Graph Theory

Judith Stumpp

6. November 2013

Bei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung Graph Theory vom Winter-semester 2011/2012. Sie wurde gehalten von Prof. Maria Axenovich Ph.D. . Der Mitschrieb erhebt wederAnspruch auf Vollständigkeit, noch auf Richtigkeit!

1

Kapitel 1

Definitions

The graph is a pair V,E. V is a finite set and E ⊆(V2

)a pair of elements in V . V is called the set of vertices

and E the set of edges.

Visualize: G = (V,E), V = {1, 2, 3, 4, 5}, E = {{1, 2}, {1, 3}, {2, 4}}

1 2

3

4

5

History: word: Sylvester (1814-1897) and Cayley (1821-1895)Euler - developed graph theory

Königsberg bridges (today Kaliningrad in Russia):

A

B

D C

A

D

B

C

Problem: Travel through each bridge once, come back to the original point.Impossible!

Notations:

• Kn= (V,(V2

)) - complete graph on n vertices |V | = n

K5 K3

v1v2

v3

v4v5

v6

C6 = v1v2v3v4v5v6

• Cn - cycle on n verticesV = {v1, v2, . . . , vn}, E = {{v1, v2}, {v2, v3}, . . . , {vn−1, vn}, {vn, v1}}

• Pn - path on n vertices (Note: Pn . path on n edges (Diestel))V = {v1, v2, . . . , vn}, E = {{v1, v2}, {v2, v3}, . . . , {vn−1, vn}}

2

• Let P be a path from v1 to vn. The subpath of P from vi to vj is viPvj and the subpath from vi+1 tovj is

◦viPvj .

• En= (V, ∅), |V | = n isolated vertices

E10

m

n

Km,n

• Kn,m= (A ∪B,A×B), A ∩B = ∅ complete bipartite graph

• Peterson graph: V =({1,2,3,4,5}

2

), E = {{{i, j}, {k, l} : {i, j} ∩ {k, l} = ∅}

{1, 2}

{3, 4}{3, 5}{4, 5}

{1, 3}

{2, 3}

{1, 4} {2, 4}

{1, 5}

{2, 5}

• Kneser Graph K(n, k)= ((Vk

), E)

|V | = n, E = {{A,B} : A,B ∈(Vk

)and A ∩B = ∅}.(

Vk

)is the set of k-element subsets of V, |

(Vk

)| =

(|V |k

)

• Qn - hypercube of dimension n.Qn = {2{1,2,...,n}, E}, E = {{A,B} : |A4B| = 1} (A4B := (A ∪B)− (A ∩B))V - set of binary n-tuples E - pairs of binary tuples different in 1 position

Q1 Q2 Q3

V = {∅, {1}} = {(0), (1)} V = {(0, 0), (0, 1), (1, 0), (1, 1)} V = {(0, 0, 0), . . . , (1, 1, 1)}

1

0

(1,1)

(1,0)(0,1)

(0,0)

(1,1,1)

(1,1,0) (1,0,1) (0,1,1)

(1,0,0)(0,1,0)

(0,0,1)

(0,0,0)

(1, . . . , 1)

(0, . . . , 0)

weight 1 (# 1’s in a binary tuple)weight 2

weight n-1

n

(n2

)(n3

)

(nn2

)

(n

n−1

)

Qn−1

Qn−1

001001

01

Qn :

3

Parameter: Let G = (V,E) be a graph. The order of G ist the number of vertices (|V |) and the size of Gis the number of edges (|E|).If the order of G is n, then 0 ≤ size(G) ≤

(n2

).

If e = {x, y} ∈ E, x is adjacent to y and x is incident to e.There is a n× n matrix A of G = ({v1, . . . , vn}, E) which is called the adjacent matrix.

For

v1

v2v3 v4 A =

0 1 1 01 0 1 11 1 0 00 1 0 0

.

Subgraph: H ⊆ G, H = (V ′, E′), G = (V,E), V ′ ⊆ V, E′ ⊆ Ev1

v2v3 v4

v1

v2v3

H ⊆ind

H is an induced subgraph of G if H ⊆ G and for v1, v2 ∈ V (H): {v1, v2} ∈ E(H)⇔ {v1, v2} ∈ E(G).

