MATLAB Answers

0

Permutations Cycles Counting ... speed up

Asked by Michal Kvasnicka on 20 Oct 2017
Latest activity Edited by Jan
on 23 Oct 2017
Accepted Answer by Jan
Hi, this is my code for permutation cycles counting. Any idea how to vectorize it to get some additional speedup for large set of permutations?
function count = PermCyclesCount(p)
[nR,nE] = size(p);
count = zeros(nR,1);
for j = 1:nR
i = 0;
count(j) = 0;
while i < nE
i = i + 1;
if p(j,i) > 0
flag = false; % flag is just to trick the computer into entering the while loop
first = i;
while (first ~= i) || ~flag
flag = true;
r = first;
first = p(j,first);
p(j,r) = 0; % The entries in the cycle are changed to zero to indicate that they have been `used'
end
count(j) = count(j) + 1;
end
end
end
end
Some testing data:
[~,perms] = sort(rand(1000000,60),2);
tic;c= PermCyclesCount(perms);toc
Elapsed time is 1.559758 seconds.

  0 Comments

Sign in to comment.

4 Answers

Answer by Jan
on 20 Oct 2017
Edited by Jan
on 20 Oct 2017
 Accepted Answer

% VERSION 1
[nR, nE] = size(p);
count = zeros(nR, 1);
for j = 1:nR
q = p(j, :); % <== Cut out a vector to operate on
i = 0;
while i < nE
i = i + 1;
if q(i) > 0
flag = false;
first = i;
while (first ~= i) || ~flag
flag = true;
r = first;
first = q(first);
q(r) = 0;
end
count(j) = count(j) + 1;
end
end
end
This tiny change let the code run in 1.41 instead of 2.37 sec on my computer.
And this in 1.25 sec without the flag:
% VERSION 2
[nR, nE] = size(p);
count = zeros(nR, 1);
for j = 1:nR
q = p(j, :);
for i = 1:nE
if q(i) > 0
f = q(i);
q(i) = 0;
while f ~= i
r = f;
f = q(f);
q(r) = 0;
end
count(j) = count(j) + 1;
end
end
end

  3 Comments

But your last code is about 2 times faster than mine. Thanks!
tic;c_= PermCyclesCount_(perms);toc
Elapsed time is 0.797457 seconds.
Jan
on 20 Oct 2017
This code is 20% faster, if the data a provided in columnwise order:
[~,perms] = sort(rand(60, 1000000), 1);
% VERSION 2.1
[nE, nE] = size(p); % <== Swapped
count = zeros(nR, 1);
for j = 1:nR
q = p(:, j); % <== Copy column
for i = 1:nE
if q(i) > 0
f = q(i);
q(i) = 0;
while f ~= i
r = f;
f = q(f);
q(r) = 0;
end
count(j) = count(j) + 1;
end
end
end
The C code is even 33% faster - see in the other answer.
Does the creation of perms matter also? Then:
[~,perms] = sort(rand(60, 1000000),1); % 4.88 sec
perms = Shuffle(repmat((1:60).', 1, 1000000), 1); % 2.95 sec
One again Jan, tahnk you very much for your help. Creation is not a critical part, but anyway, the Shuffle function looks very useful for me, too.

Sign in to comment.


