graph3 - SJTUbasics.sjtu.edu.cn/~chen/teaching/GR07/graph3.pdf · 2007-09-29 · GRAPH THEORY(III) Page 3 Two edges e = f are adjacentif they have an end in common. Pairwise non-adjacentvertices

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GRAPH THEORY

Yijia ChenShanghai Jiaotong University

2007/2008

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GRAPH THEORY (III) Page 2

Chapter 2. Matching Covering and Packing

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Two edges e �= f are adjacent if they have an end in common.

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Pairwise non-adjacent vertices or edges are called independent.

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Pairwise non-adjacent vertices or edges are called independent. A set of edges or of edges

is independent (or stable) if no two of its elements are adjacent.

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A set M of independent edges in a graph G = (V, E) is called a matching.

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A set M of independent edges in a graph G = (V, E) is called a matching. M is a

matching of U ⊆ V if every vertex in U is incident with an edge in M .

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matching of U ⊆ V if every vertex in U is incident with an edge in M . The vertices in U

are then called matched (by M ); vertices not incident with any edge of M are unmatched.

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A k-regular spanning subgraph is called a k-factor.

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A k-regular spanning subgraph is called a k-factor.

Thus, a subgraph H ⊆ G is a 1-factor of G if and only if E(H) is a matching of V .

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2.1 Matching in bipartite graphs

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We fix a bipartite graph G = (V, E) with bipartition {A, B}.

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Let M be an arbitrary matching in G. A path in G which starts in A at an unmatched vertex

and then contains, alternately, edges from E \ M and from M , is an alternating path with

respect to M .

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respect to M .

An alternating path that ends in an unmatched vertex of B is called an augmenting path.

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respect to M .

An alternating path that ends in an unmatched vertex of B is called an augmenting path.

An augmenting path leads to a larger matching.

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

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Theorem.[Konig 1931] The maximum cardinality of a matching in G is equal to the

minimum cardinality of a vertex cover of its edges.

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Proof.

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Proof. Let M be a matching of maximum cardinality.

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Define a set U in the following way:

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Define a set U in the following way: For every ab ∈ M , if there is an alternating path ending

in b, then b ∈ U ; otherwise a ∈ U .

�

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Theorem.[Hall 1935] G contains a matching of A if and only if |N(S)| ≥ |S| for all

S ⊆ A.

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First Proof.

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First Proof.

Let M be a matching that leaves a vertex in A unmatched.

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First Proof.

Let M be a matching that leaves a vertex in A unmatched. We will construct a augmenting

path with respect to M .

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First Proof.



Let a0, b1, a1, b2, a2, . . . be a maximal sequence of distinct vertices ai ∈ A and bi ∈ B

satisfying the following conditions for all i ≥ 1:

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First Proof.





(i) a0 is unmatched;

(ii) bi is adjacent to some vertex af(i) ∈ {a0, . . . , ai−1};

(iii) aibi ∈ M .

The sequence will end in some vertex bk ∈ B.

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First Proof.







(iii) aibi ∈ M .


Consider

P := bkaf(k)bf(k)af2(k)bf2(k)af3(k) . . . , afr(k)

with fr(k) = 0 is an alternating path.

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First Proof.







(iii) aibi ∈ M .


Consider

P := bkaf(k)bf(k)af2(k)bf2(k)af3(k) . . . , afr(k)

with fr(k) = 0 is an alternating path.

P is an augmenting path. �

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Second Proof.

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Second Proof.

Induction on |A|.

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Second Proof.

Induction on |A|.Trivial for |A| = 1.

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Second Proof.


If |N(S)| ≥ |S| + 1 for every non-empty set S � A, we pick an edge ab ∈ G and apply

induction hypothesis on the graph G′ := G − {a, b}.

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Second Proof.




If A has a non-empty proper subset A′ with |B′| = |A′| for B′ := N(A).

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Second Proof.




If A has a non-empty proper subset A′ with |B′| = |A′| for B′ := N(A).

Apply induction hypothesis on G[A′ ∪ B′] and G − (A′ ∪ B′). �

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Third Proof.

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Third Proof.

let H be a spanning subgraph of G that satisfies the marriage condition and is edge-minimal

with this property.

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Third Proof.


with this property.

By the marriage condition, dH(a) ≥ 1 for every a ∈ A.

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Third Proof.


with this property.

By the marriage condition, dH(a) ≥ 1 for every a ∈ A. Suppose there is a vertex a ∈ A

with distinct neighbours b1, b2 in H .

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Third Proof.


with this property.



By the minimality of H , the graphs H − ab1 and H − ab2 violate the marriage condition.

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Third Proof.


with this property.




So for i = 1, 2 there is a set Ai ⊆ A containing a such that |Ai| > |Bi| for

Bi := NH−abi(Ai).

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Third Proof.


with this property.




So for i = 1, 2 there is a set Ai ⊆ A containing a such that |Ai| > |Bi| for

Bi := NH−abi(Ai).

We can show∣∣NH(A1 ∩ A2 \ {a})

∣∣ ≤ |B1 ∩ B2|= |B1| + |B2| − |B1 ∪ B2|≤ |A1| − 1 + |A2| − 1 − ∣

∣A1 ∪ A2

∣∣

=∣∣A1 ∩ A2 \ {a}

∣∣ − 1.

�

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Corollary. If G is k-regular with k ≥ 1, then G has a 1-factor.

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A set of preferences for G is a family (≤v)v∈V of linear ordering ≤v on E(G).

