Maximal knotless graphs
MMAXIMAL KNOTLESS GRAPHS
LINDSAY EAKINS, THOMAS FLEMING, AND THOMAS W. MATTMAN
Abstract.
A graph is maximal knotless if it is edge maximal for theproperty of knotless embedding in R . We show that such a graph has atleast | V | edges, and construct an infinite family of maximal knotlessgraphs with | E | < | V | . With the exception of | E | = 22, we showthat for any | E | ≥
20 there exists a maxmal knotless graph of size | E | .We classify the maximal knotless graphs through nine vertices and 20edges. We determine which of these maxnik graphs are the clique sumof smaller graphs and construct an infinite family of maxnik graphs thatare not clique sums. Introduction
A graph G is maximal planar if it is edge maximal for the property ofbeing a planar graph. That is, G is either a planar complete graph, or elseadding any missing edge to G results in a non-planar graph. Maximal planargraphs are triangulations and are characterized by the number of edges: aplanar graph with | V | ≥ | E | = 3 | V | − intrinsically linked ifevery embedding of the graph in R contains a non-split link. Some earlyresults on maximal linkless (or maxnil ) graphs–those that are edge maximalfor the property of not being intrinsically linked–include a family of maximallinkless graph with 3 | V | − Q (13 ,
3) isa splitter for intrinsic linking, a property that implies it is maximal linkless[Mh]. Recently there have been several new results including families ofmaxnil graphs with 3 | V |− | V | edges [A], and with | V | edges [NPP]. Lower bounds for thenumber of edges required for a maxnil graph have been established [A], andmethods for creating new maxnil graphs via clique sum have been developed[NPP].We extend this work with what appears to be the first study of maximalknotless graphs. A graph is intrinsically knotted (IK) if every embedding in R includes a non-trivially knotted cycle, and a graph is not IK or nIK if ithas a knotless embedding, that is, an embedding in which every cycle is atrivial knot. We will call a graph that is edge maximal for the nIK propertymaximal knotless or maxnik. In Section 2, we establish a connection between maximal 2-apex graphsand maxnik graphs, specifically that a 2-apex graph is maxnik if and only a r X i v : . [ m a t h . G T ] F e b f it is maximally 2-apex. This connection is instrumental in allowing theidentification of all maxnik graphs with nine or fewer vertices, and with20 or fewer edges. We remark that there is an analogous connection be-tween maximal apex graphs and maxnil graphs that may be of independentinterest.We consider clique sums of maxnik graphs in Section 3, and are able toestablish similar, if weaker, results to those of [NPP]. Most importantly, weshow that the edge sum of two maxnik graphs G and G on an edge e ismaxnik if e is non-triangular (i.e., not part of a 3-cycle) in at least one G i .Similarly, we provide conditions that ensure that the clique sum over K oftwo maxnik graphs is again maxnik. These results are used in Section 4 toconstruct new maxnik graphs from those found in Section 2.We then turn to studying general properties of maxnik graphs in Section4. We establish a lower bound for the number of edges in a maxnik graph of | V | , and construct an infinite family of maxnik graphs with fewer than | V | edges. A maximal planar graph has | E | = 3 | V | −
6, and maximal k -apexgraphs also have a fixed number of edges depending on | V | . In contrast,the number of edges in maxnil and maxnik graphs can vary. We show that,except for | E | = 22, given any | E | ≥
20, there exists a maxnik graph of size | E | .We will call a maxnik graph composite if it is the clique sum of two smallergraphs. Otherwise we say it is prime . These terms are analogous to knots,where a knot is composite if is the connected sum of two non-trivial knots,and prime otherwise. The infinite families of maxnik graphs constructedin Section 4 are all composite, as they are clique sums of smaller maxnikgraphs. In Section 5, we classify the maxnik graphs found in Section 2 andconstruct an infinite family of prime maxnik graphs.2. Classification through order nine and size 20
Theorem 2.1.
