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Introduction to Finite Automata

FundamentalsDeterministic finite automataRepresentations of automata

Definition of a languageProof of language content

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Review:What is the difference between

S not equal to T and S T = null?

Recursive definitions use that thing being defined as part of the definition.

Example: recursive definition of a treeBasis: single node is a treeIH: assume that T1 … Tk are treesInduction: connect T1 … Tk to the node that is the root of the recursively defined tree

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Review: FA Recognizing Strings Ending in “ing”

empty Saw ii

Not i

Saw ingg

i

Not i or g

Saw inn

i

Not i or n

Start

Double circle denotes “finish” or “accepting”state

We can get the next character in any stateWhere we go depends on which character

What is wrong with this FA?

Rules for deterministic finite automata

Finite automata move from state to state in response to input controls

Deterministic finite automata (DFA) are always in a unique state

Each input determines one and only one state to which the automaton will transition from its current state

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Ready Sending

data in

done

timeout

Start

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Review: FA Recognizing Strings Ending in “ing”

empty Saw ii

Not i

Saw ingg

i

Not i or g

Saw inn

i

Not i or n

Start

Does this example obey the rules?

How many string ending in “ing” will this FA find in the input?

Any letter

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Review: FA Recognizing Strings Ending in “ing”

empty Saw ii

Not i

Saw ingg

i

Not i or g

Saw inn

i

Not i or n

Start

How do I change this FA so that it only accepts input string ending in “ing”

Any letter

Central concepts of automata theory

Alphabets: sets of symbols Strings: lists of symbols from an

alphabet Languages: sets of string from the

same alphabet

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Alphabets

Any finite set of symbols Usually denoted by Examples:

=set of all ASCII characters ={0,1} binary alphabet ={a,b,…,z} lower case letters

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Sets of Strings A set of strings over alphabet Σ is a set of

lists, each element of which is a member of Σ

e denotes the empty string (length=0) Σ* denotes the set of all strings over {0,1}*={ε,0,1,00,01,10,11,000,…} k denotes strings of length k over a

specified alphabet 0={e} for any alphabet + denotes non-empty strings

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More about Strings * =0 1 2 … and 1 give different meaning to the

same entity denotes an alphabet 1 denotes strings of unit length over alphabet

If x and y are strings, then xy is a string formed by their concatenation

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Languages A language is a subset of strings for

some alphabet L=Σ* language of all strings from L can be empty L={e} is not empty. It contains the

empty string See text p31 for examples from binary

alphabet

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Using DFAs to define languages

Elements in the formalism for defining languages:

1. A finite set of states (Q, typically).2. An input alphabet (Σ, typically).3. A transition function (δ, typically).4. A start state in Q (q0, typically).

5. A set of final states ⊆ Q (F, typically).“Final” and “accepting” are synonyms.

Language A=(Q,,,q0,F) is a “five tuple”

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The Transition Function Takes two arguments: a state and

an input symbol. δ(q, a) = the state that the DFA

goes to when it is in state q and input a is received.

In graph representation δ(q, a) = p is shown by arc from state q to state p labeled by a

The set of strings “accepted” by A=(Q,,,q0,F) its

language How do we determine if A=(Q,,,q0,F)

accepts a string? Let a1,a2,…an denote a finite string

Locate q0 in A using “start”

Find states q1,q2,…qn such that (qi-1,ai)=qi

If qn F then string accepted, otherwise string is rejected

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DFA that accepts all strings without two

consecutive 1’s

Start

1

0

A CB1

0 0,1

No 11’s and ends in 0

No 11’s and ends In 1

Consecutive1’s havebeen seen.

L=({A,B,C}, {0,1}, , A, {A,B})

3 possible terminations

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Transition Table: δ(q, a) is the element in row q of

columns a

0 1

A A BB A CC C C

Final statesstarred

**Arrow for

start state

Start

1

0

A CB1

0 0,1

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Using graph to answer Is string in language?

Start

1

0

A CB1

0 0,1

Is string 101 in the language of the DFA below?Start at A.

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Is string in language 2

Start

1

0

A CB1

0 0,1

Follow arc labeled 1.

Is string 101 in the language of the DFA below?

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Is string in language 3

Start

1

0

A CB1

0 0,1

Then arc labeled 0 from current state B.

