所以,我的妻子在Steam上玩Hammerwatch.她遇到了一个难题,我决定尝试编写一个解决方案.
以下是拼图的工作原理：
激活开关可以打开或关闭该开关,也可以切换其相邻的开关.
这是游戏中拼图的YouTube视频：
http://www.youtube.com/watch?v=OM1XD7IZ0cg
我想出了如何让拼图的机制正常工作.我终于意识到我有两个选择让电脑解决这个问题：
A)通过随机选择开关让计算机解决
…要么…
B)创建一种算法,使计算机能够更有效地解决难题.
作为一名新程序员(在CodeAcademy教程的中途,在LPTHW中途,目前正在通过麻省理工学院edX计算机科学Python课程),我觉得我的能力有限,无法解决这个问题.我来学习了！请帮忙！
请帮助：
我需要帮助找出一个更好的方法来随机解决这个问题,甚至更好,有一个算法,让计算机系统地解决这个难题.我唯一能做的就是让计算机将拼图状态存储在列表或字典中,通过跳过这些存储的状态来协助程序,将程序指向新的可能解决方案
当前程序的工作原理：
我打算允许用户使用前9个raw_input输入拼图板的当前状态.然后它进入一个循环,随机切换拼图板的开关,直到它们全部打开.
P.S.：当我注册StackOverflow帐户并输入此消息时,我的计算机一直在后台运行该程序以找到解决方案.大约一个小时,仍然没有找到解决方案,它目前正在进行~92,000,000次迭代.我觉得它不起作用……
import random
def switcheroo(x):
"""
switches 'x' to 1 if it's a 0 and vice-versa
"""
if x == 0:
x = 1
else:
x = 0
return x
# original input variables
a1 = 0
a2 = 0
a3 = 0
b1 = 0
b2 = 0
b3 = 0
c1 = 0
c2 = 0
c3 = 0
# puzzleboard
print "\n\n"
print " 1 2 3 "
print " -------------"
print "a |",a1,"|",a2,a3,"|"
print " -------------"
print "b |",b1,b2,b3,"|"
print " -------------"
print "c |",c1,c2,c3,"|"
print " -------------"
print "\n\n"
print "What's ON/OFF? (type 0 for OFF,1 for ON)"
a1 = int(raw_input("a1: "))
a2 = int(raw_input("a2: "))
a3 = int(raw_input("a3: "))
b1 = int(raw_input("b1: "))
b2 = int(raw_input("b2: "))
b3 = int(raw_input("b3: "))
c1 = int(raw_input("c1: "))
c2 = int(raw_input("c2: "))
c3 = int(raw_input("c3: "))
# for counting the iterations within the loop
iteration = 0
# to stop loop if all switches are ON
ans = a1 and a2 and a3 and b1 and b2 and b3 and c1 and c2 and c3
while ans == False:
# randomly generates number,flipping random switches
counter = random.randint(1,9)
if counter == 1:
switch = "a1"
elif counter == 2:
switch = "a2"
elif counter == 3:
switch = "a3"
elif counter == 4:
switch = "b1"
elif counter == 5:
switch = "b2"
elif counter == 6:
switch = "b3"
elif counter == 7:
switch = "c1"
elif counter == 8:
switch = "c2"
elif counter == 9:
switch = "c9"
# PUZZLE MECHANICES #
if switch == "a1":
a1 = switcheroo(a1)
a2 = switcheroo(a2)
b1 = switcheroo(b1)
if switch == "a2":
a2 = switcheroo(a2)
a1 = switcheroo(a1)
a3 = switcheroo(a3)
b2 = switcheroo(b2)
if switch == "a3":
a3 = switcheroo(a3)
a2 = switcheroo(a2)
b3 = switcheroo(b3)
if switch == "b1":
b1 = switcheroo(b1)
b2 = switcheroo(b2)
a1 = switcheroo(a1)
c1 = switcheroo(c1)
