Open Access Published by De Gruyter Open Access September 13, 2022

# On the general position number of two classes of graphs

• Yan Yao , Mengya He and Shengjin Ji
From the journal Open Mathematics

## Abstract

The general position problem is to find the cardinality of the largest vertex subset S such that no triple of vertices of S lies on a common geodesic. For a connected graph G , the cardinality of S is denoted by gp ( G ) and called the gp -number (or general position number) of G . In the paper, we obtain an upper bound and a lower bound regarding the gp -number in all cacti with k cycles and t pendant edges. Furthermore, the exact value of the gp -number on wheel graphs is determined.

MSC 2010: 05C12; 05C69; 68Q25

## 1 Introduction

In the paper, all graphs are undirected, finite, and simple. Assume that G = ( V , E ) is a connected graph with vertex set V ( G ) and edge set E ( G ) . Let u , v V ( G ) . d G ( u , v ) denotes the number of edges on a shortest ( u , v ) -path in G . A ( u , v ) -path with length d G ( u , v ) is regarded as a v , r -geodesic. The interval I G ( u , v ) of G is the set of vertices w such that there exists a u , v -geodesic, which contains w . We refer the readers to [1] for undefined terminology and notations.

The classical Dudeney’s no-three-in-line problem [2,3,4] is to determine as many as possible vertices that can be placed in the m × m grid such that no three vertices of them lie on a line. The famous problem has been further studied in several recent papers [5,6, 7,8]. Later, Froese et al. [9] extended the problem in discrete geometry to general position subset selection problem, which is to obtain a maximal subset of vertices in general position. They also showed that the general position subset selection problem is NP-hard.

Motivated by the above two problems, the general position problem of graphs was introduced by Manuel and Klavžar [10] and Payne and Wood [11]. A subset R of V ( G ) is a general position set in graph G if no three vertices of R lie on a common geodesic. A maximum general position set is named a gp -set of G . The general position number (the gp -number for short) of G is defined as the cardinality of a gp -set, denoted by gp ( G ) . Hence, the aim of the general position problem is to determine bounds or exact value of the gp -number for some classes of graphs.

Anand et al. [12] deduced a formula for the general position number of the complement of an arbitrary bipartite graph. Klavžar and Yero [13] proved that gp ( G ) ω ( G S R ) , where G S R is the strong resolving graph of a connected graph G , and ω ( G S R ) is its clique number. In [14,15], the authors determined upper bounds of the gp -number on integer lattices and Cartesian grids and showed that the general position problem is NP-complete. For more properties of the general position problem, see [10,16,17, 18,19]. Note that many researchers are highly concerned about the structural properties of the cactus and wheel graphs under graph parameters, such as [20,21,22]. Hence, it is interesting to explore the bounds of the gp -number for cactus and wheel graphs.

A connected graph G is called a cactus if its blocks consist of cycles and edges. A chain cactus is a cactus graph for which each block has at most two cut vertices and each cut vertex is shared by exactly two blocks. If a cycle of G has one cut vertex, we name it an end-block. A vertex of a cycle is said to be nontrivial if its degree is at least 3. Let v be a cut vertex of a cycle of G . If a component of G v contains an end-block C 0 (if v belongs to some cycle C , then the component does not contain vertices of C ), then C 0 is defined as an end-block of v . Moreover, the path between two cycles is a cyclic path, i.e., ( u , v ) -path is a cyclic path, see Figure 1. Let C n t , k be the class of all cacti of order n with k cycles and t pendant edges. Let C n k be the set of all cacti of order n with k cycles. The wheel graph W n is the graph obtained from the cycle C n of order n by adding a new vertex and connecting it to all vertices of C n .

Figure 1

( u , v ) -path is a acyclic path.

In the paper, we propose upper and lower bounds of the gp -number for cactus graphs, and the structure of the extremal graphs attaining the bounds is characterized. In addition, the exact value of the gp -number is also obtained for wheels.

