0 votes 0 votes A Turing Machine goes on to an infinite loop on string $\epsilon$. Is it correct? And thus Any language containing $\epsilon$ is not REL. Is it correct? thor asked Jan 9, 2017 thor 537 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply mohit chawla commented Jan 9, 2017 reply Follow Share nope it's not! why because you never know on a given input string whether a turing machine can go to infinite loop or not. 1 votes 1 votes thor commented Jan 9, 2017 reply Follow Share Is {$\epsilon$} a REL? Is {$\epsilon$} a CSL? 0 votes 0 votes thor commented Jan 9, 2017 reply Follow Share Does a TM allow $\epsilon$ transitions? 0 votes 0 votes mohit chawla commented Jan 9, 2017 reply Follow Share is {epsilon} is REL is equivalent to saying whether there exist a TM for this or not! Yes you can design a Tm which can accept epsilon if epsilon is in it's lang.! epsilon transaction is like just start with a TM and without seeing anything move head to right side and then to left side so and go to accepting state. so it is REL. 0 votes 0 votes saurabh rai commented Jan 9, 2017 reply Follow Share https://gateoverflow.in/76419/decidability 0 votes 0 votes mohit chawla commented Jan 9, 2017 reply Follow Share @saurabh, the link that u have given shows whether Tm accepts epsilon, this problem is undecidable(not even semdidecidable) but i guess here what @thor is asking is that " Any language containing ϵ is not REL " so here lang conatining epsilon and the ques is not whether it is accept epsilon or not! pls correct me if i am wrong! 0 votes 0 votes saurabh rai commented Jan 9, 2017 reply Follow Share i think u r right 0 votes 0 votes Please log in or register to add a comment.