Even numbers combinatorial game We have $2$ piles of coins-one of them with $X$ number of sheets and the other with $Y$ number of sheets, $X$ and $Y$ are larger than $0$.  
$2$ players are playing against each others , every player at his turn chooses a pile and he is allowed to take an even number ($K$) from the coins in that pile, then he returns half of $K$ to the other pile (that he didn't choose from in this turn) and keeps the other half out of the game (you can take from different piles in different turns like you can choose now from pile $1$ and then in the next turn you can choose from pile $2$).  
As the game goes the numbers $X$ and $Y$ becomes smaller.When $X$ and $Y$ are less than $2$ the game stops.The winner is the player who does the last step.
For example:
if in pile $1$ we have $2$ coins and in pile $2$ we have $1$ coin $(2,1)$ I shouldn't start because I would lose WHY?:
$(2,1)$-->I take $2$ from pile $1$ -->$(0,2)$ then he takes $2$ from pile $2$ and wins-->$(1,0)$ because $X$ and $Y$ are less than $2$ when my turn is about to start.  
I want a strategy which always wins, you can choose weather to start or not
please specify your strategy and why it wins!
Thank you


Lemma: $(X,Y)$ is winning for the starting player if and only if $|X - Y| > 1$.

Obs. The positions $(1,0)$ and $(0,1)$ are losing because the player who gets them has no moves (hence the last move was made by the winner).
Proof of the Lemma: 
We shall prove this by induction on $X + Y$.
If $X + Y = 1$, than the statement is obviously true.
Suppose now that $X + Y = n$, and that we have shown that the statement is true for any pair $(X', Y')$ satisfying $X' + Y' < n$.
Suppose $|X - Y| \leq 1$. Without loss of generality, any move will be of the form $(X, Y) \rightarrow (X - 2k, Y + k)$, and by the triangle inequality we have $|X - 2k - (Y + k)| = |X - Y - 3k| \geq 3k - |X - Y| \geq 2$. By induction, the resulting position is winning (hence the starting position has to be losing).
Suppose now $|X - Y| \geq 2$, and suppose without loss of generality that $X \geq Y$. Set $k = \left\lfloor \frac{X - Y + 1}{3} \right\rfloor$, and consider the position $(X - 2k, Y + k)$ which is one move away from our starting position.
It should be clear that 
$$ (X - Y + 1) - 2 = X - Y - 1 \leq 3k \leq X - Y + 1,$$ 
hence 
$$X - Y - (X - Y + 1) = -1 \leq X - Y - 3k \leq 1 = X - Y - (X - Y -1).$$
Therefore, $|X - 2k - (Y + k)| = |X - Y - 3k| \leq 1$, which implies that the position $(X - 2k, Y + k)$ is losing. The proof is complete.