In the upper example it is no induced subgraph.

An induced subgraph is obtained from G by deleting vertices. E.g.:

v1

v2v3 v4

v1

v2v3

v1

v2 v4 v3 v4v2v3 v4

v1

Let G = (V,E) and G′ = (V ′, E′) be graphs. Then we define G ∪ G′ := (V ∪ C ′, E ∪ E′) and G ∩ G′ :=(V ∩ C ′, E ∩ E′).G[X] := (X, {{x, y} : x, y ∈ X, {x, y} ∈ E(G)}) is called the subgraph of G induced by a vertex setX ⊆ V (G). E.g.:

4 1

2

3 5 4 1

2

3

G G[{1, 2, 3, 4}]

A degree d(v) = deg v of a vertex is the number of edges incident to that vertex.v1

v2v3 v4 deg v1 = 2, deg v2 = 3, deg v3 = 2, deg v4 = 1In this example the degree sequence is (2, 3, 2, 1), the minimum degree δ(G) is 1 and the maximum degree∆(G) is 3.

Apparently |E(G)| = 12

n∑i=1

deg vi is true.

Thusn∑i=1

deg vi is even and therefore the number of vertices with odd degree is even.

d(G) := 1n

n∑i=1

deg vi is called the average degree of G.

4

Extremal graph theorem: We’ll prove that if G has n vertices and >⌈n2

4

⌉edges ⇒ G has a

triangle.

Let A,B ⊆ V, A∩B = ∅. P is an A-B-path if P = v1 . . . vk, V (P )∩A = {v1} and V (P )∩B = {vk}.A graph is connected if any two vertices are linked by a path. A maximal connected subgraph of agraph is a connected component.

A connected graph without cycles is called a tree. A graph without cycles (acyclic graph) is called aforest.

Other „special named“ graphs:star caterpillar spider broom

Proposition: If a graph G has a minimum degree δ(G) ≥ 2 then G has a path of length δ(G) anda cycle with at least δ(G) + 1 vertices.

proof: Let P = (x0, . . . , xk) be a longest path in G. Then all neighbors of xk are in V (P ) (y is aneighbor of x if {x, y} ∈ E). In particular k ≥ δ(G).Let i = min{j ∈ {0, . . . , k} : {xk, xj} ∈ E}. Then xixkxk−1 . . . xi is a cycle of length at least δ + 1.

x0xixk

≥ δ

�

The girth of a graph G is the length of a smallest cycle in G.

The distance dG(v, w) of v, w ∈ G is the length of the smallest path between them (min ∅ =∞).The diameter of G is max{dG(v, w) : v, w ∈ G}.

Proposition: Every nontrivial tree T has a leaf.

proof: Assume T has no leaves. T has no isolated vertices ⇒ δ(T ) ≥ 2⇒ Cn ⊆ T �

- A tree T of order n ≥ 1 has n− 1 edges.

proof: T = K1 X

Assume it holds for all trees of order < n.

Let v be a leaf of T , T ′ := T − v.⇒ |T ′| = n− 1 < n.T ′ is acyclic.Let v′, w ∈ T ′. ∃P v′ = v0, v1, . . . , vn = w ⊆ T .

To show: vi 6= v for all i = 0, . . . , n

v0, vn 6= v because v0, vn ∈ T ′, v 6∈ T ′vi 6= v (i = 1, . . . , n− 1) because dT (vi) ≥ 2, vi is not a leaf.

5

⇒ P ⊆ T ′ connecting v0 and w ⇒ T ′ connected ⇒ T is a tree.

With induction hypothesis T ′ has (n− 1)− 1 edges. Thus T has (n− 1)− 1 + 2 = n− 1 edges. �

A walk is an alternating sequence v0e0v1e1 . . . vn of vertices and edges so that ei = vivi+1 for alln = 0, . . . , n − 1. Compared to a path it is allowed to pass edges and vertices more than once. Ifv0 = vn, then the walk is a closed walk.

