2Trees: Structural Insights and the study of Hamiltonian Paths
Abstract
For a connected graph, a path containing all vertices is known as Hamiltonian path. For general graphs, there is no known necessary and sufficient condition for the existence of Hamiltonian paths and the complexity of finding a Hamiltonian path in general graphs is NPComplete. We present a necessary and sufficient condition for the existence of Hamiltonian paths in 2trees. Using our characterization, we also present a lineartime algorithm for the existence of Hamiltonian paths in 2trees. Our characterization is based on a deep understanding of the structure of 2trees and the combinatorics presented here may be used in other combinatorial problems restricted to 2trees.
1 Introduction
Hamiltonian path (cycle) problem is one of the most extensively studied problem, that looks for a spanning path (cycle) in a connected graph. Interestingly, such a problem has many applications in real life, related to medical genetic studies[9], for chromosome studies, in physics [3] and operational research [14]. Hamiltonian problem is one among the NPcomplete problems in general graphs[39].
For a graph, the fundamental research question is to find a necessary and sufficient condition for the existence of a Hamiltonian path (Hamiltonian cycle). Surprisingly, there is no known necessary and sufficient condition despite many attempts from several researchers [12, 11, 29, 6, 46, 35, 2, 33, 48, 21]. However, there are wellknown necessary conditions and sufficient conditions.
Necessary condition by V.Chvatal [8] states that if a connected graph has a Hamiltonian cycle, then for each nonempty subset , the graph has at most components.
Sufficient condition looks for structural conditions for a graph to have a Hamiltonian cycle, mostly
for the presence of higher degree vertices in a graph. Sufficient conditions based on vertex degree has been proposed in the literature [12, 11, 29, 6, 46, 35, 2, 33]. Other sufficient conditions based on graph closure, independence number and connectivity have also been formulated [48, 21].
Interestingly, several variants of Hamiltonian path (Hamiltonian cycle) have been looked at in the past by imposing appropriate constraints. A graph is said to be homogeneously traceable, if there exist a Hamiltonian path beginning at every vertex of . A hypoHamiltonian graph is a nonHamiltonian graph such that is Hamiltonian for every vertex, . Existence of homogeneously traceable graph and hypoHamiltonian graph [7, 50, 13] were studied in the literature. A graph is ordered Hamiltonian if for every ordered sequence of vertices, there exists a Hamiltonian cycle that encounters the vertices of the sequence in the given order. If there exist a Hamiltonian path between every pair of vertices then the graph is called Hamiltonian connected. A pancyclic graph on vertices is a graph which has every cycle of length . Sufficient conditions for the existence of ordered Hamiltonian, Hamiltonian connected, and pancyclic graphs, similar to Ore’s and Dirac’s results have also been proposed in the literature [26, 16, 34, 32].
On the algorithmic front, it is wellknown that Hamiltonian path (Hamiltonian cycle) is NPcomplete. When a combinatorial problem is NPcomplete in general graphs, it is natural to study the complexity on restricted graph classes or special graph classes. The popular graph classes studied in the literature are chordal, interval, grid, chordal bipartite, distance hereditary, circular arc, cubic, and planar. It is proved that Hamiltonian problem is NPcomplete on various restricted graph classes like chordal [1], grid [49], chordalbipartite [19], planar [28], bipartite [45], directed path graph [15] and rooted directed path graph [4]. On the other side, nice polynomialtime algorithms for the same has been found on interval [23, 41], circular arc [42, 51], proper interval [5, 25], distance hereditary[40], and specific sub class of grid graphs [24]. Nice structural characterization for the existence of Hamiltonian cycle in claw free graphs [44, 10, 38, 18, 30, 31, 22] has been studied in the past as well. A detailed survey on the Hamiltonian properties has been compiled by Broersma and Gould [17, 36, 37].
