create a 99x99 matrix with ones on both diagonals and zeros everywhere otherwise

Antworten (4)

Image Analyst
Image Analyst am 15 Okt. 2021

1 Stimme

Another way, even more compact:
A = eye(99) | fliplr(eye(99))
As long as it's not your homework you can use my code.
Chunru
Chunru am 15 Okt. 2021

0 Stimmen

n = 9; %99
A = eye(n);
A(n:n-1:n*n-1) = 1; % anti-diagonal
A
A = 9×9
1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1
Image Analyst
Image Analyst am 15 Okt. 2021

0 Stimmen

As long as it's not your homework you can use my code:
A = min(1, eye(99) + fliplr(eye(99)))
Chunru
Chunru am 15 Okt. 2021

0 Stimmen

% For time comparison:
n = 1000;
timeit(@() bidiag1(n))
ans = 4.1086e-04
timeit(@() bidiag2(n))
ans = 0.0027
timeit(@() bidiag3(n))
ans = 0.0019
function bidiag1(n)
a = eye(n);
a(n:n-1:n*n-1) = 1;
end
function bidiag2(n)
a = eye(n) | fliplr(eye(n));
end
function bidiag3(n)
a = min(1, eye(n) + fliplr(eye(n)));
end

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am 15 Okt. 2021

Beantwortet:

am 15 Okt. 2021

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