decrease num of for loops

Hi

Can you help me to write that 4 for loops with just 1 for loop?

I really need to speed up my program and that is the biggest problem.

for u=0:7,
        for v=0:7,
            tmp = 0;
            for x=0:7,
                for y=0:7,
                  tmp = tmp + vhod(y+1, x+1) * cosd( (2*x + 1)*u*180/16 ) * cosd( (2*y + 1)*v*180/16 ); 
                end
            end
            izhod(v+1, u+1) = round(0.25 * C(u+1) * C(v+1) * tmp);
        end
    end

Answer 1

[Matt Fig]

This should be much faster:

D = (2*(0:7).'+1)*180/16;  % No need to do this over and over.
for u=0:7
    for v=0:7
        tmp = vhod.*(cosd(D*v) * cosd(D.'*u));
        izhod2(v+1, u+1) = C(u+1) * C(v+1) * sum(tmp(:));
    end
end
izhod2 = round(0.25 *izhod2);  %compare to izhod... equal!

[Matt Fig]

Uros comments ( again as an answer!!)

"thanks, that's great. it works much faster.

can you help me out with this similar function (idct actually):

        for x=0:7,
            for y=0:7,
                tmp = 0;
                for u=0:7,
                    for v=0:7,
                      tmp = tmp +  C(u+1) * C(v+1) * vhod(v+1, u+1) * cosd( (2*x + 1)*u*180/16 ) * cosd( (2*y + 1)*v*180/16 ); 
                    end
                end
                izhod(y+1, x+1) = ceil(0.25 * tmp);
            end
        end

that's the last thing i need to speed up in my program"

[Matt Fig]

It looks so similar that I think you can go through the code I gave you and just apply what you learn.

[Uros]

i'm trying but no success for now.

getting error at line 9 (inner matrix dimensions must agree)

      D1 = (0:7);
      D2 = (1:8);
      for u=0:7,
          for v=0:7
              tmp = vhod.*( cosd((2*u + 1)*D1*180/16) * cosd((2*v + 1)*D1*180/16) * C(D2) * C(D2.'));
              izhod(v+1, u+1) = sum(tmp(:));
          end
      end
      izhod = ceil(0.25 * izhod);

[Uros]

:( i can't do it, in this case there is also C related to u and v and i'm a bit confused.

btw this is the formula i need to speed up: http://shrani.si/f/3j/bC/gP6Ixdg/brez-naslova.png

with: C = [1 / sqrt(2), 1, 1, 1, 1, 1, 1, 1];

[Uros]

still no success

current code:

      for u=0:7,
          for v=0:7
              tmp = vhod.*(cosd((2*u+1)*(0:7)*180/16) * cosd((2*v+1)*(0:7)'*180/16) ).*(C(1:8) * C(1:8)');
              izhod(v+1, u+1) = sum(tmp(:));
          end
      end
      izhod = ceil(0.25 * izhod);

:(

Answer 2

[José-Luis]

old_v = 0;  
for ii = 1:8*8*8*8 
      [u v x y] = ind2sub([8 8 8 8],ii);
       u = u-1; v = v-1; x = x-1; y = y-1;
      if old_v ~= v
          tmp = 0;
      end
      %do your stuff
      old_v = v;

end

[Matt Fig]

Uros comments:

"hm i can't really understand how it works

now i have:

      tmp = 0;

      old_v = 0;  
      for ii = 1:8*8*8*8 
            [u v x y] = ind2sub([8 8 8 8],ii);
             u = u-1; v = v-1; x = x-1; y = y-1;
            if old_v ~= v
                tmp = 0;
            end
            tmp = tmp + vhod(y+1, x+1) * cosd( (2*x + 1)*u*180/16 ) * cosd( (2*y + 1)*v*180/16 ); 

            old_v = v;
      end

but i don't know where and how to put that line:

izhod(v+1, u+1) = round(0.25 * C(u+1) * C(v+1) * tmp);"