On the General Position Number of Mycielskian Graphs

The general position problem for graphs was inspired by the no-three-in-line problem from discrete geometry. A set $S$ of vertices of a graph $G$ is a \emph{general position set} if no shortest path in $G$ contains three or more vertices of $S$. The \emph{general position number} of $G$ is the number of vertices in a largest general position set. In this paper we investigate the general position numbers of the Mycielskian of graphs. We give tight upper and lower bounds on the general position number of the Mycielskian of a graph $G$ and investigate the structure of the graphs meeting these bounds. We determine this number exactly for common classes of graphs, including cubic graphs and a wide range of trees.


Introduction
A graph G consists of a set of vertices V (G) connected by edges, which are unordered pairs of vertices.All graphs used here are simple, connected and finite.We write u ∼ v if the vertices u, v ∈ V (G) are adjacent.A path P ℓ+1 of length ℓ in G is a sequence v 0 , v 1 , . . ., v ℓ of distinct vertices such that v i v i+1 is an edge for 0 ≤ i ≤ ℓ − 1.If P is a path v 0 , v 1 , . . ., v ℓ , then the reverse path P of P is the path v ℓ , . . ., v 1 , v 0 .The distance d G (u, v) between vertices u and v of G is the length of a shortest path from u to v. A shortest path from u to v is also called a u, v-geodesic.The neighbourhood N G (u) of u ∈ V (G) is the set {v ∈ V (G) : v ∼ u} of all vertices adjacent to u.More generally, for t ≥ 0, N t (u) is the set of all vertices at distance t from u, so that N G (u) = N 1 (u).The degree deg(u) is the number |N G (u)| of neighbours of u.A leaf of G is a vertex with degree one in G and we define the leaf number ℓ(G) to be the number of leaves in G.A vertex adjacent to a leaf is a support vertex.A vertex of degree n − 1 in a graph with order n is universal.

Preprint submitted to Discrete Applied Mathematics
A cycle C ℓ of length ℓ is a sequence of vertices v 0 , v 1 , . . ., v ℓ−1 such that v i ∼ v i+1 for 0 ≤ i ≤ ℓ − 2 and also v 0 ∼ v ℓ−1 .A connected graph with no cycles is a tree.The girth g(G) of G is the length of a shortest cycle in G (if G is acyclic, then we adopt the convention that g(G) = ∞).For U ⊆ V (G) we write the subgraph induced by U as U .An independent set in G is a set of mutually non-adjacent vertices, i.e. a set of vertices that induces an empty subgraph of G, and the cardinality α(G) of a largest independent set in G is the independence number of G.By contrast, a clique in a graph is a set of mutually adjacent vertices.The complete graph K n is the graph with order n such that every pair of vertices is adjacent, i.e. the whole vertex set is a clique.A matching M in G is an independent set of edges, i.e. if e 1 , e 2 are two distinct edges of M, then e 1 and e 2 do not have an endvertex in common.The size of the largest matching in G is the matching number of G and is denoted by ν(G).For any subsets V 1 , V 2 ⊆ V (G) we denote the set of edges of For any graph-theoretical terminology not defined here we take [6] as our standard.
The general position problem originated in Dudeney's no-three-in-line problem and the general position subset selection problem from discrete geometry [11,13,24].The general position problem was generalised to graph theory independently in [7] and [20] as follows.A set S of vertices in a graph G is a general position set, or is in general position, if no shortest path in G contains three or more vertices of G.A largest general position set of G is called a gp-set and its cardinality gp(G) is the general position number of G (or gp-number for short).The structure of general position sets was characterised in [2].In particular, it was shown in this article that any general position set induces an independent union of cliques, i.e. a disjoint union of one or more cliques W 1 , . . ., W t , t ≥ 1, with no edges between cliques W i and W j if 1 ≤ i < j ≤ t.These cliques must also have the distance-constant property, meaning that for any cliques W i and W j in this collection and any vertices w 1 , w ′ 1 ∈ W i , w 2 , w ′ 2 ∈ W j , we have d G (w 1 , w 2 ) = d G (w ′ 1 , w ′ 2 ).We make use of two variants of general position sets.The largest number of vertices in a general position set that is also an independent set is called the independent position number ip(G) and was studied in [27].The article [17] defined a general d-position set to be a subset S ⊆ V (G) such that no shortest path of length at most d contains three or more vertices of S. The largest number of vertices in a general d-position set is denoted by gp d (G).Other types of position sets have also been investigated, including monophonic position sets [28], mutual visibility sets [8], edge general position sets [21], Steiner position sets [16], mobile position sets [15], vertex position sets [26], lower general position sets [9] and others.Some of the most recent papers on this subject include [3,18,29,19,22].
In this paper we discuss the general position numbers of the Myclieskian of a graph.This construction was introduced by Mycielski in [23] in order to produce triangle-free graphs with arbitrarily large chromatic number.Properties of the Mycielskian construction have been investigated extensively, for example dominator colouring number [1], connectivity [4], energy [5], L(2, 1)-labelling number [10], circular chromatic number [12], Gromov hyperbolicity [14], wide diameter [25] and Italian domination number [30], amongst many others.
Let G be a graph with vertex set V = {u 1 , . . ., u n } and edge set E. The Mycielskian of G is the graph M(G) defined as follows.The vertex set of In other words, the edge set of M(G) consists of the edges of G, and for any u ∈ V we join We note that the Mycielskian is more often denoted by µ(G), but we avoid this notation here to avoid a clash with the notation used for the mutual visibility number, another position type invariant.We call the vertex u ′ the M-twin of u (and conversely u is the M-twin of u ′ ).For any set X = {x 1 , . . ., x t } of vertices of G we denote the set of M-twins of the vertices in X by X ′ = {x ′ 1 , . . ., x ′ t }.The plan of this paper is as follows.In Section 2 we characterise general position sets of M(G) that contain the root vertex u * .In Section 3 we describe general position sets of M(G) in terms of a 'dual' partition of V (G) into four parts.In Section 4 we give upper and lower bounds for gp(M(G)) and characterise the graphs that meet our upper bound.We also give exact values of gp(M(G)) for some common families of graphs.Section 5 presents a bound on the gp-number of the Mycielskian of regular graphs, which we use to classify the gp-numbers of Mycielskians of cubic graphs and all sufficiently large regular graphs.Finally in Section 6 we determine the gp-numbers of the Mycielskians of a wide class of trees.

