1. Introduction
Suppose that we have been entrusted to color (or partition) the collection of all subsets of the set so that
the number of times each element of appears in each color class (all sets of the same color) is exactly . Such a coloring is called an factorization of . A solution for the case with 10 colors is given below.
Note that the number of times each element of appears in is . Thus, for to be factorable, it is clear that (i) must divide . In addition, a simple double counting argument shows that (ii) must divide . One may wonder if conditions (i) and (ii) are also sufficient for to be factorable. In the 18th century, Sylvester considered the case of this problem which remained open until the 1970s when Baranyai solved this 120yearold problem completely [5]. In fact, Baranyai proved a far more general result which, in particular, implies that is factorable if and only if and .
We are interested in a Sudokutype version of Baranyai’s theorem. A partial factorization of a set is a coloring of with at most colors so that the number of times each element of appears in each color class is at most . Note that a color class may be empty.
Problem 1.
Under what conditions can a partial factorization of be extended to an factorization of ?
We are given a coloring of a subset , and our task is to complete the coloring. In other words, we need to color so that the coloring of provides an factorization of . Baranyai’s theorem settles the case when .
A partial 4factorization of is given below (Here we abbreviate a set to ).
156, 248, 379, 126, 348, 579, 127, 349, 568, 124, 389, 567
148, 267, 359, 168, 279, 345, 159, 278, 346, 134, 259
128, 347, 569, 178, 249, 356, 169, 247, 358, 123
146, 239, 578, 137, 289, 456, 136, 257
129, 367, 458, 125, 368, 479, 147, 258, 369, 157
189, 246, 357, 158, 237, 469, 138, 245, 679, 139, 268
145, 236, 789, 167, 238, 459, 149, 256, 378, 135, 269, 478
It is not too difficult to extend this to the following 4factorization.
156, 248, 379, 126, 348, 579, 127, 349, 568, 124, 389, 567
148, 267, 359, 168, 279, 345, 159, 278, 346, 134, 259, 678
128, 347, 569, 178, 249, 356, 169, 247, 358, 123, 467, 589
146, 239, 578, 137, 289, 456, 136, 257, 489, 179, 235, 468
129, 367, 458, 125, 368, 479, 147, 258, 369, 157, 234, 689
189, 246, 357, 158, 237, 469, 138, 245, 679, 139, 268, 457
145, 236, 789, 167, 238, 459, 149, 256, 378, 135, 269, 478
The case of Problem 1 is closely related to completing partial Latin squares, (see Lindner’s excellent survey [16]). A special case of Problem 1 when , and the partial factorization is a 1factorization of for some , was studied by Cruse (for ) [8], Cameron [7], and Baranyai and Brouwer [6]. Baranyai and Brouwer conjectured that a 1factorization of can be extended to a 1factorization of if and only if and divides . Häggkvist and Hellgren [10] gave a beautiful proof of this conjecture. For further generalizations of HäggkvistHellgren’s result, we refer the reader to two recent papers by the author and Newman [2, 3] in which extending factorizations of to factorizations of is studied (for ).
At this point, it should be clear to the reader that the 1factorization of in the first example, can not be extended to a 1factorization of , but it can be extended to a 1factorization of .
Like most results in the literature, our primary focus is the case where (for some ). However, unlike those, here we do not require the given partial factorization to be a factorization itself. In this case, Problem 1 was settled by Rodger and Wantland over 20 years ago for [18], and recently by the author and Rodger for [4]. In this paper, we settle the cases and . The major obstacle from to stems from the natural difficulty of generalizing a graph theoretic result to hypergraphs.
Note that, in order to extend a partial factorization of to an factorization of (for ), it is clearly necessary that , . Let be the smallest such that any partial factorization of satisfying , can be extended to an factorization of . Combining the results of this paper with those of [2, 3, 4], it can be easily shown that , and .
Last but not least, we shall consider Problem 1 in the case when . In this direction, we solve a variation of the problem when we allow sets of size less than , and in our extension of the coloring we also extend the sets of size less than to sets of size .
2. Notation and Tools
A hypergraph is a pair where is a finite set called the vertex set, is the edge multiset, where every edge is itself a multisubset of . This means that not only can an edge occur multiple times in , but also each vertex can have multiple occurrences within an edge. By an edge of the form , we mean an edge in which vertex occurs times for . The total number of occurrences of a vertex among all edges of is called the degree, of in . The multiplicity of an edge in , written , is the number of repetitions of in (note that is a multiset, so an edge may appear multiple times). If is an edge in , then we abbreviate to . If are multisubsets of , then means , where the union of s is the usual union of multisets. Whenever it is not ambiguous, we drop the subscripts; for example we write and instead of and , respectively.
For , is said to be uniform if for each , and an factor in a hypergraph is a spanning regular subhypergraph. An factorization is a partition of the edge set of into factors. The hypergraph with is called a complete uniform hypergraph. A edgecoloring of is a mapping and color class of , written , is the subhypergraph of induced by the edges of color .
Let be a hypergraph, let be some finite set, and let be a surjective mapping. The map extends naturally to . For we define . Note that need not be injective, and may be a multiset. Then we define the hypergraph by taking and . We say that is an amalgamation of , and that is a detachment of . Associated with is a (number) function defined by ; to be more specific we will say that is a detachment of . Then has vertices. Note that induces a bijection between the edges of and the edges of , and that this bijection preserves the size of an edge. We adopt the convention that it preserves the color also, so that if we amalgamate or detach an edgecolored hypergraph the amalgamation or detachment preserves the same coloring on the edges. We make explicit a straightforward observation: Given , and the amalgamation is uniquely determined, but given , and the detachment is in general far from uniquely determined.
There are quite a lot of other papers on amalgamations and some highlights include [9, 11, 12, 13, 14, 15, 17, 18].
Given an edgecolored hypergraph , we are interested in finding a detachment obtained by splitting each vertex of into a prescribed number of vertices in so that (i) the degree of each vertex in each color class of is shared evenly among the subvertices in the same color class in , and (ii) the multiplicity of each edge in is shared evenly among the subvertices in . The following theorem, which is a special case of a general result in [1], guarantees the existence of such detachment (Here means ).
Theorem 2.1.
(Bahmanian [1, Theorem 4.1]) Let be a edgecolored hypergraph and let . Then there exists a detachment (possibly with multiple edges) of whose edges are all sets, with amalgamation function , being the number function associated with , such that

