On the general position number of two classes of graphs

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.□

Lemma 2.2

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

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 ℓ , T ℓ + 1 , … , T w with roots v 1 , … , v ℓ , v ℓ + 1 , … , v w , where the first ℓ roots  (resp. the last w − ℓ roots) belong to cycles (resp. cyclic paths). Let T 1 = ⋃ i = 1 ℓ { T i − v i } , T 2 = ⋃ i = ℓ + 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 ( ℓ + 1 ≤ i ≤ w ) .

Assume that there is some v i 0 ∈ S ′ for ℓ + 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 ≤ ℓ ) , 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 ℓ S ∩ ( V ( T i ) − v i ) ∣ = ∑ i = 1 ℓ ∣ S ∩ V ( T i ) ∣ = ∑ i = 1 ℓ ∣ 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 .□

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 ) ≤ ℓ 2 − 2 for even ℓ . 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 ) ≤ ℓ 2 − 1 = ℓ − 1 2 − 1 for odd ℓ . 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 ℓ 2 − 1 for even ℓ or ℓ 2 for odd ℓ . 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 ≤ ℓ 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 ) ≥ ℓ 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 ) = ℓ 2 . If ℓ is even, then ℓ 2 = ℓ 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 ℓ is odd, then ℓ 2 = ℓ − 1 2 . Using a similar way for even ℓ , it is verified that ∣ S ∩ V ( H 0 ) ∣ = 1 .

Suppose d c ( v i 0 , v j 0 ) = ℓ 2 − 1 = ℓ 2 − 1 for even ℓ . 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.□

Theorem 2.5

If G ∈ C n t , k , then

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

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 .□

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

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 ℓ 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 ≤ ℓ 1 ≤ ℓ 2 ≤ … ≤ ℓ k . In fact, e d ( G ) = 2 + ∑ i = e d ( G ) + 1 k ℓ 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. ℓ r ≤ 3 for 1 ≤ r ≤ k . In particular, ℓ r ≤ 2 if C r is even, ℓ r ≤ 3 otherwise. If ℓ 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 ℓ 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 ℓ i ′ be the number of cut vertices of C t i in G ′ for i = 1 , 2 , … , k . Evidently, ℓ i ′ = ℓ i for i ≠ r , e d ( G ) , ℓ r ′ = ℓ r − 1 ≥ 3 , and ℓ e d ( G ) ′ = ℓ 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 ℓ 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 ℓ 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 ℓ r ≤ 2 . Together (A) with ( B ) , it is deduced that G is a chain cactus.□

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.