Each tree of order 3 or more contains at least one vertex whose removal results in a disconnected graph. In fact, every vertex in a tree that is not a leaf has this property. Furthermore, the removal of every edge in a tree results in a dis- connected graph (with exactly two components). On the other hand, no vertex or edge in a nonseparable graph of order 3 or more has this property. Hence, in this sense, nonseparable graphs possess a greater degree of connectedness than trees. We now look at the two most common measures of connectedness of graphs. In the process of doing this, we will encounter some of the best known and most useful theorems dealing with the structure of graphs.
The Connectivity of a Graph
Avertex-cut of a graphGis a set S of vertices of Gsuch that G−S is disconnected. A vertex-cut of minimum cardinality inGis called aminimum vertex-cut of Gand this cardinality is called the vertex-connectivity (or, more simply, the connectivity) ofGand is denoted byκ(G). (The symbol κ is the Greek letterkappa.)
LetS be a minimum vertex-cut of a (noncomplete) connected graphGand letG1,G2, . . ., Gk (k≥2) be the components of G−S. Then the subgraphs Bi =G[V(Gi)∪S] are called the branches ofG at S or the S-branches of G. For the minimum vertex-cutS={u, v} of the graphGof Figure 2.27, the threeS-branches ofGare also shown in that figure.
Complete graphs do not contain vertex-cuts. Indeed, the removal of any proper subset of vertices from a complete graph results in a smaller complete graph. The connectivity of the complete graph of ordernis defined as n−1, that is, κ(Kn) =n−1. In general then, the connectivity κ(G) of a graph
90 CHAPTER 2. TREES AND CONNECTIVITY
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u v
x y
z
v u
w
s t
t
v u v
z w
s
G:
u
Figure 2.27: The branches of a graph atS={u, v}
Gis the smallest number of vertices whose removal from Gresults in either a disconnected graph or a trivial graph. Therefore, for every graphGof ordern,
0≤κ(G)≤n−1.
Thus a graphGhas connectivity 0 if and only if eitherG=K1orGis discon- nected; a graphGhas connectivity 1 if and only ifG=K2orGis a connected graph with cut-vertices; and a graphGhas connectivity 2 or more if and only ifGis a nonseparable graph of order 3 or more.
Often it is more useful to know that a given graphGcannot be disconnected by the removal of a certain number of vertices rather than to know the actual connectivity ofG. A graphGisk-connected, k≥1, if κ(G)≥k. That is,G isk-connected if the removal of fewer thankvertices fromGresults in neither a disconnected nor a trivial graph. In particular, to show that a graphGof order n≥k+ 1 isk-connected, it suffices to show thatG−S is connected for every setS⊆V(G) with|S|=k−1. The 1-connected graphs are then the nontrivial connected graphs, while the 2-connected graphs are the nonseparable graphs of order 3 or more.
The Edge-Connectivity of a Graph
How connected a graph G is can be measured not only in terms of the number of vertices that need to be deleted from Gto arrive at a disconnected or trivial graph but in terms of the number of edges that must be deleted from Gto produce a disconnected or trivial graph.
An edge-cut of a graph G is a subset X of E(G) such that G−X is disconnected. An edge-cut of minimum cardinality inGis aminimum edge- cut and this cardinality is theedge-connectivityof G, which is denoted by λ(G). (The symbol λ is the Greek letterlambda.) The trivial graphK1 does not contain an edge-cut but we define λ(K1) = 0. Therefore, λ(G) is the minimum number of edges whose removal fromGresults in a disconnected or
trivial graph. Since the set of edges incident with any vertex of a graph Gof ordernis an edge-cut ofG, it follows that
0≤λ(G)≤n−1. (2.4)
A graph Gis k-edge-connected, k ≥1, if λ(G)≥ k. That is, G is k-edge- connected if the removal of fewer than k edges from G results in neither a disconnected graph nor a trivial graph. Thus a 1-edge-connected graph is a nontrivial connected graph and a 2-edge-connected graph is a nontrivial con- nected bridgeless graph.
For the graphGof Figure 2.28,κ(G) = 2 and λ(G) = 3. Both{u, v1} and {u, v2}are minimum vertex-cuts, while{e1, e2, e3}is a minimum edge-cut.
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v2 G:
v1 e2
e1
e3
Figure 2.28: Connectivity and edge-connectivity
The edge-connectivity of the complete graphs is given in the next theorem.
Theorem 2.28 For every positive integern, λ(Kn) =n−1.
