# 2-D Image Transformations and Alignment

Given a set of corresponding points in two images, how can you align them? What if the pixel aspect ratios in the two images are different?

# Representation

Raster images are often represented as an array of shape $$(Y, X, C)$$ corresponding to the height, width, and number of channels (e.g., $$C = 3$$ for RGB images). The values in the array indicate intensities. By convention, a coordinate of $$(y, x) = (0, 0)$$ (0-indexed) refers to the upper-left corner of the image. [MATLAB, Python - Matplotlib, Python - Pillow]

For the purpose of computing 2-D transformations, however, we are primarily concerned with pixel coordinates. A pixel coordinate $$\vec{p}$$ can be represented as a vector

$\vec{p} = \begin{bmatrix} x \\ y \end{bmatrix}$

Note that $$x, y \in \mathbb{R}$$ are not limited to the set of non-negative integers. While a transformed coordinate does not correspond to a discrete pixel in an image, the intensities of the transformed image can be estimated by interpolation (see below).

Linear transformations (maps) such as rotation, scaling, and shear can be represented as matrix multiplication

$\begin{equation} \label{eq:linear_transformation} \vec{p}' = A \vec{p} \end{equation}$ \begin{aligned} A_\text{rotate} &= \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} & \text{rotate counterclockwise by \theta} \\ A_\text{scale} &= \begin{bmatrix} s_x & 0 \\ 0 & s_y \end{bmatrix} & \text{scale by s_x horizontally and s_y vertically} \\ A_\text{shear} &= \begin{bmatrix} 1 & c_x \\ c_y & 1 \end{bmatrix} & \text{shear by c_x horizontally and c_y vertically} \end{aligned}

Since translations

\begin{aligned} \vec{p}' &= \vec{p} + \vec{t} = \vec{p} + \begin{bmatrix} t_x \\ t_y \end{bmatrix} & \text{translate by t_x horizontally and t_y vertically} \end{aligned}

are not strictly linear transformations, they cannot be represented as matrix multiplication as in Equation \ref{eq:linear_transformation}. However, if pixels are represented in homogeneous coordinates, affine (linear map with translations) and projective transformations can also be computed using matrix multiplication.

For affine transformations, it is sufficient to work with a simple augmented representation for both the pixel coordinate vector and the transformation.

$\underbrace{\begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix}}_{\vec{p}'} = \underbrace{\begin{bmatrix} s_x \cdot \cos(\theta) & -c_x \cdot \sin(\theta) & t_x \\ c_y \cdot \sin(\theta) & s_y \cdot \cos(\theta) & t_y \\ 0 & 0 & 1 \end{bmatrix}}_A \underbrace{\begin{bmatrix} x \\ y \\ 1 \end{bmatrix}}_{\vec{p}}$

For projective transformations, however, the first two elements of the last row of the transformation matrix $$A$$ may be nonzero. Consequently, the “augmented element” of the transformed vector $$\vec{p}'$$ is not necessarily $$1$$ and is a general example of a homogeneous coordinate:

Given a point (x, y) on the Euclidean plane, for any non-zero real number Z, the triple (xZ, yZ, Z) is called a set of homogeneous coordinates for the point. By this definition, multiplying the three homogeneous coordinates by a common, non-zero factor gives a new set of homogeneous coordinates for the same point. In particular, (x, y, 1) is such a system of homogeneous coordinates for the point (x, y). [Wikipedia]

The Cartesian coordinate $$(\tilde{x}, \tilde{y})$$ corresponding to a homogeneous coordinate $$\vec{p}' = \begin{bmatrix} x' & y' & z \end{bmatrix}^\top$$ is then given by

$(\tilde{x}, \tilde{y}) = \left(\frac{x'}{z}, \frac{y'}{z} \right)$

## Types of 2-D transformations

A general 2-D transformation can be represented as a 3x3 matrix

$T = \begin{bmatrix} a_0 & a_1 & a_2 \\ b_0 & b_1 & b_2 \\ c_0 & c_1 & c_2 \end{bmatrix}$

Different constraints on $$T$$ give rise to special named transformations with recognizable geometries.

