Small Cycle Systems Chris Rodger Auburn University

Hamilton Cycle 4-cycle

Graph Decompositions - Paths Partition the edges of your favorite graph so that each element of the partition induces something interesting. K 7 = Use colors on the edges to denote the partition. Maybe you like paths, but with variety!

Graph Decompositions - Cycles Partition the edges of your favorite graph so that each element of the partition induces something interesting. K 7 = Use colors on the edges to denote the partition. Or perhaps you like small cycles. This is a C 3 -decomposition of K 7. Steiner Triple System

Existence Problems When do 4-cycle systems exist? Clearly the number of edges must be divisible by 4 And the degree of each vertex must be even. So n must be congruent to 1 modulo Each edge has an associated “difference”.

One Construction 4-cycle decompositions are easy to make now that we have a decomposition of K 9. Let’s try K 17 Put 4-cycle decompositions of K 9 on each of these two sets of 9 vertices. Pair the vertices and place a 4-cycle between pairs in different columns.

It Solves the Existence Problem Add as many columns as you like!

Another 4-cycle system of K 17 Sometimes we need something more complicated Rotate left to right Pure Difference 4 Mixed Difference 4 What’s missing?? ∞ Mixed differences 2 and -2

Other Cycle Lengths Of course that was just for 4-cycles. The existence of m-cycle systems of order n has been solved after a long history. Clearly – the number of edges must be divisible by m – the degree of each vertex must be even, and – n must be at least m, or n = 1. Alspach, Gavlas, Šajna ( Hoffman, Lindner, Rodger )

Partial Cycle Systems A set of edge disjoint 4-cycles in K n is said to be a partial 4-cycle system of order n. This is a partial 4-cycle system of order 11 that is EQUITABLE.

Equitable Partial Cycle Systems Equitable: for each pair of vertices u and v, the number of cycles containing u differs by at most one from the number of cycles containing v. These are VERY useful! 3-cycles: Andersen, Hilton and Mendelsohn 4-cycles and 5-cycles: Raines and Staniszló Any mixture of cycle lengths! Bryant, Horsley and Maenhaut

Embeddings A 4-cycle system P of λK v is said to be embedded in a 4-cycle system Q of λK v+u if P is a sub- multiset of Q. P with λ = 1Q

We Have An Embedding Already This is an embedding of a 4-cycle system of K 9 (left vertices) into a 4-cycle system of K 17.

Embeddings - History The Lindner problem of embedding a partial 3- cycle system of order n into an STS(v) has been solved! (Bryant and Horsley) A necessary condition requires that v ≥ 2n+1 For 4-cycles the situation is messier, but recent progress has been dramatic: Necessarily v ≥ n+n 1/2 -1 Lindner had the best result of 2n+15 until recently: n /2 n 3/4 + o(n 3/4 ) (Lindner and Hilton) n + n 1/2 + o(n 1/2 )(Füredi and Lehel)

Embeddings - History Partial 3-cycle system of order n into an STS(v) v ≥ 2n+1 (Bryant, Horsley) Partial 4-cycles systems: n + n 1/2 + o(n 1/2 )(Füredi, Lehel) Partial 2k-cycle systems Around kn(Hoffman, Lindner,Rodger) Partial 2k+1-cycle systems Around (4k+2)n(Lindner, Rodger, Stinson)

Enclosings A 4-cycle system P of λK v is said to be enclosed in a 4-cycle system Q of (λ+μ)K v+u if P is a sub- multiset of Q. P with λ = 1Q with λ+μ = 2

Theorem A 4-cycle system of λK v can be enclosed in a 4-cycle system of (λ+μ)K v+u if and only if 1.(v+u-1)(λ+μ) is even, 2.The number of new edges is divisible by 4, 3.If u = 1 then μ(v-1)/2 ≥ λ + μ, and 4.If u = 2 then μv(v-1)/2 ≥ λ + μ. (Newman and Rodger)

The biggest case: u ≥ 3 So let’s move on to the more interesting cases! Use existing results on maximum partial 4-cycle systems. It is easy to use the edges that join 2 new vertices! Add at least 3 vertices

The neatest case: u = 2 Settling this case involves: Equitable partial 4-cycle systems, Directed Euler Tours, and Expanding nearly-regular graphs into copies of K 2,2. But first we check the necessary condition.

Recall: If u = 2 then μv(v-1)/2 ≥ λ + μ In here λ edges between each pair of vertices are already used. Eventually μ more edges are used between each pair of vertices in here. So μv(v-1)/2 ≥ (λ + μ) There are λ + μ edges joining the 2 added vertices. Add u = 2 vertices.

