meljun cortes - automata with solution and answer

7/31/2019 MELJUN CORTES - AUTOMATA with SOLUTION and ANSWER

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MELJUN P. CORTES--------> Theory of Automata and Formal Language

2012

SET

It is any well defined collection of objects.

It is a collection of distinct objects, without repetition and without ordering. It is denoted by capital letter.

Example:A = { , ,0} C = {1, 2, 3}B = {a, b, c, d} D = {Len, Joseph}

1. Roster / Tabular Form

Listing or enumerating all the elements of a given set.

Example: A = {a, b, c}

2. Set-Builder Form Elements have properties in common.

Elements must satisfy a given rule or condition.Example: A = {x | x is a positive integer < 4}

A = {x | x < 5}

It is a member of a given set or an object in the collection.

It is denoted by small letter/s.Example:

A = {a, b, c} ; a A; b A; c A; d A

1. elements can be counted.

Example:A = {a, b, c}

2. elements cannot be counted.

Example:A = {x | x is a positive integer}

1. Order of elements is not important2. Repetition must be ignored

Prepared by: Marilyn M. Sanchez2003


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If A is a subset of B, A is contained in B or every element of A is in B.

Example:A = {a, b}

B = {a, b, c}

Every set is a subset of itself.

An empty set is a subset of every set.

If A B then there is at least one element in A that is not in B.

Example: Let A = {1, 2, 3}

= {{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, {}}

To check:|A| = 2n = 8

A is a proper subset of B if A B and A B

It is the total number of elements in a given set

Example:A = {1, 2, 3}

|A| = 3

It is a set containing no element

Example : A = { } or A = {}

A = {} is not an empty set

It is a set containing only one element.e.g. A = {1}



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At least one set is a subset of another set.

Conditions so that A & B are comparable:

1. A B & B A

Example: A = {1, 2, 3, 4}B = {4, 3, 2, 1}

2. A B & B A

Example: A = {1, 2}B = {1, 2, 3}

3. A B & B AExample: A = {a, b, c}

B = {a, b}

No set is a subset of the other set. (A B & B A & A B)

Example: A = {l, e, n}B = {s, e, p, h}

Sets having exactly the same cardinality and kind of elements.

Example: A = {l, e, n}B = {n, l, e}

Set having the same cardinality.

Example: A = {1, 2, 3}B = {4, 5, 6}

Every element of A is associated or paired to every element in B.

Example: A = {1, 2, 3}B= {l, e, n}

Sets having no common elements

Example A = {1, 4, 5}B = {3, 2, 6}


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It is the biggest set under investigation wherein the other sets are subsets of this givenset.

Example: Let U= {1, 2, 3, 4, 5, 6}A = 1, 2, 3}B = {4, 5, 6}C = {5, 6}

It is a collection of set. All elements of the given set are sets.

Example: A = {A, B, C}A = {{1, 2, 3}, {4, 5, 6}, {5, 6}}

B = {A, B, C, D, 0, 2}C = {A, B, C, D, z}

new set of elements that belong to U but not in A (with respect to U)

Example: Let U= {1, 2, 3, 4, 5, 6}

A = {1, 2, 3}A = {4, 5, 6}

new set of elements that belong to B but not in A (with respect to B)

Example: Let B = {a, b, c, d, e}

A = {a, b, c}A = {d, e}

It is a family of all subsets of a given set.

It is the set of all subsets.

Example: Let A = {1, 2, 3}P (A) = {{1}, {2}, {3}, {1, 2}, {1, 3}, {}, {1, 2, 3}}


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1.

It is a new set of elements that belong to A or to B or to both A and B.

Example: Let A = {1, 2, 3, 4}B = {2, 4, 6, 7}

AB = {1, 2, 3, 4, 5, 6, 7}

2.

It is a new sets of elements that belong to both A and B.

Example: Let A = {1, 2, 3, 4}B = {2, 4, 6, 7}

AB = {2, 4}

3.

It is a new set of elements that belong to A but not to B.

Example: Let A = {1, 2, 3, 4, 5}

B = {1, 2, 4, 6, 7}A - B = {3, 5}B - A = {6, 7}

4.

It is a new set of elements that belongs to U but not to A.

