我发现了一个编程问题网站https://www.ohjelmointiputka.net/postit/tehtava.php?tunnus=muslam，并且存在一个我无法解决的问题。对于普通笔记本电脑而言，什么样的算法能够以最小的移动速度解决以下问题：
我有一个圆形的n个灯，并由L_1，...，L_n枚举。其中一些已打开，其中一些已关闭。我也得到了一个正整数m。每转一圈，我选择一盏灯L_i然后灯L_ {i-m}，...，L_ {i + m}将改变它们的状态，我的意思是如果灯L_j被关闭然后现在它被打开，反之亦然。索引是模n，因此灯L_ {n-1}，L_ {n}，L_1，L_2是连续的。
如果灯的初始状态（从L_1到L_n）如下表所示，那么关闭所有灯的最小转数是多少以及必须按哪些开关（列数是房间名称，数字）对于灯具来说，开关变化到一个方向的最近的灯数是多少，所以总共2n + 1个灯具开关有效，并且灯的初始状态为：
================================================================================
B 6 1 101101
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C 10 2 1011010110
--------------------------------------------------------------------------------
D 20 1 11111011101010111111
--------------------------------------------------------------------------------
E 30 7 011100001010011011100001010011
--------------------------------------------------------------------------------
F 39 6 110100111111101000011000100110111100010
--------------------------------------------------------------------------------
G 53 9 0101100101111100100011100111101001001010
0010000010110
--------------------------------------------------------------------------------
H 120 7 1010110110000100000101011001011111010111
1010011101001100000010001010011010110000
0000100110010100010010110111000000010110
--------------------------------------------------------------------------------
I 220 27 1110111111101000100110011001100110100000
0010100011000111101100111111000001010000
1010110110011100100010011011010111100011
0101101000010000100110111101001001011010
1101001001110110001100011010111101001100
11010111110101010100
--------------------------------------------------------------------------------
J 500 87 1010001101101001110001101001000101010100
0001111111001101011000000011001111111011
1001110011010111111011010100010011011001
1001101110011011100001000111110101011111
1100111100001100110011101110101100001111
1100010010011010001111000000101110101101
1010100001100011111000111001000101101000
1011111111101111000000011111010001000000
1110011110111101010010011000000100010100
0011101011010011010110011110111000010010
0111100100011010010110001000011100101001
1110111010001001011001111011111011010110
10101101111011101110
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K 1002 83 0010100100100101000000110101111111101011
1101000101111110001110000110110110010101
1110110011011101100110111001110110010011
1101111010110011110101100001101010100011
1110001100011111110100011110100111111100
0011001011100110101100001101000001110010
0110100000100100100000011010000010111100
1110001110011110101001100111101101010000
0101010000011010011110101001001001000000
0011000100011011011001111010001101111000
0100001011010011001010111001111100110001
0011111110101101001100111101110000000000
1101100100000011000010010100010101001000
1100001000101001100110010100001000001101
1101000100001010011000101001101000100010
0011010001011101010100011101001101101100
0111110100110011001111000000001001001001
1001111001011111000010110000110010101000
1011001100111101000101000110000111010100
0010011011010111001101011001111000001011
1110101010101101111011111110100001100110
1000101100110011010000110000011011110011
0010000010000000111101101000001111101111
0100111110010101100011101001111101010000
1111100010011001110111111000101000000101
01
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L 2107 108 0111110100011000011111101110010101100011
1001111011101001001110111110001100011001
1001010100101011101101001000010111111111
1001101010111011110100100101000101100011
1110100010010010101110100000111100101000
0111101011111100010010110000100110100100
0100110101110010110011110010101101100111
1110010011000110110111010110010100101110
0111111101110000111001111100100010010001
1010110011000101100111111001011110101110
0111010110111110110101000101100100011000
1011000011011110001111100110100010100101
1101111100110011001110010010001010101111
1000001001000110011110010011011101110100
1011111100110010011000010110010110101010
0110101000011011110001010000010001000110
1001110101001001110110111111010011010111
1111011001000110111001000011101101110001
0000011111101000010101011111011011000011
1111000000011100010011011001011000110101
1101011111100001100010110010110011000000
0001001111100101110100100011011010011100
0000001111010101000111011000110110100001
1010110011100110111010111110110000010000
1000101001111001000110000101010000010111
