# Nested for-loops too slow

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Niels 2011 年 4 月 18 日
Hello,
I'm having trouble to compute a summation with four indices. My code below is very time consuming. The problem is have to compute for every i,j,k,l the three smallest elements of A. Any other suggestions that run a bit faster than my code? (N is approximately 156)
sum = 0;
for i=1:N
for j=1:N
for k=1:N
for l=1:N
A = [N-(i-1),N-(j-1),N-(k-1),N-(l-1)];
B = sort(A);
sum = sum + B(1)*B(2)*B(3);
end
end
end
end
Thank you very much!

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### 採用された回答

Egon Geerardyn 2011 年 4 月 18 日
You might want to work more on that code manually before implementing it like that. As the only thing that really matters is N, I think it should be possible to derive a more closed-form formula for the total sum and at least it should be possible to reduce the number of iterations by factoring out some constants.
E.g. I note [i,j,k,l] as your complete index: going from [1,1,1,1] to [1,1,1,X], you know what the results will be for any X (larger than 1): you already know the smallest values so you are wasting time iterating of all other values, sorting, multiplying and summing while you could just see that in that case you can add (N-1) * 1 * 1 * 1
Also remember that your case is quite symmetric towards your indexes: [i,j,k,l] will produce the same output as e.g. [j,i,k,l] and [i,j,l,k] (and so forth). When you think about it, you will see that there are 24 possible permutations for the most general case (where [i,j,k,l] are all different). So by taking that into account, I assume it should at least be possible to limit the iteration range you took by calculating the output once and adding 24*sum for the general case. You will get different situations for [i,i,k,l] (12 cases for every i); [i,i,i,l] (in 4 cases for every i) and [i,i,i,i] (1 case for every i). You will need to account for that in your code.
When doing something like this, I would suggest you first leave out some dimensions, start to work with a 2 dimensional loop, first of all that will allow you to use this crude code and secondly that also allows you to study a simpler case first, find a faster solution and generalize that solution to your exact case.
For the 2D case, [i,j] represents a matrix and you will see that you can see that [i,j] and [j,i] gives the same output. However, a special case exists [i,i] only occurs once.
So for 2D, a naive implemtation might be something like:
sum = 0;
N=156;
for i=1:N
for j=1:N
A = N - [i,j] + 1;
sum = sum + min(A);
end
end
sum
An implementation with fewer iterations (approximately half as many, if I'm right):
sum = 0;
N=156;
for i=1:N
A = N - [i,i] + 1;
sum = sum + min(A);
for j=(i+1):N
A = N - [i,j] + 1;
sum = sum + 2*min(A);
end
end
sum
You can of course combine this with Robert Cumming's answer, precalculate the parts you already know.
##### 3 件のコメント表示非表示 2 件の古いコメント
Niels 2011 年 4 月 18 日
@Egon: Thanks to your answer I finally found a much quicker method. I'd really expected to get answers that could help me with my real problem. :)
Thank you!

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### その他の回答 (2 件)

Teja Muppirala 2011 年 4 月 18 日
You really shouldn't use "sum" as a variable name. That is the name of a very useful built-in function. I'll call your variable "S" instead.
Looking at the code, you can deduce that S should be a polynomial function of N. Doing a little bit of curve fitting you can indeed get an analytic expression: S(N) =
(1/420)*(30*N.^7 + 105*N.^6 + 147*N.^5 + 105*N.^4 + 35*N.^3 - 2*N)
So If were to type into MATLAB:
S = @(N)(1/420)*(30*N.^7 + 105*N.^6 + 147*N.^5 + 105*N.^4 + 35*N.^3 - 2*N);
S(uint64(156))
S = 164235165014258
##### 3 件のコメント表示非表示 2 件の古いコメント
Teja Muppirala 2011 年 4 月 18 日
That should be:
A 7th order polynomial fit to "8 points" not "7 points"

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Robert Cumming 2011 年 4 月 18 日
Pulling out the i, j and k elements only when they changewill speed it up:
sum = 0;
A = zeros(4,1);
for i=1:N
A(1) = N-(i-1);
for j=1:N
A(2) = N-(j-1);
for k=1:N
A(3) = N-(k-1);
for l=1:N
A(4) = N-(l-1);
B = sort(A);
sum = sum + B(1)*B(2)*B(3);
end
end
end
end
##### 1 件のコメント表示非表示 なし
Niels 2011 年 4 月 18 日
Thank you for pointing this out. Significant change in speed!

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