Chia, G. L.

Thomassen, Carsten

Robertson ([5]) and independently, Bondy ([1]) proved that the generalized Petersen graph P(n, 2) is non-hamiltonian if n equivalent to 5 ( mod 6), while Thomason [7] proved that it has precisely 3 hamiltonian cycles if n equivalent to 3 (mod 6). The hamiltonian cycles in the remaining generalized Petersen graphs were enumerated by Schwenk [6]. In this note we give a short unified proof of these results using Grinberg's theorem

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L.; Thomassen, CarstenPublished in:Ars CombinatoriaPublication date:2011Document VersionPublisher's PDF, also known as Version of recordLink back to DTU OrbitCitation (APA):Chia, G. L., & Thomassen, C. (2011). Grinberg's Criterion Applied to Some Non-Planar Graphs. ArsCombinatoria, 100, 3-7.Grinberg’s Criterion Applied to SomeNon-Planar GraphsG.L. Chiaa and Carsten ThomassenbaInstitute of Mathematical Sciences, University Malaya,50603 Kuala Lumpur, MalaysiabDepartment of Mathematics, Technical University of Denmark,DK-2800, Lyngby, DenmarkAbstractRobertson ([5]) and independently, Bondy ([1]) proved that thegeneralized Petersen graph P (n, 2) is non-hamiltonian if n ≡ 5 (mod6), while Thomason [7] proved that it has precisely 3 hamiltoniancycles if n ≡ 3 (mod 6). The hamiltonian cycles in the remaininggeneralized Petersen graphs were enumerated by Schwenk [6]. In thisnote we give a short unified proof of these results using Grinberg’stheorem.A celebrated result of Grinberg (see [4]) concerning planar hamiltoniangraphs states that if a planar graph G has a hamiltonian cycle C whichpartitions its fi faces of degree i into f′i (respectively f′′i ) faces of degree iin the interior (respectively exterior) of C, then∑i≥3(i− 2)(f′i − f′′i ) = 0.If there is precisely one natural number i not congruent to 2(mod 3)such that fi > 0, then Grinberg’s equation cannot be satisfied, and hencethe graph is non-hamiltonian. But even if the equation can be satisfied, itis still possible, in special cases, to use the criterion to prove that a graphis non-hamiltonian. Thus Thomassen [8] used the criterion to describe aninfinite class of cubic planar hypohamiltonian graphs (all of which have aface partition that satisfies Grinberg’s equation), and also the Tutte graphcan be shown to be non-hamiltonian using the Grinberg criterion, see forexample, [2] p.166 and [3] Chapter 6. In this note we apply the criterionto a class of non-hamiltonian graphs, namely some generalized Petersengraphs.1Suppose n and k are two integers such that 1 ≤ k ≤ n − 1 and n ≥ 5.The generalized Petersen graph P (n, k) is defined to have vertex-set {ui, vi :i = 0, 1, . . . , n − 1} and edge-set {uiui+1, uivi, vivi+k : i = 0, 1, . . . , n − 1with subscripts reduced modulo n}.Let Fm denote the m-th Fibonacci number defined by F1 = F2 = 1,and Fm = Fm−1 + Fm−2 for m > 2. It is easy to prove, by induction on n,that the number of matchings (including the empty matching) of the pathwith n vertices is Fn+1. Hence the number of matchings of the cycle withn vertices is Fn+1 + Fn−1.Theorem 1 Let n be a natural number, n ≥ 5. Then(i) ([1], [5]) P (n, 2) is non-hamiltonian if n ≡ 5 (mod 6),(ii) ([7]) P (n, 2) has precisely three hamiltonian cycles if n ≡ 3 (mod6), and(iii) ([6]) the number of hamiltonian cycles in P (n, 2) isn if n ≡ 1 (mod 6)2(Fn2+1 + Fn2−1 − 1) if n ≡ 0, 2 (mod 6)n+ 2(Fn2+1 + Fn2−1 − 1) if n ≡ 4 (mod 