## 2 The upper bound of cacti regarding gp -number

From [10], we know that gp ( C 3 ) = 3 , gp ( C 4 ) = 2 , and gp ( C n ) = 3 for n 5 .

## Lemma 2.1

Let G be a cactus with k ( 2 ) cycles, and S be a gp -set of G. If C 0 is a cycle of G, then V ( C 0 ) S 2 .

## Proof

Let C 1 and C 2 be two cycles of G connected by a path P (maybe trivial). Assume that V ( C 1 ) V ( P ) = u and V ( C 2 ) V ( P ) = r . Let u 1 and u 2 , u 3 and u 4 be the adjacent vertices of u and r in C 1 and C 2 , respectively. Note that R = { u 1 , u 2 , u 3 , u 4 } is a general position set of G . Therefore, gp ( G ) 4 .

Suppose V ( C 0 ) S = 3 and let V ( C 0 ) S = { x , y , z } . This means that I ( x , y ) I ( y , z ) I ( z , x ) = V ( C 0 ) and I ( x , y ) I ( y , z ) = y , I ( y , z ) I ( z , x ) = z , I ( x , y ) I ( z , x ) = x . Since G is a cactus, the cycle C 0 possesses at least one nontrivial cut vertex, marked as v . Let H 0 be a nontrivial subgraph of G such that V ( H 0 ) V ( C 0 ) = v . Without loss of generality, assume that v I ( x , y ) . For some v 0 V ( H 0 ) , every path connecting the vertex v 0 and all vertices in C 0 goes through v . Therefore, x (or y ) I ( v 0 , z ) , and we deduce that S V ( H 0 ) = 0 . Then, gp ( G ) = 3 , a contradiction.□

Manuel and Klavžar [10] determined the the gp -number of trees. We present it here.

## Lemma 2.2

If L is the set of leaves of a tree T, then gp ( T ) = L .

Suppose G is cactus and contains a cycle C t and a cyclic path P r . Set v a vertex of C t (or P r ) with degree more than two and its two neighbors as v 1 and v 2 . G v denotes the subgraph of G v v 1 v v 2 containing v . T k denotes the maximal subgraph induced by the vertices in acyclic subgraph of G v v together with v . Clearly, T k is a tree, we call it a pendant tree associated with v , and v is the root of T k . A root vertex v is referred as a nontrivial cut vertex of G .

## Lemma 2.3

If G C n t , k and T is the subgraph formed from pendant trees by removing their roots in G, then there exists a gp -set S such that S V ( T ) = t .

## Proof

Let G C n t , k has w pendant trees labeled as T 1 , , T , T + 1 , , T w with roots v 1 , , v , v + 1 , , v w , where the first roots  (resp. the last w roots) belong to cycles (resp. cyclic paths). Let T 1 = i = 1 { T i v i } , T 2 = i = + 1 w { T i v i } and T = T 1 T 2 . Let L i ( 1 i w ) be the set of leaves of the pendant tree T i and X i ( 1 i w ) be the set of vertices with degree one in G . Obviously, L i = X i . Let S be a gp -set of G .

Claim 1. S does not contain any root v i ( + 1 i w ) .

Assume that there is some v i 0 S for + 1 i 0 w . As shown in Figure 2, v i 0 is a nontrivial cut vertex of G , which, besides T i 0 , has p ( 2 ) components containing cycles labeled as H 1 , H 2 , , H p . For every triple including v i 0 of S marked as { x , y , v i 0 } , we conclude that x and y simultaneously belong to some component (or the pendant tree) of v i 0 , assume that H 1 . If not, suppose that x V ( H j ) and y V ( H k ) (or V ( T i 0 v i 0 ) ), then any x , y -geodesic is via v i 0 , contradicted with S . By the way, we obtain a new general position set S = S V ( H 1 ) { u 2 , u 3 , , u p , u p + 1 } . Obviously, S S + 2 , which is a contradiction.

We now consider the relationship between S and the remaining roots of G .

