NLP Homework 03 ——冯煜博
题目描述
（盒子和球模型）假设有3个盒子，每个盒子里装有红白两种颜色的球，盒子里的红白球有下表列出，初始状态分布。
解答
1. 给出HMM模型 (mu=(A, B,pi))
A，“状态转移矩阵”如下：
[egin{array}{c|lcr}
A & ext{box1} & ext{box2} & ext{box3} \
hline
box1 & 0.5 & 0.2 & 0.3 \
box2 & 0.3 & 0.5 & 0.2 \
box3 & 0.2 & 0.3 & 0.5
end{array}
]
B，“符号发射概率”如下矩阵：
[egin{array}{c|lcr}
B & ext{red} & ext{white} \
hline
box1 & frac{1}{2} & frac{1}{2} \
box2 & frac{2}{5} & frac{3}{5} \
box3 & frac{7}{10} & frac{3}{10}
end{array}
]
(pi)，“初始状态的概率分布”如下矩阵
[egin{array}{c|cr}
& ext{box1} & ext{box2} & ext{box3} \
hline
pi & 0.2 & 0.4 & 0.4
end{array}
]
2. 试用前向算法计算观测序列概率：{红，白，红}
O = {红， 白， 红}
step 1 初始化
(a_{红}(box1) = pi_{box1} b_{box1}(红) = 0.2 imes frac{1}{2} = 0.1)
(a_{红}(box2) = pi_{box2} b_{box2}(红) = 0.4 imes frac{2}{5} = 0.16)
(a_{红}(box3) = pi_{box3} b_{box3}(红) = 0.4 imes frac{7}{10} = 0.28)
step 2 归纳计算
[egin{align}
a_{红白}(box1) & = left[ sum_{i=1}^{3} a_{红}(box_{i})a_{box_{i} box_{1}}
ight] b_{box_{1}}(白) \
& =left[ a_{红}(box1) imes 0.5 + a_{红}(box2) imes 0.3 + a_{红}(box3) imes 0.2
ight] imes frac{1}{2} \
& = 0.077
end{align}
]
[egin{align}
a_{红白}(box2) & = left[ sum_{i=1}^{3} a_{红}(box_{i})a_{box_{i} box_{2}}
ight] b_{box2}(白) \
& = left[ a_{红}(box1) imes 0.2 + a_{红}(box2) imes 0.5 + a_{红}(box3) imes 0.3
ight] imes frac{3}{5} \
& = 0.1104
end{align}
]
[egin{align}
a_{红白}(box3) & = left[ sum_{i=1}^{3} a_{红}(box_{i})a_{box_{i} box_{3}}
ight] b_{box_{3}}(白) \
& = left[ a_{红}(box1) imes 0.3 + a_{红}(box2) imes 0.2 + a_{红}(box3) imes 0.5
ight] imes frac{3}{10} \
& = 0.0606
end{align}
]
[egin{align}
a_{红白红}(box1) & = left[ sum_{i=1}^{3} a_{红白}(box_{i}) a_{box_{i}box1}
ight] b_{box1}(红) \
& = left[ a_{红白}(box1) imes 0.5 + a_{红白}(box2) imes 0.3 + a_{红白}(box3) imes 0.2
ight] imes frac{1}{2} \
& = 0.04187
end{align}
]
[egin{align}
a_{红白红}(box2) & = left[ sum_{i=1}^{3} a_{红白}(box_{i}) a_{box_{i}box2}
ight] b_{box2}(红) \
& = left[ a_{红白}(box1) imes 0.2 + a_{红白}(box2) imes 0.5 + a_{红白}(box3) imes 0.3
ight] imes frac{2}{5} \
& = 0.035512
end{align}
]
[egin{align}
a_{红白红}(box3) & = left[ sum_{i=1}^{3} a_{红白}(box_{i}) a_{box_{i}box3}
ight] b_{box3}(红) \
& = left[ a_{红白}(box1) imes 0.3 + a_{红白}(box2) imes 0.2 + a_{红白}(box3) imes 0.5
ight] imes frac{7}{10} \
& = 0.052836
end{align}
]
step 3 求和终结
[egin{align}
P(O | mu) & = sum_{i=1}^{3}a_{红白红} (box_{i}) \
& = a_{红白红}(box1) + a_{红白红}(box2) + a_{红白红}(box3) \
& = 0.04187 + 0.035512 + 0.052836 \
& = 0.130218
end{align}
]
3. 试用维特比算法，求解观测序列{红，白，红}的最优状态序列。 （要求给详细计算过程）
假设从盒子中取球是放回抽样概率模型。
step 1 初始化
(delta_{红}(box1) = pi_{box1}b_{box1}(红) = 0.2 imes frac{1}{2} = 0.1)
(delta_{红}(box2) = pi_{box2}b_{box2}(红) = 0.4 imes frac{2}{5} = 0.16)
(delta_{红}(box3) = pi_{box3}b_{box3}(红) = 0.4 imes frac{7}{10} = 0.28)
(varphi_{红}(box1) = 0)
(varphi_{红}(box2) = 0)
(varphi_{红}(box3) = 0)
step 2 归纳计算 1
[egin{align}
delta_{红白}(box1) & = max left[ delta_{红}(box1)a_{box1box1} , delta_{红}(box2)a_{box2box1} , delta_{红}(box3)a_{box3box1}
ight] b_{box1}(白) \
& = 0.28 imes 0.2 imes frac{1}{2} \
& = 0.028
end{align}
]
[egin{align}
