placing values in the form of a triangle in a matrix

If you want it to be a matrix, then I don't think there is a way to do it without adding 0's in the lower part of the triangle. The code below will work.

n = 3; %input
A = hankel(1:n);
c = A(n,1);
for i = 2:n
    for j = 1:n-1
        A(j,i) = c+1;
        c = c + 1;
    end
    n = n-1;
end

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Duncan 2014-8-11

thanks for this

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Answer 2

Patrik Ek 2014-8-11

0
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https://ww2.mathworks.cn/matlabcentral/answers/147118-placing-values-in-the-form-of-a-triangle-in-a-matrix#answer_148127

编辑：Patrik Ek 2014-8-11

在 MATLAB Online 中打开

Try this,

n = 5;
A0 = fliplr( triu(ones(n)) );
A = cumsum( A0(:) );
A = reshape(A,n,n);
A(A0<=0) = 0

The idea is to create an upper diagonal matrix of ones which you flip, then accumulate the sum of all elements (after reshaping the elements to a column matrix with (:) ). That means that the sum increase for every element, represented in the final matrix, but not for any other element. Then reshape the matrix again and set all elements which are not the flipped upper diagonal to zero to zeros.