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

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2 Definition: H 1 is a large hypergraph, H 2 is a smaller hyeprgraph, an H 2 -decomposition of H 1 is a partition of the edges of H 1 into induced subhypergraphs which are isomorphic to H 2. Trivial requirement: e(H 2 ) divides e(H 1 ). Goal: Find nontrivial requirements which H 1 must satisfy, and which guarantee an H 2 decomposition. Known results: Mainly in the graph-theoretic case: Explicit designs (H 2 is a small clique and H 1 is complete). Wilson’s Theorem. (H 2 is any graph, H 1 is a large complete graph). Gustavsson [91]. (H 2 is any graph, H 1 is a very large and very dense graph). Yuster [97]. (H 2 is any tree, H 1 is any Dirac graph of moderate size. Asymptotically optimal).

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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. (H 1 is a complete k-uniform hypergraph and H 2 is a smaller k-uniform hypergraph). In this talk we present a general decomposition result where H 2 is a simple hypertree. Definition of a simple hyperforest: Any two edges intersect in at most one vertex For any sequence e 1... e r of edges, either they have a common endpoint, or for some j, e j and e j+1 are vertex disjoint (r+1=1). If the creature is connected, it is called a simple hypertree. Note: 2-uniform hyperforests are forests.

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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.

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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:

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6 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. |E 1 |=m |E 2 |=m |E t |=m Let T be a k-uniform simple hypertree with t edges. Let H be a k-uniform hypergraph with tm edges which is 9k 3 t 5 n k-4/3 edge- expanding. Then H has a T-decomposition.

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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 e 1... e t and find an ordering of T such that: Each edge except the header of e 1 is a nonheader in exactly one edge. The header of e 1 appears only in e 1. For i > 1 The header of e i appears as a nonheader in some e j where j < i. Example: e 1 =(13,12,3) e 2 =(3,2,1) e 3 =(2,6,7) e 4 =(2,10,11) e 5 =(1,4,5) e 6 =(5,8,9). This ordering defines a parent-child relation where the parent of e i is e j if e j contains as a nonheader the header of e i. 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 e i in e p(i) is denoted s i. So, s 2 =3, s 3 =2, s 4 =2, s 5 =3, s 6 =3.

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8 Step 3: Ordering H. An ordering of H is called a T-homomorphic decomposition if it satisfies: |d p(i) (s i,v)=d i (1,v)| for all v and i > 1. |d i (s,v) - d i (v)/k| < tn 4/3 for all i,s,v. D i (s,v) denotes the set of edges of E i which contain v at position s. Thus, |D i (s,v)|= d i (s,v) The crucial lemma: H has a T-homomorphic decomposition. The proof is rather tricky. It uses the expansion of each E i together with Hall’s Theorem, and some probabilistic arguments.

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9 Step 4: Partitioning the edges of H into subsets which are homomorphic to T. A homomorphism between H 1 and H 2 is a mapping v(H 1 ) to v(H 2 ) which preserves edges in both directions. If it is a bijection, then it is an isomorphism. The fact that |d p(i) (s i,v)=d i (1,v)| for all v and i > 1 means that there are d i (1,v)! Different ways to take a perfect matching between D i (1,v) and D p(i) (s i,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 E i and each equivalence class is homomorphic to T. Important: The equivalence classes are not necessarily isomorphic to T. Example: e 1 =(1,2,3) e 2 =(2,4,5) e 3 =(5,6,7) It may be that some edge (a,b,c) in E 1 is matched to the edge (b,d,e) in E 2 and the edge (b,d,e) is matched in to the edge (e,f,a) in E 3. These three edges form an equivalence class S, but S is not isomorphic to T since 1 and 7 are both mapped to a.

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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 E i bad if it has a nonheader which already appears in some prior edge of S belonging to E j where j < i. Clearly: S is isomorphic to T iff S has no bad edges.

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11 Step 5: With high probability, each equivalence class has O(n k-4/3 ) bad edges Step 6: With high probability, any pair of distinct vertices appear together in O(n k-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.

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12 Concluding Remarks The proof yields a randomized polynomial time algorithm for generating the desired decomposition. The n k-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.

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