We look at 3-transversal designs with all groups of size 2 or 3.

The three parameters are a, the number of groups of size 2 (factors at two levels), b, the number of groups of size 3 (factors at three levels), and N, the number of blocks (runs).

For a+b = 3 these designs are trivial (the complete design repeated some number of times). So we take a+b > 3.

N must be divisible by 2, 4, 8, 3, 9, 27 when a is at least 1, 2, 3 or b is at least 1, 2, 3, respectively.

For b=0 we have a <= N/2, and equality is achieved (by systems derived from Hadamard matrices).

We give tables with horizontally a, vertically b. The entries give the number of nonisomorphic designs, and link to files with the actual designs. In the files (and below), the ternary coordinates are given first. Solutions are called nonextendible when it is impossible to add a binary column.

There is a unique design with N=8 a=4 b=0:

4 | ||

0 | 1 |

This unique design can be described as consisting of the binary words xyzw with x+y+z+w = 0.

All solutions:

4 | 5 | 6 | 7 | 8 | ||

0 | 2 | 2 | 1 | 1 | 1 | |

Nonextendible solutions:

4 | 5 | 6 | 7 | 8 | ||

0 | 0 | 1 | 0 | 0 | 1 | |

The unique nonextendible design with a=5 and b=0 can be described as consisting of the binary words xyzuv with x+y+z+u+v = 0.

The unique design with a=8 and b=0 can be described as the [8,4,4] extended Hamming code.

All solutions:

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | ||

0 | 1 | 2 | 1 | 2 | 1 | 1 | 1 | 1 | 1 | 1 | |

1 | 2 | 3 | 0 | ||||||||

Nonextendible solutions:

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | ||

0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | |

1 | 0 | 3 | 0 | ||||||||

The three solutions for a=4 b=1 can be described as consisting of words txyzw with ternary coordinate t and binary coordinates xyzw satisfying the conditions below.

- Solution #1:
0xyzw (x+y+z=0) 1xyzw (x+y+z+w=0) 2xyzw (x+y+z=1)

- Solution #2:
txyzw (x+y+z+w=0)

- Solution #3:
0xyzw (x+y+z+w=0) 1xyzw (x+y+z+w=0) 2xyzw (x+y+z+w=1)

The single solution for a=4 b=0 that cannot be extended by adding a binary column, extends to Solution #2 for a=4 b=1.

0 | ||

4 | 1 | |

5 | 0 | |

This unique system can be described as consisting of the ternary words qrst with q+r+s+t = 0.

All solutions:

4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | ||

0 | 3 | 5 | 10 | 17 | 33 | 34 | 32 | 22 | 23 | 12 | 10 | 5 | 5 | |

Nonextendible solutions:

4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | ||

0 | 0 | 0 | 0 | 0 | 0 | 9 | 2 | 0 | 0 | 0 | 0 | 0 | 5 | |

2 | ||

2 | 3 | |

The three solutions can be described as follows.

- Solution #1: Take the twenty words 02**, 20**, 12**, 21**, 22** each once, and the eight binary words of even weight each twice.
- Solution #2: Take the twelve words 01**, 10**, 22** each once, and the twelve words 00bb, 12bb, 21bb, 02bc, 20bc, 11bc (where bb are two equal bits and bc are two different bits) each twice.
- Solution #3: The trivial design ****: take everything once.

All solutions:

4 | 5 | 6 | 7 | 8 | 9 | ... | 20 | ||

0 | 3 | 3 | 9 | 25 | 105 | 213 | ... | 3 | |

Nonextendible solutions:

4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | ||

0 | 1 | 0 | 0 | 0 | 4 | 0 | 2 | 2 | 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 3 | |

We see that given a partial Hadamard matrix consisting of 13 mutually orthogonal plusminus vectors of length 20, there exists an extension to a full Hadamard matrix. More generally one can ask how many rows of a partial Hadamard matrix one needs to be guaranteed of the existence of an embedding into a full Hadamard matrix.

All solutions:

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ... | 24 | ||

0 | 1 | 4 | 10 | 45 | 397 | 8383 | 166081 | >0 | ... | 60 | |

1 | 4 | 21 | 134 | 938 | 3056 | 5018 | 3 | 0 | |||

Nonextendible solutions:

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ... | 24 | ||

0 | 0 | 0 | 1 | 1 | 1 | 81 | 64 | ? | ... | 60 | |

1 | 0 | 0 | 5 | 14 | 115 | 5010 | 3 | 0 | |||

0 | 1 | ||

3 | 1 | 5 | |

4 | 7 | 33 | |

5 | 4 | 4 | |

6 | 0 | 0 | |

4 | 5 | 6 | 7 | ... | 28 | ||

0 | 4 | 7 | 86 | 4049 | ... | 487 | |

4 | 5 | 6 | ... | 32 | ||

0 | 5 | 19 | 358 | ... | > 3578005 | |

2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | ... | 36 | ||

0 | 1 | 1 | 5 | 15 | 906 | >0 | >0 | >0 | >0 | >0 | >0 | >0 | ... | > 4745356 | |

1 | 1 | 6 | 89 | 27730 | >0 | >0 | >0 | >0 | >0 | >0 | >0 | ? | |||

2 | 9 | 465 | 380253 | >0 | >0 | >1956 | >382 | >157 | >67 | >14 | >7 | ? | |||

Examples with b = 0 exist with a up to 36.

An example of N=72 a=12 b=2.

Using the LP bound one sees that no example with N=72 a=21 b=2 exists. (Hedayat, Sloane, Stufken, p.203.)

Man V. M. Nguyen used exhaustive search to find all 379 nonisomorphic arrays with N=72 a=8 b=2, and then showed by successive extension that there are 5 systems with N=72 a=12 b=2 (with groups of order 8, 8, 16, 64, 96) and no system with N=72 a=13 b=2. However, the construction below seems to contradict these results.

Examples with a=12 b=2 may be constructed as follows:
Take the unique Hadamard matrix H of order 12. Arbitrarily partition
its rows into three groups of 4 rows. Construct a OA(24,2^{12},3)
matrix M (with symbol alphabet +,-) by putting H above -H. We now have
three groups of 8 rows such that each column has four +'s and four -'s
in each group. Construct a OA(72,3^{2}.2^{12},3)
by writing three copies of M above each other and writing
00,01,02,12,10,11,21,22,20 in front of the nine groups of 8 rows.
In this way one obtains three examples with groups of orders
288, 864, 2304. The explicit example (on file) given above
has a group of size 8.

Examples with b = 0 exist with a up to 40.

Starting with the [12,6,6] ternary quadratic residue code, we obtain a ternary [10,4,6] code with dual distance 4 by shortening twice. This yields an example with a=0 b=10 N=81. Hedayat-Sloane-Stufken, p. 22, says that Seiden (1955) proved that no example with a=0 b=11 N=81 exists.

Examples with b = 0 exist with a up to 44.

All solutions:

3 | 4 | 5 | ... | ||

0 | 1 | 7 | 56 | ... | |

1 | 9 | 487 | >0 | ... | |

Examples with b = 0 exist with a up to 48.

There exists an example with a = 16, b = 1.

Nonextendible solutions:

3 | 4 | 5 | ... | ||

0 | 1 | 0 | ? | ... | |

1 | 0 | ? | ? | ... | |

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