Consider the following sentence: a Markov model tags easilyAssume that based on a HMM, we have the following probabilities: Emission:P1(a|DET) = 0.1, P1(easily|ADV) = 0.1, P1(Markov|N) = 0.1, P1(model|N) = 0.095, P1(model|V) =0.005, P1(tags|N) = 0.080, P1(tags|V) = 0.020,Transition probabilities: P(Y|X) Y DET N V ADJ ADV X DET 0 0.55 0 0.02 0.03 N 0.01 0.1 0.08 0.01 0.02 V 0.16 0.11 0.06 0.08 0.08 ADJ 0.01 0.65 0 0.05 0 ADV 0.08 0.02 0.09 0.04 0.04 Initial probabilitiesP3(DET) = 0.20, P3(N) = 0.06, P3(V) = 0.08, P3(ADV) = 0.07, P3(ADJ) = 0.02.What are the possible tag(s) of the sentence?Assume you want to use the Viterbi algorithm to decode the sentence, write down the expression to be evaluated at the initial step.Write down the expression for the second iteration if the first tag is DET, for a second tag of ADJ
Question
Consider the following sentence: a Markov model tags easilyAssume that based on a HMM, we have the following probabilities: Emission:P1(a|DET) = 0.1, P1(easily|ADV) = 0.1, P1(Markov|N) = 0.1, P1(model|N) = 0.095, P1(model|V) =0.005, P1(tags|N) = 0.080, P1(tags|V) = 0.020,Transition probabilities: P(Y|X) Y DET N V ADJ ADV X DET 0 0.55 0 0.02 0.03 N 0.01 0.1 0.08 0.01 0.02 V 0.16 0.11 0.06 0.08 0.08 ADJ 0.01 0.65 0 0.05 0 ADV 0.08 0.02 0.09 0.04 0.04 Initial probabilitiesP3(DET) = 0.20, P3(N) = 0.06, P3(V) = 0.08, P3(ADV) = 0.07, P3(ADJ) = 0.02.What are the possible tag(s) of the sentence?Assume you want to use the Viterbi algorithm to decode the sentence, write down the expression to be evaluated at the initial step.Write down the expression for the second iteration if the first tag is DET, for a second tag of ADJ
Solution
The sentence "a Markov model tags easily" consists of four words. The possible tags for each word based on the given emission probabilities are:
- "a" - DET
- "Markov" - N
- "model" - N, V
- "tags" - N, V
- "easily" - ADV
For the Viterbi algorithm, we start with the initial step. The expression to be evaluated at the initial step for each word is:
- P1(a|DET) * P3(DET) = 0.1 * 0.20
- P1(Markov|N) * P3(N) = 0.1 * 0.06
- P1(model|N) * P3(N) = 0.095 * 0.06, P1(model|V) * P3(V) = 0.005 * 0.08
- P1(tags|N) * P3(N) = 0.080 * 0.06, P1(tags|V) * P3(V) = 0.020 * 0.08
- P1(easily|ADV) * P3(ADV) = 0.1 * 0.07
For the second iteration, if the first tag is DET and the second tag is ADJ, the expression to be evaluated is:
P1(Markov|N) * max(P(Y|X)) * P1(a|DET) * P3(DET)
where max(P(Y|X)) is the maximum transition probability from DET to any other tag. In this case, it would be 0.55 (the transition probability from DET to N). So the expression becomes:
0.1 * 0.55 * 0.1 * 0.20
Similar Questions
Consider the following sentence: a Markov model tags easilyAssume that based on a HMM, we have the following probabilities: Emission:P1(a|DET) = 0.1, P1(easily|ADV) = 0.1, P1(Markov|N) = 0.1, P1(model|N) = 0.095, P1(model|V) =0.005, P1(tags|N) = 0.080, P1(tags|V) = 0.020,Transition probabilities: P(Y|X) Y DET N V ADJ ADV X DET 0 0.55 0 0.02 0.03 N 0.01 0.1 0.08 0.01 0.02 V 0.16 0.11 0.06 0.08 0.08 ADJ 0.01 0.65 0 0.05 0 ADV 0.08 0.02 0.09 0.04 0.04 Initial probabilitiesP3(DET) = 0.20, P3(N) = 0.06, P3(V) = 0.08, P3(ADV) = 0.07, P3(ADJ) = 0.02.What are the possible tag(s) of the sentence?Assume you want to use the Viterbi algorithm to decode the sentence, write down the expression to be evaluated at the initial step.Write down the expression for the second iteration if the first tag is DET, for a second tag of ADJ
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