text stringlengths 1 1.11k | source dict |
|---|---|
turing-machines, halting-problem, correctness-proof
Title: Proof by Reduction: From Empty Language to Halting Problem on Empty Input Question:
Let language $E$ = {$\langle M \rangle$ | $M$ accepts no inputs whatsoever}
Let language $H$ = { $\langle M \rangle$ | $M$ halts on an empty string input}.
Is it possible to show that $H$ is undecidable by reducing $E$ to it? (You can take it as a given that we know $E$ to be undecidable.) If so, show the work. If not, explain why not. | {
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"tags": "turing-machines, halting-problem, correctness-proof",
"url": null
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A\cap \overline{C} \\ Be careful with the other operations. of propositional logic, and 7 - 11 also follow immediately from them as illustrated below. &= \{x\mid x\notin (A\cup B)\} \\ With similar proofs, we could prove these things: When doing set operations we often need to define a. ( cf. ) \[\bigcup_{i=1}^{n} S_i\,. There is no logical version of set difference, or set version of exclusive or (at least as far as we have defined). x \end{align*}. Since A Additional properties: Then there is an element x that is in , i.e. For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: A. The if part can be proven similarly. Hence . If and , then , Also since , . the commutativity of Proof: Suppose for contradiction that there is an element $$x\in | {
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"tags": null,
"url": "https://ecom.ablele.net/hqi53g6/ss9ng.php?c394cf=set-operations-proofs"
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general-relativity, lagrangian-formalism, tensor-calculus, action, variational-calculus
Consider infinitesimal transformations in the form of Lie derivatives
$$ \delta \phi ~=~ {\cal L}_{\varepsilon}\phi~=~\varepsilon^i \partial_i\phi~=~ \varepsilon_i \partial^i\phi, \qquad
\delta g_{ij}~=~{\cal L}_{\varepsilon}g_{ij}~=~\nabla_i\varepsilon_j+(i\leftrightarrow j) \tag{A}, $$
where $\varepsilon$ is an infinitesimal vector field.
The action
$$ S[\phi,g] ~=~\int d^nx~{\cal L}, \tag{B}$$
(where ${\cal L}$ denotes the Lagrangian density) is assumed to be diffeomorphism invariant:
$$0~=~{\cal L}_{\varepsilon}S~=~ \delta S
~=~\int d^nx \left(E~\delta\phi +E^{ij}~\delta g_{ij}\right) + \text{boundary terms}$$
$$~\stackrel{(A)}{=}~\int d^nx \left(E~\partial^i\phi -2\nabla_jE^{ji}\right)\varepsilon_i + \text{boundary terms}. \tag{C}$$
OP's sought-for identity
$$ E~\partial^i\phi ~=~2\nabla_jE^{ji}\tag{1.8}$$
follows since eq. (C) should hold for all vector fields $\varepsilon$. | {
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"tags": "general-relativity, lagrangian-formalism, tensor-calculus, action, variational-calculus",
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motor, differential-drive
You first need to determine the pitch of the gear you want to match, the 8-tooth gear. If it has a diameter of say 10 mm, then take the circumference for your new gear (πd = 31.4 mm) and divide by the number of teeth (8) to get a pitch of 3.925 mm. Note that I am using the root circle to calculate diameter, rather than outside circle.
Knowing the pitch, multiply by 33 to get the circumference: 129.525 mm. We can now determine the diameter (d = c / π), 41.25 mm.
You'll need to use the same tooth shape, that is, the angles for the faces and flanks of the teeth, so that the gears mesh well. | {
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quantum-mechanics, thermodynamics, cosmology, space-expansion, cosmological-inflation
Since the density fluctuations of these components evolve differently during the RD and MD universe, we cannot expect perfect homogeneity and isotropy. Also, currently, the universe is not perfectly homogeneous or isotropic. We are just making an approximation or assumption.
Since this is a complex topic, I cannot explain the details of it here. However, If you want to learn about how these fluctuations evolve I recommend you to read these books and articles.
1 - Cosmology: The Origin and Evolution of Cosmic Structure 1st Edition
by Prof Peter Coles, Francesco Lucchin
2 - Christos G Tsagas. Cosmological perturbations (https://arxiv.org/abs/astro-ph/0201405)
3 - David Tong. Structure formation, 2019. http://www.damtp.cam.ac.uk/user/tong/cosmo.html
4 - Malcolm S. Longair. Galaxy Formation (Astronomy and Astrophysics Library) | {
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where $$\zeta=\mathop {\max }\limits_{1 \leqslant k \leqslant N} \left| {{a_k} - \ell } \right|$$
Now, let $n_0$ be such that if $n>n_0$,
\eqalign{ & \frac{{N\zeta }}{n} < {\epsilon} \cr & \frac{N}{n} < {1} \cr} Then we get
$$\frac{N}{n}\zeta - \frac{N}{n}\frac{\epsilon }{2} + \frac{\epsilon }{2} < \epsilon - \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon$$
How far is this OK? Do you think there is an easier way to go about proving it? I now remember that by Stolz Cesàro:
$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = \ell$$ | {
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"url": "http://math.stackexchange.com/questions/210681/if-a-n-to-ell-then-hat-a-n-to-ell"
} |
newtonian-mechanics, energy, potential-energy, harmonic-oscillator, dissipation
$$E=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2$$
where the potential is
$$U=\frac{1}{2}kx^2$$
I read that if instead we are working with a damped oscillator the expression for potential energy would be the same. I understand that the expression for position and velocity have the decaying exponential so the total energy will be decreasing as expected but why doesn't the potential energy expression need to be changed in this the case? When getting an expression for potential energy wouldn't we need to also consider the resistive force as well? A good old-fashioned definition of potential energy is the amount of work a body can do by virtue of its position. So when a body goes between positions A and B its change in potential energy is the same, independently of the route it takes or how fast it goes. If it goes from A to B and back to A its net change in potential energy is zero. | {
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"tags": "newtonian-mechanics, energy, potential-energy, harmonic-oscillator, dissipation",
"url": null
} |
statistical-mechanics, condensed-matter, solid-state-physics, temperature, thermal-conductivity
\begin{equation}
\Gamma_{ij}=\gamma_0\exp(-\frac{r_{ij}}{\xi}-\frac{|E_i-E_j|}{T})
\end{equation}
where $\gamma_0$ is another parameter of the theory, that depends of phonon DOS and electron-phonon coupling. From this equation you may see how $\Gamma$ (and hence conductivity) depends on $T$.
The hopping theory was first introduced to describe electron transport in disordered semiconductors. A notable person is F. N. Mott. Nowadays, it is used also, for example, for organic materials. For metal, it might be applied, if it is disordered enough, in a sense of wave function localization as said above. | {
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"tags": "statistical-mechanics, condensed-matter, solid-state-physics, temperature, thermal-conductivity",
"url": null
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waves
Title: Formula to calculate sound volume loss per feet travelled in air I would like to know the formula to find what percentage/ratio of dB is lost per feet traveled. The medium we know is a normal air mixture with 45% humidity and a temperature if 73°F. Is there a formula to calculate how many decibels are lost? The rate at which sound volume is lost is not constant with respect to distance. Rather, it depends on the distance from the source of the sound: the sound distance does not drop as much when moving from 100 m to 101 m away from the source as it does when moving from 1 meter to 2 meters away from the source. | {
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performance, vba, excel
For i = 2 To 1511 'Loop over all values from total dataset
For j = 2 To 288503 'Loop over all values from IBES file
If Worksheets("Data").Cells(i, 3) = Worksheets("IBES").Cells(j, 1) Then
If Worksheets("Data").Cells(i, 7) = Worksheets("IBES").Cells(j, 6) Then
If Worksheets("Data").Cells(i, 10) = Worksheets("IBES").Cells(j, 9) Then
If Worksheets("Data").Cells(i, 13) = Worksheets("IBES").Cells(j, 11) Then
If Worksheets("Data").Cells(i, 8) = Worksheets("IBES").Cells(j, 7) Then
If Worksheets("Data").Cells(i, 14).Text = Worksheets("IBES").Cells(j, 13).Text Then
Worksheets("Data").Cells(i, 12) = Worksheets("IBES").Cells(j, 10).Text
Worksheets("Data").Cells(i, 18) = Worksheets("IBES").Cells(j, 16).Text
End If | {
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"tags": "performance, vba, excel",
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filters, window-functions
Where do we use these windowing functions & Filters in practical world?
Do I need to use both for better signal quality, if so which one come first(Filters => Window or Window => Filters)? Filters and windows complement each other, in a way. This is due to the convolutional lemma of the fourier transform.
Let me expand on that. A filter would be a convolution of a signal with another. In a technical sense, a window has a fixed (supposedly short) size and is seen as "cutting out" a short part of the signal. Often, a DFT would follow.
In the fourier domain, convolution and point-wise multiplication trade places, so a convolution in the time domain is a pw. multiplication in the fourier domain and vice-versa.
A filter scales the spectrum with a known frequency response. A window will filter the spectrum instead. | {
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gazebo
key2 = anim2->CreateKeyFrame(10.0);
key2->SetTranslation( math::Vector3(0,3,0) );
key2->SetRotation( math::Quaternion(0,0,0) );
_parent->GetEntity("box_model")->SetAnimation(anim);
_parent->GetEntity("box_model2")->SetAnimation(anim2, anim);
}
};
// Register this plugin with the simulator
GZ_REGISTER_WORLD_PLUGIN(AnimationTest)
}
Thank you so much! | {
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linear-algebra, permutations, matrix-product
Title: the product of a matrix and a permutation matrix Can a permutation matrix (P) be used to change the rank of another matrix (M)?
