The Doyen-Wilson theorem for 3-sun systems††thanks: Supported
by P.R.I.N. and I.N.D.A.M.
(G.N.S.A.G.A.)
**Giovanni Lo Faro
**Dipartimento di Scienze Matematiche e Informatiche,
Scienze Fisiche e Scienze della Terra
Università di Messina, Messina, Italia
email:
[email protected]
**Antoinette Tripodi
**Dipartimento di Scienze Matematiche e Informatiche,
Scienze Fisiche e Scienze della Terra
Università di Messina, Messina, Italia
email:
[email protected]
Abstract
A solution to the existence problem of G-designs with given subdesigns is known when G is a triangle with p=0,1, or 2 disjoint pendent edges: for p=0, it is due to Doyen and Wilson, the first to pose such a problem for Steiner triple systems; for p=1 and p=2, the corresponding designs are kite systems and bull designs, respectively. Here, a complete solution to the problem is given in the remaining case where G is a 3-sun, i.e. a graph on six vertices consisting of a triangle with three pendent edges which form a 1-factor.
Keywords: 3-sun systems;
embedding; difference set.
Mathematics Subject
Classification(2000): 05B05, 05B30.
1 Introduction
If G is a graph, then let V(G) and E(G) be the vertex-set and edge-set of G, respectively. The graph Kn denotes the complete graph on n vertices. The graph Km∖Kn has vertex-set V(Km) containing a distinguished subset H of size n; the edge-set of Km∖Kn is E(Km) but with the (2n) edges between the n distinguished vertices of H removed. This graph is sometimes referred to as a complete graph of order m with a hole of size n.
Let G and Γ be finite graphs. A G-design of Γ is a pair (X,B) where X=V(Γ) and B is a collection of isomorphic
copies of G (blocks), whose edges
partition E(Γ). If Γ=Kn, then we refer to such a design as a G-design of order n.
A G-design (X1,B1) of order n is said to be embedded in a
G-design (X2,B2) of order m provided X1⊆X2 and B1⊆B2 (we also say that (X1,B1) is a
subdesign (or subsystem) of (X2,B2) or (X2,B2) contains (X1,B1) as subdesign). Let N(G) denote the
set of integers n such that there exists a G-design of order
n. A natural question to ask is: given n,m∈N(G), with
m>n, and a G-design (X,B) of order n, does exists a
G-design of order m containing (X,B) as subdesign? Doyen
and Wilson were the first to pose this problem for G=K3
(Steiner triple systems) and in 1973 they showed that given
n,m∈N(K3)={v≡1,3(mod6)}, then any
Steiner triple system of order n can be embedded in a Steiner
triple system of order m if and only if m≥2n+1 or m=n
(see [5]). Over the years, any such problem has come to be
called a “Doyen-Wilson problem” and any solution a
“Doyen-Wilson type theorem”. The work along these lines is
extensive ([3], [6]-[10], [12], [13], [15]) and the interested
reader is referred to [4] for a history of this problem.
In particular, taking into consideration the case where G is a triangle with p=0,1,2, or 3 mutually disjoint pendent edges, a solution to the Doyen-Wilson problem is known when p=0 (Steiner triple systems, [5]), p=1 (kite systems, [12, 13]) and p=2 (bull designs, [6]). Here, we deal with the remaining case (p=3)
where G is a 3-sun, i.e. a graph on six vertices consisting of a triangle with three pendent edges which form a 1-factor, by giving a complete solution to the Doyen-Wilson problem for G-designs where G is a 3-sun (3-sun systems).
2 Notation and basic lemmas
The 3-sun consisting of the triangle (a,b,c) and the three disjoint pendent edges {a,d},{b,e},{c,f} is denoted by (a,b,c;d,e,f). A 3-sun system of order n (briefly, 3SS(n)) exsits if and only if n≡0,1,4,9(mod12) and if (X,S) is a 3SS(n), then ∣S∣=12n(n−1) (see [16]).
Let n,m≡0,1,4,9(mod12), with m=u+n, u≥0. The Doyen-Wilson problem for 3-sun systems is equivalent to the existence problem of decompositions of Ku+n∖Kn into 3-suns.
