You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
164 lines
5.1 KiB
164 lines
5.1 KiB
/* fp2rat.c (convert floating-point number to rational number) */
|
|
|
|
/***********************************************************************
|
|
* This code is part of GLPK (GNU Linear Programming Kit).
|
|
*
|
|
* Copyright (C) 2000, 2013 Andrew Makhorin, Department for Applied
|
|
* Informatics, Moscow Aviation Institute, Moscow, Russia. All rights
|
|
* reserved. E-mail: <mao@gnu.org>.
|
|
*
|
|
* GLPK is free software: you can redistribute it and/or modify it
|
|
* under the terms of the GNU General Public License as published by
|
|
* the Free Software Foundation, either version 3 of the License, or
|
|
* (at your option) any later version.
|
|
*
|
|
* GLPK is distributed in the hope that it will be useful, but WITHOUT
|
|
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
|
|
* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public
|
|
* License for more details.
|
|
*
|
|
* You should have received a copy of the GNU General Public License
|
|
* along with GLPK. If not, see <http://www.gnu.org/licenses/>.
|
|
***********************************************************************/
|
|
|
|
#include "env.h"
|
|
#include "misc.h"
|
|
|
|
/***********************************************************************
|
|
* NAME
|
|
*
|
|
* fp2rat - convert floating-point number to rational number
|
|
*
|
|
* SYNOPSIS
|
|
*
|
|
* #include "misc.h"
|
|
* int fp2rat(double x, double eps, double *p, double *q);
|
|
*
|
|
* DESCRIPTION
|
|
*
|
|
* Given a floating-point number 0 <= x < 1 the routine fp2rat finds
|
|
* its "best" rational approximation p / q, where p >= 0 and q > 0 are
|
|
* integer numbers, such that |x - p / q| <= eps.
|
|
*
|
|
* RETURNS
|
|
*
|
|
* The routine fp2rat returns the number of iterations used to achieve
|
|
* the specified precision eps.
|
|
*
|
|
* EXAMPLES
|
|
*
|
|
* For x = sqrt(2) - 1 = 0.414213562373095 and eps = 1e-6 the routine
|
|
* gives p = 408 and q = 985, where 408 / 985 = 0.414213197969543.
|
|
*
|
|
* BACKGROUND
|
|
*
|
|
* It is well known that every positive real number x can be expressed
|
|
* as the following continued fraction:
|
|
*
|
|
* x = b[0] + a[1]
|
|
* ------------------------
|
|
* b[1] + a[2]
|
|
* -----------------
|
|
* b[2] + a[3]
|
|
* ----------
|
|
* b[3] + ...
|
|
*
|
|
* where:
|
|
*
|
|
* a[k] = 1, k = 0, 1, 2, ...
|
|
*
|
|
* b[k] = floor(x[k]), k = 0, 1, 2, ...
|
|
*
|
|
* x[0] = x,
|
|
*
|
|
* x[k] = 1 / frac(x[k-1]), k = 1, 2, 3, ...
|
|
*
|
|
* To find the "best" rational approximation of x the routine computes
|
|
* partial fractions f[k] by dropping after k terms as follows:
|
|
*
|
|
* f[k] = A[k] / B[k],
|
|
*
|
|
* where:
|
|
*
|
|
* A[-1] = 1, A[0] = b[0], B[-1] = 0, B[0] = 1,
|
|
*
|
|
* A[k] = b[k] * A[k-1] + a[k] * A[k-2],
|
|
*
|
|
* B[k] = b[k] * B[k-1] + a[k] * B[k-2].
|
|
*
|
|
* Once the condition
|
|
*
|
|
* |x - f[k]| <= eps
|
|
*
|
|
* has been satisfied, the routine reports p = A[k] and q = B[k] as the
|
|
* final answer.
|
|
*
|
|
* In the table below here is some statistics obtained for one million
|
|
* random numbers uniformly distributed in the range [0, 1).
|
|
*
|
|
* eps max p mean p max q mean q max k mean k
|
|
* -------------------------------------------------------------
|
|
* 1e-1 8 1.6 9 3.2 3 1.4
|
|
* 1e-2 98 6.2 99 12.4 5 2.4
|
|
* 1e-3 997 20.7 998 41.5 8 3.4
|
|
* 1e-4 9959 66.6 9960 133.5 10 4.4
|
|
* 1e-5 97403 211.7 97404 424.2 13 5.3
|
|
* 1e-6 479669 669.9 479670 1342.9 15 6.3
|
|
* 1e-7 1579030 2127.3 3962146 4257.8 16 7.3
|
|
* 1e-8 26188823 6749.4 26188824 13503.4 19 8.2
|
|
*
|
|
* REFERENCES
|
|
*
|
|
* W. B. Jones and W. J. Thron, "Continued Fractions: Analytic Theory
|
|
* and Applications," Encyclopedia on Mathematics and Its Applications,
|
|
* Addison-Wesley, 1980. */
|
|
|
|
int fp2rat(double x, double eps, double *p, double *q)
|
|
{ int k;
|
|
double xk, Akm1, Ak, Bkm1, Bk, ak, bk, fk, temp;
|
|
xassert(0.0 <= x && x < 1.0);
|
|
for (k = 0; ; k++)
|
|
{ xassert(k <= 100);
|
|
if (k == 0)
|
|
{ /* x[0] = x */
|
|
xk = x;
|
|
/* A[-1] = 1 */
|
|
Akm1 = 1.0;
|
|
/* A[0] = b[0] = floor(x[0]) = 0 */
|
|
Ak = 0.0;
|
|
/* B[-1] = 0 */
|
|
Bkm1 = 0.0;
|
|
/* B[0] = 1 */
|
|
Bk = 1.0;
|
|
}
|
|
else
|
|
{ /* x[k] = 1 / frac(x[k-1]) */
|
|
temp = xk - floor(xk);
|
|
xassert(temp != 0.0);
|
|
xk = 1.0 / temp;
|
|
/* a[k] = 1 */
|
|
ak = 1.0;
|
|
/* b[k] = floor(x[k]) */
|
|
bk = floor(xk);
|
|
/* A[k] = b[k] * A[k-1] + a[k] * A[k-2] */
|
|
temp = bk * Ak + ak * Akm1;
|
|
Akm1 = Ak, Ak = temp;
|
|
/* B[k] = b[k] * B[k-1] + a[k] * B[k-2] */
|
|
temp = bk * Bk + ak * Bkm1;
|
|
Bkm1 = Bk, Bk = temp;
|
|
}
|
|
/* f[k] = A[k] / B[k] */
|
|
fk = Ak / Bk;
|
|
#if 0
|
|
print("%.*g / %.*g = %.*g",
|
|
DBL_DIG, Ak, DBL_DIG, Bk, DBL_DIG, fk);
|
|
#endif
|
|
if (fabs(x - fk) <= eps)
|
|
break;
|
|
}
|
|
*p = Ak;
|
|
*q = Bk;
|
|
return k;
|
|
}
|
|
|
|
/* eof */
|