Languages

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‫‪2ndclass‬‬ ‫‪Computation Theory‬‬ ‫النظرية االحتسابية‬ ‫برهجيات‪ -‬نظن هعلوهات‬ ‫استاذة الواده‪ :‬م‪.‬م‪ .‬نهى جويل ابراهين‬

_Computation Theory __________________________________________________________MSC Nuha J Ibrahim _

Reference: Introduction to Computer Theory/ Cohen

Algorithms for Compiler Design / O.G. Kakde .

Languages Till half of this century several people define the language as a way of understanding between the same group of beings, between human beings, animals, and even the tiny beings, this definition includes all kinds of understanding, talking, special signals and voices. This definition works till the mathematician called Chomsky said that: the language can be defined mathematically as: 1. A set of letters which called Alphabet. This can be seen in any natural language, for example the alphabet of English can be defined as: E = {a, b,…, z} 2. By concatenate letters from alphabet we get words. 3. All words from the alphabet make language. Language can be classified into two types as follows: Natural Languages Languages Formal Languages The definition above used with natural language and formal language. As in the natural language not all concatenations make permissible words, the same things happen with the formal languages. Note: formal language deals with form not meaning. You can easily distinguish from this definition that: Alphabet: finite Words: finite Language: infinite There is one thing that we must not forget it, is the alphabet could be a set of an empty set (or null string) which is a string of no letters, denoted by (Λ).

_Computation Theory __________________________________________________________MSC Nuha J Ibrahim__

Definitions 1.

Function concatenation: we can concatenate the word xxx with the word xx we can obtain the word xxxxx Xn concatenated to Xm = Xn+m Example let a=xx, b=xxx then ab=xxxxx

2. Function length: used to find the length of any word in a language. Example lets a=xxxx, b=543, c= Λ then length(a)=4, length(b)=3, length(c)=0 Note: in case where parentheses are letters of the alphabet: S={x ( ) } Then length(xxxxx)=5 But length((xx)(xxx))=9 3.

Function reverse: if a is a word in some language, then reverse of a is the same string of letter spelled backward. lets a=xxxx, b=543, c= aab Example then reverse(a)=xxxx, reverse(b)=345, reverse(c)=baa. Here we want to mention that if we apply this function on words some times the result does not satisfied with the definition of the language. Example

Let A an alphabet of the language L1 be {0 1 2 3 4 5 6 7 8 9} Let L1 = {all words that does not start with zero} And c=210 Then reverse(c)=012, which is not in L1.

4. Palindrome : is a language that={ Λ, all strings x such that reverse(x)=x} aba, aabaa, bab, bbb,… Example 5. Kleene star * : Given an alphabet ∑ we wish to define a language in which any string of letters from ∑ is a word, even the null string, this language we shall call the closure of the alphabet. if ∑={a,b,c} Example Then ∑*={Λ a b c aa ab ac ba bb bc ca cb cc aaa aab aac bba bbb bbc cca ccb ccc aaaa aaab …} Lets S= alphabet of language S*= closure of the alphabet


Example

S= {x} S*= { Λ, xn |n>=1} To prove a certain word in the closure language S* we must show how it can be written as a concatenation of words from the set S. Let S={a,ab} Example To find if the word abaab is in S* or not, we can factor it as follows: (ab)(a)(ab) Every factor in this word is a word in S* so as the whole word abaab. In the above example, there is no other way to factor this word that we called unique. While, some times the word can be factored in different ways. Example Lets S={x} n S*={ Λ, x | n>=1} If we factor the word xxxxxxxx, it can be: (xx) (xx) (xx) (xx), or (x) (xx) (xxx) (xx), or (xxx) (xxx) (xx), … We can obviously say that the method of proving that something exists by showing how to create it is called proof by constructive algorithm. If S=ø then S*= { Λ} This is not the same If S= { Λ} Then S*={ Λ} If S= {w1 w2 w3} Then S*= { Λ w1 w2 w3 w1w1w1 w1w1w2 w1w2w1…} And S+= { w1 w2 w3 w1w1w1 w1w1w2 w1w2w1…} Which is mean that S+=S* except the Λ Theorem1 S*= S** Proof Every word in S** is made up of factors from S*. Every factor from S* is made up of factors from S. Therefore, every word in S** is made up of factors from S. Therefore, every word in S** is also a word in S*. we can write this as: S** S* Or S* S** This mean S*=S**.


