graph4 - sjtubasics.sjtu.edu.cn/~chen/teaching/gr08/graph4.pdf · 2008. 10. 8. · graph theory(iv)...

GRAPH THEORY

Yijia ChenShanghai Jiaotong University

2008/2009

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GRAPH THEORY (IV) Page 2

Chapter 3. Connectivity

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G is called k-connected (for k ∈ N) if |G| > k and G − X is connected for every set

X ⊆ V with |X | < k.

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G is called k-connected (for k ∈ N) if |G| > k and G − X is connected for every set

X ⊆ V with |X | < k.

A graph is k-connected if and only if any two (distinct) vertices can be joined by k

independent paths.

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3.1 2-Connected graphs and subgraphs

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A maximal connected subgraph without a cutvertex is called a block.

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Thus, every block of a graph G is either a maximal 2-connected subgraph, or a bridge (with

its ends), or an isolated vertex.

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its ends), or an isolated vertex. Conversely, every such subgraph is a block.

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By their maximality, different blocks of G overlap in at most one vertex, which is then a

cutvertex of G.

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By their maximality, different blocks of G overlap in at most one vertex, which is then a

cutvertex of G.

Hence, every edge of G lies in a unique block, and G is the union of its blocks.

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If {V1, V2} is a partition of V , the set E(V1, V2) of all the edges of G crossing this partition

is called a cut (or cocycle).

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is called a cut (or cocycle). Recall that for V1 = {v} this cut is denoted by E(v).

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A minimal non-empty cut in G is a bond.

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A minimal non-empty cut in G is a bond.

Lemma. (i) The cycles of G are precisely the cycles of its blocks.

(ii) The bonds of G are precisely the minimal cuts of its blocks.

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Let A denote the set of cutvertices of G, and B the set of its blocks.

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The block graph of G is a bipartite graph on A ∪ B with partition {A, B} and edges aB

with a ∈ B.

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The block graph of G is a bipartite graph on A ∪ B with partition {A, B} and edges aB

with a ∈ B.

Proposition. The block graph of a connected graph is a tree.

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Proposition. A graph is 2-connected if and only if it can be constructed from a cycle by

successively adding H-paths to graphs H already constructed.

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3.2 The structure of 3-connected graphs

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Lemma. If G is 3-connected and |G| > 4, then G has an edge e such that G/e ia again

3-connected.

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Lemma. If G is 3-connected and |G| > 4, then G has an edge e such that G/e ia again

3-connected.

Recall, G/e is a graph (V ′, E′) with vertex set V ′ := (V \ {x, y}) ∪ {ve} (where ve is the

‘new’ vertex, i.e., ve /∈ V ∪ E) and edge set

E′ :={

vw ∈ E∣∣ {v, w} ∩ {x, y} = ∅

vew∣∣ xw ∈ E \ {e} or yw ∈ E \ {e}

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Proof.

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Proof.

Suppose there is no such edge e. Then, for every edge xy ∈ G, the graph G/xy contains a

separator S of at most 2 vertices, i.e., G/xy − S is disconnected.

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Proof.

Since κ(G) ≥ 3, the contracted vertex vxy of G/xy lies in S and |S| = 2, i.e. G has a vertex

z /∈ {x, y} such that {vxy, z} separates G/xy.

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Proof.

Then any two vertices separated by {vxy, z} in G/xy are separated in G by T := {x, y, z}.

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Proof.

Then any two vertices separated by {vxy, z} in G/xy are separated in G by T := {x, y, z}.

Since no proper subset of T separates G, every vertex in T has a neighbour in every

component C of G − T .

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Proof. (Cont’d)

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Proof. (Cont’d)

We choose the edge xy, the vertex z, and the component C so that |C| is as small as

possible, and pick a neighbour v of z in C.

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Proof. (Cont’d)

By assumption, G/zv is again not 3-connected, so again there is a vertex w such that

{z, v, w} separates G, and as before every vertex in {z, v, w} has a neighbour in every

component of G − {z, v, w}.

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Proof. (Cont’d)

As x and y are adjacent, G \ {z, v, w} has a component D such that D ∩ {x, y} = ∅.

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Proof. (Cont’d)

Then every neighbour of v in D lies in C (since v ∈ C), so D ∩ C �= ∅, and hence D � C

by the choice of D.

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Proof. (Cont’d)

Then every neighbour of v in D lies in C (since v ∈ C), so D ∩ C �= ∅, and hence D � C

by the choice of D.

