# New Sufficient Conditions for Hamiltonian Paths

Authors: M. Sohel Rahman; M. Kaykobad; Jesun Sahariar Firoz
Journal: The Scientific World Journal
 (2014)

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## Abstract

A Hamiltonian path in a graph is a path involving all the vertices of the graph. In this paper, we revisit the famous Hamiltonian path problem and present new sufficient conditions for the existence of a Hamiltonian path in a graph.


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## Body

## 1. Introduction

Hamiltonian paths and cycles are named after William Rowan Hamilton who invented the puzzle that involves finding a Hamiltonian cycle in the edge graph of the dodecahedron. Although Hamilton solved this particular puzzle, finding Hamiltonian cycles or paths in arbitrary graphs is proved to be among the hardest problems of computer science [1]. As a result, instead of complete characterization, most researchers aimed to find sufficient conditions for a graph to possess a Hamiltonian cycle or path. In this paper, we focus on degree based sufficient conditions for the existence of Hamiltonian paths in a graph.To the best of our knowledge, the quest for good sufficient degree based conditions for Hamiltonian cycles or paths dates back to 1952 when Dirac presented the following theorem, whereδ(G) denotes the degree of the minimum degree vertex of the graph G.Theorem 1 (see [2]).
IfG is a simple graph with n vertices, where n≥3 and δ(G)≥n/2, then G contains a Hamiltonian cycle.Later Ore in 1960 presented a highly celebrated result where a lower bound for the degree sum of nonadjacent pairs of vertices was used to force the existence of a Hamiltonian cycle. In particular, Ore proved the following theorem, wheredu denotes the degree of the vertex u.Theorem 2 (see [3]).
LetG be a simple graph with n vertices and u, v distinct nonadjacent vertices of G with du+dv≥n. Then, G has a Hamiltonian cycle.A graph satisfying Ore’s condition has a diameter of only two [4], where the diameter of a graph is the longest distance between two vertices. But if a sufficient condition can be derived for a graph with diameter more than two, Hamiltonian path or cycle may be found with fewer edges. With this motivation, Rahman and Kaykobad [5] proposed a sufficient condition to find a Hamiltonian Path in a graph involving the parameter δ(u,v), which denotes the length of the shortest path between u and v.Theorem 3 (see [5]).
LetG=(V,E) be a connected graph with n vertices such that for all pairs of distinct nonadjacent vertices u,v∈V one has du+dv+δ(u,v)≥n+1. Then, G has a Hamiltonian path.In some subsequent literature, the condition “du+dv+δ(u,v)≥n+1, where u, v are distinct nonadjacent vertices of a graph having n vertices,” is referred to as the “Rahman-Kaykobad” condition. A number of interesting results were achieved extending and using the “Rahman-Kaykobad” condition as listed below.Theorem 4 (see [6]).
LetG be a 2-connected graph which satisfies the “Rahman-Kaykobad” condition. If G contains a Hamiltonian path with endpoints at distance 3, then G contains a Hamiltonian cycle.Theorem 5 (see [7]).
LetG be a connected graph which satisfies the “Rahman-Kaykobad” condition. Then, either G contains a Hamiltonian cycle or G belongs to some specific classes of graphs.Theorem 6 (see [8]).
LetG be a 2-connected graph with n≥3 vertices. If du+dv≥n-1 for every pair of vertices u and v with δ(u,v)=2, then G contains a Hamiltonian cycle, unless n is odd and G belongs to some specific classes of graphs.The sufficient conditions of Theorems4, 5, and 6 can be seen as incremental improvements over the result of Rahman and Kaykobad [5]. To the best of our knowledge, the latest and the best (so far) result of “Rahman-Kaykobad” condition was reported in [9]. In particular, in [9], the authors show that “Rahman-Kaykobad” condition is “almost” sufficient to make a graph pancyclic, where a graph is pancyclic if it contains a cycle of length k for 3≤k≤n.Theorem 7 (see [9]).
LetG be a 2-connected graph of order n≥6, which satisfies the “Rahman-Kaykobad” condition. Then, either G is pancyclic or G belongs to some specific classes of graphs.In [10], the authors proved the following traceability analogue of the famous Fan’s condition for hamiltonicity [11].Theorem 8.
LetG be a connected graph of order n. If max⁡{d(u),d(v)}≥(n-1)/2 for each pair of vertices u,v∈V(G) at distance 2, then G is traceable.By this theorem, the existence of a Hamiltonian path for each pair of vertices of only distance two is ensured. In this paper, we mention a more generalized version of Theorem8. In particular, we present new sufficient conditions for a graph to possess a Hamiltonian path and Theorem 8 can be seen as a special case of our sufficient conditions. This time, we achieve a lower bound for the degree sum of nonadjacent pairs of vertices that is 2 lesser than Ore’s condition. However, this condition cannot guarantee a Hamiltonian path for all graphs and we present such examples as well. The rest of the paper is organized as follows. In Section 2, we present some preliminary notations and results needed throughout the paper. Section 3 presents our main results. Finally, we very briefly conclude in Section 4.


