## A card trick, 4

Here is Donald Preece’s solution to his problem, taken from his paper `Some partly cyclic 13×4 Youden “squares” and a balanced arrangement for a pack of cards’, in Utilitas Math. 22 (1982), 255–263. (I hope I haven’t made any typing mistakes!)

 A♠ J♣ Q♥ K♦ 5♠ 6♠ 7♠ 2♦ 3♣ 4♥ 8♦  9♣ 10♥ 2♥ A♥ 6♣ 10♠ J♦ Q♦ 4♦ 9♥ 8♥ K♠ 7♣ 3♠ 5♣ 3♦ 8♠A♦ 7♥ 2♣ K♣ Q♣ J♠ 10♦ 9♦ 6♥ 5♥ 4♠ 4♣ 5♦ 9♠ A♣ K♥ 3♥ J♥ 10♣ Q♠ 8♣ 2♠ 7♦ 6♦

[Note: When I changed the WordPress style, I didn’t realise that it was going to screw up the tables. Apologies for the poor alignment here, but I haven’t found any other fix!]

In the preceding post on this problem, I described a simple method of Youdenization of a design which has a cyclic automorphism: write a block in the first column and develop cyclically. The design here (the projective plane of order 3) has 1728 cyclic automorphisms, and so this many different Youdenizations can be produced in this way; but no two of them are balanced with respect to each other, so the layout cannot be produced using this simple method. Donald found it using a symmetry of order 3.

$\begin{array}{*{13}{r}} A\spadesuit & J\clubsuit & Q\heartsuit & K\diamondsuit & 5\spadesuit & 6\spadesuit & 7\spadesuit & 2\diamondsuit & 3\clubsuit & 4\heartsuit & 8\diamondsuit & 9\clubsuit & 10\heartsuit \\ 2\heartsuit & A\heartsuit & 6\clubsuit & 10\spadesuit & J\diamondsuit & Q\diamondsuit & 4\diamondsuit & 9\heartsuit & 8\heartsuit & K\spadesuit & 7\clubsuit & 3\spadesuit & 5\clubsuit \\ 3\diamondsuit & 8\spadesuit & A\diamondsuit & 7\heartsuit & 2\clubsuit & K\clubsuit & Q\clubsuit & J\spadesuit & 10\diamondsuit & 9\diamondsuit & 6\heartsuit & 5\heartsuit & 4\spadesuit \\ 4\clubsuit & 5\diamondsuit & 9\spadesuit & A\clubsuit & K\heartsuit & 3\heartsuit & J\heartsuit & 10\clubsuit & Q\spadesuit & 8\clubsuit & 2\spadesuit & 7\diamondsuit & 6\diamondsuit \end{array}$