A maxnik graph is -connected. If | V | ≥ , then δ ( G ) ≥ .If | V | ≥ , then ≤ | E | ≤ n − .Proof. Suppose G is maxnik. If G is not connected, then, in a knotlessembedding, add an edge e to connect two components. This is a knotlessembedding of G + e , contradicting G being maximal knotless.Suppose G has connectivity one with cut vertex v . Label the two compo-nents of G \ v as A and B . Let a be a neighbor of v in A and b be a neighborof v in B . These must exist as G is connected. We will argue that G + ab is also nIK, a contradiction.Form an embedding of G as follows: Embed A and B so that they areknotless and disjoint. Embed v on a plane separating them. Embed edgesfrom v to A on the A side of the plane, edges from v to B on the B side, sothat the embedding remains knotless. Now, isotope the rest of A (and B )until edge va (and vb ) is embedded in the plane. Next add edge ab so thatthe triangle abv bounds a disk. ny cycle contained in A (or in B ) is an unknot. Any cycle c that usesvertices from both A and B must use at least two vertices in the triangle abv . Since abv bounds a disk, this means the cycle c is a connected sum ofa cycle in A and a cycle in B . Since those are unknots, c must be as well.This shows G + ab is nIK, contradicting G being maxnik. So a maxnik graphcannot have connectivity one and must be 2-connected.Suppose G is a maxnik graph with | V | ≥
3. Since G is connected, δ ( G ) >
0. If v ∈ V ( G ) has degree one, let u be the neighbor of v and w (cid:54) = v adifferent neighbor of u . In a knotless embedding of G , we can introduce theedge vw that closely follows the path v, u, e . This gives a knotless embeddingof G + vw , contradicting the maximality of G .Suppose G is maxnik with | V | ≥
7. The lower bound on size is a conse-quence of the observation [JKM, Mt] that an IK graph has at least 21 edges.The upper bound follows as a graph with | E | ≥ | V | −
14 has a K minorand is therefore IK [Md, CMOPRW]. (cid:3) In Theorem 4.3 below, we construct an infinite family of maxnik graphs,each with δ ( G ) = 2. Theorem 2.2. A -apex graph is maxnik if and only if it is maximal -apex.Proof. Let G be 2-apex. If G is not maximal 2-apex, then there is an edge e so that G + e is 2-apex, hence nIK [BBFFHL, OT]. This shows that G is notmaxnik. Conversely, if G is maximal 2-apex there are two cases, dependingon | V | . If n = | V | <
7, then K n is 2-apex, so G = K n . But, K n is also nIKand therefore maxnik. If | V | ≥
7, then | E | = 5 | V | −
15. Since G is 2-apex,it is nIK. Adding any edge e , we have G + e with 5 | V | −
14 edges. It followsthat G has a K minor and is IK [Mt, CMOPRW]. This shows that G ismaxnik. (cid:3) A similar result, with essentially the same proof, holds for maxnil.
Theorem 2.3.
An apex graph is maxnil if and only if it is maximal apex.
Theorem 2.4.
For | V | = n ≤ , K n is the only maxnik graph. The onlymaxnik graphs for n = 7 and are the three -apex graphs derived fromtriangulations on five and six vertices.Proof. In [Mt, Proposition 1.4] it’s shown that every nIK graph of order 8or less is 2-apex. So, the maxnik graphs are the maximal 2-apex graphs.For n ≤
6, all graphs are 2-apex, so K n is the only maximal knotless graph.For n = 7, the maximal 2-apex graph is K − , formed by adding two verticesto the unique graph with a planar triangulation on five vertices, K − . Thetwo maximal planar graphs on 8 vertices are formed by adding two verticesto the two triangulations on six vertices, the octahedron and a graph whosecomplement is a 3-path. We will call these graphs K − K − P . (cid:3) et E (called N in [HNTY]) be the nIK nine vertex graph in the Hea-wood family. Figure 2 in Section 4 below shows a knotless [Mt] embeddingof E . Theorem 2.5.
The graph E is maxnik.Proof. That E is nIK is established in [Mt]. Up to symmetry, there aretwo types of edges that may be added. One type yields the graph E + e ,shown to be IK (in fact minor minimal IK or MMIK) in [GMN]. The otherpossible addition yields a graph that has as a subgraph F in the Heawoodfamily. Kohara and Suzuki [KS] established that F is MMIK. (cid:3) Figure 1.