## 2. Preliminaries

We consider only simple graphs and hence neither self-loop nor multiedges are present. Suppose we have a graphG=(V,E) with n vertices. We sometimes use the notations V[G]=V and E[G]=E. Two vertices, u,v∈V are said to be adjacent/neighbours to each other if (u,v)∈E; otherwise, they are nonadjacent. The set of neighbours of a vertex u in G is denoted by Nu. If G′ is a subgraph of G and u∈V[G′], then Nu[G′] denotes the set of neighbours of u (confined) in G′. Now, du=|Nu| and du[G′]=|Nu[G′]|. We use G′¯ to denote the complimentary graph of G′ with respect to G; that is, V[G′¯]=V-V[G′] and E[G′¯]=E-E[G′]. A path P of G is defined by a sequence of vertices 〈u=x1,x2,…,xk=v〉 such that V[P]={xi∣1≤i≤k}⊆V, E[P]={(xi,xi+1)∣1≤i<k}⊆E. Sometimes, P is referred to as a u, v-path and u and v are referred to as the end vertices or endpoints of P. Also, sometimes we use the notation |P| to denote the length of P. So, by our definition, |P|=k-1. If we have (x1,xk)∈E, then the graph C=(V[P],E′) such that E′=E[P]⋃{(x1,xk)} is called a cycle. In what follows, we only consider simple paths and simple cycles. A path P (cycle C) is called a Hamiltonian path (cycle) if V[P]=V (V[C]=V). Given a path P of G as defined above, assume that dx1[P]≠0 and dxk[P]≠0. Two edges (x1,xi),(xk,xj)∈E, 1≤i≠j≤k, are said to be crossover edges if and only if j=i-1. For example, in Figure 1, the pair of edges (x1,x5) and (x8,x4) are crossover edges. The following is a well-known fact.Figure 1
Crossover edges.Fact 1.
SupposeP=〈x1,x2,…,xk〉 is a path of G. If there exist crossover edges (x1,xi), (xk,xi-1), then there is a cycle C=(V[P],E′) in G.Proof.
We easily get a cycleC as follows:
(1)C=〈x1,x2,…,xi-1,xk,xk-1,…,xi,x1〉.In what follows, we extensively use the following result.Lemma 9 (see [5]).
LetG=(V,E) be a connected graph with n vertices and P a longest path in G. If P is contained in a cycle then P is a Hamiltonian path.An independent set of a graphG=(V,E) is a set of vertices V′⊆V such that all pairs of vertices u,v∈V′ are nonadjacent in G. A graph can be decomposed into independent sets in the sense that the entire vertex set of the graph can be partitioned into pairwise disjoint independent subsets. Such independent subsets are called partite sets or simply parts. A graph is said to be a k-partite graph, if its vertex set can be decomposed into k partite sets but not fewer. So, a bipartite graph is a graph that can be decomposed into two partite sets but not fewer. Similarly, a tripartite graph is a graph that can be decomposed into three partite sets but not fewer. A 1-partite graph is the same as an independent set or an empty graph.One often writesG=(A⋃B,E) to denote a bipartite graph whose partite sets are A and B. If |A|=|B|, that is, if the two partite sets have equal cardinality, then G is called a balanced bipartite graph. On the other hand, if ||A|-|B||≤1, then we say that G is a semibalanced bipartite graph. Note that, by definition, a balanced bipartite graph is also a semibalanced bipartite graph. It is easy to see that, for a bipartite graph G to possess a Hamiltonian path, G must be semibalanced. Similar to the notation used for bipartite graphs, a tripartite graph with partite sets A, B, and C may be denoted by G=(A⋃B⋃C,E).