Let r and s be integers with r<s, define [r,s]={r,r+1,...,s} and [s,r]=∅. Let Zu=[0,u−1] and H={∞1,∞2,…,∞t}, H∩Zu=∅. If S=(a,b,c;d,e,f) is a 3-sun whose vertices belong to Zu∪H and i∈Zu, let S+i=(a+i,b+i,c+i;d+i,e+i,f+i), where the sums are modulo u and ∞+i=∞, for every ∞∈H. The set (S)={S+i:i∈Zu} is called the
orbit of S under Zu and S is a base block of (S).
To solve the Doyen-Wilson problem for 3-sun systems we use the difference method (see [11], [14]). For every pair of distinct elements i,j∈Zu, define ∣i−j∣u= min{∣i−j∣,u−∣i−j∣} and set Du={∣i−j∣u:i,j∈Zu}={1,2,…,⌊2u⌋}. The elements of Du are called differences of Zu.
For
any d∈Du, d=2u, we can form a single
2-factor {{i,d+i}:i∈Zu}, while if u is even and d=2u, then we can form a 1-factor {{i,i+2u}:0≤i≤2u−1}. It is also worth remarking that
2-factors obtained from distinct differences are disjoint from
each other and from the 1-factor.
If D⊆Du, denote by ⟨Zu∪H,D⟩
the graph with vertex-set V=Zu∪H and the edge-set E={{i,j}:∣i−j∣u=d, d∈D}∪{{∞,i}:∞∈H, i∈Zu}. The graph ⟨Zu∪H,Du⟩ is the complete graph Ku+t∖Kt based on Zu∪H and having H as hole. The elements of H are called infinity points.
Let X be a set of size n≡0,1,4,9(mod12). The aim of the paper is to decompose the graph ⟨Zu∪X,Du⟩ into 3-suns. To obtain our main result the ⟨Zu∪X,Du⟩ will be regarded as a union of suitable edge-disjoint subgraphs of type ⟨Zu∪H,D⟩ (where H⊆X may be empty, while D⊆Du is always non empty) and then each subgraph will be decomposed into 3-suns by using the lemmas given in this section.
From here on suppose u≡0,1,3,4,5,7,8, 9,11(mod12).
Lemmas 2.1 - 2.4 give decompositions of subgraphs of type ⟨Zu∪H,D⟩ where D contains particular differences, more precisely, D={2}, D={2,4} or D={1, 3u}.
Lemma 2.1
Let u≡0(mod4), u≥8. Then the graph ⟨Zu∪{∞1,∞2}, {2}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
for i=0,1,…,4u−1. □
Lemma 2.2
Let u≡0(mod12). Then the graph ⟨Zu∪{∞1,∞2,∞3, ∞4},{2}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
for i=0,1,…,12u−1. □
Lemma 2.3
The graph ⟨Zu∪{∞1,∞2,∞3, ∞4},{2,4}⟩, u≥7, u=8, can be decomposed into 3-suns.
Proof. Let u=4k+r, with r=0,1,3, and consider the 3-suns
[TABLE]
for i=0,1,…,k−3, k≥3, plus the following blocks as the case may be.
If r=0,
[TABLE]
If r=1,
[TABLE]
If r=3,
[TABLE]
With regard to the difference 4 in Z7, note that ∣4∣7=3 and the seven distinct blocks obtained for k=1 and r=3 gives a decomposition of ⟨Z7∪{∞1,∞2,∞3, ∞4},{2,3}⟩ into 3-suns.
□
Lemma 2.4
Let u≡0(mod3), u≥12. Then the graph ⟨Zu∪{∞1,∞2, …,∞8},{1, 3u}⟩
can be decomposed into 3-suns.
Proof.
If u≡0(mod6) consider the 3-suns:
[TABLE]
If u≡3(mod6) consider the 3-suns:
[TABLE]
□
Lemmas 2.5 - 2.9 allow to decompose ⟨Zu∪H,D⟩ where u is even and D contains the difference 2u.
Lemma 2.5
Let u be even, u≥8. Then the graph ⟨Zu∪{∞1,∞2,∞3},{1, 2u}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
□
Lemma 2.6
Let u≡0(mod12). Then the graph ⟨Zu∪{∞1,∞2,∞3,∞4}, {1,2u}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
for i=0,1,…,12u−1. □
Lemma 2.7
Let u be even, u≥8. Then the graph ⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
□
Lemma 2.8
Let u≡0(mod12). Then the graph ⟨Zu∪{∞1, ∞2,…,∞7},{1,2u}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
for i=0,1,…,12u−1. □
Lemma 2.9
Let u≡0(mod12). Then the graph ⟨Zu∪{∞1,∞2,∞3}, {1,2,2u}⟩
can be decomposed into 3-suns.
