Nim Game - Problem Statement

$n$ stones in a pile. Each turn, remove 1, 2, or 3 stones. Whoever takes the last stone wins. $dp[i]$ = true if the player facing $i$ stones can win. Base: $dp[0] = false$ (no stones = you lose). Transition: $dp[i] = \lnot dp[i-1] \lor \lnot dp[i-2] \lor \lnot dp[i-3]$.

If any move leads to opponent losing, you win. Pattern emerges: $dp[i] = (i \mod 4 \neq 0)$. Multiples of 4 are losing positions. This is the classic Nim result.