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Forbidden subgraphs and the existence of a spanning tree without small degree stems
Discrete Mathematics 313 (2013) 2206–2212
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Forbidden subgraphs and the existence of a spanning tree without small degree stems Michitaka Furuya, Shoichi Tsuchiya ∗ Department of Mathematical Information Science, Tokyo University of Science, 1-3 Kagurazaka, Shinjuku-ku, Tokyo 162-8601, Japan
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Article history: Received 25 March 2012 Received in revised form 27 February 2013 Accepted 28 May 2013 Available online 19 June 2013 Keywords: [2, k]-spanning tree ([2, k]-ST) Homeomorphically irreducible spanning tree (HIST) HIST-extendability Forbidden subgraph
abstract A spanning tree with no vertices of degree from 2 to k of a graph is called a [2, k]-spanning tree ([2, k]-ST ) of the graph. In this paper, we consider sets of forbidden subgraphs that imply the existence of a [2, k]-ST in a connected graph of sufficiently large order, and give the characterization of such sets. We also focus on connected graphs that contain no path of order 4 as an induced subgraph, and characterize such graphs having a [2, 2]-ST. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper, all graphs are finite, simple and undirected. Let G be a graph. We let V (G) and E (G) denote the vertex set and the edge set of G, respectively. For v ∈V (G), we let d(v) and N (v) denote the degree of v and the neighborhood of v , respectively. For X ⊆ V (G), we let N (X ) = v∈X N (v). For v ∈ V (G) and a non-negative integer i, we let Ni (v) = {u ∈ V (G) | the distance between v and u is i}. We let ∆(G) = max{d(u) | u ∈ V (G)} and δ(G) = min{d(u) | u ∈ V (G)}, and for k ≥ 0, let Vk (G) = {v ∈ V (G) | d(v) = k}, V≥k (G) = {v ∈ V (G) | d(v) ≥ k} and V≤k (G) = {v ∈ V (G) | d(v) ≤ k}. For X ⊆ V (G), we let G[X ] be the subgraph of G which is induced by X . For S ⊆ V (G) and u ∈ S, we let pn(u, S ) = {v | N (v) ∩ S = {u}}. We let Kn denote the complete graph of order n. For terms and symbols not defined here, we refer the reader to [3]. In graph theory, it is a fundamental problem deciding whether a graph has spanning trees of particular types. For example, the Hamiltonian path problem, that is to find a spanning tree with all but two vertices of degree 2, is a famous problem. In this paper, we consider a spanning tree which is a class antithetical to the Hamiltonian path in point of a degree condition. If a spanning tree T of G has no vertex of degree 2, then T is called a homeomorphically irreducible spanning tree (HIST ) of G. Some properties of the HIST were explored in, for example, [1,4,10]. For a set H of connected graphs, a graph G is said to be H -free if G contains no member of H as an induced subgraph. We also say that the members of H are forbidden subgraphs. If G is {H }-free, then G is simply said to be H-free. For two sets H1 and H2 of forbidden subgraphs, we write H1 ≤ H2 if for every H2 ∈ H2 , there exists H1 ∈ H1 such that H1 is an induced subgraph of H2 . Note that if H1 ≤ H2 , then an H1 -free graph is H2 -free (see [6]). Forbidden subgraphs have appeared in many areas of graph theory (see, for example, [2,7,8]). Our first aim is to give a characterization of the sets H such that every connected H -free graph of sufficiently large order has a HIST. We now extend the definition of the HIST. For k ≥ 2, if a spanning tree T of G has no vertex x with
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Corresponding author. E-mail addresses: [email protected] (M. Furuya), [email protected] (S. Tsuchiya).
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.05.020
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Fig. 1. The graphs Yn , Zn and Fn .
Fig. 2. The 1-subdivided K4 .
2 ≤ dT (x) ≤ k, then T is called a [2, k]-spanning tree ([2, k]-ST ) of G. Note that a [2, 2]-ST is exactly a HIST. We define five graphs as follows (see Fig. 1): Let Pn denote a path of order n. Let Km,n denote a complete bipartite graph with partite sets of size m and n. Let Yn be a graph obtained from K1,n+1 by subdividing one edge, that is to say V (Yn ) = {x1 , . . . , xn , y1 , y2 , y3 } and E (Yn ) = {xi y1 | 1 ≤ i ≤ n} ∪ {y1 y2 , y2 y3 }. Let Zn be a graph obtained from Kn ∪ P2 by adding an edge that joins a vertex of Kn to an endvertex of P2 , that is to say V (Zn ) = {x1 , . . . , xn , y1 , y2 } and E (Zn ) = {xi xj | i ̸= j} ∪ {x1 y1 , y1 y2 }. Let Fn be the graph obtained from Kn by adding a pendant edge to each vertex, that is to say V (Fn ) = {x1 , . . . , xn , y1 , . . . , yn } and E (Fn ) = {xi xj | i ̸= j} ∪ {xi yi | 1 ≤ i ≤ n}. Our main result is as follows: Theorem 1.1. Let k ≥ 2 be an integer, and let H be a set of connected graphs. Then there exists an integer n0 = n0 (k, H ) such that every connected H -free graph of order at least n0 has a [2, k]-ST if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Surprisingly, every set of forbidden subgraphs that essentially implies the existence of a [2, k]-ST does not depend on k. Therefore we get the following result as a corollary. Corollary 1.2. Let H be a set of connected graphs. Then there exists an integer n0 = n0 (H ) such that every connected H -free graph of order at least n0 has a HIST if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. When we study the relationship between a certain graph property and forbidden subgraphs, we often restrict the number of forbidden subgraphs (see, for example, [5]). Next we restrict the cardinality of the set H in Corollary 1.2. Let H be a set of connected graphs with |H | ≤ 2, and suppose that every connected H -free graph of sufficiently large order has a HIST. Then by Corollary 1.2, we can check that H ≤ {P4 }. Therefore, one may expect that many connected P4 -free graphs have a HIST. Further, we introduce graphs having a HIST such that a specified vertex is a stem of the HIST. A vertex of a tree is called a stem if it has degree at least 2 in the tree. Note that every stem of [2, k]-ST has degree at least k + 1. A graph is HIST-extendable if V≥3 (G) ̸= ∅ and, for each v ∈ V≥3 (G), G has a HIST T such that v is a stem of T . We characterize the connected P4 -free HIST-extendable graphs (without the condition ‘‘sufficiently large order’’). Theorem 1.3. Let G be a connected P4 -free graph of order n ≥ 4. Then G is HIST-extendable if and only if (i) n = 4 and ∆(G) = 3, (ii) n = 5, ∆(G) = 4 and G has no vertex of degree 3, or (iii) n ≥ 6 and G is not isomorphic to K2,n−2 . A graph obtained from K4 by subdividing one edge is called a 1-subdivided K4 (see Fig. 2). By inspection, we can see that a connected P4 -free graph G of order n ∈ {4, 5} has a HIST if and only if G is neither a K2,n−2 nor a 1-subdivided K4 . Furthermore, for n ≥ 4, K2,n−2 has no HIST (see Section 2). Therefore we get a relationship between P4 -free graphs and the existence of a HIST by Theorem 1.3. Corollary 1.4. Let G be a connected P4 -free graph of order n ≥ 4. Then G has a HIST if and only if G is neither a K2,n−2 nor a 1-subdivided K4 . We prove Theorem 1.1 in Section 2 and Theorem 1.3 in Section 3. In Section 4, we give a result concerning the HISTextendability similar to Theorem 1.1.
