Section 2.1 Matching in bipartite graphs in Graph Theory presentation

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Section 2.1 Matching in bipartite graphs in Graph Theory

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Exercise 1. Suppose that M is a matching not maximal. ... Choose one end a or b from each edge ab M as follows: ... b' U since ab' M. Since b' is selected, ... –

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Title: Section 2.1 Matching in bipartite graphs in Graph Theory

1
Section 2.1 Matching in bipartite graphsin
Graph Theory

Handout for reading seminar

2
Matching

Task Find independent edges as many as possible.
Independent edges edges not sharing any vertex.
Q Are there independent edges containing all
vertices?
Matching M ? E of U given G(V,E) and U ? V
Every vertex in U is incident with an edge in M.
A set M ? E of independent edges in graph G.

v1
v2
v3
v4
Vertices in V are matched by M2.
Vertices in U are matched by M1.
G
unmatched
Uv1, v2, v3, v4
3
k-factor

k-factor a k-regular spanning subgraph
k-regular
graph any of whose vertex has the same degree k
spanning subgraph
one having all vertices of the original graph
The followings are equivalent.
A subgraph H ? G(V,E) is a 1-factor of G.
E(H) (a set of edges in H) is a matching of V.

G
2-regular
M2 (matched)
unmatched
4
Notation in this section

G (V, E) bipartite graph with bipartition A,
B
VA ? B, A ? B ?.
a, a, represent vertices in A. No edge between
a, a.
b, b, represent vertices in B. No edge between
b, b.

b
ab
a
A
B
A
B
A
B
Examples of matchings
G
5
Matching and augmenting paths (1/3)

How can we find a (large) matching?
By expanding a small one with augmenting paths.
Augmenting path
Starts an unmatched vertex in A, and ends at an
unmatched one in B.
Vertices except 1st and last in the path are
attached to edges in M.
In finding an augmenting path, the followings are
repeated
Move from A to B through an edge in E M.
Move from B to A through an edge in M.

A
B
A
B
Augmenting path P
Matching M
6
Matching and augmenting paths (2/3)

How can we find a (large) matching?
By expanding a small one with augmenting paths.
Suppose M and an augmenting path P are given.
(E(P) ? M) ?(M ? E(P)) is also a matching.
Each vertex on P is incident with just one edge
in E(P) ? M. Then E(P) ? M is a matching.
M ? E(P) is obviously a matching.
These sets do not share any vertex, so the union
is also a matching.

A
B
A
B
A
B
Matching M
Augmenting path P
Matching M
7
Matching and augmenting paths (3/3)

How can we find a matching?
By expanding a small one with augmenting paths.
Two new vertices (the first and the last vertices
in the path) are included in the new matching
with one expansion.
Alternating path
Augmenting one w/o the condition on the end
vertex.
The end is not necessarily an unmatched vertex in
B.

A
B
A
B
A
B
Matching M
Augmenting path P
Matching M
8
Exercise 1

Suppose that M is a matching not maximal.
M has less edges than some other matching M in
G.
Then G contains an augmenting path w.r.t. M.
Proof
A edge e in M M is an augmenting path.
What happens when M is not maximum but maximal?
(HW)

A
B
A
B
A
B
A
B
G
Matching M
Aug. path w.r.t. M
New matching M
9
Vertex cover

A set U ? E is a vertex cover of E if every edge
of G is incident with a vertex in U.

not a v. cover
v. cover
A
B
A
B
A
B
A
B
v. cover
not a v. cover
G
G
10
Theorem 2.1.1 (König 1931) (1/6)

Let M and U be a maximum matching and a minimum
vertex cover in G(A ? B, E).
Then, M U.
Its proof consists of three parts.
Show that M ? U, where U is any vertex
cover.
Construct U from M, where UM.
If U is a vertex cover, it is the minimum since
M ? U.
Show that U is a (minimum) vertex cover.

11
Theorem 2.1.1 (König 1931) (2/6)

Proof for M ? U (showing M?T?U)
www.cs.mcgill.ca/kaleigh/teaching/360_tutorials/t
utorial_07.html
All the endpoints of the edges in M are distinct.
-- (1)
By definition of matching.
A vertex cover T of M requires at least one end
for each edge in M. -- (2)
T then has to have at least M vertices
(M?T).
G contains at least the same edges of M.
To cover edges in M, U (vc of G) must be at
least T.

T
U
(1)
(2)
G
12
Theorem 2.1.1 (König 1931) (3/6)

Construction of U from M
We will construct a vertex cover U from M as
follows
Choose one end a or b from each edge ab ? M as
follows
Choose b in B if some alternating path ends in
the vertex.
Choose a in A otherwise.

