1. 2007 Kézdy André Kézdy Department of Mathematics University of Louisville * Preliminary report.  More -valuations for trees via the combinatorial nullstellensatz* 0 10 8 6 2 3 3 5 1 42 2 1 5 4 3 0 1 2 3 4 5 6 7 8 9 K 11 T

2. 2007 Kézdy Decompositions A decomposition of a graph G is a partition of E ( G ) into pairwise edge-disjoint subgraphs. If all of the subgraphs are isomorphic to a given graph H, then we say that H decomposes G. Examples… K4K4 P3P3 P 3 cyclically decomposes K 4 K6-eK6-eK 1, 3 K 1,3 does not decomposes K 6 - e

3. 2007 Kézdy Numerous unresolved conjectures! CONJECTURE G. Ringel, 1963 Any tree with n edges decomposes K n, n. CONJECTURES R. Häggkvist and R. Graham,1984? Any tree with n edges decomposes any 2 n -regular graph. Any tree with n edges decomposes any n -regular bipartite graph. CONJECTURE A. Kotzig, 1963 Any tree with n edges cyclically decomposes K 2 n +1.

4. 2007 Kézdy Some recent results Any tree with n edges and radius r decomposes K t for some THEOREM Shiue and Fu 2006 Any tree with n edges and radius r decomposes K 2 hn,2 hn for some THEOREM A. Lladó and S.C. López 2004 Suppose T is a tree with edges. There is a cyclic packing of copies of T into K n. THEOREM Brandt and Woźniak 2004 roughly half of what might be possible slightly improves bounds by Kézdy and Snevily, 2002

5. 2007 Kézdy - valuations A  -valuation of a graph G ( V, E ) with n edges is an injection such that, if then, for all i, 10 8 0 3 6 2 12 3 4 5 THEOREM Rosa, 1967 Any tree with n edges cyclically decomposes K 2 n +1 if and only if it has a  -valuation. or

6. 2007 Kézdy Cyclic decomposition 0 10 8 6 2 3 3 5 1 42 2 1 5 4 3 0 1 2 3 4 5 6 7 8 9 K 11 T

7. 2007 Kézdy Alon, N. (1999) Combinatorial Nullstellensatz, Combinatorics, Probability, and Computing 8 7-29. Main tool THEOREM F an arbitrary field where t i 2 Z º 0 If the coefficient of in g is nonzero then, for any subsets S 1, S 2,…, S n of F satisfying | S i | > t i, there are elements s 1 2 S 1, s 2 2 S 2,…, s n 2 S n such that Alon, 1999

8. 2007 Kézdy stunted trees stunted trees are trees that admit a linear ordering of edges NOTE: If T is a stunted tree with n edges, then e 1, e 2,…, e n such that and, for all j = 3,…, n, e1e1 e2e2 e3e3 e4e4 e5e5 a stunted tree: So long paths, for example, are not stunted trees.

9. 2007 Kézdy A theorem If T is a stunted tree on n edges and p = 2 n +1 is a prime, then T has a  -valuation. THEOREM Kézdy, 2006 Suppose T is a stunted tree with n edges ordered e 1, e 2, …, e n. Proof. For all vertices v i, define Apply the combinatorial nullstellensatz working over Z p [ x 1, x 2, …, x n ]. assign variables x 1, x 2, …, x n V ( T ) = { v 0, v 1, v 2, …, v n } root: order vertices “chronologically” NOTE:

10. 2007 Kézdy consider this polynomial… degree guarantees that edge labels are distinct in absolute value modulo p guarantees that induced vertex mapping is an injection into K p f T is homogeneous of degree We’ll show that the coefficient of this monomial has absolute value 1. a carefully selected factor

11. 2007 Kézdy ‘complementary’ monomials Vandermonde’s identity this sum has exactly one term of the form… CLAIM: Basis: m = 1 every edge is a factor in so  (1) = 1 follows from inspection of Use induction on m to prove that

12. 2007 Kézdy induction step Induction step: every edge e m, e m +1,…, e 2m is in m factors of inspection of QED reveals that  ( m ) · 2 m So,  ( m ) 2 { m, m +1, m +2,…,2 m } Because T is stunted: Tree induced by first m -1 edges eheh

13. 2007 Kézdy Some consequences… COROLLARY Suppose 2 n +1 is prime and T is a stunted tree with edges ordered e 1,…, e n. If S 1,…, S n µ Z 2n+1 such that | S i | = i, then T has a  -valuation in which e i avoids labels in S i. COROLLARY Adding sufficiently many leaves to an arbitrary vertex of a tree eventually produces a tree with a  -valuation.

14. 2007 Kézdy What about extensions? No luck with unique monomials in other trees. (except trees that are ‘nearly stunted’: an edge near the root can be contracted to form a stunted tree) One way to extend is via a permanent related to a similar polynomial… recall.. zero or constant (absolute value) on all inputs! Data produced using polynomial-time, unique f-factor algorithm (Gabow, Kaplan, Tarjan, 2001) shows that the only small trees with ‘nice’ factors are stunted or nearly stunted trees.

15. 2007 Kézdy Scheim’s Lemma LEMMA Suppose that n is a positive integer, s 1,…, s n nonnegative integers and P ( x 1,…, x n ) (over the reals) of degree s 1 +  + s n. TFAE: Scheim, David E. (1974) Discrete Mathematics, 377-382 The number of edge 3-coloring of a planar cubic graph as a permanent. IMPORTANT because it relates coefficients with evaluations of the polynomal

16. 2007 Kézdy The sign of new labelings g T will be nonzero only on something between a  and a graceful-labeling of T What is the interpretation for the coefficient of this monomial? g T is homogeneous of degree parity of sign of permutation parity of ‘sign’ of a nonzero evaluation of g T is the ‘product’ of these signs Using Scheim lemma this coefficient is a nonzero constant times the difference between number of even and odd evaluations of g T (modulo p). Answer recall:

17. 2007 Kézdy Approach via n th roots of unity

18. 2007 Kézdy Question: linear transformation Does there exist an element of the range whose preimage has an odd number of elements in S ? S y p an odd prime p = 2 n +1 f f -1 ( y ) f has the matrix form: C is an ( n +1) £ ( n +1) unipotent upper triangular matrix

19. 2007 Kézdy THANKS!