Proof. Consider the 3-suns
[TABLE]
for i=0,1,…,12u−1. □
The following lemma “ combines ” one infinity point with one difference d=2u,3u such that gcd(u,d)u≡0(mod3) (therefore, u≡0(mod3)).
Lemma 2.10
Let u≡0(mod3) and d∈Du∖{2u,3u} such that
p=gcd(u,d)u≡0(mod3). Then the graph ⟨Zu∪{∞},{d}⟩
can be decomposed into 3-suns.
Proof. The subgraph ⟨Zu,{d}⟩ can be decomposed into pu cycles of length p=3q, q≥2.
If q>2, let
(x1,x2,…,x3q) be a such cycle and consider the 3-suns
[TABLE]
for i=0,1,…,q−1 (where the sum is modulo 3q).
If q=2, let (x1(j),x2(j),x3(j),x4(j),x5(j),x6(j)), j=0,1,…,6u−1, be the 6-cycles decomposing ⟨Zu,{d}⟩ and consider the 3-suns
[TABLE]
for j=0,1,…,6u−1 (where the sums are modulo 6u). □
Subsequent Lemmas 2.14 - 2.11 allow to decompose ⟨Zu∪H,D⟩, where ∣H∣=1,2,3,5, ∣D∣=6−∣H∣ and 2u∈D; here, u and D are any with the unique condition that if D contains at least three differences d1,d2,d3, then d3=d2−d1 or d1+d2+d3=u.
Lemma 2.11
Let d1,d2,d3,d4,d5∈Du∖{2u} such that
d3=d2−d1 or d1+d2+d3=u. Then the graph ⟨Zu∪{∞},{d1,d2,d3,d4,d5}⟩ can be decomposed into 3-suns.
Proof. If d3=d2−d1, consider the orbit of (d1,d2,0;∞,d2+d5,d4)
(or (d1,d2,0;∞,d2+d5,−d4), if
d2+d5=d4) under Zu. If d1+d2+d3=u, consider the orbit of (−d1,d2,0;∞,d2+d5,d4) (or (−d1,d2,0;∞,d2+d5,−d4), if
d2+d5=d4) under Zu. □
Lemma 2.12
Let d1,d2,d3,d4∈Du∖{2u} such that
d3=d2−d1 or d1+d2+d3=u. Then the graph ⟨Zu∪{∞1,∞2},{d1,d2,d3,d4}⟩ can be decomposed into 3-suns.
Proof. Consider the orbit of (d1,d2,0;∞1,∞2,d4) or (−d1,d2,0;∞1, ∞2,d4) under Zu when, respectively, d3=d2−d1 or d1+d2+d3=u. □
Lemma 2.13
Let d1,d2,d3∈Du∖{2u} such that
d3=d2−d1 or d1+d2+d3=u. Then the graph ⟨Zu∪{∞1,∞2,∞3},{d1,d2,d3}⟩ can be decomposed into 3-suns.
Proof. Consider the orbit of (d1,d2,0;∞1,∞2,∞3) or (−d1,d2,0;∞1, ∞2,∞3) under Zu when, respectively, d3=d2−d1 or d1+d2+d3=u. □
Lemma 2.14
Let d∈Du∖{2u}, the graph ⟨Zu∪{∞1,∞2, ∞3,∞4,∞5},{d}⟩
can be decomposed into 3-suns.
Proof. The subgraph ⟨Zu,{d}⟩ is regular
of degree 2 and so can be decomposed into l-cycles, l≥3. Let
(x1,x2,…,xl) be a such cycle. Put l=3q+r, with r=0,1,2, and consider the 3-suns with the sums modulo l
[TABLE]
for i=0,1,…,q−2, q≥2, plus the following blocks as the case may be.