Sufficiency with u = 2: μv(v-1)/2 ≥ λ + μ In P, μ edges between each pair of vertices need to be used in 4-cycles. We just saw: there are λ + μ edges joining the 2 added vertices, each of which must be in 4-cycles like this. So exactly μv(v-1)/2 – (λ + μ) edges “must” be in 4-cycles joining vertices in P. P

Sufficiency with u = 2: μv(v-1)/2 ≥ λ + μ In P, μ edges between each pair of vertices need to be used in 4-cycles. So start with an equitable partial 4-cycle system C 1 of μK v with exactly μv(v-1)/2 – (λ + μ) edges! P It turns out that this number is divisible by 4. And it is not negative!

Sufficiency with u = 2: μv(v-1)/2 ≥ λ + μ In P, μ edges between each pair of vertices need to be used in 4-cycles. P Now look at the complement in μK v of C 1. It has exactly λ + μ edges! All vertices have even degree, so form a directed Euler tour. End Start Let C 2 be the Set of these 4-cycles. In P, μ edges between each pair of vertices ARE NOW used in 4-cycles. v

Sufficiency with u = 2: μv(v-1)/2 ≥ λ + μ P E S The remaining edges induce a bipartite graph B between P and {S, E}. Since C 1 is equitable, each vertex v in P has degree 2s or 2s+2 in B (for some s). Form a graph on V(P) in which each vertex v has degree d B (v)/2. For each edge add a 4-cycle. v Also note that s or s+1edges in B respectively join v to each of S and E.

The last necessary condition: If u = 1 then μ(v-1)/2 ≥ λ + μ In here λ edges between each pair of vertices are already used. Add λ + μ edges going to each of the v vertices. Eventually μ more edges between pairs of vertices are used in here. So μv(v-1)/2 ≥ (λ + μ)v

A typical case: v = 4x+1; μ even (so 4 | λ+μ) (v-1)/2-1 (v-1)/2 µ/ (v-1)/2-1 (v-1)/2 Select ( λ+μ)/4 of the (μ/2)(v-1)/4 cells and form cycles like these. 2i- 1 2i 0 V-1V-1 2i- 1 Recall the necessary condition: μ(v-1)/2 ≥ λ + μ ∞ is the added vertex ∞ 2i-1 2i

v = 4x+1; μ even (v-1)/2-1 (v-1)/2 2i-1 2i µ/ (v-1)/2-1 (v-1)/2 Select the remaining cells and form cycles like these. 2i- 1 2i 2i- 1 0 V -1 ∞

v ≡ 2 (mod 4) v+1 is (λ+μ)-admissible So 4 divides (λ + μ) v(v+1)/2 (v+1 is (λ+μ)-admissible) So 4 divides (λ + μ) And 4 divides λv(v-1)/2 (v is λ-admissible) So 4 divides λ. So 4 divides μ. The half difference and the 0 mixed difference are now the difficulty.

Another typical case: v ≡ 2 (mod 4); so 4 | λ,μ λ/2 differences for these We have pure and mixed differences 1,2, …, (v-2)/4, and mixed difference 0. 0 v/2 -1 ∞ s s and -s s

Another typical case: v ≡ 2 (mod 4); so 4 | λ,μ λ/2 differences for these We have pure and mixed differences 1,2, …, (v-2)/4, and mixed difference 0. μ/4 differences for these s s This uses 4 times: mixed difference 0; ∞ to both sides 0 v/2 -1 ∞

Another typical case: v ≡ 2 (mod 4); so 4 | λ,μ λ/2 differences for these We have pure and mixed differences 1,2, …, (v-2)/4, and mixed difference 0. μ/4 differences for these Use each of the remaining differences in these 4-cycles 0 v/2 -1 ∞

v ≡ 0 (mod 4) Really Tricky! Try μ ≡ 3 (mod 4) λ is even (v is λ-admissible) S o λ + μ is odd 4 divides (λ + μ) v(v+1)/2 (v+1 is (λ+μ)-admissible) S o v ≡ 0 (mod 8) μ (v-1)/2 = (4x+3)(4y+3)/2 = 4z + 1/2 λ + μ ≤ μ(v-1)/2 - 3/2 So 1.5 parallel classes must avoid ∞!

v ≡ 0 (mod 8); μ ≡ 3 (mod 4) 0 v/2 -1 ∞ 1 copy of these (using: one v/8 difference; one mixed 0 difference; ∞ to both sides) The orange edges provide the required 1.5 parallel classes That avoid ∞ This vertex structure allows us to find the 1.5 parallel classes avoiding ∞ in a natural way. v /8 v /4

Do you remember this? Sometimes we need something more complicated Rotate left to right