Is there any literature to this effect, or to the contrary?
I've tried a few small examples and the resulting matrix (M2) seems to always have the same rank as the input matrix (M)
M2 = M P This probably doesn't belong on the TCS stack exchange, but I'll answer anyways.
No, multiplication by a permutation matrix will never change the rank of the matrix. Permutation matrices are orthogonal, so if matrix M has an SVD:
$$ M = U \Sigma V^* $$
Then the product $MP$ has an SVD:
$$ MP = U \Sigma V^* P = U \Sigma W^* $$
Recall that the rank is the number of non-zero singular values. Because $M$ and $MP$ have the same singular values $\Sigma$, they must have the same rank (and a whole bunch of other stuff). | {
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-
Assuming that $f$ is bounded, you can prove that it is constant by a discrete analogue of the Borel-Caratheodory inequality, as shown in this equivalent question.
The key fact is such a function attains its maximum on a ball on the boundary, so it suffices to provide lower bounds for
$$\Gamma_N = \max_{|x|+|y|=N}f(x,y).$$
- | {
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newtonian-mechanics, forces, vectors
How do the force vectors end up rolling the bike up and over vertical when the rear wheel loses contact with the road? I would have expected the bike to slide such that my left side would hit the ground instead of the right side.
Did dropping my head to peek under my left arm pit to see what happened affect the forces and help flip me to the right? i.e., would my chances of counteracting the flip have improved had I kept looking forward? Here is how this happens. I'll begin with the example of a skidding car, which is analogous and easier to visualize. | {
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ros
Title: Rviz2 requests data ahead of system clock -> requires extrapolation into the future
This problem is similar to, but not the same as:
https://answers.ros.org/question/382207/rviz2-requests-data-ahead-of-system-clock-requires-extrapolation-into-the-future/
I am having synchronization issues in Rviz2 when trying to visualize my transforms. The transformation is updated using the following code:
rclcpp::Time now = this->get_clock()->now();
geometry_msgs::msg::TransformStamped t;
// Read message content and assign it to
// corresponding tf variables
t.header.stamp = now;
t.header.frame_id = "world";
t.child_frame_id = msg->child_frame_id;
t.transform.translation.x = msg->pose.pose.position.x;
t.transform.translation.y = msg->pose.pose.position.y;
t.transform.rotation = msg->pose.pose.orientation;
// Send the transformation
tf_broadcaster_->sendTransform(t); | {
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nanoscience
Maybe the reason you are thinking of UHV is because the related technique of scanning tunneling microscopy (STM) on the other hand is better performed in UHV. This is because STM because is an electronic technique, is highly sensitive to surface adsorbates/contamination, and has difficulty getting atomic resolution in the presence of air currents. Moreover, it's fairly common to cool to cryogenic temperatures in STM as well to remove thermal broadening.
For comparison, air currents are less of an issue for AFM since the technique is usually done in an AC type mode and has much less lateral spatial resolution compared to STM. | {
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quantum-field-theory, commutator, causality, propagator, klein-gordon-equation
Isn't this an issue? Don't we want this to be zero outside the light cone in order to preserve causality?
He then claims that what we really have to do is calculate $[\phi(x),\phi(y)]$. But then, what was the point of the previous calculation? What is my takeaway here? I am really confused about this. A propagator like
$\langle 0 | \phi(x) \phi(y)|0 \rangle$
(or a time-ordered correlation) | {
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statistical-mechanics, phase-transition, spin-models
In both papers, the approach is to show that their model becomes close enough (in a strong sense) to the mean-field Potts model once the interaction range is sufficiently large (in the first case), or the rate of exponential decay sufficiently slow (in the second case). It is of course easy to see that $q_c=2$ in the mean-field model.
Note also that the fact that, in both works, it was necessary to take a large enough range (or slow enough exponential decay) of the interaction is expected to be only a technical issue, originating from the need to compare with the mean-field model. What is expected is that the relevant feature of the model required to have $q_c=4$ in two dimensions is planarity. In particular, the conjecture is indeed that adding a next-nearest neighbor (translation invariant, ferromagnetic) interaction to your nearest neighbor model should change $q_c$ from $4$ to $2$. This has not yet been proved, unfortunately. | {
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python, beginner, tkinter
def get_names(self, roster, i):
words = roster[i]
self.Results.config(text=words)
def draw_names(self, roster, count):
if count == 0:
self.mixed = sample(roster, len(roster))
self.get_names(self.mixed, count)
count +=1
elif 0 < count < len(roster):
self.get_names(roster, count)
count +=1
else:
count = 0
self.mixed = sample(roster, len(roster))
self.get_names(self.mixed, self.count)
self.count_s7 +=1
def s7_Names(self):
s7 = ['Name1', 'Name2', 'Name3', 'Name4', 'Name5', 'Name6','Name7',
'Name 8', 'Name9','Name10', 'Name11','Name12','Naome13']
self.draw_names(s7, self.count_s7)
def s8_Names(self):
s8 = ['Kid1','Kid2','Kid3','Kid4','Kid5','Kid6','Another kid']
self.draw_names(s8, self.count_s8) | {
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image-processing
In signal processing, you'd like the interpolating function $f(m,n)$ to be the ideal low-pass filter. However, its frequency response requires infinite support and is useful only for bandlimited signals. Most images are not bandlimited and in image processing there are other factors to consider (such as how the eye interprets images. What's mathematically optimal might not be visually appealing). The choice of an interpolating function, much like window functions, depends very much on the specific problem at hand. I have not heard of Connes, Welch and Parzen (perhaps they're domain specific), but the others should be the 2-D equivalents of the mathematical functions for a 1-D window given in the Wikipedia link above.
Just as with window functions for temporal signals, it is easy to get a gist of what an image interpolating kernel does by looking at its frequency response. From my answer on window functions: | {
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c++, recursion, template, c++20, constrained-templates
recursive_reduce_string function test with std::deque<std::wstring>:
0123
recursive_reduce_string function test with std::deque<std::deque<std::wstring>>:
0123012301230123
recursive_reduce_string function test with std::vector<std::u8string>:
€2.00€2.00€2.00€2.00
recursive_reduce_string function test with std::vector<std::vector<std::u8string>>:
€2.00€2.00€2.00€2.00€2.00€2.00€2.00€2.00€2.00€2.00€2.00€2.00
recursive_reduce_string function test with std::vector<std::pmr::string>:
123123123
Computation finished at Thu Oct 26 08:44:11 2023
elapsed time: 0.00210049
Godbolt link is here.
All suggestions are welcome.
The summary information:
Which question it is a follow-up to?
A recursive_reduce Template Function with Unwrap Level Implementation in C++ and
A recursive_reduce_string Template Function Implementation in C++
What changes has been made in the code since last question?
Another approach of recursive_reduce_string template function implementation is proposed in this post. | {
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distributed-systems
the collection of all messages labeled with some particular proposal $n$.
It is important to distinguish between the definition of "round" here and the "communication round". In this sense, a round is dedicated to a single proposal (and belongs to a single proposer), because different proposals have globally unique timestamps.
Q2: I found a number of sources which indicate there are three phases in Paxos. Which is true?
As explained in James Aspnes' lecture notes,
Note that acceptance is a purely local phenomenon; additional messages are needed to detect which if any proposals have been accepted by a majority of acceptors. Typically this involves a fourth round, where acceptors send accepted(n, v) to all learners.
In the "Paxos Made Simple" paper, Lamport mentions the "learning" phase in Section 2.3 (see Page 6). Notice that it is a fourth phase in Aspnes' lecture note. | {
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"openwebmath_score": null,
"tags": "distributed-systems",
"url": null
} |
java, beginner, array, design-patterns
}
} catch (InputMismatchException e) { // If input is not an integer.
// Skip to finally block
} finally {
invalidValuePrompt(); // Remind user to select correct value.
menuMethod(); // Recursive call gives user option to create PIN, Password or EXIT.
}
}
} | {
"domain": "codereview.stackexchange",
"id": 38843,
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"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "java, beginner, array, design-patterns",
"url": null
} |
to the nearest tenth as needed. When a coin is tossed, there are two possible outcomes: heads (H) or ; tails (T) We say that the probability of the coin landing H is ½. meeting a person at a party). Probability Darts 3 - Displaying top 8 worksheets found for this concept. Probability. 0009, 10 percent. 31% check Approved by eNotes Editorial. The numbers from 1-30 are in a bag. Some of the worksheets for this concept are Probability darts, Probability statistics work 5, 11 2 theoretical and experimental probability, 1 samplespaceand probability, Lesson practice b 11 2 experimental probability, Work geometric probability, Name period geometry unit 10 work 9 geometric. Chemistry Scarsdale High School Electron Probability Objective: Determine the probability of finding an electron near the nucleus of an atom. Find the probability of earning each score. You keep throwing. Note also that the square also consists of$8$parts each housing one part of the region. It is given that AC=26, BC=24, | {
"domain": "clodd.it",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.986151392226466,
"lm_q1q2_score": 0.8085389568582907,
"lm_q2_score": 0.8198933381139645,
"openwebmath_perplexity": 835.8275025840575,
"openwebmath_score": 0.5500837564468384,
"tags": null,
"url": "http://clodd.it/cfop/probability-darts-answers.html"
} |
Escalator questions tend to confuse people but the point to note here is that they are just like Boats and Streams questions. If you are going in the direction of the escalator, it is like going downstream and your effective speed will be your speed + escalator speed. When you are going against the escalator, it is like going upstream and your effective speed will be your speed - escalator speed. The total distance to be traveled is the total number of steps on the escalator.
