\mathbf{r}_A-\mathbf{r}_C)$ which makes the centrifugal term
$$ \mathbf{a}_A = \boldsymbol{\omega} \times \left( \boldsymbol{\omega} \times (\mathbf{r}_A-\mathbf{r}_C)\right)$$
A centrifugal force is only defined at the center of mass G, and it is not a vector field like centrifugal acceleration. | {
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"tags": "newtonian-mechanics, reference-frames, centrifugal-force, coriolis-effect, conservative-field",
"url": null
} |
java, socket, stream, network-file-transfer, aes
private byte[] hex2byte(String s) {
return DatatypeConverter.parseHexBinary(s);
}
public static void main(String [] args) throws Exception {
FileSender fileSender = new FileSender(FILE, "localhost", 8081);
fileSender.sendFile();
}
} try-with-resources | {
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"url": null
} |
reinforcement-learning, papers, reward-design, reward-shaping, potential-reward-shaping
The issue is actually worse than you describe---it's more general than $s = s'$. In particular, the issue presents itself differently based on the sign of your potential function. For example, in your case it looks like the potential function is positive: $P(s) > 0$ for all $s$. The issue (as you have found) is that an increase in potential (regardless of whether $s = s'$) might not be enough to overcome the multiplication by $\gamma$, and thus the PBRS term may be negative. In particular, only when the fold-change in $P$ is large enough ($\frac{P(s')}{P(s)} > \frac{1}{\gamma}$) will the PBRS term actually be positive.
The situation changes when the potential function is negative, i.e. if $P(s) < 0$ for all $s$. In this case, you can actually get a positive PBRS signal even when there is a decrease in potential! In particular, only when the fold-change in $P$ is large enough (same inequality as before) will the PBRS term actually be negative. | {
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an argument that establishes an algebraic fact by relying on counting principles. There are also Investigate! We do not do that in the proofs of Theorems 2.5 and 2.6, noting that this work has already been done in [2] and [4]. | {
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homework-and-exercises, electromagnetism, capacitance
If the plates were circles it would be trivial, but being squares, I'm sure I'm missing something. You want to know $$\int_0^a\int_0^a rdxdy$$ with $$r=\sqrt{x^2+y^2}$$ via substitution $x=\alpha a$ and $y=\beta a$, you can bring it into the form
$$ a \int_0^1\int_0^1 \sqrt{\alpha^2+\beta^2}d\alpha d\beta.$$
For the remaining intgral, Wolfram alpha is your friend. With some more work you can show that in the exact result is $$ \int_0^a\int_0^a rdxdy=\frac{a}{3}\left[\sqrt{2}+\log \left(1+\sqrt{2}\right)\right]. $$ | {
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Let's assume for simplicity that $I$ is finitely generated, so that we know that $A^\wedge$ is complete.
To get such an isomorphism, the first thing we want is section $\sigma : A/I \to A^\wedge$ to the canonical map $A^\wedge \to A/I$. Besides $\sigma$ we want a map $\tau : I/I^2 \to A^\wedge$ which is $A/I$-linear (here we are using $\sigma$), maps into the kernel $IA^\wedge$ of the canonical map $A^\wedge \to A/I$ and then induces an isomorphism $I/I^2 \to IA^\wedge/I^2A^\wedge$. Given $\sigma$ and $\tau$ we can at least define an $R$-algebra map $$A/I[[I/I^2]] \longrightarrow A^\wedge$$ In general the existence of $\sigma$ and $\tau$ is not good enough to imply that this map is an isomorphism. But if you assume for example that $I$ is a quasi-regular ideal then it is an isomorphism (follows more or less immediately from the definition of quasi-regular ideals). | {
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power-engineering, electrical-grid
3) It would be an easier target for malicious activity since it would be easier to get to without the need for heavy machinery. However, cable maps aren't very difficult to get, and someone could just as easily target shallow cables using the current system.
Ultimately, does anyone have any criticisms of the potential solution? Or does anyone know of this idea having been implemented but not worked, or was determined to not be feasible? Thank you very much! Your four drawbacks are very good reasons for not laying cables on the ground and just covering them with soil or "stuff".
Erosion will be a major issue. Exposed cables would be a safety hazard to humans and wildlife. It would also increase the risk of cable degradation and failures. Countering the affects of erosion, but covering the cable, will also add to the operating cost of the cable because the uncovered parts of the cable would need to be covered over again, repeatedly. | {
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python, performance, python-3.x, numpy, matplotlib
# Keep count of position in colour cycle
colour_count = 0
ax = add_subplot(111)
axscatter = ax.scatter
# Work out normalising function
chn = chainfrom_iterable(grouping_dict.values())
flattened = list(chn)
std = nanstd(flattened)
mean = nanmean(flattened)
if std:
two_ops = lambda x: (x - mean) / std
v_norm = vectorize(two_ops)
else:
one_op = lambda x: x - mean
v_norm = vectorize(one_op)
# Keep track of total number of values plotted this iteration
total_length = 0 | {
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statistical-mechanics, electric-circuits, semiconductor-physics, perpetual-motion
The resolution here is the same- if the diode is at the same temperature as the rest of the circuit, it has its own noise and is equally likely to pass current in the forward direction as the backwards direction.
If the diode is not at the same temperature as the rest of the circuit, then you can extract some energy from the thermal noise. But there's nothing wrong with that- there are two reservoirs at different temperatures and it becomes just a fancy heat engine. | {
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machine-learning, neural-network, deep-learning, recommender-system, software-recommendation
Title: How to draw neural network diagrams with this particular style? I would like to draw a neural network architecture with the follow style. Do you know which tool can be used to do this? The paper is Operation-aware Neural Networks for User Response Prediction. I asked me something similar as well as I thought that a lot of papers use rather high quality images but it seems that all the authors generate them individually: https://graphicdesign.stackexchange.com/questions/114886/software-to-draw-schemes-quickly
For NNs, there is already an answered question: How to draw Deep learning network architecture diagrams?
Have a look at https://www.quora.com/What-tools-are-good-for-drawing-neural-network-architecture-diagrams and https://softwarerecs.stackexchange.com/questions/47841/drawing-neural-networks | {
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ros, rviz, urdf, ros-melodic, xacro
`
`
(I don´t know why the format of the code is wrong)
Launch the urdf file with robot_state_publisher package as normal.
With this I can use my model of FreeCAD to see it in rviz, without create a complex urdf file.
Best regards.
Alessandro
Originally posted by Alessandro Melino with karma: 113 on 2020-06-19
This answer was ACCEPTED on the original site
Post score: 0
Original comments
Comment by gvdhoorn on 2020-06-19:
This gets you a static mesh. It visualises in RViz yes, but that's not what people typically want when they ask how to convert <insert CAD program here> to URDF.
Comment by Alessandro Melino on 2020-06-22:
Yes, I know is a static mesh, but for my application is enough because my robot just have the wheels as mobile part, so I just want to visualize the robot. Thank you for the information by the way. | {
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php, object-oriented, mvc, ajax, pdo
$core = new market();
/* articles $_POST requests */
if(isset($_POST['action']) && $_POST['action'] === 'insertArticle'){
$code = filter_var($_POST['code'],FILTER_SANITIZE_STRING);
$barcode = filter_var($_POST['barcode'],FILTER_SANITIZE_NUMBER_INT);
$qty = filter_var($_POST['quantity'],FILTER_SANITIZE_STRING);
$brand = filter_var($_POST['brand'],FILTER_SANITIZE_STRING);
$article_name = filter_var($_POST['artName'],FILTER_SANITIZE_STRING);
$article_type = filter_var($_POST['artType'],FILTER_SANITIZE_STRING);
$price = filter_var($_POST['price'],FILTER_SANITIZE_NUMBER_INT);
$promo_stats = filter_var($_POST['promoStats']);
$note = filter_var($_POST['note'],FILTER_SANITIZE_STRING); | {
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quantum-field-theory, particle-physics, dirac-equation, magnetic-moment, protons
$$
-{e\over 2M} \left (\frac{1}{2} F^{\mu\nu} \bar \psi \sigma _{\mu\nu}\psi\right )=i{e\over 2M} \left ( \partial^\nu A^\mu \bar \psi {[\gamma_\mu,\gamma_\nu]\over 4}\psi\right ),
$$
by integrating by parts,
$$
=-i{eA^\mu\over 8M} \partial^\nu \left ( \bar \psi [\gamma_\mu,\gamma_\nu] \psi\right ),
$$
and shape your coupling vertices into your capricious language. Recall to also include the Gordon decomposition canonical part, and adjust m and M to get the ultimate moment you wish. As I insisted, the effective form I recommended is normally adhered to in effective Lagrangeanese today... But you might like this.
