透视变换从原理到应用

透视变换从原理到应用 一，理论

略

二，计算

已知从原图位置 ( u , v , w ) (u,v,w) (u,v,w)到目标图像位置 ( x ′ , y ′ , z ′ ) (x',y',z') (x′,y′,z′)的变换矩阵为 A A A
A = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] A=left[ begin{array}{ccc} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ end{array} right] A=⎣⎢⎢⎢⎢⎢⎢⎡​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​⎦⎥⎥⎥⎥⎥⎥⎤​
即 ( x ′ , y ′ , z ′ ) = ( u , v , w ) ∗ A = ( a 11 u + a 21 v + a 31 w , a 12 u + a 22 v + a 32 w , a 13 u + a 23 v + a 33 w ) (x',y',z')=(u,v,w)*A=(a_{11}u+a_{21}v+a_{31}w,a_{12}u+a_{22}v+a_{32}w,a_{13}u+a_{23}v+a_{33}w) (x′,y′,z′)=(u,v,w)∗A=(a11​u+a21​v+a31​w,a12​u+a22​v+a32​w,a13​u+a23​v+a33​w)
a 33 = 1 , w = 1 a_{33}=1,w=1 a33​=1,w=1,从而有

x = x ′ z ′ = a 11 u + a 21 v + a 31 a 13 u + a 23 v + 1 x=frac{x'}{z'}=frac{a_{11}u+a_{21}v+a_{31}}{a_{13}u+a_{23}v+1} x=z′x′​=a13​u+a23​v+1a11​u+a21​v+a31​​,

y = y ′ z ′ = a 12 u + a 22 v + a 32 a 13 u + a 23 v + 1 y=frac{y'}{z'}=frac{a_{12}u+a_{22}v+a_{32}}{a_{13}u+a_{23}v+1} y=z′y′​=a13​u+a23​v+1a12​u+a22​v+a32​​.---------（1）

一个点有两个方程，解 A A A的8个参数至少需要4个点，设原图的四个点为 ( u 1 , v 1 ) , ( u 2 , v 2 ) , ( u 3 , v 3 ) , ( u 4 , v 4 ) (u_1,v_1),(u_2,v_2),(u_3,v_3),(u_4,v_4) (u1​,v1​),(u2​,v2​),(u3​,v3​),(u4​,v4​)，目标图像四个点为 ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) , ( x 4 , y 4 ) (x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4) (x1​,y1​),(x2​,y2​),(x3​,y3​),(x4​,y4​)，则有
x i = a 11 u i + a 21 v i + a 31 a 13 u i + a 23 v i + 1 x_i=frac{a_{11}u_i+a_{21}v_i+a_{31}}{a_{13}u_i+a_{23}v_i+1} xi​=a13​ui​+a23​vi​+1a11​ui​+a21​vi​+a31​​

y i = a 12 u i + a 22 v i + a 32 a 13 u i + a 23 v i + 1 ， i = 1 , 2 , 3 , 4 y_i=frac{a_{12}u_i+a_{22}v_i+a_{32}}{a_{13}u_i+a_{23}v_i+1}，i=1,2,3,4 yi​=a13​ui​+a23​vi​+1a12​ui​+a22​vi​+a32​​，i=1,2,3,4。展开得到