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2. Characterizing forbidden families for the [2, k ]-ST In this section, we prove Theorem 1.1. First we consider the ‘‘only if’’ part. We start with a useful lemma. Lemma 2.1. Let G be a graph. (i) If G has a cut set S with S ⊆ V2 (G), then G has no HIST. (ii) The graph obtained from G by adding a pendant edge to each vertex has no HIST. Proof. (i) Let A and A′ be two components of G − S. Suppose that G has a HIST T . Since T is connected, there exists a path P on T which joins A and A′ . Then P contains a vertex x ∈ V2 (G) as an internal vertex. On the other hand, every vertex in V2 (G) is a leaf of T , which is a contradiction. Thus G′ has no HIST. (ii) Let G′ be the graph obtained from G by adding a pendant edge to each vertex. Suppose that G′ has a HIST T . Since every vertex in V (G) is adjacent to a vertex in V1 (G′ ), the set V (G) is equal to the set of stems of T . In particular, δ(T [V (G)]) ≥ 2, which contradicts the fact that T has no cycle. Thus G′ has no HIST. Let H1 = {Pl | l ≥ 3}, H2 = {K2,m | m ≥ 2}, H3 = {Yn | n ≥ 1}, H4 = {Zp | p ≥ 2} and H5 = {Fq | q ≥ 2}. By Lemma 2.1(i), for each 1 ≤ i ≤ 4, every graph in Hi has no HIST. By Lemma 2.1(ii), every graph in H5 has no HIST. Furthermore, for each 1 ≤ i ≤ 5, Hi is an infinite set. Therefore every set H of connected graphs such that every connected H -free graph of sufficiently large order has a [2, k]-ST satisfies that H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. This completes the proof of the ‘‘only if’’ part of Theorem 1.1. We next show the ‘‘if’’ part. A clique of a graph is a set of pairwise adjacent vertices, and an independent set is a set of pairwise nonadjacent vertices. For two positive integers s and t, the Ramsey number R(s, t ) is the minimum positive integer R such that any graph of order at least R contains either a clique of cardinality s or an independent set of cardinality t. The Ramsey number R(s, t ) exists for all positive integers s and t (see, for example, [3]). For positive integers k, l, m, n, p and q, let
α(k, n, p, q) = 3(k − 1)R(q, (p + q − 1)n + 1) + 2p + q and
β(k, m, n, q) = (k − 1)R(q, n + 1) + 2n + m. By the definition of the relation ‘‘≤’’, it suffices to show the following proposition. Proposition 2.2. Let k ≥ 2, l ≥ 4, m ≥ 2, n ≥ 1, p ≥ 2 and q ≥ 2 be positive integers. Then every connected {Pl , K2,m , Yn , Zp , Fq }-free graph of order at least (R(α, β) − 1)l−2 + 1 has a [2, k]-ST, where α = α(k, n, p, q) and β = β(k, m, n, q). Proof. Let G be a connected {Pl , K2,m , Yn , Zp , Fq }-free graph with |V (G)| ≥ (R(α, β) − 1)l−2 + 1. Since G is Pl -free, the diameter of G is at most l − 2. Hence ∆(G) ≥ R(α, β). Let x be a vertex with d(x) = ∆(G). Then G[N (x)] contains either a clique of cardinality α or an independent set of cardinality β . Case 1: G[N (x)] contains a clique of cardinality α . Let U1 be a clique of cardinality α in G[N (x)]. For each u ∈ N2 (x), we define vu ∈ N (u) as follows: if u ̸∈ N (U1 ), let vu ∈ N (u) ∩ N (x); if u ∈ N (U1 ), let vu ∈ N (u) ∩ U1 . We choose vu so that |{vu | u ∈ N2 (x)}|(=|{vu | u ∈ N2 (x) − N (U1 )}| + |{vu | u ∈ N2 (x) ∩ N (U1 )}|) is as small as possible. Let U2 = {vu | u ∈ N2 (x) − N (U1 )} and U3 = {vu | u ∈ N2 (x) ∩ N (U1 )} (see Fig. 3). By the minimality of |U2 | + |U3 |, pn(v, U2 ) ∩ (N2 (x) − N (U1 )) ̸= ∅ for each v ∈ U2 , and pn(v, U3 ) ∩ (N2 (x) ∩ N (U1 )) ̸= ∅ for each v ∈ U3 . For each v ∈ U2 , fix a vertex uv ∈ pn(v, U2 ) ∩ (N2 (x) − N (U1 )). For each v ∈ U3 , fix a vertex uv ∈ pn(v, U3 ) ∩ (N2 (x) ∩ N (U1 )). For v ∈ U2 , if |N (v) ∩ U1 | ≤ α − p + 1, then G[(U1 − N (v)) ∪ {x, v, uv }] contains a Zp as an induced subgraph, a contradiction. Thus
|N (v) ∩ U1 | ≥ α − p + 2 for each v ∈ U2 .
(2.1)
Claim 2.1. (i) For u ∈ N2 (x) − N (U1 ), if u′ ∈ V (G) − (N (x) ∪ N (U1 )) is adjacent to u, then vu u′ ∈ E (G). (ii) No vertex in N2 (x) − N (U1 ) is adjacent to a vertex in N3 (x). (iii) For v, v ′ ∈ U2 , uv uv ′ ̸∈ E (G). Proof. Let u ∈ N2 (x) − N (U1 ) and u′ ∈ V (G) − (N (x) ∪ N (U1 )) with uu′ ∈ E (G). By (2.1) and the definition of α, |N (vu ) ∩ U1 | ≥ α − p + 2 ≥ p − 1. If vu u′ ̸∈ E (G), then G[(N (vu ) ∩ U1 ) ∪ {vu , u, u′ }] contains a Zp as an induced subgraph, a contradiction. Consequently (i) holds. By (i), (ii) clearly holds. For v ∈ U2 , since uv ∈ pn(v, U2 ), no vertex in U2 − {v} is adjacent to uv . This together with (i) leads to (iii). Claim 2.2. |U2 | < R(q, n + 1).
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Fig. 3. U1 , U2 , U3 and U4 .
Proof. Suppose that |U2 | ≥ R(q, n + 1). Then G[U2 ] contains either a clique of cardinality q or an independent set of cardinality n + 1. Suppose that G[U2 ] contains a clique X of cardinality q. Then by Claim 2.1(iii), X ∪ {uv | v ∈ X } induces an Fq in G, a contradiction. Thus G[U2 ] contains an independent set X ′ of cardinality n + 1. Let v ∈ X ′ . Then X ′ ∪ {x, uv } induces a Yn in G, a contradiction. Claim 2.3. |U3 | < p + q. Proof. Suppose that |U3 | ≥ p + q. Since G is Fq -free, G[U3 ∪ {uv | v ∈ U3 }] contains no Fq as an induced subgraph. Then there exists v, v ′ ∈ U3 such that uv uv ′ ∈ E (G). Since |U3 | ≥ p + 1, G[(U3 − {v ′ }) ∪ {uv , uv ′ }] contains a Zp as an induced subgraph, a contradiction. For each u ∈ N3 (x), let wu ∈ N (u) ∩ N2 (x). We choose wu so that |{wu | u ∈ N3 (x)}| is as small as possible. Let U4 = {wu | u ∈ N3 (x)}. By Claim 2.1(ii), U4 ⊆ N2 (x) ∩ N (U1 ). By the minimality of |U4 |, pn(w, U4 ) ∩ N3 (x) ̸= ∅ for each w ∈ U4 . For each w ∈ U4 , fix a vertex uw ∈ pn(w, U4 ) ∩ N3 (x). If there exists w ∈ U4 such that |N (w) ∩ U1 | ≤ α − p + 1, then G[(U1 − N (w)) ∪ {vw , w, uw }] contains a Zp as an induced subgraph, a contradiction. Thus
|N (w) ∩ U1 | ≥ α − p + 2 for each w ∈ U4 .