Proof. Since the edge-connectivity ofK1 is defined to be 0, we may assume thatn≥2. We have observed in (2.4) thatλ(Kn)≤n−1. LetXbe a minimum edge-cut of Kn. Then |X| = λ(Kn) and G−X consists of two components, sayG1andG2. Suppose thatG1 has orderk. ThenG2 has ordern−k. Thus
|X|=k(n−k). Sincek≥1 andn−k≥1, it follows that (k−1)(n−k−1)≥0 and so
(k−1)(n−k−1) =k(n−k)−(n−1)≥0, which implies that
λ(Kn) =|X|=k(n−k)≥n−1.
Therefore,λ(Kn) =n−1.
Relationship of Connectivities and Minimum Degree
That the edge-connectivity of a graph is bounded above by its minimum degree and bounded below by its connectivity was first observed by Hassler Whitney [238].
92 CHAPTER 2. TREES AND CONNECTIVITY Theorem 2.29 For every graphG,
κ(G)≤λ(G)≤δ(G).
Proof. Let G be a graph of order n. If G is disconnected, then κ(G) = λ(G) = 0; while if Gis complete, then κ(G) =λ(G) =δ(G) =n−1. Thus the desired inequalities hold in these two cases. Hence we may assume thatGis a connected graph that is not complete.
SinceG is not complete, δ(G) ≤n−2. Let v be a vertex of Gsuch that degv =δ(G). If the edges incident with v are deleted fromG, then a discon- nected graph is produced. Henceλ(G)≤δ(G)≤n−2.
It remains to show thatκ(G)≤λ(G). LetX be a minimum edge-cut ofG. Then|X|=λ(G)≤n−2. Necessarily,G−X consists of two components, say G1 andG2. Suppose that the order ofG1 isk. Then the order of G2 isn−k, where k≥1 andn−k≥1. Also, every edge in X joins a vertex ofG1 and a vertex of G2. We consider two cases.
Case1. Every vertex of G1 is adjacent to every vertex ofG2. Then|X|= k(n−k). Sincek−1≥0 andn−k−1≥0, it follows that
(k−1)(n−k−1) =k(n−k)−(n−1)≥0 and so
λ(G) =|X|=k(n−k)≥n−1.
This, however, contradictsλ(G)≤n−2 and so Case 1 cannot occur.
Case 2. There exist a vertex uin G1 and a vertex v inG2 such that uv /∈ E(G). We now define a set U of vertices ofG. Lete∈X. Ifeis incident with u, say e=uv′, then the vertex v′ is placed in the setU. If e is not incident withu, saye=u′v′ whereu′ is inG1, then the vertexu′is placed inU. Hence, for every edgee∈X, one of its two incident vertices belongs toU butu, v /∈U. Thus|U| ≤ |X|andU is a vertex-cut. Therefore,
κ(G)≤ |U| ≤ |X|=λ(G), as desired.
The connectivity of a graphGof a given ordernand sizemcan only be so large. For example, ifm < n−1, thenGis disconnected and soκ(G) = 0. On the other hand, ifm ≥n−1, then there is a sharp upper bound for κ(G) in terms of the average degree ofG, which we present next.
Theorem 2.30 If Gis a graph of ordern and sizem≥n−1, then κ(G)≤
2m n
.
Proof. Since the sum of the degrees of the vertices ofG is 2m, the average degree of the vertices of G is 2m/n and so δ(G) ≤ 2m/n. Since δ(G) is an integer,δ(G)≤ ⌊2m/n⌋. By Theorem 2.29, κ(G)≤ ⌊2m/n⌋.
The observation stated in Theorem 2.30 is due to Frank Harary. In the second book ever written on graph theory,Th´eorie des Graphes et Ses Appli- cations, the author Claude Berge [15] wrote (translated into English):
What is the maximum connectivity of a graph with n vertices and medges?
In 1962 Harary [109] answered Berge’s question (see Exercise 13) by showing that for every pairn, m of integers with 2≤n−1 ≤m ≤ n
2
, there exists a graphGof ordernand sizemwithκ(G) =2m
n
.Of course, ifm=n−1≥2, every treeT of ordernhas the desired property since
κ(T) = 2m
n
=
2n−2 n
= 1.
We observed thatκ(G) = 2 andλ(G) = 3 for the graph Gof Figure 2.28.
Since δ(G) = 4, this graph shows that the two inequalities stated in Theo- rem 2.29 can be strict. The first of these inequalities cannot be strict for cubic graphs, however.
Theorem 2.31 For every cubic graphG, κ(G) =λ(G).
Proof. For a cubic graphG, it follows thatκ(G) =λ(G) = 0 if and only if Gis disconnected. Ifκ(G) = 3, then λ(G) = 3 by Theorem 2.29. So two cases remain, namely κ(G) = 1 and κ(G) = 2. Let U be a minimum vertex-cut of G. Then|U|= 1 or|U|= 2. SoG−U is disconnected. LetG1 andG2be two components ofG−U. SinceGis cubic, for eachu∈U, at least one ofG1 and G2 contains exactly one neighbor ofu.
