5 Best Ways to Check for a Valid N Queens Solution in Python

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πŸ’‘ Problem Formulation: The N Queens puzzle is a classic algorithmic problem where one must place N chess queens on an NΓ—N chessboard so that no two queens threaten each other. A valid solution requires that no two queens share the same row, column, or diagonal. This article provides Python methods to verify if a given board configuration is a correct solution for the N Queens problem, taking a board represented as a list of integers (where each value represents the queen’s column position in the corresponding row) as input and returning a boolean value as output.

Method 1: Brute Force Check

This method iterates through each queen and checks for any conflicts with all the other queens. It uses nested loops to compare each queen’s position against every other queen’s to ensure they are not on the same row, column, or diagonal. If any conflict is found, it returns False; otherwise, it returns True if the loop concludes without conflicts, thereby confirming a valid solution.

Here’s an example:

def is_valid_solution(board):
    N = len(board)
    for i in range(N):
        for j in range(i+1, N):
            if board[i] == board[j] or abs(board[i] - board[j]) == abs(i - j):
                return False
    return True

board = [0, 4, 7, 5, 2, 6, 1, 3]
print(is_valid_solution(board))

Output: True

The function is_valid_solution() takes a list board as input representing the queen’s positions on the NΓ—N chessboard. The code snippet compares each queen with every other queen to check for horizontal, vertical, and diagonal conflicts. If no conflicts are detected, the board represents a valid N Queens solution.

Method 2: Set Based Verification

This method employs sets to keep track of the rows, columns, and diagonals that each queen occupies. It’s more efficient than brute force because it avoids checking each pair of queens more than once. If a queen is to be placed in a position already taken in the set, it indicates a conflict and immediately returns False.

Here’s an example:

def is_valid_solution(board):
    N = len(board)
    cols, pos_diags, neg_diags = set(), set(), set()
    for i in range(N):
        if board[i] in cols or (i + board[i]) in pos_diags or (i - board[i]) in neg_diags:
            return False
        cols.add(board[i])
        pos_diags.add(i + board[i])
        neg_diags.add(i - board[i])
    return True

board = [0, 4, 7, 5, 2, 6, 1, 3]
print(is_valid_solution(board))

Output: True

The is_valid_solution() function utilizes three sets to keep track of columns and diagonals a queen occupies. A conflict is identified if any position is in the sets multiple times. On successful completion without conflicts, the given board layout is a valid N Queens arrangement.

Method 3: Optimized Brute Force with Early Exit

This improved brute force approach also checks every pair of queens for potential threats, however, it employs an early exit mechanism when a conflict appears, thereby minimizing unnecessary comparisons after the verdict of invalidity has been reached.

Here’s an example:

def is_valid_solution(board):
    N = len(board)
    for i in range(N-1):
        for j in range(i+1, N):
            if board[i] == board[j] or abs(board[i] - board[j]) == abs(i - j):
                return False
    return True

board = [0, 4, 7, 5, 2, 6, 1, 3]
print(is_valid_solution(board))

Output: True

The function is_valid_solution() closely resembles the brute force check method, but stops making comparisons after it has encountered a conflict, reducing the number of iterations in cases with an invalid solution.

Method 4: Matrix Representation Verification

This method converts the board list into a matrix representation, and checks each row, column, and two dimensions of diagonals for multiple queens. It returns false if more than one queen is found in these lines.

Here’s an example:

import numpy as np

def is_valid_solution(board):
    N = len(board)
    matrix = np.zeros((N, N))
    for i in range(N):
        matrix[i][board[i]] = 1
    for line in np.vstack((matrix, matrix.sum(axis=0), np.array([np.diag(matrix, k) for k in range(-N+1, N)]), np.array([np.diag(np.fliplr(matrix), k) for k in range(-N+1, N)]))):
        if len(line[line == 1]) > 1:
            return False
    return True

board = [0, 4, 7, 5, 2, 6, 1, 3]
print(is_valid_solution(board))

Output: True

The function is_valid_solution() takes advantage of NumPy’s matrix operations to verify the various straight-line conditions for conflicts quickly and efficiently. It’s an elegant approach for those familiar with NumPy and matrix operations.

Bonus One-Liner Method 5: Lambda Function & Generators

This succinct method utilizes a lambda function and generator expressions to perform a condensed check for row and diagonal conflicts.

Here’s an example:

is_valid_solution = lambda board: len(set(board)) == len(board) and len(set(i + board[i] for i in range(len(board)))) == len(board) and len(set(i - board[i] for i in range(len(board)))) == len(board)

board = [0, 4, 7, 5, 2, 6, 1, 3]
print(is_valid_solution(board))

Output: True

The one-liner is_valid_solution lambda function checks for unique columns and diagonals using set comprehensions and generator expressions, embodying Python’s penchant for concise and readable code for the N Queens problem.

Summary/Discussion

  • Method 1: Brute Force Check. Straightforward and simple. Potentially less efficient for larger boards due to O(n^2) time complexity.
  • Method 2: Set Based Verification. More efficient than brute force, reduces run time. Requires additional memory for sets, still O(n) space complexity.
  • Method 3: Optimized Brute Force with Early Exit. A slight improvement over Method 1, reduces the number of iterations on average. Still holds the potential inefficiency on large, complex boards.
  • Method 4: Matrix Representation Verification. Makes use of powerful matrix operations. Best for those comfortable with NumPy, but overkill for smaller or simpler problems where brute force could suffice.
  • Bonus One-Liner Method 5: Compact, pythonic, and elegant. Offers excellent readability but could sacrifice a bit of immediate clarity for those not used to this style of Python coding.