Graph Theory. Judith Stumpp. 6. November 2013


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1 Graph Theory Judith Stumpp 6. November 2013
2 ei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung Graph Theory vom Wintersemester 2011/2012. Sie wurde gehalten von Prof. Maria xenovich Ph.D.. Der Mitschrieb erhebt weder nspruch auf Vollständigkeit, noch auf Richtigkeit! 1
3 Kapitel 1 Definitions The graph is a pair V, E. V is a finite set and E ( V 2) 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}} History: word: Sylvester ( ) and Cayley ( ) Euler  developed graph theory Königsberg bridges (today Kaliningrad in Russia): D C D C Problem: Travel through each bridge once, come back to the original point. Impossible! Notations: K n = (V, ( V 2) )  complete graph on n vertices V = n v 2 v 6 v 3 K 5 K 3 v 5 v 4 C 6 = v 2 v 3 v 4 v 5 v 6 C n  cycle on n vertices V = {, v 2,..., v n }, E = {{, v 2 }, {v 2, v 3 },..., {v n 1, v n }, {v n, }} P n  path on n vertices (Note: P n. path on n edges (Diestel)) V = {, v 2,..., v n }, E = {{, v 2 }, {v 2, v 3 },..., {v n 1, v n }} 2
4 Let P be a path from to v n. The subpath of P from v i to v j is v i P v j and the subpath from v i+1 to v j is v i P v j. E n = (V, ), V = n isolated vertices m n E 10 K m,n K n,m = (, ), = complete bipartite graph Peterson graph: V = ( ) {1,2,3,4,5}, E = {{{i, j}, {k, l} : {i, j} {k, l} = } {1, 2} 2 {4, 5} {3, 5} {1, 3} {2, 5} {1, 4} {2, 4} {3, 4} {2, 3} {1, 5} Kneser Graph K(n, k)= ( ( V k), E) V = n, E = {{, } :, ( V k) and = }. ( V ) ( k is the set of kelement subsets of V, V ( k) = V ) k Q n  hypercube of dimension n. Q n = {2 {1,2,...,n}, E}, E = {{, } : = 1} ( := ( ) ( )) V  set of binary ntuples E  pairs of binary tuples different in 1 position Q 1 Q 2 Q 3 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) (1,0) (1,1,0) (1,0,0) (1,0,1) (0,1,0) (0,1,1) (0,0,1) Q n : ( n n 1 ) 0 (1,..., 1) weight n1 (0,0) (0,0,0) ( n ) n 2 ( n 3) ( n 2) n (0,..., 0) weight 2 weight 1 (# 1 s in a binary tuple) Q n 1 Q n 1 3
5 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 of G = ({,..., v n }, E) which is called the adjacent matrix For v 3 v 2 v 4 = Subgraph: H G, H = (V, E ), G = (V, E), V V, E E v 3 v 2 v 4 v 3 v 2 H ind H is an induced subgraph of G if H G and for, v 2 V (H): {, v 2 } E(H) {, v 2 } E(G). In the upper example it is no induced subgraph. n induced subgraph is obtained from G by deleting vertices. E.g.: v 2 v 3 v 4 v 3 v 2 v 2 v 4 v 3 v 2 v 4 v3 v 4 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 set X V (G). E.g.: G 2 G[{1, 2, 3, 4}] degree d(v) = deg v of a vertex is the number of edges incident to that vertex. v 3 v 2 v 4 deg = 2, deg v 2 = 3, deg v 3 = 2, deg v 4 = 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 pparently E(G) = 1 2 deg v i is true. i=1 Thus n deg v i is even and therefore the number of vertices with odd degree is even. i=1 d(g) := 1 n n deg v i is called the average degree of G. i=1 4
6 n Extremal graph theorem: We ll prove that if G has n vertices and > 2 edges G has a 4 triangle. Let, V, =. P is an path if P =... v k, V (P ) = { } and V (P ) = {v k }. graph is connected if any two vertices are linked by a path. maximal connected subgraph of a graph is a connected component. connected graph without cycles is called a tree. 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 = (x 0,..., x k ) be a longest path in G. Then all neighbors of x k 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} : {x k, x j } E}. Then x i x k x k 1... x i is a cycle of length at least δ + 1. δ x k x i x 0 The girth of a graph G is the length of a smallest cycle in G. The distance d G (v, w) of v, w G is the length of the smallest path between them (min = ). The diameter of G is max{d G (v, w) : v, w G}. Proposition: Every nontrivial tree T has a leaf. proof: ssume T has no leaves. T has no isolated vertices δ(t ) 2 C n T  tree T of order n 1 has n 1 edges. proof: T = K 1 ssume 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 = v 0,,..., v n = w T. To show: v i v for all i = 0,..., n v 0, v n v because v 0, v n T, v T v i v (i = 1,..., n 1) because d T (v i ) 2, v i is not a leaf. 5
