Tutorial 6 of CSCI2110
Bipartite Matching
Tutor: Zhou Hong (周宏)
hzhou@cse.cuhk.edu.hk
Outline
• Maximum Bipartite Matching
• B-Matching
• Tic-Tac-Toe (Optional)
• Midterm Review
Maximum Matching
The bipartite matching problem:
Find a matching with the maximum number of edges.
B
A
E
G=(A∪B,E)
In the following, we assume
|A|=|B| for simplicity.
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.
Oracle of Perfect Matching
output
input
Perfect
Perfect
No Perfect Matching
Our Goal:
find a maximum matching by querying perfect matching
oracle as few as possible
Bipartite Graph Contains Perfect Matching
If G already has a perfect matching M, then M must be a
maximum matching. We done with only one query!
A
B
However, what shall we do if G does not contain a perfect
matching?
Matching of Size n-1
To get an ideal of the reduction step, first, we assume G has a
maximum matching of size n-1.
A
B
First Attempt
• 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.
A
B
Drawback of This Approach?
We need to try different pair of vertices, in the worst
case, O(n2) queries are required.
A Better Solution
• 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
B
A matching of size n-1 in original graph guarantees a perfect
matching in new graph
A Better Solution
• 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
B
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
B-Matching
Suppose we have a bipartite graph G=(A∪B,E), for each vertex v,
there is a degree bound bv.
𝑣
A
𝑏𝑣
B
A b-matching is a subset of edges so that each vertex v∈A∪B
has degree at most bv.
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 ′
𝑏𝑢1 = 2
𝑏𝑢2 = 1
𝑏𝑣1 = 1
𝑢1
𝑣1
𝑢2
𝑣2
𝑏𝑣2 = 2
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 ′
𝑏𝑢1 = 2
𝑏𝑢2 = 1
𝑏𝑣1 = 1
𝑢1
𝑣1
𝑢2
𝑣2
𝑏𝑣2 = 2
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
3
1
2
3
4
5
6
7
8
9
4
5
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
…
…
3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Note that, player 1 can never
mark both squares in a pair.
4
…
…
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?