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To solve this problem, we need to count the number of valid Latin squares that can be formed from a given partial grid. A Latin square of order ( n ) is an ( n \times n ) grid where each row and each column contains all ( n ) distinct symbols exactly once.

Approach

1. Validity Check: First, we check if the given partial grid is valid. A valid partial grid should not have any duplicate symbols in any row or column.
2. Backtracking with Minimum Remaining Values (MRV) Heuristic: To efficiently count the valid Latin squares, we use backtracking with the MRV heuristic. This heuristic selects the empty cell with the fewest possible valid symbols to fill next, which minimizes the number of recursive calls and prunes invalid paths early.

Solution Code

def is_valid(grid):
    n = len(grid)
    # Check rows for duplicates
    for row in grid:
        seen = set()
        for c in row:
            if c != '.' and c in seen:
                return False
            seen.add(c)
    # Check columns for duplicates
    for col in range(n):
        seen = set()
        for row in range(n):
            c = grid[row][col]
            if c != '.' and c in seen:
                return False
            seen.add(c)
    return True

def count_latin_squares(grid):
    n = len(grid)
    if not is_valid(grid):
        return 0

    # Collect all empty cells
    empty_cells = []
    for i in range(n):
        for j in range(n):
            if grid[i][j] == '.':
                empty_cells.append((i, j))

    if not empty_cells:
        return 1

    # Determine the symbol set (assuming uppercase letters A-Z)
    symbols = [chr(ord('A') + i) for i in range(n)]

    # Find the cell with the minimum remaining valid symbols (MRV heuristic)
    min_candidates = n + 1
    best_cell = None
    best_candidates = []

    for (i, j) in empty_cells:
        row_used = set()
        for k in range(n):
            if grid[i][k] != '.':
                row_used.add(grid[i][k])
        col_used = set()
        for k in range(n):
            if grid[k][j] != '.':
                col_used.add(grid[k][j])

        # Calculate possible symbols for this cell
        possible = [s for s in symbols if s not in row_used and s not in col_used]

        if len(possible) < min_candidates:
            min_candidates = len(possible)
            best_cell = (i, j)
            best_candidates = possible

            # Early exit if no valid symbols for this cell
            if min_candidates == 0:
                break

    if min_candidates == 0:
        return 0

    i, j = best_cell
    total = 0

    # Try all valid symbols for the best cell
    for symbol in best_candidates:
        # Place the symbol in the cell
        grid[i][j] = symbol
        # Recursively count valid Latin squares from this state
        total += count_latin_squares(grid)
        # Backtrack: remove the symbol from the cell
        grid[i][j] = '.'

    return total

def solve(n, grid):
    return count_latin_squares(grid)

# Example usage:
# n = 2
# grid = [['A', '.'], ['.', 'A']]
# print(solve(n, grid))  # Output: 1

Explanation

1. Validity Check: The is_valid function ensures that the partial grid does not have duplicates in any row or column, which is a prerequisite for forming a valid Latin square.
2. MRV Heuristic: The count_latin_squares function uses the MRV heuristic to select the next cell to fill. This heuristic helps reduce the number of recursive calls by choosing the cell with the least number of valid symbols, thus pruning invalid paths early.
3. Backtracking: For each valid symbol in the selected cell, we place the symbol and recursively count the valid Latin squares from the new state. After the recursive call, we backtrack by removing the symbol to explore other valid options.

This approach efficiently counts the valid Latin squares by leveraging the MRV heuristic to minimize the search space and prune invalid paths quickly. It works well for small to medium values of ( n ) (up to ( 10 ) or ( 12 )) depending on the number of pre-filled cells.

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