- Email: [email protected]
Non-separating trees in connected graphs
Discrete Mathematics 309 (2009) 5235–5237
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Note
Non-separating trees in connected graphs Ajit A. Diwan ∗ , Namrata P. Tholiya Department of Computer Science and Engineering, Indian Institute of Technology Bombay, Powai, Mumbai 400076, India
article
info
Article history: Received 3 October 2008 Received in revised form 3 February 2009 Accepted 21 March 2009 Available online 19 April 2009
abstract Let T be any tree of order d ≥ 1. We prove that every connected graph G with minimum degree d contains a subtree T 0 isomorphic to T such that G − V (T 0 ) is connected. © 2009 Elsevier B.V. All rights reserved.
Keywords: Non-separating tree Isomorphic subtree Minimum degree
1. Introduction A subgraph H of a connected graph G is said to be non-separating if G − V (H ) is a connected non-empty graph. It is well-known that every non-trivial connected graph G contains a vertex v such that G − v is connected. It is also known that every connected graph with minimum degree 2 contains a non-separating edge [4] and that every connected graph with minimum degree 3 contains a non-separating induced cycle [6]. A non-trivial connected graph G is said to be k-cohesive if for any two distinct vertices u and v , d(u) + d(v) + d(u, v) ≥ k, where d(u) is the degree of u and d(u, v) is the distance between u and v . Locke [2] conjectured that for k ≥ 3, every connected 2k-cohesive graph contains a non-separating copy of every tree of order k, and proved it for paths [3]. Abreu and Locke [1] proved that every connected (2k + 2)-cohesive graph contains a non-separating copy of every tree of order k and diameter at most 4. We show that every connected graph of minimum degree d contains a non-separating copy of every tree of order d. The graph mKd−1 ∨ K1 , for m ≥ 3, shows that the degree bound is tight for any tree of order d. Our result may be considered to be a partial step toward Locke’s conjecture. The proof is based on a technique used by Mader to prove a completely different result. Mader [5] showed that every graph with minimum degree d contains an edge uv such that there are d internally vertex-disjoint paths between u and v . We extend this technique slightly to obtain our result. The proof also leads to a polynomial-time algorithm for finding a non-separating copy of any given tree of order d in a connected graph of minimum degree d. The notation used is largely standard and follows, for example, [7]. 2. Main result Theorem 1. Let G be a connected graph with minimum degree d ≥ 1. Then for any tree T of order d, G contains a subtree T 0 isomorphic to T such that G − V (T 0 ) is connected.
∗
Corresponding author. E-mail address: [email protected] (A.A. Diwan).
0012-365X/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2009.03.037
5236
A.A. Diwan, N.P. Tholiya / Discrete Mathematics 309 (2009) 5235–5237
Before proving the theorem, we introduce some definitions. An ordered clique K in a graph G is a complete subgraph of G with an ordering imposed on the vertices of K . Let K be an ordered clique in a graph G with the ordering v1 , v2 , . . . , vk of its vertices. A subtree T of G is said to be consistent with the ordered clique K if every vertex vi ∈ V (K ) ∩ V (T ) has at most one neighbor in T that is not contained in {v1 , v2 , . . . , vi−1 }. We consider ordered pairs of the form (G, K ), where K is an ordered clique in a graph G. We assume throughout that d ≥ 1 is an integer and T is any fixed tree of order d. Let u1 , u2 , . . . , ud be an ordering of the vertices of T such that ui is adjacent to exactly one vertex uj with j > i, for 1 ≤ i < d. We prove a statement stronger than Theorem 1 by induction. Lemma 2. Let (G, K ) be an ordered pair such that K is a non-empty ordered clique in a connected graph G, and K is a proper subgraph of G. Let v1 , v2 , . . . , vk be the ordering of the vertices of K . If every vertex in V (G) \ V (K ) has degree at least d in G, then there exists a subtree T 0 of G satisfying the following properties. (1) (2) (3) (4)
v1 6∈ V (T 0 ).
T 0 is isomorphic to T . T 0 is consistent with K . Every connected component of G − (V (K ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|.