Example: Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}A = {1, 2, 3, 4}B = {2, 4, 6}

A = {5, 6, 7, 8, 9}

B = {1, 3, 5, 7, 8, 9}

5. It is a new set of elements that belong to A or B but not to A and B.

Example: Let A = {1, 2, 3, 4, 5}B = {0, 1, 2, 6, 7, 8}

A B = {0, 3, 4, 5, 6, 7, 8}


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It deals with idealized computer device.

It tells how to design and use special language.

3.

It tells whether the problem is algorithmically solvable or not.

It tells how to design and use special language.

It is a finite set of symbols.

It can be letters from a to z or digits from 0-9.

It is denoted by .

It is a finite sequence of an alphabet.

It is a set of strings generated from some alphabets.

Examples:

Let {a,b}

Write the strings generated from the given sets.

1. L1 = A set of strings that begins and ends with .

Answer:

L1 ={aa, aba, abbbba, aaaaa, aaaaba, }

2. L2 = A set of strings divisible by 2.

Answer:

L2 ={aa, bb, abba, bbaa, bababa, }


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It is any number of leading symbols of a string.

Example: W ={star}Prefix of a String = {e, s, st, sta, star}

It is any number of leading symbols other than the string itself.

Example: W ={star}Proper Prefix of a String = {e, s, st, sta}

It is any number of trailing symbols of a string.

Example: W ={star}Suffix of a String = {e, r, ar, tar, star }

It is any number of trailing symbols other than the string itself.

Example: W ={star}

Proper Suffix of a String = {e, r, ar, tar}

It is string formed by writing the first string followed by the second string with nointervening spaces.

It is denoted by ( ) concatenation operator.

Example:

automata theory = automatatheory

* That is if w & x are strings, then wx is the concatenation of these two strings.

It is denoted by WR.

It is a string that spelled backwards.

Example:marilynR = nyliram


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Prepared by: Marilyn M. Sanchez

2003

It is a mathematical model of a system with discrete inputs and outputs.

It consist of finite set of a sets and a set of transitions from state to state the occur on

input symbol chosen from alphabet ().

1. states qs and ps

2. q0 initial state

3. 0, 1, a, b input symbols

4. w, x, y, z strings of input symbols.

Finite Automata

are limited in strength but they are a thoroughly understood subclass of more powerfulcomputational models.

a further reason for studying FA is their applicability to the design of several common types ofcomputer algorithms and programs.

Example:Lexical analysis phase of Compilers is often based on the simulation of a automatation.

Vertices denotes the store

Edges transition function

Initial State Final State

( set of states, set of input alphabet )


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Example:State Diagram

qo q1

State Table

x y

q0 q1 q1q1 q0 q0

Transition Function

(q0,x) = q0

(q0,y) = q0

(q1,x) = q1

(q1,y) = q1

1. For each input symbol in , there is exactly one transition of each state (possibly back to

the state itself).

It do not accept empty strings.

It is a quintuple where M = (Q, , q0, , F)

where:Q = finite set of all sets


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= finite set of input symbolsq0 = initial state

= transition functionF = set of final state


Examples:

1. Diagrams that ends with 10.

qo

q1 q21

0 1

0

1

0

Transition Table

x y

q0 q1 q1

q1 q2 q1q2 q3 q1

Strings that can be derived:

10

010

0110

01010

Transition Function

(q0,0) = q0

(q0,1) = q1

(q1,0) = q2 (q1,1) = q1

(q2,0) = q0

(q2,1) = q1

2. Draw a transition diagram that will end with 001.


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qo

q1 q30

1 0

0q2

1

01

1


* In FA or DFA, for each input there is exactly one transition output of each state.

Lets try to modify the finite automaton by alowing zero, one or more transitions from astate on the same input symbol. The modified model is called Non-Deterministic FiniteAutomaton.

If for some ofQ, a, (q,a) does either to a unique state or several states or notstates at all, then the FA is a NFA.

A Non-Deterministic Finite Automaton is a quintuple M=(Q,,q0,,F)

where: Q = set of states

= set of input alphabets

= transition function

---> QX to 2Qq0 = initial stateF = set of final state

It allows zero, one or more transitions for every inputs symbol.

