在MATLAB中从4个点计算2D均匀透视变换矩阵 [英] Calculate a 2D homogeneous perspective transformation matrix from 4 points in MATLAB

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问题描述

我有2D的4个点的坐标，它们形成一个矩形，并且在应用了透视变换之后它们的坐标.

透视变换是在同构坐标中计算的，并由3x3矩阵 M 定义.如果矩阵未知，如何从给定点计算出来?

一分的计算公式为:

  |M11 M12 M13 ||P1.x ||w * P1'.x ||M21 M22 M23 |* |P1.y |= |w * P1'.y ||M31 M32 M33 ||1 ||w * 1 |

要同时计算所有点，我将它们一起写在一个矩阵 A 中，并类似地将变换后的点写在矩阵 B 中:

  |P1.x P2.x P3.x P4.x |A = |P1.y P2.y P3.y P4.y ||1 1 1 1 |

因此等式为 M * A = B ，这可以通过 M = B/A 或 M =(A'\ B')'.

但这不是那么容易.我知道变换后点的坐标，但我不知道确切的 B，因为存在因子 w 并且在齐次变换后不需要 1.因为在同构坐标中，向量的每个倍数都是同一点，所以我不知道会得到哪个倍数.

考虑到这些未知因素，我将等式写为 M * A = B * W 其中 W 是一个对角矩阵，对角线上 B 中每个点的因子均为w1 ... w4.所以 A 和 B 现在完全已知，我必须为 M 和 W 求解这个方程.>

如果我可以将等式重新排列为 x * A = B 或 A * x = B 的形式，其中 x 类似于 M * W 我可以解决它，而知道 M * W 的解决方案可能已经足够了.但是，尽管尝试了所有可能的重排，但我都没有设法这样做.直到我不敢封装(M * W)，因为一个是3x3矩阵，另一个是4x4矩阵.在这里，我被卡住了.

也 M * A = B * W 没有针对 M 的单一解决方案，因为 M 的每一个倍数都是相同的转换.将其写为线性方程组系统，只需将 M 的条目之一固定即可获得单个解决方案.此外，可能有些输入根本无法解决 M ，但是暂时不用担心.

我实际上想要实现的是某种矢量图形编辑程序，用户可以在内部计算变换矩阵的同时拖动形状边界框的角进行变换.

实际上我在JavaScript中需要它，但是如果我什至不能在MATLAB中解决这个问题，我就会完全陷入困境.

解决方案

应该是一个简单的问题.那么如何将 M * A = B * W 转换为可解决的形式?这只是矩阵乘法，因此我们可以将其写为线性方程组.您会知道: M11 * A11 + M12 * A21 + M13 * A31 = B11 * W11 + B12 * W21 + B13 * W31 + B14 * W41 .每个线性方程组都可以用 Ax = b 的形式编写，或者避免与我的问题中已使用的变量混淆: N * x = y .就是这样.

根据我的问题的示例:我生成一些输入数据，这些输入数据具有已知的 M 和 W :

  M = [1 2 3;4 5 6;7 8 1];A = [0 0 1 1;0 1 0 1;1 1 1 1];W = [4 0 0 0;0 3 0 0;0 0 2 0;0 0 0 1];B = M * A *(W ^ -1);

然后我忘记了 M 和 W .这意味着我现在有13个要解决的变量.我将 M * A = B * W 重写为线性方程组，并从那里重写为 N * x = y .在 N 中，每一列都有一个变量的因数:

  N = [A(1,1)A(2,1)A(3,1)0 0 0 0 0 0 -B(1,1)0 0 0;0 0 0 A(1,1)A(2,1)A(3,1)0 0 0 -B(2,1)0 0 0;0 0 0 0 0 0 0 A(1,1)A(2,1)A(3,1)-B(3,1)0 0 0;A(1,2)A(2,2)A(3,2)0 0 0 0 0 0 0 -B(1,2)0 0;0 0 0 A(1,2)A(2,2)A(3,2)0 0 0 0 -B(2,2)0 0;0 0 0 0 0 0 A(1,2)A(2,2)A(3,2)0 -B(3,2)0 0;A(1,3)A(2,3)A(3,3)0 0 0 0 0 0 0 0 -B(1,3)0;0 0 0 A(1,3)A(2,3)A(3,3)0 0 0 0 0 -B(2,3)0;0 0 0 0 0 0 A(1,3)A(2,3)A(3,3)0 0 -B(3,3)0;A(1,4)A(2,4)A(3,4)0 0 0 0 0 0 0 0 0 -B(1,4);0 0 0 A(1,4)A(2,4)A(3,4)0 0 0 0 0 0 -B(2,4);0 0 0 0 0 0 A(1,4)A(2,4)A(3,4)0 0 0 -B(3,4);0 0 0 0 0 0 0 0 1 0 0 0 0];

y 是:

  y = [0;0;0;0;0;0;0;0;0;0;0;0;1];

请注意，根据 y ， N 中最后一行描述的方程的解为1.这就是我在问题中提到的内容，您必须修复 M 的条目之一才能获得单个解决方案.(我们之所以这样做，是因为 M 的每一个倍数都是相同的变换.)并且用这个等式，我说M33应该是1.

我们为 x 解决了此问题:

  x = N \ y

并获得:

  x = [1.00000;2.00000;3.00000;4.00000;5.00000;6.00000;7.00000;8.00000;1.00000;4.00000;3.00000;2.00000;1.00000]

是 [M11，M12，M13，M21，M22，M23，M31，M32，M33，w1，w2，w3，w4]的解决方案

在JavaScript中执行此操作时，我可以使用数字JavaScript 库，该库具有所需的功能解决以解决Ax = b.

