Finding A Chessboard: An Introduction To Computer Vision

Martin C. Martin

March 9, 2005 6

Motivation & Inspiration

• A spare time project, because I thought it would be fun

• Project: Chess against the computer, where board and your pieces are real, it’s pieces are projected

• Working so far: very robust localizing of chessboard (full 3D location and orientation)

March 9, 2005 7

Requirements

• Interaction should be as natural as possible– E.g. pieces don’t have to be centered in their squares, or even

completely in them

• Should be easy to set up and give demonstrations– Little calibration as possible– Work in many lighting conditions– Although only with this board and pieces

• Camera needs to be at angle to board• Board made by my father just before he met my mother

– My brother and I learned to play on it

• I’m not a big chess fan, I just like project

March 9, 2005 8

Computer Vision ’70s to ’80s: Feature Detection

• Initial Idea: Corners are unique, look for them• Compute the “cornerness” at each pixel by

adding the values of some nearby pixels, and subtracting others, in this pattern:

+ + + - - -+ + + - - -+ + + - - -- - - + + +- - - + + +- - - + + +

• Constant Image (e.g. middle of square): output = 0

• Edge between two regions (e.g. two squares side-by-side): output = 0

• Where four squares come together: output max (+ or -)

March 9, 2005 9

Corner Detector In Practice

• Actually, the absolute value of the output

March 9, 2005 10

Problems With Corner Detection

• Edge effects

• Strong response for some non-corners

• Easily obscured by pieces, hand

• How to link them up when many are obscured?

• Go back to something older: Find edges

March 9, 2005 11

Computer Vision ’60s toEarly ’70s: Line Drawings

March 9, 2005 12

Edge Finding

• Very common in early computer vision– Early computers didn’t have much power– Very early: enter lines by hand– A little later: extract line drawing from image

• Basic idea: for vertical edge, subtract pixels on left from pixels on right

• Similarly for horizontal edge

- - - + + +- - - + + +- - - + + +- - - + + +- - - + + +- - - + + +

March 9, 2005 13

Edge Finding

• Need separate mask for each orientation?• No! Can compute intensity gradient from horizontal &

vertical gradients– Think of intensity as a (continuous) function of 2D position:

f(x,y)– Rate of change of intensity in direction (u, v) is– Magnitude changes as cosine of (u, v)– Magnitude maximum when (u, v) equals (∂f/∂x, ∂f/∂y)

• (∂f/∂x, ∂f/∂y) is called the gradient – Magnitude is strength of line at this point– Direction is perpendicular to line

vy

fux

f

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Gradient

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Localizing Line• How do we decide where

lines are & where they aren’t?

• One idea: threshold the magnitude– Problem: what threshold to

use? Depends on lighting, etc.

– Problem: will still get multiple pixels at each image location

March 9, 2005 16

Laplacian

• Better idea: find the peak– i.e. where the 2nd derivative

crosses zero

2

2

2

2

:y

f

x

fLaplacian

• Image shows magnitude• One ridge is positive, one negative

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Gradient: Before And After Suppression

March 9, 2005 18

Extracting Whole Lines

• So Far: intensity image “lineness” image• Next: “lineness” image list of lines• Need to accumulate contributions from across

image– Could be many gaps

• Want to extract position & orientation of lines– Boundaries won’t be robust, so consider lines to run

across the entire image

March 9, 2005 19

Hough Transform

• Parameterize lines by angle and distance to center of image

• Discretize these and create a 2D grid covering the entire range

• Each unsuppressed pixel is part of a line– Perpendicular to the

gradient– Add it’s strength to the

bin for that line

θ

d

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Hough Transform

AngleDis

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ente

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Take The 24 Biggest Lines

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Demonstration

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Coordinate Systems

• In 2D (u, v) (i.e. on the screen):– Origin at center of screen– u horizontal, increasing to the right– v vertical, increasing down– Maximum u and v determined by field of view.

• In 3D (x, y, z) (camera’s frame):– Origin at eye– Looking along z axis– x & y in directions of u & v respectively

March 9, 2005 25

2D 3D

• Perspective projection– (u, v): image coordinates (2D)– (x, y, z): world coordinates (3D)– Similar Triangles

• Free to assume virtual screen at d = 1

– Relation: u = x/z, v = y/z

d

v

z

y

March 9, 2005 26

2D 3D

p = (u, v) p = (x, y, z)

u = x/z, v=y/z

March 9, 2005 27

Lines in 3D map to lines on screen

• Proof: the 3D line, plus the origin (eye), form a plane.

• All light rays from the 3D line to the eye are in this plane.

• The intersection of that plane and the image plane form a line

March 9, 2005 28

2D 3D

p = (u, v) p = (x, y, z)

u = x/z, v=y/z

Line Line

March 9, 2005 29

From 2D Lines To 3D Lines

• If a group of lines are parallel in 3D, what’s the corresponding 2D constraint?