If S does not contain any root v i ( 1 i ) , we take S = S . Every vertex with degree 1 cannot lie on the geodesics of any other pair pendant vertices. Hence, S V ( T 1 ) = i = 1 S ( V ( T i ) v i ) = i = 1 S V ( T i ) = i = 1 X i = i = 1 w L i by Lemma 2.2.

We hence suppose that S includes some root v i 0 , which belongs to a cycle C 0 . Evidently, S = ( S V ( G T i 0 ) ) ( S ( V ( T i 0 ) { v i 0 } ) ) { v i 0 } . We claim that T i 0 is a path. Note that every path from the vertices in V ( G T i 0 ) to the vertices of V ( T i 0 ) { v i 0 } is via v i 0 , which implies S V ( G T i 0 ) = or S ( V ( T i 0 ) { v i 0 } ) = . First, suppose that S ( V ( G T i 0 ) ) = . Mark the two adjacent vertices of v i 0 as x and y in C 0 . Then, we obtain a new general position set S = { x , y } ( S ( V ( T i 0 ) { v i 0 } ) ) . This means that S > S , which is a contradiction with the maximum of S . That is to say, S ( V ( T i 0 ) { v i 0 } ) = . If X i 0 2 , we have a new general position set S = ( S V ( G T i 0 ) ) X i 0 . Then S > S , contradicted with the maximum of S . So X i 0 = 1 , which implies that T i 0 is a path. Let S = ( S { v i 0 } ) X i 0 . Clearly, S = S and S is a gp -set of G .

Assume now that S contains some of these roots. Then, repeating the aforementioned process, we obtain a new gp -set S of G such that all roots are not its elements.

By Lemma 2.2, we conclude that S V ( T ) = i = 1 w S V ( T i ) = i = 1 w X i = i = 1 w L i = t .□

Figure 2

The two graphs G and G 0 used in Lemma 2.3 and Theorem 2.4.

Let C be the set of all cycles and T be the subgraph formed from pendant trees by removing their roots in G . From Lemmas 2.1 and 2.3, we deduce gp ( G ) = S V ( C ) + S V ( T ) 2 k + t . In order to demonstrate that the bound is best, we next exhibit the structure of cacti such that their g p -number equal 2 k + t .

For convenience, we introduce some notations. Let G C n t , k and C = u 1 u 2 u u 1 be a cycle of G . For two vertices u i and u j , clearly, C consists of two ( u i , u j ) -paths. If all cut-vertices of C is located on one ( u i , u j ) -path, then this path is referred to as a ( u i , u j ) -cut-path and use d c ( u i , u j ) to denote the number of edges in the ( u i , u j ) -cut-path. Let D c be min u i , u j V ( C ) d c ( u i , u j ) of C . We call cycle C a good cycle if D c is no more than either 2 2 for even or 2 1 for odd . Trivially, the cycle as an end-block of cactus is a good cycle. In addition, if D c is at least either 2 for even or 2 + 1 for odd , then the cycle is regarded as a bad cycle. Furthermore, C is referred to a normal cycle if D c equals to 2 1 for even or 2 for odd . Let H 0 = v 1 v 2 v v 1 be a cycle of G 0 (see Figure 1). Assume that H i ( 1 i ) is a connected subgraph with V ( H i ) V ( H 0 ) = { v i } and V ( H i ) = n i for i { 0 , 1 , , } .

## Theorem 2.4

Let G 0 C n t , k . gp ( G 0 ) = 2 k + t if only all cycles of G 0 are good.

## Proof

Let G 0 C n t , k with gp ( G 0 ) = 2 k + t . Suppose S is a gp -set of G 0 . Hence, S = 2 k + t and S V ( H 0 ) = 2 for every cycle H 0 of G 0 by Lemma 2.3. To verify the conclusion, it is sufficient to show S V ( H 0 ) = 2 if and only if H 0 is a good cycle.

