how to get sum of elements of lower right triangular matrix.? i have tried this one, can anyone please correct this one
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vamshidhar Reddy Peruvala
2016 年 12 月 2 日
回答済み: Ibrahim Abouemira
2019 年 5 月 19 日
function [sum] = halfsum( a )
[m, n]=size(a); sum=0;
for i= m:-1:1
for j=n:-1:1
if i==j || j<=i
sum=sum+a(i,j);
end
end
end
end
output: halfsum([1 2 3;4 5 6;7 8 9])
ans =
34
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Image Analyst
2016 年 12 月 2 日
Don't use sum() as the name of your variable! You're destroying a built-in function. I wish we could make this a sticky on the forum because it happens so often, probably more often than people confusing (row,column) with (x,y).
Use tril and fliplr like this:
m = magic(9) % Sample matrix.
lowerRight = fliplr(tril(fliplr(m)))
theSum = sum(lowerRight(:))
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Image Analyst
2016 年 12 月 4 日
The j loop should start at i, not n. And use theSum instead of sum. And the "if" is unnecessary.
Ibrahim Abouemira
2019 年 5 月 19 日
Hello,
Here's a proposed answer to your question.
The function takes as input at the most two-dimensional array. It computes the sum of the elements of that are in the lower right triangular part(counter-diagonal elements).
For example, if the input is [1 2; 3 4; 5 6; 7 8], then the function would return 21.
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