• Equation of line in 2D: Au + Bv + C = 0• Substituting in our formula for u & v:

– Ax/z + By/z + C = 0– Ax + By + Cz = 0

• A 3D plane through the origin containing the 3D line

– Let L = (A, B, C) & p = (x, y, z). Then L•p = 0

March 9, 2005 30

2D 3D

p = (u, v) p = (x, y, z)

u = x/z, v=y/z

Line Line

Line Au + Bv + C = 0

Plane

Ax + By + Cz = 0

L•p = 0

March 9, 2005 31

Recovering 3D Direction

• Let’s represent the 3D line parametrically, i.e. as the set of p0+td for all t, where d is the direction.

• The 9 parallel lines on the board have the same d but different p0.

• For all t: L•(p0+td) = L•p0 + t L•d = 0• Since p0 is on the line, L•p0 = 0.• Therefore, L•d = 0, i.e. Axd + Byd + Czd = 0• That is, the point (xd, yd, zd) (which is not necessarily on

the 3D line) projects to a point on the 2D line• d is the same for all 9 lines in a group; so it must be the

common intersection point, i.e. the vanishing point– Knowing the 2D vanishing point gives us the full 3D direction!

March 9, 2005 33

Vanishing Point

March 9, 2005 34

Ames Room

Viewing Direction

March 9, 2005 35

2D 3D

p = (u, v) p = (x, y, z)

u = x/z, v=y/z

Line Line

Line Au + Bv + C = 0

Plane

Ax + By + Cz = 0

L•p = 0

Common Intersection Parallel Lines

Vanishing Point Common Direction

March 9, 2005 36

Finding The Vanishing Point

• Want to find a group of 9 2D lines with a common intersection point

• For two lines A1u+B1v+C1 = 0 and A2u+B2v+C2 = 0, intersection point is on both lines, therefore satisfies both linear equations:

• In reality, only approximately intersect at a common location

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1

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C

v

u

BA

BA

March 9, 2005 37

Representing the Vanishing Point

• Problem: If camera is perpendicular to board, 2D lines are parallel no solution

• Problem: If camera is almost perpendicular to board, small error in line angle make for big changes in intersection location

• Euclidean distance between points is poor measure of similarity

March 9, 2005 38

Desired Metric

• Sensitivity Analysis for Intersection Point:

• Of the form u’/w, v’/w. If -1 ≤ A,B,C ≤ 1, then -2 ≤ u’, v’, w ≤ 2

• So, the only way for the intersection point to be large is if w is small; the right metric is roughly 1/w

• A representation with that property is…

1221

2112

1221

2112 ,BABA

CACAv

BABA

CBCBu

March 9, 2005 39

Homogeneous Coordinates

• Introduced by August Ferdinand Möbius• An example of projective geometry• Represent a 2D point using 3 numbers• The point (u, v, w) corresponds to u/w, v/w• Formula for perspective projection means any

3D point is already the homogeneous coordinate of its 2D projection

• Multiplying by a scalar doesn’t change the point:– (cu, cv, cw) represents same 2D point as (u, v, w)

March 9, 2005 40

Homogeneous Coordinates

(u, v, 1)

w=1

March 9, 2005 41

Project Intersection Point Onto Unit Sphere

March 9, 2005 42

Lines Become “Great Circles”

March 9, 2005 43

Clustering: Computer Vision In The Nineties

• 60s and 70s: Promise of human equivalence right around the corner

• 80s: Backlash against AI– Like an “internet startup” now

• 90s: Extensions of existing engineering techniques– Applied statistics: Bayesian Networks– Control Theory: Reinforcement Learning

March 9, 2005 44

K Means

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2D to 3D: Distance• How do we get the distance to the board? From the

spacing between lines.• While 3D lines map to 2D lines, points equally spaced

along a 3D line AREN’T equally spaced in 2D:

March 9, 2005 52

Distance To Board

• Let c = (0, 0, zc) be the point were the z axis hits the board

• For each line in group 1, we find the 3D perpendicular distance from c to the line, along d2

• These should be equally spaced in 3D

• Find the common difference, like Millikan oil drop expr.

March 9, 2005 53

Distance To Board

• Let Li = (Ai, Bi, Ci) be the ith line in group 1.• We want to find ti such that the point c + ti d2

is on Li, i.e.• Li•(c + ti d2) = 0• Li•c + ti Li•d2 = 0• ti = - Li•c / Li•d2 = - Cizc / Li•d2

• zc is the only unknown, so put that on the left• Let si = ti/zc = -Ci / Li•d2

• If we choose our 3D units so that the squares are 1x1, then the common difference is 1/zc

March 9, 2005 54

Distance To Board

• So, given the si, we need to find t0 and zc such that:

• si = ti/zc = (i + t0) / zc, i = 0…8

• But, we have outliers & occasional omission

• So we use robust estimation

March 9, 2005 55

Robust Estimation

• Many existing parameter estimation algorithms optimize a continuous function– Sometimes there’s a closed form (e.g. MLE of center

of Gaussian is just the sample mean)– Sometimes it’s something more iterative (e.g.

Newton-Rhapson)

• However, these are usually sensitive to outliers– Data cleaning is often a big part of the analysis– The reason why decision trees (which are extreemly

robust to outliers) are the best all-round “off-the-shelf” data mining technique

March 9, 2005 56

Robust Estimation

• Find distance (and offset) to maximize score:

– Involves evaluating on a fine grid– Isn’t time consuming here, since the number

of data points is small