a 11 u i + a 12 0 − a 13 u i x i + a 21 v i + a 22 0 − a 23 v i x i + a 31 + a 32 0 = x i a_{11}u_i+a_{12}0-a_{13}u_ix_i+a_{21}v_i+a_{22}0-a_{23}v_ix_i+a_{31}+a_{32}0=x_i a11​ui​+a12​0−a13​ui​xi​+a21​vi​+a22​0−a23​vi​xi​+a31​+a32​0=xi​
a 11 0 + a 12 u i − a 13 u i y i + a 21 0 + a 22 v i − a 23 v i y i + a 31 0 + a 32 = y i , i = 1 , 2 , 3 , 4 a_{11}0+a_{12}u_i-a_{13}u_iy_i+a_{21}0+a_{22}v_i-a_{23}v_iy_i+a_{31}0+a_{32}=y_i,i=1,2,3,4 a11​0+a12​ui​−a13​ui​yi​+a21​0+a22​vi​−a23​vi​yi​+a31​0+a32​=yi​,i=1,2,3,4。即
[ u 1 0 − u 1 x 1 v 1 0 − v 1 x 1 1 0 0 u 1 − u 1 y 1 0 v 1 − v 1 y 1 0 1 u 2 0 − u 2 x 2 v 2 0 − v 2 x 2 1 0 0 u 2 − u 2 y 2 0 v 2 − v 2 y 2 0 1 u 3 0 − u 3 x 3 v 3 0 − v 3 x 3 1 0 0 u 3 − u 3 y 3 0 v 3 − v 3 y 3 0 1 u 4 0 − u 4 x 4 v 4 0 − v 4 x 4 1 0 0 u 4 − u 4 y 4 0 v 4 − v 4 y 4 0 1 ] . [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 ] = [ x 1 y 1 x 2 y 2 x 1 y 1 x 2 y 2 ] left[ begin{array}{ccc} u_1 &0&-u_1x_1 &v_1&0&-v_1x_1&1&0 \ 0&u_1&-u_1y_1&0&v_1&-v_1y_1&0&1\ u_2 &0&-u_2x_2 &v_2&0&-v_2x_2&1&0 \ 0&u_2&-u_2y_2&0&v_2&-v_2y_2&0&1\ u_3 &0&-u_3x_3 &v_3&0&-v_3x_3&1&0 \ 0&u_3&-u_3y_3&0&v_3&-v_3y_3&0&1\ u_4 &0&-u_4x_4 &v_4&0&-v_4x_4&1&0 \ 0&u_4&-u_4y_4&0&v_4&-v_4y_4&0&1\ end{array} right].left[ begin{array}{ccc} a_{11}\ a_{12}\ a_{13}\ a_{21}\ a_{22}\ a_{23}\ a_{31}\ a_{32}\ end{array} right]=left[ begin{array}{ccc} x_1\ y_1\ x_2\ y_2\ x_1\ y_1\ x_2\ y_2\ end{array} right] ⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​u1​0u2​0u3​0u4​0​0u1​0u2​0u3​0u4​​−u1​x1​−u1​y1​−u2​x2​−u2​y2​−u3​x3​−u3​y3​−u4​x4​−u4​y4​​v1​0v2​0v3​0v4​0​0v1​0v2​0v3​0v4​​−v1​x1​−v1​y1​−v2​x2​−v2​y2​−v3​x3​−v3​y3​−v4​x4​−v4​y4​​10101010​01010101​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​.⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​a11​a12​a13​a21​a22​a23​a31​a32​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​x1​y1​x2​y2​x1​y1​x2​y2​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​
从而有，
[ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 ] = [ u 1 0 − u 1 x 1 v 1 0 − v 1 x 1 1 0 0 u 1 − u 1 y 1 0 v 1 − v 1 y 1 0 1 u 2 0 − u 2 x 2 v 2 0 − v 2 x 2 1 0 0 u 2 − u 2 y 2 0 v 2 − v 2 y 2 0 1 u 3 0 − u 3 x 3 v 3 0 − v 3 x 3 1 0 0 u 3 − u 3 y 3 0 v 3 − v 3 y 3 0 1 u 4 0 − u 4 x 4 v 4 0 − v 4 x 4 1 0 0 u 4 − u 4 y 4 0 v 4 − v 4 y 4 0 1 ] − 1 . [ x 1 y 1 x 2 y 2 x 1 y 1 x 2 y 2 ] left[ begin{array}{ccc} a_{11}\ a_{12}\ a_{13}\ a_{21}\ a_{22}\ a_{23}\ a_{31}\ a_{32}\ end{array} right]=left[ begin{array}{ccc} u_1 &0&-u_1x_1 &v_1&0&-v_1x_1&1&0 \ 0&u_1&-u_1y_1&0&v_1&-v_1y_1&0&1\ u_2 &0&-u_2x_2 &v_2&0&-v_2x_2&1&0 \ 0&u_2&-u_2y_2&0&v_2&-v_2y_2&0&1\ u_3 &0&-u_3x_3 &v_3&0&-v_3x_3&1&0 \ 0&u_3&-u_3y_3&0&v_3&-v_3y_3&0&1\ u_4 &0&-u_4x_4 &v_4&0&-v_4x_4&1&0 \ 0&u_4&-u_4y_4&0&v_4&-v_4y_4&0&1\ end{array} right]^{-1}.left[ begin{array}{ccc} x_1\ y_1\ x_2\ y_2\ x_1\ y_1\ x_2\ y_2\ end{array} right] ⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​a11​a12​a13​a21​a22​a23​a31​a32​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​u1​0u2​0u3​0u4​0​0u1​0u2​0u3​0u4​​−u1​x1​−u1​y1​−u2​x2​−u2​y2​−u3​x3​−u3​y3​−u4​x4​−u4​y4​​v1​0v2​0v3​0v4​0​0v1​0v2​0v3​0v4​​−v1​x1​−v1​y1​−v2​x2​−v2​y2​−v3​x3​−v3​y3​−v4​x4​−v4​y4​​10101010​01010101​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​−1.⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡​x1​y1​x2​y2​x1​y1​x2​y2​​⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤​
最终得到变换矩阵，不过这个八阶矩阵求逆直接劝退我自己写代码的想法了，后来看到这个思路不错，其思路是两个四边形变换之间用一个小的正方形过度一下，这个正方形的坐标不是0就是1大大简化了计算，后面的代码也是基于这个思路做的。

