(b) Let Gand H be graphs withV(G) ={v1, v2, . . . , vn} and V(H) = {u1, u2, . . . , un},n≥3. Two verticesui anduj are adjacent inH if and only if vi and vj belong to a common cycle inG. Characterize those graphsGfor whichH is complete.
8. Prove that ifGis a graph of ordern≥3 with the property that degu+ degv ≥ n for every pair u, v of nonadjacent vertices of G, then G is nonseparable.
9. Prove or disprove: IfB is a block of order 3 or more in a connected graph G, then there is a cycle inB that contains all the vertices ofB.
10. A connected graphGcontains k blocks andℓ cut-vertices. What is the relationship between kandℓ?
11. Prove or disprove: IfGis a connected graph with cut-vertices anduand v are antipodal vertices ofG, then no block ofGcontains bothuandv. 12. (a) Show that for every positive integerk, there exists a connected graph
Gand a non-cut-vertexuofGsuch that rad(G−u) = rad(G) +k.
(b) Prove for every nontrivial connected graph G and every non-cut- vertex vofGthat rad(G−v)≥rad(G)−1.
(c) LetGbe a nontrivial connected graph with rad(G) =r. Among all connected induced subgraphs of Ghaving radiusr, let H be one of minimum order. Prove that rad(H−v) = r−1 for every non-cut- vertex vofH.
62 CHAPTER 2. TREES AND CONNECTIVITY
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G:
v1 v2 v3
v4 v5 v6
v7 v8 v9
v10 v11
v12 v13
Figure 2.4: Bridges and cut-vertices in a graph
Theorem 2.10 An edge ein a graph G is a bridge ofGif and only if e lies on no cycle inG.
Proof. We may assume thatGis connected, for otherwise we can consider a component ofGcontaininge.
First, suppose thate=uvis an edge ofGthat is not a bridge. SinceG−e is connected, there is au−vpathP inG−e. ThenP together witheproduces a cycle inGcontaininge.
For the converse, assume thate=uvis an edge ofGbelonging to a cycle of G. Sinceelies on a cycle ofG, there is au−v pathP′ inGnot containinge.
We show thatG−eis connected and, consequently, thateis not a bridge. Let xandy be two vertices of G. SinceGis connected,Gcontains anx−y path Q. If edoes not lie onQ, then Qis anx−y path inG−eas well. If, on the other hand, elies on Q, then replacing ein Qby the u−v pathP′ produces anx−y walk inG−e. By Theorem 1.16,G−econtains anx−y path. Thus G−eis connected.
We are now prepared to discuss one of the best known and most useful classes of graphs. An acyclic graph has no cycles. A tree is a connected acyclic graph. By Theorem 2.10, a tree is a connected graph, every edge of which is a bridge. Each graph in Figure 2.5 is a tree.
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Figure 2.5: Three trees
Origin of Trees
Trees appeared implicitly in the 1847 work [133] of the German physicist Gustav Kirchhoff in his study of currents in electrical networks, while Arthur Cayley [34] used trees in 1857 to count certain types of chemical compounds.
Trees are important to the understanding of the structure of graphs and are used to systematically visit the vertices of a graph. Trees are also widely used in computer science as a means to organize and utilize data.
The simplest organic chemical molecules are thealkanes. Alkanes are hy- drocarbons and so their molecules consist only of carbon and hydrogen atoms, denoted by the symbols C and H, respectively. The valency of each carbon atom is 4 and of each hydrogen atom is 1, so if an alkane molecule has ncar- bon atoms, then there must be 2n+ 2 hydrogen atoms, producing the formula CnH2n+2 for each alkane molecule. The five simplest alkanes along with their chemical formulas are shown in Figure 2.6. In earlier days, the degree of a vertexv in a graph was sometimes referred to as thevalencyofv.
