Haskell自我娱乐帖

无心人

条件12 let filter12 (x1, x2, x3, x4, x5, x6) = not （xor (x4 == 'B') (x6 == 'B')） 条件13： let filter13 (x1, x2, x3, x4, x5, x6) = not （xor (x3 == 'G') (x6 == 'R')） 所有简单条件罗列完 let t1 = filter filter1 t6 let t2 = filter filter2 t1 let t3 = filter filter3 t2 let t5 = filter filter5 t3 let t6_1 = filter filter6 t5 let t7 = filter filter7 t6_1 let t8 = filter filter8 t7 let t9 = filter filter9 t8 let t12 = filter filter12 t9 let t13 = filter filter13 t12 length t13 输出： 144

无心人

条件4 绿房子在白房子左面 即绿房子序号 小于等于白房子 目前只能说白房子序号大于等于2 let filter4_1 (x1, x2, x3, x4, x5, x6) = ((x2 == 'W') && (x1 >= 2)) || not (x2 == 'W') 且绿房子序号小于5 let filter4_2 (x1, x2, x3, x4, x5, x6) = ((x2 == 'G') && (x1 < 5)) || not (x2 == 'G') let t4 = filter filter4_2 (filter filter4_1 t13) length t4 134组

无心人

目前各位置候选数 1、15 2、37 3、15 4、37 5、33

无心人

中午考虑了 把10个简单条件筛选后的144个结果 就应该按编号分5个表分开了 而这5个表现在如果进行组合的话 理论的组合数大概在1亿内 但考虑相互之间存在很多冲突 考虑冲突情况 此时的实际数字应该是很小的 而余下的五个条件的利用 我想也应该是在组合了5个表的基础上应用 不过，haskell如何组合5个表，且使符合不冲突的条件 我还是不会作的 哈

无心人

let is1 (x1, x2, x3, x4, x5, x6)= (x1 == 1) let is2 (x1, x2, x3, x4, x5, x6)= (x1 == 2) let is3 (x1, x2, x3, x4, x5, x6)= (x1 == 3) let is4 (x1, x2, x3, x4, x5, x6)= (x1 == 4) let is5 (x1, x2, x3, x4, x5, x6)= (x1 == 5) let l1 = filter is1 t13 let l2 = filter is2 t13 let l3 = filter is3 t13 let l4 = filter is4 t13 let l5 = filter is5 t13 分解后元素数量 l1 21 l2 37 l3 12 l4 37 l5 37

无心人

let filter4_G (x1, x2, x3, x4, x5, x6) = not (x2 == 'G') let filter4_W (x1, x2, x3, x4, x5, x6) = not (x2 == 'W') let l11 = filter filter4_W l1 let l51 = filter filter4_G l5 length l11 15 length l51 33 应用条件4得到的结论 1不能有白房子 5不能有绿房子

无心人

应用条件14 2是蓝房子 let filter2_B (x1, x2, x3, x4, x5, x6) = (x2 == 'B') let l21 = filter filter2_B l2 length l21 11

无心人

条件10 抽混合烟的人和养猫的不是一个 let filter10 (x1, x2, x3, x4, x5, x6) = not ((x4 == 'C') && (x6 == 'U')) 条件11 养马的和抽DUNHILL烟的不是一个 let filter11 (x1, x2, x3, x4, x5, x6) = not ((x4 == 'H') && (x6 == 'D')) 条件15 抽混合烟的和喝矿泉水的不是一个 let filter15 (x1, x2, x3, x4, x5, x6) = not ((x5 == 'W') && (x6 == 'U'))

