Getting number of variables whose sum is lower than certain number

Hi all,

I have a problem because I don't know how to get the number of X's whose sum is lower than number N. Condition that must be fulfilled is that loop should repeat until the sum of X's reaches or exceeds number N. In case the sum exceeds number N, *it should give me 'There are 3 numbers X.' which is 25, 36 and 16. Number 49 should not be taken into consideration since when it is included in the sum (126) it exceeds number N (100).*

Here is my try but obviously there should be something more:

N=100;
r=[5 6 4 7 3 2];
s=0;
for i=1:1:length(r)
    X=r(i).^2;
    sprintf('X = %.1f',X)
    s=s+X;
    if (s > N || s == N)
  break
    end
end
sprintf('Sum = %.1f',s)
sprintf('There are %d numbers X.'length(X))

Output

ans =

X = 25.0

ans =

X = 36.0

ans =

X = 16.0

ans =

X = 49.0

ans =

Sum = 126.0

ans =

There are 1 numbers X.

I hope I explained as clearer as possible. Thank you for your help!

Answer 1

[Star Strider]

Your for loop increments the counter until you exit it, so ‘i’ tracks the quantity of numbers you used to complete the loop.

Try this:

sprintf('There are %d numbers X.', i)

[Banjo]

That is ok, it now shows 4. But how to solve it so it doesn't consider last X in the sum since it exceeds number 100. Thanks!

[Star Strider]

My pleasure.

Simply save the previous values, and use them:

N=100;
r=[5 6 4 7 3 2];
s=0;
for i=1:1:length(r)
    X=r(i).^2;
    sprintf('X = %.1f',X)
    s=s+X;
    if (s >= N)
  break
    end
    s_prev = s;                                 % <== ADDED
    i_prev = i;                                 % <== ADDED
end
sprintf('Sum = %.1f',s_prev)
sprintf('There are %d numbers X.', i_prev)

[Banjo]

Thank you very much, sir! That solved the problem.

[Star Strider]

As always, my pleasure!

[Banjo]

Hello, I have one more question. How could I add one more function so it sums up respectively with the sum of the function x=r^2 until it reaches or exceeds number N. Example:

N=100;
r=[5 6 4 7];
s=0; m=0; k=0;
for i=1:1:length(r)

      X=r(i).^2;
      sprintf('X = %.1f',X)
      s=s+X;

      Y=r(i)+5;
      sprintf('Y = %.1f',Y)
      m=m+Y;

      k=k+(s+m);

      if (k >= N)
    break
      end
      k_prev = k;
      i_prev = i;  % This should be only number of X's (not sum with Y)
  end
sprintf('Sum = %.1f',k_prev)
sprintf('There are %d numbers X.', i_prev)

The output which I look for:

ans =

X = 25.0     % s=25

ans =

Y = 10.0     % m=10

ans =

X = 36.0     % s=61

ans =

Y = 11.0

ans =

X = 16.0

ans =

Y = 9.0

ans =

X = 49.0

ans =

Y = 12.0

ans =

Sum = 96.0

ans =

There are 2 numbers X.

When I try with the code above it gives me result which I don't want. Thank you!

[Star Strider]

My pleasure.

I ran your code looking at the values of the various summations. I do not understand how you get the output you want from the code you wrote.

Changing ‘k’ to:

k = (s+m)

is the only change I can think of that will give you anything close to the result you want. It produces a sum of 82, not 96.

[Banjo]

Hello, I managed to get sum of 82 which is ok. Number 96 was just an example, not a real number.

I was wondering if it is possible to use two for loops, one for X and one for Y since I want to get how many X's and Y's are included in the sum. For the code above it should be 'There are 2 numbers X.' and 'There are 2 numbers Y'. Thanks!

[Star Strider]

My pleasure.

If I remember correctly, that’s what I got as well.

A double loop would likely be overkill. You could put in a separate counter for ‘Y’, however since you are summing each of them, adding the sums, and then using that sum as your exit criterion, the counters for both would be the same. (There is no condition on calculating ‘Y’ that depends on ‘X’, ‘s’, ‘m’, or ‘k’, so the same single iteration counter ‘i’ applies to all of them.)

I would just leave it as is, unless you also want to add a separate fprintf statement for ‘Y’, using the same counter.