python 数独求解器

Posted 2021-05-09

# Sudoku Solver - Peter A. Norvig

digits = '123456789'
rows = 'ABCDEFGHI'
cols = digits

def cross(A, B):
    return [a+b for a in A for b in B]

squares = cross(rows, cols)

# unit - collection of nine squares (col, row or box)
# peers - squares that share a unit

unitlist = ([cross(rows, c) for c in cols] + [cross(r, cols) for r in rows] \
+ [cross(rs, cs) for rs in ('ABC', 'DEF', 'GHI') for cs in ('123', '456', '789')])

units = dict((s, [u for u in unitlist if s in u]) for s in squares)

peers = dict((s, set(sum(units[s], []))-set([s])) for s in squares)

# values - a dict with squares as keys, the value of each key will be the possible digits for that squares
# grid - string format of the puzzle

def parse_grid(grid):
    """Convert grid to a dict of possible values, {square: digits}, or
    return False if a contradiction is detected."""
    ## To start, every square can be any digit; then assign values from the grid.
    values = dict((s, digits) for s in squares)

    for s, d in grid_values(grid).items():
        if d in digits and not assign(values, s, d):
            return False
    return values

def grid_values(grid):
    "Convert grid into a dict of {square: char} with '0' or '.' for empties"
    chars = [c for c in grid if c in digits or c in '0.']
    assert len(chars) == 81
    return dict(zip(squares, chars))

# Constraint propagation: use the general strategies for reducing the number of
# possible values for a square
# 1. IF a square has only one possible value, eliminate that value from the square's peers
# 2. If a unit has only one possible place for a value, then put the value there.

def assign(values, s, d):
    """Eliminate all the other values (except d) from values[s] and propagate.
    Return values, except return False if a contradiction is detected."""
    other_values = values[s].replace(d, '')
    if all(eliminate(values, s, d2) for d2 in other_values):
        return values
    else:
        return False

def eliminate(values, s, d):
    """Eliminate d from values[s]; propagate when values or places <= 2.
    Return values, except return False if a contradiction is detected."""

    if d not in values[s]:
        return values # Already eliminated

    values[s] = values[s].replace(d, '')

    ## Strategy 1
    if len(values[s]) == 0:
        return False
    elif len(values[s]) == 1:
        d2 = values[s]
        if not all(eliminate(values, s2, d2) for s2 in peers[s]):
            return False

    ## Strategy 2
    for u in units[s]:
        dplaces = [s for s in u if d in values[s]]
        if len(dplaces) == 0:
            return False
        elif len(dplaces) == 1:
            if not assign(values, dplaces[0], d):
                return False
    return values

def display(values):
    "Display these values as a 2-D grid."
    width = 1+max(len(values[s]) for s in squares)
    line = '+'.join(['-'*(width*3)]*3)
    for r in rows:
        print (''.join(values[r+c].center(width)+('|' if c in '36' else '')
                      for c in cols))
        if r in 'CF': print (line)
    print()

# DFS
def search(values):
    if values is False:
        return False
    if all(len(values[s]) == 1 for s in squares):
        return values
    n, s = min((len(values[s]), s) for s in squares if len(values[s]) > 1)

    return some(search(assign(values.copy(), s, d)) for d in values[s])

def solve(grid):
    return search(parse_grid(grid))

def some(seq):
    for e in seq:
        if e:
            return e
    return False 


grid = '003020600900305001001806400008102900700000008006708200002609500800203009005010300'

display(parse_grid(grid))