Is string 101 in the language of the DFA below?

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Is string in language 4

Start

1

0

A CB1

0 0,1

Finally arc labeled 1 from current state A. Resultis an accepting state, so 101 is in the language.

Is string 101 in the language of the DFA below?

Extended Transition FunctionDelta-hat

We defined δ(q, a) as the state that the DFA goes to when it is in state q and input a is received.

We want an extended transition function (q, w) defined as the state a DFA is in after it processes string w starting from state q

This extended transition function allows a language to be defined by L={w|(q0,w)F}

21˄

˄

˄

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Recursive definition of Delta-hat

Write w=xa where a is last symbol in w Basis: δ(q, ε) = q, where e is empty string Basis: δ(q, a) = δ(q, a) Induction: δ(q,xa) = δ(δ(q,x),a) If δ(q,x)=p then δ(q,w) = δ(p,a)

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˄

˄

˄

˄

˄

˄

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Use δ(q,xa) = δ(δ(q,x),a) to decompose delta-hat to nested

delta’s0 1

A* A BB* A CC C C

δ(B,011) = δ(δ(B,01),1) = δ(δ(δ(B,0),1),1) =

δ(δ(A,1),1) = δ(B,1) = C011 not accepted

˄ ˄

For convenience of slide preparation, I will use delta-underline to mean delta-hat

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Quiz #2 Monday 9-8-14on material in lecture “finite automata 2”and text pp28-52

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Provej=1 to n j = n(n+1)/2

Base case n=1j=1 to 1 j = 1(1+1)/2

Review: What method is used in this proof?If S(?) then S(?)

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Review: Induction on integers

• Methods if S(n) then S(n+1) and if S(n-1) then S(n) are equivalent but if S(n+1) then S(n) is inconsistent with induction principal.

• What you assume is your IH. Make sure it is true.

• I’m not accepting the “identity” method (text pp20-21) because it does not generalize to other forms of induction.

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Review: Language of a DFA

Automata of all kinds define languages. If A=(Q,,,q0,F) is an automaton, L(A) is its

language. L(A) = the set of strings w such that δ(q0, w) is in F.˄

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0 1

A* A BB* A CC C C

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Formal proof of language content

Delta_hat and transition table make it easy to test if a string is in L(DFA)

Formal proof of language content is a problem in equality of sets

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S = L(DFA)T = strings of 0’s and 1’s with no consecutive 1’s

To prove S=T, we need to prove If w is in S, then w is in T If w is in T, then w is in S

Proof of “If w is in S, then w is in T” is by induction on length of the string |w|

Proof of language content

Start

1

0

A CB 1

0 0,1

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Since DFA has 2 accepting states, if w is in S=L(DFA) has 2 cases

•If δ(A, w) = A (string ends with 0)•If δ(A, w) = B (string ends with 1)

Start

1

0

A CB1

0 0,1

No 11’s and ends in 0

No 11’s and ends in 1

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Prove: if w in L(DFA) then w no 11’s

Basis: |w| = 0; i.e., w = ε δ(A, e)=A by definition (e in S=L(DFA)) since ε has no 1’s, e is in T Basis is true

Induction: |w| >0, write w = xa, where a is the last symbol of w.

inductive hypothesis: if x in L(DFA) then x no 11’s

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IH: if x in L(DFA) then x no 11’s has 2 cases

Inductive hypothesis: if δ(A, x) = A, then x has no

consecutive 1’s and ends in 0. if δ(A, x) = B, then x has no

consecutive 1’s and ends in a single 1.

Start

1

0

A CB 10 0,1

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If δ(A, xa) = A then δ(A, x) must be A or B and a must be 0 (see DFA above). By the IH, x has no 11’s and ends in 0 or 1. Thus, w has no 11’s and end in 0.

If δ(A, xa) = B then δ(A, x) must be A and a must be 1 (see DFA above). By IH (case 1), x has no 11’s and ends in 0. Thus, w has no 11’s and ends in 1

Start

1

0

A CB 10 0,1Induction

w=xa

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Contrapositive : If w is not accepted by DFA (i.e. (A,w)=C) then w has 11’s. If δ(A,w)=C then w = x1y, where δ(A,x)=B

and y is the tail of w that follows what gets to C the first time.

