Why is this simple loop taking so much time?

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Mohamad Al Hassan
评论： Walter Roberson 2020-4-3
Why is this simple loop taking me so much time?
n=2^13;
idx=cell(2*n-1,1);
for i=1:2*n-1
temp=zeros(i,2);
for j=1:i
if j>n
idx1=0;
else
idx1=j;
end
if i-j+1>n
idx2=0;
else
idx2=i-j+1;
end
temp(j,:)=[idx1,idx2];
end
idx{i}=temp;
end

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回答（1 个）

Walter Roberson
Walter Roberson 2020-4-3
When i is a particular value, say M, then you do M loop iterations total over the j. If we were to say that each loop iteration took time "1", then M of them takes time M. When i was one fewer, M-1, then you did M-1 operations for that, and one fewer still, M-2, you did M-2 operations for that. The total number of operations by the time of doing i = M is 1+2+3+4...+M, which is M*(M+1)/2 .
So by the end of 2^13 operations, you are doing 2^13 * (2^13+1) / 2 = 2^12 * (2^13+1) = 33558528 operations.
... and that is going to take a while.
On my system, n = 2^10 took roughly 1/4 second, for what would have been 2^10*(2^10+1)/2 = 524800 operations. If each loop was constant time we could predict 33558528 / 524800 * 0.25 = about 15 seconds.
Elapsed time is 14.064167 seconds.
Close enough.
  2 个评论
Mohamad Al Hassan
Mohamad Al Hassan 2020-4-3
Is there any way to make it faster? Without having to resort to a parfor loop.
Walter Roberson
Walter Roberson 2020-4-3
Vectorize with logical tests.
Or recognize that you can split the code into sections:
for i=1:n
for j = 1 : i
j is never greater than n in this case so you do not need to test it
end
end
for i=n+1:2*n-1
for j = 1 : n
j is never greater than n in this case so you do not need to test it
end
for j = n+1:i
j is always greater than n in this case so you do not need to test it
end
end
Or you could do
for i=1:2*n-1
temp=zeros(i,2);
for j=1:i
if i-j+1>n
idx2=0;
else
idx2=i-j+1;
end
temp(j,:)=[j,idx2]
end
temp(n+1:end,1) = 0;
idx{i}=temp;
end
That is, you let j be assigned for the first index always and then later in vectorized form you zap part of it to be 0.

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