- If G has a u-v-walk (between vertices u, v) ⇒ G has a u-v-path.

proof: Consider the shortest walk between u and v isW . ThenW is a path. If not,W has a repeatedvertex W = ue0v1e1 . . .︸ ︷︷ ︸

=:W1

vi . . .︸ ︷︷ ︸=:W

vi . . . v︸ ︷︷ ︸=:W2

, then W ′ = W1W2 is a shorter u-v-walk. �

- If G has an odd closed walk (i.e. odd # edges) then G has an odd cycle.

proof: If there are no repeated vertices (except for first and last) ⇒ we have an odd cycle.

If there is a repeated vertex vi, W = v0e0v1 . . .︸ ︷︷ ︸1’st part of W2

vi . . . vi︸ ︷︷ ︸W1

. . . vn = v0︸ ︷︷ ︸2’nd part of W2

.

W is a union of two closed walks W1 and W2. Either W1 or W2 is an odd closed walk

⇒ by induction it contains an odd cycle. �

- If G has a closed walk with a non-repeated edge W = v0e0v1 . . . ei . . . ei is unique, then Gcontains a cycle.

proof: Induction on # vertices.

Basis:

Step: W = v0e0v1 . . .︸ ︷︷ ︸1’st part of W2

vi . . . vi︸ ︷︷ ︸W1

. . . vn = v0︸ ︷︷ ︸2’nd part of W2

(note, there is a repeated vertex vj, otherwise W is a cycle)So, W is a union of two closed walks W1 and W2 and either W1 or W2 has a non-repeated edge.By induction, that walk contains a cycle. �

Definition: An Eulerian tour is a closed walk containing all edges of a graph and repeating no edge.

e.g.: Eulerian tour v1e1v2e2 . . . e8v9 = v1 in

e1

v1 = v5 = v9

v2e2

e3 e4

e5

e6

e7

e8v4

v6 = v3

v7 v8

Theorem: A connected graph G has an Eulerian tour iff (i.e. if and only if) each degree of vertexin G is even.

6

proof:

„⇒“ : If there is an Eulerian tour then clearly the number of edges entering the vertex is the numberof edges leaving the vertex.

„⇐“ : Assume that each degree is even.Consider a walk with longest number of edges and no repeated edge, W = v0 . . . vk. Thus,there is no edge incident to v0 that is not in W . Since deg v0 is even, v0 must be vn, i.e. Wis a closed walk.If all edges are in W , done. Otherwise, there is an edge e, not in W . Since G is connected,there is such e incident to a vertex in W . Say e = viu. Then W ′ = ueviWvi is a longer walkwith no repeated edges.

Other idea: all edges in G are even, δ(G) ≥ 2 ⇒ G has a cycle C. Delete C from G(problem: G− C maybe isn’t connected).

C C

G G

�

Connectivity:We say that a Graph G is vertex k-connected if |V (G)| > k and deleting any (k − 1) vertices doesnot disconnect the graph.

Any connected graph is 1-connected. If a graph is 2-connected then there exists no cut-vertex whichis a vertex whose deletion disconnects a graph. Trees are not 2-connected.

If G is connected, X ⊆ V, G−X disconnected ⇒ X is called a cut-set.

κ(G) = max{k : G is k-connected}

e.g.: κ(v1

v2v3 v4) = 1, κ(Cn) = 2, κ(Kn,m) = min{m,n}.G is called l-edge connected if G 6= En and G does not become disconnected after deleting any (l−1)edges.

λ(G) (= κ′(G)) = max{l : G is l-edge-connected}e.g.: λ(tree) = 1, λ(Cn) = 2.

If λ(G) = 1 there exists a so called bridge (cut edge) .

Clearly λ(G) ≤ δ(G). But it could be that 1 = λ(G) << δ(G) = 99K100K100

.

Lemma: For any connected G: κ(G) ≤ λ(G) ≤ δ(G).

proof: Idea: want to find the set of at most λ := λ(G) vertices that disconnects the graph.

7

Let E be a set of λ edges disconnecting G. Then E is a cut, i.e. ∃S ⊆ V : ∀e ∈ E, one endpoint ofe is in S, another is in S := V − S.

E

SS

If in G there are all edges between S and S. λ = |E| = |S| · |S| ≥ |V (G)| − 1 ≥ κ(G).

Otherwise ∃x ∈ S, y ∈ S, x 6∼ y (i.e. xy 6∈ E(G)).

