李航《统计学习方法》（第二版）习题10.1python实现

题目要求使用后向算法计算，程序中分别使用了前向算法和后向算法，最后得到的结果是一样的。

前向算法和后向算法：

输入是：
A: 状态转移概率矩阵
B: 观测概率矩阵
p_i: 初始状态概率向量
O: 观测序列
T: 观测序列的个数
N: 状态序列的个数
输出是：
P: 此模型参数下，得到此观测序列的概率

前向算法：

def fore_algorithm(A, B, p_i, O, T, N):
    alpha = np.zeros((T, N))
    o = np.zeros(T, np.int)

    for i in range(T):
        if O[i] == '白':
            o[i] = 1
        else:
            o[i] = 0

    for i in range(N):
        h = o[0]
        alpha[0][i] = p_i[i] * B[i][h]

    for t in range(T-1):
        h = o[t+1]
        for i in range(N):
            a = 0
            for j in range(N):
                a += (alpha[t][j] * A[j][i])
            alpha[t+1][i] = a * B[i][h]

    P = 0
    for i in range(N):
        P += alpha[T-1][i]
    return P

后向算法：

def back_algorithm(A, B, p_i, O, T, N):
    P = 0
    o = np.zeros(T, np.int)
    beta = np.ones((T, N))

    for i in range(T):
        if O[i] == '白':
            o[i] = 1
        else:
            o[i] = 0

    for t in range(T-1):
        t = T - t - 2
        h = o[t + 1]
        h = int(h)

        for i in range(N):
            beta[t][i] = 0
            for j in range(N):
                beta[t][i] += A[i][j] * B[j][h] * beta[t+1][j]

    for i in range(N):
        h = o[0]
        h = int(h)
        P += p_i[i] * B[i][h] * beta[0][i]
    return P

根据题目的要求，已知条件如下：

T = 4
N = 3
O = ['红', '白', '红', '白']
A = [[0.5, 0.2, 0.3],
     [0.3, 0.5, 0.2],
     [0.2, 0.3, 0.5]]
B = [[0.5, 0.5],
     [0.4, 0.6],
     [0.7, 0.3]]
p_i = [0.2, 0.4, 0.4]

将已知条件输入到函数中：

PF = fore_algorithm(A, B, p_i, O, T, N)
print('前向算法所得结果: ')
print(PF)
PB = back_algorithm(A, B, p_i, O, T, N)
print('后向算法所的结果: ')
print(PB)

运行后得到的结果为：

前向算法所得结果: 
0.06009079999999999
后向算法所的结果: 
0.06009079999999999

欢迎大神批评指正！