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How to convert floats to human-readable fractions
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/* | |
** 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])); | |
} |
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How to convert floats to human-readable fractions?
https://stackoverflow.com/questions/95727/how-to-convert-floats-to-human-readable-fractions