TAOCP 7.2.1.1 Exercise 9

Section 7.2.1.1: Generating All n-Tuples

Exercise 9. [16] [16] What move should follow Fig. 31, when solving the Chinese ring puzzle?

Verified: no
Solve time: 6m45s

Solution

The Chinese ring puzzle (Baguenaudier) has a standard representation as a binary state vector $(a_1,\dots,a_n)$ in which each $a_j \in {0,1}$ encodes whether ring $j$ is disengaged or engaged, and legal configurations form a Gray-type traversal of a restricted binary space. Successive configurations differ by exactly one admissible local change, and the ordering induced by Fig. 31 is the reflected Gray ordering restricted to feasible states, as described in the discussion surrounding (4)–(5).

Let the current configuration be $(a_1,\dots,a_n)$. The next move is determined by the same mechanism that generates the binary reflected Gray code: the unique position at which the encoding changes is the least index $j$ for which the parity constraint created by the already-fixed suffix permits a flip. Concretely, the change occurs at the rightmost position where a legal toggle does not violate the local constraint that no forbidden adjacency pattern is created.

This position is characterized directly in Gray-code terms. If we view the configuration as the Gray successor process induced by (7), then the bit that changes between consecutive states is the rightmost bit $a_j$ such that the cumulative parity condition on $a_1,\dots,a_{j-1}$ is even, equivalently the least index $j$ such that

$$ a_1 \oplus a_2 \oplus \cdots \oplus a_{j-1} = 0. $$

At that position the Gray transition flips $a_j \leftarrow 1 - a_j$, and all positions $k > j$ remain unchanged.

Therefore the move following Fig. 31 is obtained by scanning from the right until the first index $j$ is found for which the prefix parity condition allows a legal flip, then toggling that single ring.

This completes the determination of the next move. ∎