Fraction Algorithm

[Fishstick]

I need an algorithm to add two fractions. Its easy to do with your head but when I went to write it i was all like HTF?

So yeah, its for C++ by the way so if you feel like writing it do so.

I have to add:
[u]frac1.numerator[/u] + [u]frac2.numerator[/u]
frac1.denominator frac2.denominator

[ThePuska]

frac1.numerator * frac2.denominator + frac2.numerator * frac1.denominator
-----------------------------------------------------------------------------------------
frac1.denominator * frac2.denominator

[nullsquared]

[QUOTE=ThePuska;19533166]frac1.numerator * frac2.denominator + frac2.numerator * frac1.denominator
-----------------------------------------------------------------------------------------
frac1.denominator * frac2.denominator[/QUOTE]

Yup, just find a common denominator by multiplying the two denominators (is not necessarily the smallest common multiple, but it works)

1/2 + 1/4
(1*4)/(2*4) + (1*2)/(4*2)
4/8 + 2/8
6/8

Then you can simplifying by looking for common factors from 2 to the numerator, so we get

6/8 (check from 2 to 6)
(2*3)/(2*4)
3/4

If the numerator is bigger than the denominator, we want to divide to find the amount of wholes, and then leave the remainder as a fraction

whole = numerator / denominator
remaining_numerator = numerator % denominator

[ThePuska]

[QUOTE=nullsquared;19533512]Yup, just find a common denominator by multiplying the two denominators (is not necessarily the smallest common multiple, but it works)

1/2 + 1/4
(1*4)/(2*4) + (1*2)/(4*2)
4/8 + 2/8
6/8

Then you can simplifying by looking for common multiples from 2 to the numerator, so we get

6/8 (check from 2 to 6)
(2*3)/(2*4)
3/4[/QUOTE]

If you want to simplify it, it might be best to factorize the original numerators and denominators

[editline]10:10PM[/editline]

And simply remove all factors that occur both in either of the numerators and either of the denominators

edit: Wait no I don't know what I'm doing, that'd only apply if you were multiplying them

[nullsquared]

[QUOTE=ThePuska;19533580]If you want to simplify it, it might be best to factorize the original numerators and denominators

[editline]10:10PM[/editline]

And simply remove all factors that occur both in either of the numerators and either of the denominators[/QUOTE]

That's essentially what I meant, I accidentally said multiples instead of factors.

[Mrs Affinity]

Haha, thats pretty funny, I had to do this for a lab in my Computer Sciences class. It was pretty easy though.

[ThePuska]

[QUOTE=nullsquared;19533721]That's essentially what I meant, I accidentally said multiples instead of factors.[/QUOTE]

Ignore my post, you need to factorize them after adding anyway

[Fishstick]

I don't need to simplify for this assignment, its just practice using structures which we are just learning about. Thanks for the help though.

[PvtCupcakes]

Or if you want some fun just divide the fractions and add them to get a decimal.
(1/2) + (1/4) = 0.75
Then convert 0.75 back to a fraction.

I did this a long time ago, I think it was something like this:

double dec = 0.75;

//this should get rid of the decimal
//if not, just truncate it
int mod = 1000000; //mod will be our denomenator too
double num = dec * mod;
int numerator = (int) num;

int gcd = gcd(numerator, mod);
numerator = numerator/gcd;
mod = mod/gcd;

printf("%d/%d", numerator, mod);

//Euclid's Algorithm
int gcd(int a, int b)
{
  if(b == 0)
    return a;
  else
    return gcd(b, a % b);
}

I haven't tested this though. But I did it about a year and a half ago.

[turby]

[QUOTE=PvtCupcakes;19545681]Or if you want some fun just divide the fractions and add them to get a decimal.
(1/2) + (1/4) = 0.75
Then convert 0.75 back to a fraction.[/QUOTE]

All fun and games until binary floating point loses it's precision

[PvtCupcakes]

[QUOTE=turby;19546174]All fun and games until binary floating point loses it's precision[/QUOTE]

Thats true, but it usually loses it precision by sticking a 1 or something in the last decimal place which I think is the 6th place. So you get a number like 1.000001

If you make that function I posted only accurate to the first 5 decimals (multiply by 10^5) you shouldn't have that problem because it'll just truncate that 1.

[nullsquared]

[QUOTE=PvtCupcakes;19546344]Thats true, but it usually loses it precision by sticking a 1 or something in the last decimal place which I think is the 6th place. So you get a number like 1.000001

If you make that function I posted only accurate to the first 5 decimals (multiply by 10^5) you shouldn't have that problem because it'll just truncate that 1.[/QUOTE]

Or, you can just user integer addition and multiplication for the fractions and have practically infinite precision :downs: