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EFFICIENT BITMAP EFFICIENT BITMAP RESEMBLENCE UNDER RESEMBLENCE UNDER

TRANSLATIONTRANSLATION

Klara Kedem & Daniel Cohen

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Introduction

“Whether two point set A and B resemble each other ?”

minimum Hausdorff distance under translation

Use of the graphic hardware in order to speed-up run-times of the existing algorithms

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The algorithms

Huttenlocher, Kedem and Sharir Huttenlocher, Klanderman and Rucklidge Chew and Kedem

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Graphic Hardware

Pixel planeWord for each pixel Processor for each pixelOperations that supported are restricted

(or, and, …) Parallel execution

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In order to use graphic hardware the input for the algorithms will be in raster form, i.e., matrices of pixels, each having value 0 or 1, standing for white or black color of the image

110010011000100100110000101010

1111101111

Bitmaps

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Automatic navigation.In automatic navigation two successive

images are taken by a camera attached to an autonomous vehicle, with a short time interval between them. We need to find the matching features in the images in order to determine the exact motion and the new location of the vehicle

The application

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Assume that we have two consecutive images and we have a vague knowledge of motion between them, which consist of a translation value and a tolerance

We compute a rasterized approximation of the exact motion within window W

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Example

The map is known, the motion is roughly known

We need to find the exact motion

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Theoretical Background

A, B –point sets(a,b) – any metric (for example L1, L2, L)Directed Hausdorf distance from a A to B

)1(),(min),( baBadBb

d(a, B)

a

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Theoretical Background

Directed Hausdorff distance from A to B

It measures the distance of the point of A which is farthest from the set B

Attention: h(A, B) is not symetric

)2(),(minmax),( badBAhBbAa

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Theoretical Background

h(A, B) h(B, A)

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Theoretical Background Symmetric Hausdorff distance from A to B

)3()],(),,(max[),( ABhBAhBAH

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Theoretical BackgroundThe measure of resemblance D(A, B), between the sets A and B - the minimum value of the Hausdorff distance under translation.

A is fixed, B is allowed to translate in the plane

where B + x is the set B translated by x

),(min),(2

xBAHBADRx

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Theoretical Background

we discuss the directed Hausdorff distance only

Consider the directed distances in (*) as function that depends on the translation x of the set B.

(*)),(minmax),( axbdAxBhAaBb

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Voronoi Surface

Consider the function

The graph of this function

is the Voronoi surface of A

)6(),(min)( axdxgAa

A

}|))(,{( 2Rxxgx A

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Voronoi Surface in one dimention

1a 2a 3a

||)( 3 xaxg A )(xg A

x x

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Voronoi Surface on the plane

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Upper Envelope

For each bB, we shift by –b f(x) is the upper envelope of these shifts

)(xg A

bAaBbbA

Bbaxdxgxf

),(minmax)(max)(

),(minmax),(minmax axbdbaxdAaBbAaBx

),( AxBh

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)(xf

Upper Envelope

x

Goal – is to find the translation x that minimize f(x)

- this is directed minimum Hausdorff distance

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Algorithm 1:Huttenlocher, Kedem and Sharir

(i) Initialization: Generate ; Set f(x) to zero;

(ii) For each b B do f(x) = max (f(x), )), for all x W.

(iii) Find the minimum of f(x) for all x W.

)(xg A

)( bxg A

minf

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Speed-ups

Encode the distance values in the distance transform gA(x) into words in the pixel plane

distance i = gA(x) present by its unary representation U(i) = 2i-1

min(U(i), U(j) = and (U(i), U(j)) max(U(i), U(j) = or (U(i), U(j))

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Speed-ups

If the length of a word in the pixel plane is 6 bits, then gA(x) = 1 encoded by 000001

gA(x) = 6 encoded by 111111

Restricts the distance we can represent to be smaller then the size (in bits) of words in the pixel plane

truncate gA(x)

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Speed-upsk –size of a word in the pixel plane in bits

first – transform gA(x) to k-gA(x)

truncate all the negative values of k-gA(x) => get

k-gA(x ) if gA(x) <= k g’A(x) =

0 otherwise

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)(xg A

)(xgk A

)(' xg A

Horizontal mirroring of

)(xg ATruncated horizontal mirroring of

)(xg A

k

k

25Inverted distance truncated distance

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By the transformation above we compute a truncated lower envelope f’(x) of g’A(x+b)

Maximum of f’(x) is desired minimum Hausdorff distance (under k-bit limitation)

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Algorithm 3: The Huttenlocher, Klanderman and Rucklidge

1. Initialization: Compute ; Initialize

2. For each x W do

(a) For each b B do d = max(d, );

(b) = min( , d);

)(xg A minf

)( bxg A

minf minf

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Search of the minimum is done in the inner loop Speed-up: as soon as for some b the computed

distance d is grater that the current value of fmin

leave the inner loop since x will not yield a better Hausdorff distance

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The framework of Chew and Kedem examines certain cross-sections of Voronoi surfaces.

