Tournaments - BUKU GRAPHS & DIGRAPH PDF

156 CHAPTER 4. DIGRAPHS 11. Prove that ifDis a connected digraph such thatP

v∈V(D)|odv−idv|= 2t, wheret≥1, thenE(D) can be partitioned into subsetsEi, 1≤i≤t, so that the subgraphG[Eⁱ] induced byEi is an open trail for eachi.

12. Let D be a connected digraph of ordern with V(D) = {v₁, v₂, . . . , vn}. Prove that if odvi≥idvi for 1≤i≤n, thenD is Eulerian.

13. A vertexv in a digraphD is said to bereachable from a vertexuinD ifD contains a u−v path. LetD be a digraph and for each vertexuof D, letR(u) be the set of vertices reachable fromuand letr(u) =|R(u)|. Since u∈R(u) for every vertexuof D, it follows thatr(u)≥1. Prove that if r(x)6=r(y) for every two distinct verticesxand y of D, thenD contains a Hamiltonian path.

14. Corollary 4.9 states: If Dis a strong digraph of ordernsuch thatdegv ≥ n for every vertex v of D, then D is Hamiltonian. Show that if the digraphDis not required to be strong, thenDneed not be Hamiltonian.

15. Show for infinitely many positive integers n that there exists a digraph D of order nsuch that odv≥(n−1)/2 and idv≥(n−1)/2 for every vertex vof Dbut D is not Hamiltonian.

16. Prove that ifD is a nontrivial strong digraph, thenL(D) is strong.

17. Give an example of a Hamiltonian digraph whose line digraph is not Hamiltonian.

18. Give an example of an Eulerian digraph whose line digraph is not Eulerian.

19. Give an example of a digraphDsuch thatDis not Eulerian andL(L(D)) is Hamiltonian.

20. Prove Theorem 4.11:LetDbe a nontrivial connected digraph. ThenL(D) is Hamiltonian if and only if D is Eulerian.

u, vof distinct vertices, exactly one of (u, v) and (v, u) is an arc. A tournament T then models a round robin tournament in which no ties are permitted. The vertices ofT are the teams in the round robin tournament and (u, v) is an arc in T if teamudefeats teamv.

Figure 4.4 shows two tournaments of order 3. In fact, these are the only two tournaments of order 3. The number of non-isomorphic tournaments increases sharply with their orders, however. For example, there is only one tournament of order 1 and one of order 2. As we just observed, the tournamentsT₁andT₂in Figure 4.4 are the only two tournaments of order 3. There are four tournaments of order 4, 12 of order 5, 56 of order 6 and over 154 billion of order 12.

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T₁: T₂:

Figure 4.4: The tournaments of order 3 Since the size of a tournament of ordernis ⁿ₂

, it follows from Theorem 1.23 that

v∈V(T)

odv= X

v∈V(T)

idv= n

Transitive Tournaments

A tournamentT istransitive if whenever (u, v) and (v, w) are arcs ofT, then (u, w) is also an arc ofT. The tournamentT₂ of Figure 4.4 is transitive whileT₁is not. The following result gives an elementary property of transitive tournaments. Anacyclic digraphhas no cycles.

Theorem 4.12 A tournament is transitive if and only if it is acyclic.

Proof. LetTbe an acyclic tournament and suppose that (u, v) and (v, w) are arcs ofT. SinceT is acyclic, (w, u)∈/ E(T). Therefore, (u, w)∈E(T) andT is transitive.

Conversely, suppose thatT is a transitive tournament and assume thatT contains a cycle, say C = (v₁, v₂, . . . , vk, v₁), where k ≥3. Since (v₁, v₂) and (v₂, v₃) are arcs of the transitive tournamentT, it follows that (v₁, v₃) is also an arc of T. Similarly, (v₁, v₄), (v₁, v₅), . . ., (v₁, vk) are arcs of T. However, this contradicts the fact that (vk, v₁) is an arc ofT. Thus,T is acyclic.