Then, you can very easily make the equations made by Bunuel and Sarang above:
$$\frac{y}{x+4}=15 and \frac{y}{4-x}=45$$ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9385759633816046,
"lm_q1q2_score": 0.8103586038588099,
"lm_q2_score": 0.8633916011860785,
"openwebmath_perplexity": 3706.1072975835295,
"openwebmath_score": 0.5227562189102173,
"tags": null,
"url": "http://gmatclub.com/forum/an-absent-minded-scientist-walks-on-an-escalator-at-a-rate-104469.html"
} |
php, time-limit-exceeded, pdo
require_once('../config/database.php');
require_once('../config/functions.php');
$keywords = preg_split("/MyWebSite\//", getcwd());
$folder = explode('/', $keywords[1]);
if (!isset($_POST['login']) || !isset($_POST['email']) || !isset($_POST['password']) || !isset($_POST['password_check']))
{
echo "missing some fields, did you try to edit my html?";
die();
}
if (strlen($_POST['login']) < 5 || strlen($_POST['email']) == 0 || strlen($_POST['password']) < 5 || strlen($_POST['password_check']) < 5)
{
echo "fieds badly filled";
die();
}
if ($_POST['password'] != $_POST['password_check'])
{
echo "passwords field aren't same";
die();
}
try
{
$db = new PDO($DB_DSN, $DB_USER, $DB_PASSWORD);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // to get an exception when caught an error :)
} catch (PDOException $e) {
print "Erreur !: " . $e->getMessage() . "<br/>";
die();
} | {
"domain": "codereview.stackexchange",
"id": 25938,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "php, time-limit-exceeded, pdo",
"url": null
} |
python
print("--- %s seconds ---" % (time.time() - start_time))
print("Other Function:")
from collections import defaultdict
start_time = time.time()
def dict_with_indices(dta):
""" Returns a dictionary with a list of indices for each item in dta
Args:
dta (list): list of data to be indexed
Returns:
dict: dict with list of indices as values
Examples:
>>> dict_with_indices([1,2,3,3,1,0,5]]
{1: [0, 4], 2: [1], 3: [2, 3], 0: [5], 5: [6]}
"""
result = defaultdict(list)
for idx, val in enumerate(dta):
result[val].append(idx)
return result
dict_with_indices(sample_data)
print("--- %s seconds ---" % (time.time() - start_time))
Times for this new tests:
First Function:
--- 1.9537413120269775 seconds ---
My Function:
--- 0.7061352729797363 seconds ---
Other Function:
--- 0.6053769588470459 seconds ---
New Solution seems to be the better one, thanks to @agtoever :
from collections import defaultdict | {
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "python",
"url": null
} |
turing-machines, finite-automata, pushdown-automata, computation-models, nondeterminism
A deterministic Turing machine decider has two kinds of halting states, accepting and rejecting, and defines a partial function as follows: if $M$ halts on $x$ at an accepting state, then $M(x)=1$; if it halts at a rejecting state, then $M(x)=0$; if it doesn't halt, then $M(x)=\bot$. If $M$ always halts then we say that it accepts the language $L = \{x : M(x) = 1\}$.
A non-deterministic Turing machine (which is always a decider) is allowed to "branch" (have several possible options at any given point in time), and has the following semantics: | {
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"tags": "turing-machines, finite-automata, pushdown-automata, computation-models, nondeterminism",
"url": null
} |
or from. Enters when you try to combine them Siméon Denis Poisson in 1837 gain insights from different of. Trials or the probability ( Poisson probability would equal 0.368 + 0.368 or.... Distribution tutorial x > x given ) articles – Poisson distribution be anything - a unit of time a. Is constant ; the occurrence of one accident Reply Cancel Reply as in the hour... Be anything - a unit of time you take the simple example calculating! Be used for calculating or creating new math problems: x is the Euler ’ consider! Ends 2.016 vs. 0.653 - this is an average of 1 phone call in the number of successes result. Functions on a 2-hour time period success in a given number of times the event is to occur symbolized...: probability mass f this Calculator is used to find Poisson … Poisson distribution.. Most n successes in a period of time with a known average rate ;... That a single success will occur during a short interval is small { 2.04^ { 10 to a. Below the form you to understand Poisson | {
"domain": "com.br",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9843363480718235,
"lm_q1q2_score": 0.824499779769548,
"lm_q2_score": 0.8376199673867852,
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"openwebmath_score": 0.6509507894515991,
"tags": null,
"url": "http://goldenbowl.com.br/kitchenaid-toaster-gyu/poisson-distribution-calculator-2d01da"
} |
Let $\delta= \psi(a)-\ln \mu$. Then:
\begin{align} \int_0^\infty x^{a-1} e^{-\mu x} \ln x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^2\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^2+\zeta(2,a) \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^3\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^3+3\zeta(2,a)\delta-2\zeta(3,a) \right\} \\ \int_0^\infty x^{a-1} e^{-\mu x} \ln^4\!x \,dx &=\frac{\Gamma(a)}{\mu^a}\left\{ \delta^4+6\zeta(2,a)\delta^2-8\zeta(3,a)\delta + 3\zeta^2(2,a)+6\zeta(4,a)) \right\} \end{align}
where $\zeta(z,q)=\sum_{n=0}^\infty \frac{1}{(n+q)^z}$ is the Hurwitz zeta function (the Riemann zeta function is the special case $q=1$).
Now on to the moments of the log of a gamma random variable.
Noting firstly that on the log scale the scale or rate parameter of the gamma density is merely a shift-parameter, so it has no impact on the central moments; we may take whichever one we're using to be 1. | {
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"openwebmath_score": 0.9848909378051758,
"tags": null,
"url": "https://stats.stackexchange.com/questions/312803/skewness-of-the-logarithm-of-a-gamma-random-variable"
} |
performance, comparative-review, swift, combinatorics
// constant copy of the elements to pass to the iterator on its creation.
let elements: [Element]
init(startingFrom elements: [Element]) {
self.elements = elements
}
init<S : Sequence>(_ elements: S) where S.Iterator.Element == Element {
self.elements = elements.sorted()
}
struct Iterator : IteratorProtocol {
private var current: [Element]
private var firstIteration = true
// you should create a PermutationSequence rather than invoking this
// initialiser directly.
fileprivate init(startingFrom elements: [Element]) {
self.current = elements
}
mutating func next() -> [Element]? {
var continueIterating = true
// if it's the first iteration, we avoid doing the permute() and reset the flag
if firstIteration {
firstIteration = false
} else {
continueIterating = current.permute()
} | {
"domain": "codereview.stackexchange",
"id": 24964,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "performance, comparative-review, swift, combinatorics",
"url": null
} |
optics, astrophysics, astronomy, spectroscopy
The picture below shows an echellogram for the bHROS echelle spectrograph on the Gemini S telescope. The cross-disperser is a set of fused silica prisms. These provide an order separation of at least 3mm. The detector is represented by the grey rectangle - you can basically position that to record any part of the echellogram. The width of each of the echellogram orders (order number labelled on the left hand side) is governed by the length of the slit. You cannot make the slit length too large or the orders will overlap with the orders above and below. | {
"domain": "physics.stackexchange",
"id": 62600,
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"tags": "optics, astrophysics, astronomy, spectroscopy",
"url": null
} |
• @PontusS It is correct. I got to the part where I need to prove that if $z$ is a convex linear combination of $x_1, x_2$ then the ball with center $z$ and radius $d(z-x_0)$ is a subset of the union of the balls with centers $x_1, x_2$ and radii $d(x_1 - x_0), d(x_2-x_0)$. I can't say whether this statement is correct right off the bat. But if you can prove this, then the proof for convexity of $M$ is straightforward. The proof in the accepted answer is a lot more elegant though. – lightxbulb Feb 22 at 15:45 | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9711290922181331,
"lm_q1q2_score": 0.8050065944815276,
"lm_q2_score": 0.8289388104343892,
"openwebmath_perplexity": 138.53903448794583,
"openwebmath_score": 0.9746814370155334,
"tags": null,
"url": "https://math.stackexchange.com/questions/3122592/prove-convexity-of-set-with-triangular-inequality"
} |
ros-melodic
Original comments
Comment by sisko on 2021-04-03:
Thanks @Tryan.
I did not realise slam does not work well in an environment of long straight corridors. Infact, I had assumed the opposite. Those lines are actually the kerbs of sidewalks of a simulated city as detected by the laser(s) or lidar on my robot model.
As the environment is a city, the lidar does detect buildings and other infrastructure.
So, I wonder, is it possible to combine multiple sensor data to create better slam data ?
My thinking is perhaps combining odometry, laser and lidar would provide better data for slam.
