decomposeProjectionMatrix gives unexpected result - opencv

I have the following projection matrix P:
-375 0 2000 262500
-375 2000 0 262500
-1 0 0 700
This projection matrix projects 3D points in mm on a detector in px (with 1px equals to 0.5mm) and is built from the intrinsic matrix K and the extrinsic matrix [R|t] (where R is a rotation matrix and t a translation vector) according the relation P = K [R|t].
2000 0 375 0 0 1 0
K = 0 2000 375 R = 0 1 0 t = 0
0 0 1 -1 0 0 700
For some reasons I need to decompose P back into these matrices. When I use decomposeProjectionMatrix I get this as a rotation matrix:
0 0 0
0 0 0
-1 0 0
Which doesn't look like a rotation matrix to me.
Moreover when I build back the projection matrix from the Open CV decomposition I get this matrix:
-375 0 0 262500
-375 0 0 262500
-1 0 0 700
Looks similar but it is not the same.
I'm wondering if I'm doing something wrong or if I'm unlucky and that was one of the rare cases where this function fails.
Note that I did the decomposition by myself and I get coherent results but I would rather use Open CV functions as much as possible.

The problem seems to be in the RQ decomposition used by decomposeProjectionMatrix.
Even though the first square of the matrix P is non singular, the RQDecomp3x3 function gives incorrect results:
0 0 375 0 0 0
R = 0 0 375 Q = 0 0 0
0 0 1 -1 0 0
So a work around is to use a homemade function (here written in Python) based on the section 2.2 of Peter Sturm's lectures:
def decomposeP(P):
import numpy as np
M = P[0:3,0:3]
Q = np.eye(3)[::-1]
P_b = Q # M # M.T # Q
K_h = Q # np.linalg.cholesky(P_b) # Q
K = K_h / K_h[2,2]
A = np.linalg.inv(K) # M
l = (1/np.linalg.det(A)) ** (1/3)
R = l * A
t = l * np.linalg.inv(K) # P[0:3,3]
return K, R, t
I use the anti-identity matrix Q to build the non conventional Cholesky decomposition U U* where U is upper triangular.
This method differs slightly from the Peter Sturm's one as we use the relation P = K[R|t] while in Peter Sturm's lectures the relation used is P = K[R|-Rt].
A C++ implementation using only Open CV is trickier as they don't really expose a function for Cholesky decompostion:
void chol(cv::Mat const& S, cv::Mat& L)
{
L = cv::Mat::zeros(S.rows, S.rows, cv::DataType<double>::type);
for (int i = 0; i < S.rows; ++i) {
for (int j = 0; j <= i ; ++j) {
double sum = 0;
for (int k = 0; k < j; ++k)
sum += L.at<double>(i,k) * L.at<double>(j,k);
L.at<double>(i,j) = (i == j) ? sqrt(S.at<double>(i,i) - sum) : (S.at<double>(i,j) - sum) / L.at<double>(j,j);
}
}
}
void decomposeP(cv::Mat const& P, cv::Mat& K, cv::Mat& R, cv::Mat& t)
{
cv::Mat M(3, 3, cv::DataType<double>::type);
for (int i = 0; i < 3; ++i)
for (int j = 0; j < 3; ++j)
M.at<double>(i, j) = P.at<double>(i ,j);
cv::Mat Q = cv::Mat::zeros(3, 3, cv::DataType<double>::type);
Q.at<double>(0, 2) = 1.0;
Q.at<double>(1, 1) = 1.0;
Q.at<double>(2, 0) = 1.0;
cv::Mat O = Q * M * M.t() * Q;
cv::Mat C;
chol(O, C);
cv::Mat B = Q * C * Q;
K = B / B.at<double>(2,2);
cv::Mat A = K.inv() * M;
double l = std::pow((1 / cv::determinant(A)), 1/3);
R = l * A;
cv::Mat p(3, 1, cv::DataType<double>::type);
for (int i = 0; i < 3; ++i)
p.at<double>(i, 0) = P.at<double>(i ,3);
t = l * K.inv() * p;
}

Related

Convolution in Golang

I want execute a convolution product on an image.