A knotless embedding of G , . Theorem 2.6.
There are seven maxnik graphs of order nine.Proof.
The seven graphs are the five maximal 2-apex graphs with 30 edges, E , and the graph G , , shown in Figure 1. Note that G , is the comple-ment of K (cid:116) K (cid:116) C . Theorems 2.2 and 2.5 show that six of these sevengraphs are maxnik. To see that G , is as well, note that the embeddingshown in Figure 1, due to Ramin Naimi [N], is knotless. Up to symmetry,there are two ways to add an edge to the graph. In either case, the newgraph has a K minor and is IK.It remains to argue that no other graphs of order nine are maxnik. Weknow that order nine graphs with size 21 or less are either IK, the graph E ,or else 2-apex, see [Mt, Propositions 1.6 and 1.7]. Using Theorem 2.2, thiscompletes the argument for graphs with | E | ≤
21. Suppose G is maxnik oforder nine with | E | ≥
22. By Theorem 2.1, we can assume | E | ≤
30. If G is 2-apex, by Theorem 2.2, it is one of the five maximal 2-apex graphs. So,we can assume G is not 2-apex. The minor minimal not 2-apex (MMN2A)graphs through order nine are classified in [MP]. With a few exceptions hese graphs are also MMIK. If G has an IK minor (including a MMIKminor) it is IK and not maxnik. So, we can assume G has as a minor agraph that is MMN2A, but not MMIK. There are three such graphs. Oneis E , the other two, G and G , have 26 and 27 edges. In Theorem 2.5,we showed that E is maxnik. The other two are subgraphs of G , . Tocomplete the proof, we observe that any order nine graph that contains G is either a subgraph of G , or else IK and similarly for G . In fact, forthose that are IK, we can verify this by finding a MMIK minor, either inthe K or K , , , family, or else the graph G , described in [GMN]. (cid:3) Theorem 2.7.
The only maxinik graph of size 20 is K − . There are sevenmaxnik graphs with at most 20 edges.Proof. Work above establishes this through order nine. The seven maxnikgraphs with at most 20 edges are the seven on seven or fewer vertices.Suppose G of order ten or more and size 20 is maxinik. By [Mt, Theorem2.1], G is 2-apex and therefore maximal 2-apex. But this means | E | =5 | V | − ≥
35, a contradiction. (cid:3)
Remark 2.8.
A computer search suggests that E is the only maxnik graphof size 21. The search makes use of the 92 known MMIK graphs of size 22,see [FMMNN] . Clique sums of maxnik graphs
Clique sums of maxnil graphs were studied in [NPP], and we will showsimilar, if weaker, versions in the case of maxnik graphs. These results areused in Section 4.
Lemma 3.1.
For t ≤ , the clique sum over K t of nIK graphs is nIK.Proof. Let G and G be nIK graphs, and let Γ( G ) denote the set of allcycles in G . Let G be the clique sum of G i over a clique of size t . Let f i bean embedding of G i that contains no non-trivial knot.Suppose t = 1. We may extend the f i to an embedding of G by embedding f ( G ) in 3-space with z >
0, and f ( G ) with z < G = G ∪ v G , soby isotoping vertex v from each G i to the plane z = 0 and identifying themthere, we have an embedding f ( G ). A closed cycle in G must be containedin a single G i , and hence given c ∈ Γ( G ), then c ∈ Γ( G i ) for some i . As theembeddings f i ( G i ) contain no nontrivial knot, c must be the unknot, andhence G is nIK.Suppose t = 2. We may extend the f i to an embedding of G by embedding f ( G ) in 3-space with z >
0, and f ( G ) with z < G = G ∪ e G , so byshrinking the edge e in each G i and then isotoping them to the plane z = 0and identifying them there, we have an embedding f ( G ). A closed cycle c ∈ Γ( G ) must either be an element of Γ( G i ), or c = c c , with c i ∈ Γ( G i ).As the embeddings f i ( G i ) contain no nontrivial knot, in the first case c isthe unknot, and in the second, it is the connect sum of unknots and henceunknotted. Thus, G is nIK. (cid:3) or H , H , . . . , H k subgraphs of graph G , let (cid:104) H , H , . . . , H k (cid:105) G denotethe subgraph induced by the vertices of the subgraphs. Lemma 3.2.