(2.2)
Claim 2.4. (i) For each u ∈ N3 (x), if u′ ∈ V (G) − N2 (x) is adjacent to u, then wu u′ ∈ E (G). (ii) N4 (x) = ∅. (iii) For w, w ′ ∈ U4 , uw uw′ ̸∈ E (G). Proof. Let u ∈ N3 (x) and u′ ∈ V (G)−N2 (x) with uu′ ∈ E (G). By (2.2) and the definition of α, |N (wu )∩U1 | ≥ α−p+2 ≥ p−1. If wu u′ ̸∈ E (G), then G[(N (wu ) ∩ U1 ) ∪ {wu , u, u′ }] contains a Zp as an induced subgraph, a contradiction. Consequently (i) holds. By (i), (ii) clearly holds. For w ∈ U4 , since uw ∈ pn(w, U4 ), no vertex in U4 − {w} is adjacent to uw . This together with (i) leads to (iii). Claim 2.5. |U4 | < R(q, (p + q − 1)n + 1). Proof. Suppose that |U4 | ≥ R(q, (p + q − 1)n + 1). Then G[U4 ] contains either a clique of cardinality q or an independent set of cardinality (p + q − 1)n + 1. Suppose that G[U4 ] contains a clique X of cardinality q. Then by Claim 2.4(iii), X ∪{uw | w ∈ X } induces an Fq in G, a contradiction. Thus G[U4 ] contains an independent set X ′ of cardinality (p + q − 1)n + 1. By Claim 2.3, there exists a vertex v ∈ U3 which is adjacent to at least n + 1 vertices in X ′ . Let w ∈ N (v) ∩ X ′ . Then G[(N (v) ∩ X ′ ) ∪ {v, uw }] contains a Yn as an induced subgraph, a contradiction. Now we construct a [2, k]-ST T of G such that {x} ∪ ( 2≤i≤4 Ui ) is the set of stems of T . By (2.1), Claims 2.2, 2.3 and the definition of α , for each v ∈ U2 , |N (v) ∩ (U1 − U3 )| ≥ (α − p + 2) − (p + q − 1) = 3(k − 1)R(q, (p + q − 1)n + 1) + 3 ≥ (k − 1)(R(q, n + 1) − 1) ≥ (k − 1)|U2 |. Hence there exist (k − 1)|U2 | vertices av,i ∈ U1 − U3 (v ∈ U2 , 1 ≤ i ≤ k − 1) such that {av,i | 1 ≤ i ≤ k − 1} ⊆ N (v) for each v ∈ U2 . Let L1 = {av,i | v ∈ U2 , 1 ≤ i ≤ k − 1}. By (2.2), Claims 2.3, 2.5 and the definition of α , for each w ∈ U4 , |N (w) ∩ (U1 − (U3 ∪ L1 ))| ≥ (α − p + 2) − (p + q − 1) − (k − 1)(R(q, n + 1) − 1) = 3(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) + 3 ≥ (k − 1)(R(q, (p + q − 1)n + 1) − 1) ≥ (k − 1)|U4 |. Hence there exist (k − 1)|U4 | vertices bw,i ∈ U1 − (U3 ∪ L1 ) (w ∈ U4 , 1 ≤ i ≤ k − 1) such that {bw,i | 1 ≤ i ≤ k − 1} ⊆ N (w) for each w ∈ U4 . Let L2 = {bw,i | w ∈ U4 , 1 ≤ i ≤ k − 1}. By Claim 2.3 and the definition of
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Fig. 4. A desired [2, k]-ST of G.
α, |U1 − (U3 ∪ L1 ∪ L2 )| ≥ α − (p + q − 1) − (k − 1)(R(q, n + 1) − 1) − (k − 1)(R(q, (p + q − 1)n + 1) − 1) = 2(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) + p + 1 ≥ (k − 1)(p + q − 1) ≥ (k − 1)|U3 |. Hence there exist (k − 1)|U3 | vertices cv,i ∈ U1 −(U3 ∪ L1 ∪ L2 ) (v ∈ U3 , 1 ≤ i ≤ k − 1). Let L3 = {cv,i | v ∈ U3 , 1 ≤ i ≤ k − 1}. By the definition of α, |U1 − (L1 ∪ L2 ∪ L3 )| ≥ α − (k − 1)(R(q, n + 1) − 1) − (k − 1)(R(q, (p + q − 1)n + 1) − 1) − (k − 1)(p + q − 1) = 2(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) − (k − 1)(p + q − 2) + 2p + q ≥ k + 1. For each v ∈ U2 , let Ev = {v u | u ∈ N2 (x) − N (U1 ), vu = v} ∪ {v av,i | 1 ≤ i ≤ k − 1}. For each v ∈ U3 , let Ev = {v u | u ∈ N2 (x) ∩ N (U1 ), vu = v} ∪ {v cv,i | 1 ≤ i ≤ k − 1}. For each w ∈ U4 , let Ew = {w u | u ∈ N3 (x), wu = w} ∪ {wbw,i | 1 ≤ i ≤ k − 1}. Then the spanning tree T of G with E (T ) = {xx | x ∈ N (x) − (L1 ∪ L2 ∪ L3 )} ∪ ′
′
Ey
y∈U2 ∪U3 ∪U4
is a desired [2, k]-ST of G (see Fig. 4). Case 2: G[N (x)] contains an independent set of cardinality β . Let U1 be an independent set of cardinality β in G[N (x)]. Claim 2.6. N2 (x) ∩ N (U1 ) = ∅. Proof. Suppose that N2 (x) ∩ N (U1 ) ̸= ∅, and let y ∈ U1 and y′ ∈ N2 (x) such that yy′ ∈ E (G). If |N (y′ ) ∩ U1 | ≤ β − n, then G[(U1 − N (y′ )) ∪ {x, y, y′ }] contains a Yn as an induced subgraph, a contradiction. Thus |N (y′ ) ∩ U1 | ≥ β − n + 1. Then by the definition of β, |N (y′ ) ∩ U1 | ≥ m. Then G[(N (y′ ) ∩ U1 ) ∪ {x, y′ }] contains a K2,m as an induced subgraph, a contradiction. For each u ∈ N2 (x), let vu ∈ N (u) ∩ N (x). We choose vu such that |{vu | u ∈ N2 (x)}| is as small as possible. Let U2 = {vu | u ∈ N2 (x)}. By Claim 2.6, U2 ⊆ N (x) − U1 . By the minimality of |U2 |, pn(v, U2 ) ∩ N2 (x) ̸= ∅ for each v ∈ U2 . For each v ∈ U2 , fix a vertex uv ∈ pn(v, U2 ) ∩ N2 (x). If there exists v ∈ U2 such that |N (v) ∩ U1 | ≤ β − n, then G[(U1 − N (v)) ∪ {x, v, uv }] contains a Yn as an induced subgraph, a contradiction. Thus
|N (v) ∩ U1 | ≥ β − n + 1 for each v ∈ U2 .
(2.3)
Claim 2.7. (i) For each u ∈ N2 (x), if u′ ∈ V (G) − N (x) is adjacent to u, then vu u′ ∈ E (G). (ii) N3 (x) = ∅. (iii) For v, v ′ ∈ U2 , uv uv ′ ̸∈ E (G). Proof. Let u ∈ N2 (x) and u′ ∈ V (G) − N (x) with uu′ ∈ E (G). By Claim 2.6, u is adjacent to no vertex in U1 . By (2.3) and the definition of β, |N (vu ) ∩ U1 | ≥ β − n + 1 ≥ n. If vu u′ ̸∈ E (G), then G[(N (vu ) ∩ U1 ) ∪ {vu , u, u′ }] contains a Yn as an induced subgraph, a contradiction. Thus (i) holds. By (i), (ii) clearly holds. For v ∈ U2 , since uv ∈ pn(v, U2 ), no vertex in U2 − {v} is adjacent to uv . This together with (i) leads to (iii). Claim 2.8. |U2 | < R(q, n + 1). Proof. Suppose that |U2 | ≥ R(q, n + 1). Then G[U2 ] contains either a clique of cardinality q or an independent set of cardinality n + 1. Suppose that G[U2 ] contains a clique X of cardinality q. Then by Claim 2.7(iii), X ∪ {uv | v ∈ X } induces an
M. Furuya, S. Tsuchiya / Discrete Mathematics 313 (2013) 2206–2212
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Fig. 5. A desired [2, k]-ST of G.