delta_{红白}(box2) & = max left[ delta_{红}(box1)a_{box1box2} , delta_{红}(box2)a_{box2box2} , delta_{红}(box3)a_{box3box2}
ight] b_{box1}(白) \
& = 0.28 imes 0.3 imes frac{3}{5} \
& = 0.0504
end{align}
]
[egin{align}
delta_{红白}(box3) & = max left[ delta_{红}(box1)a_{box1box3} , delta_{红}(box2)a_{box2box3} , delta_{红}(box3)a_{box3box3}
ight] b_{box3}(白) \
& = 0.28 imes 0.5 imes frac{3}{10} \
& = 0.042
end{align}
]
记忆回退路径 1
[egin{align}
varphi_{红白}(box1) & = argmax left[ delta_{红}(box1)a_{box1box1} , delta_{红}(box2)a_{box2box1} , delta_{红}(box3)a_{box3box1}
ight] b_{box1}(白) \
& = frac{1}{2} delta_{红}(box3)a_{box3box1}
end{align}
]
[egin{align}
varphi_{红白}(box2) & = argmax left[ delta_{红}(box1)a_{box1box2} , delta_{红}(box2)a_{box2box2} , delta_{红}(box3)a_{box3box2}
ight] b_{box2}(白) \
& = frac{3}{5} delta_{红}(box3) a_{box3box2}
end{align}
]
[egin{align}
delta_{红白}(box3) & = argmax left[ delta_{红}(box1)a_{box1box3} , delta_{红}(box2)a_{box2box3} , delta_{红}(box3)a_{box3box3}
ight] b_{box3}(白) \
& = frac{3}{10} delta_{红}(box3)a_{box3box3} \
end{align}
]
归纳计算 2
[egin{align}
delta_{红白红}(box1) & = max left[ delta_{红白}(box1)a_{box1box1} , delta_{红白}(box2)a_{box2box1} , delta_{红白}(box3)a_{box3box1}
ight] b_{box1}(红) \
& = 0.0504 imes 0.3 imes frac{1}{2} \
& = 0.0756
end{align}
]
[egin{align}
delta_{红白红}(box2) & = max left[ delta_{红白}(box1)a_{box1box2} , delta_{红白}(box2)a_{box2box2} , delta_{红白}(box3)a_{box3box2}
ight] b_{box1}(红) \
& = 0.0504 imes 0.5 imes frac{2}{5} \
& = 0.01008
end{align}
]
[egin{align}
delta_{红白红}(box3) & = max left[ delta_{红白}(box1)a_{box1box3} , delta_{红白}(box2)a_{box2box3} , delta_{红白}(box3)a_{box3box3}
ight] b_{box3}(红) \
& = 0.042 imes 0.5 imes frac{7}{10} \
& = 0.0147
end{align}
]
记忆回退路径 2
[egin{align}
varphi_{红白红}(box1) & = argmax left[ delta_{红白}(box1)a_{box1box1} , delta_{红白}(box2)a_{box2box1} , delta_{红白}(box3)a_{box3box1}
ight] b_{box1}(红) \
& = frac{1}{2} delta_{红}(box2)a_{box2box1}
end{align}
]
[egin{align}
varphi_{红白红}(box2) & = argmax left[ delta_{红白}(box1)a_{box1box2} , delta_{红白}(box2)a_{box2box2} , delta_{红白}(box3)a_{box3box2}
ight] b_{box2}(红) \
& = frac{2}{5} delta_{红白}(box2) a_{box2box2}
end{align}
]
[egin{align}
delta_{红白红}(box3) & = argmax left[ delta_{红白}(box1)a_{box1box3} , delta_{红白}(box2)a_{box2box3} , delta_{红白}(box3)a_{box3box3}
ight] b_{box3}(白) \
& = frac{7}{10} delta_{红白}(box3)a_{box3box3} \
end{align}
]
step 3 终结
[egin{align}
hat{Q}_{红白红} & = argmax left[ delta_{红白红}(box1) , delta_{红白红}(box2) , delta_{红白红}(box3)
ight] \
& = delta_{红白红}(box3)
end{align}
]
[egin{align}
hat{P}(hat{Q}_{红白红}) & = max left[ delta_{红白红}(box1) , delta_{红白红}(box2) , delta_{红白红}(box3)
ight] \
& = 0.0147
end{align}
]
step 4 路径（状态序列)回溯
[egin{align}
hat{q}_{红白} = varphi_{红白红}(hat{q}_{红白红}) = varphi_{红白红} left[ delta_{红白红}(box3)
ight]
end{align}
]
[egin{align}
hat{q}_{红} = varphi_{红白}(hat{q}_{红白}) = varphi_{红白} left[ delta_{红白}(box2)
ight]
end{align}
]
[egin{align}
hat{q} = varphi_{红}(hat{q}_{红}) = varphi_{红} left[ delta_{红}(box3)
ight]
end{align}
]