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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 \(y\).

Is \(\text{Nim}(7,11,15)\) a winning position or a losing position? If it is a winning position, what is the optimal move?

Show that \(\text{Nim}(x_1,\dots,x_k)\) is an losing position if and only if \(x_1 \oplus \dots \oplus x_k = 0\). \(x \oplus y\) denotes the nim-sum of \(x\) and \(y\).