Powers of Graphs and Line Graphs - BUKU GRAPHS & DIGRAPH PDF

A number of graph operations have been defined and studied that have led to several results dealing with Hamiltonian and Eulerian properties. One of the simplest operations is that of the subdivision graph of a graph. The subdivision graph S(G) of a graphGis that graph obtained fromGby replacing each edgee=uvofGby a new vertexweand the two new edgesuweandvwe. The subdivision graph ofK₄ is shown in Figure 3.15.

IfGis a graph of ordernand size m, then the order ofS(G) isn+mand its size is 2m. Furthermore, S(G) is a bipartite graph with partite sets V(G) andV(S(G))−V(G).

138 CHAPTER 3. EULERIAN AND HAMILTONIAN GRAPHS

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K₄: S(K₄) :

Figure 3.15: The subdivision graph of a graph

The Square and Cube of a Graph

Associated with each connected graph of ordernand diameterdis a class of graphs defined in terms of distance. For each positive integer k, the kth powerG^k of a graphGis that graph withV(G^k) =V(G) anduv∈E(G^k) if and only if 1≤dG(u, v)≤k. ThusG¹ =GandG^k =Kn ifk≥d. The graph G² is also called thesquare ofGwhile G³ is called the cubeof G. A graph with its square and cube are shown in Figure 3.16.

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G³: G²:

Figure 3.16: A graph and its square and cube

If G is a nontrivial connected graph of order n and k is an integer with 1≤k≤n−1, thenG^k isk-connected (see Exercise 4). Since thekth powerG^k (k≥2) of a connected graphGcontainsGas a subgraph (as a proper subgraph if G is not complete), it follows that G^k is Hamiltonian if G is Hamiltonian.

WhetherGis Hamiltonian or not, for a connected graphGof order at least 3 and for a sufficiently large integerk, the graphG^k is Hamiltonian sinceG^d is complete ifGhas diameterd. It is therefore natural to ask for the minimumk for whichG^k is Hamiltonian. Certainly, for connected graphs in general,k= 2 will not suffice since ifG=S(K₁,3) is the graph of Figure 3.16, thenG² is not Hamiltonian. The graphG³isHamiltonian, however.

In fact, the cube of every connected graph of order at least 3 is Hamiltonian.

Indeed, a stronger result exists, discovered by Milan Sekanina [202] and later, but independently, by Jerome Karaganis [129].

Theorem 3.26 IfGis a connected graph, thenG³is Hamiltonian-connected.

Proof. IfH is a spanning tree ofGandH³ is Hamiltonian-connected, then G³is Hamiltonian-connected. Hence it suffices to prove that the cube of every tree is Hamiltonian-connected. To show this, we proceed by induction on n, the order of the tree. For small values ofnthe result is obvious.

Assume for all treesHof order less thannthatH³is Hamiltonian-connected and letT be a tree of ordern. Letuandvbe any two vertices ofT. We consider two cases.

Case1. uandvare adjacent inT. Lete=uv, and consider the forestT−e. This forest has two components, one tree Tu containing u and the other tree Tv containingv. By hypothesis,Tu³andTv³are Hamiltonian-connected. Letu₁ be any vertex ofTu adjacent tou, and let v₁ be any vertex ofTv adjacent to v. IfTu orTv is trivial, we defineu₁=uor v₁=v, respectively. Note thatu₁ andv₁are adjacent inT³sincedT(u₁, v₁)≤3. LetPube a Hamiltonianu−u₁ path (which may be trivial) ofTu³ and letPv be a Hamiltonianv₁−v path of Tv³. The path formed by beginning withPufollowed by the edgeu₁v₁ and then the pathPv is a Hamiltonianu−vpath ofT³.

Case 2. u and v are not adjacent in T. Since T is a tree, there exists a unique path between every two of its vertices. LetP be the uniqueu−vpath of T, and let f = uw be the edge of P incident with u. The graph T −f consists of two trees, one treeTu containinguand the other treeTwcontaining w. By hypothesis, there exists a Hamiltonian w−v path Pw in Tw³. Let u₁ be a vertex ofTu adjacent to u, or let u₁ =uifTu is trivial, and let Pu be a Hamiltonianu−u₁path inTu³. BecausedT(u₁, w)≤2, the edgeu₁wis present in T³. Hence the path formed by starting withPu followed by u₁w and then Pwis a Hamiltonianu−vpath ofT³.

