# Write a function called halfsum that takes as input an at most two-dimensional matrix A and computes the sum of the elements of A that are in the diagonal or are to the right of it. example, the input is [1 2 3; 4 5 6; 7 8 9],the ans is 26

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Deepanshu Parihar on 28 May 2015
Commented: DGM on 25 Jul 2022
function s = halfsum(A)
[row col] = size(A);
if row ~= col
error('Expecting a square matrix here people...');
end
s = 0;
for ii = 1:row
for jj = ii:col
s = s + A(ii,jj);
end
end
Nadeem U Rehman on 10 Dec 2020
thank you @Rik

Buwaneka Dissanayake on 21 Jun 2020
function summa = halfsum(M)
[a b] = size(M);
if a>1
for n = 1:a;
for m = 1:b;
if n>m;
M(n,m) = 0;
summa = sum(sum(M));
end
end
end
else
summa = sum(M);
end
end
##### 2 CommentsShowHide 1 older comment
DGM on 25 Jul 2022
Edited: DGM on 25 Jul 2022
Of all the inefficient loop-based approaches on this page, this appears to be the most wasteful by far. For every single element of the array, you calculate the sum of the entire array and then discard the result. The only sum that doesn't get discarded is the last one. As a consequence, the time required grows rapidly as the matrix size increases -- all for nothing.
Try feeding this a large array. For a 400x400 matrix, this takes roughly 1000x as much time as the other loop-based examples, and nearly 6000x as much time as @Sean de Wolski's concise and efficient answer. Would you dare to feed it a 4000x4000 matrix?

Sean de Wolski on 28 May 2015
f = @(x)sum(sum(triu(x))) % make function
f(magic(3)) % use it

Joseph Cheng on 28 May 2015
your input should be
input = [1 2 3;4 5 6;7 8 9]
and not
input = [1 2 3 4 5 6 7 8 9]
##### 1 CommentShowHide None
Joseph Cheng on 28 May 2015
which if you read carefully the problem statement and the code the input is a 3x3 square matrix. the input you provided it was a 1x9 array.

Pragyan Dash on 19 Sep 2020
%this worked for me. Happy to help!
function summa = halfsum(M)
[row col] = size(M)
summa = 0;
for ii = 1:row;
for jj = 1:col;
if jj >= ii;
summa = summa + M(ii, jj);
end
end
end
##### 1 CommentShowHide None
very clean thanks

Srishti Saha on 7 Apr 2018
This code works perfectly for me:
%function to compute sum of lower most right side triangle in an X*2 matrix
function u = halfsum(P)
u1 = P(end:-1:1, 1:end);
u2 = triu(u1);
u = sum(u2(:));
end

Ajith Bharadwaj on 3 Feb 2020
function summa = halfsum(A)
[row col] = size(A);
for ii = 1:row
for jj = ii:col
summa = summa + A(ii,jj);
end
end
Amit Jain on 3 Nov 2020
I hope this clears your doubt. You did not tell function what to if there is n ==1 means there is column vector. I have made this correction. Now function works fine. Thanks

ERTIZA HOSSAIN SHOPNIL on 21 May 2020
function summa=halfsum(A)
t=triu(A);
list=sum(t);
s=0;
for n=list
s=s+n;
end
summa=s;
end
Rik on 24 Aug 2020
That means you will have to copy another answer to cheat on your homework. Or think yourself.

saurav Tiwari on 25 Jun 2020
function summa=halfsum(a)
[m,n]=size(a)
for i=1:m
j=1:n
x(i)=sum(a(i,j))
end
summa=sum(x)
end
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 25 Jun 2020
Incorrect code. This calculates sums of an entire row, not the sum of the diagonal and right of the diagonal.

youssef boudhaouia on 27 Jul 2020
A solution with double For-loop:
function summa=halfsum(M)
summa=0;
s=size(M);
for i=1:s(1)
for j=1:s(2)
if j>=i
summa=summa+M(i,j);
else
summa=summa;
end
end
end
Walter Roberson on 4 Aug 2020
You do not need the if condition if you use
for j=i:s(2)

Abdul Quadir Khan on 18 Oct 2020
function summa = halfsum(x)
[row,col] = size(x);
allsum=0;
for n=1:row
for c=n:col
allsum=x(n,c)+allsum;
end
end
summa=allsum;

Muhammad on 25 Jul 2022
function summa=halfsum(M)
[row,col]=size(M);
total1=0;
total2=0;
for m=1:row
if m==1
for n=1:col
P=M(m,n)+total1;
total1=P;
end
elseif m>1 && m<=n
for n=m:col
Q= M(m,n)+total2;
total2=Q;
end
end
end
summa=total1+total2;
##### 1 CommentShowHide None
DGM on 25 Jul 2022
Compared to the other loop-based approaches, what is the particular benefit of this one? It's not particularly fast or concise in comparison.
Posting reference answers is a good thing, but it's best to explain why your answer takes the form it has. This is particularly important when there are many similar answers in the same thread.
Maybe describe the problem-solving approach that brought you to the solution. Maybe you asserted certain goals or constraints that led you to solve it differently than others. Explain why a reader might want to follow your example -- or even the cases in which they might want to follow one of the others. It doesn't have to be much description, but something is better than nothing.