Disjoint skolem sequences and related disjoint structures

Disjoint Skolem Sequences and Related Disjoint Structures*

C. Baker and N. Shalaby Department of Mathematics and Computer Science, Mount Allison University, Sackville, New Brunswick, Canada EOA 3CO

ABSTRACT

A Skolem sequence of order n is a sequence S = (sl, s2 . . . , szn) of 2n integers satisfying the following conditions: (1) for every k E {1, 2 , . . . ,n} there exist exactly two elements si,sj such that si = sj = k ; (2) If si = sj = k, i < j then j - i = k. In this article we show the existence of disjoint Skolem, disjoint hooked Skolem, and disjoint near-Skolem sequences. Then we apply these concepts to the existence problems of disjoint cyclic Steiner and Mendelsohn triple systems and the existence of disjoint 1-covering designs. 0 1993 John Wiley & Sons, Inc.

1. SEQUENCES AND CYCLIC STS

A Steiner triple system of order v, STS(v), is a pair of sets ( V , B ) , where IVl = v and B consists of 3-subsets (triples or blocks) of V such that any two elements of V occur in exactly one triple. An STS(v) exists if and only if v = 1 or 3 (mod 6). Two Steiner triple systems on the same set V are disjoint if they have no blocks in common. In [8,9,10,20], i t is shown that the maximum number of painvise disjoint Steiner triple systems on a v-element set is v - 2 for all v > 7. The maximum is 2 for v = 7.

A STS(v) is cyclic if its automorphism group contains a v-cyclic. A cyclic STS(v) exists for all v = 1 or 3 (mod 6) except for v = 9 [13]. This question of existence is equivalent to finding solutions to Heffter’s difference problems 161:

* Research partially supported by grants from NSERC Canada.

0 1YY3 John Wiley & Sons, Inc. Journal of Combinatorial Designs, Vol. 1, No. 5 (1993)

330 BAKER AND SHALABY

Heffter’s first difference problem: Can the set (1,. . . , 3 n } be partitioned into TI

ordered triples [ai, bi, ci], i = 1,. . . , n , such that in each triple either ai + hi + ci or ai + bi - ci = 0 (mod 6n + I)? Heffter’s second difference problem: Can the set {1,2,. . . , 2n , 2n + 2 , . . . , 3 n + I} be partitioned into n ordered triples [ai, bi, ci], i = 1,. . . , n, such that in each triple either ai + bi + c; or ai + hi - ci = 0 (mod 6n + 3)?

Cyclic STS(v) with v = 1 (mod 6) [3(mod 6 ) ] are generated by solutions to Heffter’s first [second] problem in the following way. Each triple [a , , h, , c,] in the partition gives a starter block (0, a , , a , + b,), plus, when v = 6n + 3, the additional “short starter block” (0,2n + 1,4n + 2). These blocks cycled modulo v are the blocks in the cyclic STS(v) . For example, {[l, 5,6], [2,11,13], [3,7, lo], [4,8,12]} is a solution to Heffter’s second problem with n = 4. This solution gives the base blocks ((0, 1,6), (0,2,13), (0,3, lo), (0,4,12)} (mod 27); plus (0,9,18) (mod 27), of a cyclic STS(27). In this way, we construct a STS(v ) on the set {O, ..., v - 1) that admits the automorphism (0,1,. . . , v - 1). We note that two cyclic STS(v) are disjoint if their sets of base blocks are disjoint. However, if v = 3 (mod 6), all such STS(v ) contain the short starter block and hence are not disjoint; cf. [3,4]. Let n, (v ) represent the maximum number of pairwise disjoint cyclic STS(v) constructed over (0,. . . , v - 1) that admit the automorphism (0,1,. . . , v - 1). Rosa [14] uses the notation d, , (2 ,3 , v ) and Colbourn n ( v ) for n,(v) . Then n,(6n + 3 ) = 1. Hence for disjoint cyclic STS(v) , we need only consider v = 1 (mod 6).

While studying Steiner triple systems, Skolem [18] asked whether it is possible to partition the set {1,2,. . . , 2 n } into n pairs (uk, vk), where vk - uk = k , for k = 1,. . . , n . He showed that such a partition is possible if and only if n = 0 or 1 (mod 4). Nickerson 11 I ] was the first to write this as a sequence of length 2 4 in which each k = 1, . . . , n is placed in the two positions uk and vk = uk + k . Such a partition or sequence is called a Skolem sequence of order n .

The set { [ k , uk + n , uk + k + n ] I k = 1,. . . , n } , where uk is defined as above, is also a solution to Heffter’s first problem. Hence the two sets of starter blocks { (O ,k ,uk + k + n ) I k = 1, . . . , n } and {(o,uk + n , uk + k + n ) I k = 1, . . . , n } give two disjoint STS(6n + 1) where n = 0 or 1 (mod 4). These two disjoint STS are related in exactly the same way as the two given in [5,Theorem 41.

To deal with the case where n = 2 or 3 (mod 4), Skolem [18] considered partitioning

O’Keefe [12] proved that such a partition did exist if and only if n = 2 or 3 (mod 4). Using Nickerson’s notation, we have a sequence (called a hooked Skolem sequence of order n ) of length 2n + 1 with a 0 (called the hook) in the 2nth position and each k = 1,. . . , n placed in the two positions uk and vk = uk + k . The two sets of starter blocks ((0, k , U k + k + n ) I k = 1,. . . , a } and ( (0 ,uk + n , U k + k + n ) I k = 1, . . . , n > again give two disjoint STS(6n + l), where n = 2 or 3 (mod 4).

For example, the set { I , . . . , S} is partitioned into the pairs (1,2), (4,6), (5, S), and (3,7). Written in Nickerson’s notation: 1, 1 ,4 ,2 ,3 ,2 ,4 ,3 is a Skolem sequence of order 4. This sequence gives the solution for the first Heffter problem for n = 4: [1,5,6], [2,8, 101, [3,9,12], [4,7,11]. This will give the base blocks for two disjoint cyclic STS(25):

the Set {1,2,.. . , 2 n - 1,2n + 1} into n pairs ( U k , V k ) , SO that Vk - U k = k = I , . . . , n .

1. 2. Therefore, 2 5 n , ( v ) . By [3], n, (v) 5 v - 5 for v = 1 (mod 6).

(0,1,6), (072, lo), (0,3,12), (0,4,11) (mod 251, (0,5,6), (0,8, lo), (0,9,12), (0,7,11) (mod 25).

DISJOINT SKOLEM SEQUENCES 331

In this article we introduce the concept of disjoint Skolem sequences and apply it to several unsolved problems in design theory concerning the existence of disjoint cyclic STS(v), disjoint cyclic Mendelsohn triple systems and disjoint A-covering designs when repeated elements are allowed.

2. DISJOINT SKOLEM AND HOOKED SKOLEM SEQUENCES

Formally then, a Skolem sequence of order n is an integer sequence S = ( ~ 1 , . . . , ~ 2 ~ )

such that i. every integer k = 1,. . . , n occurs in exactly two positions;

ii. if s, = s, = k , i < j , then j - i = k. A hooked Skolem sequence of order n is an integer sequence S = ( ~ 1 , . . . , s2,,+1)

satisfying properties (i) and (ii) for Skolem sequences plus 111. s2,, = 0 (szn is called the hook).

Both sequences can be represented as sets of disjoint pairs of integers, { ( i , j ) I s, = s,}. Two Skolem (or hooked Skolem) sequences S and S’ of order n are disjoint if

s, = s, = k = si = s: implies that { i , j } # { t , u}, for all k = 1 , . . . , n . Given a Skolem sequence, S = ( s l , . . . ,s2, ,) , the reverse S, = ( s z , , , . . . ,s1) is also

a Skolem sequence. If S and S, are disjoint, then S is reverse-disjoint. We give some examples to illustrate these definitions:

Two disjoint Skolem sequences of order 4 are 1 1 4 2 3 2 4 3 and 2 3 2 4 3 1 1 4. Note that both are reverse disjoint.

Two disjoint hooked Skolem sequences of order 7 are 5 7 1 1 6 5 3 4 7 3 6 4 2 0 2 a n d 6 1 1 5 7 2 6 2 5 3 4 7 3 0 4 .

