ラケットの学習を開始し、任意のアリティ ツリーと生成再帰を使用して、特定のボードから可能なすべてのバージョンのボードを取得しています。
では、このボードがあるとしましょう。 false は空を意味します。
(list "X" "X" "O" "O" "X" "O" "X" #false "X")
要件によると、これに対する解決策は次のようになります。
(list
(list "X" "X" "O" "O" "X" "O" "X" "X" "X")
(list "X" "X" "O" "O" "X" "O" "X" "O" "X"))
ラケットのソリューションはうまく機能します。Python で同じことを試しましたが、期待どおりに動作しません。
次のような出力が得られます。
[['X', 'X', 'O', 'O', 'X', 'O', 'X', 'X', 'X'], [['X', 'X', 'O', 'O', 'X', 'O', 'X', 'O', 'X'], []]]
また
['X', 'X', 'O', 'O', 'X', 'O', 'X', 'X', 'X', 'X', 'X', 'O', 'O', 'X', 'O', 'X', 'O', 'X']
またはさらに悪い。
私が望む出力を得ることができないようです。
他に何も機能しない場合は、出力に対して後処理を行うことを考えていましたが、それは避けたいと思います。
私が必要とするのはこれです：
[['X', 'X', 'O', 'O', 'X', 'O', 'X', 'X', 'X'], ['X', 'X', 'O', 'O', 'X', 'O', 'X', 'O', 'X']]
いずれにせよ、お手伝いできることがあればお知らせください。
ここに私のpythonコードがあります：
"""
Brute force solution for tic-tac-toe.
"""
"""
Data definitions:
;; Value is one of:
;; - false
;; - "X"
;; - "O"
;; interp. a square is either empty (represented by false) or has and "X" or an "O"
;; Board is (listof Value)
;; a board is a list of 9 Values
"""
#
## CONSTANTS
#
B0 = [False for i in range(9)]
B1 = [
False, "X", "O",
"O", "X", "O",
False, False, "X"
]
B2 = [
"X", "X", "O",
"O", "X", "O",
"X", False, "X",
]
B3 = [
"X", "O", "X",
"O", "O", False,
"X", "X", False,
]
"""
PROBLEM 1
In this problem we want you to design a function that produces all
possible filled boards that are reachable from the current board.
In actual tic-tac-toe, O and X alternate playing. For this problem
you can disregard that. You can also assume that the players keep
placing Xs and Os after someone has won. This means that boards that
are completely filled with X, for example, are valid.
"""
def fill_board(index, bd):
"""
Returns a list of 2 board versions with
the index filled with "X" and "O"
:param index: int; index of position in list to be filled
:param bd: Board
:return: (listof Board)
"""
return [
bd[:index] + ["X"] + bd[index+1:],
bd[:index] + ["O"] + bd[index + 1:],
]
assert fill_board(0, B1) == [
[
"X", "X", "O",
"O", "X", "O",
False, False, "X"
],
[
"O", "X", "O",
"O", "X", "O",
False, False, "X"
],
]
assert fill_board(5, B3) == [
[
"X", "O", "X",
"O", "O", "X",
"X", "X", False,
],
[
"X", "O", "X",
"O", "O", "O",
"X", "X", False,
],
]
def find_blank(bd):
"""
Return the index of the
first empty (False) value
in the board.
ASSUME: there is at least one
empty cell.
:param bd: Board
:return: Index
"""
return bd.index(False)
assert find_blank(B0) == 0
assert find_blank(B2) == 7
assert find_blank(B3) == 5
def next_boards(bd):
"""
Produce the next version of initial board.
Finds the first empty (False) cell, and produces
2 versions of the board; one with X and one with O
:param bd: Board
:return: (listof Board)
"""
return fill_board(find_blank(bd), bd)
assert next_boards(B0) == [
["X"] + B0[1:],
["O"] + B0[1:],
]
assert next_boards(B3) == [
[
"X", "O", "X",
"O", "O", "X",
"X", "X", False,
],
[
"X", "O", "X",
"O", "O", "O",
"X", "X", False,
],
]
def solve(bd):
"""
Produce all possible filled boards that are
reachable from the current board.
:param bd: Board (listof Value)
:return: (listof Board)
"""
def is_full(bd):
"""
Returns true if board is full; meaning
if every value on the board is a string.
:param bd: Board (listof Value)
:return: Boolean
"""
return all(type(i) is str for i in bd)
def solve__bd(bd):
"""
Mutually refential function with
solve__lobd. This is where all the actual
computation takes place.
The two functions are responsible for
generating and operating on the tree.
The tree (arbitraty arity tree) represents
another version of the board filled with an
additional X or O.
:param bd: Board (listof Value)
:return: (listof Board)
"""
if is_full(bd):
return list(bd)
return solve__lobd(next_boards(bd))
def solve__lobd(lobd):
"""
Helper function of solve, alongside solve__bd
:param lobd: (listof Board)
:return: (listof Board)
"""
if not lobd:
return []
return [solve__bd(lobd[0]), solve__lobd(lobd[1:])]
return solve__bd(bd)
assert solve(B2) == [
[
"X", "X", "O",
"O", "X", "O",
"X", "X", "X",
],
[
"X", "X", "O",
"O", "X", "O",
"X", "O", "X",
],
]
assert solve(B3) == [
[
"X", "O", "X",
"O", "O", "X",
"X", "X", "X",
],
[
"X", "O", "X",
"O", "O", "X",
"X", "X", "O",
],
[
"X", "O", "X",
"O", "O", "O",
"X", "X", "X",
],
[
"X", "O", "X",
"O", "O", "O",
"X", "X", "O",
],
]