I am new to OpenVX, learning from the document that OpenVX uses a row-major storage. And the below matrix access example illustrate it, just like the ordinary row-major access pattern as we used in plain C code.
Then I go to the vx_matrix and vxCreateMatrix document page. The former has such statements:
VX_MATRIX_ROWS - The M dimension of the matrix [REQ-1131]. Read-only [REQ-1132]. Use a vx_size parameter.
VX_MATRIX_COLUMNS - The N dimension of the matrix [REQ-1133]. Read-only [REQ-1134]. Use a vx_size parameter.
While the latter said:
vx_matrix vxCreateMatrix(
vx_context c,
vx_enum data_type,
vx_size columns,
vx_size rows);
So according to my comprehension, in OpenVX world, when i said an MxN matrix, M refers to the row size and N refers to the column size. And the vxCreateMatrix declaration just follow what the row-major storage said, parameter column first and then row.
However, it really confuses me when i reach Warp Affine page, it said:
This kernel performs an affine transform with a 2x3 Matrix M with this method of pixel coordinate translation [REQ-0498]:
And the C declartion:
// x0 = a x + b y + c;
// y0 = d x + e y + f;
vx_float32 mat[3][2] = {
{a, d}, // 'x' coefficients
{b, e}, // 'y' coefficients
{c, f}, // 'offsets'
};
vx_matrix matrix = vxCreateMatrix(context, VX_TYPE_FLOAT32, 2, 3);
vxCopyMatrix(matrix, mat, VX_WRITE_ONLY, VX_MEMORY_TYPE_HOST);
If the M is a 2x3 matrix, according to the previous section, it should has 2 row and 3 column. Then why should it be declared as mat[3][2] and createMatrix accept column=2 and row=3 as argument? Is my comprehension totally wrong?
This would be a good start and help for your implementation
https://software.intel.com/content/www/us/en/develop/documentation/sample-color-copy/top/color-copy-pipeline/color-copy-pipeline-the-scan-pre-process-openvx-graph.html
I have a list L = [[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]] Ii. That represents my matrix. The size can change dynamic, so the blocksize can be different, 4x4 = 4 elements, 9x9= 9 elements
I want to obtain the 4 squares that compose the List.(In this case it's a matrix 4 by 4). If I have that matrix:
5 6 7 8
10 11 12 13
1 2 3 4
14 15 16 17
The result should be:
R = [5,6,10,11],[7,8,12,13],[1,2,14,15],[3,4,16,17].
Any suggestions are welcomed. Thanks
The first thing you need is really a lever for turning a list of lists into a matrix. What distinguishes a 2-dimensional matrix from a list of lists? The idea of a coordinate system. So you need a way to relate a coordinate pair with the corresponding value in the matrix.
at(Matrix, X, Y, V) :- nth0(X, Matrix, Row), nth0(Y, Row, V).
This predicate makes it possible to index the matrix at (X,Y) and get the value V. This turns out to be, IMO, a massive demonstration of what makes Prolog powerful, because once you have this one, simple predicate, you gain:
The ability to obtain the value at the point supplied:
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], 1,3, V).
V = 13.
The ability to iterate the entire matrix (only instantiate Matrix and leave the other arguments as variables):
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V).
X = Y, Y = 0,
V = 5 ;
X = 0,
Y = 1,
V = 6 ;
...
X = 3,
Y = 2,
V = 16 ;
X = Y, Y = 3,
V = 17.
The ability to search the matrix for values:
?- at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, 14).
X = 3,
Y = 0 ;
false.
So this is a pretty useful lever! In a conventional lanugage, you'd need three different functions to do all these things, but this is different, because in Prolog we just have to define the relationship between things (in this case, a data structure and a coordinate pair) and Prolog can do quite a bit of the heavy lifting.
It's easy to see how we could produce a particular submatrix now, by just defining the sets of X and Y values we'd like to see. For instance, to get the upper-left matrix we would do this:
?- between(0,1,X), between(0,1,Y),
at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V).
X = Y, Y = 0,
V = 5 ;
X = 0,
Y = 1,
V = 6 ;
X = 1,
Y = 0,
V = 10 ;
X = Y, Y = 1,
V = 11.
