Viterbi Table
start 1.00000 0.00000 0.00000 0.00000 0.00000
v 0.00000 0.15000 0.03000 0.00810 0.00000
n 0.00000 0.20000 0.04050 0.00810 0.00000
end 0.00000 0.00000 0.00000 0.00000 0.00405
START ground control station END
Backtrace
start -1 -1 -1 -1 -1
v -1 Start n n -1
n -1 Start v v -1
end -1 -1 -1 -1 v
START ground control station END
===============================================================
Tags (answer 1):
start v n v end
Tags (answer 2):
start v v n end
There are two equal probability paths (tag sequences).
Notice that doesn't mean the model does nothing.
There are 8 possible tag sequences, and
6 of them are eliminated.
In the calculations below v2(1) stands for the Viterbi
value at time 2 for the state indexed 1 in the Viterbi
table. Looking at the order of the rows in the Viterbi table,
using 0-based indexing, that's v. So v2(1) is the viterbi
value of the v-state at time t = 2.
================================================================
t = 1
V1(1) = max(0.30000 * 1.00000 * 0.50000,0.30000 * 0.00000 * 0.10000,0.30000 * 0.00000 * 0.50000,0.30000 * 0.00000 * 0.00000)
V1(1) = max(0.15000,0.00000,0.00000,0.00000)
V1(1) = 0.15
# The terms in the max are supplied by the states we may have come from
# on this best path and they are listed in the same order as the
# states in the backtrace table, so the state supplying the amx path
# prob is Start
backtrace[v][1] = Start
V1(2) = max(0.40000 * 1.00000 * 0.50000,0.40000 * 0.00000 * 0.90000,0.40000 * 0.00000 * 0.50000,0.40000 * 0.00000 * 0.00000)
V1(2) = max(0.20000,0.00000,0.00000,0.00000)
V1(2) = 0.2
backtrace[n][1] = Start
---------------------------------------------------------------------
t = 2
V2(1) = max(0.30000 * 0.00000 * 0.50000,0.30000 * 0.15000 * 0.10000,0.30000 * 0.20000 * 0.50000,0.30000 * 0.00000 * 0.00000)
V2(1) = max(0.00000,0.00450,0.03000,0.00000)
V2(1) = 0.03
backtrace[v][2] = v
V2(2) = max(0.30000 * 0.00000 * 0.50000,0.30000 * 0.15000 * 0.90000,0.30000 * 0.20000 * 0.50000,0.30000 * 0.00000 * 0.00000)
V2(2) = max(0.00000,0.04050,0.03000,0.00000)
V2(2) = 0.0405
backtrace[n][2] = v
----------------------------------------------------------------------
t = 3
V3(1) = max(0.40000 * 0.00000 * 0.50000,0.40000 * 0.03000 * 0.10000,0.40000 * 0.04050 * 0.50000,0.40000 * 0.00000 * 0.00000)
V3(1) = max(0.00000,0.00120,0.00810,0.00000)
V3(1) = 0.0081
backtrace[v][3] = n
V3(2) = max(0.30000 * 0.00000 * 0.50000,0.30000 * 0.03000 * 0.90000,0.30000 * 0.04050 * 0.50000,0.30000 * 0.00000 * 0.00000)
V3(2) = max(0.00000,0.00810,0.00607,0.00000)
V3(2) = 0.0081
backtrace[n][3] = v
-------------------------------------------------------------------------
t = 4
V4(3) = max(1.00000 * 0.00000 * 0.00000,1.00000 * 0.00810 * 0.50000,1.00000 * 0.00810 * 0.50000,1.00000 * 0.00000 * 0.00000)
V4(3) = max(0.00000,0.00405,0.00405,0.00000)
V4(3) = 0.00405
backtrace[end][4] = v
(Of course it's a tie with n)