• No se han encontrado resultados

4. Complejidad: Según Byrne (1995), la mayoría de las acciones presentadas experimentalmente eran demasiado simples y propuso aumentar su complejidad en futuras

1.2 Justificación de los estudios de imitación

LetT be a tree on n vertices, rooted at t1, letH ⊆ V (T ), and let k be a positive integer. For any vertexx ∈ T , there is a unique path in T from x to t1; letPx denote the set of the firstk vertices of this path, starting fromx. Let H1 =S

x∈HPx, and then for eachi > 1 let Hi+1be formed fromHi by adding the vertices ofP

x for anyx ∈ Hi with at least two children in Hi. After at mostn steps we must have Hi = Hi+1, when we terminate the process. We refer to this finalHiasH with leading paths included, denotedP

k(H). So H ⊆ Pk(H)⊆ V (T ). Note thatPk(H) depends on both the value of k and the root t1ofT .

Next we prove two results which enable us to make use of this definition. The first shows that ifH is small thenPk(H) is small, and the second shows that if H is small then it is possible to embed any componentT0ofT [Pk(H)] in a regular and dense cycle of cluster tournaments such that the vertices ofV (T0)∩ H are embedded in the first cluster and the ‘root’ of H is embedded

in a given cluster.

Proposition 2.23 Letk be any positive integer, let T be a tree on n vertices, rooted at some

t1 ∈ T, and letH⊆ V (T ). Then|Pk(H)| 6 3k|H|.

Proof. Consider any componentT0 of T [Pk(H)], and let t01 be the unique vertex of T0 with minimald(t1, t01). Then every vertex of T0 lies on the path from some vertex ofH to t1, and so T0 is precisely the set of vertices in paths between t0

1 and vertices of H ∩ V (T0). Thus onlyt0

1 and vertices ofH can be leaves of T0. It follows thatT [Pk(H)] has at most 2|H| leaves. Since T [Pk(H)] is a forest, it follows that the number of vertices of T [Pk(H)] with at least two children inT [Pk(H)] is also at most 2|H|. Furthermore, any vertex x ∈ T for which the vertices ofPx were added to Pk(H) at any stage is either a member of H or has at least two children inPk(H). This is true for at most 3|H| vertices x, and for each such vertex at most k

Lemma 2.24 Suppose that 1/m  1/k  ε  d. LetT be a directed tree rooted at some

t1 ∈ T. LetH ⊆ V (T )be of size |H| 6 m/10k, letT0 be a component ofT [Pk(H)]which

does not containt1, and lett01 be the unique vertex ofT0 with minimald(t01, t1). Let Gbe an

ε-regulard-dense cycle of cluster tournaments on clustersV1, . . . , Vk, each of sizem. Then for

anyj ∈ [k],Gcontains a copy ofT0 with the vertext0

1 corresponding to some vertex ofVj, and

every vertex inV (T0)∩ Hcorresponding to some vertex ofV 1.

Proof. Informally, from the perspective of t0

1, T0 begins with a path of lengthk − 1 (from t01 tot, say) before possibly branching out. So we find a copy of T0 inG by first embedding the vertices of this path so thatt0

1is embedded inVj andt is embedded in V1. We then embed all of the remaining vertices ofT0 inV

1.

More formally, note that for each 0 6 s 6 k − 1 there is precisely one vertex xs of T0 with

d(t01, xs) = s (so x0 = t01, and xi ∈ H for any i < k − 1). Let F ⊆ [k − 1] be the set of/ thoses such that xs−1 → xs, and let B ⊆ [k − 1] be the set of s such that xs−1 ← xs. Then

|F | + |B| = k − 1, so either |F | > k − j or |B| > j − 1. Suppose first that |B| > j − 1. Then

chooseB0 ⊆ B of size j − 1. We allocate the vertices of T0 to the clustersV

1, . . . , Vj. Begin by allocatingx0 toVj. Then for eachs ∈ [k − 1] in turn, let Vi be the cluster to which xs−1 was allocated, and allocatexstoViifs /∈ B0, or toVi−1ifs∈ B0. Then since|B0| = j − 1, xk−1 is assigned toV1. Finally, allocate all other vertices ofT0toV1. Then every edge ofT0is allocated either canonically or within a cluster.

Next we embed T0 in G so that every vertex is embedded within the cluster to which it is allocated. To begin, by a standard regularity argument we may choose for eachi a set V0

i ⊆ Vi so that |Vi0| > 9m/10 and every vertex v ∈ Vi0 has at least dm/2 outneighbours in V0

i+1. Let G0 = G[V0

1 ∪ · · · ∪ Vk0]. Now, for each i, let Si be the set of vertices of T0 allocated to

Vi. So |S2|, . . . , |Sk| 6 k − 1 and |S1| 6 |T0|. Then by Proposition 2.23, 3|S1| 6 3|T0| 6

9k|H| 6 |V0

have successfully embeddedT0[S1 ∪ · · · ∪ Si−1] in G[V10 ∪ · · · ∪ Vi0−1] for some i 6 j. Then precisely one vertext ∈ Si has a neighbour t0 ∈ Si−1, and t0 has already been embedded to somev0 ∈ V0

i−1. Nowv0has at leastdm/2 > 3|Si| outneighbours in Vi0, and so by Theorem 1.2 we may embedT0[S

i] among these outneighbours. Let v be the vertex to which t is embedded; then since v is an outneighbour of v0, we have extended our embedding to an embedding of

T0[S

1∪ · · · ∪ Si] in G[V10∪ · · · ∪ Vi0]. Continuing in this manner we obtain an embedding of T0 inG, with t01 embedded inVj andV (T0)\ {x0, . . . , xk−2} ⊇ V (T0)∩ H embedded into V1, as desired. A similar argument achieves this if|F | > k − j.