on the complexity of affine image matching stacs 2007, aachen university of lbeck institute for...

On the Complexity ofAffine Image Matching

STACS 2007, Aachen

University of LübeckInstitute for Theoretical Computer ScienceLübeck, Germany

Thursday, February 22th 2007

Christian Hundt and Maciej Liśkiewicz

Foto: Andreas Herrmann

What is Image Matching?On the Complexity of Affine Image Matching

? ? ?

OverviewOn the Complexity of Affine Image Matching

1. Notation and the Problem

2. Structure of the Searchspace

3. Polynomial Time Algorithm

Section 1On the Complexity of Affine Image Matching

Notation and the Problem

Notation and the ProblemR2-Functions and Images

On the Complexity of Affine Image Matching

A: R R R



A: Z Z R

Discretization of A ( step I ):



A: Z Z R



A: Z Z Z

14 19 12 14

18 21 13 17

16 18 13 15

14 19 12 14

Discretization of A ( step II ):

Notation and the ProblemImage Metric


BA

: R R R ( e.g. (a,b) = |a-b| )

(A, B) = ( A(x,y), B(x,y) ) dx dy- -

Notation and the ProblemImage Metric


BA

(A, B) = ( A(x,y), B(x,y) ) N

N

Discretization of for Images:

14N2

y=-N x=-N

Notation and the ProblemTransformations


f(A)(x,y) = A( f -1(x,y) )

A f(A)

f

f -118

18



f(A)(x,y) = A( f -1(x,y) )

A f(A)

f -1

????????????

and with Images?



A f(A)



A f(A)

Nearest Neighbour Interpolation!!!



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst

Discretization of f:

(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst


(3,1)(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst


(3,1)(2,1)(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst


(3,1)(2,1)(1,1)(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst


(3,1)(2,1)(1,1)(0,1)(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



f (i, j) =[ f -1(i, j) ], if in (-N ... N)2

sonst


(3,1)(2,1)(1,1)(0,1)

(0,2)

(0,3)

(1,2)

(1,3)

(2,2)

(2,3)

(3,2)

(3,3)

(0,0)

(0,1)

(0,2)

(0,3)

(1,0) (2,0) (3,0) (4,0)

(1,1) (2,1) (3,1) (4,1)

(1,2) (2,2) (3,2) (4,2)

(1,3) (2,3) (3,3) (4,3)



Discretization of F:

N(F) = { f | f F and f injecive }

Notation and the ProblemImage Matching


BA

f ?

Input: Image A, distorted image B, set of admissible transformations F

Solution:Transformation f in F

Goal: min (f(A), B) = min ( A( (x,y) ), B(x,y) )14N2 y=-N x=-N

N N

f F N(F)

Notation and the ProblemHow to ...


1. enumerate N(F) ?

2. estimate |N(F)| ?

Notation and the ProblemElastic Image Matching


f

f(A)A

Set of transformation F: ||(x,y)-(x‘,y‘)|| 1 ||f(x,y)-f(x‘,y‘)||

[Keysers; Unger; 2002] Elastic Image Matching is NP-complete.

Notation and the ProblemProjective Image Matching


f(A)A

Set of transformation F: f(x,y) = (a/c, b/c) and (a,b,c) = M(x,y,1)

Open Problem.

f

Notation and the ProblemAffine Image Matching


f(A)A

Set of transformation F: f(x,y) = M(x, y) + t

Our result: Affine Image Matching can be solved in polynomial time.

f

Notation and the ProblemMore Image Matching Classes


f

f

f

f

f

Translation:

Scaling:

Rotation:

Rotation &Scaling:

LinearTransformation:

f(x,y) = +1 0

0 1

s 0

0 s

cos sin

-sin cos

s cos s sin

-s sin s cos

x

y

t1

t2

f(x,y) =x

y

f(x,y) =x

y

f(x,y) =x

y

f(x,y) =a1 a2

a3 a4

x

y


Structure of the Search Space

Structure of the Search SpaceAffine Transformation vs. R6


f(x,y) = +a1 a2

a3 a4

xy

t1

t2

Transformation:

Vector in R6: v = ( )T



f(x,y) = +a2

a3 a4

xy

t1

t2

Transformation:

Vector in R6: a1, v = ( )T



f(x,y) = +

a3 a4

xy

t1

t2

Transformation:

Vector in R6: a1, a2, v = ( )T

v = ( )T



f(x,y) = +xyTransformation:

Vector in R6: a1, a2, t1,

a3 a4

t2

v = ( )T



f(x,y) = +

a3 a4

xyTransformation:

Vector in R6:

t2

a1, a2, t1,

v = ( )T




Vector in R6:

t2

a1, a2, t1, a3,

a4

v = ( )T



f(x,y) = +

a4

xyTransformation:

Vector in R6:

t2

a1, a2, t1, a3,

v = ( )T




Vector in R6:

t2

a1, a2, t1, a3, a4,

v = ( )T



f(x,y) = +xy

t2

Transformation:

Vector in R6: a1, a2, t1, a3, a4,

v = ( )T




Vector in R6: a1, a2, t1, a3, a4, t2

v = ( )T




Vector in R6:

a1 a2

a3 a4

t1

t2

a1, a2, t1, a3, a4, t2

v = f -1

Structure of the Search SpaceHyperplanes


Xiji‘: i x1 + j x2 + x3 - i‘ + 0.5 0

Yijj‘: i x4 + j x5 + x6 - j‘ + 0.5 0

Let HN be a set of hyperplanes in R6 which contains for all i, j, i‘, j‘ [-N,N] the hyperplanes:

A(HN) is the set of cells in R6 defined by HN.

Structure of the Search SpaceCentral Property


Lemma 1. For two vectors u, v R6 it holds that u = v iff if for all i, j, i‘ [-N,N] u and v belong to the same half-subspace according to the partition of R6 with the hyperplane Xiji‘.

The same holds for Yijj‘.

Structure of the Search SpaceMain Result

Theorem 1. Let F be the set of affine transformations, then N(F) A(HN).


Corollary 1. |A(HN)| = O(N18).


Polynomial Time Algorithm

Polynomial Time AlgorithmEfficient Representation

Lemma 2. Each convex cell of A(HN) contains a vector coordinates of which can be encoded by rational numbers of lenght O(log N).Moreover the representatives can be computed efficiently.


Polynomial Time AlgorithmThe Algorithm


Theorem 2. [Edelsbrunner, 1987] Any arrangement of n hyperplanes in R6 can be constructed in O(n6) time.

Polynomial Time AlgorithmThe Algorithm


Algorithm ImageMatching( )Input: Images A, B of size (2N+1) (2N+1)Output: Transformation f = arg min (f‘(A), B)

1. Construct A(HN).2. Set f = Id and = .3. For all C A(HN) do4. compute the representative g for C,5. compute ‘ = (g-1(A), B),6. if (‘ < ) then set = ‘ and f = g-1.7. Return f.

f‘ F

Polynomial Time AlgorithmRunning Time


Lemma 3. The algorithm runs in time O(N20).

Theorem 3. The Affine Image Matching Problem can be solved in polynomial time.

Polynomial Time AlgorithmExtensions


1. Many subclasses of Affine Transformation

2. Images with more than two Dimensions.

3. Linear Interpolation

Summary

1. The Search Space of discretized inverse affine Transformation has a regular structure.

2. The Affine Image Matching Problem can be solved in polynomial time.


Thank you for your attention!

Any Questions?


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