How I can find a code to find (m) number form (1, n) and their sum must be different at each time.

Ex : if the n = 5 and m = 4, which mean I want 4 number from (1,5) . The numbers appear are 1,2,3,4 and the sum is : 1+2=3 1+3=4 1+4=5 2+3=5 2+4=6 3+4=7 Clearly that the sum 5 is repeated so these number are rejected.

Answer 1

[Image Analyst]

I know it seems obvious but did you try a simple for loop? I assume you did and got something like

n = 5;
m = 4;
numbers = linspace(1, n-1, m) % 4 numbers from 1 to m-1 (4)
for k = 1 : m
	num1 = numbers(k);
	for k2 = k+1 : m
		num2 = numbers(k2);
		fprintf('%.1f + %.1f = %f\n', num1, num2, num1+num2);
	end
end

numbers =

1 2 3 4

1.0 + 2.0 = 3.000000

1.0 + 3.0 = 4.000000

1.0 + 4.0 = 5.000000

2.0 + 3.0 = 5.000000

2.0 + 4.0 = 6.000000

3.0 + 4.0 = 7.000000

but I'm wondering why this was no good. Did you need something more "sophisticated" for some reason?

[Rami Ahmed]

What I need is no repeated sum which mean when I sum the number that appear the sum must be diiferent, 5 as a sum was repeated

[Image Analyst]

You can use ismember() or unique():

n = 5;
m = 4;
numbers = linspace(1, n-1, m) % 4 numbers from 1 to m-1 (4)
index = 1;
theSum = [];
for k = 1 : m
    num1 = numbers(k);
    for k2 = k+1 : m
        num2 = numbers(k2);
        if index == 1 || ~ismember(num1+num2, theSum)
            theSum(index) = num1 + num2;
            fprintf('%.1f + %.1f = %f\n', num1, num2, theSum(index));
            index = index + 1;
        end
    end
end

[Rami Ahmed]

But I need to sum all these number with each other

[Image Analyst]

Do you mean like

overallSum = sum(theSum);

[Rami Ahmed]

To be honest I dont know what that means

[Image Analyst]

You said "sum all these number with each other". That means you use the sum() function. So what do we put into the sum function? All these numbers. Well, all the numbers that were the result of adding two numbers are in theSum. So I put theSum vector into sum. Does it make sense now? Otherwise give explicit description of what "all these number" means to you.

[Rami Ahmed]

Its means the numbers that appear when I run, I want to generate numbers when I sum any of these numbers with any other numbers that appear the sum must be new, which means no sum repeated if I do it for all numbers that appear.

But I can't exclude any sum, 4 numbers mean 6 different sums.

Please if you can help me...

[Image Analyst]

Your output in the original post gave 6 sums but then you said 5 was repeated (1+4 and 2+3) so you did not want to include that (I exclude the second pair that gives a repeat). That leaves 5 sums, not 6. My code provides the 5 sums.

1.0 + 2.0 = 3.000000

1.0 + 3.0 = 4.000000

1.0 + 4.0 = 5.000000

2.0 + 4.0 = 6.000000

3.0 + 4.0 = 7.000000

What is the missing 6th sum?

This isn't your homework is it? Because I'm not allowed to give you homeworls solutions where you could turn in my code as your own.

[Rami Ahmed]

The wrong is with the result appear when I run the code, where the numbers must be 1,2,3,5 these numbers have 6 differnet sum. 1+2=3, 1+3=4, 1+5=6, 2+3=5, 2+5=7, 3+5=8,

[Image Analyst]

So now they go out to 5 not 4? Did you change m or n?

Is m always n-1 or can it be any integer at all?

What happened to 1+4, 2+4, 3+4, and 4+5? Why are they not to be included?

[Rami Ahmed]

No, m =4 and n=5 that means 4 numbers from 1to 5 I want to choose, and m can be any integer but must be less than n. I didnt choose 4 becsuse I want 4 numbers from 1 to 5 so I choose 1,2,3,5 and the sum of 4 with other numbers not necessary or I dont need it..

[Image Analyst]

So if you have 4 numbers from 1 to 5, inclusive, then m is

n = 5;
m = 4;
numbers = linspace(1, n, m)

numbers = 1×4

1.0000 2.3333 3.6667 5.0000

Is that what you want?

[Rami Ahmed]

Yes, but I want an integer number

[Image Analyst]

Which integers? Do you just want to round and take unique ones?

n = 5;
m = 4;
numbers = unique(round(linspace(1, n, m)))

numbers = 1×4

1 2 4 5

So 3 will not be done? Or do you have some other algorithm?

What is the use case for this? No more help until you say if it's homework or not.

[Rami Ahmed]

I swear its not homework its personal issue I want to solve because It will help me in my school, because I'm a math teacher. Above 1+5=6 and 2+4=6 so 1,2,4,5 are rejected.

[Image Analyst]

Try this:

n = 5;
m = 5;
numbers = linspace(1, n, m)
index = 1;
theSums = [];
for k = 1 : length(numbers)
    for k2 = 1 : length(numbers)
        num1(index) = numbers(k);
        num2(index) = numbers(k2);
        theSums(index) = num1(index) + num2(index);
        fprintf('%.4f + %.4f = %.4f\n', num1(index), num2(index), theSums(index));
        index = index + 1;
    end
end
[uniqueSums, indexes] = unique(theSums)
% Extract only the unique numbers:
num1 = num1(indexes)
num2 = num2(indexes)

% Double check.
for k = 1 : length(indexes)
    fprintf('%.4f + %.4f = %.4f\n', num1(k), num2(k), uniqueSums(k));
end

[Rami Ahmed]

I can try any n and m or just for m=n=5

[Image Analyst]

It should work for any n and m. Did you try any?

[Rami Ahmed]

Yes, but its gave me not an integer number

[Image Analyst]

Again, you can call round(). But you must realize that you cannot divide a range of n long into m segments without the segments being fractional, except for certain special cases. Like I said:

n = 5;
m = 4;
numbers = linspace(1, n, m)

numbers = 1×4

1.0000 2.3333 3.6667 5.0000

numbers = unique(round(linspace(1, n, m)))

numbers = 1×4

1 2 4 5