T := (N(x) ∩ S) ∪ ({z ∈ S : z ∼ S} − {x})

T is a vertex cut, in particular after deleting T , x and y are in different connected components.We have |T | ≤ |E| = λ because

|N(x)| ≤ #(edges incident to x) and |{z ∈ S : z ∼ S} − {x}| ≤ #(edges incident to this set).

�

Definition: A graph G is d-degenerate if there is a vertex order v1, v2, . . . , vn:

|N(vi) ∩ {vi+1, . . . , vn}| ≤ d.

I.e. we eliminate the graph by deleting a vertices sequence, s.t. at most d edges are gone at a time.

v1 v2 vnv3

≤ d ≤ d ≤ d

Let T be a graph. T is a tree if it is connected and acyclic.

• T is a tree iff T is connected and has |V (T )| − 1 edges.

• T is 1-degenerate.

• A leaf in a nontrivial tree is a vertex of degree 1.

• If G is a graph with δ(G) ≥ |V (T )| − 1 (T tree) then G contains T as a subgraph.

G T

δ(G) ≥ 6

8

Lemma: A graph is bipartite if and only if it has no odd cycles.

proof:

„⇒“ : Let G be a bipartite graph, then any cycle has a form u1v1u2v2 . . . ukvku1, where ui ∈ U, vi ∈V, 1 ≤ i ≤ k, U, V are partite sets of G.

„⇐“ : Assume that G is connected and has no odd cycles. We shall prove that G is bipartite withpartite sets U, V defined as follows.Fix x ∈ V (G).Let U = {u : dist(x, u) is even}, V {v : dist(x, v) is odd}We need to verify that G[U ], G[V ] are empty graphs.Assume that u, u′ ∈ U and {u, u′} ∈ E(G).Consider a walk formed by shortest x-u-path, shortest x-u′-path and u, u′.

u u′

This is an odd closed walk that contains an odd cycle, a contradiction.Thus G[U ] is an empty graph.Similarly G[V ] is an empty graph.

�

Matchings:A matching is a graph that is a disjoint (vertex) union of edges.

Philip Hall (Apr. 1904 - Dec. 1982) Cambridge, UK

Recall that N(S) for a set S of vertices is a set of neighbors of vertices in S.

Hall’s matching theorem 1935: Let G be a bipartite graph with partite sets A,B. Then G hasa matching containing all vertices of A if and only if |N(S)| ≥ |S| for any S ⊆ A.

S

N(S)

S

N(S)

A

B

bad

proof:

9

„⇒“ : obvious

S

N(S)

A

B

„⇐“ : Assume that |N(S)| ≥ |S| for any S ⊆ A.We shall proof that there is a matching containing all elements of A by induction on |A|.If |A| = 1, clear.Assume that |A| > 1

Case 1: |N(S)| ≥ |S|+ 1, for any S ⊂ A, S 6= A.Let {x, y} =: e ∈ E(G). Consider G′ = G− {x, y}.|NG′(S)| ≥ |NG(S)| − 1 ≥ |S|+ 1− 1 = |S|, for any S ⊆ A− {y}.

S

A

B x

y

Thus, Hall’s condition is true for G′, and there is a matchingM ′, containing all elementsof A− {y}, by induction.So, M ′ ∪ {x, y} is a matching saturating A in G.

A

B x

yM ′

Case 2: ∃S ⊂ A, S 6= A such that |N(S)| = |S|.

S

N(S)

A

B

S′

N(S′)

By induction, there is a matching containing all vertices of S. Let apply induction toG[A− S,B −N(S)].Assume that there is S ′ ⊆ A− S such that |N(S ′) ∩ (B −N(S))| < |S ′|.Then |N(S ′ ∪ S)| = |N(S) ∪ (N(S ′) ∩ (B −N(S)))|<6= |S|+ |S ′|.A contradiction to Hall’s condition applied to S ∪ S ′.Thus for any S ′ ⊆ A−S, |N(S ′)∩(B−N(S))| ≥ |S ′|, and there is a matching saturatingA− S in G[A− S,B −N(S)]. Together with a matching between S and N(S), it givesa matching saturating A.

