Embedding handcuffed designs in D

Embedding handcuffed designs in D-designs,
where D is the triangle with attached edge ∗
Gaetano Quattrocchi
Dipartimento di Matematica e Informatica, Universita’ di Catania
viale A. Doria, 6 – 95125 Catania, Italia
[email protected]
Dedicated to Professor Alex Rosa
on the occasion of his 65th birthday.
Abstract
Let D be the triangle with attached edge (i. e. D is the graph having vertices
{a0 , a1 , a2 , a3 } and edges {a0 , a1 }, {a0 , a2 }, {a1 , a2 }, {a0 , a3 }). J.C. Bermond and
J. Schönheim [1], proved that a D-design of order n exists if and only if n ≡ 0 or 1
(mod 8).
Let (W, C) be a nontrivial D-design of order n, n ≥ 8, and let V ⊂ W ,
|V | = v < n. We say that a handcuffed design (V, P) of order v and block size s
([3]) is embedded in (W, C) if there is an injective mapping f : P → C such that
B is a subgraph of f (B) for every B ∈ P.
For each n ≡ 0 or 1 (mod 8), we determine the spectrum of all the integers
v such that there is a nontrivial handcuffed design of order v and block size s
embedded in a D-design of order n.
AMS classification: 05B05.
Keywords: Graph design; Embedding; Path.
1
Introduction
Let G be a subgraph of Kn , the complete undirected graph on n vertices. A G-design
of Kn is a pair (W, C), where W is the vertex set of Kn and C is an edge-disjoint
∗
Supported by MIUR and by C.N.R. (G.N.S.A.G.A.), Italy.
1
decomposition of Kn into copies of the graph G. Usually we say that C is a block of
the G-design if C ∈ C, and C is called the block-set.
Let D = [a1 , a2 , a0 ✶ a3 ] be the triangle with attached edge (i. e. D is the
graph having vertices {a0 , a1 , a2 , a3 } and edges {a0 , a1 }, {a0 , a2 }, {a1 , a2 }, {a0 , a3 }).
J.C. Bermond and J. Schönheim [1], proved that a D-design of order n exists if and
only if n ≡ 0 or 1 (mod 8).
A 4-cycle system of order n is a C4 -design of Kn , where C4 is the 4-cycle (cycle of length 4), i.e. C4 is the graph having vertices {a0 , a1 , a2 , a3 } and edges
{a0 , a1 }, {a1 , a2 }, {a2 , a3 }, {a0 , a3 }. It is well-known that a 4-cycle system of order
n exists if and only if n ≡ 1 (mod 8).
Note that D and C4 are the only two graphs having 4 vertices and 4 edges.
Let Ps = (a1 , a2 , . . . , as ) be the simple path with s − 1 edges and s vertices, i. e. Ps
is the graph having vertices {a1 , a2 , . . . , as } and edges {a1 , a2 }, {a2 , a3 }, . . . , {as−1 , as }.
A handcuffed design of order v and block size s H(v, s, 1) [3], is a Ps -design of Kv
(V, P) having the property that each vertex of V belongs to the same number of
blocks of P.
S. H. Y. Hung and N. S. Mendelsohn [4] proved that a handcuffed design H(v, 2h+
1, 1) (h ≥ 1) exists if and only if v ≡ 1 (mod 4h), and a handcuffed design H(v, 2h, 1)
(h ≥ 2) exists if and only if v ≡ 1 (mod 2h − 1).
Let G1 be a subgraph of G2 , and let V ⊆ W . We say that a G1 -design (V, P) of
order v is embedded in a G2 -design (W, C) of order n if there is a injective function
f : P → C such that B is a subgraph of f (B) for every B ∈ P.
The mapping f is called the embedding of (V, P) in (W, C).
In the case that G1 = Kv and G2 = Kn , the above definition is well-known in the
literature. The most famous example is the embedding of any affine space of order v
in a projective space of order v.
Related to this example is the embedding for resolvable balanced incomplete block
designs.
If G1 = G2 then we obtain the usual embedding definition for G-designs.
The following spectrum problem arises: Let G1 be a subgraph of G2 . For each
admissible n, determine the set of all the integers v such that there exists a G1 -design
of order v embedded in a G2 -design of order n.
Recentely, S. Milici and the author [5] considered the embedding of a handcuffed
design in a 4-cycle system, by giving a complete solution to the related spectrum
problem.
As we pointed out, there is only one other graph having 4 vertices and 4 edges.
The aim of this paper is to study the spectrum problem in this case. More precisely
we determine, for each n ≡ 0 or 1 (mod 8), n ≥ 8, the set
SH(Ps , D, n)
of all the integers v such that there is a nontrivial handcuffed design H(v, s, 1) embedded in a D-design of order n.
2
Example 1 Let V = {0, 1, 2, 3}, W = V ∪{a0 , a1 , a2 , a3 , a4 }, P = {(0, 1), (0, 2), (0, 3),
(1, 2), (1, 3), (2, 3)} and C = {[a1 , 1, 0 ✶ a4 ], [a0 , 2, 0 ✶ a2 ], [a3 , 0, 3 ✶ a2 ],
[a2 , 1, 2 ✶ a4 ], [a0 , 1, 3 ✶ a4 ], [a1 , 3, 2 ✶ a3 ], [a3 , a1 , a2 ✶ a0 ], [a4 , a1 , a0 ✶ a3 ],
[a3 , 1, a4 ✶ a2 ]}. It is easy to see that (V, P) is a handcuffed design H(4, 2, 1) embedded
in the D-design (W, C) of order 9.
Example 2 Let V = {0, 1, 2, 3, 4}, W = V ∪ {∞, a0 , b0 }, P = {(1, 2, 4), (3, 0, 2),
(0, 1, 4), (3, 4, 0), (1, 3, 2)} and C = {[b0 , 1, 2 ✶ 4], [b0 , 3, 0 ✶ 2], [a0 , 0, 1 ✶ 4],
[a0 , 3, 4 ✶ 0], [∞, 1, 3 ✶ 2], [∞, 4, b0 ✶ a0 ], [a0 , 2, ∞ ✶ 0]}. It is easy to see that (V, P)
is a handcuffed design H(5, 3, 1) embedded in the D-design (W, C) of order 8.
Let (V, P) be a handcuffed design H(v, s, 1) embedded in a D-design (W, C) of
order n. It is either s = 2 or s = 3. We will consider these cases in section 2 and
3 respectively. Moreover note that whenever V and C are known, the path set P is
uniquely determined.
Example 3 Let V = {0, 1, . . . , 7} and W = {∞1 , ∞2 , a0 , a1 , . . . , a6 }. Put
C1 = {[0, 1, a0 ✶ a2 ], [2, 3, a0 ✶ a3 ], [a0 , 5, 4 ✶ ∞2 ], [6, 7, a0 ✶ a4 ], [0, 2, a1 ✶ a2 ],
[1, 3, a1 ✶ a3 ], [4, 6, a1 ✶ a4 ], [a1 , 5, 7 ✶ ∞1 ], [0, 3, a2 ✶ a3 ], [1, 2, a2 ✶ a4 ],
[4, 7, a2 ✶ a5 ], [5, 6, a2 ✶ a6 ], [0, 4, a3 ✶ a4 ], [a3 , 1, 5 ✶ ∞2 ], [a3 , 2, 6 ✶ ∞2 ],
[a3 , 3, 7 ✶ ∞2 ], [a4 , 5, 0 ✶ ∞2 ], [a4 , 6, 1 ✶ ∞2 ], [a4 , 7, 2 ✶ ∞2 ], [a4 , 4, 3 ✶ ∞2 ],
[a5 , 0, 6 ✶ ∞1 ], [7, 1, a5 ✶ a6 ], [a5 , 2, 4 ✶ ∞1 ], [a5 , 3, 5 ✶ ∞1 ], [a6 , 7, 0 ✶ ∞1 ],
[a6 , 4, 1 ✶ ∞1 ], [a6 , 5, 2 ✶ ∞1 ], [a6 , 6, 3 ✶ ∞1 ]} and
C2 = {[a2 , ∞1 , ∞2 ✶ a0 ], [∞1 , a0 , a1 ✶ ∞2 ], [∞1 , a3 , a5 ✶ a1 ],
[∞2 , a3 , a6 ✶ a1 ], [∞2 , a4 , a5 ✶ a0 ], [∞1 , a4 , a6 ✶ a0 ]}}.
Clearly (W, C1 ∪ C2 ) is a D-design of order 17. For each C ∈ C1 there is exactly
one edge having vertices in V . Let P be the set of such edges. Then (V, P) is the
H(8, 2, 1) embedded in (V, C1 ∪ C2 ).
Refer to [2] for results and definitions omitted in the present paper.
2
SH(P2, D, n)
We begin this section by giving the following necessary condition.
Theorem 1 Let (V, P) be a H(v, 2, 1), v ≥ 2, embedded in a D-design (W, C) of
order n. Then n ≥ 2v.
Proof. Let f : P → C be the embedding. Since √f is injective, it follows that
2
|P| ≤ |C|. So n2 − n − 4v 2 + 4v ≥√0. Therefore n ≥ 1+ 16v2−16v+1 . The result follows
2
from the inequalities 2v − 1 ≤ 1+ 16v2−16v+1 ≤ 2v. ✷
The remaining part of this section is devoted to prove that the necessary condition
given in Theorem 1 is also sufficient, i.e. we will prove that
SH(P2 , D, n) = {v | 2 ≤ v ≤
3
n
, v integer}.
2
At first we give some useful notation. Let v = 4k, k ≥ 1. We denote by F =
{F0 , F1 , . . . , F4k−2 } a one-factorization of the complete graph Kv on point set V = Z4k ,
such that for each t = 2k − 1, 2k, . . . , 4k − 2 the one-factor Ft has edges {i, i + t + 1},
i = 0, 1, . . . , 2k − 1 and i + t + 1 is reduced to the range {2k, 2k + 1, . . . , 4k − 1},
(mod 2k) (see Example 4). It is well-known that such a one-factorization exists for
each k ≥ 1.
For i = 0, 1, . . . , 4k − 2, denote by ai Fi the set of triangles
{{ai , α, β} | {α, β} ∈ Fi }.
Let t1 , t2 ∈ {2k − 1, 2k, . . . , 4k − 2}, t1 = t2 . Define
B(∞, t1 , t2 ) = {[at1 , i + t1 + 1, i ✶ ∞] | {i, i + t1 + 1} ∈ at1 Ft1 }∪
{[at2 , i, i + t2 + 1 ✶ ∞] | {i, i + t2 + 1} ∈ at2 Ft2 }.
Lemma 1 For every v ≡ 0
D-design of order 2v.
(mod 4), v ≥ 4, there exists a H(v, 2, 1) embedded in a
Proof. Let v = 4k, k ≥ 1, V = Z4k and W = V ∪ {∞, a0 , a1 , . . . , a4k−2 }. Define F,
ai Fi , i = 0, 1, . . . , 4k − 2, and B(∞, 4k − 3, 4k − 2) as above.
Now we construct a D-design (W, C) of order 2v. Put in C the following blocks.
Step 1. All the blocks of B(∞, 4k − 3, 4k − 2).
Step 2. For j = 0, 1, . . . , k − 1, the blocks [∞, a2k+2j , a2k+2j−1 ✶ a2k+2j−2 ].
Now we form blocks of C by attaching to each triangle of ∪4k−4
i=0 ai Fi a pendant
edge. Remark that these pendant edges must cover the edges of the complete graph
K{∞,a0 ,a1 ,...,a4k−2 } not covered by blocks in Step 2. If k = 1 then form the blocks
[0, 1, a0 ✶ ∞] and [2, 3, a0 ✶ a2 ] and in this case the proof is completed. For k ≥ 2,
we proceed in the following way (see the following Example 4).
Step 3. For i = 0, 1, . . . , 2k − 2, attach to three triangles of ai Fi the pendant edges
{∞, ai }, {a4k−3 , ai } and {a4k−2 , ai }.
Step 4. For i = 0, 1, . . . , 2k − 3, take 2k − 3 triangles of ai Fi that are not used to form
the blocks in Step 3 and attach to them the pendant edges {ai+j+1 , ai },
j = 0, 1, . . . , 2k − 4.
Remark that the blocks given in Steps 3 and 4 cover all the triangles of ai Fi ,
i = 0, 1, . . . , 2k − 3, and three triangles of a2k−2 F2k−2 .
Step 5. For i = 2k − 1, 2k, . . . , 4k − 4, attach to triangles of ai Fi the following pendant
edges: {ai+j+2 , ai }, j = 0, 1, . . . , 4k − i − 4 and {aρ , ai }, ρ = 0, 1, . . . , i − 2k + 2.