Suppose that a tournamentT of order nwith vertex setV(T)={v₁,v₂,. . ., vn}represents a round robin tournament involving competition amongnteams v₁, v₂, . . . , vn. If team vi defeats team vj, then (vi, vj) is an arc of T. The number of victories by team vi is the outdegree of vi. For this reason, the

158 CHAPTER 4. DIGRAPHS outdegree of the vertexviin a tournament is also referred to as thescoreofvi. A sequences₁, s₂, . . . , sn of nonnegative integers is called ascore sequence of a tournamentif there exists a tournamentT of ordernwhose vertices can be labeledv₁, v₂, . . . , vn such that odvi=si fori= 1,2, . . . , n.

Figure 4.5 shows transitive tournaments of ordernforn= 3,4,5. The score sequence of every transitive tournament has an interesting property. The following result describes precisely which sequences are score sequences of transitive tournaments.

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Figure 4.5: Transitive tournaments of orders 3, 4, 5

Theorem 4.13 A nondecreasing sequence π of n nonnegative integers is a score sequence of a transitive tournament of order n if and only if π is the sequence 0,1, . . . , n−1.

Proof. First we show thatπ: 0,1, . . . , n−1 is a score sequence of a transitive tournament of order n. Let T be the tournament with vertex set V(T) = {v₁, v₂, . . . , vn} and arc setE(T) ={(vi, vj) : 1≤i < j ≤n}. We claim that T is transitive. Let (vi, vj) and (vj, vk) be arcs of T. Then i < j < k. Since i < k, (vi, vk) is an arc ofT and soT is transitive. For 1≤i≤n, odvi=n−i. Therefore, a score sequence ofT isπ: 0,1, . . . , n−1.

Next, we show that ifT is a transitive tournament of ordern, then 0,1, . . . , n−1 is a score sequence of T. This is equivalent to showing that every two vertices ofT have distinct scores. Letuandw be two vertices ofT. Assume, without loss of generality, that (u, w) is an arc ofT. LetW be the set of vertices of T to which w is adjacent. Therefore odw =|W|. For eachx∈ W, (w, x) is an arc of T. SinceT is transitive, (u, x) is also an arc ofT. However then, odu≥ |W|+ 1 and so odu6= odw.

The proof of Theorem 4.13 shows that the structure of a transitive tournament is uniquely determined.

Corollary 4.14 For every positive integer n, there is exactly one transitive tournament of order n.

Combining this corollary with Theorem 4.12, we arrive at yet another corollary.

Corollary 4.15 For every positive integern, there is exactly one acyclic tournament of order n.

Although there is only one transitive tournament of each ordern, in a certain sense, which we now explore, every tournament has the structure of a transitive tournament. Let T be a tournament. We define a relation on V(T) by u is related tovif there is both au−vpath and av−upath inT. This relation is an equivalence relation and, as such, this relation partitionsV(T) into equivalence classes V₁, V₂, . . . , Vk (k ≥ 1). Let Si = T[Vⁱ] for i = 1,2, . . . , k. Then each subdigraph Si is a strong tournament and, indeed, is maximal with respect to the property of being strong. The subdigraphsS₁, S₂, . . . , Sk are called the strong componentsof T. So the vertex sets of the strong components ofT produce a partition ofV(T).

LetT be a tournament with strong componentsS₁, S₂, . . . , Sk, and letTede- note that digraph whose verticesu₁, u₂, . . . , ukare in one-to-one correspondence with these strong components (whereuicorresponds toSi,i= 1,2, . . . , k) such that (uⁱ, uj) is an arc ofTe, i6=j, if and only if some vertex ofSi is adjacent to some vertex of Sj. If (ui, uj) is an arc of Te, then because Si and Sj are distinct strong components ofT, it follows that every vertex of Si is adjacent to every vertex of Sj. Hence, Te is obtained by identifying the vertices of Si

fori= 1,2, . . . , k. A tournamentT and its associated digraphTe are shown in Figure 4.6.

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Te: S₁

S₂

S₃

u₂

u₃ u₁

T :

Figure 4.6: A tournamentT and its associated transitive tournamentTe Observe that for the tournamentT of Figure 4.6, Te is itself a tournament, indeed a transitive tournament. That this always occurs follows from Theo- rem 4.16. (See Exercise 5.)

