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 weder Anspruch auf Vollständigkeit, noch auf Richtigkeit!

1 Kapitel 1

Deﬁnitions

The graph is a pair V,E. V is a ﬁnite set and E V a pair of elements in V . V is called the set of vertices ⊆ 2 and E the set of edges. 1 2 5 Visualize: G = (V,E),V = 1, 2, 3, 4, 5 ,E = 1, 2 , 1, 3 , 2, 4 3 { } {{ } { } { }} 4

History: word: Sylvester (1814-1897) and Cayley (1821-1895) Euler - developed graph theory Königsberg bridges (today Kaliningrad in Russia): A A

D C D C

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

Notations:

V Kn= (V, ) - complete graph on n vertices V = n • 2 | | v1 v2 v6 v3

v5 v4 K5 K3 C6 = v1v2v3v4v5v6

Cn - cycle on n vertices • V = v1, v2, . . . , vn ,E = v1, v2 , v2, v3 ,..., vn−1, vn , vn, v1 { } {{ } { } { } { }} n Pn - path on n vertices (Note: P . 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 viP vj and the subpath from vi+1 to • ◦ vj is viP vj.

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

n E10 Km,n

Kn,m= (A B,A B),A B = complete bipartite graph • ∪ × ∩ ∅ Peterson graph: V = {1,2,3,4,5} ,E = i, j , k, l : i, j k, l = • 2 {{{ } { } { } ∩ { } ∅} 1, 2 { }

4, 5 3, 5 3, 4 { } { } { } 1, 3 2, 5 { } { } 1, 4 2, 4 { } { }

2, 3 1, 5 { } { } Kneser Graph K(n, k)= ( V ,E) • k V = n, E = A, B : A, B V and A B = . | | {{ } ∈ k ∩ ∅} V is the set of k-element subsets of V, V = |V | k | k | k Q n- hypercube of dimension n. • {1,2,...,n} Qn = 2 ,E ,E = A, B : A B = 1 (A B := (A B) (A B)) { } {{ } | 4 | } 4 ∪ − ∩ 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) {∅ { }} { } { } { (1,1,1) }

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

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

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

Qn : (1,..., 1) n n 1 weight n-1 −

n n 2 1001 0001 n 3 Qn 1 n weight 2 Qn 1 − 2 weight 1 (# 1’s in a binary tuple) − n (0,..., 0)

3 Parameter: Let G = (V,E) be a graph. The order of G ist the number of vertices ( V ) and the size of G | | is the number of edges ( E ). | | If the order of G is n, then 0 size(G) n . ≤ ≤ 2 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. × { } v1 0 1 1 0 1 0 1 1 For v3 v2 v4 A =  . 1 1 0 0 0 1 0 0     Subgraph: H G, H = (V 0,E0),G = (V,E),V 0 V,E0 E ⊆ ⊆ ⊆ v1 v1 v3 v2 v4 v3 v2

H H is an induced subgraph of G if H G and for v1, v2 V (H): v1, v2 E(H) v1, v2 E(G). ind⊆ ⊆ ∈ { } ∈ ⇔ { } ∈ In the upper example it is no induced subgraph. An induced subgraph is obtained from G by deleting vertices. E.g.:

v1 v1 v1 v1

v v3 v2 v4 v3 2 v2 v4 v3 v2 v4 v3 v4

Let G = (V,E) and G0 = (V 0,E0) be graphs. Then we deﬁne G G0 := (V C0,E E0) and G G0 := ∪ ∪ ∪ ∩ (V C0,E E0). ∩ ∩ G[X] := (X, x, y : x, y X, x, y E(G) ) is called the subgraph of G induced by a vertex set {{ } ∈ { } ∈ } X V (G). E.g.: ⊆ G 2 G[ 1, 2, 3, 4 ] 2 { }

4 1 3 5 4 1 3 A degree d(v) = deg v of a vertex is the number of edges incident to that vertex. v1 v v3 2 v4 deg v1 = 2, deg v2 = 3, deg v3 = 2, deg v4 = 1 In this example the degree sequence is (2, 3, 2, 1), the minimum degree δ(G) is 1 and the maximum degree ∆(G) is 3. n 1 Apparently E(G) = 2 deg vi is true. | | i=1 n P Thus deg vi is even and therefore the number of vertices with odd degree is even. i=1 P n 1 d(G) := n deg vi is called the average degree of G. i=1 P

