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Need math function (or algorithm) for reducing fractions

Posted on 1998-04-19
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I need to find a C++ math library function for reducing fractions, but have not been able to find one. Is there one available? If not, can you help me with an algortihm to do this i.e. how to get an answer of 3/5 when the input was 6/10?
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Question by:englm
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q2guo earned 150 total points
ID: 1162143
The following function is not the most efficient funciton for doing fraction reduction.  But its the only one I can come
up with in 2 minutes.

The argument d and n represent denominator and neminator respectively.


void Reduce(int &d, int &n) {
    for (int i = n/2; i > 1; i--) {
        if ( (n%i) == 0) {
            if ( (d%i) == 0) {
                 n = n / i;
                 d = d / i;
                 break;
            }
        }
    }
}
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Expert Comment

by:ozo
ID: 1162144
//Heres one Euclid came up with over 2 millennia ago
int gcd(int a,int b){
  while( b ){
    int c;
    c = a%b;
    a = b;
    b = c;
  }
  return a;
}

main(int argc,char *argv[]){
   int d,n,f;
   d = atoi(argv[1]);
   n = atoi(argv[2]);
   f = gcd(d,n);
   printf("%d/%d\n",d/f,n/f);
}

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Expert Comment

by:ozo
ID: 1162145
(If n was supposed to represent the numerator, I think I got my variable names backwards)

But it looks like Reduce(4,4) will only reduce to 2/2
and Reduce(4,-4) seems to have an even worse problem.
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Expert Comment

by:Ready4Dis
ID: 1162146
Why (q2guo) do you start at n/2? Shouldn't you check to see which is heigher?  What if the fraction is 10/2.? It will return 10/2 which can be 5/1...
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Expert Comment

by:q2guo
ID: 1162147
Ready4Dis, I assumed that the fraction is a pure one, meaning that the numerator is always smaller that the denominator.
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Expert Comment

by:ozo
ID: 1162148
Then if n is assumed to be smaller than d, shouldn't you have started with d/2?
(I think that's what confused my labels:-) fortunately, the gcd approach doesn't depend such assumptions.
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Expert Comment

by:q2guo
ID: 1162149
ozo: why would you start with d/2?  Since n is smaller than d
all i(s) between d/2 and n/2 are just extra work.
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Expert Comment

by:ozo
ID: 1162150
Given a fraction like 2/10, with the numerator 2 smaller than the denominator 10,
starting i=n/2 i.e. i=1 will fail to reduce it to 1/5
(with a fraction like 10/2, since n=10 and d=2, i=n/2 would start at 5, so Reduce(2,10) would find (10/2)/(2/2) = 5/1)

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Expert Comment

by:q2guo
ID: 1162151
Ok, here is the function again

void Reduce(int &d, int &n) {
        int temp = (d >= n ? d : n);
        for (int i = temp; i > 1; i--) {
            if ( (n%i) == 0) {
                if ( (d%i) == 0) {
                     n = n / i;
                     d = d / i;
                     break;
                }
            }
        }
}
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Expert Comment

by:ozo
ID: 1162152
Still has a problem with -10/2
Here's my (Euclid's) function again with the variables renamed:

int gcd(int a,int b){
  while( b ){
    int c = a%b;
    a = b;
    b = c;
  }
  return a;
}

main(int argc,char *argv[]){
   int n,d,i;
   n = atoi(argv[1]);
   d = atoi(argv[2]);
   i = gcd(n,d);
   printf("%d/%d\n",n/i,d/i);
}


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