Proof. Suppose there exists a counterexample. Let (G, K ) be a counterexample that minimizes |V (G)| + |V (G) \ V (K )|. If |V (G)\ V (K )| = 1 then let w be the only vertex in V (G)\ V (K ). Since the degree of w is at least d in G, |NG (w)∩ V (K )| ≥ d. Let vi1 , vi2 , . . . , vid be vertices in V (K ) that are adjacent to w , such that i1 < i2 < · · · < id . Now let the vertex uj of T correspond to the vertex vij+1 in G for 1 ≤ j < d, and let the vertex ud of T correspond to the vertex w in G. This gives a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . Since G − (V (K ) ∪ V (T 0 )) is empty, the fourth property in Lemma 2 is trivially satisfied. This contradicts the assumption that (G, K ) is a counterexample. Suppose |V (G) \ V (K )| > 1. We consider two cases. Case 1. There exists a vertex v ∈ V (G) \ V (K ) such that V (K ) ⊆ NG (v). Let K 0 be the ordered clique v1 , v2 , . . . , vk , v in G. Then (G, K 0 ) satisfies the hypothesis of Lemma 2, and by the minimality of the counterexample (G, K ), (G, K 0 ) is not a counterexample. Hence, there exists a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K 0 , and every connected component of G − (V (K 0 ) ∪ V (T 0 )) contains a vertex w with |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )|. We show that T 0 satisfies all four properties in Lemma 2, contradicting the assumption that (G, K ) is a counterexample. Clearly, v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . It remains to show that every connected component of G − (V (K ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|. If v ∈ V (T 0 ) then any component of G − (V (K ) ∪ V (T 0 )) is a component of G − (V (K 0 ) ∪ V (T 0 )). Also, for any vertex w in such a component, |NG (w) ∩ V (K )| ≥ |NG (w) ∩ V (K 0 )| − 1. Since |V (K ) ∩ V (T 0 )| = |V (K 0 ) ∩ V (T 0 )| − 1 and every component of G − (V (K 0 ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )|, it follows that for the same vertex w, |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|. If v 6∈ V (T 0 ) then for the component of G − (V (K ) ∪ V (T 0 )) that contains v , |NG (v) ∩ V (K )| > |V (K ) ∩ V (T 0 )|, since v is adjacent to all vertices in V (K ) and v1 6∈ V (T 0 ). Any other component of G − (V (K ) ∪ V (T 0 )) is also a component of G − (V (K 0 ) ∪ V (T 0 )). For any vertex w in such a component, |NG (w) ∩ V (K )| = |NG (w) ∩ V (K 0 )| since w is not adjacent to v . Hence, any such component contains a vertex w such that |NG (w) ∩ V (K )| = |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )| = |V (K ) ∩ V (T 0 )|. Case 2. No vertex in V (G) \ V (K ) is adjacent to all vertices in V (K ). We must have |V (K )| = k > 1, since G is connected and K is a proper subgraph of G. We construct a new graph G0 and an ordered clique K 0 in G0 as follows. Let X = NG (vk ) \ V (K ) be the set of neighbors of vk that are not contained in V (K ). For each vertex w ∈ X , let π (w) be the largest index i, such that w is not adjacent to vi ∈ V (K ). Note that π (w) < k and is well-defined, since w is adjacent to vk and is not adjacent to at least one vertex in V (K ), by assumption. Let V (G0 ) = V (G −vk ) and E (G0 ) = E (G −vk )∪{wvπ (w) |w ∈ X }. Let K 0 = K − vk with the ordering v1 , v2 , . . . , vk−1 of its vertices. We call the edges {wvπ (w) |w ∈ X } bad edges in G0 . This construction ensures that G0 is connected and the degree of every vertex in V (G0 )\ V (K 0 ) is at least d in G0 . Also, K 0 is a non-empty, proper subgraph of G0 ; hence (G0 , K 0 ) satisfies the hypothesis of Lemma 2. Since |V (G0 ) \ V (K 0 )| = |V (G) \ V (K )| and |V (G0 )| < |V (G)|, by the minimality of the counterexample (G, K ), (G0 , K 0 ) is not a counterexample. Let T 00 be a subtree of G0 such that v1 6∈ V (T 00 ), T 00 is isomorphic to T and consistent with K 0 , and every connected component of G0 − (V (K 0 ) ∪ V (T 00 )) contains a vertex w with |NG0 (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 00 )|. We modify T 00 to find a subtree T 0 of G satisfying the four properties in Lemma 2, thus contradicting the assumption that (G, K ) is a counterexample. If none of the bad edges in G0 is contained in T 00 then let T 0 be the same as T 00 . Suppose wvi is a bad edge contained in T 00 for some w ∈ X and vi ∈ V (K 0 ). Since T 00 is consistent with K 0 and v1 6∈ V (T 00 ), all other neighbors of vi in T 00 are contained in {v2 , . . . , vi−1 }; hence T 00 can contain at most one bad edge incident with vi . Also, from the construction, there is exactly one bad edge incident with a vertex w ∈ X . This implies that the bad edges contained in T 00 are independent.
