Distance from point to point: dist = sqrt(dx * dx + dy * dy);
But sqrt is too slow and I can't accept that. I found a method called Taylor McLaughlin Series to estimate the distance of two points on the book. But I can't comprehend the following code. Thanks for anyone who helps me.
#define MIN(a, b) ((a < b) ? a : b)
int FastDistance2D(int x, int y)
{
// This function computes the distance from 0,0 to x,y with 3.5% error
// First compute the absolute value of x, y
x = abs(x);
y = abs(y);
// Compute the minimum of x, y
int mn = MIN(x, y);
// Return the distance
return x + y - (mn >> 1) - (mn >> 2) + (mn >> 4);
}
I have consulted related data about McLaughlin Series, but I still can't comprehend how the return value use McLaughlin Series to estimate the value. Thanks for everyone~
This task is almost duplicate of another one:
Very fast 3D distance check?
And there was link to great article:
http://www.azillionmonkeys.com/qed/sqroot.html
In the article you can find different aproaches for approximation of root. For example maybe this one is suitable for you:
int isqrt (long r) {
float tempf, x, y, rr;
int is;
rr = (long) r;
y = rr*0.5;
*(unsigned long *) &tempf = (0xbe6f0000 - *(unsigned long *) &rr) >> 1;
x = tempf;
x = (1.5*x) - (x*x)*(x*y);
if (r > 101123) x = (1.5*x) - (x*x)*(x*y);
is = (int) (x*rr + 0.5);
return is + ((signed int) (r - is*is)) >> 31;
}
If you can calculate root operation fast, then you can calculate distance in regular way:
return isqrt(a*a+b*b)
And one more link:
http://www.flipcode.com/archives/Fast_Approximate_Distance_Functions.shtml
u32 approx_distance( s32 dx, s32 dy )
{
u32 min, max;
if ( dx < 0 ) dx = -dx;
if ( dy < 0 ) dy = -dy;
if ( dx < dy )
{
min = dx;
max = dy;
} else {
min = dy;
max = dx;
}
// coefficients equivalent to ( 123/128 * max ) and ( 51/128 * min )
return ((( max << 8 ) + ( max << 3 ) - ( max << 4 ) - ( max << 1 ) +
( min << 7 ) - ( min << 5 ) + ( min << 3 ) - ( min << 1 )) >> 8 );
}
You are right sqrt is quite a slow function. But do you really need to compute the distance?
In a lot of cases you can use the distance² instead.
E.g.
If you want to find out what distance is shorter, you can compare the squares of the distance as well as the real distances.
If you want to check if a 100 > distance you can as well check 10000 > distanceSquared
Using the ditance squared in your program instead of the distance you can often avoid calculating the sqrt.
It depends on your application if this is an option for you, but it is always worth to be considered.
Related
I'm looking for some nice C code that will accomplish effectively:
while (deltaPhase >= M_PI) deltaPhase -= M_TWOPI;
while (deltaPhase < -M_PI) deltaPhase += M_TWOPI;
What are my options?
Edit Apr 19, 2013:
Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea:
static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348;
static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696;
// Floating-point modulo
// The result (the remainder) has same sign as the divisor.
// Similar to matlab's mod(); Not similar to fmod() - Mod(-3,4)= 1 fmod(-3,4)= -3
template<typename T>
T Mod(T x, T y)
{
static_assert(!std::numeric_limits<T>::is_exact , "Mod: floating-point type expected");
if (0. == y)
return x;
double m= x - y * floor(x/y);
// handle boundary cases resulted from floating-point cut off:
if (y > 0) // modulo range: [0..y)
{
if (m>=y) // Mod(-1e-16 , 360. ): m= 360.
return 0;
if (m<0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(106.81415022205296 , _TWO_PI ): m= -1.421e-14
}
}
else // modulo range: (y..0]
{
if (m<=y) // Mod(1e-16 , -360. ): m= -360.
return 0;
if (m>0 )
{
if (y+m == y)
return 0 ; // just in case...
