The winning strategy in (a) is easily found: all piles are even in number so Player 2 maintains this all-even status by parroting Player 1's every move. This eventually guarantees a win for Player 2.

Configuration (b) is trickier. We will see that there is a guaranteed win for Player 1. We first mathematically encode the game as a system operating on the set $S$ of sequences $s = (s_i)_i^\infty$ of non-negative integers with only finitely many non-zero entries.

For each $i \in \mathbb{N}$, the function $f_i$ is defined on the subset of $S$ of sequences $s$ such that $s_i > 0$: $f_i(s) = t$, where $t_i = s_i - 1$ and $t_i = s_i$ otherwise. The function $g: S \to S$ is defined by $g(s) = t$, where $t_j = \max\{0, s_j - 1\}$, $j \in \mathbb{N}$. Below, we write $\mathbb{F}$ for the set of functions consisting of $g$ and every $f_i$.

A particular game is given by a sequence $(s^k)_{k=0}^\infty$ of elements of $S$, where $s_{k+1} = f(s^k)$ for some $f \in \mathbb{F}$. The winner is determined by the parity of the minimal $k$ for which $s^k$ is the zero element, denoted $z$ below.

Given $s \in S$, the notation $f(s)$ will always refer to $f(s)$ for some $f \in \mathbb{F}$; the number of positive and odd entries in $s$ are denoted by $n(s)$ and $o(s)$, respectively. We call $W \subset S$ a strategy if the following conditions hold:
(1) If $s \in W$, and if $f(s)$ is defined, then $f(s) \notin W$, but there exists $f' \in \mathbb{F}$ such that $f'(f(s)) \in W$.
(2) $z \in W$.
(3) $W$ is rearrangement invariant: if $w \in W$, then every rearrangement of $w$ lies in $W$.

(1) says that a player should strive to leave a configuration in $W$ after his/her move. If this is achieved, every move the other player makes gives a configuration not in $W$; the first player can then move it back to $W$, and eventually win because of (2). (3) just encodes the fact that a winning strategy is readily adjusted to handle swapping of piles.

A strategy $W$ tells us how to win from certain configurations, but it might not tell us how to proceed from every winnable configuration. $W$ is $k$-complete if for each $s \in S$, $n(s) \le k$, either $s \in W$ or $f(s) \in W$ for some $f \in \mathbb{F}$. (A $k$-complete strategy tells us how to win from every configuration involving at most $k$ piles.)

The previously given solution to Configuration (a) is essentially equivalent to the fact that
$$W_0 := \{s \in S \mid o(s) = 0\}$$
is a strategy. To verify (1), it suffices to note that if $s \in W_0 \setminus \{z\}$ and if $f(s)$ is defined, then $o(f(s)) \in \{1, n(s)\} \subset \mathbb{N}$, but $f(f(s)) \in W$.

Clearly, $W_0$ is 2-complete, but it is not 3-complete. We will enlarge $W_0$ to first make it 3-complete, and then 4-complete. In general, if we have a strategy $W$, and a rearrangement invariant $A \subset S$, then $A \cup W$ is also a strategy if for each $s \in A$:
* there is no $f \in \mathbb{F}$ with $f(s) \in W$, but
* if $f(s)$ is defined, then there exists $f' \in \mathbb{F}$ such that $f'(f(s)) \in W$.

In the above situation, we call $A \cup W$ a one-step augmentation (of $W$). The sole reason that $W_0$ is not 3-complete is because of $s$ with $n(s) = 3$ and $o(s) = 2$. However, if we define
$$A := \{s \in S \mid n(s) = 3 \text{ and } o(s) = 2\},$$
then $W_1 := A \cup W_0$ is a one-step augmentation and also a 3-complete strategy. (Note that if $o(s) = 2$, then $o(f(s))$ is either 3 or 1. In the former case, $g(f(s)) \in W_0$; in the latter, $f(f(s)) \in W_0$, where $i$ is the index of the odd entry.)

Finally, let $W_2 := W_1 \cup B$, where $B$ consists of all $s$ such that $n(s) = 4$ and either
* $o(s) = 3$ and the odd entries of $s$ all exceed 1, or
* $o(s) = 2$ and the odd entries of $s$ both equal 1.
We call the former sequences Type 1 and the latter Type 2.

We now prove that $W_2$ is a strategy. (It is not a one-step augmentation of $W_1$, so the proof requires a little more effort.)

First, suppose $s$ is of Type 1. Since $o(g(s)) = 1$, we have $f_i(g(s)) \in W_0$, where $i$ is the index of the odd entry in $g(s)$. Alternatively, $o(f_i(s)) \in \{2, 4\}$ whenever $f_i(s)$ is defined. If $o(f_i(s)) = 4$, then $g(f_i(s)) \in W_0$, as desired. If instead $o(f_i(s)) = 2$, then the next step depends on whether or not both of the even entries equal 2. If they do, then $g(f_i(s))$ is of Type 2, as desired. If instead the $j$th entry is even and exceeds 2, then $f_j(f_i(s))$ is of Type 1.

Suppose instead that $s$ is of Type 2. First, $n(g(s)) = 2$ and $o(g(s)) = 2$, so $g(g(s)) \in W_0$. Next, assume that $f_i(s)$ is defined. It might be that $s_i = 1$, in which case $o(f_i(s)) = 1$ and $n(f_i(s)) = 3$. Writing $j$ for the index of the single odd entry in $f_i(s)$, we have $f_j(f_i(s)) \in W_0$, as required. Suppose instead that $s_i \neq 1$, and so $s_i$ is even. Then $o(f_i(s)) = 3$. Taking $j$ so that $s_j = 1$, we see that $n(f_j(f_i(s))) = 3$ and $o(f_j(f_i(s))) = 2$, so $f_j(f_i(s)) \in W_1$, as required.

Using $W_2$, we see that Player 1 has a guaranteed win in Configuration (b); the first move should be to remove a stone from one of the even piles.