The magnetic moments of baryons cannot be calculated from first principles (except perhaps on the lattice), but they can be easily calculated in the effective constituent quark model, indeed one of its early triumphs! You'll find the calculation in most decent books (e.g. D Perkins') on it. | {
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ai-design, agi, ai-safety, neo-luddism, asimovs-laws
Another promising area is only including text data and having artificial judges. This wouldn't immediately mean that AIs would obey laws but it is the start of AIs understanding the word of law. | {
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classical-mechanics
Title: When an object is moving at a constant velocity,would the normal force and $mg$ be equal? Does the object's normal force and $mg$ cancel out, resulting in the two force becoming equal, or would one force be greater than the other? Thank you!
Edit: Also would the $mg$ be considered weight or would it be more correct to just refer to the force as $mg$? If the object is moving on a horizontal frictionless surface, then the normal force equals $mg$.
Yes, $mg$ is the weight of the object. | {
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If the number of R moves is 0, then every move must be a V move, and there is only 1 such route.
If the number of R moves is 1, then there are 2 diagonal moves, first an R and later an L. There are ${7\choose2} = 21$ such routes.
If the number of R moves is 2, then there are 4 diagonal moves, and there are two orders in which they can occur, namely RRLL or RLRL. Any other order would mean the king going off the left side of the chessboard. So in this case there are ${7\choose4}\times2 = 56$ possible routes.
Finally, if the number of R moves is 3, then there are 6 diagonal moves, and there are five orders in which they can occur. You can list these five possible orders if you want, but the significance of this number 5 is that it is the Catalan number C(3) (see property 5 in the list on that link page to understand why Catalan numbers are relevant here). So in this case there are ${7\choose6}\times5 = 35$ possible routes. | {
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python, performance, python-3.x, tree
def main():
tree = AutoCompleteTree()
while True:
text = input(">>> ")
tree.store_input(text)
tree.print_tree()
if __name__ == '__main__':
main() All in all I think your implementation is well done. Clearly, if I didn't have at least a few comments, I wouldn't be taking the trouble to post an answer. Admittedly some (hopefully not all) of my comments could be considered "nitpicking."
Improved Data Structures
The Node.nodes attribute is a list when, as I believe you now know, making it a dict that maps a letter to a Node instance would permit faster searching. Likewise for the AutoCompleteTree.roots attribute which should also be a dictionary that maps a letter to a Node instance.
Type Hints
If you are going to do use type hints, there might be some room for improvement. Function __has_letter_node is defined as follows:
def __has_letter_node(self, letter: str) -> object | None: | {
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c#, wpf, mvvm, system.reactive, reactive-programming
private string __sFull;
public string Full
{
get { return __sFull; }
set { this.RaiseAndSetIfChanged(ref __sFull, value); }
}
readonly ObservableAsPropertyHelper<string> __oapTitle;
public string Title { get { return __oapTitle.Value; } }
readonly ObservableAsPropertyHelper<string> __oapFirst;
public string First { get { return __oapFirst.Value; } }
readonly ObservableAsPropertyHelper<string> __oapMiddle;
public string Middle { get { return __oapMiddle.Value; } }
readonly ObservableAsPropertyHelper<string> __oapLast;
public string Last { get { return __oapLast.Value; } }
//NAME PARSING FUNCTIONALITY BELOW THIS LINE | {
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$So, F(-1) = 0.5$ and
$F(1) = P(-1) + 0 + 0 + ... + P(1)$
= $0.5 + 0.5 = 1$
Correct Answer: $C$
by
Sir IN the GATECSE PROBABILITY 2 links that are FORMULA FOR DISTRIBUTIONS AND NOTES FOR DISTRIBUTIONS ARE NOT opening.
Sir I m confused about f(x) at c -1 is .5 why???? @Arjunsir
Probability
x=-1 0.5
x=1 0.5
F(X) is the cumulative distribution function and let P is the probability and given X is random variable so
The formula for cumulative distribution function is F(X)=P(X<=x)
F(-1)=P(X<=-1)
= P(x=-1)
=0.5
and F(1)=P(X<=1)
=P(x=-1)+P(x=1)
=0.5 + 0.5
=1
1
9,231 views | {
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"tags": null,
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formal-languages, formal-grammars, reference-request, books
Title: Textbook for understanding formal grammars I am looking to understand the Chomsky Hierarchy. I've read some textbooks that touch on formal grammars (textbooks on computability, which relate automata to specific sets of formal grammars, notably context-free grammars and regular grammars), but not one that specifically deals with formal grammars as an object.
Is there a good textbook for this? Probably the most-used textbook today is:
Hopcroft, Motwani, & Ullman, Introduction to Automata Theory, Languages, and Computation (3rd edition).
A couple of other common ones are:
Sipser, Introduction to the Theory of Computation.
Linz, An Introduction to Formal Languages and Automata.
And a couple of older ones: | {
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surface-chemistry, reaction-mechanism, polymers
Title: What is the mechanism of APTES mono-layer formation on glass substrates? The reagent APTES is a fairly common "ink" for microcontact printing, a technique that forms covalent bonds between the silanols found on the surface of the glass and the silane in the the APTES. It's also been demonstrated that the mono-layer then polymerizes somewhat, forming bonds between neighboring silanols.
What is the mechanism by which the printing and polymerization occur?
What is the reason for using APTES instead of a more reactive trichlorosilane? Here's two links (1, 2) that show the simple scheme on forming monolayers. The essential figure is also shown here.
For the second part of your question... I can only point out this article that describes which properties change, if you make the molecules more bulky or add chloride, etc. I would guess that APTES has well known properties and they are "neutral" regarding common applications. | {
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photography, hubble-telescope
have easily saturated Hubble's CCD. So even if Hubble did spend 10 continuous days observing this patch of the sky, it would have been broken into many, many smaller exposures. This means, even if the Moon did get in the way during part of such a 10-day exposure, they could have just not made any observations during that time. | {
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java, object-oriented
Here are some excerpt of how your final code may looks like:
class Branch {
private final ArrayList<Customer> customers;
private final String name;
public Branch(String name) {
this.customers = new ArrayList<>();
this.name = name;
}
public String getBranchName() {
return this.name;
}
public ArrayList<Customer> getCustomers() {
return new ArrayList<>(customers);
}
public void addCustomer(Customer customer) {
if (!isValidCustomer(customer)) {
throw new InvalidCustomerException( customer);
}
customers.add(customer);
}
private boolean isValidCustomer(Customer customer) {
return customer!=null &&
!contains(customer);
} | {
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"tags": "java, object-oriented",
"url": null
} |
quantum-field-theory, general-relativity, dark-matter, dark-energy, theory-of-everything
Even if we succeed to fit GR and QFT in one framework, we still won't have theory of everything, for there would be other unexplained things.
Of course, having both GR-like and QFT-like properties is only a necessary condition, not a sufficient condition. Even ignoring gravity, there are a countless number of different models that satisfy the general principles of QFT, and most of them are completely unrealistic, so clearly just satisfying those general principles is not enough. A similar comment applies to GR even if we ignore quantum effects; by itself, GR would let us have pretty much any spacetime metric we wanted if we ignored the question of what type of matter it would require, so clearly just satisfying the general principles of GR (even if only approximately) is not enough. | {
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"tags": "quantum-field-theory, general-relativity, dark-matter, dark-energy, theory-of-everything",
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proteins, homework, protein-binding, protein-folding, protein-interaction
I am specifically stuck on explaining the effect of the 0.5M urea vs 5M urea on the soluble and insoluble proteins represented by the pellet and supernatant respectively. My thinking is that at low urea concentrations (0.5M) the pellet protein is destabilised by urea and solubility increases showing a band. However, soluble proteins are denatured and show no band. However, at 5M urea, the pellet proteins are reversibly soluble and therefore do not show as a band whereas the soluble proteins are OK and so show a 'normal' band. I don't know if this is the right explanation, so could someone please explain it to me?? What you need to do is compare the relative amounts of the protein in the insoluble (pellet) fraction to the soluble (supernatant) one. This way you can determine how soluble the RMAS has become through the effect of the indicated compound. | {
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"tags": "proteins, homework, protein-binding, protein-folding, protein-interaction",
"url": null
} |
experimental-physics, dark-matter
More minor(?) problems are that the axion-electron coupling constant estimated by Fraser et al. lies an order of magnitude higher than an upper limit found by new laboratory work and that this high coupling constant is inconsistent with a low-mass axion that is capable of being treated as cold dark matter. | {
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python, python-3.x, sqlite, tkinter
def _task_combobox_selection_changed(self, event):
if self.task_combobox.get() != '':
self.button.state(['!disabled'])
def _button_click(self):
if (state):
self.button.configure(text='Start')
self._stop_timer()
else:
self.button.configure(text='Stop')
self._start_timer()
def _get_tasks_for_project(self, project):
selected_project = Project.get_project_by_name(project)
db.get_tasks_for_project(self.selected_project.get())
self.task_combobox['values'] = selected_project.get_tasks()
def _start_timer(self):
global state
state = True
self.project_combobox.configure(state='disabled')
self.task_combobox.configure(state='disabled') | {
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kinetics, enzyme-kinetics
Title: Are good leaving groups good in both directions? In biology, phosphate groups are often used, good leaving groups.