else
return y+m; // Mod(-106.81415022205296, -_TWO_PI): m= 1.421e-14
}
}
return m;
}
// wrap [rad] angle to [-PI..PI)
inline double WrapPosNegPI(double fAng)
{
return Mod(fAng + _PI, _TWO_PI) - _PI;
}
// wrap [rad] angle to [0..TWO_PI)
inline double WrapTwoPI(double fAng)
{
return Mod(fAng, _TWO_PI);
}
// wrap [deg] angle to [-180..180)
inline double WrapPosNeg180(double fAng)
{
return Mod(fAng + 180., 360.) - 180.;
}
// wrap [deg] angle to [0..360)
inline double Wrap360(double fAng)
{
return Mod(fAng ,360.);
}
One-liner constant-time solution:
Okay, it's a two-liner if you count the second function for [min,max) form, but close enough — you could merge them together anyways.
/* change to `float/fmodf` or `long double/fmodl` or `int/%` as appropriate */
/* wrap x -> [0,max) */
double wrapMax(double x, double max)
{
/* integer math: `(max + x % max) % max` */
return fmod(max + fmod(x, max), max);
}
/* wrap x -> [min,max) */
double wrapMinMax(double x, double min, double max)
{
return min + wrapMax(x - min, max - min);
}
Then you can simply use deltaPhase = wrapMinMax(deltaPhase, -M_PI, +M_PI).
The solutions is constant-time, meaning that the time it takes does not depend on how far your value is from [-PI,+PI) — for better or for worse.
Verification:
Now, I don't expect you to take my word for it, so here are some examples, including boundary conditions. I'm using integers for clarity, but it works much the same with fmod() and floats:
Positive x:
wrapMax(3, 5) == 3: (5 + 3 % 5) % 5 == (5 + 3) % 5 == 8 % 5 == 3
wrapMax(6, 5) == 1: (5 + 6 % 5) % 5 == (5 + 1) % 5 == 6 % 5 == 1
Negative x:
Note: These assume that integer modulo copies left-hand sign; if not, you get the above ("Positive") case.
wrapMax(-3, 5) == 2: (5 + (-3) % 5) % 5 == (5 - 3) % 5 == 2 % 5 == 2
wrapMax(-6, 5) == 4: (5 + (-6) % 5) % 5 == (5 - 1) % 5 == 4 % 5 == 4
Boundaries:
wrapMax(0, 5) == 0: (5 + 0 % 5) % 5 == (5 + 0) % 5 == 5 % 5 == 0
wrapMax(5, 5) == 0: (5 + 5 % 5) % 5 == (5 + 0) % 5== 5 % 5 == 0
wrapMax(-5, 5) == 0: (5 + (-5) % 5) % 5 == (5 + 0) % 5 == 5 % 5 == 0
Note: Possibly -0 instead of +0 for floating-point.
The wrapMinMax function works much the same: wrapping x to [min,max) is the same as wrapping x - min to [0,max-min), and then (re-)adding min to the result.
I don't know what would happen with a negative max, but feel free to check that yourself!
If ever your input angle can reach arbitrarily high values, and if continuity matters, you can also try
atan2(sin(x),cos(x))
This will preserve continuity of sin(x) and cos(x) better than modulo for high values of x, especially in single precision (float).
Indeed, exact_value_of_pi - double_precision_approximation ~= 1.22e-16
On the other hand, most library/hardware use a high precision approximation of PI for applying the modulo when evaluating trigonometric functions (though x86 family is known to use a rather poor one).
Result might be in [-pi,pi], you'll have to check the exact bounds.
Personaly, I would prevent any angle to reach several revolutions by wrapping systematically and stick to a fmod solution like the one of boost.
There is also fmod function in math.h but the sign causes trouble so that a subsequent operation is needed to make the result fir in the proper range (like you already do with the while's). For big values of deltaPhase this is probably faster than substracting/adding `M_TWOPI' hundreds of times.
deltaPhase = fmod(deltaPhase, M_TWOPI);
EDIT:
I didn't try it intensively but I think you can use fmod this way by handling positive and negative values differently:
if (deltaPhase>0)
deltaPhase = fmod(deltaPhase+M_PI, 2.0*M_PI)-M_PI;
else
deltaPhase = fmod(deltaPhase-M_PI, 2.0*M_PI)+M_PI;
The computational time is constant (unlike the while solution which gets slower as the absolute value of deltaPhase increases)
I would do this:
double wrap(double x) {
return x-2*M_PI*floor(x/(2*M_PI)+0.5);
}
There will be significant numerical errors. The best solution to the numerical errors is to store your phase scaled by 1/PI or by 1/(2*PI) and depending on what you are doing store them as fixed point.