We now assume that H 0 is a good cycle. Suppose d c ( v i 0 , v j 0 ) 2 2 for even . We take the two vertices that are adjacent to v i 0 and v j 0 on another ( v i 0 , v j 0 ) -path in H 0 , and mark them as v i 0 and v j 0 , respectively. It is clear that the two vertices have degree two. For an arbitrary vertex u 0 in G V ( H 0 ) , v i 0 is not on a ( u 0 , v j 0 ) -shortest path and v j 0 is also not on a ( u 0 , v i 0 ) -shortest path. If S V ( H 0 ) < 2 , then let S be the set obtained from S by removing all the elements of its subset S V ( H 0 ) and adding the two vertices v i 0 and v j 0 . We deduce that S is a general position set with S S + 1 , which contradicts the maximum of S . Hence, S V ( H 0 ) = 2 .

Suppose d c ( v i 0 , v j 0 ) 2 1 = 1 2 1 for odd . Let v i 0 and v j 0 denote the two vertices of degree two, which are adjacent to v i 0 and v j 0 on another ( v i 0 , v j 0 ) -path, respectively. Hence, for an arbitrary vertex w 0 in G V ( H 0 ) , v i 0 is not on a ( w 0 , v j 0 ) -shortest path and v j 0 is also not on a ( w 0 , v i 0 ) -shortest path. By a similar way as mentioned earlier, we obtain S V ( H 0 ) = 2 .

Therefore, we obtain that S V ( H 0 ) = 2 if H 0 is a good cycle.

On the contrary, G includes at least a cycle H 0 , which is not a good cycle. Note that end-blocks of G 0 are good. We thus assume that H 0 is not an end-block as shown in Figure 2. (It includes at least two cut vertices.) Hence, its D c is no less than either 2 1 for even or 2 for odd . Let v i 0 and v j 0 be two cut-vertices of H 0 for which D c = d c ( v i 0 , v j 0 ) .

We first consider the special case that H 0 contains only two cut vertices. Then, D c 2 , and we deduce that S V ( H 0 ) 1 . Suppose now that there are at least three cut vertices in V ( H 0 ) . If d c ( v i 0 , v j 0 ) 2 + 1 , then there exists a cut vertex v r which belongs to the ( v i 0 , v j 0 ) -cut-path of the cycle H 0 , such that one vertex is not on the geodesic of the other two vertices in the triple { v i 0 , v j 0 , v r } . Let now v k 0 ( v i 0 , v j 0 , v r ) be an arbitrary vertex of H 0 . Assume that v k 0 is one vertex of the shortest ( v i 0 , v j 0 ) -path. Then, it is easy to deduce that every shortest path from the vertices in H v i 0 to the vertices in H v j 0 goes through v k 0 . If S V ( H 0 ) = 2 , then at least one of S V ( H v i 0 ) , S V ( H v j 0 ) and S V ( H v r ) is equal to zero, so gp ( G 0 ) < 2 k + t by Lemmas 2.1 and 2.3. Hence, S V ( H 0 ) 1 .

Assume that d c ( v i 0 , v j 0 ) = 2 . If is even, then 2 = 2 . For a vertex u of H v i 0 v i 0 and a vertex w of H v j 0 v j 0 , all vertices of H 0 lie on some ( u , w ) -geodesic. If S V ( H 0 ) 1 . Then, S V ( H v i 0 ) = 0 or S V ( H v j 0 ) = 0 . According to Lemmas 2.1 and 2.3, gp ( G 0 ) < 2 k + t . Hence, S V ( H 0 ) = 0 If is odd, then 2 = 1 2 . Using a similar way for even , it is verified that S V ( H 0 ) = 1 .

Suppose d c ( v i 0 , v j 0 ) = 2 1 = 2 1 for even . We conclude that any vertex, which lies on a ( v i 0 , v j 0 ) -cut-path is not an element of S . Let P 0 be another ( v i 0 , v j 0 ) -path in H 0 . Clearly, S ( V ( P 0 ) \ { v i 0 , v j 0 } ) = 1 .

Therefore, the proof is complete.□

By means of Theorem 2.4, we deduce that there are a lot of graphs belonging to C n t , k for which the gp -number is equal to 2 k + t and these graphs have different shapes.