If I had an opposite case, where a phosphate group remains bound to an enzyme while the main substrate is removed (e.g. E-P04-S => E-PO4 + S) is the substrate considered a good leaving group?
Put another way, does the fact that Phosphate is a good leaving group affect the ease at which S is removed? While you are correct that phosphate is a good leaving group, it does not play that role in the reaction you described. | {
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पढें-Railway RPF Ki Taiyari Kaise Kare Puri Jankari Exam, Fees; Railway RPF Gk Questions In Hindi PDF; 500 Synonyms and Antonyms MCQs In PDF For Other Examinations; Biology Book Download In Hindi PDF; CCC Notes In Hindi PDF Download. Maths Formulas allow students for hands-on practice and assist them to score high both in class exam and board exam. Mathematics Formula Sheet Perimeter The perimeter of a polygon is equal to the sum of the lengths of its sides. 433 psi 1 square foot. The diameter of its circular base is 12 inches. Google "ACT Math Formulas" and you get a grab-bag of subpar results, including an old PDF from 1996 and two popular but glaringly incomplete lists of ACT Math Formulas. Download Algebra formula PDF: Algebra formula PDF Chart is available here to download. To increase 500 by 25% you can either find 25% of 500, then add it on or find 125% of 500. For example, lease rental payments on real estate. formula it is located there. more hyperlinks within the handbook and | {
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"openwebmath_score": 0.6267150044441223,
"tags": null,
"url": "http://wgml.podistiagnadello.it/math-formulas-pdf.html"
} |
python, image, steganography
Parameters:
image_path (str or None): Path to the image.
Depending on if the path or filename is not provided,
the code will automatically fill in the blanks with
default values.
allow_print (bool): If printing the current progress of the
code execution is allowed.
Set to False to totally disable all printing.
"""
self.defaults = ISGlobals().get
self.allow_print = allow_print | {
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} |
c++, algorithm, boost, numerical-methods
template<class Real>
static void BM_tanh_sinh_runge_move_pole(benchmark::State& state)
{
auto f = [](Real t) { Real tmp = t/5; return 1/(1+tmp*tmp); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) -1, (Real) 1));
}
}
template<class Real>
static void BM_tanh_sinh_exp_cos(benchmark::State& state)
{
auto f = [](Real t) { return exp(t)*cos(t); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) 0, half_pi<Real>()));
}
} | {
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"tags": "c++, algorithm, boost, numerical-methods",
"url": null
} |
c, graph
void min_spantree(int no_vert, int no_edge, int root)
{
int u, v, i;
struct queue *q = init_q();
for (i = 0; i < no_vert; i++) {
key[i] = MAX_INT;
parent[i] = -1;
}
key[root] = 0;
for (i = 0; i < no_vert; i++) {
q = enqueue(q, i);
}
/* printf(" Key Parent\n");
for (i = 0; i < no_vert; i++) {
printf("%6d%6d\n", key[i], parent[i]);
}*/
while (!is_empty(q)) {
// traverse(q);
u = extract_min(&q, no_vert);
// printf("U: %d\n", u);
find_adj(u, no_vert);
for (i = 0; i < adj_top; i++) {
v = adj[i];
if (isin_q(q, v) && graph[u][v] < key[v]) {
parent[v] = u;
key[v] = graph[u][v];
}
}
adj_top = 0;
}
free_q(&q);
}
int main(int argc, char *argv[])
{
int no_vert, no_edge, root_vert;
int i, from, to; | {
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} |
asymptotics, landau-notation, lower-bounds
$$3n^2+15n-5\leq 3n^2+15n^2=18n^2$$
But when looking for the constant for the lower bound I find myself typically resorting to looking at limits. Is there any kind of short cut to finding the lower bound constant like there is for the upper bound constant? You can codify your method in the following lemma.
Lemma. If $f(n)/g(n) \rightarrow C$, where $C > 0$, then $f(n) = \Theta(g(n))$.
The proof is the same as the one you gave. After you prove this lemma once and for all, you can use it forever. That's actually a good way of verifying $f(n) = \Theta(g(n))$.
Note that the converse to the lemma isn't true. For example, let $f(n) = n$ and let $g(n) = \exp\lfloor\log n\rfloor$. The ratio $f(n)/g(n)$ moves inside the interval $[1,e)$, and in particular does not tend to a constant limit. | {
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ros, ros-melodic
Title: Image publisher gives numpy error
Hi, I am running the code from the directory of the file. After initiating both the publisher and subscriber the code works perfectly for 15sec and then give the error shown bellow.
Code:
#!/usr/bin/env python
import cv2
import rospy
from sensor_msgs.msg import Image
from cv_bridge import CvBridge, CvBridgeError
def read_video():
video_publisher = rospy.Publisher("tennis_ball_image", Image, queue_size=100)
video_capture = cv2.VideoCapture("video/tennis-ball-video.mp4")
while not rospy.is_shutdown():
_,rgb_frame = video_capture.read()
ros_image = CvBridge().cv2_to_imgmsg(rgb_frame, "bgr8")
video_publisher.publish(ros_image)
cv2.waitKey(10)
video_capture.release()
if __name__ == '__main__':
try:
rospy.init_node("tennis_ball_publisher", anonymous=True)
read_video()
rospy.spin()
except rospy.ROSInterruptException:
rospy.loginfo("node terminated.") | {
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thermodynamics, temperature, equilibrium
Now, for simplicity let us suppose $m_1c_1=m_2c_2$, such that, the total entropy variation of thermal equilibrium is
$\Delta S_1+\Delta S_2= mc \ln \frac{T_e^2 }{T_1 T_2}$.
It is easy to prove that the total variation will be a maximum when $T_e= \frac{T_1+T_2}{2}$.
In short: the thermal equilibrium is the state of maximal entropy of the system.
(This can be demonstrated also when $m_1c_ \neq m_2c_2$ with a bit more of algebra...). | {
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python, web-scraping, tkinter, beautifulsoup
should not declare a global. Simply return path.
Separation of UI and logic
download_fanfic is a big old ball of yarn. You have calls to requests beside calls to messagebox. Separate out the actual downloading logic and parsing logic into their own separate functions that do not have any tk code in them at all. Errors can be signalled via exceptions, and progress can be signalled via a generic callback function.
Logging
Try converting this:
set_status("Writing FanFiction to " ...
into a call to the standard logging framework, and adding your own logging handler to do what set_status currently does. It will make your application much more flexible - you could fairly easily flip one switch to have the entire thing run in console-only mode.
Requests check
if response.status_code == 200:
should be replaced with:
if response.ok:
or better yet
response.raise_for_status()
Threading | {
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One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions that are not far from the definition of topological spaces. The same can be said about hereditarily separable and hereditarily Lindelof spaces. Out of these basic ingredients come the notion of S-spaces and L-spaces, the existence of which is one of the key motivating questions in set-theoretic topology in the twentieth century. The study of S and L-spaces is a body of mathematics that had been developed for nearly a century. It is a fruitful area of research at the boundary of topology and axiomatic set theory. | {
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"openwebmath_score": 0.9994206428527832,
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"url": "https://dantopology.wordpress.com/tag/point-set-topology/"
} |
php, file-system
}
}
}
?>
Very messy I know, I am sure there is a better way to do this just not sure how. You might want to try using the scandir function to build your arrays of directories and files, makes it a bit more future proof.
