重新回顾：
前向变量αt(i):在时刻t,在已知模型μ=（A,B,π）的条件下，状态处于si,输出序列为O102...Ot,前向变量为αt(i)
后向变量βt(i):在时刻t,在已知模型μ=（A,B,π）和状态处于si的条件下，输出序列为Ot+1Ot+2...OT,后向变量为βt(i)
公式推导：
P(O,qt=si|μ） = P(O1O2...OT, qt=si|μ）
=P(O1O2...Ot, qt=si,Ot+1Ot+2...OT|μ)
=P(O1O2...Ot, qt=si|μ) * P(Ot+1Ot+2...OT|O1O2...Ot, qt=si,μ)
=P(O1O2...Ot, qt=si|μ) * P(Ot+1Ot+2...OT|qt=si,μ)
=αt(i) * βt(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
本文转自jihite博客园博客，原文链接：http://www.cnblogs.com/kaituorensheng/archive/2012/12/05/2803182.html，如需转载请自行联系原作者
