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Maximum Number of Moves to Kill All Pawns

Difficulty: Hard

Problem Description

There is a 50 x 50 chessboard with one knight and some pawns on it. You are given two integers kx and ky where (kx, ky) denotes the position of the knight, and a 2D array positions where positions[i] = [xi, yi] denotes the position of the pawns on the chessboard. Alice and Bob play a turn-based game, where Alice goes first. In each player's turn, the player selects a pawn and captures it with the knight in the fewest possible moves. The knight can capture any pawn, but only the selected pawn can be captured in that turn. Alice aims to maximize the total number of moves made by both players until there are no more pawns on the board, while Bob tries to minimize them. Return the maximum total number of moves that Alice can achieve, assuming both players play optimally.

Key Insights

  • The knight has 8 possible moves, and its movement patterns must be considered for calculating pawn capture distances.
  • The game is turn-based, with Alice trying to maximize and Bob minimizing the total moves.
  • A breadth-first search (BFS) can be used to compute the minimum moves for the knight to capture each pawn.
  • The problem can be modeled as a game-theoretic scenario where each player has a best response to the other's actions.

Space and Time Complexity

Time Complexity: O(2^n * n^2) where n is the number of pawns (at most 15), due to the exploration of all combinations of pawn captures. Space Complexity: O(n) for storing the distances and the state of pawns.

Solution

To solve the problem, we will use a combination of breadth-first search (BFS) to find the shortest paths for the knight to each pawn, and dynamic programming (DP) to simulate the turn-based game. We will create a distance matrix to store the number of moves required for the knight to reach each pawn from its current position. The game will be modeled using a bitmask to represent the state of pawns on the board, allowing us to efficiently explore all possible capture orders.

Code Solutions

from collections import deque

def knight_moves(kx, ky):
    directions = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]
    queue = deque([(kx, ky, 0)])
    visited = set()
    visited.add((kx, ky))
    distances = {}

    while queue:
        x, y, moves = queue.popleft()
        distances[(x, y)] = moves

        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < 50 and 0 <= ny < 50 and (nx, ny) not in visited:
                visited.add((nx, ny))
                queue.append((nx, ny, moves + 1))

    return distances

def max_moves(kx, ky, positions):
    n = len(positions)
    distances = knight_moves(kx, ky)
    distance_matrix = []

    for pawn in positions:
        distance_matrix.append(distances.get((pawn[0], pawn[1]), float('inf')))

    dp = [-1] * (1 << n)
    dp[0] = 0

    for mask in range(1 << n):
        for i in range(n):
            if (mask & (1 << i)) == 0:  # If the pawn i is not captured
                next_mask = mask | (1 << i)
                moves = distance_matrix[i]
                if dp[next_mask] == -1:
                    dp[next_mask] = dp[mask] + moves
                else:
                    dp[next_mask] = max(dp[next_mask], dp[mask] + moves)

    return dp[(1 << n) - 1]

# Example usage
print(max_moves(1, 1, [[0, 0]]))  # Output: 4
print(max_moves(0, 2, [[1, 1], [2, 2], [3, 3]]))  # Output: 8
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