for each , each and ,

for distinct and with for ,
Let be the hypergraph obtained by adding a new vertex and new edges to so that
In other words, is an amalgamation of , obtained by identifying an arbitrary set of vertices in .
An immediate consequence of Theorem 2.1 is the following.
Corollary 2.2.
Let . A partial factorization of can be extended to an factorization of if and only if the new edges of can be colored so that
(1) 
Proof.
First, suppose that a partial factorization of can be extended to an factorization of . By amalgamating the new vertices of into a single vertex , we clearly obtain . The edgecoloring of (in which each color class is an factor) induces a edgecoloring in that satisfies (1).
The following observation will be quite useful throughout the paper.
Proposition 2.3.
For every with ,
(2) 
(3) 
Proof.
In order to avoid trivial cases, throughout the rest of this paper we assume that .
3. Arbitrary
If we replace every edge of a hypergraph by copies of , then we denote the new hypergraph by . For hypergraphs with the same vertex set , we define their union, written , to be the hypergraph with vertex set and edge set . For a hypergraph and , let be the hypergraph whose vertex set is and whose edge set is .
Let be an arbitrary subset of vertices in with . Then . A partial factorization of is a coloring of the edges of with at most colors so that for each color , for each vertex of (Note that has singleton edges). In the next result, we completely settle the problem of extending a partial factorization of to an factorization of . Note that here we are not only extending the coloring, but also the edges of size less than to edges of size . The case was solved in [4].
Theorem 3.1.
For with , any partial factorization of can be extended to an factorization of if and only if , , and for all ,
(4) 
(5) 
Proof.
To prove the necessity, suppose that a given partial factorization of is extended to an factorization of . For to be factorable, the two divisibility conditions are clearly necessary. By extending an edge of size () in to an edge of size in , the color of does not change, and so (4) is necessary. Since the number of edges in each color class of is exactly , the necessity of (5) is implied.
To prove the sufficiency, suppose that a partial factorization of is given, , , and that (4), (5) are satisfied. Let , and let . For , an edge of type in is an edge in containing but not containing . Note that there are edges of type in .
There is a clear onetoone correspondence between the edges of size in and the edges of type in for each . We color the edges of type in with the same color as the corresponding edge in for . By Corollary 2.2, if we can color the remaining edges of (edges of type ) so that the following condition is satisfied, then we are done.
(6) 
Let be the number of edges of type in , for . Note that for . We color the edges of type so that for ,
Since , is an integer for . The following shows that for .
Now we show that all edges of the type will be colored, or equivalently that, .
To complete the proof, we show that for . We have
∎
For a hypergraph and , let be the hypergraph whose vertex set is and whose edge set is .
Let with , and let . An edge is of type , if (for ). Let be a partial factorization of . Then a partial factorization of is said to be friendly if

the color of each edge of type 0 is the same in and , and

the number of edges of type and color is the same in and for each and each color .
We are interested in finding the conditions under which a partial factorization of can be extended to an factorization of .
Lemma 3.2.
For with , if a partial factorization of can be extended to an factorization of , then

,

,

for each , and ,

for ,
where .
It remains an open question whether these conditions are sufficient. Here we prove a weaker result.
Corollary 3.3.
Let with , and let be a partial factorization of , and assume that (N1)–(N4) are satisfied. Then there exists a friendly partial factorization of that can be extended to an factorization of .
Proof.
By eliminating all the vertices in , and shrinking the edges containing vertices in , we obtain . The rest of the proof follows from Theorem 3.1. ∎
4.
Theorem 4.1.
For , any partial factorization of can be extended to an factorization of if and only if and .
Proof.
For the necessary conditions, see the previous section. To prove the sufficiency, we need to show that the edges of can be colored with colors so that (6) is satisfied.
First we color the edges in of the form where and . We color these edges greedily so that for each and . We claim that this coloring can be done in such a way that all edges of this type are colored. Suppose by contrary that there is an edge in of the form with that can not be colored. This implies that for each either or or . Thus for each , . On the one hand, , and on the other hand, . Thus, we have
which is equivalent to . Now, we show that since and , we have , which is a contradiction, and therefore, all edges in of the form where and can be colored using the greedy approach described above.
First, note that for , both and are positive. Therefore,
Now we greedily color all the edges of the form where and , so that for each and . We show that this is possible. If by contrary, some edge with remains uncolored, then for each either or , and so . We have
which is equivalent to . Using Mathematica (Wolfram Alpha) it can be shown that this inequality does not have any real solution under the constraints . Therefore, all edges of the form where and , can be colored.
Since for each ,
we can color all the edges of the form where so that for each , there are edges of this type colored incident with for each . Note that after this coloring,
(7) 
For , let be the number of edges colored of the form where , respectively. We color the edges of the form so that there are exactly
edges of this type colored for . Since , and , is a positive integer for . We claim that all edges of the form will be colored, or equivalently, .
To complete the proof, we show that for . First note that for , . Therefore,
Combining this with (7) implies that (