Decomposing Hypergraphs with Hypertrees

1. Decomposing Hypergraphs with Hypertrees Raphael Yuster University of Haifa - Oranim

2. Definition: H1is a large hypergraph, H2 is a smaller hyeprgraph, an H2-decomposition of H1 is a partition of the edges of H1 into induced subhypergraphs which are isomorphic to H2. Trivial requirement: e(H2) divides e(H1). Goal: Find nontrivial requirements which H1 must satisfy, and which guarantee an H2 decomposition. Known results: Mainly in the graph-theoretic case: • Explicit designs (H2 is a small clique and H1 is complete). • Wilson’s Theorem. (H2 is any graph, H1 is a large complete graph). • Gustavsson [91]. (H2 is any graph, H1 is a very large and very dense graph). • Yuster [97]. (H2 is any tree, H1 is any Dirac graph of moderate size. Asymptotically optimal).

3. There are practically no general decomposition results for hypergraphs. There are, however: • Explicit small designs (cf. Colbourn+Dinitz) • Exact Algebraic constructions (e.g. Alon) • Decomposing complete k-uniform hypergraphs into delta-systems (Lonc) • General packing and covering result of Rdl. (H1 is a complete k-uniform hypergraph and H2 is a smaller k-uniform hypergraph). In this talk we present a general decomposition result where H2is a simple hypertree. Definition of a simple hyperforest: • Any two edges intersect in at most one vertex • For any sequence e1 . . . er of edges, either they have a common endpoint, or for some j, ej and ej+1 are vertex disjoint (r+1=1). • If the creature is connected, it is called a simple hypertree. Note: 2-uniform hyperforests are forests.

4. Example of a 3-uniform simple hypertree The main result Let T be a k-uniform simple hypertree with t edges. If H is a k-uniform hypergraph with tm edges and with Then H has a T-decomposition.

5. The result is asymptotically best possible since for every T with t>1 edges, it is easy to construct a hypergraph H with tm edges and with minimum degree which does not have a T-decomposition. In fact, we are able to prove a stronger edge-expansion type result, from which the main theorem follows: A hypergraph with n=|V| vertices is called r edge-expanding if for every XV with |X| n/2, there are at least r|X| edges intersecting X and V \ X. An easy exercise: Every k-uniform hypergraph with minimum degree: is r edge-expanding. Thus, the main theorem follows from the following theorem:

6. Let T be a k-uniform simple hypertree with t edges. Let H be a k-uniform hypergraph with tm edges which is 9k3t5nk-4/3 edge- expanding. Then H has a T-decomposition. The proof has some similarities with the corresponding proof for trees, but is more difficult since hypergraphs and hypertrees are more complex objects than graphs and trees. Outline of the proof: Step 1: Partition the tm edges of H into t sets of exactly m edges each such that any edge set is still a good expander, and such that the degree of v in each set is close to 1/t of the original. |E1|=m |E2|=m |Et|=m

7. Step 2: Ordering T An ordered hypergraph is a family of ordered sets. We call the first vertex in each edge the header of the edge. An important property of simple hypertrees is the following: We can always label the edges e1 . . . et and find an ordering of T such that: • Each edge except the header of e1 is a nonheader in exactly one edge. • The header of e1 appears only in e1. • For i > 1 The header of ei appears as a nonheader in some ej where j < i. Example:e1=(13,12,3) e2=(3,2,1) e3=(2,6,7) e4=(2,10,11) e5=(1,4,5) e6=(5,8,9). This ordering defines a parent-child relation where the parent of ei is ej if ej contains as a nonheader the header of ei. In the last example: p(2)=1 p(3)=2 p(4)=2 p(5)=2 p(6)=5. The corresponding position of the header of ei in ep(i) is denoted si. So, s2=3, s3=2, s4=2, s5=3, s6=3.

8. Step 3: Ordering H. An ordering of H is called a T-homomorphic decomposition if it satisfies: • |dp(i)(si,v)=di(1,v)| for all v and i > 1. • |di(s,v) - di(v)/k| < tn4/3 for all i,s,v. Di(s,v) denotes the set of edges of Ei which contain v at position s. Thus, |Di(s,v)|= di(s,v) The crucial lemma: H has a T-homomorphic decomposition. The proof is rather tricky. It uses the expansion of each Ei together with Hall’s Theorem, and some probabilistic arguments.

9. Step 4: Partitioning the edges of H into subsets which are homomorphic to T. A homomorphism between H1 and H2 is a mapping v(H1) to v(H2) which preserves edges in both directions. If it is a bijection, then it is an isomorphism. The fact that |dp(i)(si,v)=di(1,v)| for all v and i > 1 means that there are di(1,v)! Different ways to take a perfect matching between Di(1,v) and Dp(i)(si,v) for each v and each i>1. Thus, we have a “being matched” relation defined on e(H), which is a symmetric relation, so by taking the transitive closure of this relation we get an equivalence relation, whose equivalence classes are subsets of e(H), with one edge from each Ei and each equivalence class is homomorphic to T. Important: The equivalence classes are not necessarily isomorphic to T. Example: e1=(1,2,3) e2=(2,4,5) e3=(5,6,7) It may be that some edge (a,b,c) in E1 is matched to the edge (b,d,e) in E2 and the edge (b,d,e) is matched in to the edge (e,f,a) in E3. These three edges form an equivalence class S, but S is not isomorphic to T since 1 and 7 are both mapped to a.

10. Let L* be an equivalence relation. There are exactly: different ways to create L*. We need to show that in at least one of these ways, all the equivalence classes are isomorphic to T. Note: Any sequential process for selecting the perfect matchings can get stuck! We shall therefore select each perfect matching randomly, each of the n(t-1) matchings is selected independently. One cannot prove that with positive probability all the equivalence classes are isomorphic to T. We overcome this difficulty as follows: Let S be an equivalence class. We call the unique edge of S belonging to Ei bad if it has a nonheader which already appears in some prior edge of S belonging to Ej where j < i. Clearly: S is isomorphic to T iff S has no bad edges.

11. Step 5: With high probability, each equivalence class has O(nk-4/3) bad edges Step 6: With high probability, any pair of distinct vertices appear together in O(nk-4/3) equivalence classes. The proofs of Step 5 and Step 6 involve a rather technical computation of conditional probabilities. Hence, we can pick an equivalence relation L’ satisfying the properties in Step 5 and Step 6. Step 7: The properties possessed by L’ enable us to mend L’ into an equivalence relation L with no bad edges. This mending process also involves probabilistic arguments.

12. Concluding Remarks • The proof yields a randomized polynomial time algorithm for generating the desired decomposition. • The nk-4/3 term in the required expansion can be improved, but we do not know how to eliminate the dependency on k. We conjecture, however, that the minimum degree requirement in the main theorem is of the form where c(T) is a suitable constant only depending on T.