Fq in G, a contradiction. Thus G[U2 ] contains an independent set X ′ of cardinality n + 1. Let v ∈ X ′ . Then X ′ ∪ {x, uv } induces a Yn in G, a contradiction. Now we construct a [2, k]-ST T of G such that {x} ∪ U2 is the set of stems of T . By (2.3), Claim 2.8 and the definition of β , for each v ∈ U2 , |N (v) ∩ U1 | ≥ β − n + 1 = (k − 1)R(q, n + 1) + n + m + 1 ≥ (k − 1)(R(q, n + 1) − 1) ≥ (k − 1)|U2 |. Hence there exist (k − 1)|U2 | vertices av,i ∈ U1 (v ∈ U2 , 1 ≤ i ≤ k − 1) such that {av,i | 1 ≤ i ≤ k − 1} ⊆ N (v) for each v ∈ U2 . Let L1 = {av,i | v ∈ U2 , 1 ≤ i ≤ k − 1}. By the definition of β, |U1 − L1 | ≥ β − (k − 1)(R(q, n + 1) − 1) = k − 1 + 2n + m ≥ k + 1. For each v ∈ U2 , let Ev = {v u | u ∈ N2 (x), vu = v} ∪ {v av,i | 1 ≤ i ≤ k − 1}. Then the spanning tree T of G with
E (T ) = {xx′ | x′ ∈ N (x) − L1 } ∪
Ev
v∈U2
is a desired [2, k]-ST of G (see Fig. 5). This completes the proof of Proposition 2.2.
Remark. By the argument in the proof of Proposition 2.2, we also see that the radius of a connected {Pl , K2,m , Yn , Zp , Fq }-free graph of sufficiently large order is at most 3. 3. A HIST of a P4 -free graph In this section, we prove Theorem 1.3. In [9], a theorem which gives a characterization of P4 -free graphs was proved. By the theorem, we obtain the following lemma which is well known when we deal with P4 -free graphs. Lemma 3.1. Let G be a connected P4 -free graph, and let S be a minimum cut set of G. Then every vertex in S is adjacent to every vertex in V (G) − S. Proof of Theorem 1.3. We first consider a connected graph G of order n ∈ {4, 5}. Let v ∈ V≥3 (G). If there exists a HIST T of G such that v is a stem of T , then v is the unique stem of T , and hence d(v) = n − 1. On the other hand, if d(v) = n − 1, then there clearly exists a HIST of G such that v is a stem of T . This implies that the case n ∈ {4, 5} of Theorem 1.3 holds. In the argument below, we consider only the graphs of order at least 6. Furthermore, in Section 2, we have already checked that K2,n−2 has no HIST (and so K2,n−2 is not HIST-extendable). Thus it suffices to show the ‘‘only if’’ part of Theorem 1.3 under the condition n ≥ 6. Let G be a connected P4 -free graph of order n ≥ 6 with G ̸≃ K2,n−2 . By Lemma 3.1, there exists a partition of {X1 , X2 } of V (G) such that every vertex in X1 is adjacent to every vertex in X2 . Claim 3.1. Suppose that X1 ∩ V≥3 (G) ̸= ∅ and X2 ∩ V≥3 (G) ̸= ∅. Then, for every x1 ∈ X1 ∩ V≥3 (G) and every x2 ∈ X2 ∩ V≥3 (G), there exists a HIST T of G such that x1 and x2 are the stems of T . Proof. Without loss of generality, we may assume that |X1 | ≥ |X2 |. By the definition of {X1 , X2 }, X2 ⊆ N (x1 ) and X1 ⊆ N (x2 ). Since dG (x1 ) ≥ 3, there exists a subset Y1 of N (x1 ) such that |Y1 | ≥ 3 and X2 ⊆ Y1 . We choose Y1 so that |Y1 | is as small as possible. By the minimality of Y1 , |X1 ∩ Y1 | = max{3 − |X2 |, 0}. If |X1 ∩ Y1 | = 3 − |X2 |, then |X1 − Y1 | = |X1 | − |X1 ∩ Y1 | = |X1 | − (3 − |X2 |) = |V (G)| − 3 ≥ 3; if |X1 ∩ Y1 | = 0, then |X1 − Y1 | = |X1 | − |X1 ∩ Y1 | = |X1 | ≥ |V (G)|/2 ≥ 3. Hence in either case, we have |X1 − Y1 | ≥ 3. Then the spanning tree T of G with E (T ) = {x1 u | u ∈ Y1 } ∪ {x2 u | u ∈ X1 − Y1 } is a desired HIST of G.
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Let v ∈ V≥3 (G). It suffices to show that there exists a HIST of G such that v is a stem of T . We may assume that v ∈ X1 . If |X1 | = 1, then d(v) = n − 1, and hence there exists a desired HIST of G. Thus we may assume that |X1 | ≥ 2. If X2 ∩ V≥3 (G) ̸= ∅, then by Claim 3.1, there exists a desired HIST of G. Thus we may assume that X2 ∩ V≥3 (G) = ∅. By the definition of {X1 , X2 }, this implies that |X1 | = 2 and X2 is an independent set of G. Write X1 = {v, v ′ }. Since G ̸≃ K2,n−2 , vv ′ ∈ E (G), and hence d(v) = n − 1. Then G has a spanning star with center v , as desired. This completes the proof of Theorem 1.3.
4. The [2, k ]-ST-extendability In this section, we extend the HIST-extendability to [2, k]-ST, and give a result concerning it similar to Theorem 1.1. For k ≥ 2, a graph G is [2, k]-ST-extendable if V≥k+1 (G) ̸= ∅ and, for every v ∈ V≥k+1 (G), G has a [2, k]-ST T such that v is a stem of T . We show the following proposition. Proposition 4.1. Let k ≥ 2 be an integer. If a graph G has a [2, 2k]-ST, then G is [2, k]-ST-extendable. Proof. Let v ∈ V≥k+1 (G). It suffices to show that there exists a [2, k]-ST T of G such that v is a stem of T . Let T ′ be a [2, 2k]-ST of G. Since T ′ is a [2, k]-ST of G, if v is a stem of T ′ , then T ′ is a desired HIST. Thus we may assume that dT ′ (v) = 1. Let U be a subset of NG (v) − NT ′ (v) with |U | = k. For each u ∈ U, let Pu be the unique v − u path in T ′ , and let eu be the unique edge of Pu which is incident with u. Then every component of F = T ′ − {eu | u ∈ U } is a tree with no vertices of degree from 2 to k. Furthermore, dF (u) = 0 if and only if u ∈ U ∩ V≤1 (F ). Therefore F ∪ {v u | u ∈ U } is a desired [2, k]-ST. This together with Theorem 1.1 implies the following results. Corollary 4.2. Let k ≥ 2 be an positive integer, and let H be a set of connected graphs. Then there exists an integer n0 = n0 (k, H ) such that every connected H -free graph of order at least n0 is [2, k]-ST-extendable if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Corollary 4.3. Let H be a set of connected graphs. Then there exists an integer n0 = n0 (H ) such that every connected H -free graph of order at least n0 is HIST-extendable if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Acknowledgments The authors are grateful to referees for their helpful comments. References [1] M.O. Albertson, D.M. Berman, J.P. Hutchinson, C. Thomassen, Graphs with homeomorphically irreducible spanning trees, J. Graph Theory 14 (1990) 247–258. [2] M.D. Barrus, M. Kumbhat, S.G. Hartke, Graph classes characterized both by forbidden subgraphs and degree sequences, J. Graph Theory 57 (2008) 131–148. [3] G. Chartrand, L. Lesniak, P. Zhang, Graphs & Digraphs, fifth ed., Chapman and Hall/CRC, Boca Raton, FL, 2010. [4] G. Chen, H. Ren, S. Shan, Homeomorphically irreducible spanning trees in locally connected graph, Combin. Probab. Comput. 21 (2012) 107–111. [5] R.J. Faudree, R.J. Gould, Characterization forbidden pairs for Hamiltonian properties, Discrete Math. 173 (1997) 45–60. [6] S. Fujita, K. Kawarabayashi, C.L. Lucchesi, K. Ota, M. Plummer, A. Saito, A pair of forbidden subgraphs and perfect matchings, J. Combin. Theory Ser. B 96 (2006) 315–324. [7] R.J. Gould, J.M. Harris, Forbidden triples and traceability: a characterization, Discrete Math. 203 (1999) 101–120. [8] K. Ota, G. Sueiro, Forbidden induced subgraphs for perfect matchings, Graphs Combin. 29 (2013) 289–299. [9] D. Seinsche, On a property of the class of n-colorable graphs, J. Combin. Theory Ser. B 16 (1974) 191–193. [10] S. Tsuchiya, Rooted HIST property on planar triangulations, Ars Combin. (2013) in press.
