# Solution

(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.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@H-Dudeney

### References

#### Project Gutenberg

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

This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.