Decomposing K-n UP into triangles

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Decomposing K

n

∪ P into triangles

Chin-Mei Fu

a

_{, Hung-Lin Fu}

b

_{, C.A. Rodger}

c

a_{Department of Mathematics, Tamkang University, Tamsui, Taipei Shien, Taiwan ROC} b_{Department of Applied Mathematics, National Chiao Tung University, Hsin Chu, Taiwan ROC} c_{Department of Discrete and Statistical Sciences, 235 Allison Lab, Auburn University AL 36849, USA}

Received 26 September 2002; received in revised form 10 February 2003; accepted 22 April 2003 In honor of Curt Lindner on the occasion of his 65th birthday

Abstract

In this paper, we extend the work on minimum coverings of Kn with triangles. We prove that when P is any forest on n vertices the multigraph G = Kn∪ P can be decomposed into triangles if and only if three trivial necessary conditions

are satis4ed: (i) each vertex in G has even degree, (ii) each vertex in P has odd degree, and (iii) the number of edges in G is a multiple of 3. This result is of particular interest because the corresponding packing problem where the leave is any forest is yet to be solved. We also consider some other families of packings, and provide a variation on a proof by Colbourn and Rosa which settled the packing problem when P is any 2-regular graph on at most n vertices.

c

1. Introduction

A Steiner triple system of order v isa pair (S; t) where S is a v-set and t is a collection of 3-element subsets of S such that each pair of elements in S occur together in a triple exactly once. It is well known that a Steiner triple system of order v, STS(v), exists if and only if v ≡ 1 or 3 (mod 6) [5]. In terms of graph decompositions, an STS(v) can be viewed as a partition of the edges of Kv, each element of which induces a triangle C3; we denote such a decomposition

by C3| Kv.

For each value of v for which there is no STS(v), results in the literature establish how close one can come by omitting some pairs from triples. A packing of a graph G with triangles is a partition of the edge set of a subgraph H of G, each element of which induces a triangle; the remainder graph of this packing, also known as the leave, is the subgraph G − H formed from G by removing the edges in H. If the remainder graph is minimum in size (that is, has the least number of edges among all possible leaves of G), then the packing is called a maximum packing. The following result is also well-known (a simple proof of this result can be found in [6], for example). A graph is said to be odd if each vertex has odd degree.

Theorem 1.1 (Hanani [4]). The remainder graph L for the maximum packings of Kn with triangles are as follows:

n

0 1 2 3 4 5

L F Ø F Ø F1 C4

(mod 6)

F is a l-factor, F1 is an odd spanning forest with n=2 + 1 edges (tripole), and C4 is a cycle of length 4.

E-mail address:rodgec1@mail.auburn.edu(C.A. Rodger).

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It is natural to ask for which subgraphs H of Kn; C3| Kn− H. When n is odd and H is 2-regular graph, the following result has been obtained by Colbourn and Rosa [1]:

Theorem 1.2. Let n be an odd positive integer. Let H be a 2-regular subgraph of Kn. If n = 9 then suppose that H = C4∪ C5. Then C3| Kn− H if and only if the number of edges in Kn− H is a multiple of 3.

For convenience, in what follows, we call a graph G 3-su9cient if each vertex of G is of even degree and 3 | |E(G)|. In the hope of determining for which H it is possible that C3| Kn−H, the following conjecture deserves to be mentioned: Conjecture 1 (Nash-Williams). Let H be a subgraph of Kn(n = 9) such that Kn− H is 3-suDcient and (H) 61

4(n − 1). Then C3| Kn− H.

The exceptional case when n = 9 is required since when H is a disjoint union of a 4-cycle and a 5-cycle there is no decomposition of Kn− H into triangles. Clearly, if the conjecture is proved to be true, then Theorem1.2 would become an easy corollary. So far, this conjecture is not yet settled.

It is also worth mentioning another problem of interest which seems possible to solve, yet still remains open.