Step 6. It is easy to see that 2k − 3 triangles of a2k−2 F2k−2 remain to be covered by
2k − 3 blocks. Attach to them the pendant edge {a0 , a2k−2 } and, if k ≥ 3, {aj , a2k−2 },
j = 2k, 2k + 1, . . . , 4k − 5.
It is easy to see that (V, P), P = ∪4k−2
i=0 Fi , is a H(v, 2, 1) embedded in the D-design
(W, C) of order 2v. ✷
4
As an example of the above Lemma 1, we produce a H(12, 2, 1) embedded in a
D-design of order 24.
Example 4 Let V = {0, 1, . . . , 11} and W = V ∪ {∞, a0 , a1 , . . . , a10 }. Take the
following one-factorization of K{0,1,...,11} :
F0 = {(0, 1), (2, 3), (4, 5), (6, 7), (8, 9), (10, 11)},
F1 = {(0, 2), (1, 4), (3, 5), (6, 8), (7, 10), (9, 11)},
F2 = {(0, 3), (1, 5), (2, 4), (6, 9), (7, 11), (8, 10)},
F3 = {(0, 4), (1, 3), (2, 5), (6, 10), (7, 9), (8, 11)},
F4 = {(0, 5), (1, 2), (3, 4), (6, 11), (7, 8), (9, 10)},
F5 = {(0, 6), (1, 7), (2, 8), (3, 9), (4, 10), (5, 11)},
F6 = {(0, 7), (1, 8), (2, 9), (3, 10), (4, 11), (5, 6)},
F7 = {(0, 8), (1, 9), (2, 10), (3, 11), (4, 6), (5, 7)},
F8 = {(0, 9), (1, 10), (2, 11), (3, 6), (4, 7), (5, 8)},
F9 = {(0, 10), (1, 11), (2, 6), (3, 7), (4, 8), (5, 9)},
F10 = {(0, 11), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}.
Put P = ∪10
i=0 Fi . The block set C is given by the following blocks.
Step 1. The blocks of B(∞, 9, 10) = {[a9 , 10, 0 ✶ ∞], [a9 , 11, 1 ✶ ∞], [a9 , 6, 2 ✶ ∞],
[a9 , 7, 3 ✶ ∞], [a9 , 8, 4 ✶ ∞], [a9 , 9, 5 ✶ ∞]} ∪ {[a10 , 0, 11 ✶ ∞], [a10 , 1, 6 ✶ ∞],
[a10 , 2, 7 ✶ ∞], [a10 , 3, 8 ✶ ∞], [a10 , 4, 9 ✶ ∞], [a10 , 5, 10 ✶ ∞]}.
Step 2. The blocks [∞, a6 , a5 ✶ a4 ], [∞, a8 , a7 ✶ a6 ], [∞, a10 , a9 ✶ a8 ].
Step 3.
[0, 1, a0
[3, 5, a1
[1, 3, a3
Form the blocks:
✶ ∞], [2, 3, a0 ✶ a9 ], [4, 5, a0 ✶ a10 ], [0, 2, a1 ✶ ∞], [1, 4, a1 ✶ a9 ],
✶ a10 ], [0, 3, a2 ✶ ∞], [1, 5, a2 ✶ a9 ], [2, 4, a2 ✶ a10 ], [0, 4, a3 ✶ ∞],
✶ a9 ], [2, 5, a3 ✶ a10 ], [0, 5, a4 ✶ ∞], [1, 2, a4 ✶ a9 ], [3, 4, a4 ✶ a10 ].
Step 4. Form the blocks:
[6, 7, a0 ✶ a1 ], [8, 9, a0 ✶ a2 ], [10, 11, a0 ✶ a3 ], [6, 8, a1 ✶ a2 ], [7, 10, a1 ✶ a3 ],
[9, 11, a1 ✶ a4 ], [6, 9, a2 ✶ a3 ], [7, 11, a2 ✶ a4 ], [8, 10, a2 ✶ a5 ], [6, 10, a3 ✶ a4 ],
[7, 9, a3 ✶ a5 ], [8, 11, a3 ✶ a6 ].
Step 5. Form the blocks:
[0, 6, a5 ✶ a7 ], [1, 7, a5 ✶ a8 ], [2, 8, a5 ✶ a9 ], [3, 9, a5 ✶ a10 ], [4, 10, a5 ✶ a0 ],
[5, 11, a5 ✶ a1 ], [0, 7, a6 ✶ a8 ], [1, 8, a6 ✶ a9 ], [2, 9, a6 ✶ a10 ], [3, 10, a6 ✶ a0 ],
[4, 11, a6 ✶ a1 ], [5, 6, a6 ✶ a2 ], [0, 8, a7 ✶ a9 ], [1, 9, a7 ✶ a10 ], [2, 10, a7 ✶ a0 ],
[3, 11, a7 ✶ a1 ], [4, 6, a7 ✶ a2 ], [5, 7, a7 ✶ a3 ], [0, 9, a8 ✶ a10 ], [1, 10, a8 ✶ a0 ],
[2, 11, a8 ✶ a1 ], [3, 6, a8 ✶ a2 ], [4, 7, a8 ✶ a3 ], [5, 8, a8 ✶ a4 ].
Step 6. Form the blocks:
[6, 11, a4 ✶ a0 ], [7, 8, a4 ✶ a6 ], [9, 10, a4 ✶ a7 ].
Lemma 2 For every v ≡ 0
D-design of order 2v + 1.
(mod 4), v ≥ 4, there exists a H(v, 2, 1) embedded in a
5
Proof. Let v = 4k, k ≥ 1, V = Z4k and W = V ∪ {∞1 , ∞2 , a0 , a1 , . . . , a4k−2 }.
Define F, ai Fi , i = 0, 1, . . . , 4k − 2, B(∞1 , 4k − 5, 4k − 4) and B(∞2 , 4k − 3, 4k − 2)
as at the beginning of this section. The proof for k = 1, 2 follows from Examples 1
and 3 respectively.
Case k = 3. Let C1 = B(∞1 , 7, 8) ∪ B(∞2 , 9, 10).
Let C2 = {[a0 , a1 , a2 ✶ a7 ], [a2 , a4 , a3 ✶ a7 ], [a6 , a8 , a7 ✶ a0 ], [a6 , a10 , a9 ✶ a1 ],
[a6 , ∞1 , ∞2 ✶ a1 ], [a7 , a9 , ∞1 ✶ a4 ], [∞2 , a10 , a7 ✶ a1 ], [∞2 , a8 , a9 ✶ a0 ],
[∞1 , a10 , a8 ✶ a0 ]}.
Form the block set C3 by attaching a pendant edge to the triangles of ai Fi ,
i = 0, 1, . . . , 6, in the following way:
For i = 0, attach {∞1 , a0 }, {∞2 , a0 }, {at , a0 }, t = 3, 4, 5, 10.
For i = 1, attach {∞1 , a1 }, {at , a1 }, t = 3, 4, 5, 8, 10.
For i = 2, 3, attach {∞1 , ai }, {∞2 , ai }, {at , ai }, t = 5, 8, 9, 10.
For i = 4, attach {∞2 , a4 }, {at , a4 }, t = 5, 7, 8, 9, 10.
For i = 5, attach {∞1 , a5 }, {∞2 , a5 }, {at , a5 }, t = 7, 8, 9, 10.
For i = 6, attach {at , a6 }, t = 0, 1, 2, 3, 4, 5.
It is easy to see that (W, C1 ∪ C2 ∪ C3 ) is the required D-design of order 25.
Case k = 4. Let C1 = B(∞1 , 11, 12) ∪ B(∞2 , 13, 14).
Let C2 = {[a1 , a2 , a0 ✶ a13 ], [a2 , a3 , a4 ✶ ∞1 ], [a4 , a5 , a6 ✶ a9 ],
[a7 , a8 , a6 ✶ ∞1 ], [a8 , a9 , a0 ✶ a14 ], [a10 , a11 , a12 ✶ a3 ], [a10 , a13 , a14 ✶ a2 ],
[a10 , ∞1 , ∞2 ✶ a2 ], [a11 , a13 , ∞1 ✶ a1 ], [a11 , a14 , ∞2 ✶ a1 ], [a12 , a13 , ∞2 ✶ a0 ],
[a12 , a14 , ∞1 ✶ a0 ]}.
Form the block set C3 by attaching a pendant edge to the triangles of ai Fi ,
i = 0, 1, . . . , 6, in the following way:
For i = 0, attach {at , a0 }, t = 3, 4, 5, 6, 7, 10, 11, 12.
For i = 1, attach the edges {at , a1 }, t = 7, 8, 9, 10, 11, 12, 13, 14.
For i = 2, attach {∞1 , a2 }, {at , a2 }, t = 5, 7, 8, 9, 11, 12, 13.
For i = 3, attach {∞1 , a3 }, {∞2 , a3 }, {at , a3 }, t = 1, 7, 9, 11, 13, 14.
For i = 4, attach {∞2 , a4 }, {at , a4 }, t = 1, 8, 9, 11, 12, 13, 14.
For i = 5, attach {∞1 , a5 }, {∞2 , a5 }, {at , a5 }, t = 1, 3, 11, 12, 13, 14.
For i = 6, attach {∞2 , a6 }, {at , a6 }, t = 1, 2, 3, 11, 12, 13, 14.
For i = 7, attach {∞1 , a7 }, {∞2 , a7 }, {at , a7 }, t = 4, 5, 11, 12, 13, 14.
For i = 8, attach {∞1 , a8 }, {∞2 , a8 }, {at , a8 }, t = 3, 5, 11, 12, 13, 14.
For i = 9, attach {∞1 , a9 }, {∞2 , a9 }, {at , a9 }, t = 5, 7, 11, 12, 13, 14.
For i = 10, attach {at , a10 }, t = 2, 3, 4, 5, 6, 7, 8, 9.
It is easy to see that (W, C1 ∪ C2 ∪ C3 ) is the required D-design of order 33.
Case k ≥ 5. In order to construct a D-design of order 2v + 1 (W, C), put in C the
following blocks.
Step 1. The blocks of B(∞1 , 4k − 5, 4k − 4) ∪ B(∞2 , 4k − 3, 4k − 2).
Step 2. The blocks [∞1 , ∞2 , a4k−6 ✶ a4k−7 ], [a4k−5 , a4k−4 , ∞1 ✶ a4k−7 ],
[∞1 , a4k−2 , a4k−3 ✶ a4k−7 ], [a4k−5 , a4k−3 , ∞2 ✶ a4k−7 ], [∞2 , a4k−4 , a4k−2 ✶ a4k−7 ],
6
[a4k−6 , a4k−2 , a4k−5 ✶ a4k−7 ], [a4k−6 , a4k−3 , a4k−4 ✶ a4k−7 ].
Now we form the remaining blocks by attaching to each triangle of ∪4k−6
i=0 ai Fi a
pendant edge. It is easy to see that these pendant edges must cover the graphs
K{a0 ,a1 ,...,a3k−8 } , K{a3k−7 ,a3k−6 ,...,a4k−7 } , K{a0 ,a1 ,...,a3k−8 },{a3k−7 ,a3k−6 ,...,a4k−7 } and
K{a0 ,a1 ,...,a4k−8 },{∞1 ,∞2 ,a4k−6 ,a4k−5 ,...,a4k−2 } .
Step 3. For i = 0, 1, . . . , k − 1, attach to k − i triangles of a4k−7−i F4k−7−i the pendant
edges {a4k−7−i , a3k−7+ρ }, ρ = 0, 1, . . . , k − 1 − i. These pendant edges cover the
complete graph K{a3k−7 ,a3k−6 ,...,a4k−7 } .
Step 4. Let k = 5. Attach the pendant edges {a13 , aρ }, ρ = 2, 4, 5, 6, 7, to the
remaining 5 triangles of a13 F13 not used in Step 3.
Let k ≥ 6. For i = 2k − 7, 2k − 6, . . . , 3k − 8, attach the pendant edges {a4k−7 , ai }
to the remaining k triangles of a4k−7 F4k−7 not used in Step 3.
, i(i+1)
+ 1, . . . , i(i+1)
+ i},
Step 5. Let k = 5. Put Λi = {(1 − x) (mod 8) | x = i(i+1)
2
2
2
i = 0, 1, 2, 3. Then attach to i+1 triangles of a11−i F11−i the pendant edges {a11−i , aρ },
ρ ∈ Λi .
Let k ≥ 6. Put Γi = {(2k − 8 − x) (mod 3k − 7) | x = i(i+1)
+ (k − 6)i, i(i+1)
+
2
2
(i+1)(i+2)
(k − 6)i + 1, . . . ,
+ (k − 6)(i + 1) − 1}, i = 0, 1, . . . , k − 1. Then attach to
2
i + k − 5 triangles of a4k−8−i F4k−8−i the pendant edges a4k−8−i aρ , ρ ∈ Γi .