The original image is:
So I test the convolution with gimp. With this matrix:
1 1 1
1 1 1
1 1 1
and the divider 9
I obtain
When I execute my algorithm I obtain:
My algorithm is:
func Convolution(img *image.Image, matrice [][]int) *image.NRGBA {
imageRGBA := image.NewNRGBA((*img).Bounds())
w := (*img).Bounds().Dx()
h := (*img).Bounds().Dy()
sumR := 0
sumB := 0
sumG := 0
var r uint32
var g uint32
var b uint32
for y := 0; y < h; y++ {
for x := 0; x < w; x++ {
for i := -1; i <= 1; i++ {
for j := -1; j <= 1; j++ {
var imageX int
var imageY int
imageX = x + i
imageY = y + j
r, g, b, _ = (*img).At(imageX, imageY).RGBA()
sumG = (sumG + (int(g) * matrice[i+1][j+1]))
sumR = (sumR + (int(r) * matrice[i+1][j+1]))
sumB = (sumB + (int(b) * matrice[i+1][j+1]))
}
}
imageRGBA.Set(x, y, color.NRGBA{
uint8(min(sumR/9, 255)),
uint8(min(sumG/9, 255)),
uint8(min(sumB/9, 255)),
255,
})
sumR = 0
sumB = 0
sumG = 0
}
}
return imageRGBA
}
Where are the error ?
Thank you for your help.
r, g, and b are uint32 values, and they contain 16bits of color information which is always greater than 255 if started as a non-zero 8 bit value.
You then can't operate on the RGBA values and truncate them to a uint8; that gives you a useless result because the least significant bits are just fractional parts of the 8bit values.
Compare the candidate integer value with the max 16bit value 65535, and shift it 8 bits before truncating it to get the 8 most significant bits.
uint8(min(sumR/9, 0xffff) >> 8),
uint8(min(sumG/9, 0xffff) >> 8),
uint8(min(sumB/9, 0xffff) >> 8),

Gradient descent values not correct

I'm attempting to implement gradient descent using code from :
Gradient Descent implementation in octave
I've amended code to following :
X = [1; 1; 1;]
y = [1; 0; 1;]
m = length(y);
X = [ones(m, 1), data(:,1)];
theta = zeros(2, 1);
iterations = 2000;
alpha = 0.001;
for iter = 1:iterations
theta = theta -((1/m) * ((X * theta) - y)' * X)' * alpha;
end
theta
Which gives following output :
X =
1
1
1
y =
1
0
1
theta =
0.32725
0.32725
theta is a 1x2 Matrix but should'nt it be 1x3 as the output (y) is 3x1 ?
So I should be able to multiply theta by the training example to make a prediction but cannot multiply x by theta as x is 1x3 and theta is 1x2?
Update :
%X = [1 1; 1 1; 1 1;]
%y = [1 1; 0 1; 1 1;]
X = [1 1 1; 1 1 1; 0 0 0;]
y = [1 1 1; 0 0 0; 1 1 1;]
m = length(y);
X = [ones(m, 1), X];
theta = zeros(4, 1);
theta
iterations = 2000;
alpha = 0.001;
for iter = 1:iterations
theta = theta -((1/m) * ((X * theta) - y)' * X)' * alpha;
end
%to make prediction
m = size(X, 1); % Number of training examples
p = zeros(m, 1);
htheta = sigmoid(X * theta);
p = htheta >= 0.5;
You are misinterpreting dimensions here. Your data consists of 3 points, each having a single dimension. Furthermore, you add a dummy dimension of 1s
X = [ones(m, 1), data(:,1)];
thus
octave:1> data = [1;2;3]
data =
1
2
3
octave:2> [ones(m, 1), data(:,1)]
ans =
1 1
1 2
1 3
and theta is your parametrization, which you should be able to apply through (this is not a code, but math notation)
h(x) = x1 * theta1 + theta0
thus your theta should have two dimensions. One is a weight for your dummy dimension (so called bias) and one for actual X dimension. If your X has K dimensions, theta would have K+1. Thus, after adding a dummy dimension matrices have following shapes:
X is 3x2
y is 3x1
theta is 2x1
so
X * theta is 3x1
the same as y

How can I get ellipse coefficient from fitEllipse function of OpenCV?

I want to extract the red ball from one picture and get the detected ellipse matrix in picture.
Here is my example:
I threshold the picture, find the contour of red ball by using findContour() function and use fitEllipse() to fit an ellipse.