Conjecture 2. Let H be a spanning odd forest of a complete graph of even order n. Then C3|Kn−H if and only if Kn−H is 3-suDcient.

When no STS(v) exists, a second approach to approximating one has been to use some pairs more than once. A covering of a graph G with triangles is a collection of triangles, p, such that each edge of G occurs in at least one triangle in p. So, if G(p) is the multigraph formed by joining each pair of vertices u and v with x edges if and only if p contains x triples that contain both u and v, then clearly C3|G(p). The multigraph G(p) − G is called the excess graph of G; it is

also known as the padding of the covering p of G. A covering with smallest excess graph (in size) is called a minimum covering, and these have been found as the following result describes (a simple proof of this result can be found in [6], for example).

Theorem 1.3 (Hanani [4]). The excess graph P for the minimum coverings of Kn with triangles are as follows:

n ₀ ₁ ₂ ₃ ₄ ₅

P F Ø F1 Ø F1

(mod 6)

As with the packing problem, we would like to go further and know what are the possible excess graphs of Kn. For example, if n ≡ 0(mod 6) then Theorem 1.3 shows that a 1-factor can be an excess graph of Kn. In this paper

we generalize such results by showing that the excess graph P can in fact be any spanning odd forest as long as the multigraph Kn∪ P (formed by adding the edges of P to Kn) is 3-suDcient (see Theorem2.5). Clearly for such a result

it is necessary that n be even so that Kn∪ P is 3-suDcient. This result is of particular interest in that the corresponding

packing problem (see Conjecture2) still remains unsolved. It also provides a companion result to the following theorem of Colbourn and Rosa [2]. Of course, if n is odd then one would require all vertices in P to have even degree.

Theorem 1.4 (Colbourn and Rosa [2]). Let H be any 2-regular (not necessarily spanning) graph de:ned on V (Kn). Then C3| Kn∪ H if and only if Kn∪ H is 3-su9cient.

In this paper we provide a new proof of Theorem 1.4, presented in Section3. Both proofs are similar in that they rely on Theorem 1.2, but are certainly diGerent in that the original proof considers the possible congruence classes of v in turn, whereas here the argument is based on the possible lengths of the cycles in the padding.

In what follows, we assume that all the packings and coverings are done using triangles. 2. Forest paddings

Before we prove the main results, we need a couple of lemmas. The following is easy to see, but useful enough to list separately:

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Lemma 2.1. Let G be packed with remainder graph L. If there exists a graph P de:ned on V (G) such that C3| L ∪ P,

then G can be covered with excess graph P.

Proof. The union of the 3-cycles in the given packing of G and the 3-cycles in the decomposition of L ∪ P provides the result.

The next lemma is of some interest in its own right.

Lemma 2.2. Let H be an odd tree. Then H can be decomposed into |V (H)|=2 − 1 paths of length 2 and a path of length one. Moreover, this can be done so that the end vertices of these paths are all distinct, and so that the path of length one can be any edge of the tree.

Proof. The proof is by induction on the number of vertices. Let f be the edge in the path of length one. Since all vertices have odd degree, H must have an even number 2n of vertices, so has an odd number 2n − 1 of edges. Clearly the result is true if H is a star, so we can assume that n ¿ 6 and that if R = (v1; v2; : : : ; vx) is a longest path in H then it has length at least 3. Since R is a longest path, and since v2 and vx−1 have odd degree, each of these 2 vertices is adjacent to at least two leaves in H. So at least one of the two, say v2 is adjacent to two leaves, neither of which is incident with the edge f. Deleting the two leaves incident with v2 produces an odd tree with two fewer vertices, so by induction it has a decomposition into one path of length 1 containing the edge f and the remaining edges being in paths of length 2, as described in the lemma. Adding the path of length 2 that joins the two deleted leaves in H completes the proof, since clearly these leaves cannot be the ends of any 2-path produced by induction.

Using a similar idea, the following decomposition can also be obtained. A graph is unicyclic if it contains exactly one cycle.

