CS计算机代考程序代写 python “””

“””
CSC148, Summer 2021
Assignment 3: Automatic Puzzle Solver
==============================
This code is provided solely for the personal and private use of
students taking the CSC148 course at the University of Toronto.
Copying for purposes other than this use is expressly prohibited.
All forms of distribution of this code, whether as given or with
any changes, are expressly prohibited.

Authors: Diane Horton, Jonathan Calver, Sophia Huynh,
Maryam Majedi, and Jaisie Sin.

All of the files in this directory are:
Copyright (c) 2021 Diane Horton, Jonathan Calver, Sophia Huynh,
Maryam Majedi, and Jaisie Sin.

This module is adapted from the CSC148 Winter 2021 A2 with permission from
the author.

=== Module Description ===

This module contains the sudoku puzzle class.

Note: Some of the provided code makes use of list comprehensions
and can be a bit hard to understand if you aren’t familiar with
them. You do NOT need to understand HOW the provided code works,
but you need to understand WHAT services it is providing, so that you
can make use of the provided private methods to help you with the task of
implementing fail_fast.
“””

from __future__ import annotations
from typing import List, Set
from puzzle import Puzzle

EMPTY_CELL = ‘ ‘

class SudokuPuzzle(Puzzle):
“””
A Sudoku puzzle.

=== Private Attributes ===
_n: the number of rows/columns in this puzzle’s grid
_grid: the grid representing this puzzle; each sublist
represents one row of the grid
_symbol_set: the set of all symbols that each row/column/subsquare must have
exactly one of, for this puzzle to be solved

=== Representation Invariants ===
_n is a positive, square integer >= 4 (e.g. 4, 9, 16)
“””
_n: int
_grid: List[List[str]]
_symbol_set: Set[str]

def __init__(self, n: int, grid: List[List[str]],
symbol_set: Set[str]) -> None:
“””
Create a new n x n SudokuPuzzle with symbols
from and the specified .

Note:
– Grid symbols are represented as letters or numerals
and must be single characters.
– In , an empty square is represented by the constant
EMPTY_CELL
– a copy of grid is NOT made

Preconditions:
– n is a positive, square integer, n >= 4 (e.g. 4, 9, 16)
– there are n symbols in the given symbol_set
– there are n lists in grid, and each list has n symbols, each of which
is either an EMPTY_CELL or in the symbol_set.
“””

self._n, self._grid, self._symbol_set = n, grid, symbol_set

def __eq__(self, other: SudokuPuzzle) -> bool:
“””
Return True if this SudokuPuzzle is equivalent to , False
otherwise.

Two SudokuPuzzles are equivalent if they have equal grids
and equal symbol sets.

>>> r1 = [“A”, “B”, “C”, “D”]
>>> r2 = [“D”, “C”, “B”, “A”]
>>> r3 = [” “, “D”, ” “, ” “]
>>> r4 = [” “, ” “, ” “, ” “]
>>> s1 = SudokuPuzzle(4, [r1, r2, r3, r4], {“A”, “B”, “C”, “D”})
>>> r1_2 = [“A”, “B”, “C”, “D”]
>>> r2_2 = [“D”, “C”, “B”, “A”]
>>> r3_2 = [” “, “D”, ” “, ” “]
>>> r4_2 = [” “, ” “, ” “, ” “]
>>> s2 = SudokuPuzzle(4, [r1_2, r2_2, r3_2, r4_2], {“A”, “B”, “C”, “D”})
>>> s1 == s2
True
>>> r1_3 = [“A”, “B”, “C”, “D”]
>>> r2_3 = [“D”, “C”, “B”, “A”]
>>> r3_3 = [” “, “D”, ” “, ” “]
>>> r4_3 = [” “, “A”, ” “, ” “]
>>> s3 = SudokuPuzzle(4, [r1_3, r2_3, r3_3, r4_3], {“A”, “B”, “C”, “D”})
>>> s1 == s3
False
“””
return (self._grid == other._grid
and self._symbol_set == other._symbol_set)

def __str__(self) -> str:
“””
Return a human-readable string representation of this SudokuPuzzle.

