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°.

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    $\begingroup$ Of course $M_{12}$ has a subgroup of order $11$; this has twelve orbits on the hexads. Could you pick the suits to each consist of three orbits? Of course, with this construction the fixed point of the subgroup has a special status. $\endgroup$ – Angina Seng Sep 8 at 18:30
  • $\begingroup$ Thank you — I hadn’t thought of that at all. Would it give any ‘playable’ symmetries on the cards? That is, any properties of quadruples or triples or pairs, which could be used by somebody devising a playable game? Also, alas, my undergraduate maths degree is three decades stale: back then I was competent at number theory; less so at group theory. So the question “Could you … ?” is, alas, probably not. If it is easy for you to describe steps to be taken, please say more. Would I need to construct a list of $M_{12}$ elements (yikes!)? And how should the twelve orbits be split into three? $\endgroup$ – jdaw1 Sep 8 at 18:54
  • $\begingroup$ More about qualities useable by a game designer. The hoped-for symmetry in the question: “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”, would allow a Poker variant: three-card hands; best hand is three cards of the same suit with a quad in common; not as good is suits not same; quads shared pairwise but not collectively would be even less good; etc. In contrast, less useful are invisible qualities hidden to the players. They might be the best available, but I’m hoping for visible qualities. $\endgroup$ – jdaw1 Sep 8 at 19:28
  • $\begingroup$ Lovely question! Might be tempted to get some cards, they look beautiful! $\endgroup$ – linguo Sep 8 at 19:33
  • $\begingroup$ Question edited to allow a number of suits other than four. Also, design comments have been received — updated designs at jdawiseman.com/papers/games/steiner_hexads/steiner_hexads.html $\endgroup$ – jdaw1 Sep 11 at 8:15

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.

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)

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).

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
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  • $\begingroup$ This is excellent work — thank you. Absent a major improvement from somebody else, in a few days it will be marked as the answer. (Small aside: another reason for pairing suits is because I’d like six suits of three colours, and would prefer the implied suit pairings to be non-arbitrary.) $\endgroup$ – jdaw1 Sep 22 at 13:29
  • $\begingroup$ FTR, a small change, effectively a relabelling of the suits, such that they are ordered by sum-square card totals, as this removes one arbitrariness. Obviously the first suit, comprising just 21s and 45s, is by far the largest at 11×(21²+45²) = 27126. Suit values are [27126 24502 24410 24306 24282 24270]. $\endgroup$ – jdaw1 Oct 10 at 8:57
  • $\begingroup$ From Professor Peter Keevash, and posted here with his permission. ¶1 These look like they will be excellent cards, and I will be happy to acquire a set myself! $\endgroup$ – jdaw1 Oct 23 at 16:47
  • $\begingroup$ From Peter Keevash, ¶2 I do know the solution to the question in the stackexchange answer, which could perhaps be incorporated into the design if it is not too late. There is a natural way, discovered by Walecki in the 19th century, to partition K_{12} (the complete graph on 12 vertices) into 6 paths of length 11. The wikipedia page “Hamiltonian decomposition” illustrates his construction as applied to the related problem of partitioning K_9 into 4 cycles of length 9 (“Hamiltonian cycles”). $\endgroup$ – jdaw1 Oct 23 at 16:47
  • $\begingroup$ From Peter Keevash, ¶2 On deleting the top vertex from that picture one obtains a partition of K_8 into 4 paths of length 7. Similarly one obtains a partition of K_{13} into 6 cycles of length 13 or a partition of K_{12} into 6 paths of length 11. The sequence of shifts mod 12 in each path is +1,−2,+3,−4,+5,±6,−5,+4,−3,+2,−1. $\endgroup$ – jdaw1 Oct 23 at 16:48

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