多视图几何---基础矩阵

「爱情、让人受尽委屈。」 2022-02-19 16:41 625阅读 0赞

**目录**

一、基础矩阵(Fundamental Matrix)

1、对极几何

2、基础矩阵推导

二、实现结果

三、实现代码

四、问题

--------------------

## 一、基础矩阵(**Fundamental Matrix)** ##

这时候就要推荐大家一首歌了**?**[The fundamental matrix song][]，相信这首歌对了解基础矩阵有所帮助~

基础矩阵体现了两视图几何（对极几何，epipolar geometry）的内在射影几何（projective geometry）关系，基础矩阵可以由两照相机的投影变换矩阵（相对旋转R和平移t）表示。基础矩阵的秩小于等于2，自由度为7。

### 1、对极几何 ###

我们先了解一下什么是对极几何

**对极几何**是视图几何理论的基础，对极几何（Epipolar Geometry）描述了同一场景两幅图像之间的视觉几何关系。

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70][]

空间中的点X在 两个像平面中的投影分别为 x, x’，则X与C, C', x, x'共面

对极几何关系中主要包含以下元素：

对极点=基线与像平面相交点（光心在另一幅图像中的投影）

对极平面 = 包含基线的平面

对极线 = 对极平面与像平面的交线

### 2、基础矩阵推导 ###

给定一对图像，对于图像上的每点x，在另一幅图像中存在一条对应的对极线l'。在第二幅图像上，任何与该点x匹配的点x'必然在对极线l'上。该对极线试过点x与第一个摄像机中心C的射线在第二副图像上的投影。因此，存在一个从一幅图像上的点到另一幅图像上与之对应的对极线的映射![x\\rightarrow l'][x_rightarrow l],这个映射是一个对射，即是由基本矩阵F标识的从点到直线的射影映射。

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 1][]

设X在C，C'坐标系中的相对坐标分别p，p'，则有p = Rp'+T

根据三线共面，有

![(p-T)^\{T\}(T\\times p)=0 \\Rightarrow (R^\{T\}p')^\{T\}(T\\times p)=0][p-T_T_T_times p_0 _Rightarrow _R_T_p_T_T_times p_0]

![T\\times p=Sp][T_times p_Sp]

![S=\\begin\{bmatrix\} 0 & -T\_\{z\}& T\_\{y\} \\\\ T\_\{z\} & 0& -T\_\{x\}\\\\ -T\_\{y\}& T\_\{x\}& 0 \\end\{bmatrix\}][S_begin_bmatrix_ 0 _ -T_z_ T_y_ _ T_z_ _ 0_ -T_x_ -T_y_ T_x_ 0 _end_bmatrix]

所以有

![(R^\{T\}p')^\{T\}(Sp)=0][R_T_p_T_Sp_0]

![（p'^\{T\}R)(Sp)=0][p_T_R_Sp_0]![(p'^\{T\}R)(Sp)=0][p_T_R_Sp_0 1]

![p'^\{T\}Ep=0][p_T_Ep_0]

E为本质矩阵，描述了空间中的点在两个坐标系中的坐标对应关系。

K和K'分别为两个相机的内参矩阵，则有

![p=K^\{-1\}x][p_K_-1_x]  和   ![p'=K'^\{-1\}x'][p_K_-1_x 1]

![(K^\{-1\}x')E(K^\{-1\}x)=0][K_-1_x_E_K_-1_x_0]

得![x'^\{T\}Fx=0][x_T_Fx_0]

F则为基础矩阵，可用于简化匹配和去除错配特征。

接下来让我们来尝试计算图相对的特征匹配，并估计基础矩阵和计算照相机矩阵

## 二、实现结果 ##

**(1) 两幅图像**

室内图像的基础矩阵(F)和投影矩阵(P)

![20190417142155808.png][]

sift特征匹配结果：                              投影点：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 2][]        ![20190417132841516.png][]          ![20190417132859603.png][]

保留照相机前面的点的匹配

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 3][]              ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 4][]

室外图像的基础矩阵(F)和投影矩阵(P)

![20190417145025168.png][]

sift特征匹配结果：                              投影点：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 5][]               ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 6][]        ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 7][]

保留照相机前面的点的匹配：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 8][]              ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 9][]

