我有一项人工智能练习要做。目标是在屏幕上打印一个CNF(n,m,k)。这是一个布尔公式，包含n个字母，m个子句，每个子句必须有k个字母。我有一个规则，大写字母表示真，小写字母表示假。此外，我还需要从子句中删除冗余和重言式。例如，CNF(6,3,3)可能看起来像这样：
[[A,true]or[b,false]or[D,true]] and[[B,true]or[d,false]or[a,false]] and [[d,false]or[A,true]or[B,true]]你可以看到这里有3个子句，每个子句恰好有3个字母，字母表是6个字母（6个大写和6个小写）。这里没有冗余或重言式。冗余是像[[A,true]or[A,true]]这样的情况。重言式是像[[A,true]or[A,false]]这样的情况。现在我发布我的Python代码，它能工作，但只删除了冗余。正如我在问题中所写，我需要删除重言式。这里是代码
import randomimport stringimport sysimport os#这个函数创建元组列表# n = 符号数量, m = 子句数量, k = 子句长度def createTuple(n): if n > 26: print("不可能有超过25个字母") sys.exit() # 创建并打乱25个数字的列表 my_list = [0] * (2 * n) for x in range(0, 2 * n): my_list[x] = x random.shuffle(my_list) # 创建小写或大写符号的列表 symbols = [0] * (2 * n) x = 0 for count in range(0, n): symbols[count] = string.ascii_lowercase[count] for count in range(n, 2 * n): symbols[count] = string.ascii_uppercase[count - n] symbols_rand = [0] * (2 * n) for count in range(0, 2 * n): symbols_rand[count] = symbols[my_list[count]] list_tuple = [0] * (2 * n) for counts in range(0, (2 * n)): if symbols_rand[counts].islower(): list_tuple[counts] = [(symbols_rand[counts]), False] if symbols_rand[counts].isupper(): list_tuple[counts] = [(symbols_rand[counts]), True] return list_tuple#这个函数创建CNFdef createCnf(n, m, k): list_tuple = createTuple(n) stot = "" ki = "" x = 0 i = 0 countx = 0 for count in range(0, m): ktot = "{" temple_tuple = createTuple(n) temple_tuple2 = [] for index in range(0, k): #冗余控制 i = index while(temple_tuple[i] == None): i += 1 countx = 0 #重言式控制（仅在两个元素相邻时有效） while countx < len(temple_tuple2): x = temple_tuple[i][0] if temple_tuple2[countx][0].isupper(): c = temple_tuple2[countx][0].lower() else: c = temple_tuple2[countx][0].upper() if (x == c): i += 1 print("发现重言式") countx = 0 else: countx = countx + 1 ki = temple_tuple[i] temple_tuple2.append(ki) temple_tuple[i] = None if index != (k - 1): ktot += str(ki) + " or " elif index == (k - 1): ktot += str(ki) # 重言式控制结束 ktot += "}" if count != (m - 1): stot += ktot + " and " elif count == (m - 1): stot += ktot print(stot)lista_tuple = createTuple(6)createCnf(6,3,3)回答：
这是答案
import randomimport stringimport sysimport os #这个函数创建元组列表 # n = 符号数量, m = 子句数量, k = 子句长度def createTuple(n):if n > 26: print("不可能有超过25个字母") sys.exit()# 创建并打乱25个数字的列表my_list = [0] * (2 * n)for x in range(0, 2 * n): my_list[x] = xrandom.shuffle(my_list)# 创建小写或大写符号的列表symbols = [0] * (2 * n)x = 0for count in range(0, n): symbols[count] = string.ascii_lowercase[count]for count in range(n, 2 * n): symbols[count] = string.ascii_uppercase[count - n]symbols_rand = [0] * (2 * n)for count in range(0, 2 * n): symbols_rand[count] = symbols[my_list[count]]list_tuple = [0] * (2 * n)for counts in range(0, (2 * n)): if symbols_rand[counts].islower(): list_tuple[counts] = [(symbols_rand[counts]), False] if symbols_rand[counts].isupper(): list_tuple[counts] = [(symbols_rand[counts]), True]return list_tuple#这个函数创建CNFdef createCnf(n, m, k):list_tuple = createTuple(n)stot = ""ki = ""x = 0i = 0countx = 0for count in range(0, m): ktot = "{" temple_tuple = createTuple(n) temple_tuple2 = [] for index in range(0, k): #冗余控制 i = index while(temple_tuple[i] == None): i += 1 countx = 0 #重言式控制（仅在两个元素相邻时有效） while countx < len(temple_tuple2): x = temple_tuple[i][0] if temple_tuple2[countx][0].isupper(): c = temple_tuple2[countx][0].lower() else: c = temple_tuple2[countx][0].upper() if (x == c): i += 1 print("发现重言式") countx = 0 else: countx = countx + 1 ki = temple_tuple[i] temple_tuple2.append(ki) temple_tuple[i] = None if index != (k - 1): ktot += str(ki) + " or " elif index == (k - 1): ktot += str(ki) # 重言式控制结束 ktot += "}" if count != (m - 1): stot += ktot + " and " elif count == (m - 1): stot += ktotprint(stot)lista_tuple = createTuple(6)createCnf(6,3,3)