Two disjoint Skolem sequences S and S’ each produce two disjoint sets of starter blocks, B1 = {(O,k,i + k + n)} and B2 = { (O , i + n , i + k + n)} for S and B{ =

{ (O,k , t + k + n)} and B: = ( (0 , t + n , t + k + n)} for S’, where i and t represent the location of the first occurrence of k in S and S’ respectively. If B1 n B: # 4, then some (i, j ) = ( t , u) ; a contradiction. Also t + n > n 2 k, so B1 n B: = 4 . Hence these four sets of starter blocks are disjoint.

Since the sequence 1 1 4 2 3 2 4 3 is reverse-disjoint, its reverse gives another two cyclic STS (25) disjoint from those given in Section 1:

4. (0,11,12), (0,7,9), (0,5,8), (0,6,10) (mod 25). Since each Skolern (or hooked Skolem) sequence must have an n in two places, n

positions apart, there are only n distinct ways of selecting the pair of positions to contain n. Therefore we obtain the following result.

...

3. (0,1,12), (0,2,9), (0,3,8), (0,4910) (mod 25),

Lemma 1. quences of order n is at most n.

The maximum number of mutually disjoint Skolem (hooked Skolem) se-

Theorem 2. Skolem sequences of order n.

For all n = 0 , l (mod 4), n 2 4, there exist at least four mutually disjoint

ProoJ: Skolerri sequences.

In each case, we present the partitions equivalent to two disjoint reverse-disjoint


Case 1. The first partition consists of the pairs: (4s, 8s + 2), (4s + 1,12s + l), (12s + 2,12s

((r, 8s - r + 1) I r = 1,. . . ,4s - l}, ((8s + 2r - 1,16s - 2r + 1) I r = 1,. . . , s - l}, ((8s + 2r + 2,16s - 2r + 2)1 r = 1 ,..., s-l}, ((10s + r + 1,14s - r + 3)lr =

1,. . . ,2s - l} and the second, (3,4), (10,12), (2, S), (9,13), (11,16), (1,7), (8,15), (6,14) if n = 8 and (8s - 1,16s - 2), (10s - 2,lOs - l), (12s, 16s), (16s - 3,16s - l), (4s - 1,12s - l), (4s,12s - 2), ((r,8s - r - 1)lr = 1 ,..., 4s - 2}, ((8s + r - 1,16s - r - 3) I r = 1,. . . ,2s - 2}, ((10s + r - 1,14s - r - 1) I r = 1,. . . ,2s - 2) otherwise. Case 2. n = 4 (mod 8). If n > 4, let n = 8s + 4 and the first partition is (4s + 2,8s + 6), (4s + 3,12s + 7), (12s + 5,12s + 6), (14s + 8,14s + lo), {(r, 8s - r + 5) I r = 1,. . . ,4s + l}, ((8s + 2r + 3,16s - 2r + 9) I r = 1,. . . ,s}, ((8s + 2r + 6,16s - 2r + 10) I r = 1,. .. , s - l}, ((10s + r + 4,143 - r + 8) I r

the second is (1,4s + 3), (4s + 4,12s + 8), (10s + 6,lOs + 7), ((r + 1,8s - r + 6)lr = 1 ,..., 4s + 1},((8s + r + 5,16s - r + 9)lr = 1 ,..., 2s}, ((10s + 7 + r, 14s - r + 9) I r = 1,. . . ,2s}. If n = 4, the partitions are (1,2), (4,6), (5,8), (3,7) and (2,3), (6,8), (4,7), and (1,s). Case 3. n = 1 (mod 4), n > 1. Let n = 4s + 1. Then the first partition is (2s, 6s + l), (2s + 1,4s + l), (7s + 1, 7s + 2), ((r,4s - r + 1) I r = 1,. . . ,2s - l}, ((4s + r + 1,8s - r + 3)lr = 1 ,..., s}, ((5s + r + I, 7s - r + 1)lr = 1 ,..., s - I}; the second is (1,2), (7,9), (3,6), (4,8), (5,lO) if n = 5, and (2s + 1,4s + 3),(2s + 2,6s + 2),(7s + 3,7s + 4),((r,4s - r + 3)lr = 1 ,...,a}, ((4s + r + 3,8s - r + 3 ) ( r = 1 ,..., s - 2},((5s + r + 1,7s - r + 3)lr = 1 ,..., s}, otherwise.

To see that these solutions are reverse-disjoint, we check that all the values of (i, j ) and their new positions in the reverse sequence (2n + 1 - j,2n + 1 - i) are distinct componentwise. For example, in Case 3, (r,4s - r + 3) becomes (8s + 3 - (4s - r + 3), 8s + 3 - r) in the reverse sequence and the two pairs are distinct componentwise for all s > 0. 0

n = 0 (mod S), n > 0. Let n = 8s.

+ 3), (10s - 1,lOs + l),

= l , . . .,2s};

Corollary 3. For all v 2 25, v = 1 or 7 (mod 24), n,(v) 2 8.

We note that n,(7) = 2 [14].

Theorem 4. hooked Skolem sequences of order n.

For all n = 2,3 (mod 4), n 2 6,there are at least 3 mutually disjoint

Proot Case 1. disjoint hooked sequences:

n = 2 (mod 4),n 2 6. For n = 6, we give the following three mutually

5 2 4 2 6 5 4 1 1 3 6 0 3 1 1 2 5 2 4 6 3 5 4 3 0 6 2 3 2 6 3 5 1 1 4 6 5 0 4 .


For n = 10, the following two disjoint hooked sequences are also disjoint from the one constructed in (iii) below:

5 3 8 1 0 3 5 2 7 2 6 8 9 4 1 0 7 6 4 1 1 0 9 9 6 1 0 1 1 5 7 6 8 9 5 4 1 0 7 3 4 8 3 2 0 2 .

For n > 10, let n = 4s + 2. The sequences are given in (i)-(iv). i. The first sequence is constructed from the pairs [12]:

(2s + Y,6s + 2), (4s + 2,6s + 3), (4s + 3,8s + 5), (7s + 3,7s + 4), ((4s + r + 3,8s - r + 4), (5s + r + 2,7s - r + 3)11 5 r 5 s - I}, {(r, 4s .- r + 2) I 1 5 r 5 2s).

ii. We append the hooked Skolem sequence of order 2 to the end of the Langford sequence with d = 3, m = 4s constructed in [17], to obtain the pairs

(2s + 1 , 4 ~ + l), ( 3 ~ , 7 ~ + l), (2s - 1,6s + l), (2~ , 6s), (8s + 1,8s + 2), (8s + 3,8s + 5) , ((s - r + 1,3s + r) I 1 5 r 5 s}, ((5s - r + 1,7s + r + l), (6s - r,6s + r + 1)1 1 5 r 5 s - l}, ((2s - r - 1,2s + r + 1)11 5 r 5 s - 2).

(1,4t + 3), (4t + 2,12t + 4), (lot + 2,lOt + 4), (12t + 5,12t + 6), ( ( r + 1,8t - r + 4) 11 5 r 5 4t}, ((8t + 2r + 2,16t - 2r + 4) 11 5 r 5 t - l}, ( (8t + 2r + 3,16t - 2r + 7) 11 5 r 5 t},((lOt + r + 4,14t - r + 6), 1 5 r 5

iii. If n = 2 (mod 8), let n = 8t + 2. Use the pairs

2t - 11. iv. If n = 6 (mod 8), let n = 8t + 6 ( t 2 1). Use the pairs

(1,4t + 5),(4t + 4,12t + 10),(12t + 8,12t + 9),(14t + 11,14t + 13), ((r + 1,8t - r + 8)l l 5 r 5 4t + 2},((10t + r + 7,14t - r + 11)l 1 5 r 5 2t}, ((8t + 2r + 6,16t - 2r + 12),(8t + 2r + 7,16t - 2r + 15) I 1 5 r 5 t } . Case 2. n = 3 (mod 4). For n = 7, we give the following mutually disjoint hooked sequences:

3 6 2 3 2 7 5 6 1 1 4 5 7 0 4 4 2 6 2 4 5 3 7 6 3 5 1 1 0 7 6 4 7 5 3 4 6 3 5 7 1 1 2 0 2

For n 3’ 7, let n = 4s + 3 (s > 1). The sequences are given in (i)-(iii). i. This sequence, due to O’Keefe [12], is given by the pairs:

( s , ~ s -t- 3 ) , ( ~ + 1,s + 2) , (2~ + 2,4s + 3),(2~ + 3,6s + 5), (4s + 4,8s + 7), {(r,4s - r + 3),(s + r + 2,3s - r + 3)1 1 5 r 5 s - I}, ((4s + r + 4,8s - r + 6) 11 5 r 5 2s).