$baseDir = ABSPATH."/uploads";
$calendarinfo = cal_info(0); //built-in function to generate calendar info, e.g. month names
$years = scandir($baseDir);
//You might want to ignore certain folders, e.g. on UNIX systems:
$ignore = [".","..",".DS_Store"];
foreach ($years as $year) {
if (!in_array($year, $ignore)) { //the folder is not in the ignore array
//then scan the year directory for user folders
$users = scandir($baseDir."/".$year);
foreach ($users as $user) { | {
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algorithm, genetic-algorithms
There is evidence that evolution occurs more rapidly in Wright’s model but the evidence is inconclusive. The theory concentrates on the probability that a mutation will occur and then become dominant in a population. In a small population a beneficial mutation is more likely to be selected for reproduction, but premature convergence is a serious danger. While in a larger population a mutant is less likely to be removed from the population during reproduction. I would strongly recommend a read of Games of life by Karl Sigmund. | {
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laserscan, transform
Title: tranformation using static publisher and listener
Hi,
If I have two frames I want to do transformation.. in my code i have the following:
void laserCallback (const sensor_msgs::LaserScan::ConstPtr& scan_in)
{
double t1 = ros::Time::now().toSec() ;
//std::cout<<"LASER" << scan_in->header.frame_id << std::endl ;
if(!listener_.waitForTransform(scan_in->header.frame_id,
"/uav/baselink_ENU",
scan_in->header.stamp + ros::Duration().fromSec(scan_in->ranges.size()*scan_in->time_increment),
ros::Duration(1.0)))
{
std::cout << "RETURN" << std::endl ;
return;
}
sensor_msgs::PointCloud msg;
// Which one of the follwoing should I use ?
//projector_.projectLaser(*scan_in, msg);
//projector_.transformLaserScanToPointCloud("/uav/baselink_ENU",*scan_in, msg,listener_); | {
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meteorology, clouds
In that chart, how should I interpret the terms "clear", "scattered", "broken", "overcast", and "obscured"? I understand what they mean in English, but what do they mean in terms of my ability to see the eclipse? The mean is probably the average cloud cover in percentages: 41% cloud coverage on average over time. The categories you mentioned: 'clear', 'scattered', 'broken' and 'overcast' are defined in different ways sometimes. Basically there is a human observing the sky analyzing the coverage in oktas or tenths (e.g. 6/8), or more likely there is a sounding device looking through the sky and getting a value in percentages of the cloud cover (e.g. 75.0%). To get one of these categories from the device data, the percentages are turned into oktas or tenths. Some data is lost in this transformation and thus from the categorized data it is impossible to get calculate the actual mean. | {
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I suppose you're right, @Aaron, but would you happen to have an explicit example of a "rootable" nilpotent on hand, just so I can check my intuition? :) – J. M. Sep 20 '11 at 16:46
@J.M. Consider the $4\times 4$ matrix with two $2\times 2$ blocks. This is the square of a matrix conjugate to a $4\times 4$ Jordan block, unless I've done my calculations wrong. – Aaron Sep 20 '11 at 16:49 | {
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string-theory
A single string has mass going like $\sqrt{N}$ so this is also its energy $p^0$ in the rest frame where $\vec p=0$. However, in the light cone gauge,
$$p^- = \frac{(p^i)^2+m^2}{2p^+} $$
note that this is just the usual $p_\mu p^\mu = m^2$ solved for $p^-$ in the light cone gauge, so that the $m^2$ in the numerator goes like $N$ (the total excitation of the string) rather than $\sqrt{N}$.
Also, the energy may mean the world sheet energy, the generator of translations in the world sheet temporal coordinate $\tau$. Then, for an excited string, the energy goes like $H\sim L_0+\tilde L_0\sim N$ much like in the light cone gauge. However, there are different coefficients in these two interpretations of the energy and one has to treat the nonzero modes and ghosts differently.
I am confident that all the standard textbooks make it clear from the context which "energy" they are talking about. | {
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# Limit of this expression $\lim \limits_{n\to \infty }\left( \frac{15n+7}{n \log\left( n\right) } \right)$
The limit $$\lim \limits_{n\to \infty }\left( \frac{15n+7}{n \log\left( n\right) } \right)$$ Is it valid to do
$\frac{15 + 7/n}{\log n}$
which makes the numerator go to $15$ as the bottom goes to infinity, therefore the whole thing goes to $0$? How would I show more formally that this is true?
Your reasoning is perfectly fine, and can be made fully rigorous as follows. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$
$\lim_{n\to\infty} \dfrac{15n+7}{n \ln(n)}$
$\ = \lim_{n\to\infty} \dfrac{15+7/n}{\ln(n)}$ [the division is valid because $n \ne 0$ and it does not change the value]
$\ = \dfrac{\lim_{n\to\infty}( 15+7/n )}{\lim_{n\to\infty} \ln(n)}$ [since these two limits exist in $\overline{\rr}$ and the ratio is defined in $\overline{\rr}$] | {
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I wanted to know if my solution is correct, something to improve or maybe some other. Thank you
• I'm not familiar with the notation $X'$. Would you mind explaining what it means? – Robert Shore Feb 15 '19 at 0:09
• Is the set of accumulation points of $X$. – Juan Daniel Valdivia Fuentes Feb 15 '19 at 0:10
• It seems obvious that if $Y \subseteq X$, then any accumulation point of $Y$ must also be an accumulation point of $X$. – Robert Shore Feb 15 '19 at 0:14 | {
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} |
navigation, mapping, pose, scan, hector-localization
Comment by gleb on 2016-09-28:
Do you mean a custom amcl-odom-model-type?
Comment by Henrique on 2016-09-29:
No, amcl, infere its new position by odom pose of the robot (the position in the map that the robot think it is), plus laserscan and instant movement (in odom too). So what you need to do is to make your odom publish correct. (make a topic that reads the joint states and pubblish the correct odom) | {
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pressure
would that be like the iron up here on
the surface or does the pressure
change its structure?
One difference to "surface iron" is obvious: the innermost core is solid, in spite
of the temperature. And one has to be aware that the iron in the core is not
pure, there will be some other metals and carbon in the melt of outer core.
At the border of outer core and mantle sometimes sulfides are assumed. | {
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• Not really the case. For example, consider $x^4-y^4=1$ which only has $x=1, y=0$ as a solution. The answers will usually be zero or a finite number of solutions, or rather, just a finite number of solutions. If the numbers can be real and irrational numbers, then you have infinite solutions in most cases. If you extend it to imaginary numbers, you have infinite answers always. – user1952500 Jul 26 '17 at 16:29 | {
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machine-learning, classification, random-forest, decision-trees
Does it mean that at each node in the tree, we randomly select m variables from the variable set which is fixed for that tree? Or from the global variable set of the training dataset? And then from the selected set of variables we select 1 variable heuristically (e.g. whichever variable maximises information gain) -- is that a correct statement? To train a tree in a random forest, at each split m variables are randomly selected from the global variable set, and the variable that maximizes information gain is chosen to be split on. Source: https://www.stat.berkeley.edu/~breiman/randomforest2001.pdf (the original random forest paper). | {
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ros, roslib
Running $ cd /opt/ros/electric/ros/core/roslib/src;python -c 'import roslib; print "success"' results in;
success
user@user-K53SV:/opt/ros/electric/ros/core/roslib/src$
Any ideas on what is wrong? Also why the PYTHONPATH looks as it does? Should it?
Thanks!
raymachira
Originally posted by raymachira on ROS Answers with karma: 1 on 2012-08-13
Post score: 2
Original comments
Comment by raymachira on 2012-08-13:
I have tried sourcing /opt/ros/electric/setup.sh While this command works, I still get the error
Comment by Lorenz on 2012-08-13:
Please provide the complete backtrace. Maybe it contains something important.
Comment by raymachira on 2012-08-13:
Backtrace of what exactly?
Comment by Lorenz on 2012-08-13:
The python backtrace that shows you the import error. Please always copy-paste the exact error you are getting. See http://www.ros.org/wiki/Support
Comment by raymachira on 2012-08-13:
Sure. Here it is: >>Traceback (most recent call last): | {
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java, performance, primes, sieve-of-eratosthenes
Now what really bugging me is the indexing i did in primeSum() method for this optimized sieve
Sure, the 2 needs a special handling and then you should make sure that m and n are odd. I guess that you code is wrong and I'd try something like this instead:
private static int primeSum(int n, int m) {
if (m < 2) {
return 0; // no primes below 2
}
if (n < 2) {
n = 2; // no primes below 2
}
// Add 2 if in range.
int sum = n <= 2 && 2 <= m ? 2 : 0;
// Round up to next odd in a very hacky way.
n += ~n & 1;
// Round down to next odd.
m -= ~m & 1;
for (int i = numberToIndex(n), p = n; i<=numberToIndex(m); ++i, p+=2) {
sum += isPrimeIndex[i] ? p : 0;
}
} | {
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needs of today's applied calculus student. 8) Sep 30, 2003 · 100 Calculus Projects: Complete Set of Projects These student projects have been developed by the mathematics department of IUPUI for their introductory calculus sequence. application of vector calculus in engineering field ppt 2) If is always Solenoidal Vector. Course Description. This lesson contains the following Essential Knowledge (EK) concepts for the *AP Calculus course. Extraordinary original hand-woven creations made in Santa Fe, New Mexico. Review Precalculus 2. The following steps are usually easy to carry out and give important clues as to the shape of a curve: Determine the $x$– and $y$-intercepts of the curve. 2T 10 Total MODULE - II DIFFERENTIAL CALCULUS – 2 1. Posted on January 16, 2021 Written by No Comments on application of calculus in engineering ppt. Calculus Lesson Plan Objectives Coach Gordon Stephens PPT. This is just one of the solutions for you to be successful. Laws of Limits 5. How to use the | {
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machine-learning, linear-regression, machine-learning-model, gradient-descent, feature-scaling
Title: Does Feature Normalization affect Gradient Descent | Linear Regression am new to datascience and i want to learn linear regression so i coded linear regression from scratch and performed gradient descent to find the best $w_\theta$ and $b_\theta$ values using a tutorial. And it went just fine i was able to find the best $w_\theta$ , $b_\theta$ values and i ploted the line-of-best-fit (below).