Theorem 4.16 IfT is a tournament with exactlyk strong components, then Teis the transitive tournament of order k.

Since for every tournament T, the tournament Te is transitive, it follows that ifT is a tournament that is not strong, thenV(T) can be partitioned as {V₁, V₂, . . . , Vk} (k≥2) such that T[Vi] is a strong tournament for eachi, and ifvi∈Viandvj ∈Vj, wherei > j, then (vi, vj)∈E(T). This decomposition is often useful when studying the properties of tournaments that are not strong.

We already noted that there are four tournaments of order 4. Of course, one of these is transitive, which consists of four trivial strong components

160 CHAPTER 4. DIGRAPHS S₁, S₂, S₃, S₄, where the vertex ofSiis adjacent to the vertex ofSjif and only if i > j. There are two tournaments of order 4 containing two strong components S₁ and S₂, depending on whether S₁ or S₂ is the strong component of order 3. (No strong component has order 2.) Since there are four tournaments of order 4, there is exactly one strong tournament of order 4. These tournaments are depicted in Figure 4.7. The arcs not drawn in the tournamentsT₁, T₂ and T₃ that are not strong are all directed downward, as indicated by the double arrow.

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S₂

S₁ S₂ S₃ S₄

S₂

S₁

⇓

T₁

⇓ ⇓

T₂ T₃ T₄

S₁

Figure 4.7: The four tournaments of order 4

We also stated that there are 12 tournaments of order 5. There are six tournaments of order 5 that are not strong, shown in Figure 4.8. Again all arcs that are not drawn are directed downward. Thus there are six strong tournaments of order 5.

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T₁ T₂ T₃ T₄ T₅ T₆

Figure 4.8: The six tournaments of order 5 that are not strong

Score Sequences of Tournaments

Theorem 4.13 characterizes score sequences of transitive tournaments. We next investigate score sequences of tournaments in general. We begin with a theorem similar to Theorem 1.12.

Theorem 4.17 A nondecreasing sequence π : s₁, s₂, . . . , sn (n ≥2) of nonnegative integers is a score sequence of a tournament if and only if the sequence π₁:s₁, s₂, . . . , ss_n, ss_n+1−1, . . . , sn−1−1 is a score sequence of a tournament.

Proof. Assume thatπ₁is a score sequence of a tournament. Then there exists a tournamentT₁of ordern−1 havingπ₁as a score sequence. Hence the vertices ofT₁ can be labeled asv₁, v₂, . . . , vn−1 such that

odvi=

si for 1≤i≤sn

si−1 fori > sn.

We construct a tournamentT by adding a vertexvn toT₁wherevn is adjacent to vi if 1 ≤ i≤ sn and vn is adjacent from vi otherwise. The tournamentT then hasπas a score sequence.

For the converse, we assume thatπ is a score sequence. Hence there exist tournaments of ordernwhose score sequence isπ. Among all such tournaments, let T be one such that V(T) = {v₁, v₂, . . . , vn}, odvi = si for i = 1,2, . . . , n and the sum of the scores of the vertices adjacent from vn is minimum. We claim that vn is adjacent to vertices having scoress₁, s₂, . . . , ss_n. Assume, to the contrary, that vn is not adjacent to vertices having scores s₁, s₂, . . . , ss_n. Necessarily, then, there exist vertices vj and vk with j < k andsj < sk such that vn is adjacent to vk and vn is adjacent from vj. Since the score of vk

exceeds the score of vj, there exists a vertex vtsuch that vk is adjacent to vt, andvt is adjacent tovj (Figure 4.9(a)). Thus, a 4-cycleC = (vⁿ, vk, vt, vj, vn) is produced. If we reverse the directions of the arcs of C, a tournamentT^′ is obtained also havingπas a score sequence (Figure 4.9(b)). However, inT^′, the vertex vn is adjacent tovj rather thanvk. Hence the sum of the scores of the vertices adjacent fromvn is smaller inT^′ than inT, which is impossible. Thus, as claimed,vn is adjacent to vertices having scoress₁, s₂, . . . , ssn. ThenT−vn

is a tournament having score sequenceπ₁.