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Forbidden subgraphs and the existence of a spanning tree without small degree stems Michitaka Furuya, Shoichi Tsuchiya ∗ Department of Mathematical Information Science, Tokyo University of Science, 1-3 Kagurazaka, Shinjuku-ku, Tokyo 162-8601, Japan
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Article history: Received 25 March 2012 Received in revised form 27 February 2013 Accepted 28 May 2013 Available online 19 June 2013 Keywords: [2, k]-spanning tree ([2, k]-ST) Homeomorphically irreducible spanning tree (HIST) HIST-extendability Forbidden subgraph
abstract A spanning tree with no vertices of degree from 2 to k of a graph is called a [2, k]-spanning tree ([2, k]-ST ) of the graph. In this paper, we consider sets of forbidden subgraphs that imply the existence of a [2, k]-ST in a connected graph of sufficiently large order, and give the characterization of such sets. We also focus on connected graphs that contain no path of order 4 as an induced subgraph, and characterize such graphs having a [2, 2]-ST. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper, all graphs are finite, simple and undirected. Let G be a graph. We let V (G) and E (G) denote the vertex set and the edge set of G, respectively. For v ∈V (G), we let d(v) and N (v) denote the degree of v and the neighborhood of v , respectively. For X ⊆ V (G), we let N (X ) = v∈X N (v). For v ∈ V (G) and a non-negative integer i, we let Ni (v) = {u ∈ V (G) | the distance between v and u is i}. We let ∆(G) = max{d(u) | u ∈ V (G)} and δ(G) = min{d(u) | u ∈ V (G)}, and for k ≥ 0, let Vk (G) = {v ∈ V (G) | d(v) = k}, V≥k (G) = {v ∈ V (G) | d(v) ≥ k} and V≤k (G) = {v ∈ V (G) | d(v) ≤ k}. For X ⊆ V (G), we let G[X ] be the subgraph of G which is induced by X . For S ⊆ V (G) and u ∈ S, we let pn(u, S ) = {v | N (v) ∩ S = {u}}. We let Kn denote the complete graph of order n. For terms and symbols not defined here, we refer the reader to [3]. In graph theory, it is a fundamental problem deciding whether a graph has spanning trees of particular types. For example, the Hamiltonian path problem, that is to find a spanning tree with all but two vertices of degree 2, is a famous problem. In this paper, we consider a spanning tree which is a class antithetical to the Hamiltonian path in point of a degree condition. If a spanning tree T of G has no vertex of degree 2, then T is called a homeomorphically irreducible spanning tree (HIST ) of G. Some properties of the HIST were explored in, for example, [1,4,10]. For a set H of connected graphs, a graph G is said to be H -free if G contains no member of H as an induced subgraph. We also say that the members of H are forbidden subgraphs. If G is {H }-free, then G is simply said to be H-free. For two sets H1 and H2 of forbidden subgraphs, we write H1 ≤ H2 if for every H2 ∈ H2 , there exists H1 ∈ H1 such that H1 is an induced subgraph of H2 . Note that if H1 ≤ H2 , then an H1 -free graph is H2 -free (see [6]). Forbidden subgraphs have appeared in many areas of graph theory (see, for example, [2,7,8]). Our first aim is to give a characterization of the sets H such that every connected H -free graph of sufficiently large order has a HIST. We now extend the definition of the HIST. For k ≥ 2, if a spanning tree T of G has no vertex x with
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Corresponding author. E-mail addresses: [email protected] (M. Furuya), [email protected] (S. Tsuchiya).
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.05.020
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Fig. 1. The graphs Yn , Zn and Fn .
Fig. 2. The 1-subdivided K4 .
2 ≤ dT (x) ≤ k, then T is called a [2, k]-spanning tree ([2, k]-ST ) of G. Note that a [2, 2]-ST is exactly a HIST. We define five graphs as follows (see Fig. 1): Let Pn denote a path of order n. Let Km,n denote a complete bipartite graph with partite sets of size m and n. Let Yn be a graph obtained from K1,n+1 by subdividing one edge, that is to say V (Yn ) = {x1 , . . . , xn , y1 , y2 , y3 } and E (Yn ) = {xi y1 | 1 ≤ i ≤ n} ∪ {y1 y2 , y2 y3 }. Let Zn be a graph obtained from Kn ∪ P2 by adding an edge that joins a vertex of Kn to an endvertex of P2 , that is to say V (Zn ) = {x1 , . . . , xn , y1 , y2 } and E (Zn ) = {xi xj | i ̸= j} ∪ {x1 y1 , y1 y2 }. Let Fn be the graph obtained from Kn by adding a pendant edge to each vertex, that is to say V (Fn ) = {x1 , . . . , xn , y1 , . . . , yn } and E (Fn ) = {xi xj | i ̸= j} ∪ {xi yi | 1 ≤ i ≤ n}. Our main result is as follows: Theorem 1.1. Let k ≥ 2 be an integer, and let H be a set of connected graphs. Then there exists an integer n0 = n0 (k, H ) such that every connected H -free graph of order at least n0 has a [2, k]-ST if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Surprisingly, every set of forbidden subgraphs that essentially implies the existence of a [2, k]-ST does not depend on k. Therefore we get the following result as a corollary. Corollary 1.2. Let H be a set of connected graphs. Then there exists an integer n0 = n0 (H ) such that every connected H -free graph of order at least n0 has a HIST if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. When we study the relationship between a certain graph property and forbidden subgraphs, we often restrict the number of forbidden subgraphs (see, for example, [5]). Next we restrict the cardinality of the set H in Corollary 1.2. Let H be a set of connected graphs with |H | ≤ 2, and suppose that every connected H -free graph of sufficiently large order has a HIST. Then by Corollary 1.2, we can check that H ≤ {P4 }. Therefore, one may expect that many connected P4 -free graphs have a HIST. Further, we introduce graphs having a HIST such that a specified vertex is a stem of the HIST. A vertex of a tree is called a stem if it has degree at least 2 in the tree. Note that every stem of [2, k]-ST has degree at least k + 1. A graph is HIST-extendable if V≥3 (G) ̸= ∅ and, for each v ∈ V≥3 (G), G has a HIST T such that v is a stem of T . We characterize the connected P4 -free HIST-extendable graphs (without the condition ‘‘sufficiently large order’’). Theorem 1.3. Let G be a connected P4 -free graph of order n ≥ 4. Then G is HIST-extendable if and only if (i) n = 4 and ∆(G) = 3, (ii) n = 5, ∆(G) = 4 and G has no vertex of degree 3, or (iii) n ≥ 6 and G is not isomorphic to K2,n−2 . A graph obtained from K4 by subdividing one edge is called a 1-subdivided K4 (see Fig. 2). By inspection, we can see that a connected P4 -free graph G of order n ∈ {4, 5} has a HIST if and only if G is neither a K2,n−2 nor a 1-subdivided K4 . Furthermore, for n ≥ 4, K2,n−2 has no HIST (see Section 2). Therefore we get a relationship between P4 -free graphs and the existence of a HIST by Theorem 1.3. Corollary 1.4. Let G be a connected P4 -free graph of order n ≥ 4. Then G has a HIST if and only if G is neither a K2,n−2 nor a 1-subdivided K4 . We prove Theorem 1.1 in Section 2 and Theorem 1.3 in Section 3. In Section 4, we give a result concerning the HISTextendability similar to Theorem 1.1.