**(2) 三幅图像(可以更多图像)**

图像1和图像2的sift匹配                       移除不在照相机前面点之后的剩余匹配

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 10][]       ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 11][]        ![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 12][]

图一和图二的三维点：

![20190417165625890.png][]       ![20190417165638309.png][]

基础矩阵

![20190417155024230.png][]

这是第二副图像的照相机矩阵

![20190417154957988.png][]

这是第三幅图像的照相机矩阵

![20190417154452188.png][]

## 三、实现代码 ##

**（1）照相机和三维结构的计算**

# coding: utf-8
    
    from PIL import Image
    from numpy import *
    from pylab import *
    import numpy as np
    
    from PCV.geometry import camera
    from PCV.geometry import homography
    from PCV.geometry import sfm
    from PCV.localdescriptors import sift
    from imp import reload
    
    camera = reload(camera)
    homography = reload(homography)
    sfm = reload(sfm)
    sift = reload(sift)
    
    
    im1 = array(Image.open('E:/study_work/python/image/test1.jpg'))
    sift.process_image('E:/study_work/python/image/test1.jpg', 'im1.sift')
    l1, d1 = sift.read_features_from_file('im1.sift')
    
    im2 = array(Image.open('E:/study_work/python/image/test2.jpg'))
    sift.process_image('E:/study_work/python/image/test2.jpg', 'im2.sift')
    l2, d2 = sift.read_features_from_file('im2.sift')
    
    # 匹配特征
    matches = sift.match_twosided(d1, d2)
    ndx = matches.nonzero()[0]
    x1 = homography.make_homog(l1[ndx, :2].T)
    ndx2 = [int(matches[i]) for i in ndx]
    x2 = homography.make_homog(l2[ndx2, :2].T)
    
    d1n = d1[ndx]
    d2n = d2[ndx2]
    x1n = x1.copy()
    x2n = x2.copy()
    print(len(ndx))
    
    figure(figsize=(16,16))
    sift.plot_matches(im1, im2, l1, l2, matches, True)
    show()
    
    def F_from_ransac(x1, x2, model, maxiter=5000, match_threshold=1e-6):
        """ Robust estimation of a fundamental matrix F from point
        correspondences using RANSAC (ransac.py from
        http://www.scipy.org/Cookbook/RANSAC).
    
        input: x1, x2 (3*n arrays) points in hom. coordinates. """
    
        from PCV.tools import ransac
        data = np.vstack((x1, x2))
        d = 5 # 20 is the original
        # 计算F，并返回正确点索引
        F, ransac_data = ransac.ransac(data.T, model, 8, maxiter, match_threshold, d, return_all=True)
        return F, ransac_data['inliers']  # 返回最佳基础矩阵
    
    # 使用RANSAC方法估计E
    model = sfm.RansacModel()
    F, inliers = F_from_ransac(x1n, x2n, model, maxiter=5000, match_threshold=1e-5)
    # 计算照相机矩阵
    P1 = array([[1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0]])
    P2 = sfm.compute_P_from_fundamental(F)  # 从基础矩阵中计算第二个照相机矩阵
    
    print 'P2', P2
    print 'F', F
    
    # 三角剖分正确点，并已初步在所有照相机前面的点
    X = sfm.triangulate(x1n[:, inliers], x2n[:, inliers], P1, P2)
    # plot the projection of X
    cam1 = camera.Camera(P1)
    cam2 = camera.Camera(P2)
    x1p = cam1.project(X)    # 投影点
    x2p = cam2.project(X)
    figure()
    imshow(im1)
    gray()
    # 输出X
    plot(x1p[0], x1p[1], 'o')
    # plot(x1[0], x1[1], 'r.')  # 基础矩阵的点
    axis('off')
    figure()
    imshow(im2)
    gray()
    plot(x2p[0], x2p[1], 'o')
    # plot(x2[0], x2[1], 'r.')
    axis('off')
    show()
    
    
    figure(figsize=(16, 16))
    imj = sift.appendimages(im1, im2)
    imj = vstack((imj, imj))
    
    imshow(imj)
    
    cols1 = im1.shape[1]
    rows1 = im1.shape[0]
    for i in range(len(x1p[0])):
        if (0<= x1p[0][i]<cols1) and (0<= x2p[0][i]<cols1) and (0<=x1p[1][i]<rows1) and (0<=x2p[1][i]<rows1):
            plot([x1p[0][i], x2p[0][i]+cols1],[x1p[1][i], x2p[1][i]],'c')
    axis('off')
    show()
    
    
    d1p = d1n[inliers]
    d2p = d2n[inliers]
    
