Moser asked for a construction of explicit tournaments on $n$ vertices having
at least $(\frac{n}{3e})^n$ Hamilton cycles. We show that he could have asked
for rather more

Calkin, Neil J.

Novick, Beth

Ushijima-Mwesigwa, Hayato

English

arXiv

Moser asked for a construction of explicit tournaments on n vertices having at least Hamilton cycles. We show that he could have asked rather more

Clemson University: TigerPrints

Clemson UniversityTigerPrintsPublications School of Mathematical and Statistical Sciences4-2016What Moser Could Have Asked: CountingHamilton Cycles in TournamentsNeil J. CalkinClemson University, calkin@clemson.eduBeth NovickHayato Ushijima-MwesigwaFollow this and additional works at: https://tigerprints.clemson.edu/mathematics_pubsPart of the Mathematics CommonsThis Article is brought to you for free and open access by the School of Mathematical and Statistical Sciences at TigerPrints. It has been accepted forinclusion in Publications by an authorized administrator of TigerPrints. For more information, please contact kokeefe@clemson.edu.Recommended CitationPlease use publisher's recommended citation. http://www.maa.org/press/periodicals/american-mathematical-monthly/american-mathematical-monthly-april-2016arXiv:1506.00699v1  [math.CO]  1 Jun 2015What Moser Could Have Asked: CountingHamilton Cycles in TournamentsNeil J. Calkin, Beth Novick and Hayato Ushijima-MwesigwaJune 3, 2015AbstractMoser asked for a construction of explicit tournaments on n verticeshaving at least ( n3e)n Hamilton cycles. We show that he could have askedfor rather more.1 Introduction. . . the cycle has taken us upthrough forests.Robert M. PirsigIn his classic book on tournaments, Moon [4, Section 10] discusses the ques-tion of exhibiting tournaments with a large number of Hamilton cycles. Heposes the question (Exercise 4, attributed to Moser), of constructing a tour-nament on n vertices having at least ( n3e)n Hamilton cycles. Presumably, theintended construction is to take three tournaments, T1, T2, T3, onn3 vertices,and construct a new tournament C3(T1, T2, T3) by orienting all edges from T1to T2, T2 to T3, and T3 to T1 (See Figure 1 ). The number of Hamilton cyclesT2T3 T1Figure 1: C3(T1, T2, T3)AMS Subject Classification: 05C20, 05C30, 05A161in C3(T1, T2, T3) which do not use any edges internal to T1, T2, or T3 is(n3 )!3n3∼√8pi3n3( n3e)n>( n3e)n.In this note, we show that this construction has many more Hamilton cycles.Indeed, if T1, T2, and T3 are all transitive, we show that the number of Hamiltoncycles is asymptotic to 1(1−log 2)(n−1)!(3 log 2)n .2 Background and DefinitionsA tournament is an oriented, complete graph. A Hamilton cycle or path in atournament T , is a spanning directed cycle or directed path in T . A tournamentwith no directed cycles is called transitiveCounting Hamilton paths and cycles in tournaments is a very old problem,dating back to the 1940’s: in one of the first applications of the probabilisticmethod, Szele [6] showed that the expected number of Hamilton paths in arandom tournament is n!2n−1 , therefore showing that there exists a tournamenton n vertices with at least this many Hamilton paths. The same argument showsthat there exists a tournament with at least (n−1)!2n Hamilton cycles. Moonobserved that it seems difficult to give explicit tournaments with at least thismany Hamilton cycles.Deep results of Cuckler [3] show that every regular tournament on n verticeshas at least n!(2+o(1))n Hamilton cycles.Given tournaments T1, T2, T3, we can construct a tournament C3(T1, T2, T3)by orienting all edges from T1 to T2, T2 to T3, and T3 to T1. We will callsuch tournaments triangular. Wormald [8] showed that if T1, T2, T3 are randomtournaments, then the expected number of Hamilton cycles is 2 (n−1)!2n .We show that all triangular tournaments have a relatively large number ofHamilton cycles, even in the extreme case when they constructed from transitivetournaments.Let S(m, k) denote the Stirling number of the second kind, that is, S(m, k)is the number of set partitions of {1, 2, . . .m} into exactly k parts.3 Main ResultTheorem 1. Let T1, T2, and T3 be any tournaments on m1,m2, and m3 verticesrespectively. Then the number H of Hamilton cycles in C3(T1, T2, T3) is at leastH ≥min{m1,m2,m3}∑k=1S(m1, k)S(m2, k)S(m3, k)k!3k, (1)with equality when T1, T2, and T3 are transitive.2Corollary 2. If T1, T2, and T3 are transitive tournaments onn3 vertices, thenumber of Hamilton cycles in C3(T1, T2, T3) is asymptotic to1(1− log 2)(n− 1)!