## 3. Sufficient Conditions

In this section we present our main results. First, we present the following useful lemma.Lemma 10.
LetP=〈x1,x2,…,xp〉 be a longest path of G such that p≥4. If dx1+dxp≥p-1+l, l≥1, then there exists at least l crossover edges.Proof.
SinceP is a longest path, (x1,xp)∉E(Lemma 9) and all the neighbors of x1 and xp must lie within P. Assume that N(x1)={xi∣(x1,xi)∈E} and N+(xp)={xi+1∣(x1,xi)∈E,i+1≤p}. Now, it is clear that x1,xp∉N(x1) and x1,x2∉N+(xp). Therefore, we must have |N(x1)⋃N+(xp)|≤p-1. Now we have the following:
(2)|N(x1)⋂N+(xp)|=|N(x1)|+|N+(xp)|-|N(x1)⋃N+(xp)|=dx1+dxp-|N(x1)⋃N+(xp)|≥dx1+dxp-p+1≥p-1+l-p+1=l.
Therefore, we have |N(x1)⋂N+(xp)|≥l and we are done.Remark 11.
Note carefully that, in Lemma10, when we talk about multiple crossover edges, they may not necessarily be disjoint.Now, we present the following sufficient condition.Theorem 12.
LetG=(V,E) be a connected graph with n≥5 vertices, such that n is odd. If for all pairs of nonadjacent vertices u, v one has du+dv≥n-2, then G has a Hamiltonian path.Proof.
LetP=〈x1,x2,…,xk〉 be a longest path of G of length k-1. If P is a Hamiltonian path, we are done. So, assume otherwise. Clearly, (x1,xk)∉E because, otherwise, by Lemma 9, we are done. Now, we must have dx1[P¯]=dxk[P¯]=0, since, otherwise, P is part of a longer path, a contradiction. Therefore, we must have dx1=dx1[P] and dxk=dxk[P].
Now, sincex1, xk are nonadjacent, we must have dx1+dxk≥n-2. Now, if we assume that dx1+dxk<k-1, then we must have k-2≥n-2; that is, k≥n, implying that P is a Hamiltonian path. Therefore, assume that dx1+dxk≥k-1. Since we cannot allow a crossover edge, if we want to maximize the total degree sum of dx1+dxk, we can have configurations similar to one of three configurations, namely, Config-1, Config-2, and Config-3 shown in Figures 2 to 5. It can be verified easily that any other configuration will result in dx1+dxk<k-1 and according to the argument presented above P then would have to be a Hamiltonian path. Note carefully that in both Config-1 and Config-2 we have dx1+dxk≥k-1. The main properties of Config-1 and Config-2 are listed as the following facts. And Config-3 is a combination of Config-1 and Config-2.
Fact
2. In Config-1, we have Nx1={xj∣1<j≤r}, Nxk={xj∣r≤j<k}, such that Nx1⋂Nxk={xr}.
Fact
3. In Config-2, if k is odd, then we have Nx1=Nxk={xj∣jiseven}. On the other hand, if k is even, then we have Nx1={xj∣jisevenandj≠k} and Nxk=Nx1⋃{xk-1}.
Now, we consider two cases as follows.
Case
1  (dx1+dxk>k-1). From Figures 2 to 4 it is easy to see that if dx1+dxk≥k, then we definitely will have a crossover edge resulting in a cycle containing another path P′ such that |P|=|P′|. Therefore, by Lemma 9, it follows that the length of a longest path cannot be k-1, a contradiction.
Case
2  (dx1+dxk=k-1). Since (x1,xk)∉E, we have dx1+dxk≥n-2. Therefore, k-1≥n-2⇒k≥n-1. Now, if k>n-1, we are done. So, assume that k=n-1. Then we have a vertex y such that y∈V-V[P]. Now, for y, we cannot have (y,x1),(y,xk)∈E because, then, we have a longer path than P, a contradiction. Now we must have dy+dx1≥n-2 and dy+dxk≥n-2. Then, we have the following:(3)dx1+dxk+2×dy≥2×(n-2)⟹2×dy+k-1≥2n-4⟹2×dy+n-1-1≥2n-4⟹2×dy+n≥2n-2⟹2×dy≥n-2⟹dy≥(n-2)2⟹dy≥n2-1.
