wk 2/3

Question 1

What is a first derivative operator

Answer 1

A filter or function used for edge detection by using the first derivative wrt intensity change of an image

Question 2

What is a second derivative operator

Answer 2

An operator which aims to find edges by finding the second derivative of the image then applying zero crossing to find edges

Question 3

What is the intuitive approach to finding the first derivative edges of an image. what is the first problem that appears and how is it solved (despite other shortcomings)

Answer 3

Approximate the gradient by e.g. f[ i, j+1] - f[ i, j].
These can be put into convolutional filter form [-1, 1] and [ [ 1 ], [ -1 ] ].

These however find the gradient at interpolated points of i, j + 1/2 and i +1/2, j respectively. Consequently we combine these filters into 2d filters to find gradient at the same location of i + 1/2 and j + 1/2

[[ -1, 1 ], [ -1, 1 ] ] = G_x
[[ 1, 1 ], [ -1, -1 ] ] = G_y

Question 4

What are the 4 steps to edge detection

Answer 4

Filtering:
-removing noise e.g. Gaussian

Enhancement:
-methods to increase edge detector performance e.g. laplacian or normalization of image, increasing intensity etc

Detection:
-Thresholding or zero crossing etc

Localization: determining the actual position of an edge

Question 5

Compare Roberts and Sobel Filter

Answer 5

Roberts:
- 2x2 filter
- calculates interpolated gradient at i +1/2, j +1/2 rather than the gradient at the actual pixel location
-does not take the entire neighbourhood of the pixels into account

Sobel:
- 3x3 filter
- calculates the approximated gradient at exact location i,j
- takes the full neighbourhood of the pixel into account
- emphasises the pixels nearer the centre of the filter

Question 6

What are second derivative filters

Answer 6

• take the second derivative of the images intensity map
• the result of the second derivative are two zero crossed lines at areas of the greatest intensity change (local maxima)
  -points at zero crossing are more or less considered edges

Question 7

What is the Laplacian operator

Answer 7

it is derived from taking the second derivative of the x and y magnitude matrices with respect to x and y components respectively. This gives the 3x3 Laplacian operator

To find edges it can be convolved with the image to be edged

Question 8

What is a problem with the laplacian

Answer 8

It is very sensitive to noise and can therefore create many false edges

Question 9

what are the 2 solutions to laplacians noise problem

Answer 9

-Threshold zerocrossing
-Image smoothing e.g. Gaussian / Laplacian of Gaussian

Question 10

what is the gaussian filter, what are its key advantages

Answer 10

-It is a smoothing filter which fits surrounding pixels to a normal distribution depending on parameters
-noise is likely to be on the edges of the images normaly distributed pixels so it is very effective at removing noise
-it is seperable to 1 dimensions from 2d so it can be implemented very efficiently
-it is symmetric meaning its rotation wont make a difference and it smoothes in all directions equally

Question 11

How does laplacian of gaussian implement the 4 steps to edge detection

Answer 11

1) filtering -> Gaussian convolved
2) enhancement -> Laplacian in 2 dimensions accentuates local maxima
3) detection: zero crossing
4) localization: sub-pixel reconstruction and linear interpolation

Question 12

How do you make the LoG operator

Answer 12

convolve a 2d gaussian with a Laplacian

Question 13

Using LoG how do we do edge detection explicitly

Answer 13

• create LoG filter by convolving the gaussian with the Laplacian
  -Convolve the LoG operator with the image
  -Zero cross the resulting image

Question 14

What is a more efficient way to apply smoothing when dealing with a large gaussian filter, large image or both

Answer 14

Decompose the Gaussian filter into x and y components ( 1 dimensional Gaussians) and convolve the image one by one with the x and y Gaussian components. This is more efficient then convolving with 2-dimensional Gaussians

Question 15

What is the difference between how 1st derivative operators work compared to 2 derivative

Answer 15

1st derivative operators find first derivative, choose some threshold, and assume all magnitudes of first derivative image map above the threshold are edges

2 derivative operators find only points with a local maxima (by zero crossing) and consider them edges

Question 16

Why are second derivative filters better than first-derivative filters

Answer 16

-Second-derivative operators find edges corresponding to peaks not only in the first derivative but also zero-crossed ones in the second derivative
-Zero crossing is far better at localisation
-better edge response, as it only finds local maxima edges, they wont be as dummy thicc as first derivative operators who will display gradual intensity changes unlike second derivative operators