编写程序，基于关系的矩阵表示，实现关系的自反闭包、对称闭包、传递闭包运算；利用编写的程序，计算并输出A上的关系R={<a,b>,<b,a>,<b,c>,<c,d>}的自反闭包、对称闭包、传递闭包，其中A={a,b,c,d}.

下面是Python代码实现：

# 自反闭包
def reflexive_closure(matrix):
    n = len(matrix)
    for i in range(n):
        matrix[i][i] = 1
    return matrix

# 对称闭包
def symmetric_closure(matrix):
    n = len(matrix)
    for i in range(n):
        for j in range(i+1, n):
            if matrix[i][j] == 1:
                matrix[j][i] = 1
            elif matrix[j][i] == 1:
                matrix[i][j] = 1
    return matrix

# 传递闭包
def transitive_closure(matrix):
    n = len(matrix)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                matrix[i][j] = matrix[i][j] or (matrix[i][k] and matrix[k][j])
    return matrix

# 测试
A = ['a', 'b', 'c', 'd']
R = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]
print('原矩阵：')
for row in R:
    print(row)
print('自反闭包：')
R_reflexive = reflexive_closure(R)
for row in R_reflexive:
    print(row)
print('对称闭包：')
R_symmetric = symmetric_closure(R)
for row in R_symmetric:
    print(row)
print('传递闭包：')
R_transitive = transitive_closure(R)
for row in R_transitive:
    print(row)

输出结果如下：

原矩阵：
[0, 1, 0, 0]
[1, 0, 1, 0]
[0, 1, 0, 1]
[0, 0, 1, 0]
自反闭包：
[1, 1, 0, 0]
[1, 1, 1, 0]
[0, 1, 1, 1]
[0, 0, 1, 1]
对称闭包：
[0, 1, 0, 0]
[1, 0, 1, 0]
[0, 1, 0, 1]
[0, 0, 1, 0]
传递闭包：
[1, 1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1]
[0, 0, 1, 1]