Written Homework 3 Solutions

CSSE 461 - Computer Vision

Solutions

Problem 1: Reddish Tint

Given a 3-channel color image with width width and height height stored in a 3-dimensional array F, write pseudocode to give the image a reddish tint. Assume that F[r, c, i] is the syntax to access the value of the ith color channel (where 0 is red, 1 is green, 2 is blue) of the pixel at the rth row and cth column. Your answer can, but does not need to involve any color space transformations.

Solution

One approach is to simply increase the red channel:

F[:,:,0] += some_constant

However, this approach has limitations. For example, if you start with a perfectly blue image, scaling up red won’t effectively tint it red.

A better approach is to blend the image with a pure red image:

red = np.zeros_like(F)
red[:,:,0] = 1
F = alpha * F + (1-alpha) * red

This works well because it can move towards red by reducing other channels in addition to adding to the red channel.

Problem 2: Contrast

Given a grayscale image \(f(x, y)\), how could you increase the contrast? In other words, how could you make the bright stuff brighter and dark stuff darker? As above, your approach should not allow values to go outside their original range from 0 to 1.

Solution

One approach is to shift the midpoint, scale, then shift back:

# First temporarily shift the "middle" of grayscale space to make values <0.5 negative
F[r, c] = F[r, c] - 0.5
# Next scale by the amount of contrast we want, making all values <0.5 darker and >0.5 brighter
F[r,c] = F[r,c] * contrast_factor
# Finally move the midpoint back into regular grayscale space
F[r,c] = F[r,c] + 0.5

This works well for non-extreme adjustments.

Another approach used in practice is to use some kind of nonlinear s-shaped curve (see [https://en.wikipedia.org/wiki/Sigmoid_function]) to modify the values so you don’t have as much detail getting clipped to 0 or 1.

Problem 3: Shift Left

In terms of an input image \(f(x, y)\), write a mathematical expression for a new image \(g\) that is shifted four pixels to the left.

Solution

\(g(x, y) = f(x + 4, y)\)

Problem 4: Scale Up

In terms of an input image \(f(x, y)\), write a mathematical expression for a new image \(g\) that is twice as big (i.e., larger by a factor of two in both \(x\) and \(y\)).

Solution

\(g(x, y) = f(x/2, y/2)\)

To make \(g\) bigger, you need \(x/2, y/2\) instead of \(2x, 2y\). Think of it as a lookup - to become bigger, \(g\) needs to look up its value from smaller coordinates in \(f\).

Problem 5: Cross-Correlation with Zero Padding

Recall that \(f \otimes w\) means cross-correlating image \(f\) with filter \(w\). Compute the following cross-correlation using same output size and zero padding. \[ \begin{bmatrix} 0 & 1 & 0\\ 0 & 1 & 0\\ 0 & 1 & 0 \end{bmatrix} \otimes \begin{bmatrix} 1 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 1 \end{bmatrix} \]

Solution

\[ \begin{bmatrix} 3 & 6 & 3\\ 4 & 8 & 4\\ 3 & 6 & 3 \end{bmatrix} \]

Problem 6: Cross-Correlation with Repeat Padding

Perform the same convolution as above, but use repeat padding.

Solution

\[ \begin{bmatrix} 4 & 8 & 4\\ 4 & 8 & 4\\ 4 & 8 & 4 \end{bmatrix} \]

Problem 7: Describe Filter Effect

Describe in words the result of applying the following filter using cross-correlation. If you aren’t sure, try applying it to the image above to gain intuition.

\[ \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \]

Solution

This shifts things one pixel to the left. If we were using convolution (instead of cross-correlation), it would shift right.

Problem 8: Convolution

Compute the following convolution, assuming repeat padding and same output size. \[ \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix} * \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & -1\\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \ & \ & \ \\ \ & \ & \ \\ \hspace{1em} & \hspace{1em} & \hspace{1em} \end{bmatrix} \]

Solution

\[ \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix} * \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & -1\\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} \]

Problem 9: Blur then Sharpen

Does blurring then sharpening an image yield the original, unfiltered image?

Solution

No. The information lost when blurring is truly gone and cannot be recovered by sharpening.

Problem 10: Convolution Filters

For each of the following, decide whether it’s possible to design a convolution filter that performs the given operation.

1. Max filter: the output pixel is the maximum value among the pixels in the input window

2. Threshold: the output pixel is

   • 1.0 if the input pixel is > 0.5

   • 0.0 otherwise

3. \(y\) partial derivative: the output is a finite-differences approximation of the input image’s vertical derivative \(\frac{\partial}{\partial y} f(x, y)\).

Solution

1. No - convolution is a linear operation, max is not
2. No - convolution is linear, threshold is not
3. Yes - this can be done with a finite difference filter

Problem 11: Blur and Derivative Filter

Compute the following convolution, which results in a new filter kernel, and describe the effect of this new kernel in words. \[ \begin{bmatrix} 1 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 1 \end{bmatrix} * \begin{bmatrix} 0 & 0 & 0\\ 1 & 0 & -1\\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \ & \ & \ \\ \ & \ & \ \\ \hspace{1em} & \hspace{1em} & \hspace{1em} \end{bmatrix} \]

Solution

\[ \begin{bmatrix} 2 & 0 & -2\\ 4 & 0 & -4\\ 2 & 0 & -2 \end{bmatrix} \]

or scaled by 1/2 if using repeat padding.

It gives the derivative of a blurred image, making our derivative estimate a little more robust to noise. This filter highlights only vertical edges - a transposed version of the same filter would work on horizontal edges.

Problem 12: Structure Tensors

Compute the structure tensor for each of the following image patches. I have it on good authority that these images are noise-free, so we can safely skip the Sobel filter and compute gradients using 3x1 and 1x3 centered finite difference filters and repeat padding. \[ \begin{bmatrix} 2 & 2 & 2\\ 2 & 2 & 2\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 & 2\\ 0 & 2 & 2\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 2 & 2 & 2\\ 0 & 2 & 2\\ 0 & 0 & 2 \end{bmatrix} \]

Solution

Gradients: \[ X: \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 2 & 2 & 0\\ 2 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0\\ 2 & 2 & 0\\ 0 & 2 & 2 \end{bmatrix} \\ Y: \begin{bmatrix} 0 & 0 & 0\\ 2 & 2 & 2\\ 2 & 2 & 2 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0\\ 0 & 2 & 2\\ 0 & 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0\\ 2 & 2 & 0\\ 0 & 2 & 0 \end{bmatrix} \]

Structure tensor: \[ \begin{bmatrix} 0 & 0 \\ 0 & 24 \end{bmatrix} \begin{bmatrix} 16 & 4 \\ 4 & 16 \end{bmatrix} \begin{bmatrix} 16 & 12 \\ 12 & 16 \end{bmatrix} \]

Problem 13: Eigenvalues of Structure Tensors

Using software of your choice (e.g., np.linalg.eigvals, or use the formula described here), compute the smallest eigenvalue of each of the structure tensors you computed in the prior problem.

Solution

0, 12, 4