But what I want is to get coefficient of this ellipse. Because the fitEllipse() return a rotation rectangle (RotatedRect), so I need to re-write this function.
One Ellipse can be expressed as Ax^2 + By^2 + Cxy + Dx + Ey + F = 0; So I want to get u=(A,B,C,D,E,F) or u=(A,B,C,D,E) if F is 1 (to construct an ellipse matrix).
I read the source code of fitEllipse(), there are totally three SVD process, I think I can get the above coefficients from the results of those three SVD process. But I am quite confused what does each result (variable cv::Mat x) of each SVD process represent and why there are three SVD here?
Here is this function:
cv::RotatedRect cv::fitEllipse( InputArray _points )
{
Mat points = _points.getMat();
int i, n = points.checkVector(2);
int depth = points.depth();
CV_Assert( n >= 0 && (depth == CV_32F || depth == CV_32S));
RotatedRect box;
if( n < 5 )
CV_Error( CV_StsBadSize, "There should be at least 5 points to fit the ellipse" );
// New fitellipse algorithm, contributed by Dr. Daniel Weiss
Point2f c(0,0);
double gfp[5], rp[5], t;
const double min_eps = 1e-8;
bool is_float = depth == CV_32F;
const Point* ptsi = points.ptr<Point>();
const Point2f* ptsf = points.ptr<Point2f>();
AutoBuffer<double> _Ad(n*5), _bd(n);
double *Ad = _Ad, *bd = _bd;
// first fit for parameters A - E
Mat A( n, 5, CV_64F, Ad );
Mat b( n, 1, CV_64F, bd );
Mat x( 5, 1, CV_64F, gfp );
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
c += p;
}
c.x /= n;
c.y /= n;
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
p -= c;
bd[i] = 10000.0; // 1.0?
Ad[i*5] = -(double)p.x * p.x; // A - C signs inverted as proposed by APP
Ad[i*5 + 1] = -(double)p.y * p.y;
Ad[i*5 + 2] = -(double)p.x * p.y;
Ad[i*5 + 3] = p.x;
Ad[i*5 + 4] = p.y;
}
solve(A, b, x, DECOMP_SVD);
// now use general-form parameters A - E to find the ellipse center:
// differentiate general form wrt x/y to get two equations for cx and cy
A = Mat( 2, 2, CV_64F, Ad );
b = Mat( 2, 1, CV_64F, bd );
x = Mat( 2, 1, CV_64F, rp );
Ad[0] = 2 * gfp[0];
Ad[1] = Ad[2] = gfp[2];
Ad[3] = 2 * gfp[1];
bd[0] = gfp[3];
bd[1] = gfp[4];
solve( A, b, x, DECOMP_SVD );
// re-fit for parameters A - C with those center coordinates
A = Mat( n, 3, CV_64F, Ad );
b = Mat( n, 1, CV_64F, bd );
x = Mat( 3, 1, CV_64F, gfp );
for( i = 0; i < n; i++ )
{
Point2f p = is_float ? ptsf[i] : Point2f((float)ptsi[i].x, (float)ptsi[i].y);
p -= c;
bd[i] = 1.0;
Ad[i * 3] = (p.x - rp[0]) * (p.x - rp[0]);
Ad[i * 3 + 1] = (p.y - rp[1]) * (p.y - rp[1]);
Ad[i * 3 + 2] = (p.x - rp[0]) * (p.y - rp[1]);
}
solve(A, b, x, DECOMP_SVD);
// store angle and radii
rp[4] = -0.5 * atan2(gfp[2], gfp[1] - gfp[0]); // convert from APP angle usage
if( fabs(gfp[2]) > min_eps )
t = gfp[2]/sin(-2.0 * rp[4]);
else // ellipse is rotated by an integer multiple of pi/2
t = gfp[1] - gfp[0];
rp[2] = fabs(gfp[0] + gfp[1] - t);
if( rp[2] > min_eps )
rp[2] = std::sqrt(2.0 / rp[2]);
rp[3] = fabs(gfp[0] + gfp[1] + t);
if( rp[3] > min_eps )
rp[3] = std::sqrt(2.0 / rp[3]);
box.center.x = (float)rp[0] + c.x;
box.center.y = (float)rp[1] + c.y;
box.size.width = (float)(rp[2]*2);
box.size.height = (float)(rp[3]*2);
if( box.size.width > box.size.height )
{
float tmp;
CV_SWAP( box.size.width, box.size.height, tmp );
box.angle = (float)(90 + rp[4]*180/CV_PI);
}
if( box.angle < -180 )
box.angle += 360;
if( box.angle > 360 )
box.angle -= 360;
return box;
}
The source code link: https://github.com/Itseez/opencv/blob/master/modules/imgproc/src/shapedescr.cpp
The function fitEllipse returns a RotatedRect that contains all the parameters of the ellipse.