General position sets of M(G) containing the root u *
Firstly, we introduce a convention that we use throughout the paper.Given a set S of vertices of a graph, we call a geodesic containing at most two vertices of S sound and a shortest path containing at least three vertices of S unsound.
It is useful to disregard gp-sets of M(G) that contain the root vertex.In this section we classify the graphs such that M(G) has a general position set of cardinality at least n + 1 that does not contain u * .We also show that for any graph that is not a complete graph there is always a gp-set of M(G) that does not contain the root u * .Lemma 1.Let G = (V, E) be a connected graph with order n ≥ 3. Then there is a general position set S of M(G) with |S| ≥ n + 1 that contains u * if and only if G is a complete graph K n or the join of K 1 with a disjoint union of cliques, at least one of which is K 1 .
Proof.Suppose that S is a gp-set of M(G) with |S| ≥ n + 1 that contains the root u * .Firstly, since S ∩ V (G) cannot contain an induced path P 3 , the subgraph S ∩ V (G) of G induced by S ∩ V (G) must be an independent union of cliques.Also, S can contain at most one vertex of V ′ , since any two vertices of V ′ are connected by a shortest path through u * .Suppose that there is a vertex u ∈ V (G) such that u, u ′ ∈ S. By our previous observation, u ′ is the only vertex of V ′ in S. Also, no neighbour of u in G lies in S ∩ V (G), since if u ∼ v, then u, v, u ′ would be an unsound path.We conclude that |S| ≤ n+2−deg G (u). Hence u is a leaf of G and S ∩V (G) = V (G) \ {v}, where v is the support vertex of u.Let W be any clique of S ∩ V (G) apart from the leaf u.As G is connected, there must an edge from some vertex w of W to a vertex outside W ; as u is adjacent only to v and S ∩ V (G) is an independent union of cliques, it must be that w ∼ v. Since d M(G) (u, w) = 2, by the distance-constant property u must be at distance two to every vertex of W . Thus v is a universal vertex and G has the claimed structure.It is easily verified that {u * , u ′ } ∪ (V (G) \ {v}) is in general position.An example is shown on the right of Figure 2.1.Now suppose that there is no vertex u such that u, u ′ ∈ S. Then we have |S| = n + 1 and for any vertex v of G, S contains exactly one of v and v ′ .If S ∩ V ′ = ∅, then V (G) ⊂ S and by our earlier observation G is a clique, in which case V (G) ∪ {u * } is in general position.Otherwise S contains a vertex w ′ ∈ V ′ and so S = {u * , w ′ } ∪ (V (G) \ {w}).However, as G is connected, w has a neighbour z in G, and u * , w ′ , z would be an unsound path, a contradiction.
Notice that the general position sets exhibited in the proof of Lemma 1 are not necessarily largest possible.In Corollary 16 we determine the gp-numbers of the Mycielskian of the join of K 1 with a disjoint union of cliques, thereby classifying those graphs for which the Mycielskian has a gp-set containing u * .Corollary 2. For any graph with order n ≥ 3, the root vertex u * of M(G) belongs to every gp- Proof.If gp(M(G)) = n, then V ′ is a gp-set of M(G) that does not contain u * , so we can confine our attention to graphs G with gp(M(G)) ≥ n + 1.By Lemma 1, if a graph is such that every gp-set of M(G) contains u * , then G is either a complete graph or the join of K 1 with an independent union of cliques, at least one of which consists of a single vertex (moreover, the general position sets exhibited in the proof would have to be gp-sets).If G has the latter structure and u is a leaf with universal support vertex, then V ′ ∪ {u} is a general position set with the same order.Thus, whether or not the general position sets containing u * are gp-sets, we see that u * is not contained in every gp-set of such a graph.Now suppose that G is a complete graph.It follows from the proof of Lemma 1 that the only general position set of M(G) containing u * is V (G) ∪ {u * }, which has cardinality n + 1.
Suppose that S is a general position set of M(G) that does not contain u * .Then there must be a vertex u of K n such that u, u ′ ∈ S. Now if v is any vertex of K n other than u, the path u, v, u ′ is a geodesic, so v ∈ S. It follows that S = V ′ ∪ {u}.However, in this case if v, w ∈ V (K n ) \ {u}, the path v ′ , u, w ′ would be unsound, a contradiction.Thus V (G) ∪ {u * } is the unique gp-set of M(G).
The gp-set for M(K 5 ) is shown on the left of Figure 2.1.Notice that gp(M(K n )) = n + 1 is true for n = 2, as M(K 2 ) ∼ = C 5 , but the root vertex u * is not contained in every gp-set.
It follows from Lemma 1 that the general position sets containing u * can easily be found in polynomial time and that for any non-complete graph G there is a gp-set of M(G) that does not contain.Thus in the remainder of this article we can safely ignore gp-sets containing u * .