(2s + ;!,6s + 5),(6s + 6,8s + 7),(5s + 4,5s + 5), ((r,4s - r + 4)1 1 5 r 5 2s + 1},{(4s + r + 3,8s - r + 6)1 1 5 r 5 s}, ( ( 5 s + r + 5 , 7 s - r + 6 ) 1 1 5 r 5 s - l } .

(1,2s t- 2),(3s + 2,3s + 3),(2s + 1,6s + 4), ((1 + r,4s - r + 4),(4s + 2r + 2,8s - 2r + 6)11 5 r 5 s}, ((s + r + 1,3s - r + 2)11 5 r 5 s - 1},((4s + 2r + 3,8s - 2r + 9)11 5 r 5

ii. ZJse the pairs:

iii. LJse the pairs:

s + 1). 0

334 BAKER A N D SHALABY

Corollary 5. For all v 2 37,v = 13 or 19 (mod 24), n,(v) 2 6.

We note that there are two disjoint hooked Skolem sequences of order 3: 1 1 2 3 2 0

These results can also be used to provide information about Mendelsohn triple systems. AMendelsohn triple system of order v,MTS(v), is a pair (V, B ) of sets where IV( = v

and B consists of cyclic triples on V (i.e., ( a , b, c ) is the triple containing the ordered pairs (a , b) , (b , c), and (c , a ) ) such that every ordered pair of distinct elements from V occurs in exactly one triple. Cyclic Mendelsohn triple systems are defined analagously to cyclic STS. We observe that a MTS(v) can be constructed from a STS(v) by replacing each triple (a, b, c) in the STS(v) by two cyclic triples ( a , b, c) and ( a , c, b). Clearly, in the cyclic case it suffices to replace the starter blocks in this way. Let m,(v) denote the maximum number of disjoint cyclic MTS(v)’s. By [4], m,(v) 5 v - 5 and this bound is attained when v = 7 or 13. In addition, 12 5 mc(19) 5 14 and 17 5 ~ ~ ~ ( 2 . 5 ) 5 20.

3 and 3 1 1 3 2 0 2; so n,(19) 2 4. In fact, n,(19) = 8 and n,(13) = 2 [ 3 ] .

Corollary 6. (to Theorem 2). For all v 2 25 and v = 1, 7 (mod 24), m,(v) 2 8.

Corollary 7. (to Theorem 4). For all v 2 37 and v = 13,19 (mod 24), m,(v) 2 6.

3. DISJOINT NEAR-SKOLEM SEQUENCES

Suppose some integer m(1 5 m 5 n ) is omitted from {1 ,2 , . . . , n}. The remaining n - 1 integers can be used to construct a sequence of length 2n - 2, so that each integer k occurs exactly twice in locations k positions apart. Such a sequence is called a near- Skolem sequence (we give a precise definition below), and is an obvious generalization of a Skolem sequence. In this section, we prove the existence of reverse-disjoint near-Skolem sequences of order n.

Let m and n be integers, m 5 n. An m-near-Skolem sequence of order n (and defect m) is an integer sequence S = ( ~ 1 , . . . , szn-2) satisfying

i. every k E {1,2,. . . , m - 1, m + 1 , . . . , n } occurs in exactly two positions in S; ii. if si = s j = k , i < j , then k = j - i.

As with Skolem sequences, near-Skolem sequences can be represented as a set of disjoint integer pairs (this time, a partition of the set {1,2,. . . ,2n - 2) into n - 1 pairs). Hooked near-Skolem sequences can be defined analagously.

Theorem 8. [ 161 An m-near Skolem sequence of order n , m 5 n, exists if and only if n = 0, l(mod 4) and m is odd, or n = 2 , 3 (mod 4) and m is even. A hooked m-near Skolem sequence of order n, m 5 n exists if and only if n = 0,1 (mod 4) and m is even or n = 2 , 3 (mod 4) and m is odd.

Disjoint near-Skolem sequences are defined in the usual way. In order to prove the existence of reverse-disjoint m-near-Skolem sequences, we first introduce Langford sequences which are generalizations of Skolem sequences.

Let d , p be positive integers. Simpson in [17] defines a set of p consecutive positive integers ( d , d + 1,. . . , d + p - 1) to be perfect Langford if the integers (1,. . . ,2p} can be arranged in disjoint pairs { (a i , bi) 11 5 i 5 p } so that {bi - ai 11 5 i 5 p } = { d , d + 1,. . . , d + p - l}. (In the original problem [7], bi - ai = i + 1 which


corresponds to d = 2 in this formulation.) Using Nickerson’s form, this is equivalent to a sequence S = ( ~ 1 , . . . , s+), with entries from {d , d + 1 , . . . , d + p - l}, where ai and bi give the positions in S of bi - ai = d + i - 1,l 5 i 5 p . Let the order of such a sequence be n = d + p - 1 , the largest entry in the sequence. For example, 4 2 3 2 4 3 is perfect Langford with d = 2, p = 3 and order 4. Hooked Langford sequences can be defined analagously.

Lemma 9. sequence of order n with d = 2.

For n = 0 , l (mod 4), there exists a reverse-disjoint perfect Langford

Pro05 Case 1. (s, 5 s - l ) , ( 2 ~ , 6 ~ ) , (4s - 1 , 6 ~ - l), {(r ,4s -- r - I ) , ( s + r ,3s - r) , (4s + r - 1,8s - r - I ) , (5s + r - I,7s - r ) I I 5 r 5: s}. Case 2. (14s, 14s + 2),(8s + 4,12s + 2)*,((8s + 2r + 4,16s - 2r + 2) 1 1 5 r 5 s - l}, ((8s + 2r + 1,16s - 2r + 1 ) 11 5 r 5 s},{(lOs + r + 2,14s - r ) 1 1 5 r 5 2s -

((4s + r , 12s - r + 2) I r = 1,2},{(r, 8s - r + 3) I 1 5 r 5 4s}, where * is omitted if s = 1 (i.e., if n = 9). Case 3. n = 5 (mod 8). Let n = 8s + 5. Use the pairs (2,4s + 2)*, (4s + 1,12s + 6) , (4s + 3,12s + 7),(6s + 2,6s + 4), ((2r - 1,8s - 2r + 5)1 1 5 r 5 s + 1}*,{(2r + 2,8s - 2r + 6)1 1 5 r 5 s}, ((2s + r + 2,6s - r + 2) I 1 5 r 5 2s - 2}, ( (8s + r + 4,16s - r + 9) I 1 5 r 5

n = 0 (mod 4). Let n = 4s > 0. Use the pairs

n = 1 (mod S), n # 1. Let n = 8s + 1. Use the pairs

31,

4s + l } , where * are omitted if s = 0 (i.e., if n = 5).

Theorem 10. reverse-disjoint rn-near-Skolern sequence of order n.

For all n 2 4, rn, n satisfying the conditions of Theorem 8, there exists a

Pro05 If rn = 1, then n = 0 , l (mod 4). A 1-near-Skolem sequence is a perfect Langford sequence with d = 2, so the required sequences are given in the previous lemma. In the remaining cases rn > 1.

Case 1. For n 5 1,2,4,5,6 or 7 (mod 8) and rn > 1 (rn odd or even as appropriate), the sequences given in (16, Theorem 31 are reverse-disjoint except when n = 10 and 4 5 rn 5 10. Any reverse-disjoint Skolem sequence of order 9 given in Theorem 2 is a 10-near Skolem sequence of order 10. We present the following sequences for the other cases.

n = 10 and rn = 4

8 6 7 2 1 0 2 9 6 8 7 1 1 5 3 1 0 9 3 5

n = 10 and rn = 6 8 9 4 1 0 1 1 4 7 8 3 9 5 3 1 0 7 2 5 2

n = 10 and rn = 8 7 9 4 10 1 1 4 7 6 3 9 5 3 10 6 2 5 2 .


