Turing Machine Variants Chuck Cusack B ased very

Turing Machine Variants Chuck Cusack B ased (very heavily) on notes by Edward O kie (Radford University) and “Introduction to the Theory of Computation”, 3 rd edition, Michael Sipser. (Original version at http: //www. radford. edu/~nokie/classes/420/)

Equivalence of models �Two variants of a TM are equivalent if each one can be simulated on the other. �Recall that a language is Turing-recognizable if some single-tape Turing machine recognizes it. �Thus, if a TM variant is equivalent to a single-tape TM, then a language is Turing-recognizable if and only if some Turing machine of that type recognizes it.

Main Results � Theorem 3. 13: Every multitape Turing machine has an equivalent single-tape Turing machine. � Theorem 3. 16: Every nondeterministic Turing machine has an equivalent deterministic Turing machine. � Theorem 3. 21: A language is Turing-Recognizable iff some enumerator enumerates it. (We won’t focus on this one)

Stay-Put TMs �One of the simplest variations is a TM that allows the head to stay put as well as more left or right �Notice that a stay-put TM can simulate a regular TM �In fact, you can view a regular TM as a stay-put TM that just happens to not use the stay-put feature �You can simulate a stay-put TM on a TM by replacing every transition that stays with two transitions: � Write the symbol as normal and go right � Read and write the same symbol and go left �Thus, these two variants are equivalent

Multitape Turing Machine �Multiple tapes �Transition function � Reads from all heads � Writes to all heads � Each head goes left, right or stays �We can simulate it with a single-tape TM as illustrated to the right. �Let’s look a little closer. Image from Sipser’s book

Proof Idea for Theorem 3. 13 � Theorem 3. 13: Every multitape Turing machine has an equivalent single-tape Turing machine. � Simulate an n-tape machine using a one tape machine: � Store information from all tapes on the single tape. � Use new symbol (e. g. #) to separate information from each tape � Use a dotted symbol to represent the location of each tape head � When the tape head moves to a location, replace the symbol there (e. g. symbol x) with the equivalent dotted symbol (i. e. with dotted x) � When the tape head moves from a location, write a regular (i. e. nondotted) symbol at that location � Requires adding dotted symbols to tape alphabet � Develop new transition function: � A step of the n-tape machine will represent steps on all n of the parts of the single tape � A right movement onto the # results in shifting all of the symbols that are to the right of the # one symbol to the right

Equivalence of Multi- and single-tape TMs �Corollary 3. 15: A language is Turing-recognizable if and only if some multitape Turing machine recognizes it. �Proof: � �If L is Turing-recognizable, then some single-tape TM recognizes it. �A single-tape TM is a multitape TM (with just one tape), so some multitape Turing machine recognizes it. � �If some multitape Turing machine recognizes L, Theorem 3. 13 implies that some single-tape Turing machine recognizes L. �Therefore L is Turing recognizable (by definition).

Nondeterministic Turing Machine �Transition function: δ: Q x Γ → P(Q x Γ x {L, R}) � As expected, multiple (0 r no) transitions allowed �Machine N accepts w if some branch accepts w �Theorem 3. 16: Every nondeterministic Turing machine has an equivalent deterministic Turing machine. �Let’s sketch a proof

Theorem 3. 16: How to Prove? � How did we prove DFA and NFA were equivalent? � Defined an algorithm for creating a DFA that is equivalent to an given NFA � The states of the DFA were power sets of the NFA states � The DFA accepted the same strings as the NFA � How to prove that a Nondeterministic TM has an equivalent deterministic TM? � Create a deterministic TM that executes the possible computations on the nondeterministic TM � Basic Approach: � Specify possible nondeterministic choices made by the NTM � Continue trying more and more choices � If the NTM accepts, the DTM will find the accepting computation � If the NTM rejects, the DTM continues forever, looking for an accepting computation � If the NTM loops, the DTM will continuing trying computations forever

Representing Nondeterministic Choices �Let N be our NTM �Let b be the largest number of choices from N’s transition function. �Let Σb = {1, 2, . . . , b} �Represent a sequence of n choices with a string from Σb of length n �Example: 534 represents taking choice 5, then choice 3, then choice 4

Simulating a Nondeterministic Machine � Simulate N with a 3 -tape deterministic machine D as follows: � Tape 1 of D is initialized with a copy of the input tape for N and never changes after initialization. � Tape 2 of D has a copy of what N's tape would be at that point of the computation that is currently specified by tape 3 � Tape 3 of D successively contains a sequence of numbers that represent every possible path through N. � Each path represents (part of) a computation in N � Each number represents a choice at a nondeterministic branch in the computation � Tape 3 successively contains the numbers in lexicographic order (i. e. in alphabetic order, shortest first) � Example: if b=4 (i. e. 4 is the maximum number of choices), then tape 3 would hold 1, 2, 3, 4, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44, 111, . . . , 114, 121, 122, 123, 124, . . . , 443, 444, 1111, . . .

Simulating N on D � Initialize Tape 1. � Initialize Tape 3 with the number for the first possible computation N � Loop � Copy Tape 1 to Tape 2 � Execute N using Tape 2 as input and following the computation choices specified in Tape 3 � If sequences of choices represent an accepting computation for N (i. e. N enters an Accept state), then D halts and accepts � Else (i. e. a number represents an infeasible or rejecting computation) continue with the next step � Initialize Tape 3 with the string that represents the next possible sequence of choices of N � Repeat loop

N and D are Equivalent �N and D accept the same language � That is, N accepts L iff D accepts L because �If N accepts w, then D will accept w (eventually) �If N rejects or loops on w, then D will loop (as it continues trying more and more choices of the transition function) �Can we avoid looping in D? � If N always accepts or rejects every string, then D can be modified to accept or reject every string � If N loops on some strings, then D will loop on the strings on which N rejects or loops

Simulation order �Why is the order we simulate the NTM important? �Here we are essentially doing a breadth-first search of the tree of possible choices the NTM can make �What if we did a depth-first search instead?