计算将二维四边形转换为另一个四边形的矩阵 [英] Computing a matrix which transforms a quadrangle to another quadrangle in 2D
问题描述
在下图中，目标是计算单应矩阵H，该矩阵将点a1 a2 a3 a4转换为其对应的b1 b2 b3 b4. 那是:
In the figure below the goal is to compute the homography matrix H which transforms the points a1 a2 a3 a4 to their counterparts b1 b2 b3 b4. That is:
[b1 b2 b3 b4] = H * [a1 a2 a3 a4]
您建议哪种方法是计算H(3x3)的最佳方法. a1 ... b4是2D中的点，它们以齐次坐标系表示(即[a1_x a1_y 1]'，...). 编辑: 对于这些类型的问题，我们使用SVD，所以我想看看如何在Matlab中简单地做到这一点.
What way would you suggest to be the best way to calculate H(3x3). a1...b4 are points in 2D which are represented in homogeneous coordinate systems ( that is [a1_x a1_y 1]', ...). EDIT: For these types of problems we use SVD, So i would like to see how this can be simply done in Matlab.
编辑:
这是我最初尝试使用Maltlab中的svd(H = Q/P)解决此问题的方法.考虑给定示例的以下代码
Here is how I initially tried to solve it using svd (H=Q/P) in Maltlab. Cosider the following code for the given example
px=[0 1 1 0]; % a square
py=[1 1 0 0];
qx=[18 18 80 80]; % a random quadrangle
qy=[-20 20 60 -60];
if (DEBUG)
fill(px,py,'r');
fill(qx,qy,'r');
end
Q=[qx;qy;ones(size(qx))];
P=[px;py;ones(size(px))];
H=Q/P;
H*P-Q
answer:
-0.0000 0 0 0 0
-20.0000 20.0000 -20.0000 20.0000 0.0000
-0.0000 0 0 0 -0.0000
我期望答案是一个空矩阵，但事实并非如此！...这就是为什么我在StackOverflow中问这个问题. 现在，我们都知道这是一个投影变换，显然不是欧几里得.但是，最好知道在一般情况下是否仅使用4个点即可计算出这样的矩阵.
I am expecting the answer to be a null matrix but it is not!... and that's why I asked this question in StackOverflow. Now, we all know it is a projective transformation not obviously Euclidean. However, it is good to know if in general care calculating such matrix using only 4 points is possible.
推荐答案
使用您发布的数据:
P = [px(:) py(:)];
Q = [qx(:) qy(:)];
计算转换:
H = Q/P;
应用转换:
Q2 = H*P;
比较结果:
err = Q2-Q
输出:
err =
7.1054e-15 7.1054e-15
-3.5527e-15 -3.5527e-15
-1.4211e-14 -2.1316e-14
1.4211e-14 1.4211e-14
对于所有意图和目的，均为零.
which is zeros for all intents and purposes..
因此，正如您在评论中指出的那样，以上方法将不会计算3x3单应性矩阵.它只用提供的点数来解决方程组:
So as you pointed out in the comments, the above method will not compute the 3x3 homography matrix. It simply solves the system of equations with as many equations as points provided:
H * A = B --> H = B*inv(A) --> H = B/A (mrdivide)
否则，MATLAB在图像处理工具箱中具有CP2TFORM功能.这是应用于所示图像的示例:
Otherwise, MATLAB has the CP2TFORM function in the image processing toolbox. Here is an example applied to the image shown:
%# read illustration image
img = imread('http://i.stack.imgur.com/ZvaZK.png');
img = imcomplement(im2bw(img));
%# split into two equal-sized images
imgs{1} = img(:,fix(1:end/2));
imgs{2} = img(:,fix(end/2:end-1));
%# extract the four corner points A and B from both images
C = cell(1,2);
for i=1:2
%# some processing
I = imfill(imgs{i}, 'holes');
I = bwareaopen(imclearborder(I),200);
I = imfilter(im2double(I), fspecial('gaussian'));
%# find 4 corners
C{i} = corner(I, 4);
%# sort corners in a consistent way (counter-clockwise)
idx = convhull(C{i}(:,1), C{i}(:,2));
C{i} = C{i}(idx(1:end-1),:);
end
%# show the two images with the detected corners
figure
for i=1:2
subplot(1,2,i), imshow(imgs{i})
line(C{i}(:,1), C{i}(:,2), 'Color','r', 'Marker','*', 'LineStyle','none')
text(C{i}(:,1), C{i}(:,2), num2str((1:4)'), 'Color','r', ...
'FontSize',18, 'Horiz','left', 'Vert','bottom')
end
在检测到角点之后，现在我们可以获得空间变换:
With the corners detected, now we can obtain the spatial transformation:
%# two sets of points
[A,B] = deal(C{:});
%# infer projective transformation using CP2TFORM
T = cp2tform(A, B, 'projective');
%# 3x3 Homography matrix
H = T.tdata.T;
Hinv = T.tdata.Tinv;
%# align points in A into B
X = tformfwd(T, A(:,1), A(:,2));
%# show result of transformation
line(X([1:end 1],1), X([1:end 1],2), 'Color','g', 'LineWidth',2)
结果:
>> H = T.tdata.T
H =
0.74311 -0.055998 0.0062438
0.44989 -1.0567 -0.0035331
-293.31 62.704 -1.1742
>> Hinv = T.tdata.Tinv
Hinv =
-1.924 -0.42859 -0.0089411
-2.0585 -1.2615 -0.0071501
370.68 39.695 1
我们可以自己确认计算结果
We can confirm the calculation ourselves:
%# points must be in Homogenous coordinates (x,y,w)
>> Z = [A ones(size(A,1),1)] * H;
>> Z = bsxfun(@rdivide, Z, Z(:,end)) %# divide by w
Z =
152 57 1
219 191 1
62 240 1
92 109 1
映射到B中的点:
%# maximum error
>> max(max( abs(Z(:,1:2)-B) ))
ans =
8.5265e-14
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