Turing Machines Variants

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# Turing Machines Variants - PowerPoint PPT Presentation

Turing Machines Variants. Zeph Grunschlag. Announcement. Midterms not graded yet Will get them back Tuesday. Agenda. Turing Machine Variants Non-deterministic TM’s Multi-Tape. Input-Output Turing Machines.

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### Turing Machines Variants

Zeph Grunschlag

Announcement
• Will get them back Tuesday
Agenda
• Turing Machine Variants
• Non-deterministic TM’s
• Multi-Tape
Input-Output Turing Machines

Input/output (or IO or transducing) Turing Machines, differ from TM recognizers in that they have a neutral halt state qhalt instead of the accept and reject halt states. The TM is then viewed as a string-function which takes initial tape contents u to whatever the non blank portion of the tape is when reaching qhalt . If v is the tape content upon halting, the notation fM (u) = v is used.

If M crashes during the computation, or enters into an infinite loop, M is said to be undefined on u.

Input-Output Turing Machines

When fM crashes or goes into an infinite loop for some contents, fM is a partial function

If M always halts properly for any possible input, its function f is total (i.e. always defined).

TM Notations

There are three ways that Sipser uses to describe TM algorithms.

• High level –pseudocode which explains how algorithm works without the technical snafoos of TM notation
• Implementation level –describe how the TM operates on its tape. No need to mention states explicitly.
• Low-level description. One of:
• Set of complete “goto” style instructions
• State diagram
• Formal description: spell out the 7-tuple
High-Level TMExample

Let's for example describe a Turing Machine M which multiplies numbers by 2 in unary:

M = "On input w = 1n

For each character c in w

copy c onto the next available b blank space"

Implementation-Level TMExample

The idea is to carry out the high level description by copying each character to the end. We also need to keep track of which characters have already been copied (or were copies themselves) by distinguishing these characters. One way is to use a different character, say X.

EG: Let’s see how 11111 is transformed.

Implementation-Level TMExample

So round by round, tape transformed as follows:

• 11111
• X1111X
• XX111XX
• XXX11XXX
• XXXX1XXXX
• XXXXXXXXXX
• 1111111111
Implementation-Level TMExample

Implementation level describes what algorithm actually looks like on the Turing machine in a way that can be easily turned into a state-diagram

Some useful subroutines:

• fast forward
• move to the right while the given condition holds
• rewind
• move to the left while the given condition holds
• May need to add extra functionality
• Add \$ if need to tell when end of tape is
Implementation-Level TMExample

M = "On input w = 1n

• HALT if no input
• Write \$ in left-most position
• Sweep right and write X in first blank
• Sweep left through X-streak and 1-streak
• Go right
• If read X, go right and goto 9. Else, replace 1 by X, move right.
• If read X [[finished original w]] goto 8 Else, goto 3
• Sweep to the right until reach blank, replace by X
• Sweep left replacing everything non-blank by 1
• HALT
Implementation-Level TMExample

At the low level the Turing Machine is completely described, usually using a state diagram:

9

\$R

1L

XL

1\$,R

X,L

1L

0

2

3

4

\$|XR

1|XR

1R

1X,R

X1,L

6

5

R

X1,L

XR

\$1,L

1,L

XR

9

8

halt

Non-Deterministic TM’s

A non-Deterministic Turing Machine N allows more than one possible action per given state-tape symbol pair.

A string w is accepted by N if after being put on the tape and letting N run, N eventually enters qacc onsome computation branch.

If, on the other hand, given any branch, N eventually enters qrej or crashes or enters an infinite loop on, w is not accepted.

Symbolically as before:

L(N) = { x  S* |  accepting config. y, q0 x *y }

(No change needed as  need not be function)

Non-Deterministic TM’sRecognizers vs. Deciders

N is always called a non-deterministicrecognizer and is said to recognizeL(N); furthermore, if in addition for all inputs and all computation branches, N always halts, then N is called a non-deterministic decider and is said to decide L(N).