Instead of working in radians, use angles scaled by 1/(2π) and use modf, floor etc. Convert back to radians to use library functions.
This also has the effect that rotating ten thousand and a half revolutions is the same as rotating half then ten thousand revolutions, which is not guaranteed if your angles are in radians, as you have an exact representation in the floating point value rather than summing approximate representations:
#include <iostream>
#include <cmath>
float wrap_rads ( float r )
{
while ( r > M_PI ) {
r -= 2 * M_PI;
}
while ( r <= -M_PI ) {
r += 2 * M_PI;
}
return r;
}
float wrap_grads ( float r )
{
float i;
r = modff ( r, &i );
if ( r > 0.5 ) r -= 1;
if ( r <= -0.5 ) r += 1;
return r;
}
int main ()
{
for (int rotations = 1; rotations < 100000; rotations *= 10 ) {
{
float pi = ( float ) M_PI;
float two_pi = 2 * pi;
float a = pi;
a += rotations * two_pi;
std::cout << rotations << " and a half rotations in radians " << a << " => " << wrap_rads ( a ) / two_pi << '\n' ;
}
{
float pi = ( float ) 0.5;
float two_pi = 2 * pi;
float a = pi;
a += rotations * two_pi;
std::cout << rotations << " and a half rotations in grads " << a << " => " << wrap_grads ( a ) / two_pi << '\n' ;
}
std::cout << '\n';
}}
Here is a version for other people finding this question that can use C++ with Boost:
#include <boost/math/constants/constants.hpp>
#include <boost/math/special_functions/sign.hpp>
template<typename T>
inline T normalizeRadiansPiToMinusPi(T rad)
{
// copy the sign of the value in radians to the value of pi
T signedPI = boost::math::copysign(boost::math::constants::pi<T>(),rad);
// set the value of rad to the appropriate signed value between pi and -pi
rad = fmod(rad+signedPI,(2*boost::math::constants::pi<T>())) - signedPI;
return rad;
}
C++11 version, no Boost dependency:
#include <cmath>
// Bring the 'difference' between two angles into [-pi; pi].
template <typename T>
T normalizeRadiansPiToMinusPi(T rad) {
// Copy the sign of the value in radians to the value of pi.
T signed_pi = std::copysign(M_PI,rad);
// Set the value of difference to the appropriate signed value between pi and -pi.
rad = std::fmod(rad + signed_pi,(2 * M_PI)) - signed_pi;
return rad;
}
I encountered this question when searching for how to wrap a floating point value (or a double) between two arbitrary numbers. It didn't answer specifically for my case, so I worked out my own solution which can be seen here. This will take a given value and wrap it between lowerBound and upperBound where upperBound perfectly meets lowerBound such that they are equivalent (ie: 360 degrees == 0 degrees so 360 would wrap to 0)
Hopefully this answer is helpful to others stumbling across this question looking for a more generic bounding solution.
double boundBetween(double val, double lowerBound, double upperBound){
if(lowerBound > upperBound){std::swap(lowerBound, upperBound);}
val-=lowerBound; //adjust to 0
double rangeSize = upperBound - lowerBound;
if(rangeSize == 0){return upperBound;} //avoid dividing by 0
return val - (rangeSize * std::floor(val/rangeSize)) + lowerBound;
}
A related question for integers is available here:
Clean, efficient algorithm for wrapping integers in C++
A two-liner, non-iterative, tested solution for normalizing arbitrary angles to [-π, π):
double normalizeAngle(double angle)
{
double a = fmod(angle + M_PI, 2 * M_PI);
return a >= 0 ? (a - M_PI) : (a + M_PI);
}
Similarly, for [0, 2π):
double normalizeAngle(double angle)
{
double a = fmod(angle, 2 * M_PI);
return a >= 0 ? a : (a + 2 * M_PI);
}
In the case where fmod() is implemented through truncated division and has the same sign as the dividend, it can be taken advantage of to solve the general problem thusly:
For the case of (-PI, PI]:
if (x > 0) x = x - 2PI * ceil(x/2PI) #Shift to the negative regime
return fmod(x - PI, 2PI) + PI
And for the case of [-PI, PI):
if (x < 0) x = x - 2PI * floor(x/2PI) #Shift to the positive regime
return fmod(x + PI, 2PI) - PI
[Note that this is pseudocode; my original was written in Tcl, and I didn't want to torture everyone with that. I needed the first case, so had to figure this out.]
