kandi X-RAY | Factorization Summary
kandi X-RAY | Factorization Summary
Top functions reviewed by kandi - BETA
- Update the conductor
- Update the circuit count
- Returns an iterable containing the members of this charge
- Create or join the conductor set
- Draws the boxes
- Returns true if any of the boxes have been set
- Draws the render data
- Deplete the current value
- Removes the given number of items from the list
- Invalidate the conductor
- Gets a mutable list
- Increase value
- Tests whether the given collection contains all elements of the given collection
- Adds a charge to the conductor set
- Removes the first element from the list
- Serialize this ID into an internal map
- Removes all elements in the given collection
- Adds all elements in the specified collection to this set
- Tries to take a value from the queue
- Removes the element at the specified position
- Add line
- Removes all elements from this collection that are contained in the given collection
- Iterate over the elements in this map
- Return the Charge density reading from a given IChargeConductor
- Clears all the boxes for the next tick
- Returns the first argument in the given advancer
Factorization Key Features
Factorization Examples and Code Snippets
Trending Discussions on Factorization
I have a problem with non-negative matrix factorization in octave. I try to estimate synergies from Emg-data, but octave only lets me do this for two or more synergies, but not for one. I was able to reproduce the problem with the following code. nmf_bpas is from the linear-algebra pkg from octave-forge....
ANSWERAnswered 2021-May-20 at 17:34
This seems to be a bug in
From what I can tell, the bug is on line 373. Change
I am trying to execute
radial basis interpolation in
I have developed the following code:...
ANSWERAnswered 2021-Apr-19 at 03:17
Your training data contains only a single sampling point with multiple values. Replace with more sound data and your code works:
I'm trying to parallelize this piece of code that search for a max on a column. The problem is that the parallelize version runs slower than the serial
Probably the search of the pivot (max on a column) is slower due the syncrhonization on the maximum value and the index, right?...
ANSWERAnswered 2021-Mar-18 at 20:16
There is at least two things wrong with your parallelization
I have written a code which performs LU-factorization with OpenMP. I feel like the code is performing too slow. The computer I am working on have 16 cores and the time for 16 threads is 12.5s and for 4 threads 14.3s. This is the section of the code where the parallelization is happening. I am new to C programming and feel like I am missing something that is slowing the multithreading down....
ANSWERAnswered 2021-Mar-10 at 18:18
Based on the code that you have provided it seems to me that you do not need any lock whatsoever, try the following parallel version:
I am trying to define an operator overloads in C# for 2 classes:
On the outside, they act the same, but internally
PrimeFactorization (stores number as a list of prime factors) and
int for storing the nominator and denominator.
They would follow this interface (if I use JSDoc type definitions):...
ANSWERAnswered 2021-Feb-21 at 14:52
You want a discriminated union type. C# does not support them natively as a first-class language feature but you can hack-it using OneOf: https://github.com/mcintyre321/OneOf
So you would end-up with:
I have some issues in python console app. I coded an factorization console app. It works well, but when I input a number that is power of 2 like 4, 8, 16, the while loop doesn't stop. So it doesn't print anything and the code doesn't stop. Bellow is my code....
ANSWERAnswered 2021-Feb-18 at 11:06
The problem is that you don't change
m inside the
while loop. So, for example, for input
k comes out to be
n/m=2 which satisfies
k%m==0 and since neither
m changes so it runs forever.
You can simplify the code by modifying
n in the while
loop to keep decreasing if it is divisible by the current divisor
m. You can't do the same for
k is reset to
n again with the line
k = n and it will start with the original number giving incorrect output.
Here is a bit modified version of the code with the outer
I am new to cuda and cuBlas, and recently I am trying to use batched cuBlas API to solve multiple systems of linear equations. Here's my code:
The size of the matrix is N, and the number of matrices (batch size) is numOfMat....
ANSWERAnswered 2021-Feb-16 at 03:03
I think you need to look more closely at your data. If I run a modification of your code on Google Colab (Tesla T4) I get this:
Which looks largely like your figure. But look more closely (log scales help):
You can clearly see that up to a certain point, the runtime is largely independent of the number of matrices (around 2^8 = 64), but then scaling is linear as sizes increase. That is the transition from being able to parallelize the workload to reaching parallel capacity and having to schedule many parallel groups of operations to execute the workload. You might infer that for this particular GPU, the GPU run out of parallel capacity at between 64 and 128 concurrent operations (The T4 has 40 SM, so it might well be 80 if an SM could accommodate 2 operations per SM concurrently), after which runtime scales with multiples of that limiting size.
This is completely normal behaviour for any parallel computation architecture I am familiar with.
the xlearn predict function gives a different mse than what you get by looking at the predictions and calculating it yourself. Here is code to do this; you can run it by cloning the xlearn repository and copying the below code in
demo/regression/house_price in the repository
ANSWERAnswered 2021-Feb-09 at 23:58
A lot of people use 1/2 MSE for the loss because it makes the derivative "easier". Given that they use the word "loss" rather than "MSE" or something like that, I'd bet this is what's going on.
For clarity, if your loss is
1/2n * [(y_1 - p_1)^2 + ... + (y_n - p_n)^2]
then the derivative (wrt p) would be
-1/n * [(y_1 - p_1) + ... + (y_n - p_n)]
The 2 goes away because you end up multiplying by 2 for the power rule.
pardon the formatting... I don't know how to do math stuff here.
I have working Python code (below) that cracks an RSA key and times the prime factorization. How do I put this code into a module that takes a list of integers representing n for creating an n-bit prime number for range 0...2^n? I want to pass in a list of integers and return a table of the n-values and run times like this:...
ANSWERAnswered 2021-Jan-17 at 22:57
There are several issues here. To start with, your factor function regenerates a new pq value, ignoring the one you produced earlier. This counts the time to get pq twice. Also, measuring time on a single execution for a process that uses a random number generator doesn't give you a good idea of the performance. You need to make several runs to get an average.
I cleaned up your code a bit and changed the way time is measured to get an average over 100 runs:
Note: I am not that experienced in Python, therefore my code may not be as good as it could/should be.
I am attempting to create a tool to facilitate calculating the algebraic factors of a certain form of number (see https://en.wikipedia.org/wiki/Aurifeuillean_factorization). This is mostly as a test/learning experience, however I have run into a problem when attempting to calculate the parameter "c", which is defined as 2^(2k+1)+1. The addition step does not work for me. I am simply getting the returned value as 2^129, instead of 2^129+1 as I am looking to get. Is this an issue with Python itself, or am I making some sort of mistake in this.
ANSWERAnswered 2021-Jan-14 at 01:54
k = (exponent - 2) / 4 makes
float, which means you potentially introduce numerical error in computations down the line. Use integer division to stay in
int world from the start:
No vulnerabilities reported
You can use Factorization like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Factorization component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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