Nim-Style Counting Game: Optimal Strategy and Losing Positions

Worked Solution

How to Think About It: This is a classic combinatorial game theory problem. The key is to think backward from the end: who controls the game? If you can force your opponent to always face a multiple of 4, they can never win -- because whatever move they make (1, 2, or 3), you can always make the pair sum to 4, marching them toward the next multiple of 4. The insight is that the move set $\{1, 2, 3\}$ sums to 4, and that symmetry is the entire game.

Before any formal proof, here is the gut check: position 0 is a loss for the player to move (the game is already over if you are asked to move from 0 -- actually, 0 is not reachable mid-game; the real base case is position $N$ itself). Start from $N$ and count backward. $N$ is a win (the previous player just moved there and won, so the player to move has no move to make -- but more precisely, any total from which you can reach $N$ in one move is a winning position). The losers cluster at $N-4$, $N-8$, ..., and these are exactly $\{0, 4, 8, 12, \ldots\}$ relative to the start.

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Part (a) -- Proof: Losing Positions Are Multiples of 4

We classify every position $k \in \{0, 1, \ldots, N\}$ as either a P-position (Previous player wins, i.e., the player to move loses) or an N-position (Next player wins, i.e., the player to move wins).

Claim: $k$ is a P-position if and only if $4 \mid k$.

Proof by strong induction:

*Base cases:* - $k = N$: The player to move has already won (the game ends when the total *reaches* $N$). Actually, we should think of it this way: a position $k < N$ is a winning position if you can move to $N$ (i.e., $N - k \in \{1, 2, 3\}$), otherwise you need to move to a P-position. Position $N$ itself is a terminal winning state -- the player who just moved there has won, so $N$ is not a position any player "faces."

*Inductive step:* Assume the claim holds for all positions $> k$.

• If $4 \mid k$: Every move from $k$ adds $m \in \{1, 2, 3\}$, landing at $k + m$. Since $m \in \{1, 2, 3\}$, none of $k+1, k+2, k+3$ is divisible by 4. By the inductive hypothesis, all of these are N-positions (winning for the player who faces them). So every move from $k$ leads to a position where the *opponent* wins -- meaning $k$ is a P-position. $\checkmark$

• If $4 \nmid k$: Let $r = k \bmod 4 \in \{1, 2, 3\}$. Choose move $m = r$. Then $k + m = k + r \equiv 0 \pmod{4}$, which is a P-position by the inductive hypothesis -- the player who faces $k + r$ loses. So from $k$, the current player can move to a P-position, making $k$ an N-position. $\checkmark$

This completes the induction. The P-positions are $\{0, 4, 8, 12, \ldots\}$.

Boundary check at $N$: If $4 \mid N$, then $N$ is a P-position -- but $N$ is the winning terminal state. The resolution: position $N$ is a terminal win for the player who *arrives* there. The player who *faces* total $N$ (if this could happen) has already lost. Since $N$ is reachable from $N - 1$, $N - 2$, $N - 3$ (all N-positions when $4 \mid N$), the induction is consistent.

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Part (b) -- O(1) Algorithm

From part (a), the first player faces position 0, which is a P-position (a loss) if and only if $4 \mid 0$ -- but position 0 is the start, not the terminal. Re-read: the player to move from total $k$ loses iff $4 \mid k$.

The first player moves from total 0. Since $4 \mid 0$, position 0 is a P-position -- the first player loses if and only if $4 \mid N$? No: we need to be careful. The P-positions were derived relative to the winning condition at $N$. The correct statement is:

• Compute $r = N \bmod 4$.
• If $r = 0$: The second player wins. No optimal first move exists for player 1.
• If $r \in \{1, 2, 3\}$: The first player wins by moving $r$ on the first turn (bringing the total to $r$, which satisfies $4 \mid r$? No -- $r \in \{1,2,3\}$ means $r$ is an N-position for player 2, i.e., a P-position for the one who faces it).

Let me state it cleanly:

$$\text{First player wins} \iff N \not\equiv 0 \pmod{4}$$

If the first player wins, the optimal first move is $m = N \bmod 4 \in \{1, 2, 3\}$. This brings the total to $m$, and since $4 \mid (N - m)$... actually the optimal move is $m^{*} = N \bmod 4$, which leaves a total of $m^{*}$. From total $m^{*}$, the remaining distance to $N$ is $N - m^{*}$, which is divisible by 4. Now it is player 2's turn to face a game where the remaining distance is a multiple of 4 -- a losing position by part (a).

Algorithm (pseudocode): ``` function analyze(N): r = N mod 4 if r == 0: return ("Second player wins", None) else: return ("First player wins", r) # optimal first move is r ```

This runs in $O(1)$ time and $O(1)$ space.

Example: $N = 14$. $14 \bmod 4 = 2$. First player wins; optimal first move is 2, bringing total to 2. Remaining distance is 12, a multiple of 4. Whatever player 2 adds ($m_2 \in \{1,2,3\}$), player 1 responds with $4 - m_2$, maintaining the invariant that the remaining distance is always a multiple of 4 after player 1's turn.

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Part (c) -- Generalization and Trading Relevance

The reasoning above is a special case of Sprague-Grundy theory for impartial combinatorial games (games where both players have the same available moves from any position). The key steps always are:

1. Identify terminal positions (wins or losses).
2. Classify positions by backward induction: a position is a P-position (loss for the player to move) if every move leads to an N-position; it is an N-position (win) if at least one move leads to a P-position.
3. Find the pattern -- in many games the P-positions have a clean modular or Grundy-value structure.

For sums of independent games (e.g., multiple simultaneous Nim piles), Sprague-Grundy assigns a *Grundy value* (nimber) to each position, and the combined game is a P-position iff the XOR of all Grundy values is 0.

Adversarial trading relevance: The analogy to trading is not perfect, but the structural lesson is real. In adversarial settings -- competitive market making, options auctions, or sequential bidding -- an optimal player reasons backward from the endgame. If you know your counterparty is rational, you can identify the states where they are "stuck" (P-position analog: no good move, forced to make a bad trade) and the states where they hold the advantage. Market makers who understand the structure of a repeated game can set quotes that force the informed trader into a sequence of moves that always leaves the MM in a favorable position -- the same logic as maintaining the invariant that the remaining distance is a multiple of 4.