Unsolved Problems in Graph Decompositions Hung-Lin Fu (傅恆霖) 國立交通大學應用數學系

Motivation The study of graph decomposition has been one of the most important topics in graph theory and also play an important role in the study of combinatorial designs. We are looking for more modern applications in recent years.

Preliminaries A graph G is an ordered pair (V,E) where V the vertex set is a nonempty set and E the edge set is a collection of subsets of V. In the collection E, a subset (an edge) is allowed to occur many times, such edges are called multi-edges. If both V and E of G are finite, the graph G is a finite graph. G is an infinite graph otherwise. If E contains subsets which are not 2-element subsets, then G is a hyper-graph.

Continued … A simple graph is a 2-uniform hyper-graph without multi-edges. A multi-graph is a 2-uniform hyper-graph. A complete simple graph on v vertices denoted by Kv is the graph (V,E) where E contains all the 2-element subsets of V. Hence, Kv has v(v-1)/2 edges. We shall use Kv to denote the complete multi-graph with multiplicity  , I.e. each edge occurs  times.

Graph Decomposition We say a graph G is decomposed into graphs in H if the edge set of G, E(G), can be partitioned into subsets such that each subset induces a graph in H. On the other hand, we may say : A decomposition of a graph G is a list of graphs G1, G2, ...., Gt such that each edge of G occurs in exactly one graph of the list. Now, H is the set of graphs in the list. For simplicity, we say that G has an H-decomposition and Gi’s are the members of the decomposition. If H = {H}, then G has an H-decomposition denoted by H|G.

Balanced Incomplete Block Designs (BIBD) A BIBD or a 2-(v,k,) design is an ordered pair (X,B) where X is a v-set and B is a collection of k-element subsets (blocks) of X such each pair of elements of X occur together in exactly  blocks of B. A Steiner triple system of order v, STS(v), is a 2-(v,3,1) design and it is well-known that an STS(v) exists iff v is congruent to 1 or 3 modulo 6.

Another point of view The existence of an STS(v) is equivalent to the existence of a K3-decomposition of Kv, i.e. decomposing Kv into triangles.

More General The existence of a 2-(v,k,) design can be obtained by finding a Kk-decomposition of Kv. Example: 2K4 can be decomposed into 4 triangles (1,2,3), (1,2,4), (1,3,4) and (2,3,4). A 2-(4,3,2) design exists and its blocks are: {1,2,3}, {1,2,4}, {1,3,4} and {2,3,4}.

Pairwise Balanced Designs If Kv can be decomposed into complete subgraphs of order in a prescribed set K, then we have a 2-(v,K,) design, also known as a (v,K,) pairwise balanced design(PBD). A (22,{4,7},1) PBD exists. A pair of orthogonal latin squares of order 22 can be constructed from this PBD!

Cycle Systems A cycle is a connected 2-regular graph. We use Ck to denote a cycle with k vertices and therefore Ck has k edges. If G can be decomposed into Ck’s, then we say G has a k-cycle system and denote it by Ck | G. Clearly, if the decomposition is possible, then we have (1) |G|  k, (2) G is an even graph and (3) |E(G)| is a multiple of k. (k-sufficient) If Ck | Kv, then we say a k-cycle system of order v exists.

Known Results Ck | Kv if and only if Kv is k-sufficient. Let v be even and I is a 1-factor of Kv. Then Ck | Kv – I if and only if Kv – I is k-sufficient. After more than 40 years effort, the above two theorems have been proved following the combining results of B. Alspach et al. (2001, JCT(B))

An Idea of Decomposition A mapping  from V(G) into {0, 1, 2, …, |E(G)|} is an -labeling if {|(u) - (v)| : uv is an edge of G} = {1, 2, 3, …, |E(G)|} and there exists a  such that for each uv in E(G), either (u)   < (v) or (v)   < (u). C4 has an -labeling. (See it?) So are the cycles of length 4k. A labeling without the second condition is called a -labeling or a graceful labeling.