Empty string accepted

A sequence of symbols, say a1, a2, ...an is accepted by NFA if there exists a sequence of

transition corresponding to the input sequence that leads from the initial states to thefinal state.

Example:Construct a NFA that accepts string ending with 011.


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qo

q1 q30

1

1q2

1

0

(q0,011)--> q3To check if 1011 is a valid string

q0q0q1q2q2

q0q0q0


The reason for the word nondeterministic is that we are in a state where there aremultiple outgoing edges all having the same input symbol we have a choice of next state.

The only difference between a NFA & DFA is that in DFA the next state function takesus to a uniquely defined state whereas in NFA the next state takes us to a set of states.

Example :

qo

q1

q4

01

0

q2

q3

q2

q00

1

1

1,0

0

1

0 1

q0 {q0,q3} {q0,q1}

q1 0 {q2}

q2 {q2} {q2}

q3 {q4} 0

q4 {q4} {q4}


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Let the input be 01001

(q0, ) =

(q0, 0) = ((q0, )0) = (q0,0)=

(q1, 01) = ((q0,0)1) =(q0,1) U (q3,1)={q0,q1}U{ }=

(q1, 010) = ((q0,01)0)=(q0,0) U (q1,0)={q0,q3}U{ }=

(q2, 0100) = ((q0,010)0)=(q0,1) U (q3,1) U (q3,0)={q0,q3} U

{q4}=

(q2, 01001) = ((q0,1) U (q3,1) U (q4,1) = {q0,q1} U {} U {q4}=


Theorem: For every NFA model, there exist one and only one equivalent DFA

Let M be the equivalent DFA of a NFA M.

M=(Q, ,,q0,F)

Example:Transform the given NFA to DFA

qo

q20

q11

000,1

={{},q0,q1,q2,{q0,q1},{q0,q1},{q0,q2},{q1,q2},{q0,q1,q2}}

=q2 ={q2,{q0,q2},{q1,q2},{q0,q1,q2}}

0 1

q0 {q0} {q0,q1}

q1 {q1,q2} {}

q2 {q2} {}


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({},0) = {}

({},1) = {}

(q0,0) = q0

(q0,1) = {q0,q1}

(q1,0) = {q1,q2} (q1,1) = {}

(q2,0) = q2

(q2,1) = {}

((q0,q1),0) = {q0,0} U (q1,0)

= q0 U {q1,q2} = {q0,q1,q2}

((q0,q1),1) = {q0,1) U (q1,1)

= {q0,q1} U {} = { q0,q1}

((q0,q2),0) = {q0,0) U (q2,0)= {q0} U {q2} = {q0,q2}


((q0,q2),1) = {q0,1) U (q2,1)= {q0,q1} U {q2} = {q0,q1,q2}

((q1,q2),0) = {q1,0) U (q2,0)

= {q1,q2} U {q2} = {q1,q2} ((q0,q1,q2),0) = {q0,0) U (q1,0) U (q2,0)

= {q0} U {q1,q2} U {q2} = {q0,q1,q2}

((q0,q1,q2),1) = {q0,1) U (q1,1) U (q2,1)

= {q0,q1} U {} U {} = {q0,q1}

0 1

q0 {q0} {q0,q1}

q1 {q1,q2} {}

q2 {q2} {}q0,q1 {q0,q1,q2} {q0,q1}

q0,q2 {q0,q2} {q0,q1}

q1,q2 {q1,q2} {}

q0,q1,q2 {q0,q1,q2} {q0,q1}

{} {} {}


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{}

q1

q2

q1q2

10

0

0

1

1

0,1


It is quintuple M = {Q,,,q0,F} where the transition functions maps.

:Q x ( U {})2Q

(q,a)={p | (q,a) p and a= or a

will consist of all states p such that there is a transition labeled a from q to p where a is

either or a symbol in the.

-closure(q)= { p | , a path from q to p marked }-closure(p)= Uqp -closure where p is a set of states

: Q x 2Q

(q,w) { p | (q,w) contains p including edges labeled (q,) = -closure (q)

* (q,w) will be all states p such that one can go from q to p doing a path labeled w,perhaps including edges labeled .