Lemma 2.3. Let H be a connected unicyclic odd graph. Then H can be decomposed into |V (H)|=2 paths of length two. Moreover, all the end vertices of these length two paths are distinct.

Proof. Let C =(v1; v2; : : : ; v3) be the cycle in H. Since all vertices have odd degree, for 1 6 i 6 x the vertex vi is adjacent to at least one vertex wi that is not in C. Deleting the edges in C leaves a forest with exactly x components, each of which is an odd tree. So we can apply the previous lemma to each tree with the path of length one containing the edge joining vi to wi for 1 6 i 6 x. Adding the 2-paths (wi; vi; vi+1) for 1 6 i 6 x produces the required decomposition.

Now, we can consider the excess graphs which are either unicyclic odd graphs or spanning odd forests, beginning with the easier unicyclic graphs.

Theorem 2.4. Let H be a spanning odd subgraph of Kn in which each component is unicyclic. Then C3| Kn∪ H if and only if Kn∪ H is 3-su9cient.

Proof. The necessity is obvious so we prove the suDciency. Since each vertex of H is of odd degree, Kn∪ H being

3-suDcient means that n must be even. Since each component of H is a unicyclic odd graph, H can be decomposed into paths pi for 1 6 i 6 n=2 of length two such that all the end vertices are distinct; let ai and bi be the end vertices of pi. Let L = {{ai; bi}|1 6 i 6 n=2}. Thus L is a 1-factor of Kn. Furthermore, H has n edges, so since Kn∪ H is 3-suDcient,

it follows that 3|(n

2) + n so n ≡ 0 or 2 (mod 6). Therefore, Kn has a maximum packing with leave a 1-factor, which we can assume is L. So adding to this maximum packing the triangles de4ned on the three vertices in pi for 1 6 i 6 n=2 shows that Kn∪ H can be decomposed into triangles.

We now turn to the most interesting case, where H is a forest. Since every non-trivial tree has a vertex of degree 1, it is clear that if H is a forest then for each vertex in Kn∪ H to have even degree it is necessary that n is even, that each

vertex in H is odd, and that H spans Kn.

Theorem 2.5. Let H be an odd spanning forest of Kn. Then C3| Kn∪ H if and only if Kn∪ H is 3-su9cient.

Proof. First, we consider n = 10. Suppose that H has u components. Apply Lemma 2.2 to each component in H to decompose H into paths p1; p2; : : : ; ps of length two and u paths of length one. Let the ends of pi be ai and bi for

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Fig. 1.

By Lemma 2.2, we know that n = 2s + 2u. Let L1 be the set of edges {{b2; a3}; {b3; a4}; : : : ; {bs−1; as}; {bs; a2}} and

L2 be {{ai; bi}|i = 2; 3; : : : ; s}.

Note that if s 6 1, then H is the excess graph of a minimum covering, so the result follows from Theorem 1.3. Also if s = 2 then |E(H)| = 2 + n=2, so |E(Kn∪ H)| = 2 + n2=2 which is not divisible by 3, so this case cannot arise since

Kn∪ H is 3-suDcient. So we can assume that s ¿ 3, so L1∪ L2 is an even cycle of length at least 4. Let c1 and d1 be

in a component of size greater than 2; so we can assume that d1 is the center of one of the paths of length 2, say the

2-path (a1; d1; b1).

Let be the collection of triangles obtained in the maximum packing of G = (Kn− c1) − (L1∪ L2). Since n = 10, since

Kn− c1 is in fact a complete graph of order n − 1, and since L1∪ L2 is 2-regular, by Theorem1.2the above maximum

packing has an empty leave as long as the number of edges in G=(Kn−c1)−(L1∪L2) is a multiple of 3. This is in fact the

case since the 3-suDcient graph Kn∪H =G∪L1∪L2∪H ∪{c1ci; c1dj; c1ak; c1bh| 2 6 i 6 u; 1 6 j 6 u; 1 6 k ; h 6 s} and since

(Kn∪ H) − G can be decomposed into a set of triangles " = {(c1; d1; a1); (c1; d1; b1); (c1; b2; a3); (c1; b3; a4); : : : ; (c1; bs; a2);

{a2; b2} ∪ p2; : : : ; {as; bs} ∪ ps; (c1; c2; d2); (c1; c3; d3); : : : ; (c1; cu; du)}. Now, ∪ " gives the decomposition of Kn∪ H into

triangles.