Step 6. Let k = 5. For i = 0, 1, 3, attach the pendant edges {ai , a12 } and {ai , a13 } to
two triangles of ai Fi , and for i = 2, 4, 5, 6, 7 attach the pendant edge {ai , a12 } to one
triangle of ai Fi .
Let k ≥ 6. For i = 0, 1, . . . , 2k − 8, attach the pendant edge {ai , a4k−7 } to one
triangle of ai Fi .
Step 7. Let k = 5. For i = 0, 1, 2, 3 let Λi be the set defined in Step 5. Then for each
t ∈ {0, 1, . . . , 7} − Λi , attach the pendant edge {at , a11−i } to one triangle of at Ft .
Let k ≥ 6. For i = 0, 1, . . . , k − 1, let Γi be the set defined in Step 5. Then for
each t ∈ {0, 1, . . . , 3k − 8} − Γi , attach the pendant edge {at , a4k−8−i } to one triangle
of at Ft .
Note that the pendant edges used in Steps 4, 5, 6 and 7 cover the complete bipartite
graph K{a0 ,a1 ,...,a3k−8 },{a3k−7 ,a3k−6 ,...,a4k−7 } .
Step 8. Attach the pendant edges {a4k−6 , ai }, i = 2k−7, 2k−6, . . . , 4k−8, to triangles
of a4k−6 F4k−6 .
For i = 0, 1, . . . , k − 1, attach to six triangles of a4k−8−i F4k−8−i the pendant edges
{a4k−8−i , ∞1 }, {a4k−8−i , ∞2 }, {a4k−8−i , a4k−5 }, {a4k−8−i , a4k−4 }, {a4k−8−i , a4k−3 },
{a4k−8−i , a4k−2 }.
For i = 0, 1, . . . , 2k − 8, attach to six triangles of ai Fi the pendant edges {ai , ∞2 },
{ai , a4k−6 }, {ai , a4k−5 }, {ai , a4k−4 }, {ai , a4k−3 }, {ai , a4k−2 }.
For i = 2k − 7, 2k − 6, . . . , 3k − 8, attach to five triangles of ai Fi the pendant edges
{ai , ∞2 }, {ai , a4k−5 }, {ai , a4k−4 }, {ai , a4k−3 }, {ai , a4k−2 }.
7
The pendant edges used in Steps 2-7 cover the graph
K{a0 ,a1 ,...,a4k−8 },{∞1 ,∞2 ,a4k−6 ,a4k−5 ,...,a4k−2 } − K{∞1 },{a0 ,a1 ,...,a3k−8 } . So only the edges of
K{a0 ,a1 ,...,a3k−8 } ∪ K{∞1 },{a0 ,a1 ,...,a3k−8 } remain to be utilized.
Step 9. Form blocks [a1+ρ , a3+ρ , aρ ✶ ∞1 ] for ρ = 0, 1, . . . , 3k − 8, where the suffices
are (mod 3k − 7).
Step 10. In this step we suppose that k is even, the case where k is odd will be
considered in the next one.
Let k = 6. Put in C the following blocks.
(1) For i = 0, 1, 2, 3, 4, attach the pendant edge {ai , ai+4 } to one triangle of ai Fi .
(2) For i = 5, attach the pendant edges {a5 , a0 } and {a5 , a9 } to two triangles of a5 F5 .
(3) For i = 6, attach the pendant edges {a6 , a10 }, {a6 , a0 }, {a6 , a1 }, to three triangles
of a6 F6 .
(4) For i = 7, 8, 9, 10, attach the pendant edges {ai , ai−7+ρ }, ρ = 0, 1, 2, to three
triangles of ai Fi .
It is easy to see that for k = 6 the proof is completed.
Let k = 2h, h ≥ 4. Put in C the following blocks.
(1) For each i = 0, 1, . . . , 2h−1, exactly 3h−8 triangles of ai Fi remain to be completed.
To do this attach the pendant edges {ai , ai+4+ρ }, ρ = 0, 1, . . . , 3h − 9.
(2) For each i = 2h, 2h + 1, . . . , 4h − 8 exactly 3h − 7 triangles of ai Fi remain to be
completed. We can proceed in the following way.
If h ≥ 5 then for i = 2h, 2h + 1, . . . , 3h − 5, attach the pendant edges {ai , a6h−8 } and
{ai , ai+ρ+4 }, ρ = 0, 1, . . . , 3h − 9.
For i = 3h−4, 3h−3, . . . , 4h−8 (remark that for h = 4 we have 2h = 3h−4), attach the
pendant edges {ai , ai+4+ρ }, ρ = 0, 1, . . . , 6h−i−13, and {ai , at }, t = 0, 1, . . . , i−3h+4.
(3) For each i = 4h − 7, 4h − 6, . . . , 6h − 8, exactly 3h − 6 triangles of ai Fi remain to
be completed. We can proceed in the following way.
For i = 4h − 7, 4h − 6, . . . , 6h − 12, attach the pendant edges
{ai , ai+4+ρ }, ρ = 0, 1, . . . , 6h − i − 12, and {ai , at }, t = 0, 1, . . . , i − 3h + 4.
For i = 6h − 11, 6h − 10, 6h − 9, attach the pendant edges
{ai , at }, t = i − 6h + 11, i − 6h + 12, . . . , i − 3h + 4.
For i = 6h − 8, attach the pendant edges
{ai , at } for t = 3, 5, . . . , 2h − 1 and for t = 3h − 4, 3h − 3, . . . , 4h − 8.
This completes the proof for even k.
Step 11. Let k = 2h + 1, h ≥ 2.
If h = 2, then for i = 4, 5, 6, 7, attach the pendant edge {ai , ai−4 }, to the last
“free” triangle of ai Fi . This completes the proof for k = 5.
Suppose now h ≥ 3. Put in C the following blocks.
(1) For each i = 0, 1, . . . , 4h−6, exactly 3h−6 triangles of ai Fi remain to be completed.
We proceed in the following way.
For i = 0, 1, . . . , 6 and h = 3 attach the pendant edges {ai , ai+4+ρ }, ρ = 0, 1, 2.
For i = 0, 1, . . . , 2h + 1 and h ≥ 4 then attach the pendant edges {ai , ai+4+ρ },
ρ = 0, 1, . . . , 3h − 7.
8
For i = 2h + 2, 2h + 3, . . . , 3h − 3 and h ≥ 5, attach the pendant edges {ai , at } for
t = 6h − 5 and t = i + 4, i + 5, . . . , i + 3h − 4.
For i = 3h − 2, 3h − 1, . . . , 4h − 6 and h ≥ 4, attach the pendant edges {ai , at } for
t = 0, 1, . . . , i − 3h + 2 and t = i + 4, i + 5, . . . , 6h − 6.
(2) For each i = 4h − 5, 4h − 4, . . . , 5h − 2, exactly 3h − 5 triangles of ai Fi remain to
be completed. We proceed in the following way.
If h = 3 then attach the following pendant edges:
{a7 , at } for t = 0, 11, 12, 13;
{a8 , at } for t = 0, 1, 12, 13;
{a9 , at } for t = 0, 1, 2, 13;
{a10 , at } for t = 0, 1, 2, 3;
{a11 , at } for t = 1, 2, 3, 4;
{a12 , at } for t = 2, 3, 4, 5;
{a13 , at } for t = 3, 4, 5, 6.
The proof for h = 3 is completed.
If h = 4 then attach the following pendant edges:
{ai , at } for i = 11, 12, 13, 14, 15, t = 0, 1, . . . , i − 10 and t = i + 4, i + 5, . . . , 19;
{a16,at } for t = 0, 1, . . . , 6;
{a17,at } for t = 1, 2, . . . , 7;
{a18,at } for t = 2, 3, . . . , 8.
If h = 5 then attach the following pendant edges:
{ai , at } for i = 15, 16, . . . , 21, t = 0, 1, . . . , i − 13 and t = i + 4, i + 5, . . . , 25;
{a22,at } for t = 0, 1, . . . , 9;
{a23,at } for t = 1, 2, . . . , 10.
If h = 6 then attach the following pendant edges:
{ai , at } for i = 19, 20, . . . , 27, t = 0, 1, . . . , i − 16 and t = i + 4, i + 5, . . . , 31;
{a28,at } for t = 0, 1, . . . , 12.
If h ≥ 7 then attach the following pendant edges:
{ai , at } for i = 4h − 5, 4h − 4, . . . , 5h − 2, t = 0, 1, . . . , i − 3h + 2 and
t = i + 4, i + 5, . . . , 6h − 5.
(3) Let h ≥ 4. For each i = 5h − 1, 5h, . . . , 6h − 5, exactly 3h − 4 triangles of ai Fi
remain to be completed. We proceed in the following way.
If h = 4 then attach the pendant edges {a19 , at }, t = 3, 4, . . . , 10. For h = 4 the
proof is completed.
If h = 5 then attach the following pendant edges:
{a24 , at } for t = 2, 3, . . . , 12;
{a25 , at } for t = 3, 4, . . . , 11 and t = 13, 14.
For h = 5 the proof is completed.
If h = 6 then attach the following pendant edges:
{a29 , at } for t = 1, 2, . . . , 14;
{a30 , at } for t = 2, 3, . . . , 15;
{a31 , at } for t = 3, 4, . . . , 13 and t = 16, 17, 18.
For h = 6 the proof is completed.
If h ≥ 7 then attach the following pendant edges:
{ai , at } for i = 5h−1, 5h, . . . , 6h−9, t = 0, 1, . . . , i−3h+3 and t = i+4, i+5, . . . , 6h−5.
9
{a6h−8 , at } for t = 0, 1, . . . , 3h − 5;
{a6h−7 , at } for t = 1, 2, . . . , 3h − 4;
{a6h−6 , at } for t = 2, 3, . . . , 3h − 3;
{a6h−5 , at } for t = 3, 4, . . . , 2h + 1 and t = 3h − 2, 3h − 1, . . . , 4h − 6.
For h ≥ 7 the proof is completed. ✷
Theorem 2 For each n ≡ 0 or 1
n
, v integer}.
2
(mod 8), n ≥ 8, SH(P2 , D, n) = {v | 2 ≤ v ≤
Proof. The proof follows from Lemmas 1 and 2 and from the fact that if w ≤ v,
then any H(w, 2, 1) is embedded in any H(v, 2, 1). ✷
3
SH(P3, D, n)
In this section we want to determine, for each n ≡ 0 or 1 (mod 8), n ≥ 8, the
spectrum SH(P3 , D, n) of the integers v such that there is a nontrivial handcuffed
design H(v, 3, 1) embedded in a D-design of order n.
Theorem 3 Suppose that (V, P) is a H(v, 3, 1), v ≡ 1
in a D-design (W, C). Then n > 3v−1
.
2
(mod 4), v ≥ 5, embedded
Proof. Let f be the embedding of (V, P) in (W, C), and let f (P) = {f (B) | B ∈ P}.
Clearly each C ∈ f (P) cannot cover any edge of KW −V . It follows that |C − f (P)| ≥
(n−v)(n−v−1)
. Take a vertex x ∈ W − V . Then |{x, y} | y ∈ V | = v. Suppose
8
that |C − f (P| = (n−v)(n−v−1)
. Then the edges {x, y}, y ∈ V , must be covered
8
by blocks C ∈ f (P) and so v = |{{x, y} | y ∈ V }| is an even integer. But v ≡ 1
(mod 4), therefore |C − f (P)| > (n−v)(n−v−1)
. So n(n−1)
= v(v−1)
+ |C − f (P|. Therefore
8
8
4
n(n−1)
v(v−1)
(n−v)(n−v−1)
− 4 >
from which the result follows. ✷
8
8
Theorem 4 For each n ≡ 0
2n−9
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 24, SH(P3 , D, n) = {v | 5 ≤ v ≤
. Then
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−9
3
it is sufficient to prove that {v | 5 ≤ v ≤ 2n−9
,
v
≡
1
(mod
4)}
⊆
SH(P
3 , D, n).
3
Put n = 24 + 24k, k ≥ 0.
