# Theory of Computation

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1 Theory of Computation Lecture #10 Sarmad Abbasi Virtual University Sarmad Abbasi (Virtual University) Theory of Computation 1 / 43

2 Lecture 10: Overview Linear Bounded Automata Acceptance Problem for LBAs Emptiness Problem for LBAs Computation Histories Emptiness Problem for CFGs An undecidable Problem for CFGs Post Correspondence Problem Sarmad Abbasi (Virtual University) Theory of Computation 2 / 43

3 Linear Bounded Automata An LBA L is a restricted type of TM. Its tape head is not permitted to move off the portion of the tape containing the input. Here is a diagram of an LBA: Sarmad Abbasi (Virtual University) Theory of Computation 3 / 43

4 Linear Bounded Automata As you can see the input portion of the tape is the only tape that the LBA is allowed to move its head on. Sarmad Abbasi (Virtual University) Theory of Computation 4 / 43

5 Acceptance Problem for LBAs Let us define the acceptance problem for LBAs. A lba = { M, w : M is a LBA that accepts w}. Is A lba decidable? Sarmad Abbasi (Virtual University) Theory of Computation 5 / 43

6 Acceptance Problem for LBAs Theorem Let M be an LBA with q states and g tape symbols. There are at most qng n distinct configurations of M for a tape of length n. Recall that a configuration is completely given by 1 The state of the machine. 2 The position of its head. 3 The contents of the tape. Sarmad Abbasi (Virtual University) Theory of Computation 6 / 43

7 Acceptance Problem for LBA In this case we have 1 q possibilities for the state. 2 n possibilities for the position of the head. 3 g possibilities for each tape cell. Since, there are n of them hence there are g n total possibilities. Sarmad Abbasi (Virtual University) Theory of Computation 7 / 43

8 Acceptance Problem for LBA Example: Suppose M has 7 states and 5 tape symbols. How many distinct configuration can we have for a tape of length 3. 1 The machine can be in any 7 states. 2 Its head can be scanning any tape cell. 3 possibilities. 3 1 Cell 1 can have any symbol (5 possibilities). 2 Cell 2 can have any symbol (5 possibilities). 3 Cell 2 can have any symbol (5 possibilities). So there are a total of distinct configurations. Sarmad Abbasi (Virtual University) Theory of Computation 8 / 43

9 Acceptance Problem for LBA Theorem Let M be an LBA with q states and g tape symbols. If M takes more then qng n steps on an input,x, of length n then the M does not halt on x. Note that in that case one of the configuration of the LBA has repeated. Hence, M is stuck in an endless loop. Sarmad Abbasi (Virtual University) Theory of Computation 9 / 43

10 Acceptance Problem for LBA Theorem A lba is decidable. We can simulate an LBA on its given input only for a finite amount of time to decide if it accepts the input or not. Sarmad Abbasi (Virtual University) Theory of Computation 10 / 43

11 Acceptance Problem for LBA Let L be a decider given as follows: 1 Input M, w where M is an LBA and w is a string. 2 Let n = w 3 Simulate M on w for qng n steps. 4 If M accepts accept. Else reject. Note that L is a decider. This is because if M does not accept w in qng n steps then it is in an endless loop and it will never accept w. Sarmad Abbasi (Virtual University) Theory of Computation 11 / 43

12 Acceptance Problems Summary A DFA = { D, w : D is a DFA that accepts w}. A DFA is decidable. Reason: we have to simulate D on w only for n steps. A REX = { R, w : R is a regular expression that generates w}. A REX is decidable. We convert the regular expression into a dfa and then use the decider for A REX to decide A REX. A CFG = { G, w : G is a context free grammar that generates w}. A CFG is decidable. Since we can convert G into Chomsky normal form and list all derivations of length 2n 1. A PDA = { P, w : P is Pda that accepts w}. This is decidable. Since, we can convert a PDA into a equivalent grammar and then use the decider for A PDA. Sarmad Abbasi (Virtual University) Theory of Computation 12 / 43

13 Emptiness Problem for LBA Let us define the emptiness problem for LBAs. A REX = { M : M is a LBA and L(M) = }. Is A REX undecidable? Sarmad Abbasi (Virtual University) Theory of Computation 13 / 43

