1

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 complete).
is a small clique and H
1
• Wilson’s Theorem. ( H
2 large complete graph).
is any graph, H
1 is a
• Gustavsson [91]. ( H
2 is any graph, very large and very dense graph).
H
1 is a
• Yuster [97]. ( H
2 is any tree, H
1 is any Dirac graph of moderate size.
Asymptotically
2 optimal).

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 is a smaller k -uniform hypergraph).
H
2
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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Example of a 3-uniform simple hypertree
Let T be a k -uniform simple hypertree with t edges. If H is a k -uniform hypergraph with tm edges and with

( H )

2 k

1 n k

1
( k

1 )!
( 1
 o ( 1 ))
Then H has a T -decomposition.
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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

( H )



 n / 2
 k

1

1



1 which does not have a T -decomposition.
In fact, we are able to prove a stronger edgeexpansion 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:



  k
 is r edge-expanding.
1

1




( k

1 ) r
Thus, the main theorem follows from the following theorem:
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Let T be a k -uniform simple hypertree with t edges. Let H be a k -uniform hypergraph with tm edges which is 9 k 3 t 5 n k-4/3 edgeexpanding. 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.
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.
1
2
t
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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 find an ordering of T such that:
. . . e t and
• Each edge except the header of e
1 nonheader in exactly one edge.
is a
•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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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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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 edge ( b,d,e ) in E
2 is matched to the 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 .

Let L * be an equivalence relation. There are exactly: i t 

2 v

V d i
( 1 , v ) 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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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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• 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


 n / 2
 k

1

1


 c ( T ) where c ( T ) is a suitable constant only depending on T .
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