And the gradient descent code i used to find the $w_\theta$ , $b_\theta$ values is below .
def step_gradient_descent(data,m,b,learning_rate=0.0001):
b_gradient= m_gradient = 0
N=float(len(data))
for i in range(len(data)):
[x,y]=data[i]
y_=(m*x)+b
m_gradient+= - (2/N)*(x*(y-y_))
b_gradient+= -(2/N)*(y-y_)
#print("m ={}, b ={}".format(m_gradient,b_gradient))
m_new = m-(learning_rate*m_gradient)
b_new = b-(learning_rate*b_gradient)
return (m_new,b_new) | {
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m can take multiple values from 1 to 5. But since value of n is not provided this is insufficient. Statement 2 : n=12. Now since value of m not provided => insufficient. Combining statement 1 and 2 : m<6 and n=12. still this is not sufficient. Lets say m=5 and n=12 then mean will not be an integer. and for values m=4 and n=12, mean will be an integer. Thus both together are not sufficient. => option E. _________________ Please take a moment to hit Kudos if you like my post Manager Joined: 09 Jul 2015 Posts: 58 Re: Math Revolution and GMAT Club Contest! There is a sequence, 4(10^n), [#permalink] ### Show Tags 21 Dec 2015, 18:48 Answer is E. 4(10^n), 4(10^(n-1)), ..., 4(10^(n-m)) ; sum of these numbers can be written as 4*10^n + 4*(10^n)/10 + 4*(10^m)/10^2+... Simplifying. 4*10^n(1+1/10+1/10^2+... 1/10^m)... 4*10^n ((10^m+10^m-1+10^m-2+...+1)/10^m))... Average of the above = (4*10^n ((10^m+10^m-1+10^m-2+...+1)/10^m)))/m .. simplifyig, A*(4*10^n)/(m*10^m) where A = | {
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filters, infinite-impulse-response
this is LTI so then the exponential function is an eigenfunction for the natural response.
for the natural solution, try
$$ y_\text{n}[n] = y_\text{n}[0] a^n $$
then
$$\begin{align}
y[n] &= x[n] + (1-Q) y[n-1] \\
y_\text{n}[0] a^n &= 0 + (1-Q) y_\text{n}[0] a^{n-1} \\
a^n &= (1-Q) a^{n-1} \\
&= (1-Q) a^n a^{-1} \\
1 &= (1-Q) a^{-1} \\
\end{align}$$
so i guess $a = 1-Q$.
for steady state, you posit a solution that is like the input plus all of its finite differences. since the input is constant and all finite differences of a discrete constant function are zero, then the steady state output is a constant. you sum the scaled steady state with the scaled transient response for the complete response.
$$\begin{align}
y[n] &= x[n] + (1-Q) y[n-1] \\
C + y_\text{n}[0] a^n &= D + (1-Q)(C + y_\text{n}[0] a^{n-1}) \\
C + y_\text{n}[0] (1-Q)^n &= D + (1-Q)(C + y_\text{n}[0] (1-Q)^{n-1}) \\ | {
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navigation, ekf, odometry, ros-kinetic, robot-localization
Originally posted by positron on ROS Answers with karma: 3 on 2018-02-19
Post score: 0
Is this a correct setup?
Yes. Only one node may publish a given transform, so if you want the EKF to publish odom->base_footprint, then you need to make sure no other nodes are publishing that node.
Do I need to set some other frame_id
there?
No. If you have the world_frame set to odom for the EKF, it'll attempt to transform all future pose data into the odom frame before it uses the data. If it's already in the odom frame, you're golden. If, for example, you had some other world-fixed coordinate frame called rotated_odom that was rotated from the odom frame by a fixed yaw value, you'd use a static_transform_publisher to publish rotated_odom->odom, and the EKF would use that to transform the messages into the odom frame before using them.
Won't these nodes mess up data when
the odom transform is published by
ekf_localization but not themselves? | {
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IEEE Transactions on Image Processing Vol. , chemical reactions) and are widely used to describe pattern-formation phenomena in variety of biological, chemical and physical sys-tems. % Set up 1D domain from 0. | {
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a. K3 b. a 2-regular simple graph c. simple graph with = 5 & = 3 d. simple disconnected graph with 6 vertices e. graph that is Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Relevance. Since G is disconnected, there exist 2 vertices x, y that do not belong to a path. Graph Complement, Cliques and Independent Sets16 Chapter 3. Once the graph has been entirely traversed, if the number of nodes counted is equal to the number of nodes of G, the graph is connected; otherwise it is disconnected. A cut point for a graph G is a vertex v such that G-v has more connected components than G or disconnected. If every vertex is linked to every other by a single edge, a simple graph is said to be complete. In graph theory, the degreeof a vertex is the number of connections it has. An A graph is disconnected if at least two vertices of the graph are not connected by a path. We say that a graph can be embedded in the | {
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stability, proteins, structural-biology
So what exactly is it that imparts such high thermal stability to some proteins? Is it just a vastly intricate network of the earlier mentioned interactions, or is it because of something else? Szilágyi and Závodszky published an article in the journal Structure which analyses a number of different structural parameters of proteins of moderately thermophilic ($45~\mathrm{^\circ C} < \vartheta_\mathrm{opt} < 80~\mathrm{^\circ C}$) and hyperthermophilic ($\vartheta_\mathrm{opt} \approx 100~\mathrm{^\circ C}$) organisms compared to homologous mesophilic organisms.[1] I shall summarise their results in this answer. | {
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Page 4 of 18 Limits as x approaches ∞ For rational functions, examine the x with the largest exponent, numerator and denominator. 2, and Miscellaneous Extra Questions NCERT Solutions are extremely helpful while doing your homework or while preparing for the exam. Throughout Swill denote a subset of the real numbers R and f: S!R will be a real valued function de ned on S. com - Bethany Kochan. One Bernard Baruch Way (55 Lexington Ave. Solve the problem. 1 Limit of a Function Suppose f is a real valued function de ned on a subset Dof R. If not, explain why. Use the drop menus below to access exercises in other Chapters and Sections. Best answer: Presumably Walter White is dead (and almost certainly Cranston is busy making another movie right now), so it's no surprise he's not in this movie. Many statements we make about functions are only true over intervals where the function is continuous. Limits and Continuity. A hands-on exercise designed to aid you in preventing a disease outbreak | {
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astrochemistry, infrared, sofia
In this metal-free and low-density environment, neutral helium atoms formed the Universe’s first molecular bond in the helium hydride ion HeH+ through radiative association with protons. As recombination progressed, the destruction of HeH+ created a path to the formation of molecular hydrogen. Why was helium hydride (HeH+) the universe's first molecule? Why was it the hydrogen helium cation rather than the dihydrogen cation? At the risk of self-plagiarism: (https://physics.stackexchange.com/a/532568/43351 with a bit added).
Molecular chemistry in the early universe requires species with bound electrons. Helium hydride is the first molecule to form because neutral helium atoms, formed about 120,000 years after the big bang, could combine with plentiful protons; but it was another 260,000 years until significant numbers of neutral hydrogen atoms formed, and it is only once these are present that there is a route to forming H$_2$.