If δ(A,x)=B then x = z1 for some z. Thus w = z11y and has 11.

Start

1

0

A CB1

0 0,1

Prove: if w no 11’s then w in L(DFA)

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Exercise 2.2.2 p53: prove (q, xy)=((q, x), y)

Recall: xy is the concatenation of strings x and y

When a is last character in string, true by definition of delta_hat: (q, xa)=((q, x), a)

Base case is true Complete proof by induction on |y|

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Let y=za((q,x),y)=((q,x),za) let(q,x)=p

=(p,za) =((p,z),a) def of txt p49 =(((q,x),z)a) substitution for p

Assume (q,xz)=((q,x),z) for strings shorter than ySince |z|<|y|((q,x),y)=(((q,x),z)a)=((q,xz),a)

=(q,xza) def of txt p49 =(q,xy) substitution for za

prove (q,xy)=((q,x),y) by induction on |y|

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Review: Assignment 2Due 9-10-14Exercise 2.2.9a, text p 54Given: A=(Q,,,q0,{qf}) and (q0, a)=(qf, a) for all a in Prove: (q0, w) = (qf, w) for all w with |w| > 0 using induction on the length of the string.  Your proof must include the following:(1) truth of base case(2) statement of inductive hypothesis(3) application of inductive hypothesis

Hint: let w=xa and used definition of delta_hat 

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Exercise 2.2.9b, text p54Given: A = (Q, , , q0, {qf}) and (q0, w) = (qf, w) iff w in L(A) with |w| > 0 Prove by induction on integers that if |x| > 0 and x is in L(A) then xk (k > 0 concatenations of x) is also in L(A) Your proof must include the following:(1) truth of base cases k=1 and k=2(2) statement of inductive hypothesis k>3(3) application of inductive hypothesis to show that if S(k-1) then S(k)

Note: the proof is about membership in L(A) Hint: use the result from exercise 2.2.2 in text p53

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Review: designing finite automataExercise 2.2.4a: A=({A,B,C},{0,1},,A,{C})

L(A)=set of input strings ending in 00Graph A

Exercise 2.2.10: A= 0 1->A A B

*B B A

L(A)=set input strings with an odd number of 1’s

Prove by induction on length of string that an input string with an odd number of 1’s is accepted by A

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Assignment #3 due 9-19-14Exercise 2.2.10: DFA= 0 1

->A A B *B B A

L(DFA)=set input strings with an odd number of 1’s

Prove by induction on length of string that (A,w)=A if w contains an even number of 1’s(A,w)=B if w contains an odd number of 1’s

Hints: Basis includes |w|=0 and |w|=1if w=xa, then a can be zero or oneif w=xa and a=0 then x and w contain the

same number of 1’s

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More hints on Assignment #3Exercise 2.2.10: DFA= 0 1

->A A B *B B A

Define parity(w) as whether w contains and even or odd number of ones.

Prove by induction on |w|If parity(w)=even, then (A,w)=AIf parity(w)=odd, then (A,w)=B

Basis includes w={e,0,1}. For each case, your proof must ask “what is parity(w)?” and “what is (A,w)?”.

Induction with w=xa has 4 cases: party(x)=even, a={0,1} and party(x)=odd, a={0,1}. For each case, your proof must ask the same questions as in the Basis.

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Regular Languages

A language L is regular if it consist entirely of strings accepted by some DFA. The DFA must accept only the strings in

L, no others.

Some languages LNR are not regular. No DFA exist that accepts all the string

in LNR

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Example of LNR

L1 = {0n1n | n ≥ 1}

L1 is the set of strings consisting of n 0’s followed by n 1’s, such that n is at least 1.

Thus, L1 = {01, 0011, 000111,…}

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Another LNR

L2 = {w | w in {(, )}* and w is balanced } alphabet consists of the parenthesis

symbols ’(’ and ’)’. Balanced parentheses are those that can

appear in an arithmetic expression. (), ()(), (()), (()()), etc

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Many Languages are Regular

We will discuss 4 ways to describe a Regular Language DFA’s Non-deterministic FA’s Non-deterministic FA’s with -

transtions Regular expressions

Will show at all are equivalent

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