An ellipse is defined by 5 parameters:
xc : x coordinate of the center
yc : y coordinate of the center
a : major semi-axis
b : minor semi-axis
theta : rotation angle
You can obtain these parameters like:
RotatedRect e = fitEllipse(points);
float xc = e.center.x;
float yc = e.center.y;
float a = e.size.width / 2; // width >= height
float b = e.size.height / 2;
float theta = e.angle; // in degrees
You can draw an ellipse with the function ellipse using the RotatedRect:
ellipse(image, e, Scalar(0,255,0));
or, equivalently using the ellipse parameters:
ellipse(res, Point(xc, yc), Size(a, b), theta, 0.0, 360.0, Scalar(0,255,0));
If you need the values of the coefficients of the implicit equation, you can do like (from Wikipedia):
So, you can get the parameters you need from the RotatedRect, and you don't need to change the function fitEllipse.
The solve function is used to solve linear systems or least-squares problems. Using the SVD decomposition method the system can be over-defined and/or the matrix src1 can be singular.
For more details on the algorithm, you can see the paper of Fitzgibbon that proposed this fit ellipse method.
Here is some code that worked for me which I based on the other responses on this thread.
def getConicCoeffFromEllipse(e):
# ellipse(Point(xc, yc),Size(a, b), theta)
xc = e[0][0]
yc = e[0][1]
a = e[1][0]/2
b = e[1][1]/2
theta = math.radians(e[2])
# See https://en.wikipedia.org/wiki/Ellipse
# Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 is the equation
A = a*a*math.pow(math.sin(theta),2) + b*b*math.pow(math.cos(theta),2)
B = 2*(b*b - a*a)*math.sin(theta)*math.cos(theta)
C = a*a*math.pow(math.cos(theta),2) + b*b*math.pow(math.sin(theta),2)
D = -2*A*xc - B*yc
E = -B*xc - 2*C*yc
F = A*xc*xc + B*xc*yc + C*yc*yc - a*a*b*b
coef = np.array([A,B,C,D,E,F]) / F
return coef
def getConicMatrixFromCoeff(c):
C = np.array([[c[0], c[1]/2, c[3]/2], # [ a, b/2, d/2 ]
[c[1]/2, c[2], c[4]/2], # [b/2, c, e/2 ]
[c[3]/2, c[4]/2, c[5]]]) # [d/2], e/2, f ]
return C

Difference between real fft and complex fft with imaginary part of zero in fftw?

I have a real 2d matrix. I am taking its fft using fftw. But the result of using a real to complex fft is different from a complex ( with imaginary part equal to zero) to complex fft.
real matrix
0 1 2
3 4 5
6 7 8
result of real to complex fft
36 -4.5+2.59808i -13.5+7.79423i
0 -13.5-7.79423i 0
0 0 0
Code:
int r = 3, c = 3;
int sz = r * c;
double *in = (double*) malloc(sizeof(double) * sz);
fftw_complex *out = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * sz);
fftw_plan p = fftw_plan_dft_r2c_2d(r, c, in, out, FFTW_MEASURE);
for ( int i=0; i<r; ++i ){
for ( int j=0; j<c; ++j ){
in[i*c+j] = i*c + j;
}
}
fftw_execute(p);
using a complex matrix with imaginary part of zero
complex matrix
0+0i 1+0i 2+0i
3+0i 4+0i 5+0i
6+0i 7+0i 8+0i
result of complex to complex fft
36 -4.5 + 2.59808i -4.5 - 2.59808i
-13.5 + 7.79423i 0 0
-13.5 - 7.79423i 0 0
Code:
int r = 3, c = 3;
int sz = r * c;
fftw_complex *out = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * sz);
fftw_complex *inc = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * sz);
p = fftw_plan_dft_2d( r,c, inc, out, FFTW_FORWARD,FFTW_MEASURE);
for ( int i=0; i<r; ++i ){
for ( int j=0; j<c; ++j ){
inc[i*c+j][0] = i*c+j;
inc[i*c+j][1] = 0;
}
}
fftw_execute(p);
I am after the result of complex to complex fft. But the real to complex fft is much faster and my data is real. Am I making a programming mistake or the result should be different?
