Anti-Ramsey hypergraph numbers

The anti-Ramsey number arn(H) of an r-uniform hypergraph is the maximum number of colors that can be used to color the hyperedges of a complete r-uniform hypergraph on n vertices without producing a rainbow copy of H . In this paper, we determine anti-Ramsey numbers for paths of length 2, certain stars and complete hypergraphs, and the complete 3-uniform hypergraph of order 4 with a single hyperedge removed.


Introduction
First defined by Erdős, Simonovits, and Sós [4] in 1973, the anti-Ramsey number of an runiform hypergraph H (r) determines the number of colors needed for the hyperedges of a complete r-uniform hypergraph to guarantee the existence of a rainbow subhypergraph isomorphic to H (r) . While much work has been done on the evaluation of such numbers in the case of graphs (e.g., see [1], [5], [6], and [8]), little work has considered more general uniformities. One exception is the recent work ofÖzkahya and Young [7], where anti-Ramsey numbers for hypergraph matchings were studied. In this article, we determine the anti-Ramsey numbers for r-uniform paths of length 2, certain stars and complete r-uniform hypergraphs, and the complete 3-uniform hypergraph of order 4 missing a single hyperedge.
We begin with the relevant definitions. Suppose throughout that r ≥ 2. An r-uniform hypergraph H (r) = (V, E) consists of a nonempty set V of vertices and a set E consisting of unordered r-tuples of distinct vertices. Elements of E are called hyperedges. When we wish to specify the underlying hypergraph, we write V (H (r) ) and E(H (r) ) in place of V and E, respectively. We refer to such a hypergraph by just H (r) , including the superscript (r) to emphasize the specific uniformity being considered. If the uniformity is dropped from the notation, it should be assumed that r = 2. The complete r-uniform hypergraph K (r) n contains n vertices and every r-tuple of distinct vertices forms a hyperedge. If a single hyperedge is removed, we write K (r) n − e to denote the resulting hypergraph.
The other two types of hypergraph that we will consider fall under the category of trees when r = 2: paths and stars. Let 1 ≤ t < r. The t-tight r-uniform path of length k, denoted P (r) k,t , consists of distinct vertices x 1 , x 2 , . . . , x r+(k−1)(r−t) and hyperedges  and r = 2, we replace P k,1 with the usual notation P k . The t-tight r-uniform star of order p, denoted S  p,t consists of all hyperedges that include all vertices in C and some selection of r − t vertices from U . The set C is called the center of the star S (r) p,t . Figure 2 provides some examples of stars with varying parameters. In the special case where t = 1 and r = 2, we replace S p,1 with the usual (bipartite) notation K 1,p−1 . www.ejgta.org A k-coloring is called exact if it is surjective. Denote the set of all exact k-colorings of K If we assume that H (r) is an r-uniform hypergraph of order p that lacks isolated vertices, then for n ≥ p, the anti-Ramsey number ar n (H (r) ) is defined to be the maximum k such that some exact k-coloring in C k (K (r) n ) lacks a rainbow subhypergraph isomorphic to H (r) . Observe that every n rcoloring of K (r) n results in a rainbow copy of H (r) and |E(H (r) )| colors are required to produce a rainbow H (r) . It follows that The anti-Ramsey number ar n (H (r) ) is related to the rainbow number rb n (H (r) ), defined to be the minimum k such that every exact k-coloring of K (r) n contains a rainbow H (r) (c.f. Section 11.4 of [3]). It follows that rb n (H (r) ) = ar n (H (r) ) + 1, and the reader should be aware that the definition of a rainbow number is sometimes given as the definition of an anti-Ramsey number.
In Section 2, we prove several results for general uniformity. In particular, we show that We also provide a theorem relating anti-Ramsey numbers for r-uniform complete hypergraphs to anti-Ramsey numbers for certain (n − r)-uniform stars. This result allows us to use some known results on graphs to obtain anti-Ramsey numbers for certain hypergraphs. The section is concluded with the determination of lower bounds for the anti-Ramsey numbers ar r+2 (K (r) r+1 ). The anti-Ramsey number for K (3) 4 − e (the complete 3-uniform hypergraph of order 4 with a single hyperedge removed) is the focus of Section 3. We prove a general upper bound and give exact evaluations in the cases n = 5, 6. Section 4 concludes with a few directions for future inquiry.

