我目前正在试验眼动追踪我已经成功地使用带有轮廓和霍夫变换的 OpenCV 构建了一个虹膜追踪算法。但下一步对我来说还不清楚。我想知道我正在做的计算对于将眼睛的中心转换到屏幕上是否正确。用户的头部位置固定。
我想要的是一种适用于所有眼睛偏离路线的算法。有角度计算吗？那么当用户更多地向右看时，线性？
我现在要做的是：首先我让用户查看特定点，并使用 RANSAC 来检测最接近屏幕位置的虹膜位置。我用屏幕和虹膜上的四个 2D 点来做这件事。我为此使用 Homography 来获得正确的计算。
void gaussian_elimination(float *input, int n){
// ported to c from pseudocode in
// http://en.wikipedia.org/wiki/Gaussian_elimination
float * A = input;
int i = 0;
int j = 0;
int m = n-1;
while (i < m && j < n){
// Find pivot in column j, starting in row i:
int maxi = i;
for(int k = i+1; k<m; k++){
if(fabs(A[k*n+j]) > fabs(A[maxi*n+j])){
maxi = k;
}
}
if (A[maxi*n+j] != 0){
//swap rows i and maxi, but do not change the value of i
if(i!=maxi)
for(int k=0;k<n;k++){
float aux = A[i*n+k];
A[i*n+k]=A[maxi*n+k];
A[maxi*n+k]=aux;
}
//Now A[i,j] will contain the old value of A[maxi,j].
//divide each entry in row i by A[i,j]
float A_ij=A[i*n+j];
for(int k=0;k<n;k++){
A[i*n+k]/=A_ij;
}
//Now A[i,j] will have the value 1.
for(int u = i+1; u< m; u++){
//subtract A[u,j] * row i from row u
float A_uj = A[u*n+j];
for(int k=0;k<n;k++){
A[u*n+k]-=A_uj*A[i*n+k];
}
//Now A[u,j] will be 0, since A[u,j] - A[i,j] * A[u,j] = A[u,j] - 1 * A[u,j] = 0.
}
i++;
}
j++;
}
//back substitution
for(int i=m-2;i>=0;i--){
for(int j=i+1;j<n-1;j++){
A[i*n+m]-=A[i*n+j]*A[j*n+m];
//A[i*n+j]=0;
}
}
}
ofMatrix4x4 findHomography(ofPoint src[4], ofPoint dst[4]){
ofMatrix4x4 matrix;
// create the equation system to be solved
//
// from: Multiple View Geometry in Computer Vision 2ed
// Hartley R. and Zisserman A.
//
// x' = xH
// where H is the homography: a 3 by 3 matrix
// that transformed to inhomogeneous coordinates for each point
// gives the following equations for each point:
//
// x' * (h31*x + h32*y + h33) = h11*x + h12*y + h13
// y' * (h31*x + h32*y + h33) = h21*x + h22*y + h23
//
// as the homography is scale independent we can let h33 be 1 (indeed any of the terms)
// so for 4 points we have 8 equations for 8 terms to solve: h11 - h32
// after ordering the terms it gives the following matrix
// that can be solved with gaussian elimination:
float P[8][9]={
{-src[0].x, -src[0].y, -1, 0, 0, 0, src[0].x*dst[0].x, src[0].y*dst[0].x, -dst[0].x }, // h11
{ 0, 0, 0, -src[0].x, -src[0].y, -1, src[0].x*dst[0].y, src[0].y*dst[0].y, -dst[0].y }, // h12
{-src[1].x, -src[1].y, -1, 0, 0, 0, src[1].x*dst[1].x, src[1].y*dst[1].x, -dst[1].x }, // h13
{ 0, 0, 0, -src[1].x, -src[1].y, -1, src[1].x*dst[1].y, src[1].y*dst[1].y, -dst[1].y }, // h21
{-src[2].x, -src[2].y, -1, 0, 0, 0, src[2].x*dst[2].x, src[2].y*dst[2].x, -dst[2].x }, // h22
{ 0, 0, 0, -src[2].x, -src[2].y, -1, src[2].x*dst[2].y, src[2].y*dst[2].y, -dst[2].y }, // h23
{-src[3].x, -src[3].y, -1, 0, 0, 0, src[3].x*dst[3].x, src[3].y*dst[3].x, -dst[3].x }, // h31
{ 0, 0, 0, -src[3].x, -src[3].y, -1, src[3].x*dst[3].y, src[3].y*dst[3].y, -dst[3].y }, // h32
};
gaussian_elimination(&P[0][0],9);
matrix(0,0)=P[0][8];
matrix(0,1)=P[1][8];
matrix(0,2)=0;
matrix(0,3)=P[2][8];
matrix(1,0)=P[3][8];
matrix(1,1)=P[4][8];
matrix(1,2)=0;
matrix(1,3)=P[5][8];
matrix(2,0)=0;
matrix(2,1)=0;
matrix(2,2)=0;
matrix(2,3)=0;
matrix(3,0)=P[6][8];
matrix(3,1)=P[7][8];
matrix(3,2)=0;
matrix(3,3)=1;
return matrix;
}