Chordal graphs are one among the restricted graph classes possessing nice structural characteristics. A graph is said to be chordal if every cycle of length more than three has a chord. A chord is an edge joining two nonconsecutive vertices of a cycle. Given that chordal graphs have polynomialtime algorithm on various classical combinatorial problems such as vertex cover, clique, it is natural to investigate the complexity of Hamiltonian problems on chordal graphs. As already mentioned, Hamiltonian cycle problem on chordal graphs is NPcomplete, this brings our focus on some subclasses of chordal graphs. Interestingly, interval graphs, a quite popular subclass of chordal graphs have a polynomialtime algorithm for Hamiltonian problem. Similarly, other special graph classes like properinterval graphs and circular arc graphs also possess polynomialtime algorithms. To the best of our knowledge, these are the only polynomialtime results for Hamiltonian cycle problem on the sub class of chordal graphs.
The objective of this paper is two fold. First, we present structural insights on 2trees. Further, we present a necessary and sufficient condition for the existence of Hamiltonian paths, and using the characterization, a polynomialtime algorithm to obtain Hamiltonian paths in 2trees is also presented.
Our Approach: Given a 2tree , we perform a series of computations to obtain a Hamiltonian path. We first check whether is pyramid free. If so, we output a Hamiltonian path. We next check whether is pyramid free and contains exactly one pyramid. If so, contains a Hamiltonian path. If is pyramid free and contains at least two pyramids, then we first perform a pruning of the 2tree by removing 2degree vertices iteratively satisfying some structural condition. During pruning, we also color the edges, in particular if an edge in is colored blue during pruning, it indicates that there is a pyramid free sub 2tree with as the base 2tree. We also observe that the first level pruning yields a pyramid free 2tree with some edges are colored blue. On this pruned 2tree we identify five sets of edges (nonblue edges) which will be removed from . The existence of Hamiltonian path in is determined based on some structural conditions on this simplified graph. We also highlight that each pruning step is a solution preserving step and indeed guarantees a Hamiltonian path.
Road Map: We next present graph preliminaries. In Section 2, we present a necessary and sufficient condition for a 2tree to have Hamiltonian paths and Hamiltonian cycles. The algorithm for finding a Hamiltonian path in a 2tree is presented in Section 2.4.
1.1 Graph Preliminaries
Notation is as per [20]. In this paper we work with simple, connected, unweighted graphs.
For a graph the vertex set is and the edge set is and is adjacent to in and .
The neighborhood of vertex is .
The degree of a vertex is .
denotes the maximum degree in .
For a vertex , close(u).
For an edge , close(e) .
A 2tree G can be inductively constructed as follows.
An edge is a 2tree. If G is a 2tree on vertices, then select an edge and add a vertex to such that ; is also a 2tree on vertices.
We call a 2tree , if it has vertices and an edge such that . A pyramid is shown in Figure 1. We call a 2tree , pyramid free if contains no pyramid as an induced subgraph.
denotes a complete graph on vertices.
A vertex is called a simplicial vertex if induces a complete subgraph of [27].
Perfect vertex Elimination Ordering (PEO) is an ordering of the vertices of a graph as () such that each is a simplicial vertex of the induced subgraph on vertices . Note that by definition 2trees are chordal.
An (s,t)Hamiltonian path is a Hamiltonian path from to . If , then the induced subgraph is represented as . For , we also use . is free if does not contain as an induced subgraph. An edgeinduced subgraph of is formed on the edge set and edge is incident on the vertex . denotes the number of connected components in the graph . For a connected graph , =. is a vertex separator if . A cut vertex is a vertex such that . represents the component of a disconnected graph containing a vertex . A connected component is a component without a cut vertex. A block is a maximal 2connected component of a graph. Path from vertex to , is represented as , where vertices are termed as internal vertices of . We use to represent and hence . Blue path is a path with all its edges blue.
2 Structural insights into 2trees
In this section we shall present some insights into the structure of 2trees. Below observation is a wellknown characteristics of any 2tree.
Observation 1