If r=0,
[TABLE]
If r=1,
[TABLE]
If r=2,
[TABLE]
□
Finally, after settling the infinity points by using the above lemmas, if u is large we need to decompose the subgraph ⟨Zu,L⟩, where L is the set of the differences unused (difference leave). Since by applying Lemmas 2.1-2.13 it could be necessary to use the differences 1, 2 or 4, while Lemma 2.14 does not impose any restriction, it is possible to combine infinity points and differences in such a way that the difference leave L is the set of the “ small ” differences, where 1, 2 or 4 could possibly be avoided.
Lemma 2.15
Let α∈{0,4,8} and u, s be positive integers such that u>12s+α. Then there exists a decomposition of ⟨Zu,L⟩ into 3-suns, where:
α=0* and L=[1,6s];*
α=4* and L=[3,6s+2];*
α=8* and L=[3,6s+4]∖{4,6s+3}.*
Proof.
Consider the orbits (Sj) under Zu, where
Sj=(5s+1+j, 5s−j, 0;3s, s,u−2−2j), j=0,1,…,s−1.
Consider the orbits in i), where (S0) is replaced with the orbit of (6s+1, 4s, 0;s, 9s,6s+2).
Consider the orbits in i), where the orbits (S0) and (S1) are replaced with the orbits of (6s+1, 4s, 0;s, 9s,6s+4) and (5s+2, 5s−1, 0;3s, s,6s+2). □
3 The main result
Let (X,S) be a 3-sun system of order n and m≡0,1,4,9(mod12).
Lemma 3.1
If (X,S) is embedded in a 3-sun system of order
m>n, then m≥57n+1.
Proof. Suppose (X,S) embedded in (X′,S′), with
∣X′∣=m=n+u (u positive integer). Let ci be the number of 3-suns of S′ each of which contains
exactly i edges in X′∖X. Then ∑i=16i×ci=(2u) and ∑i=15(6−i)ci=u×n, from which it follows
6c2+12c3+18c4+24c5+30c6=2u(5u−2n−5) and so u≥52n+1 and m≥57n+1. □
By previous Lemma:
-
if n=60k+5r, r=0,5,8,9, then m≥84k+7r+1;
2. 2.
if n=60k+5r+1, r=0,3,4,7, then m≥84k+7r+3;
3. 3.
if n=60k+5r+2, r=2,7,10,11, then m≥84k+7r+4;
4. 4.
if n=60k+5r+3, r=2,5,6,9, then m≥84k+7r+6;
5. 5.
if n=60k+5r+4, r=0,1,4,9, then m≥84k+7r+7.
In order to prove that the necessary conditions for embedding a 3-sun system (X,S) of order n in a 3-sun system of order m=n+u, u>0 are also sufficient, the graph ⟨Zu∪X,Du⟩ will be expressed as a union of edge-disjoint subgraphs ⟨Zu∪X,Du⟩=⟨Zu∪X,D⟩∪⟨Zu,L⟩, where L=Du∖D is the difference leave, and ⟨Zu∪X,D⟩ (if necessary, expressed itself as a union of subgrapphs) will be decomposed by using Lemmas 2.1-2.14, while if L=∅, ⟨Zu,L⟩ will be decomposed by Lemma 2.15. To obtain our main result we will distiguish the five cases 1.−5. listed before by giving a general proof for any k≥0 with the exception of a few cases for k=0, which will be indicated by a star ⋆ and solved in Appendix. Finally, note that:
u≡0,1,4, or 9(mod12), if n≡0(mod12);
u≡0,3,8, or 11(mod12), if n≡1(mod12);
u≡0,5,8, or 9(mod12), n≡4(mod12);
u≡0,3,4, or 7(mod12), if n≡9(mod12).
Proposition 3.1
For any n=60k+5r, r=0,5,8,9, there exists a decomposition of Kn+u∖Kn into 3-suns for every admissible u≥24k+2r+1.
Proof. Let X={∞1,∞2,…, ∞60k+5r}, r=0,5,8,9, and u=24k+2r+1+h, with h≥0. Set h=12s+l, 0≤l≤11 (l depends on r), and distinguish the following cases.
Case 1: r=0,5,8,9 and l=0 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪X,D⟩∪⟨Zu,L⟩, where D=[6s+1,12k+r+6s], ∣D∣=12k+r, and L=[1,6s], and apply Lemmas 2.14 and 2.15.
Case 2: r=0,9 and l=8 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{2,6s+3,6s+5}⟩∪⟨Zu∪{∞4},{1}⟩∪⟨Zu∪{∞5},{6s+4}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+4], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.13, 2.10, 2.14 and 2.15.
