How to create this patterned matrix?
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采纳的回答
N = 8;
M = toeplitz(0:N-1,[0,N:-1:1]);
M(N,:) = N:-1:0
更多回答（2 个）
Davide Masiello
2022-3-7
编辑：Davide Masiello
2022-3-7
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 个评论
Burger King
2022-3-7
Davide Masiello
2022-3-7
Anytime! I guess you indeed wanted that row to be skipped. @Stephen's answer will do that very efficiently.
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]
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