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His point is that the language of all inputs on which a Turing Machine holds after at most N steps is regular. So for any given Language L that a Turing Machin

by ablob 4y ago

His point is that the language of all inputs on which a Turing Machine holds after at most N steps is regular. So for any given Language L that a Turing Machine TM decides, there is a family of languages: L_n = { w | TM accepted the word w in at most n steps } where n is a natural number such that every L_n is a regular language. And as you already said, a regular language can be decided by a finite state machine. Edit: It was never stated that a Turing Machine could be simulated completely, but that one can choose an N that is large enough for a given need. If that N wasn't large enough for a given input, then so be it - the input is not accepted (or whatever you defined it to be).

Kranar 4y ago
>It was never stated that a Turing Machine could be simulated completely, but that one can choose an N that is large enough for a given need. What I read was that an FSM can be used to simulate Turing Machines that always halt which was then edited to claim that an FSM can simulate a Turing Machine that halts for a given input. Both those claims are false for reasons I explained. You have further revised the claim to state that for a given bound on the length of an input, there exists an FSM that can simulate a Turing Machine for that same input. That's technically true but that's no more insightful than just saying that FSMs are a proper subset of Turing Machines. For example you wouldn't say that the FSM corresponding to some regular expression, for example "a*b*", is a simulator for a Turing Machine that's a decider for strings that start with zero or more "a"s followed by zero or more "b"s. If you put a fixed bound on the length of the input then you can just take a notebook, list all inputs in lexicographical order along with whether that input maps to true or false. Since presumably the length of the input is bounded by some fixed constant N, then the number of inputs is also bounded by 2^N (assuming a binary alphabet). The confusion that really needs to be avoided is the idea that there's a subset of Turing Machines that can be simulated by FSMs on the basis of whether or not the TM halts, as though TMs that halt can be simulated by an FSM, and that the key differentiator between an FSM and a TM has to do with infinite loops. It's totally possible for a TM to halt on all inputs and yet there is no corresponding FSM.
- bdowling 4y ago
  Hi Kranar. See my other post with the reference to linear bounded automata. I think you're right on all points. Also, I agree that "simulates" is probably the wrong word. There would be a one-to-one mapping between FSM states and restricted-TM (or Linear Bounded Automaton) states and it would "simulate" the TM by stepping through mapped states until it reached a terminal state, but the number of states required and the transition table would be ridiculous. I don't think it would exactly be a look-up table, but it's close. (i.e., it would be O(2^N), for a binary alphabet, to construct the transition table, but looking up the terminal state for each input state would also be O(2^N)).

by ablob 4y ago

Kranar 4y ago

>It was never stated that a Turing Machine could be simulated completely, but that one can choose an N that is large enough for a given need. What I read was that an FSM can be used to simulate Turing Machines that always halt which was then edited to claim that an FSM can simulate a Turing Machine that halts for a given input. Both those claims are false for reasons I explained. You have further revised the claim to state that for a given bound on the length of an input, there exists an FSM that can simulate a Turing Machine for that same input. That's technically true but that's no more insightful than just saying that FSMs are a proper subset of Turing Machines. For example you wouldn't say that the FSM corresponding to some regular expression, for example "a*b*", is a simulator for a Turing Machine that's a decider for strings that start with zero or more "a"s followed by zero or more "b"s. If you put a fixed bound on the length of the input then you can just take a notebook, list all inputs in lexicographical order along with whether that input maps to true or false. Since presumably the length of the input is bounded by some fixed constant N, then the number of inputs is also bounded by 2^N (assuming a binary alphabet). The confusion that really needs to be avoided is the idea that there's a subset of Turing Machines that can be simulated by FSMs on the basis of whether or not the TM halts, as though TMs that halt can be simulated by an FSM, and that the key differentiator between an FSM and a TM has to do with infinite loops. It's totally possible for a TM to halt on all inputs and yet there is no corresponding FSM.

bdowling 4y ago
Hi Kranar. See my other post with the reference to linear bounded automata. I think you're right on all points. Also, I agree that "simulates" is probably the wrong word. There would be a one-to-one mapping between FSM states and restricted-TM (or Linear Bounded Automaton) states and it would "simulate" the TM by stepping through mapped states until it reached a terminal state, but the number of states required and the transition table would be ridiculous. I don't think it would exactly be a look-up table, but it's close. (i.e., it would be O(2^N), for a binary alphabet, to construct the transition table, but looking up the terminal state for each input state would also be O(2^N)).