Non-Deterministic TMExample

Consider the non-deterministic method:

void nonDeterministicCrossOut(char c)

while()

else

cross out, go right, return

OR go right // even when reading c

OR go left // even when reading c

Non-Deterministic TMExample

Using randomCross() put together a non-deterministic program:

1. while(some character not crossed out)

nonDeterministicCrossOut(‘0’)

nonDeterministicCrossOut(‘1’)

nonDeterministicCrossOut(‘2’)

2. ACCEPT

Q: What language does this non-deterministic program recognize ?

Non-Deterministic TMExample

A: {x  {0,1,2}* | x has the same no.

of 0’s as 1’s as 2’s }

Q: Suppose q is the state of the TM while running inside nonDeterministicCrossOut(‘1’) and q’ is the state of the TM inside nonDeterministicCrossOut(‘2’).

Suppose that current configuration is

u = 0XX1Xq12X2

For which v do we have u v ?

Non-Deterministic TMExample

A: 0XX1Xq12X2 

0XX1qX12X2 | 0XX1X1q2X2 | 0XX1XXq’ 2X2

These define 3 branches of computation tree:

Q: Is this a non-deterministic TM decider?

0XX1Xq12X2

0XX1qX12X2

0XX1X1q2X2

0XX1XXq’ 2X2

Non-Deterministic TMExample

A: No. This is a TM recognizer, but not a decider. nonDeterministicCrossOut() often enters an infinite branch of computation since can see-saw from right to left to right, etc. ad infinitum without ever crossing out anything. I.e., computation tree is infinite!

Note: If you draw out state-diagrams, you will see that the NTM is more compact, than TM version so there are some advantages to non-determinism! Later, will encounter examples of “efficient” nondeterministic programs for practically important problems, with no known efficient counterpart: The P vs. NP Problem.

NTM’sKonig’s Infinity Lemma

For Problem 3.3 in Sipser the following fact is important:

If a NTM is a decider then given any input, there is a number h such that all computation branches involve at most h basic steps. I.e., computation tree has height h. Follows from:

Konig’s Infinity Lemma: An infinite tree with finite branching at each node must contain an infinitely long path from the root.

Or really, the contrapositive is used: A tree with no infinite paths, and with finite branching must itself be finite.

Konig’s Infinity LemmaProof Idea

Idea is to “smell-out” where the infinite part of the tree is and go in that direction:

Konig’s Infinity LemmaProof Idea

Idea is to “smell-out” where the infinite part of the tree is and go in that direction:

Konig’s Infinity LemmaProof Idea

Idea is to “smell-out” where the infinite part of the tree is and go in that direction:

Konig’s Infinity LemmaProof Idea

Idea is to “smell-out” where the infinite part of the tree is and go in that direction:

Konig’s Infinity LemmaProof Idea

Idea is to “smell-out” where the infinite part of the tree is and go in that direction:

Konig’s Infinity LemmaProof

Proof. Given an infinite tree with finite branching construct an infinite path inductively:

Vertex v0 : Take the root.

Edge vn  vn+1 : Suppose v0v1…vn-1vn has been constructed and that the subtree from vn is infinite. Then one of vn’s finite no. of children, call it vn+1, must have an infinite subtree, so add the edge vn  vn+1. 

Multi-tape TM’s

Often it’s useful to have several tapes when carrying out a computations. For example, consider a two tape I/O TM for adding numbers (we show only how it acts on a typical input)

Multitape TM’sFormal Notation

NOTE: Sipser’s multitape machines cannot pause on one of the tapes as above example. This isn’t a problem since pausing 1-tape machines can simulate pausing k-tape machines, and non-pausing 1-tape machines can simulate 1-tape pausing machines by adding dummy R-L moves for each pause.

Formally, the d-function of a k-tape machine:

Multitape TM’sConventions
• Input always put on the first tape
• If I/O machine, output also on first tape
• Can consider machines as “string-vector” generators. E.g., a 4 tape machine could be considered as outputting in (S*)4