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# Finding the sum of all numbers between two negative numbers

I have to create a robust algorithm that calculates the sum of two numbers. I've been at this all night (not that question, but performing tasks) and this one has got my brain falling out a little.

I previously already had the algorithm for two positive numbers working great. And for two negative numbers, I use the same solution. However first I make the two numbers positive, then at the end I convert them back to a negative again ( total = total - (total * 2); ) since adding two negative results in a negative number. This works well enough.

However, where I'm stuck is finding the sum of all the numbers between a negative and a positive. For instance, the sum of all numbers between -50 and 50.

I'm not looking for code as this is an assignment but perhaps some hints on how to achieve this would be great! Maths isn't my super strong point.
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If I've got what you want to do right, I think you're trying to sum an arithmetic series - http://mathcentral.uregina.ca/QQ/database/QQ.02.06/jo1.html

For a negative sequence, swap the signs so S=( N(N-1) ) / 2
Edited by BorisTheSpider - 12/12/12 at 4:57pm
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Just a small nit pick, a better way to turn an integer negative is to multiply by -1
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Quote:
Originally Posted by BorisTheSpiderÂ

If I've got what you want to do right, I think you're trying to sum an arithmetic series - http://mathcentral.uregina.ca/QQ/database/QQ.02.06/jo1.html
For a negative sequence, swap the signs so S=( N(N-1) ) / 2

Oh my, that's actually quite handy. For one, I wasn't using that. I was using a while loop with this:
Code:
``````public void Sum(int a, int b)
{
int total = 0;
int value = a;

while(value < b)
{
total += value;
System.out.println(total);
value++;
}```
```

The code now looks like:
Code:
``````public int sum2(int a, int b)
{
int sum = b * (b + a) / 2;
return sum;
}```
```

Which is much shorter!

I'll test the negatives shortly.

I wonder though, how do you do a positive and a negative together?
Quote:
Originally Posted by tompsonnÂ

Just a small nit pick, a better way to turn an integer negative is to multiply by -1

Again, that's handy as well!
Edited by Rubers - 12/12/12 at 5:26pm
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The sum of every integer between and including a and b is (a+b)*(a-b+1)/2
I THINK, early here. Not sure if it will work with a negative and positive integer... try it

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Quote:
Originally Posted by tompsonnÂ

The sum of every integer between and including a and b is (a+b)*(a-b+1)/2
I THINK, early here.

Ok, strange results with these two.
Code:
````sum = b * (b + a) / 2;`
```
gives 5050 where a = 1 and b =100.

but
Code:
````sum = b * (b - a) / 2;`
```

gives -4949 where a = -1 and b =-100.

thompsonn's
Code:
````sum = (a + b) * (a - b + 1) /2;`
```

gives -4949 wherea = 1 and b = 100.

but gives -5050 where a = -1 and b = -100!
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I think it could be: for the sum of every integer between a and b inclusive: ((a+b)/2)(a+b)
based on: (n/2)(a(sub 1) + a(sub n))
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Yeah I buggered something there I think
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It's ok, I got this:
Code:
``````public int sum2(int a, int b)
{
int sum = 0;

if(a > 0 && b > 0)
{
sum = b * (b + a) / 2;
}

if(a < 0 && b < 0)
{
a = -a;
b = -b;
sum = b * (b + a) / 2;
sum = sum * -1;
}
return sum;
}```
```

Or, there's this for the negatives. I'm not sure which is better:
Code:
``````       if(a < 0 && b < 0)
{
sum = b * (b - a) / 2;
sum = sum - (a - b * -1);
sum = sum * -1;
}```
```
I think I like the first better, though. It seems a bit cleaner even if it is one more line.
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S = n[2a+(n-1)d]/2 is entirely general, with the following provisos:

1. Always start at the smaller number - that is to say, if summing negative numbers, start with the most negative, and if crossing from -ve to +ve, start with the -ve one.

2. If the sequence has zero as a term, make sure you count that when calculating n

3. Don't make the first or last term zero - start with the first nonzero term and end with the last nonzero term

In pig-pseudo-c:

a = insert a
b = insert b
d = insert difference between terms

// make d +ve

d=abs(d)

// set a < b

if b < a
{
temp=a
a=b
b=temp
}

// Fix a=0

if a==0
{
if b<0 then a=a-d
else
a=a+d
}

// fix b=0

if b==0
{
b=b-d
}

// get the number of terms

n = abs( (b-a)/d ) + 1

// sum them

sum = n(2a+(n-1)d)/2

Needs testing - may contain traces of bogosity
Edited by BorisTheSpider - 12/12/12 at 6:47pm
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