, 2023

If $n$ is a power of 2, then the minimum $P = 1$; otherwise, the minimum $P = 2$.

When $n = 2^k$, we only need to repeatedly take any two piles with the fewest walnuts, take all their walnuts and combine them, and eventually we can end up with a single pile, which is clearly the smallest possible value of $P$.

Now consider $2^k < n < 2^{k+1}$. Below, we use $t$-pile to denote a pile with $t$ walnuts. Consider the following operations:

1. First, from the initial $n$ piles we select $2^k$ piles, then repeatedly take any two piles among them with the fewest walnuts, take all their walnuts and combine them, and eventually we can end up with one $2^k$-pile together with $m = n - 2^k$ 1-piles. Call this $2^k$-pile the XL pile.

2. Next, from the XL pile and a 1-pile we each take one walnut, forming a 2-pile. If $m < 2^k - 1$, then we again take one walnut each from the XL pile and the 2-pile. Repeat this action until the XL pile has $m$ walnuts left. At this point we have one $m$-pile, one 2-pile, and $n - m - 2 = 2^k - 2$ 1-piles.

3. We then pair up and combine all the 1-piles, thereby obtaining $2^{k-1}$ 2-piles. Then, from these piles, repeatedly take the two smallest piles and combine them, and eventually we will obtain a $2^k$-pile. At this point we are left with one $m$-pile and one $2^k$-pile, so $P = 2$.

We only need to prove that when $n$ is not a power of 2, it is impossible to reduce the operations to a single pile. First note that if we select an $a$-pile and a $b$-pile, and each time take $c \le \min(a, b)$ walnuts, then we have
$$a \to a - c$$
$$b \to b - c$$
$$0 \to 2c$$
Also note that if there exists an odd integer $q$ dividing $\gcd(a-c,b-c,2c)$, then we must necessarily have $q|a$ and $q|b$.

Now, since $n$ is not a power of 2, it must have an odd factor $q > 1$. If we finally end up with a single $n$-pile, then according to the discussion above, the number of walnuts in every initial 1-pile must be divisible by $q$, which is obviously a contradiction. Hence when $n$ is not a power of 2, the minimum value of $P$ is 2.