Let A 1 be a cactus in which all cycles and all pendant trees share a common cut vertex v . Let R be the vertex set formed by all neighbors of v in cycles and the pendant vertices in A 1 . In fact, R is a general position set, and R = 2 k + t . Hence, gp ( A 1 ) 2 k + t . Conversely, Lemmas 2.1 and 2.3 lead to gp ( A 1 ) 2 k + t . Thus gp ( A 1 ) = 2 k + t .

In addition, let A 2 denote a chain cactus with the following two properties: (a) the length of every cycle in A 2 is at least 5 and its two cut vertices (if they exist) are adjacent, (b) there is one cut vertex of some cycle possessing pendant tree. Evidently, take two neighbors of the two cut vertices different with themselves for each cycle and all pendant vertices forming a general position set of A 2 , denote it by R . Then, R = 2 k + t and gp ( A 2 ) 2 k + t . So gp ( A 2 ) = 2 k + t by Lemmas 2.1 and 2.3.

We now give two concrete examples. Let B 1 C n t , k with k = 3 , t = 3 , n = 17 , and B 2 C n t , k with k = 4 , t = 2 , n = 19 , see Figure 3. We have their general position sets S and S that consists of all solid vertices, respectively. Evidently, S = 9 = 2 k + t and S = 10 = 2 k + t .

Figure 3

B 0 is used in Theorem 2.5, B 1 and B 2 attain the bound in Theorem 2.4.

By means of Theorem 2.4, the main result holds.

## Theorem 2.5

If G C n t , k , then

gp ( G ) max { 3 , 2 k + t } ,

where the equality holds if and only if all cycles of G are good.

## Proof

Case 1. k = 1 and G with at most one pendant vertex. Note that G is a unicyclic graph, mark its unique cycle as C 0 . Clearly, C 0 is a good cycle. We check that gp ( G ) 3 . In addition, gp ( G ) = 3 means that G C n for n 4 or G contains a pendant vertex.

Case 2. k 2 or k = 1 and G with at least two pendant vertices. Observe that 3 2 k + t . Lemmas 2.1 and 2.3 result in gp ( G ) 2 k + t . Moreover, Theorem 2.4 implies that gp ( G ) = 2 k + t if and only if each cycle in G is good.

Together with the aforementioned two cases, the proof is complete.□

## Corollary 2.6

Let G C n k , we have

gp ( G ) max { 3 , n 1 } ,

where equality holds if and only if G C 3 or G B 0 .

## Proof

Suppose G C n k . If G is isomorphic to C 3 , then gp ( G ) = 3 . We now assume that G C 3 . Let t be the number of pendant edges in G . Clearly, t n 2 k 1 . Since k is bounded, we deduce that gp ( G ) 2 k + t 2 k + n 2 k 1 = n 1 .

gp ( G ) = n 1 leads to that t = n 2 k 1 . That is to say, every cycle of G is a triangle, and all triangles share a common vertex. Thus, G B 0 , see Figure 3. Actually, it is easy to check that all vertices except for the cut vertex form a general position set of B 0 .□

## 3 A lower bound for cacti

In this section, we will prove some lower bounds for cacti. Note that for a graph G C n t , k , its pendant vertices consist of a general position set. Hence, there exist an obvious lower bound, gp ( G ) t . It is interesting to determine all graphs attaining this bound. In addition, we also consider cactus graph with t = 0 . In particular, if G C n 0 , 1 , then G C n , and hence, gp ( G ) = 3 for n = 3 or n 5 or gp ( G ) = 3 for n = 4 . Moreover, if G C n 0 , 2 , then gp ( G ) = 4 . We now consider G C n 0 , k for k 3 .

## Theorem 3.1

Suppose k and k 1 are two integers with k 3 and k 1 0 . If G C n 0 , k has k 1 odd cycles, then

gp ( G ) k 1 + 2 k 1 3 , 4 o t h e r w i s e .