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vj vk

(a) (b)

Figure 4.9: A step in the proof of Theorem 4.22 As an illustration of Theorem 4.17, we consider the sequence

π: 1,2,2,3,3,4.

In this case,sn(actuallys₆) has the value 4; thus, we delete the last term, repeat the firstsn= 4 terms, and subtract 1 from the remaining terms, obtaining

162 CHAPTER 4. DIGRAPHS π₁^′ : 1,2,2,3,2.

Rearranging, we have

π₁: 1,2,2,2,3.

Repeating this process twice more, we have π^′₂ : 1,2,2,1 π₂ : 1,1,2,2 π₃ : 1,1,1.

The sequenceπ₃is clearly a score sequence of a tournament. By Theorem 4.17, π₂ is as well, as are π₁ and π. We can use this information to construct a tournament with score sequenceπ. The sequence π₃ is the score sequence of the tournament T₃ of Figure 4.10. Proceeding from π₃ to π₂, we add a new vertex to T₃ and join it to two vertices of T₃ and from the other, producing a tournament T₂ with score sequence π₂. To proceed from π₂ to π₁, we add a new vertex to T₂ and join it tovertices having scores 1, 2 and 2 and from the remaining vertex of T₂, producing a tournament T₁ with score sequence π₁. Continuing in the same fashion, we finally produce a desired tournamentT with score sequenceπ by adding a new vertex to T₁ and joining ittovertices having scores 1, 2, 2 and 3, and joining itfromthe other vertex.

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1 1

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2 T₂: 1

3 2

1 T₃:

T₁: T :

Figure 4.10: Construction of a tournament with a given score sequence The sociologist Hyman Garshin Landau [143] characterized those sequences of nonnegative integers that are score sequences of tournaments. The proof we present of his theorem is due to Carsten Thomassen [215].

Theorem 4.18 A nondecreasing sequence π : s₁, s₂, . . . , sn of nonnegative integers is a score sequence of a tournament if and only if for each integer k with 1≤k≤n,

i=1

si≥ k

, (4.1)

with equality holding whenk=n.

Proof. Suppose first thatπ:s₁, s₂, . . . , snis a score sequence of a tournament of order n. Then there exists a tournament T with V(T) = {v₁, v₂, . . . , vn} such that odvi =si for i = 1,2, . . . , n. For an integer k with 1 ≤k≤n and S ={v₁, v₂, . . . , vk}, the subdigraph T₁ =T[S] induced by S is a tournament of orderk and size ^k₂

. Since od^Tvi≥od^T1vi for 1≤i≤k, it follows that

i=1

si=

i=1

od^Tvi ≥

i=1

od^T1vi = k

We now verify the converse. Suppose that the converse is false. Then among all counterexamples for which n is minimum, let π : s₁, s₂, . . . , sn be one for whichs₁ is minimum. Suppose first that there exists an integerkwith 1≤k≤n−1 such that

i=1

si= k

. (4.2)

Sincek < n, it follows thatπ₁:s₁, s₂, . . . , skis a score sequence of a tournament T₁ of orderk.

Let τ : t₁, t₂, . . . , tn−k be the sequence, where ti = sk+i −k for i = 1,2, . . . , n−k. Since

k+1

i=1

si≥ k+ 1

it follows from (4.2) that sk+1=

k+1

i=1

si−

i=1

si≥ k+ 1

− k

=k.

Sinceπis a nondecreasing sequence,

ti=sk+i−k≥sk+1−k≥0

for i = 1,2, . . . , n−k and so τ is a nondecreasing sequence of nonnegative integers. We now show thatτ satisfies (4.1).

For each integerr with 1≤r≤n−k, we have

i=1

ti=

i=1

(sk+i−k) =

i=1

sk+i−rk=

r+k

i=1

si−

i=1

si−rk.

164 CHAPTER 4. DIGRAPHS Since

r+k

i=1

si≥ r+k

and

i=1

si= k

, it follows that _r

i=1

ti≥ r+k

− k

−rk= r

with equality holding for r =n−k. Thus τ satisfies (4.1). Sincen−k < n, there is a tournamentT₂of ordern−khaving score sequenceτ.

