使用 R 的 Motion 的简单正交结构——确定度量约束

Question

我想根据 Tomasi 和 Kanade [1992] 从运动程序构建一个简单的 结构。这篇文章可以在下面找到：

https://people.eecs.berkeley.edu/~yang/courses/cs294-6/papers/TomasiC_Shape%20and%20motion%20from%20image%20streams%20under%20orthography.pdf

这种方法看起来优雅而简单，但是，我在计算上述参考的等式 16 中概述的 度量约束 时遇到了问题。

我正在使用 R 并在下面概述了我的工作：

给定一组图像

我想跟踪三个柜门的角和一张图片（图片上的黑点）。首先我们读入点作为矩阵 w where

最终，我们想要将 w 分解为描述 3 维点的旋转矩阵 R 和形状矩阵 S。我会尽可能多地保留细节，但可以从 Tomasi 和 Kanade [1992] 的论文中收集到完整的数学描述。

我提供以下w：

w.vector=c(0.2076,0.1369,0.1918,0.1862,0.1741,0.1434,0.176,0.1723,0.2047,0.233,0.3593,0.3668,0.3744,0.3593,0.3876,0.3574,0.3639,0.3062,0.3295,0.3267,0.3128,0.2811,0.2979,0.2876,0.2782,0.2876,0.3838,0.3819,0.3819,0.3649,0.3913,0.3555,0.3593,0.2997,0.3202,0.3137,0.31,0.2718,0.2895,0.2867,0.825,0.7703,0.742,0.7251,0.7232,0.7138,0.7345,0.6911,0.1937,0.1248,0.1723,0.1741,0.1657,0.1313,0.162,0.1657,0.8834,0.8118,0.7552,0.727,0.7364,0.7232,0.7288,0.6892,0.4309,0.3798,0.4021,0.3965,0.3844,0.3546,0.3695,0.3583,0.314,0.3065,0.3989,0.3876,0.3857,0.3781,0.3989,0.3593,0.5184,0.4849,0.5147,0.5193,0.5109,0.4812,0.4979,0.4849,0.3536,0.3517,0.4121,0.3951,0.3951,0.3781,0.397,0.348,0.5175,0.484,0.5091,0.5147,0.5128,0.4784,0.4905,0.4821,0.7722,0.7326,0.7326,0.7232,0.7232,0.7119,0.7402,0.7006,0.4281,0.3779,0.3918,0.3863,0.3825,0.3472,0.3611,0.3537,0.8043,0.7628,0.7458,0.7288,0.727,0.7213,0.7364,0.6949,0.5789,0.5491,0.5761,0.5817,0.5733,0.5444,0.5537,0.5379,0.3649,0.3536,0.4177,0.3951,0.3857,0.3819,0.397,0.3461,0.697,0.671,0.6821,0.6821,0.6719,0.6412,0.6468,0.6235,0.3744,0.3649,0.4159,0.3819,0.3781,0.3612,0.3763,0.314,0.7008,0.6691,0.6794,0.6812,0.6747,0.6393,0.6412,0.6235,0.7571,0.7345,0.7439,0.7496,0.7402,0.742,0.7647,0.7213,0.5817,0.5463,0.5696,0.5779,0.5761,0.5398,0.551,0.5398,0.7665,0.7326,0.7439,0.7345,0.7288,0.727,0.7515,0.7062,0.8301,0.818,0.8571,0.8878,0.8766,0.8561,0.858,0.8394,0.4121,0.3876,0.4347,0.397,0.38,0.3631,0.3668,0.2971,0.912,0.8962,0.9185,0.939,0.9259,0.898,0.8887,0.8571,0.3989,0.3781,0.4215,0.3725,0.3612,0.3461,0.3423,0.2782,0.9092,0.8952,0.9176,0.9399,0.925,0.8971,0.8887,0.8571,0.4743,0.4536,0.4894,0.4517,0.446,0.4328,0.4385,0.3706,0.8273,0.8171,0.8571,0.8878,0.8766,0.8543,0.8561,0.8394,0.4743,0.4554,0.4969,0.4668,0.4536,0.4404,0.4536,0.3857) w=matrix(w.vector,ncol=16,nrow=16,byrow=FALSE)

然后根据等式 2 创建注册测量矩阵 wm 作为

来自

wm = w - rowMeans(w)

我们可以将 wm 分解为“2FxP”矩阵 o1 对角线 'PxP' 矩阵 e 和 'PxP' 矩阵 o2使用奇异值分解。

svdwm <- svd(wm) o1 <- svdwm$u e <- diag(svdwm$d) o2 <- t(svdwm$v) ## dont forget the transpose!

但是，由于噪音，我们只关注 o1 的前 3 列、e 的前 3 个值和 o2 的前 3 行：

o1p <- svdwm$u[,1:3] ep <- diag(svdwm$d[1:3]) o2p <- t(svdwm$v)[1:3,] ## dont forget the transpose!