Equality holds if and only if G is a chain cactus for which, except for two end-blocks, D c of every even cycle equals to the half number of its vertices and D c of every odd cycle equals to the floor of the half number of its vertices, or G is a cactus such that at least one cycle owns three cut vertices, which includes three properties: there are at least one odd cycle with three cut vertices, all even cycles have two cut vertices, and all end-blocks are odd.

## Proof

Let k and k 1 are two integers and k 3 and k 1 0 . Suppose G is a cactus in C n 0 , k having the minimal gp -number. Let S be a gp -set of G .

To exhibit clearly, we now introduce some notations. Let C t 1 , C t 2 , , C t k be the k cycles of G . e d ( G ) denotes the number of end-blocks in G . Let i be the number of cut vertices on C t i for i = 1 , 2 , , k . Assume that e d 2 ( G ) denotes the number of the cycles with two cut vertices in G . Without loss of the generality, suppose 1 1 2 k . In fact, e d ( G ) = 2 + i = e d ( G ) + 1 k i 2 . Let v e d be a cut vertex of C t e d ( G ) . In addition, set v e d the vertex on C t e d ( G ) such that d c ( v e d , v e d ) is as big as possible and d c ( v e d , v e d ) t e d 2 . To show the conclusion, we first show the claim.

Claim 1 G has the following properties.

1. Each vertex in cyclic paths of G has degree two.

2. For each vertex v on cycles of G , G v has at most two components.

3. r 3 for 1 r k . In particular, r 2 if C r is even, r 3 otherwise. If r = 3 , then C t r is odd and end-blocks of G are odd.

4. If G has an even end-block, then G is a chain cactus.□

## Proof

(A) On the contrary, suppose that there exists a vertex u 0 in a cyclic path of G with d G ( u 0 ) 3 . Set its one neighbor u 1 such that u 1 isn’t on the shortest ( u 0 , v e d ) -path. Let G be the new graph from G by deleting the edge u 0 u 1 and joining u 1 v e d . Set S a gp -set of G . Clearly, S S = V ( C t e d ) S V ( C t e d ) S 2 1 = 1 , is a contradiction.

Using the same way, we can verify (B); thus, the proof is omitted.

(C) On the contrary, suppose r 4 with r e d ( G ) + 1 as small as possible. Thus, there exists three cut vertices as v r , v r 0 , and v r 1 of C t r such that v r is a vertex in the ( v r 0 , v r 1 ) -cut-path with length D c . G is the new graph from G by removing the subgraph from v r to v e d . Let S be a gp -set of G . Let i be the number of cut vertices of C t i in G for i = 1 , 2 , , k . Evidently, i = i for i r , e d ( G ) , r = r 1 3 , and e d ( G ) = e d ( G ) + 1 = 2 . Note that V ( C t r ) S = V ( C t r ) S . Hence, S S = V ( C t e d ) S V ( C t e d ) S 2 1 = 1 , which contradicts with the choice of G .

Suppose there is an even cycle C t r with r = 3 . Let u 0 is a cut vertex of C t r . G denotes the graph obtained from G by shifting the subgraph from u 0 to an end-block C t h ( h e d ( G ) ) and adjusting the positions of cut vertices in two cycles for which C t r is bad and C t h is not good. Let S is a gp -set in G . Observe that V ( C t r ) S = V ( C t r ) S = 0 . Thus, S S = V ( C t h ) S V ( C t h ) S 2 1 = 1 , which is a contradiction.

If r = 3 , by the aforementioned discussion, we have C t r is odd. All end-blocks of G are odd. If not, using the same way,we also obtain a contradiction.