LetT be the tournament withV(T) =V(T₁)∪V(T₂) and E(T) =E(T₁)∪E(T₂)∪ {(u, v) :u∈V(T₂), v∈V(T₁)}.

Thenπis a score sequence forT, contrary to our assumption. Consequently,

i=1

si>

k 2

fork= 1,2, . . . , n−1. In particular,s₁>0.

We now consider the sequenceπ^′ :s₁−1, s₂, s₃, . . . , sn−1, sn+ 1. Then π^′ is a nondecreasing sequence of nonnegative integers satisfying (4.1). By the minimality ofs₁, there is a tournamentT^′ of ordernhaving score sequenceπ^′. Let xand y be vertices ofT^′ such that od^T^′x=sn+ 1 and od^T^′y =s₁−1.

Since odT′x≥odT′y+ 2, there is a vertex w6=x, y such that (x, w)∈E(T^′) and (w, y)∈E(T^′). Thus P = (x, w, y) is a path inT^′.

LetT be a tournament obtained from T^′ by reversing the directions of the arcs inP. Thenπis a score sequence forT, producing a contradiction.

Frank Harary and Leo Moser [111] obtained a related characterization of sequences of nonnegative integers that are score sequences of strong tournaments (see Exercise 10).

Theorem 4.19 A nondecreasing sequence π : s₁, s₂, . . . , sn of nonnegative integers is a score sequence of a strong tournament if and only if

i=1

si>

k 2

for 1≤k≤n−1 and

i=1

si= n

Furthermore, every tournament whose score sequence satisfies these conditions is strong.

Distance in Tournaments

Recall that ifuandv are vertices of a digraphD such thatD contains at least one u−v path, then the length of a shortest u−v path is called the directed distance fromutov and is denoted byd(u, v).~

Theorem 4.20 Ifuis a vertex of maximum score in a nontrivial tournament T, then for every vertex v ofT,d~(u, v)≤2.

Proof. Assume that odu= k. Necessarily,k ≥1. Let v₁, v₂, . . . , vk denote the vertices of T adjacent from u. Then d(u, v~ ⁱ) = 1 for i = 1,2, . . . , k. If V(T) ={u, v₁, v₂, . . . , vk}, then the proof is complete.

Assume, then, that V(T)− {u, v₁, v₂, . . . , vk} is nonempty, and let v ∈ V(T)− {u, v₁, v₂, . . ., vk}. Ifv is adjacent from some vertex vi, 1≤i ≤k, then d(u, v) = 2, producing the desired result. Suppose that this is not the~ case. Then v is adjacent to all of the verticesv₁, v₂, . . . , vk, as well as tou, so odv≥1 +k= 1 + odu. However, this contradicts the fact thatuis a vertex of maximum score.

If a tournament represents a competition among teams where ties are not permitted, then Theorem 4.20 says that ifAis a team with the greatest number of victories and B is any other team, then either A has defeated B or A has defeated a team that has defeatedB.

Landau discovered Theorem 4.20 first during a study of pecking orders and domination among chickens. In the case of chickens, the theorem says that if chickencpecks the largest number of chickens, then for every other chickend, eithercpecksd, orcpecks some chicken that pecksd. Thuscdominates every other chicken either directly or indirectly in two steps. Such a chicken is called a king chicken. In a nontrivial tournamentT, a vertexv is called aking if for every vertex udistinct from v, eitherv is adjacent to uor v is adjacent to a vertex that is adjacent to u. By Theorem 4.20, every nontrivial tournament contains a king.

In the case of nontrivial strong tournaments, there are always at least three kings, as we are about to see. Let D be a strong digraph. Recall that the eccentricitye(v) of a vertexvofD is defined as

e(v) = max{d(v, w) :~ w∈V(D)}. The radius ofD is

radD= min{e(v) : v∈V(D)}

and the center Cen(D) ofDis defined as the induced subdigraph D[{v: e(v) = rad(D)}].

