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@rayheberer
Created December 20, 2017 21:53
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from utils import *
import ast
row_units = [cross(r, cols) for r in rows]
column_units = [cross(rows, c) for c in cols]
square_units = [cross(rs, cs) for rs in ('ABC','DEF','GHI') for cs in ('123','456','789')]
# TODO: Update the unit list to add the new diagonal units
left_diag_units = [[r+c for (r, c) in zip(rows, cols)]]
right_diag_units = [[r+c for (r, c) in zip(rows, cols[::-1])]]
unitlist = row_units + column_units + square_units + left_diag_units + right_diag_units
units = dict((s, [u for u in unitlist if s in u]) for s in boxes)
peers = dict((s, set(sum(units[s],[]))-set([s])) for s in boxes)
def naked_twins(values):
"""Eliminate values using the naked twins strategy.
Parameters
----------
values(dict)
a dictionary of the form {'box_name': '123456789', ...}
Returns
-------
dict
The values dictionary with the naked twins eliminated from peers
Notes
-----
Your solution can either process all pairs of naked twins from the input once,
or it can continue processing pairs of naked twins until there are no such
pairs remaining -- the project assistant test suite will accept either
convention. However, it will not accept code that does not process all pairs
of naked twins from the original input. (For example, if you start processing
pairs of twins and eliminate another pair of twins before the second pair
is processed then your code will fail the PA test suite.)
The first convention is preferred for consistency with the other strategies,
and because it is simpler (since the reduce_puzzle function already calls this
strategy repeatedly).
"""
twins = {}
for unit_id, unit in enumerate(unitlist):
twins[unit_id] = []
candidate_twins = [box for box in unit if len(values[box]) == 2]
for i in range(len(candidate_twins)-1):
for j in range(i+1, len(candidate_twins)):
if sorted(values[candidate_twins[i]]) == sorted(values[candidate_twins[j]]):
twins[unit_id].append(([candidate_twins[i], candidate_twins[j]], values[candidate_twins[i]]))
for unit in twins.keys():
for twin in twins[unit]:
digit1 = twin[1][0]
digit2 = twin[1][1]
for box in unitlist[unit]:
if box not in twin[0]:
values[box] = values[box].replace(digit1, '')
values[box] = values[box].replace(digit2, '')
return values
def eliminate(values):
"""Apply the eliminate strategy to a Sudoku puzzle
The eliminate strategy says that if a box has a value assigned, then none
of the peers of that box can have the same value.
Parameters
----------
values(dict)
a dictionary of the form {'box_name': '123456789', ...}
Returns
-------
dict
The values dictionary with the assigned values eliminated from peers
"""
solved_values = [box for box in values.keys() if len(values[box]) == 1]
for box in solved_values:
digit = values[box]
for peer in peers[box]:
values[peer] = values[peer].replace(digit, '')
return values
def only_choice(values):
"""Apply the only choice strategy to a Sudoku puzzle
The only choice strategy says that if only one box in a unit allows a certain
digit, then that box must be assigned that digit.
Parameters
----------
values(dict)
a dictionary of the form {'box_name': '123456789', ...}
Returns
-------
dict
The values dictionary with all single-valued boxes assigned
Notes
-----
You should be able to complete this function by copying your code from the classroom
"""
for unit in unitlist:
for digit in '123456789':
dplaces = [box for box in unit if digit in values[box]]
if len(dplaces) == 1:
values[dplaces[0]] = digit
return values
def reduce_puzzle(values):
"""Reduce a Sudoku puzzle by repeatedly applying all constraint strategies
Parameters
----------
values(dict)
a dictionary of the form {'box_name': '123456789', ...}
Returns
-------
dict or False
The values dictionary after continued application of the constraint strategies
no longer produces any changes, or False if the puzzle is unsolvable
"""
solved_values = [box for box in values.keys() if len(values[box]) == 1]
stalled = False
while not stalled:
solved_values_before = len([box for box in values.keys() if len(values[box]) == 1])
values = naked_twins(values)
values = eliminate(values)
values = only_choice(values)
solved_values_after = len([box for box in values.keys() if len(values[box]) == 1])
stalled = solved_values_before == solved_values_after
if len([box for box in values.keys() if len(values[box]) == 0]):
return False
return values
def search(values):
"""Apply depth first search to solve Sudoku puzzles in order to solve puzzles
that cannot be solved by repeated reduction alone.
Parameters
----------
values(dict)
a dictionary of the form {'box_name': '123456789', ...}
Returns
-------
dict or False
The values dictionary with all boxes assigned or False
Notes
-----
You should be able to complete this function by copying your code from the classroom
and extending it to call the naked twins strategy.
"""
values = reduce_puzzle(values)
if values is False:
return False
if all(len(values[s]) == 1 for s in boxes):
return values
n, s = min((len(values[s]), s) for s in boxes if len(values[s]) > 1)
for value in values[s]:
new_sudoku = values.copy()
new_sudoku[s] = value
attempt = search(new_sudoku)
if attempt:
return attempt
def solve(grid):
"""Find the solution to a Sudoku puzzle using search and constraint propagation
Parameters
----------
grid(string)
a string representing a sudoku grid.
Ex. '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'
Returns
-------
dict or False
The dictionary representation of the final sudoku grid or False if no solution exists.
"""
values = grid2values(grid)
values = search(values)
return values
if __name__ == "__main__":
diag_sudoku_grid = '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'
display(grid2values(diag_sudoku_grid))
result = solve(diag_sudoku_grid)
display(result)
try:
import PySudoku
PySudoku.play(grid2values(diag_sudoku_grid), result, history)
except SystemExit:
pass
except:
print('We could not visualize your board due to a pygame issue. Not a problem! It is not a requirement.')
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