7 P T connecting v 0 and w T connected T is a tree. With induction hypothesis T has (n 1) 1 edges. Thus T has (n 1) = n 1 edges. walk is an alternating sequence v 0 e 0 e 1... v n of vertices and edges so that e i = v i v i+1 for all n = 0,..., n 1. Compared to a path it is allowed to pass edges and vertices more than once. If v 0 = v n, then the walk is a closed walk.  If G has a uvwalk (between vertices u, v) G has a uvpath. proof: Consider the shortest walk between u and v is W. Then W is a path. If not, W has a repeated vertex W = ue 0 e 1... v }{{} i... v }{{} i... v, then W }{{} = W 1 W 2 is a shorter uvwalk. =:W 1 =:W 2 =: W  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 v i, W = v 0 e 0... v }{{} i... v i... v }{{} n = v 0. }{{} 1 st part of W 2 W 1 2 nd part of W 2 W is a union of two closed walks W 1 and W 2. Either W 1 or W 2 is an odd closed walk by induction it contains an odd cycle.  If G has a closed walk with a nonrepeated edge W = v 0 e 0... e i... e i is unique, then G contains a cycle. proof: Induction on # vertices. asis: Step: W = v 0 e 0... v }{{} i... v i... v }{{} n = v }{{} 0 1 st part of W 2 W 1 2 nd part of W 2 (note, there is a repeated vertex v j, otherwise W is a cycle) So, W is a union of two closed walks W 1 and W 2 and either W 1 or W 2 has a nonrepeated edge. y induction, that walk contains a cycle. Definition: n Eulerian tour is a closed walk containing all edges of a graph and repeating no edge. e.g.: Eulerian tour e 1 v 2 e 2... e 8 v 9 = in e 2 v 2 e 1 = v 5 = v 9 v 6 = v 3 e 5 e 6 e 3 e 4 v 4 e 8 e 7 v 7 v 8 Theorem: connected graph G has an Eulerian tour iff (i.e. if and only if) each degree of vertex in G is even. 6
8 proof: : : If there is an Eulerian tour then clearly the number of edges entering the vertex is the number of edges leaving the vertex. ssume that each degree is even. Consider a walk with longest number of edges and no repeated edge, W = v 0... v k. Thus, there is no edge incident to v 0 that is not in W. Since deg v 0 is even, v 0 must be v n, 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, there is such e incident to a vertex in W. Say e = v i u. Then W = uev i W v i 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 kconnected if V (G) > k and deleting any (k 1) vertices does not disconnect the graph. ny connected graph is 1connected. If a graph is 2connected then there exists no cutvertex which is a vertex whose deletion disconnects a graph. Trees are not 2connected. If G is connected, X V, G X disconnected X is called a cutset. e.g.: κ( v 3 v 2 κ(g) = max{k : G is kconnected} v 4) = 1, κ(c n ) = 2, κ(k n,m ) = min{m, n}. G is called ledge connected if G E n and G does not become disconnected after deleting any (l 1) edges. λ(g) (= κ (G)) = max{l : G is ledgeconnected} e.g.: λ(tree) = 1, λ(c n ) = 2. If λ(g) = 1 there exists a so called bridge (cut edge). Clearly λ(g) δ(g). ut it could be that 1 = λ(g) << δ(g) = 99 K 100 K 100. 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
9 Let Ẽ be a set of λ edges disconnecting G. Then Ẽ is a cut, i.e. S V : e Ẽ, one endpoint of e is in S, another is in S := V S. Ẽ S S If in G there are all edges between S and S. λ = Ẽ = S S V (G) 1 κ(g). Otherwise x S, y S, x y (i.e. xy 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 Ẽ = λ because N(x) #(edges incident to x) and {z S : z S} {x} #(edges incident to this set). Definition: graph G is ddegenerate if there is a vertex order, v 2,..., v n : N(v i ) {v i+1,..., v n } 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 v 2 v 3 v n 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 1degenerate. 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