Alternating paths
Vertex cover
A
B
A
B
A
B
A
B
13
Theorem 2.1.1 (König 1931) (4/6)

Proof that U is a vertex cover
For any edge ab?E(G), show a?U or b?U (that is,
ab is covered by U) ().

a
b
b
ab ? M for some b
Any matching incident with a or b ?
ab ? M
ab ? M for some a
a
ab ? M and ab ? M for any a and b
a
ab ? E
ab ? M?
b
a
b
ab?M is a matching. This contradicts with
that M is maximum.
ab ? M
By the construction of U, a?U or b?U ().
14
Theorem 2.1.1 (König 1931) (5/6)

Proof that U is a vertex cover
For any edge ab?E(G), show a?U or b?U (that is,
ab is covered by U) ().

ai
bj
ak
b
P
b
b?U since ab?M. Since b is selected, there
must be an alt. path P ending at b by
definition of U.
a
b
a?U
ab?M for some b
impossible
a?U?
ab ? E
Any matching incident with a or b ?
()
a?U
ab is an alt. path. Since ab?M, b?U by def. of
U. ()
a
b
a
ab ? M for some a
15
Theorem 2.1.1 (König 1931) (6/6)

Proof that U is a vertex cover
For any edge ab?E(G), show a?U or b?U (that is,
ab is covered by U) ().

b
PPb
(if b?P)
ai
bj
ak
a
b
b
b
P
ai
bj
ak
ai
bj
ak
a
b
b
b
P
(if b?P)
PPbab
b?U since ab?M. b is selected, then there must
be an alt. path P ending at b by definition of
U.
From P, we can create an alt. path P ending
at b.
P is an aug. path ending at b.
A matching larger than M can be obtained with
P.
a?U
M is not maximum. Then, a must be in U.
ab ? E
a?U?
16
Conditions for the existence of 1-factor

Consider a matching containing all vertices in A.
Necessary condition
N(S)?S for all S ? A (the marriage condition
MC), where
S any subset of A
N(S) neighbors of vertices in S
The condition is in fact a sufficient condition.

Matching of A unavailable
S
N(S) 3 2 2
S 3 2 1
N(S) 3 1 1
S 3 2 1
a, a, a
a, a, a
a, a
a, a
a
a
17
Theorem 2.1.2 (Hall 1935) (1/5)

G contains a matching M of A iff N(S)?S for
all S ? A.
(?)
a ? A has at least one neighbor, i.e.,
N(a)?a1.
For any a?A, there is an edge ab in M for some
b?B.
Neighbors of a and a in M can be distinct if a ?
a.
We can choose the other ends of a, a in M as the
neighbors.
For any S, at least S distinct neighbors can be
found.
That is, N(S)?S for all S ? A.

18
Theorem 2.1.2 (Hall 1935) (2/5)

G contains a matching iff N(S)?S for all S ?
A.
We follow here the third proof in the textbook.
(?)
Consider a spanning subgraph H of G that
satisfies the marriage condition (MC), and
is edge-minimal with MC.
Note dH(a)NH(a)?a1for every a?A by MC.
We show dH(a) 1 for every a?A.
Then, E(H) forms a matching of A.
MC guarantees that no two such edges can share a
vertex in B.

a
dH(a)1
a1,a22, N(a1,a2)1.
dH(a)1
a
19
Theorem 2.1.2 (Hall 1935) (3/5)

G contains a matching iff N(S)?S for all S ?
A.
(?)
Proof by contradiction on H
Suppose dH(a) ?2, and the neighbors are b1 and
b2.
Since the edge of H is minimal for MC, H ab1
and H ab2 should violate MC for some A1, A2 ?
A.
Bi NH abi(Ai).

b1
b1
A1
A1 NH (A1)
A2 NH (A2)
b2
b2
a
a
A1gt NH ab1(A1) B1
A2gt NH ab2(A2) B2
A2
B1
B2
20
Theorem 2.1.2 (Hall 1935) (4/5)

G contains a matching iff N(S)?S for all S ?
A.
(?)
NH(A1?A2 a) ? NH-ab1-ab2(A1?A2)
? NH-ab1-ab2(A1 )
?NH-ab1-ab2(A2)
B1 ? B2
B1 B2 B1
? B2

b1
b1
A1
b2
b2
B1?B2
B1? B2
B1
B2
a
a

A2
NH(A1?A2 a)
B1?B2
21
Theorem 2.1.2 (Hall 1935) (5/5)

G contains a matching iff N(S)?S for all S ?
A.
(?)
NH( A1?A2 a) ?
? B1 B2 B1
? B2
? A1 1 A2
1 A1 ? A2
By the definition of Ai and Bi (Ai gt Bi ?
Ai 1 ? Bi)
A1 A2 A1
? A2 2
A1?A2 2
A1?A2 a 1.
Hence H violates the marriage condition, contrary
to the assumption.
Therefore, dH(a) 1 for every a?A.

22
Corollary 2.1.3

If G is k-regular with k? 1, then G has a
1-factor.
k-factor a k-regular spanning subgraph
k-regular
graph any of whose vertex has the same degree k
spanning subgraph
one having all vertices of the original graph
A 1-factor is also called a perfect matching.