如上图所示，从甲到到乙直接地变换是 A A A,甲到丙的变换是 A 1 A_1 A1​，丙到乙的变换是 A 2 A_2 A2​。先考虑从正方形丙到四边形乙的变换。其中，正方形四个点的坐标为 1 : ( 0 , 1 ) , 2 : ( 1 , 1 ) , 3 : ( 1 , 0 ) , 4 : ( 0 , 0 ) 1:(0,1),2:(1,1),3:(1,0),4:(0,0) 1:(0,1),2:(1,1),3:(1,0),4:(0,0)，乙的四个点的坐标为 1 : ( x 1 , y 1 ) , 2 : ( x 2 , y 2 ) , 3 : ( x 3 , y 3 ) , 4 : ( x 4 , y 4 ) 1:(x_1,y_1),2:(x_2,y_2),3:(x_3,y_3),4:(x_4,y_4) 1:(x1​,y1​),2:(x2​,y2​),3:(x3​,y3​),4:(x4​,y4​)，两边的四个点分别对应，由公式（1），有
x 1 = a 21 + a 31 a 23 + 1 x_1=frac{a_{21} + a_{31}}{a_{23}+1} x1​=a23​+1a21​+a31​​
y 1 = a 22 + a 32 a 23 + 1 y_1=frac{a_{22}+a_{32}}{a_{23}+1} y1​=a23​+1a22​+a32​​
x 2 = a 11 + a 21 + a 31 a 13 + a 23 + 1 x_2=frac{a_{11}+a_{21}+a_{31}}{a_{13}+a_{23}+1} x2​=a13​+a23​+1a11​+a21​+a31​​
y 2 = a 12 + a 22 + a 32 a 13 + a 23 + 1 y_2=frac{a_{12}+a_{22}+a_{32}}{a_{13}+a_{23}+1} y2​=a13​+a23​+1a12​+a22​+a32​​
x 3 = a 11 + a 31 a 13 + 1 x_3=frac{a_{11}+a_{31}}{a_{13}+1} x3​=a13​+1a11​+a31​​
y 3 = a 12 + a 32 a 13 + 1 y_3=frac{a_{12} + a_{32}}{a_{13}+1} y3​=a13​+1a12​+a32​​
x 4 = a 31 x_4=a_{31} x4​=a31​
y 4 = a 32 y_4=a_{32} y4​=a32​
将 x 4 , y 4 x_4,y_4 x4​,y4​带入并化简得