10 Lemma: 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 u 1 u 2 v 2... u k v k u 1, where u i U, v i V, 1 i k, U, V are partite sets of G. ssume 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. ssume that u, u U and {u, u } E(G). Consider a walk formed by shortest xupath, shortest xu 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: matching is a graph that is a disjoint (vertex) union of edges. Philip Hall (pr 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,. Then G has a matching containing all vertices of if and only if N(S) S for any S. S N(S) S bad N(S) proof: 9
11 S : obvious N(S) : ssume that N(S) S for any S. We shall proof that there is a matching containing all elements of by induction on. If = 1, clear. ssume that > 1 Case 1: N(S) S + 1, for any S, S. Let {x, y} =: e E(G). Consider G = G {x, y}. N G (S) N G (S) 1 S = S, for any S {y}. x S y Thus, Hall s condition is true for G, and there is a matching M, containing all elements of {y}, by induction. So, M {x, y} is a matching saturating in G. x M y Case 2: S, S such that N(S) = S. N(S) N(S ) S S y induction, there is a matching containing all vertices of S. Let apply induction to G[ S, N(S)]. ssume that there is S S such that N(S ) ( N(S)) < S. Then N(S S) = N(S) (N(S ) ( N(S))) < S + S. contradiction to Hall s condition applied to S S. Thus for any S S, N(S ) ( N(S)) S, and there is a matching saturating S in G[ S, N(S)]. Together with a matching between S and N(S), it gives a matching saturating. 10
12 Corollaries of Hall s theorem: 1) Let G be bipartite with partite sets,, such that N(S) S d for all S, and some fixed positive integer d. Then G contains a matching of size at least d. 2) kregular bipartite graph has a perfect matching, i.e. matching containing all vertices of a graph. Here kregular is a graph with all degrees equal to k. G has partite sets, : E(G) = #edges incident to = k = #edges incident to = k = 3) k regular bipartite graph has a proper kedge coloring. proof: 1) Construct G. C C = d, add all edges between and C. In G N G (S) N G (S) + d S d + d = S. y Hall s theorem, there is a matching in G saturating, 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? #edges from S to is S k = #edges between S and N(S) = q #edges from N(S) to is N(S) k #edges between S and N(S) = q. N(S) k q = S k N(S) S. 11
13 Nonbipartite graphs: kfactor in a graph is a spanning (containing each vertex) subgraph in which each vertex has degree k. perfect matching = 1factor 2factor Denes König (Sep Oct. 1944) Gyula König (Dec pr. 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 alternating path: starts with an unmatched vertex of M (alternating one point in and one in ). Take the longest alternating path. vertex cover: for any element of {a, b} E(M), a, 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). ssume 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
14 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 u N(v) {u}, such that 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 contains vertices of M only and u / V (M)). Thus W contains v. So, W covers f too. Thus W covers G. 1 v 2 f v 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: (ill Tutte May May 2002) graph G has a perfect matching (1factor) 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 1factor. dd edge onebyone, so the resulting graph G is no 1factor. 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 M 1, M 2 are perfect matchings in G, M 1 M 2 = (M 1 M 2 ) (M 1 M 2 ) are only cycles. 13
