透视变换

2024-05-07 本文已影响0人 leon_tly

透视变换原理及求解推导

透视变换（Perspective Transformation）是将二维的图片投影到一个三维视平面上，然后再转换到二维坐标下，所以也称为投影映射（Projective Mapping）。简单来说就是二维 → 三维 → 二维的一个过程。
矩阵形式表示
$\begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ \end{bmatrix} * \begin{bmatrix} x\\ y\\ 1\\ \end{bmatrix} = \begin{bmatrix} X\\ Y\\ Z\\ \end{bmatrix}$
其中 $\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{bmatrix}$ 表示线性变换
其中 $\begin{bmatrix} a_{31}&a_{32} \end{bmatrix}$ 表示透视变换
其余表示平移变换
通过除以 $Z$ 轴将三维坐标转换为二维坐标 $Z=a_{31}x + a_{32}y + a_{33}$
$x^{'} = \frac{X}{Z} = \frac{a_{11}x + a_{12}y + a_{13}}{a_{31}x + a_{32}y + a_{33}}$
$y^{'} = \frac{Y}{Z} = \frac{a_{21}x + a_{22}y + a_{23}}{a_{31}x + a_{32}y + a_{33}}$
其中 $a_{33} = 1$
有8个未知数，因此需要原图上的4个点和目标图上的4个点来构成8个方程来求解变换矩阵。
$x^{'} = \frac{X}{Z} = \frac{a_{11}x + a_{12}y + a_{13}}{a_{31}x + a_{32}y + 1}$
$({a_{31}x + a_{32}y + 1}) x^{'} = a_{11}x + a_{12}y + a_{13}$
$a_{11}x + a_{12}y + a_{13} - {a_{31}xx^{'} - a_{32}yx^{'}} = x^{'}$
同理得到
$a_{21}x + a_{22}y + a_{23} - {a_{31}xy^{'} - a_{32}yy^{'}} = y^{'}$

当有 $(x_{0}, y_{0})$ , $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ 和 $(x^{'}_{0}, y^{'}_{0})$ , $(x^{'}_{1}, y^{'}_{1})$ , $(x^{'}_{2}, y^{'}_{2})$ , $(x^{'}_{3}, y^{'}_{3})$ 8个点时可以得到方程

$\begin{cases} a_{11}x_{0} + a_{12}y_{0} + a_{13} - {a_{31}x_{0}x_{0}^{'} - a_{32}y_{0}x_{0}^{'}} = x_{0}^{'}\\ a_{21}x_{0} + a_{22}y_{0} + a_{23} - {a_{31}x_{0}y_{0}^{'} - a_{32}y_{0}y_{0}^{'}} = y_{0}^{'}\\ a_{11}x_{1} + a_{12}y_{1} + a_{13} - {a_{31}x_{1}x_{1}^{'} - a_{32}y_{1}x_{1}^{'}} = x_{1}^{'}\\ a_{21}x_{1} + a_{22}y_{1} + a_{23} - {a_{31}x_{1}y_{1}^{'} - a_{32}y_{1}y_{1}^{'}} = y_{1}^{'}\\ a_{11}x_{2} + a_{12}y_{2} + a_{13} - {a_{31}x_{2}x_{2}^{'} - a_{32}y_{2}x_{2}^{'}} = x_{2}^{'}\\ a_{21}x_{2} + a_{22}y_{2} + a_{23} - {a_{31}x_{2}y_{2}^{'} - a_{32}y_{2}y_{2}^{'}} = y_{2}^{'}\\ a_{11}x_{3} + a_{12}y_{3} + a_{13} - {a_{31}x_{3}x_{3}^{'} - a_{32}y_{3}x_{3}^{'}} = x_{3}^{'}\\ a_{21}x_{3} + a_{22}y_{3} + a_{23} - {a_{31}x_{3}y_{3}^{'} - a_{32}y_{3}y_{3}^{'}} = y_{3}^{'}\\ \end{cases}$
写成矩阵形式则有
$\begin{bmatrix} x_{0} & y_{0} & 1 & 0 & 0 & 0 & -x_{0}x_{0}^{'} & -y_{0}x_{0}^{'}\\ 0 & 0 & 0 & x_{0} & y_{0} & 1 & -x_{0}y_{0}^{'} & -y_{0}y_{0}^{'}\\ x_{1} & y_{1} & 1 & 0 & 0 & 0 & -x_{1}x_{1}^{'} & -y_{1}x_{1}^{'}\\ 0 & 0 & 0 & x_{1} & y_{1} & 1 & -x_{1}y_{1}^{'} & -y_{1}y_{1}^{'}\\ x_{2} & y_{2} & 1 & 0 & 0 & 0 & -x_{2}x_{2}^{'} & -y_{2}x_{2}^{'}\\ 0 & 0 & 0 & x_{2} & y_{2} & 1 & -x_{2}y_{2}^{'} & -y_{2}y_{2}^{'}\\ x_{3} & y_{3} & 1 & 0 & 0 & 0 & -x_{3}x_{3}^{'} & -y_{3}x_{3}^{'}\\ 0 & 0 & 0 & x_{3} & y_{3} & 1 & -x_{3}y_{3}^{'} & -y_{3}y_{3}^{'}\\ \end{bmatrix} * \begin{bmatrix} a_{11} \\ a_{12}\\ a_{13}\\ a_{21}\\ a_{22}\\ a_{23}\\ a_{31}\\ a_{32}\\ \end{bmatrix} = \begin{bmatrix} x_{0}^{'} \\ y_{0}^{'}\\ x_{1}^{'}\\ y_{1}^{'}\\ x_{2}^{'}\\ y_{2}^{'}\\ x_{3}^{'}\\ y_{3}^{'}\\ \end{bmatrix}$

