Problems
Explain why a position \(g\) is a winning position if there is a move that turns \(g\) into a losing position. On the other hand, explain why a position is a losing position if all moves turns it into a winning position.
A technique that can be used to completely solve certain games is drawing game graphs. Given a game \(G\), we draw an arrow pointing from a position \(g\) to a position \(h\) if there is a move from \(g\) to \(h\).
As a simple example, the game graph of \(\text{Nim}(2)\) is shown below.
Draw the game graph of \(\text{Nim}(2,2)\). Is \(\text{Nim}(2,2)\) a winning position or losing position?
Let \(x,y\) be nonnegative integers. Determine when \(\text{Nim}(x,y)\) is a losing position and when it is a winning position.
When we write 137 in decimal, we mean \(1 \cdot 10^2 + 3 \cdot 10 + 7 \cdot 1\). If we write it instead using powers of 2, we have \(137 = 1 \cdot 2^7 + 0 \cdot 2^6 + 0 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0\). To tell apart binary representation from decimals, we can use the following notation: \(137 = (10001001)_2\).
What is the number 273 in binary? Note that using binary is useful for finding whether a particular Nim game is a winning position or a losing position.
Let us define XOR (or addition mod 2). XOR is defined for 0 and 1 only. Here is a table recording the values of XOR:
| XOR | 0 | 1 |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 1 | 0 |
Now we define the important concept of nim-sum. Given two natural numbers \(x\) and \(y\), we first convert them into binary representations and then compute XOR on individual digits. The resulting number, denoted \(x \oplus y\), is the nim-sum of \(x\) and \(y\). Here is an example.
| 1 | 0 | 1 | 1 | 0 | |
| XOR | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
This is simply saying \(22 \oplus 5 = 19\). Note that \(22=(10110)_2\) and \(5=(00101)_2\).
Verify \((x \oplus y) \oplus z = x \oplus (y \oplus z)\), so we can speak of \(x \oplus y \oplus z\) with no ambiguity.
Show that \(x \oplus y = 0\) if and only if \(x = y\). Remember that \(x \oplus y\) denotes the nim-sum of \(x\) and \(y\).
Show that \(\text{Nim}(x,y,z)\) is a losing position if and only if \(x \oplus y \oplus z = 0\). Remember that \(x \oplus y\) denotes the nim-sum of \(x\) and \(y\).