if switch == "b2":
b2 = switcheroo(b2)
a2 = switcheroo(a2)
b1 = switcheroo(b1)
b3 = switcheroo(b3)
c2 = switcheroo(c2)
if switch == "b3":
b3 = switcheroo(b3)
b1 = switcheroo(b1)
b2 = switcheroo(b2)
c3 = switcheroo(c3)
# Edit 1
if switch == "c1":
c1 = switcheroo(c1)
c2 = switcheroo(c2)
b1 = switcheroo(b1)
if switch == "c2":
c2 = switcheroo(c2)
c1 = switcheroo(c1)
c3 = switcheroo(c3)
b2 = switcheroo(b2)
if switch == "c3":
c3 = switcheroo(c3)
c2 = switcheroo(c2)
b3 = switcheroo(b3)
if switch == "stop":
break
# prints puzzle-board state at end of loop iteration
print "\n\n"
print " 1 2 3 "
print " -------------"
print "a |","|"
print " -------------"
print "b |","|"
print " -------------"
print "c |","|"
print " -------------"
print "\n\n"
# prints which # was randomly generated
print "random #: ",counter
# tracks loop iteration
iteration += 1
print "iteration",iteration
if ans == True:
print "I figured it out!"
解决方法
假设您正在解决3×3问题,并且变量设置如下：
x_1 x_2 x_3 x_4 x_5 x_6 x_7 x_8 x_9
这个初始状态是：
a_1 a_2 a_3 a_4 a_5 a_6 a_7 a_8 a_9
现在,您可以写下解决方案必须满足的一些方程式(算术模2).它基本上编码关于哪些开关导致特定灯光切换的规则.
a_1 = x_1 + x_2 + x_4 a_2 = x_1 + x_2 + x_3 + x_5 ... a_5 = x_2 + x_4 + x_5 + x_6 + x_8 ... a_9 = x_6 + x_8 + x_9
现在你可以使用高斯消元法来解决这组联立方程.因为你在算术模2中工作,它实际上比实数上的联立方程更容易.例如,要删除第二个等式中的x_1,只需将第一个等式添加到其中即可.
a_1 + a_2 = (x_1 + x_2 + x_4) + (x_1 + x_2 + x_3 + x_5) = x_3 + x_4 + x_5
具体来说,这是算术模2中的高斯消元算法：
>选择一个带有x_1的等式.将其命名为E_1.
>将E_1添加到其他每个未命名的等式中,其中包含x_1.
>重复x_2,x_3,….,x_n.
现在,E_n是仅包含x_n的等式.您可以将从此处获得的x_n值替换为先前的等式.重复E_ {n-1},E_1.
总的来说,这解决了O(n ^ 3)操作中的问题.
这是一些代码.
class Unsolvable(Exception):
pass
def switches(n,m,vs):
eqs = []
for i in xrange(n):
for j in xrange(m):
eq = set()
for d in xrange(-1,2):
if 0 <= i+d < n: eq.add((i+d)*m+j)
if d != 0 and 0 <= j+d < m: eq.add(i*m+j+d)
eqs.append([vs[i][j],eq])
N = len(eqs)
for i in xrange(N):
for j in xrange(i,N):
if i in eqs[j][1]:
eqs[i],eqs[j] = eqs[j],eqs[i]
break
else:
raise Unsolvable()
for j in xrange(i+1,N):
if i in eqs[j][1]:
eqs[j][0] ^= eqs[i][0]
eqs[j][1] ^= eqs[i][1]
for i in xrange(N-1,-1,-1):
for j in xrange(i):
if i in eqs[j][1]:
eqs[j][0] ^= eqs[i][0]
eqs[j][1] ^= eqs[i][1]
return [(i//m,i%m) for i,eq in enumerate(eqs) if eq[0]]
print switches(4,3,([1,0],[0,1,1],0]))
您可以为其指定开关阵列的高度和宽度,以及一次一行的初始状态.它返回您需要按下的开关以关闭所有灯.
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