import functoolsclass Relation(object):def __init__(self, sets, rel):# rel为sets上的二元关系assert not(len(sets)==0 and len(rel) > 0)  # 不允许sets为空而rel不为空assert sets.issuperset(set([x[0] for x in rel]) | set([x[1] for x in rel]))  # 不允许rel中出现非sets中的元素self.rel = relself.sets = setsdef __str__(self):relstr = '{}'setstr = '{}'if len(self.rel) > 0:relstr = str(self.rel)if len(self.sets) > 0:setstr = str(self.sets)return 'Relation: ' + relstr + ' on Set: ' + setstrdef __eq__(self, other):return self.sets == other.sets and self.rel == other.reldef diagonalRelation(self):# 返回一个给定集合(self.sets)的对角线关系对象return Relation(self.sets, set([(a, a) for a in self.sets]))def __mul__(self, other):  # 返回合成的结果，为一个Relation对象assert self.sets == other.sets  # 必须满足self和other的A->B的映射空间相同# 注意序偶return Relation(self.sets, set([(x, y1) for x, y in other.rel for x1, y1 in self.rel if y == x1]))def __pow__(self, power, modulo=None):assert power >= -1  # 必须满足幂次>=-1if power == -1:  # 求反return Relation(self.sets, set([(x, y) for y, x in self.rel]))elif power == 0:  # 求对角线return self.diagonalRelation()return self ** (power - 1) * self  # 递归：*已经重载def __add__(self, other):assert self.sets == other.sets  # 必须满足元素相同return Relation(self.sets, self.rel)def toMatrix(self):"""将序偶集合形式的关系转换为矩阵"""matrix = []elems = sorted(list(self.sets))  # 为保证矩阵的唯一性，需对self.sets中的元素先排序line = [0] * len(self.sets)for elem in elems:for x, y in self.rel:if x == elem:line[elems.index(y)] = 1matrix.append(line)line = [0] * len(self.sets)return matrixdef isReflexive(self):"""判断是否自反"""for x in self.sets:if (x, x) not in self.rel:return Falsereturn Truedef isIrreflexive(self):"""判断是否 反自反"""for x in self.sets:if (x, x) not in self.rel:return Falsereturn Truedef isSymmetric(self):"""判断是否 对称"""for x, y in self.rel:if (y, x) not in self.rel:return Falsereturn Truedef isAsymmetric(self):"""判断是否 非对称"""for x, y in self.rel:if (y, x) in self.rel:return Falsereturn Truedef isAntiAsymmetric(self):"""判断是否 非对称"""for x, y in self.rel:if (y, x) in self.rel and x != y:return Falsereturn Truedef isTransitive(self):"""判断是否 传递"""for x, y in self.rel:for x1, y1 in self.rel:if x1 == y:if (x, y1) not in self.rel:return Falsereturn Truedef reflexiveClosure(self):"""计算自反闭包"""return self + self.diagonalRelation()def symmetricClosure(self):"""对称闭包"""return self + self ** -1def transitiveClosure(self):"""传递闭包"""closure = selffor i in range(2, len(self.sets) + 1):if closure.isTransitive():  # 反正是或运算，可以省此判断，或许还会更快算得breakclosure = closure + self ** ireturn closuredef transitiveClosure3(self):# 该方法利用Roy-Warshall计算传递闭包# 现将关系转换为矩阵，再调用__warshall函数m = self.toMatrix()return self.__warshall(m)def __warshall(self, a):  # 参数a为关系矩阵assert (len(row) == len(a) for row in a)n = len(a)# Roy-Warshall求传递闭包的算法for r in range(n):for c in range(len(a[r])):if a[c][r] == 1:for k in range(n):a[c][k] = a[c][k] | a[r][k]  # .......return adef isEquivalenceRelation(rel: Relation):"""判断等价关系"""return rel.isReflexive() and rel.isSymmetric() and rel.isTransitive()def createPartition(rel: Relation):  """对给定的Relation对象rel，求其决定的rel.sets上的划分"""# 如果rel不是等价关系，返回空集if not isEquivalenceRelation(rel):print("The given relation is not an Equivalence Relation")return set([])# 如rel是等价关系，求划分、商集partition = set([])for elem in rel.sets:  # 给个元素的等价类partition.add(frozenset(y for x, y in rel.rel if x == elem))return partitiondef createEquivalenceRelation(partition, A):# 根据所给集合A的划分partition创建等价关系assert functools.reduce(lambda x, y: x.union(y), partition) == A  # 断言保证 ...# 每个等价类的元素与元素之间、元素与自身存在关系return Relation(A, set((a,b) for part in partition for a in part for b in part))def isPartialOrder(rel: Relation):"""判断是否 半序 = 自反 + 反对称 + 传递"""return (rel ** (-1)).rel.intersection(rel.rel) == set((x, x) for x in rel.sets)# return rel.isReflexive() and rel.isAntiSymmetric() and rel.isTransitive()def isQuasiOrder(rel):"""判断是否 拟序 = 反自反 + 反对称 + 传递"""# return (rel ** (-1)).rel.intersection(rel.rel) == set()return rel.isIrreflexive() and rel.isAntiSymmetric() and rel.isTransitive()def isLinearOrder(rel: Relation):"""判断是否 全序 = 半序+ = (自反+) + 反对称 + 传递"""if not isPartialOrder(rel):return Falseelse:# 由于带有传递性：所以第二个==可以不用return set(x for x, y in rel.rel if x != y) == rel.sets == set(y for x, y in rel.rel if x != y)def join(rel1: Relation, rel2: Relation):"""实现关系矩阵的join运算"""assert rel1.sets == rel2.sets# 首先得到二者的矩阵M1 = rel1.toMatrix()M2 = rel2.toMatrix()m = len(M1)n = mM = []  # 实现关系矩阵的join运算，结果存于M中for r in range(m):row = [0] * nfor c in range(n):row[c] = M1[r][c] | M2[r][c]  # 或M.append(row)return Mdef meet(rel1: Relation, rel2: Relation):"""实现关系矩阵的meet运算"""assert rel1.sets == rel2.sets# 首先得到二者的矩阵M1 = rel1.toMatrix()M2 = rel2.toMatrix()m = len(M1)n = mM = []  # 实现关系矩阵的meet运算，结果存于M中for r in range(m):row = [0] * nfor c in range(n):row[c] = M1[r][c] & M2[r][c]  # 与M.append(row)return Mdef booleanProduct(rel1: Relation, rel2: Relation):assert rel1.sets == rel2.sets  # 保证了矩阵大小一致# 首先得到二者的矩阵M1 = rel1.toMatrix()M2 = rel2.toMatrix()m = len(M1)n = mM = []  # 结果存于M中for r in range(m):row_M1_r = M1[r]  # 准备M1矩阵的行row_M_r = [0] * nfor c in range(n):col_M2_c = [M2[r_2][c] for r_2 in range(m)]  #  准备M2矩阵的列# 计算r行c列的结果row_M_r[c] = 1 if sum([x & y for x, y in zip(row_r, col_M2_c)]) else 0M.append(row_M_r)return M