Case 3: r=5,8 and l=4 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{2,4}⟩∪⟨Zu∪{∞5},{1}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+r+6s+2]∖{6s+4}, ∣D′∣=12k+r−1, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.3, 2.10, 2.14 and 2.15.
Case 4: r=0,8 and l=3 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4,∞5},{2}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+r+6s+1], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.5, 2.1, 2.14 and 2.15.
Case 5: r=0 and l=11 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2,2u}⟩∪⟨Zu∪{∞4,∞5},{4,6s+3,6s+5,6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+6s+5]∖{6s+7}, ∣D′∣=12k−1, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.9, 2.12, 2.14 and 2.15.
Case 6: r=5 and l=1 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7,∞8,∞9, ∞10},{2}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞10}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+5], ∣D′∣=12k+3, and L=[3,6s+2], and apply Lemmas 2.7, 2.2, 2.14 and 2.15.
Case 7: r=5,9 and l=9 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4,∞5},{2,6s+3,6s+4,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+4], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.5, 2.12, 2.14 and 2.15.
Case 8: r=8 and l=7 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2,2u}⟩∪⟨Zu∪{∞4},{4}⟩∪⟨Zu∪{∞5},{6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+11]∖{6s+4,6s+5}, ∣D′∣=12k+7, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.9, 2.10, 2.14 and 2.15.
Case 9: r=9 and l=5 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4},{2}⟩∪⟨Zu∪{∞5},{4}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4,∞5}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+11]∖{6s+4}, ∣D′∣=12k+8, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.5, 2.10, 2.14 and 2.15. □
Proposition 3.2
For any n=60k+5r+1, r=0,3,4,7, there exists a decomposition of Kn+u∖Kn into 3-suns for every admissible u≥24k+2r+2.
Proof. Let X={∞1,∞2,…, ∞60k+5r+1}, r=0,3,4,7,
and u=24k+2r+2+h, with h≥0. Set h=12s+l, 0≤l≤11, and distinguish the following cases.
Case 1: r=0,3 and l=1 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞},{6s+2}⟩∪⟨Zu∪(X∖{∞}),D′⟩∪⟨Zu,L⟩, where D′=[6s+1,12k+r+6s+1]∖{6s+2}, ∣D′∣=12k+r, and L=[1,6s], and apply Lemmas 2.10, 2.14 and 2.15.
Case 2: r=0,3,4,7 and l=9 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,6s+3,6s+4}⟩∪⟨Zu∪{∞4,∞5, ∞6},{2,6s+5,6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+5]∖{6s+7}, ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.13, 2.14 and 2.15.
Case 3: r=4⋆,7 and l=5 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{2,4}⟩∪⟨Zu∪{∞5},{1}⟩∪⟨Zu∪{∞6},{6s+8}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+r+6s+3]∖{6s+4,6s+8}, ∣D′∣=12k+r−1, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.3, 2.10, 2.14 and 2.15.
Case 4: r=0,4 and l=6 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4,∞5,∞6},{2,6s+3,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+4,12k+r+6s+3]∖{6s+5}, ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.5, 2.13, 2.14 and 2.15.
Case 5: r=0 and l=10 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7,∞8,∞9, ∞10},{2}⟩∪⟨Zu∪{∞11},{4,6s+3,6s+5,6s+6,6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞11}),D′⟩∪⟨Zu,L⟩, where D′=[6s+8,12k+6s+5], ∣D′∣=12k−2, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.7, 2.2, 2.11, 2.14 and 2.15.
Case 6: r=3,7 and l=0 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪(X∖{∞1,∞2,…, ∞6}),D′⟩∪⟨Zu,L⟩, where D′={2}∪[6s+3,12k+r+6s], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.7, 2.14 and 2.15.
Case 7: r=3 and l=4 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪{∞5},{2}⟩∪⟨Zu∪{∞6},{6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+5]∖{6s+5}, ∣D′∣=12k+2, and L=[3,6s+2], and apply Lemmas 2.6, 2.10, 2.14 and 2.15.
Case 8: r=4 and l=2 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪{∞5,∞6},{2}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+5], ∣D′∣=12k+3, and L=[3,6s+2], and apply Lemmas 2.6, 2.1, 2.14 and 2.15.