1011100001000110001100010000001011101110
1001111110100010010000011000100101010101
1001001001110110101000001001001100001011
0011011100011111100111001110101101110001
0111010000010011110110011011000011101001
1111011010010000101111000010000001100110
1001011101001000010101001001011111111011
1000111000100001101100101110100011111100
1011001111101111110110101111101111011111
1001111100110101110101111110010010101101
1111111111000100100111100011101110110100
0100011011001010110100101101000000110010
0010010001001110110100011111100011111101
0100110111101101010101010100110110011011
0001111111000100000111011010101011000010
0011011110110110110100011001101111001000
1000000011110011100111100000001010010011
1000011101111100000101010101010010100101
1010001011010100011011001110110010100000
1000111101111000010111111101010110110111
0110001111100011001110000100100101001111
0000111111100010011001010000010110111000
1000110110001000001100110000001011000010
1000101101110000101100100010101111100011
1000010010111101000010000110011010000001
0010001100001000001100110111110100100111
1001100110001000100101011111001011001111
110001011111001101010101001
================================================================================
其中1表示相应的灯在开头打开，0表示相应的灯在开始时关闭。
我修改了https://ask.sagemath.org/question/39335/how-to-find-a-minimum-number-of-switch-presses-to-shut-down-all-lamps/中的代码。它根据场Z / 2Z上的问题做了矩阵方程。它使用了一些Sage自己的线性代数函数，如nonzero_positions，http://doc.sagemath.org/html/en/reference/matrices/sage/matrix/matrix0.html，right_kernel，http://doc.sagemath.org/html/en/prep/Quickstarts/Linear-Algebra.html和solve_right，http://doc.sagemath.org/html/en/tutorial/tour_linalg.html。
我能够解决一些案例，比如H：
input = '101011011000010000010101100101111101011110100111010011000000100010100110101100000000100110010100010010110111000000010110'
m = 7
d = len(input)
F = GF(2)
L_init = vector(F,input)
M = matrix(F,d,d)
for i in range(d):
for j in range(-m,m+1):
M[i,(i+j) % d] = 1
I = M.solve_right(L_init)
K = M.right_kernel();
m = min([len((I+k).nonzero_positions()) for k in K])
print (m)
一段时间后输出46但有些情况需要花费太多时间，比如
input ='1110111111101000100110011001100110100000001010001100011110110011111100000101000010101101100111001000100110110101111000110101101000010000100110111101001001011010110100100111011000110001101011110100110011010111110101010100'
m = 27
d = len(input)
F = GF(2)
L_init = vector(F,input)
M = matrix(F,d,d)
for i in range(d):
for j in range(-m,m+1):
M[i,(i+j) % d] = 1
I = M.solve_right(L_init)
K = M.right_kernel();
m = min([len((I+k).nonzero_positions()) for k in K])
print (m)
那么如何才能找到一种更快的算法来以最少的开关次数关闭所有灯？
我试着编写自己的Gauss-Jordan实现：
def swap(m,i,j):
for k in range(8):
temp = m[i][k]
m[i][k]=m[j][k]
m[j][k]=temp
return m
def addrow(m,i,j):
for k in range(8):
m[i][k] += m[j][k]
m[i][k] %= 2
return m
w = 8;
m = [[0 for x in range(w)] for y in range(w)]
for i in range(w):
for j in range(w):
if max(i,j)-min(i,j) < 3:
m[i][j] = 1
else:
m[i][j] = 0
print(m)
m = swap(m,0,1)
print(m)
m = addrow(m,2,3)
print(m)
for i in range(0,8):
if m[i][i] == 0:
j = i+1
while j<8:
if m[j][i] == 1:
m = swap(m,j,i)
if m[j][i] == 1:
m = addrow(m,i,j)
j += 1
print(m)
但看起来矩阵方程总是不会给出最佳的压力集https://www.quora.com/What-is-the-algorithm-for-solving-lights-out-puzzle-in-minimum-number-of-moves-in-Java。这可以通过将不同的灯位置视为图形节点并将开关按下作为顶点来解决，然后使用Thorup的算法（https://dl.acm.org/citation.cfm?doid=316542.316548）来找到最小的印刷机组吗？但是，我认为图形太大而无法安装到计算机内存中，因此我认为该方法需要某种方法将图形拆分为较小的部分，并多次使用该算法来找到最佳解决方案。
https://ask.sagemath.org/question/39335/how-to-find-a-minimum-number-of-switch-presses-to-shut-down-all-lamps/中有一个算法，但该算法无法在我的计算机上运行。
F = GF(2)
CASES = (
( 6, 1, '101101' ),
( 10, 2, '1011010110' ),
( 20, 1, '11111011101010111111' ),
( 30, 7, '011100001010011011100001010011' ),
( 39, 6, '110100111111101000011000100110111100010' ),
( 53, 9,
'''01011001011111001000111001111010010010100010000010110''' ),
( 120, 7,
'''10101101100001000001010110010111110101111010011101001100000010001010011010110000
0000100110010100010010110111000000010110''' ),
( 220, 27,
'''11101111111010001001100110011001101000000010100011000111101100111111000001010000
10101101100111001000100110110101111000110101101000010000100110111101001001011010
110100100111011000110001101011110100110011010111110101010100''' ),
( 500, 87,
'''10100011011010011100011010010001010101000001111111001101011000000011001111111011
10011100110101111110110101000100110110011001101110011011100001000111110101011111
