I want to use KDE with the Gaussian Kernel. If I'm correct, the sum of all f(x) must be 1 ( ~ rounding ) ?
My Implementation looks like this:
float K( float const& val)
{
const float p=1.0 / std::sqrt( 2.0 * M_PI);
float result = 0.5 * (val*val);
result = p * std::exp(- result);
return result;
};
std::vector< std::pair<float, float> kde( float *val, int len float h)
{
std::vector< std::pair<float, float>> density( len );
const float p = 1.0 / (h * len );
for(int r=0;r<len;r++)
{
float sum = 0;
for(int i=0;i<len;i++)
sum += k( (val[r] - val[i]) / h );
density[r] = std::make_pair( val[r], p*sum );
}
return density;
}
And I choosed h > 0. Am i right that p*sum is the probability for the value val[r] ? The sum over all probability is > 1 ( but looks ok for me ).
You misinterpreted the assumptions on the probability density here. The density integrates to one, whereas its values at certain points are definitely not 1.
Let's discuss it using the following formula from the linked Wikipedia article which you seem to use:
This formula provides the density f_h(x) evaluated at point x.
From my review, your code correctly evaluates this quantity. Yet, you misinterpreted the quantity which should be one. As a density, the integral over the complete space should yield one, i.e.
This property is called normalization of the density.
Moreover, being a density itself, each summand of f_h(x) should yield 1/n when integrated over the whole space, when one also includes the normalization constant. Again, there's no guarantee on the values of the summands.
In one dimension, you can easily confirm the normalization by using the trapezoidal rule or another quadrature scheme (--if you provide a working example, I can try to do that.)
Related
I have such a function that calculates weights according to Gaussian distribution:
const float dx = 1.0f / static_cast<float>(points - 1);
const float sigma = 1.0f / 3.0f;
const float norm = 1.0f / (sqrtf(2.0f * static_cast<float>(M_PI)) * sigma);
const float divsigma2 = 0.5f / (sigma * sigma);
m_weights[0] = 1.0f;
for (int i = 1; i < points; i++)
{
float x = static_cast<float>(i)* dx;
m_weights[i] = norm * expf(-x * x * divsigma2) * dx;
m_weights[0] -= 2.0f * m_weights[i];
}
In all the calc above the number does not matter. The only thing matters is that m_weights[0] = 1.0f; and each time I calculate m_weights[i] I subtract it twice from m_weights[0] like this:
m_weights[0] -= 2.0f * m_weights[i];
to ensure that w[0] + 2 * w[i] (1..N) will sum to exactly 1.0f. But it does not. This assert fails:
float wSum = 0.0f;
for (size_t i = 0; i < m_weights.size(); ++i)
{
float w = m_weights[i];
if (i == 0) {
wSum += w;
} else {
wSum += (w + w);
}
}
assert(wSum == 1.0 && "Weights sum is not 1.");
How can I ensure the sum to be 1.0f on all platforms?
You can't. Floating point isn't like that. Even adding the same values can produce different results according to the cpu used.
All you can do is define some accuracy value and ensure that you end up with 1.0 +/- that value.
See: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
Because the precision of float is only 23 bits (see e.g. https://en.wikipedia.org/wiki/Single-precision_floating-point_format ), rounding error quickly accumulates therefore even if the rest of code is correct, your sum becomes something like 1.0000001 or 0.9999999 (have you watched it in the debugger or tried to print it to console, by the way?). To improve precision you can replace float with double, but still the sum will not be exactly 1.0: the error will just be smaller, something like 1e-16 instead of 1e-7.
The second thing to do is to replace strict comparison to 1.0 with a range comparison, like:
assert(fabs(wSum - 1.0) <= 1e-13 && "Weights sum is not 1.");
Here 1e-13 is the epsilon within which you consider two floating-point numbers equal. If you choose to go with float (not double), you may need epsilon like 1e-6 .
Depending on how large your weights are and how many points there are, accumulated error can become larger than that epsilon. In that case you would need special algorithms for keeping the precision higher, such as sorting the numbers by their absolute values prior to summing them up starting with the smallest numbers.
