modify a program to obtain the same result from right to left

I have this matrix:

H=[0 1 0 0 0 0 0 0 0 0 0 0;

0 0 0 1 0 0 0 0 0 0 0 0;

0 0 0 0 0 1 0 0 0 0 0 0;

0 0 0 0 0 0 0 1 0 0 0 0;

0 0 0 0 0 0 0 0 0 1 0 0;

0 0 0 0 0 0 0 0 0 0 0 1]

I want use but from right to left. It means I want obtain this matrix

H=[0 1 0 0 0 0 0 0 0 0 1 0;

0 0 0 1 0 0 0 0 1 0 0 0;

0 0 0 0 0 1 1 0 0 0 0 0;

0 0 0 0 1 0 0 1 0 0 0 0;

0 0 1 0 0 0 0 0 0 1 0 0;

1 0 0 0 0 0 0 0 0 0 0 1]

[Rik]

I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is considered rude.

modify a program to obtain the same result from right to left

I have this MATLAB program

dr = 6; dc = 3; N = 12; M=6; r=gcd(M,N); c=N/r; b=M/r;

H=zeros(M,N);

%Full rank (left)

vCol = c:c:N;

for iCol = 1:numel(vCol )

iRow = (iCol - 1) * b + 1;

H(iRow:iRow + b - 1, vCol(iCol)) = 1;

end

Using this program I obtain the result of this matrix:

H=[0 1 0 0 0 0 0 0 0 0 0 0;

0 0 0 1 0 0 0 0 0 0 0 0;

0 0 0 0 0 1 0 0 0 0 0 0;

0 0 0 0 0 0 0 1 0 0 0 0;

0 0 0 0 0 0 0 0 0 1 0 0;

0 0 0 0 0 0 0 0 0 0 0 1]

I want use the same part of Full rank in this program but from right to left. It means I want obtain this matrix

H=[0 1 0 0 0 0 0 0 0 0 1 0;

0 0 0 1 0 0 0 0 1 0 0 0;

0 0 0 0 0 1 1 0 0 0 0 0;

0 0 0 0 1 0 0 1 0 0 0 0;

0 0 1 0 0 0 0 0 0 1 0 0;

1 0 0 0 0 0 0 0 0 0 0 1]

Notice: It's not an identity matrix because it depends on the rate b and c.

Answer 1

[Rik]

Your loop can probably be replaced by a vectorized operation, but I will leave that for you to figure out. I suspect fliplr and or() are all you need:

dr = 6; dc = 3; N = 12; M=6; r=gcd(M,N); c=N/r; b=M/r;
H=zeros(M,N);
vCol = c:c:N;
for iCol = 1:numel(vCol )
    iRow = (iCol - 1) * b + 1;
    H(iRow:iRow + b - 1, vCol(iCol)) = 1;
end

H = double( H | fliplr(H) );
disp(H)

     0     1     0     0     0     0     0     0     0     0     1     0
     0     0     0     1     0     0     0     0     1     0     0     0
     0     0     0     0     0     1     1     0     0     0     0     0
     0     0     0     0     1     0     0     1     0     0     0     0
     0     0     1     0     0     0     0     0     0     1     0     0
     1     0     0     0     0     0     0     0     0     0     0     1

[high speed]

What do you mean by vectorized operation please

[Rik]

By vectorized opeation I mean something like this:

M=6;N=12;
H=zeros(M,N);
H( (M+1):(2*M+1):end )=1;
disp(H)

     0     1     0     0     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     0     0     1     0     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0
     0     0     0     0     0     0     0     0     0     0     0     1

This way you generate the entire array at once. Careful construction of the vector of indices might avoid a loop. With the exceptions of functions that hide a loop (e.g. cellfun and arrayfun), code without a loop will almost always be faster.