A horizontal shear is a matrix of the form \( S^H_{\alpha} =\begin{bmatrix}1&\alpha\\ 0&1 \end{bmatrix}\) , with its parameter \( \alpha\) being a given real number.

Consider the column vector \( X=\begin{bmatrix} x\\ y \end{bmatrix}\) , and prove that \( Y=S^H_{\alpha} X\) has the same ordinate as \( X\) .

2.1.1 Solution

Consider the horizontal shears \( S^H_{\alpha} =\begin{bmatrix}1&\alpha\\ 0&1 \end{bmatrix}\) of parameter \( \alpha\) , and \( S^H_{\alpha'} =\begin{bmatrix}1&\alpha'\\ 0&1 \end{bmatrix}\) of parameter \( \alpha'\) , and prove that \( S^H_{\alpha}S^H_{\alpha'}=S^H_{\alpha+\alpha'}\) .

2.2.1 Solution

A vertical shear is a matrix of the form \( S^V_{\beta} =\begin{bmatrix}1&0\\ \beta&1 \end{bmatrix}\) , with its parameter \( \beta\) being a given real number.

Consider the column vector \( X=\begin{bmatrix} x\\ y \end{bmatrix}\) , and prove that \( Y=S^V_{\beta} X\) has the same abscissa as \( X\) .

2.3.1 Solution

Consider the vertical shears \( S^V_{\beta} =\begin{bmatrix}1&0\\ \beta&1 \end{bmatrix}\) of parameter \( \beta\) , and \( S^V_{\beta} =\begin{bmatrix}1&0\\ \beta&1 \end{bmatrix}\) of parameter \( \beta'\) , and prove that \( S^V_{\beta}S^V_{\beta'}=S^V_{\beta+\beta'}\) .

2.4.1 Solution

3.1.1 Solution

Consider the following formulae.

• \( \tan(\theta/2)=\frac{\sin(\theta/2)}{\cos(\theta/2)}\) , that is well defined because \( -\frac{\pi}{2}<\frac{\theta}{2}<\frac{\pi}{2}\) , so that \( \cos(\theta/2)\ne 0\) .

• \( \sin(\theta)=2\sin(\theta/2)\cos(\theta/2)\) .

• \( \cos(\theta)=1-2\sin^2(\theta/2)\) .

• \( \cos^2(\theta)+\sin^2(\theta)=1\) .

Prove the following formulae (in that order).

1. \( 1+\alpha\beta=\cos(\theta)\) .

2. \( \alpha(2+\alpha\beta)=-\sin(\theta)\) .

3.2.1 Solution

Use the fact that \( R_{\theta}= \begin{bmatrix}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta) \end{bmatrix}\) .

3.3.1 Solution

For an image, the rotation of \( \pi/2\) involves both row and column indexes together, making the parallelisation impossible.

As the horizontal (resp. vertical) shears act only on the row (resp. column) indexes, making it possible to parallelise the process.

So that it is interesting to decompose the rotation into shears, for the sake of fastness of the computation.

4.1.1 NB

In Python, a \( 512\times 512\) image is a \( 512\times 512\times 3\) tensor with \( 3\) layers.

For an image in colors, each layer is an RGB component of a color, red, green and blue.

For a grayscale image, as we have, only the layer \( 0\) conains information, and it is the level of gray of each pixel.

To import it, we need the function ’imread’ in the library ’imageo’.

Then we extract the layout ’0’ to obtain a \( 512\times 512\) matrix.

That matrix can then be visualized we the function ’imshow’ of the library ’matplotlib.pyplot’.

We shall rotate by \( \pi/2\) the image, both directely and with the help of 3 sucessive shears.

The matrix of the rotation of angle \( \pi/2\) is \( R=\begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}\) .

To rotate the image ’lena’, each vector ’array([i,j])’, with i and j calculated from the center of the image, must be multiplied to the right by the matrix \( R\) . Indeed, the indexes (i,j) are analog to the vectors of the canonical basis in the plane of the image.

As \( \tan(\pi/4)=1\) , the matrix of the horizontal shear is \( HS=\begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix}\) .