Case k = 0. Let V0 = {χ0 , χ1 , χ2 , χ3 , χ4 } ∪ {1, 2, . . . , 8}, A0 = {a0 , b0 , c0 , d0 }, and
W0 = V0 ∪ A0 ∪ {∞1 , ∞2 , . . . , ∞7 }. Let D0 = { [∞1 , χ0 , χ1 ✶ χ4 ], [∞2 , χ1 , χ2 ✶ χ0 ],
[∞7 , χ2 , χ3 ✶ χ1 ], [∞2 , χ3 , χ4 ✶ χ2 ], [∞4 , χ4 , χ0 ✶ χ3 ], [∞1 , ∞3 , ∞2 ✶ χ0 ],
[∞2 , ∞6 , ∞4 ✶ ∞5 ], [∞2 , ∞5 , ∞7 ✶ ∞1 ], [χ2 , ∞5 , ∞3 ✶ χ1 ], [χ1 , ∞6 , ∞5 ✶ χ0 ],
[χ3 , ∞3 , ∞6 ✶ χ0 ], [χ1 , ∞4 , ∞7 ✶ χ4 ], [χ0 , ∞3 , ∞7 ✶ ∞6 ], [χ2 , ∞1 , ∞4 ✶ χ3 ],
[χ4 , ∞1 , ∞6 ✶ χ2 ], [χ3 , ∞1 , ∞5 ✶ χ4 ], [3, ∞4 , ∞3 ✶ χ4 ]} and let
H0 = { [∞7 , 4, 1 ✶ χ0 ], [∞1 , 1, 2 ✶ χ0 ], [∞2 , 2, 3 ✶ 1], [∞5 , 2, 4 ✶ 3], [a0 , 3, χ0 ✶ 4],
10
[∞1 , 8, 5 ✶ χ0 ], [∞4 , 5, 6 ✶ χ0 ], [∞7 , 6, 7 ✶ 5], [∞6 , 7, 8 ✶ 6], [c0 , 7, χ0 ✶ 8],
[b0 , 1, χ1 ✶ 2], [b0 , 2, χ2 ✶ 1], [b0 , 3, χ3 ✶ 4], [b0 , 4, χ4 ✶ 3], [c0 , χ3 , 1 ✶ χ4 ],
[c0 , χ4 , 2 ✶ χ3 ], [c0 , χ1 , 3 ✶ χ2 ], [c0 , χ2 , 4 ✶ χ1 ], [a0 , 5, χ1 ✶ 6], [a0 , 6, χ2 ✶ 5],
[a0 , 7, χ3 ✶ 8], [a0 , 8, χ4 ✶ 7], [d0 , χ3 , 5 ✶ χ4 ], [d0 , χ4 , 6 ✶ χ3 ], [d0 , χ1 , 7 ✶ χ2 ],
[d0 , χ2 , 8 ✶ χ1 ], [∞2 , 5, 1 ✶ 6], [∞6 , 6, 2 ✶ 5], [∞7 , 8, 3 ✶ 7], [∞1 , 7, 4 ✶ 8],
[∞5 , 3, 5 ✶ 4], [∞1 , 3, 6 ✶ 4], [∞3 , 1, 7 ✶ 2], [∞3 , 2, 8 ✶ 1], [a0 , ∞1 , b0 ✶ c0 ],
[c0 , ∞1 , d0 ✶ χ0 ], [b0 , 6, ∞2 ✶ a0 ], [d0 , 4, ∞2 ✶ 7], [b0 , d0 , ∞3 ✶ 4], [c0 , 6, ∞3 ✶ a0 ],
[d0 , 2, ∞4 ✶ 1], [a0 , 4, ∞4 ✶ c0 ], [b0 , 8, ∞5 ✶ 6], [d0 , 1, ∞5 ✶ 7], [a0 , 1, ∞6 ✶ 4],
[d0 , 3, ∞6 ✶ c0 ], [a0 , 2, ∞7 ✶ b0 ], [c0 , 5, ∞7 ✶ d0 ], [c0 , ∞5 , a0 ✶ d0 ], [∞6 , b0 , 5 ✶ ∞3 ],
[∞2 , c0 , 8 ✶ ∞4 ], [7, ∞4 , b0 ✶ χ0 ]}.
Put C0 = D0 ∪ H0 . Then (W0 , C0 ) is a D-design of order 24 with an embedded
H(13, 3, 1) based on point set V0 and three embedded H(v, 3, 1) for v = 5, 9.
In order to prove the theorem for k ≥ 1, we construct (using the well-known
embedding v → v + 4 [4]) a H(13 + 16k, 3, 1) (V, P) having an embedded H(v, 3, 1)
for each admissible v, 5 ≤ v ≤ 13 + 16k. Then we embed (V, P) in a D-design (W, C)
of order 24 + 24k.
Put Vi = {χ0 , χ1 , χ2 , χ3 , χ4 }∪{1+8i, 2+8i, 3+8i, 4+8i, 5+8i, 6+8i, 7+8i, 8+8i},
Ai = {ai , bi , ci , di }, and Wi = Vi ∪ Ai ∪ {∞1 , ∞2 , . . . , ∞7 }, i = 0, 1, . . . , 2k. Let
V = ∪2k
i=0 Vi .
For each i = 1, 2, . . . , 2k, let (Wi , Ci ) be the D-design of order 24 isomorphic to
(W0 , C0 ), based on the point set Wi . Denote by (Vi , Pi ) the H(13, 3, 1) embedded in
(Wi , Ci ).
It is well-known that there is a near one-factorization F = {F0 , F1 , . . . , F2k } of the
complete graph K1+2k on point set Z1+2k , satisfying the following properties:
(1) for each i = 0, 1, . . . , 2k, i is the missing vertex in Fi ;
(2) F0 = {{2ρ − 1, 2ρ} | ρ = 1, 2, . . . , k};
(3) if k ≥ 2 then there is a one-factor G of K2k based on point set {1, 2, . . . , 2k} and
suborthogonal to F (i.e. |F0 ∩ G| = 0 and |Fi ∩ G| ≤ 1 for each i = 1, 2, . . . , 2k).
Let {x, y}, x < y, be an edge of Fi . Define
P (i, x, y) = { (1 + 8y, 1 + 8x, 2 + 8y), (1 + 8y, 2 + 8x, 2 + 8y), (3 + 8y, 3 + 8x, 4 + 8y),
(3+8y, 4+8x, 4+8y), (5+8y, 5+8x, 6+8y), (5+8y, 6+8x, 6+8y), (7+8y, 7+8x, 8+8y),
(7+8y, 8+8x, 8+8y), (3+8x, 1+8y, 4+8x), (3+8x, 2+8y, 4+8x), (1+8x, 3+8y, 2+8x),
(1+8x, 4+8y, 2+8x), (7+8x, 5+8y, 8+8x), (7+8x, 6+8y, 8+8x), (5+8x, 7+8y, 6+8x),
(5+8x, 8+8y, 6+8x), (5+8y, 1+8x, 6+8y), (5+8y, 2+8x, 6+8y), (7+8y, 3+8x, 8+8y),
(7+8y, 4+8x, 8+8y), (1+8y, 5+8x, 2+8y), (1+8y, 6+8x, 2+8y), (3+8y, 7+8x, 4+8y),
(3+8y, 8+8x, 4+8y), (3+8x, 5+8y, 4+8x), (3+8x, 6+8y, 4+8x), (1+8x, 7+8y, 2+8x),
(1+8x, 8+8y, 2+8x), (7+8x, 1+8y, 8+8x), (7+8x, 2+8y, 8+8x), (5+8x, 3+8y, 6+8x),
(5 + 8x, 4 + 8y, 6 + 8x)}.
Let P = ∪2k
i=0 {Pi ∪ (∪{x,y}∈Fi P (i, x, y)}. Then (V, P) is the required
H(13 + 16k, 3, 1).
Now we embed (V, P) in a D-design (W, C) of order 24 + 24k. Let W = ∪2k
i=0 Wi .
Step 1. Put in C the blocks of C0 and the blocks of Ci − Di , i = 1, 2, . . . , 2k.
11
Step 2. Let {x, y}, x < y, be an edge of the near one-factor Fi . Define
B(i, x, y) = { [ai , 1 + 8y, 1 + 8x ✶ 2 + 8y], [ai , 2 + 8y, 2 + 8x ✶ 1 + 8y],
[ai , 3 + 8y, 3 + 8x ✶ 4 + 8y], [ai , 4 + 8y, 4 + 8x ✶ 3 + 8y], [ai , 5 + 8y, 5 + 8x ✶ 6 + 8y],
[ai , 6 + 8y, 6 + 8x ✶ 5 + 8y], [ai , 7 + 8y, 7 + 8x ✶ 8 + 8y], [ai , 8 + 8y, 8 + 8x ✶ 7 + 8y],
[bi , 3 + 8x, 1 + 8y ✶ 4 + 8x], [bi , 4 + 8x, 2 + 8y ✶ 3 + 8x], [bi , 1 + 8x, 3 + 8y ✶ 2 + 8x],
[bi , 2 + 8x, 4 + 8y ✶ 1 + 8x], [bi , 7 + 8x, 5 + 8y ✶ 8 + 8x], [bi , 8 + 8x, 6 + 8y ✶ 7 + 8x],
[bi , 5 + 8x, 7 + 8y ✶ 6 + 8x], [bi , 6 + 8x, 8 + 8y ✶ 5 + 8x], [ci , 5 + 8y, 1 + 8x ✶ 6 + 8y],
[ci , 6 + 8y, 2 + 8x ✶ 5 + 8y], [ci , 7 + 8y, 3 + 8x ✶ 8 + 8y], [ci , 8 + 8y, 4 + 8x ✶ 7 + 8y],
[ci , 1 + 8y, 5 + 8x ✶ 2 + 8y], [ci , 2 + 8y, 6 + 8x ✶ 1 + 8y], [ci , 3 + 8y, 7 + 8x ✶ 4 + 8y],
[ci , 4 + 8y, 8 + 8x ✶ 3 + 8y], [di , 3 + 8x, 5 + 8y ✶ 4 + 8x], [di , 4 + 8x, 6 + 8y ✶ 3 + 8x],
[di , 1 + 8x, 7 + 8y ✶ 2 + 8x], [di , 2 + 8x, 8 + 8y ✶ 1 + 8x], [di , 7 + 8x, 1 + 8y ✶ 8 + 8x],
[di , 8 + 8x, 2 + 8y ✶ 7 + 8x], [di , 5 + 8x, 3 + 8y ✶ 6 + 8x], [di , 6 + 8x, 4 + 8y ✶ 5 + 8x]}.
Put in C the blocks of ∪2k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
Step 3. For ρ = 1, 2, . . . , k, put in C the blocks (the suffices are
(mod 1 +
2k)) [d2ρ+σ , bρ+σ , aσ ✶ cρ+σ ], [d2ρ−1+σ , bk+ρ+σ , aσ ✶ aρ+σ ], [b2ρ+σ , bσ , cρ+σ ✶ cσ ],
[d2ρ+σ , dσ , cρ+σ ✶ a2ρ+σ ], σ = 0, 1, . . . , 2k.
Note that the blocks in C cover all the edges of KW except the following: {3 +
8i, ∞3 }, {3 + 8i, ∞4 }, i = 1, 2, . . . , 2k.
Step 4. Since F0 = {{2ρ − 1, 2ρ} | ρ = 1, 2, . . . , k}, then
[a0 , 3 + 8(2ρ), 3 + 8(2ρ − 1) ✶ 4 + 8(2ρ)] ∈ C. Replace these blocks with
[∞4 , 3 + 8(2ρ), 3 + 8(2ρ − 1) ✶ 4 + 8(2ρ)].
Note that the blocks in C cover all the edges of KW except the following: {3+8i, a0 },
{3 + 8i, ∞3 }, i = 1, 2, . . . , 2k.
Step 6. Suppose k = 1. Then remove from C the block [c0 , 6, ∞3 ✶ a0 ] and form the
following blocks [c0 , 6, ∞3 ✶ 19], [11, ∞3 , a0 ✶ 19].
It is easy to see that the for k = 1 the proof is completed. From now on we suppose
that k ≥ 2.
Step 7. For ρ = 1, 2, . . . , k, let σρ ∈ {1, 2, . . . , 2k} such that Fσρ ∩G = {τ1 , τ2 }, τ1 < τ2 ,
σρ ∈ {1, 2, . . . , 2k}. For each ρ, remove from C the block [aσρ , 3+8τ2 , 3+8τ1 ✶ 4+8τ2 ].
Since by definition G is a one-factor of the complete graph on point set
{1, 2, . . . , 2k}, then ∪kρ=1 Fσρ ∩ G = {1, 2, . . . , 2k}. It follows that
∪{τ1 ,τ2 }∈G {3 + 8τ1 , 3 + 8τ2 } = {3 + 8i | i = 1, 2, . . . , 2k}
Then we can form the blocks [∞3 , 3 + 8τ2 , 3 + 8τ1 ✶ 4 + 8τ2 ].