a 21 = a 23 x 1 + x 1 − x 4 a_{21}=a_{23}x_1+x_1-x_4 a21​=a23​x1​+x1​−x4​ ----(2)
a 22 = a 23 y 1 + y 1 − y 4 a_{22}=a_{23}y_1+y_1-y_4 a22​=a23​y1​+y1​−y4​ ----(3)
a 11 = a 13 x 3 + x 3 − x 4 a_{11}=a_{13}x_3+x_3-x_4 a11​=a13​x3​+x3​−x4​ ----(4)
a 12 = a 13 y 3 + y 3 − y 4 a_{12}=a_{13}y_3+y_3-y_4 a12​=a13​y3​+y3​−y4​ ----(5)
a 11 + a 21 − a 13 x 2 − a 23 x 2 = x 2 − x 4 a_{11}+a_{21}-a_{13}x_2-a_{23}x_2=x_2-x_4 a11​+a21​−a13​x2​−a23​x2​=x2​−x4​ ----(6)
a 12 + a 22 − a 13 y 2 − a 23 y 2 = y 2 − y 4 a_{12}+a_{22}-a_{13}y_2-a_{23}y_2=y_2-y_4 a12​+a22​−a13​y2​−a23​y2​=y2​−y4​ ----(7)

将上面(2、3、4、5)带入到（6、7）进行消元得到关于 a 13 , a 23 a_{13},a_{23} a13​,a23​的等式
a 13 ( x 3 − x 2 ) + a 23 ( x 1 − x 2 ) = x 2 − x 1 + x 4 − x 3 a 13 ( y 3 − y 2 ) + a 23 ( y 1 − y 2 ) = y 2 − y 1 + y 4 − y 3 a_{13}(x_3-x_2)+a_{23}(x_1-x_2)=x_2-x_1+x_4-x_3\ a_{13}(y_3-y_2)+a_{23}(y_1-y_2)=y_2-y_1+y_4-y_3 a13​(x3​−x2​)+a23​(x1​−x2​)=x2​−x1​+x4​−x3​a13​(y3​−y2​)+a23​(y1​−y2​)=y2​−y1​+y4​−y3​
即
[ x 3 − x 2 x 1 − x 2 y 3 − y 2 y 1 − y 2 ] . [ a 13 a 23 ] = [ x 2 − x 1 − x 3 + x 4 y 2 − y 1 + y 4 − y 3 ] left[ begin{array}{ccc} x_3-x_2 &x_1-x_2 \ y_3-y_2&y_1-y_2\ end{array} right].left[ begin{array}{ccc} a_{13}\ a_{23}\ end{array} right]=left[ begin{array}{ccc} x_2-x_1-x_3+x_4\ y_2-y_1+y_4-y_3\ end{array} right] [x3​−x2​y3​−y2​​x1​−x2​y1​−y2​​].[a13​a23​​]=[x2​−x1​−x3​+x4​y2​−y1​+y4​−y3​​]
当最左边矩阵可逆时
[ a 13 a 23 ] = [ x 3 − x 2 x 1 − x 2 y 3 − y 2 y 1 − y 2 ] − 1 . [ x 2 − x 1 − x 3 + x 4 y 2 − y 1 + y 4 − y 3 ] left[ begin{array}{ccc} a_{13}\ a_{23}\ end{array} right]=left[ begin{array}{ccc} x_3-x_2 &x_1-x_2 \ y_3-y_2&y_1-y_2\ end{array} right]^{-1}.left[ begin{array}{ccc} x_2-x_1-x_3+x_4\ y_2-y_1+y_4-y_3\ end{array} right] [a13​a23​​]=[x3​−x2​y3​−y2​​x1​−x2​y1​−y2​​]−1.[x2​−x1​−x3​+x4​y2​−y1​+y4​−y3​​]
左边矩阵不可逆时容易计算得到1，2，3三点共线，将透视矩阵代入可得四点共线，也就是该映射退化成将平面映射成一条线了，自己用的时候基本上用不到这种情况，这里不考虑该情况了，得到 a 13 , a 23 a_{13},a_{23} a13​,a23​后带入(2),(3),(4),(5)，可将所有参数计算得出，求出 A 2 A_2 A2​后,再求丙到甲的变换矩阵 A 1 − 1 A_1^{-1} A1−1​,最终得到 A = A 1 ∗ A 2 A=A_1*A_2 A=A1​∗A2​