15 Let S be a set of vertices of degree V (G) 1. We shall show that S is bad in G. Claim: ll components of G S are complete. ssume not, i.e. there is a noncomplete component in G 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 (G ) 1, there is d / {a, b, c}, such that b d. y maximality of G, there is a perfect matching M, in G {{a, c}}, there is a perfect matching M 2 in G {{b, d}}. Note ac E(M 1 ), bd E(M 2 ). We shall create a perfect matching of G. Consider M 1 M 2, ac, bd E(M 1 M 2 ). If ac, bd belong to different cycles of M 1 M 2 : ac bd Take the edges of M 2 in a component containing ac, take edges of M 1 in a component with bd, otherwise take edges of M 1. If ac, bd belong to the same cycle of M 1 M 2, then b d a c or d b a c contradiction, since G has no 1factor, so all components of G 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 G has no perfect matching. Thus S is bad in G. 14
16 even even odd S odd odd odd G is obtained from G by deleting edges, so q G (S) q G (S) > S. 1factor 2 1 ffactor 3factor 3 4 kfactor  spanning subgraph, all degrees = k ffactor: If f : V N, an ffactor is a spanning subgraph H of G such that deg H (v) = f(v). Let e(v)= deg(v) f(v) 0 (excess). (v) e(v) Replace each vertex of G with (v) d(v) K e(v), d(v) For adjacent u and v, put an edge between (u) and (v), such that these edges form a matching. n ffactor, 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 deg H (v) = f(v). 1factor or matching ffactor, f v 2 () (v1) (v 2) (v 2) f( ) = 3, f(v 2 ) = 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 (v) (v), (v) = deg(v), (v) = deg(v) f(v), and for adjacent vertices u, v placing an edge between (u) and (v), so that these edges are disjoint, and placing a 15
17 complete bipartite graph between (u) (u) for each vertex u. Claim: G has an ffactor if and only if T (G, f) has 1factor. proof: ssume that M is an ffactor of G, to create a 1factor in T, take the edges corresponding to M, and take missing edges between (u) and (u) u V. ssume that M is a 1factor in T, create an ffactor in G by deleting (u), u V (G), contracting (u) into a single vertex, u V (G). Hfactor: Given a graph G, and a graph H, such that V (G) : V (H) ( : = divisible). n Hfactor of G is a spanning subgraph of G that is a vertexdisjoint union of copies of H. H = G = H = K 2 Hfactor perfect matching. Hajnal & Szemerédi 70: If G satisfies δ(g) k 1n, n :k, then G has a K k kfactor. K k K k K k K k lonyuster 95: If G satisfies δ(g) χ(h) 1 χ(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 vertexdisjoint.... 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. G χ(g) = 3 χ(k k ) = k, χ(c 3 ) = 3, χ(c 4 ) = 2, χ(k m,n ) = 2, χ(c 2k+1 ) = 3 There are graphs with large V (G) and small χ(g). 16
18 Connectivity:, V (G), path P is a path v 0,,..., v k such that V (P ) = {v 0 }, V (P ) = {v k }. C V E, we say that X separates and if each path contains an element of X. v v is an path v 2 e 1 e 2 e 5 u 1 e 3 u e 4 2 u 3 v 3 sep. set : {v 2, u 2 } {e 1, e 5, e 4 } {e 1, u 1 } Note that a separating set must contain. Note and X separates and X separates and. = Menger s theorem (1927): (Karl Menger Jan Oct. 1985) Let G be a graph,, V (G). Min #vertices separating and = Max #vertexdisjoint paths. proof: ssume that =. Let k = k(g;, ) = min #vertices separating and, k(g;, ) max # vertexdisjoint path (easy). We shall prove that max # vertexdisjoint path k(g;, ) = k by stronger induction: If P is any set of less than k disjoint paths then there is a set Q of disjoint paths that includes the endpoints of P and Q = P + 1. P Lets prove this by induction on V (G). asis: V (G) = 0. P < k There is an edge between and, not adjacent to vertices of P, otherwise V (P ) < k is 17
19 a vertex separating and. Step: We have P, a set of less than k path, vertex disjoint. There is an vpath for v \(V (P )), otherwise V (P ) is a set of less than k vertices separating and, call it R. P x R v Let x be the last vertex of R that also belongs to a path in P call it P. Let = (V (xp ) V (xr)). P = P \{P } {P x}. note k(g;, ) k(g;, ). y induction, there is a larger set of  paths, Q, Q P + 1, Q contains endpoints of P. P y Let y be an endpoint of a path in Q in that is not an endpoint of P. Case : 1 Case: 2 Case: 3 y Q y Q 1 y, Case 1: y : Take Q = Q {Q} }{{} path containing x {Q xp }. Case 2: y xp : Take Q = Q {Q} {Q xr} {Q 1 } }{{} path containing y {Q 1 yp }. Case 3: y xr: Take Q = Q {Q} {Q xp } {Q 1 } {Q 1 yr}. If G = (V, E) a graph, then a line graph L(G) of G is a graph L(G) = (E, E ), E = {{e, ẽ} : e, ẽ E and e, ẽ are adjacent}. 18
