本文介绍了利用HMM与Viterbi算法来纠正印刷错误的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我想用HMM与Viterbi算法来纠正印刷错误，我计算出所需的概率，但是当我申请Viterbi算法我得到了很糟糕的结果，我检查线路code线，我无法找到错误
公共ForwardViterbi（字符串[]的状态，字符串[]的意见，双[] startProbability，双[，] transitionProbability，双[，] emissionProbability，双比例因子）
{
this.states =状态;
this.observations =观测;
this.startProbability = startProbability;
this.transitionProbability = transitionProbability;
this.emissionProbability = emissionProbability;
this.scaleFactor =比例因子;
}
// ------------------------------------------------ ----------------------
//方法
公共无效处理（INT []的问题）
{
双[，] T =新的双[states.Length，3]。 //我们将存储概率序列的维特比路径
VPATH =新INT [problem.Length]
vProbs =新的双[problem.Length]
//初始化Ť
// ------------------------------------------------ ------------------
对于（INT状态= 0;状态＆LT; states.Length;状态++）
{
T [状态，0] = startProbability [状态];
T [状态1] =状态;
T [状态，2] = startProbability [状态];
}
对于（INT输出= 0;输出＆LT; problem.Length;输出++）
{
双[，] U =新的双[states.Length，3]。 //我们将使用这个数组来计算未来的可能性
Console.WriteLine（\ nTesting假设{0}（{1}），输出的观察[问题[输出]）;
双最高= 0;
对于（INT nextState = 0; nextState＆LT; states.Length; nextState ++）
{
双总= 0;
双argMax = 0;
双valMax = 0;
Console.WriteLine（估计概率为未来的状态{0}（{1}），nextState，各国[nextState]）;
对于（INT状态= 0;状态＆LT; states.Length;状态++）
{
Console.WriteLine（测试状态为{0}（{1}），规定[状态]，状态）;
双概率= T [状态，0];
双v_path = T [状态1]。
双v_prob = T [状态，2]。
双P = emissionProbability [状态，问题[输出] * transitionProbability [状态，nextState] *比例因子;
概率* = P;
v_prob * = P;
共有+ =概率;
如果（v_prob＆GT; valMax）
{
valMax = v_prob;
argMax = nextState;
}
Console.WriteLine（中VProbability {0} {1}具有规模{2} ^ {3}，规定[nextState]，v_prob，比例因子，输出+ 1）;
如果（v_prob＆GT;最高）
{
最高= v_prob;
VPATH [输出] = nextState;
vProbs [输出] = v_prob;
}
}
U [nextState，0] =总;
U [nextState，1] = argMax;
U [nextState，2] = valMax;
}
T = U;
Console.WriteLine（最高的概率为{0}在{2} ^ {3}）状态{1}（比例因子，最高，状态[VPATH [输出]，比例因子，输出+ 1）;
}
//应用SumMax
双总= 0;
双ValMax = 0;
对于（INT状态= 0;状态＆LT; states.Length;状态++）
{
双概率= T [状态，0];
双v_path = T [状态1]。
双v_prob = T [状态，2]。
共有+ =概率;
如果（v_prob＆GT; ValMax）
{
ValMax = v_prob;
}
}
Console.WriteLine（\ nAnalysis：总概率（所有路径的总和）为给定的状态，:: {0} \ n此维特比路径概率为:: {1}，道达尔，ValMax）;
Console.WriteLine（以上的结果psented含{0} ^ {1}的一个比例因子$ P $，比例因子，problem.Length）;
}
解决方案
我刚刚检查过这实现和的。这其中似乎并没有工作。在一个在维基百科目前的工作。如果你想 - 你可以对它们进行比较，但我懒得做这种
（我已经实现解决方案，正是您的问题，描述的）
I want to use HMM with Viterbi Algorithm to correct typographical errors, I calculated the required probability but when I apply Viterbi algorithm I got very bad results, I checked the code line by line and I couldn't find the error
public ForwardViterbi(string[] states, string[] observations, double[] startProbability, double[,] transitionProbability, double[,] emissionProbability, double scaleFactor)
{
this.states = states;
this.observations = observations;
this.startProbability = startProbability;
this.transitionProbability = transitionProbability;
this.emissionProbability = emissionProbability;
this.scaleFactor = scaleFactor;
}
//----------------------------------------------------------------------
//The Methods
public void Process(int[] problem)
{
double[,] T = new double[states.Length, 3]; //We will store the probability sequence for the Viterbi Path
vPath = new int[problem.Length];
vProbs = new double[problem.Length];
//initialize T
//------------------------------------------------------------------
for (int state = 0; state < states.Length; state++)
{
T[state, 0] = startProbability[state];
T[state, 1] = state;
T[state, 2] = startProbability[state];
}
for (int output = 0; output < problem.Length; output++)
{
double[,] U = new double[states.Length, 3]; //We will use this array to calculate the future probabilities
Console.WriteLine("\nTesting hypothesis {0} ({1})", output, observations[problem[output]]);
double highest = 0;
for (int nextState = 0; nextState < states.Length; nextState++)
{
double total = 0;
double argMax = 0;
double valMax = 0;
Console.WriteLine(" Estimating probability for future state {0} ({1})", nextState, states[nextState]);
for (int state = 0; state < states.Length; state++)
{
Console.WriteLine(" The testing state is {0} ({1})", states[state], state);
double prob = T[state, 0];
double v_path = T[state, 1];
double v_prob = T[state, 2];
double p = emissionProbability[state, problem[output]] * transitionProbability[state, nextState] * scaleFactor;
prob *= p;
v_prob *= p;
total += prob;
if (v_prob > valMax)
{
valMax = v_prob;
argMax = nextState;
}
Console.WriteLine(" VProbability of {0} is {1} with scale {2}^{3}", states[nextState], v_prob, scaleFactor, output + 1);
if (v_prob > highest)
{
highest = v_prob;
vPath[output] = nextState;
vProbs[output] = v_prob;
}
}
U[nextState, 0] = total;
U[nextState, 1] = argMax;
U[nextState, 2] = valMax;
}
T = U;
Console.WriteLine("The highest probability was {0} in state {1} (scale factor of {2}^{3})", highest, states[vPath[output]], scaleFactor, output + 1);
}
//Apply SumMax
double Total = 0;
double ValMax = 0;
for (int state = 0; state < states.Length; state++)
{
double prob = T[state, 0];
double v_path = T[state, 1];
double v_prob = T[state, 2];
Total += prob;
if (v_prob > ValMax)
{
ValMax = v_prob;
}
}
Console.WriteLine("\nAnalysis: Total probability (sum of all paths) for the given state is :: {0}\nThe Viterbi Path Probability is :: {1}", Total, ValMax);
Console.WriteLine("The above results are presented with a scale factor of {0}^{1}", scaleFactor, problem.Length);
}
解决方案
I've just checked this implementation and the one published on wikipedia.This one doesn't seem to work. The one on wikipedia does work.If you want - you can compare them, but I'm lazy to do this.
(I've implemented solution for exactly your problem as described here)
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