General position sets of M(G) in terms of partitions of V (G)
In this section we demonstrate a duality between general position sets of M(G) that do not contain u * and certain partitions of V (G) into four (possibly empty) parts.This will allow us to calculate gp(M(G)) working entirely inside G. First we observe the connection between shortest paths in the base graph G and the shortest paths in the Mycielskian M(G).

Definition 3. The projection of a path
if and only if it is an expansion of a shortest path in G.
With any general position set S of M(G) that does not contain u * we associate the partition π Then we have gp(M(G)) = n + n 1 − n 4 , where n is the order of the graph and n i = |V i | for i = 1, 2, 3, 4. We now identify the essential properties of such a partition that guarantee it has the form π(S), where S is a general position set of M(G).We call such a partition an MGP-partition (short for 'Mycielskian general position').
into four (possibly empty) sets is an MGP-partition if and only if it satisfies the following three conditions: , then either u 0 , u ℓ ∈ V 2 and ℓ ≥ 3, or else P or P has one of the following forms: Note the following three important properties of MGP-partitions.
Proof.We verify the three properties from Definition 5 in turn.
1 , w would be an unsound path, where w 1 is any neighbour of w in G.
Condition 3: Suppose that P is a geodesic u 0 , u 1 , . . ., u ℓ in G with length ℓ ≤ 4 that has both endpoints and an internal vertex in , then P has an unsound expansion that is a geodesic in M(G), so we can assume that the initial vertex u 0 of P is in V 2 .Suppose firstly that ℓ = 2. Then we must have We now complete our characterisation of gp-sets of Mycielskians of graphs.
Proof.We need to verify that all shortest paths between vertices of S = σ(π) are sound.First let u, v ∈ V 1 ∪V 3 .By Condition 3 of Definition 5, no expansion of a shortest u, v-path in G with length at most four contains a third vertex of S. Furthermore, if d G (u, v) ≥ 4, then by Observation 4 the only other geodesics to consider are the paths of the form u, w ′ 1 , u * , w ′ 2 , v, where u ∼ w 1 and v ∼ w 2 in G.By Condition 1 of Definition 5 neither of the vertices w 1 or w 2 is in V 1 .Hence we can assume that w 1 ∈ V 2 .Then by Condition 2 we would have has length two and P is unsound only if it passes through a common neighbour of u and v in V 1 ∪ V 3 .By Condition 2 and Lemma 6 this is impossible.
We can thus assume that P is an unsound geodesic with one endpoint in V 1 ∪ V 3 and the other in are the paths u, w, u ′ , where u ∼ w in G.By Lemma 6 the vertex w does not lie in S. Hence the endpoints of P are not M-twins of each other.Let us denote the endpoints by u, v ′ , where u ∈ If ℓ(P ) = 2, then by Condition 1 of Definition 5 we must have u ∈ V 3 , v ∈ V 2 and the internal vertex of P lies in V 3 .However, this form of shortest path is not allowed by Condition 3.
Hence ℓ(P ) = 3.By Observation 4, the u, v ′ -geodesics in M(G) are either expansions of shortest paths of length three in G, or else have the form u, w ′ , u * , v ′ , where u ∼ w in G.By Condition 3 the former geodesics are sound and the latter are sound by Condition 2. This shows that all shortest paths in M(G) are sound and thus S is in general position.
Since both maps π and σ are one-to-one, this completes the proof of the claimed duality between general position sets of M(G) not containing u * and MGP-partitions of G.
Corollary 9. Let G be a non-complete graph with order n.