Example:

q0q0,q1

q0,q1,q2

q0,q2


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qo

q3

q1 a

a

q2

transition: (q0,a) q0

(q0,) q1

(q1,) q2

(q2,a) q3

-closure:

q0

q0 q1

q0 q1 q2


For each NFAs there is one and only one equivalent NFA.

NFA or NFAs: M = {Q, , , q0, F}

Let Equivalent NFA be M = {Q, , 1, q0, F1}

where: F = {F U closure (qo)1 l closure (qo) contains state of F }

Example:

1. Convert the given NFA to NFA.

qo

q3

q1

0

q2

1 2

-closure (q0) = {q0,q1,q2}-closure (q1) = {q1,q2}-closure(p) = -closure of {q0,q1}

= -closure {q0} U -closure {q1}= {q0,q1,q2} U {q1,q2}= {q0,q1,q2}


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0 1 2 q0 {q0} {} {} q1

q1 {} q1 {} q2

q2 {} {} q2 {}

0 1 2

q0 {q0,q1,q2} {q1,q2} {q2}

q1 0 {q1,q2} {q2}

q2 {} {} {q2}

= {F U closure (q0)}= {q0,q1,q2}

qo

q3q1

0

q2

1 2

0,1,2

0,11,2


2. Convert the given NFA to NFA.

q0

q4 q5

q1 q2 q2q3

q5

a,b

a a

a,b

b

b

F = {q3,q5} -closure (q0) = {q0,q1,q4}q0 = {q0}

= {a,b}

a b E


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q0 {} {} {q1,q4}

q1 {q1,q2} q1 {}

q2 {q3} {} {}

q3 {q3} {q3} {}

q4 {} {q4} {}

q5 {} {q5} {}

a b

q0 {q1,q2} {q1,q5}

q1 {q1,q2} {q1}

q2 {q3} 0

q3 {q3} {q3}

q4 {} {q5}

q5 {} {q5}


F = {q3,q5}

q0

q4

q5

q1

q2 q2q3

q5

a,b

a

a

a,b

b

b

b

a,b

a


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DEFINITION: is an alphabet

A regular expression over is recursively defined as:

1. 0 is a regular expression denoting the set of {}.

2. is a regular expression denoting the set of {}.

3. For any a , a is a regular expression denoting the set {a}.4. If r,s,t are reg. exp. Denoting sets R,S,T respectively, then

(r t s) denotes R u S

rs denotes RSr* denotes R*

finite set of symbols

L1,L2,L: set of strings from


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Examples:

a. L1L2 = { xyl xL1 ^ YL2} is the concatenation of L1 & L2e.g = {a,b}

L1 = {a,b} {}={}

L2 = {ba,a} ={}

L1L2 = {a} {ba}{b} {a }

= {aba,aa,bba,ba}

b. L0 = {}Li = LL i1

e.g. L = {0,1}

L0 = {}

Find L3

L1 = LL0 = {0,1},{}= {0,1}

L2 = LL1 ={0,1}{0,1}

= {00,01,10,11}L3 = LL2 = {0,1}{00,01,10,11}

= {000,001,010,011,100,101,110,111}


of L is defined asL* = U

Li

i = 0

of L is defined asL+ = U

Li

i = 0

Example:L = {0,1}L* = {e,0,1,00,11,101,01,10,111,000,101,..}L+ = {0,1,00,11,101,01,10,111,000,101,.}

L = {1}L* = {e,1,11,111,1111,..}

Examples: Let = {0,1}


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1. 0 = {0}

2. 01 = {01}

3. 0 + 1 = {0,1}

4. 1* = {e,1,11,111,1111,}5. 0* + 1 = {e,0,1,00,000,..}

6. 1*00 = {00,100,1100,11100,.}7. (0+1)* = {e,0,1,00,11,000,111,10,01,101,0001.}8. (00+1*)* = {e,00,1,11,1111,00001,100.}9. (010+101) = {e,010,101,1011,10111,101111..}

10. (0+1)*101 = set of strings ending with 101.11. (0+1)*11(0+1)* = set of strings with 11 as a substring.

