Further results on maximal nontraceable graphs of smallest size

Let g ( n ) denote the minimum number of edges of a maximal nontraceable (MNT) graph of order n . In 2005 Frick and Singleton (Lower bound for the size of maximal nontraceable graphs, Electronic Journal of Combinatorics , 12(1) R32, 2005) proved that g ( n ) = (cid:100) 3 n − 2 2 (cid:101) for n ≥ 54 as well as for n ∈ I , where I = { 12 , 13 , 22 , 23 , 30 , 31 , 38 , 39 , 40 , 41 , 42 , 43 , 46 , 47 , 48 , 49 , 50 , 51 } and they determined g ( n ) for n ≤ 9 . We determine g ( n ) for 18 of the remaining 26 values of n , showing that g ( n ) = (cid:100) 3 n − 2 2 (cid:101) for n ≥ 54 as well as for n ∈ I ∪{ 18 , 19 , 20 , 21 , 24 , 25 , 26 , 27 , 28 , 29 , 32 , 33 } and g ( n ) = (cid:100) 3 n 2 (cid:101) for n ∈ { 10 , 11 , 14 , 15 , 16 , 17 } . We give results based on “analytic” proofs as well as computer searches.


Introduction
We consider only simple, finite graphs G and denote the vertex set, the edge set and the size (number of edges) of G by V (G), E(G) and e(G), respectively.We denote the complement of G by G.The degree of a vertex v in a graph G is denoted by d G (v).If no confusion can result we simply write d(v).
If S is a nonempty subset of V (G), then S denotes the subgraph of G induced by S.
A graph G is hamiltonian if it has a Hamilton cycle (a cycle containing all the vertices of G), and traceable if it has a Hamilton path (a path containing all the vertices of G).A graph G is maximal nonhamiltonian (MNH) if G is not hamiltonian, but G + e is hamiltonian for each e ∈ E(G).A graph G is maximal nontraceable (MNT) if G is not traceable, but G + e is traceable for each e ∈ E(G).Let x and y be distinct vertices in G. Then G is xy-traceable if G is traceable between x and y, i.e. has a Hamilton path with x and y as endvertices.Also, G is xy-maximal nontraceable (xy-MNT) if G is not xy-traceable (xy-nontraceable), but G + e is xy-traceable for each e ∈ E(G).A graph G is z-traceable if it is traceable from z (i.e.there exists a Hamilton path with z as an endvertex), and it is z-maximal nontraceable (z-MNT) if G is not z-traceable, but G + e is z-traceable for each e ∈ E(G).A graph G is hypohamiltonian if it is not hamiltonian, but G − v is hamiltonian for every v ∈ V (G).
Many MNT graphs have either xy-MNT graphs or z-MNT graphs as subgraphs and thus we have a section devoted to xy-MNT graphs and z-MNT graphs of smallest size.