As for the case n=10, the necessary conditions imply that H can only be a spanning tree or a forest with 4 components. The latter case is a minimum covering and the decomposition is known by Theorem 1.3. In the former case we have s = 4 and the proof in the general case works here too since L1∪ L2 is a cycle of length 6 in this case, and by Theorem

1.2C3| K9− C6.

3. 2-Regular paddings

Next, we consider the situation where the excess graph is a 2-regular graph de4ned on n vertices, so n must be odd. We allow a double edge (a 2-cycle) to be a part of the 2-regular graph. Recall that the join of two graphs G1 and G2,

G1∨ G2, is the graph de4ned on V (G1) ∪ V (G2) such that E(G1∨ G2) = E(G1) ∪ E(G2) ∪ E(K|V (G1)|;|V (G2)|). Let O2 denote

the graph with two vertices and no edges.

Theorem 3.1. Let H be any 2-regular (not necessarily spanning) graph de:ned on V (Kn). Then C3|Kn∪ H if and only

if Kn∪ H is 3-su9cient.

Proof. The necessity is trivial, so we prove the suDciency. Since Kn∪ H is 3-suDcient, n must be odd. We prove the

theorem by using a similar technique to that used to prove the last theorem. We split the proof into four cases. We assume that n = 11 in the 4rst three cases, then deal with n = 11 at the end of the proof.

(i) Suppose that H contains an odd cycle D1.

Let D1= (a0; a1; a2; : : : ; a2m), and D2; D3; : : : ; Ds be the remaining cycles in H. Here, we consider a double edge as a

2-cycle. If there are an odd number of vertices in Kn that are in no cycle in H then let one of them be named ∞, and in

any case let the remaining vertices that are in no cycle in H be e1; e2; : : : ; e2x (so possibly x = 0). By arranging the cycles

as in Fig.1, we can convert the covering problem of Kn to the packing problem of Kn−2where we can use Theorem1.2.

First, we de4ne the triangles between {a1; a3} and {a0; a2; ai; ej| 4 6 i 6 2m; 1 6 j 6 2x}. Let =

{(a1; a0; a2m); (a1; a2; a3); (a0; a1; a2); (a2; a3; a4); (ao; a3; a4); (a1; a2i+2; a2i+3); (a3; a2i+3; a2i+4) | i = 1; 2; : : : ; m − 2}. If

x = 0 then replace the 3-cycle (a0; a3; a4) in with the 3-cycles in {(a3; a4; e1); (a3; e2i; e2i+1); (a3; e2x; a0);

(a1; e2j−1; e2j)|1 6 i 6 x − 1; 1 6 j 6 x}. Notice that partitions into triangles the edges in (K2 ∨ D1) ∪

(a0; a2; a4; e1; e2; : : : ; e2x) (the edges in the cycle (a0; a2; a4; e1; e2; : : : ; e2x) form part of the leave in Kn−2 when we

