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Trouble with handling values in anti-diagonal of a square matrix

6 vues (au cours des 30 derniers jours)
Minh Hoang
Minh Hoang le 8 Mar 2024
Commenté : Dyuman Joshi le 8 Mar 2024
I am trying to create a square matrix whose diagonal and antidiagonal have the same value (let say, 8). However the antidiagonal values in the matrix I create do not act in the way I expect. Here is my code:
function outXmatrix = sqrtmatx(s) %a function whose input is the dimension of the square matrix
nrows = s;
A = zeros(nrows, nrows);
for c = 1:nrows
for r = 1:nrows
for i = 0:(nrows - 1)
if r == c
A(r,c) = 8; %Assign value 8 to the diagonal
elseif r == i+1 && c == nrows-i %Assign value 8 to the antidiagonal
A(r,c) = 8;
else %Assign value 1 to the other elements
A(r,c) = 1;
end
end
end
outXmatrix = A;
end
For example, s = 5
My expected matrix is
A = [8 1 1 1 8,
1 8 1 8 1,
1 1 8 1 1,
1 8 1 8 1,
8 1 1 1 8]
The matrix my function produced is
sqrtmatx(5)
I hope someone could point out the problem in my code. Thanks a lot!

Réponse acceptée

Dyuman Joshi
Dyuman Joshi le 8 Mar 2024
1 - Preallocate using ones() instead of zeros() and remove the else part.
2 - Update the condition for checking the anti-diagonal element, and remove the 3rd for loop.
y = sqrtmatx(5)
y = 5×5
8 1 1 1 8 1 8 1 8 1 1 1 8 1 1 1 8 1 8 1 8 1 1 1 8
y = sqrtmatx(8)
y = 8×8
8 1 1 1 1 1 1 8 1 8 1 1 1 1 8 1 1 1 8 1 1 8 1 1 1 1 1 8 8 1 1 1 1 1 1 8 8 1 1 1 1 1 8 1 1 8 1 1 1 8 1 1 1 1 8 1 8 1 1 1 1 1 1 8
function outXmatrix = sqrtmatx(s) %a function whose input is the dimension of the square matrix
nrows = s;
A = ones(nrows, nrows);
for c = 1:nrows
for r = 1:nrows
if r == c
A(r,c) = 8; %Assign value 8 to the diagonal
%% updated check for antidiagonal term
elseif r + c == s+1 %Assign value 8 to the antidiagonal
A(r,c) = 8;
end
end
outXmatrix = A;
end
end
  2 commentaires
Minh Hoang
Minh Hoang le 8 Mar 2024
It worked! Thank you so much!
Dyuman Joshi
Dyuman Joshi le 8 Mar 2024
You're welcome!
You could vectorize your code as well. See - eye, flip

Connectez-vous pour commenter.

Plus de réponses (1)

Chuguang Pan
Chuguang Pan le 8 Mar 2024
diagVal=8;
otherVal=1;
nrows=8;
if mod(nrows,2) % odd
A=(diagVal-otherVal)*eye(nrows) + (diagVal-otherVal)*fliplr(eye(nrows))+...
otherVal*ones(nrows);
A(ceil(nrows/2),ceil(nrows/2))=A(ceil(nrows/2),ceil(nrows/2))-(diagVal-otherVal);
else
% even
A=(diagVal-otherVal)*eye(nrows) + (diagVal-otherVal)*fliplr(eye(nrows))+...
otherVal*ones(nrows);
end
disp(A)
8 1 1 1 1 1 1 8 1 8 1 1 1 1 8 1 1 1 8 1 1 8 1 1 1 1 1 8 8 1 1 1 1 1 1 8 8 1 1 1 1 1 8 1 1 8 1 1 1 8 1 1 1 1 8 1 8 1 1 1 1 1 1 8

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