43_päges_maximum matching algorithm

8/3/2019 43_pges_maximum matching algorithm

1/43

Maximum Weighted

Matching Algorithm

Christina Witwer

Wien 2002

Institute for Theoretical Biochemistry

University of Vienna


2/43

Some Definitions(I)

A graph G consists of a set V of vertices (nodes) and

a set E of edges

G = (V, E)

A weighted graph has a weight wij assigned to each

edge [vi, vj] E.

24

24

32

32

12

24

0

1

2

3

4

5

12

12

Matching: A subset M E of edges is called a mat-

ching, if each vertex is contained in no more than one

edge.

24

24

32

32

12

24

0

1

2

3

4

5

12 12


3/43

Some Definitions(II)

Maximum cardinality matching: contains the maxi-

mum number of edges of a graph that form a mat-

ching.

Maximum weighted matching: The sum of the weights

of the edges forming the matching is maximal.


4/43

Motivation

View a RNA-sequence with a list of possible basepairs

as a graph:Nucleotides Vertices

Basepairs Edges

Sequence: GUCAUCGA

V = {1, 2, 3, 4, 5, 6, 7, 8}

Possible basepairs:

E = {(1, 3), (1, 5), (1, 6), (2, 4), (2, 7), (2, 8), (3, 7), (5, 7), (5, 8

G1

G7

A8 U2

A4U5

C6

C3

G1

G7

A8 U2

A4U5

C6

C3

C

A

G

UC

G AU


5/43

Some Definitions(III)

Alternating path with respect to a matching M: Or-

dered list of vertices, so that

each vertex is contained only once

the edges connecting the vertices are alternately

in M and not in M.

1

2

34

5

6


6/43

Some Definitions(IV)

Augmenting path with respect to a matching M: An

alternating path that starts and ends with an unmat-

ched edge.

Symmetric Difference: S1 S2 = (S1 S2) (S1 S2)

M = {(2, 4), (3, 5)}

P = {(1, 2), (2, 4), (4, 3), (3, 5), (5, 6)}

1

2

3

4

5

6

Augment: M = M P

1

2

3

4

5

6

M = {(1, 2), (4, 3), (5, 6)}


7/43

Berges Lemma: A matched graph (G, M) has an aug-

menting path if and only if M is not a maximum car-

dinality matching.

Sketch for an algorithm

k-Matching

search for an augmenting path

augment

(k+1)-Matching

If no more augmenting path exists, the matching is

maximum


8/43

Maximum Cardinality Matching

Search for an augmenting path

Initially all unmatched vertices of the graph are labe-led even (+), all matched vertices are unlabeled (*).

Starting with an unmatched vertex r+ a tree T is

grown.

Let v / T be adjacent to any vertex u+ T. T is

extended by taking the unmatched edge uv and the

matched edge vw to T. v is labeled odd (-) and w is

labeled even (+).

Grow step

r

v w+ * *

r

v w+ +

When an even vertex u+ / T is adjacent to any vertex

u+ T, an augmenting path p = (v , u , . . . , r) with

respect to M has been found.

r n u v

+ + +r n u v

_+ ++


9/43

Blossoms

r 1 b

6

3

4

5

7 8

+

+

+

+ +

+

Shrink step: G = (V, E)

V = (V \ B) {b}

E = (V \ B) {ub : uv (B) and u B}

r 1 b

6

3

4

5

7 8

b+

+

+ +


10/43

More Definitions

(S) = {uv E : u S and v S}


1

5

23

4

6

7

B

8

(S) = {(3, 7), (4, 6)}

(S) = {(1, 2), (2, 3), (3, 4), (4, 5), (1, 5)}


11/43

Blossoms can be nested

0 1

2

3

4 5

67

8

910

11

r

13

14

B1B2

Each vertex is a (trivial) blossom.

B1 is a subblossom of B2.

B2 is a (non-trivial) surface blossom.


12/43

Search for an augmenting path (I)

r 1 2

3 4

5 6

7 8

9 10 11 +

*

* *

* *

+ * *

*

* *

grow step

r 1 2

3 4

5 6

7 8

9 10 11

+ +

* *

* *

* *

* * +

grow steps


13/43

Search for an augmenting path (II)

r 1 2

3 4

5 6

7 8

9 10 11

++

+

+

* *

+

+

shrink step

r 1 2(b) 7 8

9 10 11

+ + +

+* *

grow steps


14/43

Search for an augmenting path (III)

r 1 2(b) 7 8

9 10 11+

+

+

++

expand step

r 1 2

3 4

5 6

7 8

9 10 11

Augmenting path!