现在我们可以求解方程(14)中的rhat和shat

来自

rhat <- o1p%*%ep^(1/2) shat <- ep^(1/2) %*% o2p

然而，这些结果并不是唯一的，我们仍然需要通过等式（15）求解R和S

通过使用等式 (16) 的度量约束

现在我需要找到 Q。我相信有两种可能的方法，但不清楚如何使用。

方法 1 涉及求解 B，其中 B=Q%*%solve(Q) 然后使用 Cholesky 分解求 Q。方法 1 似乎是文献中的常见选择，但是，细节不多给出了如何实际求解线性系统。很明显，B 是 6 个未知数的“3x3”对称矩阵。但是，考虑到度量约束（方程式 16），我不知道如何在给定 3 个方程式的情况下求解 6 个未知数。我是不是忘记了属性个对称矩阵？

方法 II 涉及使用非线性方法来估计 Q，并且在运动文献的结构中不太常用。

任何人都可以就如何解决这个问题提供一些建议吗？提前致谢，如果我需要更清楚地回答我的问题，请告诉我。

Answer 1

$\hat{i}_{f}$ 可以写成 ${a_if b_if c_if}$ 。

$\hat{j}_{f}$ 可以写成 ${a_jf b_jf c_jf}$ .

$QQ^T$ 可以写成 $\begin{pmatrix} q_{11}&q_{12}&q_{13}\\ q_{12}&q_{22}&q_{23}\\ q_{13}&q_{23}&q_{33}\\ \end{pmatrix}$ .

所以我们的方程是：

$\begin{pmatrix} a_{if}&b_{if}&c_{if} \end{pmatrix} \cdot \begin{pmatrix} q_{11}&q_{12}&q_{13}\\ q_{12}&q_{22}&q_{23}\\ q_{13}&q_{23}&q_{33}\\ \end{pmatrix} \cdot$ $\begin{pmatrix} a_{if}\\ b_{if}\\ c_{if}\\ \end{pmatrix} = 1$

$\begin{pmatrix} a_{jf}&b_{jf}&c_{jf} \end{pmatrix} \cdot \begin{pmatrix} q_{11}&q_{12}&q_{13}\\ q_{12}&q_{22}&q_{23}\\ q_{13}&q_{23}&q_{33}\\ \end{pmatrix} \cdot$ $\begin{pmatrix} a_{jf}\\ b_{jf}\\ c_{jf}\\ \end{pmatrix} = 1$

$\begin{pmatrix} a_{if}&b_{if}&c_{if} \end{pmatrix} \cdot \begin{pmatrix} q_{11}&q_{12}&q_{13}\\ q_{12}&q_{22}&q_{23}\\ q_{13}&q_{23}&q_{33}\\ \end{pmatrix} \cdot$ $\begin{pmatrix} a_{jf} \\ b_{jf} \\ c_{jf} \end{pmatrix} = 0$

所以第一个等式可以写成：

$a_{if}\cdot a_{if}\cdot q_{11}+ a_{if}\cdot b_{if}\cdot q_{12}+ a_{if}\cdot c_{if}\cdot q_{13}$ $+ b_{if}\cdot a_{if}\cdot q_{12}+ b_{if}\cdot b_{if}\cdot q_{22}+ b_{if}\cdot c_{if}\cdot q_{23}$ $+ c_{if}\cdot a_{if}\cdot q_{13}+ c_{if}\cdot b_{if}\cdot q_{23}+ c_{if}\cdot c_{if}\cdot q_{33}=1$

$\Leftrightarrow a_{if}\cdot a_{if}\cdot q_{11}+ 2\cdot a_{if}\cdot b_{if}\cdot q_{12}+ 2\cdot a_{if}\cdot c_{if}\cdot q_{13}+$ $b_{if}\cdot b_{if}\cdot q_{22}+ 2\cdot b_{if}\cdot c_{if}\cdot q_{23}+ c_{if}\cdot c_{if}\cdot q_{33}=1$

相当于

$\begin{pmatrix} a_{if}a_{if}& 2a_{if}b_{if}& 2a_{if}c_{if}& b_{if}b_{if}& 2b_{if}c_{if}& c_{if}c_{if} \end{pmatrix}\cdot\\[35pt].$ $\begin{pmatrix} q_{11}\\ q_{12}\\ q_{13}\\ q_{22}\\ q_{23}\\ q_{33}\\ \end{pmatrix} = 1$

为了简短起见，我们现在定义：

$V_{x,y}\;:=\begin{pmatrix} a_{x}a_{y}& a_{x}b_{y}+a_{y}b_{x}& a_{x}c_{y}+a_{y}c_{x}& b_{x}b_{y}& b_{x}c_{y}+b_{y}c_{x}& c_{x}c_{y} \end{pmatrix}$ （我知道间距非常小，但是，是的，这是一个 Vector...）

所以对于所有不同帧 f 中的所有方程，我们可以写一个大方程：

$\begin{pmatrix} V_{i1,i1}\\ V_{j1,j1}\\ V_{i1,j1}\\ V_{i2,i2}\\ V_{j2,j2}\\ V_{i2,j2}\\ \vdots\\ \end{pmatrix}\cdot$ $\begin{pmatrix} q_{11}\\ q_{12}\\ q_{13}\\ q_{22}\\ q_{23}\\ q_{33}\\ \end{pmatrix} = \begin{pmatrix} 1\\ 1\\ 0\\ 1\\ 1\\ 0\\ \vdots\\ \end{pmatrix}$

（对不起，丑陋的公式...）现在您只需要使用 Cholesky 分解或其他方法来求解 $QQ^T$ -矩阵...

使用 R 的 Motion 的简单正交结构——确定度量约束

Simple Orthographic Structure from Motion using R -- Determining Metric Constraints

r

opencv

image-processing

non-linear-regression

structure-from-motion