Details: | {
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cosmology, space-expansion
$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$
where $\rho$ is the average density of matter (both normal + dark matter) and $p$ is the pressure. The pressure is effectively zero for our universe, and since right now $a=1$ the equation simplifies to:
$$ \ddot{a} = -\frac{4\pi G\rho}{3} + \frac{\Lambda c^2}{3} $$
If you take your two equal masses $m$ separated by a distance $\ell$ then the gravitational force between them is:
$$ F = -\frac{Gmm}{\ell^2} $$
where the minus seen means the force is attractive. The acceleration $A$, given by $F/m$, is:
$$ A = -\frac{Gm}{\ell^2} $$
And all we have to do to answer your question is to set this gravitational acceleration $A$ equal to minus the universe acceleration $-\ell \ddot{a}$ so they balance out. This gives us:
$$ \frac{Gm}{\ell^2} = \ell\left(\frac{-4\pi G\rho}{3} + \frac{\Lambda c^2}{3}\right) \tag{2} $$ | {
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c#, game, user-interface, uwp, checkers-draughts
{Piece.whiteKing.Value, new Uri("ms-appx:///../Assets/WhiteKing.png", UriKind.Absolute)},
{Piece.blackChecker.Value, new Uri("ms-appx:///../Assets/BlackChecker.png", UriKind.Absolute)},
{Piece.blackKing.Value, new Uri("ms-appx:///../Assets/BlackKing.png", UriKind.Absolute)}
}; | {
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c++, algorithm, recursion, template, c++20
std::vector<std::u8string> u8string_vector1{};
for(int i = 0; i < 4; ++i)
{
u8string_vector1.push_back(u8"\u20AC2.00");
}
assert(recursive_any_of<1>(u8string_vector1, [](auto&& i) { return i == u8"\u20AC2.00"; }));
assert(recursive_any_of<1>(u8string_vector1, [](auto&& i) { return i == u8"\u20AC1.00"; }) == false);
std::pmr::string pmr_string1 = "123";
std::vector<std::pmr::string> pmr_string_vector1 = {pmr_string1, pmr_string1, pmr_string1};
assert(recursive_any_of<1>(pmr_string_vector1, [](auto&& i) { return i == "123"; }));
assert(recursive_any_of<1>(pmr_string_vector1, [](auto&& i) { return i == "456"; }) == false);
std::cout << "All tests passed!\n";
return;
}
void recursive_none_of_tests()
{
auto test_vectors_1 = n_dim_container_generator<4, int, std::vector>(1, 3);
test_vectors_1[0][0][0][0] = 2;
assert(recursive_none_of<4>(test_vectors_1, [](auto&& i) { return i % 2 == 0; }) == false); | {
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resource-request, experimental-realization
With respect to cryptography, NIST is currently sponsoring the development of post-quantum cryptography standards that would possibly be secure against quantum attacks even if a large quantum computer were made available in the near term. Google has even begun implementing a variation of one of these protocols (HRSS) for key exchange. However, claims that a protocol is "post-quantum" should generally be taken with a grain of salt since its entirely unknown whether a method to break these protocols (classical or quantum) exists. | {
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kinematics
Now, sunlight is light and not rain. This means that sunlight comes down from the sky at the speed of light. This is much much faster than a person can run so there is no worry that by running too fast the person is going to "run into the light" and get sun on their face for example. This is like the second scenario. This means it makes sense for the person in sunlight to hold their umbrella vertically whether they are standing or running.
I'll just point out that there's a more mathematical answer where you can consider the problem in the reference frame of the running person and see that the velocity vector of the rain changes directions as the person runs faster. This could also be analyzed using special relativity to see if the angle of the sun rays changes as the person runs. In my answer I'm neglecting all relativistic effects ($v\ll c$). I'll leave the relativistic/vector answer to someone else. | {
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electromagnetism, gravity, interactions, weak-interaction, strong-force
It could be argued that the region around an uncharged, non-magnetic black hole with no surrounding matter would fit the specifications of your gravitation-only region of space.
The big challenge would be to find a place in the universe that contains exactly zero photons, neutrinos, etc. There probably is no such place-- only places where the density of such things is greatly reduced. | {
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filters, adaptive-filters, least-squares
Title: Least Mean Squares (LMS) Filter Weight Update I have a general question regarding Least mean squares adaptive filters.
Using the example of noise cancellation, I understand that if you have a set of reference signals (S) and corrupted signals (S+N) then you can perform minimisation of the error to find the optimal weights for the set of inputs.
I am wondering if this is normally performed 'offline' or if this can be performed on the fly with the weights changing over time? Adaptive means that you try to find a "best" filter, locally, according to some criterion. Using it online is not mandatory. Indeed on images the requirements for "onlinity" or causality are often less stringent than for signals.
However, for efficiency, it can be beneficial to update coefficients every time, or even more perform some approximation. | {
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ros, rosserial-client
Templates - templates MUST be defined in header files due to the nature of how C++ instantiates templates
In general, you should try to keep as much code as possible in the cpp files so that only a single compiled version of your functions/methods/classes decreasing compile time and the size of your executable targets. If inlining is required, modern C++ compilers perform very aggressive inlining and in most cases will automatically inline functions better than you can. Also, this prevents users from having to recompile everything that touches your library if the implementation changes without the interface changing -- only a relink is required.
I also personally like keeping implementations out of header files as in C/C++ they are useful for referencing the API. Putting the code in header files can sometimes make them hard to read.
Originally posted by mirzashah with karma: 1209 on 2013-01-23
This answer was ACCEPTED on the original site
Post score: 4 | {
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>Choose any vertex from the graph and put it in set A. Follow every edge from that vertex and put all vertices at the other end in set B. Erase all the vertices you used. >Now for every vertex in B, follow all edges from each and put the vertices on the other end in A, erasing all the vertices you used. Jul 11, 2018В В· A sphere has no edges. Square Pyramid. A square pyramid has 4 lateral triangular faces and 1 square base. A square pyramid has 5 vertices. A square pyramid 8 edges. Triangular Prism. A triangular prism has 3 rectangular lateral faces and 2 triangular bases. A triangular prism has 6 vertices. A triangular prism has 9 edges. | {
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c#, .net-core
Log.Logger = new LoggerConfiguration()
.ReadFrom.Configuration(builder.Build())
.Enrich.FromLogContext()
//.WriteTo.Console() // activate to send the logging to the console
.WriteTo.File("LogFiles/apploggings.txt") // activate to send the logging to a file
.CreateLogger();
var host = Host.CreateDefaultBuilder()
.ConfigureServices((context, services) =>
{
services.AddLogging();
services.Configure<UserInformation>(builder.Build().GetSection("UserInformation"));
services.Configure<ApplicationInformation>(builder.Build().GetSection("ApplicationInformation"));
services.AddTransient<IMainView, MainView>();
services.AddTransient<IMainViewModel, MainViewModel>();
services.AddTransient<IAppSettingProvider, AppSettingProvider>();
services.AddTransient<ISettingView, SettingView>();
services.AddTransient<ISettingViewModel, SettingViewModel>();
})
.UseSerilog()
.Build(); | {
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quantum-mechanics, homework-and-exercises, angular-momentum, wavefunction, probability
Title: Determine the state $|\psi \rangle$ The question is:
The angular momentum components of an atom prepared in the state $|\psi\rangle$ are measured and the following experimental probabilities are obtained:
\begin{equation} P(+\hat{z}) = 1/2, P(−\hat{z}) = 1/2,
\end{equation}
\begin{equation}
P(\hat{x}) = 3/4, P(−\hat{x}) = 1/4,
\end{equation}
\begin{equation}
P(+\hat{y}) = 0.067, P(−\hat{y}) = 0.933.
\end{equation}
From this experimental data, determine the state $|\psi \rangle$. Note that in performing the measurements, the state $|\psi \rangle$ is prepared again and again. | {
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ngs, illumina, 10x-genomics, barcode
Around one million.
Conclusion
This analysis doesn't give conclusive evidence for the four hypotheses above, but it seems like there is a noise-generating process in 10X data that causes there to be many barcodes with only a few readpairs. Your initial guess is almost certainly correct. I don't know about the linked read libraries, but in the 10X single cell sequencing protocol, separating real barcodes from noise barcodes is an important and sensitive step in the analysis pipeline.
The current analysis pipeline is to look at a plot, like the one you produced and find the "knees". The plot you produced has two knees, on at around 10 and one at around 200. I'd guess that the ones below 10 are what we call "error barcodes" generated by sequencing is PCR errors from real barcodes, and the ones between 10 and 200 are what we'd call "background" barcodes - in scRNAseq they represent things like GEMs that didn't encapsulate a cell, but did encapsulate some non-cellular DNA or RNA. | {
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python, algorithm, complexity
The reason for this is the copying in the recursive calls. In Python, the list slice items[mid:] has to be copied, and the time taken to do this is proportional to the number of items in the copy. See the time complexity page on the Python wiki.