Theorem 4.20 provides an immediate result dealing with the radius of a strong tournament (see Exercise 12).

166 CHAPTER 4. DIGRAPHS Corollary 4.21 Every nontrivial strong tournament has radius2.

We now show that every nontrivial strong tournament contains at least three kings.

Theorem 4.22 The center of every nontrivial strong tournament contains at least three vertices.

Proof. Let T be a nontrivial strong tournament. Then by Corollary 4.21, radT = 2. Letwbe a vertex having eccentricity 2. SinceT is strong, there are vertices adjacent to w; let v be one of these having maximum score. Among the vertices adjacent tov, letube one of maximum score. We show that both uandvhave eccentricity 2, which will complete the proof.

Assume, to the contrary, that one of the vertices u and v does not have eccentricity 2. Suppose, then, thatx∈ {u, v}ande(x)≥3. Hence, there exists a vertex y in T such that d(x, y)~ ≥ 3. Thus, y is adjacent to x. Moreover, y is adjacent to every vertex adjacent from x. These observations imply that ody >odx.

Suppose thatx=v. Sincexis adjacent tow, it follows that y is adjacent to w. However, ody > odv, which contradicts the defining property of v.

Therefore,x6=v and sox=u. Herexis adjacent tov so thaty is adjacent to v, but ody >odu. This contradicts the defining property ofu. Hence, x6=u and the proof is complete.

Hamiltonian Tournaments

The large number of arcs that a tournament has often produces a variety of paths and cycles. We now investigate paths and cycles in tournaments. We begin with perhaps the most basic result of this type, a property of tournaments first observed by László Rédei [181] in 1934. A path in a digraphDcontaining every vertex ofD is aHamiltonian path.

Theorem 4.23 Every tournament contains a Hamiltonian path.

Proof. LetT be a tournament of order n and let P = (v₁, v₂, . . . , vk) be a longest path inT. IfPis not a Hamiltonian path ofT, then 1≤k < nand there is a vertexv of T not onP. SinceP is a longest path, (v, v₁),(v^k, v)∈/ E(T), and so (v₁, v),(v, v^k)∈E(T). This implies that there is an integeri(1≤i < k) such that (vⁱ, v)∈E(T) and (v, vⁱ₊₁)∈E(T) (see Figure 4.11). But then

(v₁, v₂, . . . , vi, v, vi+1, . . . , vk)

is a path whose length exceeds that ofP, producing a contradiction.

A simple but useful consequence of Theorem 4.23 concerns transitive tournaments.

.. .. .. . . . . . .. .. .

...... ...................... ...................... ...................... ........................ ...................... ......................

p p p p p p ^.^.^.^.................................................................................................................................................................................. ......................

...................... ..

....................... . ...................... . ......................

. .. . ....... ......................................

.. . ......... . ........................ .. .. .. .. .. . ..

.................................................................................

.... ........................ . . ... .

......................

v₂ v₃ vk−1 vk

v vi vi+1

P : v₁

Figure 4.11: A step in the proof of Theorem 4.23

Corollary 4.24 Every transitive tournament contains exactly one Hamilto- nian path.

The preceding corollary is a special case of a result found independently by R´edei [181] and Tibor Szele [213], who showed that every tournament contains an odd number of Hamiltonian paths.

While every tournament contains a Hamiltonian path, certainly not every tournament contains a Hamiltonian cycle. Indeed, by Theorem 4.12, every transitive tournament is acyclic. If a tournament T contains a Hamiltonian cycle, then T is strong by Theorem 4.2. Paul Camion [32] showed that the converse is true as well.

Theorem 4.25 A nontrivial tournamentT is Hamiltonian if and only if T is strong.

Proof. We have already seen that every Hamiltonian tournament is strong.

For the converse, assume that T is a nontrivial strong tournament. Thus T contains cycles. Let C be a cycle of maximum length in T. IfC contains all of the vertices of T, then C is a Hamiltonian cycle. So assume that C is not Hamiltonian, say

C= (v₁, v₂, . . . , vk, v₁),

where 3≤k < n. IfT contains a vertexvthat is adjacent to some vertex ofC and adjacent from some vertex ofC, then there must be a vertexvi ofC that is adjacent tov such thatvi+1 is adjacent from v. In this case,

C^′ = (v₁, v₂, . . . , vi, v, vi+1, . . . , vk, v₁)

is a cycle whose length is greater than that of C, producing a contradiction.