Case 2. (12s + t - 2,12s + t - l), (2,4~)*, (6s - 2,6s),

{ ( 2 r - 1 , 8 s - 2 r - 1 ) ~ 1 ~ r ~ s } , { ( 2 s + r , 6 s - r - 2 ) J l ~ r ~ 2 s - 3 } , { ( 1 2 ~ - t + r - 2 , 1 2 ~ + t - r - 3 ) 1 0 5 r 5 t - 2 ) , { ( 8 s + r - 1 , 1 6 s - r - 2 )

where * is omitted when s = 1. Case 3.

and p = n - 2 (such a sequence exists by [7; Theorem 13).

(12s + t + 2,12s + t + 3), (8s,12s + 3), (6s + 1,6s + 3), ((4s + r - 2,12s - r + 3) I r = 1,2},{(2r,8s - 2r + 3) 11 5 r 5 s - I}, {(2r - 1,8s - 2r) 11 5 r 5 s},{(2s + r - 1,6s - r ) 11 5 r 5 2s - l}, ((12s - t + r + 2,123 + t - r + 2 ) J 1 5 r 5 t - 2}, ((8s + r + 2 , 1 6 ~ - I + 4) 10 5 r 5 4s - t}.

If n = 0 (mod 8), let n = 8s and rn = 2t + l ;s , t > 0. Use the pairs

((4s + r - 2,12s - r - 2)lr = 0,1},{(2r + 2,8s - 2r)J 1 5

10 5 r 5 4s - t - 2},

s - l},

If n = 3 (mod 8), let n = 8s + 3 and rn = 2t. For n 2 11 and rn = 2, adjoin the sequence 1,l to the Langford sequence with d = 3

If n 5 11 and rn > 2, use the pairs

4. A-COVERINGS AND SKOLEM SEQUENCES

In this section, we use disjoint Skolem, hooked Skolem, and near-Skolem sequences to construct disjoint 1-coverings.

In [2], Billington defines vA(t,k,v) to be the minimum number of blocks of size k (with repeated entries permitted) such that each collection of elements of size t from a v-set V occurs at least h times. She calls such a set of blocks a lambda covering of V and vA(t, k , v), the covering number. For example, (O,O,l), (1,1,2), (2,2,0) is a 1-covering of the 3-element set {0,1,2}. In this article, Billington proves that V1(2,3, v) = [v(v + 3)/61, for all v by constructing suitable 1-coverings of triples.

Two lambda coverings of a set are disjoint if they have no common blocks. For a set of size v, let maxA(t,k,v) be the maximum number of disjoint lambda coverings and denote max1(2,3, v) by max,(v). We note that there are only two disjoint 1-coverings of the set {0,1,2}: (O,O, l), (1,1,2), (2,2,0) and (O,O, 2), (1,1,0), (2,2, l), so maxl(3) = 2.

-

Lemma 11.

v - 1 i f v = 0 or 3 (mod 6) { V otherwise.

max,(v) 5

Pro05 If v = 0 or 3 (mod 6), then the minimum number of pairs in a 1-covering (with repeated elements) is (;) + 2v = v(v + 3)/2 = 3[v(v + 3) /6 ] . Therefore, no triples of the form xyx can occur in such a 1-covering and so each 1-covering must contain a triple of the form xxa, for each x. For a fixed x, only v - 1 choices of a are possible. Hence maxl(v) 5 v - 1.

We use the results of Sections 2 and 3 to derive some lower bounds for maxl(v). In our constructions, disjoint Skolem, hooked Skolem or near-Skolem sequences are used to generate sets of starter blocks as in Section 1. These starter blocks are cycled modulo some specified number to generate part of the block set. The remainder of the block set is then given as a completion. Most of these constructions generalize those

For other values of v, since xyx is permitted, maxl(v) 5 v.


given by Billington [2]. In some of the cases, we employ special constructions using combinations of Skolem, hooked Skolem, and near-Skolem sequences to produce the necessary disjoint starter blocks.

In the case where v = 4 (mod 6), Billington uses a STS(v - 1) in her construction. However, all cyclic STS(v - 1) with automorphism (0,1,2,. . . , v - 1) intersect, so a generalization of the construction in this case would require some noncyclic STS and would not use disjoint Skolem sequences and is beyond the scope of this article. (It is likely that the method used for v = 2 (mod 6), but with four disjoint STS(v - 1) will work.)

Theorem 12. For all v > 3,v $ 4 (mod 6), maxl(v) 2 4.

Proof: Case 1.

Let LI = 0 or 1 (mod 4). If a 2 4, there exists a reverse-disjoint 1-near Skolem sequence of order a, by Lemma 9, which generates four disjoint sets of starter blocks. All four sets are cycled modulo 6a - 3, but in the last two 0 , l and 2 are written as 6a - 3, 6a - 2 and 6a - 1, respectively. The four block sets are then extended to 1-coverings by adding each of (1) and (2) below to one of the first two sets and each of (3), (4) to one of the third and fourth sets.

and {(3j,3j,6a - 2),(3j + 1,3j + 1,6a - 1),(3j + 2,3j + 2,6a - 3 ) l j = 0 ,..., 2a - 2}, {(3j,3j + 1,6a - 3),(3j + 1,3j + 2,6a - 2),(3j + 2,3j + 3,6a - 1) 1 j = 0,. . . ,2a - 2}, where (6a - 4,6a - 3,6a - 1) is replaced by (6a - 4,0,6a -

v = 0 (mod 6). Let v = 6a.

1. ( 6 ~ - 3 , 6 ~ - 3 , 6 ~ - 2 ) , ( 6 ~ - 2 , 6 ~ - 2 , 6 ~ - 1 ) , ( 6 ~ - 1 , 6 ~ - 1 , 6 ~ - 3)

1). 2. ( 6 ~ - 3,6a - 3,6a - 1 ) , ( 6 ~ - 2 , 6 ~ - 2 , 6 ~ - 3 ) , ( 6 ~ - 1 , 6 ~ - 1 , 6 ~ - 2)

plus the last 12a - 6 blocks of 1) with 6a - 2,6a - 1,6a - 3 replaced by 6a - 1,6a - 3,6a - 2, respectively.

3. (O,O, l), (1,1,2), (2,2,0) and {(3j,3j, l), (3j + 1,3j + 1,2), (3j + 2,3j + 2,O) I j = 1 , ..., 2 a - 1},{(3j,3j+ 1,0),(3j+ 1 , 3 j + 2 , 1 ) , ( 3 j + 2 , 3 j + 3 , 2 ) l j = 1,. . . ,2a - l} where (6a - 1,6a, 2) is replaced by (6a - 1,3,2).

by 2,0,1, respectively. 4.

If a = 1, the starter set is empty and the above completions form the 1-coverings. If a == 2, the four 1-coverings are: 1’. (0,1,4) cycled modulo 9,(9,9, lo), (10,10, l l) , (11,11,9), {(i, i,9) I i = 0,3,6},

{(i, i, 10)i I i = 1,4,7}, {(i, i, 11) I i = 2,5, 8}, (0,2, lo), (6,8, lo), (3,5, lo), (2,4,9),

2’. (0,2,5) cycled modulo 9,(9,9, l l ) , (10,10,9), (11,11, lo), { ( i , i, 11) I i = 0,3,6},

(O,O, 2), (1,1,0), (2,2,1) plus the last 12a - 6 blocks of (3) with 1,2,0 replaced

(8,L 9), (5,7,9), (4,6,11), (1,3,11), (7,0,11);

{(i, i, 9) I i = 1,4,7}, {( i , i, 10) I i = 2,581, (0,L lo), (3,4, lo), (6,7,10>, (1,2,11), (4,5,11), (7,8,11), (2,3,9), (5,6,9), (8,0, 9);

8,111, (9,11, I), (6,8,1), (3,5, I), (11,4, O), (8,10, O), (5,7,0), (4,6,2), (10,3,2), (7,9,2); (0,3,5) cycled modulo 9, but with 0,1,2 written as 9,10,11, respectively,

plus (0,0, 2), (1,1,0), (2,2, l ) , {(i, i, 1) I i = 3,6,9}, {(i, i, 2) I i = 4,7, lo}, {(i, i ,O) I i =

5,8,11>, (9,10,0), (3,4, O), (6,7,0), (10,11, I), (4,5,1), (7,8, I), (1 1,3,2), (5,6,2), (8,9,2).