deltaPhase -= floor(deltaPhase/M_TWOPI)*M_TWOPI;
The way suggested you suggested is best. It is fastest for small deflections. If angles in your program are constantly being deflected into the proper range, then you should only run into big out of range values rarely. Therefore paying the cost of a complicated modular arithmetic code every round seems wasteful. Comparisons are cheap compared to modular arithmetic (http://embeddedgurus.com/stack-overflow/2011/02/efficient-c-tip-13-use-the-modulus-operator-with-caution/).
In C99:
float unwindRadians( float radians )
{
const bool radiansNeedUnwinding = radians < -M_PI || M_PI <= radians;
if ( radiansNeedUnwinding )
{
if ( signbit( radians ) )
{
radians = -fmodf( -radians + M_PI, 2.f * M_PI ) + M_PI;
}
else
{
radians = fmodf( radians + M_PI, 2.f * M_PI ) - M_PI;
}
}
return radians;
}
If linking against glibc's libm (including newlib's implementation) you can access
__ieee754_rem_pio2f() and __ieee754_rem_pio2() private functions:
extern __int32_t __ieee754_rem_pio2f (float,float*);
float wrapToPI(float xf){
const float p[4]={0,M_PI_2,M_PI,-M_PI_2};
float yf[2];
int q;
int qmod4;
q=__ieee754_rem_pio2f(xf,yf);
/* xf = q * M_PI_2 + yf[0] + yf[1] /
* yf[1] << y[0], not sure if it could be ignored */
qmod4= q % 4;
if (qmod4==2)
/* (yf[0] > 0) defines interval (-pi,pi]*/
return ( (yf[0] > 0) ? -p[2] : p[2] ) + yf[0] + yf[1];
else
return p[qmod4] + yf[0] + yf[1];
}
Edit: Just realised that you need to link to libm.a, I couldn't find the symbols declared in libm.so
I have used (in python):
def WrapAngle(Wrapped, UnWrapped ):
TWOPI = math.pi * 2
TWOPIINV = 1.0 / TWOPI
return UnWrapped + round((Wrapped - UnWrapped) * TWOPIINV) * TWOPI
c-code equivalent:
#define TWOPI 6.28318531
double WrapAngle(const double dWrapped, const double dUnWrapped )
{
const double TWOPIINV = 1.0/ TWOPI;
return dUnWrapped + round((dWrapped - dUnWrapped) * TWOPIINV) * TWOPI;
}
notice that this brings it in the wrapped domain +/- 2pi so for +/- pi domain you need to handle that afterward like:
if( angle > pi):
angle -= 2*math.pi
I did use the findcontours() method to extract contour from the image, but I have no idea how to calculate the curvature from a set of contour points. Can somebody help me? Thank you very much!