This contradicts the choice of xy, z and C. �

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3.3 Menger’s Theorem

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Theorem.[Menger 1927] Let G = (V, E) be a graph and A, B ⊆ V . Then the minimum

number of vertices separating A from B in G is equal to the maximum number of disjointA-B paths in G.

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Whenever G, A,B are given as in the theorem, we denote by

k = k(G, A, B)

the minimum number of vertices separating A from B in G.

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Whenever G, A,B are given as in the theorem, we denote by

k = k(G, A, B)

the minimum number of vertices separating A from B in G.

Clearly, G cannot contain more than k disjoint A-B paths; our task will be to show that k

such paths exist.

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First Proof.

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First Proof.

We apply induction on ‖G‖ = |E(G)|.

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First Proof.

We apply induction on ‖G‖ = |E(G)|.If G has no edge, then |A ∩ B| = k and we have k trivial A-B paths.

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First Proof.

So we assume that G has an edge e = xy.

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First Proof.

If G has no k disjoint A-B paths, then neither does G/e; here, we count the contracted

vertex ve as an element of A (resp. B) in G/e if in G at least one of x, y lies in A (resp. B).

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First Proof.

By the induction hypothesis, G/e contains an A-B separator Y of fewer than k vertices.

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First Proof.

Among these must be the vertex ve, since otherwise Y ⊆ V (G) would be an A-B separator

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First Proof.

Among these must be the vertex ve, since otherwise Y ⊆ V (G) would be an A-B separator

Then X := (Y \ {ve}) ∪ {x, y} is an A-B separator in G of exactly k vertices.

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Proof. (Cont’d)

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Proof. (Cont’d)

We now consider the graph G − e.

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Proof. (Cont’d)

Since x, y ∈ X , every A-X separator in G − e is also an A-B separator in G and hence

contains at least k vertices.

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Proof. (Cont’d)

So by induction there are k disjoint A-X paths in G − e, and similarly there are k disjoint

X − B paths in G − e.

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Proof. (Cont’d)

So by induction there are k disjoint A-X paths in G − e, and similarly there are k disjoint

X − B paths in G − e.

As X separates A from B, these two path systems do not meet outside X , and can thus be

combined to k disjoint A-B paths. �

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Let P be a set of disjoint A-B paths, and let Q be another such set.

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We say that Q exceeds P if the set of vertices in A that lie on a path in P is a proper subset

of the set of vertices in A that lie on a path in Q, and likewise for B.

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We say that Q exceeds P if the set of vertices in A that lie on a path in P is a proper subset

of the set of vertices in A that lie on a path in Q, and likewise for B. Then, in particular,

|Q| ≥ |P| + 1.

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Second Proof.

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Second Proof.

CLAIM. If P is any set of fewer than k disjoint A-B paths in G, then G contains a set of

|P| + 1 disjoint A-B paths exceeding P .

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Second Proof.

Keeping G and A fixed, we let B vary and apply induction on |⋃P|. Let R be an A-B path

that avoids the (fewer than k) vertices of B that lie on a path in P .

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Second Proof.

If R avoids all the paths in P , then P ∪ {R} exceeds P , as desired.

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Second Proof.

If R avoids all the paths in P , then P ∪ {R} exceeds P , as desired. (This will happen when

P = ∅, so the induction starts.)

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Second Proof.

If R avoids all the paths in P , then P ∪ {R} exceeds P , as desired. (This will happen when

P = ∅, so the induction starts.)

If not, let x be the last vertex of R that lies on some P ∈ P .

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Proof. (Cont’d)

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Proof. (Cont’d)

Put B′ := B ∪ V (xP ∪ xR) and P ′ := (P \ {P}) ∪ {Px}.

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Proof. (Cont’d)

Then |P ′| = |P| (but |⋃P ′| < |⋃P|) and k(G, A, B′) ≥ k(G, A, B).

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Proof. (Cont’d)

By the induction hypothesis there is a set Q of |P′| + 1 disjoint A-B′ paths exceeding P′.

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Proof. (Cont’d)

Then Q′ contains a path Q ending in x, and a unique path Q′ whose last vertex y is not

among the last vertices of the paths in P′.

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Proof. (Cont’d)

If y /∈ xP , we let Q be obtained from Q′ by adding xP to Q, and adding yR to Q′ if y /∈ B.

