2 2 votes Consider the following languages : L1 : {< M, q >|M is a Turing Machine that visits state q on some input within 15 steps}. L2 : {< M >|M is a Turing Machine, |M|< 200 where |M|is number of states in machine}. Which of the following is correct ? L1 is decidable but L2 is not decidable L2 is decidable but L1 is not decidable Both L1 and L2 is decidable Neither L1 nor L2 is decidable Theory of Computation + – Anirban Biswas 2.1k views answer comment Share Follow Print See 1 comment 1 1 comment reply Gate Mission 1 commented Dec 31, 2016 reply Follow flag Both decidable ? 0 0 replyShare Please log in or register to add a comment.
Best answer 6 6 votes L1: Since steps are finite, configurations are also finite. Within finite configurations we can know whether that particular state has been encountered or not. Hence decidable. L2: Number of states can be known just looking at encoding of inputted TM by/in UTM. Hence decidable. Therefore, both decidable. Aghori answered Dec 31, 2016 • selected Jan 9, 2017 by vijaycs Aghori comment Share Follow See all 5 Comments 5 5 Comments reply Show 2 previous comments Aghori commented Jan 4, 2017 reply Follow flag Yeah. Whatever the length be, we need to check till bounded steps only. 1 1 replyShare ST commented Jan 6, 2017 i edited by ST Jan 9, 2017 reply Follow flag L1 is partially decidable. 2 2 replyShare Nitinkumar.097 commented Nov 23, 2020 reply Follow flag Regarding L1, why we didn't used Rice theorem and say it is undecidable. Why This approach is giving the wrong answer? 0 0 replyShare Please log in or register to add a comment.