Question

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character ‘.’.

Note:

• The given board contain only digits 1-9 and the character ‘.’.
• You may assume that the given Sudoku puzzle will have a single unique solution.
• The given board size is always 9x9.

Solution

There are 27 constraints: 9 rows, 9 columns, and 9 sub-boxes. Use a set for each constraint to represent possible values so far. Use backtracking to try each number. Before backtracking, iterate through positions where only one possible option can be found.

class Solution:
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        # Populate unused values for each constraint
        # 0-8 is row, 9-17 is column, and 18-26 is sub-box
        options = [set(range(1, 10)) for i in range(27)]
        pos = [] # Empty positions

        for i in range(0, 9):
            for j in range(0, 9):
                c = board[i][j]
                if c != '.':
                    num = int(c)
                    options[i].remove(num)
                    options[9 + j].remove(num)
                    options[18 + (i//3)*3 + j//3].remove(num)
                else:
                    pos.append((i, j))

        self.populateSingleOption(pos, options, board)
        answer = []
        self.backtracking(pos, answer, options)
        for k in range(0, len(pos)):
            i, j = pos[k]
            num = answer[k]
            board[i][j] = str(num)

        return

    def populateSingleOption(self, pos, options, board):
        min_options = 1 # Track whether there are single options populated in last iteration
        while min_options == 1:
            min_options = 10 # max value
            for k in range(len(pos) - 1, -1, -1): # Because items are removed during iteration, do so from the the end
                i, j = pos[k]
                option = set.intersection(options[i], options[9+j], options[18 + (i//3)*3 + j//3])
                if not option:
                    return
                if len(option) < min_options:
                    min_options = len(option)
                if len(option) == 1:
                    num = option.pop()
                    board[i][j] = str(num)
                    pos.pop(k) # This position is populated, remove
                    options[i].remove(num)
                    options[9+j].remove(num)
                    options[18 + (i//3)*3 + j//3].remove(num)

    def backtracking(self, pos, answer, options):
        if len(answer) == len(pos):
            return

        i, j = pos[len(answer)]
        option = set.intersection(options[i], options[9+j], options[18 + (i//3)*3 + j//3])

        if len(option) == 0:
            return
        else:
            for num in option:
                # Try this number
                answer.append(num)
                options[i].remove(num)
                options[9+j].remove(num)
                options[18 + (i//3)*3 + j//3].remove(num)

                self.backtracking(pos, answer, options)
                if len(answer) == len(pos):
                    break

                # This number does not satisfy requirement, restore to previous state
                answer.pop()
                options[i].add(num)
                options[9+j].add(num)
                options[18 + (i//3)*3 + j//3].add(num)