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2012-08-28T10:09:22.660
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OK, the code below will improve the timing significantly. It's not perfect yet, it can all be optimized a lot further. But, before I do so: I think what you want is fundamentally impossible. So you want <ul> <li>all rows contain the numbers 1 through 12, in a random permutation</li> <li>any value between 1 and 4 must be present either twice or not at all in any column</li> </ul> I have a hunch this is impossible (that's why your code never completes), but let me think about this a bit more. Anyway, my 5-minute-and-obvious-improvements-only-version: <pre><code>clc clear all jum_kel = 8; jum_bag = 12; uk_pop = 10; A = jum_kel; % renamed to make language independent B = jum_bag; % and a lot shorter for readability C = uk_pop; krom = zeros(A, B, C); for ii = 1:C; for a = 1:A krom(a,:,ii) = randperm(B); end end gab1 = sum(krom == 1); gab2 = sum(krom == 2); gab3 = sum(krom == 3); gab4 = sum(krom == 4); gabh1 = sum( gab1 ~= 2 & gab1 ~= 0 ); gabh2 = sum( gab2 ~= 2 & gab2 ~= 0 ); gabh3 = sum( gab3 ~= 2 & gab3 ~= 0 ); gabh4 = sum( gab4 ~= 2 & gab4 ~= 0 ); tot = gabh1+gabh2+gabh3+gabh4; for ii = 1:C ii while tot(:,:,ii) ~= 0 for a = 1:A krom(a,:,ii) = randperm(B); end gabb1 = sum(krom(:,:,ii) == 1); gabb2 = sum(krom(:,:,ii) == 2); gabb3 = sum(krom(:,:,ii) == 3); gabb4 = sum(krom(:,:,ii) == 4); gabbh1 = sum(gabb1 ~= 2 & gabb1 ~= 0) gabbh2 = sum(gabb2 ~= 2 & gabb2 ~= 0); gabbh3 = sum(gabb3 ~= 2 & gabb3 ~= 0); gabbh4 = sum(gabb4 ~= 2 & gabb4 ~= 0); tot(:,:,ii) = gabbh1+gabbh2+gabbh3+gabbh4; end end </code></pre>
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1. POHow to improve execution time of the following Matlab code
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1. USRody Oldenhuis
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1. COI think so, it's imposible to do. but when I tried to be used only for value "1". it was work fast. hahaha so i think, it is still have chance to solve right.
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 USFebri Dwi Laksono
2. CObut I have to do make this matrix. Maybe you know another suggestion code mr. and of course it is faster than current code. because i just know code like this to make matrix like what i want.
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