高斯消元法求解代码

struct Point{
    float x;
    float y;
};


static std::vector<std::vector<float>> getPerspectiveTransform(const std::vector<Point>& src, const std::vector<Point>& dst) 
{
    std::vector<std::vector<float>> A(8, std::vector<float>(8, 0));
    std::vector<float> b(8, 0);
    
    // 矩阵赋初始值
    for (int i = 0; i < 4; ++i) {
        A[i * 2][0] = src[i].x;
        A[i * 2][1] = src[i].y;
        A[i * 2][2] = 1.0;
        A[i * 2][6] = -src[i].x * dst[i].x;
        A[i * 2][7] = -src[i].y * dst[i].x;

        A[i * 2 + 1][3] = src[i].x;
        A[i * 2 + 1][4] = src[i].y;
        A[i * 2 + 1][5] = 1.0;
        A[i * 2 + 1][6] = -src[i].x * dst[i].y;
        A[i * 2 + 1][7] = -src[i].y * dst[i].y;

        b[i * 2] = dst[i].x;
        b[i * 2 + 1] = dst[i].y;
    }

    std::vector<float> x(8);
    for (int i = 0; i < 8; ++i) {
        x[i] = b[i];
    }

    for (int i = 0; i < 8; ++i) {
        int pivot = i;
        for (int j = i + 1; j < 8; ++j) {
            if (std::abs(A[j][i]) > std::abs(A[pivot][i])) {
                pivot = j;
            }
        }
        // 按照列循环，将该列中值比较大的行向上交换
        std::swap(A[i], A[pivot]);
        std::swap(x[i], x[pivot]);
        // 消除主元
        for (int j = i + 1; j < 8; ++j) {
            float factor = A[j][i] / A[i][i];
            for (int k = i; k < 8; ++k) {
                A[j][k] -= factor * A[i][k];
            }
            x[j] -= factor * x[i];
        }
    }
    // 求解
    std::vector<float> H(9);
    for (int i = 7; i >= 0; --i) {
        float sum = 0;
        for (int j = i + 1; j < 8; ++j) {
            sum += A[i][j] * H[j];
        }
        H[i] = (x[i] - sum) / A[i][i];
    }
    H[8] = 1.0;

    std::vector<std::vector<float>> transform(3, std::vector<float>(3));
    for (int i = 0; i < 3; ++i) {
        for (int j = 0; j < 3; ++j) {
            transform[i][j] = H[i * 3 + j];
        }
    }

    return transform;
}

透视变换

透视变换原理及求解推导

高斯消元法求解代码

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