Solution
Knaves Never Stand Together
Show the problem again
Twenty islanders sit around a round table, each a knight (always truthful) or a knave (always lying). Every one of them says: "At least one of my two neighbours is a knave." What is the largest possible number of knights at the table? Give a single integer.
Worked solution
A knave's statement is false, so both of a knave's neighbours are knights. Hence no two knaves are adjacent. A knight's statement is true, so every knight has at least one knave neighbour. Let $k$ be the number of knaves. Each knave has two neighbours, so it can serve as the required knave neighbour for at most two knights; therefore the number of knights is at most $2k$. Total seats: $20 = \text{knights} + k \le 3k$, so $k \ge 7$ and the number of knights is at most $20 - 7 = 13$. This is attainable. Seat six blocks K N K followed by a final block K N around the circle: every knave sits between two knights, and every knight is adjacent to a knave, giving 13 knights and 7 knaves. An exhaustive search over all $2^{20}$ seatings confirms that 13 is the maximum. Answer: 13.
Source: Original extremal variant of a standard knights-and-knaves round-table exercise. Statement written for AxiomIQ.