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

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

[Deepanshu Parihar]

but i am getting the error

Problem 3 (halfsum): Feedback: Your program made an error for argument(s) [1 2 3 4 5 6 7 8 9 10] \please help

[Joseph Cheng]

please reformat your question using the {}code button and view the preview section so it is easily readable.

[Deepanshu Parihar]

see now Mr Cheng

[Moses Patson]

This code may fail the row and column vector test.

Answer 1

[Sean de Wolski]

f = @(x)sum(sum(triu(x))) % make function
f(magic(3)) % use it

Answer 2

[Buwaneka Dissanayake]

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

[Arshdeep Singh]

can you explain the logic of this function

[DGM]

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?

Answer 3

[Pragyan Dash]

%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

[THIERNO AMADOU MOUCTAR BALDE]

very clean thanks

Answer 4

[DGM]

Okay, it's my turn. Let's try something absolutely ridiculous.

% some test arrays
A = magic(1000); % square
B = [A A]; % wide
C = [A; A]; % tall

% Sean de Wolski's example as a reference
f = @(x)sum(sum(triu(x))); % a sensible choice
ref = [f(A) f(B) f(C)];

% logical addressing using a polygonal mask
testme = [halfsum_poly(A) halfsum_poly(B) halfsum_poly(C)];

% check the results
isgood = isequal(testme,ref)

isgood = logical
   1

% literally just draw a triangle over the values you want
function summa = halfsum_poly(A)
    sz = size(A);
    mxsz = max(sz);
    x = [0 1 1 0]*mxsz; % use max() to support non-square inputs
    y = [0 0 1 0]*mxsz + 0.5; % offset to capture main diagonal
    mk = poly2mask(x,y,sz(1),sz(2)); % generate logical mask
    summa = sum(A(mk)); % apply the mask
end

Wait, as ridiculous as that is, it actually works? Of course it works. These are basic image processing tools. It doesn't use triu(), and it even works for non-square inputs. I don't think anyone would expect this silliness to be fast, but it doesn't have the worst complexity out of the examples on this page. For large inputs (1000x1000), it's comparable in speed to many of the examples here.

Would your TA accept this if you turned it in for your homework? Wanna gamble?

Are there more sensible ways to solve the problem using logical indexing? Yes.

Are there ways which might seem wasteful, but might have advantages? Yes.

Are there ways to do it with loops that are both concise and fast? Yes.

What's the lesson here? If you're going to post an answer among many others, try to post something that adds to the information present. Test it to make sure it works (you can run it right here in the editor). Describe what your code does (comments and otherwise). Does your answer provide particular benefits? Does it have relative drawbacks?

[Rik]

I'm tempted to create a new account so I can give you two upvotes. (and as a TA I would personally accept this answer)

You missed the golden opportunity to use the length function.

[DGM]

I think the cautionary warning was more to imply that copying a curiously-uncommon answer is a good way to invite scrutiny, especially if your other efforts in the class are comparably lazy.

After I posted that, I was tempted to change it to calculate the sum of the lower triangular matrix instead, just to keep everyone honest (and attentive).

I kind of wanted to post all the silly examples I tested to emphasize that there's plenty of opportunity to try something different, but I figured that would be too much free stuff. I don't know. Maybe if I flip those it'd be fair.

Yeah, I avoid length() so much that I'd have never thought of that.

Answer 5

[Joseph Cheng]

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]

[Joseph Cheng]

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.

Answer 6

[Srishti Saha]

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

[Muhammad Amir Javed]

How to sum upper triangle of matrix without using triu?

[Image Analyst]

@Muhammad Amir Javed you could use a double for loop. The starting value for the inner for loop would be the current value of the outer for loop.

Answer 7

[ERTIZA HOSSAIN SHOPNIL]

function summa=halfsum(A)
    t=triu(A);
    list=sum(t);
    s=0;
    for n=list
        s=s+n;
    end
    summa=s;
    
end

[Saurov Baidya]

its brings error at a certain step

[Meron Pawlos]

It restircts the use of keyword triu

[Rik]

That means you will have to copy another answer to cheat on your homework. Or think yourself.

[Rik]

@Bengisu, Regarding your flag ("this user constantly bullies people on the matlab answers page, claiming them to be cheaters and should be restricted in my opinion."):

What legitimate purpose do you think this thread serves? If there is some sort of system that restricts the use of 'triu', that is either a homework submission system, or a system like Cody. In either case I would think it is fair to call copying answers cheating. If there is a legitimate use case I would expect some sort of reply. Can you come up with a reason why they should not be called out?

I never intended to bully anyone, but I will call out people who seem to attempt to cheat on their homework. Do you have any suggestions for how I should have phrased my comment?

[Rik]

There is a fine line (or overlap) between rudeness and clarity/unabiguity. I will have followed up those questions with something along the lines of 'these are not rhetorical questions' (or at least I should have).

My questions to you were actually serious, and not at all intended as bullying. If you see a legitimate purpose to this thread, I would welcome the explanation. If I'm missing something, I honestly would like to know.

Same goes for my wording. If you have a suggestion for me that has the same level of transparency but conveys the message a bit (or a lot) nicer, please feel free to suggest it. For this kind of interaction I use canned responses, so a suggestion now will improve many posts to come.

I hope you do reply if you have any constructive/actionable feedback. I hope my record speaks for itself and that you can help me improve it further.

[DGM]

I hate to continue, but this is one of the threads I try to keep ... well I wouldn't call it "tidy", but you get the idea. I think there's a side to these interactions that's being overlooked.

@Bengisu: At least as far as my own similar remarks are concerned, asking "why did you post this" in cluttered threads like these is often in part a response to the ubiquity of answers which are:

• not notably unique from existing answers
• not explained or commented (or formatted)
• not correct in their results
• incomplete or result in errors
• verbatim copies of other answers on the same page
• not answers at all or are irrelevant to the question

How many of the answers on this thread describe their conceptual approach to the solution? How many explain what makes their approach better or different than other similar answers? Would a novice reader know if there is relative benefit in yet another uncommented loop-based example? Are there any? Disregarding ethics, it's poor communication, and lots of it.

From my own perspective, we're not merely trying to help people with MATLAB; sometimes we also try to help people write good questions and answers. Some of us are just trying to keep things organized. Maybe some of us are trying to help people improve in ways that are more important than this week's MATLAB homework. In that effort, some of us have the patience of a saint, but the rest of us are mere humans.

Considering that students do get in trouble for offloading their work on the forum, I think calling this sort of thing cheating is a fair generalization, despite the arguable existence of exceptions. That said, I think the ship has already sailed with regards to providing answers to this simple assignment. I personally wouldn't mind if someone else posted a different answer and had something to say about it, though others are free to disagree.

Answer 8

[youssef boudhaouia]

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

[Purushottam Shrestha]

Great

[Saurov Baidya]

can you please explain the solution in detail.... i m still confused about it

[youssef boudhaouia]

First we initialize the output summa. Second we assign s as the size of the matrix M . In fact , size(M) returns a vector with two values. The first value s(1) is the row number and the second value s(2) is the column number. The variable i is for the row indexes( from 1 to s(1)) and the variable j is for column indexes(from 1 to s(2)) . In every time where we find a value of the matrix which has a column index ( j ) higher or equal to the row index ( i ), we add this value to summa, and so over, until we find the total sum. If the condition (j>=i) is not true, the variable summa stays the same and doesn't change, that means : summa=summa .

I hope it's clear now . you can ask if there's something u r not convinced with

[Walter Roberson]

You do not need the if condition if you use

for j=i:s(2)