How can I ensure the sum to be 1.0f on all platforms?
As the other answers (and comments) have stated, you can't achieve this, due to the inexactness of floating point calculations.
One solution is that, instead of using double, use a fixed point or multi-precision library such as GMP, Boost Multiprecision Library, or one of the many others out there.
Here is a part of my code:
double tmp = OP.innerProduct(OQ);
double tmp2 = -1;
and the value of tmp and tmp2 is: (in binary)
tmp = 0b1011111111110000000000000000000000000000000000000000000000000001
tmp2= 0b1011111111110000000000000000000000000000000000000000000000000000
If I used acos(tmp), it will return nan.
I don't want the nan value, and I would like to ignore the small error to keep tmp in the range [-1,1].
How to do so?
EDIT:
I have two points given in spherical coordinate. ( for example, (r,45,45) (r,225,-45) )
Then I need to change them to cartesian coordinate. (a small error occur here!)
Then I want to compute the angle between two points.
The analytical solution is different to computer solution(since the small error).
I would like to make the two solutions same.
Are you trying to prevent branching? I usually make a little helper when I'm doing anything like this:
template<typename T>
inline T Clamp( T val, T low, T high ) {
return val < low ? low : (val > high ? high : val);
}
And then:
double result = acos( Clamp(tmp, -1.0, 1.0) );
If you're trying to write highly optimized code without branching, this won't help. Depending on the accuracy you require, you might consider making an acos lookup table and just put an extra value at each end to handle error-induced overflow.
[edit] I've just had a play around with a [-1,1] clamp without branching. Of course, this only cures inaccuracies. If you call it with a number that is grossly outside the range, it will bomb:
inline double safer_acos (double val)
{
double vals[] = {-1.0, val, val, 1.0};
return acos( vals[int(2.0 + val)] );
}
From this question: Random number generator which gravitates numbers to any given number in range? I did some research since I've come across such a random number generator before. All I remember was the name "Mueller", so I guess I found it, here:
Box-Mueller transform
I can find numerous implementations of it in other languages, but I can't seem to implement it correctly in C#.
This page, for instance, The Box-Muller Method for Generating Gaussian Random Numbers says that the code should look like this (this is not C#):
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
double gaussian(void)
{
static double v, fac;
static int phase = 0;
double S, Z, U1, U2, u;
if (phase)
Z = v * fac;
else
{
do
{
U1 = (double)rand() / RAND_MAX;
U2 = (double)rand() / RAND_MAX;
u = 2. * U1 - 1.;
v = 2. * U2 - 1.;
S = u * u + v * v;
} while (S >= 1);
fac = sqrt (-2. * log(S) / S);
Z = u * fac;
}
phase = 1 - phase;
return Z;
}
Now, here's my implementation of the above in C#. Note that the transform produces 2 numbers, hence the trick with the "phase" above. I simply discard the second value and return the first.
public static double NextGaussianDouble(this Random r)
{
double u, v, S;
do
{
u = 2.0 * r.NextDouble() - 1.0;
v = 2.0 * r.NextDouble() - 1.0;
S = u * u + v * v;
}
while (S >= 1.0);
double fac = Math.Sqrt(-2.0 * Math.Log(S) / S);
return u * fac;
}
My question is with the following specific scenario, where my code doesn't return a value in the range of 0-1, and I can't understand how the original code can either.
u = 0.5, v = 0.1
S becomes 0.5*0.5 + 0.1*0.1 = 0.26
fac becomes ~3.22
the return value is thus ~0.5 * 3.22 or ~1.6
That's not within 0 .. 1.
What am I doing wrong/not understanding?
If I modify my code so that instead of multiplying fac with u, I multiply by S, I get a value that ranges from 0 to 1, but it has the wrong distribution (seems to have a maximum distribution around 0.7-0.8 and then tapers off in both directions.)
Your code is fine. Your mistake is thinking that it should return values exclusively within [0, 1]. The (standard) normal distribution is a distribution with nonzero weight on the entire real line. That is, values outside of [0, 1] are possible. In fact, values within [-1, 0] are just as likely as values within [0, 1], and moreover, the complement of [0, 1] has about 66% of the weight of the normal distribution. Therefore, 66% of the time we expect a value outside of [0, 1].
