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Add arrays/sudoku_solver.py (#10623)
* Create Sudoku_Solver Each of the digits 1-9 must occur exactly once in each row. Each of the digits 1-9 must occur exactly once in each column. Each of the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid. The '.' character indicates empty cells. * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Rename Sudoku_Solver to sudoku_solver.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Update sudoku_solver.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Christian Clauss <cclauss@me.com>
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Anubhavpandey27
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data_structures/arrays/sudoku_solver.py
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data_structures/arrays/sudoku_solver.py
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"""
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Please do not modify this file! It is published at https://norvig.com/sudoku.html with
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only minimal changes to work with modern versions of Python. If you have improvements,
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please make them in a separate file.
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"""
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import random
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import time
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def cross(items_a, items_b):
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"Cross product of elements in A and elements in B."
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return [a + b for a in items_a for b in items_b]
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digits = "123456789"
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rows = "ABCDEFGHI"
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cols = digits
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squares = cross(rows, cols)
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unitlist = (
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[cross(rows, c) for c in cols]
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+ [cross(r, cols) for r in rows]
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+ [cross(rs, cs) for rs in ("ABC", "DEF", "GHI") for cs in ("123", "456", "789")]
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)
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units = {s: [u for u in unitlist if s in u] for s in squares}
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peers = {s: set(sum(units[s], [])) - {s} for s in squares}
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def test():
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"A set of unit tests."
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assert len(squares) == 81
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assert len(unitlist) == 27
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assert all(len(units[s]) == 3 for s in squares)
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assert all(len(peers[s]) == 20 for s in squares)
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assert units["C2"] == [
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["A2", "B2", "C2", "D2", "E2", "F2", "G2", "H2", "I2"],
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["C1", "C2", "C3", "C4", "C5", "C6", "C7", "C8", "C9"],
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["A1", "A2", "A3", "B1", "B2", "B3", "C1", "C2", "C3"],
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]
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# fmt: off
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assert peers["C2"] == {
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"A2", "B2", "D2", "E2", "F2", "G2", "H2", "I2", "C1", "C3",
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"C4", "C5", "C6", "C7", "C8", "C9", "A1", "A3", "B1", "B3"
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}
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# fmt: on
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print("All tests pass.")
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def parse_grid(grid):
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"""Convert grid to a dict of possible values, {square: digits}, or
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return False if a contradiction is detected."""
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## To start, every square can be any digit; then assign values from the grid.
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values = {s: digits for s in squares}
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for s, d in grid_values(grid).items():
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if d in digits and not assign(values, s, d):
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return False ## (Fail if we can't assign d to square s.)
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return values
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def grid_values(grid):
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"Convert grid into a dict of {square: char} with '0' or '.' for empties."
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chars = [c for c in grid if c in digits or c in "0."]
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assert len(chars) == 81
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return dict(zip(squares, chars))
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def assign(values, s, d):
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"""Eliminate all the other values (except d) from values[s] and propagate.
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Return values, except return False if a contradiction is detected."""
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other_values = values[s].replace(d, "")
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if all(eliminate(values, s, d2) for d2 in other_values):
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return values
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else:
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return False
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def eliminate(values, s, d):
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"""Eliminate d from values[s]; propagate when values or places <= 2.
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Return values, except return False if a contradiction is detected."""
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if d not in values[s]:
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return values ## Already eliminated
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values[s] = values[s].replace(d, "")
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## (1) If a square s is reduced to one value d2, then eliminate d2 from the peers.
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if len(values[s]) == 0:
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return False ## Contradiction: removed last value
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elif len(values[s]) == 1:
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d2 = values[s]
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if not all(eliminate(values, s2, d2) for s2 in peers[s]):
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return False
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## (2) If a unit u is reduced to only one place for a value d, then put it there.
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for u in units[s]:
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dplaces = [s for s in u if d in values[s]]
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if len(dplaces) == 0:
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return False ## Contradiction: no place for this value
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elif len(dplaces) == 1:
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# d can only be in one place in unit; assign it there
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if not assign(values, dplaces[0], d):
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return False
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return values
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def display(values):
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"Display these values as a 2-D grid."
