三元(2D)算法的实现

Question

我正在尝试在2D中实现三元化过程。与此相关的维基百科文章：倾斜

我在这个网站上发现了一个很好的问题，在这里，算法得到了很好的解释：人工智能

毕竟，我试图在c++中实现该算法。不幸的是我遇到了一些问题。让我们看看我的实现。它只是一个函数:第一个输入是三个向量，每个向量用X，Y坐标表示一个2D点。另一个输入变量(r1、r2、r3)表示每个点的距离/半径。

#include <iostream>
#include <fstream>
#include <sstream>
#include <math.h> 
#include <vector>
using namespace std;

std::vector<double> trilateration(double point1[], double point2[], double point3[], double r1, double r2, double r3) {
    std::vector<double> resultPose;
    //unit vector in a direction from point1 to point 2
    double p2p1Distance = pow(pow(point2[0]-point1[0],2) + pow(point2[1]-point1[1],2),0.5);
    double exx = (point2[0]-point1[0])/p2p1Distance;
    double exy = (point2[1]-point1[1])/p2p1Distance;
    //signed magnitude of the x component
    double ix = exx*(point3[0]-point1[0]);
    double iy = exy*(point3[1]-point1[1]);
    //the unit vector in the y direction. 
    double eyx = (point3[0]-point1[0]-ix*exx)/pow(pow(point3[0]-point1[0]-ix*exx,2) + pow(point3[1]-point1[1]-iy*exy,2),0.5);
    double eyy = (point3[1]-point1[1]-iy*exy)/pow(pow(point3[0]-point1[0]-ix*exx,2) + pow(point3[1]-point1[1]-iy*exy,2),0.5);
    //the signed magnitude of the y component
    double jx = eyx*(point3[0]-point1[0]);
    double jy = eyy*(point3[1]-point1[1]);
    //coordinates
    double x = (pow(r1,2) - pow(r2,2) + pow(p2p1Distance,2))/ (2 * p2p1Distance);
    double y = (pow(r1,2) - pow(r3,2) + pow(iy,2) + pow(jy,2))/2*jy - ix*x/jx;
    //result coordinates
    double finalX = point1[0]+ x*exx + y*eyx;
    double finalY = point1[1]+ x*exy + y*eyy;
    resultPose.push_back(finalX);
    resultPose.push_back(finalY);
    return resultPose;
}

正如我提到的，我跟踪了这的文章。我认为问题在于计算y坐标的部分。我也不确定最后一部分，我计算finalX，finalY.

我的主要职能如下：

int main(int argc, char* argv[]){
    std::vector<double> finalPose;
    double p1[] = {4.0,4.0};
    double p2[] = {9.0,7.0};
    double p3[] = {9.0,1.0};
    double r1,r2,r3;
    r1 = 4;
    r2 = 3;
    r3 = 3.25;
    finalPose = trilateration(p1,p2,p3,r1,r2,r3);
    cout<<"X:::  "<<finalPose[0]<<endl;
    cout<<"Y:::  "<<finalPose[1]<<endl; 
    //x = 8, y = 4.1

}

结果应该在X~8和Y~4.1左右，但我得到X= 13.5542和Y=-5.09038。

所以我的问题是:对于x和y的计算，我有问题，我想我可以把算法解到x，然后我有计算y的问题。

Y= (r12 - r32 + i2 + j2) / 2j - ix /j

我不知道在这里应该使用哪一个i和j，因为我有两个i (ix，iy)和两个j(jx，jy)。正如你所看到的，我使用了iy和jy，但是在行的末尾，我使用了ix，因为与x相乘。谢谢！

Answer 1

我用了一些辅助变量，但效果很好.

#include <iostream>
#include <fstream>
#include <sstream>
#include <math.h> 
#include <vector>
using namespace std;

struct point 
{
    float x,y;
};

float norm (point p) // get the norm of a vector
{
    return pow(pow(p.x,2)+pow(p.y,2),.5);
}

point trilateration(point point1, point point2, point point3, double r1, double r2, double r3) {
    point resultPose;
    //unit vector in a direction from point1 to point 2
    double p2p1Distance = pow(pow(point2.x-point1.x,2) + pow(point2.y-   point1.y,2),0.5);
    point ex = {(point2.x-point1.x)/p2p1Distance, (point2.y-point1.y)/p2p1Distance};
    point aux = {point3.x-point1.x,point3.y-point1.y};
    //signed magnitude of the x component
    double i = ex.x * aux.x + ex.y * aux.y;
    //the unit vector in the y direction. 
    point aux2 = { point3.x-point1.x-i*ex.x, point3.y-point1.y-i*ex.y};
    point ey = { aux2.x / norm (aux2), aux2.y / norm (aux2) };
    //the signed magnitude of the y component
    double j = ey.x * aux.x + ey.y * aux.y;
    //coordinates
    double x = (pow(r1,2) - pow(r2,2) + pow(p2p1Distance,2))/ (2 * p2p1Distance);
    double y = (pow(r1,2) - pow(r3,2) + pow(i,2) + pow(j,2))/(2*j) - i*x/j;
    //result coordinates
    double finalX = point1.x+ x*ex.x + y*ey.x;
    double finalY = point1.y+ x*ex.y + y*ey.y;
    resultPose.x = finalX;
    resultPose.y = finalY;
    return resultPose;
}

int main(int argc, char* argv[]){
    point finalPose;
    point p1 = {4.0,4.0};
    point p2 = {9.0,7.0};
    point p3 = {9.0,1.0};
    double r1,r2,r3;
    r1 = 4;
    r2 = 3;
    r3 = 3.25;
    finalPose = trilateration(p1,p2,p3,r1,r2,r3);
    cout<<"X:::  "<<finalPose.x<<endl;
    cout<<"Y:::  "<<finalPose.y<<endl; 
}

$产出如下：

X:::  8.02188
Y:::  4.13021