Jon's Topics

Question 1

Robust, Repeatable, Invariance and Constraints

Answer 1

Robust and Repeatable computer vision is achieved through engineered Invariance and applied Constraints. (We wants Robust and Repeatable, we achieve using Invariance and Constraints)

Question 2

Feature Space

Answer 2

Abstract space defined by a feature extraction process. Transforming raw data into feature of some fixed number of elements (dimensionality of the space).

Question 3

Distance Measures

Answer 3

Euclidean (L2) Distance: 2rt[SUMi=1->n (q_i-p_i)^2]

L1 Distance: SUMi=1->n ( | p_i-q_i | ). Where p,q = points

Question 4

Supervised ML: Classification

Answer 4

Binary - Linear: Learn a hyperplane that separates two classes with a minimum error. Non-Linear: Can learn curved lines to fit data, lose generality by overfitting.
Multiclass - KNN

Question 5

Unsupervised ML: Clustering

Answer 5

Aims to group data without any prior knowledge. K-Means: Guaranteed to find optimal results, doesn’t perform consistently. Trial and error to find best k value.

Question 6

Image Features (4 types)

Answer 6

Global, Grid-based, Region-based, Local

Question 7

Global Features

Answer 7

Extracted contents from entire image, outputs single Feature Vector. Image Histogram is common. Joint Colour Histogram usually considers all pixel colour components together.

Question 8

Segmentation

Answer 8

Used for segmenting image into regions for Region based features. Using: Thresholding - Basic, Otsu, and Local or Adaptive Thresholding. K-Means

Question 9

Otsu Thresholding

Answer 9

Otsu: Assumes image contains two classes of pixels (fore/background). Calculates optimum threshold to separate two classes with minimal intra-class variance.

Question 10

Local or Adaptive Thresholding

Answer 10

Computes a different threshold for each pixel, based on values of neighbouring pixels. E.g. Mean adaptive thresh sets pixel to BG if it’s value is less than mean of it’s neighbours plus an offset.

Question 11

Segmentation using K-Means

Answer 11

Simple method: Cluster colour vector of all pixels, assign each pixel based on closest cluster centroid. Could spatially encode FV to help base segments on relative position of pixels, to avoid loner different region pixels.

Question 12

Connected Components

Answer 12

Type of segment in which all pixels are reachable to each other through a path of spatially adjacent pixels. Commonly 4-connectivity or 8-connectivity

Question 13

Connected Component Labelling

Answer 13

Takes binary image and produces a set of connected components. Uses 2 pass algorithm

Question 14

Simple Scalar Features

Answer 14

Area: #pixels in component
Perimeter: Length around component
Inner Border: Inner pixels forming edge
Outer Border: Outer pixels of edge
Compactness: Measure how tightly packed pixels are
Dispersion: How spread out shape is

Question 15

Moments

Answer 15

Describes Distribution of pixels in a shape. Standard 2D Cartesian Moments can be used as shape descriptors. Different shapes have different moments. Not invariant to scaling, translation, and rotation.

Question 16

Central Moments + Normalised Central Moments + Hu Moments

Answer 16

Central: Translation invariant as moments computed about centroid of component.
Normalised Central: Translation and Scale Invariant.
Hu: 7 scale, rotation and invariant moments.

Question 17

Chain Codes

Answer 17

Simple way of encoding boundary of an object. Walk around boundary and encode direction on each step as a number. Cyclically shift code so it forms smallest integer value. Invariant to start point.

Question 18

Chain Codes Properties

Answer 18

Make rotation invariant by encoding differences in direction. Make scale invariant by resampling component to fixed size. Perimeter for 8-con chain code = #(Even numbers in CC) + Rt2 * #(Odd numbers in CC). Not good for shape matching due to noise, resampling and no good distance metric.

Question 19

Fourier Descriptor

Answer 19

Encode shape info by decomposing boundary into small set of frequency components. 1: Define representation of the curve (boundary). 2: Expand representation using Fourier Theory.

Question 20

Region Adjacency Graph

Answer 20

Build graph from set of connected components. Each node = component. Nodes connected if they share a border. Invariant to distortion, not to occlusion. Good for creating markers like QR Codes.

Question 21

Point Distribution Models

Answer 21

Learning a low dimensional parametric model of how the shape of an object can vary. Shape represented by number of fixed 2D points at distinguishable locations on the object. Needs training set of images with same points marked.