While the theory behind Gombat's answer is correct, there are some errors in the code as well as in the formulae (the denominator t+n-x should be t+n-t). I have made several changes:
use symmetric derivatives to get more precise locations of curvature maxima
allow to use a step size for derivative calculation (can be used to reduce noise from noisy contours)
works with closed contours
Fixes:
* return infinity as curvature if denominator is 0 (not 0)
* added square calculation in denominator
* correct checking for 0 divisor
std::vector<double> getCurvature(std::vector<cv::Point> const& vecContourPoints, int step)
{
std::vector< double > vecCurvature( vecContourPoints.size() );
if (vecContourPoints.size() < step)
return vecCurvature;
auto frontToBack = vecContourPoints.front() - vecContourPoints.back();
std::cout << CONTENT_OF(frontToBack) << std::endl;
bool isClosed = ((int)std::max(std::abs(frontToBack.x), std::abs(frontToBack.y))) <= 1;
cv::Point2f pplus, pminus;
cv::Point2f f1stDerivative, f2ndDerivative;
for (int i = 0; i < vecContourPoints.size(); i++ )
{
const cv::Point2f& pos = vecContourPoints[i];
int maxStep = step;
if (!isClosed)
{
maxStep = std::min(std::min(step, i), (int)vecContourPoints.size()-1-i);
if (maxStep == 0)
{
vecCurvature[i] = std::numeric_limits<double>::infinity();
continue;
}
}
int iminus = i-maxStep;
int iplus = i+maxStep;
pminus = vecContourPoints[iminus < 0 ? iminus + vecContourPoints.size() : iminus];
pplus = vecContourPoints[iplus > vecContourPoints.size() ? iplus - vecContourPoints.size() : iplus];
f1stDerivative.x = (pplus.x - pminus.x) / (iplus-iminus);
f1stDerivative.y = (pplus.y - pminus.y) / (iplus-iminus);
f2ndDerivative.x = (pplus.x - 2*pos.x + pminus.x) / ((iplus-iminus)/2*(iplus-iminus)/2);
f2ndDerivative.y = (pplus.y - 2*pos.y + pminus.y) / ((iplus-iminus)/2*(iplus-iminus)/2);
double curvature2D;
double divisor = f1stDerivative.x*f1stDerivative.x + f1stDerivative.y*f1stDerivative.y;
if ( std::abs(divisor) > 10e-8 )
{
curvature2D = std::abs(f2ndDerivative.y*f1stDerivative.x - f2ndDerivative.x*f1stDerivative.y) /
pow(divisor, 3.0/2.0 ) ;
}
else
{
curvature2D = std::numeric_limits<double>::infinity();
}
vecCurvature[i] = curvature2D;
}
return vecCurvature;
}
For me curvature is:
where t is the position inside the contour and x(t) resp. y(t) return the related x resp. y value. See here.
So, according to my definition of curvature, one can implement it this way:
std::vector< float > vecCurvature( vecContourPoints.size() );
cv::Point2f posOld, posOlder;
cv::Point2f f1stDerivative, f2ndDerivative;
for (size_t i = 0; i < vecContourPoints.size(); i++ )
{
const cv::Point2f& pos = vecContourPoints[i];
if ( i == 0 ){ posOld = posOlder = pos; }
f1stDerivative.x = pos.x - posOld.x;
f1stDerivative.y = pos.y - posOld.y;
f2ndDerivative.x = - pos.x + 2.0f * posOld.x - posOlder.x;
f2ndDerivative.y = - pos.y + 2.0f * posOld.y - posOlder.y;
float curvature2D = 0.0f;
if ( std::abs(f2ndDerivative.x) > 10e-4 && std::abs(f2ndDerivative.y) > 10e-4 )
{
curvature2D = sqrt( std::abs(
pow( f2ndDerivative.y*f1stDerivative.x - f2ndDerivative.x*f1stDerivative.y, 2.0f ) /
pow( f2ndDerivative.x + f2ndDerivative.y, 3.0 ) ) );
}
vecCurvature[i] = curvature2D;
posOlder = posOld;
posOld = pos;
}
It works on non-closed pointlists as well. For closed contours, you may would like to change the boundary behavior (for the first iterations).
UPDATE:
Explanation for the derivatives:
A derivative for a continuous 1 dimensional function f(t) is:
But we are in a discrete space and have two discrete functions f_x(t) and f_y(t) where the smallest step for t is one.
The second derivative is the derivative of the first derivative:
Using the approximation of the first derivative, it yields to:
There are other approximations for the derivatives, if you google it, you will find a lot.
