0 votes 0 votes L1={w|starts with a and ends with b} L2={w|starts with b and ends with a} then L1 ⋂ L2 will be: 1. Finite 2.CFL but not regular 3.none of the above. vishal singh asked Feb 11, 2017 vishal singh 775 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes I tried in this way. All strings in L1 start with a while strings in L2 start with b. So their intersection will be EMPTY. Answer,= Finite :) sh!va answered Feb 11, 2017 selected Feb 11, 2017 by Arjun sh!va comment Share Follow See 1 comment See all 1 1 comment reply Arjun commented Feb 11, 2017 reply Follow Share That should be the only way :O How was your GATE? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes L1 is regular so finite L2 is also regular so finite regular lang are closed under intersection so finite Sankha Narayan Bose answered Feb 11, 2017 Sankha Narayan Bose comment Share Follow See all 2 Comments See all 2 2 Comments reply Arjun commented Feb 11, 2017 reply Follow Share Are you saying all regular languages are finite? 0 votes 0 votes Debasmita Bhoumik commented Feb 11, 2017 reply Follow Share @sankha All finite languages are regular. It is not true that all regular languages are finite. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes L1 always starts with 'a' and L2 always starts with 'b' So the intersection will be € ie empty string € is a finite set So option 1. Finite set is correct bhargav9873 answered Feb 12, 2017 edited Feb 12, 2017 by bhargav9873 bhargav9873 comment Share Follow See all 4 Comments See all 4 4 Comments reply Arjun commented Feb 12, 2017 reply Follow Share € is never used to denote empty set. It represents empty string. 0 votes 0 votes bhargav9873 commented Feb 12, 2017 reply Follow Share Yup... My mistake It's an empty string not empty But the logic for the answer still remains the same Btw thanks arjun for pointing out my wrong use of words 0 votes 0 votes Arjun commented Feb 12, 2017 reply Follow Share nopes, it is still not correct. It becomes correct only if both $L_1$ and $L_2$ also have empty string in them- which is not the case here. So, the intersection is empty set and not a set containing empty string. 0 votes 0 votes bhargav9873 commented Feb 12, 2017 reply Follow Share So is empty set finite?? 0 votes 0 votes Please log in or register to add a comment.