Now, since n≥5 and dy cannot be a fractional value, we must have dy≥2. Now we have two cases.
Case
2.a (Config-1). In this case we have a configuration similar to Figure 2. Now, let xi,xj∈Ny. Assume without loss of generality that j>i. If j=i+1, then we easily get a Hamiltonian path P′=〈x1,x2,…,xi,y,xj,xj+1,…,xk〉 and we are done. So, assume that j=i+l, l>1.
Now, recall that, in this case, there exists a vertexxr,r∉{1,k} such that (xr,x1),(xr,xk)∈E (Fact 2). Now, we have three subcases.
Case
2.a.1
(
r
≤
i
<
j
<
k
). From Fact 2, it is clear that (xk,xi+1)∈E. Therefore, we get a Hamiltonian path
(4)P′′=〈x1,x2,…,xi,y,xj,xj+1,…,xk,xi+1,xi+2,…,xj-1〉
and we are done.
Case
2.a.2
(
1
<
i
<
j
≤
r
). This is symmetrical to Case 2.a.1.
Case
2.a.3
(
1
<
i
<
r
<
j
). Again, from Fact 2, it is clear that (xk,xj-1)∈E. Therefore, we get a Hamiltonian path
(5)P′′′=〈x1,x2,…,xi,y,xj,xj+1,…,xk,xj-1,xj-2,…,xi+1〉,
and we are done.
Case
2.b (Config-2). Since n is odd, in this case, we have k is even. Hence, we have a configuration similar to Figure 3. Recall that we have dy≥(n/2)-1. Since n is odd and dy cannot be a fractional number, we must have dy≥((n+1)/2)-1. In other words, we have dy≥((k+2)/2)-1=k/2.
First of all, if we have either of the edges(y,x1),(y,xk)∈E, we are done. So assume otherwise. Furthermore, if we have (y,xk-1)∈E, then we get a Hamiltonian path P′=〈x1,x2,…,xk-2,xk,xk-1,y〉 and we are done. So, assume otherwise. So, x1,xk-1,xk∉Ny. Therefore, we have k-3 vertices as candidates for membership in Ny.
Now, sincen≥5, we have k=n-1≥4. Therefore, we must have dy≥2. Now, let xi,xj∈Ny. Without the loss of generality assume that j>i. Clearly, if j=i+1, we are done, since we get a Hamiltonian path P′=〈x1,x2,…,xi,y,xj,xj+1,…,xk〉. So, assume that j=i+l, l>1. Since k-3 is an odd number, it follows that dy≤(k-2)/2=(k/2)-1. This contradicts our deduction above that dy≥k/2.
Case
2.c (Config-3). As mentioned above Config-3 is a combination of Config-1 and Config-2. Recall that in this case our assumption is dx1+dxk=k-1. Hence we have dy≥(n/2)-1. Clearly, x1,xk∉Ny because then we get a longer path, a contradiction. It is easy to verify that this would force two consecutive vertices xi,xi+1∈Ny for 2≤i≤n-2. Then, again we get a longer path simply including the subpath 〈xi,y,xi+1〉, which leads us to a contradiction.
And this completes our proof.Figure 2
Config-1.Figure 3
Config-2:k is even (the dotted edge is nonexistent).Figure 4
Config-2:k is odd.Figure 5
Config-3: a combination of Config-1 and Config-2.Note that in the proof of Theorem12, the condition that n is odd is assumed only in Case 2.b where Config-2 is considered. So, based on the proof of Theorem 12, we have the following two corollaries.Corollary 13.
Suppose one has a graphG=(V,E) with n≥5 vertices such that for all pairs of nonadjacent vertices u, v one has du+dv≥n-2. If one has configurations similar to Config-1, then G must possess a Hamiltonian path.Corollary 14.
Suppose one has a graphG=(V,E) with n≥5 vertices such that for all pairs of nonadjacent vertices u, v one has du+dv≥n-2. If one has configurations similar to Config-2 and n is odd, then G must possess a Hamiltonian path.We now present another sufficient condition.Theorem 15.