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H
H H
propane (C3H8)
H
C C C C C C
H
H H
H H
H H
H H H
H H H
H
H H
H
H H H
H
H H
C C
C H H
H H
H H H H
H H
C C C C
C
methane (CH4) ethane (C2H6)
butane (C4H10) isobutane (C4H10) Figure 2.6: The five simplest alkanes
All of the structures in Figure 2.6 are trees. Observe that both butane and isobutane have the same chemical formulaC4H10 but they have a differ- ent structure. (They are not isomorphic.) So it is possible for two different alkanes to have the same chemical formula. In fact, there are three different alkanes having the formulaC5H12and 1, 117, 743, 651, 746, 953, 270 (over one quintillion) different alkanes having the formulaC50H102.
64 CHAPTER 2. TREES AND CONNECTIVITY
Special Trees
There are several well-known classes of trees. For example, the pathsPn
and starsK1,n−1 are trees of ordern≥2. Fort≥2, only one vertex of a star is not a leaf; the vertex of degreet in K1,t is thecentral vertex of K1,t. A tree containing exactly two vertices that are not leaves (which are necessarily adjacent) is called a double star. Thus a double star is a tree of diameter 3. A caterpillaris a tree T of order 3 or more, the removal of whose leaves produces a path (which is called thespineofT). Thus every path, every star (of order at least 3) and every double star is a caterpillar. Figure 2.7 shows four treesT1, T2, T3andT4. The treeT1is a star,T2 is a double star andT3is a caterpillar that is not a double star, whileT4is not a caterpillar.
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Figure 2.7: Four trees
Properties of Trees
Since every tree is connected, every two vertices are connected by a path.
In fact, even more can be said.
Theorem 2.11 A graphGis a tree if and only if every two vertices ofGare connected by a unique path.
Proof. First, suppose thatGis a tree and thatuandvare two vertices ofG.
SinceG is connected,Gcontains at least one u−v path. On the other hand, ifG were to contain twou−v paths, then Gwould contain a cycle, which is impossible. Therefore,Gcontains exactly one u−vpath.
Conversely, letGbe a graph in which every two vertices are connected by a unique path. Certainly then, G is connected. If Gwere to contain a cycle C, then every two vertices on C would be connected by two paths. Thus G contains no cycle and soGis a tree.
While every vertex of degree 2 or more in a tree is a cut-vertex, the vertices of degree 1 (the leaves) are not. These observations provide a corollary of Theorem 2.1.
Corollary 2.12 Every nontrivial tree contains at least two leaves.
For a cut-vertexv ofT, there are degvbranches ofT at v. In the treeT of Figure 2.8, the four branches ofT atv are shown in that figure. In the treeT of Figure 2.8 and in each of the trees T1, T2, T3 and T4 of Figure 2.7, the size of the tree is one less than its order. This is not only true of these trees, it is true of all trees.
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Figure 2.8: The branches of a tree at a vertex
There are several ways of constructing a new tree from a given tree. For example, ifv is a leaf in a treeT, thenT −v is also a tree. If a new vertex is added toT and joined to any vertex of T, then another tree results.
Theorem 2.13 IfT is a tree of order nand sizem, thenm=n−1.
Proof. We proceed by induction on the order of a tree. There is only one tree of order 1, namely K1, and it has no edges. Thus the basis step of the induction is established. Assume that the size of every tree of ordern−1≥1 isn−2 and letT be a tree of ordernand sizem. By Corollary 2.12,T has at least two leaves. Letv be one of them. As we observed,T−vis a tree of order n−1. By the induction hypothesis, the size ofT−v isn−2. Thus the size of T ism= (n−2) + 1 =n−1.
Suppose that a treeTof ordern≥3, sizemand maximum degree ∆(T) = ∆ hasni vertices of degreei(1≤i≤∆). Then
X
v∈V(T)
degv=
∆
X
i=1
ini= 2m= 2n−2 = 2
∆
X
i=1
ni−2. (2.1) Solving (2.1) forn1, we have the following.
Theorem 2.14 LetT be a tree of ordern≥3 having maximum degree∆and containingni vertices of degreei(1≤i≤∆). Then the numbern1of leaves of T is given by
n1= 2 +n3+ 2n4+· · ·+ (∆−2)n∆.
66 CHAPTER 2. TREES AND CONNECTIVITY If s : d1, d2, . . . , dn is the degree sequence of a tree of order n ≥ 2, then necessarilyPn
i=1di= 2n−2.Here too, more can be said.
