Table 1 The process of solving the optimal path by Viterbi algorithm

Initialization:

δ₁(j)=w · F₁(y₀=start,y₁=j,x), j=1,2,...,m

Recurrence for I = 2,3,..., n:

\({\delta _{i}}(l) = \underset {1 \le j \le m}{\max } \left \{{\delta _{i - 1}}(j) + w\; \cdot \;{F_{i}}\left ({y_{i - 1}} = j,{y_{i}} = l,x\right)\right \},\;\;\;l = 1,2,...,m\)

\({\varphi _{i}}(l) \!= \arg \underset {1 \le j \le m}{\max } \left \{\! {\delta _{i - 1}}(j) + w\; \cdot \;{F_{i}}\left ({y_{i - 1}}= j,{y_{i}} = l,x\right)\right \},\;\;\;l = 1,2,...,m\)

Termination when i = n:

\(\underset {y}{\max } \left (w\; \cdot \;F(y,x)\right) = \underset {i \le j \le m}{\max } {\delta _{n}}(j)\)

\(y_{n}^{*} = \arg \underset {i \le j \le m}{\max } {\delta _{n}}(j)\)

Return path:

\(y_{1}^{*} = {\varphi _{i + 1}}\left (y_{i + 1}^{*}\right),\;\;\;i = n - 1,n - 2,...,1\)