It is, of course, an immediate consequence of Theorem 3.26 that for any connected graph G of order at least 3, its cube G³ is Hamiltonian. Although it is not true that the square of every connected graph of order at least 3 is Hamiltonian, it was conjectured independently by Crispin Nash-Williams and Michael D. Plummer that for 2-connected graphs this is the case. In 1974, Her- bert Fleischner [82] verified this conjecture. More recent proofs of this theorem have been given by by Stanislav ˇRiha [183] in 1991 and Agelos Georgakopoulos [93] in 2009.

Theorem 3.27 If G is a2-connected graph, thenG² is Hamiltonian.

A variety of results strengthening (but employing) Fleischner’s work have been obtained. For example, Gary Chartrand, Arthur Hobbs, Heinz Jung, S. F.

Kapoor and Crispin Nash-Williams [37] showed that the square of a 2-connected graph is Hamiltonian-connected.

Theorem 3.28 If G is a2-connected graph, thenG²is Hamiltonian-connected.

Proof. SinceG is 2-connected, G has order at least 3. Let uand v be any two vertices ofG. LetG₁, G₂, . . . , G₅be five distinct copies ofGand letuiand

140 CHAPTER 3. EULERIAN AND HAMILTONIAN GRAPHS vi (i= 1,2, . . . ,5) be the vertices inGi corresponding touandvinG. Form a new graphF by adding to the unionG₁+G₂+· · ·+G₅two new verticesw₁and w₂and ten new edgesw₁ui andw₂vi(i= 1,2, . . . ,5). The graphF is shown in Figure 3.17. Clearly, neitherw₁norw₂is a cut-vertex ofF. Furthermore, since each graphGi is 2-connected and contains two vertices adjacent to vertices in V(F)−V(Gⁱ), no vertex ofGi is a cut-vertex ofF. HenceF is 2-connected.

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G₁ G₂ G₃ G₄ G₅

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u₄

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Figure 3.17: The graphF in the proof of Theorem 3.28

By Theorem 3.27,F²has a Hamiltonian cycleC, which, of course, contains w₁ andw₂. At least one of the graphsGi, sayG₁, contains no vertex adjacent tow₁orw₂ onC. Sinceu₁andv₁are the only vertices ofG₁ adjacent onCto vertices not inG₁, it follows thatChas au₁−v₁path containing all vertices of G₁. ThusG²₁ has a Hamiltonianu₁−v₁ path, which implies that G²contains a Hamiltonianu−v path.

The Line Graph of a Graph

The most familiar graph operation of a graph is that of the line graph. The line graphL(G) of a graphGis that graph whose vertices can be put in one-to- one correspondence with the edges ofGin such a way that two vertices ofL(G) are adjacent if and only if the corresponding edges ofGare adjacent. A graph and its line graph are shown in Figure 3.18 where the vertexui (1≤i≤6) of L(G) corresponds to the edgeei ofG.

It is relatively easy to determine the number of vertices and the number of edges in the line graphL(G) of a graphGin terms of easily computed quantities in G. Indeed, if G is a graph of order n and size m with degree sequence d₁, d₂, . . . , dn andL(G) is a graph of ordern^′ and sizem^′, thenn^′ =m and

m^′=

i=1

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u₆ u₃ e₃

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Figure 3.18: A graph and its line graph

since each edge of L(G) corresponds to a pair of adjacent edges of G. For each nontrivial connected graph G, its line graph L(G) is also connected (see Exercise 10). For the setGof all connected graphs and the setG^′of all nonempty connected graphs, we may think of L as a function then, namelyL :G^′ → G. Hassler Whitney [238] showed that this function is very nearly injective.

Theorem 3.29 Let G₁ and G₂ be nontrivial connected graphs. If L(G₁) ∼

= L(G₂), then G₁∼

=G₂ unless one ofG₁ andG₂ isK₃ and the other isK₁,3. The function L : G^′ → G is not close to being surjective, however. The graphK₁,3is one of many graph that is not isomorphic to the line graph of any graph. Suppose that there is a graphH such that L(H) =K₁,3. Then H is a graph of size 4 containing an edge that is adjacent to the other three edges, no two of which are adjacent to each other. Such a graph H does not exist and soK₁,3is not the line graph of any graph. Indeed, Lowell W. Beineke [10]

obtained the following result.