15/43

Algorithm: Search for an Augmenting Path

let r be the only vertex of T

while no edge uv with u+ T and v+ / T exists {

if an edge uv with u+ T and v / T exists:grow step

else if an edge uv with u+ T and v+ T exists:

shrink step

else terminate,

T is abandoned since no augmenting path

for r exists}

an edge uv with u+ T and v+ / T exists

expand step augmenting path.


16/43

Algorithm: Maximum Cardinality Matching

M: Abitrary matching in G

label all free vertices even, unlabel all matched vertices

for each vertex r in G {

if r is matched continue with an other vertexsearch for an augmenting path

if an augmenting path has been found {

augment M = M P

unlabel all vertices contained in T

delete all blossoms of T

destroy T

}

else T has been abandoned

continue with another vertex

}

M is a maximum cardinality matching


17/43

Linear programming

Primal Problem:

Maximizeq

j=1

cjxj

subject toq

j=1

aijxj b(i) for all i = 1, . . . , p ,

xj 0 for all j = 1, . . . , q .

Dual Problem:

Minimizep1=1

b(i)(i)

subject topi=1

aij(i) cj for all j = 1, . . . , q ,

(i) 0 for all i = 1, . . . , p .

Complementary Slackness Conditions: A pair (x, ) of

the primal and dual feasible solutions are optimal, if

they satisfy the following conditions:

(i) b(i) q

j=1

aijxj = 0 for each i = 1, . . . , p .xj

p

i=1

aij(i) cj

= 0 for each j = 1, . . . , q .


18/43

Linear programming (Example)Maximize 200x1 + 400x2

subject to 1/40x1 + 1/60x2 11/50x1 + 1/50x2 1xi 0 i = 1, 2

x2

x1

Feasible Region


19/43

LP Formulation of MWM (I)

integer linear program:

MaximizeeEw

exe

subject to

e(u)

xe 1 for all u V

xe {0, 1} for all e E

xe = 0 if e M1 ife M

1 8r 2

3 4

5 6

7

9

3224

28

24

28

28

16

20

24

28

we = w(u, v), xe = x(u, v)

w(r, 1) = 28, w(1, 2) = 32, . . .

x(1, 2) = 1, x(3, 4) = 1, xe = 0 for the other edgeseE

wexe = 60


20/43

ad LP Formulation of MWM(II)

Weighthed graph:

1

2 3

4

5

2

3

3

2

2

Optimal Solution of the integer linear program:edge 1 2 2 3 3 4 4 5 1 5xe 1 0 0 1 0

eE

wexe = 5

1

2 3

4

5

2

3

3

2

2

Optimal Solution of the linear programing relaxation:edge 1 2 2 3 3 4 4 5 1 5xe 0.5 0.5 0.5 0.5 0.5

eE

wexe = 6


21/43

LP Formulation of MWM(II)

integer linear program:

MaximizeeE

wexe

subject to

e(u)

xe 1 for all u V

xe {0, 1} for all e E

linear programing relaxation:

MaximizeeE

wexe

subject to

e(u)

xe 1 for all u V

xe 0 for all e E

linear program:

MaximizeeE

wexe

subject to

e(u)

xe 1 for all u Ve(B)

xe |B|/2 for all B O

xe 0 for all e E

O = {B V : |B| is odd and |B| 3}


22/43

More Definitions



1

5

23

4

6

7

B

8

(S) = {(3, 7), (4, 6)}

(S) = {(1, 2), (2, 3), (3, 4), (4, 5), (1, 5)}


23/43

Primal-Dual Method for MWM

Primal:

Maximize uvE

wuvxuv

subject to

uv(u)

xuv 1 for all u Vuv(B)

xuv |B|/2 for all B O

xuv 0 for all uv E

Dual:

MinimizeuV

yu +BO

|B|/2zB

subject to yu 0 for all u V

zB 0 for all B O

yu + yv+

BOuv(B)

zB wuv for all uv E

Complementary Slackness Conditions:

xuv > 0 = uv = 0 for all uv E

yu > 0 =

uv(u)

xuv = 1 for all u V

zB > 0 = uv(B)

xuv = |B|/2 for all B O

Reduced Cost: uv = yu + yv wuv +BO

uv(B)

zB


24/43

Initial Solution

xe = 0 for each edge e E(M = )

yu = max{we/2 : e (u)} for each vertex u V

zB = 0 for each B O

Feasible Solution of the Primal:

(P1)

uv(u)

xuv 1 for all u V

(P2) uv(B)

xuv |B|/2 for all B O

(P3) xuv 0 for all uv E

Feasible Solution of the Dual:

(D1) yu 0 for all u V

(D2) zB 0 for all B O

(D3) yu + yv+BO

uv(B)

zB wuv for all uv E

Complementary Slackness Conditions:

(CS1) xuv > 0 = uv = 0 for all uv E

(CS2) yu > 0 =

uv(u)xuv = 1 for all u V

(CS3) zB > 0 =

uv(B)



25/43

Reducing Violations of CS 2

CS 2: yu > 0 = uv(u)

xuv = 1 for all u V

Let r be a vertex that violates CS 2

(i.e. r is unmatched and yr > 0.)

match r

or

adjust the dual solution (y, z) such that y = 0


26/43

Matching a free vertex r

Search for an augmenting path, but only use tight

edges (where uv = yu + yv wuv +

BOuv(B)

zB = 0).

Otherwise CS 1 ( xuv > 0 = uv = 0 ) would be

violated.

case 1: found an augmenting path

augment

case 2: T is abandoned (no further thight edges are

incident to any vertex u+ T)

dual adjustment


27/43

Requirements to the Dual Adjustment

The dual solution (y, z) gets adjusted to (y, z) such

that

R1: the objective value (uV

yu +BO

|B|/2zB) of the

dual problem strictly decreases

R2: the solutions are still feasible to the Primal

R3: the solutions are still feasible to the Dual

R4: (CS1) and (CS3) remain true for (y, z)

R5: yr strictly decreases


28/43

Performing a Dual Adjustment

> 0

yv = yv for all v+ T,

yv = yv + for all v T,

yv = yv for all v{|+} / T,

zB = zB + 2 for all B+ T,

zB = zB 2 for all B T,

zB = zB for all B{|+} / T,

zB is only adjusted when B is a non-trivial surface

blossom.


29/43

Satisfaction of the Requirements (I)

R1: the objective value (f = uV yu+ BO|B|/2zB) ofthe dual problem strictly decreases

f = f f

v+ T : fv+ =

v T : fv =

B+ T :

fB+ = |B|()+|B|/2(2) = |B|+(|B|1) =

B T :

fB = |B| + |B|/2(2) = |B| (|B| 1) =

n+ = n + 1

f = n+() + n() =

.


30/43

Satisfaction of the Requirements (II)

R2: the solutions are still feasible to the Primal

x is not altered

R3: the solutions are still feasible to the Dual

restrictions on the value of :

(D1) : yu 0 yu for all u+ T

(D2) : zB 0 zB/2 for all B T

.


31/43

Satisfaction of the Requirements (III)

(D3) : uv = yu + yv wuv +

BOuv(B)

zB 0

e = uv : u T or v T

Case 1: uv = yu + yv wuv 0 e / (B)

Case 1a: u+ T and v T:

yu = yu , yv = yv +

uv = uv

Case 1b: u+ T and v+ T:

yu = yu , yv = yv

uv = uv 2 uv/2

Case 1c: u+ T and v{|+} / T:

yu = yu , y

v = yv

uv = uv uv

Case 1d: u T and v T:

yu = yu + , yv = yv +

uv = uv + 2

Case 1e: u T and v{|+} / T:

yu = yu + , yv = yv

uv = uv +


32/43

Satisfaction of the Requirements (IV)

(D3) : uv = yu + yv wuv +

BOuv(B)

zB 0

e = uv : u T or v T

Case 2: e = uv (B)

Case 2a: B+ T :

yu = yu , yv = yv , z

B = zB + 2

uv = uv

Case 2b: B T :

yu = yu + , yv = yv + , z

B = zB 2

uv = uv

.


33/43

Satisfaction of the Requirements (V)

R4: (CS1) and (CS3) remain true for (y, z)

(CS1): xuv > 0 uv = 0

uv = uv

(CS3) : zB > 0 uv(B)


R5: yr strictly decreases

r+ T

.