REGULAR EXRESSION NFA

0 or {}qo q1

= {}* w/o inputqoq0

a = {a}q

oq

1a


If r & s are regular expression, assume that there is a NFAe corresponding to r &

s.

q0 qf L (r) q0 qf L (s)

a. r + s ( r or s)

q0 q2qf

q0 qf L (r)

q0 qf L (s)


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b. rs (r concatenates s or concatenation of rs)

q0 qf L (s)q0 qf L (r)

c. r* (r closure or r kleene closure)

q0 q2qf

q0 qf L (r)

Examples:

Construct the equivalent NFAe of the ff, reg. expression:

1. 0 W/ ANSWER na ni..

qo

q1

0

Prepared by: Marilyn M. Sanchez

2003

2. 01

q2qo

q11

q3

0

3. 0+1


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q2

qo

q1

1q4q3

q5

0

4. 1*

q2q

oq1

1q3

5. 11+0

q1

1

0

0q7

q2

q2q5

0

q2q

oq1 q3

111

1q3q1 q2 q4

1

q6

q0


Context-free Grammars describe context-free languages.

Regular set is an example of Context Free language.

Context-Free grammar (CFG) is a quadruple

G = (V,T,P,S) where V set of variables {S,B}


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T set of terminals (a,b,c)

P Production or rule (How to form the string)S the starting symbol

Examples :

Construct CFG for the following:

1. ab*c = {ac,abc,abbc,abbbc,.}

given: S aBc

B bB

B

S aBc S aBc S aBc

ac abBc abbBc

ac abc abbbBc

abc abbbbBc

abbbbbBc

abbbbbc

abbbbbc

2. Construct a CFG for anbn; n>0

= {ab,aabb,aaabbb..}V=SS=ST={a,b}

P: S aSb

S

S aSb S aSb ab aaSbb

ab aabb

aabb


3. Construct a CFG for the following: W/ ANSWER na ni..a. 0*1 = {1,01,001,..}

V=S,B

T={0,1}S=S


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P: S B1

B0B

B

S B1 S B1

1 0B1 1 00B1

000B1

0001

0001b. (0+1)*

S 1s|0s

S

01 111001

S 0s S 1s

01s 11s

01

111s 01 1110s

11100s

111001s

111001

111001

c. ambn | m>= n

S aMb

M aMb | aM |

aabb aaabb

S aMb S aMb

aaMbb aaMbb aabb aaaMbb

aabb aaabb

aaabb


d. S | 0 | 1

S 1s1 | 0s0


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S 0s0 S 1s1 S 0s0

00 101 00s00

00 00100

e. zanzbnz | n>=0

S aSb S aSb

ab aaSbb

ab aabb

aabb

f. set of integers (+ or -)

S +N| -N

S 0N | | 9N | 0 | | 9

S +N S -N

+1N -9N

+10 -98N

-987

g. {anbncmdm | n,m > 0}

S abcd| aMbcNd

M aMb |

N aNb |

aabbcd aabbccdd

S aMbcNd S aMbcNd

aaMbbcd aaMbbccNdd aabbcd aabbccdd

aabbcd aabbccdd


S MN

M aMb |ab


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N cNd | cd

aabbcd

S MN

aMbcd

aabbcd

h. {xmyxm| m >= 0} S xSx |y

S xSx S xSx

xyx xxSxx

xxxSxxx

xxxyxxx



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Context Free Grammars machine counterpart.(Just as the RE have an equivalent

autometon-FA).

Non deterministic device.(Deterministic version is only a subset of all CFLs).

Have input tape, finite control and a stack

Essentially a FA with control of both input and stack FILO list.

is a string of symbols from some alphabet. The leftmost symbol of the stack isconsidered to be the tape of the stack.

1. An input symbol is used.

Depending on the input symbol, the top symbol on the stack, and state of thefinite control, a number of choices are possible. Each choice consist of next statefor the finite control and a (possibly empty) string of symbols to replace the topstack symbol. After selecting a choice, the input head is advanced one symbol.

2. Input symbol is not used (-move).

Similar to the first except that the input symbol is not used and the input head isnot advanced after the move.

