1 Linear Bounded Automata LBAs

2 Linear Bounded Automata (LBAs) are the same as Turing Machines with one difference: The input string tape space is the only tape space allowed to use

3 Left-end marker Input string Right-end marker Working space in tape All computation is done between end markers Linear Bounded Automaton (LBA)

4 We define LBA’s as NonDeterministic Open Problem: NonDeterministic LBA’s have same power with Deterministic LBA’s ?

5 Example languages accepted by LBAs: LBA’s have more power than NPDA’s LBA’s have also less power than Turing Machines

6 The Chomsky Hierarchy

7 Unrestricted Grammars: Productions String of variables and terminals String of variables and terminals

8 Example unrestricted grammar:

9 A language is recursively enumerable if and only if is generated by an unrestricted grammar Theorem:

10 Context-Sensitive Grammars: and: Productions String of variables and terminals String of variables and terminals

11 The language is context-sensitive:

12 A language is context sensistive if and only if is accepted by a Linear-Bounded automaton Theorem: There is a language which is context-sensitive but not recursive Observation:

13 Non-recursively enumerable Recursively-enumerable Recursive Context-sensitive Context-free Regular The Chomsky Hierarchy

14 Decidability

15 Consider problems with answer YES or NO Examples: Does Machine have three states ? Is string a binary number? Does DFA accept any input?

16 A problem is decidable if some Turing machine decides (solves) the problem Decidable problems: Does Machine have three states ? Is string a binary number? Does DFA accept any input?

17 Turing Machine Input problem instance YES NO The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem

18 The machine that decides (solves) a problem: If the answer is YES then halts in a yes state If the answer is NO then halts in a no state These states may not be final states

19 YES states NO states Turing Machine that decides a problem YES and NO states are halting states

20 Difference between Recursive Languages and Decidable problems The YES states may not be final states For decidable problems:

21 Some problems are undecidable: which means: there is no Turing Machine that solves all instances of the problem A simple undecidable problem: The membership problem

22 The Membership Problem Input:Turing Machine String Question: Does accept ?

23 Theorem: The membership problem is undecidable Proof:Assume for contradiction that the membership problem is decidable (there are and for which we cannot decide whether )

24 Thus, there exists a Turing Machine that solves the membership problem YES accepts NO rejects

25 Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input

26 accepts ? NO YES accept Turing Machine that accepts and halts on any input reject

27 Therefore,is recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Since is chosen arbitrarily, every recursively enumerable language is also recursive

28 Therefore, the membership problem is undecidable END OF PROOF

29 Another famous undecidable problem: The halting problem

30 The Halting Problem Input:Turing Machine String Question: Does halt on input ?

31 Theorem: The halting problem is undecidable Proof:Assume for contradiction that the halting problem is decidable

32 There exists Turing Machine that solves the halting problem YEShalts on doesn’t halt on NO

33 Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input

34 halts on ? YES NO Run with input reject accept reject Turing Machine that accepts and halts on any input Halts on final state Halts on non-final state

35 Thereforeis recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Since is chosen arbitrarily, every recursively enumerable language is also recursive

36 Therefore, the halting problem is undecidable END OF PROOF