Case 9: r=7 and l=8 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2,2u}⟩∪⟨Zu∪{∞4,∞5,∞6},{4, 6s+3,6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),D′⟩∪⟨Zu,L⟩, where D′=[6s+5,12k+6s+11]∖{6s+7}, ∣D′∣=12k+6, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.9, 2.13, 2.14 and 2.15.
□
Proposition 3.3
For any n=60k+5r+2, r=2,7,10,11, there exists a decomposition of Kn+u∖Kn into 3-suns for every admissible u≥24k+2r+2.
Proof. Let X={∞1,∞2,…, ∞60k+5r+2}, r=2,7,10,11,
and u=24k+2r+2+h, with h≥0. Set h=12s+l, 0≤l≤11, and distinguish the following cases.
Case 1: r=2,11 and l=3 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1},{6s+2}⟩∪⟨Zu∪{∞2},{6s+4}⟩∪⟨Zu∪(X∖{∞1,∞2}),D′⟩∪⟨Zu,L⟩, where D′=[6s+1,12k+r+6s+2]∖{6s+2,6s+4}, ∣D′∣=12k+r, and L=[1,6s], and apply Lemmas 2.10, 2.14 and 2.15.
Case 2: r=2,7,10,11 and l=7 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2},{1,2,6s+3,6s+4}⟩⟨Zu∪(X∖{∞1,∞2}), D′⟩∪⟨Zu,L⟩, where D′=[6s+5,12k+r+6s+4], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.12, 2.14 and 2.15.
Case 3: r=7,10 and l=11 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,6s+3,6s+4}⟩∪⟨Zu∪{∞4,∞5, ∞6},{2,6s+5,6s+7}⟩∪⟨Zu∪{∞7},{6s+8}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}), D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+6]∖{6s+7,6s+8}, ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.13, 2.10, 2.14 and 2.15.
Case 4: r=2 and l=6 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪{∞5,∞6,∞7},{2, 6s+3,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}),D′⟩∪⟨Zu,L⟩, where D′=[6s+4,12k+6s+5]∖{6s+5}, ∣D′∣=12k+1, and L=[3,6s+2], and apply Lemmas 2.6, 2.13, 2.14 and 2.15.
Case 5: r=2,10 and l=10 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7},{2,6s+3,6s+4,6s+5,6s+6}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}),D′⟩∪⟨Zu,L⟩, where D′=[6s+7,12k+r+6s+5], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.7, 2.11, 2.14 and 2.15.
Case 6: r=7,11 and l=4 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4,∞5,∞6,∞7}, {2,4}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+r+6s+2]∖{6s+4}, ∣D′∣=12k+r−1, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.5, 2.3, 2.14 and 2.15.
Case 7: r=7 and l=8 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪{∞4,∞5,∞6},{2,6s+3,6s+5}⟩⟨Zu∪{∞7},{6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}),D′⟩∪⟨Zu,L⟩, where D′=[6s+4,12k+6s+11]∖{6s+5,6s+7}, ∣D′∣=12k+6, and L=[3,6s+2], and apply Lemmas 2.5, 2.13, 2.10, 2.14 and 2.15.
Case 8: r=10 and l=2 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7},{2}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞7}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+11], ∣D′∣=12k+9, and L=[3,6s+2], and apply Lemmas 2.7, 2.10, 2.14 and 2.15.
Case 9: r=11 and l=0 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞7},{1,2u}⟩∪⟨Zu∪(X∖{∞1,∞2, …,∞7}),D′⟩∪⟨Zu,L⟩, where D′={2}∪[6s+3,12k+6s+11], ∣D′∣=12k+10, and L=[3,6s+2], and apply Lemmas 2.8, 2.14 and 2.15. □
Proposition 3.4
For any n=60k+5r+3, r=2,5,6,9, there exists a decomposition of Kn+u∖Kn into 3-suns for every admissible u≥24k+2r+3.
Proof. Let X={∞1,∞2,…, ∞60k+5r+3}, r=2,5,6,9,
and u=24k+2r+3+h, with h≥0. Set h=12s+l, 0≤l≤11, and distinguish the following cases.