Here's a python implementation mainly based on Philipp's C++ code. For those interested, more details on the derivation can be found in Chapter 10.4.2 of:
Klette & Rosenfeld, 2004: Digital Geometry
def getCurvature(contour,stride=1):
curvature=[]
assert stride<len(contour),"stride must be shorther than length of contour"
for i in range(len(contour)):
before=i-stride+len(contour) if i-stride<0 else i-stride
after=i+stride-len(contour) if i+stride>=len(contour) else i+stride
f1x,f1y=(contour[after]-contour[before])/stride
f2x,f2y=(contour[after]-2*contour[i]+contour[before])/stride**2
denominator=(f1x**2+f1y**2)**3+1e-11
curvature_at_i=np.sqrt(4*(f2y*f1x-f2x*f1y)**2/denominator) if denominator > 1e-12 else -1
curvature.append(curvature_at_i)
return curvature
EDIT:
you can use convexityDefects from openCV, here's a link
a code example to find fingers based in their contour (variable res) source
def calculateFingers(res,drawing): # -> finished bool, cnt: finger count
# convexity defect
hull = cv2.convexHull(res, returnPoints=False)
if len(hull) > 3:
defects = cv2.convexityDefects(res, hull)
if type(defects) != type(None): # avoid crashing. (BUG not found)
cnt = 0
for i in range(defects.shape[0]): # calculate the angle
s, e, f, d = defects[i][0]
start = tuple(res[s][0])
end = tuple(res[e][0])
far = tuple(res[f][0])
a = math.sqrt((end[0] - start[0]) ** 2 + (end[1] - start[1]) ** 2)
b = math.sqrt((far[0] - start[0]) ** 2 + (far[1] - start[1]) ** 2)
c = math.sqrt((end[0] - far[0]) ** 2 + (end[1] - far[1]) ** 2)
angle = math.acos((b ** 2 + c ** 2 - a ** 2) / (2 * b * c)) # cosine theorem
if angle <= math.pi / 2: # angle less than 90 degree, treat as fingers
cnt += 1
cv2.circle(drawing, far, 8, [211, 84, 0], -1)
return True, cnt
return False, 0
in my case, i used about the same function to estimate the bending of board while extracting the contour
OLD COMMENT:
i am currently working in about the same, great information in this post, i'll come back with a solution when i'll have it ready
from Jonasson's answer, Shouldn't be here a tuple on the right side too?, i believe it won't unpack:
f1x,f1y=(contour[after]-contour[before])/stride
f2x,f2y=(contour[after]-2*contour[i]+contour[before])/stride**2
I need to make a simple bandpass audio filter.
Now I've used this simple C++ class: http://www.cardinalpeak.com/blog/a-c-class-to-implement-low-pass-high-pass-and-band-pass-filters
It works well and cut off the desired bands. But when I try to change upper or lower limit with small steps, on some values of limit I hear the wrong result - attenuated or shifted in frequency (not corresponding to current limits) sound.
Function for calculating impulse response:
void Filter::designBPF()
{
int n;
float mm;
for(n = 0; n < m_num_taps; n++){
mm = n - (m_num_taps - 1.0) / 2.0;
if( mm == 0.0 ) m_taps[n] = (m_phi - m_lambda) / M_PI;
else m_taps[n] = ( sin( mm * m_phi ) -
sin( mm * m_lambda ) ) / (mm * M_PI);
}
return;
}
where
m_lambda = M_PI * Fl / (Fs/2);
m_phi = M_PI * Fu / (Fs/2);
Fs - sample rate (44.100)
Fl - lower limit
Fu - upper limit
And simple filtering function:
float Filter::do_sample(float data_sample)
{
int i;
float result;
if( m_error_flag != 0 ) return(0);
for(i = m_num_taps - 1; i >= 1; i--){
m_sr[i] = m_sr[i-1];
}
m_sr[0] = data_sample;
result = 0;
for(i = 0; i < m_num_taps; i++) result += m_sr[i] * m_taps[i];
return result;
}
Do I need to use any window function (Blackman, etc.)? If yes, how do I do this?
I have tried to multiply my impulse response to Blackman window:
m_taps[n] *= 0.42 - 0.5 * cos(2.0 * M_PI * n / double(N - 1)) +
0.08 * cos(4.0 * M_PI * n / double(N - 1));
but the result was wrong.
And do I need to normalize taps?
