By Kuhn D., Osthus D.

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**Extra resources for A Note on Complete Subdivisions in Digraphs of Large Outdegree**

**Example text**

R}, dH (x) = dG (x) − |NG (x) ∩ Vj | ≥ r−1 dG (x). r Corollary 2 (Caro and Roditty [2]). Let r, k be positive integers and G a graph of order n and minimum degree δ ≥ (r + 1)k/r − 1. Then r n. γk (G) ≤ r+1 Proof. Let r = r + 1 and let V1 , V2 , . . , Vr and H be like in Corollary 1 such that |V1 | ≥ |V2 | ≥ · · · ≥ |Vr |. Then, together with the hypothesis on δ, it follows r r r that dH (x) ≥ r r−1 dG (x) = r+1 dG (x) ≥ r+1 δ ≥ k − r+1 and hence, since dH (x) is an integer, dH (x) ≥ k for all x ∈ V (G).

Lemma 1. Every graph has a unique minimal shrinkage. Proof. For a vertex x of degree at most 2 in some graph H and a vertex x let H(x) denote the graph obtained from H by suppressing x. Suppose (reductio ad absurdum) that the set G of all shrinkages of G had two distinct minimal elements. We choose two distinct minimal elements H1 , H2 plus a common upper bound H of them from G in such a way that |V (H)| is minimal. For i ∈ {1, 2}, Hi < H holds, so H contains a vertex xi of degree at most 2 such that Hi ≤ H(xi ) ≤ H.

R} and each u ∈ Vi . Therefore, each Vi is a k-dependent set of G for 1 ≤ i ≤ r. Since n βk (G) ≥ max{|Vi | : 1 ≤ i ≤ r} ≥ , r the desired bound follows. 3. NEW BOUNDS ON γk (G) Corollary 6. Let G be a graph of order n and minimum degree δ. If k ≤ δ is an integer, then γk (G) ≤ δ n. 2δ + 1 − k for every positive integers a and b and since the Proof. Since ba ≤ a+b−1 b x function x+1 is increasing for x positive, Corollary 3 implies γk (G) ≤ δ k/(δ + 1 − k) δ/(δ + 1 − k) n= n. n≤ δ/(δ + 1 − k) + 1 2δ + 1 − k k/(δ + 1 − k) + 1 A simple calculation yields 2k − δ − 1 δ ≤ 2δ + 1 − k 2k − δ for (δ + 4)/2 ≤ k ≤ δ − 1 and thus δ ≥ 6.

### A Note on Complete Subdivisions in Digraphs of Large Outdegree by Kuhn D., Osthus D.

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