11001111000011001100111011101011000011111100010010011010001111000000101110101101
10101000011000111110001110010001011010001011111111101111000000011111010001000000
11100111101111010100100110000001000101000011101011010011010110011110111000010010
01111001000110100101100010000111001010011110111010001001011001111011111011010110
10101101111011101110''' ),
( 1002, 83,
'''00101001001001010000001101011111111010111101000101111110001110000110110110010101
11101100110111011001101110011101100100111101111010110011110101100001101010100011
11100011000111111101000111101001111111000011001011100110101100001101000001110010
01101000001001001000000110100000101111001110001110011110101001100111101101010000
01010100000110100111101010010010010000000011000100011011011001111010001101111000
01000010110100110010101110011111001100010011111110101101001100111101110000000000
11011001000000110000100101000101010010001100001000101001100110010100001000001101
11010001000010100110001010011010001000100011010001011101010100011101001101101100
01111101001100110011110000000010010010011001111001011111000010110000110010101000
10110011001111010001010001100001110101000010011011010111001101011001111000001011
11101010101011011110111111101000011001101000101100110011010000110000011011110011
00100000100000001111011010000011111011110100111110010101100011101001111101010000
111110001001100111011111100010100000010101''' ),
( 2107, 108,
"""01111101000110000111111011100101011000111001111011101001001110111110001100011001
10010101001010111011010010000101111111111001101010111011110100100101000101100011
11101000100100101011101000001111001010000111101011111100010010110000100110100100
01001101011100101100111100101011011001111110010011000110110111010110010100101110
01111111011100001110011111001000100100011010110011000101100111111001011110101110
01110101101111101101010001011001000110001011000011011110001111100110100010100101
11011111001100110011100100100010101011111000001001000110011110010011011101110100
10111111001100100110000101100101101010100110101000011011110001010000010001000110
10011101010010011101101111110100110101111111011001000110111001000011101101110001
00000111111010000101010111110110110000111111000000011100010011011001011000110101
11010111111000011000101100101100110000000001001111100101110100100011011010011100
00000011110101010001110110001101101000011010110011100110111010111110110000010000
10001010011110010001100001010100000101111011100001000110001100010000001011101110
10011111101000100100000110001001010101011001001001110110101000001001001100001011
00110111000111111001110011101011011100010111010000010011110110011011000011101001
11110110100100001011110000100000011001101001011101001000010101001001011111111011
10001110001000011011001011101000111111001011001111101111110110101111101111011111
10011111001101011101011111100100101011011111111111000100100111100011101110110100
01000110110010101101001011010000001100100010010001001110110100011111100011111101
01001101111011010101010101001101100110110001111111000100000111011010101011000010
00110111101101101101000110011011110010001000000011110011100111100000001010010011
10000111011111000001010101010100101001011010001011010100011011001110110010100000
10001111011110000101111111010101101101110110001111100011001110000100100101001111
00001111111000100110010100000101101110001000110110001000001100110000001011000010
10001011011100001011001000101011111000111000010010111101000010000110011010000001
00100011000010000011001101111101001001111001100110001000100101011111001011001111
110001011111001101010101001""" )
)
def getA( N, M, delay=0 ):
R = range(N)
lists_for_M = [ [ (j+k+delay) % N for k in range(M) ]
for j in R ]
return matrix( F, N, N, [ [ F(1) if k in lists_for_M[j] else F(0)
for k in R ]
for j in R ] ).transpose()
def showVector( vec, separator='' ):
return separator.join( [ str( entry ) for entry in vec ] )
def exchange( wblocks, wpattern, b, k1, k2 ):
blocks, pattern = wblocks[:], wpattern[:] # copy the information
for block in blocks:
block[k1] = F(1) - block[k1]
block[k2] = F(1) - block[k2]
pattern[k1] = b - pattern[k1]
pattern[k2] = b - pattern[k2]
return blocks, pattern
def blocksInformation( blocks ):
print "Blocks:"
for block in blocks:
print showVector( block )
print
# main search for all cases...
def searchLampSwitches( N, m, lamps
, showBlocks=False
, returnSolution=False ):
"""N, m, lamps is an entry in CASES..."""