>>> r1 = [“A”, “B”, “C”, “D”]
>>> r2 = [“D”, “C”, “B”, “A”]
>>> r3 = [” “, “D”, ” “, ” “]
>>> r4 = [” “, ” “, ” “, ” “]
>>> s = SudokuPuzzle(4, [r1, r2, r3, r4], {“A”, “B”, “C”, “D”})
>>> print(s)
——-
|AB|CD|
|DC|BA|
——-
| D| |
| | |
——-
>>> s = SudokuPuzzle(9, \
[[” “, ” “, ” “, “9”, ” “, “2”, ” “, ” “, ” “], \
[” “, “9”, “1”, ” “, ” “, ” “, “6”, “3”, ” “], \
[” “, “3”, ” “, ” “, “7”, ” “, ” “, “8”, ” “], \
[“3″, ” “, ” “, ” “, ” “, ” “, ” “, ” “, “8”], \
[” “, ” “, “9”, ” “, ” “, ” “, “2”, ” “, ” “], \
[“5″, ” “, ” “, ” “, ” “, ” “, ” “, ” “, “7”], \
[” “, “7”, ” “, ” “, “8”, ” “, ” “, “4”, ” “], \
[” “, “4”, “5”, ” “, ” “, ” “, “8”, “1”, ” “], \
[” “, ” “, ” “, “3”, ” “, “6”, ” “, ” “, ” “]], \
{“1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”})
>>> print(s)
————-
| |9 2| |
| 91| |63 |
| 3 | 7 | 8 |
————-
|3 | | 8|
| 9| |2 |
|5 | | 7|
————-
| 7 | 8 | 4 |
| 45| |81 |
| |3 6| |
————-
“””
rslt = ”
sqn = round(self._n ** (1 / 2))

div = “-” * (self._n + sqn + 1) + “\n”

for i in range(self._n):
if i % sqn == 0:
rslt += div
row = ‘|’
for j in range(sqn):
row = (row
+ “”.join(self._grid[i][j * sqn: (j + 1) * sqn]) + ‘|’)
rslt += row + ‘\n’
rslt += div
return rslt.rstrip()

def is_solved(self) -> bool:
“””
Return True if this SudokuPuzzle is solved, False otherwise.

>>> r1 = [“A”, “B”, “C”, “D”]
>>> r2 = [“C”, “D”, “A”, “B”]
>>> r3 = [“B”, “A”, “D”, “C”]
>>> r4 = [“D”, “C”, “B”, “A”]
>>> grid = [r1, r2, r3, r4]
>>> s = SudokuPuzzle(4, grid, {“A”, “B”, “C”, “D”})
>>> s.is_solved()
True
>>> r1 = [“A”, “B”, “C”, “D”]
>>> r2 = [“C”, “D”, “A”, “B”]
>>> r3 = [“B”, “D”, “A”, “C”] # “D” and “A” in this row are not valid.
>>> r4 = [“D”, “C”, “B”, “A”]
>>> grid = [r1, r2, r3, r4]
>>> s = SudokuPuzzle(4, grid, {“A”, “B”, “C”, “D”})
>>> s.is_solved()
False
“””
# check if there are any EMPTY_CELLs left
if any(EMPTY_CELL in row for row in self._grid):
return False
# check that all rows and columns have correct symbols
for i in range(self._n):
for j in range(self._n):
# row
if self._row_set(i) != self._symbol_set:
return False
if self._column_set(j) != self._symbol_set:
return False
# check that all subsquares have correct symbols
sqn = round(self._n ** 0.5)
for i in range(0, self._n, sqn):
for j in range(0, self._n, sqn):
if self._subsquare_set(i, j) != self._symbol_set:
return False
return True

def extensions(self) -> List[SudokuPuzzle]:
“””
Return list of extensions of SudokuPuzzle self.

Note: For sudoku puzzles, we define the extensions to be puzzles
where the first empty square (starting from the top left corner
reading left to right and then top to bottom) is filled in
with a symbol not already occuring in that row, column, or
subsquare.

Precondition: Assume the puzzle is correct so far.

>>> r1 = [“A”, “B”, “C”, “D”]
>>> r2 = [“C”, “D”, “A”, “B”]
>>> r3 = [“B”, “A”, “D”, “C”]
>>> r4 = [“D”, “C”, “B”, ” “]
>>> grid = [r1, r2, r3, r4]
>>> s = SudokuPuzzle(4, grid, {“A”, “B”, “C”, “D”})
>>> L1 = list(s.extensions())
>>> grid[-1][-1] = “A”
>>> L2 = [SudokuPuzzle(4, grid, {“A”, “B”, “C”, “D”})]
>>> len(L1) == len(L2) == 1
True
>>> all([s in L2 for s in L1])
True
>>> all([s in L1 for s in L2])
True
“””
# temporary variables to give convenient names to each attribute
symbols, symbol_set, n = self._grid, self._symbol_set, self._n
if not any(EMPTY_CELL in row for row in symbols):
return []
# get position of first empty position
r = 0 # row with first empty position
while EMPTY_CELL not in symbols[r]:
r += 1
c = symbols[r].index(EMPTY_CELL) # column with first empty position