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Proof. (Cont’d)

Otherwise y ∈◦x P , and we let Q be obtained from Q′ by adding xR to Q and adding yP to

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Proof. (Cont’d)

Otherwise y ∈◦x P , and we let Q be obtained from Q′ by adding xR to Q and adding yP to

In both cases Q exceeds P , as desired. �

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Applied to a bipartite graph, Mengers theorem specializes to the assertion of Konig’s

theorem.

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Applied to a bipartite graph, Mengers theorem specializes to the assertion of Konig’s

theorem.

The third proof is based on the alternating paths, thus algorithmic in nature.

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Let again G, A, B be given, and let P be a set of disjoint A-B paths in G.

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Let us say that an A-B separator X ⊆ V lies on P if it consists of a choice of exactly one

vertex from each path in P .

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If we can find such a separator X , then clearly k ≤ |X | = |P|, and Menger’s theorem will

be proved.

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If we can find such a separator X , then clearly k ≤ |X | = |P|, and Menger’s theorem will

be proved.

V [P] :=⋃ {

V (P ) | P ∈ P}

E[P] :=⋃ {

E(P ) | P ∈ P}.

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Let a walk W = x0e0x1e1 . . . en−1xn in G with ei �= ej for i �= j be said toalternate with respect to P

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Let a walk W = x0e0x1e1 . . . en−1xn in G with ei �= ej for i �= j be said toalternate with respect to P if it starts in A \ V [P] and

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Let a walk W = x0e0x1e1 . . . en−1xn in G with ei �= ej for i �= j be said toalternate with respect to P if it starts in A \ V [P] and the following three conditions holdfor all i < n :

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(i) if ei = e ∈ E[P], then W traverses the edge e backwards, i.e. xi+1 ∈ P◦xi for some

P ∈ P;

(ii) if xi = xj with i �= j, then xi ∈ V [P];

(iii) if xi ∈ V [P], then {ei−1, ei} ∩ E[P] �= ∅, where e−1 := e0.

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P ∈ P;

By (ii), any vertex outside V [P] occurs at most once on W .

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P ∈ P;

Since the edges ei of W are all distinct, (iii) implies that any vertex v ∈ V [P] occurs at mosttwice on W .

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P ∈ P;

Since the edges ei of W are all distinct, (iii) implies that any vertex v ∈ V [P] occurs at mosttwice on W .

For v �= xn, this can happen in exactly the following two ways. If xi = xj with0 < i < j < n, then

either ei−1, ej ∈ E[P] and ei, ej−1 /∈ E[P]

or ei, ej−1 ∈ E[P] and ei−1, ej /∈ E[P].

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Lemma. If an alternating walk W ends in B \ V [P], then G contains a set of disjoint A-B

paths exceeding P .

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paths exceeding P .

Proof. We may assume that W has only its first vertex in A \ V [P] and only its last vertex in

B \ V [P].

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paths exceeding P .

B \ V [P].

Let H be the graph on V (G) whose edge set is the symmetric difference of E[P] with

{e0, . . . , en−1}.

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paths exceeding P .

B \ V [P].

Let H be the graph on V (G) whose edge set is the symmetric difference of E[P] with

{e0, . . . , en−1}.

In H , the ends of the paths in P and of W have degree 1 (or 0, if the path or W is trivial),

and all other vertices have degree 0 or 2.

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For each vertex a ∈ (A ∩ V [P]) ∪ {x0}, therefore, the component of H containing a is apath, P = v0 . . . vk say, which starts in a and ends in A or B.

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Using conditions (i) and (iii), one easily shows by induction on i = 0, . . . , k − 1 that P

traverses each of its edges e = vivi+1 in the forward direction with respect to P or W .

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(Formally: if e ∈ P ′ with P ′ ∈ P , then vi ∈ P ′ ◦vi+1; if e = ej ∈ W , then vi = xj and

vi+1 = xj+1.)

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vi+1 = xj+1.)

Hence, P is an A-B path.

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vi+1 = xj+1.)

Similarly, for every b ∈ (B ∩ V [P]) ∪ {xn} there is an A-B path in H that ends in b.

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vi+1 = xj+1.)

Similarly, for every b ∈ (B ∩ V [P]) ∪ {xn} there is an A-B path in H that ends in b.

The set of A-B paths in H therefore exceeds P .

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Lemma. If no alternating walk W as above ends in B \ V [P], then G contains an A-Bseparator on P .

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Proof. Let

A1 := A ∩ V [P] and A2 := A \ A1,

and B1 := B ∩ V [P] and B2 := B \ B1.