Note that the blocks in C cover all the edges of KW except the following: {3+8i, a0 },
i = 1, 2, . . . , 2k, and {3 + 8τ1 , aσρ }, {3 + 8τ2 , aσρ }, {τ1 , τ2 } ∈ Fσρ ,
ρ = 1, 2, . . . , k.
Step 8. Consider the following two one-factors of the complete graph K2k on vertex
set {3 + 8i | i = 1, 2, . . . , 2k}, Γ1 = {{3 + 8i, 3 + 8(i + 1)} | i = 1, 3, 5, . . . , 2k − 1}
and Γ2 = {{3 + 8τ1 , 3 + 8τ2 } | {τ1 , τ2 } ∈ G}.
12
Since |G ∩ F0 | = 0, then any edge of Γ1 has at most one vertex in common with
an edge of Γ2 . Let Λ be the bipartite graph having as vertices the edges of Γ1 ∪ Γ2 ,
and edges the pairs {E1 , E2 } such that E1 is an edge of Γ1 , E2 is an edge of Γ2 and
|E1 ∩E2 | = 1. It is easy to see that Λ is a regular graph of degree 2, decomposable into
even cycles. Then it is possible to decompose Λ into two disjoint one-factors. Take
one of these one-factors. It is easy to see that this one-factor induces a one-to-one
mapping g from the edge set of Γ1 to the edge set of Γ2 such that E1 and g(E1 ) have
exactly one vertex in common, for each edge E1 of Γ1 .
Let E1 = {3 + 8(2ρ − 1), 3 + 8(2ρ)}. Put g({3 + 8(2ρ − 1), 3 + 8(2ρ)}) =
{3 + 8τ1 , 3 + 8τ2 }. Suppose that 3 + 8(2ρ − 1) = 3 + 8τ1 and 3 + 8(2ρ) = 3 + 8τ2
(similarly we can prove the theorem in the other cases). Now we form blocks by using
the noncovered edges of KW (see Step 7).
Suppose σρ ≤ k. Remove from C the block [d2σρ −1 , bk+σρ , a0 ✶ aσρ ]. Using the
edges of this block and the edges {a0 , 3 + 8(2ρ − 1)}, {a0 , 3 + 8(2ρ)}, {aσρ , 3 + 8τ1 } =
{aσρ , 3 + 8(2ρ − 1)}, {aσρ , 3 + 8τ2 }, form the following blocks:
[d2σρ −1 , bk+σρ , a0 ✶ 3 + 8(2ρ)] and [3 + 8(2ρ − 1), a0 , aσρ ✶ 3 + 8τ2 ].
Suppose σρ > k. Remove from C the block [d2k−σρ , bk , aσρ ✶ a0 ] and replace with
the following: [d2k−σρ , bk , aσρ ✶ 3 + 8τ2 ] and [3 + 8(2ρ − 1), aσρ , a0 ✶ 3 + 8(2ρ)]. ✷
Theorem 5 For each n ≡ 1
2n−11
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 25, SH(P3 , D, n) = {v | 5 ≤ v ≤
. Then
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−11
3
it is sufficient to prove that {v | 5 ≤ v ≤ 2n−11
,
v
≡
1
(mod
4)}
⊆
SH(P
,
3 D, n).
3
Put n = 25 + 24k, k ≥ 0.
Case k = 0. Let V0 = {χ0 , χ1 , χ2 , χ3 , χ4 } ∪ {1, 2, . . . , 8}, A0 = {a0 , b0 , c0 , d0 }, and
W0 = V0 ∪ A0 ∪ {∞1 , ∞2 , . . . , ∞8 }. Let D0 = { [∞1 , χ0 , χ1 ✶ χ4 ], [∞2 , χ1 , χ2 ✶ χ0 ],
[∞3 , χ2 , χ3 ✶ χ1 ], [∞4 , χ3 , χ4 ✶ χ2 ], [∞5 , χ4 , χ0 ✶ χ3 ], [∞6 , ∞4 , ∞5 ✶ χ1 ],
[∞1 , ∞4 , ∞7 ✶ χ2 ], [∞2 , ∞5 , ∞8 ✶ χ3 ], [∞3 , ∞4 , ∞8 ✶ χ1 ], [∞3 , ∞5 , ∞7 ✶ χ1 ],
[χ2 , ∞8 , ∞6 ✶ χ4 ], [∞2 , ∞4 , χ0 ✶ ∞3 ], [∞7 , ∞8 , χ0 ✶ ∞6 ], [∞3 , ∞6 , χ1 ✶ ∞4 ],
[∞1 , ∞5 , χ2 ✶ ∞4 ], [χ4 , ∞7 , ∞2 ✶ ∞3 ], [χ3 , ∞7 , ∞6 ✶ ∞2 ], [∞1 , ∞2 , χ3 ✶ ∞5 ],
[∞8 , ∞1 , χ4 ✶ ∞3 ], [a0 , ∞1 , ∞3 ✶ b0 ], [c0 , ∞6 , ∞1 ✶ 2]}, and let
H0 = { [∞7 , 4, 1 ✶ χ0 ], [∞8 , 1, 2 ✶ χ0 ], [∞2 , 1, 3 ✶ 2], [∞4 , 3, 4 ✶ 2], [b0 , 4, χ0 ✶ 3],
[∞8 , 8, 5 ✶ χ0 ], [∞5 , 5, 6 ✶ χ0 ], [∞6 , 5, 7 ✶ 6], [∞7 , 7, 8 ✶ 6], [a0 , 8, χ0 ✶ 7],
[a0 , 1, χ1 ✶ 2], [a0 , 2, χ2 ✶ 1], [a0 , 3, χ3 ✶ 4], [a0 , 4, χ4 ✶ 3], [c0 , χ3 , 1 ✶ χ4 ],
[c0 , χ4 , 2 ✶ χ3 ], [c0 , χ1 , 3 ✶ χ2 ], [c0 , χ2 , 4 ✶ χ1 ], [b0 , 5, χ1 ✶ 6], [b0 , 6, χ2 ✶ 5],
[b0 , 7, χ3 ✶ 8], [b0 , 8, χ4 ✶ 7], [d0 , χ3 , 5 ✶ χ4 ], [d0 , χ4 , 6 ✶ χ3 ], [d0 , χ1 , 7 ✶ χ2 ],
[d0 , χ2 , 8 ✶ χ1 ], [∞4 , 5, 1 ✶ 6], [∞4 , 6, 2 ✶ 5], [∞5 , 7, 3 ✶ 8], [∞2 , 8, 4 ✶ 7],
[∞1 , 3, 5 ✶ 4], [∞1 , 4, 6 ✶ 3], [∞3 , 1, 7 ✶ 2], [∞3 , 2, 8 ✶ 1], [a0 , ∞2 , 5 ✶ ∞3 ],
[a0 , ∞7 , 6 ✶ ∞6 ], [a0 , ∞4 , 7 ✶ ∞2 ], [c0 , 5, ∞7 ✶ b0 ], [c0 , 6, ∞3 ✶ 4], [c0 , 7, ∞8 ✶ 4],
[c0 , ∞4 , 8 ✶ ∞1 ], [b0 , 1, ∞1 ✶ 7], [b0 , 2, ∞5 ✶ 4], [b0 , ∞8 , 3 ✶ ∞6 ], [d0 , ∞5 , 1 ✶ ∞6 ],
[d0 , ∞7 , 2 ✶ ∞2 ], [d0 , ∞3 , 3 ✶ ∞7 ], [d0 , 4, ∞6 ✶ 8], [a0 , c0 , ∞5 ✶ 8], [a0 , b0 , ∞6 ✶ 2],
[a0 , d0 , ∞8 ✶ 6], [c0 , b0 , ∞2 ✶ 6], [c0 , χ0 , d0 ✶ ∞2 ], [b0 , ∞4 , d0 ✶ ∞1 ]}.
Put C0 = D0 ∪ H0 . Then (W0 , C0 ) is a D-design of order 25 with an embedded
H(13, 3, 1) based on point set V0 and three embedded H(v, 3, 1), v = 5, 9.
13
In order to prove the theorem for k ≥ 1, we construct (using the well-known
embedding v → v + 4 [4]) a H(13 + 16k, 3, 1) (V, P) having an embedded H(v, 3, 1)
for each admissible v, 5 ≤ v ≤ 13 + 16k. Then we embed (V, P) in a D-design (W, C)
of order 24 + 24k.
Put Vi = {χ0 , χ1 , χ2 , χ3 , χ4 }∪{1+8i, 2+8i, 3+8i, 4+8i, 5+8i, 6+8i, 7+8i, 8+8i},
Ai = {ai , bi , ci , di }, and Wi = Vi ∪ Ai ∪ {∞1 , ∞2 , . . . , ∞8 }, i = 0, 1, . . . , 2k. Let
V = ∪2k
i=0 Vi .
For each i = 1, 2, . . . , 2k, let (Wi , Ci ) be the D-design of order 25 isomorphic to
(W0 , C0 ), based on the point set Wi . Denote by (Vi , Pi ) the H(13, 3, 1) embedded in
(Wi , Ci ).
Let F = {F0 , F1 , . . . , F2k } be a near one-factorization of the complete graph K1+2k
on point set Z1+2k , such that
(1) for each i = 0, 1, . . . , 2k, i is the missing vertex in Fi ;
(2) F0 = {{2ρ − 1, 2ρ} | ρ = 1, 2, . . . , k}.
For each {x, y} ∈ Fi , x < y, define P (i, x, y) as in Theorem 4.
Let P = ∪2k
i=0 {Pi ∪ (∪{x,y}∈Fi P (i, x, y)}. Then (V, P) is the required
H(13 + 16k, 3, 1).
Now we embed (V, P) in a D-design (W, C) of order 25 + 24k. Let W = ∪2k
i=0 Wi .
Step 1. Put in C the blocks of C0 and the blocks of Ci − Di , i = 1, 2, . . . , 2k.
Step 2. For each {x, y} ∈ Fi , x < y, define B(i, x, y) as in Theorem 4.
Put in C the blocks of ∪2k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
For ρ = 1, 2, . . . , k, replace the blocks [a0 , 2 + 8(2ρ), 2 + 8(2ρ − 1) ✶ 1 + 8(2ρ)] with
[∞1 , 2 + 8(2ρ), 2 + 8(2ρ − 1) ✶ 1 + 8(2ρ)].
Step 3. Let
θ(k) =
and
k
2
k−1
2
τ (k) =
if k is even
if k is odd, k ≥ 3
k
2
k+1
2
if k is even
if k is odd.
Put in C the following blocks (the suffices are (mod 1 + 2k)):
(4) [d2ρ−1+σ , bk+ρ+σ , aσ ✶ aρ+σ ], [b2ρ+σ , bσ , cρ+σ ✶ cσ ], ρ = 1, 2, . . . , k, σ = 0, 1, . . . , 2k;
(5) [d4ρ−2+σ , b2ρ−1+σ , aσ ✶ c2k−2ρ+2+σ ], [d4ρ−2+σ , dσ , c2ρ−1+σ ✶ aσ ], ρ = 1, 2, . . . , τ (k),
σ = 0, 1, . . . , 2k;
(6) for k ≥ 2, [d4ρ+σ , b2ρ+σ , aσ ✶ c2ρ+σ ], [d4ρ+σ , dσ , c2ρ+σ ✶ a4ρ+σ ], ρ = 1, 2, . . . , θ(k),
σ = 0, 1, . . . , 2k.
Note that the blocks in C cover all the edges of KW except the following: {∞1 , ai },
{∞3 , ai }, {∞3 , bi }, {∞1 , ci }, {∞6 , ci }, {a0 , 2 + 8i}, i = 1, 2, . . . , 2k.
Step 4. Let
µ(t) =
2t + 1
2k − 2t
if t = 0, 1, . . . , k−1
2
k−1
k−1
if t = 2 + 1, 2 + 2, . . . , k − 1, k ≥ 2.
14
Let χ(t) be the element of Z2k+1 such that χ(t)µ(t) = k − t.
For t = 0, 1, . . . , k−1
replace the blocks [d2µ(t)−1 , bk+µ(t) , a0 ✶ aµ(t) ] and
2
[d(2+j)µ(t)−1 , bk+(1+j)µ(t) , ajµ(t) ✶ a(1+j)µ(t) ], j = 1, 2, . . . , χ(t), with
[d2µ(t)−1 , bk+µ(t) , a0 ✶ 2 + 8(t + 1)] and [d(2+j)µ(t)−1 , bk+(1+j)µ(t) , ajµ(t) ✶ a(j−1)µ(t) ] respectively.
Note that in the process we use the edges {a0 , 2 + 8(t + 1)} and we release
{aχ(t)µ(t) , a(χ(t)+1)µ(t) } = {ak−t , ak+t+1 }.
