How to create this patterned matrix?

 Accepted Answer

N = 8;
M = toeplitz(0:N-1,[0,N:-1:1]);
M(N,:) = N:-1:0
M = 8×9
0 8 7 6 5 4 3 2 1 1 0 8 7 6 5 4 3 2 2 1 0 8 7 6 5 4 3 3 2 1 0 8 7 6 5 4 4 3 2 1 0 8 7 6 5 5 4 3 2 1 0 8 7 6 6 5 4 3 2 1 0 8 7 8 7 6 5 4 3 2 1 0

More Answers (2)

Generalizing for any nxn matrix
clear,clc
n = 9;
A = zeros(n);
for i = 1:n
A(i,i+1:n) = flip(i:n-1);
if i > 1
A(i,1:i) = flip(0:i-1);
end
end
which yields
A =
0 8 7 6 5 4 3 2 1
1 0 8 7 6 5 4 3 2
2 1 0 8 7 6 5 4 3
3 2 1 0 8 7 6 5 4
4 3 2 1 0 8 7 6 5
5 4 3 2 1 0 8 7 6
6 5 4 3 2 1 0 8 7
7 6 5 4 3 2 1 0 8
8 7 6 5 4 3 2 1 0
However, in your example, the row starting with 7 is missing. I am not sure whether that's intentional or just a typo.
In the first instance, the code above can be arranged to remove rows starting with a certain value.

3 Comments

Stephen23
Stephen23 on 7 Mar 2022
Edited: Stephen23 on 7 Mar 2022
"Generalizing for any nxn matrix"
The OP requested an 8x9 matrix (which is not square).
Check the last rows of the OP's matrix too, you will find that they do not match your answer.
I didnt even realize it was "5 6 8". Thanks a lot man
Anytime! I guess you indeed wanted that row to be skipped. @Stephen's answer will do that very efficiently.

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try this:
x=[0,8,7,6,5,4,3,2,1];
A1=circshift(x,1);
A2=circshift(x,2);
A3=circshift(x,3);
A4=circshift(x,4);
A5=circshift(x,5);
A6=circshift(x,6);
A7=circshift(x,8);
Matrix=[x;A1;A2;A3;A4;A5;A6;A7]
Matrix = 8×9
0 8 7 6 5 4 3 2 1 1 0 8 7 6 5 4 3 2 2 1 0 8 7 6 5 4 3 3 2 1 0 8 7 6 5 4 4 3 2 1 0 8 7 6 5 5 4 3 2 1 0 8 7 6 6 5 4 3 2 1 0 8 7 8 7 6 5 4 3 2 1 0

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