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2. Characterizing forbidden families for the [2, k ]-ST In this section, we prove Theorem 1.1. First we consider the ‘‘only if’’ part. We start with a useful lemma. Lemma 2.1. Let G be a graph. (i) If G has a cut set S with S ⊆ V2 (G), then G has no HIST. (ii) The graph obtained from G by adding a pendant edge to each vertex has no HIST. Proof. (i) Let A and A′ be two components of G − S. Suppose that G has a HIST T . Since T is connected, there exists a path P on T which joins A and A′ . Then P contains a vertex x ∈ V2 (G) as an internal vertex. On the other hand, every vertex in V2 (G) is a leaf of T , which is a contradiction. Thus G′ has no HIST. (ii) Let G′ be the graph obtained from G by adding a pendant edge to each vertex. Suppose that G′ has a HIST T . Since every vertex in V (G) is adjacent to a vertex in V1 (G′ ), the set V (G) is equal to the set of stems of T . In particular, δ(T [V (G)]) ≥ 2, which contradicts the fact that T has no cycle. Thus G′ has no HIST. Let H1 = {Pl | l ≥ 3}, H2 = {K2,m | m ≥ 2}, H3 = {Yn | n ≥ 1}, H4 = {Zp | p ≥ 2} and H5 = {Fq | q ≥ 2}. By Lemma 2.1(i), for each 1 ≤ i ≤ 4, every graph in Hi has no HIST. By Lemma 2.1(ii), every graph in H5 has no HIST. Furthermore, for each 1 ≤ i ≤ 5, Hi is an infinite set. Therefore every set H of connected graphs such that every connected H -free graph of sufficiently large order has a [2, k]-ST satisfies that H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. This completes the proof of the ‘‘only if’’ part of Theorem 1.1. We next show the ‘‘if’’ part. A clique of a graph is a set of pairwise adjacent vertices, and an independent set is a set of pairwise nonadjacent vertices. For two positive integers s and t, the Ramsey number R(s, t ) is the minimum positive integer R such that any graph of order at least R contains either a clique of cardinality s or an independent set of cardinality t. The Ramsey number R(s, t ) exists for all positive integers s and t (see, for example, [3]). For positive integers k, l, m, n, p and q, let
α(k, n, p, q) = 3(k − 1)R(q, (p + q − 1)n + 1) + 2p + q and
β(k, m, n, q) = (k − 1)R(q, n + 1) + 2n + m. By the definition of the relation ‘‘≤’’, it suffices to show the following proposition. Proposition 2.2. Let k ≥ 2, l ≥ 4, m ≥ 2, n ≥ 1, p ≥ 2 and q ≥ 2 be positive integers. Then every connected {Pl , K2,m , Yn , Zp , Fq }-free graph of order at least (R(α, β) − 1)l−2 + 1 has a [2, k]-ST, where α = α(k, n, p, q) and β = β(k, m, n, q). Proof. Let G be a connected {Pl , K2,m , Yn , Zp , Fq }-free graph with |V (G)| ≥ (R(α, β) − 1)l−2 + 1. Since G is Pl -free, the diameter of G is at most l − 2. Hence ∆(G) ≥ R(α, β). Let x be a vertex with d(x) = ∆(G). Then G[N (x)] contains either a clique of cardinality α or an independent set of cardinality β . Case 1: G[N (x)] contains a clique of cardinality α . Let U1 be a clique of cardinality α in G[N (x)]. For each u ∈ N2 (x), we define vu ∈ N (u) as follows: if u ̸∈ N (U1 ), let vu ∈ N (u) ∩ N (x); if u ∈ N (U1 ), let vu ∈ N (u) ∩ U1 . We choose vu so that |{vu | u ∈ N2 (x)}|(=|{vu | u ∈ N2 (x) − N (U1 )}| + |{vu | u ∈ N2 (x) ∩ N (U1 )}|) is as small as possible. Let U2 = {vu | u ∈ N2 (x) − N (U1 )} and U3 = {vu | u ∈ N2 (x) ∩ N (U1 )} (see Fig. 3). By the minimality of |U2 | + |U3 |, pn(v, U2 ) ∩ (N2 (x) − N (U1 )) ̸= ∅ for each v ∈ U2 , and pn(v, U3 ) ∩ (N2 (x) ∩ N (U1 )) ̸= ∅ for each v ∈ U3 . For each v ∈ U2 , fix a vertex uv ∈ pn(v, U2 ) ∩ (N2 (x) − N (U1 )). For each v ∈ U3 , fix a vertex uv ∈ pn(v, U3 ) ∩ (N2 (x) ∩ N (U1 )). For v ∈ U2 , if |N (v) ∩ U1 | ≤ α − p + 1, then G[(U1 − N (v)) ∪ {x, v, uv }] contains a Zp as an induced subgraph, a contradiction. Thus
|N (v) ∩ U1 | ≥ α − p + 2 for each v ∈ U2 .
(2.1)
Claim 2.1. (i) For u ∈ N2 (x) − N (U1 ), if u′ ∈ V (G) − (N (x) ∪ N (U1 )) is adjacent to u, then vu u′ ∈ E (G). (ii) No vertex in N2 (x) − N (U1 ) is adjacent to a vertex in N3 (x). (iii) For v, v ′ ∈ U2 , uv uv ′ ̸∈ E (G). Proof. Let u ∈ N2 (x) − N (U1 ) and u′ ∈ V (G) − (N (x) ∪ N (U1 )) with uu′ ∈ E (G). By (2.1) and the definition of α, |N (vu ) ∩ U1 | ≥ α − p + 2 ≥ p − 1. If vu u′ ̸∈ E (G), then G[(N (vu ) ∩ U1 ) ∪ {vu , u, u′ }] contains a Zp as an induced subgraph, a contradiction. Consequently (i) holds. By (i), (ii) clearly holds. For v ∈ U2 , since uv ∈ pn(v, U2 ), no vertex in U2 − {v} is adjacent to uv . This together with (i) leads to (iii). Claim 2.2. |U2 | < R(q, n + 1).
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Fig. 3. U1 , U2 , U3 and U4 .
Proof. Suppose that |U2 | ≥ R(q, n + 1). Then G[U2 ] contains either a clique of cardinality q or an independent set of cardinality n + 1. Suppose that G[U2 ] contains a clique X of cardinality q. Then by Claim 2.1(iii), X ∪ {uv | v ∈ X } induces an Fq in G, a contradiction. Thus G[U2 ] contains an independent set X ′ of cardinality n + 1. Let v ∈ X ′ . Then X ′ ∪ {x, uv } induces a Yn in G, a contradiction. Claim 2.3. |U3 | < p + q. Proof. Suppose that |U3 | ≥ p + q. Since G is Fq -free, G[U3 ∪ {uv | v ∈ U3 }] contains no Fq as an induced subgraph. Then there exists v, v ′ ∈ U3 such that uv uv ′ ∈ E (G). Since |U3 | ≥ p + 1, G[(U3 − {v ′ }) ∪ {uv , uv ′ }] contains a Zp as an induced subgraph, a contradiction. For each u ∈ N3 (x), let wu ∈ N (u) ∩ N2 (x). We choose wu so that |{wu | u ∈ N3 (x)}| is as small as possible. Let U4 = {wu | u ∈ N3 (x)}. By Claim 2.1(ii), U4 ⊆ N2 (x) ∩ N (U1 ). By the minimality of |U4 |, pn(w, U4 ) ∩ N3 (x) ̸= ∅ for each w ∈ U4 . For each w ∈ U4 , fix a vertex uw ∈ pn(w, U4 ) ∩ N3 (x). If there exists w ∈ U4 such that |N (w) ∩ U1 | ≤ α − p + 1, then G[(U1 − N (w)) ∪ {vw , w, uw }] contains a Zp as an induced subgraph, a contradiction. Thus
|N (w) ∩ U1 | ≥ α − p + 2 for each w ∈ U4 .