Other Enclosings For 3-cycle systems: The problem remains open. There are earlier results by Colbourn and Hamm, and also with Rosa (this conference in 1985!). There are several recent results by Hurd, Munson and Sarvate that consider small enclosings. One of the necessary conditions is quadratic. Enclosings do not exist in the interval: Recent result approach this gap from both sides (Newman and Rodger Nothing appears to be known for longer cycles. (v+1)(1 –(1-(4mv)/(v-1) 2 ( λ+m) 2 ) 1/2, (v+1)(1 +(1-(4mv)/(v-1) 2 ( λ+m) 2 ) 1/2

Spouse Avoiding Dinners Try to find a way for 4 couples to sit at 2 tables, each seating 4 people so that each sits next to each other person exactly once. Friday Men Women W4W1 W3W2 M1 M3M4 M2W4W3 W2W1 M3 M2M4 M1W4W2 W1W3 M2 M1M4 M3 Saturday Sunday Not the spouses!

Spouse Avoiding Dinners Try to find a way for 4 couples to sit at 2 tables, each seating 4 people so that each sits next to each other person exactly once. Friday Men Women Saturday Sunday Not the spouses! Can you do this so that each table has 2 men and 2 women?

Must one avoid one’s spouse?? Cycle systems of graphs other than K n are also interesting. Join vertices in the same group with λ 1 edges and vertices in different groups with λ 2 edges λ 1 = 2 and λ 2 = 1 Pure and Mixed Edges No! You now have an excuse for another dinner!

Cycle Systems with 2 Associate Classes K(a,p; λ 1,λ 2 ) Suppose a is even. There exists a C 4 -factorization of K(a,p; λ 1,λ 2 ) if and only if 1.4 divides ap 2.λ 1 is even, and 3.If a ≡ 2 (mod 4) then λ 2 a(p-1) ≥ λ 1. (Billington, Rodger) 12p 1 2 a λ1λ1 λ2λ2

Cycle Systems with 2 Associate Classes 12p 1 2 a K(a,p; λ 1,λ 2 ) λ1λ1 λ2λ2 Suppose a ≡ 1 (mod 4). There exists a C 4 -factorization of K(a,p; λ 1,λ 2 ) if and only if 1.4 divides p 2.λ 2 > 0 and is even, and 3.λ 2 a(p-1) ≥ λ 1, (Rodger, Tiemeyer) except possibly if a = 9 and λ 1 is odd.

What about a ≡ 3 (mod 4)? Looks difficult from my point of view!

Why must λ 2 a(p-1) ≥ λ 1 ? K(a,p; λ 1,λ 2 ) 1 2 a =6 Suppose a ≡ 2 (mod 4). Consider one C 4 -factor. Every part must contain at least 2 vertices incident with mixed edges. So each C 4 - factor must contain at least p mixed edges! 1p

4-cycle systems of K(a,p; λ 1,λ 2 ) There exists a 4-cycle system of K(a,p; λ 1,λ 2 ) if and only if 1.Each vertex has even degree, 2.The number of edges is divisible by 4, 3.If a = 2 then λ 2 > 0, and λ 1 ≤ 2(p-1) λ 2 4.If a = 3 then λ 2 > 0, and λ 1 ≤ 3(p-1) λ 2 /2 if λ 2 is even, and λ 1 ≤ 3(p-1) λ 2 /2 if λ 2 is odd. (Hung Lin Fu, Rodger) For 3-cycles: Fu, Rodger, Sarvate For block designs: Bose and Shimamoto – 1952! - (p-1)/9

Why is λ 1 ≤ 3(p-1) λ 2 /2 when a = 3? Every 4-cycle must use at least 2 mixed edges. So 3pλ 1 ≤ 9p(p-1) λ 2 /2 1p K(a,p; λ 1,λ 2 ) Each of these uses an even number of mixed edges.

Is λ 1 ≤ 3(p-1) λ 2 /2 –(p-1)/9 when λ 2 is odd? There are an odd number of edges between each pair of parts! So some 4-cycles must use at least 3 mixed edges So 3pλ 1 ≤ 9p(p-1) λ 2 /2 – (p-1)/9 1p K(a,p; λ 1,λ 2 ) Each of these uses an even number of mixed edges.

Plenty More! 4-cycle systems that cover 2-paths (Heinrich and Nonay, Cox and Rodger, Kobayashi and Nakamura) Resolvable versions (Kobayashi and Nakamura) 4-cycle systems of line graphs of K n and of line graphs of complete multipartite graphs (Rodger and Sehgal) 4-cycle systems of K n minus any graph with – maximum degree 3 – One vertex of any degree, all others of degree at most 2 (Fu, Fu and Rodger, Sehgal, Ash)

Thanks for listening!