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Now we consider the edges between {a1; a3} and V (Di); i = 2; 3; : : : ; s. If V (Di) = {b1; b2; : : : ; bt} where t is even, then let "i = {(a1; b2i−1; b2i); (a3; b2i; b2i+1)|1 6 i 6 t=2}. Clearly, "i partitions O2 ∨ Di in this case. Next, pair up all except possibly one, say D2 of the odd cycles. If V (Di) = {b1; b2; : : : ; b2l+1} is paired with Dj = {c1; c2; : : : ; c2h+1} then de4ne )i;j= {(a1; b2i−1; b2i); (a3; b2i; b2i+1); (a1; c2j; c2j+1); (a3; c2j−1; c2j)|i = 1; 2; : : : ; l; j = 1; 2; : : : ; h} ∪ {(a1; b2l+1; c1); (a3; c2h+1; b1)} ∪ {(b1; b3; b2l+1); (c1; c3; c2h+1)}. Then )i;j partitions the edge set of {O2 ∨ (Di ∪ Dj)} ∪ {(b1; b3; b2l+1; c1; c3; c2h+1)}. Note that as in previous case, (b1; b3; b2l+1; c1; c3; c2h+1) will become a part of the leave in Kn−2 when Theorem1.2is applied.

If we have an even number of odd cycles, then one, say D2, remains unpaired. Clearly H is of even size in

this case, so there exists the vertex ∞ in V (Kn) which is not in V (H). Let D2 = (d1; d2; : : : ; d2k+1). De4ne * =

{(a1; d2i−1; d2i); (a3; d2i; d2i+1); (a1; d2k+1; ∞); (a3; ∞; d1); (d1; d3; d2k+1) | i = 1; 2; : : : ; k}. Then * partitions (O2 ∨ D2) ∪

{(d1; d3; d2k+1; ∞)}. Again, (d1; d3; d2k+1; ∞) will become part of the remainder graph in Kn−2.

In each subcase we are able to place the edges in the cycles D1 to Ds into triangles, as well as the edges between

{a1; a3} and V (Kn)\{a1; a3}. This implies that after taking away all the edges in triangles de4ned by ; "i; )i;j and * as

the case may be, the remaining edges induce the graph Kn−2− H where H is a 2-regular graph de4ned on V (Kn−2).

(More precisely, the components of H _{are a cycle of length 2x + 3, one 6-cycle for each of the paired odd length cycles}

in H, and a 4-cycle if H has an even number of odd length cycles.) Since 3 E(Kn∪H)| and since Kn−2−His obtained

by taking away edges in triangles from Kn∪ H, it follows that Kn−2− H is also 3-suDcient. By Theorem1.2, the proof

follows since n = 11.

(ii) Suppose that H contains only even cycles and that |V (H)| = n − 1. Let the vertex in V (Kn)\V (H) be ∞, and the

cycles in H be Di= (b(i)1 ; b(i)2 ; : : : ; b(i)f(i)), i = 1; 2; : : : ; s. Since f(i) is even for each i, by deleting the edges in the triangles

in {(∞; b(i)

2j−1; b(i)2j)|1 6 j 6 f(i)=2}, the multi-graph that remains is the union of Kn−1 and a 1-factor. By Theorem 1.3

the edges in this graph can be partitioned into triangles, so the proof in this case follows.

(iii) Suppose that H contains only even cycles and that |V (H)| ¡ n−2. Since n is odd, |V (H)| 6 n−3 and n−|V (H)| is odd.

First, if n − |V (H)| = 3 then let the vertices in {a1; a2; a3} be the vertices in Kn that are not in H. Since Kn∪

H is 3-suDcient it immediately follows that n must be equivalent to 5 (mod 6). So, by Theorem 1.1 there exists a packing T of the graph formed from Kn by deleting vertices in {a1; a2; a3} and then deleting the edges in a 1-factor

F1. Also, H has a 1-factorization (since all cycles in H have even length) into 2 1-factors F2 and F3. Then R ∪

{(ai; b; c) | 1 6 i 6 3; {b; c} ∈ Fi} provides the required covering.