Preliminary results
For easy reference we include certain results from [8].
then the neighbours of v are adjacent.Also, one of the neighbours has degree at least 4 and the other neighbour has degree 2 or at least 4. Theorem 2.2 [8] Suppose G is a connected MNT graph without vertices of degree 1 or adjacent vertices of degree 2. If G has order n ≥ 7 and m vertices of degree 2, then e(G) ≥ 1 2 (3n + m).We include the main steps of the proof of the following theorem, as the proofs of our main results relate to the construction and cases used in this proof.
for n ≥ 10.Thus we assume that G is connected.We need to prove that the sum of the degrees of the vertices of G is at least 3n − 2. In view of Theorem 2.2, we let The remaining vertices of degree 2 can be dealt with simultaneously with the vertices of degree 1.We let Let m = |M | and s = |S|.If S = ∅, then it follows from Theorem 2.2 that e(G) ≥ 1 2 (3n + m).Thus we assume S = ∅.
We observe that, if H is a component of the graph S , then either An example of such a graph G is depicted in Figure 1.
Since G is MNT the graph S has at most three components.We consider three cases: Case 1. S has exactly three components, say H 1 , H We note that no Hamilton cycle in G − S contains y 1 y 2 , otherwise G would be traceable.Also Case 2.2.2.G − S is nonhamiltonian, i.e.G − S is MNH: We Case 3. S has exactly one component, say H: We introduce the following terminology which we require in proving certain results in order to show that g(10) = 15.
A vertex v in a subgraph H of a graph G is an attachment vertex of H if v has a neighbour in V (G) − V (H).The circumference c(G) of a graph G is the order of a longest cycle in G.We suppose that a cycle C has an orientation.If u, v ∈ V (C) we denote the path on C from u to v (with same orientation as C) by C[u, v] and the other path on C from u to v by C[u, v].The paths obtained by deleting the endvertices u, v are denoted by C(u, v) and C(u, v), respectively.Lemma 2.5 Suppose C is a circumference cycle of a graph G and G − V (C) has a path P = yLz with endvertices y and z, where we also consider V (L) = ∅ as well as The following lemma is a direct result of Lemma 2.5.
We also require the following lemma in order to prove g(10) = 15.
Lemma 2.7 Suppose G is a graph of order 10 as described in Case 3 of Theorem 2.3, with Proof: Suppose G − S has a cut vertex y and that G has the structure as depicted in Figure 2. Consider G + zy.Since G + zy is traceable it follows that B is traceable from y.If B is not complete, then A is traceable between x and y making G traceable.Hence B is complete.
If B = K 3 , then G has two vertices of degree 2, and if B = K 4 , then d G (y) > 3, both of which lead to contradictions.

xy-MNT and z-MNT graphs
In this section we investigate some properties of xy-MNT and z-MNT graphs and show how these graphs can be used in constructing MNT graphs.We also present some xy-MNT graphs of small order and smallest size, as well as some cubic xy-MNT graphs of larger orders. Proof: Then F +xz has a Hamilton path xzQy and consequently F −x has a Hamilton path zQy.Hence F − x is traceable from y.
Then F + ux has a Hamilton path xuP y, and hence uP is a Hamilton path of F − {x, y}, i. e. F − {x, y} is traceable from u.A similar proof holds if uy / ∈ E(F ).
2 Lemma 3.2 Suppose F is an xy-MNT graph with n ≥ 6, where neither x nor y is a universal vertex and (a) Then w is nonadjacent to x and y. Proof: (a) Suppose w is adjacent to x and u / ∈ N (x).Then F + xu is not xy-traceable, a contradiction.A similar argument holds if w is adjacent to y.(a) Then w is nonadjacent to z. Proof: (a) Suppose w is adjacent to z and v, and u / ∈ N (z).Then F + zu has a Hamilton path of the form zuP vw.But then uP vwz is a Hamilton path in F , a contradiction.
z is complete.Then F + zw has a Hamilton path of the form zwv i v j P or zwv i Qv j , where {i, j} = {1, 2}.But then zv i wv j P or zv i Qv j w is a Hamilton path in F , a contradiction.
We consider the following construction.
Construction 1: Suppose F is an xy-MNT graph and H x ∼ = K n and H y ∼ = K m , for positive integers n and m.Suppose the vertex sets V (F ), V (H x ) and V (H y ) are disjoint.Then [F, H x , H y ] denotes the graph obtained by joining each vertex of H x to x and each vertex of H y to y.
Theorem 3.4 Suppose F is an xy-MNT graph and neither x nor y is a universal vertex of Proof: Since F is not xy-traceable it follows that G is not traceable.We now prove that G + uv is traceable for all u, v ∈ V (G), where uv / ∈ E(G).Suppose that u, v ∈ V (F ).Since F + uv has a Hamilton path between x and y it follows that G + uv is traceable.
Suppose u ∈ V (H x ) ∪ {x} and v ∈ V (H y ).Then, since x is not a universal vertex it follows from Lemma 3.1(b) that there is a Hamilton path in G + uv.
Suppose u ∈ V (H x ) and v ∈ V (F ) − {x, y}.Then, from Lemma 3.1(c), there is a Hamilton path in G + uv.
All other cases for u and v follow from symmetry. 2 We now consider z-MNT graphs.It follows from the definition that if F z is a z-MNT graph, then Construction 2: Suppose F z is a z-MNT graph and H z ∼ = K n .Suppose the vertex sets V (F z ) and V (H z ) are disjoint.Then [F z , H z ] denotes the graph obtained by joining each vertex of H z to z.

