# Create binary matrix with some conditions (Matlab)

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high speed am 8 Apr. 2021
Kommentiert: high speed am 9 Apr. 2021
I have this matrix of dimensions (6x12)
H=[1 1 1 1 1 1 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]
The first row must still as it is (1 1 1 1 1 1 0 0 0 0 0 0)
1) As a first condition, I want change the second row in order to get it (0 0 0 1 0 0 0 1 1 1 1 1) it means we change the zeros from 8 to 12 column of the second row to ones.
2) As a second condition, from the third to sixth row I must add ones in order to get a matrix with 6 ones in each row and 3 ones in each column without changing the place of ones of this initially matrix.
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### Akzeptierte Antwort

DGM am 9 Apr. 2021
Well, I'm not exactly the guy to ask for elegant solutions to anything, but I'll take a stab at it.
clc
H=[1 1 1 1 1 1 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];
H(2,8:end)=1;
% H(1:2,:) can't change
% remaining rows to be arranged
% such that all(sum(H,1)==3) && all(sum(H,2)==6)
% i'm going to assume no further rules apply
% blindly just fill the bulk of the area
for r=3:size(H,1)
hasroom=((3-sum(H,1))>0) & ~H(r,:);
H(r,find(hasroom,5))=1;
end
% clean up any underruns
% for other initial conditions or more general use, i can't guarantee that
% this won't need to become an iterative process
leancol = find(sum(H,1)<3,1);
leanrow = find(sum(H,2)<6,1);
openrows = [0; 0; ~H(3:end,leancol)]; % these are places in leancol where a 1 can be placed
opencols = ~H(leanrow,:); % these are places in leanrow where a 1 can be placed
swappable = bsxfun(@and,openrows,opencols) & H; % these are ones that can be moved
[r c] = find(swappable,1);
H([r leanrow],[leancol c])=1;
H(r,c)=0;
% dump the result and show the criteria are met
H
all(sum(H,1)==3) && all(sum(H,2)==6)
Make of that what you will. I have a feeling that this problem is intended to be solved generally with an iterative process, so my simplistic single-pass code may not work for other scenarios. I'm not sure of anything except that someone else can find a better way.
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high speed am 9 Apr. 2021
It works. Thank you so much

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