2-regular
3-regular
Its 1-factor
Its 1-factor
23
Proof of Corollary 2.1.3 (1/2)

AB.
Since G is k-regular, dA?v?Ad(v)kA and
dB?v?Bd(v)kB.
Since G is bipartite, each edge connects a vertex
in A to a vertex in B. Then, dA dB, that is,
kAkB.
We will see that N(S)?S for all S ? A.
By Theorem 2.1.2, G then has a 1-factor (perfect
matching).

dA 34
dB 34
24
Proof of Corollary 2.1.3 (2/2)

We will see that N(S)?S for all S ? A.
Every set S ? A is joined to N(S) by kS edges.
These kS edges ( ) are among the kN(S)
edges ( ) of G incident with N(S).
Then, kS ? kN(S) for any S, which satisfies
the condition of Theorem 2.1.2.
Since AB, every vertex is included in the
matching.

kN(S) edges
kS edges
S
N(S)
S
N(N(S))
N(S)
N(S)
N(N(S))
25
Stable matching (1/2)

Suppose a graph G and its preference are given.
For example, the following matching is unstable.
Because choosing pair (A, M) as a matching makes
both A and M happier.
A and M might be going to break the current
matching (engagement) for their preference.
A matching is called stable if it has no such
pair.

ranks of preference
?
?
Alice (A)
Kathy (K)
A K M N B L K C L M D M N L
K B A L D B C M A C D N D A
Bill (B)
Lucy (L)
Charles (C)
Mary (M)
David (D)
Nancy (N)
This example was taken from a transcript of a
lecture http//www.misojiro.t.u-tokyo.ac.jp/iwat
a/dmi/dmi03j.pdf
26
Stable matching (2/2)

A stable matching M is defined as follows
For any e (a, b) in E M, there exists some f ?
M s.t
f ?a e or f ?b e,
where f ?a e denotes the order of preference for
edges of a.
In the example below, (A, K) ?A (A, M) ?A (A, N).

For every set of preferences, G has a stable
matching.
Its proof is based on an algorithm called the
Gale-Shapley algorithm to find a stable matching
for G.
The algorithm finds a male-optimal matching.
By interchanging a set of males and a set of
females, it is also able to find a female-optimal
matching.
Note that a stable matching is not unique.
There might be another stable matching having
them happier.
It depends on the definition of happiness.

28
Gale-Shapley algorithm

Algorithm GS(G) G(A,B, E)
MuA
while Mu??
for each a?Mu a is an unengaged man
Mu proposal(a, b) s.t. b?B is highest-ranked
not yet proposed to for a)
end-for
end-while
function proposal(a, b)
if b is engaged with someone, say, a
if a ltb a then b is engaged with a MuMu ?
a - a
else b keeps the engagement with a
else b is engaged with a MuMu - a
return Mu

29
Example of results by the G-S algorithm
?
?
A K M N B L K C L M D M N L
K B A L D B C M A C D N D A
Each of A, B, C, and D proposed to 1st-ranked
female. C failed to find his fiance.
?
?
C proposed to the highest-ranked female not yet
proposed to. M changed her fiance, and D then
lost his fiance.
A K M N B L K C L M D M N L
K B A L D B C M A C D N D A
D proposed to the highest- ranked female not yet
proposed to. Then, each of them has found
her/his fiance.
30
Proof for Theorem 2.1.4 (1/2)

We will see G-S finds a stable matching by
following the transcript.
It terminates in finite steps because the number
of candidates for proposal is monotonically
decreased in the while loop of the G-S algorithm.
The matching the G-S algorithm finds is stable.
Suppose that there is an unstable pair in the
result of G-S.
The pair (a, b), where (a, b) ?a (a, b), (a,
b) ?b (a, b).
In the matching by G-S, (a, b) and (a, b) are
selected.

a
b
?
?
a b b a b
b a b a a
a
b
31
Proof for Theorem 2.1.4 (2/2)

Suppose that there is an unstable pair in the
result of G-S.
The pair (a, b), where (a, b) ?A (a, b), (a,
b) ?b (a, b).
In the matching by G-S, (a, b) and (a, b) were
selected.
By the construction of G-S, it must be that the
proposed to b by a must be taken place before
the proposal to b by a.
This case can be occurred only when (1) or (2).
(1) b refused the proposal from a.
(1) never happens because (a, b) ?b (a, b),
and then b also should refuse the proposal from
a.
(2) b accepted the proposal from a by breaking
the relation with a.
(2) never happens because b should engage with
a, not a when a proposed to b.
From these observations, any matching by G-S is
always stable.

?
?
a b b a b
b a b a a
32
Corollary 2.1.5

Every regular graph G of positive even degree 2k
has a 2-factor.
Proof consists of three steps.
G contains an Euler tour.
Every vertex in G is replaced by a pair (v, v-).
The resulting graph is k-regular bipartite graph.
By Corollary 2.1.3, a k-regular bipartite graph
has a 1-factor.
Every pair is collapsed into a single vertex.
The 1-factor graph is turned into a 2-factor
graph.

33
Proof of Corollary 2.1.5 (1/3)

G contains an Euler tour by Theorem 1.8.1.
Theorem 1.8.1 A connected graph is Eulerian iff
every vertex has even degree.
If G is not a connected graph, we consider each
connected subgraph for the corollary.

Former and latter halves of an Euler tour
4-regular
34
Proof of Corollary 2.1.5 (2/3)