本文整理汇总了Python中Board.swap_row方法的典型用法代码示例。如果您正苦于以下问题：Python Board.swap_row方法的具体用法？Python Board.swap_row怎么用？Python Board.swap_row使用的例子？那么恭喜您, 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类Board的用法示例。
在下文中一共展示了Board.swap_row方法的1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__
# 需要导入模块: import Board [as 别名]
# 或者: from Board import swap_row [as 别名]
class Generator:
# # constructor for generator, reads in a space delimited
# def __init__(self, starting_file):
# # opening file
# f = open(starting_file)
# # reducing file to a list of numbers
# numbers = filter(lambda x: x in '123456789',list(reduce(lambda x,y:x+y,f.readlines())))
# print (numbers)
# numbers = map(int,numbers)
# ['1', '2', '3', '4', '5', '6', '7', '8', '9', '4', '5', '6', '7', '8', '9', '1', '2', '3', '7', '8', '9', '1', '2', '3', '4', '5', '6', '2', '1', '4', '3', '6', '5', '8', '9', '7', '3', '6', '5', '8', '9', '7', '2', '1', '4', '8', '9', '7', '2', '1', '4', '3', '6', '5', '5', '3', '1', '6', '4', '2', '9', '7', '8', '6', '4', '2', '9', '7', '8', '5', '3', '1', '9', '7', '8', '5', '3', '1', '6', '4', '2']
# # closing file
# f.close()
# # constructing board
# self.board = Board(numbers)
# constructor for generator, reads in a space delimited
def __init__(self):
numbers = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '4', '5', '6', '7', '8', '9', '1', '2', '3', '7', '8', '9', '1', '2', '3', '4', '5', '6', '2', '1', '4', '3', '6', '5', '8', '9', '7', '3', '6', '5', '8', '9', '7', '2', '1', '4', '8', '9', '7', '2', '1', '4', '3', '6', '5', '5', '3', '1', '6', '4', '2', '9', '7', '8', '6', '4', '2', '9', '7', '8', '5', '3', '1', '9', '7', '8', '5', '3', '1', '6', '4', '2']
numbers = map(int,numbers)
self.board = Board(numbers)
# function randomizes an existing complete puzzle
def randomize(self, iterations):
# not allowing transformations on a partial puzzle
if len(self.board.get_used_cells())==81:
# looping through iterations
for x in range(0, iterations):
# to get a random column/row
case = random.randint(0, 3)
# to get a random band/stack
block = random.randint(0, 2) * 3
# in order to select which row and column we shuffle an array of
# indices and take both elements
options = range(0,3)
random.shuffle(options)
piece1, piece2 = options[0],options[1]
# pick case according to random to do transformation
if case == 0:
self.board.swap_row(block + piece1, block + piece2)
elif case == 1:
self.board.swap_column(block + piece1, block + piece2)
elif case == 2:
self.board.swap_stack(piece1, piece2)
elif case == 3:
self.board.swap_band(piece1, piece2)
else:
raise Exception('Rearranging partial board may compromise uniqueness.')
# method gets all possible values for a particular cell, if there is only one
# then we can remove that cell
def reduce_via_logical(self, cutoff = 81):
cells = self.board.get_used_cells()
random.shuffle(cells)
for cell in cells:
if len(self.board.get_possibles(cell)) == 1:
cell.value = 0
cutoff -= 1
if cutoff == 0:
break
# method attempts to remove a cell and checks that solution is still unique
def reduce_via_random(self, cutoff=81):
temp = self.board
existing = temp.get_used_cells()
# sorting used cells by density heuristic, highest to lowest
new_set = [(x,self.board.get_density(x)) for x in existing]
elements= [x[0] for x in sorted(new_set, key=lambda x: x[1], reverse=True)]
# for each cell in sorted list
for cell in elements:
original = cell.value
# get list of other values to try in its place
complement = [x for x in range(1,10) if x != original]
ambiguous = False
# check each value in list of other possibilities to try
for x in complement:
# set cell to value
cell.value = x
# create instance of solver
s = Solver(temp)
# if solver can fill every box and the solution is valid then
# puzzle becomes ambiguous after removing particular cell, so we can break out
if s.solve() and s.is_valid():
#.........这里部分代码省略.........