Let be a 2tree. forbids , and as an induced subgraph.
Lemma 1
Let be a 2tree and . If , then .
Proof
We use induction on . The claim is immediate for . For , let be a simplicial vertex in such that . From the induction hypothesis, in , for every , . Clearly, for every , , and . This completes the induction.
Theorem 2.1 (Chvatal [8])
If a graph has a Hamiltonian cycle, then for every , .
Theorem 2.1 is a wellknown necessary condition for Hamiltonicity in general graphs. Also there is no necessary and sufficient condition for hamiltonicity in general graphs. In Theorem 2.2, we present a necessary and sufficient condition for the existence of Hamiltonian cycles in 2trees. Further, we show that Theorem 2.1 is indeed sufficient for 2trees, which we establish using Theorem 2.2, and Theorem 2.3.
Observation 2
For every , any pyramid free 2tree is also a pyramid free 2tree.
Theorem 2.2
Let G be a 2tree. G has a Hamiltonian cycle if and only if is pyramid free.
Proof
Necessity:
Assume for a contradiction that has a pyramid. This implies there exist such that . By Lemma 1, . Further, by Theorem 2.1, has no Hamiltonian cycle, a contradiction to the premise.
Sufficiency:
For any pyramid free 2tree on more than two vertices, the unique Hamiltonian cycle of is obtained by using the edge set .
Theorem 2.3
Let G be a 2tree. For every , if and only if G is pyramid free.
Proof
Corollary 1
For a 2tree , has a Hamiltonian cycle if and only if for every , .
Proof follows from Theorem 2.2 and 2.3.
It is easy to see that graphs with Hamiltonian cycles contain
Hamiltonian paths as well. However, the converse is not true always. Like
Hamiltonian cycle problem there is no known necessary and sufficient
condition for the existence of Hamiltonian paths in general graphs. We
below recall a necessary condition on graphs having Hamiltonian paths.
Lemma 2
Let be a connected graph. If has a Hamiltonian path, then for every , .
Proof
Suppose to the contrary assume that in there exist at least components. Any Hamilton path switches between different components at least times each time using a different element of , a contradiction.
Lemma 3
Let be a 2tree. If contains a pyramid as an induced subgraph, then has no Hamiltonian path.
Proof
The converse of the above lemma is not true and a counter example is illustrated in Figure 2. The example highlights the fact that there exist 2trees with no pyramid and contain pyramids, yet it does not have Hamiltonian paths. We shall now focus our structural analysis on 2trees containing pyramids. In Lemma 4, we show that pyramid free 2trees having exactly one pyramid has a Hamiltonian path.
Lemma 4
Let be a pyramid free 2tree. If contains exactly one pyramid as an induced subgraph, then there exist a Hamiltonian path in .
Proof
Let the pyramid is on the edge . Note that . By Lemma 1, and let and be those components. Let , , and . Consider the graphs induced on , respectively. Clearly each is a 3pyramid free 2tree. By Theorem 1, each has a Hamiltonian cycle and hence a Hamiltonian path. The Hamiltonian path of and are , , and , respectively. The path is a Hamiltonian path in .
We next present some combinatorial observations on pyramid free 2trees with at least two pyramids for the existence of Hamiltonian paths. We also observe that not all such 2trees possess Hamiltonian paths. From now on we shall work with such 2trees for our discussion.
2.1 A Simplification (Vertex Pruning)
We now present an approach that transforms a pyramid free 2tree with pyramids into a 2tree without pyramids. Intuitively, for such a pyramid with base edge , there are three 2trees growing out of . While pruning, out of the three 2trees we retain two and prune the other. While doing so, to remember the pruned 2tree, we introduce coloring and labeling as part of our approach. Coloring of signifies that there is a 2tree growing from and signifies the vertices of .
For a 2tree , by vertex pruning we remove vertices of degree 2 satisfying some property and color some of the edges in , based on the closeness property. In particular, a vertex of degree 2 is pruned if its close edge, is not colored and on pruning , is colored blue.
Let , and on deleting , we color the vertices blue and also the edge blue.
We remember the pruned vertices using a label associated with . Initially all the edges are unlabeled,
i.e., (empty string) for every .
On deleting , we label as follows:

