How to improve the speed of computing trace in my code?

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Xiaohan Du
Xiaohan Du on 20 Jun 2018
Commented: Xiaohan Du on 20 Jun 2018
Hi all,
In my code there is a key function which cost the majority of computational power due to the large number of repetitive computations. Imagine I have 2 cell arrays 'respi' and 'respj' which contains results of SVD vectors:
>> respi
respi =
1×10 cell array
Columns 1 through 9
{3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell}
Column 10
{3×1 cell}
>> respi{1}
ans =
3×1 cell array
{242×5 double}
{ 5×5 double}
{ 3×5 double}
I have a function which takes respi and respj, compute and store the trace as follows
function otpt = uTuPartDemo(respi, respj)
otpt = zeros(length(respi), length(respj));
for iTr = 1:length(respi)
u1 = respi{iTr}; % find the ith cell element.
for jTr = 1:length(respj)
u2 = respj{jTr}; % find the jth cell element
otpt(iTr, jTr) = ... % compute the trace.
trace((u2{3}' * u1{3}) * u1{2}' * (u1{1}' * u2{1}) * u2{2});
And I need to run this function repetitively for many times as follows:
for i = 1:n % very large number, say n = 100000
otpt = uTuPartDemo(respi, respj);
How can I improve the speed and efficiency of function uTuPartDemo? Many thanks!

Answers (1)

Pieter Hamming
Pieter Hamming on 20 Jun 2018
Presumably the trace calculation takes most time. Since you're only interested in the diagonals anyway, why not use the approach explained here. It should reduce your computational strain.
A minor other issue: you don't need to define u1 and u2. Instead of
u1 = respi{iTr};
Just use
It will save you a minor amount of memory.
  1 Comment
Xiaohan Du
Xiaohan Du on 20 Jun 2018
Hi Pieter,
Before trying this approach, I'd like to know if I need to recover the matrix before SVD (i.e. computing A = u1{1} * u1{2} * u1{3}', B = u2{1} * u2{2} * u2{3}'), then compute sum(A.*B',2)?

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