Remaining Solutions Previous Homework Decidable languages are closed

Remaining Solutions Previous Homework • Decidable languages are closed under concatenation – Why the idea of “bridging the transitions” doesn’t work – Proof by splitting input in parts • Why this proof does not work to prove that Turingrecognizable languages are closed under concatenation – How to “repair” the proof • {an bn cn: n = 1, 2 , …} is decidable • K-Stack pushdown automata

Multi-tape Turing Machines Control head 1 a 1 a 2 … Tape 1 head 2 a 1 a 2 … … Tape 2 We add a fixed number of tapes

Multi-tape Turing Machines: Transitions NEW SLIDE Transactions in a multi-tape Turing machine allows the head to stay put where they are. So the general for a 2 -Tape Turing machine are: Q ( ) ({L, R, S} {L, R, S}) So for example the transition: ((q, (a, b)), (q’, (a, a), (L, S))) Is moving the head of the first tape to the left while keeping the head of the second tape where it was

Multi-tape Turing Machines vs Turing Machines M 2 a 1 a 2 … ai … b 1 b 2 … bj … We can simulate a 2 -tape Turing machine M 2 in a Turing machine M: • we can represent the contents of the 2 tapes in the single tape by using special symbols • We can simulate one transition from the M 2 by constructing multiple transitions on M • We introduce several (but fixed) new states into M

Using States to “Remember” Information Configuration in a 2 -tape Turing Machine M 2: Tape 1 a b State: s Tape 2 b b a State in the Turing machine M: “s+b+1+a+2” Which represents: • M 2 is in state s • Cell pointed by first header in M 2 contains b • Cell pointed by second header in M 2 contains an a

Using States to “Remember” Information (2) State in the Turing machine M: “s+b+1+a+2” How many states are there in M? (# states in M 2) * | or or S| * | or or S| Yes, we need large number of states for M but it is finite!

Configuration in a 2 -tape Turing Machine M 2: Tape 1 a b State in M 2: s Tape 2 b b a Equivalent configuration in a Turing Machine M: 1 a b 2 3 b b a 4 e State in M: s+b+1+a+2

Simulating M 2 with M • The alphabet of the Turing machine M extends the alphabet 2 from the M 2 by adding the separator symbols: 1, 2, 3 , 4 and e, and adding the mark symbols: and • We introduce more states for M, one for each 5 -tuple p+ +1+ +2 where p in an state in M 2 and +1+ +2 indicates that the head of the first tape points to and the second one to • We also need states of the form p+ +1+ +2 for control purposes

Simulating transitions in M 2 with M 1 a b 2 3 b b a 4 e State in M: s+b+1+a+2 • At the beginning of each iteration of M 2, the head starts at e and both M and M 2 are in an state s • We traverse the whole tape do determine the state p+ +1+ +2, Thus, the transition in M 2 that is applicable must have the form: ((p, ( , )), (q, ( 1, 2) , ( , S))) in M 2 ((p+ +1+ +2, e), (q+ 1+ +S+2, e, ) in M

Simulating transitions in M 2 with M (2) • To check if the transformation (q, ( , ), …) is applicable, we go forwards from the first cell. • If transformation is (or ) we move the marker to the right (left): i i 1 … • To overwrite characters, M must first determine the correct position

state: s 1 a b a b 2 3 b b a 4 e a b 2 1 3 b b a 4 e b a b 2 1 a 3 b b a 4 e a b 2 1 a b 3 b b a 4 e Output: state: s+b+1 1 a b 2 3 b b a 4 e

Multi-tape Turing Machines vs Turing Machines (final) • We conclude that 2 -tape Turing machines can be simulated by Turing machines. Thus, they don’t add computational power! • Using a similar construction we can show that 3 -tape Turing machines can be simulated by 2 -tape Turing machines (and thus, by Turing machines). • Thus, k-tape Turing machines can be simulated by Turing machines

Implications • If we show that a function can be computed by a k-tape Turing machine, then the function is Turing-computable • In particular, if a language can be decided by a k-tape Turing machine, then the language is decidable Example: Since we constructed a 2 -tape TM that decides L = {anbn : n = 0, 1, 2, …}, then L is Turing-computable.