As indicated in FFTW documentation
Then, after an r2c transform, the output is an n0 × n1 × n2 × … × (nd-1/2 + 1) array of fftw_complex values in row-major order
In other words, the output for your real-to-complex transform of your sample real matrix really is:
36 -4.5+2.59808i
-13.5+7.79423i 0
-13.5-7.79423i 0
You may notice that these two columns match exactly the first two columns of your complex-to-complex transform. The missing column is omitted from the real-to-complex transform since it is redundant due to symmetry. As such, the full 3x3 matrix including the missing column could be constructed using:
fftw_complex *outfull = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * sz);
int outc = (c/2+1);
for ( int i=0; i<r; ++i ){
// copy existing columns
for ( int j=0; j<outc; ++j ){
outfull[i*c+j][0] = out[i*outc+j][0];
outfull[i*c+j][1] = out[i*outc+j][1];
}
// generate missing column(s) from symmetry
for ( int j=outc; j<c; ++j){
int row = (r-i)%r;
int col = c-j;
outfull[i*c+j][0] = out[row*outc+col][0];
outfull[i*c+j][1] = -out[row*outc+col][1];
}
}

Multi otsu(multi-thresholding) with openCV

I am trying to carry out multi-thresholding with otsu. The method I am using currently is actually via maximising the between class variance, I have managed to get the same threshold value given as that by the OpenCV library. However, that is just via running otsu method once.
Documentation on how to do multi-level thresholding or rather recursive thresholding is rather limited. Where do I do after obtaining the original otsu's value? Would appreciate some hints, I been playing around with the code, adding one external for loop, but the next value calculated is always 254 for any given image:(
My code if need be:
//compute histogram first
cv::Mat imageh; //image edited to grayscale for histogram purpose
//imageh=image; //to delete and uncomment below;
cv::cvtColor(image, imageh, CV_BGR2GRAY);
int histSize[1] = {256}; // number of bins
float hranges[2] = {0.0, 256.0}; // min andax pixel value
const float* ranges[1] = {hranges};
int channels[1] = {0}; // only 1 channel used
cv::MatND hist;
// Compute histogram
calcHist(&imageh, 1, channels, cv::Mat(), hist, 1, histSize, ranges);
IplImage* im = new IplImage(imageh);//assign the image to an IplImage pointer
IplImage* finalIm = cvCreateImage(cvSize(im->width, im->height), IPL_DEPTH_8U, 1);
double otsuThreshold= cvThreshold(im, finalIm, 0, 255, cv::THRESH_BINARY | cv::THRESH_OTSU );
cout<<"opencv otsu gives "<<otsuThreshold<<endl;
int totalNumberOfPixels= imageh.total();
cout<<"total number of Pixels is " <<totalNumberOfPixels<< endl;
float sum = 0;
for (int t=0 ; t<256 ; t++)
{
sum += t * hist.at<float>(t);
}
cout<<"sum is "<<sum<<endl;
float sumB = 0; //sum of background
int wB = 0; // weight of background
int wF = 0; //weight of foreground
float varMax = 0;
int threshold = 0;
//run an iteration to find the maximum value of the between class variance(as between class variance shld be maximise)
for (int t=0 ; t<256 ; t++)
{
wB += hist.at<float>(t); // Weight Background
if (wB == 0) continue;
wF = totalNumberOfPixels - wB; // Weight Foreground
if (wF == 0) break;
sumB += (float) (t * hist.at<float>(t));
float mB = sumB / wB; // Mean Background
float mF = (sum - sumB) / wF; // Mean Foreground
// Calculate Between Class Variance
float varBetween = (float)wB * (float)wF * (mB - mF) * (mB - mF);
// Check if new maximum found
if (varBetween > varMax) {
varMax = varBetween;
threshold = t;
}
}
cout<<"threshold value is: "<<threshold;
To extend Otsu's thresholding method to multi-level thresholding the between class variance equation becomes:
Please check out Deng-Yuan Huang, Ta-Wei Lin, Wu-Chih Hu, Automatic
Multilevel Thresholding Based on Two-Stage Otsu's Method with Cluster
Determination by Valley Estimation, Int. Journal of Innovative
Computing, 2011, 7:5631-5644 for more information.