General r-Uniform Results
In this section, we focus on proving a few results that are independent of uniformity. First, we consider the case of paths of length 2.
Proof. Observe that 2 colors are needed to produce a rainbow path of length 2. So, ar n (P n . Any hyperedge that includes vertices from both e 1 and e 2 must be colored red or blue, creating a nontrivial overlap between a red hyperedge and a blue hyperedge. It remains to be shown that for any 1 ≤ t ≤ r − 2, there exists a rainbow path P (=⇒) Assume there exists a t-tight rainbow path of length 2 with blue hyperedge e 1 and red hyperedge e 2 . Without loss of generality, suppose that e 1 = y 1 y 2 · · · y r−t x 1 x 2 · · · x t and e 2 = x 1 x 2 · · · x t z 1 z 2 · · · z r−t (see the first image in Figure 3). If no (t + 1)-tight rainbow path of length 2 exists, then the hyper- edge e 3 = y r−t x 1 x 2 · · · x t z 1 z 2 · · · z r−t−1 must be blue and the hyperedge e 4 = y 2 y 3 · · · y r−t x 2 x 3 · · · x t z 1 z r−t must be red (see the second image in Figure 3). However, e 3 and e 4 now form a (t + 1)tight rainbow path of length 2.
(⇐=) Assume there exists a (t + 1)-tight rainbow path of length 2 with blue hyperedge e 1 and red hyperedge e 2 . Without loss of generality, suppose that e 1 = y 1 y 2 · · · y r−t−1 x 1 x 2 · · · x t+1 and e 2 = x 1 x 2 · · · x t+1 z 1 z 2 · · · z r−t−1 (see the first image in Figure 4). Note that since n ≥ 2r − t, there exists some vertex w that is not contained in e 1 or e 2 . If no t-tight rainbow path of length 2 exists, then the hyperedge e 3 = wx 2 x 3 · · · x t+1 z 1 z 2 · · · z r−t−1 must be blue and the hyperedge e 4 = xy 1 y 2 · · · y r−t−1 x 1 x 2 · · · x t must be red (see the second image in Figure 4). However, e 3 and e 4 now form a t-tight rainbow path of length 2.
Hence, at most 1 color can be used without creating a rainbow path of any given tightness, concluding the proof.
One particularly useful trick to working with anti-Ramsey numbers of hypergraphs is to construct a natural bijection between exact k-colorings of K where the vertex sets of K n,n−m (for example, see Figure 5). Applying Theorem 2.2 to known anti-Ramsey numbers on graphs, we can obtain select results on hypergraphs. Erdős, Simonovits, and Sós [4] proved that The same principal of using induced colorings leads us to the following theorem involving complete hypergraphs.
Proof. Using the concepts introduced in Theorem 2.2, the bijection r+1 to a rainbow K 1,r+1 . One way to maximize the number of colors used in a rainbow K 1,r+1 -free coloring of K r+2 is to color edges by maximizing the number of disjoint small cycles, which are each given distinct colors. By the Division Algorithm, there exists unique q, t ∈ Z such that r + 2 = 3q + t, where 0 ≤ t ≤ 2.
When t = 0, we can partition the vertices in K r+2 into q K 3 -subgraphs, with each K 3 receiving its own color. The remaining edges each get their own color, producing a K 1,r+1 -free coloring of K r+2 using a total of colors. When t = 1, we partition the vertices into q − 1 K 3 -subgraphs and a single cycle C 4 of length four. Giving each cycle and all remaining edges distinct colors gives a total of q + r + 2 2 − (3q + 1) colors. When t = 2, we partition the vertices into q − 1 K 3 -subgraphs and a single cycle C 5 of length five. Following the same color scheme as before, we obtain a total of colors. In all three cases, we have produced a coloring of K r+2 that lacks a rainbow K 1,r+1 using q + r + 2 2 − (r + 2) colors. Observing that q = r+2 3 completes the proof of the theorem. having vertex set {x 1 , x 2 , x 3 , x 4 }, one can construct an exact k-colored K (3) k+2 that lacks a rainbow K

The Anti-Ramsey Number for K
4 − e by assigning color s (with s = 3, 4, . . . , k) to all hyperedges of the form x i x j x s+2 whenever i < s + 2 and j < s + 2. It follows that (1) The following theorem proves that this bound is exact when n = 5.
Theorem 3.1. ar 5 (K 4 − e) = 3. Proof. It remains to be shown that every exact 4-coloring of K (using say, red, blue, green, and purple) contains a K 4 spanned by hyperedges using at least 3 colors. We will prove this by contradiction. Consider a 4-coloring of K 4 . Without loss of generality, assume that it has at least as many red hyperedges as it has blue hyperedges. The remaining six hyperedges in the K 5 all contain x 5 and at least one of them is green and another is purple. If any green hyperedge has a vertex from S in common with a purple hyperedge, then a 3colored K 4 is produced. This forces there to only be a single green hyperedge and a single purple hyperedge whose intersection in S is empty. One of these hyperedges must include two vertices from {x 2 , x 3 , x 4 }. Suppose it is the green hyperedge, which is given by x 2 x 3 x 5 . Then x 1 x 4 x 5 is purple and at least one of x 1 x 2 x 4 and x 1 x 3 x 4 is red. Suppose the former case. At this point, we have Figure 6. Observe that the subhypergraph induced by {x 1 , x 2 , x 4 , x 5 } is red/purple and the subhypergraph induced by {x 2 , x 3 , x 4 , x 5 } is blue/green. This gives us a contradiction since no coloring of the hyperedge x 2 x 4 x 5 satisfies these conditions. It follows that every exact 4-coloring of K    The remaining hyperedges are those that contain x n and exactly two vertices from {x 1 , x 2 , . . . , x n−1 }. If n is even, then n−1 is odd, and adding in n 2 additional colors forces two such hyperedges to have some vertex in common from {x 1 , x 2 , . . . , x n−1 }. Without loss of generality, suppose that x 1 x 2 x n and x 2 x 3 x n have two of the additional colors. Then the subhypergraph induced by {x 1 , x 2 , x 3 , x n } contains a rainbow K colors. Similarly, If n is odd, then n−1 is even, and adding in n+1 2 additional colors forces two such hyperedges to have some vertex in common from {x 1 , x 2 , . . . , x n−1 }, forming a rainbow K 4 − e.