It is easy to see that the graph
Suppose F is an xy-MNT graph and F z is a z-MNT graph, where x and z are universal vertices and [F, H x , H y ] and [F z , H z ] are MNT graphs of order n ≥ 10.Then V (F ) ∪ V (H y ) − x and F z − z are themselves MNT.If each of H x , H y and H z is either a K 1 or K 2 , then by using Theorem 2.3 and results for the minimum size of MNT graphs of orders less than 10 given in [8] it is not difficult to check that [F, H x , H y ] and [F z , H z ] have size greater than 3n 2 .Thus, since smallest MNT graphs of a specific order may arise from xy-MNT or z-MNT graphs, we searched, with the aid of a computer, for xy-MNT and z-MNT graphs of given order and smallest size in which neither x nor y nor z is a universal vertex.The algorithm for xy-MNT graphs F of order n starts with an initial graph of order n that is necessarily a subgraph of any xy-MNT graph of order n.From Lemmas 3.1(a) and (b) we choose an n-path of the form xyQ as the initial graph.The algorithm then performs a depth-first tree search, systematically adding all possible combinations of edges, but keeping the graph xy-nontraceable.When a predetermined upper bound on the number of edges is reached, that branch is terminated.
We could find all xy-MNT graphs of minimum size of order 5 up to 13 in reasonable computing time.These graphs are depicted in the Figures 3-6.We also established, by computer, that the smallest MNH graphs G 14 of order 14 and size 22, and G 15 of order 15 and size 24 (see Figure 7) given in [9] are in fact xy-MNT.A similar search for z-MNT graphs of minimum size was undertaken for orders up to 12.We noted that for 7 ≤ n ≤ 12 all z-MNT graphs of minimum size can be constructed from xy-MNT graphs of minimum size in one or both of the following ways: (1) Attach a K 1 to x (or y) of a smallest xy-MNT graph of order n − 1.
(2) Attach a K 2 to x (or y) of a smallest xy-MNT graph of order n − 2. If K i , i = 1, 2 is attached to x (or y), then z is y (or x).
We therefore do not include sketches of these graphs.[9].Each of the MNH graphs of orders 12 and 13 have one more edge than the corresponding MNH graphs of smallest size listed in [9] (b) N Fj (x) ∩ N Fj (y) = ∅ for j ∈ {5, 6, 6 * , 7, 7 * , 9, 12 * , 13, 13 * }.In these cases it is easy to check that [F j , H x , H y ] is MNT.
(c) The graph F * 7 is also a z-MNT graph (but not of smallest size), where z is the universal vertex.
(e) In F 10 and in G 15 , x and y can be any two adjacent vertices of the graphs.
A cubic xy-MNT graph with N (x) ∩ N (y) = ∅ of order n can be used to construct an MNT graph of order n = n + 2 or n = n + 3 by attaching a K 1 to x and a K 1 or K 2 to y.These MNT graphs realise the lower bound 3n −2
(b) The graphs Q 16 , Q 18 are hamiltonian.This is to be expected since the smallest MNH graphs of orders 16 and 18 listed in [9] have 25 and 28 edges respectively.
(c) In Q 30 , x and y can be any two adjacent vertices of the graph and in Q 22 , Q 24 and Q 26 , where applicable, x, y can be x , y or x , y .
(d) There are no cubic xy-MNT graphs of order 14.