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A matching M is stable if for every edge e ∈ E \ M there exists an edge f ∈ M such that e

and f have a common vertex v with e <v f .

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Given a matching M , a vertex a ∈ A is acceptable to a vertex b ∈ B if e = ab ∈ E \ M and

any edge f ∈ M at b satisfies f <b e.

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A vertex a ∈ A is happy with M if a is unmatched or its matching edge f ∈ M satisfies

f >a e for all edges e = ab such that a is acceptable to b.

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Lemma. 1. If a matching M is stable, then every vertex a ∈ A is happy with M .

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Lemma. 1. If a matching M is stable, then every vertex a ∈ A is happy with M .

2. If every matched vertex a ∈ A is happy with M and every unmatched a ∈ A is not

acceptable to any b ∈ B, then M is stable.

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Theorem.[Gale and Shapley 1962] For every set of preferences, G has a stable matching.

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Proof.

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Proof.

Starting with the empty matching, construct a sequence of matchings that each keeps all the

vertices in A happy.

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Proof.



Given such a matching M , consider a vertex a ∈ A that is unmatched but acceptable to

some b ∈ B. (If no such a exists, terminate the sequence.)

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Proof.





Add to M the ≤a-maximal edge ab such that a is acceptable to b, and discard from M any

other edge at b.

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Proof.






other edge at b.

The new matching M ′ is better than the matching M : every vertex b in an edge f ∈ M is

incident also with some f ′ ∈ M ′ with f ≤b f ′.

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Proof.






other edge at b.

The new matching M ′ is better than the matching M : every vertex b in an edge f ∈ M is

incident also with some f ′ ∈ M ′ with f ≤b f ′. The improvement cannot continue forever.

�

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Corollary.[Peterson 1891] Every regular graph (not necessarily bipartite) of positive even

degree has a 2-factor.

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2.2 Matching in general graphs

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Let G be a graph. CG denotes the number of its components, and q(G) denotes its set of

odd components, those of odd order.

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Theorem.[Tutte 1947] A graph G has a 1-factor if and only if q(G − S) ≤ |S| for all

S ⊆ V (G).

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Theorem.[Tutte 1947] A graph G has a 1-factor if and only if q(G − S) ≤ |S| for all

S ⊆ V (G).

Proof. (=⇒) Every odd component in G − S must send a factor edge to S.

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Proof. (Cont’d)

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Proof. (Cont’d)

(⇐=) Let G = (V, E) be a graph without a 1-factor. Our task is to find a bad set S ⊆ V ,

one that violates Tutte’s condition.

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Proof. (Cont’d)



We can assume G is edge-maximal without a 1-factor. (Why?)

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Proof. (Cont’d)




Then for a bad S we have:

all the components of G − S are complete

and every vertex s ∈ S is adjacent to all the vertices of G − s.(1)

Otherwise, by adding the missing edges, S is still bad and hence G still has no 1-factor.

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Proof. (Cont’d)








But also conversely, if a set S ⊆ V satisfies (1) then either S or the empty set must be bad:

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Proof. (Cont’d)








But also conversely, if a set S ⊆ V satisfies (1) then either S or the empty set must be bad: if

S is not bad we can join the odd components of G − S disjointly to S and pair up all the

remaining vertices, unless |G| is odd, in which case ∅ is bad.

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Proof. (Cont’d)

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Proof. (Cont’d)

Let S be the set of vertices that are adjacent to every other vertex. If this set S does not

satisfy (1), then some component of G − S has nonadjacent vertices a, a′.

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Proof. (Cont’d)



Let a, b, c be the first three vertices on a shortest a-a′ path in this component; then

ab, bc ∈ E but ac /∈ E.

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Proof. (Cont’d)




ab, bc ∈ E but ac /∈ E. Since b /∈ S, there is a vertex d ∈ V such that bd /∈ E.

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Proof. (Cont’d)




ab, bc ∈ E but ac /∈ E. Since b /∈ S, there is a vertex d ∈ V such that bd /∈ E.

By the maximality of G, there is a matching M1 of V in G + ac, and a matching M2 of V in

G + bd.

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Proof. (Cont’d)

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Proof. (Cont’d)

Let P = d . . . v be a maximal path in G starting at d with an edge from M1 and containing

alternately edges from M1 and M2.

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Proof. (Cont’d)



If the last edge of P lies in M1, then v = b, since otherwise we could continue P .

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Proof. (Cont’d)



If the last edge of P lies in M1, then v = b, since otherwise we could continue P . Let us

then set C := P + bd.

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Proof. (Cont’d)





If the last edge of P lies in M2, then by the maximality of P the M1-edge at v must be ac,

so v ∈ {a, c};

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Proof. (Cont’d)






so v ∈ {a, c}; then let C be the cycle dPvbd.

In each case, C is an even cycle with every other edge in M2, and whose only edge not in E

is bd.

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Proof. (Cont’d)






so v ∈ {a, c}; then let C be the cycle dPvbd.

In each case, C is an even cycle with every other edge in M2, and whose only edge not in E

is bd.

Replacing in M2 its edges on C with the edges of C − M2, we obtain a matching of V

contained in E, a contradiction.

�

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Corollary.[Petersen 1891] Every bridgeless cubic graph has a 1-factor.

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Documents

graph3 - SJTUbasics.sjtu.edu.cn/~chen/teaching/GR07/graph3.pdf · 2007-09-29 · GRAPH THEORY(III) Page 3 Two edges e = f are adjacentif they have an end in common. Pairwise non-adjacentvertices