3’. (0,3,4) cycled modulo 9, but with 0,1,2 written as 9,10,11, respectively, plus (O,O, l), (1,1,2), (2,2,0), {(i, i, 0) I i = 3,6,9}, {(i, i, 1) I i = 4,7, lo}, { ( i , i, 2) I i = 5,

4’.

BAKER AND SHALABY 338

If a = 3, the four 1-coverings are: 1’. (0,1,6), (0,3,7) cycled modulo 15, (15,15,16), (16,16,17), (17,17, Is), {(i, i,

( i + l , i + 1 ,16) , ( i+2 , i+2 ,17)1 i=0 ,3 ,6 ,9 ,12} , (0,2,16), (6,8,16), (12,14,16), (3,5,16), (9,11,16), (2,4,15), (8,10, Is), (14,1, Is), (5,7,15), (11,13,15), (4,6,17), (10,12,17), (1,3,17), (7,9,17), (13,0,17);

2/. (0,2,6), (0,3,8) cycled modulo 15, (15,15,17), (16,16,15), (17,17,16), {(i, i, 1% (i + 1,i + 1,17), (i + 2,i + 2,15)li = 0,3,6,9,12}, (0,1,15), (3,4,15), (6,7,15), (9,10, IS), (12,13,15), (1,2,16), (4,5,16), (7,8,16), (10,11,16), (13,14,16), (2,3,17), (5,6,17), (8,9,17), (11,12,17), (14,0,17).

tively, plus (O,O, I), (1,L 2), (2,2, 01, {(i,i,O), (i + 1,i + l , l) , (i + 2,i + 2,2)li = 3,6,9,12,15}, (15,17, I), (68 , I), (12,14,1), (3,5,1), (9,11,1), (1794, O), (8,10, O), (14,169 O), 6 7 , O), (11,13,0), (4,6,2), (10,12,2), (16,3,2), (7,9,2), (13,15,2);

tively, plus (O,O, 2), (1,1,0), (2,2, I), {(i,i,2), (i + 1,i + 1,0), (i + 2,i + 2 , l ) l i = 3,6,9,12,15}, (15,16,1), (3,4,1), (67 , I), (9,10,1), (12,13, I), (16,173 2), (4,5, 21, (7,8,2)7 (10,117 2), (13,14,2), (17,3,0), (5,6,0), (8,9,0), (11,127 01, (14215,O).

3’. (0,5,6), (0,4,7) cycled modulo 15, but with 0,1,2 written as 15,16,17, respec-

4’. (0,4,6), (0,5,8) cycled modulo 15, but with 0,1,2 written as 15,16,17, respec-

We note that the starters for a = 2,3 come from the 2-near Skolem sequence of order a (1 and 3 above) and the 1-near hooked Skolem sequence of order a (2 and 4 above).

If a = 2 or 3 (mod 4), a > 4, then we use the reverse-disjoint 2-near Skolem sequence of order a (Theorem 10) in the following adaptation of the above constructions:

1”. the completion in (1) with the last set of 6a - 3 blocks replaced by {(6jY,2(3j + 1)*,6a - 1),(2(3j + 1)*,2(3j + 2 ) * , 6 ~ - 2),(2(3j + 2)*,2(3j + 3)*, 6a - 3) I j = 0,. . . ,2a - 2}, where all computations * are modulo 6a - 3;

2”. is obtained from (1”) as (2) is obtained from (1); 3“. the completion in (3) with the last set of 6a - 3 blocks replaced

by {(6j*,2(3j + 1)*,2),(2(3j + 1)*,2(3j + 2)*, 1),(2(3j + 2)*,2(3j + 3)*,0) I j = 1,. . . ,2a - l}, where * indicates multiplication modulo 6a - 3 with 0,1,2 written as 6a - 3,6a - 2,6a - 1, respectively;

4”. is obtained from (3”) as (4) comes from (3). Case 2. v = 1 (mod 6). Let v = 6a + 1.

First, consider a = 0 or 1 (mod 4). If a 2 4, there exists a reverse-disjoint 1-near Skolem sequence, S, of order a, by Theorem 10. This can be used to generate four sets, each containing a - 1 starter blocks. (If a = 1, these sets of starter blocks are void and this method still works.) Each set of starter blocks is cycled modulo v - 2 = 6a - 1. The four disjoint coverings are each completed by adding one of the following four disjoint sets.

1. [2] (O,O, 3a - 1) cycled modulo 6a - 1; plus 6a + 1 blocks (6a - 1,6a - 1,6a), (6a,6a, 0), (6a - 2,0,6a - 1); {(2j, 2 j + 1,6a - 1) I j = 0,. . . ,3a - 2}, {(2j - 1 , 2 j , 6 ~ ) I j = 1,. . . ,3a - 1).


2. (0,0,3a) cycled modulo 6a - 1 and the last 6a + 1 blocks of (1) with 6a and 6a + I interchanged.

3. (O,O, 1) cycled modulo 6a - 1; (6a - 1,6a - 1,3a + l), (6a,6a, l), (6a - 1, 6a, 1); plus the 6a - 2 blocks ((1 + 2j(3a - l), 1 + (2j + 1) (3a - 1),6a - 1) I j =

0 ,..., 3a - 2}, ((1 + (2j - 1) (3a - l), 1 + 2j(3a - 11, 6a) Ij = 1 ,..., 3a - l}, where all multiplications are performed modulo 6a - 1.

4. (0,0,6a - 2) cycled modulo 6a - 1; (6a - 1,6a - 1, l), (6a, 6a, 3a + l), (6a - 1,6a, 3a + 1); plus the last 6a - 2 blocks of (3) with 6a and 6a - 1 interchanged.

Now suppose a = x (mod 4), where x = 2 or 3 and a > 3. If x = 2, then there exists a reverse-disjoint Skolem sequence of order a - 1, by Theorem 2; if x = 3, there exist two mutually disjoint hooked Skolem sequences of order a - 1. In each case, we get four disjoint sets of starter blocks, S1,. . . , S,. The four l-coverings are:

{ ( 2 j ( 3 ~ - I), (2j + 1) ( 3 ~ - l), 6~ - l)l j = 0,. . . , 3 ~ - l}, (((2j - 1) (3a - l), 2j(3a - l), 6a) I j = 1,. . . ,3a - l}, where all multiplications are performed modulo 6a - 1; plus (6a - 1,6a - 1,6a),(6a,6a,O);

2. S2 and (3a - x;3a - x,O) cycled modulo 6a - 1; plus the last 6a + 1 blocks of (1) with 6a and 6a - 1 interchanged;

3. S, and (0,0,3a - 1) cycled modulo 6a - 1 but with 0 and 1 written as 6a - 1 and 6a, respectively; {(2j(3a - x), (2j + 1) (3a - x), 0) I j = 0,. . . ,3a - I}, {((2j - 1) (3a - x), 2j(3a - x), 1) I j = 1,. . . ,3a - l} where all multiplications are performed modulo 6a - 1 and every 0 or 1 arising from the multiplications is written as 6a - 1 or 6a, respectively; plus (0,0, l), (1,1,6a - 1);

4. S4 and (3a - 1,3a - 1,O) cycled modulo 6a - 1 but with 0 and 1 written as 6a - 1 and 6a, respectively; plus the last 6a + 1 blocks of (3) with 0 and 1 interchanged.