Also, I think this is not the Box-Mueller transform, but is actually the Marsaglia polar method.
I am no mathematician, or statistician, but if I think about this I would not expect a Gaussian distribution to return numbers in an exact range. Given your implementation the mean is 0 and the standard deviation is 1 so I would expect values distributed on the bell curve with 0 at the center and then reducing as the numbers deviate from 0 on either side. So the sequence would definitely cover both +/- numbers.
Then since it is statistical, why would it be hard limited to -1..1 just because the std.dev is 1? There can statistically be some play on either side and still fulfill the statistical requirement.
The uniform random variate is indeed within 0..1, but the gaussian random variate (which is what Box-Muller algorithm generates) can be anywhere on the real line. See wiki/NormalDistribution for details.
I think the function returns polar coordinates. So you need both values to get correct results.
Also, Gaussian distribution is not between 0 .. 1. It can easily end up as 1000, but probability of such occurrence is extremely low.
This is a monte carlo method so you can't clamp the result, but what you can do is ignore samples.
// return random value in the range [0,1].
double gaussian_random()
{
double sigma = 1.0/8.0; // or whatever works.
while ( 1 ) {
double z = gaussian() * sigma + 0.5;
if (z >= 0.0 && z <= 1.0)
return z;
}
}
I used to work with MATLAB, and for the question I raised I can use p = polyfit(x,y,1) to estimate the best fit line for the scatter data in a plate. I was wondering which resources I can rely on to implement the line fitting algorithm with C++. I understand there are a lot of algorithms for this subject, and for me I expect the algorithm should be fast and meantime it can obtain the comparable accuracy of polyfit function in MATLAB.
This page describes the algorithm easier than Wikipedia, without extra steps to calculate the means etc. : http://faculty.cs.niu.edu/~hutchins/csci230/best-fit.htm . Almost quoted from there, in C++ it's:
#include <vector>
#include <cmath>
struct Point {
double _x, _y;
};
struct Line {
double _slope, _yInt;
double getYforX(double x) {
return _slope*x + _yInt;
}
// Construct line from points
bool fitPoints(const std::vector<Point> &pts) {
int nPoints = pts.size();
if( nPoints < 2 ) {
// Fail: infinitely many lines passing through this single point
return false;
}
double sumX=0, sumY=0, sumXY=0, sumX2=0;
for(int i=0; i<nPoints; i++) {
sumX += pts[i]._x;
sumY += pts[i]._y;
sumXY += pts[i]._x * pts[i]._y;
sumX2 += pts[i]._x * pts[i]._x;
}
double xMean = sumX / nPoints;
double yMean = sumY / nPoints;
double denominator = sumX2 - sumX * xMean;
// You can tune the eps (1e-7) below for your specific task
if( std::fabs(denominator) < 1e-7 ) {
// Fail: it seems a vertical line
return false;
}
_slope = (sumXY - sumX * yMean) / denominator;
_yInt = yMean - _slope * xMean;
return true;
}
};
Please, be aware that both this algorithm and the algorithm from Wikipedia ( http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_line ) fail in case the "best" description of points is a vertical line. They fail because they use
y = k*x + b
line equation which intrinsically is not capable to describe vertical lines. If you want to cover also the cases when data points are "best" described by vertical lines, you need a line fitting algorithm which uses
A*x + B*y + C = 0
line equation. You can still modify the current algorithm to produce that equation:
y = k*x + b <=>
y - k*x - b = 0 <=>
B=1, A=-k, C=-b
In terms of the above code:
B=1, A=-_slope, C=-_yInt
And in "then" block of the if checking for denominator equal to 0, instead of // Fail: it seems a vertical line, produce the following line equation:
x = xMean <=>
x - xMean = 0 <=>
A=1, B=0, C=-xMean
I've just noticed that the original article I was referring to has been deleted. And this web page proposes a little different formula for line fitting: http://hotmath.com/hotmath_help/topics/line-of-best-fit.html
double denominator = sumX2 - 2 * sumX * xMean + nPoints * xMean * xMean;
...
