Two players set several piles of chips. The piles are arranged in a row from left to right. By turns each player takes one chip from one of the piles and adds at will as many chips as he/she wishes (can be none) to any piles placed to the left of the pile from which the chip was taken. Assuming that the game ever finishes, the player that takes the last chip wins.
Obviously (if there are initially at least two piles) a player can make the duration of the game arbitrarily long by taking a chip from the rightmost pile and adding a large enough number of chips to the left of it, but prove that, no matter how they play, the game will eventually end after finitely many steps.
Find a winning strategy.
Next, assume that instead of one row of piles we have several rows of piles (not necessarily with the same number of piles each). Let $(m,n)$ denote the position of the $n$-th pile of row $m$. By turns each player takes one chip from one of the piles, say the one in position $(m,n)$, and then adds as many chips as he/she wishes (can be none) to any piles in positions $(m',n')$, where either $m'<m$, or $m'=m$ and $n'<n$. Additionally the player can add any number of piles with any number of chips each to rows $1,\dots,m-1$. Prove that also in this case the game always ends in finitely many steps, and find a winning strategy.
Generalize the problem to multi-dimensional arrangements of piles (and solve it) - the idea would be to have the piles represented with tuples of positive integers $(n_1,\dots,n_d)$, each with $N(n_1,\dots,n_d)$ chips, where $N(n_1,\dots,n_d) = 0$ for all but finitely many tuples. Each player picks a tuple $(n_1,\dots,n_d)$ such that $N(n_1,\dots,n_d) > 0$, takes one chip from it, and adds as many chips as he/she wishes (can be none) to piles placed at positions $(n'_1,\dots,n'_d) <_{\text{lex}} (n_1,\dots,n_d)$, where $<_{\text{lex}}$ is the (strict) lexicographical order in $(\mathbb{Z}^+)^d$. Again, prove that the game always finishes, and find a winning strategy.