Theorem 2.15 A sequence s:d1, d2, . . . , dn of n≥2 positive integers is the degree sequence of a tree of order nif and only if
n
X
i=1
di= 2n−2.
Proof. First, let T be a tree of order n and size m with degree sequence s:d1, d2, . . . , dn. Then, as observed,
n
X
i=1
di= 2m= 2(n−1) = 2n−2.
We verify the converse by induction onn. When n= 2, the only sequence of two positive integers whose sum is 2n−2 = 2 is 1,1, which is the degree sequence of the treeK2.
Assume for a given integern≥3 that any sequence ofn−1 positive integers whose sum is 2(n−2) = 2n−4 is the degree sequence of a tree of ordern−1.
Now letd1, d2, . . . , dnbe a sequence ofnpositive integers whose sum is 2n−2. We show that this is the degree sequence of a tree of ordern. Suppose thatd1≥ d2≥ · · · ≥dn. Since each term is positive andPn
i=1di= 2n−2, it follows that dn−1=dn= 1 and that 2≤d1≤n−1. Henced1−1, d2, . . . , dn−1is a sequence ofn−1 positive integers whose sum is 2(n−1)−2 = 2n−4. By the induction hypothesis, there is a tree T′ of order n−1 with V(T′) = {v1, v2, . . . , vn−1} such that degT′v1=d1−1 and degT′vi =di for 2≤i≤n−1. LetT be the tree of ordern obtained fromT′ by adding a new vertexvn and joining it to v1. The tree T then has the degree sequenced1, d2, . . . , dn.
A graph without cycles is aforest. That is, a forest is an acyclic graph.
Thus each tree is a forest and every component of a forest is a tree. All of the graphsF1,F2 andF3 in Figure 2.9 are forests but none are trees.
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F1: F2: F3:
Figure 2.9: Forests
The following is an immediate corollary of Theorem 2.13.
Corollary 2.16 The size of a forest of ordernhavingkcomponents isn−k.
By Theorem 2.13, if G is a graph of order n and size m such that G is connected and has no cycles (that is,Gis a tree), thenm=n−1. It is easy to see that the converse of this statement is not true. However, if we were to add to the hypothesis of the converse either of the two defining properties of a tree, then the converse would be true.
Theorem 2.17 Let G be a graph of order n and size m. If G has no cycles andm=n−1, thenGis a tree.
Proof. It remains only to show thatG is connected. Suppose that the com- ponents of G are G1, G2, . . . , Gk, where k ≥ 1. Let ni be the order of Gi
(1≤i≤k) and mi the size ofGi. Since each graphGi is a tree, it follows by Theorem 2.13 thatmi=ni−1 and by Corollary 2.16 that m=n−k. Hence
n−1 =m=
k
X
i=1
mi =
k
X
i=1
(ni−1) =n−k.
Thusk= 1 and soGis connected. Therefore,Gis a tree.
Theorem 2.18 Let G be a graph of order n and size m. If G is connected andm=n−1, thenGis a tree.
Proof. Assume, to the contrary, that there exists some connected graph of ordernand sizem=n−1 that is not a tree. Necessarily then,Gcontains one or more cycles. By successively deleting an edge from a cycle in each resulting subgraph, a tree of ordernand size less thann−1 is obtained. This contradicts Theorem 2.13.
Combining Theorems 2.13, 2.17 and 2.18, we have the following.
Theorem 2.19 Let G be a graph of order n and size m. If G satisfies any two of the following three properties, then Gis a tree:
(1) Gis connected, (2) Ghas no cycles, (3) m=n−1.
As one would anticipate, graphs often contain many subgraphs that are trees. In fact, for each tree of a fixed order, every graph whose vertices have sufficiently large degree contains a subgraph that is isomorphic to the tree.
Theorem 2.20 Let T be a tree of order k. If Gis a graph for whichδ(G)≥ k−1, thenGcontains a subgraph that is isomorphic to T.
Proof. We proceed by induction onk. The result is obvious fork= 1 since K1 is a subgraph of every graph and fork= 2 sinceK2is a subgraph of every nonempty graph.