_slope = (sumXY - sumY*xMean - sumX * yMean + nPoints * xMean * yMean) / denominator;
The formulas are identical because nPoints*xMean == sumX and nPoints*xMean*yMean == sumX * yMean == sumY * xMean.
I would suggest coding it from scratch. It is a very simple implementation in C++. You can code up both the intercept and gradient for least-squares fit (the same method as polyfit) from your data directly from the formulas here
http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_line
These are closed form formulas that you can easily evaluate yourself using loops. If you were using higher degree fits then I would suggest a matrix library or more sophisticated algorithms but for simple linear regression as you describe above this is all you need. Matrices and linear algebra routines would be overkill for such a problem (in my opinion).
Equation of line is Ax + By + C=0.
So it can be easily( when B is not so close to zero ) convert to y = (-A/B)*x + (-C/B)
typedef double scalar_type;
typedef std::array< scalar_type, 2 > point_type;
typedef std::vector< point_type > cloud_type;
bool fit( scalar_type & A, scalar_type & B, scalar_type & C, cloud_type const& cloud )
{
if( cloud.size() < 2 ){ return false; }
scalar_type X=0, Y=0, XY=0, X2=0, Y2=0;
for( auto const& point: cloud )
{ // Do all calculation symmetric regarding X and Y
X += point[0];
Y += point[1];
XY += point[0] * point[1];
X2 += point[0] * point[0];
Y2 += point[1] * point[1];
}
X /= cloud.size();
Y /= cloud.size();
XY /= cloud.size();
X2 /= cloud.size();
Y2 /= cloud.size();
A = - ( XY - X * Y ); //!< Common for both solution
scalar_type Bx = X2 - X * X;
scalar_type By = Y2 - Y * Y;
if( fabs( Bx ) < fabs( By ) ) //!< Test verticality/horizontality
{ // Line is more Vertical.
B = By;
std::swap(A,B);
}
else
{ // Line is more Horizontal.
// Classical solution, when we expect more horizontal-like line
B = Bx;
}
C = - ( A * X + B * Y );
//Optional normalization:
// scalar_type D = sqrt( A*A + B*B );
// A /= D;
// B /= D;
// C /= D;
return true;
}
You can also use or go over this implementation there is also documentation here.
Fitting a Line can be acomplished in different ways.
Least Square means minimizing the sum of the squared distance.
But you could take another cost function as example the (not squared) distance. But normaly you use the squred distance (Least Square).
There is also a possibility to define the distance in different ways. Normaly you just use the "y"-axis for the distance. But you could also use the total/orthogonal distance. There the distance is calculated in x- and y-direction. This can be a better fit if you have also errors in x direction (let it be the time of measurment) and you didn't start the measurment on the exact time you saved in the data. For Least Square and Total Least Square Line fit exist algorithms in closed form. So if you fitted with one of those you will get the line with the minimal sum of the squared distance to the datapoints. You can't fit a better line in the sence of your defenition. You could just change the definition as examples taking another cost function or defining distance in another way.
There is a lot of stuff about fitting models into data you could think of, but normaly they all use the "Least Square Line Fit" and you should be fine most times. But if you have a special case it can be necessary to think about what your doing. Taking Least Square done in maybe a few minutes. Thinking about what Method fits you best to the problem envolves understanding the math, which can take indefinit time :-).
Note: This answer is NOT AN ANSWER TO THIS QUESTION but to this one "Line closest to a set of points" that has been flagged as "duplicate" of this one (incorrectly in my opinion), no way to add new answers to it.
The question asks for:
Find the line whose distance from all the points is minimum ? By
distance I mean the shortest distance between the point and the line.
The most usual interpretation of distance "between the point and the line" is the euclidean distance and the most common interpretation of "from all points" is the sum of distances (in absolute or squared value).
