CSCI 3130: Formal languages and automata theory

CSCI 3130: Formal languages and automata theory

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

The Chinese University of Hong Kong

Turing Machines

Fall 2011

What is a computer?

A computer is a machine that manipulates data according to a list of instructions.

computerinput

programoutput

What is a computer?

world

void hello( String name){ print( “Hello, “, name);} Hello, world

E2-E4

computer

deep blue

?

?

?

Turing Machines

…a b b ☐ ☐

input blanks

controlhead

Can both read from and write to the tape

Head can move both left and right

Tape is infinite

Has two special states accept and reject

Example

L1 = {w#w: w ∈{a, b}*}

Strategy:

Read and remember the first symbol

Cross it off (x)

Read the first symbol past #

If they don’t match, reject

abbaa#abbaa

xbbaa#abbaa

xbbaa#abbaa

If they do, cross it off xbbaa#xbbaa

Example

L1 = {w#w: w ∈ {a, b}*}

Strategy:

Look for the first uncrossed symbol

Cross it off (x)

Read the first uncrossed symbol past #

xbbaa#xbbaa

xxbaa#xbbaa

xxbaa#xbbaa

If they match, cross it off, else reject xxbaa#xxbaa

At the end, there should be only xs and #sxxxxx#xxxxx

If not, reject; otherwise, accept.

How Turing Machines operate

…a b a ☐q1 q2

a/bL

Replace a with b, and move head left

current state

…a b b ☐q1 q2

a/bL

new state

Formal Definition

A Turing Machine is (Q, , , , q0, qacc, qrej):– Q is a finite set of states; is the input alphabet not containing the blank

symbol ☐ is the tape alphabet ( ) including ☐

– q0 in Q is the start state;

– qacc, qrej in Q are the accepting and rejecting state is the transition function

: (Q – {qacc, qrej}) → Q {L, R}

Turing Machines are deterministic

Configurations

• A configuration consists of the current state, the head position, and tape contents

…a b a ☐q1 abq1a

q1 qacca/bR

abbqacc…a b b ☐

qacc

configuration

Configurations

• We say configuration C yields C ’ if the TM can go from C to C ’ in one step

• The start configuration of the TM on input w is q0w

• An accepting configuration is one that contains qacc;A rejecting configuration is one that contains qrej

abq1a abbqaccyields

The language of a Turing Machine

• We say M accepts x if there exists a sequence of configurations C0, C1, ..., Ck where

C0 is starting Ci yields Ci+1Ck is accepting

The language recognized by M is the set of all strings that M accepts

Looping

• Something strange can happen in a Turing Machine:

• Inputs can be divided into three types

q0

qacc

qrej

= {0, 1}

input: ☐/☐R

0/0R

1/1RThis machinenever halts

qacc qrej

accept reject loop

Halting

• We say M halts on x if there exists a sequence of configurations C0, C1, ..., Ck where

C0 is starting Ci yields Ci+1

Ck is acceptingor rejecting

A TM M is a decider if it halts on every inputLanguage L is decidable if it is recognized by a TM that halts on every input

Programming Turing Machines

L1 = {w#w: w ∈ {a, b}*}Description of Turing Machine:

Until you reach #

Read and remember entry

Move right past # and past all xsIf this entry is different, reject

Otherwise

Move left past # and to right of first x

If you see only xs followed by ☐, accept

xbbaa#xbbaa

xxbaa#xxbaa

Write x xxbaa#xbbaaxxbaa#xbbaa

Write x xxbaa#xxbaa

1

2

3

4

5

6

7

8


L1 = {w#w: w ∈ {a, b}*}

q0

qa1 qa2

q2

qb1 qb2

q3 qrejq1 qacc

a/xR

#/#R

#/#R

b/xL

b/xR

a/xL

a/aRb/bR

a/aRb/bR

x/xR

x/xR

x/xR

a/aLb/bLx/xL

☐/☐R#/#R #/#L

a/aLb/bL

x/xR

everything else

1

23

23

4

4

56

56

78


q0

qa1 qa2

q2

qb1 qb2

q3q1 qacc

a/xR

#/#R

#/#R

b/xL

b/xR

a/xL

a/aRb/bR

a/aRb/bR

x/xR

x/xR

x/xR

a/aLb/bLx/xL

☐/☐R#/#R #/#L

a/aLb/bL

x/xR

1

23

23

4

4

56

56

78

input: aab#aab

configurations:

q0aab#aabxqa1ab#aabxaqa1b#aabxabqa1#aabxab#qa2aabxabq2#xabxaq3b#xabxq3ab#xabq3xab#xabxq0ab#xab


L2 = {aibjck: i j = k and i, j, k > 0 }

High-level description of TM:

For every a:

Cross off the same number of bs and cs

Uncross the crossed bs (but not the cs)

Cross off this a

If all as and cs are crossed off, accept.

aabbcccc

aabbcccc

= {a, b, c}


L2 = {aibjck: i j = k and i, j, k > 0 }

Low-level description of TM:

Scan input from left to right to check it looks like aa*bb*cc*

Move the head to the first symbol of the tape

For every a:

Cross off the same number of bs and cs

Restore the crossed of bs (but not the cs)

Cross off this a

If all as and cs are crossed off, accept.

how do we know?

how to do this?