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Proof. Let

A1 := A ∩ V [P] and A2 := A \ A1,

and B1 := B ∩ V [P] and B2 := B \ B1.

For every path P ∈ P , let xP be the last vertex of P that lies on some alternating walk; if no

such vertex exists, let xP be the first vertex of P .

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Proof. Let

A1 := A ∩ V [P] and A2 := A \ A1,

and B1 := B ∩ V [P] and B2 := B \ B1.

CLAIM.

X := {xP |P ∈ P}meets every A-B path in G;

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Proof. Let

A1 := A ∩ V [P] and A2 := A \ A1,

and B1 := B ∩ V [P] and B2 := B \ B1.

CLAIM.

X := {xP |P ∈ P}meets every A-B path in G; i.e., X is an A-B separator on P .

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Proof. (Cont’d)

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Proof. (Cont’d)

Suppose there is an A-B path Q that avoids X .

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Proof. (Cont’d)

Q meets V [P], as otherwise it would be an alternating walk ending in B2.

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Proof. (Cont’d)

Now the A-V [P] path in Q is either an alternating walk or consists only of the first vertex of

some path in P .

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Proof. (Cont’d)

some path in P .

Therefore Q also meets the vertex set V [P′] of

P ′ :={PxP | P ∈ P}

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Proof. (Cont’d)

some path in P .

Therefore Q also meets the vertex set V [P′] of

P ′ :={PxP | P ∈ P}

Let y be the last vertex of Q in V [P′], say y ∈ P ∈ P , and let x := xP .

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Proof. (Cont’d)

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Proof. (Cont’d)

As Q avoids X and hence x, we have y ∈ P◦x.

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Proof. (Cont’d)

Hence, x = xP is not the first vertex of P , and so there is an alternating walk W ending at

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Proof. (Cont’d)

Then W ∪ xPyQ is a walk from A2 to B.

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Proof. (Cont’d)

Then W ∪ xPyQ is a walk from A2 to B.

If this walk alternates and ends in B2, we have our desired contradiction.

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Proof. (Cont’d)

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Proof. (Cont’d)

How could W ∪ xPyQ fail to alternate?

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Proof. (Cont’d)

For example, W might already use an edge of xPy.

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Proof. (Cont’d)

For example, W might already use an edge of xPy. But if x′ is the first vertex of W on

xP◦y, then W ′ := Wx′Py is an alternating walk from A2 to y. (By Wx′ we mean the

initial segment of W ending at the first occurrence of x′ on W ; from there, W ′ follows P

back to y.)

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Proof. (Cont’d)

back to y.)

Even our new walk W ′yQ need not yet alternate: W ′ might still meet◦y Q.

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Proof. (Cont’d)

back to y.)

By definition of P ′ and W , however, and the choice of y on Q, we have

V (W ′) ∩ V [P] ⊆ V [P ′] and V (◦y Q) ∩ V [P ′] = ∅.

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Proof. (Cont’d)

back to y.)

By definition of P ′ and W , however, and the choice of y on Q, we have

V (W ′) ∩ V [P] ⊆ V [P ′] and V (◦y Q) ∩ V [P ′] = ∅.

Thus, W ′ and◦y Q can meet only outside P .

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Proof. (Cont’d)

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Proof. (Cont’d)

If W ′ does indeed meet◦y Q, we let z be the first vertex of W ′ on

◦y Q and set

W ′′ := W ′zQ.

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Proof. (Cont’d)

◦y Q and set

W ′′ := W ′zQ. Otherwise we set W ′′ := W ′ ∪ yQ.

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Proof. (Cont’d)

◦y Q and set

In both cases W ′′ alternates with respect to P′, because W ′ does and◦y Q avoids V [P ′].

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Proof. (Cont’d)

◦y Q and set

W ′′ satisfies condition (iii) at y in the second case even if y occurs twice on W ′, because

W ′′ then contains the entire walk W ′ and not just its initial segment W ′y.

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Proof. (Cont’d)

◦y Q and set

By definition of P ′, therefore, W ′′ avoids V [P] \ V [P ′].

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Proof. (Cont’d)

◦y Q and set

By definition of P ′, therefore, W ′′ avoids V [P] \ V [P ′]. Thus W ′′ also alternates with

respect to P and ends in B2, contrary to our assumptions. �

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Third Proof of Menger’s Theorem.

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Let P contain as many disjoint A-B paths in G as possible.

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Then by the first lemma, no alternating walk ends in B \ V [P].

