eecs 274 computer vision model fitting. fitting choose a parametric object/some objects to represent...

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EECS 274 Computer Vision

Model Fitting

Fitting

• Choose a parametric object/some objects to represent a set of points

• Three main questions:– what object represents this set of points best?– which of several objects gets which points?– how many objects are there?(you could read line for object here, or circle, or

ellipse or...)

• Reading: FP Chapter 15

Fitting and the Hough transform• Purports to answer all

three questions– in practice, answer isn’t

usually all that much help

• We do for lines only• A line is the set of points

(x, y) such that

• Different choices of , d>0 give different lines

• For any (x, y) there is a one parameter family of lines through this point, given by

• Each point gets to vote for each line in the family; if there is a line that has lots of votes, that should be the line passing through the points

0)(sin)(cos ryx

0)(sin)(cos ryx

tokens votes

•20 points •200 bins in each direction•# of votes is indicated by the pixel brightness •Maximum votes is 20•Note that most points in the vote array are very dark, because they get only one vote.

r

Hough transform

• Construct an array representing , r

• For each point, render the curve (, r) into this array, adding one at each cell

• Difficulties– Quantization error: how

big should the cells be? (too big, and we cannot distinguish between quite different lines; too small, and noise causes lines to be missed)

– Difficulty with noise

• How many lines?– count the peaks in the

Hough array

• Who belongs to which line?– tag the votes

• Hardly ever satisfactory in practice, because problems with noise and cell size defeat it

points votes

r

•Add random noise ([0,0.05]) to each point.•Maximum vote is now 6

points votes

As noise increases, # of max votes decreases difficult to use Hough transform less robustly

•As noise increase, # of max votes in the right bucket goes down, and it is more likely to obtain a large spurious vote in the accumulator•Can be quite difficult to find a line out of noise with Hough transform as the # of votes for the line may be comparable with the # of vote for a spurious line

Choice of model

b

a

x

xx

y

y

baxyi

ii

1

)(

22

2

Least squares but assumes error appears only in y Total least squares

b

a

b

a

yyyyxxy

yxxyxxx

ybxac

b

a

c

b

a

yx

yyxy

xxyx

bacbyaxi

ii

2

2

2

2

222

0

2

2

1

1 s.t. )(

),( vu

0 cbyax

1 if is line to

),( from distancelar perpendicu22 bacbvau

vu

Who came from which line?

• Assume we know how many lines there are - but which lines are they?– easy, if we know who came from which

line

• Three strategies– Incremental line fitting– K-means– Probabilistic (later!)

linesl

ix

ji

i j

xldist

lineth to due data

2),(

Fitting curves other than lines• In principle, an easy

generalisation– The probability of

obtaining a point, given a curve, is given by a negative exponential of distance squared

• In practice, rather hard– It is generally difficult to

compute the distance between a point and a curve

Implicit curves

• (u,v) on curve, i.e., ϕ(u,v)=0

• s=(dx,dy)-(u,v) is normal to the curve

0

04 where,0

04 where,0

04 where,0

022

0

22

222

222

222

22222

feydxcybxyaxconic

acbfeydxcybxyaxparabolae

acbfeydxcybxyaxhyperbolae

acbfeydxcybxyaxellipse

rbabyaxyxcircle

cbyaxline

0))(,())(,(),;,(

0

,

udvux

udvuy

ddvu

sT

yxT

yxyx

Robustness

• As we have seen, squared error can be a source of bias in the presence of noise points– One fix is EM - we’ll do this shortly– Another is an M-estimator

• Square nearby, threshold far away

– A third is RANSAC• Search for good points

Missing data

• So far we assume we know which points belong to the line

• In practice, we may have a set of measured points– some of which from a line, – and others of which are noise

• Missing data (or label)

Least squares fits the data well

Single outlier (x-coordinate is corrupted) affects the least-squares result

Single outlier (y-coordinate is corrupted) affects the least-squares result

Bad fit

Heavy tail, light tail

• The red line represents a frequency curve of a long tailed distribution.

• The blue line represents a frequency curve of a short tailed distribution.

• The black line is the standard bell curve..

M-estimators

• Often used in robust statistics

out flattensfunction thehow controls ,),(

residue :),( ,parameters model :

));,((

22

2

u

uu

xr

xr

ii

iii

22

2

),(x

xx

A point that is severalaway from the fitted curve will have no effect on the coefficients

Other M-estimators

• Defined by influence function

• Nonlinear function, solved iteratively• Iterative strategy

– Draw a subset of samples randomly – Fit the subset using least squares– Use the remaining points to see fitness

• Need to pick a sensible σ, which is referred as scale

• Estimate scale at each iteration

0));,((

iii xr

|);(|median 4826.1 )1()()( ni

nii

n xr

Appropriate σ

small σ

large σ

Matching features

What do we do about the “bad” matches?

Szeliski

RAndom SAmple Consensus

Select one match, count inliers

RAndom SAmple Consensus

Select one match, count inliers

Least squares fit

Find “average” translation vector

RANSAC

• Random Sample Consensus

• Choose a small subset uniformly at random

• Fit to that• Anything that is close to

result is signal; all others are noise

• Refit• Do this many times and

choose the best

• Issues– How many times?

• Often enough that we are likely to have a good line

– How big a subset?• Smallest possible

– What does close mean?• Depends on the

problem– What is a good line?

• One where the number of nearby points is so big it is unlikely to be all outliers

Richard Szeliski Image Stitching 34

Descriptor Vector• Orientation = blurred gradient• Similarity Invariant Frame

– Scale-space position (x, y, s) + orientation ()

RANSAC for Homography

RANSAC for Homography

RANSAC for Homography

Probabilistic model for verification

Finding the panoramas

Finding the panoramas

Finding the panoramas

Results

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