summing #ers and differencing #ers

My code below basically does this:

The FL Function:
- User types in one number (lets say 2)
- User types in second number (lets say 4)
- The computer caclulates and outputs what 4+3+2 is.

The FG Function:
- User types in one number (lets say 4)
- User types in second number (lets say 2)
- The computer caclulates and outputs what 4-3-2 is.

The FG function works fine until I type something in like:
First number: 4
Second number: 3

If they are one apart they don't come out with the right answer. I have narrowed this down to the n/2 which makes in all situations (where the first and second number are one apart) n/2 equal to one. I need a revise my formula but can't think of a way. Here is my code:


#include <iostream>
using namespace std;
float fg();
float fl();

float first;
float second;

int main()
{

cout << "Type in A1\n";
cin >> first;
cout << "Type in An\n";
cin >> second;

if (first < second)
{
fl();
}
else if (first > second)
{
fg();
}


return 0;
}

float fl()
{
float n;
float final;

n = ((second + 1) - first);
final = n/2 * (first + second);
cout << "The sum between " << first << " and " << second << " is, "<< final << "\n";
return 0;
}

float fg()
{
float n;
float final;

n = ((first + 1) - second);
final = n/2 * (first + second);
cout << "The difference between " << first << " and " << second << " is, -"<< final << "\n";
return 0;
}


P.S. I realize I could do this easier with arrays..But I want to try it with a custom formula

Your formulae are wrong. By using this well-known formula (was it Gauss who proposed it, or Euler? I forget who it was):
Sum(1..n) = n(n+1)/2

I derived the formulae where the first number is not 1, to be (assuming second is larger than first, of course):

FL = (second + first) (second - first + 1) / 2
FG = (2 * second + (first - second)(first + second - 1))/2

This works for all cases, whether the numbers are one apart or not. A discussion of how I derived these formulae is probably best discussed in the algorithms forum.

Hmm..my first formula works in all cases. My second is flawed and I will use yours although I don't understand it 100% yet. Thanks alot though for your quick reply. Regards.

-andy

That's because your FL() formula is the same as my FL() formula (rearrange the symbols and you'll see). However, your FG() formula is not the same as mine.

[edit]
BTW my first try at FG was wrong too.
FG() = second + ((first - second)*(first + second - 1)/2)
[/edit]

hmm..your formula:
final = (2 * second + (first - second) * (first + second - 1))/2;

doesn't seem to work. Here is my results for a practice run:

Type in A1
4
Type in An
2
The difference between 4 and 2 is, -7

It should be -9? 4-3-2 = -9

EDIT: Nope..your new formula:
final = second + ((first - second)*(first + second - 1)/2);

Still results in -7

Hehe, check out my corrected formula.

fixed it..it should be:


final = (2 * second + (first - second) * (first + second + 1))/2;

Thanks for pointing me in the right direction again scorp.

-andy

My formulae are correct for FG(), both first and second try are the same thing. BTW 4-3-2 = -1, not -9

if you put second = 4 and first = 2 and try out my formula, it works correctly.

Doh! Im stuck on negatives..lol. BRB to try your method out.

final = second + ((first - second)*(first + second - 1)/2);

Output:

Type in A1
5
Type in An
3
5-4-3 = -10

hmm..

If you check the first reply in my thread, I said second was assumed to be the greater number in my formula. If you put second = 5, first = 3, then you get

final = 5 + ((3-5) * (3 + 5- 1)/2)
= 5 + ((-2) * (7)/2)
= 5 + (-7)
= -2

5-4-3 = -2

Therefore my formula works correctly. The reason you're getting -10 is because you reversed the values of first and second.

Thanks alot man. Take it easy.

-andy