#1025

Easy

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Divisor Game

Divisor Game is a game theory problem where players take turns subtracting divisors of a number n until one player loses.

math dynamic programming brainteaser game theory

Problem Statement

In the Divisor Game, Alice and Bob take turns subtracting divisors of a given number n. Alice always starts, and on each turn, a player subtracts a divisor of the current number from it. The player who cannot make a valid move loses the game.

If n is even, Alice can subtract 1, turning the number odd, which gives her a strategic advantage. If n is odd, Alice must subtract an odd divisor, making the number even again. The game continues until no valid move can be made.

Examples

Example 1

Input: n = 2

Output: true

Alice chooses 1, and Bob has no more moves.

Example 2

Input: n = 3

Output: false

Alice chooses 1, Bob chooses 1, and Alice has no more moves.

Constraints

1 <= n <= 1000

Solution Approach

State Transition with Dynamic Programming

This problem is best approached by simulating the game states using dynamic programming. Create an array where each index i represents whether the position i is a winning or losing position. Transition between states based on whether the current number is odd or even.

Game Theory Insight

The problem can be analyzed using game theory by considering the positions that are winning or losing based on the parity of n. If n is even, Alice can always subtract 1 to make it odd. If n is odd, Alice’s only option is to subtract an odd number, giving Bob a chance to make the number even.

Minimizing Computational Effort

Instead of trying to find all divisors for each turn, optimize by focusing on the state transitions and memoizing the results. This allows us to compute the game outcomes efficiently without redundant calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on how efficiently we calculate divisors and transition states. The space complexity depends on the dynamic programming array size, which is O(n).

What Interviewers Usually Probe

Look for understanding of game theory and dynamic programming concepts.
Check if the candidate can optimize the approach beyond brute force.
Assess the candidate’s ability to recognize patterns in state transitions.

Common Pitfalls or Variants

Common pitfalls

Not considering the impact of parity (even/odd) in game decisions.
Overcomplicating the problem by calculating all divisors instead of focusing on state transitions.
Failing to optimize space usage with dynamic programming.

Follow-up variants

Extend the problem to multiple players instead of just Alice and Bob.
Consider the case where players can subtract any divisor instead of only the proper ones.
Introduce a restriction on the number of moves a player can make.

FAQ

What is the core idea behind the Divisor Game?

The core idea is to subtract divisors from n, and the winner is the one who forces the opponent into a losing position by making optimal moves.

How can dynamic programming help solve this problem?

Dynamic programming is used to track winning and losing positions for each possible value of n, allowing us to efficiently compute the game's outcome.

Why is the parity of n important in the Divisor Game?

The parity determines the type of move a player can make. Alice can always subtract 1 when n is even, but when n is odd, Alice must subtract an odd divisor.

What would happen if there were more players involved?

With more players, the game would become more complex. Each player's strategy would need to be adapted to ensure a winning position, with new state transitions for each additional player.

What other patterns or algorithms are related to this problem?

This problem relates to game theory, dynamic programming, and state transitions. It is also linked to problems involving turn-based games and optimal strategies.

terminal

Solution

Solution 1: Mathematical Induction

- When $n=1$, the first player loses.

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class Solution:
    def divisorGame(self, n: int) -> bool:
        return n % 2 == 0

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