(D) Suppose G includes an even end-block. From (C), we deduce that r 2 . Together (A) with ( B ) , it is deduced that G is a chain cactus.□

If 1 r 2 for r k , then G is a chain cactus. Note that, except for two end-blocks, all others internal cycles contain two cut vertices. Furthermore, we discuss the contribution of internal cycles to S , as an bad even cycle C 0 , S V ( C 0 ) = 0 , and as a normal odd cycle C 0 , S V ( C 0 ) = 1 . Since G has the minimum with respect to the gp -number, odd cycles should be end-blocks as many as possible. Recall that the condition G owns k 1 odd cycles and k k 1 even cycles, we hence arrive at gp ( G ) = k 1 2 + 4 = k 1 + 2 for k 1 3 and gp ( G ) = 4 for k 1 = 0 , 1 , 2 . If r = r + 1 = = k = 3 , then all end-blocks are odd and e d ( G ) = 2 + i = r k i 2 = 2 + ( k r + 1 ) . Therefore, gp ( G ) = 2 e d ( G ) + k 1 ( k r + 1 ) e d ( G ) = k 1 + 2 .

Based on the aforementioned conclusion, we deduce the following two results.

## Corollary 3.2

If G C n 0 , k has k odd cycles, then gp ( G ) k + 2 with equality if and only if G is either a chain cactus for which, except for two end-blocks, D c of each cycle equals to the floor of the half number of its vertices, or a cactus such that each cycle owns at most three cut vertices, which includes three properties: there are odd cycles with three cut vertices, and all even cycles have two cut vertices and all end-blocks are odd.

## Corollary 3.3

If G C n 0 , k has k even cycles, then gp ( G ) 4 with equality if and only if G is a chain cactus for which, except for two end-blocks, D c of each cycle equals to the half number of its vertices.

## Theorem 3.4

If G C n t , k , then gp ( G ) t with equality if and only if the cycles are bad.

## Proof

Suppose G C n t , k , S is the set of pendant vertices in G . By the property of the set of pendant vertices, gp ( G ) t . If equality holds, then S is the gp -set of G by Lemma 2.3. This means that the vertices of all cycles do not contribute to S . In other words, for any cycle C a of G , V ( C a ) S = 0 . Let h be the number of cut vertices of C a . Clearly, h 2 . (On the contrary, C a is an end-block. So V ( C a ) S = 2 , a contradiction.)

Assume that h = 2 , we deduce that D c a 2 for an even a , D c a 2 1 otherwise. Observe that V ( C a ) S = 0 results in C a is bad, which deduces that a is even and D c = a 2 . We next assume that h 3 . V ( C a ) S = 0 infers that C a is bad from the definition of the bad cycle. By the choice of C a , we finish the proof.□

## 4 Wheel graphs

In this section, we will deduce the gp -number of wheels. For convenience, we now introduce some notation. The vertex with degree n in W n is called the center of W n and denoted by w . Let C n = W n w , and S denote a gp -set of W n . Let G 0 = C n S for short.

## Theorem 4.1

If n 3 , then gp ( W n ) = 4 , if n = 3 , 3 , if n = 4 , 5 , 2 3 n , if n 6 .

## Proof

Let C n = v 1 v 2 v n v 1 . If n = 3 , then W 3 K 4 . Clearly, { v 1 , v 2 , v 3 , w } forms a general position set of W 3 , so gp ( W n ) 4 . Hence, gp ( W n ) = 4 by V ( W 3 ) = 4 . Suppose n = 4 , 5 . Evidently, { v 1 , v 2 , w } is a general position set of W n . In addition, S V ( W n ) with S 4 , W n [ S ] leads to an induced P 3 . Thus, gp ( W n ) 3 , which infers gp ( W n ) = 3 . We next assume that n 6 . Let S be a gp -set of W n .□

Claim 1. If n 6 , then the center w S .

## Proof of Claim 1

If w S , then S = 3 . Since n 6 , there are two edges for which their distance is no less than 2 in C n , and we denote their ends as x 1 , x 2 and x 3 , x 4 , respectively. Then S = { x 1 , x 2 , x 3 , x 4 } is a general position set, which implies that S > S . So w S .□

Note that C n [ S ] contains no path P 3 . Hence, it consists of some K 2 and K 1 .

Claim 2. G 0 contains at most one K 2 .

