04/12/2005Tucker, Sec. 4.41 Applied Combinatorics, 4 th Ed. Alan Tucker Section 4.4 Algorithmic Matching Prepared by Joshua Schoenly and Kathleen McNamara.

04/12/2005 Tucker, Sec. 4.4 1

Applied Combinatorics, 4th Ed.Alan Tucker

Section 4.4

Algorithmic Matching

Prepared by Joshua Schoenly and Kathleen McNamara

04/12/2005 Tucker, Sec. 4.4 2

Some Definitions

X-Matching - All vertices in X are used

Maximal Matching - The largest possible number of independent edges

Note: an X-matching is necessarily maximal.

Matching - a set of independent edges

X Y

Note: independent edges refer to edges that do not share a common vertex

04/12/2005 Tucker, Sec. 4.4 3

More Definitions

Edge Cover

A set of vertices so that every edge is incident to

at least one of them

R(A)

Range of a set A of vertices

R(A) is the set of vertices adjacent to at least one vertex in A

A = red, R(A) = green

04/12/2005 Tucker, Sec. 4.4 4

Matching Network

za

All edges infinite capacity

All edges capacity 1

By converting the bipartite graph to a network, we can use network flow techniques to find matchings.

04/12/2005 Tucker, Sec. 4.4 5

Matching Network

Matching a – z flow

za

X – matching saturation at a

za

04/12/2005 Tucker, Sec. 4.4 6

Matching Network

Maximal matching Maximal flow

za

04/12/2005 Tucker, Sec. 4.4 7

Matching Network

Edge cover Finite a – z cut

za

A = red on left, B = red on right

S = red vertices, an edge cover

P = a, black on left, red on right

P = z, red on left, black on right

Edge cover means that all edges have at least one red endpoint:

P to P, to , to P.

Thus only uncovered edges, would go from P to .

P P P

P

04/12/2005 Tucker, Sec. 4.4 8

Matching Network

Not an edge cover Infinite a – z cut

za

A = red on left, B = red on right

S = red vertices, NOT an edge cover

P = a, black on left, red on right

P = z, red on left, black on right

An infinite cut would go through an edge with two black endpoints, which corresponds to an uncovered edge.

04/12/2005 Tucker, Sec. 4.4 9

Theorem 1Recall Corollary 2a from Section 4.3:

The size of a maximal flow is equal to the capacity of a minimal cut.

Since….

Then,

The size of a maximal matching is equal

to the size of a minimal edge cover.

Edge cover Finite a – z cut

Matching a – z flow

and

04/12/2005 Tucker, Sec. 4.4 10

Finding Matchings

za

Take any matching, convert to a network, and then use the augmenting flow algorithm to find a maximal flow, hence a

maximal matching.

04/12/2005 Tucker, Sec. 4.4 11

An easy way to think about it

1. Start with any matching.

2. From an unmatched vertex in X, alternate between non-matching and matching edges until you hit an unmatched vertex in Y.

3. Then switch between the non-matching and matching edges along to path to pick up one more matching edge.

4. Continue this until no unmatched vertex in X leads to an unmatched vertex in Y.

04/12/2005 Tucker, Sec. 4.4 12

Example

Start with any matching

Find an alternating path

Start at an unmatched vertex

in X

End at an unmatched vertex

in Y

Switch matching to non-matching and vice versa

A maximal matching!

Note: if this does not result in a maximal matching, start from step one and do it again.

04/12/2005 Tucker, Sec. 4.4 13

Theorem 2Hall’s Marriage Theorem

A bipartite graph has an X-matching if and only if for every subset A of X, the number of vertices in R(A) is greater

than or equal to the number of vertices in A.

X – matching

True for all A in X

A is the set of red vertices

R(A) is the set of blue vertices

04/12/2005 Tucker, Sec. 4.4 14

Proof ofHall’s Marriage Theorem

“ ”: An X-matching implies that for every subset A of X, |R(A)| is greater than or equal to |A|.

X – matching For any A, the X – matching gives

at least one vertex in R(A) for every vertex in A.

04/12/2005 Tucker, Sec. 4.4 15

Proof ofHall’s Marriage Theorem

“ ”: If |R(A)| is greater than or equal to |A| for every subset A of X, then there is an X-matching.

• First note that if M is a maximal matching, then .

• Taking A = X, we have that since the range of X is contained in Y. Thus .

• Also, by Theorem 1, if S is a minimal edge cover, then .

• Note that if , then M must be an X – matching.

• Thus, it suffices to show that for all edge covers S.

min X , YM

M S

X X YR XM

XM

X S

04/12/2005 Tucker, Sec. 4.4 16

Since , .

Now let A be the vertices in X but not in S, so that .

Thus, .

Now, S is an edge-cover, so if it doesn’t contain on the X side, it must contain all the vertices a goes to on the Y side in order to cover the edges in between. Thus , so .

Thus,

But , hence ,

so as needed. QED

Proof of Hall’s Marriage Theorem“ ”: Continued… we need to show that for all

edge covers S. X S

S S X S Y S S X S Y

S X X A

S S X S Y X A S Y

a A

R A S Y R A S Y

S X A S Y X A R A

0A R A A R A

S X A R A X

A

S

R(A)

04/12/2005 Tucker, Sec. 4.4 17

Class ProblemProblem #7

In the following table of remaining games, is it possible for the Bears to be champions (or co-champions) if they win all remaining games? Build the appropriate network model.

Team Wins to Date

Games to Play

With Bears

With Lions

With Tigers

With Vampires

Bears 26 14 -- 4 4 6

Lions 34 5 4 -- 0 1

Tigers 32 8 4 0 -- 4

Vampires 29 11 6 1 4 --

04/12/2005 Tucker, Sec. 4.4 18

Class ProblemProblem #7

In the following table of remaining games, is it possible for the Bears to be champions (or co-champions) if they win all remaining games? Build the appropriate network model.

Team Wins to Date

Games to Play

With Bears

With Lions

With Tigers

With Vampires

Bears 26 14 -- 4 4 6

Lions 34 1 4 -- 0 1

Tigers 32 4 4 0 -- 4

Vampires 29 5 6 1 4 --

04/12/2005 Tucker, Sec. 4.4 19

Solution to Class Problem

6,1

8,0

11,4

∞,0

1,1

4,4

0,0∞,1 ∞,0

∞,0 ∞,0

∞,4

a z

TV

LV

LTL

T

V

04/12/2005 Tucker, Sec. 4.4 20

Solution to Class Problem

6,1

8,0

11,4

∞,0

1,1

4,4

0,0∞,1 ∞,0

∞,0

∞,0

∞,4

a z

TV

LV

LTL

T

V

Yay! There is still hope for the Bears!