三，代码

#include 
#include 


typedef struct _M_POINT_ {
	int x;
	int y;
}M_POINT;

typedef struct _POINTS_ {
	M_POINT p1;
	M_POINT p2;
	M_POINT p3;
	M_POINT p4;
}POINTS;


class Persperctive_transform {
public:
	Persperctive_transform() {};
	~Persperctive_transform() {};

	void persperctive_transformation(POINTS p_ori, POINTS p_goal, float T[9])
	{
		float A1[9];
		float A2[9];
		float A1_1[9];

		transformT(p_ori, A1);
		transformT(p_goal, A2);

		mat_inverse_3d(A1, A1_1);

		Mat_multy(A1_1, A2, T);


	}

private:

	void transformT(POINTS p, float A[9])
	{
		//float A[9];

		float det = (p.p3.x - p.p2.x) * (p.p1.y - p.p2.y) - (p.p1.x - p.p2.x) * (p.p3.y - p.p2.y);
		A[2] = (p.p1.y - p.p2.y) * (p.p2.x - p.p1.x + p.p4.x - p.p3.x) + (p.p2.x - p.p1.x) * (p.p2.y - p.p1.y + p.p4.y - p.p3.y);
		A[2] /= det;
		A[0] = A[2] * p.p3.x + p.p3.x - p.p4.x;
		A[1] = A[2] * p.p3.y + p.p3.y - p.p4.y;

		A[5] = (p.p2.y - p.p3.y) * (p.p2.x - p.p1.x + p.p4.x - p.p3.x) + (p.p3.x - p.p2.x) * (p.p2.y - p.p1.y + p.p4.y - p.p3.y);
		A[5] /= det;
		A[3] = A[5] * p.p1.x + p.p1.x - p.p4.x;
		A[4] = A[5] * p.p1.y + p.p1.y - p.p4.y;

		A[6] = p.p4.x;
		A[7] = p.p4.y;
		A[8] = 1;

	}int pow(int x, int n)
	{
		int t = 1;
		for (int i = 0; i < n; i++)
		{
			t *= x;
		}
		return t;
	}

	void mat_inverse_3d(float ori[9], float res[9])
	{


		float det = 0;


		float A11, A12, A13, A21, A22, A23, A31, A32, A33;

		A11 = pow(-1, 2) * (ori[4] * ori[8] - ori[5] * ori[7]);
		A12 = pow(-1, 3) * (ori[3] * ori[8] - ori[5] * ori[6]);
		A13 = pow(-1, 4) * (ori[3] * ori[7] - ori[4] * ori[6]);

		A21 = pow(-1, 3) * (ori[1] * ori[8] - ori[2] * ori[7]);
		A22 = pow(-1, 4) * (ori[0] * ori[8] - ori[2] * ori[6]);
		A23 = pow(-1, 5) * (ori[0] * ori[7] - ori[1] * ori[6]);