Implementation details:

Cross off the same number of bs and cs:

Move right until you see a c

If any bs are left, repeat

Put a special marker on top of first a

Move the head to the first symbol of the tape

aabbcccc

aabbcccc

aaqbbcccc

= {a, b, c}


L3 = {#x1#x2...#xl : xi ∈ {0, 1}* and xi ≠ xj for each i ≠ j}

#01#0011#1High-level description of TM:

On input w,

For every pair of blocks xi and xj in w,

Compare the blocks xi and xjIf they are the same, reject.Otherwise, accept.



#01#0011#1Low-level description:

Place a mark on the leftmost #(i.e. replace # by #) and move right

1.

0.If input is , or has exactly one #, accept.

#01#0011#1

Place another mark on next unmarked #(If there are no more #, accept)2. #01#0011#1



current state:

Repeat Step 35.

#01#0011#1

#01#0011#1

#01#0011#1

#01#0011#1

Compare the two strings to the rightof marked #. If there are same, reject3.

4.Move the right # to the right If not possible, move the left # to the next # and put the right # on the next If not possible, accept



#01#0011#1

#01#0011#1

accept4.Move the right # to the right If not possible, move the left # to the next # and put the right # on the next If not possible, accept

Compare the two strings to the rightof marked #. If there are same, reject3.

How to describe Turing Machines?

• Unlike for DFAs, NFAs, PDAs, we rarely give complete state diagrams of Turing Machines

• Usually we give a high-level description:A recipe about the workings of the Turing Machine

Just like in cooking, practice makes perfect!


L4 = { 〈 G 〉 : G is a connected undirected graph}

Q: How do we feed a graph into a Turing Machine?

A: We represent it by a string, e.g.

1 2

3 4

(1,2,3,4)((1,4),(2,3),(3,4)(4,2))

Convention for describing graphs:

(nodes)(edges)

no node must repeatedges are pairs (node1,node2)



On input 〈 G 〉 ,

0.Verify that 〈 G 〉 is the description of a graph (no vertex repeats; edges only go between nodes)

1.Mark the first node of G

2.Repeat until no new nodes are marked:

For each node, mark it if it is attached to an already marked node

3.If all nodes are marked accept, otherwise reject.

1 2

3 4

x

x

x

x



(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2)) etc.

1 2

3 4

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

(1,2,3,4)((1,4)(2,3)(3,4)(4,2))

What’s so great aboutTuring Machines?

Hilbert’s list of 23 problems

• Leading mathematician in the 1900s

• At the 1900 International Congressof Mathematicians, proposed a list of23 problems for the 20th century

• Hilbert’s 10th problem:David Hilbert

(1862-1943)

Find a method to tell if an equation like xyz – 3xy + 6xz + 2 = 0 has integer solutions

A brief history of computing devices

Z3 (Germany, 1941) ENIAC (Pennsylvania, 1945)

PC (1980) MacBook Air (2008)

Computation is universal

In principle, all thesehave the same problem-solving ability

The Church-Turing Thesis

Turing Machine

If an algorithm can be implemented on any realistic computer, then it can be implemented on a Turing Machine.


Turing Machinequantum computing

DNA computing

cosmic computing

Alan Turing

• Invented the Turing Test to tellhumans and computers apart

• Broke German encryption machines in World War II

• The Turing Award is the “Nobel prize of Computer Science”

Alan Turing (1912-1954)

Turing’s motivation: By studying Turing Machines, we can understand the limitations of real computers.

Undecidability

• What Hilbert meant to ask was:

• Building on work of Gödel, Church, Turing, Davis, Putnam, Robinson, in 1970 Matijasievič showed:

Find a Turing Machine to tell if an equation like xyz – 3xy + 6xz + 2 = 0 has integer solutions

There is no such Turing Machine!

Computer program analysis

public static void main(String args[]) { System.out.println("Hello World!"); }

What does this program do?

How about this one?

public static void main(String args[]) { int i = 0; for (j = 1; j < 10; j++) { i += j; if (i == 28) { System.out.println("Hello World!"); } } }

Computers cannot analyze programs!

No Turing Machine can do this:

input: The code of a java program P

Accept if P prints “hello, world”

Reject if not

Significance: It is impossible for computer to predict what a computer program will do!

How do you argue things like that?

To argue what computers cannot do,we need to have a precise definitionof what a computer is.

“On Computable Numbers, with an Application to the Entscheidungsproblem”

1936:

Section 1. Computing Machines

Turing’s answer: A computer is just a Turing Machine.


“On Computable Numbers, with an Application to the Entscheidungsproblem”

1936:

Section 9. The extent of the computable numbers

All arguments [for the CT Thesis] which can be given are bound to be, fundamentally, appeals to intuition, and for this reason rather unsatisfactory mathematically. The arguments which I shall use are of three kinds: 1. A direct appeal to intuition 2. A proof of the equivalence of two definitions (In case the new definition has greater intuitive appeal) 3. Giving examples of large classes of numbers which are computable.

CSCI 3130: Formal languages and automata theory

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