I've got coordinates of 4 points in 2D that form a rectangle and their coordinates after a perspective transformation has been applied.

The perspective transformation is calculated in homogeneous coordinates and defined by a 3x3 matrix M. If the matrix is not known, how can I calculate it from the given points?

The calculation for one point would be:

| M11 M12 M13 |   | P1.x |   | w*P1'.x |
| M21 M22 M23 | * | P1.y | = | w*P1'.y |
| M31 M32 M33 |   | 1    |   | w*1     |

To calculate all points simultaneously I write them together in one matrix A and analogously for the transformed points in a matrix B: 

    | P1.x P2.x P3.x P4.x |
A = | P1.y P2.y P3.y P4.y |
    | 1    1    1    1    |

So the equation is M*A=B and this can be solved for M in MATLAB by M = B/A or M = (A'\B')'.

But it's not that easy. I know the coordinates of the points after transformation, but I don't know the exact B, because there is the factor w and it's not necessary 1 after a homogeneous transformation. Cause in homogeneous coordinates every multiple of a vector is the same point and I don't know which multiple I'll get.

To take account of these unknown factors I write the equation as M*A=B*W where W is a diagonal matrix with the factors w1...w4 for every point in B on the diagonal. So A and B are now completely known and I have to solve this equation for M and W.

If I could rearrange the equation into the form x*A=B or A*x=B where x would be something like M*W I could solve it and knowing the solution for M*W would maybe be enough already. However despite trying every possible rearrangement I didn't managed to do so. Until it hit me that encapsulating (M*W) would not be possible, since one is a 3x3 matrix and the other a 4x4 matrix. And here I'm stuck.

Also M*A=B*W does not have a single solution for M, because every multiple of M is the same transformation. Writing this as a system of linear equations one could simply fix one of the entries of M to get a single solution. Furthermore there might be inputs that have no solution for M at all, but let's not worry about this for now.

What I'm actually trying to achieve is some kind of vector graphics editing program where the user can drag the corners of a shape's bounding box to transform it, while internally the transformation matrix is calculated.

And actually I need this in JavaScript, but if I can't even solve this in MATLAB I'm completely stuck.

解决方案

Should have been an easy question. So how do I get M*A=B*W into a solvable form? It's just matrix multiplications, so we can write this as a system of linear equations. You know like: M11*A11 + M12*A21 + M13*A31 = B11*W11 + B12*W21 + B13*W31 + B14*W41. And every system of linear equations can be written in the form Ax=b, or to avoid confusion with already used variables in my question: N*x=y. That's all.

An example according to my question: I generate some input data with a known M and W:

M = [
    1 2 3;
    4 5 6;
    7 8 1
];
A = [
    0 0 1 1;
    0 1 0 1;
    1 1 1 1
];
W = [
    4 0 0 0;
    0 3 0 0;
    0 0 2 0;
    0 0 0 1
];
B = M*A*(W^-1);

Then I forget about M and W. Meaning I now have 13 variables I'm looking to solve. I rewrite M*A=B*W into a system of linear equations, and from there into the form N*x=y. In N every column has the factors for one variable:

N = [
    A(1,1) A(2,1) A(3,1)      0      0      0      0      0      0 -B(1,1)       0       0       0;
         0      0      0 A(1,1) A(2,1) A(3,1)      0      0      0 -B(2,1)       0       0       0;
         0      0      0      0      0      0 A(1,1) A(2,1) A(3,1) -B(3,1)       0       0       0;
    A(1,2) A(2,2) A(3,2)      0      0      0      0      0      0       0 -B(1,2)       0       0;
         0      0      0 A(1,2) A(2,2) A(3,2)      0      0      0       0 -B(2,2)       0       0;
         0      0      0      0      0      0 A(1,2) A(2,2) A(3,2)       0 -B(3,2)       0       0;
    A(1,3) A(2,3) A(3,3)      0      0      0      0      0      0       0       0 -B(1,3)       0;
         0      0      0 A(1,3) A(2,3) A(3,3)      0      0      0       0       0 -B(2,3)       0;
         0      0      0      0      0      0 A(1,3) A(2,3) A(3,3)       0       0 -B(3,3)       0;
    A(1,4) A(2,4) A(3,4)      0      0      0      0      0      0       0       0       0 -B(1,4);
         0      0      0 A(1,4) A(2,4) A(3,4)      0      0      0       0       0       0 -B(2,4);
         0      0      0      0      0      0 A(1,4) A(2,4) A(3,4)       0       0       0 -B(3,4);
         0      0      0      0      0      0      0      0      1       0       0       0       0
];

And y is:

y = [ 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 1 ];

Notice the equation described by the last row in N whose solution is 1 according to y. That's what I mentioned in my question, you have to fix one of the entries of M to get a single solution. (We can do this because every multiple of M is the same transformation.) And with this equation I'm saying M33 should be 1.

We solve this for x:

x = N\y

and get:

x = [ 1.00000; 2.00000; 3.00000; 4.00000; 5.00000; 6.00000; 7.00000; 8.00000; 1.00000; 4.00000; 3.00000; 2.00000; 1.00000 ]

which are the solutions for [ M11, M12, M13, M21, M22, M23, M31, M32, M33, w1, w2, w3, w4 ]

When doing this in JavaScript I could use the Numeric JavaScript library which has the needed function solve to solve Ax=b.

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