Let G be a maxnik graph with a vertex cut set S = { x, y } ,and let G , G , . . . , G r denote the connected components of G \ S . Then xy ∈ E ( G ) and (cid:104) G i , S (cid:105) G is maxnik for all ≤ i ≤ r .Proof. As G is 2-connected by Theorem 2.1, each of x and y has at leastone neighbor in each G i . Suppose xy / ∈ G . Form G (cid:48) = G + xy and let G (cid:48) i = (cid:104) G i , S (cid:105) G (cid:48) . For each i , edge xy ∈ G (cid:48) i . But G (cid:48) i is a minor of G , as thereexists G j with i (cid:54) = j since S is separating, and there exists a path from x to y in G j as G j is connected. Thus in (cid:104) G i , G j , S (cid:105) G , we may contract G j to x to obtain a graph isomorphic to G (cid:48) i . Thus, G (cid:48) i is nIK. So, by Lemma 3.1, G (cid:48) = G (cid:48) ∪ xy G (cid:48) ∪ xy . . . ∪ xy G (cid:48) r is nIK. This contradicts the fact that G ismaxnik, and hence xy ∈ E ( G ).Suppose that one or more of the G i are not maxnik. Then add edgesas needed to each G i to form graphs H i that are maxnik. Then the graph H = H ∪ xy H ∪ xy . . . ∪ xy H r is nIK by Lemma 3.1 and contains G as asubgraph. As G is maxnik, G = H and hence G i = H i for all i , so every G i is maxnik as well. (cid:3) Lemma 3.3.
Let G and G be maxnik graphs. Pick an edge in each G i and label it e . Then G = G ∪ e G is maxnik if e is non-triangular in atleast one G i .Proof. Suppose that e is non-triangular in G and has endpoints x, y . Addan edge ab to the graph G . The graph G is nIK by Lemma 3.1. If both a, b ∈ G i for some i , then G + ab is IK, as the G i are each maxnik. Thus, wemay assume that a ∈ G and b ∈ G . The edge e is non-triangular in G ,so vertex a is not adjacent to both endpoints of e . We may assume that a is not adjacent to x . As G is connected, we construct a minor G (cid:48) of G + ab by contracting the whole of G to vertex x . Note that as b ∈ G , we havethe edge ax in G (cid:48) , and in fact G (cid:48) = G + ax . As G is maxnik, G (cid:48) is IK andso is G + ab . Thus, G is maxnik. (cid:3) Lemma 3.4.
For i = 1 , , let G i be maxnik, containing a -cycle C i , andadmitting to a knotless embedding such that C i bounds a disk whose interioris disjoint from the graph. Then the clique sum G over K formed by iden-tifying C and C is nIK. Moreover, G is maxnik if C i is not part of a K in at least one G i .Proof. Let f i be the knotless embedding of G i . Embed the f i ( G i ) so thatthey are separated by a plane. We may then extend this to an embedding f ( G ) by isotoping the C i to the separating plane and identifying them there.Let Γ( G ) denote the set of all cycles in G . As the cycles C i bound a diskin f ( G ), if a closed cycle c ∈ Γ( G ) is not contained in one of the f i ( G i ), then c = c c , with c i ∈ Γ( G i ). As the embeddings f i ( G i ) contain no nontrivialknot, in the first case c is the unknot, and in the second, it is the connectsum of unknots and hence unknotted. Thus, G is nIK. uppose C is not contained in a 4-clique in G . We will show G + ab is IK, and hence G is maxnik. As the G i are maxnik, we may assume that a ∈ G and b ∈ G , as otherwise G + ab is IK. As C is not contained in a4-clique in G , there exists a vertex x in C that is not adjacent to a . As G is connected, there is a path from b to x . Contract G to a . This graphcontains G + ax as a minor, and hence is IK, as G is maxnik and does notcontain edge ax . Thus, G is maxnik. (cid:3) Bounds on maximal knotless graphs
We now consider maximal knotless graphs in general and establish boundson the possible number of edges, and the maximal and minimal degrees. Wefirst show a lemma that will be useful for establishing a lower bound. Asimilar result holds for maximal linkless graphs as well.
