Thread Subject: Vecorizing for speed(?)

Hi all,

I have a function in which I calculate a tensor using nested
loops. I'm trying to vectorize some of the computations for
speed, but I'm finding some odd behaviour which I hope
somebody can explain.

Inside 4 'for' loops cycling through i,j,alpha,beta, when I
replace:

tempsum=0;
for l=1:n
   tempsum = tempsum + ...
      summation_kernel(i,l,alpha)*...
      summation_kernel(j,l,beta);
end output_c(i,alpha,j,beta) = tempsum/n;

with:

output_c(i,alpha,j,beta) = ...
   summation_kernel(i,:,alpha)*...
   summation_kernel(j,:,beta)'/n;

and check using tic toc around these expressions then the
vectorized version is much faster, as expected. But the
strange thing is that the function as a whole runs SLOWER
with this change. Anybody got any ideas why this should happen?

Many thanks, S

"Simon Preston" <preston.simon+mathsworks@gmail.com> wrote
in message <flmall$suc$1@fred.mathworks.com>...
> Hi all,
>
> I have a function in which I calculate a tensor using nested
> loops. I'm trying to vectorize some of the computations for
> speed, but I'm finding some odd behaviour which I hope
> somebody can explain.
>
> Inside 4 'for' loops cycling through i,j,alpha,beta, when I
> replace:
>
> tempsum=0;
> for l=1:n
> tempsum = tempsum + ...
> summation_kernel(i,l,alpha)*...
> summation_kernel(j,l,beta);
> end output_c(i,alpha,j,beta) = tempsum/n;
>
> with:
>
> output_c(i,alpha,j,beta) = ...
> summation_kernel(i,:,alpha)*...
> summation_kernel(j,:,beta)'/n;
>
> and check using tic toc around these expressions then the
> vectorized version is much faster, as expected. But the
> strange thing is that the function as a whole runs SLOWER
> with this change. Anybody got any ideas why this should
happen?
>
> Many thanks, S
>

Does anybody have any thoughts on this?

Your explaination is not clear to understand your problem. Before giving any explaination to your problem, I think it is necessary to understand your problem. You please explain your problem in a little elaborate way with full codes.

Simon Preston:

How many elements do you have in the second dimension?
Try with n greater than 10 elements.
You can feel the difference.

Anh Huy Phan
RIKEN - BSI

Yumnam Kirani Singh <kirani.singh@gmail.com> wrote in
message
<7625549.1200313848871.JavaMail.jakarta@nitrogen.mathforum.org>...
> Your explaination is not clear to understand your problem.
Before giving any explaination to your problem, I think it
is necessary to understand your problem. You please explain
your problem in a little elaborate way with full codes.

Maybe including the code in the original post was what made
it unclear - I'll try again.

Suppose OLD CODE and NEW CODE calculate the same quantity
but in a different way. If I do:

function
...
tic;
OLD CODE;
toc;
...
end

function
...
tic;
NEW CODE;
toc;
...
end

then NEW CODE is better. But if I do:

tic;
function
...
OLD CODE;
...
end
toc;

tic;
function
...
NEW CODE;
...
end
toc;

then OLD CODE is better. Why is this?

NEW CODE is a vectorized version of OLD CODE, using vector
products in place of loops.

Thanks, S

"Simon Preston" <preston.simon+mathsworks@gmail.com> wrote
in message <fmfnce$760$1@fred.mathworks.com>...
> Yumnam Kirani Singh <kirani.singh@gmail.com> wrote in
> message
>
<7625549.1200313848871.JavaMail.jakarta@nitrogen.mathforum.org>...
> > Your explaination is not clear to understand your problem.
> Before giving any explaination to your problem, I think it
> is necessary to understand your problem. You please explain
> your problem in a little elaborate way with full codes.
>
> Maybe including the code in the original post was what made
> it unclear - I'll try again.
>
> Suppose OLD CODE and NEW CODE calculate the same quantity
> but in a different way. If I do:
>
> function
> ...
> tic;
> OLD CODE;
> toc;
> ...
> end
>
> function
> ...
> tic;
> NEW CODE;
> toc;
> ...
> end
>
> then NEW CODE is better. But if I do:
>
> tic;
> function
> ...
> OLD CODE;
> ...
> end
> toc;
>
> tic;
> function
> ...
> NEW CODE;
> ...
> end
> toc;
>
> then OLD CODE is better. Why is this?
>
> NEW CODE is a vectorized version of OLD CODE, using vector
> products in place of loops.
>
> Thanks, S
>
>
>

and everything represented by the ellipses (...) is exactly
the same between the old and new code?

Jos

>
> and everything represented by the ellipses (...) is exactly
> the same between the old and new code?
>
> Jos
>

Yep, that's right.

Hi Simon,

Are you sure you are getting the exact same results from
both your code.

try this:

a = rand(1,200);
a*a' == sum(a.*a)

Usually they are not equal at the least significant bit. If
the rest of your code has precision sensitive decision
nodes, it may be executing in a whole different path with
your vectorized code.

Volkan

"Volkan " <volkan@buyukgungor.gmail.com> wrote in message
<fmg5ri$4vd$1@fred.mathworks.com>...
> Hi Simon,
>
> Are you sure you are getting the exact same results from
> both your code.
>
> try this:
>
> a = rand(1,200);
> a*a' == sum(a.*a)
>
> Usually they are not equal at the least significant bit. If
> the rest of your code has precision sensitive decision
> nodes, it may be executing in a whole different path with
> your vectorized code.
>
> Volkan

Hi Volkan,

Good thinking, but no, none of the code in the ellipsis runs
conditionally on the value calculated in OLD CODE/NEW CODE

Thanks for the suggestion. But it's still a mystery

In article <fmg9ji$ck8$1@fred.mathworks.com>,
preston.simon+mathsworks@gmail.com says...
> "Volkan " <volkan@buyukgungor.gmail.com> wrote in message
> <fmg5ri$4vd$1@fred.mathworks.com>...
> > Hi Simon,
> >
> > Are you sure you are getting the exact same results from
> > both your code.
> >
> > try this:
> >
> > a = rand(1,200);
> > a*a' == sum(a.*a)
> >
> > Usually they are not equal at the least significant bit. If
> > the rest of your code has precision sensitive decision
> > nodes, it may be executing in a whole different path with
> > your vectorized code.
> >
> > Volkan
>
> Hi Volkan,
>
> Good thinking, but no, none of the code in the ellipsis runs
> conditionally on the value calculated in OLD CODE/NEW CODE
>
> Thanks for the suggestion. But it's still a mystery
>
>

are you reusing variables in the second code so that the memory
allocation step is avoided. assign to new variables each time for a
real comparison perhaps. or preallocate for both before the timing.

--
Loren
http://blogs.mathworks.com/loren/