Fibonacci Nim Strategy

1. Start with a non-Fibonacci number, and pick first: then you're guaranteed to win.

   

   where

   

   The fact of the matter is that we can do this:

   Theorem 13.4: Any positive integer N can be expressed as a sum of distinct Fibonacci numbers, no two of which are consecutive; that is,

   

   where and for (the Zeckendorf representation).

   Proof: By the second principle of induction.

2. You pick first, and take the number of sticks indicated by the smallest in the Zeckendorf representation, .

   In practice, this will leave our opponent unable to follow the same strategy we are using, since

   

   for all . The only troublesome case is the case of m=1: but the pair is never necessary, since it can be replace by , since

   

   If was in the sum, then you might protest that we now have consecutive Fibonaccis, but the fact of the matter is that we can always pass the problem up, until it disappears. The worst cases are these:

   

   

   So this only happens when we started with a Fibonacci. Otherwise, we will ultimately pass the problem up until a gap exists, and we have another representation that is non-consecutive Fibonacci numbers.

3. Hence Player 2 faces

   

   and can't choose to remove - she or he must take some portion thereof; hence we split , as follows:

   

   or

   

   Player 2 must take something, but can't take all of , leaving a succession of non-consecutive Fibonacci numbers for us, Bwahahahaaaa!

4. As the play continues, we ultimately get down to where we end up with something like

   

   to us; we take , and they're stuck. They take 1, we take 2 and win; they take 2, we take 1 and win.