In order to avoid copying in find, it's best to keep indexes into the part of the sequence which might contain the item we're looking for:
def find(x, items):
"""Return True if x is an element of the sequence items, assuming
items is sorted.
"""
lo = 0
hi = len(items)
while lo < hi:
mid = (lo + hi) // 2
if x == items[mid]: return True
elif x < items[mid]: hi = mid
else: lo = mid + 1
return False
Python has a built-in module bisect for binary search of sorted sequences, and using this module it's possible to write find like this:
from bisect import bisect_left
def find(x, items):
"""Return True if x is an element of the sequence items, assuming
items is sorted. | {
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c++, machine-learning, neural-network
//Bias calcs
for (size_type i = 0; i < layers_size(); ++i)
{
for (size_type j = 0; j < layer_size(i); ++j)
{
m_neuron_weights[i][j].back() += m_learningRate * m_neuron_layer_errors[i][j] * m_activation.derivative(m_unactivated_neuron_layers[i][j]);
}
}
}
void train(const batch inputs, const batch targets, const std::string& learningResultsFilePath = "test_results.csv")
{
std::ofstream results(learningResultsFilePath);
results.clear();
results << "Errors, Iterations" << std::endl;
for (size_type i = 0; i < inputs.size(); ++i)
{
forward(inputs[i].data());
backpropagate(targets[i].data());
if (i % 16 == 0)results << mse() << ',' << i << std::endl;
}
results.close();
} | {
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classical-mechanics, chaos-theory, non-linear-systems
The middle phase is when the chaos is actually growing and the distances are growing exponentially.
At the end, at very long timescales, the maximum possible distance is basically reached and equilibrium is reached so the distance has nowhere to grow further. | {
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javascript, strings, regex
return str.replace(new RegExp('^[' + result + ']+|['+ result +']+$', 'g'), '');
}
The function can be used as follows:
trim(' - hello** +', '+*- ');
The escaping is required as + * . etc. would be used in the regular expression. Is there a better way to trim any character in JavaScript? Or a better way to do the escaping, as not every character needs it? There are many different ways to tackle trimming strings, a regex is probably the most popular, and there are lots of blogs and performance tests out there, (one such blog), that you can look at and decide which is best.
So continuing along the lines that you have demonstrated, then this may be of some help to you.
Javascript
/*jslint maxerr: 50, indent: 4, browser: true */
var trim = (function () {
"use strict";
function escapeRegex(string) {
return string.replace(/[\[\](){}?*+\^$\\.|\-]/g, "\\$&");
} | {
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$\Delta =x-d=n\lambda \Rightarrow x=\left(d+n\lambda \right)$
squaring both sides, we obtain
$x^2={\left(d+n\lambda\right)}^2=d^2+n^2\lambda^2+2n\lambda d$
Now as shown in the figure, use the Pythagorean theorem to write $x^2=d^2+4^2$. Substitute this into the relation above
$d^2+16=d^2+{\left(n\lambda\right)}^2+2\left(n\lambda\right)d\Rightarrow d=\frac{16-{\left(n\lambda\right)}^2}{2\left(n\lambda\right)}$
$d=\frac{16-{\left(\frac{1}{2}\lambda\right)}^2}{2\left(\frac{1}{2}\lambda\right)}=\frac{16-\frac{{\left(3.85\right)}^2}{4}}{3.85}=3.19\ {\rm m}$
If $n=3/2\$then $d=1.50$ so the largest value of $d$ is $3.19\ {\rm m}$. | {
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standard-model, potential-energy, quarks, gluons, quark-gluon-plasma
Title: Where did all the negative potential energy in gluons went to? I have watched PBS Space Time that gluons could be viewed as negative potential energy (tension between quarks) and that they actually are 99% of all the mass in protons and neutrons.
Now the way I understand negative potential energy is like 'borrowing' it from somewhere and using it up and then if you want to get to zero value state you need to add energy to get there, since to get to negative potential that energy had to used by something. Example would be when gas that got compressed and turned hot, but that energy was used so it cooled down to room temperature and then if uncompressed it freezes because energy is gone.
So what was it that took all that gluon energy and used it up (where did it go)?
Or am I misunderstanding this?
and that they actually are 99% of all the mass in protons and neutrons. | {
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c++, performance, programming-challenge, hash-map
You're reading each account number, storing them then going back through the data to get the counts. The map class allows for a key that doesn't exist to create an entry on the fly. Using this will eliminate the second loop:
std::string temp = "";
std::map<std::string, int> count;
for (i = 0; i < n; ++i) {
std::getline(std::cin, temp);
count[temp]++;
}
std::cout << "\n"; | {
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water, corrosion
Some materials for supercritical water include Inconel and Hastelloy, nickel-chromium steels.
BTW, if you need an observation port or a non-conductive feed-through, silica-based glasses won't do, since they dissolve in supercritical water. Perhaps sapphire ($\ce{Al2O3}$, corundum) might survive for a time, or not -- supercritical water can etch it under certain conditions.
You might research the Benson boiler. | {
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waves, wavelength, superposition
Consider a Slinky. One can make it swing in the fundamental mode or in modes with one or two nodes by adding energy from your hands in just the right intervals. You will feel that, that gives a positive feedback to that mode. That also is what happens at the lip of an organ pipe etc. The standing waves are resonances.
Or one can look at what happens when you make a pulse on the Slinky. Or on this Java simulation by Falstad. The pulse travels back and forth, getting reflected at both ends. So there is a periodic signal, with many harmonics of the fundamental frequency.
Falstad's simulation also has the option of adding a driving force with arbitrary frequency. It is like pushing a swing at arbitrary times: it won't add energy to the system. | {
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waves, acoustics, education
$$\frac{A}{x_0}e^{-ikx_0}e^{i\omega t-ikx}$$
This is a plane wave with constant amplitude $A/x_0$ and an additional phase shift of $e^{-ikx_0}$. | {
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quantum-mechanics, electromagnetism, wavefunction, quantum-optics, cavity-qed
In this general case, however, dynamics such as described by the OP are conceivable.
Scenario 4: Classical charge oscillation in the field
For classical charge oscillation, we do not have a notion of quantized orbitals or the state structure defined by the atom. In a complicated environment as the birefringent cavity, a rotation of the charge oscillation can happen. Even a separation into a multi-modal oscillation is possible.
Summary | {
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python, python-3.x, game, wordle
updating the board and
updating the colors and
deciding whether the game is over
Split this method into 4 smaller methods and it'll be much simpler to read, understand and debug.
Magic numbers
checkRight() has a few constants that are related to colors. We call such numbers "magic numbers", because they appear out of nothing with no semantics. The color comment is far from the usage (at least 4 lines).
We typically give them names. Since they are constants, we agree to make them upper case:
GREEN = 32
YELLOW = 33
GREY = 30 | {
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java, linked-list
return j;
}
Which you can use like
int j = findNext(twoDmatrix[i], 0);
and
j = findNext(twoDmatrix[i], j + 1);
The second one allows you to change the enclosing for loop to a while loop.
Simplifying
for (int i = 0; i < twoDmatrix.length; i++) {
for (int j = 0; j < twoDmatrix[i].length; j++) {
if (i % 2 == 0 && j % 2 == 0) {
twoDmatrix[i][j] = 1;
} else if (i % 2 == 0 && j % 2 != 0) {
twoDmatrix[i][j] = 0;
} else if (i % 2 != 0 && j % 2 == 0) {
twoDmatrix[i][j] = 0;
} else {
twoDmatrix[i][j] = 1;
}
}
} | {
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c, strings, formatting, pointers
// ---------------------------------------------------
byte counter = 0;
void loop() {
Serial.print(counter);
Serial.print(" ");
delay(1000);
if (++counter > 9) { counter = 0; }
}
The above can be run in an online emulator at https://wokwi.com/arduino/projects/315730913481720385
My function last3Digits() seems clumsy to me. I imagine it can be made smaller without losing readability, and I imagine I must have made some other rookie mistakes.
overwriting a string constant (what is the better way to allocate space?)
redundant setting of string terminator (felt safer)
? Well, there are some concerning points about your code:
You are not using const where appropriate. It is there to help you avoid obvious bugs, like overwriting constants. Crank up the warning-level, and the compiler warns about assigning string-constants to mutable pointers.
If you need a small persistent space, ask for it. Don't misuse string-constants:
static char result[4] = ""; | {
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formal-languages, kleene-star
Title: Kleene star Empty language I had a test a couple of days ago and one of the question had 2 statements :
$L^+ = L^\ast$
$L$ contains $\varepsilon$
I had to say does 1 imply 2, does 2 imply 1 or do they both imply each other.
