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assignments = []
def assign_value(values, box, value):
Please use this function to update your values dictionary!
Assigns a value to a given box. If it updates the board record it.
# Don't waste memory appending actions that don't actually change any values
if values[box] == value:
return values
values[box] = value
if len(value) == 1:
return values
def cross(A, B):
"Cross product of elements in A and elements in B."
return [s+t for s in A for t in B]
def grid_values(grid):
Convert grid into a dict of {square: char} with '123456789' for empties.
grid(string) - A grid in string form.
A grid in dictionary form
Keys: The boxes, e.g., 'A1'
Values: The value in each box, e.g., '8'. If the box has no value, then the value will be '123456789'.
assert len(grid) == 81, "Input grid must be a string of length 81 (9x9)"
return { boxes[i] : elem if elem !='.' else '123456789' for i, elem in enumerate(grid) }
def display(values):
Display the values as a 2-D grid.
values(dict): The sudoku in dictionary form
def naked_twins(values):
"""Eliminate values using the naked twins strategy.
values(dict): a dictionary of the form {'box_name': '123456789', ...}
the values dictionary with the naked twins eliminated from peers.
# Find all instances of naked twins
for unit in unitlist:
## create a new dictionary where key is options for a box in the unit
## and the corresponding value as the an array of boxes in the unit that hold the key
new_dict = {}
for elems in unit:
new_dict.setdefault(values[elems], []).append(elems)
for v in new_dict.keys():
## find two boxes that contain only two options
if len(new_dict[v]) == 2 and len(v) == 2:
for key in unit:
## skip any box with assigned value
if len(values[key])==1:
## for other boxes, remove the twin values from their options
if not (key == new_dict[v][0] or key == new_dict[v][1]):
val = values[key].replace(v[0], "")
val = val.replace(v[1], "")
assign_value(values, key, val)
# Eliminate the naked twins as possibilities for their peers
return values
def peers(value):
"""Returns the peers of a particular box.
value(string): A box value
array of box values that are peers of the given value
row_peers = row_units[rows.index(value[0])]
col_peers = column_units[cols.index(value[1])]
sqr_peers = []
for rs in square_units:
if value in rs:
sqr_peers += rs
## add all normal peers
ans = row_peers + col_peers + sqr_peers
## add peers for diagonal sudoku
## if central box, both diagonals are peers
if value == "E5":
return ans + left_diag_peers + right_diag_peers
## value belongs to left diagonal
elif value in left_diag_peers:
return ans + left_diag_peers
## value belongs to right diagonal
elif value in right_diag_peers:
return ans + right_diag_peers
return ans
def eliminate(values):
"""Eliminate values from peers that are assigned to a box
values(dict): a dictionary of the form {'box_name': '123456789', ...}
the values dictionary with the assigned values eliminated from peers.
for key in values:
if len(values[key])==1:
for peer in peers(key):
## check that peer is not key, since peers array contains the original value also
if not peer == key:
assign_value(values, peer, values[peer].replace(values[key], ''))
return values
def only_choice(values):
"""Assign a value to the box if only that box contains that option amongst all the peers in that unit
values(dict): a dictionary of the form {'box_name': '123456789', ...}
the values dictionary with the only choice strategy implemented.
for unit in unitlist:
all_values = ''.join([ values[key] for key in unit ]) ## join all the options in all the boxes of the unit
singles = []
for k in '123456789': ## count instances of 1-9 in the all_Values string
if all_values.count(k) == 1:
## find the box that contains the singleton value and assign it to that box
for key in unit:
if( k in values[key] ):
assign_value(values, key, k)
return values
def reduce_puzzle(values):
"""Reduce the puzzle using different strategies
values(dict): a dictionary of the form {'box_name': '123456789', ...}
the values dictionary with the reduced options
stalled = False
while not stalled:
# Check how many boxes have a determined value
solved_values_before = len([box for box in values.keys() if len(values[box]) == 1])
# Your code here: Use the Eliminate Strategy
# Your code here: Use the Only Choice Strategy
# add the naked-twins
# Check how many boxes have a determined value, to compare
solved_values_after = len([box for box in values.keys() if len(values[box]) == 1])
# If no new values were added, stop the loop.
stalled = solved_values_before == solved_values_after
# Sanity check, return False if there is a box with zero available values:
if len([box for box in values.keys() if len(values[box]) == 0]):
return False
return values
def search(values):
"Using depth-first search and propagation, try all possible values."
# First, reduce the puzzle using the previous function
values = reduce_puzzle(values)
if values is False:
return False ## Failed earlier
if all(len(values[s]) == 1 for s in boxes):
return values ## Solved!
# Choose one of the unfilled squares with the fewest possibilities
n,s = min((len(values[s]), s) for s in boxes if len(values[s]) > 1)
# Now use recurrence to solve each one of the resulting sudokus, and
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 grid.
grid(string): a string representing a sudoku grid.
Example: '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'
The dictionary representation of the final sudoku grid. False if no solution exists.
values = grid_values(grid)
for key in values:
assign_value(values, key, values[key])
return search(values)
if __name__ == '__main__':
#diag_sudoku_grid = '2.............62....1....7...6..8...3...9...7...6..4...4....8....52.............3'
diag_sudoku_grid ='9.1....'
from visualize import visualize_assignments
except SystemExit:
print('We could not visualize your board due to a pygame issue. Not a problem! It is not a requirement.')
cols = '123456789'
rows = 'ABCDEFGHI'
boxes = cross('ABCDEFGHI', '123456789')
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')]
left_diag_peers = [ rows[i] + cols[i] for i in range(0, 9) ]
right_diag_peers = [ rows[8-i] + cols[i] for i in range(0, 9) ]
unitlist = row_units + column_units + square_units + [left_diag_peers] + [right_diag_peers]
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