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Then by the first lemma, no alternating walk ends in B \ V [P].

By the second lemma, this implies that G has an A-B separator X on P , giving

k ≤ |X | = |P| as desired. �

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A set of a-B paths is called an a-B fan if any two of the paths have only a in common.

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A set of a-B paths is called an a-B fan if any two of the paths have only a in common.

Corollary. For B ⊆ V and a ∈ V \ B, the minimum number of vertices �= a separating a

from B in G is equal to the maximum number of paths forming an a-B fan in G.

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Corollary. Let a and b be two distinct vertices of G.

(i) If ab /∈ E, then the minimum number of vertices �= a, b separating a from b in G is equal

to the maximum number of independent a-b paths in G.

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(ii) The minimum number of edges separating a from b in G is equal to the maximum

number of edge-disjoint a-b paths in G.

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(ii) The minimum number of edges separating a from b in G is equal to the maximum

number of edge-disjoint a-b paths in G.

Proof. For (ii) we consider :

The line graph L(G) of G is the graph on E in which x, y ∈ E are adjacent as

vertices if and only if they are adjacent as edges in G.

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Theorem.[Global Version of Mengers Theorem]

(i) A graph is k-connected if and only if it contains k independent paths between any two

vertices.

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Theorem.[Global Version of Mengers Theorem]

(i) A graph is k-connected if and only if it contains k independent paths between any two

vertices.

(ii) A graph is k-edge-connected if and only if it contains k edge-disjoint paths between any

two vertices.

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3.5 Linking pairs of vertices

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Let G be a graph, and let X ⊆ V (G) be a set of vertices.

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We call X linked in G if whenever we pick distinct vertices s1, . . . , s�, t1, . . . , t� in X we

can find disjoint paths P1, . . . , P� in G such that each Pi links si to ti and has no inner

vertex in X .

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vertex in X .

If |G| ≥ 2k and every set of at most 2k vertices is linked in G, then G is k-linked.

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vertex in X .

If |G| ≥ 2k and every set of at most 2k vertices is linked in G, then G is k-linked.

This is equivalent to requiring that k-linked disjoint paths Pi = si . . . ti exist for every

choice of exactly 2k vertices s1, . . . , sk, t1, . . . , tk.

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Clearly, every k-linked graph is k-connected.

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Clearly, every k-linked graph is k-connected.

Theorem.[Jung 1970; Larman and Mani 1970] There is a function f : N → N such that

every f(k)-connected graph is k-linked, for all k ∈ N.

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Lemma. There is a function h : N → N such that every graph of average degree at least

h(r) contains Kr as a topological minor, for every r ∈ N.

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Lemma. There is a function h : N → N such that every graph of average degree at least

h(r) contains Kr as a topological minor, for every r ∈ N.

RECALL:

If we replace the edges of X with independent paths between their ends (so that none

of these paths has an inner vertex on another path or in X), we call the graph G obtained

a subdivision of X and write G = TX .

If G = TX is a subgraph of another graph Y , then X is a topological minor of Y .

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Proof.

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Proof.

For r ≤ 2, the assertion holds with h(r) = 1.

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Proof.

Assume that r ≥ 3. We show by induction on m = r, . . . ,(r2

)that every graph G with

average degree d(G) ≥ 2m has a topological minor X with r vertices and m edges;

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Proof.

average degree d(G) ≥ 2m has a topological minor X with r vertices and m edges; for

m =(r2

)this implies the assertion with h(r) = 2(

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Proof.

m =(r2

RECALL:

Proposition. Every graphs G with at least one edge has a subgraph with δ(H) >

ε(H) ≥ ε(G) = 12d(G).

Proposition. Every graph G contains a path of length δ(G) and a cycle of length at

least δ(G) + 1 (provided that δ(G) ≥ 2).

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Proof.

m =(r2

RECALL:

Proposition. Every graphs G with at least one edge has a subgraph with δ(H) >

ε(H) ≥ ε(G) = 12d(G).

Proposition. Every graph G contains a path of length δ(G) and a cycle of length at

least δ(G) + 1 (provided that δ(G) ≥ 2).

If m = r then G contains a cycle of length at least ε(G) + 1 ≥ 2r−1 + 1 ≥ r + 1, and the

assertion follows with X = Cr.

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Proof. (Cont’d)

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Proof. (Cont’d)

Now let r < m ≤ (r2

), and assume the assertion holds for smaller m.

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Proof. (Cont’d)

Consider G with d(G) ≥ 2m, hence ε(G) ≥ 2m−1.

