0 votes 0 votes Find minimal states in DFA accepting {w1aw2, |w1| = 2, |w2| >=3, w1,w2 $\in$ $(a+b)^*$}? Is it $7$ or $8$? thor asked Feb 11, 2017 thor 627 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes There will a total of 8 states 1 for start 2 for w1 language ie 1st 2 elements 1 for 'a' 3 for w2 language ie next 3 elements min 1 for dead or trap stage Total = 8 states bhargav9873 answered Feb 12, 2017 selected Feb 12, 2017 by Tendua bhargav9873 comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes 8 is the answer . it is dfa there must be a trap state yankur9 answered Feb 12, 2017 yankur9 comment Share Follow See 1 comment See all 1 1 comment reply amitnitk commented Feb 12, 2017 reply Follow Share Yes 8 is the answer including dead state 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes Here is the dfa. In dfa we need to define the transition for every alphabet at every state otherwise it will not be a dfa. Hemant Parihar answered Feb 12, 2017 Hemant Parihar comment Share Follow See all 0 reply Please log in or register to add a comment.