if , then

if , and then

if , and then

otherwise
For example, if the blue edges and are labeled and , respectively, then the label of the new blue edge will be .
For any 2tree , we define a sub 2tree of , which is obtained by recursively pruning 2degree vertices of such that is uncolored. Note that if is a pyramid free 2tree, then is pyramid free.
Further, for every 2degree vertex in , is blue. Since is pyramid free, contains a Hamiltonian cycle, and hence a Hamiltonian path as well. However, our objective is to find a Hamiltonian path in containing all the blue edges, as labels of blue edges records the pruned vertices. Further, such a Hamiltonian path can be easily extended to a Hamiltonian path in using the labels. Given this observation, we would like to investigate to get some more insights. We call as the vertex pruned 2tree of . An expanded 2tree of is a 2tree obtained by growing each blue edge in with a pyramid free 2tree corresponding to the label of the blue edge.
We define the Blue graph of as a sub graph induced on the blue edges of .
For the next lemma we consider a pyramid free 2tree with at least two pyramids and let be the vertex pruned 2tree of and be the blue graph of .
Lemma 5
If has a Hamiltonian path, then the following hold:
(i) has exactly two vertices of degree .
(ii)
(iii) For such that and , at most one of has degree in
Proof
(i) Clearly, there are at least two vertices of degree in as has at least two pyramids. Assume for a contradiction that there exist at least three vertices of degree in such that , , (see Figure 3).
Clearly, has a Hamiltonian cycle and hence a Hamiltonian path. Now we claim that any longest path in (which is a Hamiltonian path in ) cannot be transformed to a Hamiltonian path in . Note that one of and has a specific order of appearence in . In particular, either or or appear consecutively in . Without loss of generality, let appear consecutively in . While extending to , we must include , thus we get . However in this extension is unvisited in . Therefore, is not a Hamiltonian path. This shows that any can not be extended to any Hamiltonian path in , a contradiction to the premise.
(ii) If , then there exist such that . Since is pyramid free and contains a Hamiltonian cycle, is connected. Further, there exist a path in and for every , and are in different components of . Note that , and by Lemma 2, has no Hamiltonian path, a contradiction.
(iii) Assume for a contradiction that . Note that the edge is a blue edge.
If there exist a blue edge , such that , then , and by Lemma 2, has no Hamiltonian path, a contradiction. Hence we can assume that there exist two edges such that , , and are blue as shown in Figure 4. Symmetric argument holds for the vertex . We now show by case analysis that any longest path in can not be transformed into any Hamiltonian path in . Since is a Hamiltonian path, must contain the vertex . Depending on the position of in we see various possibilities for as follows. , , , , . Now we shall show that each of the above Hamiltonian paths in can not be extended to any Hamiltonian path in .
We present the detailed case analysis in Table 1.
Case  Justification 