+1, k−1
+2, . . . , k−1 and j = 1, 2, . . . , χ(t)−1, replace
If k ≥ 2, then for t = k−1
2
2
the blocks [d2µ(t)−1 , bk+µ(t) , a0 ✶ aµ(t) ] and [d(2+j)µ(t)−1 , bk+(1+j)µ(t) , ajµ(t) ✶ a(1+j)µ(t) ]
with [d2µ(t)−1 , bk+µ(t) , a0 ✶ 2 + 8(t + 1)] and [d(2+j)µ(t)−1 , bk+(1+j)µ(t) , ajµ(t) ✶ a(j−1)µ(t) ]
respectively.
Note that in the process we use the edges {a0 , 2 + 8(t + 1)} and we release
{a(χ(t)−1)µ(t) , aχ(t)µ(t) } = {ak+t+1 , ak−t }.
It is easy to see that {{ak+t+1 , ak−t } | t = 0, 1, . . . , k − 1} is a one-factor of the
complete graph K2k on point set {a1 , a2 , . . . , a2k }. Then, for t = 0, 1, . . . , k − 1, we
can form the blocks [ak+t+1 , ak−t , ∞3 ✶ bt+1 ].
Note that the blocks in C cover all the edges of KW except the following: {∞1 , ai },
{∞1 , ci }, {∞6 , ci } for i = 1, 2, . . . , 2k, and {a0 , 2 + 8i}, {∞3 , bi }
for i = k + 1, k + 2, . . . , 2k.
Step 5. Replace the blocks [b2+σ , bσ , c1+σ ✶ cσ ], with the following:
[b2 , b0 , c1 ✶ ∞1 ] for σ = 0;
[b2+σ , bσ , c1+σ ✶ c2+σ ], σ = 1, 2, . . . , 2k.
It is easy to see that in the process we use the edge {∞1 , c1 } and we release {c1 , c2 }.
Step 6. If k = 1, then jump to Step 7. Suppose k ≥ 2. For σ = 2, 4, 6, . . . , 2k − 2,
replace the blocks [b2+σ , bσ , c1+σ ✶ c2+σ ] (note that we constructed these blocks in
Step 5), with [b2+σ , bσ , c1+σ ✶ ∞1 ].
It is easy to see that in the process we use the edges {∞1 , c1+σ } and we release
{c1+σ , c2+σ }, σ = 2, 4, 6, . . . , 2k − 2.
Step 7. Replace the blocks [d4ρ−2 , b2ρ−1 , a0 ✶ c2k−2ρ+2 ] (for ρ = 1, 2, . . . , τ (k)) and, if
k ≥ 2, [d4ρ , b2ρ , a0 ✶ c2ρ ] (for ρ = 1, 2, . . . , θ(k)), with
[d4ρ−2 , b2ρ−1 , a0 ✶ 2 + 8(k + 2ρ − 1)] (for ρ = 1, 2, . . . , τ (k)) and
[d4ρ , b2ρ , a0 ✶ 2 + 8(k + 2ρ)] (for ρ = 1, 2, . . . , θ(k)) respectively.
Note that in the process we use the edges {a0 , 2 + 8i}, i = k + 1, k + 2, . . . , 2k, and
we release {a0 , c2ρ }, ρ = 1, 2, . . . , k.
Step 8. For σ = 1, 3, 5, . . . , 2k − 1, replace the blocks [d2+σ , dσ , c1+σ ✶ aσ ], with
[d2+σ , dσ , c1+σ ✶ a0 ].
It is easy to see that in the process we use the edges {a0 , c2ρ }, ρ = 1, 2, . . . , k, and
we release {aσ , c1+σ }, σ = 1, 3, 5, . . . , 2k − 1.
Step 9. For σ = 1, 3, 5, . . . , 2k − 1, put in C the blocks [aσ , c1+σ , ∞1 ✶ a1+σ ].
Note that the blocks in C cover all the edges of KW except the following: {∞6 , ci },
i = 1, 2, . . . , 2k, {∞3 , bi }, i = k + 1, k + 2, . . . , 2k, {c2i−1 , c2i }, i = 1, 2, . . . , k.
15
Step 10. Replace the block (constructed in Step 6) [b2k , b2k−2 , c2k−1 ✶ ∞1 ], with
[c2k−1 , b2k−2 , b2k ✶ ∞3 ].
If k ≥ 2, then for ρ = 2, 3, . . . , k, replace the blocks (constructed in Step 3)
[b2k−ρ+1 , b2k−3ρ+1 , c2k−2ρ+1 ✶ c2k−3ρ+1 ], with [c2k−2ρ+1 , b2k−3ρ+1 , b2k−ρ+1 ✶ ∞3 ].
Note that the blocks in C cover all the edges of KW except the following:
{∞6 , ci }, i = 1, 2, . . . , 2k, {c2i−1 , c2i }, i = 1, 2, . . . , k, {∞1 , c2k−1 } and, if k ≥ 2,
{c2k−3ρ+1 , c2k−2ρ+1 }, ρ = 2, 3, . . . , k.
Step 11. Put in C the block [∞6 , c2k , c2k−1 ✶ ∞1 ] and, if k ≥ 2, the following
[∞6 , c2k−2ρ+2 , c2k−2ρ+1 ✶ c2k−3ρ+1 ], ρ = 2, 3, . . . , k. ✷
Theorem 6 For each n ≡ 8
2n−1
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 8, SH(P3 , D, n) = {v | 5 ≤ v ≤
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−1
. Then
3
2n−1
it is sufficient to prove that {v | 5 ≤ v ≤ 3 , v ≡ 1 (mod 4)} ⊆ SH(P3 , D, n).
Put n = 8 + 24k, k ≥ 0. The case k = 0 follows from Example 2. In order to prove
the theorem for k ≥ 1, we construct (using the well-known embedding v → v + 4 [4])
a H(5 + 16k, 3, 1) (V, P) such that for each admissible v, 5 ≤ v ≤ 5 + 16k, there is a
H(v, 3, 1) embedded in (V, P). Then we embed (V, P) in a D-design (W, C) of order
8 + 24k.
Put Vi = {0, 1 + 4i, 2 + 4i, 3 + 4i, 4 + 4i} and Ai = {ai , bi }, i = 0, 1, . . . , 4k.
Let (Wi , Ci ) be the D-design of order 9 isomorphic to the one constructed in Example 2 based on the point set Wi = Ai ∪ Vi ∪ {∞}. Denote by (Vi , Pi ) the H(5, 3, 1)
embedded in (Wi , Ci ).
Let F = {F0 , F1 , . . . , F4k } be a near one-factorization of the complete graph K1+4k
on point set Z1+4k , such that i is the missing vertex in Fi for each i = 0, 1, . . . , 4k.
Let {x, y}, x < y, be an edge of Fi . Define
P (i, x, y) = { (1 + 4x, 1 + 4y, 2 + 4x), (1 + 4x, 2 + 4y, 2 + 4x), (3 + 4x, 3 + 4y, 4 + 4x),
(3+4x, 4+4y, 4+4x), (3+4y, 1+4x, 4+4y), (3+4y, 2+4x, 4+4y), (1+4y, 3+4x, 2+4y),
(1 + 4y, 4 + 4x, 2 + 4y)}.
4k
Let V = ∪4k
i=0 Vi and P = ∪i=0 {Pi ∪(∪{x,y}∈Fi P (i, x, y)}. Then (V, P) is the required
H(5 + 16k, 3, 1).
Now we embed (V, P) in a D-design (W, C) of order 8 + 24k. Let W = ∪4k
i=0 Wi .
Step 1. Put in C the blocks of C0 and of Ci − {[ai , 2 + 4i, ∞ ✶ 0]} for i = 1, 2, . . . , 4k
(note that [ai , 2 + 4i, ∞ ✶ 0] ∈ Ci ). Let Ti = {ai , 2 + 4i, ∞}.
Step 2. Let {x, y}, x < y, be an edge of the near one-factor Fi . Define
B(i, x, y) = { [ai , 1 + 4x, 1 + 4y ✶ 2 + 4x], [ai , 2 + 4x, 2 + 4y ✶ 1 + 4x],
[ai , 3 + 4x, 3 + 4y ✶ 4 + 4x], [ai , 4 + 4x, 4 + 4y ✶ 3 + 4x), [bi , 3 + 4y, 1 + 4x ✶ 4 + 4y],
[bi , 4 + 4y, 2 + 4x ✶ 3 + 4y], [bi , 1 + 4y, 3 + 4x ✶ 2 + 4y], [bi , 2 + 4y, 4 + 4x ✶ 1 + 4y]}.
Put in C the blocks of ∪4k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
Step 3. Put in C the following blocks (the suffices are (mod 1 + 4k)):
(1) For each even ρ, 0 ≤ ρ ≤ k − 1, [a2+4ρ+σ , b1+2ρ+σ , aσ ✶ a1+2ρ+σ ],
16
[b4+4ρ+σ , a2+2ρ+σ , bσ ✶ b2+2ρ+σ ], σ = 0, 1, . . . , 4k.
(2) If k ≥ 2, then for each odd ρ, 0 ≤ ρ ≤ k − 1, [a2+4ρ+σ , b1+2ρ+σ , aσ ✶ a2+2ρ+σ ],
[b4+4ρ+σ , a2+2ρ+σ , bσ ✶ b1+2ρ+σ ], σ = 0, 1, . . . , 4k.
The triangles Ti and the blocks given in Steps 1, 2 and 3 cover all the edges of
the complete graph KW . In order to complete the proof we will remove from C some
blocks. Then we rearrange their edges and the triangles Ti in order to form new
blocks of C.
Step 4. Let
θ(k) =
and
τ (k) =
k−2
k−1
k−1
k−2
if k is even
if k is odd
if k is even
if k is odd, k ≥ 3
Remove from C the blocks [a2+4ρ , b1+2ρ , a0 ✶ a1+2ρ ] for ρ = 0, 2, 4, . . . , θ(k). If k ≥ 2,
then remove also the following blocks [a2+4ρ , b1+2ρ , a0 ✶ a2+2ρ ], ρ = 1, 3, 5, . . . , τ (k).
Using the edges {a0 , a1+2ρ } and the triangles T1+2ρ , ρ = 0, 2, 4, . . . , θ(k), form the
blocks [∞, 2 + 4(1 + 2ρ), a1+2ρ ✶ a0 ].
Using the edges {a0 , a2+4ρ } and the triangles T2+4ρ , ρ = 0, 1, 2, . . . , k − 1, form the
blocks [∞, 2 + 4(2 + 4ρ), a2+4ρ ✶ a0 ].
If k ≥ 2 then, using the edges {a0 , a2+2ρ } and the triangles T2+2ρ ,
ρ = 1, 3, 5, . . . , τ (k), form the blocks [∞, 2 + 4(2 + 2ρ), a2+2ρ ✶ a0 ].
Step 5. Remove from C the blocks [a0 , b4k−2ρ , a4k−4ρ−1 ✶ a4k−2ρ ], ρ = 0, 2, 4, . . . , θ(k).
Using their edges, the triangles T4k−2ρ and the edges {a0 , b1+2ρ }, form the blocks
[a4k−4ρ−1 , b4k−2ρ , a0 ✶ b1+2ρ ], [∞, 2+4(4k−2ρ), a4k−2ρ ✶ a4k−4ρ−1 ], ρ = 0, 2, 4, . . . , θ(k).
Step 6. Remove from C the blocks [a4+4ρ , b3+4ρ , a2+4ρ ✶ a3+4ρ ], ρ = 0, 1, 2, . . . , k − 1.
Using their edges, the triangles T3+4ρ and the edges {a2+4ρ , b1+2ρ }, form the following
blocks: [a4+4ρ , b3+4ρ , a2+4ρ ✶ b1+2ρ ], [∞, 2 + 4(3 + 4ρ), a3+4ρ ✶ a2+4ρ ],
ρ = 0, 1, 2, . . . , k − 1.
It is easy to see that for k = 1 the proof is completed.
Step 7. (I) Let k be an odd integer, k ≥ 3. Remove from C the blocks
[a0 , b4k−2ρ , a4k−4ρ−1 ✶ a4k−2ρ+1 ], [a4k−2ρ+1 , b4k−2ρ−2 , a4k−2ρ−5 ✶ a4k−2ρ−1 ],
ρ = 1, 3, 5, . . . , k − 2. Using their edges, the triangles T4k−2ρ−1 and the edges
{a0 , b1+2ρ }, form the following blocks [a4k−4ρ−1 , b4k−2ρ , a0 ✶ b1+2ρ ],
[a4k−2ρ−5 , b4k−2ρ−2 , a4k−2ρ+1 ✶ a4k−4ρ−1 ], [∞, 2 + 4(4k − 2ρ − 1), a4k−2ρ−1 ✶ a4k−2ρ−5 ],
ρ = 1, 3, 5, . . . , k − 2.
