An Improved An Improved Degree Based Condition Degree Based Condition

for for Hamiltonian CyclesHamiltonian Cycles

November 22, 2005November 22, 2005

ContentsContents• Introduction

– Hamiltonian Path and Cycle– Hamiltonian Graph

• Motivation• Proposed Condition for Hamiltonicity• Proof• Significance and Conclusion

IntroductionIntroduction• A Hamiltonian cycle is a

spanning cycle in a graph, i.e., a cycle through every vertex and a Hamiltonian path is a spanning path.

• A graph containing a Hamiltonian cycle is said to be Hamiltonian.

• It is clear that every graph with a Hamiltonian cycle has a Hamiltonian path but the converse is not necessarily true

MotivationMotivation• The Hamiltonian problem is generally considered to be

determining conditions under which a graph contains a spanning cycle. Named after Sir William Rowan Hamilton, this problem traces its origins to the 1850s [4].

• The problem of finding whether a graph G is Hamiltonian is proved to be NP-complete for general graphs [1].

• No easily testable characterization is known for Hamiltonian graphs. Nor there exists any such condition to test whether a graph contains a Hamiltonian path or not.

• Hence research efforts have been spent for finding the necessary and sufficient conditions for a graph to be Hamiltonian. (see [4] for detail survey)

Motivation(contd.)Motivation(contd.)• Since the problem is NP-Complete no

polynomial time characterization is possible.

• The problem has been proved to be NP-Complete even for simple special graphs like planar, cubic, 3-connected and with no face having fewer than 5 edges; bipartite graphs, square of a graph or even HP is provided.

• So there have been efforts on discovering sufficient conditions based upon different graph parameters.

Known sufficient conditionsKnown sufficient conditions• If G is a simple graph with n>=3 and

minimum degree >=n/2 then G is Hamiltonian. (Dirac)

• If G is simple and for every pair of non-adjacent vertices u,v d(u)+d(v)>=n then G is Hamiltonian iff G+uv is Hamiltonian. (Ore)

• If G is simple then G is Hamiltonian iff its closure is Hamiltonian.(Bondy and Chvatal)

Known Results(contd)Known Results(contd)• If d(u)+d(v)>=n for every pair of

non-adjacent vertices then G is Hamiltonian. (Ore)

• Let G=(V,E) be a connected graph on n vertices and P be a longest path in G having length k and with n vertices u and v. Then the following statements must hold:

Known Results(contd.)Known Results(contd.)• (a) Either l(u,v)>1 or P is a HP

contained in a HC.

• (b) If l(u,v)>=3, then dP(u)+ dP(v) <=k-l(u,v)+2

• © If l(u,v)=2, then either dP(u)+ dP(v) <=k or P is a HP and there is a HC. (Rahman and Kaykobad)

Known Results(Contd)Known Results(Contd)• Let G=(V,E) be a connected graph

with n vertices such that for all pairs of distinct non-adjacent vertices u,v we have d(u)+d(v)+l(u,v)>=n+1. Then G has a HP. (Rahman and Kaykobad)

• This theorem also implies Ore’s condition for Hamiltonicity.

New resultsNew results• Rahman and Kaykobad [3] introduced distance

as the new parameter.

• Sufficiency condition of Ore[2] results from Rahman and kaykobad when distance of end vertices of a HP is 2[3].

• In this paper, we are going to establish that the same condition[3] forces Hamiltonian cycle to be present in the graph excepting for the case where end points of a Hamiltonian path is at a distance greater than 2.

),( vul

Our Proposed ConditionOur Proposed Condition• Theorem 1.1 (Ore[2]) If for every pair of distinct non-

adjacent vertices u and v of G , then G is Hamiltonian.

• Theorem 1.2 (Rahman and Kaykobad [3]) Let G(V,E) be a connected graph with n vertices such that for all pairs of

distinct non-adjacent vertices we have , then G has a Hamiltonian path.

• Let us assume• Now we reformulate the theorem 1.2 in the following way to

ensure that graph G is indeed Hamiltonian.

• Theorem 1.3 Let G(V,E) be a graph without cut vertices and cut edges such that for all pairs of distinct non-adjacent vertices we have , then G is Hamiltonian

( ) ( )d u d v n

),(,21),()()( vulnvulvdud

ji ,1ji,

1),()()( nvulvdud

ProofProof• Before presenting the proof we need

to clarify the following terms in the theorem 1.3:

• Cut Vertex: A vertex whose deletion along with incident edges results in a graph with more components than the original graph. i.e. vertex a in the figure

• Cut Edge : An edge of a connected graph is called a cut-edge if removing the edge disconnects the graph. i.e. edge (a,b) is a cut edge.

b

a

(1/10)

ProofProof• Since hypothesis of Theorem 1.3 is identical to that of

Theorem 1.2, existence of a Hamiltonian path in such graphs is ensured according to [3].

• We will prove that the same graph contains a Hamiltonian cycle provided end vertices of a Hamiltonian Path is at a distance of at least 3.

• Since, sufficiency condition of Ore implies that distance of end vertices of a Hamiltonain Path is two[3], let us consider a Hamiltonian Path H(u,v) for which .

