exchange cells with concerning time

Hi I have a code. In this code p and D are matrices with 100 rows 100 columns and 4 in third dimension. D is (0-1) matrix. in my code cells one by one are selected and changed. if the value of function (here is not important what it is) is decreased, change is accepted. here I want to set the value of pp equal 0 for previous i's and j's and also previous j1's and j2's. because I didn't want to select cellS that be selected in previous selections. my code is very time consuming and I want to decrease it. My code:

    for i=1:O1
    for j=1:O2
        s=find(D(i,j,:)==1);
        pp=p(:,:,s).*(1-D(:,:,s));
        w=sum(sum(pp));
        p1=bsxfun(@rdivide,pp,w);
        r=rand;
        jj=find(r<=cumsum(p1(:)),1,'first');
   [j1 j2]=ind2sub(size(p1),jj);
   g=find(D(j1,j2,:)==1);
   x1=D;
   D(j1,j2,g)=0;
   D(i,j,g)=1;
   D(i,j,s)=0;
   D(j1,j2,s)=1;
   x2=D;
   cost1=costfunction(full(x1));
   cost2=costfunction(full(x2));
   if cost1<cost2
       D=ndSparse(x1);
   elseif cost1>cost2
       D=ndSparse(x2);
   end
    end
   D;
  end

[Guillaume]

Can you explain the meaning of the four values in the third dimension? It looks like it's used to keep track of the values that have been changed but I'm not sure.

It would certainly help in telling you how to avoid processing the same thing twice.

It would be a good idea to use meaningful names for your variable. It immediately makes it clearer what the code does.

Answer 1

[Jan]

I assume you are using FEX: ndSparse.

100x100 ist not large. Converting this array from sparse to full repeatedly is expensive. Try to stay at full matrices.

Omit the "==1" ind find(D(i,j,:)==1), if D contains 0 or 1 only.

But most of all: Use the profiler to identify the bottleneck. If 95% of the time is spent in costfunction, improving the shown code cannot increase the speed by more then 5%.

[fatema saba]

But I think the problem is related to searching all elements of matrix. some of them repeat more than one. I want to omit elements that are analyzed, but I don't know how can I do it?

Answer 2

[Guillaume]

As Jan says, a 100x100x4 matrix is not large at all, it's only 320 kB as double. In your case, you don't even need to have it as double. Since it only contains 0 or 1, you could use a logical matrix, which would only use 40 kB:

D = logical(D);

Another thing that is not needed is most of the find. Again you can use logical indexing instead. The line

s=find(D(i,j,:)==1);

can be replaced by

s = D(i,j,:) == 1; %s is a logical array

and will give the exact same result. The latter is faster. Even faster, since D is only 0 or 1 is:

s = logical(D(i,j,:));

and if D were a logical matrix, then:

s = D(i,j,:);

pp can also be calculated slightly faster, using logical operations instead of multiplication:

pp = p(:,:,s) & ~D(:,:,s); %regardless whether D is logical or not

Similarly, for g you do not need the find

g = logical(D(j1, j2, :)); %no need for logical if D is already logical

None of the above is going to vastly speed your code. You need to clarify what step you're hoping to shortcut.