Theorem 3.30 A graphGis isomorphic to the line graph of some graph if and only if none of the graphs of Figure 3.19is isomorphic to an induced subgraph of G.

We turn to the problem of determining characteristics possessed by a graph that yield certain Hamiltonian properties of its line graph. Frank Harary and Crispin Nash-Williams [112] characterized those graphs having a Hamiltonian line graph. A circuit C in a graphGis called adominating circuitif every edge ofG either belongs toC or is adjacent to an edge of C. Equivalently, a circuitCin a graphGis a dominating circuit if every edge ofGis incident with a vertex ofC.

Theorem 3.31 Let G be a graph without isolated vertices. Then L(G) is Hamiltonian if and only ifG=K₁,ℓfor someℓ≥3orGcontains a dominating circuit.

Proof. IfG=K₁,ℓ for someℓ≥3, then L(G) is Hamiltonian sinceL(G) = Kℓ. Suppose, then, thatGcontains a dominating circuit

C= (v₁, v₂, . . . , vt, v₁).

142 CHAPTER 3. EULERIAN AND HAMILTONIAN GRAPHS

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Figure 3.19: The induced subgraphs not contained in any line graph

It suffices to show that there exists an orderings:e₁, e₂, . . . , emof themedges ofGsuch thatei andei+1are adjacent edges ofG, for 1≤i≤m−1, as aree₁ andem, since such an orderingscorresponds to a Hamiltonian cycle ofL(G).

Begin the orderingsby selecting, in any order, all edges ofGincident withv₁ that are not edges of C, followed by the edgev₁v₂. At each successive vertex vi, 2≤i≤t−1, select, in any order, all edges of G incident with vi that are neither edges of C nor previously selected edges, followed by the edge vivi+1. This process terminates with the edgevt−1vt. The orderingsis completed by adding the edge vtv₁. Since C is a dominating circuit of G, every edge of G appears exactly once in s. Furthermore, consecutive edges of sas well as the first and last edges ofsare adjacent inG.

Conversely, suppose thatGis not a star butL(G) is Hamiltonian. We show that G contains a dominating circuit. Since L(G) is Hamiltonian, there is an orderings:e₁, e₂, . . . , emof themedges of G such thateiandei+1are adjacent edges ofGfor 1≤i≤m−1, as aree₁andem. For 1≤i≤m−1, letvi be the vertex ofGincident with bothei andei+1. (Note that 1≤k6=q≤m−1 does not necessarily imply that vk 6=vq.) Since Gis not a star, there is a smallest integerj₁exceeding 1 such thatvj₁ 6=v₁. The vertexvj₁−1is incident withej₁, the vertex vj1 is incident with ej1 and vj1−1 =v₁. Thus, ej1 =v₁vj1. Next, let j₂ (if it exists) be the smallest integer exceeding j₁ such that vj2 6= vj1. The vertex vj2−1 is incident with ej2, the vertex vj2 is incident with ej2 and vj2−1 =vj1. Thus, ej2 =vj1vj2. Continuing in this fashion, we finally arrive at a vertex vjt such that ejt = vjt−1vjt, wherevj_t = vm−1. Since every edge of Gappears exactly once ins and since 1< j₁< j₂ <· · ·< jt≤m−1, this

construction yields a trail

T= (v₁, ej1, vj1, ej2, vj2, . . . , vjt−1, ejt, vjt =vm−1)

inGwith the properties that (i) every edge ofGis incident with a vertex ofT and (ii) neithere₁noremis an edge ofT.

Letwbe the vertex of Gincident with both e₁ andem. We consider four possible cases.

Case1. w=v₁=vm−1. ThenT itself is a dominating circuit ofG.

Case2. w=v₁ andw6=vm−1. Sinceemis incident with bothwandvm−1, it follows thatem =vm−1w =vm−1v₁. ThusC = (T , e^m, v₁) is a dominating circuit ofG.

Case 3. w= vm−1 andw 6=v₁. Since e₁ is incident with both w and v₁, we have thate₁=wv₁ =vm−1v₁. ThusC= (T , e₁, v₁) is a dominating circuit ofG.