34/43

Satisfaction of the Requirements (Summary)

= min{1, 2, 3, 4, }, where

1 = minuV {yu : u+

T}2 = min

uvE{uv : u+ T, v{|+} / T}

3 = minuvE

{uv/2 : u+ T, v+ T}

4 = minBO

{zB/2 : B T}

Case = 1 : yu = 0, u+ T

an even length alternating path p from u to r exists:

match r by replacing M by M p

Case = 2 : the responsible edge e = uv becomes

tight and can be used to extend T

Case = 3 : the responsible edge e = uv becomes

tight and can be used to shrink a new blossom

Case = 4 : zB of B will drop to zero, B cannot

participate in another dual adjustment expand B


35/43

Expand step for an odd Blossom

B

Bj

+

+

2

3 4

6

ru

Bj

+

+

+

* *

2

3 4

6

r

u


36/43

MWM: Algorithm

Initial Solution:

let M be the empty matching

yu = max{we/2 : e E} for each vertex u G

label each vertex u g evenfor each vertex r G {

if r is matched or yr = 0 continue

let Br be the only blossom of T

repeat {

if an vertex u+ T with yu = 0 exists {let P be the alternating path from u to r

replace M by m Pelse if an edge uv with u+ T and uv = 0

exists {

case v / T: grow stepcase v+ T: shrink stepcase v+ / T: augment step

else if there exists an odd blossom B T

with zB = 0

expand step for B

else {

determine

perform dual adjustment}

} until r is matched or yr = 0

}


37/43

Example (I)

Weighthed Graph:

1 8r 2

3 4

5 6

7

9

3224

28

24

28

28

16

20

24

28

node u 1 2 3 4 5 6 7 8 9 ryu 16 16 14 14 14 14 10 10 12 14

After 4 searches:

* *

* *

* *

* *

+

1 8r 2

3 4

5 6

7

9

3224

2428

16

24

28

28

20

28

+


38/43

Example (II)

node u 1 2 3 4 5 6 7 8 9 ryu 16 16 14 14 14 14 10 10 12 14

* *

* *

* *

* *

+

1 8r 2

3 4

5 6

7

9

3224

24

28

16

24

28

28

20

28

+

starting the search with r+ T

(r+, 1) = 2

no tight edges are incident to r+ dual adjustment

1 = 14, 2 = 2, 3 = , 4 =

= 2


39/43

Example(III)

yv = yv for all v+ T

T = {r+

} = 2

node u 1 2 3 4 5 6 7 8 9 ryu 16 16 14 14 14 14 10 10 12 14yu 16 16 14 14 14 14 10 10 12 12

edge (r, 1) becomes tight grow step

*

* *

*

*

*

+

1 8r 2

3 4

5 6

7

9

32

24

24

28

16

24

28

28

20

28

+ +

(2+, 3) = 6, (2+, 5) = 6

no tight edges are incident to 2+

dual adjustment1 = 12, 2 = 6, 3 = , 4 =

= 6


40/43

Example (IV)

yv = yv for all v+ Tyv = yv + for all v

T

T = {r+, 1, 2+}

= 6

node u 1 2 3 4 5 6 7 8 9 ryu 16 16 14 14 14 14 10 10 12 12yu 22 10 14 14 14 14 10 10 12 6

edge (2+, 3) becomes tight grow step

edge (2+, 5) becomes tight grow step

* *

+

1 8r 2

3 4

5 6

7

9

32

24

24

28

16

24

28

28

20

28

+ +

+

+


41/43

Example (V)

node u 1 2 3 4 5 6 7 8 9 ryu 22 10 14 14 14 14 10 10 12 6

edge (4+, 6+) is thight shrink step

+

* *

B(2)+

9

1 8r 2

3 4

5 6

732

24

2428

16

28

28

20

28

+ +

+

+ +

24

+

(5+, 9+) = 2, (6+, 7) = 8

no tight edges are incident to (B(2))+ dual adjustment

1 = 6, 2 = 2, 3 = , 4 =

= 2


42/43

Example (VI)

yv = yv for all v+ T

yv = yv + for all v T

= 2

node u 1 2 3 4 5 6 7 8 9 ryu 22 10 14 14 14 14 10 10 12 6yu 24 8 12 12 12 12 10 10 12 4

zB = zB + 2 for all B+ T

zB(B(2)) = 4

edge (5+, 9+) is thight, 9+ / T augment step

* *

B(2)+

+

1 8r 2

3 4

5 6

732

24

24 28

16

28

28

20

28

+ +

+

+ +

24

+

9

expand the blossom and find the even length alterna-

ting path from vertex 5 to r:


43/43

Example (VII)

1 8r 2

3 4

5 6

7

9

3224

2428

16

24

28

28

20

28

uvE

wuvxuv = 108

augment

1 8r 2

3 4

5 6

7

9

3224

2428

16

24

28

28

20

28

uvEwuvxuv = 124

43_päges_maximum matching algorithm

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