From: Walter Roberson on
fckool wrote:

> c=mod(m*G,2)
> ´
> but it give me an error:
> ??? Error using ==> mtimes
> Inner matrix dimensions must agree.
>
> G it's an array calculated before this.


* is the Matrix Multiplication operator. As per usual, m*G is only possible if
the second dimension of m is the same as the first dimension of G. If you want
to multiply each corresponding element of m by each corresponding element of
G, then use m .* G .

By the way, watch out for m1 versas m .
From: dpb on
fckool wrote:
....

> c=mod(m*G,2)
> ´
> but it give me an error:
> ??? Error using ==> mtimes
> Inner matrix dimensions must agree.
>
> G it's an array calculated before this.

Well, "*" is matrix multiplication in Matlab so one must have conformant
array sizes for the operation as defined by matrix operations.

If G is same dimensions as m and you mean element-wise mulitplication,
that's ".*" in Matlab

doc times
doc mtimes

--
From: fckool on
with the code that I showed you, it works... but I cant use it.. I have to use another one..

Your code maybe not work because when you add a space (string value?) you are increasing the array dim, not?

who can I separate the bits, for a better read, without increase its size?
From: Walter Roberson on
fckool wrote:

> Your code maybe not work because when you add a space (string value?) you are increasing the array dim, not?

It does increase the dimensions, but that isn't necessarily a problem.

> who can I separate the bits, for a better read, without increase its size?

m = repmat(' ', 2^k, k);
for i = 1 : 2^k
m(i,:) = dec2bin(i-1,k);
end

mdisplay = repmat(' ', 2^k, 2*k-1);
mdisplay(:,1:2:2*k-1) = m;


Or, if you prefer an expression over efficiency,

mdisplay = cell2mat(arrayfun(@(c) [c ' '], m, 'Uniform',0))
From: fckool on
perfect, know it works.

I have to explain every line of my program can you to explain this 2 lines please?:

mdisplay = repmat(' ', 2^k, 2*k-1);
mdisplay(:,1:2:2*k-1) = m;

The efficient expression it's very good but I never gonna get it explains by myself lol..


I have another question:

how can I test the efficiency of my code? (how much time it need to run)


thank you very much, and i apology if I'm being boring.