2
$\begingroup$

In the tutorial, it is said that:

Let v range over the tags. Define U(k, v) to be the score of the best sequence of tags from 1 to k, where tag k is required to be v. This is a maximization over k −1 tags because tag number k is fixed to have value v

It is very confusing that why should the Kth tag be fixed?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ v is the STOP tag, and the sequence needs to end with a STOP tag. $\endgroup$
    – Franck Dernoncourt
    Commented Dec 28, 2016 at 18:04

1 Answer 1

Reset to default
0
$\begingroup$

I learned from here that the first and last tokens are all fixed.

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.