A.A. Diwan, N.P. Tholiya / Discrete Mathematics 309 (2009) 5235–5237
5237
Let w1 vi1 , w2 vi2 , . . . , wm vim be the bad edges contained in T 00 , such that 1 < i1 < i2 < · · · < im < k. By the definition of π (w), if wvi is a bad edge in G0 , then wvj is an edge in G for all i < j ≤ k. Construct T 0 from T 00 by replacing the vertex vij ∈ V (T 00 ) by the vertex vij+1 for 1 ≤ j ≤ m, where vim+1 is vk . This is possible since vij+1 is adjacent to wj in G and all other neighbors of vij in T 00 are contained in {v2 , v3 , . . . , vij −1 }. Since T 00 is consistent with K 0 , any such neighbor of vij will not have a bad edge incident with it in T 00 , and will be a vertex in T 0 also. Therefore T 0 is a subtree of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . If w is any vertex in V (G) \ V (K ), then |NG (w) ∩ V (K )| = |NG0 (w) ∩ V (K 0 )| since any bad edge incident with w in G0 is replaced by the edge wvk in G. Since the connected components of G − (V (K ) ∪ V (T 0 )) are also the components of G0 − (V (K 0 ) ∪ V (T 00 )), every such component contains a vertex w with |NG (w) ∩ V (K )| = |NG0 (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 00 )| = |V (K ) ∩ V (T 0 )|. Hence, T 0 is a subtree of G satisfying the properties in Lemma 2. Theorem 1 now follows from Lemma 2. Let G be any connected graph with minimum degree d ≥ 1, and let v1 be any vertex in V (G). Let K be the ordered clique in G containing only the vertex v1 . Then (G, K ) satisfies the hypothesis of Lemma 2, and there is a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and every connected component of G −(V (T 0 )∪{v1 }) contains a vertex that is adjacent to v1 . Thus T 0 is a non-separating subtree of G isomorphic to T . Acknowledgements We thank the referees for their comments and suggestions, which helped in improving the presentation. References [1] M. Abreu, S.C. Locke, Non-separating n-trees up to diameter 4 in a (2n+2)-cohesive graph, in: Proceedings of the Thirty-third Southeastern International Conference on Combinatorics, Graph Theory and Computing, Boca Raton, Florida, 2002, Congr. Numer. 154 (2002) 21–30. [2] S.C. Locke, Problem 10647, MAA Monthly 105 (1998) 176. [3] S.C. Locke, P. Tracy, H.-J. Voss, Highly cohesive graphs have long non-separating paths: Solution to Problem 10647, MAA Monthly 108 (2001) 470–472. [4] L. Lovász, Combinatorial Problems and Exercises, North Holland, 1979, p. 40, Exercise 6.6(b). [5] W. Mader, Existenz gewisser Konfigurationen in n-gesättigten Graphen und in Graphen genügend großer Kantendichte, Math. Ann. 194 (1971) 295–312. [6] C. Thomassen, B. Toft, Non-separating induced cycles in graphs, J. Combin. Theory Ser. B 31 (1981) 199–224. [7] D.B. West, Introduction to Graph Theory, second ed., Prentice-Hall, 2001.