LetG=(A⋃B,E) be a semibalanced bipartite connected graph with n≥5 vertices. If, for all pairs of nonadjacent vertices u, v one has du+dv≥n-2, then G has a Hamiltonian path.Proof. Now, the proof assumes the same hypotheses of the proof of Theorem 12. Clearly, based on Corollaries 13 and 14, it suffices to consider only Config-2 when n is even. Therefore, what follows should be treated as a continuation of the proof of Theorem 12 excluding Case 2.b. Additionally, we assume that G is a semibalanced bipartite connected graph.Sincen is even, in this case, k is odd. Hence, we have a configuration similar to Figure 4. Recall that we have dy≥(n/2)-1. In other words, we have dy≥((k+1)/2)-1=(k/2)+(1/2)-1=(k/2)-(1/2)=(k-1)/2.First of all, if we have either of the edges(y,x1),(y,xk)∈E, we are done. So assume otherwise. So, x1,xk∉Ny. Therefore, we have k-2 vertices as candidates for membership in Ny.Now, sincen is even, n≥5 implies n≥6. Hence, we have k=n-1≥5. Therefore, we must have dy≥2. Now, let xi,xj∈Ny. Without the loss of generality assume that j>i. Clearly, if j=i+1, we are done, since we get a Hamiltonian path P′=〈x1,x2,…,xi,y,xj,xj+1,…,xk〉. So, assume that j=i+l, l>1. Now, we claim the following.Claim 1.
Ifxi,xj∈Ny such that either i or j or both are odd, then we have a Hamiltonian path.Proof of Claim1.
Assume thati is even and j=i+l, l is odd. Clearly, j-1 is even and we know that xj-1 is adjacent to both x1 and xk (Fact 3). Then, we easily get a Hamiltonian path
(6)P′′=〈x1,x2,…,xi,y,xj,xj+1,…,xk,xj-1,xj-2,…xi+1〉.
Similarly we can show the existence of another Hamiltonian path ifi is odd and j is even. The case when both i and j are odd also follows easily.Now, by Claim1, if xi,xj∈Ny, such that either i or j or both are odd, then we have a Hamiltonian path and we are done. Therefore, assume otherwise. Then, since we have k-2 vertices as candidates for membership in Ny, dy≥(k-1)/2, and k is odd, we must have Ny={xj∣jiseven}. Therefore, dy=(k-1)/2.Now, ifi,j are odd, then we must have (xi,xj)∉E, because otherwise we get an odd cycle contradicting our assumption that G is a bipartite graph. Similarly, if i,j are even, then we must have (xi,xj)∉E for the same reason. Then, the two partite sets A and B of our graph G=(A⋃B,E) are defined as follows:
(7)A={xi∣iiseven},B={xi∣iisodd}⋃{y}.Then, we have|B|-|A|>1, which contradicts our assumption that G is semibalanced and this completes the proof of Theorem 15.Interestingly, we have the following theorem as well.Theorem 16.
LetG=(A⋃B⋃C,E) be a tripartite connected graph with n≥5 vertices. If, for all pairs of nonadjacent vertices u, v one has du+dv≥n-2, then G has a Hamiltonian path.Proof.
The proof of this theorem almost exactly follows the proof of Theorem15. The only difference now is that we assume G to be tripartite instead of semibalanced bipartite. Then, we can continue with the same arguments as we did in the proof of Theorem 15 and reach a position where G turns out to be a bipartite graph. This is a contradiction since G is tripartite and this completes the proof.