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Proof. (Cont’d)

Consider G with d(G) ≥ 2m, hence ε(G) ≥ 2m−1. We can assume that G is connected,

since G has a component C with ε(C) ≥ ε(G).

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Proof. (Cont’d)

since G has a component C with ε(C) ≥ ε(G). Let U ⊆ V (G) be a maximal set such that

U is connected in G and ε(G/U) ≥ 2m−1.

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Proof. (Cont’d)

U is connected in G and ε(G/U) ≥ 2m−1. We have N(U) �= ∅, as G is connected.

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Proof. (Cont’d)

Let H := G[N(U)].

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Proof. (Cont’d)

Let H := G[N(U)]. If H has a vertex v of degree dH(v) < 2m−1, we may add it to U and

obtain a contradiction to the maximality of U . Therefore d(H) ≥ δ(H) ≥ 2m−1.

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Proof. (Cont’d)

obtain a contradiction to the maximality of U . Therefore d(H) ≥ δ(H) ≥ 2m−1. By the

induction hypothesis, H contains a TY with |Y | = r and ‖Y ‖ = m − 1.

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Proof. (Cont’d)

obtain a contradiction to the maximality of U . Therefore d(H) ≥ δ(H) ≥ 2m−1. By the

induction hypothesis, H contains a TY with |Y | = r and ‖Y ‖ = m − 1. Let x, y be two

branch vertices of this TY that are non-adjacent in Y . Since x and y lie in N(U) and U is

connected in G, G contains an x-y path whose inner vertices lie in U . Adding this path to

the TY , we obtain the desired TX . �

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Proof.

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Proof.

f(k) = h(3k) + 2k,

where h is as in the previous lemma.

Shanghai

Proof.

f(k) = h(3k) + 2k,

Let G be an f(k)-connected.

Shanghai

Proof.

f(k) = h(3k) + 2k,

Let G be an f(k)-connected. Then

d(G) ≥ δ(G) ≥ κ(G) ≥ h(3k).

Shanghai

Proof.

f(k) = h(3k) + 2k,

d(G) ≥ δ(G) ≥ κ(G) ≥ h(3k).

Choose K = TK3k ⊆ G as in the previous lemma,

Shanghai

Proof.

f(k) = h(3k) + 2k,

d(G) ≥ δ(G) ≥ κ(G) ≥ h(3k).

Choose K = TK3k ⊆ G as in the previous lemma, and let U denote its set of branch

vertices.

Shanghai

Proof. (Cont’d)

Shanghai

Proof. (Cont’d)

We show G is k-linked:

Shanghai

Proof. (Cont’d)

We show G is k-linked: Let s1, . . . , sk, t1, . . . , tk ∈ V (G) be pairwise distinct.

Shanghai

Proof. (Cont’d)

Note κ(G) ≥ 2k. Hence by Menger’s theorem, G contains disjoint paths

P1, . . . , Pk, Q1, . . . , Qk, such that each Pi starts in si, each Qi starts in ti, and all these

paths end in U but have no inner vertices in U .

Shanghai

Proof. (Cont’d)

Let the set P of these paths be chosen so that their total number of edges outside E(K) is as

small as possible.

Shanghai

Proof. (Cont’d)

small as possible.

Let u1, . . . , uk be those k vertices in U that are not an end of a path in P .

Shanghai

Proof. (Cont’d)

small as possible.

For each i = 1, . . . , k, let Li be the U -path in K (i.e., the subdivided edge of the K3k) from

ui to the end of Pi in U , and let vi be the first vertex of Li on any path P ∈ P .

Shanghai

Proof. (Cont’d)

small as possible.

P has no more edges outside E(K) than PviLiui does, so viP = viLi and hence P = Pi.

Shanghai

Proof. (Cont’d)

small as possible.

Similarly, if Mi denotes the U -path in K from ui to the end of Qi in U , and wi denotes the

first vertex of Mi on any path in P , then this path is Qi.

Shanghai

Proof. (Cont’d)

small as possible.

Similarly, if Mi denotes the U -path in K from ui to the end of Qi in U , and wi denotes the

first vertex of Mi on any path in P , then this path is Qi.

Then the paths siPiviLiuiMiwiQiti are disjoint for different i and G is k-linked. �

Shanghai

graph4 - sjtubasics.sjtu.edu.cn/~chen/teaching/gr08/graph4.pdf · 2008. 10. 8. · graph theory(iv)...

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