Hi! Can I write the following code
for j = 2 : N-1
alpha(j+1,:) = A(j,:).*alpha(j,:) + B(j,:);
end
in a form like this:
J = 2:N-1;
alpha(J+1,:) = A(J,:).*alpha(J,:) + B(J,:);
I tried to use this form but the alphas are incorrectly calculated.

6 commentaires

Stephen23
Stephen23 le 1 Avr 2018
Modifié(e) : Stephen23 le 1 Avr 2018
"Recursive computation without loop"
Recursive calculations do not require loops: loops are iterative.
Why do you want to get rid of the loop: what is the problem with it?
avenior
avenior le 1 Avr 2018
To speed up these calculations. In vector form they are much faster executed.
Stephen23
Stephen23 le 1 Avr 2018
Modifié(e) : Stephen23 le 1 Avr 2018
"To speed up these calculations. In vector form they are much faster executed."
How do you know this? It looks to me like premature optimization, which is a classic programming anti-pattern:
avenior
avenior le 1 Avr 2018
Modifié(e) : avenior le 1 Avr 2018
"How do you know this?"
Experimentally, I replaced some loops in my code with vector calculations.
Walter Roberson
Walter Roberson le 1 Avr 2018
You got faster code that calculated the wrong thing.
With the loop the value of B(1,:) affects alpha(2,:), and that has an effect that changes all later output. With the vectorized version you do not get the feedback of earlier B values affecting all later values.
avenior
avenior le 1 Avr 2018
"You got faster code that calculated the wrong thing."
Yes I know. In my code there are other loops that are calculated without using the previous value. I wrote about them.

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Walter Roberson
Walter Roberson le 1 Avr 2018

1 vote

No, values are not stored into the destination until the entire right hand side finishes. Using a vector index on the output does not do an implicit iterative calculation.
If the question is about whether the calculation can be vectorized, the answer is that it can be vectorized for any given length. However the vectorized version is a bit nasty to write out and would be notably slower than the loop.

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