6)Proof: Assume first that n ≡ 3, 5 (mod 6) and n ≥ 5.If the two edges u0u1 and v0v2 are deleted from P (n, 2), the resultis a planar graph whose face-degree sequence is (5, . . . , 5, n+52, n+72) (seeFigure 1) and hence is non-hamiltonian by Grinberg’s criterion becausewhen n ≡ 3 (mod 6), n+52and n+72are 1, 2 (mod 3) respectively whilewhen n ≡ 5 (mod 6), n+52and n+72are 2, 0 (mod 3) respectively.This means that(∗) if P (n, 2) has a hamiltonian cycle C, then C must contain at leastone of the edges u0u1, v0v2.Assume now that C is a hamiltonian cycle of P (n, 2). Since C cannotcontain all the edges of the inner cycle v0v2v4 · · · vn−2v0, we may assumethat v0v2 is not an edge in C. But then this implies that the paths vn−2v0u0and u2v2v4 must be part of C.Since v0v2 is not an edge in C, the observation (∗) implies that u0u1 isan edge in C. By symmetry, u1u2 is also an edge in C. This follows becausethere is an automorphism (the reflection fixing u1, v1) of P (n, 2) which inter-changes between the edges u1u0, u2u1 and keeps v0v2 fixed. (“Reflection”here refers to the standard drawing of P (n, k) where the vertices u0, u1, . . .and also the vertices v0, v1, . . . form convex n-gons.) Hence C contains the2u1 u2u0un−1un−2un−3un−4un−5u5u4u3v1 vn−1 vn−3 vn−5 v4v2v0 vn−2 vn−4 v5 v3Figure 1: P (n, 2) with u0u1 and v0v2 deleted, n ≡ 3, 5 (mod 6)path vn−2v0u0u1u2v2v4. So C also contains the path vn−1v1v3u3u4 andtherefore C does not contain the edge v3v5. Summarizing, we have provedthat if a hamiltonian cycle C does not contain the edge v0v2, then C doesnot contain the edge v3v5 either. By repeating this argument with v3v5instead of v0v2, we conclude that C does not contain the edge v6v8 either.In fact C does not contain any of the edges v0v2, v3v5, v6v8, v9v11, . . . .Since C must contain some edge of the inner cycle v0v2v4 · · · vn−2v0, weconclude that n ≡ 0 (mod 3). Hence P (n, 2) has no hamiltonian cycle ifn ≡ 5 (mod 6). In the case that n ≡ 3 (mod 6), the argument eventuallyleads to a unique hamiltonian cycle which can be rotated to yield preciselythree hamiltonian cycles.Assume next that n ≡ 1 (mod 6). Then Grinberg’s equation is satisfiedbut only if the two faces of degrees n+52and n+72are both in the interior (orexterior) of the hamiltonian cycle on the resulting graph of Fig. 1. This ispossible only if the edge v1vn−1 is not contained in the hamiltonian cycle.Assume that C′ is such a hamiltonian cycle. Then it is easy to see (from Fig.1) that the paths vn−3vn−1un−1u0v0vn−2un−2un−3 and v4v2u2u1v1v3u3u4must be part of C′. There is a unique hamiltonian cycle containing thesepaths which can be rotated to yield n hamiltonian cycles. If it is not possibleto obtain a hamiltonian cycle by deleting a pair uiui+1, vivi+2, then we geta contradiction as in the case when n ≡ 5 (mod 6).When n ≥ 6 is even, P (n, 2) is a planar graph. Again, the abovemethod can be applied. Note that (∗) cannot be satisfied for each pairuiui+1, vivi+2 and also for each pair uiui−1, vivi−2 because the argumentin the case n ≡ 3, 5 (mod 6) would lead to a 2-factor consisting of two3cycles rather than a hamiltonian cycle. So there exists some pair of edgesuiui+1, vivi+2 (or some pair uiui−1, vivi−2), say u0u1, v0v2, which is avoidedby some hamiltonian cycle.Draw P (n, 2) in the plane such that the n2-cycle v1v3 . . . vn−1v1 is theouter face boundary. In this case, the face-degree sequence of the resultinggraph (after deleting the edges u0u1 and v0v2) is (5, . . . , 5,n2, n2+ 6) andGrinberg’s equation is satisfied. For this to be possible, any hamiltoniancycle must contain the edge v1vn−1 (which is common to both then2-faceand the (n2+ 6)-face) unless n ≡ 4 (mod 6).Assume first that n ≡ 0, 2 (mod 6). Then for any even integer 0 ≤i ≤ n2, whenever the pair of edges vivi+2, uiui+1 is deleted, then the pathsui−1uivivi−2, vi−1vi+1ui+1ui+2vi+2vi+4 and ui+4ui+3vi+3vi+5 must be partof the hamiltonian cycle. We now follow the hamiltonian cycle along thepath vi+2vi+4 . . .. If we never use an edge vjuj there is a unique way tocontinue the hamiltonian cycle. On the other hand, the first time we use anedge vjuj we repeat the previous configuration with j instead of i. Thoseedges of the cycle v0v2 . . . v0 which are not in the hamiltonian cycle clearlyform a matching. On the other hand, whenever we specify a matching onthis cycle, there is a unique hamiltonian cycle which avoids this matchingand also avoids edges of the form uiui+1, vivi+2. Therefore the numberof hamiltonian cycles in P (n, 2) avoiding some pairs of edges of the formuiui+1, vivi+2 is Fn2+1+Fn2−1−1. Note that for any such hamiltonian cycle,(∗) fails for the pairs uiui+1, vivi+2 but holds for the pairs uiui−1, vivi−2.By symmetry (more precisely, by taking the cycle u1u2...unu1 in itsreverse order) the number of hamiltonian cycles in P (n, 2) avoiding pairsof edges of the form uiui−1, vivi−2 is also Fn2+1+Fn2−1−1. Thus the totalnumber of hamiltonian cycles in P (n, 2) is 2(Fn2+1 + Fn2−1 − 1) in the casen ≡ 0, 2 (mod 6).The reason that we have counted all hamiltonian cycles is that (∗) failseither for the pairs uiui+1, vivi+2 or for the pairs uiui−1, vivi−2, as notedabove.For n ≡ 4 (mod 6), there are two types of hamiltonian cycles, namelythe 2(Fn2+1+Fn2−1−1) hamiltonian cycles which we have already counted,and also the unique hamiltonian cycle that avoids u0u1, v0v2, v1vn−1, andthose which can be obtained from this hamiltonian cycle by rotation andreflection. There are n hamiltonian cycles of the latter type. This completesthe proof. 4References[1] J.A. Bondy, Variations on the hamiltonian theme, Canad. Math.Bull. 15 (1972) 57-62.[2] M. Capobianco and J.C. Molluzzo, “Examples and Counterex-amples in Graph Theory”, North-Holland, (1978) New York.[3] G. Chartand and L.Lesniak, “Graphs and Digraphs”, Chapman &Hall/CRC, 4th Edition (2005), New York.[4] E. Grinberg, Plane homogeneous graphs of degree three withoutHamiltonian circuits, Latvian Math. Yearbook, Izdat. Zinatne Riga 4(1968) 51-58 [Russian, Latvian and English summaries.][5] G.N. Robertson, “Graphs under Girth, Valency, and ConnectivityConstraints”, Ph. D. Thesis, University of Waterloo, Ontario, Canada,1968.[6] A.J. Schwenk, Enumeration of Hamiltonian cycles in certain gener-alized Petersen graphs, J. Combinat. Theory B 47 (1989) 53-59.[7] A.G. Thomason, Cubic graphs with three hamiltonian cycles are notalways uniquely edge-colorable, J. Graph Theory 6 (1982) 219-221.[8] C. Thomassen, Planar cubic hypohamiltonian and hypotraceablegraphs, J. Combinat. Theory B 30 (1981) 36-44.[9] W.T. Tutte, On Hamiltonian circuits, J. London Math. Soc. 21(1946) 98-101.5

Grinberg's Criterion Applied to Some Non-Planar Graphs

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Grinberg's Criterion Applied to Some Non-Planar Graphs

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