Steiner(5,6,12) system: symmetrical split into four or six

Question

I am making a pack of Steiner(5,6,12) cards, and intend to make it available to others. The plan is that there will be 143 cards, 2¼″×3½″ ≈ 57mm×89mm, comprising: the 132 Steiner cards; one or two jokers of each of the four suits; two unsuited jokers; and, à la ‘bridge card’, a card giving some explanation and asking that games for these cards be tagged ‘#SteinerKirkmanCards’.

The 132 Steiner cards are to be symmetrically assigned to four suits [edit: or six — and six might be better], 33 [or 22] cards in each suit. ¿How should this assignment to suits should be done?

Each quadruple of letters (e.g., ABCD) appears on exactly four cards. Is it possible that, for every quadruple, its four cards be three of one suit and one of another? There are 495 such quadruples, so it cannot be that each suit has the same number of the ones. Can it be that one special suit has exactly one instance of each quad, and that the other three suits each have three instances of one third of the quads? Or can it be that one special suit has exactly one instance of each quad, and that the other three suits each have one instance of one third of the quads and two instances of another third (such that every quad is 1:0:1:2, with permutations of last three)? Please, what of this is possible, and how?

Or, if none of that works, can there be a different strong symmetry? Because the cards are defined by the uniqueness of the quintuples, it would be natural to suit based on quads or triples or pairs.

Each letter appears on 66 cards, so letters cannot be even across the four suits. Can the chosen symmetry be done with the letters similarly distributed across the suits (e.g., each letter’s suit frequencies being 12:15:18:21)? This is also a desirable type of symmetry.

The particular list of hexads (but improved in an answer below) currently being used is not special; if permuting letters would help gain any elegant qualities, please permute.

The ‘non-bridge card’ could include concise credit for the assignment to suits.

Edit: I’ve been asked by a different channel whether the number of suits must must be four. No. For reasons of game play, I think the number of suits must be ≥3, and ≤6, and a factor of 132: so three or four or six. Indeed, that permits an extra request: it might be that some games could work with a half pack, 66 cards, being half of the suits. If the number of suits is even, and there is a ‘natural’ split of the suits, please do say what that is. (Also a conventional deck has four suits of two colours; this deck might have four suits of two colours, or six suits of three colours.)

Though this post is not about the visual layout of the cards, it is possible that the layout might interact with the possible games which might interact with the best choice of symmetry. Hence there are low-resolution drafts, some ⟳180°.

Answer 1

Edit: The suiting (and hexads) proposed here are available here. Updated to account for Keevash's comments

if permuting letters would help gain any elegant qualities, please permute.

I suggest using the shuffle numbering from Sphere Packings, Lattices and Groups Chapter 12.

This has the nice property that if you take the sums of the hexads the break down as:

21 11
24  2
25  2
26  4
27  4
28  6
29  6
30  8
31  8
32 10
33 10
34 10
35  8
36  8
37  6
38  6
39  4
40  4
41  2
42  2
45 11

The hexads with sum 21 (light hexads) have some nice properties (enumerated in SPLaG) so one additional nice property we might like to have is all the light hexads in the same suit.

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Another property of the Steiner(5,6,12) hexads, is that they have a correspondence to octads of Steiner(5,8,24). The correspondence described in SPLaG involves extending a hexad to an octad using the shuffle number. We can therefore create a function from Steiner(5,6,12) $\to \{\{1,2\},\{1,3\},\cdots,\{1,12\},\{2,3\}, \cdots, \{11,12\}\}$ (dual duads). As a map of complementary pairs (two hexads are a complementary pair if they are disjoint) this is a bijection.

Therefore, a we might like to assign suits to (6) subsets of this set (of size 11) which would induce (6) suits (of size 22) on the hexads.

Since $\{\{1,2\},\{1,3\},\cdots,\{1,12\},\{2,3\}, \cdots, \{11,12\}\}$ is the edge set for $K_{12}$ (the complete graph on 12 vertices $\{1, \cdots, 12\}$) we could pick disjoint paths on $K_{12}$ which cover it. This would have the additional nice property that "suits" would gain an ordering.

To make this more concrete, if our first path in $K_{12}$ is $1 \to 2 \to 3 \to \cdots \to 12$, we would end up with the light / heavy hexads with an ordering:

(1, 2, 3, 4, 5, 6)
(0, 1, 2, 3, 7, 8)
(0, 1, 2, 4, 5, 9)
(0, 1, 3, 4, 6, 7)
(0, 1, 2, 3, 5, 10)
(0, 1, 2, 4, 6, 8)
(0, 2, 3, 4, 5, 7)
(0, 1, 2, 3, 6, 9)
(0, 1, 3, 4, 5, 8)
(0, 1, 2, 5, 6, 7)
(0, 1, 2, 3, 4, 11)

(The ordering is "hexads are connected to other hexads of the same suit with exactly 3 points in common)

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Now the question I have is "how do we 'nicely' choose 6 disjoint paths on $K_{12}$?". Peter Keevash in the comments proposes using a technique of Walecki we can obtain 6 paths related by rotational symmetry. (And therefore order the suits accordingly).

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Forming the suits in this way has the following properties:

1. 22 cards in each of 6 suits
2. Complementary hexads in the same suit, which implies every letter in each suit 11 times
3. There is an ordering within each suit (on complementary pairs)
4. One of the suits is formed by the light and heavy hexads
5. Cards evenly "connect" to other suits. (Where two hexads are connected if they match on exactly 3 points).

I would still like to see:

• A natural way to pick the "direction" in each ordering