A Beautiful Result Theorem (Alex Rosa, 1966) If a graph G of size q has an -labeling, then K2q+1 can be decomposed into copies of G. Proof. Use difference method! Theorem (Alex Rosa) If a graph G of size q has an -labeling, then K2pq+1 can be decomposed into copies of G. Proof. Now, we have p starters.

Alspach’s Problem It was posed by B. Alspach around 1980. If n is odd, 3  m1  m2  …  mt  n and m1 + m2 + … + mt = n(n-1)/2, then Kn can be decomposed into t cycles of lengths m1, m2, … , mt, respectively. If n is even, 3  m1  m2  …  mt  n and m1 + m2 + … + mt = n(n-1)/2 - n/2, then Kn - I can be decomposed into t cycles of lengths m1, m2, … , mt, respectively where I is a 1-factor of Kn.

Best results to date P. Balister Use closed trails instead of cycles. D. Bryant Use 2-regular graphs instead of cycles. Others: (3,4,6)-case, (3,5)-case, (4,5)-case, (n-2,n-1,n)-case, (m,n)-case, cycles of distinct lengths, … How about yours?

Path Analogue If 1  m1  m2  …  mt  n-1 and m1 + m2 + … + mt = n(n-1)/2, then Kn can be decomposed into t paths of lengths m1, m2, … , mt, respectively. If n is odd and mt  n-3, then the decomposition can be obtained without too much difficulty. (Use a special designed eulerian circuit of Kn.) Other cases are unsolved in general except for some special ones. Someone has to do something on the case when n is even!!!

Matching Analogue If 1  m1  m2  …  mt  n/2 and m1 + m2 + … + mt = n(n-1)/2, then Kn can be decomposed into t matchings of sizes m1, m2, … , mt, respectively. This problem has been solved recently by using an elegant lemma. (Regularizing Lemma, D. Bryant, 2007) This is an excellent exercise for graduate students to check if he (or she) is familiar with graph decomposition.

Bipartite Analogue Let the host graph be Kn,n whenever n is even and Kn,n – F whenever n is odd. (F is a 1-factor of Kn,n). Now, decompose the host graph into even cycles with prescribed lengths as long as the sum of lengths is equal to n2 or n2 – n respectively. There is an exception: we are not able to decompose K4,4 into one 8-cycle and two 4-cycles!

Known Results If all cycles are of length either 4 or 6, then it is an easy one. Of course, you need to have some basic idea of complete bipartite graph Kn,n, namely, representing this graph by using an nxn array. For details, please check with Prof. Kau (Tamkang University). Many researchers are involved. (4,6)-case, (4,8)-case, (4,6,8)-case, (4,6,8,10)-case, even cycles of distinct lengths, …

Decomposing Complete Bipartite Graphs into Paths Can we decompose Km,n (m  n) into paths with prescribed lengths? If 1  m1  m2  …  mt  2m-1 (or 2m whenever m < n) and m1+m2 + … +mt = mn, then Km,n can be decomposed into t paths of lengths m1, m2, … , mt, respectively. It is known that we can do it when mn = k(k+1)/2 and mi = i, i = 1,2,…,k. (Cao et al, JGT) A more general result was obtained recently by the same group of people as above.

2-factorizations A 2-factor of a graph G is a spanning 2-regular subgraph of G. If a 2-factor is connected, then it is a Hamilton cycle. If G can be decomposed into 2-factors, then we say that G has a 2-factorization. It is well known that an even regular graph has a 2-factorization. (Petersen’s Theorem)

The Oberwolfach Problem Can we decompose K2n+1 into n 2-factors each of which is isomorphic to a given 2-factor ? If the components of  are cycles of length 1, 2, 3, …, s, then the corresponding instance of the Oberwolfach problem is denoted by OP(2n+1; 1,2,3, …,s). Known: 1 = 2 = 3 = … = s = h(odd). (Ch-factors) (Alspach et al, JCT(A) 1989) 2-factor with two distinct cycle lengths.

