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

2. Outline Undirected Graphs Directed Graphs Selected Proof Ideas

3. Motivation and Definition 816 357 492 15 987 216 345 24 9 12 141318 1511 Magic SquareAntimagic Square 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 1 2 3 78 5 6 4 6 20 10 15 21

4. Some Easy Observations P n, C n (n≥3) are antimagic. Graphs of maximum degree n-1 are antimagic for n≥3 (e.g. K n, S n, W n ). 1234 1 3574 1235 4 1 3589 4 [Hartsfield, Ringel, 1990] u G-u 1. Use labels 1,2,…,m-(n-1) arbitrarily v1v1 v2v2 … v n-1 1. w(v 1 ) ≤ w(v 2 ) ≤ … ≤ w(v n-1 ) 2. Use remaining largest labels to achieve antimagic property 2. w(v 1 ) < w(v 2 ) < … < w(v n-1 ) < w(u) P 2 is not antimagic. 1 1 1

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

6. What about Directed Graphs? 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 1 2 3 78 5 6 4 -4 -14 8 11 Question 2: Given an undirected graph G, is there an antimagic orientation D(G)? Question 1: Given an undirected graph G, is every orientation D(G) antimagic? ∃ ∀

7. First Observations P 2 has an antimagic orientation. 1 1 12 There are directed graphs that are not antimagic. 12 3 2 2 Lemma: If G is an undirected bipartite antimagic graph, then G has an antimagic orientation. w(a 1 ) w(a 2 ) w(a 3 ) w(a 4 ) w(b 1 ) w(b 2 ) w(b 3 ) - - - Conjecture: Every connected directed graph with at least 4 vertices is antimagic.

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

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

10. Proof of Theorem 3 Theorem 3: Let G be a (2d+1)-regular undirected graph. Then there exists an antimagic orientation of G. Proof idea: (Almost) magic labelling + distortion 12 1. Add perfect matching M to connect different components 13 14 15 1 1 2 3 4 5 6 7 8 11 9 10 -2 -2 13 -2 13 0 3 4 9 14 -3 -6 -7 -12 -2 (extension to 2d-regular graphs is possible with slight extra conditions) G M 4 5 3. Remove matching edges and obtain an antimagic labelling 2. Orient and label along some eulerian cycle (use duplicate labels for M-edges)

11. 1 m= ( ) n 2 u rn-1-r G-u Proof of Theorem 2 (cliques) Theorem 2: Every orientation of K n (n≥4) is antimagic. Proof idea: Subset control + parity argument 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 H vrvr w(v 1 ) ≤ … ≤ w(v r ) v1v1 v2v2 … v n-1 v r+1 … w(v r+1 ) ≤ … ≤ w(v n-1 ) w(v 1 ) < … < w(v r ) w(v r+1 ) < … < w(v n-1 ) 5. Achieve antimagic property in each class and by parity among all vertices

12. Theorem [Alon, Kaplan, Lev, Roditty, Yuster, 2004] : Every graph with minimum degree  (log n) is antimagic. Proof can be adapted for every orientation of every undirected graph with minimum degree  (log n) (Theorem 1) Proof idea: Lovász Local Lemma 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 ≈ n 2 n 2 A 1, A 2,…,A s events Pr[A i ] ≤ p A i independent of all but at most r other events p(r+1) < ⅓ With positive probability no event A i holds Proof by Probabilistic Methods (1)

13. Assume G is d-regular (d=C log n) and has an even number of edges 1. Partition edges into pairs, such that no pair shares an endvertex 2. Randomly partition label set into pairs and assign to edge pairs 1 m {10,17} {3,5} {11,13} u u 2 d possible choices S(u) = {24,26,28,31,33,35} ⅛¼Pr[w(u)=k] =⅛⅛⅛¼ 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 2 d choices: Pr[w(u)=k] ≤ small 2 random phases 4. Flip coin for each edge pair which edge gets which label Pr[A(u,v)] = Pr[w(u)=k] ≤ small u v v 5. Dependencies: r = (6d+2)n ≈ 6dn  n 2 6. LLL: p(r+1) = small (6dn+1)  ⅓ Proof by Probabilistic Methods (2)

14. Questions?