Case 1. κ(G) = |U| = 1. Thus U consists of a cut-vertex u of G. Since some component ofG−U contains exactly one neighbor wof u, the edgeuw is a bridge ofGand soλ(G) =κ(G) = 1.
Case 2. κ(G) = |U| = 2. Let U = {u, v}. Assume that each of uand v has exactly one neighbor, say u′ and v′, respectively, in the same component of G−U. (This is the case that holds ifuv∈E(G).) ThenX ={uu′, vv′} is an edge-cut ofG andλ(G) =κ(G) = 2. (See Figure 2.29(a) for the situation whenuandv are not adjacent.)
Hence we may assume thatuhas one neighboru′ inG1and two neighbors in G2; while v has two neighbors in G1 and one neighbor v′ in G2 (see Fig- ure 2.29(b)). Therefore,uv /∈E(G) andX ={uu′, vv′}is an edge-cut ofG; so λ(G) =κ(G) = 2.
94 CHAPTER 2. TREES AND CONNECTIVITY
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v′ v
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(b) v
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Figure 2.29: A step in the proof of Case 2
As would be expected, the higher the degrees of the vertices of a graph, the more likely it is that the graph has large connectivity. There are several sufficient conditions of this type. We present the simplest of these, due to Gary Chartrand and Frank Harary [36].
Theorem 2.32 Let G be a graph of order n and let k be an integer with 1≤k≤n−1. If
degv≥
n+k−2 2
for every vertex v ofG, thenGis k-connected.
Proof. Suppose that this theorem is false. Then there is a graphGsatisfying the hypothesis of the theorem such thatGis notk-connected. Certainly then, G is not a complete graph. Hence there exists a vertex-cutU of Gsuch that
|U|=ℓ≤k−1. The graphG−U is therefore disconnected of ordern−ℓ.
Let G1 be a component of G−U of smallest order, say n1. Thus n1 ≤
⌊(n−ℓ)/2⌋. Let v be a vertex of G1. Necessarily, v is adjacent in Gonly to the vertices ofU or to other vertices ofG1. Hence
degv ≤ ℓ+ (n1−1)≤ℓ+⌊(n−ℓ)/2⌋ −1
= ⌊(n+ℓ−2)/2⌋ ≤ ⌊(n+k−3)/2⌋, contrary to the hypothesis.
The class ofk-edge-connected graphs is characterized in the following simple but useful theorem.
Theorem 2.33 A nontrivial graphGisk-edge-connected if and only if there exists no nonempty proper subset W of V(G) such that the number of edges joiningW andV(G)−W is less than k.
Proof. First assume that there exists no nonempty proper subsetW ofV(G) such that the number of edges joiningW andV(G)−W is less thankbut that G is not k-edge-connected. Since G is nontrivial, there exist ℓ edges, where 0< ℓ < k, such that their deletion fromGresults in a disconnected graphH.
Let H1 be a component of H. Since the number of edges joining V(H1) and V(G)−V(H1) is at mostℓ, whereℓ < k, this is a contradiction.
Conversely, suppose thatGisk-edge-connected. If there exists a nonempty proper subset W of V(G) such that j edges (j < k) join W and V(G)−W, then the deletion of these j edges produces a disconnected graph, which is impossible.
According to Theorem 2.29,λ(G)≤δ(G) for every graphG. The following theorem of Jan Plesn´ık [173] gives a sufficient condition for equality to hold in this case.
Theorem 2.34 IfGis a connected graph of diameter 2, thenλ(G) =δ(G).
Proof. LetX be an edge-cut with|X|= λ(G) = λ, and let H1 and H2 be the two components ofG−X . Now it cannot occur that bothH1andH2have vertices v1 and v2, respectively, that are not incident with any edge of X, for thend(v1, v2)≥3.
We may assume then that every vertex ofH1is incident with an edge ofX. Thus
n1=|V(H1)| ≤ |X|=λ(G)≤δ(G). (2.5) Since each vertexuinH1 is incident with at mostn1−1 edges inH1, it follows thatuis incident with at leastδ(G)−n1+ 1 edges inX. Consequently,
λ=|X| ≥n1(δ(G)−n1+ 1). (2.6) Ifn1= 1, then (2.6) givesλ(G)≥δ(G). If, however,n1>1, then multiplying the inequalityδ(G)≥n1in (2.5) byn1−1 givesδ(G)(n1−1)≥n1(n1−1) and so
n1(δ(G)−n1+ 1)≥δ(G).
Hence in both cases, we haveλ(G)≥δ(G). Thusλ(G) =δ(G).