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width = 1 + max(len(values[s]) for s in squares)
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line = "+".join(["-" * (width * 3)] * 3)
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for r in rows:
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print(
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"".join(
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values[r + c].center(width) + ("|" if c in "36" else "") for c in cols
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)
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)
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if r in "CF":
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print(line)
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print()
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def solve(grid):
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return search(parse_grid(grid))
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def some(seq):
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"Return some element of seq that is true."
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for e in seq:
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if e:
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return e
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return False
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def search(values):
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"Using depth-first search and propagation, try all possible values."
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if values is False:
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return False ## Failed earlier
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if all(len(values[s]) == 1 for s in squares):
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return values ## Solved!
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## Chose the unfilled square s with the fewest possibilities
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n, s = min((len(values[s]), s) for s in squares if len(values[s]) > 1)
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return some(search(assign(values.copy(), s, d)) for d in values[s])
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def solve_all(grids, name="", showif=0.0):
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"""Attempt to solve a sequence of grids. Report results.
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When showif is a number of seconds, display puzzles that take longer.
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When showif is None, don't display any puzzles."""
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def time_solve(grid):
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start = time.monotonic()
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values = solve(grid)
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t = time.monotonic() - start
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## Display puzzles that take long enough
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if showif is not None and t > showif:
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display(grid_values(grid))
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if values:
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display(values)
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print("(%.5f seconds)\n" % t)
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return (t, solved(values))
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times, results = zip(*[time_solve(grid) for grid in grids])
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if (n := len(grids)) > 1:
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print(
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"Solved %d of %d %s puzzles (avg %.2f secs (%d Hz), max %.2f secs)."
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% (sum(results), n, name, sum(times) / n, n / sum(times), max(times))
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)
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def solved(values):
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"A puzzle is solved if each unit is a permutation of the digits 1 to 9."
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def unitsolved(unit):
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return {values[s] for s in unit} == set(digits)
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return values is not False and all(unitsolved(unit) for unit in unitlist)
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def from_file(filename, sep="\n"):
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"Parse a file into a list of strings, separated by sep."
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return open(filename).read().strip().split(sep) # noqa: SIM115
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def random_puzzle(assignments=17):
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"""Make a random puzzle with N or more assignments. Restart on contradictions.
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Note the resulting puzzle is not guaranteed to be solvable, but empirically
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about 99.8% of them are solvable. Some have multiple solutions."""
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values = {s: digits for s in squares}
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for s in shuffled(squares):
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if not assign(values, s, random.choice(values[s])):
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break
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ds = [values[s] for s in squares if len(values[s]) == 1]
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if len(ds) >= assignments and len(set(ds)) >= 8:
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return "".join(values[s] if len(values[s]) == 1 else "." for s in squares)
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return random_puzzle(assignments) ## Give up and make a new puzzle
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def shuffled(seq):
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"Return a randomly shuffled copy of the input sequence."
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seq = list(seq)
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random.shuffle(seq)
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return seq
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grid1 = (
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"003020600900305001001806400008102900700000008006708200002609500800203009005010300"
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)
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grid2 = (
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"4.....8.5.3..........7......2.....6.....8.4......1.......6.3.7.5..2.....1.4......"
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)
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hard1 = (
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".....6....59.....82....8....45........3........6..3.54...325..6.................."
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)
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if __name__ == "__main__":
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test()
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# solve_all(from_file("easy50.txt", '========'), "easy", None)
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# solve_all(from_file("top95.txt"), "hard", None)
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# solve_all(from_file("hardest.txt"), "hardest", None)
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solve_all([random_puzzle() for _ in range(99)], "random", 100.0)
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for puzzle in (grid1, grid2): # , hard1): # Takes 22 sec to solve on my M1 Mac.
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display(parse_grid(puzzle))
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start = time.monotonic()
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solve(puzzle)
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t = time.monotonic() - start
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print("Solved: %.5f sec" % t)
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