Unsolved Problems in Graph Decompositions HungLin Fu Motivation

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

Motivation v 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. v 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. v If both V and E of G are finite, the graph G is a finite graph. G is an infinite graph otherwise. v If E contains subsets which are not 2 -element subsets, then G is a hyper-graph. v

Continued … A simple graph is a 2 -uniform hyper-graph without multi-edges. v A multi-graph is a 2 -uniform hyper-graph. v 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. v We shall use Kv to denote the complete multigraph with multiplicity , I. e. each edge occurs times. v

Graph Decomposition v v v 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 G 1, G 2, . . , 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 kelement subsets (blocks) of X such each pair of elements of X occur together in exactly blocks of B. v 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. v

Another point of view v The existence of an STS(v) is equivalent to the existence of a K 3 -decomposition of Kv, i. e. decomposing Kv into triangles.

More General v The existence of a 2 -(v, k, ) design can be obtained by finding a Kk-decomposition of Kv. v Example: 2 K 4 can be decomposed into 4 triangles (1, 2, 3), (1, 2, 4), (1, 3, 4) and (2, 3, 4). v 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 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). v A (22, {4, 7}, 1) PBD exists. v A pair of orthogonal latin squares of order 22 can be constructed from this PBD! v If

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. v 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) v If Ck | Kv, then we say a k-cycle system of order v exists. v

Known Results v Ck | Kv if and only if Kv is k-sufficient. v Let v be even and I is a 1 -factor of Kv. Then Ck | Kv – I if and only if Kv – I is ksufficient. v 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). v C 4 has an -labeling. (See it? ) So are the cycles of length 4 k. v A labeling without the secondition is called a -labeling or a graceful labeling. v

A Beautiful Result Theorem (Alex Rosa, 1966) If a graph G of size q has an -labeling, then K 2 q+1 can be decomposed into copies of G. Proof. Use difference method! v Theorem (Alex Rosa) If a graph G of size q has an -labeling, then K 2 pq+1 can be decomposed into copies of G. Proof. Now, we have p starters. v

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

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

Path Analogue v If 1 m 2 … mt n-1 and m 1 + m 2 + … + mt = n(n-1)/2, then Kn can be decomposed into t paths of lengths m 1, m 2, … , mt, respectively. v 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. ) v Other cases are unsolved in general except for some special ones. v Someone has to do something on the case when n is even!!!

Matching Analogue v If 1 m 2 … mt n/2 and m 1 + m 2 + … + mt = n(n-1)/2, then Kn can be decomposed into t matchings of sizes m 1, m 2, … , mt, respectively. v This problem has been solved recently by using an elegant lemma. (Regularizing Lemma, D. Bryant, 2007) v 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). v Now, decompose the host graph into even cycles with prescribed lengths as long as the sum of lengths is equal to n 2 or n 2 – n respectively. v There is an exception: we are not able to decompose K 4, 4 into one 8 -cycle and two 4 cycles! v

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. v For details, please check with Prof. Kau (Tamkang University). Many researchers are involved. v (4, 6)-case, (4, 8)-case, (4, 6, 8, 10)case, even cycles of distinct lengths, … v

Decomposing Complete Bipartite Graphs into Paths Can we decompose Km, n (m n) into paths with prescribed lengths? If 1 m 2 … mt 2 m-1 (or 2 m whenever m < n) and m 1+m 2 + … +mt = mn, then Km, n can be decomposed into t paths of lengths m 1, m 2, … , mt, respectively. v 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) v A more general result was obtained recently by the same group of people as above. v

2 -factorizations v. 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. v If G can be decomposed into 2 -factors, then we say that G has a 2 -factorization. v It is well known that an even regular graph has a 2 -factorization. (Petersen’s Theorem)

The Oberwolfach Problem Can we decompose K 2 n+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(2 n+1; 1, 2, 3, …, s). v Known: 1 = 2 = 3 = … = s = h(odd). (Chfactors) (Alspach et al, JCT(A) 1989) v 2 -factor with two distinct cycle lengths. v

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

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

The Hamilton-Waterloo Problem v Motivated by the Oberwolfach’s problem, the H-W problem asks for a 2 -factorization of the complete graph K 2 n+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. v 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(2 n+1; 1, 2, 3, …, s; 1, 2, 3, …, t). v Known results: 1 = 2 = 3 = … = s; 1= 2= 3= …= t. If 1 = h and 1 = k, then we denote it by “HW(2 n+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} = {2 n+1, 3}. The second one was just completely finished by J. Dinitz following earlier works of Horak et al DM(2004). v

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. v Therefore, we are looking for a 2 -factorization of K 2 n - I which consists of r Ch - factors and s Ck - factors where r + s = n - 1 and 2 n is a common multiple of h and k. v 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)}. v

An example HW*(8; 4, 8) v There are four possible cases (0, 3), (1, 2), (2, 1) and (3, 0). v Let the vertex set of K 8 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. ) v The solution for (2, 1) case is: Exercise! v How about the other two cases? Exercises!! v

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

HW*(16; 4, 8)-Problem v {(0, 3), (1, 2), (2, 1), (3, 0)} from one 1 -factor of K 4. v For the other two 1 -factors, it corresponds to the decomposition of K 4, 4 into either two C 8 factors and no C 4 -factors or two C 4 -factors or no C 8 -factors. Therefore, the combination is {(0, 3), (1, 2), (2, 1), (3, 0)} + {(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]. v 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)]. v Many more on experimental designs and Codings. v

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