Bounds, extremal cases and exact values
In this section we use Corollary 9 to derive tight upper and lower bounds for gp(M(G)) and characterise the case of equality with the upper bound.We start with the lower bound.Recall that ip(G) is the number of vertices in a largest independent set that is also in general position and gp d (G) is the number of vertices in a largest subset S ⊆ V (G) such that no shortest path with length at most d passes through three or more vertices in S. To derive bounds for gp(M(G)) we combine these concepts in the following definition.Definition 10.An independent d-position set is a subset S ⊆ V (G) such that S is an independent set and no shortest path of length at most d passes through three or more vertices of S. We denote that largest number of vertices in an independent d-position set by ip d (G).
For any graph G these parameters satisfy the inequalities ip(G) ≤ gp(G) and ip(G) ≤ ip 4 (G) ≤ α(G).Proof.The set V ′ is trivially in general position, implying that gp(M(G)) ≥ n (this is equivalent to using the MGP-partition We call any non-complete graph G with order n ≥ 3 and gp(M(G)) = max{n, 2ip 4 (G)} meagre.Otherwise G is abundant.We now show that the lower bound in Corollary 11 is tight by exhibiting two families of meagre graphs: graphs with large size, joins of K 1 with disjoint unions of cliques of cardinality at least three, and complete multipartite graphs.
Lemma 12.If G is an abundant graph with order n ≥ 3 and V 4 = ∅, then V 3 = ∅, |V 1 | = 1 and the vertex of V 1 is a leaf with support vertex that is a universal vertex.If G has a leaf with a universal support vertex, then gp(M(G)) ≥ n + 1.
Proof.Suppose that G is abundant and has MGP-partition (V 1 , V 2 , V 3 , ∅), where n 1 ≥ 1. Suppose that V 3 = ∅.As we are dealing with connected graphs, each vertex of V 1 has a path to V 3 .Therefore, by Lemma 6, the matching ( Then by Condition 2 of Definition 5 we have d(u, w) ≤ 3 for each u ∈ V 1 and w ∈ V 3 .Therefore there is a shortest path from some u ∈ V 1 to a w ∈ V 3 with length two or three with all internal vertices in V 2 .However, the existence of such a path contradicts Condition 3 of Definition 5. Thus By Condition 2 of Definition 5 any vertex of V 1 is at distance two from every vertex of V 2 , with the exception of its support vertex.Therefore any pair of vertices from V 1 are at distance three from each other, with shortest path passing through V 2 .This contradicts Condition 3 of Definition 5 unless n 1 = 1.By Condition 2 of Definition 5 it follows that the support vertex of the vertex of V 1 is universal.
Conversely, if u ∈ V (G) is a leaf with a universal support vertex, then ({u}, V (G) \ {u}, ∅, ∅) is an MGP-partition and n + n 1 − n 4 = n + 1. Theorem 13.For n ≥ 3, the largest size of a graph with order n that satisfies gp(M(G)) > n is n 2 + 1.Any abundant graphs with size n 2 + 1 is either a clique K n−1 with a leaf attached, the gem graph K 1 ∨ P 4 , or 3K 1 ∨ K 2 .The latter graph is not abundant, for the former two are.
Proof.Let G be an abundant graph with order n and size at least n 2 +1.Let (V 1 , V 2 , V 3 , V 4 ) be an MGP-partition corresponding to a gp-set of M(G).By Lemma 6 each vertex u of V 1 has degree at most 1 + n 4 if u has an edge to V 2 and degree at most n 4 otherwise.By Lemma 12, if n 1 ≥ 2, then n 1 > n 4 > 0.
Suppose that n 1 ≥ 3 and let Then, taking care not to count missing edges between u 1 , u 2 and u 3 twice, we see that there are at least missing edges in G.As n 2 ≥ t and n 1 > n 4 , we have Moreover, for equality to hold, we would need n 1 = 3, n 2 = n 3 = 0, n 4 = 2.As all other edges must be present, G is isomorphic to 3K 1 ∨ K 2 and V 1 is an independent 4-position set, so that we would have gp(M(G)) = 2ip 4 (G) and G would not be abundant.Now suppose that n 1 = 2, where since n 2 ≥ t.To have equality, we must have n 2 = t and n 3 = 0.As all other edges must be present, if t = 0, 1, then G is a clique with a leaf attached.If t = 2, then G is isomorphic to the gem graph, which is readily verified to be abundant, since its independence number is two.A gp-set of the Mycielskian of the gem graph is shown in Figure 4.1.
Finally if n 1 = 1, then we must have n 4 = 0 and Lemma 12 tells us that the vertex of V 1 is a leaf.The edges missing to the leaf account for all n − 2 missing edges, so G is a clique with a leaf attached as claimed.Theorem 14.Let K r 1 ,r 2 ,...,r k be a complete k-partite graph, where Proof.Since any partite set of G = K r 1 ,r 2 ,...,r k is an independent position set, by Corollary 11 we have gp(M(G)) = max{n, 2r 1 }.Suppose for a contradiction that (V 1 , V 2 , V 3 , V 4 ) is an MGP-partition of G with n i = |V i | for i = 1, 2, 3, 4 corresponding to a gp-set of M(G) containing more than max{n, 2r 1 } vertices.
As V 1 is an independent set, it must be contained entirely within some partite set X of G.By Lemma 6 there are no edges from V 1 to V 3 , so we have V 1 ∪ V 3 ⊆ X.If V 2 lies entirely in X, then gp(M(K r 1 ,r 2 ,...,r k )) = 2n 1 +n 2 +n 3 ≤ 2r 1 , so assume that there is a vertex y ∈ V 2 \X.Then the vertex y is adjacent to every vertex of X.As (V 1 , V 2 ) is a matching by Lemma 6, we must have n 1 = 1 and hence n 4 = 0.It now follows from Lemma 12 that the vertex of V 1 is a leaf, so there is just one vertex of G outside X and hence G is a star, i.e. a tree of order n with a universal vertex.The associated general position set of M(G) contains n + 1 vertices.However, for n ≥ 3 this would be smaller than 2r 1 = 2(n − 1), a contradiction.It follows that the graph is meagre.
Theorem 15.Let G be the join of K 1 with with a disjoint union of t ≥ 2 cliques W = ˙ t i=1 W i .Suppose that W contains t 1 cliques of order at most two and t 2 cliques of order at least three.Then gp(M(G)) = n + t 1 .
) be an MGP-partition corresponding to a gp-set S of M(G) not containing the root u * .It follows from Lemma 1 that |S| ≥ n + 1, so we can assume that n 1 > n 4 .Let u be the universal vertex of G.If u ∈ V 3 , then by Definition 5 V 1 is empty, contradicting |S| > n.If u ∈ V 1 , then by Condition 1 of Definition 5 there are no further vertices in V 1 ∪ V 3 .By Condition 3, there cannot be vertices of V 2 in two different cliques of W , so V 4 is non-empty and we would have |S| ≤ n, a contradiction.If u ∈ V 2 , then, assuming that W 1 contains a vertex v of V 1 , Definition 5 shows that G contains no further vertices of V 1 , so V 4 = ∅ and Lemma 12 shows that |W 1 | = 1 and |S| = n + 1. Hence we can assume that u ∈ V 4 .
Any clique W i contains at most one vertex of V 1 , so can contribute at most one to the difference n 1 − n 4 .By Definition 5, if any of the t 2 cliques with order at least three contains a vertex of V 1 , then it must also contain a vertex of V 4 , and so cannot contribute to n 1 − n 4 .Hence |S| ≤ n + t 1 .Conversely, setting V 4 = {u}, choosing one vertex from each of the t 1 cliques of order at most two to be in V 1 and setting all other vertices of W to belong to V 2 gives an MGP-partition.Comparing the cardinality of the general position sets obtained, including those containing the root u * from Lemma 1, we conclude that gp(M(G)) = n + t 1 .
Therefore we obtain a family of meagre graphs by letting all of the cliques in W have order at least three.