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Note
Non-separating trees in connected graphs Ajit A. Diwan ∗ , Namrata P. Tholiya Department of Computer Science and Engineering, Indian Institute of Technology Bombay, Powai, Mumbai 400076, India
article
info
Article history: Received 3 October 2008 Received in revised form 3 February 2009 Accepted 21 March 2009 Available online 19 April 2009
abstract Let T be any tree of order d ≥ 1. We prove that every connected graph G with minimum degree d contains a subtree T 0 isomorphic to T such that G − V (T 0 ) is connected. © 2009 Elsevier B.V. All rights reserved.
Keywords: Non-separating tree Isomorphic subtree Minimum degree
1. Introduction A subgraph H of a connected graph G is said to be non-separating if G − V (H ) is a connected non-empty graph. It is well-known that every non-trivial connected graph G contains a vertex v such that G − v is connected. It is also known that every connected graph with minimum degree 2 contains a non-separating edge [4] and that every connected graph with minimum degree 3 contains a non-separating induced cycle [6]. A non-trivial connected graph G is said to be k-cohesive if for any two distinct vertices u and v , d(u) + d(v) + d(u, v) ≥ k, where d(u) is the degree of u and d(u, v) is the distance between u and v . Locke [2] conjectured that for k ≥ 3, every connected 2k-cohesive graph contains a non-separating copy of every tree of order k, and proved it for paths [3]. Abreu and Locke [1] proved that every connected (2k + 2)-cohesive graph contains a non-separating copy of every tree of order k and diameter at most 4. We show that every connected graph of minimum degree d contains a non-separating copy of every tree of order d. The graph mKd−1 ∨ K1 , for m ≥ 3, shows that the degree bound is tight for any tree of order d. Our result may be considered to be a partial step toward Locke’s conjecture. The proof is based on a technique used by Mader to prove a completely different result. Mader [5] showed that every graph with minimum degree d contains an edge uv such that there are d internally vertex-disjoint paths between u and v . We extend this technique slightly to obtain our result. The proof also leads to a polynomial-time algorithm for finding a non-separating copy of any given tree of order d in a connected graph of minimum degree d. The notation used is largely standard and follows, for example, [7]. 2. Main result Theorem 1. Let G be a connected graph with minimum degree d ≥ 1. Then for any tree T of order d, G contains a subtree T 0 isomorphic to T such that G − V (T 0 ) is connected.
∗
Corresponding author. E-mail address: [email protected] (A.A. Diwan).
0012-365X/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2009.03.037
5236
A.A. Diwan, N.P. Tholiya / Discrete Mathematics 309 (2009) 5235–5237
Before proving the theorem, we introduce some definitions. An ordered clique K in a graph G is a complete subgraph of G with an ordering imposed on the vertices of K . Let K be an ordered clique in a graph G with the ordering v1 , v2 , . . . , vk of its vertices. A subtree T of G is said to be consistent with the ordered clique K if every vertex vi ∈ V (K ) ∩ V (T ) has at most one neighbor in T that is not contained in {v1 , v2 , . . . , vi−1 }. We consider ordered pairs of the form (G, K ), where K is an ordered clique in a graph G. We assume throughout that d ≥ 1 is an integer and T is any fixed tree of order d. Let u1 , u2 , . . . , ud be an ordering of the vertices of T such that ui is adjacent to exactly one vertex uj with j > i, for 1 ≤ i < d. We prove a statement stronger than Theorem 1 by induction. Lemma 2. Let (G, K ) be an ordered pair such that K is a non-empty ordered clique in a connected graph G, and K is a proper subgraph of G. Let v1 , v2 , . . . , vk be the ordering of the vertices of K . If every vertex in V (G) \ V (K ) has degree at least d in G, then there exists a subtree T 0 of G satisfying the following properties. (1) (2) (3) (4)
v1 6∈ V (T 0 ).
T 0 is isomorphic to T . T 0 is consistent with K . Every connected component of G − (V (K ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|.
