I want to give the values for a matrix parameter mat_ZZ_p A for the mat_ZZ_p type in NTL. The dimension of my vector is big. So, I am looking at a big square matrix as parameter. So, I cannot assign the values manually. One advantage here to me is that the columns of my matrix are only rotations of the first column. It is of the form
p_0 p_(n-1) p_(n-2) .... p_1
p_1 p_0 p_(n-1) .... p_2
.
.
p_(n-1) p_(n-2) p_(n-3) .... p_0
and I have a variable p which is a vector with the values p_0, p_1, ...,p_(n-1). I have assigned the 1st column of the matrix using a loop through the vector p. but I am not sure how to do the rotation for the other columns. I tried to use that the values when viewed diagonally are the same but in that case, I am not sure how to bound the loop. I tried to use the fact that there is a diagonal downward shift of elements as we move from one column to another. But again in this case, I am not able to assign the value for the 1st row, 2nd column just by referring to the previous column. Is there a standard way to do such rotation of columns?
Since I am trying to solve the system of equations in Z_p, I think the comments in this post does not help me.
Best way to solve a linear equation in code
If you refer to m[i][j] for the generic element of the matrix n x n then what you need is
m[i][j] = m[(i + n - 1) % n][j-1] for every j > 0
For a square matrix with dimensions n * n, to refer to any element not in the first column or first row, use m[i - 1][j - 1], with i and j being the row and cols.
Related
I'm trying to flatten 3D matrices of arbitrary sizes into arrays such that values can be retrieved from arrays based on the spatial indexes i, j, and k. Clearly, each array index will have to be unique. I've tried setting the array index to int idx = i + width * (j + height * k), but that's not unique because (1,0,0) and (0,1,0) would give the same index if width == 1. Does anyone know of a better way to index or flatten 3D matrices?
Your formula is OK, because if width == 1, then j must be always 0
I know how to get mean of a matrix .
it is like this . sum of all values /n elements.
But I wonder that how to get mean of every element of a matrix in c++.
kindly help me in this regards .
thanks
So it seems you have a matrix M that is varying with time ( or iterations of some kind). The mean that you want is actually a matrix of same size as M. Will the following steps should do the job?
Initialize mean matrix m of the size of M to zeros
Add the current value of M to m. So m += M
Increment counter N for number of iterations of M
Divide each element of m by N. So m = m / N.
If the matrix is representing a image, take a gray(no color) image for example, it'a easy to understand.
The element of value 0 is indicate this pixel is black, and element of value 127 tell you this position is white. 0 is darkest while 127 is brightest. The larger value, the more bright. Each pixel(matrix element) have a value indicate it's brightness. All the pixel composite a whole pic. The mean of all the element is wanting to know the average brightness of the image.
There are n points with each having two attributes:
1. Position (from axis)
2. Attraction value (integer)
Attraction force between two points A & B is given by:
Attraction_force(A, B) = (distance between them) * Max(Attraction_val_A, Attraction_val_B);
Find the summation of all the forces between all possible pairs of points?
I tried by calculating and adding forces between all the pairs
for(int i=0; i<n-1; i++) {
for(int j=i+1; j<n; j++) {
force += abs(P[i].pos - P[j].pos) * max(P[i].attraction_val, P[j].attraction_val);
}
}
Example:
Points P1 P2 P3
Points distance: 2 3 4
Attraction Val: 4 5 6
Force = abs(2 - 3) * max(4, 5) + abs(2 - 4) * max(4, 6) + abs(3 - 4) * max(5, 6) = 23
But this takes O(n^2) time, I can't think of a way to reduce it further!!
Scheme of a solution:
Sort all points by their attraction value and process them one-by-one, starting with the one with lowest attraction.
For each point you have to quickly calculate sum of distances to all previously added points. That can be done using any online Range Sum Query problem solution, like segment tree or BIT. Key idea is that all points to the left are really not different and sum of their coordinates is enough to calculate sum of distances to them.
For each newly added point you just multiply that sum of distances (obtained on step 2) by point's attraction value and add that to the answer.
Intuitive observations that I made in order to invent this solution:
We have two "bad" functions here (somewhat "discrete"): max and modulo (in distance).
We can get rid of max by sorting our points and processing them in a specific order.
We can get rid of modulo if we process points to the left and to the right separately.
After all these transformations, we have to calculate something which, after some simple algebraic transformations, converts to an online RSQ problem.
An algorithm of:
O(N2)
is optimal, because you need the actual distance between all possible pairs.
I have 2 matrices of dimension 1*280000.
I wanted to multiply one matrix with transposed second matrix using opencv.
I tried to multiply them using Multiplication operator(*).
But it is giving me error :'The total size matrix does not fit to size_t type'
As after multiplication the size will be 280000*28000 of matrix.
So,I am thinking multiplication should 32 bit.
Is there any method to do the 32bit multiplication?
Why do you want to multiply them like that? But because this is an answer, I would like to help you thinking more than just do it:
supposing that you have the two matrix: A and B (A.size() == B.size() == [1x280000]).
and A * B.t() = AB (AB is the result)
then AB = [A[0][0]*B A[0][1]*B ... A[0][279999]*B] (each column is the transposed matrix multiplied by the corresponding element of the other matrix)
AB may also be written as:
[ B[0][0]*A
B[0][1]*A
...
B[0][279999]*A]
(each row of the result will be the row matrix multiplied by the corresponding element of the column (transposed) matrix)
Hope that this will help you in what you are doing... Using a for loop you can print, or store, or what you need with the result
I have performed block SVD decomposition over image and I stored results.
Now, I need to make reconstruction from this results. I found few examples all written in Matlab, which is a mystery for me.
I only need formula from which I can reconstruct my picture, or example written in C language.
Matrix A is equal U*S*V'. How will look formula, e.g. for calculating first five singular values (product of which rows and columns)? Please provide formula with indexes in C like style. U and V' are matrices and S is vector (not matrix).
Not sure if I get your question right, but if you just need to know singular values, they are the diagonal values of the middle matrix S. S in general is a diagonal matrix, which is stored here as a vector. I mean, only the diagonal is stored, you should imagine it as a matrix if you're thinking in matrix calculations.
Those diagonal values are your singular values, if you need the first biggest singular values, just take the 5 biggest values of the vector S.
Quoting from Wikipedia:
The diagonal entries Σi,i of Σ are known as the singular values of M.
The m columns of U and the n columns of V are called the left-singular
vectors and right-singular vectors of M, respectively.
In the above quote, sigma is your S, and M is the original matrix.
You have asked for C code, yet my hope is that pseudocode will suffice (it's late, I'm tired). The target matrix A has m rows, c columns and rank rho. The variable p = min(m,n).
One strategy is to first form the the intermediate matrix product B = US. This is trivial due to the diagonal-like nature of the matrix of singular values. Assume you have rho ( = 5 ) singular values. You must enforce rho <= p.
Replace column vector u1 with s1u1.
Replace column vector u2 with s2u2.
...
Replace column vector urho with srhourho.
Replace column vector urho+1 with a zero vector of length m.
Replace column vector urho+2 with a zero vector of length m.
...
Replace column vector up with a zero vector of length m.
Next form the new image matrix A = BVT. The matrix element in row r and column c is the dot product of the rth row vector (length rho) of B with the cth column vector (length rho) of VT.
Another strategy is to jump to the form where the matrix elements of A in row r and column c are
ar,c = sum ( skur,kvc,k, { k, 1, rho } )
The row counter r runs from 1 to m; the column counter c runs from 1 to n.