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Python Alphabet.bit_pattern方法代码示例

本文整理汇总了Python中languages.Alphabet.bit_pattern方法的典型用法代码示例。如果您正苦于以下问题:Python Alphabet.bit_pattern方法的具体用法?Python Alphabet.bit_pattern怎么用?Python Alphabet.bit_pattern使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在languages.Alphabet的用法示例。


在下文中一共展示了Alphabet.bit_pattern方法的2个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: init_crosschecks

# 需要导入模块: from languages import Alphabet [as 别名]
# 或者: from languages.Alphabet import bit_pattern [as 别名]
    def init_crosschecks(self):
        """ Calculate and return a list of cross-check bit patterns for the indicated axis """

        # The cross-check set is the set of letters that can appear in a square
        # and make cross words (above/left and/or below/right of the square) valid
        board = self._autoplayer.board()
        # Prepare to visit all squares on the axis
        x, y = self.coordinate_of(0)
        xd, yd = self.coordinate_step()
        # Fetch the default cross-check bits, which depend on the rack.
        # If the rack contains a wildcard (blank tile), the default cc set
        # contains all letters in the Alphabet. Otherwise, it contains the
        # letters in the rack.
        all_cc = self._autoplayer.rack_bit_pattern()
        # Go through the open squares and calculate their cross-checks
        for ix in range(Board.SIZE):
            cc = all_cc # Start with the default cross-check set
            if not board.is_covered(x, y):
                if self.is_horizontal():
                    above = board.letters_above(x, y)
                    below = board.letters_below(x, y)
                else:
                    above = board.letters_left(x, y)
                    below = board.letters_right(x, y)
                query = u'' if not above else above
                query += u'?'
                if below:
                    query += below
                if len(query) > 1:
                    # Nontrivial cross-check: Query the word database for words that fit this pattern
                    matches = Wordbase.dawg().find_matches(query, sort = False) # Don't need a sorted result
                    bits = 0
                    if matches:
                        cix = 0 if not above else len(above)
                        # Note the set of allowed letters here
                        bits = Alphabet.bit_pattern([wrd[cix] for wrd in matches])
                    # Reduce the cross-check set by intersecting it with the allowed set.
                    # If the cross-check set and the rack have nothing in common, this
                    # will lead to the square being marked as closed, which saves
                    # calculation later on
                    cc &= bits
            # Initialize the square
            self._sq[ix].init(self._autoplayer, x, y, cc)
            # Keep track of empty squares within the axis in a bit pattern for speed
            if self._sq[ix].is_empty():
                self._empty_bits |= (1 << ix)
            x += xd
            y += yd
开发者ID:magnussig,项目名称:Netskrafl,代码行数:50,代码来源:skraflplayer.py

示例2: __init__

# 需要导入模块: from languages import Alphabet [as 别名]
# 或者: from languages.Alphabet import bit_pattern [as 别名]
    def __init__(self, state):

        # List of valid, candidate moves
        self._candidates = []
        self._state = state
        self._board = state.board()
        # The rack that the autoplayer has to work with
        self._rack = state.player_rack().contents()

        # Calculate a bit pattern representation of the rack
        if u'?' in self._rack:
            # Wildcard in rack: all letters allowed
            self._rack_bit_pattern = Alphabet.all_bits_set()
        else:
            # No wildcard: limits the possibilities of covering squares
            self._rack_bit_pattern = Alphabet.bit_pattern(self._rack)
开发者ID:borgar,项目名称:Skrafl,代码行数:18,代码来源:skraflplayer.py


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