For v = 13, the l-coverings are: 1. (0,1,3) and (0,0,5) cycled modulo 11, (11,11,12), (12,12,0), { (8 j * , (2j + 1)4*,

1 l ) l j = O ,..., 5},{(2j - 1)4*,8j*,12)lj = 1 ,..., 5); 2. (0,2,3) and (O,O, 4) cycled modulo 11, (11,11, S), (12,12, 11), {( lOj*, (2j + 1)5*,

12) I j = 3,. . . ,8},{(2j - l)5*, l O j * , 11) I j = 4,. . . ,a}; 3. (0, L!, 5) and (0, 0,7) cycled modulo 11 with 0 , l replaced by 11,12, respectively,

(0,0, 1),(1, 1,12),{(2j**,2j + I**, 1)lj = 1,. . . ,5},{(2J - 1**,2j**,O) I j = 1,. . . ,6}; 4. (0,3,5), (0, 0 , l ) cycled modulo 11 with 0,1 replaced by 11,12, respectively,

(O,O,ll),(l,l,O),{(8j**,(2j + 1)4**,l)l j = 0 ,..., 5},{(2j - 1)4*",8j**,O)lj =

1,. . . ,5}, * computed modulo 11; ** computed modulo 11 with 0 , l replaced by 11,12, respectively.

For v == 19, the hooked Skolem sequence, 1 1 2 0 2 and its reverse (which is not a hooked Skolem sequence) provide starter blocks. The 1 -coverings are:

1. (0,1,4), (0,2,7), and (0,0,6) cycled modulo 17, plus (17,17,18), (18,18,0), {(16j*, (2j + 1)8*, 17) I j = 0,. . . ,8},{(2j - 1)8*, 16j*, 18) 1 j = 1,. . . , 8};

2. (0,3,4), (0,5,7), and (0,0,8) cycled modulo 17, plus (17,17,6), (18,18,17), {(12j*,(2j + 1)6*, 17) I j = 1,. . . ,8},{((2j - 1)6*, 12j*, 18) I j = 1,. . . ,9};

3. (0,1,7), (0,2,5), (0,0, 4) cycled modulo 17, but with 0,1 replaced by 17,18, respectively; plus (0,0, 17),(1,1,0),{(16j**, (2j + 1)8**, 1) I j = 0,. . . ,8},{((2j - 1)8**, 16j**,O) 1.j = 1,. . . ,8};

spectively; plus (0,0, 1),(1, 1,7),{(8j**, (2j + 1)4**,0) I j = 3,. . . , 11},{(2j - 1)4**, 8 j * * , 1) I j = 4,.. . , ll}.

1. S1 and (0,0,3a - x) cycled modulo 6a - 1;

4. (0,6,7), (0,3,5), (O,O, 9) cycled modulo 17, but with 0,1 replaced by 17,18, re-

BAKER A N D SHALABY 340

All * multiplications are computed modulo 17; all ** multiplications are modulo 17 but with 0 , l written as 17,18 respectively.

Case 3. v = 2 (mod 6). Let v = 6a + 2, v > 14. Then there exist four pairwise disjoint cyclic STS(v - l), say S1,. . . , S4, on the vertices 0,. . . , v - 2. We first construct three sets of blocks S:, S:, Si.

To create Si replace each 0 in Sz by v - 1. S: is clearly disjoint from S1. Sz contains exactly 3a blocks of the form (O,x, y), where x, y E ( 1 , . . . , v - 2). Each of these (x, y) pairs occurs in exactly one block of S3 and different pairs must be in different blocks (since each x or y occurs in only one such pair). Therefore, exactly 3a blocks of S3 are required. Let A = { z E (1,. . . , v - 2) I ( z , x, y) is a block of S3 and (0, x, y ) is a block of SZ}. Then IAI I 3a. Take any w E (1,. . . , v - 2)b . To form S:, replace each w in S3 by v - 1. Then (v - 1, b, c) is a block of S: * (w, b, c) is a block of S3 CJ (0, b, c) is not a block of Sz e (v - 1, b, c ) is not a block of Si, so S: and S i are disjoint. We use the same method as above to construct Si. Let B1 = {z E (0,. . . , v - 2} I ( z , x , y ) is a block of S4 and (O,x,y) is a block of Sz) and BZ = {z E (0, . . . , v - 2) I ( z , r , s) is a block of S4 and (w, I, s) is a block of S3). Clearly lBIIIIBzl 5 3a. Since S4 is disjoint from SZ and S3 ,O @ B1 and w @ Bz. If there exists a u E (0,. . . , v - 2)\(B1 U B z ) , u # 0, w, then replace such u in S4 with v - 1 to create S i which is still disjoint from S1,S:, and S:.

The four 1-coverings are:

1. S1 plus the blocks {(d, d, v - 1) Id = 0,. . . , v - 2) and (v - 1, v - l ,x), any

2. S; plus the blocks {(d,d,O) I d = 1,. . . , v - 1) and (O,O,x) any x # 0, w , u, v- 1; 3. S i plus the blocks {(d, d, w ) I d = 0,. . . , v - 1; d # w} and (w, w,x), any x #

4. S i plus{(d,d,u)ld = 0 ,..., v - 1; d # u)and ( u , u , x ) , x # O,w,u,v - 1.

We now show that if we use the four cyclic STS constructed from the first partition in each case of Theorem 2 or the first two sequences in Theorem 4, such a u exists. It is sufficient to show that IBll 5 3a - 2.

Suppose a = 8s, s 2 1. Then SZ contains the blocks {(0,8s + r , 16s - r + 1) I r =

1,. . . ,4s - 1) and S4 has {(0,16s + r,24s + 1 - I) I r = 1,. . . ,4s - l} within its set of starter blocks. Therefore, the blocks in S4 containing the pairs ((8s + r , 16s - r + 1)lr = 1 ,..., 4s - 1) are ((40s + 1,8s + r,16s + 1 - r ) l r = 1 ,..., 4s - 1). So 40s + 1 is on 4s - 1 of the 3a blocks used to generate B1; i.e., 1B11 5 3a - (4s -

Similarly, if a = 8s + 4, s 2 1,S2 contains the blocks {(0,8s + r + 4,16s - r + 9)Ir = 1 ,..., 4s + 1) and S4 {(0,16s + r + 8,24s + 13 - r ) I r = 1 ,..., 4s + 1). So the blocks ((40s + 21,8s + r + 4,16s - r + 9) I r = 1 ,..., 4s + 1) in S4 are among those used to generate B1. Since 48s + 21 is on 4s + 1 of the blocks used to generate B1, IB1 I 5 3a - 4s 5 3a - 4.

For a = 4, S2 contains (among others) the blocks (0,15,16), (0,9,12), (0,8, lo), (0,7,11). The blocks in S4 containing the corresponding nonzero pairs are: (4,15,16),

Suppose a = 4s + 1, s 2 2, then S2 contains the blocks {(0,4s + r + 1,8s - r + 2) I r = 1,. . . ,2s - 1) and S4 contains {(0,8s + r + 3,12s + 4 - r ) I r = 1,. . . ,2s - 1). So the blocks ((20s + 5,4s + r + 1,8s - r + 2) I r = 1 ,..., 2s - l} in S4 are used in the construction of B I . Hence 1B11 5 3a - (2s - 2) 5 3a - 2.

x # o,w,u;

o,w,u,v - 1;

2) 5 3a - 2.

(4,9,12), (1,8, lo), (1,7,11), so lB11 5 3a - 2.

341 DISJOINT SKOLEM SEQUENCES

For = 5 , S2 contains the blocks (0,7,12), (0,2,23), (0,16,27). The blocks in S4 containing the corresponding nonzero pairs are: (29,7,12), (29,2,23), (29,16,27), so

Suppose a = 4s + 3,s 2 2. Then S2 contains the blocks {(0,8s + r + 7,12s - r + 9) I r = 1,. . . ,2s} and S4 the starters {(0,4s + r + 4,8s - r + 6 ) I r = 0,. . . ,2s}. The 2s nonzero pairs from these blocks of S2 occur in the blocks ((4s + 3,8s + r + 7,12s - r + 9) I r = 1 ,..., 2s) in S4 and hence, 1B11 I 3a - (2s - 1) I 3a - 3.

For 12 = 7, S2 contains (0,13,20), (0,1,27), (0,2,33) and S4 has (41,13,20), (41,1,27), (41,2,33), SO IBlI 5 3a - 2.

For a = 3, the two disjoint hooked Skolem sequences 3 1 1 3 2 0 2 and 1 1 2 3 2 0 3 can be used to generate four disjoint STS(19). Then IBll = 8. Let w = 1. Then 1B;~l = 7 since S3 contains (1,15,16), (1,3,9), (1,7,18), (1,8, 17) and S4 contains (11,15,16), (11,3,9), (5,7,18), (5,8,17).