Main Results
In this section we determine g(n) for n = 10, 11, 14 − 21, 24 − 29, 32, 33.From inequality (2) of Section 2 it follows that the only value of s that may produce an MNT graph G of order 10 and 14 edges is s = 2.
According to [9] the smallest MNH graph of order 8 has 15 edges and thus e(G) > 14.Thus we only consider s = 1, i. e. H ∼ = K 1 .Then, according to Lemma 2.7, G − S is 2-connected.Let z denote the vertex adjacent to H in G, and F = G − S. Then d F (z) = 2 and the other 8 vertices of F must have degree 3 each.Note that F is not traceable from z, otherwise G will be traceable.We show that such a graph G does not exist by attempting to construct F by considering circumference cycles of F .
Let C be a circumference cycle of F .We define, respectively, the degree deficiency, dd(C), of C, and the degree deficiency, dd( and determine possible degree sequences for such graphs.In most cases, for the sake of brevity, we will only state the degree sequence(s) of G and the relevant lemma(s) used for the computation.When determining the degree sequences for G, we use Lemmas 2.1, 3.  Suppose G =[F, H z ] is an MNT graph of order 14 and size 20.Then from inequality (3) of Section 2 we can only have m = 0. We consider subcases according to possible values of s.
(i) m = 0 and s = 1: The only possible degree sequence is 13 13 and we use Lemma 4.4(a) in our computation.
(ii) m = 0 and s = 2: It follows from Table 1 that the minimum number of edges is 21.
We have shown that g(14   We have thus shown that g(16) = 24 and that [G 14 , K 1 , K 1 ] is an MNT graph with order 16 and size 24 (see Figure 7).Obviously, there can be some doubt about the correctness of an algorithm when no graphs are found.The fact that MNH graphs of certain orders that were found in [9] were also found with this algorithm is reassuring.Furthermore, since the algorithm does not take into account isomorphisms of graphs, it is unlikely that a graph would be missed since the algorithm has usually more than one chance to find a graph if it exists.The source code for the algorithm can be viewed at [4].

Conclusion
The smallest size of an MNT graph of order n has now been determined for all except 8 values of n, namely n ∈ K = {34, 35, 36, 37, 44, 45, 52, 53}.The minimum size of MNT graphs of order n ≥ 18, where n / ∈ K, is 3n−2 2 and graphs of this size are constructed from cubic xy-MNT graphs.We suspect that g(n) = 3n−2 2 also for n ∈ K and that these remaining cases could be resolved by finding cubic xy-MNT graphs of orders 32, 34, 42 and 50 either by construction or by the use of more sophisticated algorithms.

2 Lemma 3 . 3
Suppose F is a z-MNT graph with n ≥ 5, where z is not a universal vertex and w ∈ V (F ) − {z} with d(w) = 2.

Case 3 (
S has 1 component, H) According to inequality (3) of Section 2 the only possibility for a graph of order 10 to have 14 edges is when m = 0, i. e. when d G (v) = 2 for all v ∈ V (G) − S. For s = 1, e(G) = 14 ⇐⇒ d G (v) = 3 for all v ∈ V (G) − S. For s = 2, taking Lemma 2.1 into account, we have e 2(a) and 3.3(a), and the fact that d(v) ≥ 3 for v ∈ V (G) − S − M .In the case where G − S is a hamiltonian xy-MNT graph we use the fact that d G−S (x) ≥ 3 and d G−S (y) ≥ 3 (

Tab. 2 :
m s degree sequence Lemma used for computer search 0 1 13 13 4 4.4(b) if d G (z) = 3 with d(u) = 4; 4.4(a) if d G (z) Subcases for n = 15, where G − S is z-MNT We have thus shown that g(15) = 23.The graphs [ are examples of MNT graphs of order 15 and size 23.Case IV: n = 16 We show that no MNT graph G of order 16 and size 23 exists.We first consider the case where G − S is an xy-MNT graph and G − S is MNH.The smallest such MNT graph of order 16 has G 14 as subgraph and has 24 edges.We now assume G − S is hamiltonian.It follows from inequality (1) of Section 2 that m + s ≤ 2. Thus m = 0 and s = 2, and the only possible degree sequence is 113 12 44 since d G−S (x) ≥ 3 and d G−S (y) ≥ 3. Hence G−S must be cubic.Applying Lemma 4.3(a) and a computer search, no graph was found.We next consider the case where G−S is a z-MNT graph.If such a graph exists then from inequality (3) of Section 2 we have m = 0.The possible subcases are: (i) m = 0 and s = 1: The degree sequence is 13 15 and we use Lemma 4.4(a).(ii) m = 0 and s = 2: No degree sequence for G is possible.