(3 log 2)n≃ 3.25889 (n− 1)!(2.07944)n. (2)Proof of Theorem 1. Take any Hamilton cycle C, of C3(T1, T2, T3), and considerC restricted to T1, T2, and T3. Since a Hamilton cycle meets every vertex inT1, T2, and T3 exactly once, C visits each subtournament the same number oftimes, say k. Hence for each Ti, C will induce a collection of k disjoint pathsthat cover the vertices of Ti. We will refer to such a collection of k paths as ak-path cover. Similarly, given k-path covers for T1, T2, T3, we can construct aHamilton cycle by joining these k-path covers together. The number of waysof doing this is k!3/k. Thus, if P (Ti, k) denotes the number of k-path covers ofTi, then the number of Hamilton cycles of C3(T1, T2, T3) which induce k-pathcovers in T1, T2, and T3 isP (T1, k)P (T2, k)P (T3, k)k!3k. (3)It follows that the number of Hamilton cycles in C3(T1, T2, T3) ismin{m1,m2,m3}∑k=1P (T1, k)P (T2, k)P (T3, k)k!3k.For any set partition of the vertex set of Ti into k nonempty sets, each part willinduce a subtournament of Ti. Re´dei [5] showed that every tournament has aHamilton path, thus each partition into k sets will induce at least one k-pathcover of Ti. Therefore the number of Hamilton cycles in C3(T1, T2, T3) is at leastmin{m1,m2,m3}∑k=1S(m1, k)S(m2, k)S(m3, k)k!3kas claimed.In the case that each Ti is transitive, each subtournament will have exactlyone Hamilton path, hence we have equality in (1).Proof of Corollary 2. Suppose now that each Ti is a transitive tournament onm vertices, then the number of Hamilton cycles in C3(T1, T2, T3) is equal tom∑k=1S(m, k)3k!3k. (4)As with many combinatorial sums, the summands in (4) are approximatedrather well by a normal distribution. Indeed, if we letµ =12 log 2and σ =√1− log 22 log 2,3define f(m) =∑mk=1 S(m, k)k!, and write p(m, k) =S(m,k)k!f(m) , then Bender [2]shows that p(m, k) is asymptotically normal with mean µm and variance σ2m.Hence, p(m, k)3 is also proportional to a normal distribution, at least in a rangeof k close to µm. This allows us to approximate the sum∑mk=1 S(m, k)3 k!3kbyan integral, showing thatm∑k=1S(m, k)3k!3k∼ f(m)3 312 212pi123µσ2(2pi)32m2.From Wilf [7, p. 176], we know thatf(m) ∼ m!2(log 2)m+1,Therefore with n = 3m, two applications of Stirling’s approximation for n! yieldsm∑k=1S(m, k)3k!3k∼(m!2(log 2)m+1)3312 212pi123µσ2(2pi)32m2∼√2pinn(1 − log 2)(n3e log 2)n∼ 1(1 − log 2)(n− 1)!(3 log 2)nAcknowledgment. The authors are very grateful to Rod Canfield for helpfuladvice regarding the asymptotics.References[1] N. Alon. The maximum number of Hamiltonian paths in tournaments Com-binatorica (1990), 319-324.[2] E. A. Bender. Central and local limit theorems applied to asymptotic enu-meration. J. Combinatorial Theory Ser. A 15 (1973), 91111.[3] B. Cuckler. Hamiltonian Cycles in Regular Tournaments Combinatorics,Probability and Computing (2007) 16, 239249.[4] J. W. Moon. Topics on Tournaments Holt, Reinhart and Winston (1968),New York.[5] L. Redei. Ein kombinatorischer Satz. Acta Litt. Szeged, 7:3943, 1934[6] T. Szele. Kombinatorikai vizsgalatok az iranyitott teljes graffal, Kapcsolat-ban, Mt. Fiz. Lapok 50 (1943), 223-256.4[7] H. S. Wilf. Generatingfunctionology Academic Press, New York (1990).[8] N.C. Wormald,. Tournaments with many Hamilton cycles.5

What Moser Could Have Asked: Counting Hamilton Cycles in Tournaments

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What Moser Could Have Asked: Counting Hamilton Cycles in Tournaments

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