6 Exploring Peg Solitaire
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6.4 Problems¶
Nim
Exercise 6.9. Nim is a simple two-player game that is played with piles of coins or other small objects. A common version is 3,4,5-Nim in which there are three piles of 3, 4, and 5 coins:
The players alternately remove one or more coins from a single pile. For example, if the first player removes three coins from the second pile, we end up with the following position:
The player who removes the last coin wins.
What is a good strategy for playing Nim? More specifically, is there a winning strategy , that is, a strategy that leads to a guaranteed win, regardless of how the opponent plays? Let’s start our analysis with with a few simple instances of Nim. If there is a single pile of \(n\) coins, the first player obviously wins
by taking all \(n\) coins, but taking fewer coins is risky because it gives the second player a chance to take all remaining coins. Taking all coins is therefore the only winning strategy for Nim with a single pile.
Since every move in Nim leads to a position with fewer coins, we can use recursion to determine whether there is a winning strategy for the current position. Let’s define a function \(W\) that determines whether a position is winnable. Continuing with the example of
\(n,m\)-Nim, this strategy can be formulated as follows. If there are no coins left when our turn begins, we lose, so we set
\(\seteqnumber{0}{6.}{1}\)
\begin{equation*}
W(0, 0) = \mathsf {F}.
\end{equation*}
(We use the symbols \(\mathsf {T}\) and \(\mathsf {F}\) to represent true and false , respectively.) Other positions are winnable if at least one of the possible next moves leads to a position that is not winnable for the opponent. If the current state is \((n,m)\),
there are \(n+m\) possible next moves, which lead to the following states:
\(\seteqnumber{0}{6.}{1}\)
\begin{equation*}
(n-1,m), (n-2,m), \ldots , (0, m), (n, m-1), (n,m-2),\ldots , (n,0).
\end{equation*}
We can win the game if at least one of these states is not winnable (for the opponent), in other words if
\(\seteqnumber{0}{6.}{1}\)
\begin{multline}
\label {eq:pegwin} W(n, m) = \neg W(n-1, m) \lor \neg W(n-2,m)\lor \ldots \lor \neg W(0,m)\\ \lor \neg W(n, m-1)\lor \neg W(n, m-2)\lor \ldots \lor \neg W(n, 0)
\end{multline}
For example, Eq. (6.2 ) tells us that \((1,0)\) is a winning position because
\(\seteqnumber{0}{6.}{2}\)
\begin{align*}
W(1,0) &= \neg W(0,0) = \neg \mathsf {F} = \mathsf {T}.
\end{align*}
On the other hand, \((1,1)\) is not a winning position because
\(\seteqnumber{0}{6.}{2}\)
\begin{align*}
W(1, 1) &= \neg W(0, 1) \lor \neg W(1,0) = \neg \mathsf {T} \lor \neg \mathsf {T} = \mathsf {F}.
\end{align*}
Here we have used the fact that the stacks of coins are interchangeable, so \(W(0,1)=W(1,0)\). As the number of coins and piles increases, the number of terms that must be considered grows rapidly.
c) Implement a recursive function that determines whether a given Nim position is winnable. The function should work with any number of piles and coins. Is the position \(10,5,4,3\) winnable for the current player? What moves are possible? Hint: Use
normalization and memoization .
The mathematical theory behind Nim was first studied by Charles L. Bouton in 1901. Bouton showed that there is a winning strategy for any number of piles and coins and derived a simple non-recursive procedure for determining the winner. His main idea was to write the number of coins in each
pile as a binary number and add them without carry . The position is winning if and only if the result is nonzero. For example, when the number of coins is \((1,3,7)\), we obtain the sum
\(\seteqnumber{0}{6.}{2}\)
\begin{equation*}
\begin{array}{ccc@{\extracolsep {1cm}}l} 0 & 0 & 1 & =1\\ 0 & 1 & 1 & =3\\ 1 & 1 & 1 & =7\\ \midrule 1 & 0 & 1 & \text {sum, ignoring carries}\\ \end {array}
\end{equation*}
Since the result is 101 and not zero, \((1, 3, 7)\) must be a winning position. The binary sum without carries is also called the Nim sum and is equivalent to the XOR operation .
d) Are \((2,5,6)\) and \((2,5,7)\) winning positions?
Solving Sudoku
Exercise 6.10. Sudoku is a popular puzzle that is played on a \(9\times 9\) board. Some of the board’s cells already contain digits. The objective is to fill the remaining cells with digits between 1 and 9 so that every row, column, and \(3\times 3\)
block contains each digit exactly once. Sudoku puzzles are usually designed in such a way that there is exactly one solution that satisfies these conditions.
For the Sudoku board shown in Fig. 6.7 , 24 of the 81 digits are given and the values of the remaining 57 digits must be found. Some numbers are easy to determine. For instance, it’s not hard to see that the square labeled A8
must be 2: We know that this digit must occur somewhere in the top-right square, but rows B and C are blocked because there is already a 2 in squares B2 and C6, so the only remaining position is A8. Similar arguments can be used to successively fill in the remaining squares.
Even though there are specialized algorithms for solving Sudoku and similar puzzles efficiently, modern computers are fast enough that a simple backtracking approach is viable as well.