Diamond Geezer’s card puzzle reminded me of a more complicated puzzle, which was devised and solved by my colleague and co-author Donald Preece. He is very proud of this, and justifiably so. In this post I will simply set the problem; later I will describe the mathematical background, and maybe give the solution.

The problem is to take a regular pack of 52 playing cards, and lay them out in a rectangle with four rows of 13 cards, so as to satisfy the following conditions:

- Each suit is represented once in each column.
- Each face value is represented once in each row.
- Two cards with the same value never occur together in a column.
- Each row contains four cards of one suit and three of each of the others.
- Any two columns share eactly one value, and any two values occur together in exactly one column.

Some brief commentary. Condition 4 says that the suits are as near as possible to being balanced over the rows; since 4 doesn’t divide 13, we can’t have exactly the same number of cards of each suit in a row. This condition could also be stated in the equivalent form “Each suit has four cards in one row and three in each of the others.”

Condition 5 is in some sense the hardest to satisfy. You may want to try to solve this condition, concentrating just on the values and the columns, and worry about the suits and the positions of the cards in the columns later.

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About Peter Cameron

I count all the things that need to be counted.