4 2 Extremal graph theorem: We’ll prove that if G has n vertices and > n edges G has a 4 ⇒ triangle. l m

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 a graph is a connected component. A connected graph without cycles is called a tree. A graph without cycles (acyclic graph) is called a forest. 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) and a 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 a neighbor 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. { ∈ { } { } ∈ } δ ≥ xk xi x0 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 0 := T v. T 0 = n 1 < n. − ⇒T 0 is | acyclic.| − 0 0 0 Let v , w T . P v = v0, v1, . . . , vn = w T . ∈ ∃ ⊆ To show: vi = v for all i = 0, . . . , n 6 0 0 v0, vn = v because v0, vn T , v T 6 ∈ 6∈ vi = v (i = 1, . . . , n 1) because dT (vi) 2, vi is not a leaf. 6 − ≥ 5 0 0 P T connecting v0 and w T connected T is a tree. ⇒ ⊆ ⇒ ⇒ With induction hypothesis T 0 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 all n = 0, . . . , n 1. Compared to a path it is allowed to pass edges and vertices more than once. If − v0 = 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 is W . Then W is a path. If not, W has a repeated 0 vertex W = ue0v1e1 ... vi ... vi . . . v, then W = W1W2 is a shorter u-v-walk.

=:W1 =:W˜ =:W2

- If G has an| odd{z closed} | {zwalk} | {z (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 ... vi . . . vi . . . vn = v0 .

1’st part of W2 W1 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 | {z } | {z } | {z } 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 G contains a cycle. proof: Induction on # vertices. Basis:

Step: W = v0e0v1 ... vi . . . vi . . . vn = v0

1’st part of W2 W1 2’nd part of W2 (note, there is a repeated vertex vj, otherwise W is a cycle) | {z } | {z } | {z } 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 v2 e2 e1

v6 = v3 e5 v1 = v5 = v9

e3 e4 e6 e8 v4 e v7 7 v8 Theorem: A connected graph G has an Eulerian tour iﬀ (i.e. if and only if) each degree of vertex in G is even.

6 proof:

„ “: If there is an Eulerian tour then clearly the number of edges entering the vertex is the number ⇒ of 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. W is a closed walk. If all edges are in W , done. Otherwise, there is an edge e, not in W . Since G is connected, 0 there is such e incident to a vertex in W . Say e = viu. Then W = ueviW vi is a longer walk with 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).≥ ⇒ − G G

C C

Connectivity: We say that a Graph G is vertex k-connected if V (G) > k and deleting any (k 1) vertices does not disconnect the graph. | | − Any connected graph is 1-connected. If a graph is 2-connected then there exists no cut-vertex which is 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 { } v1 e.g.: κ( v3 v2 v4) = 1, κ(Cn) = 2, κ(Kn,m) = min m, n . { } G is called l-edge connected if G = En and G does not become disconnected after deleting any (l 1) edges. 6 − λ(G) (= κ0(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) .

K100 K100 Clearly λ(G) δ(G). But it could be that 1 = λ(G) << δ(G) = 99 . ≤ 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 of e is in S, another is in S¯ := V S. ∃ ⊆ ∀ ∈ − E˜

S S¯

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 y (i.e. xy E(G)). ∃ ∈ ∈ 6∼ 6∈ 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.

d ≤ d d ≤ ≤ v1 v2 v3 vn

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 with ⇐ partite 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, u0 U and u, u0 E(G). ∈ { } ∈ Consider a walk formed by shortest x-u-path, shortest x-u0-path and u, u0.

u u0

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 has a matching containing all vertices of A if and only if N(S) S for any S A. | | ≥ | | ⊆

S A S bad N(S) B N(S) proof:

9 A S „ “: obvious ⇒ B N(S) „ “: 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 = A. | | ≥ | | ⊂ 6 Let x, y =: e E(G). Consider G0 = G x, y . { } ∈ − { } NG (S) NG(S) 1 S + 1 1 = S , for any S A y . | 0 | ≥ | | − ≥ | | − | | ⊆ − { } B x