Table 1: Special transformations

Transform Projective (homography) Affine Similarity Euclidean (rigid)
Constraints $$c_2 = 1$$ Projective
$$c_1 = c_2 = 0$$
Affine
$$a_0 = b_1$$, $$b_0 = -a_1$$
Similarity
\begin{aligned} &\cos^{-1}(a_0) \\ &= \sin^{-1}(-a_1) \\ &= \sin^{-1}(b_0) \\ &= \cos^{-1}(b_1)\end{aligned}
# Free Parameters 8 6 4 3
# Defining Points 4 3 2 ?
Rotation Y Y Y Y
Translation Y Y Y Y
Scaling Y Y Y (isotropic: same in x- and y-directions) N
Shear Y Y N N
Geometric interpretation Maps lines to lines Preserves lines and parallelism Preserves shape Preserves Euclidean distance between points
• Just as how “[t]he point represented by a given set of homogeneous coordinates is unchanged if the coordinates are multiplied by a common factor” (Wikipedia), the transform represented by a projective transformation matrix is unchanged if the matrix is multiplied by a scalar: the homogeneous coordinate given by $$T_\text{projective} \cdot \vec{p}$$ represents the same point as the homogeneous coordinate $$k T_\text{projective} \cdot \vec{p}$$ for any scalar $$k \in \mathbb{R}$$. Consequently, a projective transformation is uniquely represented by a projective transformation matrix up to a scalar factor, namely $$c_2$$. This is why a projective transformation matrix can always be represented with $$c_2 = 1$$.
• Since transformations can be represented as matrix muliplications, multiple transformations can be composed (combined) together $$T_\text{composed} = T_1 T_2 \cdots$$. There are two important considerations when composing transformations:
1. Non-commutativity: because matrix multiplication does not commute in general, the order in which transformations are composed matters. One consequence of non-commutativity is that affine transformations cannot be uniquely decomposed into translation, rotation, scale, and shear transformations, unless the order of transformations is known (for example, see this StackExchange Math question). The chart below summarizes which affine transformations commute with one another; code verifying the chart can be found in the Colab notebook linked at the end of the post.

Commutativity Translation Scale Rotation Shear
Translation Yes No No No
Scale No Yes No No
Rotation No No Yes No
Shear No No No No
2. The type (Table 1) of the composed transform is generally the same as the most general type among the individual transforms being composed. For example, a Euclidean transform composed with a Euclidean transform is still a Euclidean transform, but a Euclidean transform composed with an affine transform is an affine transform but not necessarily a Euclidean transform.

# Learning transformations

Let $$S \in \mathbb{R}^{3 \times n}$$ be a set of $$n$$ augmented coordinates in the source image and $$D \in \mathbb{R}^{3 \times n}$$ be a set of corresponding augmented coordinates in the destination image.

$S = \begin{bmatrix} x_{s1} & x_{s2} & \cdots \\ y_{s1} & y_{s2} & \cdots \\ 1 & 1 & \cdots \end{bmatrix}, \quad D = \begin{bmatrix} x_{d1} & x_{d2} & \cdots \\ y_{d1} & y_{d2} & \cdots \\ 1 & 1 & \cdots \end{bmatrix}$

We want to learn the transformation matrix $$T$$ such that $$T \cdot S \approx D$$.

For each set $$i = 1, ..., n$$ of corresponding points $$s_i, d_i$$, we can explicitly write out $$T s_i = d_i$$ as

\begin{aligned} x_{di} &= \frac{a_0 x_{si} + a_1 y_{si} + a_2}{c_0 x_{si} + c_1 x_{si} + c_2} \\ y_{di} &= \frac{b_0 x_{si} + b_1 y_{si} + b_2}{c_0 x_{si} + c_1 x_{si} + c_2} \\ d_{i2} &= \frac{c_0 x_{si} + c_1 x_{si} + c_2}{c_0 x_{si} + c_1 x_{si} + c_2} = 1 \end{aligned}

The denominator $$c_0 x_{si} + c_1 x_{si} + c_2$$ ensures that $$d_{i2} = 1$$ (i.e., the last row of $$D$$ is all ones). After rearranging terms, these equations can be re-written as

\begin{aligned} 0 &= a_0 x_{si} + a_1 y_{si} + a_2 - c_0 x_{si} x_{di} - c_1 x_{si} x_{di} - c_2 x_{di} \\ 0 &= b_0 x_{si} + b_1 y_{si} + b_2 - c_0 x_{si} y_{di} - c_1 x_{si} y_{di} - c_2 y_{di} \end{aligned}

Consequently, we have a set of $$2n$$ linear equations that can be expressed in matrix form as a homogeneous system $$0_{2n} = A \cdot x$$ where