Summary of Previous Class • There are languages that are not decidable – (we have not proved this yet) • Why not extend Turing machines just as we did with finite automata and pushdown automata? • First Try: multi-tape Turing machines – More convenient than Turing machines – But multi-tape Turing machines can be simulated with Turing machines – Therefore, anything we can do with a multi-tape Turing machines can be done with Turing machines • On the bright side we can use this to our advantage! • Second Try: k-stack Pushdown automata – But k-stack Pushdown Automata are equivalent to Turing machines for k 2

Third Try: Multi-Head Turing Machines Control head 1 a 1 a 2 … headn Tape • We add a fixed number of heads for the same tape • As with Multi-tape Turing machines, heads are allowed to stay put in the transitions update

Configuration in a 2 -header Turing Machine: State: s Tape Can be represented in a 3 -tape Turing Machine a b a b And use states to remember the pointers: s+b+1+b+2

Configuration in a 2 -header Turing Machine: State: s Tape a b 1 a b 2 3 e s+b+1+b+2 Or could be represented directly in a Turing Machine

Fourth Try: Nondeterministic Turing Machines Reminder: When a word is accepted by a nondeterministic automata? Given an NFA, a string w *, is accepted by A if at least one of the configurations yielded by (s, w) is a configuration of the form (f, e) with f a favorable state (s, w) (p, w’) … (q, w’’) with w’’ e (s, w) (p, w’) … (q, e) with q F … (s, w) (p, w’) … (f, e) … We will do something similar for nondeterministic TMs

Deciding, Recognizing Languages with Nondeterministic Turing machines Definition. A nondetermistic Turing machine NM decides a language L (L is said to be decidable) if: 1. If w L then: Ø at least one possible computation terminates in an acceptable configuration Ø No computation ends in a rejecting state 2. If w L then: Ø No computation ends in an acceptable state Ø At least one possible computation terminates in a rejecting configuration Definition. A Turing machine recognizes a language L if it meets condition 1 above (L is said to be Turing-recognizable)

Example of Nondeterministic TM Find a nondeterministic TM recognizing the language a*ab*b Solution: ML: a C 0 = q 0 aab b a p b q accept qaccept apab aapb ? aaqb aabqaccept� aabq� accept ?

Nondeterministic TMs can be Simulated by TMs Theorem. If a nondeterministic Turing machine, NM, recognizes a language L, then there is a Turing machine, M, recognizing the language L Theorem. If a nondeterministic Turing machine, NM, decides a language L, then there is a Turing machine, M, deciding the language L

Nondeterministic TMs can be Simulated by TMs Idea: NM: C 11 C 22 M: C 0 Let w = h. RS C 0 = q 0 h. RS … C 12 C 13 C 1 n … … C 11 C 12 C 13 … C 1 n • Simulate all computations in NM by computing them doing breadth-first order • If w L one of the Cij is an accepting configuration: (h, …) which will also be found by M • Similar argument can be made for w L C 21 C 22 … …

Context-Free Languages are Decidable • Let L be a context-free language • Let A be a pushdown automata accepting L • We simulate the pushdown automata A using a 2 -tape nondeterministic Turing machine, M. The second tape is used for simulating the stack • The header of the first tape points to the next character to be processed and the header of the second tape points to the last element (top of the stack)

Context-Free Languages are Decidable For each transition ((q, , ), (q’, )) in A, we construct several transitions in M doing the following steps: 1. Check if the next character is is (check if head of Tape 1 points to , and move the header to the right) 2. Check if the top of the stack is (check if head of Tape 2 points to , write a blank, and move the header to the left) 3. If 1 and 2 hold, “push” on tape 2 (move header to the right and write )

Possible Extensions of Turing Machines Physical Computational Adding Pushdown Non determinism Adding Matrix several automata several tapes with 2 or “no one can be told what the heads more stacks Matrix is, you have to see it for your self” None of these extensions add computational power to the Turing Machines

Homework for Friday 1. Please provide a secret nickname • So I can post all grades on web site. 2. Proof that the function: z = x + y is decidable. Assume that x, y and z are binary numbers. Hint: Use a 3 -tape Turing-Machine as follows: Tape 1: x Tape 2: y Tape 3: z 3. Suppose that a language L is enumerated by an enumerator Turing machine (we say that L is Turing-enumerable). • Prove that L must be Turing-recognizable • Can we prove that L is decidable? Provide the proof or explain why not 4. Problem 3. 14 (note that there are 2 parts)