I found a good free implementation of FIR filter:
http://www.iowahills.com/A7ExampleCodePage.html
...This Windowed FIR Filter C Code has two parts, the first is the
calculation of the impulse response for a rectangular window (low
pass, high pass, band pass, or notch). Then a window (Kaiser, Hanning,
etc) is applied to the impulse response. There are several windows to
choose from...
y[i] = waveform[i] × (0.42659071 – 0.49656062cos(w) + 0.07684867cos(2w))
where w = (2)i/n and n is the number of elements in the waveform
Try this I got the code from:
http://zone.ni.com/reference/en-XX/help/370592P-01/digitizers/blackman_window/
I hope this helps.
I have been trying to implement a simple Gaussian blur algorithm, for my image editing program. However, I have been having some trouble making this work, and I think the problem lies in the below snippet:
for( int j = 0; j < pow( kernel_size, 2 ); j++ )
{
int idx = ( i + kx + ( ky * img.width ));
//Try and overload this whenever possible
valueR += ( img.p_pixelArray[ idx ].r * kernel[ j ] );
valueG += ( img.p_pixelArray[ idx ].g * kernel[ j ] );
valueB += ( img.p_pixelArray[ idx ].b * kernel[ j ] );
if( kx == kernel_limit )
{
kx = -kernel_limit;
ky++;
}
else
{
kx++;
}
}
kx = -kernel_limit;
ky = -kernel_limit;
A brief explanation of the code above: kernel size is the size of the kernel (or matrix) generated by the Gaussian blur formula. kx and ky are variables to be used for iterating over the kernel. i is the parent loop, that nests this one, and goes over every pixel in the image. Each value variable simply holds a float R, G, or B value, and is used afterwards to obtain the final result. The if-else is used to increase kx and ky. idx is used to find the correct pixel. kernel limit is a variable set to
(*kernel size* - 1) / 2
So I can have kx going from -1 ( with a 3x3 kernel ) to +1, and the same thing with ky. I think the problem lies with the line
int idx = ( i + kx + ( ky * img.width ));
But I am not sure. The image I get is:
As can be seen, the color is blurred in a diagonal direction, and looks more like some kind of motion blur than Gaussian blur. If someone could help out, I would be very grateful.
EDIT:
The way I fill the kernel is as follows:
for( int i = 0; i < pow( kernel_size, 2 ); i++ )
{
// This. Is. Lisp.
kernel[i] = (( 1 / ( 2 * pi * pow( sigma, 2 ))) * pow (e, ( -((( pow( kx, 2 ) + pow( ky, 2 )) / 2 * pow( sigma, 2 ))))));
if(( kx + 1 ) == kernel_size )
{
kx = 0;
ky++;
}
else
{
kx++;
}
}
Few problems:
Your Gaussian misses brackets (even though you already have plenty..) around 2 * pow( sigma, 2 ). Now you multiply by variance instead of divide.
But what your problem is, is that your gaussian is centered at kx = ky = 0, as you let it run from 0 to kernel_size, instead of from -kernel_limit to kernel_limit. This results in the diagonal blurring. Something like the following should work better
kx = -kernel_limit;
ky = -kernel_limit;
int kernel_size_sq = kernel_size * kernel_size;
for( int i = 0; i < kernel_size_sq; i++ )
{
double sigma_sq = sigma * sigma;
double kx_sq = kx * kx;
double ky_sq = ky * ky;
kernel[i] = 1.0 / ( 2 * pi * sigma_sq) * exp(-(kx_sq + ky_sq) / (2 * sigma_sq));
if(kx == kernel_limit )
{
kx = -kernel_limit;
ky++;
}
else
{
kx++;
}
}
Also note how I got rid of your lisp-ness and some improvements: use some intermediate variables for clarity (compiler will optimize them away if anyway you ask it to); simple multiplication is faster than pow(x, 2); pow(e, x) == exp(x).
