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
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

4. 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)

5. 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.

6. 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)?

7. 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)

8. 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)

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.
(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

11. 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

12. 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

13. 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
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

14. Questions?