Tab. 3 :
Case V: n = 17 We show that no MNT graph G of order 17 and size 25 exists.We first consider the case where G − S is an xy-MNT graph and G − S is MNH.The smallest such MNT graphs of order 17 have either G 14 or G 15 as subgraphs and have 26 edges.We thus need to consider the case where G − S is a hamiltonian xy-MNT graph in which case m + s ≤ 3 (inequality (1)); and the case where G − S is a z-MNT graph in which case m ≤ 1 (inequality (3)).Table 3 summarizes all possible subcases.G − S m s degree sequence Lemma used for computation hamiltonian 0 2 113 13 45 or 4.3(a) xy-MNT 0 2 113 12 444 4.3(b) with d(u) b) if d G (z) = 3 with d(u) = 4; 4.4(a) if d G (z) Subcases for n = 17, where G − S is hamiltonian xy-MNT or z-MNT Hence g(17) = 26 and [G 14 , K 1 , K 2 ] and [G 15 , K 1 , K 1 ] are examples of MNT graphs of order 17 and size 26.2

Theorem 4 . 6
Let g(n) denote the minimum number of edges of an MNT graph of order n.Then g Let N G (H i ) = y i , i = 1, 2 and y 1 = y 2 .Then y 1 y 2 ∈ E(G).Now G − S is not complete.If v and w are nonadjacent vertices in V (G − S), then (G − S) + vw has a Hamilton cycle containing y 1 y 2 .Hence G − S is either hamiltonian or MNH.We consider these two cases separately: Case 2.2.1.G − S is hamiltonian: 2 , H 3 : In this case the neighbourhoods of H 1 , H 2 , H 3 are pairwise disjoint, G − S is a complete graph of order at least 4 and hence e(G) = n − s 2 + 2s − 3 for s = 3, 4, 5 or 6.This case is a Zelinka Type II construction, cf.[14].Case 2. S has exactly two components, say H 1 , H 2 : There are two subcases.Case 2.1.N G (H 1 ) = N G (H 2 ):Then G − S is a complete graph and hence Table 1 gives the minimum sizes of xy-MNT graphs and z-MNT graphs of orders up to 15 and 13 respectively.The xy-MNT graphs F 8 , F 11 , F 12 and F 13 are hamiltonian whereas the others in Figures 3-6 are MNH.The MNH ones for orders 5, 6, 7, 9 and 10 are those listed as the MNH graphs of smallest size in see Case 2.2.1 of Theorem 2.3).For all cases, no graphs with the relevant degree sequences were found when doing a computer search and hence we do not state this in every case.Case I: n = 11 According to Table 1 the smallest MNT graphs of order 11 which can be constructed from xy-MNT or z-MNT graphs have 17 edges.Hence g(11) = 17.The graphs [F 8 , K 1 , K 2 ] and [F 9 , K 1 , K 1 ] are examples of such graphs.Case II: n = 14 From Table 1 it follows that the smallest MNT graph of order 14 that can be constructed from an xy-MNT graph has 21 edges.
are examples of MNT graphs of order 14 and size 21.Case III: n = 15From Table1it follows that the smallest MNT graph of order 15 that can be constructed from an xy-MNT graph has 23 edges.Suppose G =[F, H z ] is an MNT graph of order 15 and size 22, where F is a z-MNT graph.Then from inequality (3) of Section 2 we have m ≤ 1. Table2summarizes the possible subcases.