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
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. IEEE-754 Standard. 06:16 This is wasteful...
FAST INVERSE SQUARE ROOT 3 3. The Algorithm The main idea is Newton approximation, and the magic constant is used to compute a good initial guess. Given a oating point value 1 x > 0, we want to compute √x . Dene f (y) = y12 x. Then the value we seek is the positive root ...
Themotivationtotrysuchanalgorithmismoreclearlyexplained inEberly[4],whereheassumestheshiftcreatesalinearinterpolation totheinversesquareroot.Notethereareseveralwaystospeedup thiscode,butthisnotewillnotgointofurtheroptimizations.There arealsofastermethods,perhapstablelookuptricks,butmostofthem ...
Fast inverse square root - Lomont - 2003 () Citation Context ...e demanding than multiplications. We avoided directly implementing the dividing algorithm by approximating it with additions and multiplications. Our approach is inspired by an algorithm described in =-=[9]-=-, which provides a ...
在这种情况下,在从曲面镜反射的光上。快速反平方根用于将该计算推广到三维空间。 浮点数的倒数平方根用于计算标准化向量。程序可以使用标准化向量来确定入射角和反射角。3D图形程序必须每秒执行数百万次这些计算以模拟光照。当代码在20世纪90年代早期开发时,大多数浮点处理能力都落后于整数处理的速度。在专门用于处理变...
The initial estimate and input operand are scaled to minimize final adjustments to the result mantissa and final exponent adjustments required by the algorithm are performed concurrently with any adjustment required by rounding. A pipelined implementation of the algorithm produces a result with a latency...
fast-inverse-sqrt-node-gyp This package provides a C++ implementation of the fast inverse square root algorithm for Node.js. It's implemented in C++ and runs through node's API, thanks to Node-Gyp feature of Node. This approach provides near-native performance and can be used for high-dema...
and want to find the inverse square root: 1/i. If we make a guess "x" as the inverse root, the error between our original number and our guess "x" is:This is because x is roughly 1/i. If we square x we get 1/i, and if we take the inverse we should get something close ...
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 ...