20 e 2 v 2 e e 1 2 v 3 e 1 e 3 e 4 e 5 e 3 v 4 e 5 e 4 Corollary 1: If a, b V (G), {a, b} / E(G). min #vertices separating a and b = max #independent abpaths (here independent means that they share only a and b) a b pply Menger s theorem to = N(a) and = N(b). Corollary 2: (Global version of Menger s theorem) ny graph G is kconnected 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. ssume that a and b are not connected by k independent paths. y corollary 1 a adjacent to b. Let G = G {a, b}, then G contains (k 2) independent abpaths. a k 2 b a k 2 X b v G y 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. Edgeconnectivity 1) min #edges separating a and b in G = max #edgedisjoint abpaths. a b 19
21 pply Menger s theorem to L(G) with = {edges incident to a}, = {edges incident to b}. 2) Global Menger s theorem (edgeconnected) graph is kedgeconnected if and only if there are k edgedisjoint paths between any two vertices. κ(g) = 1 blocks blockcutvertex tree. block  either a bridge or maximal 2connected subgraph. v 2 v 3 5 v v v 2... i v j if v j V ( i ). ny two block intersect by at most 1 vertex. lockcutvertex graph is a tree. block that is a leaf in a blockcutvertex tree is a block leaf. κ(g) 2 G can be constructed using eardecomposition G is created using eardecomposition if there is a sequence of graphs G 0 G 1... G, such that G 0 is a cycle, G i+1 is created from G i by adding a G i path (ear) (i.e. a path with endpoints in G i and no other vertices in G i ). outline of proof: G i : κ(g) = 2: We have that G has a cycle. Consider the largest subgraph H of G that is built as eardecomposition. 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 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, w 1. So wx xp w is an Hear. contradiction to maximality of H, so G = H. 20
22 G H w w 1 x : Show that an eardecomposition is 2connected... 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. v 3 v 3 v 2 x y v 4 v 2 v xy v 4 v 5 v 5 Tutte s theorem 1961: graph G is 3connected if and only if it exists a sequence of graphs G 0, G 1,..., G n, such that G 0 = K 4, G n = G, G i+1 is obtained from G i : G i+1 has two vertices x, y of degree 3, x y and G i = G i+1 {x, y}. x x x y y y Lemma: If G is 3connected, then there exists an edge e such that G e is 3connected. (without proof) proof: We want to prove hat if G i is 3connected, then G i+1 is also 3connected. ssume not, i.e. G i = G i+1 {x, y} and G i+1 is not 3connected, i.e. there exists a cutset S with S 2. G 1 S G 2 Let G 1 and G 2 be connected components of G i+1 S. Observe, {x, y} = S, otherwise G i is not 3connected. ut {x, y} S, otherwise G i is not 3connected (disconnected by S). So, w.l.o.g. (without loss of generality) x S, y V (G 2 ). 21
23 S G w 2 G 1 v x y G i+1 > G i ssume that there exists a vertex v V (G 2 )\{y}, then in G i {w, v xy } separates v from V (G 1 ), a contradiction. So V (G 2 ) = {y}, so deg(y) 2, a contradiction. graph G is klinked, if for any distinct vertices s 1, s 2,..., s k, t 1, t 2,..., t k, there are vertexdisjoint s i t i paths, i = 1,..., k. s 1 t 1 s 1 t 1 = t 2 s 2 t 2 s 2 t 2 = t 3 s 3 t 3 s 3 G is klinked G is kconnected (Menger s theorem) G is f(k) connected G is klinked. (ollobásthomason 96) }{{} 22k t 3 = t 1 22
(a) (b) (c) Figure 1 : Graphs, multigraphs and digraphs. If the vertices of the leftmost figure are labelled {1, 2, 3, 4} in clockwise order from
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