Tutorial 6 of CSCI2110Bipartite Matching

Tutor: Zhou Hong (周宏 )hzhou@cse.cuhk.edu.hk

Outline

• Maximum Bipartite Matching

• B-Matching

• Tic-Tac-Toe (Optional)

• Midterm Review

The bipartite matching problem:Find a matching with the maximum number of edges.

A perfect matching is a matching in which every vertex is matched (i.e. of degree 1).

Reduce to Perfect Matching:• Once you know how to solve perfect matching, you can

also do maximum matching.

Maximum Matching

G=(AB,E)A B

EIn the following, we assume |A|=|B| for simplicity.

Perfect

Matching Oracle

Oracle of Perfect Matching

Perfect

Matching Oracle

input

output

No Perfect Matching

Our Goal: find a maximum matching by querying perfect matching oracle as few as possible

Bipartite Graph Contains Perfect Matching

A B

If G already has a perfect matching M, then M must be a maximum matching. We done with only one query!

However, what shall we do if G does not contain a perfect matching?

Matching of Size n-1

A B

To get an ideal of the reduction step, first, we assume G has a maximum matching of size n-1.

First Attempt

A B

• By adding an appropriate edge, G will contain a perfect matching. We can find it by one query.

• Then, remove new added edge from the perfect matching, we done.

Drawback of This Approach?We need to try different pair of vertices, in the worst case, O(n2) queries are required.

A Better Solution

A B

• Add two dummy vertices connecting to all vertices on the opposite.

• In the new graph, we find a perfect matching by one query.

• Remove the new added edges, we have a maximum matching of original graph.

A matching of size n-1 in original graph guarantees a perfect matching in new graph

A Better Solution

A B

• Add two dummy vertices connecting to all vertices on the opposite.

• In the new graph, we find a perfect matching by one query.

• Remove the new added edges, we have a maximum matching of original graph.

On the other hand, a perfect matching in new graph will guarantee a matching of size n-1 in original graph

Complete Reduction

No perfect matching in the new graph implies no matching of size n-1 in the original graph.

What shall we do?Repeat the previous procedure until we find a perfect matching• In each iteration

– We add two dummy vertices connecting with all vertices on the opposite.

– Make a query with the updated graph.

How many queries do we need?

At most n

Using binary search, we can improve to log(n) queries!

Residence Assignment

Assignment Requirements:

• Each student can be assigned to at most one room

• Each room can accommodate a specific number of students

Our Goal:Maximize the number of residents

Basic Idea

• For a shared room with b beds, treat it as b single rooms

• In the new setting– Applying for a shared

room becomes applying for all b duplicate single rooms

– Then, we only need to find a maximum matching

A b-matching is a subset of edges so that each vertex vAB has degree at most bv.

A B

B-Matching

Suppose we have a bipartite graph G=(AB,E), for each vertex v, there is a degree bound bv.

𝑏𝑣𝑣

Matching is a special case of b-matching, with bv=1 for all v.

Reduction to Maximum Matching• For each vertex v with degree bound bv, make bv copies of

vertex v. • Connect the duplicate vertices according to the original graph.

𝑢1 ′

𝑣2 ′

𝑏𝑢1=2

𝑏𝑢2=1

𝑏𝑣1=1

𝑏𝑣2=2

𝑢1

𝑢2

𝑣1

𝑣2

Reduction to Maximum Matching• For each vertex v with degree bound bv, make bv copies of

vertex v. • Connect the duplicate vertices according to the original graph.• Find a maximum matching in the new graph.• Map the resulting matching back to the original graph.𝑢1 ′

𝑣2 ′

𝑏𝑢1=2

𝑏𝑢2=1

𝑏𝑣1=1

𝑏𝑣2=2

𝑢1

𝑢2

𝑣1

𝑣2

Tic-Tac-Toe

• A paper-and-pencil game.– Two players (X and O)– They take turns marking spaces in a 3x3 grid

• Player who succeeds in placing three respective marks in a horizontal, vertical, or diagonal row wins the game

• Best play from both parties leads to a draw

A game won by the first player X

Winning Set of Tic-Tac-Toe

Therefore, there are totally 8 winning sets

Bipartite Graph

• Now, let’s construct a bipartite graph

– On one side, the vertices corresponding to the squares in the grid (totally 9)

– On the other side, the vertices corresponding to the winning sets (totally 8)

– An edge between a square and a winning set indicate that the square is in the winning set

1 2 34 5 67 8 9

1

5

2

3

4

6

7

8

9

Generalized Tic-Tac-Toe

• Generalize to nxn grid

– There are n2 squares

– Winning sets are still the whole horizontal, vertical or diagonal rows, totally 2n+2

B-Matching

• Consider a b-matching in the bipartite graph between squares and winning sets– Degree bound for square is 1– Degree bound for winning set is 2

• We claim that– if there exists a b-matching such that each winning set

incidents with exactly 2 edges– then player 2 has a trivial strategy to avoid losing

Example of 5x5 Grid• In case of 5x5 grid Tic-Tac-Toe, there exists

a b-matching as required– (not shown)

• Player 2 has a trivial strategy to force a draw– Pair squares in the grid according to the b-

matching– If player 1 marks a square in a pair, then marks

the other square in the pair, otherwise marks randomly

1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 25

1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 25

1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 25

3

4

……

……

Note that, player 1 can never mark both squares in a pair.

Existence of The Trivial Strategy V.S. Hall’s Theorem

• For each winning set, make a copy of it in the bipartite graph

• Then, the desired b-matching is corresponding to a matching saturating both original and duplicate winning sets

Generalized Hall’s Theorem: A bipartite graph G=(V,W;E) has a matching saturating W if and only if |N(S)| >= |S| for every subset S of W.

Player 2 does not always have such a trivial strategy to avoid losing, consider the standard (3x3) Tic-Tac-Toe!

Thank You!

Q & A ?