Case4. w6=vm−1 andw6=v₁. Sinceemis incident with bothwandvm−1, it follows thatem=vm−1w. Sincee₁is incident with both wandv₁, we have that e₁ =wv₁. Thus v₁ 6= vm−1, and C = (T , em, w, e₁, v₁) is a dominating circuit ofG.

As a consequence of Theorem 3.31, ifGis either Eulerian or Hamiltonian, then L(G) is Hamiltonian. In fact, successively taking the line graph of a connected graph has some interesting consequences.

For a nonempty graphG, we writeL⁰(G) to denoteGandL¹(G) to denote L(G). For an integer k ≥ 2, the iterated line graph L^k(G) is defined as L(L^k−¹(G)), whereL^k−¹(G) is assumed to be nonempty. The following result is due to Gary Chartrand and Curtiss E. Wall [41].

Theorem 3.32 IfGis a connected graph such thatdegv≥3 for every vertex v of G, thenL²(G)is Hamiltonian.

Proof. Let v be a vertex of G, where degv = r ≥ 3. Then in L(G), the edges incident withv give rise to a subgraphHv, whereHv ∼

=Kr. LetCv be a Hamiltonian cycle inHv. LetH be the spanning subgraph ofL(G) defined by

V(H) =V(L(G)) andE(H) =S

v∈V(G)E(C^v).

Then H is connected and every vertex of H is even. Consequently,H is Eu- lerian and so H is a dominating circuit of L(G). By Theorem 3.31, L²(G) is Hamiltonian.

If G = Pn, where n ≥ 2, then L(G) = Pn−1. Thus Lⁿ⁻¹(G) = P₁ and L^k(G) is not defined fork ≥n. If G=K₁,3, thenL(G) =C₃. If G=Cn for somen≥3, then L(G) =Cn andL^k(Cⁿ) =Cn for every nonnegative integer

144 CHAPTER 3. EULERIAN AND HAMILTONIAN GRAPHS k. Suppose then that G is a connected graph that is not a path, K₁,3 or a cycle. Then ∆(G)≥3. Suppose thatGcontains a vertexuof degree 1. Then Gcontains au−v pathP = (u=u₀, u₁, . . . , uℓ=v) of minimum lengthℓ≥1 where degv ≥ 3. This gives rise to an x₀−x path P^′ of length ℓ in L(G), where x₀ corresponds to u₀u₁, xcorresponds to uℓ−1uℓ, degx₀ = 1 if ℓ ≥ 2, and degx≥3. By successively taking the line graph of G, L(G), L²(G) and so on, we eventually arrive at a graph L^k(G) for some k where there are no end-vertices. If a connected graphH contains no end-vertices but does contain vertices of degree 2, then G contain a y −z path (or y −z cycle if y = z) Q= (y=y₀, y₁, . . . , yt=z) wheret≥2, degyi= 2 for 1≤i≤t−1, degy≥3 and degz≥3. This gives rise to aw−xpathQ^′= (w=w₁, w₂, . . . , wt=x) of lengtht−1 inL(G) wherew₁corresponds to the edgey₀y₁inG,wtcorresponds to the edgeyt−1yt, degwi= 2 for 2≤i≤t−1, degw₁≥3 and degwt≥3. By successively taking the line graph ofH,L(H), L²(H) and so on, we arrive at a graphL^r(H) in which all vertices have degree at least 3. This is illustrated for the graphGof Figure 3.20. It then follows that there is a sufficiently large integer s such that the degree of every vertex ofL^s(G) is at least 3. By the discussion above and Theorem 3.32, we then have the following.

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L²(G) : L³(G) :

L(G) :

Figure 3.20: A graph and iterated line graphs

Theorem 3.33 If Gis a connected graph that is not a path, then there exists an integern₀ such thatL^k(G)is Hamiltonian for every integerk≥n₀.

The Total Graph of a Graph

A graph operation related to the line graph is the total graph. Thetotal graph T(G) of a graph G is that graph whose vertices can be put in one- to-one correspondence with the elements of the set V(G)∪E(G) such that two vertices of T(G) are adjacent if the corresponding elements in G are two adjacent vertices, two adjacent edges or an incident vertex and edge. A graph H and its total graphT(H) are shown in Figure 3.21.