A y S Thus, Hall’s condition is true for G0, and there is a matching M 0, containing all elements of A y , by induction. − { } So, M 0 x, y is a matching saturating A in G. ∪ { } B x

y A M 0

Case 2: S A, S = A such that N(S) = S . ∃ ⊂ 6 | | | | N(S) B N(S0)

A S S0 By induction, there is a matching containing all vertices of S. Let apply induction to G[A S,B N(S)]. Assume− that− there is S0 A S such that N(S0) (B N(S)) < S0 . Then N(S0 S) = N(S⊆) −(N(S0) (B |N(S)))∩<= −S + S0 |. | | | ∪ | | ∪ ∩ − | 6 | | | | A contradiction to Hall’s condition applied to S S0. ∪ Thus for any S0 A S, N(S0) (B N(S)) S0 , and there is a matching saturating A S in G[A ⊆S,B− N|(S)]. Together∩ − with| a ≥ matching | | between S and N(S), it gives a 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 some ﬁxed 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 a graph. Here k-regular is a graph with all degrees equal to k.

G has partite sets A, B :

proof:

1) Construct G0.

B C

Non-bipartite graphs:

A k-factor in a graph is a spanning (containing each vertex) subgraph in which each vertex has degree k. 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. set of vertices such that each edge is incident to some of this vertices, i.e. a set X of vertices such that G 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

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 path ending 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 the vertex 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 u0 N(v) u , such that f := v, u0 / E(M). ∈ − { } Let W 0 be{ a cover} ∈ of G f of size ν(G f) = ν(G). Then W 0 does not contain u (W 0 contains vertices of M only and u− / V (M)). Thus−W 0 contains v. So, W 0 covers f too. Thus W 0 covers G. ∈

v v 2 1 f

w u’ u

M M = ν(G) | |

Tutte’s theorem odd

S 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 edge ⇐ one-by-one, so the resulting∀ ⊆ graph G0 is no≤ | 1-factor.| We shall show that in G0 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, M1 M2 = (M1 M2) (M1 M2) are only cycles. 4 ∪ − ∩

13 Let S be a set of vertices of degree V (G) 1. We shall show that S is bad in G0. | | − Claim: All components of G0 S are complete. − Assume not, i.e. there is a non-complete component in G0 S. c − c S b b a d a

Then there is an induced path a, b, c in this component. Since b / S, deg b < V (G0) 1, there is d / a, b, c , such that b d. ∈ | | − By maximality∈ { of }G0, there is a6∼ perfect matching M, in G0 a, c , there is a perfect 0 ∪ {{ }} matching M2 in G b, d . Note ac E(M1), bd E(M2). We shall create a perfect matching of G0. ∪ {{ }} ∈ ∈ Consider M1 M2, ac, bd E(M1 M2). If ac, bd belong to diﬀerent cycles of M1 M2: bd4 ∈ 4 4 ac

Take the edges of M2 in a component containing ac, take edges of M1 in a component with bd, otherwise take edges of M1.

If ac, bd belong to the same cycle of M1 M2, then a 4 a c b c d or d

A contradiction, since G0 has no 1-factor, so all components of G0 S are complete. Claim − even odd

even S odd

odd

If S is not bad, i.e. q(S) S , we can construct a perfect matching, a contradiction to the fact that G0 has no| perfect| ≤ matching. | | Thus S is bad in G0.

14 even

odd even

odd odd

odd

0 G is obtained from G by deleting edges, so qG(S) qG (S) > S . ≥ 0 | |

1-factor

f-factor 2 1

4 3-factor 3

k-factor - spanning subgraph, all degrees = k f-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). − ≥ B(v) e(v)

A(v) d(v) Replace each vertex of G with 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 spanning → ∪ { } ∀ ∈ ≤ subgraph H of G such that degH (v) = f(v). 1-factor or matching f-factor, f 1. ≈ B(v≡1) A(v ) 1 B(v2) 3 v1 1 v2 ∅ A(v2) 3 2

2 1

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 replacing each vertex v with vertex sets A(v) B(→v), A∪{(v)} = deg(v), B(v) = deg(v) f(v), and for adjacent vertices 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) B(u) for each vertex u. 4 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 to • M, 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 of G is a spanning subgraph of G that is a vertex-disjoint| union| | of copies| of H. H = G =