If Gis a graph of order n ≥ 3 such that degu+ degv ≥n−1 for every two nonadjacent verticesuandv, then diam(G) = 2. Therefore, the following result of Linda Lesniak [145] is a consequence of Theorem 2.34.
Corollary 2.35 IfG is a graph of ordernsuch that for all distinct nonadja- cent vertices uandv,
degu+ degv≥n−1, thenλ(G) =δ(G).
Exercises for Section 2.4
1. Determine the connectivity and edge-connectivity of each complete k- partite graph.
96 CHAPTER 2. TREES AND CONNECTIVITY 2. Letv1, v2, . . . , vk bek distinct vertices of ak-connected graphG. Let H be the graph formed from Gby adding a new vertexw of degreek that is adjacent to each ofv1, v2, . . . , vk. Show thatκ(H) =k.
3. (a) Prove that if G is a k-connected graph, then G∨K1 is (k+ 1)- connected.
(b) Prove that ifGis ak-edge-connected graph, thenG∨K1is (k+ 1)- edge-connected.
4. Let G be a graph with degree sequence d1, d2, . . . , dn where d1 ≥ d2 ≥
· · · ≥dn. Determineλ(G∨K1).
5. Show that everyk-connected graph contains every tree of orderk+ 1 as a subgraph.
6. (a) Let G be a noncomplete graph of order n and connectivity k such that degv≥(n+ 2k−2)/3 for every vertexv ofG. Show that ifSis a minimum vertex-cut ofG, thenG−Shas exactly two components.
(b) Let G be a noncomplete graph of order n and connectivity k such that degv ≥(n+kt−t)/(t+ 1) for some integert≥2. Show that if S is a vertex-cut of cardinality κ(G), then G−S has at most t components.
7. For a graph G of order n ≥ 2, define the k-connectivity κk(G) of G (2≤k≤n) as the minimum number of vertices whose removal from G results in a graph with at least k components or a graph of order less thank. (Therefore,κ2(G) = κ(G).) A graph G is defined to be (ℓ, k)- connectedifκk(G)≥ℓ. LetGbe a graph of orderncontaining a set of at leastkpairwise nonadjacent vertices. Show that if
degGv≥
n+ (k−1)(ℓ−2) k
for everyv∈V(G), then Gis (ℓ, k)-connected.
8. Let G be a graph of diameter 2. Show that if S is a set of λ(G) edges whose removal disconnectsG, then at least one of the components ofG−S is isomorphic toK1 or toKδ(G).
9. Leta, bandc be positive integers witha≤b≤c. Prove that there exists a graphGwithκ(G) =a,λ(G) =bandδ(G) =c.
10. Verify that Theorem 2.32 is best possible by showing that for each positive integer k, there exists a graph G of order n ≥ k+ 1 such that δ(G) = n+k−3
2
andκ(G)< k.
11. Verify that Theorem 2.34 is best possible by finding an infinite class of connected graphs Gof diameter 3 for which λ(G)6=δ(G).
12. Theconnection numberc(G) of a connected graphGof ordern≥2 is the smallest integer k with 2≤k≤nsuch that everyinduced subgraph of order k in G is connected. State and prove a theorem that gives a relationship between κ(G) andc(G) for a graphGof ordern.
13. For an even integerk≥2, show that the minimum size of ak-connected graph of orderniskn/2.
14. Prove or disprove: LetGbe a nontrivial graph. For every vertexv ofG, κ(G−v) =κ(G) orκ(G−v) =κ(G)−1.
15. (a) Prove that ifGis a k-connected graph ande is an edge of G, then G−eis (k−1)-connected.
(b) Prove that ifGis a k-edge-connected graph and eis an edge ofG, thenG−eis (k−1)-edge-connected.
16. (a) Show that ifGis a 0-regular graph, thenκ(G) =λ(G).
(b) Show that ifGis a 1-regular graph, thenκ(G) =λ(G).
(c) Show that ifGis a 2-regular graph, thenκ(G) =λ(G).
(d) By (a) – (c) and Theorem 2.31, ifG isr-regular, where 0≤r≤3, then κ(G) =λ(G). Find the minimum positive integerr for which there exists an r-regular graphGsuch thatκ(G)6=λ(G).
(e) Find the minimum positive integer r for which there exists an r- regular graphGsuch thatλ(G)≥κ(G) + 2.
17. For a graphG, defineκ(G) = max{κ(H)}andλ(G) = max{λ(H)}, where each maximum is taken over all subgraphsH of G. How are κ(G) and λ(G) related toκ(G) andλ(G), respectively, and to each other?
18. Let G1 and G2 be two k-connected graphs, where k ≥ 2, and let G be the set of all graphs obtained by adding k edges between G1 and G2. Determine max{κ(G) :G∈ G}.