We can of course use findall/3 to gather up the solutions in one place:
?- findall(V, (between(0,1,X), between(0,1,Y),
at([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], X,Y, V)),
Vs).
Vs = [5, 6, 10, 11].
What's left for your problem is basically some arithmetic. Let's see if we have a square matrix:
square_matrix(M, Degree) :-
length(M, Degree),
maplist(length, M, InnerDegrees),
forall(member(I, InnerDegrees), I=Degree).
This is not a perfect predicate, in that it will not generate! But it will tell us whether a matrix is square and if so, what degree it has:
?- square_matrix([[5,6,7,8],[10,11,12,13],[1,2,3,4],[14,15,16,17]], D).
D = 4.
Once you have that, what you have to do is sort of formulaic:
Make sure the degree is a perfect square
Take the square root of the degree. That's how many rows or columns you have (square root 4 = 2, 2 rows and 2 columns, square root 9 = 3, 3 rows and 3 columns).
Make a relationship between the (row,column) coordinate and a list of (x,y) coordinates for the matrix in that location. For instance in the 4x4 matrix, you have four tiles: (0,0), (0,1), (1,0) and (1,1). The coordinates for (0,0) will be (0,0), (0,1), (1,0), (1,1), but the coordinates for (1,1) will be (2,2),(2,3),(3,2),(3,3). If you do a few of these by hand, you'll see it's going to amount to adding an x and y offset to all the permutations from 0 to row/column count (minus one) for both coordinates.
Now that you have that relationship, you need to do the iteration and assemble your output. I think maplist/N will suffice for this.
Hope this helps!
I am trying to create a neighborhood of pixel by using pixel matrix. The pixel matrix is the matrix of pixels in a 1 band image. Now I have to form matrix of 3*3 keeping each element of 9*9 matrix at center and have a neighbor for each element. Thus the element at (0,0) position will have neighboring elements as
[[0 0 0],
[0 2 3],
[0 3 4]]
Same case will happen to all elements in the first and last row and column. Attached image can help understanding better.
So the resultant matrix will have the size of 81*81. It is not necessary to save the small matrix in the form of matrix.
I have tried below,
n = size[0]
z= 3
x=y=0
m =0
while all( [x<0, y<0, x>=n, y>=n]):
continue
else:
for i in range(0, n):
arcpy.AddMessage("Hello" )
for x in range(m,m+3):
temp_matrix = [ [ 0 for i in range(3) ] for j in range(3) ]
for y in range(m,m+3):
temp_matrix[x][y] = arr_Pixels[x][y]
m+=1
y+=1
temp_List.append(temp_matrix)
But I am getting error: list assignment out of index. Also it looks too lengthy and confusing. I understood the error is occurring because, there is no increment in the array temp_matrix length.
Is there any better way to implement the matrix in image? Smaller matrices can be saved into list rather than matrix. Please help me.
Update #2
n = size[0]
new_matrix = []
for i in range(0,n):
for j in range(0,n):
temp_mat = [ [ 0 for k in range(3) ] for l in range(3) ]
for k in range(i-1, i+2):
for l in range(j-1,j+2):
if any([k<0, l<0, k>n-1, l>n-1]):
temp_mat[k][l] = 0
else:
temp_mat[k][l] = arr_Pixels[k][l]
new_matrix.append(temp_mat)
I think one issue is your use of while/else. The code in else only executes after the while condition is true and the while will not repeat again. This question might be helpful.
Thus, once it enters else, it will never check again that x<=n and y<=n, meaning that x and y can increase beyond n, which I assume is the length of arr_Pixels.
One better way to do it would be to create two nested for loops that increment from 0 to n and create the temp neighborhood matrices and add them to the 9x9 matrix. Here is an rough outline for that:
new_matrix = [] //future 9x9 matrix
for i in range(0, n):
for j in range(0, n):
// create a neighborhood matrix going around (i, j)
// add temp matrix to new_matrix
This method would avoid having to check that the indexes you are accessing are less than n because it assures that i and j will always be less than n-3.
I found better way of doing it by padding the whole matrix by zero. Thus it resolves the negative indexing problems.
matrix can be padded as
pixels = np.pad(arr_Pixels, (1,1), mode='constant', constant_values=(0, 0))
It adds rows and columns of zeros along the axes.
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.