Allows the PDA to manipulate the stack without reading input symbols.

no final state empty both input

tape & stack

1. Set of all inputs for which some sequence of moves causes the PDA to empty a stack.2. Designate some states as final sates define the accepted language as the set of all inputs

for which some choice of moves causes the PDA to enter final state.

with final state empty input tape

M = (Q, , , , qo, Zo, F)

where:

Q = finite set of all sets = finite set of input symbols = stack alphabet

q0 = in Q, initial state

Zo = in , start symbol

F = Q, set of final states

= mapping Q x ( U {}) x to Q x *

(q, a, Z) = {(P1, y1)(Pm, ym)}initial state Pop Push

Input Destination State



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Example:

M = ( {q1,q2}, {0,1}, {R, B, G}, , q1, R, q2 )

1. ( q1, 0, R ) = {( q1, BR )}

2. ( q1, 1, R ) = {( q1, GR )}

3. ( q1, 0, B ) = {( q1, BB )}, {( q2, )}

4. ( q1, 0, G ) = {( q1, BG )}

5. ( q1, 1, B ) = {( q1, GB )}

6. ( q1, 1, G ) = {( q1, GG )}, {( q2, )}

7. ( q2, 0, B ) = {( q2, )}

8. ( q2, 1, G ) = {( q2, )}

9. ( q1, , R ) = {( q2, )}

10. ( q2, , R ) = {( q2, )}

Determine if the input 001100 is valid for PUSHDOWN AUTOMATA.

(q1,001100,R)

(q1,01100,BR) (q2,001100,)

(q1,1100,BBR) (q2,1100,R) (q2,1100,)

(q1,100,GBBR) (q1,0,BGGBBR)

(q1,00,GGBBR) (q2,00,BBR)

(q1,0,BGGBBR) ( q2, 0,BR ) ( q2, , R ) ( q2, , )

(q1,,BBGGBBR) (q2,,GGBBR)

? ?



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M = (Q, , , , qo, B, F)where:

Q = finite set of all sets

= finite set of allowable tape symbolsB = in , is blank

= not includingB, set of input symbols

q0 = in Q, start stateZo = in , start symbol

F = Q, set of final states

= next move function Q x to Q x {L,R}

initial state destination head movement

state

(basic model) has finite control, input tape that is divided into cells, a tape head thatscans one cell of the tape at a time.

Introduced by Allan Turing in 1936

Tape has a leftmost cell but infinite to the right

Each cell may hold exactly one of a finite number of tape symbols

Initially, the n leftmost cell (n>0) hold the input, which is a string of symbolschosen from a subset of the tape symbols called input symbols.

The remaining infinity of cells each hold blank, which is a special tape symbolthat is not input symbol.

Finite Control

In one move, the TM (depending upon the symbol scanned by th tape head & the state of finitecontrol)

1. changes state2. prints symbol on the tape cell scanned, replacing what was written there3. moves its head left or right one cell


a1 a2 a1 an B B


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Recursively enumerable strings can be enumerated or listed by a Turing

Machine.Example:

M = ({qo, q1 q4}, {0,1}, {0,1,x,y,B}, , q0, B, q4)

qo (q1,x,R) (q3,y,R)

q1 (q1,0,R) (q2,y,L) (q1,y,R)

q2 (q2,0,L) (q0,x,R) (q2,y,L)

q3 (q3,y,R) (q4,B,R)

q4

Determine if the given inputs are accepted by TM

1. 0011

State Head Position

qo 0011q1 x011q1 x011q2 x0y1q2 x0y1

qo x0y1

q1 xxy1q1 xxy1q2 xxyyq2 xxyyqo xxyy

q3 xxyyq3 xxyyBq4 xxyyBB

Therefore, accepted



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2. 0011

State Head Position

qo 0011 01q1 x01101

q1 x01101q2 x0y101q2 x0y101qo x0y101q1 xxy101q1 xxy101

q2 xxyy01

q2 xxyy01qo xxyy01q3 xxyy01q3 xxyy01

?Therefore, not accepted



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AMA COMPUTER UNIVERSITYProject 8, Quezon City

COLLEGE OF COMPUTER STUDIES

MANUALINAUTOMATA THEORY

Prepared by :MELJUN P. CORTES

MSCS STUDENT

meljun cortes - automata with solution and answer

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