Let a = 4s + 2, s 2 4. Then S2 contains the blocks {(0,8s + r + 5,12s - r + 6)1 r = 1 ,..., s - 1) and S4 the starter blocks {(0,5s - b + 3,7s + b + 2)) b =

1,. . . , s} = {(0,4s + r + 2,8s - r + 3) I r = 1,. . . , s} (using b = s + 1 - r ) . Then the s - 1 nonzero pairs from the blocks of S2 listed give the following blocks of S4: ((4s + 3,8s + r + $ 1 2 ~ - r + 6 ) l r = 1 ,..., s - l} which are used in the construction of B1. Hence IBll 5 3a - (s - 2) 5 3a - 2.

For u = 14, S2 contains (0,30,41), (0,31,40), (0,32,37), (0,33,36) and S4 (15,30,41), (15,31,40), (2,32,37), (2,33,36), SO )B1 I 5 3a - 2.

For a = 10, S2 contains (0,23,27), (0,18,25), (0,14,24) and S4 (1,23,27), (1,18,25), (1,14,24), SO IBll 5 3a - 2.

For a = 6 , S2 contains (0,9,13), (0,7,12), (0,3,21) with blocks (34,9,13), (34,7,12), (34,3,21) in S4 and IB11 5 3a - 2.

Finally, consider v = 14. There is only one hooked Skolem sequence of order 2 , l 1 2 0 2 and there are only two disjoint cyclic STS(13), so we use four noncyclic STS(13). Let S1 have blocks (O,l,ll), (0,2,3), (0,4,5), (0,6,8), (0,7,12), (0,9, lo), (1,2,12), (1,3,8), (1,4,9), (1,5,10), (L6,7), (2,4, lo), (2,5,6), (2,7,11), (2,8,9), (3,4,7), (3,5, l l) , (3,6, lo), (3,9,12), (4 ,614 , (4,8,11), (5,7,9), (5,8,12), (6,9,11), (7,& lo), (10,11,12). Take S, = P 1 - l ( S l ) where p is the permutation (0,1,. . . ,10)(11)(12); cf. 1141. If we take w = 2, then u can be either 3 or 11 and we can use the given construction. Case 4.

a cycled modulo v with

Ill11 5 3a - 2.

v = 3 (mod 6). Let v = 6a + 3. If a = 0 or 1 (mod 4), a 2 4, we use the starters from a Skolem sequence of order

1. (0,0,3a + 1) cycled modulo v; 2. (O,Cl,3a + 2) cycled modulo v;

and the starters from an m-near Skolem sequence of order a + 1 also cycled modulo v with

3. (O,O,m) cycled modulo v; 4. (O,O, v - m) cycled modulo v. The following pairs of sequences will generate the required four disjoint starter sets: a = 0 (mod 4): Let m = a - 1. If a # 4,8, use the m-near Skolem sequence of

order a + 1 given in [16, Theorem 31 plus the following Skolem sequence of order a from Theorem 2: Case 1, second partition, if a = 0 (mod 8), a > 16; Case 1, first partition, if a = 16; Case 2, first partition, if a = 4 (mod 8).


For a = 4, use 1 1 4 2 3 2 4 3 and 4 2 5 2 4 1 1 5; for a = 8,6 3 1 1 3 8 6 7 4 2 5 2 4 8 7 5 a n d 9 5 3 8 4 3 5 6 4 9 2 8 2 6 1 1 .

a = 1 (mod 4): Let m = 4. If a > 5, use the Skolem sequence of order a given in the first partition of Case 3, Theorem 2 and the reverse of the 4-near Skolem sequence of order n + 1 in [16, Theorem 31. If a = 5, use 1 1 3 4 5 3 2 4 2 5 and 6 2 3 2 5 3 6 1 1 5, (Although these last two sequences are not disjoint, they do produce disjoint sets of starters. For example, 5 occurs in the same positions in both, but (0,5,15) is in the first set of starters and (0,5,16) in the other.)

For v = 9, the 1-coverings are: 1. (0,1,3) and (0,0,4) cycled modulo 9; 2. (0,2,3) and (0,0,5) cycled modulo 9; 3. (0,2,5) and (O,O, 1) cycled modulo 9; 4. (0,3,5) and (0,0,8) cycled modulo 9. If a = 2 or 3 (mod 4), use the two starter sets from the a-near Skolem sequence of

order a + 1 given in [16, Theorem 31 plus one of (O,O,a) or (0,0,5a + 3) for each, all cycled modulo v. The remaining two 1-coverings are generated by the starters from a hooked Skolem sequence of order a plus one of (0,0,3a) or (0,0,3a + 3) again cycled modulo v. For a 5 6, use the first sequence given in Theorem 4 if a = 2 (mod 4) and the third if a = 3 (mod 4). For a = 2, use 1 1 2 0 2; for a = 3, 1 1 2 3 2 0 3. Case 5. v = 5 (mod 6). Let v = 6a + 5.

i. a = 0,l (mod 4). A1 and Az[Ai and A;] are the starter sets generated by the 1-near Skolem sequence given in Lemma 9, A: and A; [A1 and A21 by its reverse if a > 4 [if a = 41. We create A3 and A4 by replacing (O,z,y) and (0,y - z,y) in A: and A: by (0,1,3a) and (0,3a - 1,3a), respectively, where

z = 2s + 1 and y = lls, if a = 4s, a > 4; z = 4s + 3 and y = 22s + 2, if a = 8s + 1; z = 8s + 5 and y = 20s + 13, if a = 8s + 5 ; z = 7 and y = 9, if a = 4.

If a = 1, then use four empty starter sets with y = 2 and z = 3. Set u = 3a - 1 and w = y - z . The four 1-coverings are: 1. A1 and (O,O, u) cycled modulo 6a + 1; (6a + 1,6a + 1,6a + 2), (6a + 2,6a +

2,6a + 3),(6a + 3,6a + 3,6a + 1),(6a + 4,6a + 4,x), any x,O 5 x 5 6a; (0,3a,6a), (6a + 2,0,6a + 4), (6a + 3,3a, 6a + 4), (6a + 1,6a, 6a + 4); {(2j,2j + 1,6a + 1),(2j + 1,2j + 2,6a + 2)1 j = 0 ,..., 3a - 1}, {(2j(3a)*, (2 j + 1) (3a)*,6a + 3) I j = 1 , . . . ,3a}, {((2j - 1) (3a)*, 2j(3a)*, 6~ + 4) I j = 2,. . . ,3a};

2. A2 and (O,O, 6a + 1 - u ) cycled modulo 6a + 1; (6a + 1,6a + 1,6a + 4), (6a + 2,6a + 2,6a + 1),(6a + 3,6a + 3,x),(6a + 4,6a + 4,6a + 2), any x,O 5 x 5 6a; ( 6 ~ , 3 ~ - 1 , 6 ~ - l), ( 6 ~ + 2 , 6 ~ , 6 ~ + 3), ( 6 ~ + 4 , 3 ~ - 1 , 6 ~ + 3), ( 6 ~ + 1 , 6 ~ - 1,6a + 3); ( 6 ~ - 2 , 6 ~ - 1 , 6 ~ + 2),(6a,0,6a + 1),(0,3a,6a + 3); {(2j,2j + 1 , 6 ~ + 2),(2j + 1,2j + 2 , 6 ~ + 1 ) l j Y O ,..., 3a - 2}, {(2j(3a)*, (2 j + 1) (3a)*, 6a + 4) I j = 2,. . . ,3a + l}, (((2j - 1) ( 3 ~ ) * , 2 j (3~ ) * , 6~ + 3) I j = 3,. . . , 3 ~ } ;

3. A3 and (O,O,w) cycled modulo 6a + 1; (6a + 1,6a + 1,6a + 3),(6a + 2,6a + 2,x),(6a + 3,6a + 3,6a + 4),(6a + 4,6a + 4,6a + l), anyx,O 5 x 5 6a; (O,y, 2y), (6a f 3,0,6a + 2), (6a + l , y , 6a + 2), (6a + 4,2y, 6a + 2);