Comment by tryan on 2021-04-03:
Yes, the main problem with such an environment arises when using only a short-range lidar. In a hallway, the scans from the lidar at different points will look very similar (two parallel walls), making it appear as though the robot has not moved. | {
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"tags": "ros-melodic",
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} |
general-relativity, stress-energy-momentum-tensor
$$\begin{align}
c &= 1 \\
\hbar &= 1 \\
4 \pi G &= 1 \\
\epsilon_0 &= 1 \\
\end{align}$$
That would non-dimensionalize the GEM equations
$$\begin{align}
\nabla \cdot \mathbf{E}_\text{g} &= -\rho_\text{g} \\
\nabla \cdot \mathbf{B}_\text{g} &= 0 \\
\nabla \times \mathbf{E}_\text{g} &= -\frac{\partial \mathbf{B}_\text{g} } {\partial t} \\
\nabla \times \mathbf{B}_\text{g} &= -\mathbf{J}_\text{g} + \frac{\partial \mathbf{E}_\text{g}} {\partial t} \\
\end{align}$$
as well as the EM counterparts:
$$\begin{align}
\nabla \cdot \mathbf{E} &= \rho \\
\nabla \cdot \mathbf{B} &= 0 \\
\nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}} {\partial t} \\
\nabla \times \mathbf{B} &= \mathbf{J} + \frac{\partial \mathbf{E}} {\partial t} \\
\end{align}$$
In both cases of GEM radiation and EM radiation, using these L-H rationalized Planck units would normalize the speed of propagation in free space to 1 and the characteristic impedance of free space to 1. | {
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"tags": "general-relativity, stress-energy-momentum-tensor",
"url": null
} |
java, sorting, heap
public static void HEAP_SORT(int A[])
{
//max heap is built with heapSize initialised
BUILD_MAX_HEAP(A);
//starting from end loop through entire array
for(int i=A.length-1;i>=0;i--)
{
//since heap is already heapified and maintains max heap property
//the first element of the array is root of max heap
//swap it with the extreme element of the array i.e. setting the largest element to the end of array
int temp = A[0];
A[0]=A[i];
A[i]=temp;
//reduce the heap window by 1
heapSize = heapSize-1;
//call max heapify on the reduced heap
MAX_HEAPIFY(A,0);
}
}
public static void main(String[] args) { | {
"domain": "codereview.stackexchange",
"id": 4734,
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"lm_q2_score": null,
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"openwebmath_score": null,
"tags": "java, sorting, heap",
"url": null
} |
python, python-2.x, numpy, coordinate-system
def check_no_viols(locations, inters, no_points):
"""Checks if the structure has any point pairs where the constraint is violated.
Returns True if the sum of violations is zero and False otherwise.
"""
no_viols = True
for point_1_no in range(no_points):
for point_2_no in range(point_1_no):
diff = locations[point_1_no] - locations[point_2_no]
distance = np.sqrt(diff.dot(diff))
if check_viol(distance, inters[point_1_no][point_2_no]) == 1:
no_viols = False
break
return no_viols
def check_viol(distance, constraint):
"""Returns 1 if the distance between two points violates the constraints present, otherwise returns 0."""
if constraint is None:
score = 0
elif constraint[0] <= distance < constraint[1]:
score = 0
else:
score = 1
return score | {
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"id": 11718,
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"openwebmath_score": null,
"tags": "python, python-2.x, numpy, coordinate-system",
"url": null
} |
dimensional-analysis, physical-constants, si-units
Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed?
What we can say, is what really gives significance to the Planck length, is that at the energy/scale distance set by it, quantum gravity effects become non-negligible.
In quantum gravity, geometry with the usual rules doesn't work if the (proper) distances are thought of as being shorter than the Planck scale. | {
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"openwebmath_score": null,
"tags": "dimensional-analysis, physical-constants, si-units",
"url": null
} |
c, reinventing-the-wheel, windows
Title: a simple implementation of unix2dos for windows On linux there is the utility called unix2dos which converts UNIX EOLs(\n) to DOS EOLs(\r\n). However on windows there is no such tool so as a result I decided to make one.
unix2dos.c:
#include <windows.h>
#include <stdint.h>
#include <stddef.h>
#define chunksize (1 << 13)
#define nullptr ((void *)0)
uint8_t buffer[chunksize + 1] = { 0 };
int64_t newline_count(HANDLE filehandle)
{
DWORD bytes_read = 0;
int64_t result = 0;
do
{
if (ReadFile(filehandle, buffer + 1, chunksize, &bytes_read, nullptr) == 0)
{
WriteConsoleW(GetStdHandle(STD_ERROR_HANDLE), L"Error: could not read file", 26, nullptr, nullptr);
ExitProcess(GetLastError());
} | {
"domain": "codereview.stackexchange",
"id": 39270,
"lm_label": null,
"lm_name": null,
"lm_q1_score": null,
"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "c, reinventing-the-wheel, windows",
"url": null
} |
python, demodulation, scipy, fsk
Title: Demodulation of FSK signal What kind of FSK signal is this and what demodulation technique can I use to demodulate it. Why is the phase of the 1's changing like that and how does that effect how it needs to be demodulated. I will be programming the demodulation in python.
I am trying to low pass at 900Hz and then generate the envelope but I am not getting the desired result.
import numpy as np
import matplotlib.pyplot as plt
import scipy.signal as signal
from scipy.fftpack import fft, rfft, rfftfreq, irfft
import scipy.signal.signaltools as sigtool
from scipy.io import wavfile
fs, data = wavfile.read('jx3p.wav')
h = signal.firwin(numtaps = 300, cutoff = 900, fs=fs)
data = signal.lfilter(h, 1, data)
data = np.abs(sigtool.hilbert(data))
import matplotlib.pyplot as plt
plt.plot(data)
plt.show() | {
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"lm_q1q2_score": null,
"lm_q2_score": null,
"openwebmath_perplexity": null,
"openwebmath_score": null,
"tags": "python, demodulation, scipy, fsk",
"url": null
} |
cryptography
Title: Doubts on Definition of Indistinguishable Encryption in the Textbook In the classic crypto textbook "Introduction to Modern Cryptography" by Jonathan Katz and Yehuda Lindell, there is a definition for indistinguishable encryption in the presence of an eavesdropper as such that for every probabilistic polynomial time adversary A there is a negligible function negl(n) such that
$\Pr[PrivK_{A,\Pi}=1] \leq negl(n)$
where PrivK is the indistinguishability experiment and for the purpose of this question we only need to know that the experiment outcome is 1 iff the adversary makes the correct guess.
My doubts are as follows. Consider a sequence of probabilistic polynomial time adversaries $\{A_i\}_{i>=1}$ whose advantage in the indistinguishability experiment is bounded by the following sequence of negligible functions
$\Pr[PrivK_{A,\Pi}=1] \leq negl_i(n) = \frac{1}{(1+1/i)^n}$ | {
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"openwebmath_score": null,
"tags": "cryptography",
"url": null
} |
python
>>> ' '.join(genkey(i) for i in range(1, 6))
'xKFbV UkbdG hVGer Evcgc 15HhX'
"""
base = len(chars)
domain = base ** length
assert(1 <= value <= domain)
n = value * prime % domain
digits = []
for _ in xrange(length):
n, c = divmod(n, base)
digits.append(chars[c])
return ''.join(digits)
A couple of things you might want to beware of:
This pseudo-random scheme is not cryptographically strong: that is, it's fairly easy to go back from the URL to the value that produced it. This can be a problem in some applications.
The random strings produced by this scheme may include real words or names in human languages. This could be unfortunate in some applications, if the resulting words were offensive or otherwise inappropriate. | {
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"tags": "python",
"url": null
} |
This means
f(t) = {-t, -2<t<0
{t, 0<=t<2
is continous as well. But it is not differentiable at t=0 is it? How come?
6. Nov 15, 2006
### HallsofIvy
Staff Emeritus
Yes, the limit from both left and right is 0 so the limit is 0. Since the limit there exists and is equal to the value of the function, the function is continuous at 0.
Your f(t) is, of course, |t| between -2 and 2. It is well know that |x|, while continuous, does not have a derivative at x= 0. Essentially, it is not "smooth": there is a corner at x= 0 so the tangent line there is not well defined. More precisely, if h< 0 then f(0+h)= f(h)= -h so
(f(0+h)- f(0))/h= -h/h= -1 while if h> 0, (f(0+h)- f(0))/h= h/h= 1. The two one-sided limits are different so the limit itself, and therefore the derivative at t=0, does not exist. | {
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"openwebmath_score": 0.9002861380577087,
"tags": null,
"url": "https://www.physicsforums.com/threads/is-this-function-continous.143924/"
} |
php, form, authentication
}?>
</div> | {
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quantum-mechanics, particle-physics, subatomic
To make an analogy, suppose physicists discover a new element. That's interesting, but like all atoms it's made of electrons, neutrons and protons. So the new element is just another way of arranging more fundamental particles. Likewise, the new particles are just another way of arranging three quarks. So they are interesting, but not Earth shattering. | {
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navigation, gmapping
Title: Post processing gmapping generated maps to add glass wall information
The gmapping system misses glass surfaces since it makes use of laser data. Unfortunately, we have an environment that has quite a few glass walls/doors. Has anyone run into this issue? If yes, how do you folks suggest we handle it? I can always modify the map manually in an image editor. Is there a better way? | {
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navigation, ros-melodic, gmapping
if you're building your workspace, those two steps should probably be reversed.
