# sudoku solver

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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:

• cells need to be filled with numbers ranging from 0 to 9.
• each row must contain unique numbers
• each column must contain unique numbers
• each 3x3 square section must contain unique numbers

## 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:

• a DFS
• set membership checking
• backtracking.

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:

• guess a number
• perform constraints checking
• if the guess passes all constraints, then we recurse into the next cell, else we recurse up the stack.

### 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] != ".":
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


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
19                 if dfs((r, r+1)[c==8], (c+1,0)[c==8]):
20                     return True
21                 else:
22                     board[r][c] = "."
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] != ".":
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)