It is easy to see that for odd k the proof is completed.
17
(II) Let k be a positive even integer. If k = 2 then jump to Step 9. Suppose k ≥ 4.
Put
k
if k2 is odd
t(k) = k2
− 1 if k2 is even
2
For ρ = 1, 3, 5, . . . , t(k), remove from C the blocks [a0 , b4k−2ρ , a4k−4ρ−1 ✶ a4k−2ρ+1 ],
[a4k−2ρ+1 , b4k−4ρ , a4k−6ρ−1 ✶ a4k−4ρ+1 ]. Using their edges, the triangles T4k−4ρ+1 and
the edges {a0 , b2k−4ρ−1 } (where we put {a0 , b2k−1 } instead of {a0 , b−1 } if k2 is odd),
form the following blocks (for ρ = 1, 3, 5, . . . , t(k)): [a4k−4ρ−1 , b4k−2ρ , a0 ✶ b2k−4ρ−1 ],
[a4k−6ρ−1 , b4k−4ρ , a4k−2ρ+1 ✶ a4k−4ρ−1 ], [∞, 2 + 4(4k − 4ρ + 1), a4k−4ρ+1 ✶ a4k−6ρ−1 ].
Step 8. Let k be even, k ≥ 4.
(I) Let k2 be odd. For ρ = 1, 2, . . . , k−2
, remove from C the blocks
4
[a4k−8ρ−2 , b2k−8ρ−1 , a4k+1−8ρ ✶ a2k−8ρ ]. Using their edges, the triangles T4k−8ρ+1 , and
the edges {a0 , b8ρ−5 }, form the following blocks
[∞, 2 + 4(4k − 8ρ + 1), a4k−8ρ+1 ✶ a2k−8ρ ], [a4k+1−8ρ , a4k−8ρ−2 , b2k−8ρ−1 ✶ a0 ],
ρ = 1, 2, . . . , k−2
.
4
It is easy to see that in this case the proof is completed.
(II) Let k2 be even. If k = 4 then jump to Step 9. If k ≥ 6, then remove the
. Then using their
blocks [a4k−8ρ−2 , b2k−8ρ−1 , a4k+1−8ρ ✶ a2k−8ρ ], ρ = 1, 2, . . . , k−4
4
edges, the triangles T4k−8ρ+1 and the edges {a0 , b8ρ−1 }, form the following blocks
[∞, 2 + 4(4k − 8ρ + 1), a4k−8ρ+1 ✶ a2k−8ρ ], [a4k−8ρ−2 , a4k−8ρ+1 , b2k−8ρ−1 ✶ a0 ],
.
ρ = 1, 2, . . . , k−4
4
Step 9. Let k be a positive even integer. Remove the block [a0 , 2, ∞ ✶ 0]. Then using
these edges, the triangle T2k+1 , and the edge {a0 , b2k−1 }, form the following blocks
[∞, 2, a0 ✶ b2k−1 ], [a2k+1 , 2 + 4(2k + 1), ∞ ✶ 0].
It is easy to see that the proof is completed. ✷
Theorem 7 For each n ≡ 9
2n−3
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 9, SH(P3 , D, n) = {v | 5 ≤ v ≤
. Then
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−3
3
2n−3
it is sufficient to prove that {v | 5 ≤ v ≤ 3 , v ≡ 1 (mod 4)} ⊆ SP(P3 , D, n).
Put n = 9 + 24k, k ≥ 0.
Case k = 0. Let V0 = {0, 1, 2, 3, 4}, A0 = {a0 , b0 } and W0 = V0 ∪ A0 ∪ {∞1 , ∞2 }.
Let
C0 = { [a0 , 0, 2 ✶ 1], [a0 , 4, 1 ✶ 0], [b0 , 3, 0 ✶ 4], [∞1 , 1, 3 ✶ 2], [∞2 , 2, 4 ✶ 3],
[∞2 , 1, b0 ✶ 4], [∞1 , 2, b0 ✶ a0 ], [∞2 , 3, a0 ✶ ∞1 ], [0, ∞2 , ∞1 ✶ 4]}.
It is easy to see that (W0 , C0 ) is a D-design of order 9 with an embedded H(5, 3, 1)
based on point set V0 .
In order to prove the theorem for k ≥ 1, we construct (using the well-known
embedding v → v + 4 [4]) a H(5 + 16k, 3, 1) (V, P) having an embedded H(v, 3, 1) for
each admissible v, 5 ≤ v ≤ 5 + 16k. Then we embed (V, P) in a D-design (W, C) of
order 9 + 24k.
18
Put Vi = {0, 1+4i, 2+4i, 3+4i, 4+4i}, Ai = {ai , bi }, and Wi = Vi ∪Ai ∪{∞1 , ∞2 },
i = 0, 1, . . . , 4k.
For each i = 1, 2, . . . , 4k, let (Wi , Ci ) be the D-design of order 9 isomorphic to
(W0 , C0 ), based on the point set Wi . Denote by (Vi , Pi ) the H(5, 3, 1) embedded in
(Wi , Ci ).
Let V = ∪4k
i=0 Vi . Let (V, P) be the H(5 + 16k, 3, 1) given in Theorem 6. Now we
embed (V, P) in a D-design (W, C) of order 9 + 24k. Let W = ∪4k
i=0 Wi .
Step 1. Put in C the blocks of C0 and the blocks of Ci − {[0, ∞2 , ∞1 ✶ 4 + 4i]},
i = 1, 2, . . . , 4k.
Step 2. Let F = {F0 , F1 , . . . , F4k } be the near one-factorization of the complete graph
K1+4k defined in Theorem 6.
Let {x, y}, x < y, be an edge of the near one-factor Fi . Define
B(i, x, y) = { [ai , 1 + 4x, 1 + 4y ✶ 2 + 4x], [ai , 2 + 4x, 2 + 4y ✶ 1 + 4x],
[ai , 3 + 4x, 3 + 4y ✶ 4 + 4x], [ai , 4 + 4x, 4 + 4y ✶ 3 + 4x), [bi , 3 + 4y, 1 + 4x ✶ 4 + 4y],
[bi , 4 + 4y, 2 + 4x ✶ 3 + 4y], [bi , 1 + 4y, 3 + 4x ✶ 2 + 4y], [bi , 2 + 4y, 4 + 4x ✶ 1 + 4y]}.
Put in C the blocks of ∪4k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
Step 3. Put in C the following blocks (the suffices are (mod 1 + 4k)):
(1) for each even ρ, 0 ≤ ρ ≤ k − 1, [a2+4ρ+σ , b1+2ρ+σ , aσ ✶ a1+2ρ+σ ],
[b4+4ρ+σ , a2+2ρ+σ , bσ ✶ b2+2ρ+σ ], σ ∈ Z1+4k ;
(2) for each odd ρ, 0 ≤ ρ ≤ k − 1, [a2+4ρ+σ , b1+2ρ+σ , aσ ✶ a2+2ρ+σ ],
[b4+4ρ+σ , a2+2ρ+σ , bσ ✶ b1+2ρ+σ ], σ = 0, 1, . . . , 4k.
Step 4. For ρ = 0, 1, . . . , k − 1, remove from C the blocks
[∞1 , 2 + 4(1 + 4ρ), b1+4ρ ✶ a1+4ρ ], [∞1 , 2 + 4(2 + 4ρ), b2+4ρ ✶ a2+4ρ ],
[∞1 , 2 + 4(3 + 4ρ), b3+4ρ ✶ a3+4ρ ], [∞2 , 3 + 4(1 + 4ρ), a1+4ρ ✶ ∞1 ],
[∞2 , 3 + 4(2 + 4ρ), a2+4ρ ✶ ∞1 ], [a3+4ρ , b2+4ρ , a1+4ρ ✶ a2+4ρ ]. Then, using their edges,
and the edges {∞1 , 4 + 4(1 + 4ρ)}, {∞1 , 4 + 4(2 + 4ρ)}, {∞1 , 4 + 4(3 + 4ρ)} and
{∞1 , 4 + 4(4 + 4ρ)}, form the following blocks
[b1+4ρ , 2 + 4(1 + 4ρ), ∞1 ✶ 4 + 4(1 + 4ρ)], [b2+4ρ , 2 + 4(2 + 4ρ), ∞1 ✶ 4 + 4(2 + 4ρ)],
[b3+4ρ , 2 + 4(3 + 4ρ), ∞1 ✶ 4 + 4(3 + 4ρ)], [∞2 , 3 + 4(1 + 4ρ), a1+4ρ ✶ b1+4ρ ],
[∞2 , 3 + 4(2 + 4ρ), a2+4ρ ✶ b2+4ρ ], [a1+4ρ , b2+4ρ , a3+4ρ ✶ b3+4ρ ],
[a1+4ρ , a2+4ρ , ∞1 ✶ 4 + 4(4 + 4ρ)], ρ = 0, 1, . . . , k − 1. ✷
Theorem 8 For each n ≡ 16
2n−5
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 16, SH(P3 , D, n) = {v | 5 ≤ v ≤
. Then
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−5
3
2n−5
it is sufficient to prove that {v | 5 ≤ v ≤ 3 , v ≡ 1 (mod 4)} ⊆ SH(P3 , D, n).
Put n = 16 + 24k, k ≥ 0.
Case k = 0. Let V0 = {0, 1, 2, . . . , 8}, A0 = {a0 , b0 , c0 , d0 }, and
W0 = V0 ∪ A0 ∪ {∞1 , ∞2 , ∞3 }. Let C0 = { [∞2 , 4, 1 ✶ 0], [∞1 , 1, 2 ✶ 0], [d0 , 2, 3 ✶ 1],
[∞3 , 3, 4 ✶ 2], [b0 , 3, 0 ✶ 4], [∞2 , 8, 5 ✶ 0], [c0 , 5, 6 ✶ 0], [a0 , 6, 7 ✶ 5], [b0 , 7, 8 ✶ 6],
19
[a0 , 8, 0 ✶ 7], [d0 , 6, 1 ✶ 5], [b0 , 5, 2 ✶ 6], [∞1 , 7, 3 ✶ 8], [c0 , 7, 4 ✶ 8], [d0 , 4, 5 ✶ 3],
[∞2 , 3, 6 ✶ 4], [∞3 , 1, 7 ✶ 2], [∞3 , 2, 8 ✶ 1], [∞2 , 7, d0 ✶ ∞3 ], [0, d0 , c0 ✶ ∞3 ],
[b0 , d0 , a0 ✶ ∞1 ], [∞3 , 6, b0 ✶ ∞2 ], [c0 , b0 , 1 ✶ a0 ], [a0 , 3, c0 ✶ 2], [∞3 , 5, a0 ✶ 4],
[∞1 , d0 , 8 ✶ c0 ], [b0 , 4, ∞1 ✶ 5], [c0 , ∞1 , ∞2 ✶ ∞3 ], [2, a0 , ∞2 ✶ 0], [∞3 , 0, ∞1 ✶ 6].
It is easy to see that (W0 , C0 ) is a D-design of order 16 with an embedded H(9, 3, 1)
based on point set V0 and two embedded H(5, 3, 1) based on point sets {0, 1, 2, 3, 4}
and {0, 5, 6, 7, 8} respectively.
In order to prove the theorem for k ≥ 1 we construct (using the well-known
embedding v → v + 4 [4]) a H(9 + 16k, 3, 1) (V, P) having an embedded H(v, 3, 1) for
each admissible v, 5 ≤ v ≤ 9 + 16k. Then we embed (V, P) in a D-design (W, C) of
order 16 + 24k.
Put Vi = {0, 1 + 8i, 2 + 8i, 3 + 8i, 4 + 8i, 5 + 8i, 6 + 8i, 7 + 8i, 8 + 8i}, V = ∪2k
i=0 Vi ,
Ai = {ai , bi , ci , di }, and Wi = Vi ∪ Ai ∪ {∞1 , ∞2 , ∞3 }, i = 0, 1, . . . , 2k.
For each i = 1, 2, . . . , 2k, let (Wi , Ci ) be the D-design of order 16 isomorphic to
(W0 , C0 ), based on the point set Wi . Denote by (Vi , Pi ) the H(9, 3, 1) embedded in
(Wi , Ci ).
Let F = {F0 , F1 , . . . , F2k } be a near one-factorization of the complete graph K1+2k
on point set Z1+2k , such that
(1) for each i = 0, 1, . . . , 2k, let i be the missing vertex in Fi ;
(2) F0 = {{2ρ − 1, 2ρ} | ρ = 1, 2, . . . , k}.
For each {x, y} ∈ Fi , x < y, define P (i, x, y) as in Theorem 4.