H = K2 H-factor perfect matching. ≈ k−1 Hajnal & Szemere´di ’70: If G satisﬁes δ(G) n, n˙:k, then G has a Kk-factor. ≥ k Kk Kk Kk Kk

Alon-Yuster ’95: If G satisﬁes δ(G) χ(H)−1 n. Then G contains at least (1 o(1)) n (H is ≥ χ(H) − |V (H)| 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, so that 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 3 G 2

2 1 χ(G) = 3 χ(Kk) = k, χ(C3) = 3, χ(C4) = 2, χ(Km,n) = 2, χ(C2k+1) = 3 There 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 ∈ ∩ ⇒ −

v2 e e v3 A B sep. set : v2, u2 v1 1 2 − { } e5 e1, e5, e4 A { } u1 e3 e4 u B e1, u1 u2 3 { } Note that a separating set must contain A B. Note B0 B and X separates A and B0 ∩X separates A and B. ⊇ ⇒

B0 A = 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-disjoint A-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 that includes the endpoints of P and Q = P + 1. P | | | |

A B

Lets prove this by induction on V (G) B A . | − − | Basis: V (G) B A = 0. | P − − | A B < k

There is an edge between A and B, not adjacent to vertices of P , otherwise V (P ) A < k is | ∩ | 17 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 vertices separating A and B, call it R∈. \ ∩

A B P x R v

Let x be the last vertex of R that also belongs to a path in P call it P . Let B0 = B (V (xP ) V (xR)). P 0 = P P ∪ P x . ∪ note k(\{G; A,} B ∪0 {) k}(G; A, B). By induction, there≥ is a larger set of A-B0-paths, Q0, Q0 P 0 + 1, Q0 contains endpoints of P 0. | | ≥ | |

A P B y

Let y be an endpoint of a path in Q0 in B0 that is not an endpoint of P 0. Case : 1 Case: 2 Case: 3

Q y y Q1 y , Case 1: y B: Take Q =∈Q0 Q Q xP . − { } ∪{ ∪ } path containing x Case 2: y xP : ∈ 0 |{z} Take Q = Q Q Q xR Q1 Q1 yP . − { } ∪ { ∪ } − { } ∪{ ∪ } path containing y Case 3: y xR: ∈ 0 | {z } 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,E0), E0 = e, e˜ : e, e˜ E and e, e˜ are adjacent . {{ } ∈ }

18 e2

v2 e1 e2 v3 e1 e3 e3 e4 e5 v1

v4 e5 e4 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 paths between a and b. outline of proof: Suppose G contains k independent paths between any two vertices, thus we need k vertices to separate 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 G0 = G a, b , then G0 contains (k 2) independent a-b-paths. − { } ≤ − k 2 ≤ − v X a a b b k 2 ≤ −

G0 By corollary 1, we can separate a and b in G0 by k 2 vertices, X. Since V (G) > k, there is v / a, b and v / component≤ − of a in G0 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

a b

19 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 two vertices.

κ(G) = 1 blocks block-cut-vertex tree. A block - either a bridge or maximal 2-connected subgraph. v1 v2 v3 B1 B2

B1 B B3 2 B4 ... B5 v4 B6 v1 v2 v5

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 and no 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 as ⇒ ear-decomposition.

Observe H G. If u, v V (H), v H u, v G u, then add uv as a ear. If H = G x ind⊆ ∈ 6∼ ∼ 6 ⇒ ∃ ∈ 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 vertex − − of P in H, w1. So wx xP w is an H-ear. A contradiction to maximality of H, so G = H. ∪

20 G

w1 H w x

„ “: 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. ◦ v3 v3

v4 v2 y vxy v2 v4 x

v5 v1 v v5 1 Tutte’s theorem 1961: A graph G is 3-connected if and only if it exists a sequence of graphs G0,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 . ≥ ∼ ◦ { }

x0 x x00 y00 y y0 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. ◦ { } | | ≤

G1 S G2

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

21 S G w 2 G1 v x y

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-disjoint si-ti-paths, i = 1, . . . , k. s t 1 1 s1 t10 = t2 s2 t 2 s2 t20 = t3 s3 t s3 3 t30 = t1 G is k-linked G is k-connected (Menger’s theorem) ⇒ G is f(k)-connected G is k-linked. (Bolloba´s-Thomason ’96) ⇒ 22k |{z}