Lemma 4.1.
Suppose G is maxnik and contains a vertex v of degree 3.Then all neighbors of v are adjacent to each other.Proof. Label the neighbors of v as x , x , x . Let E v = { x x , x x , x x } and E = E ( G ). Delete the edges in E ∩ E v to form G Y = G \ ( E ∩ E v ).Then add back all the edges of E v to form G (cid:48) = G Y + E v . We will show G = G (cid:48) .As G is maxnik, G Y has an embedding f with no nontrivial knot. Wemay extend f to an embedding of G (cid:48) by embedding each edge x i x j so thatthe 3-cycle x i vx j bounds a disk.Let Γ( G ) denote the set of all cycles in the graph G . Suppose c is a cyclein Γ( G (cid:48) ). If c does not contain one or more edges x i x j , then c ∈ Γ( G Y ), andhence is a trivial cycle in f ( G (cid:48) ). Suppose that c does contain one or moreedges x i x j . There are three possibilities: c is a 3-cycle x i vx j and bounds adisk; c includes a path of the form x i , x j , v, x k with { i, j, k } = { , , } ; or c does not include the vertex v . In the first case c is trivial as it bounds a disk.If c does not contain v , then, since the cycles x i vx j bound disks, c is isotopicto c (cid:48) ∈ Γ( G Y ) and hence trivial. Similarly, if c includes a path x i , x j , v, x k ,using the disk x i vx j , we can isotope the path to x i , v, x k to make c isotopicto c (cid:48) ∈ Γ( G Y ) and hence trivial.Thus, G (cid:48) has an embedding with no non-trivial knot. As G is maxnik, G cannot be a proper subgraph of G (cid:48) , and hence G = G (cid:48) . (cid:3) Theorem 4.2. If G is maxnik with | V | ≥ , then | E | ≥ | V | .Proof. By Theorem 2.4, K is the only maxnik graph with order five and itsatisfies the conclusion of the theorem.Suppose H has the least number of vertices among counterexamples to thetheorem. We will consider a vertex v of minimal degree in H . If deg( v ) ≥ H has | E | ≥ | V | and hence is not a counterexample, so deg( v ) ≤ v ) ≥
2, so we need only consider v of degree 2 or 3. uppose deg( v ) = 2. We will argue that H (cid:48) = H \ v is also maxnikwith | E (cid:48) | < | V (cid:48) | , contradicting our assumption that H was a minimalcounterexample. Let N ( v ) = { w, x } and note that wx ∈ E ( H ). Otherwise,in an unknotted embedding of H , we could add the edge wx so that the 3-cycle vwx bounds a disk. This will not introduce a knot into the embeddingand contradicts the maximality of H .As a subgraph of H , H (cid:48) is nIK. Suppose it is not maxnik because there isan edge ab so that H (cid:48) + ab remains nIK. In a knotless embedding of H (cid:48) + ab ,we can add the vertex v and its two edges so the 3-cycle vwx bounds a disk.This will not introduce a knot into the embedding and shows that H + ab isalso nIK, contradicting the maximality of H . Thus, no such graph H witha vertex of degree 2 can exist.So, we may assume that deg( v ) = 3. Here we cannot apply the techniquesof [A], as Y ∇ moves do not preserve intrinsic knotting [FN]. However,Lemma 4.1 allows us to show the average degree of H is actually at least3.5, and hence H is not a counterexample.Divide the vertices of H into 3 sets: A = { vertices of degree 3 } , B = { vertices of degree > A } , and C = { allother vertices of H } . Form the graph H (cid:48) = H \ C . All vertices in C havedegree 4 or greater, so it suffices to show that the vertices in each connectedcomponent of H (cid:48) have average degree 3.5 or higher.A vertex a i of degree 3 has three neighbors, label them b i , b i , a i , where a i is a neighbor of minimal degree. If deg( a i ) = 3, we continue. If not,delete all edges incident on a i except those between a i and { a i , b i , b i } .This creates a subgraph of H (cid:48) with strictly fewer edges; we will abuse nota-tion and