10

�

Corollaries of Hall’s theorem:

1) Let G be bipartite with partite sets A,B, such that |N(S)| ≥ |S| − d for all S ⊆ A, and somefixed positive integer d.Then G contains a matching of size at least |A| − d.

2) A k-regular bipartite graph has a perfect matching, i.e. matching containing all vertices of agraph. Here k-regular is a graph with all degrees equal to k.

G has partite sets A,B :

|E(G)| = #edges incident to A = |A| · k= #edges incident to B = |B| · k

⇒ |A| = |B|

3) A k regular bipartite graph has a proper k-edge coloring.

proof:

1) Construct G′.

B

A

C

|C| = d, add all edges between A and C.In G′ |NG′(S)| ≥ |NG(S)|+ d ≥ |S| − d+ d = |S|.By Hall’s theorem, there is a matching in G′ saturating A, with at most d edges not in G.

2) Let’s verify Hall’s condition.Is it true that |N(G)| ≥ |S| for any S ⊆ A?#edges from S to B is |S| · k = #edges between S and N(S) = q

#edges from N(S) to A is |N(S)| · k ≥ #edges between S and N(S) = q.|N(S)|· 6 k ≥ q = |S|· 6 k ⇒ |N(S)| ≥ |S|.

11

�

Non-bipartite graphs:

A k-factor in a graph is a spanning (containing each vertex) subgraph in which each vertex has degreek.

perfect matching = 1-factor

2-factor

Denes König (Sep. 1884 - Oct. 1944)Gyula König (Dec. 1849 - Apr. 1913)

Let ν(G) be the size of largest matching in G and τ(G) be the size of smallest vertex cover, i.e. setof vertices such that each edge is incident to some of this vertices, i.e. a set X of vertices such thatG−X is an empty graph.

X

König’s theorem ’31: If G is a bipartite graph, then ν(G) = τ(G).

Classical approach: Given a maximal matching M and want to find a vertex cover of size |M |B

A

alternating path: starts with an unmatched vertex of M (alternating one point in A and one in B).Take the longest alternating path.

vertex cover: for any element of {a, b} ∈ E(M), a ∈ A, b ∈ B pick b if there is an alternating pathending in b, otherwise pick a.

proof: (by Romeo Rizzi ’2000)

We want to prove that τ(G) ≤ ν(G) (τ(G) ≥ ν(G) trivial).

Assume that G is the smallest counterexample (#edges, #vertices).Observe that G is connected, not a path, not a cycle, i.e. ∃v : deg(v) ≥ 3

12

Let v : deg v ≥ 3. u ∈ N(v) ,

Case 1: ν(G\u) < ν(G):Take a vertex cover X by König’s theorem of G − u of size ≤ ν(G) − 1. Then X ∪ {u} is thevertex cover of G of size ≤ ν(G).

Case 2: ν(G\u) = ν(G):Then, in G there is a maximal matching, M , not containing u. There is u′ ∈ N(v) − {u}, suchthat f := {v, u′} /∈ E(M).Let W ′ be a cover of G− f of size ν(G− f) = ν(G). Then W ′ does not contain u (W ′ containsvertices of M only and u /∈ V (M)). Thus W ′ contains v. So, W ′ covers f too. Thus W ′ covers G.

v

w

1

v

u

2

u’

f

M |M | = ν(G) �

Tutte’s theorem

S

odd

odd

odd

|S| ≥ odd components of G− S

For a subset S of vertices of G, let q(S)=#odd components of G− S.

Theorem: (Bill Tutte May 1917- May 2002)A graph G has a perfect matching (1-factor) if and only if ∀S ⊆ V (G) q(S) ≤ |S|.

proof:

„⇒“ : trivial.

„⇐“ : Consider G, such that ∀S ⊆ V (G), q(S) ≤ |S|, and assume that G has no 1-factor. Add edgeone-by-one, so the resulting graph G′ is no 1-factor.We shall show that in G′ is a „bad“ set S, q(S) > |S|.We shall show that S is also a bad set in G.Observation: If M1,M2 are perfect matchings in G, M14M2 = (M1 ∪M2)− (M1 ∩M2) areonly cycles.

13

Let S be a set of vertices of degree |V (G)| − 1. We shall show that S is bad in G′.Claim: All components of G′ − S are complete.Assume not, i.e. there is a non-complete component in G′ − S.

