Question

Let’s say we have 0.33, we need to output 1/3.
If we have 0.4, we need to output 2/5.

The idea is to make it human-readable to make the user understand ‘x parts out of y‘ as a better way of understanding data.

I know that percentages is a good substitute but I was wondering if there was a simple way to do this?

Answer 1

I have found David Eppstein’s find rational approximation to given real number C code to be exactly what you are asking for. Its based on the theory of continued fractions and very fast and fairly compact.

I have used versions of this customized for specific numerator and denominator limits.

/* ** find rational approximation to given real number ** David Eppstein / UC Irvine / 8 Aug 1993 ** ** With corrections from Arno Formella, May 2008 ** ** usage: a.out r d **   r is real number to approx **   d is the maximum denominator allowed ** ** based on the theory of continued fractions ** if x = a1 + 1/(a2 + 1/(a3 + 1/(a4 + ...))) ** then best approximation is found by truncating this series ** (with some adjustments in the last term). ** ** Note the fraction can be recovered as the first column of the matrix **  ( a1 1 ) ( a2 1 ) ( a3 1 ) ... **  ( 1  0 ) ( 1  0 ) ( 1  0 ) ** Instead of keeping the sequence of continued fraction terms, ** we just keep the last partial product of these matrices. */  #include <stdio.h>  main(ac, av) int ac; char ** av; {     double atof();     int atoi();     void exit();      long m[2][2];     double x, startx;     long maxden;     long ai;      /* read command line arguments */     if (ac != 3) {         fprintf(stderr, "usage: %s r d\n",av[0]);  // AF: argument missing         exit(1);     }     startx = x = atof(av[1]);     maxden = atoi(av[2]);      /* initialize matrix */     m[0][0] = m[1][1] = 1;     m[0][1] = m[1][0] = 0;      /* loop finding terms until denom gets too big */     while (m[1][0] *  ( ai = (long)x ) + m[1][1] <= maxden) {         long t;         t = m[0][0] * ai + m[0][1];         m[0][1] = m[0][0];         m[0][0] = t;         t = m[1][0] * ai + m[1][1];         m[1][1] = m[1][0];         m[1][0] = t;         if(x==(double)ai) break;     // AF: division by zero         x = 1/(x - (double) ai);         if(x>(double)0x7FFFFFFF) break;  // AF: representation failure     }       /* now remaining x is between 0 and 1/ai */     /* approx as either 0 or 1/m where m is max that will fit in maxden */     /* first try zero */     printf("%ld/%ld, error = %e\n", m[0][0], m[1][0],            startx - ((double) m[0][0] / (double) m[1][0]));      /* now try other possibility */     ai = (maxden - m[1][1]) / m[1][0];     m[0][0] = m[0][0] * ai + m[0][1];     m[1][0] = m[1][0] * ai + m[1][1];     printf("%ld/%ld, error = %e\n", m[0][0], m[1][0],            startx - ((double) m[0][0] / (double) m[1][0])); }