When the target is minimize the sum of squared euclidean distances, the linear regression (LST) is not the algorithm to use. In addition, linear regression can not result in a vertical line. The algorithm to be used is the "total least squares". See by example wikipedia for the problem description and this answer in math stack exchange for details about the formulation.
to fit a line y=param[0]x+param[1] simply do this:
// loop over data:
{
sum_x += x[i];
sum_y += y[i];
sum_xy += x[i] * y[i];
sum_x2 += x[i] * x[i];
}
// means
double mean_x = sum_x / ninliers;
double mean_y = sum_y / ninliers;
float varx = sum_x2 - sum_x * mean_x;
float cov = sum_xy - sum_x * mean_y;
// check for zero varx
param[0] = cov / varx;
param[1] = mean_y - param[0] * mean_x;
More on the topic http://easycalculation.com/statistics/learn-regression.php
(formulas are the same, they just multiplied and divided by N, a sample sz.). If you want to fit plane to 3D data use a similar approach -
http://www.mymathforum.com/viewtopic.php?f=13&t=8793
Disclaimer: all quadratic fits are linear and optimal in a sense that they reduce the noise in parameters. However, you might interested in the reducing noise in the data instead. You might also want to ignore outliers since they can bia s your solutions greatly. Both problems can be solved with RANSAC. See my post at:
Actually I am having several questions related to the subject given in the topic title.
I am already using Perlin functions to create lightning in my application, but I am not totally happy about my implementation.
The following questions are based on the initial and the improved Perlin noise implementations.
To simplify the issue, let's assume I am creating a simple 2D lightning by modulating the height of a horizontal line consisting of N nodes at these nodes using a 1D Perlin function.
As far as I have understood, two subsequent values passed to the Perlin function must differ by at least one, or the resulting two values will be identical. That is because with the simple Perlin implementation, the Random function works with an int argument, and in the improved implementation values are mapped to [0..255] and are then used as index into an array containing the values [0..255] in a random distribution. Is that right?
How do I achieve that the first and the last offset value (i.e. for nodes 0 and N-1) returned by the Perlin function is always 0 (zero)? Right now I am modulation a sine function (0 .. Pi) with my Perlin function to achieve that, but that's not really what I want. Just setting them to zero is not what I want, since I want a nice lightning path w/o jaggies at its ends.
How do I vary the Perlin function (so that I would get two different paths I could use as animation start and end frames for the lightning)? I could of course add a fixed random offset per path calculation to each node value, or use a differently setup permutation table for improved Perlin noise, but are there better options?
That depends on how you implement it and sample from it. Using multiple octaves helps counter integers quite a bit.
The octaves and additional interpolation/sampling done for each provides much of the noise in perlin noise. In theory, you should not need to use different integer positions; you should be able to sample at any point and it will be similar (but not always identical) to nearby values.
I would suggest using the perlin as a multiplier instead of simply additive, and use a curve over the course of the lightning. For example, having perlin in the range [-1.5, 1.5] and a normal curve over the lightning (0 at both ends, 1 in the center), lightning + (perlin * curve) will keep your ends points still. Depending on how you've implemented your perlin noise generator, you may need something like:
lightning.x += ((perlin(lightning.y, octaves) * 2.0) - 0.5) * curve(lightning.y);
if perlin returns [0,1] or
lightning.x += (perlin(lightning.y, octaves) / 128.0) * curve(lightning.y);
if it returns [0, 255]. Assuming lightning.x started with a given value, perhaps 0, that would give a somewhat jagged line that still met the original start and end points.
Add a dimension to the noise for every dimension you add to the lightning. If you're modifying the lightning in one dimension (horizontal jagged), you need 1D perlin noise. If you want to animate it, you need 2D. If you wanted lightning that was jagged on two axis and animated, you'd need 3D noise, and so on.
After reading peachykeen's answer and doing some (more) own research in the internet, I have found the following solution to work for me.
With my implementation of Perlin noise, using a value range of [0.0 .. 1.0] for the lightning path nodes work best, passing the value (double) M / (double) N for node M to the Perlin noise function.
To have a noise function F' return the same value for node 0 and node N-1, the following formula can be applied: F'(M) = ((M - N) * F(N) + N * F (N - M)) / M. In order to have the lightning path offsets begin and end with 0, you simply need to subtract F'(0) from all lightning path offsets after having computed the path.