    
    # Read features
    im3 = array(Image.open('E:/study_work/python/image/test3.jpg'))
    sift.process_image('E:/study_work/python/image/test3.jpg', 'im3.sift')
    l3, d3 = sift.read_features_from_file('im3.sift')
    
    
    matches13 = sift.match_twosided(d1p, d3)
    
    
    ndx_13 = matches13.nonzero()[0]
    x1_13 = homography.make_homog(x1p[:, ndx_13])
    ndx2_13 = [int(matches13[i]) for i in ndx_13]
    x3_13 = homography.make_homog(l3[ndx2_13, :2].T)
    
    figure(figsize=(16, 16))
    imj = sift.appendimages(im1, im3)  # 创建一个新的图像分别包含两个匹配的图像和他们之间的匹配对的连接直线
    imj = vstack((imj, imj))
    
    imshow(imj)
    
    cols1 = im1.shape[1]
    rows1 = im1.shape[0]
    for i in range(len(x1_13[0])):
        if (0<= x1_13[0][i]<cols1) and (0<= x3_13[0][i]<cols1) and (0<=x1_13[1][i]<rows1) and (0<=x3_13[1][i]<rows1):
            plot([x1_13[0][i], x3_13[0][i]+cols1],[x1_13[1][i], x3_13[1][i]],'c')
    axis('off')
    show()
    
    
    P3 = sfm.compute_P(x3_13, X[:, ndx_13])
    
    
    print 'P3', P3
    
    
    print 'P1', P1
    print 'P2', P2
    print 'P3', P3

**（2）三维重建 **

# coding: utf-8
    from PIL import Image
    from pylab import *
    from PCV.geometry import camera
    from PCV.geometry import homography
    from PCV.geometry import sfm
    from PCV.localdescriptors import sift
    # 标定矩阵
    K = array([[3000, 0, 932], [0, 2398, 628], [0, 0, 1]])
    # 载入图像，并计算特征
    
    im1 = array(Image.open('E:/study_work/python/image/test1.jpg'))
    sift.process_image('E:/study_work/python/image/test1.jpg', 'im1.sift')
    l1, d1 = sift.read_features_from_file('im1.sift')
    
    im2 = array(Image.open('E:/study_work/python/image/test2.jpg'))
    sift.process_image('E:/study_work/python/image/test2.jpg', 'im2.sift')
    l2, d2 = sift.read_features_from_file('im2.sift')
    
    # 匹配特征
    matches = sift.match_twosided(d1,d2)
    ndx = matches.nonzero()[0]
    
    # 使用齐次坐标表示，并使用 inv(K) 归一化
    x1 = homography.make_homog(l1[ndx,:2].T)
    ndx2 = [int(matches[i]) for i in ndx]
    x2 = homography.make_homog(l2[ndx2,:2].T)
    x1n = dot(inv(K),x1)
    x2n = dot(inv(K),x2)
    
    # 使用 RANSAC 方法估计 E
    model = sfm.RansacModel()
    E,inliers = sfm.F_from_ransac(x1n,x2n,model)
    # 计算照相机矩阵（P2 是 4 个解的列表）
    P1 = array([[1,0,0,0],[0,1,0,0],[0,0,1,0]])
    P2 = sfm.compute_P_from_essential(E)
    
    # 选取点在照相机前的解
    ind = 0
    maxres = 0
    for i in range(4):
        # 三角剖分正确点，并计算每个照相机的深度
        X = sfm.triangulate(x1n[:,inliers],x2n[:,inliers],P1,P2[i])
        d1 = dot(P1,X)[2]
        d2 = dot(P2[i],X)[2]
    
        if sum(d1>0)+sum(d2>0) > maxres:
            maxres = sum(d1>0)+sum(d2>0)
            ind = i
            infront = (d1>0) & (d2>0)
        # 三角剖分正确点，并移除不在所有照相机前面的点
        X = sfm.triangulate(x1n[:,inliers],x2n[:,inliers],P1,P2[ind])
        X = X[:,infront]
    
