The problem is similar the Nim problem except now for every non empty pile, either player can remove 0 items from that pile and have it count as their move; however, this move can only be performed once per pile by either player. Lets call this move as a zero move.
For example, let's say pile i initially has 2 items in it. If player A decides to use a Zero-Move on this pile , then neither A nor B can perform another Zero-Move on pile ; that said, either player is free to perform a Zero-Move on any other non-empty pile that hasn't had a Zero-Move performed on it yet.
How can I convert this problem to the standard nim problem using Grundy Numbers?
Is it solvable(telling which player will win) without using Grundy Numbers?
The answer mentioned above is incorrect.It is clear that if pile size is zero no move is possible.g(0)=0 If you try to build a solution bottom up it turns out that if n is even the grundy number is n-1 and when it is odd the grundy is n+1.
Proof: Grundy number of a state is the smallest positive integer that cannot be reached in one valid move. When the pile size is zero the grundy number is 0 as no moves including the zero move is possible. For a given pile size n we actually have two states:
N is the size but no zero move is available.This is analogous to grundy number of the standard nim with grundy number equal to n.
N is the size but zero move is available.Well,it turns out from this state you can reach the state mentioned above and all other states with a zero move remaining for size k < n.
Just try working your way bottom up.For n=1(with zero move left) you can reach n=1(no zero move left) and n=0(empty pile).Therefore,g(1)=2.For n=2 one cannot reach the state n=1(no zero move left) as both making a zero move and removing blocks simultaneously is not possible.g(2)=1.