Refer to #q252478 for an example workflow that shows the steps in the correct order.
Am i wrong to clone repos there? is there a way to do otherwise without cloning them into the ws, as that seems to be the problem?
Cloning repositories (ie: building packages from sources) does have its uses, but I would strongly suggest to only do it when absolutely needed. As I wrote in #q320046, there are quite a few steps to get right, multiple points at which things can fail, and it's always more time consuming than a simple apt install.
If you still want or need to build packages from sources, but don't want those packages in your 'main workspace', then I'd suggest to take a look at workspace overlaying. You place the packages you don't want in your main workspace in the underlay, build the underlay, source it and then continue working in your main workspace. | {
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ros, navigation, odometry, broadcaster, transform
The odom topic will have the details for odometry information - the type for that topic is nav_msgs/Odometry.
In tf there will be a new transform between base_link and odom. The transformation from odom to base_link - defines the position and orientation of base_link in the odom frame - is the same as the position/orientation part of the odometry message as well. This should be apparent from the code.
However, I am not sure how you exactly compute the velocities/distances in that tutorial.
Given that you already have the ticks, I guess all that you have to do now is this:
Convert from to ticks to distance travelled or velocity - I think distance travelled will give better accuracy.
Then simply adapt that code alongwith the tutorial to post odom information.
Once you have this you have to setup the navigation stack and you can run Monte Carlo localization.
Have a great day! | {
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ds.algorithms, graph-algorithms, random-walks
The approach you propose, which I would call Markov Chain Monte Carlo, is often a competitor to the message passing approach. If you are interested in sampling approximately uniformly at random from the set of bounded-degree, bounded-depth spanning trees of a given graph, I suggest altering your approach to use "soft" bounds. I.e. instead of rejecting an edge swap that makes the tree violate the depth bound, accept it, but with lower probability than a swap which does not violate the bound. If you have a parameter that controls how much lower this probability is, you can make the constraint violating configurations less and less probable until you arrive at a feasible solution that is nearly uniformly random.
The big question is how long do you need to run the chain. Since a spanning tree with degree at most 2 is a Hamiltonian path, you should expect any generic bound to be exponential in the size of the graph. But maybe the graphs you are interested in are special in some way. | {
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\begin{align}\mathsf E(\bar X) &= \mathsf E\left(\tfrac 1 t\sum_{i=1}^t X_i\right)\\[1ex]&=\tfrac 1t\sum_{i=1}^t\mathsf E(X_i)\end{align}
The rest is that these $$t$$ samples are identically distributed, and for all $$t$$ samples their expectation is $$p$$.
So, yeah, $$\mathsf E(\bar X)=p$$
Similarly for the variance of the mean estimator, we also use independence (and so the variance of the sum equals the sum of variance):- \begin{align}\mathsf{Var}(\bar X)&=\mathsf {Var}(\tfrac 1t\sum_{i=1}^t X_i)\\[1ex]&=\sum_{i=1}^t\mathsf{Var}(\tfrac 1{t}X_i)\\[1ex]&~~\vdots\end{align}
Recall we have for any scalar $$a$$ and random variable $$Z$$, that: $$\mathsf {Var}(aZ)=a^2\mathsf{Var}(Z)$$ | {
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and compound propositions. en.wiktionary.org. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. The connectives ⊤ … "A triangle is isosceles if and only if it has two congruent (equal) sides.". Now I know that one can disprove via a counter-example. A biconditional statement is really a combination of a conditional statement and its converse. All birds have feathers. Two line segments are congruent if and only if they are of equal length. Also, when one is false, the other must also be false. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Name. Examples. The compound statement (pq)(qp) is a conjunction of two conditional statements. Just about every | {
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navigation, ekf, ros-kinetic, robot-localization, ekf-localization-node
SymmetricMatrix sys_noise_Cov(STATE_SIZE);
sys_noise_Cov = 0.0;
sys_noise_Cov(1,1) = SIGMA_SYSTEM_NOISE_X;
sys_noise_Cov(1,2) = 0.0;
sys_noise_Cov(1,3) = 0.0;
sys_noise_Cov(2,1) = 0.0;
sys_noise_Cov(2,2) = SIGMA_SYSTEM_NOISE_Y;
sys_noise_Cov(2,3) = 0.0;
sys_noise_Cov(3,1) = 0.0;
sys_noise_Cov(3,2) = 0.0;
sys_noise_Cov(3,3) = SIGMA_SYSTEM_NOISE_THETA;
Gaussian system_Uncertainty(sys_noise_Mu, sys_noise_Cov);
// create the nonlinear system model
sys_pdf = new NonlinearSystemPdf(system_Uncertainty);
sys_model = new SystemModel<ColumnVector> (sys_pdf);
/*********************************
* NonLinear Measurement model *
********************************/ | {
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javascript, object-oriented, html, game-of-life
*/
this.cellsToFLip = new Array();
/**
* list of currently alive cells. for faster draw.
*/
this.aliveCells = new Array();
}
Grid.prototype.revive = function(x, y) {
if (x <= 0 || y <= 0 || x >= this.width - 1 || y >= this.height - 1) {
return; // frame of 1 dead cell
}
this.cells[x][y].isAlive = true;
this.aliveCells.push(this.cells[x][y]); // add to alive list
};
Grid.prototype.kill = function(x, y) {
this.cells[x][y].isAlive = false;
for (var i = 0; i < this.aliveCells.length; i++) { // remove from alive list
if (this.aliveCells[i] == this.cells[x][y]) { // same cell?
this.aliveCells.splice(i, 1); // remove cell
}
}
};
/**
* sets all cells to dead.
*/
Grid.prototype.reset = function() {
for (var i = 0; i < this.width; i++) {
for (var j = 0; j < this.height; j++) {
this.kill(i, j);
}
}
this.iteration = 0;
};
/** | {
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ros, ros-kinetic, build
Title: Error The manifest must contain exactly one "description" tags
I recently try to build a package for standalone ROS node in MATLAB Simulink. I've followed the instructions perfectly until build catkin workspace step.
dinhthong@dinhthong-Inspiron-3542:~/USV_ws2$ catkin_make
Base path: /home/dinhthong/USV_ws2
Source space: /home/dinhthong/USV_ws2/src
Build space: /home/dinhthong/USV_ws2/build
Devel space: /home/dinhthong/USV_ws2/devel
Install space: /home/dinhthong/USV_ws2/install
####
#### Running command: "cmake /home/dinhthong/USV_ws2/src -DCATKIN_DEVEL_PREFIX=/home/dinhthong/USV_ws2/devel -DCMAKE_INSTALL_PREFIX=/home/dinhthong/USV_ws2/install -G Unix Makefiles" in "/home/dinhthong/USV_ws2/build"
####
-- The C compiler identification is GNU 5.4.0
-- The CXX compiler identification is GNU 5.4.0
-- Check for working C compiler: /usr/bin/cc
-- Check for working C compiler: /usr/bin/cc -- works
-- Detecting C compiler ABI info
-- Detecting C compiler ABI info - done | {
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newtonian-mechanics, momentum, inertia
Title: What causes the skidding of the wheels of a moving car? Momentum or inertia? In a moving car, when brakes are applied suddenly the wheels skid.I have these two explanations in my mind and both seems correct to me.1. The momentum of the car must be conserved so the car continues to be in motion after the brakes are applied and the wheels skid and because of friction the car comes to rest.2. Because of the inertia of the car, the wheels will go forward with the car skidding and it will come to rest because of friction.But I was told that the correct explanation is the second one. I was told that it is related to the mass but I couldn't understand the explanation given to me. Why is momentum not the reason for skidding but inertia is? Momentum and inertia are closely related properties.
Newton's first law states that an object will continue in a straight line with constant momentum if no net external forces act on it. | {
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evaporation
A temperature of $20\ \mathrm{^\circ C}$
Which leads to a density of air of $1.204\ \mathrm{kg/m^3}$, or $0.1204 \ \mathrm{g/cm^3}$
And a partial pressure of water of $0.0231$
And a diffusion coefficient of water vapor in air of $0.242\ \mathrm{cm^2/s}$
About $6\ \mathrm{cm}$ as the distance, given that the can is on average half full and the storage space was very dry | {
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discrete-signals, signal-analysis, interpolation, radar, sinc
So:
You write some code, and then you wonder whether it is correct. The time-proven remedy for this kind of issue is the ability to test that hypothesis. In software engineering, we call such rests unit tests when applied to functional units (like your RCM implementation), and integration tests when they are used to verify a complex system of individually tested units still works together.