Case 1: r=2,5,6,9 and l=4 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,6s+3,6s+4}⟩∪⟨Zu∪(X∖{∞1,∞2, ∞3}),D′⟩∪⟨Zu,L⟩, where D′={2}∪[6s+5,12k+r+6s+3], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.13, 2.14 and 2.15.
Case 2: r=2,5 and l=8 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2},{1,6s+3,6s+4,6s+5}⟩∪⟨Zu∪{∞3},{2}⟩∪(X∖{∞1,∞2, ∞3}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+5], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.12, 2.10, 2.14 and 2.15.
Case 3: r=6,9 and l=0 (odd u).
If s=0, then write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞8},{1,3u}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞8}),D′⟩, where D′=[2,12k+r+1]∖{3u}, ∣D′∣=12k+r−1, and apply Lemmas 2.4 and 2.14.
If s>0, then write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,5s,5s+1}⟩∪⟨Zu∪{∞4,∞5,∞6},{2,6s+1,6s+3}⟩∪⟨Zu∪{∞7},{6s+2}⟩∪⟨Zu∪{∞8},{6s+4}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞8}),D′⟩∪⟨Zu,L⟩, where D′={2s+1,4s}∪[6s+5,12k+r+6s+1], ∣D′∣=12k+r−1, and L=[3,6s]∖{2s+1,4s,5s,5s+1}, and apply Lemmas 2.13, 2.10 and 2.14 to decompose the first five subgraphs, while to decompose the last one apply Lemma 2.15 i) and delete the orbit (S0).
Case 4: r=2,6 and l=1 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3}), D′⟩∪⟨Zu,L⟩, where D′={2}∪[6s+3,12k+r+6s+1], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.5, 2.14 and 2.15.
Case 5: r=2⋆ and l=5 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7,∞8,∞9, ∞10},{2}⟩∪⟨Zu∪{∞11,∞12,∞13},{4,6s+3,6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2, …,∞13}),D′⟩∪⟨Zu,L⟩, where D′=[6s+5,12k+6s+5]∖{6s+7}, ∣D′∣=12k, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.7, 2.2, 2.13, 2.14 and 2.15.
Case 6: r=5,9 and l=7 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7,∞8},{2,6s+3,6s+4,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞8}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+r+6s+4], ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.7, 2.12, 2.14 and 2.15.
Case 7: r=5 and l=11 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪{∞5,∞6},{2,6s+3,6s+5,6s+6}⟩∪⟨Zu∪{∞7},{4}⟩∪⟨Zu∪{∞8},{6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞8}),D′⟩∪⟨Zu,L⟩, where D′=[6s+8,12k+6s+11], ∣D′∣=12k+4, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.6, 2.12, 2.10, 2.14 and 2.15.
Case 8: r=6 and l=9 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4,∞5,∞6},{2,6s+3,6s+5}⟩∪⟨Zu∪{∞7},{4}⟩∪⟨Zu∪{∞8},{6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…, ∞8}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+6s+11]∖{6s+7}, ∣D′∣=12k+5, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.5, 2.13, 2.10, 2.14 and 2.15.
Case 9: r=9 and l=3 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2,2u}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3}), D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+11], ∣D′∣=12k+9, and L=[3,6s+2], and apply Lemmas 2.9, 2.14 and 2.15. □
Proposition 3.5
For any n=60k+5r+4, r=0,1,4,9, there exists a decomposition of Kn+u∖Kn into 3-suns for every admissible u≥24k+2r+3.
Proof. Let X={∞1,∞2,…, ∞60k+5r+4}, r=0,1,4,9,
and u=24k+2r+3+h, with h≥0. Set h=12s+l, 0≤l≤11, and distinguish the following cases.
Case 1: r=0,1⋆,4,9 and l=2 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{2,4}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3, ∞4}),D′⟩∪⟨Zu,L⟩, where D′={1,6s+3}∪[6s+5,12k+r+6s+2], ∣D′∣=12k+r, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.3, 2.14 and 2.15.
Case 2: r=0,9 and l=6 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,6s+3,6s+4}⟩∪⟨Zu∪{∞4},{2}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4}),D′⟩∪⟨Zu,L⟩, where D′=[6s+5,12k+r+6s+4], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.13, 2.10, 2.14 and 2.15.