Answer by Jan
on 20 Oct 2017
Edited by Jan
on 23 Oct 2017

A C-Mex has about the double speed of the VERSION 2:
// VERSION 3
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
size_t nR, nE, j, i;
double *c, *p, *pr;
int *q, f, r;
// Get input:
nR = mxGetM(prhs[0]);
nE = mxGetN(prhs[0]);
p = mxGetPr(prhs[0]);
// Create output:
plhs[0] = mxCreateDoubleMatrix((mwSize) nR, 1, mxREAL);
c = mxGetPr(plhs[0]);
q = (int *) mxMalloc((nE + 1) * sizeof(int));
for (j = 0; j < nR; j++) {
// Copy row p(j, :) to q:
pr = p++;
for (i = 1; i <= nE; i++) {
q[i] = (int) (*pr); // One based indexing for q
pr += nR;
}
for (i = 1; i <= nE; i++) {
if (q[i] != 0) {
f = q[i];
q[i] = 0;
while (f != i) {
r = f;
f = q[f];
q[r] = 0;
}
*c += 1.0;
}
}
c++;
}
mxFree(q);
return;
}
This is 33% faster, if you provide the input in columnwise order instead:
[~,perms] = sort(rand(60, 1000000),1);
// VERSION 3.1
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
// !!! FOR [nE x nR] !!!
size_t nR, nE, j, i;
double *c, *p, *pr;
int *q, f, r, ic;
// Get input:
nE = mxGetM(prhs[0]);
nR = mxGetN(prhs[0]);
p = mxGetPr(prhs[0]);
// Create output:
plhs[0] = mxCreateDoubleMatrix((mwSize) nR, 1, mxREAL);
c = mxGetPr(plhs[0]);
q = (int *) mxMalloc((nE + 1) * sizeof(int));
for (j = 0; j < nR; j++) {
// Copy row p(j, :) to q:
pr = p;
for (i = 1; i <= nE; i++) {
q[i] = (int) (*pr++); // One based indexing for q
}
ic = 0;
for (i = 1; i <= nE; i++) {
if (q[i] != 0) {
f = q[i];
q[i] = 0;
while (f != i) {
r = f;
f = q[f];
q[r] = 0;
}
ic++;
}
}
*c++ = (double) ic;
pr += nE;
}
mxFree(q);
return;
}

  1 Comment

Jan, thanks a lot for your extraordinary help!!! The codes you provide here are perfect. Especially attached MEX files.

Sign in to comment.


Answer by Jos (10584)
on 20 Oct 2017

Not vectorised, but less checking, not sure if it is faster
nR = size(p,1) ;
count2 = zeros(nR,1) ;
tf = ~isnan(p) ;
for j = 1:nR
r1 = find(tf(j,:), 1 ,'first') ; % find the first cycle
while ~isempty(r1) % is there a cycle
count2(j) = count2(j) + 1 ;
while tf(j,r1)
tf(j,r1) = false ;
r2 = p(j,r1) ;
r1 = r2 ;
end
r1 = find(tf(j,:),1,'first') ; % find the next cycle
end
end

  9 Comments

So, what is the main purpose to use profiler now?
I have two codes, doing same thing. First code is in profiler mode slower than second code. But, in non-profiler mode is the first code significantly faster than second code.
Please, let me know what the Matlab user is supposed to do in this situation? Or, what is the reason to use profiler, when its results are completely distorted?
Jan
on 20 Oct 2017
The profiler disables the JIT acceleration, which could re-order the code lines to improve speed. This cannot be allowed, if the time per line should be measured.
But the JIT accelerator is essential for the run time of such loops. Therefore the profiler has a limited use here. You can determine the commands which take the most time, but not the code blocks.
@Jan Simon Yes, you are right. The profiler has a limited use, now!!! But frankly speaking, this limitation is really terrible.

Sign in to comment.


Answer by Jos (10584)
on 20 Oct 2017
Edited by Jos (10584)
on 20 Oct 2017

This is upto 10 times faster than your code (and indeed about 100 times faster than my previous code):
[nR, nE] = size(p) ;
count3 = zeros(nR,1) ;
tf = true(nR,nE) ;
for Ri = 1:nR
for Ei = 1:nE,
if tf(Ri,Ei)
count3(Ri) = count3(Ri) + 1 ;
while tf(Ri,Ei)
tf(Ri,Ei) = false ;
r = p(Ri,Ei) ;
Ei = r ;
end
end
end
end

  2 Comments

This code is definitely still a bit slower than mine :)
[~,perms] = sort(rand(1000000,60),2);
My code:
tic;c= PermCyclesCount(perms);toc
Elapsed time is 1.552508 seconds.
Your code:
tic;c_= PermCyclesCount_(perms);toc
Elapsed time is 1.664062 seconds.
Jan
on 20 Oct 2017
An improved version of Jos' code:
[nR, nE] = size(p) ;
count = zeros(nR,1) ;
tf = true(nE, nR) ; % Transposed to operate on column
for Ri = 1:nR
q = p(Ri, :); % Vector cut out
for Ei = 1:nE
k = Ei;
if tf(k, Ri)
count(Ri) = count(Ri) + 1 ;
while tf(k, Ri)
tf(k, Ri) = false ;
r = q(k) ;
k = r ;
end
end
end
end
To my surprise using a vector tf = true(nE, 1) is not faster.

Sign in to comment.