我有一个棋盘数组，看起来像这样：
00 01 02 03 04 05 06 07
08 09 10 11 12 13 14 15
16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31
32 33 34 35 36 37 38 39
40 41 42 43 44 45 46 47
48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63
现在我试图找到一个函数 position ，板上的一个数字，它给了我向上/右对角线倒数第二个数字，例如对于 60，我试图找到 38，对于 30，我试图找到 22。我可以硬编码该死的东西，但找到一个这样做的函数真的会更好。到目前为止，它难倒了我。
我没有遇到向下/向右和对角线以及向上/向左对角线的问题，但这个让我难倒了。任何帮助，不胜感激。
下面是我正在处理的代码：
# EightQueens.py
# Joshua Marshall Moore
# thwee.abacadabra.alchemist@gmail.com
# June 24th, 2017
# The eight queens problem consists of setting up eight queens on a chess board
# so that no two queens threaten each other.
# This is an attempt to find all possible solutions to the problem.
# The board is represented as a set of 64 numbers each representing a position
# on the board. Array indexing begins with zero.
# 00 01 02 03 04 05 06 07
# 08 09 10 11 12 13 14 15
# 16 17 18 19 20 21 22 23
# 24 25 26 27 28 29 30 31
# 32 33 34 35 36 37 38 39
# 40 41 42 43 44 45 46 47
# 48 49 50 51 52 53 54 55
# 56 57 58 59 60 61 62 63
# Test:
# The following combination should yield a True from the check function.
# 6, 9, 21, 26, 32, 43, 55, 60
from itertools import combinations
import pdb
# works
def down_right_boundary(pos):
boundary = (8-pos%8)*8
applies = (pos%8)+1 > int(pos/8)
if applies:
return boundary
else:
return 64
# works
def up_left_boundary(pos):
boundary = ((int(pos/8)-(pos%8))*8)-1
applies = (pos%8) <= int(pos/8)
if applies:
return boundary
else:
return -1
def up_right_boundary(pos):
boundary = [7, 15, 23, 31, 39, 47, 55, 62][pos%8]-1
# 7: nil
# 14: 7-1
# 15: 15-1
# 21: 7-1
# 22: 15-1
# 23: 23-1
# 28: 7-1
# 29: 15-1
# 30: 23-1
# 31: 31-1
# 35: 7-1
# 36: 15-1
# 37: 23-1
# 38: 31-1
# 39: 39-1
# 42: 7-1
# 43: 15-1
# 44: 23-1
# 45: 31-1
# 46: 39-1
applies = pos%8>=pos%7
if applies:
return boundary
else:
return -1
def down_left_boundary(pos):
boundary = 64
applies = True
if applies:
return boundary
else:
return 64
def check(positions):
fields = set(range(64))
threatened = set()
# two queens per quadrant
quadrants = [
set([p for p in range(0, 28) if (p%8)<4]),
set([p for p in range(4, 32) if (p%8)>3]),
set([p for p in range(32, 59) if (p%8)<4]),
set([p for p in range(36, 64) if (p%8)>3])
]
#for q in quadrants:
# if len(positions.intersection(q)) != 2:
# return False
# check the queen's ranges
for pos in positions:
pdb.set_trace()
# threatened |= set(range(pos, -1, -8)[1:]) # up
# threatened |= set(range(pos, 64, 8)[1:]) # down
# threatened |= set(range(pos, int(pos/8)*8-1, -1)[1:]) # left
# threatened |= set(range(pos, (int(pos/8)+1)*8, 1)[1:]) # right
# down right diagonal:
# There are two conditions here, one, the position is above the
# diagonal, two, the position is below the diagonal.
# Above the diagonal can be expressed as pos%8>int(pos/8).
# In the event of a position above the diagonal, I need to limit the
# range to 64-(pos%8) to prevent warping the board into a field that
# connects diagonals like Risk.
# Otherwise, 64 suffices as the ending condition.
threatened |= set(range(pos, down_right_boundary(pos), 9)[1:]) # down right
print(pos, threatened)
pdb.set_trace()
#
# up left diagonal:
# Similarly, if the position is above the diagonal, -1 will suffice as
# the range's ending condition. Things are more complicated if the
# position is below the diagonal, as I must prevent warping, again.
threatened |= set(range(pos, up_left_boundary(pos), -9)[1:]) # up left
print(pos, threatened)
pdb.set_trace()
#
# up right diagonal:
# Above the diagonal takes on a different meaning here, seeing how I'm
# dealing with the other diagonal. It is defined by pos58>pos%7. Now I
# restrict the range to a (pos%8)*8, creating a boundary along the right
# side of the board.
threatened |= set(range(pos, up_right_boundary(pos), -7)[1:]) # up right
print(pos, threatened)
pdb.set_trace()
#
# down left diagonal:
# I reuse a similar definition to that of the diagonal as above. The
# bound for left hand side of the board looks as follows:
# ((pos%8)*7)+(pos%8)
threatened |= set(range(pos, down_left_boundary(pos), 7)[1:]) # down left
print(pos, threatened)
pdb.set_trace()
#
if len(positions.intersection(threatened)) > 0:
return False
return True
if __name__ == '__main__':
# print(check(set([55]))) # pass
# print(check(set([62]))) # pass
# print(check(set([63]))) # pass
# print(check(set([48]))) # pass
# print(check(set([57]))) # pass
# print(check(set([56]))) # pass
# print(check(set([8]))) # pass
# print(check(set([1]))) # fail
# print(check(set([0])))
# print(check(set([6])))
# print(check(set([15])))
# print(check(set([7])))
# print(check(set([6])))
# print(check(set([9])))
print(check(set([21])))
print(check(set([26])))
print(check(set([32])))
print(check(set([43])))
print(check(set([55])))
print(check(set([60])))
print(
check(set([6, 9, 21, 26, 32, 43, 55, 60]))
)
# for potential_solution in combinations(range(64), 8):
# is_solution = check(set(potential_solution))
# if is_solution:
# print(is_solution, potential_solution)
使用以下棋盘位置：
chessboard = [0,1,2,3,4,5,6,7,
8,9,10,11,12,13,14,15,
16,17,18,19,20,21,22,23,
24,25,26,27,28,29,30,31,
32,33,34,35,36,37,38,39,
40,41,42,43,44,45,46,47,
48,49,50,51,52,53,54,55,
56,57,58,59,60,61,62,63]
我写了一个与你想要的相对应的函数(注释描述代码(：
def return_diagonal(position): # with position being chessboard position
while ((position+1)%8) != 0: # while (chess position)+1 is not divisible by eight:
position -= 7 # move diagonally upwards (7 chess spaces back)
position -= 1 # when reached the right end of the chessboard, move back one position
if position < 6: # if position happens to be in the negative:
position = 6 # set position by default to 6 (smallest possible value)
return position # return the position
该函数首先询问位置是否在最后一列中。
如果没有，请返回 7 个空格，即对角线向右移动。
它再次检查，直到到达最后一列。
在那里，它返回一个空格，因此它是棋盘右端左边的一个空格。
但是，如果这个数字是负数(就像左上角的许多数字一样(，这意味着对角线完全延伸到 8x8 棋盘之外。
因此，默认情况下，答案应该是 6。
我做了几个测试
print(60,return_diagonal(60))
print(30,return_diagonal(30))
print(14,return_diagonal(14))
print(1,return_diagonal(1))
具有以下输出：
原始位置，上/右对角线中倒数第二个数字
60 38
30 22
14 6
1 6