343 DISJOINT SKOLEM SEQUENCES

{(2jz*, ( 2 j + l)z*,6a + 4) I j = 0,. . . ,3a - 1},{(2j - 1)~*,2jz*,6a + 3) I j = 1,. . . , 3a},{(2jy*,(2j + l)y*,6a + 1 ) l j = 1 ,..., 3a},{(2j - l)y*,2jy*,6a + 2 ) l j = 2 ,...,

4. A4 and (0,0,6a + 1 - w) cycled modulo 6a + 1;(6a + 1,6a + l,x), (6a + 2,6a + 2,6a + 4),(6a + 3,6a + 3,6a + 2), (6a + 4,6a + 4,6a + 3), any x,O 5 x 5 6a; (2y,3y*,4y*), (6a + 4,2y,6a + 1),(6a + 2,3y*,6a + 1),(6a + 3,4y*,6a + {(2jz*, ( 2 j + l)z*, 6a + 3) I j = 0,. . . ,3a - l}, {((2j - l)z*, 2jz*, 6a + 4) I j = 0,. . . ,3a - l}, {(2jy*, ( 2 j + l)y*,6a + 2) I j = 2,. . . ,3a + l}, {((2j - l)y*, 2jy*, 6a + 1) I j = 3,. . . ,3a + 1). All calculations marked with * are computed modulo 6a + 1 . [Note: 6az = 2y(mod 6a + l).]

ii. For a = 2,3 (mod 4), a 2 6, we use the same four 1-coverings as in (i) above with u = 3a .- 2, but we construct the starter sets A1 and A2 from the appropriate hooked 1- near SkoYem sequence, S1, and the sets A: and A: from the appropriate 2-near Skolem sequence (which is disjoint from S1 listed below). A3 and A4 are then constructed by replacing (0,5,3a - 2) and (0,3a - 7,3a - 2) by (0,2,3a + 1) and (0,3a - 1,3a + l), respectively, in A: and A:. Set y = 3a - 2,z = 5 and w = 3a - 7.

to be the hooked Langford sequence with d = 6,p = a - 5 given in [17,Theorem 23. Note this forces a 2 18. If a 2 18, take S1 to be 5 2 4 2 3 5 4 3 followed by 3?6,a-5; and S2,%:+-5 hooked to 4 0 3 5 4 3 1 1 5 (i.e., the 4 replaces the 0 in %t,a-5 and the last entry of

a = 15,

3al;

1);

Define

h

replaces the other 0); For smaller values of a, we list suitable sequences:

S1: 5’2:

3 4 2 3 2 4 5 13 14 10 8 5 11 15 12 6 9 7 8 10 13 6 14 11 7 9 12 0 15 4 1 1 3 4 15 3 12 9 7 11 14 6 13 10 8 7 9 6 12 15 11 5 8 10 14 13 5 ;

a = 14, 5’1:

S2: 10 8 13 5 3 14 12 3 5 8 10 6 11 9 7 13 4 6 12 14 4 7 9 11 2 0 2 7 3 1 1 3 11 9 7 13 8 14 12 10 6 4 9 11 8 4 6 5 13 10 12 14 5;

a = 11, S1: 8 6 1 0 3 1 1 9 3 6 8 4 7 5 1 0 4 9 1 1 5 7 2 0 2 S 2 : 4 1 1 8 4 1 0 6 11 9 7 3 8 6 3 5 107 9 11 5;

a = 10, S1: 7 5 1 0 8 4 9 5 7 4 6 3 8 1 0 3 9 6 2 0 2 S2: 3 4 7 3 9 4 1 0 8 6 7 1 1 5 9 6 8 1 0 5 ;

a = 7, S1: S2: 1 1 6 7 3 4 5 3 6 4 7 5 ;

6 7 3 4 5 3 6 4 7 5 2 0 2

a = 6, Sr: 2 5 2 4 6 3 5 4 3 0 6 S 2 : 3 4 6 3 5 4 1 1 6 5 ;

a = 3,Sl: 2 3 2 0 3 and A3 = {(0,1,6),(0,2,9)},w = 4,z = 3,y = 8. a = 2,Sl : 2 0 2 and A3 = {(0,2,6)},w = 1,z = 3,y = 5.


We now list 1-coverings for the missing small cases. If v = 5, four disjoint 1-coverings are:

012 123 230 204 234 340 014 431 004 001 000 003 113 114 112 110 222 220 224 221 330 333 331 332 441 442 443 444

5. CONCLUSIONS AND OPEN QUESTIONS

We have shown the existence of four mutually disjoint Skolem sequences, three mutually disjoint hooked Skolem sequences, and two mutually disjoint near-Skolem sequences. It may not be too difficult to improve these results by adding more disjoint sequences, but to find a general construction to show the existence of the maximum number of mutually disjoint or hooked disjoint Skolem sequences does not appear to be feasible at this point. Thus the related questions of the maximum number of disjoint cyclic STS(v), cyclic Mendelsohn triple system or A-covers are still open. However, we have found initial results to support the existence of hooked near Skolem sequences and hooked Langford sequences. Also there are further applications to different combinatorial structures such as Room squares [ 151, orthogonal symmetric Latin squares, and disjoint graph factorizations.

ACKNOWLEDGMENT

The second author would like to thank his former advisor A. Rosa for fruitful discussion concerning the disjoint Skolem sequences.

REFERENCES

[l] J.-C. Berrnond, A. E. Brouwer, and A. Gerrna, Systkmes des triplets et diffkrences associkes,

[2] E. J. Billington, On lambda coverings of pairs by triples, repeated elements allowed in triples, Colloq. CNRS, Problkmes cornbinatoires et thCorie des graphes, Orsay (1976), 35 -38,

Utilitas Math. 21C (1982), 187-203. [3] C. J. Colbourn, Disjoint cyclic Steiner triple systems, Congressus Nurnerantiurn 32 (1981),

205 -212.


[4] C. J. Colbourn and M. J. Colbourn, Disjoint cyclic Mendelsohn triple systems, Ars Comb.

[5] J. Doyen, Constructions of disjoint Steiner triple systems, Proc. AMS 32 (1972), 409-416. [6] L. Heffter, Uber Tripelsysteme, Math. Ann. 49 (1897), 101-112. [7] C.D. Langford, Problem, Math Gaz. 42 (1958), 228. [8] J. X. Lu, On large sets of disjoint Steiner triple systems I, Il, and IlI, J . Combin. Theory A 34

[91 -- , On large sets of disjoint Steiner triple systems IV, V , Vl, J . Combin. Theory A 37

(1981), 3-8.

(1983), 140-146, 147-155, 156-182.

(1984), 136-163, 164-188, 189-192. [lo] -- , On large sets of disjoint Steiner triple systems VII, unfinished manuscript. [ l l ] R.S. Nickerson, Problem E1845, Amer. Math. Monthly 73 (1966), 81; Solution 74 (1967),

[I21 E. S. O’Keefe, Verification of a conjecture of Th. Skolem, Math. Scand. 9 (1961), 80-82. [ 131 R. Peltesohn, Eine Losung der beiden Heffterschen Differenzenprobleme, Compositio Math. 6

[14] A. Rosa, “Intersection properties of Steiner systems,” in: Topics on Steiner systems, C. C.

[15] N. Shalaby, Skolem sequences: Generalizations and applications, Doctoral Thesis, McMaster

[16] -- , The existence of near-Skolem and hooked near-Skolem sequences, accepted for

[I71 J. E. Simpson, Langford sequences: Perfect and hooked, Discrete Math. 44 (1983), 97-104. [I81 Th. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand.

[19] -- , Some remarks on the triple systems of Steiner, Math. Scand. 6 (1958), 273-280. [20] L. Teirlinck, A completion of Lu’s determination of the spectrum of large sets of disjoint Steiner

591-592.

(1938), 251-257.

Lindner and A. Rosa (Editors). Ann. Discrete Math. 7 (1980), 115-128.

University, 1991.

publication in Discrete Math.

5 (1957), 57-68.

triple systems, J . Combin. Theory A 57 (1991), 302-305.

Received August 18, 1992 Accepted April 28, 1993

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Disjoint skolem sequences and related disjoint structures