S

c

b

aa

c

bd

Then there is an induced path a, b, c in this component. Since b /∈ S, deg b < |V (G′)| − 1,there is d /∈ {a, b, c}, such that b 6∼ d.By maximality of G′, there is a perfect matching M , in G′ ∪ {{a, c}}, there is a perfectmatching M2 in G′ ∪ {{b, d}}. Note ac ∈ E(M1), bd ∈ E(M2). We shall create a perfectmatching of G′.Consider M14M2, ac, bd ∈ E(M14M2). If ac, bd belong to different cycles of M14M2:ac

bd

Take the edges of M2 in a component containing ac, take edges of M1 in a component withbd, otherwise take edges of M1.If ac, bd belong to the same cycle of M14M2, then

a

cb

d

ac

b

dor

A contradiction, since G′ has no 1-factor, so all components of G′− S are complete.� Claim

S

odd

odd

odd

even

even

If S is not bad, i.e. |q(S)| ≤ |S|, we can construct a perfect matching, a contradiction to thefact that G′ has no perfect matching. Thus S is bad in G′.

14

S

odd

odd

odd

odd

even

even

G is obtained from G′ by deleting edges, so qG(S) ≥ qG′(S) > |S|.

�

3-factor

f-factor12

34

1-factor

k-factor - spanning subgraph,all degrees = kf -factor: If f : V → N, an f -factor is a spanning subgraph H of G such that degH(v) = f(v).Let e(v)= deg(v)− f(v) ≥ 0 (excess).

Replace each vertex of G withA(v)

B(v) e(v)

d(v)

Ke(v), d(v)

For adjacent u and v, put an edge between A(u) and A(v), such that these edges form a matching.

An f -factor, in a graph G, for f : V (G)→ N ∪ {0}, such that ∀v ∈ V f(v) ≤ deg(v), is a spanningsubgraph H of G such that degH(v) = f(v).

1-factor or matching ≈ f -factor, f ≡ 1.

1 v2

1

3 v1

2

3 2

∅B(v1)

A(v1) B(v2)

A(v2)

f(v1) = 3, f(v2) = 1.

For a graph G and a function f : V (G)→ N∪{0}, construct an auxiliary graph T (G, f) by replacingeach vertex v with vertex sets A(v)∪B(v), |A(v)| = deg(v), |B(v)| = deg(v)−f(v), and for adjacentvertices u, v placing an edge between A(u) and A(v), so that these edges are disjoint, and placing a

15

complete bipartite graph between A(u)4B(u) for each vertex u.

Claim: G has an f -factor if and only if T (G, f) has 1-factor.

proof:

• Assume that M is an f -factor of G, to create a 1-factor in T , take the edges corresponding toM , and take missing edges between A(u) and B(u) ∀u ∈ V .

• Assume that M is a 1-factor in T , create an f -factor in G by deleting B(u), u ∈ V (G), con-tracting A(u) into a single vertex, u ∈ V (G).

�

H-factor: Given a graph G, and a graph H, such that |V (G)|:|V (H)| (: = divisible). An H-factor ofG is a spanning subgraph of G that is a vertex-disjoint union of copies of H.

H = G =

H = K2 H-factor ≈ perfect matching.

Hajnal & Szemeredi ’70: If G satisfies δ(G) ≥ k−1kn, n:k, then G has a Kk-factor.

Kk Kk KkKk

Alon-Yuster ’95: If G satisfies δ(G) ≥ χ(H)−1χ(H)

n. Then G contains at least (1− o(1)) n|V (H)| (H is

fixed, G is large, n = |V (G)|) copies of H vertex-disjoint.

...

o(n)

χ(H)-chromatic number of a graph H := min #parts into which vertex sets can be partitioned, sothat no two adjacent vertices are in same part.

χ(G) := min # colors assigned to V (G) such that no two adjacent vertices get the same color.1

2

3

2

1

G

χ(G) = 3χ(Kk) = k, χ(C3) = 3, χ(C4) = 2, χ(Km,n) = 2, χ(C2k+1) = 3There are graphs with large |V (G)| and small χ(G).

