(related to Problem: The Round Table)

The history of this problem will be found in **The Canterbury Puzzles** (No. 90). Since the publication of that book in 1907, so far as I know, nobody has succeeded in solving the case for that unlucky number of persons, $13,$ seated at a table on $66$ occasions. A solution is possible for any number of persons, and I have recorded schedules for every number up to $25$ persons inclusive and for $33.$ But as I know a good many mathematicians are still considering the case of $13,$ I will not at this stage rob them of the pleasure of solving it by showing the answer. But I will now display the solutions for all the cases up to $12$ persons inclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators.

The solution for the case of $3$ persons seated on $1$ occasion needs no remark.

A solution for the case of $4$ persons on $3$ occasions is as follows:—

$$\begin{array}{cccc} 1&2& 3& 4\\ 1&3& 4 &2\\ 1&4& 2 &3\\ \end{array}$$

Each line represents the order for a sitting, and the person represented by the last number in a line must, of course, be regarded as sitting next to the first person in the same line, when placed at the round table.

The case of $5$ persons on $6$ occasions may be solved as follows:—

$$\begin{array}{ccccc} 1 &2& 3 &4& 5\\ 1 &2& 4 &5& 3\\ 1 &2& 5 &3& 4\\ 1 &3& 2 &5& 4\\ 1 &4& 2 &3& 5\\ 1 &5& 2 &4& 3\\ \end{array}$$

The case for $6$ persons on $10$ occasions is solved thus:—

$$\begin{array}{cccccc} 1 &2& 3 &6& 4& 5\\ 1 &3& 4 &2& 5 &6\\ 1 &4& 5 &3& 6 &2\\ 1 &5& 6 &4& 2 &3\\ 1 &6& 2 &5& 3 &4\\ 1 &2& 4 &5& 6 &3\\ 1 &3& 5 &6& 2 &4\\ 1 &4& 6 &2& 3 &5\\ 1 &5& 2 &3& 4 &6\\ 1 &6&3 &4& 5 &2\\ \end{array}$$

It will now no longer be necessary to give the solutions in full, for reasons that I will explain. It will be seen in the examples above that the $1$ (and, in the case of $5$ persons, also the $2$) is repeated down the column. Such a number I call a "repeater." The other numbers descend in cyclical order. Thus, for $6$ persons we get the cycle, $2,$ $3,$ $4,$ $5,$ $6,$ $2,$ and so on, in every column. So it is only necessary to give the two lines $1$ $2$ $3$ $6$ $4$ $5$ and $1$ $2$ $4$ $5$ $6$ $3,$ and denote the cycle and repeaters, to enable any one to write out the full solution straight away. The reader may wonder why I do not start the last solution with the numbers in their natural order, $1$ $2$ $3$ $4$ $5$ $6.$ If I did so the numbers in the descending cycle would not be in their natural order, and it is more convenient to have a regular cycle than to consider the order in the first line.

The difficult case of $7$ persons on $15$ occasions is solved as follows, and was given by me in The Canterbury Puzzles:—

$$\begin{array}{ccccccc} 1 &2& 3 &4& 5 &7& 6\\ 1 &6& 2 &7& 5 &3& 4\\ 1 &3& 5 &2& 6 &7& 4\\ 1 &5& 7 &4& 3 &6& 2\\ 1 &5& 2 &7& 3 &4& 6\\ \end{array}$$

In this case the $1$ is a repeater, and there are two separate cycles, $2,$ $3,$ $4,$ $2,$ and $5,$ $6,$ $7,$ $5.$ We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line.

A solution for $8$ persons on $21$ occasions is as follows:— $$\begin{array}{cccccccc} 1 &8&6 &3& 4 &5& 2& 7\\ 1 &8& 4 &5& 7 &2& 3& 6\\ 1 &8& 2 &7& 3 &6& 4& 5\\ \end{array}$$

The $1$ is here a repeater, and the cycle $2,$ $3,$ $4,$ $5,$ $6,$ $7, 8.$ Every one of the $3$ groups will give $7$ lines.

Here is my solution for $9$ persons on $28$ occasions:— $$\begin{array}{ccccccccc} 2 &1& 9 &7& 4 &5& 6 &3& 8\\ 2 &9& 5 &1& 6 &8& 3 &4& 7\\ 2 &9& 3 &1& 8 &4& 7 &5& 6\\ 2 &9& 1 &5& 6 &4& 7 &8& 3\\ \end{array}$$

There are here two repeaters, $1$ and $2,$ and the cycle is $3,$ $4,$ $5,$ $6,$ $7,$ $8, 9.$ We thus get $4$ groups of $7$ lines each.

The case of $10$ persons on $36$ occasions is solved as follows:— $$\begin{array}{cccccccccc} 1 &10& 8 &3& 6 &5& 4 &7&2& 9\\ 1 &10& 6 &5& 2 &9& 7 &4& 3 &8\\ 1 &10& 2 &9& 3 &8& 6 &5& 7 &4\\ 1 &10& 7 &4& 8 &3& 2 &9& 5 &6\\ \end{array}$$

The repeater is $1,$ and the cycle, $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ $9, 10.$ We here have $4$ groups of $9$ lines each.

My solution for $11$ persons on $45$ occasions is as follows:— $$\begin{array}{ccccccccccc} 2&11& 9 &4& 7& 6& 5 &1& 8 &3& 10\\ 2& 1& 11& 7& 6& 3& 10 &8 &5 &4& 9\\ 2& 11& 10& 3& 9& 4& 8 &5 &1 &7& 6\\ 2& 11 &5& 8& 1& 3& 10 &6 &7 &9& 4\\ 2& 11 &1& 10& 3& 4& 9 &6 &7& 5& 8\\ \end{array}$$

There are two repeaters, $1$ and $2,$ and the cycle is, $3, 4, 5,\ldots , 11.$ We thus get $5$ groups of $9$ lines each.

The case of $12$ persons on $55$ occasions is solved thus:— $$\begin{array}{cccccccccccc} 1& 2& 3& 12& 4& 11& 5& 10& 6 &9& 7& 8\\ 1& 2& 4& 11& 6& 9& 8& 7& 10& 5& 12& 3\\ 1& 2& 5& 10& 8& 7& 11& 4& 3 &12& 6& 9\\ 1& 2& 6& 9& 10& 5& 3& 12& 7& 8& 11& 4\\ 1& 2& 7& 8 &12 &3 &6& 9& 11& 4& 5& 10\\ \end{array}$$

Here $1$ is a repeater, and the cycle is $2,$ $ 3,$ $4,$ $ 5, \ldots ,12.$ We thus get $5$ groups of $11$ lines each.

**Dudeney, H. E.**: "Amusements in Mathematics", The Authors' Club, 1917

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