$\begin{equation} \label{eq:linear_system_2n} A = \begin{bmatrix} x_{s1} & y_{s1} & 1 & 0 & 0 & 0 & -x_{s1} x_{d1} & -y_{s1} x_{d1} & -x_{d1} \\ 0 & 0 & 0 & x_{s1} & y_{s1} & 1 & -x_{s1} y_{d1} & -y_{s1} y_{d1} & -y_{d1} \\ x_{s2} & y_{s2} & 1 & 0 & 0 & 0 & -x_{s2} x_{d2} & -y_{s2} x_{d2} & -x_{d2} \\ 0 & 0 & 0 & x_{s2} & y_{s2} & 1 & -x_{s2} y_{d2} & -y_{s2} y_{d2} & -y_{d2} \\ & & & & & \vdots \end{bmatrix}, \quad x = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ b_0 \\ b_1 \\ b_2 \\ c_0 \\ c_1 \\ c_2 \end{bmatrix} \end{equation}$

Such as a system of equations can be solved in general using methods such as singular value decomposition or gradient descent. However, simplified systems of equations exist for the special cases give in Table 1.

Simplified equations for each special transformation.

Projective: $$c_2 = 1$$

$\begin{bmatrix} x_{d1} \\ y_{d1} \\ x_{d2} \\ y_{d2} \\ \vdots \end{bmatrix} = \begin{bmatrix} x_{s1} & y_{s1} & 1 & 0 & 0 & 0 & -x_{s1} x_{d1} & -y_{s1} x_{d1} \\ 0 & 0 & 0 & x_{s1} & y_{s1} & 1 & -x_{s1} y_{d1} & -y_{s1} y_{d1} \\ x_{s2} & y_{s2} & 1 & 0 & 0 & 0 & -x_{s2} x_{d2} & -y_{s2} x_{d2} \\ 0 & 0 & 0 & x_{s2} & y_{s2} & 1 & -x_{s2} y_{d2} & -y_{s2} y_{d2} \\ & & & & & \vdots \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ b_0 \\ b_1 \\ b_2 \\ c_0 \\ c_1 \end{bmatrix}$

Affine: $$c_1 = c_2 = 0$$

$\begin{bmatrix} x_{d1} \\ y_{d1} \\ x_{d2} \\ y_{d2} \\ \vdots \end{bmatrix} = \begin{bmatrix} x_{s1} & y_{s1} & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & x_{s1} & y_{s1} & 1 \\ x_{s2} & y_{s2} & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & x_{s2} & y_{s2} & 1 \\ & & \vdots \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ b_0 \\ b_1 \\ b_2 \end{bmatrix}$

Similarity: $$a_0 = b_1$$, $$b_0 = -a_1$$

$\begin{bmatrix} x_{d1} \\ y_{d1} \\ x_{d2} \\ y_{d2} \\ \vdots \end{bmatrix} = \begin{bmatrix} x_{s1} + 0 & y_{s1} - 0 & 1 & 0 \\ 0 + y_{s1} & 0 - x_{s1} & 0 & 1 \\ x_{s2} + 0 & y_{s2} - 0 & 1 & 0 \\ 0 + y_{s2} & 0 - x_{s2} & 0 & 1 \\ & \vdots \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ b_2 \end{bmatrix} = \begin{bmatrix} x_{s1} & y_{s1} & 1 & 0 \\ y_{s1} & -x_{s1} & 0 & 1 \\ x_{s2} & y_{s2} & 1 & 0 \\ y_{s2} & -x_{s2} & 0 & 1 \\ & \vdots \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ b_2 \end{bmatrix}$

Therefore, to have a fully-determined or overdetermined system of linear equations, the number of corresponding points $$n$$ must be at least half the number of free parameters (see Table 1).

Compare this approach with solving for $$T = D \cdot S^{-1}$$. Since the last row of $$D$$ is always $$\begin{bmatrix} 1 & 1 & 1\end{bmatrix}$$, the last row of $$D \cdot S^{-1}$$ is always $$\begin{bmatrix} 0 & 0 & 1\end{bmatrix}$$. Consequently, solving for $$T$$ as $$D \cdot S^{-1}$$ can learn an affine transformation, but not a projective transformation.