Corollary 16.
A graph G has a gp-set containing the root vertex u * if and only if G is a clique or G is the join of K 1 with a disjoint union of t ≥ 2 cliques W 1 , . . ., W t , where one of the cliques has order one and all of the other cliques have order at least three.
Later in the paper we will see other examples of meagre graphs: paths (Lemma 31), all cycles apart from C 3 and C 5 (Theorem 23), all cubic graph with two exceptions (Theorem 24) and in general all sufficiently large regular graphs (Theorem 25).It also follows from Corollary 18 below that for n ≥ 2 the star graph S n is meagre (in fact for this graph the bounds in Corollaries 11 and 18 coincide).We now derive a tight upper bound on gp(M(G)) in terms of the independence number α(G).
Theorem 17.For any non-complete graph G with order n and minimum degree δ ≥ 1 we have It follows from the conditions in Definition 5 and Lemma 6 that V 1 is an independent 4-position set of G. Thus n 1 ≤ ip 4 (G).Also by Lemma 6, any vertex in V 1 has at most one neighbour in V 2 and hence has at least δ − 1 neighbours in V 4 .Thus n 4 ≥ δ − 1 and so gp(M(G) To meet this bound V 1 must be a largest independent 4-position set of G, which implies that n 1 ≥ 2, and each vertex of V 1 must have a neighbour in V 2 , so by Condition 2 of Definition 5 any pair of vertices in V 1 has a common neighbour in V 4 , implying that δ ≥ 2. Thus if δ = 1 we can improve the bound to gp(M(G)) Theorem 17 also gives an upper bound in terms of the more familiar independence number.
We now characterise the graphs that meet these upper bounds.
Theorem 19.An abundant graph G satisfies gp(M(G)) = n + ip 4 (G) − δ + 1 if and only if V (G) can be partitioned into three sets V 1 , V 2 and V 4 such that • the vertices of V 2 have degree at least δ, and Proof.Suppose that G meets the upper bound in Theorem 17.Let (V 1 , V 2 , V 3 , V 4 ) be an MGP-partition associated with a gp-set of G.The proof of Theorem 17 shows that V 1 is a largest independent 4-position set, the matching We have ip 4 (G) ≥ 2 for any non-complete graph with order n ≥ 3, so n 1 ≥ 2. By Condition 2 of Definition 5, any pair of vertices in V 1 has a common neighbour in V 4 , implying that δ ≥ 2. Thus V 4 is non-empty.
Consider the vertices of V 3 .As V 1 is a maximum independent 4-position set and (V 1 , V 3 ) = ∅, it follows for that any vertex w of V 3 there is a shortest path P of length four containing w and two vertices of V 1 .Observe that any pair of vertices from V 1 are at distance two from each other, since they have a common neighbour in V 4 .Also by Condition 2 of Definition 5 each vertex of V 1 is at distance at most three from any vertex of V 3 .It follows that there is no such path P and V 3 = ∅.Similarly, if s > r and v is a vertex of {v r+1 , . . ., v s }, then there would have to be a shortest path in G of length four that contains v and two vertices of V 1 .This is impossible, since the vertices of V 1 are at distance two apart and by Condition 2 of Definition 5 each vertex of V 1 is at distance two from v. Therefore r = s.