(2.2)
Claim 2.4. (i) For each u ∈ N3 (x), if u′ ∈ V (G) − N2 (x) is adjacent to u, then wu u′ ∈ E (G). (ii) N4 (x) = ∅. (iii) For w, w ′ ∈ U4 , uw uw′ ̸∈ E (G). Proof. Let u ∈ N3 (x) and u′ ∈ V (G)−N2 (x) with uu′ ∈ E (G). By (2.2) and the definition of α, |N (wu )∩U1 | ≥ α−p+2 ≥ p−1. If wu u′ ̸∈ E (G), then G[(N (wu ) ∩ U1 ) ∪ {wu , u, u′ }] contains a Zp as an induced subgraph, a contradiction. Consequently (i) holds. By (i), (ii) clearly holds. For w ∈ U4 , since uw ∈ pn(w, U4 ), no vertex in U4 − {w} is adjacent to uw . This together with (i) leads to (iii). Claim 2.5. |U4 | < R(q, (p + q − 1)n + 1). Proof. Suppose that |U4 | ≥ R(q, (p + q − 1)n + 1). Then G[U4 ] contains either a clique of cardinality q or an independent set of cardinality (p + q − 1)n + 1. Suppose that G[U4 ] contains a clique X of cardinality q. Then by Claim 2.4(iii), X ∪{uw | w ∈ X } induces an Fq in G, a contradiction. Thus G[U4 ] contains an independent set X ′ of cardinality (p + q − 1)n + 1. By Claim 2.3, there exists a vertex v ∈ U3 which is adjacent to at least n + 1 vertices in X ′ . Let w ∈ N (v) ∩ X ′ . Then G[(N (v) ∩ X ′ ) ∪ {v, uw }] contains a Yn as an induced subgraph, a contradiction. Now we construct a [2, k]-ST T of G such that {x} ∪ ( 2≤i≤4 Ui ) is the set of stems of T . By (2.1), Claims 2.2, 2.3 and the definition of α , for each v ∈ U2 , |N (v) ∩ (U1 − U3 )| ≥ (α − p + 2) − (p + q − 1) = 3(k − 1)R(q, (p + q − 1)n + 1) + 3 ≥ (k − 1)(R(q, n + 1) − 1) ≥ (k − 1)|U2 |. Hence there exist (k − 1)|U2 | vertices av,i ∈ U1 − U3 (v ∈ U2 , 1 ≤ i ≤ k − 1) such that {av,i | 1 ≤ i ≤ k − 1} ⊆ N (v) for each v ∈ U2 . Let L1 = {av,i | v ∈ U2 , 1 ≤ i ≤ k − 1}. By (2.2), Claims 2.3, 2.5 and the definition of α , for each w ∈ U4 , |N (w) ∩ (U1 − (U3 ∪ L1 ))| ≥ (α − p + 2) − (p + q − 1) − (k − 1)(R(q, n + 1) − 1) = 3(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) + 3 ≥ (k − 1)(R(q, (p + q − 1)n + 1) − 1) ≥ (k − 1)|U4 |. Hence there exist (k − 1)|U4 | vertices bw,i ∈ U1 − (U3 ∪ L1 ) (w ∈ U4 , 1 ≤ i ≤ k − 1) such that {bw,i | 1 ≤ i ≤ k − 1} ⊆ N (w) for each w ∈ U4 . Let L2 = {bw,i | w ∈ U4 , 1 ≤ i ≤ k − 1}. By Claim 2.3 and the definition of
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Fig. 4. A desired [2, k]-ST of G.
α, |U1 − (U3 ∪ L1 ∪ L2 )| ≥ α − (p + q − 1) − (k − 1)(R(q, n + 1) − 1) − (k − 1)(R(q, (p + q − 1)n + 1) − 1) = 2(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) + p + 1 ≥ (k − 1)(p + q − 1) ≥ (k − 1)|U3 |. Hence there exist (k − 1)|U3 | vertices cv,i ∈ U1 −(U3 ∪ L1 ∪ L2 ) (v ∈ U3 , 1 ≤ i ≤ k − 1). Let L3 = {cv,i | v ∈ U3 , 1 ≤ i ≤ k − 1}. By the definition of α, |U1 − (L1 ∪ L2 ∪ L3 )| ≥ α − (k − 1)(R(q, n + 1) − 1) − (k − 1)(R(q, (p + q − 1)n + 1) − 1) − (k − 1)(p + q − 1) = 2(k − 1)R(q, (p + q − 1)n + 1) − (k − 1)(R(q, n + 1) − 1) − (k − 1)(p + q − 2) + 2p + q ≥ k + 1. For each v ∈ U2 , let Ev = {v u | u ∈ N2 (x) − N (U1 ), vu = v} ∪ {v av,i | 1 ≤ i ≤ k − 1}. For each v ∈ U3 , let Ev = {v u | u ∈ N2 (x) ∩ N (U1 ), vu = v} ∪ {v cv,i | 1 ≤ i ≤ k − 1}. For each w ∈ U4 , let Ew = {w u | u ∈ N3 (x), wu = w} ∪ {wbw,i | 1 ≤ i ≤ k − 1}. Then the spanning tree T of G with E (T ) = {xx | x ∈ N (x) − (L1 ∪ L2 ∪ L3 )} ∪ ′
′
Ey
y∈U2 ∪U3 ∪U4
is a desired [2, k]-ST of G (see Fig. 4). Case 2: G[N (x)] contains an independent set of cardinality β . Let U1 be an independent set of cardinality β in G[N (x)]. Claim 2.6. N2 (x) ∩ N (U1 ) = ∅. Proof. Suppose that N2 (x) ∩ N (U1 ) ̸= ∅, and let y ∈ U1 and y′ ∈ N2 (x) such that yy′ ∈ E (G). If |N (y′ ) ∩ U1 | ≤ β − n, then G[(U1 − N (y′ )) ∪ {x, y, y′ }] contains a Yn as an induced subgraph, a contradiction. Thus |N (y′ ) ∩ U1 | ≥ β − n + 1. Then by the definition of β, |N (y′ ) ∩ U1 | ≥ m. Then G[(N (y′ ) ∩ U1 ) ∪ {x, y′ }] contains a K2,m as an induced subgraph, a contradiction. For each u ∈ N2 (x), let vu ∈ N (u) ∩ N (x). We choose vu such that |{vu | u ∈ N2 (x)}| is as small as possible. Let U2 = {vu | u ∈ N2 (x)}. By Claim 2.6, U2 ⊆ N (x) − U1 . By the minimality of |U2 |, pn(v, U2 ) ∩ N2 (x) ̸= ∅ for each v ∈ U2 . For each v ∈ U2 , fix a vertex uv ∈ pn(v, U2 ) ∩ N2 (x). If there exists v ∈ U2 such that |N (v) ∩ U1 | ≤ β − n, then G[(U1 − N (v)) ∪ {x, v, uv }] contains a Yn as an induced subgraph, a contradiction. Thus
|N (v) ∩ U1 | ≥ β − n + 1 for each v ∈ U2 .
(2.3)
Claim 2.7. (i) For each u ∈ N2 (x), if u′ ∈ V (G) − N (x) is adjacent to u, then vu u′ ∈ E (G). (ii) N3 (x) = ∅. (iii) For v, v ′ ∈ U2 , uv uv ′ ̸∈ E (G). Proof. Let u ∈ N2 (x) and u′ ∈ V (G) − N (x) with uu′ ∈ E (G). By Claim 2.6, u is adjacent to no vertex in U1 . By (2.3) and the definition of β, |N (vu ) ∩ U1 | ≥ β − n + 1 ≥ n. If vu u′ ̸∈ E (G), then G[(N (vu ) ∩ U1 ) ∪ {vu , u, u′ }] contains a Yn as an induced subgraph, a contradiction. Thus (i) holds. By (i), (ii) clearly holds. For v ∈ U2 , since uv ∈ pn(v, U2 ), no vertex in U2 − {v} is adjacent to uv . This together with (i) leads to (iii). Claim 2.8. |U2 | < R(q, n + 1). Proof. Suppose that |U2 | ≥ R(q, n + 1). Then G[U2 ] contains either a clique of cardinality q or an independent set of cardinality n + 1. Suppose that G[U2 ] contains a clique X of cardinality q. Then by Claim 2.7(iii), X ∪ {uv | v ∈ X } induces an
M. Furuya, S. Tsuchiya / Discrete Mathematics 313 (2013) 2206–2212
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Fig. 5. A desired [2, k]-ST of G.