16

Connectivity: A,B ⊆ V (G), A-B-path P is a path v0, v1, . . . , vk such that V (P )∩A = {v0}, V (P )∩B = {vk}.C ⊆ V ∪ E, we say that X separates A and B if each A-B-path contains an element of X.

A B A B

v ∈ A ∩B ⇒ v is an A−B path

A B

v3v2

v1

u1u2

u3

A−B sep. set : {v2, u2}{e1, e5, e4}{e1, u1}

e1 e2

e3 e4

e5

Note that a separating set must contain A ∩B.Note B′ ⊇ B and X separates A and B′ ⇒ X separates A and B.

AB

B′=

Menger’s theorem (1927): (Karl Menger Jan. 1902 - Oct. 1985)Let G be a graph, A,B ⊆ V (G). Min #vertices separating A and B = Max #vertex-disjointA-B-paths.

proof: Assume that A ∩B = ∅.Let k = k(G;A,B) = min #vertices separating A and B, k(G;A,B) ≥ max # vertex-disjoint A-B-path (easy).

We shall prove that max # vertex-disjoint A-B-path ≥ k(G;A,B) = k by stronger induction:

If P is any set of less than k disjoint A-B-paths then there is a set Q of disjoint A-B-paths thatincludes the endpoints of P and |Q| = |P |+ 1.

BA

P

Lets prove this by induction on |V (G)−B − A|.

Basis: |V (G)−B − A| = 0.

A B

P

< k

There is an edge between A and B, not adjacent to vertices of P , otherwise |V (P ) ∩ A| < k is

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a vertex separating A and B.

Step: We have P , a set of less than k A-B-path, vertex disjoint.There is an A-v-path for v ∈ B\(V (P )), otherwise V (P ) ∩ B is a set of less than k verticesseparating A and B, call it R.

A B

v

xR

P

Let x be the last vertex of R that also belongs to a path in P call it P .Let B′ = B ∪ (V (xP ) ∪ V (xR)).P ′ = P\{P} ∪ {Px}.note k(G;A,B′) ≥ k(G;A,B).By induction, there is a larger set of A-B′-paths, Q′, |Q′| ≥ |P ′| + 1, Q′ contains endpoints ofP ′.

A BP

y

Let y be an endpoint of a path in Q′ in B′ that is not an endpoint of P ′.

y

y

y

Case : 1 Case: 2 Case: 3

Q1

Q

,

Case 1: y ∈ B:Take Q = Q′ − {Q}︸︷︷︸

path containing x

∪{Q ∪ xP}.

Case 2: y ∈ xP :Take Q = Q′ − {Q} ∪ {Q ∪ xR} − {Q1}︸ ︷︷ ︸

path containing y

∪{Q1 ∪ yP}.

Case 3: y ∈ xR:Take Q = Q′ − {Q} ∪ {Q ∪ xP} − {Q1} ∪ {Q1 ∪ yR}.

�

If G = (V,E) a graph, then a line graph L(G) of G is a graph L(G) = (E,E ′),E ′ = {{e, e} : e, e ∈ E and e, e are adjacent}.

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v2v3

v4

v1

e2e1

e3 e4 e5

e1

e2

e3

e5e4

Corollary 1: If a, b ∈ V (G), {a, b} /∈ E(G).

min #vertices separating a and b = max #independent a-b-paths

(here independent means that they share only a and b)

a b

A B

Apply Menger’s theorem to A = N(a) and B = N(b).

Corollary 2: (Global version of Menger’s theorem)Any graph G is k-connected if and only if for any two vertices a, b there are k independent pathsbetween a and b.

outline of proof:Suppose G contains k independent paths between any two vertices, thus we need ≥ k vertices toseparate G. So κ(G) ≥ k.

Let κ(G) = k, in particular |(G)| > k.Assume that a and b are not connected by k independent paths. By corollary 1 a adjacent to b.Let G′ = G− {a, b}, then G′ contains ≤ (k − 2) independent a-b-paths.

ab

≤ k − 2

a b

≤ k − 2v

X

G′

By corollary 1, we can separate a and b in G′ by ≤ k − 2 vertices, X.Since |V (G)| > k, there is v /∈ {a, b} and v /∈ component of a in G′ −X.Observe that v and a are separated by X ∪ {b} in G.