        # 绘制三维图像
        from mpl_toolkits.mplot3d import axes3d
        fig = figure()
        ax = fig.gca(projection='3d')
        ax.plot(-X[0], X[1], X[2], 'k.')
        axis('off')
        # 绘制 X 的投影 import camera
        # 绘制三维点
        cam1 = camera.Camera(P1)
        cam2 = camera.Camera(P2[ind])
        x1p = cam1.project(X)
        x2p = cam2.project(X)
    
        # 反 K 归一化
        x1p = dot(K, x1p)
        x2p = dot(K, x2p)
        figure()
        imshow(im1)
        gray()
        plot(x1p[0], x1p[1], 'o')
        plot(x1[0], x1[1], 'r.')
        axis('off')
        figure()
        imshow(im2)
        gray()
        plot(x2p[0], x2p[1], 'o')
        plot(x2[0], x2[1], 'r.')
        axis('off')
        show()

## 四、问题 ##

在实现过程中可能会遇到问题，如下

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 13][]

遇到此问题可能是由于你的图片不够多的匹配点，一般来说在300以上就可以成功

看匹配点数量可以输出

print(len(ndx))

更改阈值：将match\_threshold值设置大一些

![20190417161824904.png][]

如果匹配点数大于300之后仍报错，则可更改d的值，可以将d设置的较小，也能让程序正常运行

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 14][]

[The fundamental matrix song]: https://tv.sohu.com/v/dXMvNjMzMjc3NzcvMjYzNjk1ODYuc2h0bWw=.html
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70]: /images/20220219/b48b888619114945800f503ef73d52b2.png
[x_rightarrow l]: https://private.codecogs.com/gif.latex?x%5Crightarrow%20l%27
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 1]: /images/20220219/4059916e771f4f7397b9aa9ed3819a5d.png
[p-T_T_T_times p_0 _Rightarrow _R_T_p_T_T_times p_0]: https://private.codecogs.com/gif.latex?%28p-T%29%5E%7BT%7D%28T%5Ctimes%20p%29%3D0%20%5CRightarrow%20%28R%5E%7BT%7Dp%27%29%5E%7BT%7D%28T%5Ctimes%20p%29%3D0
[T_times p_Sp]: https://private.codecogs.com/gif.latex?T%5Ctimes%20p%3DSp
[S_begin_bmatrix_ 0 _ -T_z_ T_y_ _ T_z_ _ 0_ -T_x_ -T_y_ T_x_ 0 _end_bmatrix]: https://private.codecogs.com/gif.latex?S%3D%5Cbegin%7Bbmatrix%7D%200%20%26%20-T_%7Bz%7D%26%20T_%7By%7D%20%5C%5C%20T_%7Bz%7D%20%26%200%26%20-T_%7Bx%7D%5C%5C%20-T_%7By%7D%26%20T_%7Bx%7D%26%200%20%5Cend%7Bbmatrix%7D
[R_T_p_T_Sp_0]: https://private.codecogs.com/gif.latex?%28R%5E%7BT%7Dp%27%29%5E%7BT%7D%28Sp%29%3D0
[p_T_R_Sp_0]: https://private.codecogs.com/gif.latex?%25uFF08p%27%5E%7BT%7DR%29%28Sp%29%3D0
[p_T_R_Sp_0 1]: https://private.codecogs.com/gif.latex?%28p%27%5E%7BT%7DR%29%28Sp%29%3D0
[p_T_Ep_0]: https://private.codecogs.com/gif.latex?p%27%5E%7BT%7DEp%3D0
[p_K_-1_x]: https://private.codecogs.com/gif.latex?p%3DK%5E%7B-1%7Dx
[p_K_-1_x 1]: https://private.codecogs.com/gif.latex?p%27%3DK%27%5E%7B-1%7Dx%27
[K_-1_x_E_K_-1_x_0]: https://private.codecogs.com/gif.latex?%28K%5E%7B-1%7Dx%27%29E%28K%5E%7B-1%7Dx%29%3D0
[x_T_Fx_0]: https://private.codecogs.com/gif.latex?x%27%5E%7BT%7DFx%3D0
[20190417142155808.png]: /images/20220219/4fb78e7255104947bf702b493a1c33de.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz_size_16_color_FFFFFF_t_70 2]: https://img-blog.csdnimg.cn/20190417132818548.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM3Mzc0NjQz,size_16,color_FFFFFF,t_70
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