Fq in G, a contradiction. Thus G[U2 ] contains an independent set X ′ of cardinality n + 1. Let v ∈ X ′ . Then X ′ ∪ {x, uv } induces a Yn in G, a contradiction. Now we construct a [2, k]-ST T of G such that {x} ∪ U2 is the set of stems of T . By (2.3), Claim 2.8 and the definition of β , for each v ∈ U2 , |N (v) ∩ U1 | ≥ β − n + 1 = (k − 1)R(q, n + 1) + n + m + 1 ≥ (k − 1)(R(q, n + 1) − 1) ≥ (k − 1)|U2 |. Hence there exist (k − 1)|U2 | vertices av,i ∈ U1 (v ∈ U2 , 1 ≤ i ≤ k − 1) such that {av,i | 1 ≤ i ≤ k − 1} ⊆ N (v) for each v ∈ U2 . Let L1 = {av,i | v ∈ U2 , 1 ≤ i ≤ k − 1}. By the definition of β, |U1 − L1 | ≥ β − (k − 1)(R(q, n + 1) − 1) = k − 1 + 2n + m ≥ k + 1. For each v ∈ U2 , let Ev = {v u | u ∈ N2 (x), vu = v} ∪ {v av,i | 1 ≤ i ≤ k − 1}. Then the spanning tree T of G with
E (T ) = {xx′ | x′ ∈ N (x) − L1 } ∪
Ev
v∈U2
is a desired [2, k]-ST of G (see Fig. 5). This completes the proof of Proposition 2.2.
Remark. By the argument in the proof of Proposition 2.2, we also see that the radius of a connected {Pl , K2,m , Yn , Zp , Fq }-free graph of sufficiently large order is at most 3. 3. A HIST of a P4 -free graph In this section, we prove Theorem 1.3. In [9], a theorem which gives a characterization of P4 -free graphs was proved. By the theorem, we obtain the following lemma which is well known when we deal with P4 -free graphs. Lemma 3.1. Let G be a connected P4 -free graph, and let S be a minimum cut set of G. Then every vertex in S is adjacent to every vertex in V (G) − S. Proof of Theorem 1.3. We first consider a connected graph G of order n ∈ {4, 5}. Let v ∈ V≥3 (G). If there exists a HIST T of G such that v is a stem of T , then v is the unique stem of T , and hence d(v) = n − 1. On the other hand, if d(v) = n − 1, then there clearly exists a HIST of G such that v is a stem of T . This implies that the case n ∈ {4, 5} of Theorem 1.3 holds. In the argument below, we consider only the graphs of order at least 6. Furthermore, in Section 2, we have already checked that K2,n−2 has no HIST (and so K2,n−2 is not HIST-extendable). Thus it suffices to show the ‘‘only if’’ part of Theorem 1.3 under the condition n ≥ 6. Let G be a connected P4 -free graph of order n ≥ 6 with G ̸≃ K2,n−2 . By Lemma 3.1, there exists a partition of {X1 , X2 } of V (G) such that every vertex in X1 is adjacent to every vertex in X2 . Claim 3.1. Suppose that X1 ∩ V≥3 (G) ̸= ∅ and X2 ∩ V≥3 (G) ̸= ∅. Then, for every x1 ∈ X1 ∩ V≥3 (G) and every x2 ∈ X2 ∩ V≥3 (G), there exists a HIST T of G such that x1 and x2 are the stems of T . Proof. Without loss of generality, we may assume that |X1 | ≥ |X2 |. By the definition of {X1 , X2 }, X2 ⊆ N (x1 ) and X1 ⊆ N (x2 ). Since dG (x1 ) ≥ 3, there exists a subset Y1 of N (x1 ) such that |Y1 | ≥ 3 and X2 ⊆ Y1 . We choose Y1 so that |Y1 | is as small as possible. By the minimality of Y1 , |X1 ∩ Y1 | = max{3 − |X2 |, 0}. If |X1 ∩ Y1 | = 3 − |X2 |, then |X1 − Y1 | = |X1 | − |X1 ∩ Y1 | = |X1 | − (3 − |X2 |) = |V (G)| − 3 ≥ 3; if |X1 ∩ Y1 | = 0, then |X1 − Y1 | = |X1 | − |X1 ∩ Y1 | = |X1 | ≥ |V (G)|/2 ≥ 3. Hence in either case, we have |X1 − Y1 | ≥ 3. Then the spanning tree T of G with E (T ) = {x1 u | u ∈ Y1 } ∪ {x2 u | u ∈ X1 − Y1 } is a desired HIST of G.
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Let v ∈ V≥3 (G). It suffices to show that there exists a HIST of G such that v is a stem of T . We may assume that v ∈ X1 . If |X1 | = 1, then d(v) = n − 1, and hence there exists a desired HIST of G. Thus we may assume that |X1 | ≥ 2. If X2 ∩ V≥3 (G) ̸= ∅, then by Claim 3.1, there exists a desired HIST of G. Thus we may assume that X2 ∩ V≥3 (G) = ∅. By the definition of {X1 , X2 }, this implies that |X1 | = 2 and X2 is an independent set of G. Write X1 = {v, v ′ }. Since G ̸≃ K2,n−2 , vv ′ ∈ E (G), and hence d(v) = n − 1. Then G has a spanning star with center v , as desired. This completes the proof of Theorem 1.3.
4. The [2, k ]-ST-extendability In this section, we extend the HIST-extendability to [2, k]-ST, and give a result concerning it similar to Theorem 1.1. For k ≥ 2, a graph G is [2, k]-ST-extendable if V≥k+1 (G) ̸= ∅ and, for every v ∈ V≥k+1 (G), G has a [2, k]-ST T such that v is a stem of T . We show the following proposition. Proposition 4.1. Let k ≥ 2 be an integer. If a graph G has a [2, 2k]-ST, then G is [2, k]-ST-extendable. Proof. Let v ∈ V≥k+1 (G). It suffices to show that there exists a [2, k]-ST T of G such that v is a stem of T . Let T ′ be a [2, 2k]-ST of G. Since T ′ is a [2, k]-ST of G, if v is a stem of T ′ , then T ′ is a desired HIST. Thus we may assume that dT ′ (v) = 1. Let U be a subset of NG (v) − NT ′ (v) with |U | = k. For each u ∈ U, let Pu be the unique v − u path in T ′ , and let eu be the unique edge of Pu which is incident with u. Then every component of F = T ′ − {eu | u ∈ U } is a tree with no vertices of degree from 2 to k. Furthermore, dF (u) = 0 if and only if u ∈ U ∩ V≤1 (F ). Therefore F ∪ {v u | u ∈ U } is a desired [2, k]-ST. This together with Theorem 1.1 implies the following results. Corollary 4.2. Let k ≥ 2 be an positive integer, and let H be a set of connected graphs. Then there exists an integer n0 = n0 (k, H ) such that every connected H -free graph of order at least n0 is [2, k]-ST-extendable if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Corollary 4.3. Let H be a set of connected graphs. Then there exists an integer n0 = n0 (H ) such that every connected H -free graph of order at least n0 is HIST-extendable if and only if H ≤ {Pl , K2,m , Yn , Zp , Fq } for some integers l, m, n, p and q. Acknowledgments The authors are grateful to referees for their helpful comments. References [1] M.O. Albertson, D.M. Berman, J.P. Hutchinson, C. Thomassen, Graphs with homeomorphically irreducible spanning trees, J. Graph Theory 14 (1990) 247–258. [2] M.D. Barrus, M. Kumbhat, S.G. Hartke, Graph classes characterized both by forbidden subgraphs and degree sequences, J. Graph Theory 57 (2008) 131–148. [3] G. Chartrand, L. Lesniak, P. Zhang, Graphs & Digraphs, fifth ed., Chapman and Hall/CRC, Boca Raton, FL, 2010. [4] G. Chen, H. Ren, S. Shan, Homeomorphically irreducible spanning trees in locally connected graph, Combin. Probab. Comput. 21 (2012) 107–111. [5] R.J. Faudree, R.J. Gould, Characterization forbidden pairs for Hamiltonian properties, Discrete Math. 173 (1997) 45–60. [6] S. Fujita, K. Kawarabayashi, C.L. Lucchesi, K. Ota, M. Plummer, A. Saito, A pair of forbidden subgraphs and perfect matchings, J. Combin. Theory Ser. B 96 (2006) 315–324. [7] R.J. Gould, J.M. Harris, Forbidden triples and traceability: a characterization, Discrete Math. 203 (1999) 101–120. [8] K. Ota, G. Sueiro, Forbidden induced subgraphs for perfect matchings, Graphs Combin. 29 (2013) 289–299. [9] D. Seinsche, On a property of the class of n-colorable graphs, J. Combin. Theory Ser. B 16 (1974) 191–193. [10] S. Tsuchiya, Rooted HIST property on planar triangulations, Ars Combin. (2013) in press.