print N, m, lamps
M = 2*m+1
d = gcd( N, M )
R = range( d ) # we will often loop using R
b = ZZ( N/d ) # number of blocks
bound = (b/2).floor() # critical bound for the entries in pattern
print "N=%s M=%s d=gcd(N,M)=%s b=N/d=%s bound=%s" % ( N, M, d, b, bound )
A = getA( N, M, delay=-m )
v = matrix( F, N, 1, [ F(int(s)) for s in lamps if s in '01' ] )
w = vector( A.solve_right( v ) )
if d == 1:
# then w is already the (unique, thus optimal) solution
print "UNIQUE SOLUTION"
print sum( [ 1 for entry in w if entry ] )
print showVector( w )
return w if returnSolution else None
# else we build blocks
wblocks = [ [ w[ j + d*k ] for j in R ]
for k in range(N/d) ]
wpattern = [ sum( [ 1 for wblock in wblocks if wblock[k] ] )
for k in R ]
print "Initial pattern:",
print showVector( wpattern, separator = '' if b<10 else ',' )
if showBlocks:
blocksInformation( wblocks )
wplaces = [ k for k in R if wpattern[k] > bound ] # suboptimal places in wpattern
wplaces . sort( lambda k1, k2: cmp( -wpattern[k1], -wpattern[k2] ) )
print "Suboptimal places:", wplaces
while len( wplaces ) > 1:
k1, k2 = wplaces[:2]
wplaces = wplaces[2:]
wblocks, wpattern = exchange( wblocks, wpattern, b, k1, k2 )
print "New pattern:",
print showVector( wpattern, separator = '' if b<10 else ',' )
if showBlocks:
blocksInformation( wblocks )
# we count solutions, last changes can be done - but the in case...
if wplaces:
k, = wplaces
# one last move is possible, if there are places with entries > then opposite of k'th place
# examples:
# b = 8 or 9, bound = 4, and
# ---> wpattern is 00011122233349 - then exchange 49
# ---> wpattern is 00011122233339 - then exchange 39
# ---> wpattern is 00011122222225 - no good exchange, 5 becomes 3 or 4, but the 2 (best choice) becomes > 5
goodoppositeplaces = [ kk for kk in R if kk != k and wpattern[kk] > b - wpattern[k] ]
if goodoppositeplaces:
goodoppositeplaces.sort( lambda k1, k2: -cmp( wpattern[k1], wpattern[k2] ) )
# we can still get an optimization...
kk = goodoppositeplaces[0]
print "Last exchange possible, place %s can be moved (using place %s)" % ( k, kk )
wblocks, wpattern = exchange( wblocks, wpattern, b, k, kk )
wplaces = [ kk, ] if kk > bound else []
print "New pattern:",
print showVector( wpattern, separator = '' if b<10 else ',' )
if showBlocks:
blocksInformation( wblocks )
if wplaces:
k, = wplaces
oppositeplaces = [ kk for kk in R if wpattern[kk] + wpattern[k] == b ]
if not oppositeplaces:
print "UNIQUE SOLUTION, place %s cannot be moved" % k
else:
L = len( oppositeplaces )
print ( "%s = 2^%s SOLUTIONS, obtained by interchanging even number of positions in %s"
% ( 2**L, L, str( [k, ] + oppositeplaces ) ) )
else:
if b == 2*bound+1:
print "UNIQUE SOLUTION, all suboptimal places could be paired"
else:
midplaces = [ kk for kk in R if wpattern[kk] == bound ]
L = len( midplaces )
if L == 0 :
print "UNIQUE SOLUTION, all suboptimal places could be paired, none equal %s remained" % bound
elif L == 1:
print "UNIQUE SOLUTION, one middle place remained"
else:
print ( "%s = 2^%s SOLUTIONS, obtained by interchanging even number of middle positions in %s"
% ( 2**(L-1), L-1, str(midplaces) ) )
print "OPTIMAL NUMBER OF SWITCHES: %s" % sum(wpattern)
if not returnSolution:
return
sol = []
for wblock in wblocks:
sol += wblock
return sol
for N, m, lamps in CASES:
searchLampSwitches( N, m, lamps )
print 60*'.'
输出为TypeError: unable to find a common ring for all elements。
此外，https://math.stackexchange.com/questions/2569003/how-to-find-an-algorithm-to-shut-down-all-lamps-with-minimum-number-of-moves中给出的算法似乎很难实现。