Proof. Suppose there exists a counterexample. Let (G, K ) be a counterexample that minimizes |V (G)| + |V (G) \ V (K )|. If |V (G)\ V (K )| = 1 then let w be the only vertex in V (G)\ V (K ). Since the degree of w is at least d in G, |NG (w)∩ V (K )| ≥ d. Let vi1 , vi2 , . . . , vid be vertices in V (K ) that are adjacent to w , such that i1 < i2 < · · · < id . Now let the vertex uj of T correspond to the vertex vij+1 in G for 1 ≤ j < d, and let the vertex ud of T correspond to the vertex w in G. This gives a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . Since G − (V (K ) ∪ V (T 0 )) is empty, the fourth property in Lemma 2 is trivially satisfied. This contradicts the assumption that (G, K ) is a counterexample. Suppose |V (G) \ V (K )| > 1. We consider two cases. Case 1. There exists a vertex v ∈ V (G) \ V (K ) such that V (K ) ⊆ NG (v). Let K 0 be the ordered clique v1 , v2 , . . . , vk , v in G. Then (G, K 0 ) satisfies the hypothesis of Lemma 2, and by the minimality of the counterexample (G, K ), (G, K 0 ) is not a counterexample. Hence, there exists a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K 0 , and every connected component of G − (V (K 0 ) ∪ V (T 0 )) contains a vertex w with |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )|. We show that T 0 satisfies all four properties in Lemma 2, contradicting the assumption that (G, K ) is a counterexample. Clearly, v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . It remains to show that every connected component of G − (V (K ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|. If v ∈ V (T 0 ) then any component of G − (V (K ) ∪ V (T 0 )) is a component of G − (V (K 0 ) ∪ V (T 0 )). Also, for any vertex w in such a component, |NG (w) ∩ V (K )| ≥ |NG (w) ∩ V (K 0 )| − 1. Since |V (K ) ∩ V (T 0 )| = |V (K 0 ) ∩ V (T 0 )| − 1 and every component of G − (V (K 0 ) ∪ V (T 0 )) contains a vertex w such that |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )|, it follows that for the same vertex w, |NG (w) ∩ V (K )| > |V (K ) ∩ V (T 0 )|. If v 6∈ V (T 0 ) then for the component of G − (V (K ) ∪ V (T 0 )) that contains v , |NG (v) ∩ V (K )| > |V (K ) ∩ V (T 0 )|, since v is adjacent to all vertices in V (K ) and v1 6∈ V (T 0 ). Any other component of G − (V (K ) ∪ V (T 0 )) is also a component of G − (V (K 0 ) ∪ V (T 0 )). For any vertex w in such a component, |NG (w) ∩ V (K )| = |NG (w) ∩ V (K 0 )| since w is not adjacent to v . Hence, any such component contains a vertex w such that |NG (w) ∩ V (K )| = |NG (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 0 )| = |V (K ) ∩ V (T 0 )|. Case 2. No vertex in V (G) \ V (K ) is adjacent to all vertices in V (K ). We must have |V (K )| = k > 1, since G is connected and K is a proper subgraph of G. We construct a new graph G0 and an ordered clique K 0 in G0 as follows. Let X = NG (vk ) \ V (K ) be the set of neighbors of vk that are not contained in V (K ). For each vertex w ∈ X , let π (w) be the largest index i, such that w is not adjacent to vi ∈ V (K ). Note that π (w) < k and is well-defined, since w is adjacent to vk and is not adjacent to at least one vertex in V (K ), by assumption. Let V (G0 ) = V (G −vk ) and E (G0 ) = E (G −vk )∪{wvπ (w) |w ∈ X }. Let K 0 = K − vk with the ordering v1 , v2 , . . . , vk−1 of its vertices. We call the edges {wvπ (w) |w ∈ X } bad edges in G0 . This construction ensures that G0 is connected and the degree of every vertex in V (G0 )\ V (K 0 ) is at least d in G0 . Also, K 0 is a non-empty, proper subgraph of G0 ; hence (G0 , K 0 ) satisfies the hypothesis of Lemma 2. Since |V (G0 ) \ V (K 0 )| = |V (G) \ V (K )| and |V (G0 )| < |V (G)|, by the minimality of the counterexample (G, K ), (G0 , K 0 ) is not a counterexample. Let T 00 be a subtree of G0 such that v1 6∈ V (T 00 ), T 00 is isomorphic to T and consistent with K 0 , and every connected component of G0 − (V (K 0 ) ∪ V (T 00 )) contains a vertex w with |NG0 (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 00 )|. We modify T 00 to find a subtree T 0 of G satisfying the four properties in Lemma 2, thus contradicting the assumption that (G, K ) is a counterexample. If none of the bad edges in G0 is contained in T 00 then let T 0 be the same as T 00 . Suppose wvi is a bad edge contained in T 00 for some w ∈ X and vi ∈ V (K 0 ). Since T 00 is consistent with K 0 and v1 6∈ V (T 00 ), all other neighbors of vi in T 00 are contained in {v2 , . . . , vi−1 }; hence T 00 can contain at most one bad edge incident with vi . Also, from the construction, there is exactly one bad edge incident with a vertex w ∈ X . This implies that the bad edges contained in T 00 are independent.