Hence, every vertex ofT that is not on C is either adjacent to every vertex of Cor adjacent from every vertex ofC. SinceT is strong, there must be vertices of each type.

LetU be the set of all vertices ofT that are not on C and such that each vertex ofU is adjacent from every vertex ofC, and let W be the set of those vertices ofT that are not onC such that every vertex ofW is adjacent to each vertex of C(see Figure 4.12). Then U6=∅ andW 6=∅.

SinceT is strong, there is a path from every vertex ofC to every vertex of W. Since no vertex ofC is adjacent to any vertex ofW, there must be a vertex u∈U that is adjacent to a vertexw∈W. However then,

168 CHAPTER 4. DIGRAPHS

s ^.^.^.^.^.^.^.^.^.^.^.^.^.^.^.^.^.^.^.

.............s.............................. ..

...............................................

. .. . .. .. .. . .. .. . .. . .. . .. . . .. . . . .. . . .. . . . . .. . . . .. . . . . . .. . . . . .. . . . . . . . .. . . . . . . .. . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . .. . . . . . . .. . . . . . . . .. . . . . .. . . . . . .. . . . .. . . . . .. . . .. . . . .. . . .. . .. . .. . .. . .. .. .. .. .. .. . .. .......

..........................................

s s s

...

........................ . . ...

...................... .........................

............................ ............................

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

ppppppppppppppppppppp pppppppppppppppppppp

u w

vk v₁ v₂ C

U W

Figure 4.12: A step in the proof of Theorem 4.25 C^′′= (v₁, v₂, . . . , vk, u, w, v₁)

is a cycle whose length is greater than the length ofC, a contradiction.

IfT is a Hamiltonian tournament, then of course every vertex ofT lies on every Hamiltonian cycle of T. Actually, every vertex ofT lies on a triangle of T as well.

Theorem 4.26 Every vertex in a nontrivial strong tournament belongs to a triangle.

Proof. Let v be a vertex in a nontrivial strong tournament T. By The- orem 4.25, T is Hamiltonian. Thus T contains a Hamiltonian cycle (v = v₁, v₂, . . . , vn, v₁). Since v is adjacent tov₂ and adjacent from vn, there is a vertex vi with 2 ≤ i < n such that (v, vi) and (vi+1, v) are arcs of T. Thus (v, vi, vi+1, v) is a triangle ofT containingv.

It is perhaps surprising that if a tournament is Hamiltonian, then it must possess significantly stronger properties. A digraph D of ordern ≥3 is pancyclicif it contains a cycle of every possible length, that is,D contains a cycle of lengthℓfor eachℓ= 3,4, . . . , nand isvertex-pancyclicif each vertexv of D lies on a cycle of every possible length. Frank Harary and Leo Moser [111]

showed that every nontrivial strong tournament is pancyclic, while John W.

Moon [161] went one step further by obtaining the following result. The proof given here is due to Carsten Thomassen.

Theorem 4.27 Every nontrivial strong tournament is vertex-pancyclic.

Proof. LetT be a strong tournament of ordern≥3, and letv₁ be a vertex of T. We show thatv₁ lies on anℓ-cycle for eachℓ = 3,4, . . . , n. We proceed by induction onℓ.

Since T is strong, it follows by Theorem 4.26 that v₁ lies on a 3-cycle.

Assume thatv₁ lies on anℓ-cycleC= (v₁, v₂, . . . , vℓ, v₁), where 3≤ℓ≤n−1.

We show thatv₁ lies on an (ℓ+ 1)-cycle.

Case1. There is a vertexv not on C that is adjacent to at least one vertex of C and is adjacent from at least one vertex ofC. This implies that for some

Dalam dokumen BUKU GRAPHS & DIGRAPH PDF (Halaman 169-184)