Otherwise, since 3 does not divide n − 5 + n(n − 1)=2, we can assume that n − |V (H)| ¿ 7. Let {a1; a2; : : : ; a2t+1} be

the set of vertices which are not in V (H) and the set of cycles in H be D1; D2; : : : ; Ds, where Di= (bi1; bi2; : : : ; bif(i)),

i=1; 2; : : : ; s. Now, let T1={(a1; a2; a3); (a1; a4; a5); (a1; a2i; a2i+1); (a2; a2i−1; a2i); (a2; a2t+1; a4) | 3 6 i 6 t}, and for 1 6 i 6 s

let T2(i)={(a1; bi2j−1; bi2j); (a2; bi2j; bi2j+1) | 1 6 j 6 f(i)=2}. Removing the edges from these s+1 sets from Kn∪H produces

the graph Kn−2− (a3; a4; : : : ; a2t+1) which must be 3-suDcient since it is produced from a 3-suDcient graph by removing

edges in triangles; so by Theorem1.2 it can be decomposed into triangles.

(iv) Finally, we handle the case when n = 11. Since Kn∪ H is 3-suDcient, |E(H)| = 2 (mod 3). It suDces to consider

all cases where H contains no 3-cycles, since 3-cycles can always be added to the padding. If |E(H)| = 2, then the padding is a double edge, so this case is handled by Theorem 1.3.

The remaining cases are all handled in the same fashion: a packing with an appropriate leave L is obtained, usually using Theorem1.1; then by appropriately naming the vertices in L and H, Lemma 2.1is used to complete the decomposition. We describe one case in detail, then just give the ingredients for the remaining cases, it being a simple matter to check the details.

If H is a 5-cycle, say (1,2,3,4,5), then using Theorem 1.1take a packing of K11with leave the 4-cycle (1,3,5,6). Then

{(1; 2; 3); (3; 4; 5); (1; 3; 5); (1; 5; 6)} gives the required decomposition of L ∪ H that can be used in Lemma2.1. The following table gives more possible paddings H together with the leave L that can be used to obtain the result.

H C8 C4⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ C4 C2 C6 C11 C9 C2 C7 C4 C6 C5 L C4 C5 C5 C4 C3 C7 C6 C4 C5 C5 C3 C4

Finally, we have the cases H ∼= C2 ∪ C5∪ C4, C2 ∪ C2∪ C2∪ C5, and C2 ∪ C2∪ C7 to consider. Each of these

coverings can be obtained using the construction described in case (i). This follows since each of these paddings has exactly one odd length cycle. So when the construction requires Theorem 1.2to be applied, in each case a packing of K9 that has a 3-cycle leave is needed, and this clearly exists.

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4. Concluding remarks

There are other excess graphs P for which Kn∪ P can be decomposed into triangles. For example, if n = m · r with

m; r ¿ 2 C3| Kn∪ P provided that P is a vertex disjoint union of m Krs and Kn∪ P is 3-suDcient. This decomposition is also known as a group divisible design with index (2,1), see [3]. Unfortunately, being 3-suDcient is not a suDcient condition for Kn∪P to have a triangle decomposition. For example, it is not diDcult to see that K6m∪K3m;3mis 3-suDcient

provided that m is odd, but K6m∪ K3m;3m cannot be decomposed into triangles. (An additional necessary condition is that

for all partitions {S; T} of the n vertices, the number of edges joining vertices in diGerent parts must be at most half the number of edges joining vertices in the same part.) Therefore, to determine which graphs P are the excess graph for a C3-decomposition of Kn∪ P is an interesting problem, one that is not going to be easy to solve.

References

[1] C.J. Colbourn, A. Rosa, Quadratic leaves of maximal partial triple systems, Graphs Combin. 2 (1986) 317–337. [2] C.J. Colbourn, A. Rosa, Quadratic excesses of coverings by triples, Ars Combin. 24 (1987) 23–30.

[3] Hung-Lin Fu, C.A. Rodger, Group divisible designs with two associate classes: n = 2 or m = 2, J. Combin. Theory A 83 (1998) 94–117.

[4] H. Hanani, Balanced incomplete block designs and related designs, Discrete Math. 11 (1975) 255–369. [5] Rev. T. Kirkman, On a problem in combinations, Cambridge Dublin Math. J. 2 (1847) 191–204. [6] C.C. Lindner, C.A. Rodger, Design Theory, CRC Press, Boca Raton, FL, 1997.