Actually, I don’t get what Mr. Agus said in my class yesterday since I was so sleepy 
, and what shocked me was, he gave us homework to run Viterbi Algorithm in a sheets, hikz, I have to study it alone.
First of all, Viterbi Algorithm have 4 steps to define what state is being optimal state.
1. Initialization
δt(j) = Πibi(O1), for 1≤ i ≤ N
Ψ1(i) = 0
** variable i means iteration number of states, and N means total number of states.
2. Recursion
δt(j) = max [δt-1(i)aij] bi(Ot), for 2≤ t ≤ T and 1≤ j ≤ N
1≤ j ≤ N
Ψt(j) = argmax [δt-1(i)aij], for 2≤ t ≤ T and 1≤ j ≤ N
1≤ i ≤ N
** Variable t means iteration number of times and T means total number of times.
3. Termination
p* = max [δt(i)]
1≤ i ≤ N
q* = argmax [δt(i)]
1≤ i ≤ N
4. Path (state sequence) backtracking:
qt* = Ψt+1(q*t+1) for t= T-1, T-2, ….., 1
Now, the problem is I’m losing one last state since t will never be the same as T. For the illustration, I have 4 observations, let’s say, a(t=1), b(t=2), a(t=3), a(t=4).
If your iteration has fallen into t = 4, the backtracking quotation will be:
q3* = Ψ4(q*4), if you’d like to know the q4* value, the quotation is: q4* = Ψ5(q*5) which means it’s impossible since there’s no fifth observation.
So where do i know the last path should be?
-Help me- (SOL) 
<< Updated >>