continue to call it H (cid:48) . Vertex a i now has degree 3 in H (cid:48) , and wemove it to set A .If a i had degree greater than 3 in H , then, since it has the minimal degreeamong the neighbors of a i , deg( b ij ) ≥ b i , b i ∈ B . If deg( a i ) = 3 in H , vertices a ij are adjacent only to each other and the b ij . If either of the b ij have degree 3 in H , then H can be disconnected by deleting the other b ij . This is a contradiction as H is maxnik and must be 2-connected byTheorem 2.1. Thus, the b ij are in B .Consider the connected component of v in H (cid:48) , call it H (cid:48) . We will calculatethe total degree of the vertices in H (cid:48) and divide by the number of vertices.Suppose there are n vertices from set A and m vertices from set B in H (cid:48) for a total of n + m vertices. Each vertex from set A has degree 3, so thecontribution to total degree from set A is 3 n . Each vertex in A is adjacentto exactly 2 of the b ij , so the total degree contribution for set B is at least2 n from edges to set A . Further, H (cid:48) is connected. As a ij is only adjacent to b i (cid:48) j (cid:48) if i = i (cid:48) , there must be at least m − b ij , which adds2( m −
1) to the total degree. This gives an average degree of n +2 m − n + m in H (cid:48) . However, H is 2-connected by Theorem 2.1, so there must be at least2 edges from H (cid:48) to its complement in H . So within H , these vertices must V | | E | / | V | ) 0 1/2 1 3/2 2 5/2 20/7 25/8 21/9 Table 1.
The least ratios of size to order for maxnik graphsthrough order nine.have average degree greater than or equal to n +2 mn + m . Note that 2 ≤ m ≤ n ,and n +2 mn + m attains its minimum at m = n . The minimum is , and hence H must have | E | ≥ | V | . (cid:3) Theorem 4.3.
There exist maxnik graphs with | E | < | V | edges for arbi-trarily large | V | .Proof. Let e be an edge of E connecting a degree 4 vertex to one of degree5. Edge e is non-triangular and there are five other edges symmetric to it.Using Lemma 3.3, take k copies of E glued along edge e . The resultinggraph has 7 k + 2 vertices and 20 k + 1 edges. Gluing on five K graphs ineach E on the other non-triangular edges gives an additional 5 k verticesand 10 k edges. So, for each k ≥
1, we have a graph G with n = 12 k + 2vertices and m = 30 k + 1 edges. Then m = 30( n − /
12 + 1 = n − (cid:3) These two theorems suggest the following question. In Table 1 we givethe least ratios through order nine.
Question 4.4.
What is the minimal number of edges for a maxnik graph of n vertices? For maximal planar graphs, | E | = 3 | V | −
6. Similarly, maximal k -apexgraphs have a fixed number of edges depending on | V | . In contrast, as withmaximal linkless graphs, the number of edges in a maxnik graph can vary.In fact, with the exception of | E | = 22, for any | E | ≥
20, there exists amaxnik graph of that size.
Theorem 4.5.
Let n ≥ and n (cid:54) = 22 . Then there exists a maxnik graphwith | E | = n .Proof. The graph K − is a maxnik of size 20 by Theorem 2.4. The graph E has a knotless embedding where the 3-cycle abc bounds a disk [Mt], shownin Figure 2. As no vertex in E is adjacent to all three of these vertices, wemay use Lemma 3.4 to construct a maxnik graph of size 24 by taking a cliquesum over K of E and K . So, we may assume n ≥
21 and n (cid:54)∈ { , } .The graph E has size 21 and 6 non-triangular edges. Let G i denote themaxnik graph obtained from i copies of E by gluing along non-triangularedges.Note that | E ( G i +1 ) | − | E ( G i ) | = 20, and that G i contains at least 6 non-triangular edges for any i . We will now work by induction. Suppose thatmaxnik graphs exist for size n < | E ( G i ) | and for size | E ( G i ) | + 1 and size abc Figure 2.