Oberwolfach’s Conjecture It is not difficult to check that OP(9;4,5) and OP(11;3,3,5) are two exceptional cases. Conjecture : Except for the above two cases, all the others can be done.

Generalizations of the Oberwolfach’s Problem Spouse-avoiding variant: Decomposing K2n – I into isomorphic 2-factors where I a is a 1-factor of K2n. Bipartite Analogue: Decomposing Kn,n (respectively, Kn,n – I) into isomorphic 2-factors where n is even (respectively, n is odd). Multipartite Analogue, ….

The Hamilton-Waterloo Problem Motivated by the Oberwolfach’s problem, the H-W problem asks for a 2-factorization of the complete graph K2n+1 in which r of its 2-factors are isomorphic a given 2-factor Q, and s of its 2-factors are isomorphic to a given 2-factor R, with r + s = n. So, we have assignments for Q and R respectively. It is said that in Hamilton and Waterloo the size of round tables are of distinct sizes!?

Formulation of H-W Problem If the components of Q are cycles of lengths 1, 2, 3, …, s, and the components of R are cycles of lengths 1, 2, 3, …, t, then the corresponding instance of the H-W problem is denoted by HW(2n+1; 1, 2, 3, …, s; 1, 2, 3, …, t). Known results: 1 = 2 = 3 = … = s; 1= 2= 3= …= t. If 1 = h and 1 = k, then we denote it by “HW(2n+1;h,k)-problem”. It is known that we can do it if (1) {h,k}  {3,5,15} (Adams et al, G & C, 2002) and (2) {h,k} = {2n+1,3}. The second one was just completely finished by J. Dinitz following earlier works of Horak et al DM(2004).

Spouse-Avoiding Analogue: H-W* If we consider the case when one of h and k is even, then we are in H-W* problem. Therefore, we are looking for a 2-factorization of K2n - I which consists of r Ch - factors and s Ck - factors where r + s = n - 1 and 2n is a common multiple of h and k. Notice that we have to deal with the n cases (r,s)  {(0,n-1), (1,n-2), (2,n-3), …, (n-2,1), (n-1,0)}.

An example HW*(8;4,8) There are four possible cases (0,3), (1,2), (2,1) and (3,0). Let the vertex set of K8 be {0,1, …, 7}. Now, the solution for (1,2) case is: (0,2,4,6) + (1,3,5,7), (0,1,2,…, 7) and (0,3,6,1,4,7,2,5). (The 1-factor we delete is obtained from the edges of difference 4.) The solution for (2,1) case is: Exercise! How about the other two cases? Exercises!!

HW*(2n;2k,4k)-Problem Here k is at least 2 and n is a multiple of 2k. Step 1: Settle the HW*(4k;2k,4k)-problem. Step 2: Since 4k|2n, 2n = 2p•2k. Step 3: By using the 1-factorization of K2p, we can combine the results obtained in Step 1 together nicely and conclude the proof. How?

HW*(16;4,8)-Problem {(0,3),(1,2),(2,1),(3,0)} from one 1-factor of K4. For the other two 1-factors, it corresponds to the decomposition of K4,4 into either two C8-factors and no C4-factors or two C4-factors or no C8-factors. Therefore, the combination is {(0,3),(1,2),(2,1),(3,0)} + {(2,0),(0,2)} + {(2,0),(0,2)} = {(0,7),(1,6),…,(6,1),(7,0)}. (*) A + B = {a+b: a  A, b  B}.

Applications DNA library Screening Decomposing Kn into Kr x Ks’s whenever necessary conditions hold. [ Known : (r,s) = (2,3), (3,3), (4,4) and partial results for (3,4) case]. Synchronous Optical Networks Decomposing Kn into subgraphs of size at most g (grooming rate) and minimizing the sum of their orders. [Known : g = 3, 4, 5, 6, 8 and 7(almost done)]. Many more on experimental designs and Codings.

More … It is your term to get some jobs done, good luck to you and all of us. Thank you for your patience! 圖分割