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

1 Ansicht (letzte 30 Tage)
high speed am 6 Apr. 2021
Kommentiert: Rik am 18 Jun. 2021
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]
##### 1 KommentarKeine anzeigenKeine ausblenden
Rik am 18 Jun. 2021
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.

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### Akzeptierte Antwort

Rik am 6 Apr. 2021
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
##### 2 Kommentare1 älteren Kommentar anzeigen1 älteren Kommentar ausblenden
Rik am 14 Apr. 2021
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.

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