A knotless embedding of E . | E ( G i ) | + 3. Then it suffices to show that there exist maxnik graphs of size | E ( G i ) | + k for 4 ≤ k ≤
19 and k ∈ { , , , } .Clearly a maxnik graph of size | E ( G i ) | +0 exists, as G i is maxnik. We mayform a new maxnik graph from G i by gluing a copy of K m (for 3 ≤ m ≤ G i by Lemma 3.3. As G i has at least 6 non-triangular edges, we can glue on up to 6 such graphs, each adding (cid:0) m (cid:1) − k using six or fewer addends from the set { , , , } . This isclearly possible.In the base case i = 1, we have a maxnik graph of size | E ( G ) | = 21, andwe excluded graphs of size 22 and 24 ( | E ( G ) | + 1 and | E ( G ) | + 3) above.Thus we may form maxnik graphs of size | E ( G ) | + k for the k of interestas before. (cid:3) Remark 4.6.
A computer search shows there are no size 22 maxnik graphs.Our strategy is based on the classification through size 22 of the obstructionsto -apex in [MP] . Let’s call such graphs MMN2A (minor minimal not2-apex). All but eight of the graphs in the classification are MMIK. Twoexceptions are -regular of order 11, the other six are in the Heawood family.A maximal -apex graph has n − edges where n is the number ofvertices. By Theorem 2.2 a maxnik graph G of size 22 is not -apex andtherefore has a MMN2A minor. Since G is nIK, it must have one of theeight exceptions as a minor. Using a computer, we verified that no size 22expansion of any of these eight graphs is maxnik. Theorem 4.5 implies that there are maxnik graphs of nearly every size.Note that there are maxnik graphs of any order, as there exist maximal2-apex graphs of any order and by Theorem 2.2 these graphs are maxnik. V | δ ( G ) 0 1 2 3 4 5 5 5 or 6 4 to 7∆( G ) 0 1 2 3 4 5 6 7 5 to 8 Table 2.
Maximal and minimial degrees of maxnik graphsthrough order nine.We have considered the minimal number and the possible number of edgesin a maxnik graph. We now consider other aspects of maxnik graphs’ struc-ture, in particular, the maximal and minimal degree. Since ∆( G ) = | V | − G ). Proposition 4.7.
The complete graph K is the only maxnik graph withmaximal degree two.Proof. Suppose G is maxnik with ∆( G ) = 2. Then | G | ≥ δ ( G ) = 2 and G is connected. So, G is a cycle. Now, a cycle isplanar, hence 2-apex, and by Theorem 2.2, G is maximal 2-apex. However,a cycle is not maximal 2-apex unless it is K . (cid:3) Note that Lemma 4.1 has the following two immediate corollaries.
Corollary 4.8.
If a graph G is maxnik and has ∆( G ) = 3 , then G is 3-regular. Corollary 4.9.
If a graph G is maxnik and 3-regular, then G = K . These results motivate the following question.
Question 4.10.
Do there exist regular maxnik graphs other than K n with n < ? A maximal 2-apex graph will have ∆( G ) = | V | − δ ( G ) ≤
7, so ifthere is such a regular maxnik graph with | V | ≥
7, it is not 2-apex. However,through order nine, our two examples of maxnik non 2-apex graphs are bothclose to regular, having ∆( G ) − δ ( G ) ≤
2. This suggests the answer to ourquestion is likely yes.For δ ( G ), Theorem 2.1 gives a lower bound of two that is realized bythe infinite family of Theorem 4.3. On the other hand, by starting with aplanar triangulation of minimum degree five, we can construct graphs with δ ( G ) = 7 that are maximal 2-apex, and hence maxnik. At the same time,since a graph with | E | ≥ | V |−
14 has a K minor and is IK [Md, CMOPRW],a maxnik graph must have δ ( G ) ≤
9. It seems likely that there are examplesthat realize this upper bound on δ ( G ). Table 2 records the range of degreesfor maxnik graphs through order nine.5. Prime and Composite Maxnik Graphs
We will call a graph composite if it is the clique sum of two graphs.Otherwise it is prime . These terms are analogous to knots, where a knot s composite if is the connected sum of two non-trivial knots, and primeotherwise. In this section, we classify the maxnik graphs described earlier inthis paper as prime and composite. We remark that it may be of interest tostudy other instances of prime graphs, for example, prime maximal planaror prime maxnil.The infinite families of maxnik graphs constructed in Section 4 are allcomposite, as they are clique sums of smaller maxnik graphs.Note that K n is prime, so all maxnik graphs of order 6 or less are prime. Figure 3.
