Vectorization of a double for loop

Hello all, I am trying to speed up the assembly of a matrix with a complicated structure. Right now, I have several sets of nested loops such as
for i=M+2:M+N-2
for j=i+1:M+N-1
BL1(j,i) = gamm/(gbp1*h(i)^(1-beta))*...
((j-i+3/2)^beta-2*(j-i+1/2)^beta+(j-i-1/2)^beta);
end
end
Here gamm, beta, are constants, h is a vector of length M+N and BL1 is a (dense) matrix of size (M+N-1) x (M+N-1). Is it possible to vectorize this? I've tried several things, but none seem to work.
Thanks for taking a look.

3 comentarios

dpb
dpb el 9 de Nov. de 2017
gbpi also a constant?
Geoffrey Dillon
Geoffrey Dillon el 9 de Nov. de 2017
Whoops, yes: gbpi is also a constant. Sorry!
Jan
Jan el 10 de Nov. de 2017
"gbpi"? It is gbp1.

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 Respuesta aceptada

Roger Stafford
Roger Stafford el 10 de Nov. de 2017
The following code should correct the error in my first answer. This version also avoids the repetition involved in your code for the expression
((j-i+3/2)^beta-2*(j-i+1/2)^beta+(j-i-1/2)^beta)
and hopefully it should therefore take less execution time.
BL1 = zeros(M+N-1);
T = (1:N-3)';
X = (T+3/2).^beta - 2*(T+1/2).^beta + (T-1/2).^beta;
for i = M+2:M+N-2
BL1(T(1:M+N-1-i)+i,i) = gamm/(gbp1*h(i)^(1-beta))*X(1:M+N-1-i);
end

1 comentario

Geoffrey Dillon
Geoffrey Dillon el 13 de Nov. de 2017
Wow, that did it! Thank you so much for all of your help!

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Más respuestas (1)

Roger Stafford
Roger Stafford el 9 de Nov. de 2017
Editada: Roger Stafford el 10 de Nov. de 2017
You can save unneeded repetition by computing one of the factors outside the loop.
BL1 = zeros(M+N-1);
T = (1:M+N-1)';
X = (T+3/2).^beta - 2*(T+1/2).^beta + (T-1/2).^beta;
for i=M+2:M+N-2
BL1(T+i,i) = gamm/(gbp1*h(i)^(1-beta)) * X;
end
This answer is incorrect. Read my correction in the revised answer below. RAS

5 comentarios

Jan
Jan el 10 de Nov. de 2017
EDITED: The quote was not ', but a similar character. But not ´ . Strange.
Roger Stafford
Roger Stafford el 10 de Nov. de 2017
@Jan: Thanks for the editing. I thought I typed in an ordinary, garden-variety single quote. Perhaps my computer is playing tricks on me.
Geoffrey Dillon
Geoffrey Dillon el 10 de Nov. de 2017
If M=2^5 and N=2^12, then BL1 should be 4127x4127. I placed a breakpoint right after the this new code and BL1 is now 8253x4127, which will eventually cause an error.
In addition, this new block of code takes significantly longer (10x) to execute than the double for loop!
dpb
dpb el 10 de Nov. de 2017
As written for the above M,N --> BL1 will be M+N-1 x M+N-2 --> 4127x4126 as those are the upper limits for the i,j loops respectively and the indices for BL1 are those indices identically. Every row below M+2 (=4098) will be identically zero and all columns diagonally to the left will also be zero beyond that row.
Is that the intent? But if it is to be square and 4127x4127, then the i upper index is wrong or the column subscript expression isn't correct.
Did you preallocate BL1 before timing the loop solution? Awaiting the answer to the above observation I've not looked at whether there's an efficient vectorization to be done or not; often the looping solution can be pretty effective if do the obvious.
@Geoffrey: My apologies. I mistakenly read the line
for j=i+1:M+N-1
as though it had the parentheses:
for j=i+(1:M+N-1)
I have written a second “answer” which should correct this error.

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