Case 3: r=1,4 and l=10 (odd u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2},{1,6s+3,6s+5,6s+6}⟩∪⟨Zu∪{∞3},{2}⟩∪⟨Zu∪{∞4},{6s+4}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4}),D′⟩∪⟨Zu,L⟩, where D′=[6s+7,12k+r+6s+6], ∣D′∣=12k+r, and L=[3,6s+2], and apply Lemmas 2.12, 2.10, 2.14 and 2.15.
Case 4: r=0,4 and l=5 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,2u}⟩∪⟨Zu∪{∞7,∞8,∞9},{2, 6s+3,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞9}),D′⟩∪⟨Zu,L⟩, where D′=[6s+4,12k+r+6s+3]∖{6s+5}, ∣D′∣=12k+r−1, and L=[3,6s+2], and apply Lemmas 2.7, 2.13, 2.14 and 2.15.
Case 5: r=0 and l=9 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪{∞5,∞6,∞7}, {2,6s+3,6s+5}⟩∪⟨Zu∪{∞8},{4}⟩∪⟨Zu∪{∞9},{6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞9}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+6s+5]∖{6s+7}, ∣D′∣=12k−1, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.6 , 2.13, 2.10, 2.14 and 2.15.
Case 6: r=1 and l=7 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞7},{1,2u}⟩∪⟨Zu∪{∞8,∞9}, {2,4, 6s+3,6s+5}⟩∪⟨Zu∪(X∖{∞1,∞2, …,∞9}),D′⟩∪⟨Zu,L⟩, where D′=[6s+6,12k+6s+5], ∣D′∣=12k, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.8, 2.12, 2.14 and 2.15.
Case 7: r=1,9 and l=11 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4},{2,4,6s+3,6s+5,6s+6}⟩∪⟨Zu∪(X∖{∞1,∞2,∞3,∞4}),D′⟩∪⟨Zu,L⟩, where D′=[6s+7,12k+r+6s+6], ∣D′∣=12k+r, and L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.5, 2.11, 2.14 and 2.15.
Case 8: r=4 and l=1 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{1,2u}⟩∪⟨Zu∪(X∖{∞1,∞2, ∞3,∞4}),D′⟩∪⟨Zu,L⟩, where D′={2}∪[6s+3,12k+6s+5], ∣D′∣=12k+4, and L=[3,6s+2], and apply Lemmas 2.6, 2.14 and 2.15.
Case 9: r=9 and l=3 (even u).
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3},{1,2u}⟩∪⟨Zu∪{∞4},{2}⟩∪⟨Zu∪(X∖{∞1,∞2, ∞3,∞4}),D′⟩∪⟨Zu,L⟩, where D′=[6s+3,12k+6s+11], ∣D′∣=12k+9, and L=[3,6s+2], and apply Lemmas 2.5, 2.10, 2.14 and 2.15. □
Combining Lemma 3.1 and Propositions 3.1–3.5
gives our main theorem.
Theorem 3.1
Any 3SS(n) can be embedded in a 3SS(m) if and only if m≥57n+1 or m=n.
Appendix
n=21, u=12s+15
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,∞3,∞4},{2,4}⟩∪⟨Zu∪{∞5},{1}⟩∪⟨Zu∪{∞6},{6s+7}⟩∪⟨Zu∪(X∖{∞1,∞2,…,∞6}),{6s+3,6s+5,6s+6}⟩∪⟨Zu,L⟩, where L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.3, 2.10, 2.14 and 2.15.
n=13, u=12s+12
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1,∞2,…,∞6},{1,6s+6}⟩∪⟨Zu∪{∞7, ∞8,∞9,∞10},{2}⟩∪⟨Zu∪{∞11,∞12,∞13},{4,6s+3,6s+5}⟩∪⟨Zu,L⟩, where L=[3,6s+4]∖{4,6s+3}, and apply Lemmas 2.7, 2.2, 2.13 and 2.15.
n=9, u=12s+7
Write ⟨Zu∪X,Du⟩=⟨Zu∪{∞1, ∞2,∞3,∞4},{2,4}⟩∪⟨Zu∪{∞5,∞6, ∞7,∞8,∞9},{1}⟩∪⟨Zu,L⟩, where L=[3,6s+3]∖{4}, and apply Lemmas 2.3, 2.14 and decompose ⟨Zu,L⟩ as in Lemma 2.15 iii), taking in account that ∣6s+4∣12s+7=6s+3.
