7 - Problem C: Solution

Summary

We can describe each pattern by a vector in which each component is R or B; for example, (R,R, B, B, B, B,R,B) means that n = 8 and, as we move around the circle in a clockwise direction, we see the colours red, red, blue, and so on, in this order. The starting point of this list is not well defined, so we agree that two patterns will be considered to be the same if one pattern can be rotated to the other. This means that we identify any vector v with those vectors that are obtained by cyclically permuting the components of v. Now let us consider the cases n = 2, 3, 4, 5.

The case n = 2 There are three possible patterns, namely (R,R), (R,B) and (B,B), and the system evolves as follows:

(R,B) → (B,B) → (R,R) → (R,R).

Regardless of the first pattern, after (at most) two steps we reach the pattern (R,R) and remain there.

The case n = 3 There are four possible patterns, namely (R, R,R), (R, R,B), (R,B,B) and (B, B,B), and the system evolves as follows:

(B, B,B) → (R, R,R) → (R, R,R)

(R, R,B) → (R,B,B) → (R,B,B).

Here the pattern does not change after the first step, and it is then fixed at either (R, R,R) or (R,B,B). Notice that the final outcome depends on the first pattern, and that there are two distinct systems here.

The case n = 4 Here there are six possible patterns:

s1 = (R, R, R,R), s2 = (R, R, R,B), s3 = (R, R,B,B),

s4 = (R,B, R,B), s5 = (R, B, B,B), s6 = (B, B, B,B).

and you should check the following scheme:

Here the outcome does not depend on the first pattern.

The case n = 5 There are eight possible patterns:

s1 = (R, R, R, R,R), s2 = (R, R, R, R,B),

s3 = (R, R, R,B,B), s4 = (R, R,B, R,B),

s5 = (B, B, B,R,R), s6 = (B, B,R, B,R),

s7 = (B, B, B, B,R), s8 = (B, B, B, B,B),