Corollary 20. A non-complete graph G with order
and only if G is either a clique K n−1 with a leaf attached, or has the following structure: ) is a perfect matching, and Proof.It follows from the proof of Theorem 19 that if any vertex of V 1 has degree at least three, then gp(M(G)) Hence we can assume that each vertex of V 1 has degree one or two.To have gp(M(G)) = n + α − 1, either a) V 1 is a maximum independent set and n 4 = 1, or b) V 1 is an independent set of order α − 1 and V 4 = ∅.
In case b) we have α = 2 and Lemma 12 tells us that there are vertices u, v ∈ V (G) such that u is a leaf and v is a universal vertex, and V 1 = {u}, V 2 = V (G) \ {u} and V 3 = ∅.Also, if there is any pair of vertices w, z ∈ V 2 \ {v} that are non-adjacent, then the independence number of G would be at least three.Thus G is a clique K n−1 with a leaf attached, which meets the upper bound n + α − 1 by Theorem 15.

Now consider option b). As
and G is a star.The n − 1 leaves of the star form an independent 4-position set, so by Corollary 11 the star meets the upper bound n + α − 1. Hence assume that V 2 = ∅.
We split , where • V 1,2 are the vertices of V 1 with degree two, and As the matching ( , then there is no path of length two from u to any vertex of V 1 \ {u}, so that we ) is a perfect matching, V 2 would also have just one vertex and G must be the path P 3 .Upon setting V 2 = ∅ and |V 1,1 | = 2 in the statement of the corollary, P 3 has the claimed form.Hence we can now assume that not adjacent to x, the shortest paths from x to V 1,1 would have length three and pass through V 1,2 , which is not allowed by Condition 3 of Definition 5.
Conversely, if all of the above conditions are satisfied, then the partition is an MGPpartition.
Notice that if V 2 = ∅ in the second family described in Corollary 20, then we obtain a star.