### 3.1. Discussion

In Theorem12, we have presented a sufficient condition for the existence of Hamiltonian path assuming that, n, the number of vertices of the graph, is odd. When n is even, we have shown our condition to be effective for some classes of graphs. Interestingly, we can construct a graph with even number of vertices which satisfies our condition but still does not possess a Hamiltonian path. For example, consider the graph in Figure 6. The adjacency matrix of the graph of Figure 6 is presented in  Table 1. Now, it is clear from Table 1 that for every pair of nonadjacent vertices u, v in this graph we have du+dv≥10=n-2. However, the graph in Figure 6 is in fact a bipartite graph with partite sets A={x1,x3,x5,x7,x9,x11,x12} and B={x2,x4,x6,x8,x10} such that |A|-|B|>1. Therefore, it cannot possess a Hamiltonian path. Also, note that even if we make the graph nonbipartite by adding some edges in the partite set B, still the graph will not have a Hamiltonian path. In fact, even if we add all the edges to make the partite set B a clique, the graph will not possess a Hamiltonian path. On the other hand, a single edge within the partite set A can provide us a Hamiltonian path.Table 1
Adjacency matrix of the graph of Figure6.
x
1
x
2
x
3
x
4
x
5
x
6
x
7
x
8
x
9
x
10
x
11
x
12
x
1
—
1
0
1
0
1
0
1
0
1
0
0
x
2
1
—
1
0
1
0
1
0
1
0
1
1
x
3
0
1
—
1
0
1
0
1
0
1
0
0
x
4
1
0
1
—
1
0
1
0
1
0
1
1
x
5
0
1
0
1
—
1
0
1
0
1
0
0
x
6
1
0
1
0
1
—
1
0
1
0
1
1
x
7
0
1
0
1
0
1
—
1
0
1
0
0
x
8
1
0
1
0
1
0
1
—
1
0
1
1
x
9
0
1
0
1
0
1
0
1
—
1
0
0
x
10
1
0
1
0
1
0
1
0
1
—
1
1
x
11
0
1
0
1
0
1
0
1
0
1
—
0
x
12
0
1
0
1
0
1
0
1
0
1
0
—
Degree
5
7
5
7
5
7
5
7
5
7
5
5Figure 6
A (problematic) graph with even number of vertices.


## 3.1. Discussion

In Theorem12, we have presented a sufficient condition for the existence of Hamiltonian path assuming that, n, the number of vertices of the graph, is odd. When n is even, we have shown our condition to be effective for some classes of graphs. Interestingly, we can construct a graph with even number of vertices which satisfies our condition but still does not possess a Hamiltonian path. For example, consider the graph in Figure 6. The adjacency matrix of the graph of Figure 6 is presented in  Table 1. Now, it is clear from Table 1 that for every pair of nonadjacent vertices u, v in this graph we have du+dv≥10=n-2. However, the graph in Figure 6 is in fact a bipartite graph with partite sets A={x1,x3,x5,x7,x9,x11,x12} and B={x2,x4,x6,x8,x10} such that |A|-|B|>1. Therefore, it cannot possess a Hamiltonian path. Also, note that even if we make the graph nonbipartite by adding some edges in the partite set B, still the graph will not have a Hamiltonian path. In fact, even if we add all the edges to make the partite set B a clique, the graph will not possess a Hamiltonian path. On the other hand, a single edge within the partite set A can provide us a Hamiltonian path.Table 1
Adjacency matrix of the graph of Figure6.
x
1
x
2
x
3
x
4
x
5
x
6
x
7
x
8
x
9
x
10
x
11
x
12
x
1
—
1
0
1
0
1
0
1
0
1
0
0
x
2
1
—
1
0
1
0
1
0
1
0
1
1
x
3
0
1
—
1
0
1
0
1
0
1
0
0
x
4
1
0
1
—
1
0
1
0
1
0
1
1
x
5
0
1
0
1
—
1
0
1
0
1
0
0
x
6
1
0
1
0
1
—
1
0
1
0
1
1
x
7
0
1
0
1
0
1
—
1
0
1
0
0
x
8
1
0
1
0
1
0
1
—
1
0
1
1
x
9
0
1
0
1
0
1
0
1
—
1
0
0
x
10
1
0
1
0
1
0
1
0
1
—
1
1
x
11
0
1
0
1
0
1
0
1
0
1
—
0
x
12
0
1
0
1
0
1
0
1
0
1
0
—
Degree
5
7
5
7
5
7
5
7
5
7
5
5Figure 6
A (problematic) graph with even number of vertices.


## 4. Conclusion

In this paper, we have presented new degree based sufficient conditions for a graph to contain a Hamiltonian path. It would be interesting to investigate whether our condition could force a graph to contain a Hamiltonian cycle or even to be pancyclic.



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