To shear horizontally the image ’lena’, we must first set margins of \( 255\) pixels at each side of the image, to obtain a \( 1024\times 1024\) image. Then each vector ’array([i,j])’ with customed ranges must be multiplied to the right by the matrix \( HS\) . We obtain an image called ’lena1_sheared’.

And as \( \sin(\pi/2)=1\) , the matrix of the vertical shear is \( VS=\begin{bmatrix} 1&0\\ 1&1 \end{bmatrix}\) .

To shear vertically the image ’lena1_sheared’, we must first set margins of \( 512\) pixels at each side of the image, to obtaine a \( 2048\times 2048\) image. Then each vector ’array([i,j])’ with customed ranges must be multiplied to the right by the matrix \( VS\) . After a crop to the size \( 1024\times 1024\) , we obtain an image called ’lena2_sheared’.

Then we shear horizontally the image ’lena2_sheared’ to obtain the image ’lena3_rotated’. To do that, we must first set margins of \( 512\) pixels at each side of the image, to obtain a \( 2048\times 2048\) image. Then each vector ’array([i,j])’ with customed ranges must be multiplied to the right by the matrix \( HS\) . After a crop to the size \( 512\times 512\) , we obtain an image called ’lena3_rotated’.

You are invited to copy that script in a .py file, that you will store in the same directory as the image ’lena_gray.gif’ that you have previously downloaded from the course.

Then, run and enjoy!

import imageio as iio
from matplotlib.pyplot import *
from numpy import *
# Read 'lena_gray.gif'
lena0=iio.imread('lena_gray.gif')
# Extract the first layer of the image, in gray
lena=lena0[:,:,0]
# Visualize in figure 1 the image as a matrix 512x512 pixels
imshow(lena,cmap='gray')
# Direct rotation of pi/2
R=array([[0,-1],[1,0]])
lena_rotated=zeros([512,512])
for i in arange(-255,256):
 for j in arange(-255,256):
 V=array([i,j])@R 
 lena_rotated[256+i,256+j]=lena[256+V[0],256+V[1]]
		
# Visualize in figure 2 the rotated image as a 512x512 matrix
figure()
imshow(lena_rotated,cmap='gray')
		
# margins of 256 pixels at each side
lena1=zeros([1024,1024])
lena1[256:768,256:768]=lena
		
# First horizontal shear
HS=array([[1,-1],[0,1]])
lena1_sheared=zeros([1024,1024])
for i in arange(-255,255):
	for j in arange(-255+i,255+i):
	 U=array([i,j])@HS
		lena1_sheared[512+i,512+j]=lena1[512+U[0],512+U[1]]
	 
# Visualize in figure 3 the sheared image as a 1024x1024 marix
figure()
imshow(lena1_sheared,cmap='gray')
		
# margins of 512 pixels at each side
lena2=zeros([2048,2048])
lena2[512:1536,512:1536]=lena1_sheared
		
# Vertical shear
VS=array([[1,0],[1,1]])
lena2_sheared=zeros([2048,2048])
for j in arange(-511,511):
		for i in arange(-255-j,255-j):
			W=array([i,j])@VS
		 lena2_sheared[1024+i,1024+j]=lena2[1024+W[0],1024+W[1]]
# crop to 1024 over 1024 pixels
lena2_sheared=lena2_sheared[512:1536,512:1536]

# Visualize in figure 4 the double sheared image as a 1024x1024 marix
figure()
imshow(lena2_sheared,cmap='gray')

# margins of 512 pixels at each side
lena3=zeros([2048,2048])
lena3[512:1536,512:1536]=lena2_sheared

# Second horiaontal shear
lena3=zeros([2048,2048])
lena3[512:1536,512:1536]=lena2_sheared
lena3_rotated=zeros([2048,2048])
for i in arange(-511,511):
 for j in arange(-511+i,511+i):
 U=array([i,j])@HS
 lena3_rotated[1024+i,1024+j]=lena3[1024+U[0],1024+U[1]]

# Crop to 512 over 512 pixels
lena3_rotated=lena3_rotated[768:1280,768:1280]

# Visualize in figure 5 the rotated image obtained by 3 shears as a 512x512 
matrix
figure()
imshow(lena3_rotated,cmap='gray')