Let us assign each invitee a different numerical value in modulo 2*n*
(not assigning any number to Arthur himself). The operation "+" used in the
description below will denote modulo 2*n* addition.

Let us first solve the following, simpler problem: we would like a scheme for
a party that lasts 2*n* nights, during which no person "A" will be
seated *to the right* of the same person "B", twice. (This is different
from the original question in that the original question was bidirectional.)

A solution for this riddle is as follows. Let *T* be a permutation over
the numbers 0 through 2*n*-1 and let *j* designate the date from
0 (the first night of the party) to 2*n*-1 (the last). We will seat to
Arthur's left guest *T*_{0}+*j*, followed by guest
*T*_{1}+*j*, etc., until guest
*T*_{2n-1}+*j* who will sit immediately to Arthur's
right. *T*_{0} will be set, arbitrarily,
to 0. The question is, for which permutation *T* does this
constitute a valid arrangement?

Let us suppose on the contrary that the arrangement is invalid, then there
are two nights, *j* and *k*, with *j*≠*k*, such that
for some *x* and *y*,
*T*_{x}+*j*=*T*_{y}+*k*
(=person "A") while
*T*_{x+1}+*j*=*T*_{y+1}+*k*
(=person "B").

Subtracting the two equations, we have
*T*_{x+1}-*T*_{x} =
*T*_{y+1}-*T*_{y}. A valid solution
is one in which this implies *x*=*y*. Let us denote for all *i*
between 1 and 2*n*-1
*S*_{i}=*T*_{i}-*T*_{i-1}.
The derivation so far implies that the solution is valid if and only if
*S* is a permutation over the numbers between 1 and 2*n*-1.

Constant readers of UyHiP will recognize that already in the first part of
the previous riddle (May 2008) we have shown that
a solution exists for any *n*.

Returning to the full, bidirectional version of the riddle, let us create a
solution to the unidirectional, simpler riddle in which during night *j*
the guests are arranged in the reverse order to their arrangement during
night 2*n*-*j* (If "A" was to the right of "B" on night *j*,
she will be to his left on night 2*n*-*j*).

Such a solution can easily be made into a bidirectional solution by taking the
first *n* nights only. Referring again to the
May 2008 riddle, the existence of a solution is
guaranteed by the second part of that riddle.

Q.E.D.

Bojan Bašić sends the following graphical description of the solution
(illustrated here for *n*=5):

The arrangements for the various nights can be reached by rotating the shape in multiples of 36 degrees.

As a side-note, readers seeking a way to intuit the solution should consider
what happens when 2*n*+1 is a prime, *p*. Placing guest
*i**j* (mod *p*) in table position *i* at night *j*
(*j* being counted starting from 1) is clearly an immediate solution,
because any two adjacent guests on night
*j* will be numbered *j* apart. The solution to the riddle stems
from a different method of proving that this is a valid solution, namely by
showing that the ratio of the numbers assigned to any two adjacently-seated
guests is unique in the arrangement of any particular night. This validity
criterion has the advantage that it requires the use of only one numerical
operation (multiplication), and can therefore be applied to groups of any
size, not necessarily prime. The solution to this month's riddle
can be viewed as the discrete log of this arrangement
(with Arthur playing the part of "log 0").