理性近似

- Logan 对不起Chuck但我认为你在第一次迭代中调用UB 当我们有： num = 1 value = PI 我们有： 1 / PI = 0.31830 .... （int）0.31830 == ZERO 然后你做了 num / approx你有一个divis零离子... jacob 可以试着说他们解决了一个稍微不同的问题， 但是连续分数提供了最佳近似值， 易于编码，并产生令人着迷的结果，例如。 e = 2.71828 ... is { 2; 1,2,1,1,4,1,1,6,1,1,8,1,1,10，...] 继续分数形式，并且 黄金比例为[1; 1,1,1,1,1,1,1，...]。 （此cf黄金比率的形式意味着它是真实数字 最难接近理性！黄金比例 总线调度算法取决于属性IIRC，虽然大多数comp-sci算法称为黄金比率 算法。不要。） 像往常一样，维基百科似乎提供了一个很好的讨论： http://en.wikipedia.org/wiki/Continued_fraction 不记得如何拼写M_PI，是吗？我也不能，虽然 我通常只拼写拼写：3.141592635897 James Dow Allen Here is an amusing toy, which may be of especial interest to theForthians who use rational fractions to represent real numbers.The handling of criterion may also be interesting to some./* Find best rational approximation to a double *//* by C.B. Falconer, 2006-09-07. Released to public domain */#include <stdio.h>#include <stdlib.h>#include <math.h>#include <float.h>#include <limits.h>#include <errno.h>int main(int argc, char **argv){int num, approx, bestnum, bestdenom;int lastnum = 500;double error, leasterr, value, criterion, tmp;char *eptr;value = 4 * atan(1.0);if (argc 2) lastnum = strtol(argv[2], NULL, 10);if (lastnum <= 0) lastnum = 500;if (argc 1) {tmp = strtod(argv[1], &eptr);if ((0.0 >= tmp) || (tmp INT_MAX) || (ERANGE == errno)) {puts("Invalid number, using PI");}else value = tmp;}criterion = 2 * value * DBL_EPSILON;puts("Usage: ratvalue [number [maxnumerator]]\n""number defaults to PI, maxnumerator to 500");printf("Rational approximation to %.*f\n", DBL_DIG, value);for (leasterr = value, num = 1; num < lastnum; num++) {approx = (int)(num / value + 0.5);error = fabs((double)num / approx - value);if (error < leasterr) {bestnum = num;bestdenom = approx;leasterr = error;printf("%8d / %-8d = %.*f with error %.*f\n",bestnum, bestdenom,DBL_DIG, (double)bestnum / bestdenom,DBL_DIG, leasterr);if (leasterr <= criterion) break;}}return 0;} /* main */--Chuck F (cbfalconer at maineline dot net)Available for consulting/temporary embedded and systems.<http://cbfalconer.home.att.net> 解决方案Actually, doubles are already rational numbers. The denominatoris always power of 2.Now the numbers they approximate, that''s a different story...- LoganExcuse me Chuck but I think you invoke UB in the first iterationwhen we have:num = 1value = PIwe have then:1/PI = 0.31830 ....(int) 0.31830 == ZEROThen you donum/approx and you have a division by zero...jacobOne could try to argue they solve a slightly different problem,but continued fractions provide optimal approximations, areeasy to code, and yield fascinating results, eg.e = 2.71828... is {2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,...] incontinued fraction form, andthe Golden Ratio is [1;1,1,1,1,1,1,1,...].(This c.f. form of Golden Ratio means it is the real numberhardest to approximate by a rational! The Golden Ratiobus scheduling algorithm depends on that property IIRC,though most comp-sci algorithms called "Golden Ratioalgorithm" don''t.)As is so often the case these days, Wikipedia seems tooffer a good discussion: http://en.wikipedia.org/wiki/Continued_fractionCan''t remember how to spell M_PI, eh? Neither can I, thoughI usually just spell it prosaically: 3.141592635897James Dow Allen 这篇关于理性近似的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！