To randomize the lightning path, before computing the offsets for each path node, a random offset R can be computed and added to the values passed to the noise function, so that a node's offset O = F'(N+R). To animate a lightning, two lightning paths need to be computed (start and end frame), and then each path vertex has to be lerped between its start and end position. Once the end frame has been reached, the end frame becomes the start frame and a new end frame is computed. For a 3D path, for each path node N two offset vectors can be computed that are perpendicular to the path at node N and each other, and can be scaled with two 1D Perlin noise values to lerp the node position from start to end frame position. That may be cheaper than doing 3D Perlin noise and works quite well in my application.
Here is my implementation of standard 1D Perlin noise as a reference (some stuff is virtual because I am using this as base for improved Perlin noise, allowing to use standard or improved Perlin noise in a strategy pattern application. The code has been simplified somewhat as well to make it more concise for publishing it here):
Header file:
#ifndef __PERLIN_H
#define __PERLIN_H
class CPerlin {
private:
int m_randomize;
protected:
double m_amplitude;
double m_persistence;
int m_octaves;
public:
virtual void Setup (double amplitude, double persistence, int octaves, int randomize = -1);
double ComputeNoise (double x);
protected:
double LinearInterpolate (double a, double b, double x);
double CosineInterpolate (double a, double b, double x);
double CubicInterpolate (double v0, double v1, double v2, double v3, double x);
double Noise (int v);
double SmoothedNoise (int x);
virtual double InterpolatedNoise (double x);
};
#endif //__PERLIN_H
Implementation:
#include <math.h>
#include <stdlib.h>
#include "perlin.h"
#define INTERPOLATION_METHOD 1
#ifndef Pi
# define Pi 3.141592653589793240
#endif
inline double CPerlin::Noise (int n) {
n = (n << 13) ^ n;
return 1.0 - ((n * (n * n * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0;
}
double CPerlin::LinearInterpolate (double a, double b, double x) {
return a * (1.0 - x) + b * x;
}
double CPerlin::CosineInterpolate (double a, double b, double x) {
double f = (1.0 - cos (x * Pi)) * 0.5;
return a * (1.0 - f) + b * f;
}
double CPerlin::CubicInterpolate (double v0, double v1, double v2, double v3, double x) {
double p = (v3 - v2) - (v0 - v1);
double x2 = x * x;
return v1 + (v2 - v0) * x + (v0 - v1 - p) * x2 + p * x2 * x;
}
double CPerlin::SmoothedNoise (int v) {
return Noise (v) / 2 + Noise (v-1) / 4 + Noise (v+1) / 4;
}
int FastFloor (double v) { return (int) ((v < 0) ? v - 1 : v; }
double CPerlin::InterpolatedNoise (double v) {
int i = FastFloor (v);
double v1 = SmoothedNoise (i);
double v2 = SmoothedNoise (i + 1);
#if INTERPOLATION_METHOD == 2
double v0 = SmoothedNoise (i - 1);
double v3 = SmoothedNoise (i + 2);
return CubicInterpolate (v0, v1, v2, v3, v - i);
#elif INTERPOLATION_METHOD == 1
return CosineInterpolate (v1, v2, v - i);
#else
return LinearInterpolate (v1, v2, v - i);
#endif
}
double CPerlin::ComputeNoise (double v) {
double total = 0, amplitude = m_amplitude, frequency = 1.0;
v += m_randomize;
for (int i = 0; i < m_octaves; i++) {
total += InterpolatedNoise (v * frequency) * amplitude;
frequency *= 2.0;
amplitude *= m_persistence;
}
return total;
}
void CPerlin::Setup (double amplitude, double persistence, int octaves, int randomize) {
m_amplitude = (amplitude > 0.0) ? amplitude : 1.0;
m_persistence = (persistence > 0.0) ? persistence : 2.0 / 3.0;
m_octaves = (octaves > 0) ? octaves : 6;
m_randomize = (randomize < 0) ? (rand () * rand ()) & 0xFFFF : randomize;
}