Regular graphs
It was indicated in Section 4 that the regular graphs provide a rich source of meagre graphs.In this section we show that there is a strong upper bound on gp(M(G)) for regular graphs G and that for any d there are finitely many abundant d-regular graphs.
Proof.Let G be a d-regular graph, S be a gp-set of M(G) that does not include u * and (   for n = 3 follows from Corollary 2. For n ≥ 4 let S be a gp-set of M(C n ) that does not include u * and consider the corresponding MGP-partition By Theorem 21 we have gp(M(C n )) ≤ n + 1, so we must have exactly gp(M(C n )) = n + 1.As we have equality in the bound of Theorem 21, it follows that the matching (V 1 , V 2 ) has size r = 2 and n 4 = n 1 − 1. Let the cycle has length five and all cycles with length n = 4 or n ≥ 6 are meagre.However, as can be seen in Figure 5.2, this argument does yield a gp-set of cardinality six in the Grötzsch graph M(C 5 ).Notice that C 5 is isomorphic to G(2) from Theorem 22.
Theorem 24.The graph G(3) is the unique non-complete abundant cubic graph.
Proof.Let G be a non-complete cubic graph with order n and gp(M(G)) ≥ n + 1.The proof of Theorem 21 shows that gp(M(G)) = n + 1 and the matching (V 1 , V 2 ) has size r = 3 or 4.
Hence we can assume that r = 3. Write ) and as before V 3 would be empty.
Suppose that n 1 = 5.Then as 3 ≤ n 2 ≤ 8 − n 1 = 3, we have n 2 = 3, n 3 = 0 and n 4 = 4, giving n = 12.For i = 1, 2, 3, we have Thus there is no path of length at most three from v 1 to u 4 via {v 2 , v 3 }, so the path v 1 , u 1 , w, u 4 , where w is a neighbour of u 1 in V 4 , is a shortest path of the form forbidden by Condition 3 of Definition 5, a contradiction.Now suppose that n 1 = 4, so that n 4 = 3 and n 2 ∈ {3, 4}.If n 2 = 4, then n 3 = 0, giving n = 11, whereas G must have even order.Thus n 2 = 3.By Condition 2 of Definition 5 all vertices of n 3 lie at distance at most three from u i for i = 1, 2, 3.There is just one vertex x i of N 2 (u i ) that does not belong to V 1 ∪ V 2 for i = 1, 2, 3 and x i is the only vertex of N 2 (u i ) that can have neighbours in V 3 by Condition 3 of Definition 5. Thus n 3 ≤ 3 and, by the parity of n, either n 3 = 0 or n then by Condition 3 of Definition 5 neither y 1 nor y 2 is in V 3 , contradicting the parity requirement.Thus x i ∈ V 4 and for each of u 1 , u 2 , u 3 two vertices of V 4 lie in N G (u i ) and one vertex of V 4 lies in N 2 (u i ).Write V 4 = {x 1 , w 1 , w 2 }, where N G (u 1 ) = {w 1 , w 2 , v 1 }.
If n 3 = 2, then, by Condition 3 of Definition 5, Assume without loss of generality that x 1 ∼ w 2 .Then at least one vertex of {u 2 , u 3 }, say u 2 , lies in N G (w 1 ),and we obtain the configuration in Figure 5.3 (although note that the positions of u 3 and u 4 might be swapped).But now u 2 has at most one neighbour in V 4 , contradicting our previous deduction.Therefore n 3 = 0.
Thus suppose that n 1 = 4, n 2 = n 4 = 3 and n 3 = 0. Write V 4 = {w 1 , w 2 , w 3 }.Then each of u 1 , u 2 , u 3 has one edge to V 2 and all other edges incident with V 1 are to V 4 .In particular, N G (u 4 ) ⊂ V 4 .Thus |(V 1 , V 4 )| = 9 and every vertex of V 4 has neighbourhood entirely contained in V 1 .Hence each vertex of V 2 must have one neighbour in V 1 and two neighbours in V 2 , so V 2 induces a triangle.This implies that G is isomorphic to the graph in Figure 5.4.However, this graph contains a shortest path v 1 , u 1 , w 3 , u 4 , contradicting Condition 3 in Definition 5.
Finally, let n 1 = 3, n 4 = 2.Each of u 1 , u 2 , u 3 has one edge to V 2 and two edges to V 4 , making six edges from V 1 to V 4 .Hence both vertices of V 4 have neighbourhood contained in V 1 .For each of u 1 , u 2 , u 3 to reach each vertex of V 2 in distance at most two, we see that n 2 = 3 and V 2 induces a triangle in G, so that V 3 = ∅.Hence G is isomorphic to G(3), completing the proof.
Proof.Let G be an abundant d-regular graph.We can assume that r   Moreover there are at least 2d − 2 vertices of V 1 ∪ V 2 in N 2 (u) and no such vertex can have descendants at distance three from u 1 in V 3 by Condition 3 in Definition 5. Therefore we see that Since we are assuming that G is abundant, we can bound n 4 as follows: In total then the order of G is bounded above by d 3 − 2d 2 + 2d + 1.This shows that for any value of d there are only a finite number of abundant d-regular graphs.This raises the question whether we can strengthen this result by making the same conclusion for graphs with given maximum degree.The answer is no, as evidenced by the following construction.Let the maximum degree be ∆ ≥ 4. Let P be a path u 0 , u 1 , . . ., u r , where r ≥ 3, and take the Cartesian product P K 2 to form a ladder graph.If we identify the vertices of K 2 with {0, 1}, then we can write the vertices of the ladder as (u i , j) for 0 ≤ i ≤ r and j = 0, 1. Finally add ∆ − 3 leaves to each vertex (u i , 0) for 1 ≤ i ≤ r and ∆ − 4 leaves to (u 0 , 0) and (u r , 0).By taking the set of leaves to be V 1 , V 2 = V (P ) × {1} and V 4 = V (P ) × {0} we obtain an MGP-partition that shows that the graph is abundant.By increasing r we see that the number of vertices in this family is unbounded for fixed ∆.
For degrees two and three, we have shown that our construction G(d) is the unique abundant graph.Also, for both of these graphs the general position number of the Mycielskian is just one more than the order of the graph.This suggests the following problem.V 3 = ∅.As G is triangle-free, it follows star.However, by Corollary 18, the star S gp(M(S n )) = 2n − 2, whereas in the partition described above all but one vertex of G lies in V 2 .Since S is assumed to be a gp-set, this implies that gp(M(G)) = n + 1 = 2n − 2, yielding n = 3.We conclude that (V 1 , V 2 ) = ∅.
Suppose now that there is an edge v ∼ w in G, where v ∈ V 2 and w ∈ V 3 .By Condition 2 of Definition 5, all vertices of V 1 lie in N 2 (w).By Condition 2, N G (v) ∩ V 1 = ∅.Thus, for any u ∈ V 1 , there is a path u, x, w, v for some x ∈ V (G), and by Condition 3 this cannot be a shortest path.Again this implies the existence of a cycle with length at most five if V 1 = ∅.
If gp(M(G)) = n, then the claimed bound certainly holds, so assume that gp(M(G)) > n.Let M be a maximum matching of G.If x ∼ y is any edge in M, then it follows from (V 1 , V 2 ) = (V 1 , V 3 ) = ∅ that one of x, y is in V 4 and the vertices {x, y, x ′ , y ′ } of M(G) can contribute at most two to gp(M(G)).Summing over all edges of M, we see that the vertices saturated by M and their M-twins contribute at most 2ν(G) to gp(M(G)).Assuming that every other vertex lies in V 1 , we obtain the upper bound gp(M(G)) ≤ 2ν(G)+2(n−2ν(G)) = 2n − 2ν(G).
The star S n shows that the bound of Theorem 28 is tight.
Lemma 31.Let T be any tree with order n ≥ 3 and ℓ = s, i.e. each support vertex of T is adjacent to exactly one leaf.Then gp(M(T )) = n.
Proof.Let T be as described and assume for a contradiction that gp(M(T )) > n.By Lemma 6 and Corollary 30, every vertex of V 1 has neighbourhood lying in V 4 .Consider the subgraph V 1 ∪ V 4 of T .As n 1 > n 4 , there is no matching in this subgraph that saturates V 1 and we can choose a smallest subset In particular, the subtree V 1 ∪ V 4 is connected.
If any v ∈ V 4 has just one neighbour u in V 1 , then we could delete u and v from V 1 and V 4 to obtain a smaller pair with , where u 1 , u 2 ∈ N T (w 1 ).By our assumption ℓ = s, not both of u 1 , u 2 are leaves, so there is a w 2 = w 1 with u 2 ∼ w 2 .Now by the preceding argument w 2 must have a neighbour u 3 = u 2 in V 1 .To avoid cycles, we also have u 3 = u 1 .Therefore we have constructed a path u 1 , w 1 , u 2 , w 2 , u 3 , where u 1 , u 2 , u 3 ∈ V 1 .This path is a geodesic, and so violates Condition 3 of Definition 5.This contradiction establishes the result.
Lemma 31 shows in particular that paths P n with n ≥ 4 are meagre (and P 3 is also easily seen to be meagre).
Corollary 32.If T is any tree with order n ≥ 3, leaf number ℓ and s support vertices, then gp(M(T )) ≤ n + ℓ − s.

