How to create a matrix with special block diagonal structure

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I would like to create a matrix that has the following structure ( The blue dots are ones and the rest is zeros)
the number of the dots is determined by the triangular sequence. For example, the above matrix was obtained with n = 8. So is there any way to program this for any arbitrary n?

Accepted Answer

DGM
DGM on 25 May 2022
Here's one way. I'm sure there are plenty of others.
n = 8;
sz = sum(1:n);
A = zeros(sz);
idx0 = 1;
for k = 1:n
A(idx0:idx0+k-1,idx0:idx0+k-1) = 1;
idx0 = idx0+k;
end
A = tril(A);
imshow(A)

More Answers (3)

Steven Lord
Steven Lord on 25 May 2022
n = 8;
c = cell(1, n);
for k = 1:n
c{k} = tril(ones(k));
end
B = blkdiag(c{:});
spy(B)

Torsten
Torsten on 25 May 2022
ntriangles = 4;
n = ntriangles*(ntriangles+1)/2;
A = zeros(n);
rowstart = 1;
rowend = 1;
for i = 1:ntriangles
for jrow = rowstart:rowend
A(jrow:rowend,jrow) = 1.0;
end
rowstart = rowend + 1;
rowend = rowstart + i;
end

Stephen23
Stephen23 on 25 May 2022
N = 8;
C = arrayfun(@ones,1:N,'uni',0);
B = tril(blkdiag(C{:}));
spy(B)

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