sudoku solver

🏠

Question: Given a partially filled Sudoku board, write an efficient solver to complete the board according to the rules of Sudoku.

The rules of Sudoku

The rules are very simple:

Solution

You may think of solving a Sudoku board as a highly complex problem, but as you'll see it's actually very straightforward, and only involves:

Backtracking doesn't really mean much, either, since Backtracking happens as a consequence of the DFS and set membership checking. It occurs naturally.

Constraints checking

Each row must contain unique numbers

Checking whether a number is in a set is an O(1) operation, so filling a row with unique numbers can be done in O(n).

Each column must contain unique numbers

The same rule applies to columns. Once again, filling up a column with unique numbers is a fairly trivial problem.

Each section must contain unique numbers

Likewise, filling a square with unique numbers is a trivial task. Let's depict it to be as explicit as possible.

Linear scan

We'll scan the board in a linear manner, one cell at a time.

At each cell, if it's empty, we'll:

Solving with all constraints

This solver is already in the 80% fastest on leetcode; it is fast, and due to its simplicity, it's easy to remember.

Below we can see the backtracking in action.

When the solver can no longer progress; when it reaches a deadend in the solution space; it has no choice but to recurse back up the call stack, and explore new paths.

Let's break down the solution.

1     rows, cols, triples = ddict(set), ddict(set), ddict(set)
2     for r, c in product(range(9), repeat=2):
3         if board[r][c] != ".":
4             rows[r].add(board[r][c])
5             cols[c].add(board[r][c])
6             triples[(r // 3, c // 3)].add(board[r][c])

Above, we begin by populating row, column, and section sets of the pre-existing numbers. Those allow us to check our constraints in O(1) time.

1     def dfs(r, c):
2         if r == 9:
3             return True

This is our dfs basecase, in case we are done exploring the board.

1         if A[r][c] != '.':
2             return dfs((r, r+1)[c==8], (c+1,0)[c==8])

In case we already have a digit in the current cell, we move on to the next one.

1         for dig in '123456789':
2             if dig not in rows[r] and dig not in cols[c] and dig not in triples[t]:
3                 board[r][c] = dig
4                 rows[r].add(dig)
5                 cols[c].add(dig)
6                 triples[t].add(dig)

We loop over all numbers from 1 through to 9, to test them out. The test in this case is set membership in the current row, column or section.

1                 if dfs((r, r+1)[c==8], (c+1,0)[c==8]):
2                     return True

If there are no clashes, we simply recurse into the following cell.


 1 from time import time
 2 
 3 from collections import defaultdict as ddict
 4 from itertools import product
 5 
 6 def solveSudoku(board):
 7     def dfs(r, c):
 8         if r == 9:
 9             return True
10         if A[r][c] != '.':
11             return dfs((r, r+1)[c==8], (c+1,0)[c==8])
12         t = (r // 3, c // 3)
13         for dig in '123456789':
14             if dig not in rows[r] and dig not in cols[c] and dig not in triples[t]:
15                 board[r][c] = dig
16                 rows[r].add(dig)
17                 cols[c].add(dig)
18                 triples[t].add(dig)
19                 if dfs((r, r+1)[c==8], (c+1,0)[c==8]):
20                     return True
21                 else:
22                     board[r][c] = "."
23                     rows[r].discard(dig)
24                     cols[c].discard(dig)
25                     triples[t].discard(dig)
26         return False
27     rows, cols, triples = ddict(set), ddict(set), ddict(set)
28     for r, c in product(range(9), repeat=2):
29         if board[r][c] != ".":
30             rows[r].add(board[r][c])
31             cols[c].add(board[r][c])
32             triples[(r // 3, c // 3)].add(board[r][c])
33     dfs(0, 0)
34 
35 def print_sudoku(A):
36     for r in A:
37         print(r)
38     print('')
39 
40 A = [[".",".","3","8",".",".","4",".","."],
41     [".",".",".",".","1",".",".","7","."],
42     [".","6",".",".",".","5",".",".","9"],
43     [".",".",".","9",".",".","6",".","."],
44     [".","2",".",".",".",".",".","1","."],
45     [".",".","4",".",".","3",".",".","2"],
46     [".",".","2",".",".",".","8",".","."],
47     [".","1",".",".",".",".",".","5","."],
48     ["9",".",".",".",".","7",".",".","3"]]
49 
50 
51 
52 s = time()
53 print(solveSudoku(A))
54 print_sudoku(A)
55 print(time() - s)