14 Acceptance Problem for LBA We will show once again that: If A REX is decidable then A REX is decidable. Sarmad Abbasi (Virtual University) Theory of Computation 14 / 43

15 Acceptance Problem for LBA What would we like to do? Given an input M, w where M is a TM and w is a string. We want to construct a LBA B such that B is going to accept only accepting computation history of M on w. Now, if M accepts w then there is one computation history that B will accept. On the other hand if M does not accept w then L(B) =. Sarmad Abbasi (Virtual University) Theory of Computation 15 / 43

16 Computation Histories Let M be a TM. We have the following concepts. 1 A state. This just tells us what the current state of the machine is. 2 A configuration. This tells us what the state is and all the tape contents. So, given a configuration we can figure out what will be the next configuration. A computation history is a sequence of configurations. A computation history on an input w will consist of configurations C 1, C 2,..., C l such that 1 C 1 is the initial configuration of M on w. 2 C i yields C i+1. Sarmad Abbasi (Virtual University) Theory of Computation 16 / 43

17 Computation Histories A computation history of M on w is accepting if the last configuration in the history is an accepting configuration. Note that a accepting computation history of M on w is a proof that M accepts w. We will encode a computation history as follows: C 1 #C 2 # #C l. Sarmad Abbasi (Virtual University) Theory of Computation 17 / 43

18 Computation Histories Given a TM M and w we can construct an LBA B such that B accepts the accepting computation history of M on w. We give an informal description of B. Sarmad Abbasi (Virtual University) Theory of Computation 18 / 43

19 Computation Histories 1 On input C 1 #... #C l. 2 Check if C 1 = q 0 w. 3 Check if C l has an accept state. 4 Check C i yields C i+1. 1 The transition function of M is hard coded in B. 2 B zigzags between C i and C i+1 to make sure if C i yields C i+1 Note that B never uses any extra memory. It only uses the portion of the input that is given to B. Sarmad Abbasi (Virtual University) Theory of Computation 19 / 43

20 Computation Histories B is an LBA. 1 Note that the input to B is not w but a computation history H of M. 2 In order to check if H is an computation history history of M on w it does not need more tape then the portion on which H is written. Sarmad Abbasi (Virtual University) Theory of Computation 20 / 43

21 Computation Histories Theorem Let M and w be given and B be constructed as earlier. The { =, M does not accept w; L(B), M accepts w. Sarmad Abbasi (Virtual University) Theory of Computation 21 / 43

22 Computation Histories Theorem E LBA is undecidable. Suppose R decides E LBA then consider S 1 On input M, w. 2 Construct and LBA B as described earlier. 3 Run R on B. If R accepts reject. If R rejects accepts. Note that S is a decider for A TM. A contradiction. Sarmad Abbasi (Virtual University) Theory of Computation 22 / 43

23 Context Free Grammars Let us define the emptiness problem for CFGs. E cfg = { G : G is a CFGL(G) = }. Is E CFG decidable? Sarmad Abbasi (Virtual University) Theory of Computation 23 / 43

24 Emptiness Problem for CFGs 1 On input G. 2 Mark all terminals of G. 3 While no new symbols are marked. 4 If A X 1... X r is a production with all symbols marked on the right hand side. Mark A. 5 If the start symbol is marked then reject else accept. Sarmad Abbasi (Virtual University) Theory of Computation 24 / 43

25 Context Free Grammars Let us define the another problem for CFGs. ALL CFG = { G : G is a CFGL(G) = Σ }. Is ALL CFG decidable? Sarmad Abbasi (Virtual University) Theory of Computation 25 / 43

26 ALL CFG On the surface the problem looks difficult. We have to find at least one string that is not generated by the grammar G. However there are infinitely many potential strings. But our goal is to PROVE that it is undecidable. Theorem ALL CFG is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 26 / 43

27 Computation Histories Once again we will show that If ALL CFG is decidable then A TM is decidable. Sarmad Abbasi (Virtual University) Theory of Computation 27 / 43

28 Computation Histories Given a TM M and w we can construct an PDA D such that D rejects the accepting configuration history of M on w (appropriately encoded) and accepts all other strings. We give an informal description of D. Sarmad Abbasi (Virtual University) Theory of Computation 28 / 43