• This appears to hold if the pseudoinverse of $$S$$ is used instead of its inverse.
• If the number of paired coordinates $$n > 3$$, then $$S$$ is not square, so it is not invertible.
• $$S^\top (S S^\top)^{-1}$$ gives the Moore-Penrose pseudoinverse when $$S$$ is a full-rank fat ($$n \geq 3$$) matrix.
• The affine transformation learned this way appears identical to the affine transformation learned using Equation $$\ref{eq:linear_system_2n}$$.
Details

If $$n = 3$$ (i.e., $$S$$ and $$D$$ contain 3 corresponding points), then the inverse of $$S$$ is given by

$S^{-1} = \underbrace{\frac{1}{x_{s1} y_{s2} - x_{s1} y_{s3} - x_{s2} y_{s1} + x_{s2} y_{s3} + x_{s3} y_{s1} - x_{s3} y_{s2}}}_k \begin{bmatrix} y_{s2} - y_{s3} & x_{s3} - x_{s2} & x_{s2} y_{s3} - x_{s3} y_{s2} \\ y_{s3} - y_{s1} & x_{s1} - x_{s3} & x_{s3} y_{s1} - x_{s1} y_{s3} \\ y_{s1} - y_{s2} & x_{s2} - x_{s1} & x_{s1} y_{s2} - x_{s2} y_{s1} \end{bmatrix}$

Let $$k$$ denote the constant fraction factor. Since the last row of $$D$$ is always $$\begin{bmatrix} 1 & 1 & 1\end{bmatrix}$$, we get

$T_{31} = \sum_{i=1}^3 {S^{-1}}_{i1} = k ((y_{s2} - y_{s3}) + (y_{s3} - y_{s1}) + (y_{s1} - y_{s2})) = 0$ $T_{32} = \sum_{i=1}^3 {S^{-1}}_{i2} = k ((x_{s3} - x_{s2}) + (x_{s1} - x_{s3}) + (x_{s2} - x_{s1})) = 0$ \begin{aligned} T_{33} &= \sum_{i=1}^3 {S^{-1}}_{i3} \\ &= k ((x_{s2} y_{s3} - x_{s3} y_{s2}) + (x_{s3} y_{s1} - x_{s1} y_{s3}) + (x_{s1} y_{s2} - x_{s2} y_{s1})) \\ &= \frac{x_{s2} y_{s3} - x_{s3} y_{s2} + x_{s3} y_{s1} - x_{s1} y_{s3} + x_{s1} y_{s2} - x_{s2} y_{s1}}{x_{s1} y_{s2} - x_{s1} y_{s3} - x_{s2} y_{s1} + x_{s2} y_{s3} + x_{s3} y_{s1} - x_{s3} y_{s2}} \\ &= 1 \end{aligned}

So

$T = D \cdot S^{-1} = \begin{bmatrix} ? & ? & ? \\ ? & ? & ? \\ 0 & 0 & 1 \end{bmatrix}$

This appears to hold if the pseudoinverse of $$S$$ is used instead of its inverse (analytical proof?):

import numpy as np
n = 4 # number of points; an integer >= 3
S = np.vstack((np.random.rand(2, n), np.ones(n)))
D = np.vstack((np.random.rand(2, n), np.ones(n)))
D @ np.linalg.pinv(S)
# check that the last row is [0, 0, 1], within numerical error


The affine transformation learned this way appears identical to the affine transformation learned using Equation $$\ref{eq:linear_system_2n}$$ (analytical proof?):

import numpy as np
n = 4 # number of points; an integer >= 3
S = np.vstack((np.random.rand(2, n), np.ones(n)))
D = np.vstack((np.random.rand(2, n), np.ones(n)))
A = np.zeros((2 * n, 6))
for i in range(n):
A[2 * i, 0:3] = S[:, i]
A[2 * i + 1, 3:6] = S[:, i]

T1 = D @ np.linalg.pinv(S)
T2 = np.vstack(((np.linalg.pinv(A) @ D[:2, ].T.reshape(2 * n)).reshape(2, 3), [0, 0, 1]))
np.isclose(T1, T2) # should be all True


## Pixel aspect ratios

Let $$r$$ be the pixel aspect ratio:

$r = \frac{\text{width represented by a pixel}}{\text{height represented by a pixel}}$

Usually, we work with square ($$r = 1$$) pixels. However, images (especially those generated by different modalities, such as mass spectrometry imaging, MRI, etc.) may have non-square pixels. For example, a pixel that corresponds to physical dimensions of 1 cm width x 2 cm height would have a pixel aspect ratio of 0.5.