I am trying to write a .oct function for Octave that, given a single sine wave value, between -1 and 1, and sine wave period, returns a sine wave vector of period length with the last value in the vector being the given sine wave value. My code so far is:
#include <octave/oct.h>
#include <octave/dColVector.h>
#include <math.h>
#define PI 3.14159265
DEFUN_DLD (sinewave_recreate, args, , "args(0) sinewave value, args(1) is period")
{
octave_value_list retval;
double sinewave_value = args(0).double_value ();
double period = args(1).double_value ();
ColumnVector output_sinewave(period);
double degrees_inc = 360 / period;
double output_sinewave_degrees;
output_sinewave_degrees = asin( sinewave_value ) * 180 / PI;
output_sinewave(period-1) = sin( output_sinewave_degrees * PI / 180 );
for (octave_idx_type ii (1); ii < period; ii++) // Start the loop
{
output_sinewave_degrees = output_sinewave_degrees - degrees_inc;
if ( output_sinewave_degrees < 0 )
{
output_sinewave_degrees += 360 ;
}
output_sinewave( period-1-ii ) = sin( output_sinewave_degrees * PI / 180 );
}
retval(0) = output_sinewave;
return retval;
}
but is giving patchy results. By this I mean that it sometimes recreates the sine wave quite accurately and other times it is way off. I have determined this simply by creating a given sine wave, taking the last value in time and plugging this into the function to recreate the sine wave backwards through time and then comparing plots of the two. Obviously I am doing something wrong, but I can't seem to identify what.
Lets start with some trigonometric identities:
sin(x)^2 + cos(x)^2 == 1
sin(x+y) == sin(x)*cos(y) + sin(y)*cos(x)
cos(x+y) == cos(x)*cos(y) - sin(x)*sin(y)
Given the sine and cosine at a point x, we can exactly calculate the values after a step of size d, after precalculating sd = sin(d) and cd = cos(d):
sin(x+d) = sin(x)*cd + cos(x)*sd
cos(x+d) = cos(x)*cd - sin(x)*sd
Given the initial sine value, you can calculate the initial cosine value:
cos(x) = sqrt(1 - sin(x)^2)
Note that there are two possible solutions, corresponding to the two possible square-root values. Also note that all the angles in these identities are in radians, and d needs to be negative if you're going back through the wave.
Mike's note that there are two possible solutions for cos(x) made me realise that I would need to resolve the phase ambiguity of the sine wave. My second, successful attempt at this function is:
#include <octave/oct.h>
#include <octave/dColVector.h>
#include <math.h>
#define PI 3.14159265
DEFUN_DLD (sinewave_recreate_3, args, , "args(0) sinewave value, args(1) is period, args(2) is the phase")
{
octave_value_list retval;
double sinewave_value = args(0).double_value ();
double period = args(1).double_value ();
double phase = args(2).double_value ();
ColumnVector output_sinewave(period);
double X0 = asin(sinewave_value);
if (sinewave_value < 0 & phase > 180 & phase < 270)
{
X0 = PI + (0 - X0);
}
if (sinewave_value < 0 & phase >= 270)
{
X0 = X0 + 2 * PI;
}
if (sinewave_value > 0 & phase > 90)
{
X0 = PI - X0;
}
if (sinewave_value > 0 & phase < 0)
{
X0 = X0 + PI / 2;
}
double dx = PI / 180 * (360/period);
for (octave_idx_type ii (0); ii < period; ii++) // Start the loop
{
output_sinewave(period-1-ii) = sin(X0 - dx * ii);
}
retval(0) = output_sinewave;
return retval;
}
Thanks are also due to Keynslug.
There is simple formula. Here is the example in Python:
import math
import numpy as np
# We are supposing step is equal to 1degree
T = math.radians(1.0/360.0)
PrevBeforePrevValue = np.sin(math.radians(49.0)) # y(t-2)
PrevValue = np.sin(math.radians(50.0)) # y(t-1)
ValueNowRecursiveFormula = ((2.0*(4.0-T*T))/(4.0+T*T))*PrevValue - PrevBeforePrevValue
print("From RECURSIVE formula - " + str(ValueNowRecursiveFormula))
The details can be found here:
http://howtodoit.com.ua/en/on-the-way-of-developing-recursive-sinewave-generator/
You might try an easier way to go through.
Just recall that if
y = sin(x)
then first derivative of y will be equal to
dy/dx = cos(x)
So at every step of computation you add to the current value of y some delta equal to
dy = cos(x) * dx
But that might cut your accuracy down as a side-effect. You could probe it whatever. HTH.
It seems that slightly improved equation tend to be more accurate:
dy = cos(x + dx/2) * dx
Take a look at this.