Antimagic Labellings of Graphs Torsten Mütze Joint work with Dan Hefetz and Justus Schwartz

Outline Undirected Graphs Directed Graphs Selected Proof Ideas

Motivation and Definition Magic Square Antimagic Square 15 8 1 6 3 5 7 4 9 2 24 9 12 14 13 18 15 11 9 8 7 2 1 6 3 4 5 6 20 10 15 21 1 2 3 7 8 5 6 4 Antimagic graph labelling [Hartsfield, Ringel, 1990] - Undirected graph G=(V,E), m:=|E| - Labelling of the edges with 1,2,…,m - All vertex sums distinct

Some Easy Observations [Hartsfield, Ringel, 1990] 1 1 3 5 7 4 P2 is not antimagic. 1 2 3 4 Pn, Cn (n≥3) are antimagic. 1 3 5 8 9 4 1 2 3 5 4 Graphs of maximum degree n-1 are antimagic for n≥3 (e.g. Kn, Sn, Wn). u G-u 1. Use labels 1,2,…,m-(n-1) arbitrarily 2. Use remaining largest labels to achieve antimagic property 2. w(v1) < w(v2) < … < w(vn-1) < w(u) v1 v2 … vn-1 1. w(v1) ≤ w(v2) ≤ … ≤ w(vn-1)

More Results Theorem [Hefetz, 2005]: A graph on 3k vertices that admits a K3-factor is antimagic. Theorem [Cranston, 2007]: Every regular bipartite graph with minimum degree 2 is antimagic. Theorem [Alon, Kaplan, Lev, Roditty, Yuster, 2004]: Every graph with minimum degree W(log n) is antimagic. Many other special cases are known Conjecture [Hartsfield, Ringel, 1990]: Every connected graph but P2 is antimagic.

What about Directed Graphs? -4 -14 8 -1 11 Antimagic labelling of directed graphs - Directed graph D=(V,E), m:=|E| - Labelling of the edges with 1,2,…,m - All oriented vertex sums distinct 7 8 6 3 5 2 4 1 Question 1: Given an undirected graph G, is every orientation D(G) antimagic? ∃ ∀ Question 2: Given an undirected graph G, is there an antimagic orientation D(G)?

First Observations P2 has an antimagic orientation. -1 1 1 There are directed graphs that are not antimagic. 2 -1 1 2 3 -1 2 1 2 Conjecture: Every connected directed graph with at least 4 vertices is antimagic. Lemma: If G is an undirected bipartite antimagic graph, then G has an antimagic orientation. - w(a1) w(a2) w(a3) w(a4) w(b1) w(b2) w(b3)

Our Results Theorem 1: Every orientation of every undirected graph with minimum degree W(log n) is antimagic. ∀ Theorem 2: Every orientation of every undirected graph in one of the following families is antimagic: stars Sn, wheels Wn, cliques Kn (n≥4). Theorem 3: Let G be a (2d+1)-regular undirected graph. Then there exists an antimagic orientation of G. ∃ (extension to 2d-regular graphs is possible with slight extra conditions)

Proof Strategies Explicit labellings - (Almost) magic labelling + distortion - Subset control Algebraic techniques - Combinatorial Nullstellensatz Probabilistic methods - Lovász Local Lemma

Proof of Theorem 3 Theorem 3: Let G be a (2d+1)-regular undirected graph. Then there exists an antimagic orientation of G. (extension to 2d-regular graphs is possible with slight extra conditions) Proof idea: (Almost) magic labelling + distortion 3 4 9 14 -3 -6 -7 -12 -2 -1 -2 13 12 1 -1 -2 13 M G 13 7 2 11 6 1 15 10 8 15 10 9 4 5 5 4 14 3 1. Add perfect matching M to connect different components 2. Orient and label along some eulerian cycle (use duplicate labels for M-edges) 3. Remove matching edges and obtain an antimagic labelling

Proof of Theorem 2 (cliques) Theorem 2: Every orientation of Kn (n≥4) is antimagic. Proof idea: Subset control + parity argument 1 m=( ) n 2 u r n-1-r G-u H w(v1) < … < w(vr) vr w(v1) ≤ … ≤ w(vr) v1 v2 … vn-1 vr+1 w(vr+1) ≤ … ≤ w(vn-1) w(vr+1) < … < w(vn-1) 1. Pick vertex u of maximum in-degree (r:=deg+(u), deg-(u)=n-1-r) 2. Reserve the r largest labels of the same parity and the n-1-r smallest labels of the opposite parity 3. Distribute remaining odd labels over even degree subgraph H of G-u 4. Distribute remaining even labels over remaining edges of G-u 5. Achieve antimagic property in each class and by parity among all vertices

Proof by Probabilistic Methods (1) Theorem [Alon, Kaplan, Lev, Roditty, Yuster, 2004]: Every graph with minimum degree W(log n) is antimagic. Proof can be adapted for every orientation of every undirected graph with minimum degree W(log n) (Theorem 1) Proof idea: Lovász Local Lemma A1, A2,…,As events Pr[Ai] ≤ p Ai independent of all but at most r other events p(r+1) < ⅓ With positive probability no event Ai holds A(u,v) := vertex sums of u and v are the same With positive probability no two vertex sums are the same  Antimagic labelling Naive: random permutation of the edge labels  All events are dependent, r = ( )-1 ≈ n2 n 2

Proof by Probabilistic Methods (2) Assume G is d-regular (d=C log n) and has an even number of edges 1 m {10,17} {3,5} {11,13} u u v 2d possible choices S(u) = {24,26,28,31,33,35} ⅛ ¼ Pr[w(u)=k] = 1. Partition edges into pairs, such that no pair shares an endvertex 2 random phases 2. Randomly partition label set into pairs and assign to edge pairs 3. Fix a partition from 2, such that at all vertices every value from S(u) is obtained only by a small fraction among the 2d choices: Pr[w(u)=k] ≤ small 4. Flip coin for each edge pair which edge gets which label Pr[A(u,v)] = Pr[w(u)=k] ≤ small 5. Dependencies: r = (6d+2)n ≈ 6dn  n2 6. LLL: p(r+1) = small (6dn+1)  ⅓

Questions?