## Proof of Claim 2

Suppose G 0 has at least two K 2 . Let e 1 = v 1 v 2 be an edge in G 0 . It is not difficult to find another edge e 2 = v x 1 v x E ( G 0 ) such that it is closest to e 1 in C n . We now construct a new subset S from S by the following process. If the vertex v i S ( 3 i x 2 ) , then v i 1 S , for others vertices of S , we copy them to S . Since v x 2 S we deduce v x 2 S , and add v x 1 to S . In fact, every path in W n possessing some triple of S is not geodesic. Hence, S is a general position set of W n . Then, S = S + 1 > S , which is a contradiction.□

Based on the aforementioned two claims, the gp -set S is a subset of V ( C n ) and C n [ S ] consists of some K 1 and K 2 for which every two consecutive components are separated by K 1 or K 2  (at most one). For convenience, the number of K 1 and K 2 in C n [ S ] are 1 and 2 , respectively. Hence, gp ( W n ) = S = 1 + 2 2 . In addition, if G 0 contains one K 2 , G 0 = 1 + 2 + 1 ; otherwise, G 0 = 1 + 2 . In addition, 2 1 + 3 2 n . We next take three cases to show the result.

Case 1. n = 3 k .

From the fact 2 1 + 3 2 3 k , we arrive at gp ( W n ) = S = 1 + 2 2 = 2 3 ( 2 1 + 3 2 ) 1 3 1 2 3 ( 3 k ) 1 3 1 = 2 k 1 3 1 .

On the other hand, R = { v 2 , v 3 , v 5 , v 6 , v 8 , v 9 , , v 3 k 1 , v 3 k } is a general position set of W n and R = 2 k . Hence, gp ( W n ) 2 k . That is to say, 2 k 2 k 1 3 1 , which implies 1 = 0 . Consequently, 2 = k and W n [ S ] is made up of 2 k copies of K 2 .

Case 2. n = 3 k + 1 .

Using the fact 2 1 + 3 2 3 k + 1 , we deduce that gp ( W n ) = S = 1 + 2 2 = 2 3 ( 2 1 + 3 2 ) 1 3 1 2 3 ( 3 k + 1 ) 1 3 1 = 2 k + 2 3 1 3 1 .

Now let R = { v 2 , v 3 , v 5 , v 6 , v 8 , v 9 , , v 3 k 1 , v 3 k } . As in Case 1, R is a general position set of W n and R = 2 k . So, gp ( W n ) 2 k . Hence, 2 k 2 k + 2 3 1 3 1 < 2 k + 1 . Namely, gp ( W n ) = 2 k . Moreover, R is a gp -set of W n .

Case 3. n = 3 k + 2 .

From the fact 2 1 + 3 2 3 k + 2 , we obtain that gp ( W n ) = S = 1 + 2 2 = 2 3 ( 2 1 + 3 2 ) 1 3 1 2 3 ( 3 k + 2 ) 1 3 1 = 2 k + 4 3 1 3 1 .

Take R = { v 2 , v 3 , v 5 , v 6 , v 8 , v 9 , , v 3 k 1 , v 3 k , v 3 k + 2 } . It is not difficult to verify that R is a general position with R = 2 k + 1 . So gp ( W n ) 2 k + 1 . Hence, 2 k + 1 2 k + 4 3 1 3 1 < 2 k + 2 , which implies gp ( W n ) = 2 k + 1 . Furthermore, W n [ S ] consists of one K 1 and 2 k K 2 . Evidently, R is a gp -set of W n .

Combining with the aforementioned three cases, the result holds.

## Acknowledgements

We first would particularly like to thank Professor Klavžar for his meaningful and helpful suggestions. We also would like to thank anonymous reviewers whose valuable comments greatly improved the presentation of the proof. In addition, the authors acknowledge the financial support from National Natural Science Foundation of China (No. 11561032) and Shandong Provincial Natural Science Foundation of China (No. ZR2019MA012).

1. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

2. Conflict of interest: The authors state no conflict of interest.

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