# allowed symbols at position (r, c)
# A | B == A.union(B)
allowed_symbols = (self._symbol_set
– (self._row_set(r)
| self._column_set(c)
| self._subsquare_set(r, c)))

# list of SudokuPuzzles with each legal digit at position r, c
return_lst = []
for symbol in allowed_symbols:
# NOTE: type(self)(…) means create a new SudokuPuzzle,
# we do this here so that if we were to create a subclass of
# SudokuPuzzle later, then this will work as intended
new_puzzle = type(self)(n, symbols[:r]
+ [symbols[r][:c]
+ [symbol]
+ symbols[r][c + 1:]]
+ symbols[r + 1:], symbol_set)
return_lst.append(new_puzzle)
return return_lst

# TODO (Task 1): override fail_fast
# If there is an open position with no symbols available
# (i.e. all symbols are already used in the same row, column, or subsquare),
# then the sudoku puzzle is not solvable.
#
# Hint: You may find the provided private methods below helpful.
# The helpers return sets – see the provided code for extensions
# above for an example of how they can be used.
#
# Note: You can take the union of two sets, set_a and set_b as either
# set_a | set_b or set_a.union(set_b).
# Example:
# {‘1’, ‘2’, ‘3’} | {‘2’, ‘4’, ‘5’} == {‘1’, ‘2’, ‘3’, ‘4’, ‘5’}
def fail_fast(self) -> bool:
“””
Return True if some unfilled position has no allowable symbols
remaining to choose from, and hence this SudokuPuzzle can never
be completed, and False otherwise.

You can assume that this puzzle state is valid.

>>> s = SudokuPuzzle(4, \
[ \
[“A”, “B”, “C”, “D”], \
[“C”, “D”, ” “, ” “], \
[” “, ” “, ” “, ” “], \
[” “, ” “, ” “, ” “] \
], {“A”, “B”, “C”, “D”})
>>> s.fail_fast()
False
>>> s = SudokuPuzzle(4, \
[\
[“B”, “D”, “A”, “C”], \
[“C”, “A”, “B”, “D”], \
[“A”, “B”, ” “, ” “], \
[” “, ” “, ” “, ” “]\
], {“A”, “B”, “C”, “D”})
>>> s.fail_fast()
True
“””

# some private helper methods
# Note: these return sets of symbols you may find useful
def _row_set(self, r: int) -> Set[str]:
“””Return set of symbols in row r of SudokuPuzzle self’s grid.
“””
# set of elements from grid[r]
return set(self._grid[r])

def _column_set(self, c: int) -> Set[str]:
“””Return set of symbols in column c of SudokuPuzzle self’s grid.
“””

# set of elements from grid[0][c], grid[1][c],
# … grid[len(grid)-1][c]
return set(row[c] for row in self._grid)

def _subsquare_set(self, r: int, c: int) -> Set[str]:
“””Return set of symbols in subsquare of SudokuPuzzle self’s grid
where position r, c occurs.
“””

# length of subsquares
ss = round(self._n ** (1 / 2))
# upper-left position of the subsquare containing r, c
ul_row = (r // ss) * ss
ul_col = (c // ss) * ss

subsquare_symbols = []
for i in range(ss):
for j in range(ss):
subsquare_symbols.append(self._grid[ul_row + i][ul_col + j])
return set(subsquare_symbols)

# TODO (Task 2): implement has_unique_solution
# Implement this method according to its docstring
# You may import any modules that you need when implementing this method.
def has_unique_solution(self) -> bool:
“””
Return True if the this Sudoku puzzle has exactly one unique solution,
and False otherwise.

Two “solutions” are considered to be equal if the final puzzle
state is the same.

Hint: You should find the optional parameter, seen, for the Solver
class’ solve method very useful here.

>>> s = SudokuPuzzle(4, [
… [“D”, “C”, “B”, “A”],
… [“B”, “A”, “D”, “C”],
… [“C”, ” “, “A”, ” “],
… [“A”, ” “, “C”, ” “]], {“A”, “B”, “C”, “D”}
… )
>>> s.has_unique_solution()
False
“””

if __name__ == “__main__”:
# any code you want to use for testing your code above
import python_ta

python_ta.check_all(config={
# # uncomment to disable openning pyta output in browser
# ‘pyta-reporter’: ‘ColorReporter’,
‘allowed-io’: [],
‘allowed-import-modules’: [
‘doctest’, ‘python_ta’, ‘typing’, ‘puzzle’, ‘__future__’,
‘a3_solver’
],
‘disable’: [‘E1136’],
‘max-attributes’: 15
})