I answered that only 2 implies 1 but the answer is that they both imply each other.
But I don't understand, if $L = \emptyset$ (Empty Language), then $L^\ast = \{\varepsilon\}$ and $L^+ = \{ \varepsilon \}$, they are equal and $L$ doesn't contain $\varepsilon$.
Am I wrong? You're wrong about $L^+$:
$L = \emptyset \rightarrow L^+ = \emptyset$, but $L = \emptyset \rightarrow L^* = \{\epsilon\}$ | {
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electromagnetism, magnetic-fields, electric-fields, maxwell-equations, gauss-law
As we found the divergence term, we should be able to prove $\nabla\bullet \vec E =0$.
How can I derive the rest of the problem? Electric field at any arbitrary point $\mathcal{r}$ due to a charge density $\rho$ present in a given volume $\mathcal{V}$ is given by Coulomb's law as -
$$\vec{E} = \frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{V}}\frac{\rho\ \hat{r} }{r^2} d\tau$$
We will evaluate the surface integral of the $\vec{E}$ over a sphere containing the source charges.
$$\oint \vec{E} \cdot d\vec{a} = \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho\ \hat{r} }{r^2} \cdot r^2\sin (\theta)\ d\phi\ d\theta\ \hat{r}\ d\tau$$
or, $$\oint \vec{E} \cdot d\vec{a} = \int\frac{\rho}{\epsilon_{0}}d\tau$$
Invoking Fundamental theoram of gradient -
$$\int \nabla \cdot \vec{E}\ d\tau = \oint \vec{E}\cdot d\vec{a}$$
So,$$\int \nabla \cdot \vec{E}\ d\tau = \int\frac{\rho}{\epsilon_{0}}d\tau$$
Since this is true for any given volume , thus we get -
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ | {
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feynman-diagrams, integration
Example: The easiest case I have encountered is the following. Consider the topology (called FA in the Mincer manual) | {
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python, algorithm, python-3.x, pathfinding
All of this together can look like:
from collections import namedtuple
from dataclasses import dataclass, field
from typing import * # Against best-practice but shhh
import math
Edge = namedtuple('Edge', 'distance node'.split())
class Node(namedtuple('Node', 'start end edges'.split())):
def __str__(self):
return f'{self.start} -> {self.end}'
@dataclass(order=True)
class Path:
distance: int
current: Node=field(compare=False)
previous: Node=field(compare=False)
@dataclass
class Graph:
nodes: List[Node]
def shortest_paths(self, start: Node) -> Dict[Node, Path]:
if start not in self.nodes:
raise ValueError("Graph doesn't contain start node.")
paths = {}
queue = []
for node in self.nodes:
path = Path(float('inf'), node, None)
paths[node[:2]] = path
queue.append(path) | {
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ros, bumblebee2, camera-drivers, bumblebee
So, my question is, can anyone give me an advice / comment about why using one package or the other ?
I'm using ROS-electric.
Thanks, community help is very valuable !
Originally posted by Andreu on ROS Answers with karma: 169 on 2011-10-10
Post score: 2
i have been using a slightly changed version of the bumblebee2 node.
the changed i made to the node can be found in this answer: http://answers.ros.org/question/1118/bumblebee2-640x480-under-diamondback?answer=1894#1894.
these fixes work for diamondback, im not sure if they work under electric (it probably works unless some messages have changed by a lot).
(yes i know this is quite an old question, just posting for anyone with the same question)
Originally posted by Bram van de Klundert with karma: 241 on 2011-12-12
This answer was ACCEPTED on the original site
Post score: 2
Original comments
Comment by joq on 2011-12-12:
Electric should probably work, too. I don't recall any significant camera interface changes for that release. | {
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algorithms
Title: Algorithm to check if path exists Suppose that I have a 2D array $B[i,j], 1 \le i,j \le n,$ such that $B[i,j]=X$ or $B[i,j] = O$. I want to write a program to check if there is a path that exists from $B[1,1]$ to $B[n,n]$ that only consist of $O$'s. The path must only consist of the 4 directions: North, South, East, West and does not include the diagonal direction. The algorithm is not necessarily looking for the shortest path, if such a path exists, just the existence of the path is needed.
I haven't formally studied algorithms but I think that this problem is a standard one in that area of study. I am not sure what to search in google though. Can anyone direct me to references on how to solve this problem?
Thanks! Read about depth-first search, breadth-first search, and reachability in undirected graphs. That will show you how to solve your problem. | {
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electricity, everyday-life, atmospheric-science, lightning
Why we see lightnings of different colours?
Why then sometimes lightings appear to us in a variety of colours, white, more bluish, reddish etc?
One point are well known scattering phenomena, especially those related to particulate and droplets - Mie regime. The intensity of this scattering will depends on the actual atmospheric conditions and always increase with distance (as the thickness of the atmosphere increases). Assuming a homogenous and vast enough portion of the atmosphere, lightings at horizon will be reddened (to the extent of appearing yellow) as compared to nearby ones, just to give an example.
Several specific examples are given in the last part of this webpage https://uvhero.com/what-color-is-lightning/ | {
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complexity-theory, terminology, probability-theory
If $q_2$ happened to be defined as an indicator of a "good" computation in our experiment, then, by using postselection, we would have thrown out all the "bad" computations.
Note: $\mathbb{P}\left[q_2=\lvert1\rangle\right]$ must be $>0$ or else the conditional probability measure is not well defined.
So in short, postselection gives us the ability to force quantum bits to think that another quantum bit is in a specific state we choose. This lets us filter out a lot of results that we consider to be useless to us when we measure our quantum state. As demonstrated in Aaronson's proof of $\text{PostBQP = PP}$, the ability to postselect provides significant computational advantage over vanilla models of quantum computation. | {
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php, mysql, pdo
<tr>
<td> </td>
</tr>
<?
$product_query->closeCursor();
} // End While
$customer_query->closeCursor();
$db = NULL;
?>
</table>
<br><br>
<?
$end_time = time();
echo "Number of Records: " . $customer_row_count . "<br>";
echo "Start: " . $start_time . "<br>";
echo "End: " . $end_time . "<br>";
echo "Time Elapsed: " . ($end_time - $start_time);
?>
</body>
</html> If I were you, I would try to add some sort of index to your MySQL database for this fields:
cb.customers_basket_date_added
cb.customers_id | {
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quantum-mechanics, solid-state-physics, second-quantization
Still, despite $\frac{t^2}{U}$ being a smaller quantity, it typically still corresponds to quite large magnetic fields. Even if we take the rather weak $t = .5$ eV and strong $ U = 10$ eV, we would need a magnetic field corresponding to $0.025$ eV, or $\sim 500$ Tesla! So part of the answer is: you're right, the magnetic fields needed to cause a phase transition are quite big (at least at half filling, it goes down quite a bit at lower filling, again see the above reference). Note though: that is to actually get a phase transition, this of course implies you can already start seeing appreciable effects with lower fields. | {
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Also, see how this is the null space of the RREF?
• No I do not , can you explain a bit more ? – Eng .. Abdalmonem Apr 16 '14 at 22:07
• and how you got v1=(1,0,0) ? – Eng .. Abdalmonem Apr 16 '14 at 22:07
• Judging from your "last seen" (on this site) datum, you were up WAY TOO LATE last night! ;-) – Namaste Apr 17 '14 at 11:29
• I've done the same. I actually function best (both when I was a student) and onward, when I take a long "nap" (at the end of the work day), then awaken refreshed and working late. (Early rising, late bedtime - but a nice long nap in between!) – Namaste Apr 17 '14 at 12:35 | {
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rosmake
[rosmake-0] Finished <<< roslang ROS_NOBUILD in package roslang
No Makefile in package roslang
[rosmake-0] Starting >>> rosgraph [ make ]
[rosmake-1] Finished <<< xmlrpcpp ROS_NOBUILD in package xmlrpcpp
No Makefile in package xmlrpcpp
[rosmake-1] Starting >>> roscpp [ make ]
[rosmake-0] Finished <<< rosgraph ROS_NOBUILD in package rosgraph
No Makefile in package rosgraph
[rosmake-0] Starting >>> rospy [ make ]
[rosmake-0] Finished <<< rospy ROS_NOBUILD in package rospy
No Makefile in package rospy
[rosmake-0] Starting >>> diagnostic_msgs [ make ]
[rosmake-0] Finished <<< diagnostic_msgs ROS_NOBUILD in package diagnostic_msgs
No Makefile in package diagnostic_msgs
[rosmake-0] Starting >>> rbx1_speech [ make ]
[rosmake-1] Finished <<< roscpp ROS_NOBUILD in package roscpp | {
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