		A31 = pow(-1, 4) * (ori[1] * ori[5] - ori[2] * ori[4]);
		A32 = pow(-1, 5) * (ori[0] * ori[5] - ori[2] * ori[3]);
		A33 = pow(-1, 6) * (ori[0] * ori[4] - ori[1] * ori[3]);


		det += pow(-1, 2) * ori[0] * (ori[4] * ori[8] - ori[7] * ori[5]);
		det += pow(-1, 3) * ori[3] * (ori[1] * ori[8] - ori[7] * ori[2]);
		det += pow(-1, 4) * ori[6] * (ori[1] * ori[5] - ori[2] * ori[4]);

		if (abs(det) < 0.0001)
		{
			std::cout << "no inversen";
			return;
		}

		res[0] = A11 / det;
		res[1] = A21 / det;
		res[2] = A31 / det;
		res[3] = A12 / det;
		res[4] = A22 / det;
		res[5] = A32 / det;
		res[6] = A13 / det;
		res[7] = A23 / det;
		res[8] = A33 / det;



	}

	void Mat_multy(float left[9], float right[9], float res[9])
	{
		for (int i = 0; i < 9; i++)
		{
			int row = i / 3;
			int col = i % 3;

			float value = 0;
			for (int i = 0; i < 3; i++)
			{
				value += left[row * 3 + i] * right[i * 3 + col];
			}
			if (abs(value) < 0.0001) value = 0;
			res[i] = value;

		}
	}

};


int main()
{

	Persperctive_transform PT;
	POINTS p_ori;
	POINTS p_goal;

	

	cv::Mat img=cv::imread("C:\Users\user\Desktop\tes\lena.jpg");

	cv::Mat res = cv::Mat::zeros(img.rows, img.cols, CV_8UC3);
	

	p_ori.p1.x = 0; p_ori.p1.y = 0;
	p_ori.p2.x = 0; p_ori.p2.y = img.cols;
	p_ori.p3.x = img.rows; p_ori.p3.y = img.cols;
	p_ori.p4.x = img.rows; p_ori.p4.y = 0;

	p_goal.p1.x = 150; p_goal.p1.y = 50;
	p_goal.p2.x = 0; p_goal.p2.y = img.cols;
	p_goal.p3.x = img.rows-100; p_goal.p3.y = img.cols-100;
	p_goal.p4.x = img.rows; p_goal.p4.y = 0+0;

	float T[9];

	PT.persperctive_transformation(p_ori, p_goal, T);

	for (int i = 0; i < img.rows; i++)
	{
		for (int j = 0; j < img.cols; j++)
		{
			float w = T[2] * i + T[5] * j + T[8];
			float row = T[0] * i + T[3] * j + T[6];
			float col = T[1] * i + T[4] * j + T[7];
			
			
			int R = row / w;
			int C = col / w;

			
			if (R < 0 || C < 0 || R>img.rows || C>img.cols) continue;
			res.at(R, C) = img.at(i, j);
			

		}
	}

	cv::Mat res2 = cv::Mat::zeros(img.rows, img.cols, CV_8UC3);

	p_ori.p1.x = 150; p_ori.p1.y = 50;
	p_ori.p2.x = 0; p_ori.p2.y = img.cols;
	p_ori.p3.x = img.rows-100; p_ori.p3.y = img.cols-100;
	p_ori.p4.x = img.rows; p_ori.p4.y = 0;

	p_goal.p1.x = 0; p_goal.p1.y = 0;
	p_goal.p2.x = 0; p_goal.p2.y = img.cols;
	p_goal.p3.x = img.rows - 0; p_goal.p3.y = img.cols - 0;
	p_goal.p4.x = img.rows; p_goal.p4.y = 0 + 0;


	PT.persperctive_transformation(p_ori, p_goal, T);

	for (int i = 0; i < img.rows; i++)
	{
		for (int j = 0; j < img.cols; j++)
		{
			float w = T[2] * i + T[5] * j + T[8];
			float row = T[0] * i + T[3] * j + T[6];
			float col = T[1] * i + T[4] * j + T[7];

			
			int R = row / w;
			int C = col / w;

			if (R < 0 || C < 0 || R>img.rows || C>img.cols) continue;
			res2.at(R, C) = res.at(i, j);

		}
	}

	return 0;
}