Please tell me how to resolve out of memory error of this code

count = 1; j = 1; matrix=[];

while(j+63<m) k = 1; while(k+63<n) dum=zeros(1,64*64); dum(1:(min(m-j+1,64)*(min(n-k+1,64))))=reshape(img1(j:min(j+63,m),k:min(n,k+63)),1,(min(m-j+1,64)*(min(n-k+1,64)))); matrix = [matrix;j,k,dum]; count = count+1; k = k+1; end j = j+1; end

Answer 1

[Titus Edelhofer]

Hi,

assuming that the out of memory happens in the assignment

matrix = [matrix; j,k,dum];

I would suggest to preallocate matrix. Compute the number of rows matrix will have - something like (m-63)*(n-63) - and preallocate

matrix = zeros((m-63)*(n-63), 2+64*64);

Instead of

matrix = [matrix; j,k,dum];

you write

matrix(count, :) = [j k dum];

If the preallocation works (depends on m, n of course), then the rest should work.

Titus

[shikha gautam]

I Have done this but in that case it is giving error "Maximum variable size allowed by the program is exceeded" in line matrix = zeros((m-63)*(n-63), 2+64*64);

[Titus Edelhofer]

Then I guess m and n are rather large? What are they?

[shikha gautam]

It's 500*500 image. so size of m=n=500

[shikha gautam]

This is the complete code :

img=rgb2gray(imread('16.jpg')); img1=double(img).*1/256; i=1; [m,n]=size(img); N = m*n;

%initialisation b = 64; Nbp = (sqrt(m) - sqrt(b) + 1)^(sqrt(n)-sqrt(b)+1); Nb = ceil(Nbp); epsilon = 0.01; Q = 256; Nn = 1; Nf = 64; Nd = 100; count = 1; j = 1; matrix = zeros((m-63)*(n-63), 2+64*64);

while(j+63<m) k = 1; while(k+63<n) dum=zeros(1,64*64); dum(1:(min(m-j+1,64)*(min(n-k+1,64))))=reshape(img1(j:min(j+63,m),k:min(n,k+63)),1,(min(m-j+1,64)*(min(n-k+1,64)))); matrix(count, :) = [j k dum]; count = count+1; k = k+1; end j = j+1; end

%matrix = a/Q S= sortrows(matrix,(3:64*64+2)); [r, c] = size(S); list = zeros(r,4); count = 1; for j=1:r for k= j+1:min(r,j+Nn) if(k-r<=Nn) temp = [S(j,1), S(j,2), S(k,1), S(k,2)]; list(count,:) = temp; count = count+1; end end end black = 0; for j=1:count-1 temp = list(j, :); if (64*floor(sqrt((temp(1)-temp(3))^2 + (temp(4)-temp(2))^2)) < Nd) %make the pixel values of these zero %presently seems wrong to me or maybe its right black = black+1; x1 = list(j, 1); y1 = list(j, 2); x2 = list(j, 3); y2 = list(j, 4); img(64*(x1-1)+1:min(64*x1,m),64*(y1-1)+1:min(n,64*y1))= 0; img(64*(x2-1)+1:min(64*x2,m),64*(y2-1)+1:min(n,64*y2))= 0; end end disp(black) imshow(uint8(img));

[Titus Edelhofer]

As in the new thread stated: your matrix would be (500-63)*(500-63)*(2+64*64)*8 bytes = 6GB large ...

Titus

[shikha gautam]

It requires a huge memory for this. Please help me in this code to run this .

[Titus Edelhofer]

There is not much I can offer: your result is too large to fit into memory. So you need to rethink if you need to store the result like this? What do you do with it later? You don't do the calculation for fun I guess, so the question is how to proceed without storing all the matrices...

[shikha gautam]

I am using this code to detect forgery in a image. for which I have to do so. we divide the image into non-overlapping blocks and then we lexicographically sort them. After sorting, we take lexicographically "close" blocks and compute the distance between their positions. If the distance is less than a minimum threshold, then we mark both the blocks as manipulated blocks