隐马尔可夫模型（六）——隐马尔可夫模型的评估问题(前向后向相结合算法）（转载）

 

重新回顾：

    前向变量α_t(i):在时刻t,在已知模型μ=（A,B,π）的条件下，状态处于s_i,输出序列为O₁0₂...O_t,前向变量为α_t(i)

    后向变量β_t(i):在时刻t,在已知模型μ=（A,B,π）和状态处于s_i的条件下，输出序列为O_t+1O_t+2...O_T,后向变量为β_t(i)

公式推导：

    P(O,q_t=s_i|μ） = P(O₁O₂...O_T, q_t=s_i|μ）

                         =P(O₁O₂...O_t, q_t=si,O_t+1O_t+2...O_T|μ)

                         =P(O₁O₂...O_t, q_t=si|μ) * P(O_t+1O_t+2...O_T|O₁O₂...O_t, q_t=si,μ)

                         =P(O₁O₂...O_t, q_t=si|μ) * P(O_t+1O_t+2...O_T|q_t=si,μ)

                         =α_t(i) *  β_t(i)

     P(O|μ）=

案例分析：

    

      分析：

        P(q₄=s₃|O,M) =  P(q₄=s₃, O|M)/P(O|M)

                            = P(O,q₄=s₃|M)/P(O|M)

                            = α₄(3) *  β₄(3)/  

     程序：

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
        float a[3][3] = {{0.5,0.2,0.3},{0.3,0.5,0.2},{0.2,0.3,0.5}};
        float b[3][2] = {{0.5,0.5},{0.4,0.6},{0.7,0.3}};
        float result_b[8][3];
        float result_f[8][3];
        float result, result_t;
        int list[8] = {0,1,0,0,1,0,1,1};
        result_b[7][0] = 1;
        result_b[7][1] = 1;
        result_b[7][2] = 1;
        result_f[0][0] = 0.2 * 0.5;
        result_f[0][1] = 0.4 * 0.4;
        result_f[0][2] = 0.4 * 0.7;
        //Backward
        int i,j,k, count = 7;
        for (i=6; i>=0; i--)
        {
            for(j=0; j<=2; j++)
            {
                result_b[i][j] = 0;
                for(k=0; k<=2; k++)
                {
                   result_b[i][j] += result_b[i+1][k] * a[j][k] * b[k][list[count]];
                }
            }
            count -= 1;
        }
       for (i=0; i<=7; i++)
        {
            for(j=0; j<=2; j++)
            {
                printf("b[%d][%d]= %f\n",i+1,j+1, result_b[i][j]);

            }
        }
        printf("Backward:%f\n", result_b[0][0]*0.2*0.5+result_b[0][1]*0.4*0.4+result_b[0][2]*0.4*0.7);
        //Forward
        count = 1;
        for (i=1; i<=7; i++)
        {
            for(j=0; j<=2; j++)
            {
                result_f[i][j] = 0;
                for(k=0; k<=2; k++)
                {
                    result_f[i][j] += result_f[i-1][k] * a[k][j] * b[j][list[count]];
                }
            }
            count += 1;
        }
        for (i=0; i<=7; i++)
        {
            for(j=0; j<=2; j++)
            {
                printf("a[%d][%d]= %f\n", i+1, j+1, result_f[i][j]);
            }
        }
        result = result_f[7][0] + result_f[7][1] + result_f[7][2];
        printf("Forward: %f\n", result);
        
        result_t = 0;
        for (i=0; i<=2; i++)
        {
            result_t += result_f[3][i] * result_b[3][i];
        }
        printf("Result:%f\n", result_f[3][2]*result_b[3][2]/result_t);

        return 0;
}

        运行结果