Expanding to  Since neither starts or ends in or , the vertices in remain unvisited in 
Expanding to  Since starts or ends at , on expanding, one among or the vertex remain unvisited in . 
if visits the vertices , then the vertices in remain unvisited in .  
if visits the vertices , then the vertices in remain unvisited in .  
if visits the vertices , then the vertex remain unvisited in .  
if visits the vertices , then the vertices in remain unvisited in .  
Expanding to  Since starts or ends at , on expanding, one among or the vertex remain unvisited in . 
Arguments are symmetric to that of  
Expanding to  On expanding, one among or the vertex remain unvisited in . 
if visits the vertices , then the vertex remain unvisited in .  
if visits the vertices , then the vertices in remain unvisited in .  
if visits the vertices , then the vertices in remain unvisited in .  
Expanding to  On expanding, one among or the vertex remain unvisited in . 
Arguments are symmetric to that of 
2.2 Another Simplification
In the previous section we have investigated the structure of pyramid free 2trees with at least two pyramids by introducing the notion vertex pruning. In this section, we shall obtain some more insights by introducing another simplification. Our definition of and remains the same and in this section, we do not work with arbitrary , instead, we work with satisfying the following conditions to obtain the 2tree .
(i) has exactly two vertices of degree .
(ii)
(iii) For such that and , at most one of has degree in
Note that this is precisely the conclusion of Lemma 5.
The results presented in this section are based on such restricted and its corresponding .
We define the 2tree obtained from such as follows; let be two vertices of degree in and =, and and .
We shall classify four types of Hamiltonian paths in based on .
Type 1 Hamiltonian path if , , , and .
Type 2 Hamiltonian path if and .
Type 3 Hamiltonian path if , and and .
Type 4 Hamiltonian path if , and and .
Theorem 2.4
has a Hamiltonian path if and only if has type 1 or type 2 or type 3 or type 4 Hamiltonian path containing all the blue edges of .
Proof
Sufficiency: Let be a Hamiltonian path containing all the blue edges. Replace every blue edge with where is the label of the blue edge in to get the expanded path . When is further extended by including labels, we get a path , which is a Hamiltonian path in .
Necessity: Let is a blue edge in . Assume for a contradiction that there is no Hamiltonian path in such that appear consecutively in . That is, does not contain all blue edges of . On expanding , to get a path in , clearly the does not appear in . This implies that is not a Hamiltonian path in , contradicting the premise.
Lemma 6
If has a Hamiltonian path, then .
Proof
Assume for a contradiction that there exist a vertex such that . Since there exist a Hamiltonian path in , by Theorem 2.4, there exist a Hamiltonian path in containing all the blue edges of . Clearly, must contain all three blue edges incident on . However, it is well known that a path can not contain three edges having a vertex in common, a contradiction. Therefore, no such exists.
Lemma 7
If has a Hamiltonian path, then for every vertex such that , , one of the following holds:
(1) ,
(2) ,
(3)
Proof
Note that since the edge is blue, . From Lemma 6 it follows that . So . We now show that is not possible. Assume for a contradiction, . The proof of this claim is similar to Lemma 5.(iii) with minor modification on technical details. If there exist a blue edge , such that , then , and further . By Lemma 2, has no Hamiltonian path, a contradiction. Hence we can assume that there exist an edge such that , and is blue as shown in Figure 5. Also, , and is blue in . Any longest path in (which is also a Hamiltonian path in ) must contain . That is, can be one of , , . Since has a Hamiltonian path, when is extended to a Hamiltonian path in , it will include vertices and . On such expansion path will give one of to mentioned in Lemma 5. In particular is expanded to , , to , , and to . At this point an anlaysis similar to Lemma 5 will establish that to can not be extended to any Hamiltonian path in . This shows that can not be extended to any Hamiltonian path in , a contradiction.
Although in Lemma 5 we have shown is at most , there are exactly four 2trees for which . For the rest is at most which we shall prove in the next lemma.
To present the next lemma we fix the following notation. We define four special pyramid free 2trees, and as follows. and . All the edges incident on are blue for each . Additionaly, the edge is blue in , the edge is blue in , and the edges are blue in . Note that each is a for some .
Lemma 8
If has a Hamiltonian path, and , then .
Proof
Assume for a contradiction that there exist . If , then the structure of is similar to and the only difference is the edge is blue in . Then note that and by Lemma 2, has no Hamiltonian path, a contradiction. Therefore, . Let , , and the edges are blue. Clearly, there exist paths in . If is not an edge in or is a blue edge, then , again a contradiction. Therefore, must be an edge and is not blue. Now we shall see the adjacency of vertices . Clearly, there is no such that or or . Existence of such yields pyramid in the former and in the later, a contradiction. If or , then either or , a contradiction. To complete the proof we shall focus on . Let . Note that is not blue for . Further, and by Lemma 5.(i), there exist exactly two vertices of degree 2 in . Let such that . Note that the 2tree obtained from on removing has , a contradiction to Lemma 6. Symmetric argument holds for the path , and this completes a proof.
2.3 Yet Another Simplification (Edge Pruning)
In Section 2.1 we have introduced first level pruning with the help of coloring and labeling of edges. This helps to record the pruned vertices and further we obtained nice structural results on the blue graph. It is natural to ask whether the existence of Hamiltonian path in is guaranteed (necessary and sufficient condition) using the Hamiltonian path containing all blue edges of . Surprisingly, the answer is no. However, using the second level pruning presented in Section 2.2, we can guarantee a Hamiltonian path in using a Hamiltonian path containing blue edges. Having highlighted this, it is natural to prune unnecessary (not part of any Hamiltonian path) nonblue edges from (), and this is the objective of this section.
With the definition of as before we shall introduce the following notations with respect to .
We work with a unique PEO of such that .

Separator edges =

Nonseparator edges = .

The left nonseparator edge of a vertex with is left() such that and .

The right nonseparator edge of a vertex with is right() such that and .

Star vertices = such that

A forced star refers to a star vertex with the blue left nonseparator edge. If is a forced star, then such that , is also a forced star.

A double forced star refers to a forced star vertex with the blue right nonseparator edge.

For a blue separator edge incident on a star vertex , we define left separator edge, left()= such that there is no where , .

Similarly, right separator edge, right()= such that there is no where , .
With reference to Figure 6, the left and right separator edges of the blue edge of are and , respectively. The left and right nonseparator edges of a star vertex are and , respectively.
Observation 3
For each , there exists at least three separator edges incident on , and for each , there exist exactly two nonseparator edges incident on .
As mentioned before, the objective of this section is to prune unnecessary nonblue edges in and towards this end, we define five sets of edges, and (defined in Table 2) whose removal from yields the graph . Since can not be empty, need not be a 2tree. In this section, we do not work with arbitrary , instead, we work with satisfying the following conditions to obtain .

.

For every vertex such that , , one of the following should hold

,

,


Note that this is precisely the conclusions of Lemmas 6 and 7.
Set  Definition  Intuitive justification example 
(see Figure 6)  
is not blue and or  If there exist , then  
nonblue edges incident on are  
not part of any Hamiltonian  
path.  
For and is blue,  Hamiltonian path either  
or and and is not blue .  follows  