Assume for every treeT′ of orderk−1 withk≥3 and for every graphG′ withδ(G′)≥k−2, thatG′ contains a subgraph that is isomorphic toT′. Now,
68 CHAPTER 2. TREES AND CONNECTIVITY letT be a tree of orderk and let Gbe a graph with δ(G)≥k−1. We show thatGcontains a subgraph that is isomorphic to T.
Letvbe an end-vertex ofT and letube the vertex ofT that is adjacent to v. ThenT−v is a tree of orderk−1. Since δ(G)≥k−1> k−2, it follows by the induction hypothesis thatGcontains a subgraphT′ that is isomorphic to T −v. Let u′ denote the vertex of T′ corresponding to u in T −v. Since degGu′≥k−1 and the order ofT′isk−1, the vertexu′is adjacent to a vertex v′that does not belong toT′. The tree obtained by addingv′ toT′and joining it tou′ is isomorphic toT, completing the proof.
Exercises for Section 2.2
1. LetGbe a connected graph of order 3 or more. Prove that ife=uv is a bridge ofG, then at least one ofuandv is a cut-vertex ofG.
2. A graphGof order 8 has the degree sequences: 3,3,3,1,1,1,1,1. Prove or disprove:Gis a tree.
3. (a) Give an example of a tree of order 8 containing six vertices of degree 1 and two vertices of degree 4.
(b) Find all treesT where 75% of the vertices ofT have degree 1 and the remaining 25% of the vertices have another degree (a fixed degree).
4. A certain treeT of orderncontains only vertices of degree 1 and 3. Show thatT contains (n−2)/2 vertices of degree 3.
5. Draw all forests of order 6.
6. Prove that a graphGis a forest if and only if every induced subgraph of Gcontains a vertex of degree at most 1.
7. Characterize those graphs with the property that every connected sub- graph is an induced subgraph.
8. Prove that every connected graph all of whose vertices have even degrees contains no bridges.
9. LetGbe a connected graph, and lete1 ande2 be two edges ofG. Prove thatG−e1−e2 has three components if and only if bothe1 ande2 are bridges inG.
10. Prove that a 3-regular graph Ghas a cut-vertex if and only if G has a bridge.
11. A tree is calledcentral if its center is K1 and bicentralif its center is K2. Show that every tree is either central or bicentral.
12. Let T be a tree order 3 or more, and letT′ be the tree obtained fromT by deleting its end-vertices.
(a) Show that diam(T) = diam(T′) + 2, rad(T) = rad(T′) + 1 and Cen(T) = Cen(T′).
(b) Show that a treeT is central or bicentral (see Exercise 11) according to whether diam(T) = 2 rad(T) or diam(T) = 2 rad(T)−1.
13. Let Gbe a connected graph of order nand sizemsuch that V(G)={v1, v2, . . .,vn}, wherevi belongs tob(vi) blocks (1≤i≤n).
(a) Show thatPn
i=1b(vi)≤2m. (b) Show thatPn
i=1b(vi) = 2mif and only ifGis a tree.
14. Determine all treesT such thatT is also a tree.
15. Prove that ifT is a tree of order at least 3, thenT contains a cut-vertex v such that every vertex adjacent to v, with at most one exception, is an end-vertex.
16. Let T be a tree of order nwith degree sequenced1, d2, . . . , dn such that d1 ≥ d2 ≥ · · · ≥ dn. Prove that di ≤ n−1
i
for each integer i with 1≤i≤n.
17. Let G be a connected graph that is not a tree containing two distinct verticesu and v such thatG−uand G−v are both trees. Show that degu= degv.
18. Show that there exists no tree T containing two distinct edges e1 and e2 such that the two components of T−e1 are isomorphic andthe two components ofT−e2 are isomorphic.
19. Let T be a tree of ordern. Prove thatT is isomorphic to a subgraph of Cn+2.
20. Prove that if T is a tree of order n ≥ 2 that is not a star, then T is isomorphic to a subgraph of T.
21. Find all those graphsGof order n≥4 such that the subgraph induced by every three vertices ofGis a tree, or show that no such graph exists.