Inverse Square Root Algorithm 02:09 为什么求根号?因为要求归一化的向量然后做物理引擎模拟。2024-5-25 17:41:37 02:35 求这个很简单。 求这个很困难! Approximation也行啊! 04:07 The three steps. Fixed points vs floating points. Fixed points is horrible.
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The motivation to try such an algorithm is more clearly explained in Eberly [4], where he assumes the shift creates a linear interpolation to the inverse square root. Note there are several ways to speed up this code, but this note will not go into further optimizations. There are also ...
Fast Inverse Square Root(快速倒数平方根)是一种算法,用于快速计算一个数的倒数平方根。该算法最早出现在Quake III Arena游戏引擎中,用于在计算机图形学中加速向量的归一化过程。 Fast Inverse Square Root算法的中文名称可以直译为"快速倒数平方根"。 今天看到一个很有意思的算法, 是关于快速计算1/x的. 很奇怪啊...
FASTINVERSESQUAREROOT CHRISLOMONT Abstract.Computingreciprocalsquarerootsisnecessaryinmany applications,suchasvectornormalizationinvideogames.Often, somelossofprecisionisacceptableforalargeincreaseinspeed. Thisnoteexaminesandimprovesafastmethodfoundinsource- codeforseveralonlinelibraries,andprovidestheideastoderive similar...
This note examines and improves a fast method found in sourcecode for several online libraries, and provides the ideas to derive similar methods for other functions. 1 1.Chris LomontTechC. Lomont, Fast inverse square root, Tech. Rep., Department of Mathe- matics, Purdue University, West ...
Yowza! Somehow, this code gets 1/x using only multiplication and bit-shift operations. There's no division or exponents involved -- how does it work?My Understanding: This incredible hack estimates the inverse root using Newton's method of approximation, and starts with a great initial guess....
How to edit this code to get regualr square root instead of inverse square root? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 floatq_qsqrt(floatnumber) {longi;floatx2,y;constfloatthreehalfs = 1.5f; x2 = number * 0.5f; y = number; i = * (long* ) &y; i = 0x5f3759df -...
首页 翻译 英文校对 背单词 词霸下载 用户反馈 专栏平台 登录 翻译 Fast-Inverse-Square-Root 翻译 快速逆平方根 以上结果来自机器翻译。 释义
Fast inverse square root floatInvSqrt(floatx) {floatxhalf =0.5f*x;inti = *(int*)&x;//get bits for floating valuei =0x5f3759df- (i>>1);//gives initial guess y0x = *(float*)&i;//convert bits back to floatx = x*(1.5f-xhalf*x*x);//Newton step, repeating increases ...