http://www.ijicic.org/ijicic-10-05033.pdf
Here is my C# implementation of Otsu Multi for 2 thresholds:
/* Otsu (1979) - multi */
Tuple < int, int > otsuMulti(object sender, EventArgs e) {
//image histogram
int[] histogram = new int[256];
//total number of pixels
int N = 0;
//accumulate image histogram and total number of pixels
foreach(int intensity in image.Data) {
if (intensity != 0) {
histogram[intensity] += 1;
N++;
}
}
double W0K, W1K, W2K, M0, M1, M2, currVarB, optimalThresh1, optimalThresh2, maxBetweenVar, M0K, M1K, M2K, MT;
optimalThresh1 = 0;
optimalThresh2 = 0;
W0K = 0;
W1K = 0;
M0K = 0;
M1K = 0;
MT = 0;
maxBetweenVar = 0;
for (int k = 0; k <= 255; k++) {
MT += k * (histogram[k] / (double) N);
}
for (int t1 = 0; t1 <= 255; t1++) {
W0K += histogram[t1] / (double) N; //Pi
M0K += t1 * (histogram[t1] / (double) N); //i * Pi
M0 = M0K / W0K; //(i * Pi)/Pi
W1K = 0;
M1K = 0;
for (int t2 = t1 + 1; t2 <= 255; t2++) {
W1K += histogram[t2] / (double) N; //Pi
M1K += t2 * (histogram[t2] / (double) N); //i * Pi
M1 = M1K / W1K; //(i * Pi)/Pi
W2K = 1 - (W0K + W1K);
M2K = MT - (M0K + M1K);
if (W2K <= 0) break;
M2 = M2K / W2K;
currVarB = W0K * (M0 - MT) * (M0 - MT) + W1K * (M1 - MT) * (M1 - MT) + W2K * (M2 - MT) * (M2 - MT);
if (maxBetweenVar < currVarB) {
maxBetweenVar = currVarB;
optimalThresh1 = t1;
optimalThresh2 = t2;
}
}
}
return new Tuple(optimalThresh1, optimalThresh2);
}
And this is the result I got by thresholding an image scan of soil with the above code:
(T1 = 110, T2 = 147).
Otsu's original paper: "Nobuyuki Otsu, A Threshold Selection Method
from Gray-Level Histogram, IEEE Transactions on Systems, Man, and
Cybernetics, 1979, 9:62-66" also briefly mentions the extension to
Multithresholding.
https://engineering.purdue.edu/kak/computervision/ECE661.08/OTSU_paper.pdf
Hope this helps.
Here is a simple general approach for 'n' thresholds in python (>3.0) :
# developed by- SUJOY KUMAR GOSWAMI
# source paper- https://people.ece.cornell.edu/acharya/papers/mlt_thr_img.pdf
import cv2
import numpy as np
import math
img = cv2.imread('path-to-image')
img = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
a = 0
b = 255
n = 6 # number of thresholds (better choose even value)
k = 0.7 # free variable to take any positive value
T = [] # list which will contain 'n' thresholds
def sujoy(img, a, b):
if a>b:
s=-1
m=-1
return m,s
img = np.array(img)
t1 = (img>=a)
t2 = (img<=b)
X = np.multiply(t1,t2)
Y = np.multiply(img,X)
s = np.sum(X)
m = np.sum(Y)/s
return m,s
for i in range(int(n/2-1)):
img = np.array(img)
t1 = (img>=a)
t2 = (img<=b)
X = np.multiply(t1,t2)
Y = np.multiply(img,X)
mu = np.sum(Y)/np.sum(X)
Z = Y - mu
Z = np.multiply(Z,X)
W = np.multiply(Z,Z)
sigma = math.sqrt(np.sum(W)/np.sum(X))
T1 = mu - k*sigma
T2 = mu + k*sigma
x, y = sujoy(img, a, T1)
w, z = sujoy(img, T2, b)
T.append(x)
T.append(w)
a = T1+1
b = T2-1
k = k*(i+1)
T1 = mu
T2 = mu+1
x, y = sujoy(img, a, T1)
w, z = sujoy(img, T2, b)
T.append(x)
T.append(w)
T.sort()
print(T)
For full paper and more informations visit this link.