Let P = ∪2k
i=0 {Pi ∪(∪{x,y}∈Fi P (i, x, y)}. Then (V, P) is the required H(9+16k, 3, 1).
Now we embed (V, P) in a D-design (W, C) of order 16 + 24k. Let W = ∪2k
i=0 Wi .
Step 1. Put in C the blocks of C0 and the blocks of
Ci − {[∞3 , 0, ∞1 ✶ 6 + 8i], [2 + 8i, ai , ∞2 ✶ 0], [ci , ∞1 , ∞2 ✶ ∞3 ]}, i = 1, 2, . . . , 2k.
Step 2. Put in C the blocks [2 + 8i, ai , ∞2 ✶ ci ], i = 1, 2, . . . , 2k.
Step 3. For each {x, y} ∈ Fi , x < y, define B(i, x, y) as in Theorem 4.
Put in C the blocks of ∪2k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
Step 4. For ρ = 1, 2, . . . , k, put in C the blocks (the suffices are
(mod 1 +
2k)) [d2ρ+σ , bρ+σ , aσ ✶ cρ+σ ], [d2ρ−1+σ , bk+ρ+σ , aσ ✶ aρ+σ ], [b2ρ+σ , bσ , cρ+σ ✶ cσ ],
[d2ρ+σ , dσ , cρ+σ ✶ a2ρ+σ ], σ = 0, 1, . . . , 2k.
Note that the blocks in C cover all the edges of KW except the following: {∞1 , 6 +
8i}, {∞1 , ci }, i = 1, 2, . . . , 2k.
Step 5. Since F0 = {{2ρ − 1, 2ρ} | ρ = 1, 2, . . . , k}, then
[a0 , 6 + 8(2ρ), 6 + 8(2ρ − 1) ✶ 5 + 8(2ρ)] ∈ C. Replace these blocks with
[∞1 , 6 + 16ρ, 6 + 8(2ρ − 1) ✶ 5 + 16ρ].
Note that the blocks in C cover all the edges of KW except the following: {a0 , 6+8i},
{∞1 , ci }, i = 1, 2, . . . , 2k.
Step 6. For ρ = 1, 2, . . . , k, replace the blocks
[d2ρ , bρ , a0 ✶ cρ ], and [d2ρ−1 , bk+ρ , a0 ✶ aρ ], with
20
[d2ρ , bρ , a0 ✶ 6 + 8ρ], and [d2ρ−1 , bk+ρ , a0 ✶ 6 + 8(k + ρ)] respectively.
Note that the blocks in C cover all the edges of KW except the following: {a0 , ci },
{a0 , ai }, i = 1, 2, . . . , k, and {∞1 , ci }, i = 1, 2, . . . , 2k.
Step 7. For σ = 1, 2, . . . , k, replace the blocks [d2k+σ , bk+σ , aσ ✶ ck+σ ], with
[d2k+σ , bk+σ , aσ ✶ a0 ].
Note that the blocks in C cover all the edges of KW except the following: {a0 , ci },
{ai , ck+i }, i = 1, 2, . . . , k, and {∞1 , ci }, i = 1, 2, . . . , 2k.
Step 8. Let
θ(k) =
and
τ (k) =
k+1
k
if k is odd
if k is even
k−2
k−1
if k is odd
if k is even
For σ = τ (k) + 2, τ (k) + 4, τ (k) + 6, . . . , 2k − 1, replace the blocks [b2+σ , bσ , c1+σ ✶ cσ ],
with [b2+σ , bσ , c1+σ ✶ aσ−k+1 ] (note that τ (1) = −1).
If k ≥ 2, then for σ = 1, 3, 5, . . . , τ (k), replace the blocks [b2+σ , bσ , c1+σ ✶ cσ ], with
[b2+σ , bσ , c1+σ ✶ a0 ].
Note that the blocks in C cover all the edges of KW except the following: {∞1 , ci }
for i = 1, 2, . . . , 2k, {ci , c1+i } for i = 1, 3, 5, . . . , 2k − 1, {a0 , ci−1 } for
i = 2, 4, 6, . . . , θ(k), and, for i = 1, 3, 5, . . . , τ (k), either {ai+1 , ck+i+1 } if k is odd, or
{ai , ck+i } if k is even.
Step 9. Put in C the blocks [∞1 , ci+1 , ci ✶ a0 ] (i = 1, 3, 5, . . . , τ (k)) and either
[∞1 , ck+i+2 , ck+i+1 ✶ ai+1 ] (i = 1, 3, . . . , θ(k) − 1) if k is odd, or [∞1 , ck+i+1 , ck+i ✶ ai ]
(i = 1, 3, 5, . . . , τ (k)) if k is even. ✷
Theorem 9 For each n ≡ 17
2n−7
, v ≡ 1 (mod 4)}.
3
(mod 24), n ≥ 17, SH(P3 , D, n) = {v | 5 ≤ v ≤
. Then
Proof. From Theorem 3, it follows that v ∈ SH(P3 , D, n) implies v ≤ 2n−7
3
2n−7
it is sufficient to prove that {v | 5 ≤ v ≤ 3 , v ≡ 1 (mod 4)} ⊆ SH(P3 , D, n).
Put n = 17 + 24k, k ≥ 0.
Case k = 0. Let V0 = {0, 1, 2, . . . , 8}, A0 = {a0 , b0 , c0 , d0 }, and
W0 = V0 ∪ A0 ∪ {∞1 , ∞2 , ∞3 , ∞4 }. Let C0 = { [a0 , 0, 1 ✶ 4], [∞3 , 1, 2 ✶ 0],
[a0 , 2, 3 ✶ 1], [∞4 , 2, 4 ✶ 3], [c0 , 3, 0 ✶ 4], [∞4 , 8, 5 ✶ 0], [∞3 , 5, 6 ✶ 0], [∞1 , 5, 7 ✶ 6],
[∞3 , 7, 8 ✶ 6], [b0 , 8, 0 ✶ 7], [b0 , 6, 1 ✶ 5], [c0 , 5, 2 ✶ 6], [∞2 , 7, 3 ✶ 8], [d0 , 7, 4 ✶ 8],
[∞2 , 4, 5 ✶ 3], [∞4 , 3, 6 ✶ 4], [∞4 , 1, 7 ✶ 2], [∞2 , 1, 8 ✶ 2], [a0 , c0 , 7 ✶ b0 ],
[∞4 , b0 , c0 ✶ 4], [d0 , 2, ∞1 ✶ 1], [∞1 , 4, b0 ✶ ∞3 ], [c0 , 6, ∞1 ✶ 3], [a0 , ∞1 , 8 ✶ c0 ],
[d0 , a0 , 6 ✶ ∞2 ], [b0 , 5, d0 ✶ 8], [4, ∞3 , a0 ✶ b0 ], [3, ∞3 , d0 ✶ ∞4 ], [2, ∞2 , b0 ✶ 3],
[1, c0 , d0 ✶ 0], [∞2 , ∞4 , a0 ✶ 5], [c0 , ∞3 , ∞2 ✶ d0 ], [0, ∞3 , ∞4 ✶ ∞1 ],
[0, ∞2 , ∞1 ✶ ∞3 ].
21
It is easy to see that (W0 , C0 ) is a D-design of order 17 with an embedded H(9, 3, 1)
based on point set V0 and two embedded H(5, 3, 1) based on point sets {0, 1, 2, 3, 4}
and {0, 5, 6, 7, 8} respectively.
In order to prove the theorem for k ≥ 1, we construct (using the well-known
embedding v → v + 4 [4]) a H(9 + 16k, 3, 1) (V, P) having an embedded H(v, 3, 1) for
each admissible v, 5 ≤ v ≤ 9 + 16k. Then we embed (V, P) in a D-design (W, C) of
order 17 + 24k.
Put Vi = {0, 1 + 8i, 2 + 8i, 3 + 8i, 4 + 8i, 5 + 8i, 6 + 8i, 7 + 8i, 8 + 8i}, V = ∪2k
i=0 Vi ,
Ai = {ai , bi , ci , di }, and Wi = Vi ∪ Ai ∪ {∞1 , ∞2 , ∞3 , ∞4 }, i = 0, 1, . . . , 2k.
For each i = 1, 2, . . . , 2k, let (Wi , Ci ) be the D-design of order 17 isomorphic to
(W0 , C0 ), based on the point set Wi . Denote by (Vi , Pi ) the H(9, 3, 1) embedded in
(Wi , Ci ).
As in Theorem 8, we construct a H(9 + 16k, 3, 1) (V, P) having an embedded
H(v, 3, 1) for each admissible v, 5 ≤ v ≤ 9 + 16k.
Now we embed (V, P) in a D-design (W, C) of order 17 + 24k. Let W = ∪2k
i=0 Wi .
Step 1. Put in C the blocks of C0 and the blocks of
Ci − { [∞2 , ∞4 , a0 ✶ 5], [c0 , ∞3 , ∞2 ✶ d0 ], [0, ∞3 , ∞4 ✶ ∞1 ], [0, ∞2 , ∞1 ✶ ∞3 ]},
i = 1, 2, . . . , 2k.
Step 2. Let B(i, x, y) be the block set defined as in Step 3 of Theorem 8. Put in C
the blocks of ∪2k
i=0 {∪{x,y}∈Fi B(i, x, y)}.
Step 3. For ρ = 1, 2, . . . , k, put in C the blocks (the suffices are
(mod 1 +
2k)) [c2k−ρ+1+σ , bρ+σ , aσ ✶ aρ+σ ], [b2ρ+σ , bσ , dρ+σ ✶ dσ ], [c2ρ+σ , cσ , dρ+σ ✶ aσ ],
[cρ+σ , b2k−ρ+1+σ , aσ ✶ d2k−ρ+1+σ ], σ = 0, 1, . . . , 2k.
Note that the blocks in C cover all the edges of KW except the following: {ai , 5+8i},
{ai , ∞2 }, {ai , ∞4 }, {di , ∞2 }, {ci , ∞2 }, {ci , ∞3 }, i = 1, 2, . . . , 2k.
Step 4. Replace the blocks
[c2k+σ , b1+σ , aσ ✶ a1+σ ], σ = 1, 3, 5, . . . , 2k − 1, and
[c1+σ , b2k+σ , aσ ✶ d2k+σ ], σ = 2, 4, 6, . . . , 2k,
with
[c2k+σ , b1+σ , aσ ✶ 5 + 8σ], σ = 1, 3, 5, . . . , 2k − 1, and
[c1+σ , b2k+σ , aσ ✶ 5 + 8σ], σ = 2, 4, 6, . . . , 2k, respectively.
Step 5. For i = 1, 2, . . . , k, construct the triangles
Ti = {a2i−1 , a2i , ∞4 }, Γi = {a2i , d2i−1 , ∞2 }.
Note that the blocks in C and the triangles above constructed cover all the edges
of KW except the following:
{a2i−1 , ∞2 }, {d2i , ∞2 }, i = 1, 2, . . . , k, and
{ci , ∞2 }, {ci , ∞3 }, i = 1, 2, . . . , 2k.
Step 6. For σ = 1, 2, . . . , 2k, replace the blocks [c2+σ , cσ , d1+σ ✶ aσ ], with
[d1+σ , c2+σ , cσ ✶ ∞3 ].
22
Step 7. For i = 1, 2, . . . , k, construct the triangles Λi = {a2i−1 , d2i , ∞2 }.
Note that the blocks in C and the triangles Ti , Γi and Λi , cover all the edges of
KW except the following:
{a2i , d2i+1 }, i = 1, 2, . . . , k, and
{ci , ∞2 }, i = 1, 2, . . . , 2k.
Step 8. For i = 1, 2, . . . , k, put in C the blocks [∞4 , a2i−1 , a2i ✶ d2i+1 ],
[a2i , d2i−1 , ∞2 , ✶ ci ] and [a2i−1 , d2i , ∞2 , ✶ ck+i ]. ✷
References
[1] J.C. Bermond and J. Schönheim, G-decomposition of Kn , where G has four
vertices or less, Discrete Math., 19 (1977), 113–120.
[2] C.J. Colbourn and J.H. Dinitz, CRC Handbook of Combinatorial Designs, CRC
Press, Boca Raton, FL, (1996).
[3] P. Hell and A. Rosa, Graph decompositions, handcuffed prisoners, and balanced
P -designs, Discrete Math., 2 (1972), 229–252.
[4] S.H.Y. Hung and N.S. Mendelsohn, Handcuffed designs, Aequationes Math., 18
(1974), 256–266.
[5] S. Milici and G. Quattrocchi, Embedding handcuffed designs with block size 2
or 3 in 4-cycle systems, Discrete Math., 208/209 (1999), 443-449.
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