Solution Explanation for Subtree Removal Game with Fibonacci Tree

This problem involves a game played on a Fibonacci tree. The key to solving it efficiently lies in understanding the recursive nature of the Fibonacci tree and applying a dynamic programming approach or using mathematical properties.

Understanding Fibonacci Trees

A Fibonacci tree of order n (denoted as order(n)) is defined recursively:

• order(0): Empty tree
• order(1): A single node
• order(n): A root node with left subtree order(n-2) and right subtree order(n-1).

Game Logic

The game proceeds as follows:

1. Alice starts.
2. On each turn, a player removes a node and its entire subtree.
3. The player forced to remove the root node loses.

Optimal Strategy

The optimal strategy for both players is to always remove a node that leaves the opponent with an unbalanced game state. This is where the recursive nature of the Fibonacci tree and dynamic programming comes into play.

Solution Approach: Dynamic Programming

We can use dynamic programming to determine the winner for each subtree size. Let dp[i] be a boolean value indicating whether the player who starts with a subtree of size i can win. We can recursively build up this dp array:

• dp[0] = false (empty tree, the player loses immediately).
• dp[1] = false (single node, the player loses immediately).
• dp[i] = ! (dp[i - 1] && dp[i - 2]) This line expresses the core logic: a player wins if they can force their opponent into a losing position. If the opponent can win from both the left (size i-2) and right (size i-1) subtrees, then the current player loses. Otherwise, the current player can win by choosing one of the subtrees that leads to the opponent losing.

Code (Python)

def canWin(n):
    """
    Determines if Alice wins the game on a Fibonacci tree of order n.
 
    Args:
        n: The order of the Fibonacci tree.
 
    Returns:
        True if Alice wins, False otherwise.
    """
    if n <= 1:
        return False  # Base cases: Alice loses for n=0,1
 
    dp = [False] * (n + 1) # Initialize DP array with False
    dp[1] = False
 
    for i in range(2, n + 1):
        dp[i] = not (dp[i - 1] and dp[i - 2])
 
    return dp[n]
 
# Example usage
n = 3
print(canWin(n))  # Output: True
 
n = 1
print(canWin(n))  # Output: False
 
n = 2
print(canWin(n))  # Output: True

Time and Space Complexity

• Time Complexity: O(n) - The dynamic programming solution iterates through the dp array once.
• Space Complexity: O(n) - We use an array dp of size n+1 to store the results.

Alternative Solution (Mathematical)

It's possible to observe a pattern in the winning conditions based on the Fibonacci sequence. The pattern shows that the result is based on whether n is a Fibonacci number or not, and this could lead to a more concise solution without explicit dynamic programming. However, the dynamic programming approach presented above is generally easier to understand and implement.