So, here, you'd want to write unit tests. In my experience, it's important to not start with a very complex example - start with a signal that you feed into your RCM (i.e. what would normally come out of pulse compression) that consists of all zeros but for a single one in a cell that will not migrate. Input should be output, and if that works, you probably got a lot of things right. Then calculate a point where the migration is a multiple of of full cells. Test with the same single-target signal. Works? Great! | {
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forces, acceleration
So from what I see, $g$-force tells us about our non-gravitational acceleration, and it has nothing to do with force, it is obvious. Coming back to the elevator accelerating upward at $2$ $\frac{m}{s^2}$, let's say there are two people inside the elevator, and their mass is $50$ $kg$ and $60$ $kg$, respectively. Since both are accelerating at the same rate, i.e $2$ $\frac{m}{s^2}$, it clearly means both are experiencing Normal reaction of different magnitudes, $600N$ and $720N$ respectively. But both are experiencing a $g$-force of $1.2g$. | {
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Norm Of A Vector | {
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performance. How to apply the rule. Title: SMART Notebook Author: tara. Use the Empirical Rule to determine the approximate percentage of house sizes that lie between 955. Given an approximately normal distribution what percentage of all values are within 2 standard deviations from the mean? 3. Empirical Rule… Another method is to check the 68-95-99. notebook September 10, 2014 Box and Whiskers with Outliers Outlier: An extremely high or an extremely low value in the data set when compared with the rest of the values. After taking notes, asking questions, and discussing the Empirical Rule, students will complete Exit Ticket Empirical Rule. 7 rule tells us that this area-- because it's within two standard deviations-- is 95%, or 0. Note that the definition of consistency of fn depends on the distribution P of (X,Y). 0 volts and a standard deviation of 0. The Empirical Rule 99 7% Knowing that the values in a set are approximately normally distributed allows you to get a feel for how | {
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light, night-sky
Title: Why celestial objects appear brighter in the night sky than in photographs from closer proximity? The moon, for example, appears very bright in the night sky, but quite dark from photographs on/near its surface.
Similarly, Mars and Venus appear as bright dots in the sky but in closer photographs they're quite dark. The Moon, Mars, Venus are seen against a dark sky using an eye whose aperture is adjusted to the average brightness of the visual field. So bright small sources are over-exposed and so appear bright.
Photographs of planetary surfaces are taken with aperture and exposure set for proper exposure. So appear closer to how they would look if they filled your eye's field of view and so the aperture would be set to properly expose the surface. | {
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quantum-chemistry, biochemistry, stereochemistry, molecular-structure
Title: Does anyone know what program was used to make this molecular structure picture? I am a quantum chemistry beginer, and this picture color scheme is classic. I wonder how to make such pictures by myself. The first type of image is very easy to make. I would suggest Avogadro with POV-ray. Avogadro can be used to draw the structures (or open a file containing the information, like .mol2, .sdf etc.) and then high-quality images can be generated using POV-ray.
To do this, install Avogadro, then draw a structure, and then File> Export > POV-ray. A new subwindow will open, where you can specify the location of the POV-ray executable.
In my case, the executable is pvengine64.exe, it might be different in different platforms, or in different versions.
You can also change the resolution here. Finally, click "Render", and a new windows with POV-ray should open, and it should display the image, after a few seconds. | {
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quantum-mechanics, homework-and-exercises, fourier-transform, dirac-delta-distributions, normalization
=|N|^2 \int \mathrm{d}p'\delta(x'-x'')
$$
But all I can think of is that one can say $\int \mathrm{d}p'=\hbar k$ where $k$ is the wavenumber.
What am I missing? You're close, but you seem to be saying that $\exp(ikx) = \delta(x)$, which is not true, and you're missing a $2\pi$. The correct identity is
$$\int dk\ e^{ikx} = 2\pi \delta(x)$$
Therefore, with a change of variables $p=\hbar k$:
$$\int dp\ e^{ipx/\hbar} = \hbar \int dk\ e^{ikx} = 2\pi\hbar$$ | {
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In fact, we have no way to understand that without stepping out of the plane and into the third dimension. Y1LABEL Angular Cosine Distance TITLE Angular Cosine Distance (Sepal Length and Sepal Width) COSINE ANGULAR DISTANCE PLOT Y1 Y2 X . In this article, we’ve studied the formal definitions of Euclidean distance and cosine similarity. Reply. As far as we can tell by looking at them from the origin, all points lie on the same horizon, and they only differ according to their direction against a reference axis: We really donât know how long itâd take us to reach any of those points by walking straight towards them from the origin, so we know nothing about their depth in our field of view. Cosine similarity measure suggests that OA and OB are closer to each other than OA to OC. Consider another case where the points Aâ, Bâ and Câ are collinear as illustrated in the figure 1. We will show you how to calculate the euclidean distance and construct a distance matrix. As a result, | {
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Since a billion divided by a million is one thousand. So means that an additional connection at this level results in an increase in revenue of 0.31265 thouusand dollars or $321.65. ## What Is Marginal Average Cost? In each problem below, the average cost function by dividing the cost function by the variable representing the quantity. For a cost function C(Q), the average cost function is $\displaystyle \overline{C}(Q)=\frac{C(Q)}{Q}$ The marginal average cost function is the derivative of the average cost function. Problem 1 Suppose the total cost function for a product is $\displaystyle TC(Q)=\frac{3Q+1}{Q+2}\text{ hundred dollars}$ where Q is the number of units produced. 1. Find the average cost of producing 20 units. 2. Find the average cost function. 3. Find the marginal average cost function. 4. Find and interpret the marginal average cost when 20 units are produced. This means that each of the 20 units costs an average of .1386 hundred dollars or$13.86. | {
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There is some evidence that BMR correlates with the size of certain internal organs (e.g., this study on voles). This information is not generally available for people trying to loose a few pounds and so is impractical, but it does have the interesting benefit of being a possible, if partial, explanation of what specifically may be different about individuals who burn calories at a higher rate--more specific that the vague explanation: "genetics".
My BMR is about 2030 calories based on standard online calculators.
### Activity Level | {
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thermodynamics, statistical-mechanics, entropy, observers, information
Who is right? Did entropy increase or not? Both are right. As Jaynes nicely argues in the above reference, entropy is not a mechanical property, it is only a thermodynamic property. And a given mechanical system can have many different thermodynamic descriptions. These depend on what one can --or chooses to-- measure. Indeed: if you live in a universe where there are no people and/or machines that can distinguish red from blue, there would really be no sense in saying the entropy has increased in the above process. Moreover, suppose you were color blind, arrive at the conclusion that the entropy did not increase, and then someone came along with a machine that was able to tell apart red and blue, then this person could extract work from the initial configuration, which you thought had maximal entropy, and hence you would conclude that this machine can extract work from a maximal entropy system, violating the second law. The conclusion would just be that your assumption was wrong: in | {
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homework-and-exercises, general-relativity, differential-geometry, coordinate-systems, curvature
$$
similarly, you get
$$
\nabla_Y \nabla_X Z^{\ell} \enspace = Y^j \, \nabla_j \Big(X^i \, \nabla_i Z^{\ell}\Big) = Y^j \, \nabla_j \Big(X^i ( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)\Big)=\\=Y^j\partial_j\Big(X^i( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)\Big)+Y^jX^i\,\Gamma_{jn}^\ell( \partial_i Z^n+\Gamma_{im}^n Z^m)=\\=Y^j\partial_jX^i( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)+Y^jX^i\Big(\partial_j\partial_iZ^\ell +Z^m\partial_j\Gamma_{im}^\ell+\Gamma_{im}^\ell\partial_j Z^m+\Gamma_{jn}^\ell \partial_i Z^n+\Gamma_{jn}^\ell\Gamma_{im}^n Z^m\Big).
$$
If you now subtract them, you are left with | {
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# Toronto Math Forum
## MAT244--2020F => MAT244--Lectures & Home Assignments => Chapter 2 => Topic started by: Julian on September 28, 2020, 12:45:14 PM
Title: W3L3 Exact solutions to inexact equations
Post by: Julian on September 28, 2020, 12:45:14 PM
In week 3 lecture 3, we get the example $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$. | {
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experimental-physics, specific-reference, chemostat
The other thing is that this theory isn't necessarily very accurate if the perturbation (or change to the parameters) is large. However, your system is not massively nonlinear, so I would expect it to be a decent approximation. If you're worried about that, the best thing is probably to numerically integrate the system with a few different parameter values and check it. | {
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php, object-oriented, classes, inheritance
// Key is zero-based (e.g., $proc = 0, $params = 1).
foreach( func_get_args() as $key => $parameter ) {
// Skip the $proc and $params arguments to this method.
if( $key < 2 ) continue;
$count++;
$placeholders = empty( $placeholders ) ? "?" : "$placeholders,?";
array_push( $args, $parameter );
}
$sql = "";
if( empty( $params ) ) {
// If there are no parameters, then just make a call.
$sql = "SELECT $proc( $placeholders )";
}
else if( strpos( $params, "," ) !== false ) {
// If there is a comma, select the column names.
$sql = "SELECT $params FROM $proc( $placeholders )";
}
else {
// Otherwise, select the result into the given column name.