Figure 1 . 1 :
Figure 1.1:The Mycielskian of a path of length eight

Figure 2 . 1 :
Figure 2.1: General position sets (in red) used in the proof of Lemma 1

Figure 4 . 1 :
Figure 4.1: A gp-set (in red) of the Mycielskian of the gem graph

Figure 5 . 1 :
Figure 5.1: A 5-regular graph G with gp(M(G)) v} lie in N 2 (u).There are at least 2r vertices in V 1 ∪ V 2 and at most d 2 − d vertices in N 2 (u) and so 2r ≤ d 2 − d + 2. It follows that gp(M(G)) ≤ n + d In particular, for d ≥ 3 we have gp(M(G)) ≤ n + d−1 2 .Theorem 21 suggests the following construction of abundant regular graphs.Theorem 22.For all d ≥ 2, there is an abundant d-regular graph G(d) with order n = 3d−1 and gp(M(G(d))) = n + 1. Proof.We form the graph G(d) as follows.Take three sets V 1 , V 2 , V 4 of vertices, where |V 1 | = |V 2 | = d and |V 4 | = d − 1. Form the complete bipartite graph K d,d−1 with partite sets V 1 and V 4 .Make V 2 into a clique K d by adding an edge between every pair of vertices in V 2 .Finally join V 1 and V 2 by a matching of size d.The graph G(5) is shown in Figure 5.1.As suggested by our labelling of the sets,(V 1 , V 2 , ∅, V 4 ) is an MGP-partition of G(d).As ip 4 (G(d)) = α(G(d)) = d,these graphs are abundant.We now classify the abundant 2-and 3-regular graphs.It turns out that the graphs G(2) and G(3) are the only abundant regular graphs with degree three or less.Theorem 23.For n ≥ 3, the Mycielskian of the cycle C n has general position number gp(M(C n )) = n + 1, if n = 3 or 5 n, otherwise.
Consider a breadthfirst search tree T of depth three rooted at u 1 .By Condition 2 in Definition 5, all vertices of
There are at most d 2 − d vertices at distance two from u 1 and at most d(d − 1) 2 vertices in N 3 (u 1 ).

Problem 26 .
For given d ≥ 4, what is the largest value of gp(M(G)) − n over all d-regular graphs?Is G(d) the unique abundant graph?