how to multiply matrix with vector n times?

hi guys i have this code:

M =[0 1 0 0 0 0 0 0;0 0 1 0 0 0 0 0;0 0 0 1 0 0 0 0;0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0;0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 1;1 1 0 0 0 0 0 0];

A = [ 1 0 0 0 0 0 0 0];

C =mod((A*M'),2) % multiplication by module 2

i want to repeat the C =mod((A*M'),2) many times until one of the results repeated. how to do this?

[Roger Stafford]

When you say "repeated" do you mean successive vectors, C, are the same, or do you mean that one of the C's along the line is the same as some earlier C so that a cycle is created? The first meaning is easier to accomplish and the second sounds more likely.

Answer 1

[Guillaume]

I assume you mean you want to repeat

A = mod(A*M', 2);

Otherwise, if A and M never changes C = mod(A*M', 2) will always give the same result!

Assuming you want to stop when any of the A that's been generated reappers again, build a matrix of A (concatenate rows) and use ismember(..., 'rows') to find if your new A is present in the matrix:

M =[0 1 0 0 0 0 0 0;0 0 1 0 0 0 0 0;0 0 0 1 0 0 0 0;0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0;0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 1;1 1 0 0 0 0 0 0];
A = [ 1 0 0 0 0 0 0 0];
AA = A;
A = mod(A*M', 2)
while ~any(ismember(A, AA, 'rows'))
  AA = [AA; A];
  A = mod(A*M', 2);
end

[janny]

thanks Guillaume very much, it works fine,

[janny]

you know, for some multiplication it take so much time...how to limit the calculation ? make it repeat only for 50 times ? how to add a code to limit the calculation

[Guillaume]

Just add that condition to the while loop.

%...
step = 0;
maxstep = 50;
while ~any(ismember(A, AA, 'rows')) && step < maxstep
   step = step + 1;
   %...
end

[janny]

is the code should be written like this:

M =[0 1 0 0 0 0 0 0;0 0 1 0 0 0 0 0;0 0 0 1 0 0 0 0;0 0 0 0 1 0 0 0; 0 0 0 0 0 1 0 0;0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 1;1 1 0 0 0 0 0 0];

A = [ 1 0 0 0 0 0 0 0];

AA = A;

A = mod(A*M', 2)

while ~any(ismember(A, AA, 'rows'))&& step < maxstep

AA = [AA; A];
A = mod((A*M'),2);
%...
step = 0;
maxstep = 50;
step = step + 1;
 %...

end

[Guillaume]

No. Try to understand what it does.

You have to replace the original

while ~any(ismember(A, AA, 'rows'))

by the four lines I've written, in the same order (well, you can swap the first two).

[janny]

thanks man,, it works fine...

AA = A; A = mod((A*M'), 2) step = 0; while ~any(ismember(A, AA, 'rows'))&& step < maxstep AA = [AA; A]; A = mod((A*M'),2);

maxstep = 10;
step = step + 1;

end