292. Nim Game

Description of the Problem

You are playing the following Nim Game with your friend:

• Initially, there is a heap of stones on the table.
• You and your friend will alternate taking turns, and you go first.
• On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
• The one who removes the last stone is the winner.

Given n, the number of stones in the heap, return true if you can win the game assuming both you and your friend play optimally, otherwise return false.

Example 1:

Input: n = 4
Output: false
Explanation: These are the possible outcomes:
1. You remove 1 stone. Your friend removes 3 stones, including the last stone. Your friend wins.
2. You remove 2 stones. Your friend removes 2 stones, including the last stone. Your friend wins.
3. You remove 3 stones. Your friend removes the last stone. Your friend wins.
In all outcomes, your friend wins.

Example 2:

Input: n = 1
Output: true

Example 3:

Input: n = 2
Output: true

Constraints:

• 1 <= n <= 2^31 - 1

Solution

Tags: Game Theory

Explanation

Use back-tracking to consider the situation. Let W denotes winning position, L denotes losing position.

To win the game, the number of stones at the last stage should be 1, 2 or 3. In order to have this condition, the previous number of stones should be 4. According to the above reasoning, we can fill the above table to observe the winning pattern.

By observation, the winning position contains n stones where \(n \ mod \ 4 \ne 0 \).

Code (Java)

class Solution {
    public boolean canWinNim(int n) {
        return n % 4 != 0;
    }
}