问题描述
我有两个图像，并使用筛选找到三个类似的2D点。我需要计算图像之间的仿射变换。不幸的是，我错过了讲座，那里的信息对我来说有点密集。计算这个2×3矩阵的一般方法是什么？
I have two images and found three similar 2D points using a sift. I need to compute the affine transformation between the images. Unfortunately, I missed lecture and the information out there is a little dense for me. What would the general method be for computing this 2x3 matrix?
我有一个2×3矩阵[x1 y1; x2 y2; x3 y3]的点矩阵，失去了。
感谢任何帮助。
I have the matrix of points in a 2x3 matrix [x1 y1;x2 y2;x3 y3] but I am lost from there.Thanks for any help.
推荐答案
通常，2D点的仿射变换表示为
Usually, an affine transormation of 2D points is experssed as
x' = A*x
$ b b
其中 x 是三向量 [x; y; 1] 的原始2D位置， x'是转换的点。仿射矩阵 A 是
Where x is a three-vector [x; y; 1] of original 2D location and x' is the transformed point. The affine matrix A is
A = [a11 a12 a13;
a21 a22 a23;
0 0 1]
此表格在 x 和 A 都是已知，并且您希望恢复 x' 。
但是，您可以以不同的方式表达此关系。
让
However, you can express this relation in a different way.Let
X = [xi yi 1 0 0 0;
0 0 0 xi yi 1 ]
和 code>是一个列向量
and a is a column vector
a = [a11; a12; a13; a21; a22; a23]
然后
X*a = [xi'; yi']
保存所有对对应点 x_i，x_i' / code>。
Holds for all pairs of corresponding points x_i, x_i'.
当您知道点对之间的对应关系并且您希望恢复 A 。
将所有点放在一个大矩阵中 X （每个点两行），你将有2 * n-by矩阵 X 乘以未知数的向量 a 等于堆叠的2×n乘1列向量对应点（由 x_prime 表示）：
This alternative form is very useful when you know the correspondence between pairs of points and you wish to recover the paramters of A.
Stacking all your points in a large matrix X (two rows for each point) you'll have 2*n-by-6 matrix X multiplyied by 6-vector of unknowns a equals a 2*n-by-1 column vector of the stacked corresponding points (denoted by x_prime):
X*a = x_prime
解决 a ：
a = x_prime \ X
以最小二乘意义恢复 a
Recovers the parameters of a in a least-squares sense.
祝你好运， class！
Good luck and stop skipping class!
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