$sql = "SELECT $proc( $placeholders ) AS $params";
}
$db = $this->getDataStore();
$statement = $db->prepare( $sql );
for( $i = 1; $i <= $count; $i++ ) {
$statement->bindParam( $i, $args[$i - 1] );
} | {
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biochemistry, structural-biology, biophysics, antibody
Title: Structure of RAP Antibodies (Specifically RAP-5) [EDIT] - Have just found not one but two papers that address my structure problem. However they concern RAP-1A, so I guess my question is now what is the difference in structure and function of RAP-1A and RAP-5? Does anyone know of X-ray structure analysis being used to examine RAP-5? | {
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cost is C(S,U) = S + 5U. Minimizing any function means finding the deepest valley in that function. We probably want to minimize the cost associated with the recycling activity, as we do not have any information on revenues. So, \eqref{8} conveys mathematically the intent to change the cost function (by changing the network parameters), in order to effect the intermediate values calculated in $$z’s$$, so as to minimize the differences in the final output of the network. In the case we are going to see, we'll try to find the best input arguments to obtain the minimum value of a real function, called in this case, cost function. The objective function is the function to be minimized or maximized. It's a cost function because the errors are "costs", the less errors your model give, the better your model is. For multi-objective improvements, the most generally used developmental algorithms such as NSGA-II, SPEA2 and PESA-II can be utilized. The aim of the linear regression is to find a | {
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"openwebmath_score": 0.5395734310150146,
"tags": null,
"url": "http://cgvl.foresteria-lacura.it/minimize-a-cost-function.html"
} |
newtonian-mechanics, forces, kinematics, velocity, calculus
Now you are taking downward to be positive in which case for $$v^2=2g(x_f - x_i)$$ then $x_f \gt x_i$ so that $$v^2\gt 0$$ always meaning $v\gt 0$
Your assumption that $x_f\lt x_i$ is incorrect. If for example we take $x_i=5$ then the final value for $x$ can be $x_f=20$ since the positive direction is down so that $x_f-x_i$ is positive. | {
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Hmmm well you seem to have messed up twice or something along the lines so it cancelled out or something? http://www.wolframalpha.com/input/?i=integral%20-infty%20to%20infty%20e%5E-x%5E2dx&t=crmtb01
19. freckles
yeah that is the same I got I'm just saying I put the r in too early
20. Kainui
Ohhh ok I see. Alright here's a fun one haha: $\large \int\limits_0^\infty \frac{\tan^{-1}(ax)-\tan^{-1}(x)}{x}dx \text{ when } a>0$
21. freckles
now this one I do believe is entirely new to me let me think a bit
22. freckles
so that means a can be 1 and if a is 1 then the integrand is 0 which mean the output is 0 for a=1 so should the inequality include 1?
23. freckles
you know what nevermind it is probably some kinda pattern or whatever
24. Kainui | {
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temperature, everyday-life, water
In modern refrigerators a balance is struck, having various drawers for fruits and vegetables where the temperature is higher so they retain their texture and appearance.
Freezers are used for long time storage of food that is not destroyed by freezing, either in appearance or taste or texture in order to inactivate decay processes. In countries where freezing temperatures prevail in the winter it was observed that meat and fish etc retain their food properties well when frozen and can be kept for long months. The -18 degrees must be an engineering compromise, a good temperature for long time storage but not too expensive in power consumption to retain.
In any case both refrigerator and freezer have a thermostat that could be set for higher temperatures. | {
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channelcoding
Title: Why are the minimum distances of a code and its interleaved code equal? I know this is probably a very trivial question, but I am completely stuck. Let $C$ be a linear $[n,k,d]$ Code. Then the interleaving of depth $t$ is the Code $C(t)=\{(c_{11}, \dots, c_{t1}, \dots, c_{1n}, \dots, c_{tn}) \mid (c_{i1}, \dots, c_{in}) \in C \}$. Now I am trying to understand why the minimum distance of $C(t)$ is also $d$. | {
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This is College Physics Answers with Shaun Dychko. We are going to calculate the pressure at the bottom of the Mariana trench which is 11.0 kilometers below sea level and we convert that into meters. The density of seawater we look up in our table [11.2] and it is 1.025 times 10 to the 3 kilograms per cubic meter and the pressure or the gauge pressure and that is to say the amount of pressure in excess of the atmospheric pressure is the density of the fluid times acceleration due to gravity times the height of the fluid column and so that's 1.025 times 10 to the 3 kilograms per cubic meter times 9.80 newtons per kilogram times 1.10 times 10 to the 4 meters—height— and that is this many pascals which we convert into atmospheres by multiplying by 1 atm for every 1.013 times 10 to the 5 pascals and that is 1.09 times 10 to the 3 atmospheres. So that's a 1090 atm of pressure at the bottom of the Mariana trench—that's huge! And it was okay to just ignore atmospheric pressure and just talk | {
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c#, unity3d
private void OnTouchEnter(Touch touch) {
beingTouchedByFingerNumber = touch.fingerId;
isBeingTouched = true;
shipManager.GetComponent<Rigidbody>().isKinematic = true;
Vector2 v2 = Input.GetTouch(beingTouchedByFingerNumber).position;
GetOffsetOfMouseFromShip(v2);
motionState = MotionState.drag;
meshRenderer.enabled = false;
}
public void OnTouchEnter() {
isBeingTouched = true;
shipManager.GetComponent<Rigidbody>().isKinematic = true;
Vector2 v2 = Input.GetTouch(beingTouchedByFingerNumber).position;
GetOffsetOfMouseFromShip(v2);
motionState = MotionState.drag;
meshRenderer.enabled = false;
}
Here the first method could be made smaller by calling the second method after using the Touch parameter like so
private void OnTouchEnter(Touch touch) {
beingTouchedByFingerNumber = touch.fingerId;
OnTouchEnter();
} | {
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c, game
typedef struct _GVOXEL_STATE_1_0
{
GVOXEL_ID nVoxelID; // This id will be equal to the voxel index which may be handy
GVOXEL_TYPE nVoxelType; // This number will reference the table with voxel descriptions
word nMetadata; // This was described below.
byte nLightX; // This stores how much light (from 0 to 255) is receiving the face pointing to +X
byte nLightXN; // Same from below but for -X face
byte nLightY; // etc.
byte nLightYN; // etc..
byte nLightZ; // etc...
byte nLightZN; //
} GVOXEL_STATE_1_0, GVOXEL_STATE; | {
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cell-biology
Information on the licensing factors:
The licensing factors used to create the pre-replication complex is the Minichromosome Maintenance (Mcm2-7) protein, Cdc6, and Cdt1. The Cdc6 and Cdt1 are required to load the Mcm’s onto the DNA and therefore, are highly controlled during the cell cycle. The Mcm2-7 hexamer is the actual DNA helicase that is used during DNA replication.
The way that the re-replication of DNA is prevented is that when the DNA replication starts, the Mcm helicase moves away from the ORC and the newly replicated DNA, meaning it then cannot be reinitiated. Along with the physical moving of the helicase, Cdt1’s activity is suppressed by the protein Geminin, which prevents the licensing of the helicase during times of non-replication.
Image from the article on the licensing. | {
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of such numbers are π, e, √6. Irrational Numbers. Another definition we can give as “non terminating non recurring decimal numbers are irrational numbers”. This page contains a technical definition of Irrational Number. Many people are surprised to know that a repeating decimal is a rational number. 1. a. Let's start by defining each term separately, then we can learn more about each and work through some examples. Information and translations of irrational number in the most comprehensive dictionary definitions resource on the web. Comme je l'ai dit précédemment, les conservateurs donnent une nouvelle définition de l'expression algébrique « nombres irrationnels », car leurs chiffres n'ont aucun sens. For example, real numbers like √2 which are not rational are categorized as irrational. So they can't be written as a clear fraction of 2 integers. not endowed with reason or understanding. As I said previously, the Conservatives give a new definition to the algebraic term of | {
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machine-learning, k-means, clustering
Title: How to compute the number of centroids for K-means clustering algorithm given minimal distance? I need to cluster my points into unknown number of clusters, given the minimal Euclidean distance R between the two clusters. Any two clusters that are closer than this minimal distance should be merged and treated as one.
I could implement a loop starting from the two clusters and going up until I observe the pair of clusters that are closer to each other than my minimal distance. The upper boundary of the loop is the number of points we need to cluster.
Are there any well known algorithms and approaches estimate the approximate number of centroids from the set of points and required minimal distance between centroids? | {
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haskell, functional-programming
assessLine :: [Maybe Player] -> Maybe Player
assessLine line | Nothing `notElem` line && allSame line = join $ listToMaybe line
| otherwise = Nothing
getWinner :: Board -> Maybe Winner
getWinner board = Winner <$> listToMaybe (mapMaybe assessLine (ticTacToes board))
<|> if all isJust (elems board) then Just CatScratch else Nothing
And the original version—
getWinner :: Board -> Maybe Winner
getWinner board =
let mWinner = Winner <$> find (\row@(x:_) -> all (== x) row && isJust x) (toeLines board)
mCatScratch = if all isJust (elems board) then Just CatScratch else Nothing
in mWinner <|> mCatScratch | {
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Thanks for the reply. Perhaps with the French or the Indians themselves I can maintain a true dialogue!!
16. Victor Kamat says:
Ramanujan is not human, he is a beast.
17. Lênio Fernandes Levy says:
I just want to understand how Ramanujan held its deduction. In mathematics , as we all know , resolutions are not validated by computational induction . I think I have an outcome ( achieved by myself ) using the deductive method . I wonder if my deduction is near or far from the deductive process created by Ramanujan , which was undoubtedly a brilliant mathematician .
18. Dave Tate says:
Lênio, I suspect that Ramanujan arrived at his result the way Derek illustrated above, when playing around with the fun fact that (y-1)(y+1) = y^2 – 1. | {
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beginner, c, tic-tac-toe
struct game_data {
int win;
int turn;
int grid[3][3];
}
You should remove the empty line between the comment and the beginning of the struct. As it is now, the comment reads like a general remark that applies to all the part below it and not just the struct.