29 Computation Histories We will use a trick here. Let C 1,..., C l be a computation history. We will encode it as follows: C 1 #C r 2 #C 3#C r 4 #C l We will see in a minute why we are doing this. Sarmad Abbasi (Virtual University) Theory of Computation 29 / 43

30 Computation Histories Now consider the following PDA D. The PDA will take a computation history H = C 1,..., C n and do the following. 1 If C 1 is not the initial configuration of M on w accept. 2 If C l is not a final configuration then accept. 3 If there is an i such that C i does not yield C i+1 then accepts. Note that D is a non-deterministic PDA and therefore it can guess which one of these conditions is violated. Sarmad Abbasi (Virtual University) Theory of Computation 30 / 43

31 Computation Histories Since w is hard coded in D it can check if the first configuration is not the initial configuration. Similarly, it is easy to check if the last configuration is not an accepting configuration. How does D check if C i does not yield C i+1? Sarmad Abbasi (Virtual University) Theory of Computation 31 / 43

32 Computation Histories This is why we said we will use an special encoding. We had C i #C r i+1 Thus D can push C i on its stack and pop it and check if C i yields C i+1 or not. Sarmad Abbasi (Virtual University) Theory of Computation 32 / 43

33 Computation Histories Theorem Given M and w we can construct a PDA D such that D rejects the accepting computation of M on w (properly encoded). It accepts all other strings. Sarmad Abbasi (Virtual University) Theory of Computation 33 / 43

34 Computation Histories Theorem Let M and w be given and D be constructed as earlier. Let G be a grammar which generates all the strings accepted by G. Then { = Σ L(G), M does not accept w; Σ, M accepts w. Sarmad Abbasi (Virtual University) Theory of Computation 34 / 43

35 Computation Histories Theorem ALL CFG is undecidable. If ALL CFG is decidable then there is a decider R such that L(R) = ALL CFG. Suppose R decides ALL CFG then consider S 1 On input M, w. 2 Construct and LBA D as described earlier. 3 Convert D into an equivalent CFG G. 4 Run R on G. If R accepts reject. If R rejects accepts. Note that S is a decider for A TM. A contradiction. Sarmad Abbasi (Virtual University) Theory of Computation 35 / 43

36 Post Correspondence Problem Post Correspondence Problem (PCP) is a puzzle. We are given a collection of dominos each containing two strings, on on each side. A domino looks like this: Sarmad Abbasi (Virtual University) Theory of Computation 36 / 43

37 Post Correspondence Problem This domino has a on top and ab on the bottom. A collection of dominos looks like this. 1 The first domino has b on top and ca on the bottom. 2 The second domino has a on top and ab on the bottom. 3 The third domino has ca on top and a on the bottom. 4 The fourth domino has abc on top and c on the bottom. Sarmad Abbasi (Virtual University) Theory of Computation 37 / 43

38 Post Correspondence Problem We make a list of dominos by placing the next to each other. We are allowed repetitions. This list is called a match if reading of the top symbols gives us the same string as reading of the bottom symbols. Sarmad Abbasi (Virtual University) Theory of Computation 38 / 43

39 Post Correspondence Problem For example. Here we have a match. Reading of the top symbols gives us abcaaabc and same is the case if we read the bottom. Sarmad Abbasi (Virtual University) Theory of Computation 39 / 43

40 Post Correspondence Problem Lets look at another collection. This collection has three dominos. Sarmad Abbasi (Virtual University) Theory of Computation 40 / 43

41 Post Correspondence Problem 1 The first domino has abc on top and ab on the bottom. 2 The first domino has ca on top and a on the bottom. 3 The first domino has acc on top and ba on the bottom. For this collection it is not possible to find a match. Why? The top of each domino has more letter than the bottom. So in any list the top string will be longer than the bottom string. Sarmad Abbasi (Virtual University) Theory of Computation 41 / 43

42 Post Correspondence Problem Given a collection P of Dominos. Does P have a match? PCP = { P : P has a match }. Sarmad Abbasi (Virtual University) Theory of Computation 42 / 43

43 Post Correspondence Problem Theorem PCP is undecidable. Sarmad Abbasi (Virtual University) Theory of Computation 43 / 43

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