Complements of the maximal 2-apex graphs oforder nine. Top row, L to R: Big-Y, Long-Y, Hat; Bottomrow: Pentagon-bar and House.
Proposition 5.1.
The following maxnik graphs are composite: K − , K − P ,and four of the five maximal 2-apex graphs on 9 vertices, specifically Big-Y,Long-Y, Hat and House.Proof. The graph K − is formed from two copies of K summed over a 5-clique.The graph K − P is formed from K − clique sum K over a 5-clique,where the 5-clique contains exactly one endpoint of the missing edge.Big-Y is formed from K − P clique sum K over a 5-clique, where the5-clique contains both of the terminal vertices of the 3-path.Long-Y is formed from K − K over a 5-clique.Hat is formed from K − P clique sum K over a 5-clique, where the5-clique contains one terminal vertex and one (non-adjacent) interior vertexof the 3-path.House is formed from K − P clique sum K over a 5-clique, where the5-clique contains one interior vertex of the 3-path. (cid:3) Lemma 5.2. If G c is of the form K (cid:96) H , then either G is prime, or G isthe clique sum of two copies of K n over an n − clique. roof. Call the two vertices of the K in G c v and v . Suppose that G is aclique sum of G and G over a clique C . We cannot have both v and v in C , as edge v v is in G c . Without loss of generality, we may assume that v is in G \ C . So, in G c , v must be adjacent to every vertex of G \ C .Thus G \ C is v . As the only neighbor of v in G c is v , v is adjacent toevery vertex in C . Similarly for v . Thus if G is composite, it is the cliquesum of K n and K n over an n − (cid:3) Corollary 5.3.
The following maxnik graphs are prime: Pentagon-bar, G , and K − Each of these graphs has a complement of the form K (cid:96) H . As thesegraphs are not of the form K n − a single edge, they are prime by Lemma5.2. (cid:3) Note that if G is a clique sum over a t -clique, it is not ( t + 1)-connected. Proposition 5.4.
The maxnik graph E is prime.Proof. The largest clique in E is a 3-clique, but E is 4-connected and hencemust be prime. (cid:3) Lemma 5.5. If G = H ∗ K , and G is 2-apex, then G is prime maxnik ifand only if H is prime maximal planar.Proof. As G is 2-apex, it is maxnik if and only if it is maximal 2-apex, and G is maximal 2-apex if and only if H is maximal planar.If H is composite, then H is the clique sum of H and H over a t -clique.So G is the clique sum of H ∗ K and H ∗ K over a t + 2-clique, and hence G is composite.As G is maxnik, it must be 2-connected. Hence if G is composite, it mustbe G clique sum G over a t -clique C , with t ≥
2. Label two of the verticesin C as v , v . Then H is the clique sum of G \ { v , v } and G \ { v , v } over C \ { v , v } , and thus composite. (cid:3) Corollary 5.6.
There exist prime maxnik graphs of arbitrarily large size,and of any order ≥ .Proof. The octahedron graph is max planar and 4-connected. The largestclique it contains is a 3-clique, so it is prime. New triangulations formed byrepeated subdivision of a single edge are 4-connected and maximal planar,but have no 4-clique, hence are prime as well. Thus all of these graphs giveprime maxnik examples when joined with K . (cid:3) We remark that the construction of this family of graphs is similar to themaxnil families with 3 n − n − eferences [A] M. Aires, On the number of edges in maximally linkless graphs ,arXiv:1911.08552[BBFFHL] P. Blain, G. Bowlin, T. Fleming, J. Foisy, J. Hendricks, and J. Lacombe,
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