L15 Turing machines

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Turing Machines VariantsZeph Grunschlag

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AnnouncementMidterms not graded yetWill get them back Tuesday

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AgendaTuring Machine Variants Non-deterministic TM’s Multi-Tape

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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.

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Input-Output Turing MachinesWhen 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).

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TM NotationsThere are three ways that Sipser uses to describe

TM algorithms.1) High level –pseudocode which explains how

algorithm works without the technical snafoos of TM notation

2) Implementation level –describe how the TM operates on its tape. No need to mention states explicitly.

3) Low-level description. One of:Set of complete “goto” style instructionsState diagramFormal description: spell out the 7-tuple

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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"

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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.

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So round by round, tape transformed as follows:

11111 X1111X XX111XX XXX11XXX XXXX1XXXX XXXXXXXXXX 1111111111

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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 holdsMay need to add extra functionality

Add $ if need to tell when end of tape is

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M = "On input w = 1n 1. HALT if no input 2. Write $ in left-most position3. Sweep right and write X in first blank4. Sweep left through X-streak and 1-streak5. Go right6. If read X, go right and goto 9. Else, replace 1 by X,

move right.7. If read X [[finished original w]] goto 8 Else, goto 38. Sweep to the right until reach blank, replace by X9. Sweep left replacing everything non-blank by 110.HALT

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

0

halt

�R

21$,R$|XR1|XR

3 41L

1X,R

�1,L

�X,LXL

1L

56

8

1R

XR X

R9

X1,L$1,

L

$R

9

X1,L

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Non-Deterministic TM’sA 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 on some 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 * | accepting config. y, q0 x * y }

(No change needed as need not be function)

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Non-Deterministic TM’sRecognizers vs. Deciders

N is always called a non-deterministic recognizer and is said to recognize L(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).

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Non-Deterministic TMExample

Consider the non-deterministic method:

void nonDeterministicCrossOut(char c)while()if (read blank) go leftelse if (read c)cross out, go right, return OR go right // even when reading c OR go left // even when reading c

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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 ?

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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 = 0XX1Xq12X2For which v do we have u v ?

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A: 0XX1Xq12X2 0XX1qX12X2 | 0XX1X1q2X2 | 0XX1XXq’

2X2These define 3 branches of computation

tree:

Q: Is this a non-deterministic TM decider?

0XX1XXq’ 2X2

0XX1Xq12X2

0XX1X1q2X20XX1qX12X2

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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.

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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.

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Konig’s Infinity LemmaProof Idea

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

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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 v0v1…vn-

1vn 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. �

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Multi-tape TM’sOften 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)

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1

Input string

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$ 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$ 1 0

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$ 1 0 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1 1$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0 1$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1 0$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1 $ 1$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1 $$ 1 0 1 1

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Multi Tape TMAddition Example$ 1 0 1$ 1 0 1 1

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Multi Tape TMAddition Example0 0 0 0$ 0 0 0 0

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Multi Tape TMAddition Example1 0 0 01 0 0 0 0

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Multi Tape TMAddition Example1 0 0 0 01 0 0 0 0

Outputstring

HALT!

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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 -function of a k-tape machine:kkk }RL,{:δ QQ

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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 (*)4

L15 Turing machines

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