This struct definition is the best place to document which values are valid for the win fields. There are several possible choices:
true, false
0, 1
0, 1, 2
-1, 0, -1
0, 'o', 'x'
0, '0', '1'
0, '1', '2' | {
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formal-languages, regular-languages, closure-properties
an example could be:
$\varphi(a) = \{10^{i}1 | i \in \{1,...,100\}\}$, $\varphi(b) = \{\epsilon\}$
$\varphi(bba) = \varphi(b)\varphi(b)\varphi(a) = \{\epsilon\}\{\epsilon\}\{10^{i}1 | i \in \{1,...,100\}\}=\{10^{i}1 | i \in \{1,...,100\}\}$
now I can implie that the class $REG$ is closed under finite substitution, because a concatenation of finite sets is finite and every finite set is regular.
My second question: Is this my assumption about finite substitution and the closure propierty right? You seem to have understood the operation of substitution. That maps any string into a language by concatenating the images of the letters.
Your argument that regular languages are closed under finite substitution does not work. Even if the substitution maps avery word into a finite language, the substitution of a regular language usually is infinite as the initial ragular language is infinite.
A possible approach is to use regular expressions. | {
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fft, frequency-spectrum, python
results in a frequency sweep up to 19700*2, not to 19700, but maybe you did it intentionally. If not, notice that the instantaneous frequency is the derivative of the phase. To reach a frequency of $f_{max}$ in the frequency sweep, the formulas are
$$
phase(t) = f_{min}·t + (f_{max} - f_{min})/duration·t^2/2 \\
sweep\_wave(t) = \sin(2π·phase(t)) \\
instantaneous\_freq(t) = f_{min} + ((f_{max} - f_{min})/duration)·t
$$
and you should write the code
from scipy import signal
...
fmin = 21
fmax = 19700
time = np.linspace(0,duration,samplerate*duration, endpoint= False)
Const_wave = np.sin(np.pi*2*(fmin*time + (fmax-fmin)/duration*time*time/2.0)) | {
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pathology
The problem is that in today's society, stress-causing situations do
not really require the body to use up a lot of energy. So, cortisol
ends up causing the body to refuel after stress even when it doesn't
really need to refuel. This excess fuel or glucose is converted into
fat resulting in increased storage of fat. | {
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fl.formal-languages, automata-theory, parsing
and we use any lexer and parser combination to generate a recogniser for this language, in which the lexer uses the 'maximal munch' or 'longest match first' rule.
The specification might seem to be trivially equivalent to the regular expression "$a$+", but it isn't: in fact, it recognises no strings at all. The reason is that $X$ 'eats' all the $a$ characters present in the input because of 'maximal munch', leaving none for $Y$ to consume, so the parser always rejects the input.
I'd like to know if it is decidable if such a problem is present in a given lexer and parser specification. | {
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It's helpful to think of simple toy examples.
Consider a bag with two balls: one white and one black. Suppose we take out the balls one after the other without replacement, and consider the events $$W_1=\text{the first ball is white}$$and $$W_2=\text{the second ball is white}.$$
Intuitively, it's obvious these are not independent events because if you know $$W_1$$ happened then you know $$W_2$$ cannot happen since there's only one white ball. This is what your book means by "the sample space has changed": after you remove the first ball, you alter the contents of the bag.
The probability of $$W_1$$ actually determines the probability of $$W_2$$ since $$\mathrm P(W_1)+\mathrm P(W_2)=1$$. | {
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quantum-mechanics, general-relativity, black-holes, spacetime-dimensions
Time and space don't swap places inside a black hole.
The point of all this is that the coordinates are not spacetime - they are just labels we attach to spacetime.
These are very different descriptions and I would like some clarification on this. What exactly does swap and what does not in theory? The point of the question is only clarification, there is no other answer on this site that would answer my question because it is specifically, the amount of information and the different statements that are confusing.
Question: | {
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notation, error-analysis, protons
As a commenter points out, a lot has changed in the proton-spin problem in the thirty years since the paper you're reading was published. But you can see why the puzzle was interesting: $12\pm14$ was consistent with zero of the proton's spin coming from its valence quarks. | {
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newtonian-mechanics, kinematics, projectile, drag
Title: Is motion along the $x$ and $y$ axises independent with quadratic drag? It's often stated that motion in the $x$ and $y$ axises are independent, so that changing the $x$-velocity will not influence changes in the $y$-velocity. To me it seems that with quadratic drag (drag proportional to $v^2$) this shouldn't hold true. If we increase the initial velocity in the $y$-direction, this would increase the drag. It would also decrease the percent of the drag being applied in the $x$-direction, since the drag acts opposite to the velocity. Do these effects cancel? So the x part of the drag is
$$\begin{align*}
f_x &= f \cos\theta \\
&= (k v^2) \cos\theta \\
&= k v (v \cos\theta) \\
&= k v v_x \,,
\end{align*}$$
and $v$ is dependent on the $y$ component of motion as well as the x-component.
Similar consideration, of course, apply to the y-component of drag. | {
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python, numpy, pandas, machine-learning, data-visualization
def folders_to_hdf(folder_path, new_size, threads=5):
"""
Save folders containing images to hdf format.
Args:
folder_path: Path to folder containing folders containing images.
new_size: New image size(tuple).
threads: Number of parallel threads.
Return:
None
"""
for folder_name in os.listdir(folder_path):
if folder_name != '.DS_Store':
path = ''.join([folder_path, folder_name, '/'])
folder_to_hdf(path, folder_name, new_size, threads)
def clear_hdf(folder_path):
"""
Delete .h5 files in every sub-folder in folder_path.
Args:
folder_path: A folder containing folders of images and their .h5. | {
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quantum-mechanics, angular-momentum
$N=N_1 + N_2$.
The Hamiltonian can be rewritten into $H = \alpha N + \beta \frac{2}{\hbar} J_x$. The thing is that if I have the state $\left|\Psi\right> = \frac{1}{\sqrt{2}} (\left|1, 0\right> + \left|0,1\right>)$, where inside the ket there's $\left|n_1, n_2\right>$, then, if we apply the hamiltonian to this state we'll see that it's an eigenvector with eigenvalue $\alpha + \beta$.
If we also say that $J_x$ can be substituted by $\hbar m$ in the hamiltonian we'll get that the eigenvalues of the hamiltonian are $E= \alpha n + \beta 2 m$. I say also, because I didn't explain it, but the relation of $j$ and $m$ was obtained by doing the same with $J_z$.
Okay, so back to our state, if we know substitute our values of $n$ and $m$ of the state in the formula, which are $1$ and $0$ respectively, we get that its eigenvalue doesn't have any $\beta$. There's obviously something wrong with this formula, which I think is the step of the eigenvalue for $J_z$. | {
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• (i) If ${Kaz(G,S^2,\rho) \geq \epsilon}$, show that ${Cay(G,S)}$ is a two-sided ${c}$-expander for some ${c = c(\epsilon,k) > 0}$.
• (ii) Conversely, if ${Cay(G,S)}$ is a two-sided ${\epsilon}$-expander, show that ${Kaz(G,S^2,\rho) \geq c}$ for some ${c = c(\epsilon,k) > 0}$.
One advantage of working with Kazhdan constants instead of expansion constants is that they behave well with respect to homomorphisms:
Exercise 13 Let ${G, G'}$ be locally compact groups, and suppose that there is a continuous surjective homomorphism ${\pi: G \rightarrow G'}$ from ${G}$ to ${G'}$. Let ${S}$ be a compact subset of ${G}$. Show that for any unitary representation ${\rho': G' \rightarrow U(H)}$ of ${G'}$, one has ${Kaz(G',\pi(S),\rho') = Kaz(G,S,\rho' \circ \pi)}$. Conclude that ${Kaz(G',\pi(S)) \geq Kaz(G,S)}$. In particular, if ${G}$ has property (T), then ${G'}$ does also.
As a corollary of the above results, we can use Kazhdan’s property (T) to generate expander families: | {
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let, $dA$ be the area of the segment $XOY$
if we take infinitesimally small angle $\delta\theta$ then we have, $dA=\frac{1}{2}(r^2)(\sin{\delta\theta})$
we know that $\lim_{\delta\theta ->0} \frac{\sin\delta\theta}{\delta\theta} =1$
using, the above fact, we have, $dA=\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$
Now, $\lim_{\delta\theta ->0} dA = \lim_{\delta\theta ->0}\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$
we, have, $dA=\frac{1}{2}(r^2){\delta\theta}$
Now,$$\int{dA}=\int_{0}^{2\pi}\frac{1}{2}(r^2){\delta\theta}$$
$$\int{dA}=\frac{1}{2}(r^2)\int_{0}^{2\pi}{\delta\theta}$$
$$A=\frac{1}{2}(r^2)(2\pi)$$ $$A={\pi}r^2$$
we, now have the area of a circle of radius $r$, and it is ${\pi}r^2$
- | {
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circuit-complexity, randomized-algorithms
Title: Randomness and small circuits complexity classes Let $\mathcal{C}$ be a complexity class and $\textrm{BP-}\mathcal{C}$ be the randomized counterpart of $\mathcal{C}$ defined as $\textrm{BPP}$ with respect to $\textrm{P}$. More formally we provide polynomially many random bit and we accept an input iff the probability to accept is over $\frac{2}{3}$.
It is known that for non-uniform circuits class we have $\textrm{BPAC}^0=\textrm{AC}^0$:
Miklós Ajtai, Michael Ben-Or: A Theorem on Probabilistic Constant Depth Computations STOC 1984: 471-474 | {
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