how can I write this in a compact form? can anyone suggest a single line code for it

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yita_mn = [
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1;
1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1;
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
];

Accepted Answer

John D'Errico
John D'Errico on 22 Jan 2023
Edited: John D'Errico on 22 Jan 2023
At its heart, this is just a basic circulant matrix. So use a tool that will do that. I posted such a tool on the file exchange. But you can just use gallery.
n = 8; % an 8x8 circulant matrix of that form
M = gallery('circul',[0 1,zeros(1,n-3),1])
M = 8×8
0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0
Or you can view this as a Toeplitz matrix. Toeplitz matrices are a close second cousin to circulant matrices. Your matrix is always symmetric, so a symmetric Toeplitz matrix can be formed by just supplying the first row or column.
M2 = toeplitz([0 1,zeros(1,n-3),1])
M2 = 8×8
0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0
Of course there are other ways to do it. Since your matrix is actually quite sparse, you could build it as a sparse banded matrix. The virtue of that is if n is large, (as it often will be in applications like this) then you use efficient sparse storage, as well as efficient computations thereafter with that matrix.
I'll use n=50 here, but typically n might be a number in the thousands or more, if you are really needing to use a sparse matrix. 50 is large enough that you can visualize the banded structure easily, yet not too large that you cannot see the dots. (If I made n=10000, then you might not even see the dots in the corners, or see the line down the middle as actually a pair of bands just above and below the main diagonal.)
n = 50;
M3 = spdiags(ones(n,4),[-n+1, -1, 1, n-1],n,n);
spy(M3)
whos M3
Name Size Bytes Class Attributes M3 50x50 2008 double sparse
As you can see, M3 is a sparse version of the matrix you want to build.

More Answers (3)

Bruno Luong
Bruno Luong on 22 Jan 2023
n = 8;
ismember(mod((1:n)-(1:n)',n),[1 n-1])
ans = 8×8 logical array
0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0

Sargondjani
Sargondjani on 22 Jan 2023
You can do something like:
A=[zeros(1,10);eye(9),zeros(9,1)]+ ....
Try to figure out the details yourself...

Walter Roberson
Walter Roberson on 22 Jan 2023
Edited: Walter Roberson on 22 Jan 2023
circulant matrix
Or you could add two diagonal matrices
  1 Comment
Walter Roberson
Walter Roberson on 22 Jan 2023
n = 8;
circshift(diag(ones(1,n)),1) + circshift(diag(ones(1,n)),-1)
ans = 8×8
0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0

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