无心人

对5个表过滤三个条件 let l13 = filter filter10 (filter filter11 (filter filter15 l11)) let l23 = filter filter10 (filter filter11 (filter filter15 l21)) let l33 = filter filter10 (filter filter11 (filter filter15 l3)) let l43 = filter filter10 (filter filter11 (filter filter15 l4)) let l53 = filter filter10 (filter filter11 (filter filter15 l51)) Prelude> l13 [(1,'B','N','H','B','U'),(1,'B','N','F','B','U'),(1,'Y','N','C','W','D'),(1,'Y','N','C','B','D'),(1,'Y','N','F','W','D'),(1,'Y','N','F','B','D'),(1,'G','N','H','C','U'),(1,'G','N','F','C','U')] Prelude> l23 [(2,'B','G','C','W','R'),(2,'B','G','C','B','R'),(2,'B','G','H','W','R'),(2,'B','G','H','B','R'),(2,'B','G','F','W','R'),(2,'B','G','F','B','R'),(2,'B','S','D','B','U'),(2,'B','D','H','T','U'),(2,'B','D','F','T','U')] Prelude> l33 [(3,'R','E','H','M','U'),(3,'R','E','F','M','U'),(3,'B','G','C','M','R'),(3,'B','G','H','M','R'),(3,'B','G','F','M','R'),(3,'B','S','D','M','U'),(3,'Y','S','D','M','D'),(3,'W','G','C','M','R'),(3,'W','G','H','M','R'),(3,'W','G','F','M','R'),(3,'W','S','D','M','U')] Prelude> l43 [(4,'R','E','H','B','U'),(4,'R','E','F','B','U'),(4,'B','G','C','W','R'),(4,'B','G','C','B','R'),(4,'B','G','H','W','R'),(4,'B','G','H','B','R'),(4,'B','G','F','W','R'),(4,'B','G','F','B','R'),(4,'B','S','D','B','U'),(4,'B','D','H','T','U'),(4,'B','D','F','T','U'),(4,'Y','S','D','W','D'),(4,'Y','S','D','B','D'),(4,'Y','D','C','T','D'),(4,'Y','D','F','T','D'),(4,'W','G','C','W','R'),(4,'W','G','C','B','R'),(4,'W','G','H','W','R'),(4,'W','G','H','B','R'),(4,'W','G','F','W','R'),(4,'W','G','F','B','R'),(4,'W','S','D','B','U'),(4,'W','D','H','T','U'),(4,'W','D','F','T','U'),(4,'G','G','C','C','R'),(4,'G','G','H','C','R'),(4,'G','G','F','C','R'),(4,'G','S','D','C','U')] Prelude> l53 [(5,'R','E','H','B','U'),(5,'R','E','F','B','U'),(5,'B','G','C','W','R'),(5,'B','G','C','B','R'),(5,'B','G','H','W','R'),(5,'B','G','H','B','R'),(5,'B','G','F','W','R'),(5,'B','G','F','B','R'),(5,'B','S','D','B','U'),(5,'B','D','H','T','U'),(5,'B','D','F','T','U'),(5,'Y','S','D','W','D'),(5,'Y','S','D','B','D'),(5,'Y','D','C','T','D'),(5,'Y','D','F','T','D'),(5,'W','G','C','W','R'),(5,'W','G','C','B','R'),(5,'W','G','H','W','R'),(5,'W','G','H','B','R'),(5,'W','G','F','W','R'),(5,'W','G','F','B','R'),(5,'W','S','D','B','U'),(5,'W','D','H','T','U'),(5,'W','D','F','T','U')] Prelude> Prelude> length l13 8 Prelude> length l23 9 Prelude> length l33 11 Prelude> length l43 28 Prelude> length l53 24 总理论组合数8*9*11*28*24 = 532224 现在已可穷举 谁作下穷举 [ 本帖最后由 无心人 于 2008-6-20 16:39 编辑 ]

无心人

应用条件7 黄房子人抽DUNHILL烟 应用条件11 养马人住在抽DUNHILL烟人的旁边 因为2是兰房子， 绿房子在白房子左面 可能组合是 1、G B X X W 2、G B X W X 3、G B W X X 4、X B G X W 5、X B G W X 6、X B X G W 看有可利用7, 11条件的么