So, v and a are separated by ≤ k − 1 vertices, a contradiction to the fact that κ(G) = k. �

Edge-connectivity

1) min #edges separating a and b in G = max #edge-disjoint a-b-paths.

a b

AB

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Apply Menger’s theorem to L(G) with A = {edges incident to a}, B = {edges incident to b}.2) Global Menger’s theorem (edge-connected)

A graph is k-edge-connected if and only if there are k edge-disjoint paths between any twovertices.

κ(G) = 1 blocks block-cut-vertex tree.A block - either a bridge or maximal 2-connected subgraph.

B1 B2B3 B4

B5

B6

v1 v2v3

v4

v5

B1 B2

v1 v2

...

Bi ∼ vj if vj ∈ V (Bi).Any two block intersect by at most 1 vertex.

Block-cut-vertex graph is a tree.

A block that is a leaf in a block-cut-vertex tree is a block leaf.

κ(G) ≥ 2 ⇔ G can be constructed using ear-decomposition

G is created using ear-decomposition if there is a sequence of graphs G0 ⊆ G1 ⊆ . . . G, such that G0

is a cycle, Gi+1 is created from Gi by adding a Gi-path (ear) (i.e. a path with endpoints in Gi andno other vertices in Gi).

Gi

outline of proof:

„⇒“ : κ(G) = 2: We have that G has a cycle. Consider the largest subgraph H of G that is built asear-decomposition.Observe H ⊆

indG. If u, v ∈ V (H), v 6∼H u, v ∼G u, then add uv as a ear. If H 6= G ⇒ ∃x ∈

V (G)− V (H), such that x is adjacent to a vertex w ∈ V (H).G − w is connected, so in G − w there is a path from x to H, call it P , call the first vertexof P in H, w1.So wx ∪ xPw is an H-ear. A contradiction to maximality of H, so G = H.

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H

G

w x

w1

„⇐“ : Show that an ear-decomposition is 2-connected . . .

κ(G) = 3 : |V (G)| ≥ 5.

Observation: If κ(G) = 3 then there is an edge e of G such that κ(G ◦ e) ≥ 3.

Let e = {x, y} ∈ E(G), G ◦ e is obtained from G by identifying x and y, removing (if necessary)loops and multiple edges.

x

y

v3

v2

v1

v4

v5

vxyv2

v1

v3

v4

v5

Tutte’s theorem 1961: A graph G is 3-connected if and only if it exists a sequence of graphsG0, G1, . . . , Gn, such that G0 = K4, Gn = G, Gi+1 is obtained from Gi:Gi+1 has two vertices x, y of degree ≥ 3, x ∼ y and Gi = Gi+1 ◦ {x, y}.

x

x′

x′′

yy′

y′′

Lemma: If G is 3-connected, then there exists an edge e such that G ◦ e is 3-connected.(without proof)

proof: We want to prove hat if Gi is 3-connected, then Gi+1 is also 3-connected. Assume not, i.e.Gi = Gi+1 ◦ {x, y} and Gi+1 is not 3-connected, i.e. there exists a cut-set S with |S| ≤ 2.

G1G2S

Let G1 and G2 be connected components of Gi+1 − S. Observe, {x, y} 6= S, otherwise Gi is not3-connected. But {x, y} ∩ S 6= ∅, otherwise Gi is not 3-connected (disconnected by S). So, w.l.o.g.(without loss of generality) x ∈ S, y ∈ V (G2).

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G1

G2S

x yv

w

|Gi+1| > |Gi|Assume that there exists a vertex v ∈ V (G2)\{y}, then in Gi {w, vxy} separates v from V (G1),a contradiction. So V (G2) = {y}, so deg(y) ≤ 2, a contradiction. �

A graph G is k-linked, if for any distinct vertices s1, s2, . . . , sk, t1, t2, . . . , tk, there are vertex-disjointsi-ti-paths, i = 1, . . . , k.s1s2

s3

t1

t2

t3

s1s2

s3

t′1 = t2

t′2 = t3

t′3 = t1

G is k-linked ⇒ G is k-connected (Menger’s theorem)

G is f(k)︸︷︷︸22k

-connected ⇒ G is k-linked. (Bollobas-Thomason ’96)

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