inverse

Inverse sqrt for fixed point

由 ≡放荡痞女 提交于 2019-12-09 18:20:28
问题 I am looking for the best inverse square root algorithm for fixed point 16.16 numbers. The code below is what I have so far(but basically it takes the square root and divides by the original number, and I would like to get the inverse square root without a division). If it changes anything, the code will be compiled for armv5te. uint32_t INVSQRT(uint32_t n) { uint64_t op, res, one; op = ((uint64_t)n<<16); res = 0; one = (uint64_t)1 << 46; while (one > op) one >>= 2; while (one != 0) { if (op

Rails: :inverse_of and Association extensions

由 浪尽此生 提交于 2019-12-08 17:15:48
问题 I have the following set-up class Player < ActiveRecord::Base has_many :cards, :inverse_of => :player do def in_hand find_all_by_location('hand') end end end class Card < ActiveRecord::Base belongs_to :player, :inverse_of => :cards end This means the following works: p = Player.find(:first) c = p.cards[0] p.score # => 2 c.player.score # => 2 p.score += 1 c.player.score # => 3 c.player.score += 2 p.score # => 5 But the following doesn't behave the same way: p = Player.find(:first) c = p.cards

Java inverse modulo 2**64

由 ☆樱花仙子☆ 提交于 2019-12-07 06:22:56
问题 Given an odd long x , I'm looking for long y such that their product modulo 2**64 (i.e., using the normal overflowing arithmetic) equals to 1. To make clear what I mean: This could be computed in a few thousand year this way: for (long y=1; ; y+=2) { if (x*y == 1) return y; } I know that this can be solved quickly using the extended Euclidean algorithm, but it requires the ability to represent all the involved numbers (ranging up to 2**64 , so even unsigned arithmetic wouldn't help). Using

How to do inverse real to real FFT in FFTW library

由 久未见 提交于 2019-12-07 05:19:39
问题 I'm trying to do some filtering with FFT. I'm using r2r_1d plan and I have no idea how to do the inverse transform... void PerformFiltering(double* data, int n) { /* FFT */ double* spectrum = new double[n]; fftw_plan plan; plan = fftw_plan_r2r_1d(n, data, spectrum, FFTW_REDFT00, FFTW_ESTIMATE); fftw_execute(plan); // signal to spectrum fftw_destroy_plan(plan); /* some filtering here */ /* Inverse FFT */ plan = fftw_plan_r2r_1d(n, spectrum, data, FFTW_REDFT00, FFTW_ESTIMATE); fftw_execute(plan

Reverse Spectrogram A La Aphex Twin in MATLAB

由 心已入冬 提交于 2019-12-05 20:54:10
问题 I'm trying to convert an image into an audio signal in MATLAB by treating it as a spectrogram as in Aphex Twin's song on Windowlicker. Unfortunately, I'm having trouble getting a result. Here it what I have at the moment: function signal = imagetosignal(path, format) % Read in the image and make it symmetric. image = imread(path, format); image = [image; flipud(image)]; [row, column] = size(image); signal = []; % Take the ifft of each column of pixels and piece together the real-valued

sympy.solve() doesn't give one of the solutions with LambertW

由 ℡╲_俬逩灬. 提交于 2019-12-05 19:23:11
Background: I am trying to implement a function doing an inverse transform sampling . I use sympy for calculating CDF and getting its inverse function. While for some simple PDFs I get correct results, for a PDF which CDF's inverse function includes Lambert-W function , results are wrong. Example: Consider following example CDF: import sympy as sym y = sym.Symbol('y') cdf = (-y - 1) * sym.exp(-y) + 1 # derived from `pdf = x * sym.exp(-x)` sym.plot(cdf, (y, -1, 5)) Now calculating inverse of this function: x = sym.Symbol('x') inverse = sym.solve(sym.Eq(x, cdf), y) print(inverse) Output: [

Inverse Error Function in C

由 偶尔善良 提交于 2019-12-05 17:22:35
问题 Is it possible to calculate the inverse error function in C? I can find erf(x) in <math.h> which calculates the error function, but I can't find anything to do the inverse. 回答1: Quick & dirty, tolerance under +-6e-3. Work based on "A handy approximation for the error function and its inverse" by Sergei Winitzki. C/C++ CODE: float myErfInv2(float x){ float tt1, tt2, lnx, sgn; sgn = (x < 0) ? -1.0f : 1.0f; x = (1 - x)*(1 + x); // x = 1 - x*x; lnx = logf(x); tt1 = 2/(PI*0.147) + 0.5f * lnx; tt2

Java inverse modulo 2**64

由 谁都会走 提交于 2019-12-05 10:41:04
Given an odd long x , I'm looking for long y such that their product modulo 2**64 (i.e., using the normal overflowing arithmetic) equals to 1. To make clear what I mean: This could be computed in a few thousand year this way: for (long y=1; ; y+=2) { if (x*y == 1) return y; } I know that this can be solved quickly using the extended Euclidean algorithm, but it requires the ability to represent all the involved numbers (ranging up to 2**64 , so even unsigned arithmetic wouldn't help). Using BigInteger would surely help, but I wonder if there's a simpler way, possibly using the extended

How to do inverse real to real FFT in FFTW library

由 别等时光非礼了梦想. 提交于 2019-12-05 08:24:15
I'm trying to do some filtering with FFT. I'm using r2r_1d plan and I have no idea how to do the inverse transform... void PerformFiltering(double* data, int n) { /* FFT */ double* spectrum = new double[n]; fftw_plan plan; plan = fftw_plan_r2r_1d(n, data, spectrum, FFTW_REDFT00, FFTW_ESTIMATE); fftw_execute(plan); // signal to spectrum fftw_destroy_plan(plan); /* some filtering here */ /* Inverse FFT */ plan = fftw_plan_r2r_1d(n, spectrum, data, FFTW_REDFT00, FFTW_ESTIMATE); fftw_execute(plan); // spectrum to signal (inverse FFT) fftw_destroy_plan(plan); } Am I doing all the things right? I'm

Hibernate Inverse attribute

由 这一生的挚爱 提交于 2019-12-04 08:38:56
问题 I am creating a one-to-many relationship. so, i have a parent and a child. The cascade attribute is set to all. I was wondering, if we consider the following piece of code: Parent p = (Parent) session.load(Parent.class, pid); Child c = new Child("child element"); p.addChild(c); session.flush(); Q1) If the parent owns the relationship, as in , for the parent inverse=false, then would the child element addition be updated in teh database? Q2) If the child owns the relationship, as in , for the
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