Consider a source image $$\mathcal{I}_S$$ and a destination image $$\mathcal{I}_D$$, which may be of different dimensions. Let $$r_S, r_D$$ be the aspect ratios of pixels in the source and destination images, respectively. We want to generate an image $$\mathcal{I}_{S'}$$ of the same dimension as the destination image $$\mathcal{I}_D$$ in which the source image is aligned to the destination image.

Let $$S$$ and $$D$$ be corresponding homogeneous coordinates as defined above. Let $$\tilde{r} = \frac{r_D}{r_S}$$ be the ratio of pixel aspect ratios. Then,

$\tilde{D} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\tilde{r}} & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot D = \begin{bmatrix} x_{d1} & x_{d2} & \cdots \\ \frac{y_{d1}}{\tilde{r}} & \frac{y_{d2}}{\tilde{r}} & \cdots \\ 1 & 1 & \cdots \end{bmatrix}$

transforms coordinates in the destination image to a space of the same pixel aspect ratio as in the source image.

Let $$\tilde{T} \in \mathbb{R}^{3 \times 3}$$ be a transform mapping coordinates $$S$$ to $$\tilde{D}$$:

$T \cdot S = \tilde{D}.$

We can solve for $$\tilde{T}$$ using the systems of equations discussed above, giving the transform that maps $$S$$ to $$\tilde{D}$$, but not to $$D$$ as desired. To account for the pixel aspect ratios, observe that

$\tilde{T} \cdot S = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\tilde{r}} & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot D \rightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & \tilde{r} & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \tilde{T} \cdot S = T \cdot S = D$

For projective and affine transformation matrices, the scaling matrix $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \tilde{r} & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ can be folded into the final transformation matrix

$T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \tilde{r} & 0 \\ 0 & 0 & 1 \end{bmatrix} \tilde{T}$

without violating the constraints of an affine or projective matrix. In fact, the idea of “constraining” a projective or affine transformation to respect pre-defined pixel aspect ratio $$\tilde{r}$$ is not possible. The final learned transformation matrix $$T$$ can just as well be fit on $$S$$ and $$D$$ directly, without first having to scale $$D$$.

For more constrained transformation matrices, the final transformation matrix cannot be learned directly from $$S$$ and $$D$$ without violating assumed constraints on $$T$$.

## Interpolating intensities from transformed coordinates

Once you have a transformation matrix, how do you generate the transformed image, not just the transformed coordinates?

One general approach used by many popular image libraries (Python - Pillow, Python - scikit-image, MATLAB) is inverse mapping followed by interpolation. MATLAB’s documentation of its imwarp() function nicely summarizes the procedure:

imwarp determines the value of pixels in the output image by mapping locations in the output image to the corresponding locations in the input image (inverse mapping). imwarp interpolates within the input image to compute the output pixel value.

Additional notes:

1. Visualizing rotations can be tricky due to the convention of displaying image coordinates $$(0, 0)$$ (in a 0-indexed language like Python) or $$(1, 1)$$ (in a 1-indexed language like MATLAB) in the upper left corner. A statement such as “rotate an image by $$\theta$$ radians counterclockwise” usually means to produce an output image such that, when displayed using image coordinate conventions, the image appears rotated by $$\theta$$ radians counterclockwise. However, if the rotated image is displayed with the origin in the lower-left corner, then it appears as if the image was rotated clockwise. Consequently, the rotation matrix for “rotate an image by $$\theta$$ radians counterclockwise” is

$\begin{bmatrix} \cos(-\theta) & -\sin(-\theta) & 0 \\ \sin(-\theta) & \cos(-\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

See the Colab notebook for an example.

2. In skimage.transform.warp(), the interpolation step relies on scipy.ndimage.map_coordinates().
3. When dealing with non-square pixels, the inverse transformation is given by
$T^{-1} = \left(\begin{bmatrix} 1 & 0 & 0 \\ 0 & \tilde{r} & 0 \\ 0 & 0 & 1 \end{bmatrix} \tilde{T} \right)^{-1} = \tilde{T}^{-1} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\tilde{r}} & 0 \\ 0 & 0 & 1 \end{bmatrix}$

# Additional references

• scikit-image documentation
• Difference between affine and projective transformations: StackOverflow
• Learning a projective transform matrix from 4 points: StackExchange Math
Posted: Jun 16, 2020. Last updated: Jun 21, 2020.
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