I've written an example on how otsu thresholding work in python before. You can see the source code here: https://github.com/subokita/Sandbox/blob/master/otsu.py
In the example there's 2 variants, otsu2() which is the optimised version, as seen on Wikipedia page, and otsu() which is more naive implementation based on the algorithm description itself.
If you are okay in reading python codes (in this case, they are pretty simple, almost pseudo code like), you might want to look at otsu() in the example and modify it. Porting it to C++ code is not hard either.
#Antoni4 gives the best answer in my opinion and it's very straight forward to increase the number of levels.
This is for three-level thresholding:
#include "Shadow01-1.cuh"
void multiThresh(double &optimalThresh1, double &optimalThresh2, double &optimalThresh3, cv::Mat &imgHist, cv::Mat &src)
{
double W0K, W1K, W2K, W3K, M0, M1, M2, M3, currVarB, maxBetweenVar, M0K, M1K, M2K, M3K, MT;
unsigned char *histogram = (unsigned char*)(imgHist.data);
int N = src.rows*src.cols;
W0K = 0;
W1K = 0;
M0K = 0;
M1K = 0;
MT = 0;
maxBetweenVar = 0;
for (int k = 0; k <= 255; k++) {
MT += k * (histogram[k] / (double) N);
}
for (int t1 = 0; t1 <= 255; t1++)
{
W0K += histogram[t1] / (double) N; //Pi
M0K += t1 * (histogram[t1] / (double) N); //i * Pi
M0 = M0K / W0K; //(i * Pi)/Pi
W1K = 0;
M1K = 0;
for (int t2 = t1 + 1; t2 <= 255; t2++)
{
W1K += histogram[t2] / (double) N; //Pi
M1K += t2 * (histogram[t2] / (double) N); //i * Pi
M1 = M1K / W1K; //(i * Pi)/Pi
W2K = 1 - (W0K + W1K);
M2K = MT - (M0K + M1K);
if (W2K <= 0) break;
M2 = M2K / W2K;
W3K = 0;
M3K = 0;
for (int t3 = t2 + 1; t3 <= 255; t3++)
{
W2K += histogram[t3] / (double) N; //Pi
M2K += t3 * (histogram[t3] / (double) N); // i*Pi
M2 = M2K / W2K; //(i*Pi)/Pi
W3K = 1 - (W1K + W2K);
M3K = MT - (M1K + M2K);
M3 = M3K / W3K;
currVarB = W0K * (M0 - MT) * (M0 - MT) + W1K * (M1 - MT) * (M1 - MT) + W2K * (M2 - MT) * (M2 - MT) + W3K * (M3 - MT) * (M3 - MT);
if (maxBetweenVar < currVarB)
{
maxBetweenVar = currVarB;
optimalThresh1 = t1;
optimalThresh2 = t2;
optimalThresh3 = t3;
}
}
}
}
}
#Guilherme Silva
Your code has a BUG
You Must Replace:
W3K = 0;
M3K = 0;
with
W2K = 0;
M2K = 0;
and
W3K = 1 - (W1K + W2K);
M3K = MT - (M1K + M2K);
with
W3K = 1 - (W0K + W1K + W2K);
M3K = MT - (M0K + M1K + M2K);
;-)
Regards
EDIT(1): [Toby Speight]
I discovered this bug by applying the effect to the same picture at different resoultions(Sizes) and seeing that the output results were to much different from each others (Even changing resolution a little bit)
W3K and M3K must be the totals minus the Previous WKs and MKs.
(I thought about this for Code-similarity with the one with one level less)
At the moment due to my lacks of English I cannot explain Better How and Why
To be honest I'm still not 100% sure that this way is correct, even thought from my outputs I could tell that it gives better results. (Even with 1 Level more (5 shades of gray))
You could try yourself ;-)
Sorry
My Outputs:
3 Thresholds
4 Thresholds
I found a useful piece of code in this thread. I was looking for a multi-level Otsu implementation for double/float images. So, I tried to generalize example for N-levels with double/float matrix as input. In my code below I am using armadillo library as dependency. But this code can be easily adapted for standard C++ arrays, just replace vec, uvec objects with single dimensional double and integer arrays, mat and umat with two-dimensional. Two other functions from armadillo used here are: vectorise and hist.