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. T(H) :

H :

Figure 3.21: A graph and its total graph

One might observe thatT(H)∼=G²for the graphH of Figure 3.21 and the graphGof Figure 3.16. We saw thatG²is not Hamiltonian. Thus, for the graph H of Figure 3.21,T(H) is not Hamiltonian. This, in fact, is not surprising for if F is any graph, thenT(F)∼

= [S(F)]², that is, the total graph ofF is the square of the subdivision graph ofF. Since the graphG of Figure 3.16 is isomorphic toS(K₁^,₃) and the graphH of Figure 3.21 is isomorphic toK₁,3, it follows that T(H)∼

=G², whereGis the graph of Figure 3.16.

While the total graph of a nontrivial connected graphGneed not be Hamil- tonian, the same cannot be said forT(T(G)).

Theorem 3.34 If Gis a nontrivial connected graph, then T(T(G))is Hamil- tonian.

Proof. First, we considerT(G). SinceS(G) is a nontrivial connected graph, [S(G)]² is 2-connected and T(G) ∼= [S(G)]², it follows that H = T(G) is 2- connected. HenceS(H) is 2-connected. By Theorem 3.27, [S(H)]² is Hamilto- nian. ThusT(H)∼

= [S(H)]²and so T(T(G)) is Hamiltonian.

Exercises for Section 3.3

1. Determine all those connected graphsGfor whichS(G) is Eulerian.

2. Determine all those connected graphsGfor whichS(G) is Hamiltonian.

3. Show that ifGis a connected graph of ordernand sizem, thenα(S(G)) = munless Gbelongs to a familiar class of graphs.

146 CHAPTER 3. EULERIAN AND HAMILTONIAN GRAPHS 4. Show that if G is a connected graph of ordern≥2 and k is an integer

with 1≤k≤n−1, thenG^k isk-connected.

5. Show that if G is a connected graph of diameterℓ and 1≤k ≤ℓ, then diam(G^k) =⌈ℓ/k⌉.

6. A graphH is called asquare rootof a connected graphGifH²=G. (a) Give an example of a connected graph with two non-isomorphic

square roots.

(b) Give an example of a connected graph with a unique square root.

7. Show that the graphG² of Figure 3.16 is not Hamiltonian.

8. Prove that if v is any vertex of a connected graphGof order at least 4, thenG³−v is Hamiltonian.

9. Prove that ifGis a self-complementary graph of order at least 5, thenG² is Hamiltonian-connected.

10. Prove that the line graph of every nontrivial connected graph is connected.

11. Determine a formula for the number of triangles in the line graph L(G) in terms of quantities inG.

12. Prove thatL(G) is Eulerian ifGis Eulerian.

13. (a) Find a necessary and sufficient condition for a graphGto have the property that G∼=L(G).

(b) Find a necessary and sufficient condition for a graphGto have the property that L(G)∼=L²(G).

14. For each of the following, prove or disprove.

(a) IfGis Hamiltonian, thenG² is Hamiltonian-connected.

(b) IfGis connected andL(G) is Eulerian, thenGis Eulerian.

(d) IfGhas a dominating circuit, thenL(G) has a dominating circuit.

15. Prove that ifG is a connected graph andL³(G) is Eulerian, thenL²(G) is Eulerian.

16. Give an example of a connected graph G such that degv ≥3 for every vertexvof Gbut L(G) is not Hamiltonian.

17. (a) Show that if G is a k-edge-connected graph, k ≥ 2, then L(G) is k-connected.

(b) Show that if G is a k-edge-connected graph, k ≥ 2, then L(G) is (2k−2)-edge-connected.

18. Show that there exists a graphGthat is not isomorphic to the total graph of any graph.

19. (a) Prove that ifGis a nontrivial connected graph, thenT(G²) is Hamil- tonian.

(b) Prove that if G is a nontrivial connected graph, then (T(G))² is Hamiltonian.

Chapter 4 Digraphs

While there are many concepts in digraphs that are analogues to concepts we have encountered in graphs, there are also concepts that are quite unique to digraphs. We consider many of these in this current chapter, beginning with the most studied type of connectedness for digraphs.

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