• We will first show that cannot be greater than or equal to 4

3),( vul

),( vul

(2/9)

ProofProof

v1u1 vlu

Fig. 1. A possible simple graph

j k

•If (u,u1), (v,v1) exists such that l(u,v)>3 through u1, v1 , then

l(u1,v1)=l(u,v)-2; and number of vertices between u1, v1 is

l(u1,v1)-1, which implies to be l(u,v)-3.

•So,

•But to satisfy the sufficient condition

of Hamiltonicity, we have

•So we conclude that

and u and v is connected to every vertices except l(u,v)-3 vertices between u1, v1

)3),((2)()( vulnvdud

))3),(()2(),(1)()(

1),()()(

vulnvulnvdud

nvulvdud

)3),((2)()( vulnvdud

(3/9)

ProofProof• To avoid cut vertices, we need to add more edges.

• If we add an edge (j,k), 1<j<u1 and v1<k<v, then l(u,v) reduces to 3

• So we may add edges like (j,l), (m,k). In that case also l(u,v) is reduced (see fig) and that contradicts our assumption of having l(u,v)>3.

u

Fig. 2. A possible simple graph

lu1

v1 vm

j k

(4/9)

ProofProof• We may argue to have a graph like Fig3. Let us assume that we

have such a graph which also satisfies our condition of Hamiltonicity • In this case ,

• But d(u) + d(v) + l(u,v) cannot be greater than n+1. otherwise it is a contradiction to our assumption.

• So, no such graph exists with l(u,v)>3

1),()()( nvulvdud

ul

kvm

Fig. 3. A possible simple graph with l(u,v)=4

j n

1),(27),()3),((32),()()( nvulnvulvulnvulvdud

(5/9)

u vw+1w

ProofProof3),( vul

3),( vul w V

( ) ( ) 2d u d v n

1),(2),()()( nvulnvulvdud

(6/9)

• So we will proceed with and prove that there is a Hamiltonian cycle.

• Again implies that no vertex can be adjacent to both u and v at the same time. since then would have been a path u-w-v of length 2. Now those u,v are connected to different vertices and they can be connected to at most n-2 vertices.

• But then implies

• That is, each of the internal vertices must be connected to exactly one of u or v.

ProofProof• Let us assume for clarity of arguments that u is denoted by

1 and v is denoted by n, and all vertices along Hamiltonian Path H(u,v) are denoted by 2, 3… n-1.

• Since existence of cross over edges(Fig.4) as depicted in the above (Fig. 4) picture implies existence of a Hamiltonian Cycle, we assume that there is no cross-over edges.

Fig. 4. Existence of crossover edge (1, i) and (j, n) where j=i+1

n-132 nij1

(6/9)

ProofProof• As the graph does not contain cross over edges, we assume

that k is the highest index of node which is adjacent to node 1 (Fig 5). In that case all the vertices with index i<k are adjacent to 1 and vice versa for node n.

• Here the edge (k, k+1) becomes a cut edge. So to avoid such an edge, we need to add more edges.

• In that case two situations arise (see next slides):

nk+1k1

Fig. 5. A possible simple graph without crossover edges between vertices 1 and n

(7/9)

ProofProof• Case 1: To avoid cut edge, there exists an edge (i, j) such that

1<i<k and k+1<j<n (Fig 6).• In this case we have a Hamiltonian Cycle

• Here, denotes jump and denotes increasing or decreasing natural sequence

i j nk+1k1

Fig. 6. A simple graph without crossover edges between the non-adjacent vertices 1and n may contain an edge (i,j) where i<k and j>k+1

(1 ( 1) ( 1) 1)i j n j i

(8/9)

ProofProof• Case 2: We may have two edges like (i, k+1) and (k, j)

where 1<i<k and k+1<j<n (Fig 7)• In this case we have a Hamiltonian cycle

i i+1 j-1k+1 j nk1

Fig. 7. A simple graph without crossover edges between the non-adjacent vertices 1and n may contain edges like (k, j), (k+1,i) where i < k and j>k+1

(1 ( 1) ( 1) ( 1) 1)i k j n j k i

(9/9)

ConclusionConclusion• Hence, our proposed condition, which followed

from Rahman and Kaykobad[3], ensures Hamiltonian Cycle in the graph.

• The supremacy of this condition is that this requires less number of edges than similar existing degree related conditions[2] to ensure Hamiltonicity in a graph.

• The novelty of this approach is the incorporation of distance parameter l(u,v).

ReferencesReferences• [1] Garey M.R. and Johnson D.S., Computers and

Intractability: A Guide to the Theory of NP-Completeness, W.H. Freeman and Company, New York, 1979.

• [2] Ore O., Note on Hamiltonian circuits, Amer. Math. Monthly 67(196) 55.

• [3] Rahman M. Sohel and Kaykobad M., On Hamiltonian cycles and Hamiltonian paths, Information processing Letters 94(2005) , 37-51.

• [4] Gould R. J., Advances on the Hamiltonian Problem - A Survey, Graphs and Combinatorics, Volume 19, Number 1, March 2003, 7-52.