A.A. Diwan, N.P. Tholiya / Discrete Mathematics 309 (2009) 5235–5237
5237
Let w1 vi1 , w2 vi2 , . . . , wm vim be the bad edges contained in T 00 , such that 1 < i1 < i2 < · · · < im < k. By the definition of π (w), if wvi is a bad edge in G0 , then wvj is an edge in G for all i < j ≤ k. Construct T 0 from T 00 by replacing the vertex vij ∈ V (T 00 ) by the vertex vij+1 for 1 ≤ j ≤ m, where vim+1 is vk . This is possible since vij+1 is adjacent to wj in G and all other neighbors of vij in T 00 are contained in {v2 , v3 , . . . , vij −1 }. Since T 00 is consistent with K 0 , any such neighbor of vij will not have a bad edge incident with it in T 00 , and will be a vertex in T 0 also. Therefore T 0 is a subtree of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and consistent with K . If w is any vertex in V (G) \ V (K ), then |NG (w) ∩ V (K )| = |NG0 (w) ∩ V (K 0 )| since any bad edge incident with w in G0 is replaced by the edge wvk in G. Since the connected components of G − (V (K ) ∪ V (T 0 )) are also the components of G0 − (V (K 0 ) ∪ V (T 00 )), every such component contains a vertex w with |NG (w) ∩ V (K )| = |NG0 (w) ∩ V (K 0 )| > |V (K 0 ) ∩ V (T 00 )| = |V (K ) ∩ V (T 0 )|. Hence, T 0 is a subtree of G satisfying the properties in Lemma 2. Theorem 1 now follows from Lemma 2. Let G be any connected graph with minimum degree d ≥ 1, and let v1 be any vertex in V (G). Let K be the ordered clique in G containing only the vertex v1 . Then (G, K ) satisfies the hypothesis of Lemma 2, and there is a subtree T 0 of G such that v1 6∈ V (T 0 ), T 0 is isomorphic to T and every connected component of G −(V (T 0 )∪{v1 }) contains a vertex that is adjacent to v1 . Thus T 0 is a non-separating subtree of G isomorphic to T . Acknowledgements We thank the referees for their comments and suggestions, which helped in improving the presentation. References [1] M. Abreu, S.C. Locke, Non-separating n-trees up to diameter 4 in a (2n+2)-cohesive graph, in: Proceedings of the Thirty-third Southeastern International Conference on Combinatorics, Graph Theory and Computing, Boca Raton, Florida, 2002, Congr. Numer. 154 (2002) 21–30. [2] S.C. Locke, Problem 10647, MAA Monthly 105 (1998) 176. [3] S.C. Locke, P. Tracy, H.-J. Voss, Highly cohesive graphs have long non-separating paths: Solution to Problem 10647, MAA Monthly 108 (2001) 470–472. [4] L. Lovász, Combinatorial Problems and Exercises, North Holland, 1979, p. 40, Exercise 6.6(b). [5] W. Mader, Existenz gewisser Konfigurationen in n-gesättigten Graphen und in Graphen genügend großer Kantendichte, Math. Ann. 194 (1971) 295–312. [6] C. Thomassen, B. Toft, Non-separating induced cycles in graphs, J. Combin. Theory Ser. B 31 (1981) 199–224. [7] D.B. West, Introduction to Graph Theory, second ed., Prentice-Hall, 2001.