Disc of radius is put about each point of A,creating set A (the union of these discs)

A={3, 9}, B={5, 12}

0

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Consider the set A - b; the set A shifted by –b A-b – set of all translations that map b into A F = b B A - b : set of all translations that map

all points b B into A

A={3, 9}, B={5, 12}

0

0

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Hausdorff decision problem – for a given whether the minimum Hausdorff distance under translation is less then

The answer is true if there exists a translation x such that the Hausdorff distance between A and B+x is less than or equal to => F above is not empty

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Algorithm 2:Chew and Kedem

Algorithm 2:Rasterization of the Chew and Kedem algorithm

1. Given a distance , compute a disk of this radius, C().

2. For each point a A put C() centered around a, thus

creating .

3. For each point b B shift by –b and compute the

intersection, , of the overlaid (shifted) with

former ones, over W.

4. Check whether the intersection, , is not empty

A

A

F

A

F

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Given an initial Create a disc of radius filled with 1’s and fill the rest

of the bitmap with 0’s Shift this bitmap about all point a A such that the

center of the disc is at a; or operation over all the shifts gets the union A Shifting A by –b for all b B, and intersecting these

shifted copies, generates F ; this can be computed by the and operation

Implementation of Alg.2

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we can perform k decision algorithms simultaneously (k –number of bits in the pixel plane)

1, 2, … , k – ascending sequence of the values which we chose to check simultaneously by alg.2

Number the bits in the pixel plane from 1 to k (1 and k the most significant and the least significant bits, respectively)

Relate to bit number j as to the layer representing the discs of radius j (this layer will represent Aj )

A - pixel plane with al the Aj , j = 1,…,k

Interleaving algorithms 1 and 2

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word size = k

The k-bit pixel plane for discs of various radius about

a A (the template)

klayer 2: = 3

layer 4: = 7

point a

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The intersection areas , F1, F2, … , Fk, will indicate whether A and B are closer then the corresponding ’s

The and operation that we used for radius can answer k decision problems and here yields a pixel plane that contains f’ –

all of the intersected Fi together

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If for 1, … , i the answer the decision problem is false, and for i+1, … , k is true, refine the range(i,i+1 ) by creating a new sequence of ’s in this range and we run Alg.2 again

In practice the range diminishes very quickly (this procedure runs at most twice)

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Template

Define template about the pixel a A to be the pixel plane with 0 everywhere and the unary representation the set of all words in the pixel plane which get a value other than 0 by drawing the disc about a

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Algorithm 4: a combination of 1 and 2

1. Initialization : zero f’(x) for all x W ;2. Build template: Given an ascending sequence of

possible Hausdorff distances , create a template of disks in the pixel plane for one point a A;

3.Create : for each point a A copy the template, centered at a, performing an or operation between the pixel planes;

4. for each b B shift the pixel of by –b, performing the and operation between the pixel plane containing f’ and the current shift of ,

f’ = and (f’, -b)5. For each x W , find the maximum point in

f’, by performing = or(f’(x)| x W)

i ,...,1

A

A

AA

minfminf

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in algorithm 4 the distance transform g’A is replaced by A

The word size k which restrict the largest distance that could be stored in g’A is not restricting any more

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Implementation and resultsAlg.4 and Alg.3 were implemented in C on a Silicon Graphics Indigo with 8-bits pixel-plane

Performances of Alg.4 and Alg.3 were compared under different parameters

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Implementation and resultsTwo images were taken by a moving camera, camera slightly shifted downwards

A binary images were obtained

The matching has been applied between windows in 1 and corresponding windows in 2

1 2

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Implementation and results

A comparison between the algorithms for the matching times as a function of the window size with max=20

Window size

AlgorithmFig. 1Fig. 2Fig. 3

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0.340.38

0.641.53

1.162.16

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0.490.67

1.143.90

1.993.78

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0.761.12

1.704.40

2.966.38

6043

1.001.54

2.375.33

4.139.34

7043

1.311.92

3.146.00

5.3711.48

8043

1.712.41

4.206.60

7.2612.69

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Implementation and results

The mathcing times as function of max . Window size = 80

maxAlgorithmFig. 1Fig. 2Fig. 3

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1.471.34

4.273.80

7.257.87

2043

1.502.41

4.306.60

7.1612.69

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1.483.49

4.3011.80

7.3021.22

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1.684.51

4.2915.84

7.3332.80

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1.675.71

4.2619.71

7.3444.90

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Summary

Alg.4 is faster than Alg.3 in most of the experiments

For now it is the fastest known sequential implementation of the minimum Hausdorff distance computation.