// Input parameters:
// map - input image (double matrix)
// mask - region of interest to be thresholded
// nBins - number of bins
// nLevels - number of Otsu thresholds
#include <armadillo>
#include <algorithm>
#include <vector>
mat OtsuFilterMulti(mat map, int nBins, int nLevels) {
mat mapr; // output thresholded image
mapr = zeros<mat>(map.n_rows, map.n_cols);
unsigned int numElem = 0;
vec threshold = zeros<vec>(nLevels);
vec q = zeros<vec>(nLevels + 1);
vec mu = zeros<vec>(nLevels + 1);
vec muk = zeros<vec>(nLevels + 1);
uvec binv = zeros<uvec>(nLevels);
if (nLevels <= 1) return mapr;
numElem = map.n_rows*map.n_cols;
uvec histogram = hist(vectorise(map), nBins);
double maxval = map.max();
double minval = map.min();
double odelta = (maxval - abs(minval)) / nBins; // distance between histogram bins
vec oval = zeros<vec>(nBins);
double mt = 0, variance = 0.0, bestVariance = 0.0;
for (int ii = 0; ii < nBins; ii++) {
oval(ii) = (double)odelta*ii + (double)odelta*0.5; // centers of histogram bins
mt += (double)ii*((double)histogram(ii)) / (double)numElem;
}
for (int ii = 0; ii < nLevels; ii++) {
binv(ii) = ii;
}
double sq, smuk;
int nComb;
nComb = nCombinations(nBins,nLevels);
std::vector<bool> v(nBins);
std::fill(v.begin(), v.begin() + nLevels, true);
umat ibin = zeros<umat>(nComb, nLevels); // indices from combinations will be stored here
int cc = 0;
int ci = 0;
do {
for (int i = 0; i < nBins; ++i) {
if(ci==nLevels) ci=0;
if (v[i]) {
ibin(cc,ci) = i;
ci++;
}
}
cc++;
} while (std::prev_permutation(v.begin(), v.end()));
uvec lastIndex = zeros<uvec>(nLevels);
// Perform operations on pre-calculated indices
for (int ii = 0; ii < nComb; ii++) {
for (int jj = 0; jj < nLevels; jj++) {
smuk = 0;
sq = 0;
if (lastIndex(jj) != ibin(ii, jj) || ii == 0) {
q(jj) += double(histogram(ibin(ii, jj))) / (double)numElem;
muk(jj) += ibin(ii, jj)*(double(histogram(ibin(ii, jj)))) / (double)numElem;
mu(jj) = muk(jj) / q(jj);
q(jj + 1) = 0.0;
muk(jj + 1) = 0.0;
if (jj>0) {
for (int kk = 0; kk <= jj; kk++) {
sq += q(kk);
smuk += muk(kk);
}
q(jj + 1) = 1 - sq;
muk(jj + 1) = mt - smuk;
mu(jj + 1) = muk(jj + 1) / q(jj + 1);
}
if (jj>0 && jj<(nLevels - 1)) {
q(jj + 1) = 0.0;
muk(jj + 1) = 0.0;
}
lastIndex(jj) = ibin(ii, jj);
}
}
variance = 0.0;
for (int jj = 0; jj <= nLevels; jj++) {
variance += q(jj)*(mu(jj) - mt)*(mu(jj) - mt);
}
if (variance > bestVariance) {
bestVariance = variance;
for (int jj = 0; jj<nLevels; jj++) {
threshold(jj) = oval(ibin(ii, jj));
}
}
}
cout << "Optimized thresholds: ";
for (int jj = 0; jj<nLevels; jj++) {
cout << threshold(jj) << " ";
}
cout << endl;
for (unsigned int jj = 0; jj<map.n_rows; jj++) {
for (unsigned int kk = 0; kk<map.n_cols; kk++) {
for (int ll = 0; ll<nLevels; ll++) {
if (map(jj, kk) >= threshold(ll)) {
mapr(jj, kk) = ll+1;
}
}
}
}
return mapr;
}
int nCombinations(int n, int r) {
if (r>n) return 0;
if (r*2 > n) r = n-r;
if (r == 0) return 1;
int ret = n;
for( int i = 2; i <= r; ++i ) {
ret *= (n-i+1);
ret /= i;
}
return ret;
}

Resources