Question 22

Procrustes Analysis

Answer 22

Choose a reference shape, superimpose (align) all instances to current reference shape. Compute mean shape of the current set of images. If Euclidean distance between mean shape and reference shape is above a threshold, set reference shape to mean shape and repeat.

Question 23

Local Interest Point

Answer 23

Good Local Interest Point is invariant to global & local lighting change, invariant to difference in camera’s location. Detect using corner or blob detection.

Question 24

Corner Detector: Harris + Stephens

Answer 24

Search for corners through a small window. Corner found if shift in window by a small amount results in large change in intensity.

Question 25

Structure Tensor

Answer 25

Encodes the image derivatives in the x, y and xy directions. Is the second moment matrix. Symmetric matrix with M=[{Ew(Ix(xi,yi)^2), Ew(Ix(xi,yi)Iy(xi,yi)}, {Ew(Ix(xi,yi)Iy(xi,yi), Ew(Iy(xi,yi)^2)}]

Question 26

Laplacian Of Gaussian

Answer 26

Second Derivative of Gaussian. By finding local minima or maximum, you get a blob detector

Question 27

Narrow-baseline stereo, Wide-baseline stereo

Answer 27

Narrow: 2 images are very similar - local features have only been moved by a few pixels, applicable for video tracking. Wide: 2 images are not similar at all, applicable for generic matching tasks.

Question 28

Robust Local description

Answer 28

Narrow baseline: Robust to rotation & lighting not so important. Descriptiveness can be lower as search is over a smaller area.
Wide baseline: Need to be robust to intensity change, invariant to rotation. Highly Descriptive to avoid mismatches! Robust to small localisation errors of interest points.

Question 29

Matching by correlation

Answer 29

Use pixel values to describe a local region around an interest point. Basic template matching can be used to perform matching. Works well for regions with small differences, better in narrow baseline.

Question 30

Local Intensity Histogram

Answer 30

Represent local region with a histogram of pixel values describing the region around each interest point, as opposed to comparing raw pixel values. Might not be distinctive. Only rotation invariant if circular window is used. Not invariant to illumination changes. Sensitive to interest point localisation

Question 31

Localisation Sensitivity

Answer 31

Small changes in interest point. Overcome by weighting. Pixels at edge of sampling patch have less effect, those nearer have more effect. Commonly use a Gaussian weighting.

Question 32

Local Gradient Histogram

Answer 32

Encode gradient directions within a window instead of a raw pixel values in a histogram. Compute gradient orientations and magnitudes using partial derivatives of the image. Gradient Direction ‘Theta’, magnitude ‘m’

Question 33

Local Gradient Histogram properties

Answer 33

Gradient Magnitudes are invariant to brightness change. Not rotation invariant -> Find dominant orientation and cyclically shift so it is in first bin.

Question 34

SIFT Feature

Answer 34

Like Local Gradient Histogram, incorporates spatial binning. Creates multiple gradient histograms about the interest point and appends them to one feature vector. Usually spatial 4x4 grid of histograms with 8 orientations = 128-dimensional feature. Discriminative and robust!

Question 35

Feature Distinctiveness

Answer 35

Local feature too distinctive won’t match subtle variations due to noise. If it’s not distinctive enough it will match everything.

Question 36

Geometric Mapping

Answer 36

A transform function mapping x,y co-ords of points in one image to another. Used to constrain matches between two images

Question 37

Affine transform

Answer 37

Transform that retains parallel lines, but allows translation,scaling,aspect ratio,rotation,skewing. Usually: y=Ax+b where A is a 2x2 transform matrix encoding scale,rotation&scew, b=vector encoding translation. 6 degrees of freedom.

Question 38

Planar Homography

Answer 38

Adds homogeneous co-ords to deal with perspective (i.e. an object moving further away gets smaller).

Question 39

Recovering geometric mapping transform matrix

Answer 39

Solve set of simultaneous equations. 4 point matches for Homograph, 3 for affine.

Question 40

RANSAC

Answer 40

Robust Estimation algorithm to robustly estimate inliers of the model. Assume: M data items required to estimate model T. N data items in total.

Question 41

RANSAC Pseudocode

Answer 41

1. Select M data items at random
2. Estimate model T
3. Calculate K = how many N data items fit T within a tolerance tol.
4. If K is large enough, aaccept T, or computer least squares estimates using all inliers. Exit with success
5. Repeat steps 1-4 N iteration times.
6. Fail if no good T fit of data.