Regular Expression The language-defining symbols we are about to create are called regular expressions. The languages that are associated with these regular expressions are called regular languages. Example consider the language L where L={Λ x xx xxx …} by using star notation we may write L=language(x*). Since x* is any string of x's (including Λ). Example

if we have the alphabet ∑={a,b} And L={a ab abb abbb abbbb …} Then L=language(ab*)

Example

(ab)*= Λ or ab or abab or ababab or abababab or ….

L1=language(xx*) Example The language L1 can be defined by any of the expressions: xx* or x+ or xx*x* or x*xx* or x+x* or x*x+ or x*x*x*xx* … Remember x* can always be Λ. Example

language(ab*a)={aa aba abba abbba abbbba …}

Example language(a*b*)={ Λ a b aa ab bb aaa aab abb bbb … } ba and aba are not in this language so a*b* ≠ (ab)* the following expressions both define the language L2={xodd}: x(xx)* or (xx)*x But the expression x*xx* does not since it includes the word (xx)x(x).

Example

Example

consider the language T defined over the alphabet ∑={a,b,c} T={a c ab cb abb cbb abbb cbbb abbbb cbbbb …} Then T=language((a+c)b*) T=language(either a or c then some b's)

Example consider a finite language L that contains all the strings of a's and b's of length exactly three. L={aaa aab aba abb baa bab bba bbb} L=language((a+b)(a+b)(a+b)) L=language((a+b)3)


Note from the alphabet ∑={a,b} , if we want to refer to the set of all possible strings of a's and b's of any length (including Λ) we could write (a+b)* Example we can describe all words that begins with a and end with b with the expression a(a+b)*b which mean a(arbitrary string)b Example if we have the expression (a+b)*a(a+b)* then the word abbaab can be considerd to be of this form in three ways: (Λ)a(bbaab) or (abb)a(ab) or (abba)a(b) Example

(a+b)*a(a+b)*a(a+b)* = (some beginning)(the first important a)(some middle)(the second important a)(some end) Another expressions that denote all the words with at least two a's are: b*ab*a(a+b)*, (a+b)*ab*ab*, b*a(a+b)*ab* Then we could write: language((a+b)*a(a+b)*a(a+b)*) =language(b*ab*a(a+b)*) =language((a+b)*ab*ab*) =language(b*a(a+b)*ab*) =all words with at least two a's. Note: we say that two regular expressions are equivalent if they describe the same language. Example

if we want all the words with exactly two a's, we could use the expression: b*ab*ab* which describe such words as aab, baba, bbbabbabbbb,…

Example

the language of all words that have at least one a and at least one b is: (a+b)*a(a+b)*b(a+b)*+(a+b)*b(a+b)*a(a+b)*

Note: (a+b)*b(a+b)*a(a+b)* ≠ bb*aa* since the left includes the word aba, which the expression on the right side does not. Note:

(a+b)* = (a+b)* + (a+b)* (a+b)* = (a+b)*(a+b)* (a+b)* = a(a+b)* + b(a+b)* + Λ (a+b)* = (a+b)*ab(a+b)* + b*a*

Note: usually when we employ the star operation we are defining an infinite language. We can represent a finite language by using the plus alone.


Example

L={abba baaa bbbb} L=language(abba + baaa + bbbb)

Example

L={ Λ a aa bbb} L=language(Λ + a + aa + bbb)

L={Λ a b ab bb abb bbb abbb bbbb …} Example We can define L by using the expression b* + ab* Definition The set of regular expressions is defined by the following rules: Rule1: every letter of ∑ can be made into a regular expression, Λ is a regular expression. Rule2: if r1 and r2 are regular expressions, then so are: (r1) r1r2 r1+r2 r1*. Rule3: nothing else is a regular expression. Remember that r1+=r1r1* Definition If S and T are sets of strings of letters (whether they are finite or infinite sets), we define the product set of strings of letters to be: ST={all combination of a string from S concatenated with a string from T} Example if S={a aa aaa} T={bb bbb} Then ST={abb abbb aabb aabbb aaabb aaabbb} (a+aa+aaa)(bb+bbb)=abb+abbb+ aabb+aabbb+aaabb+aaabbb) Example if P={a bb bab} Q={Λ bbbb} Then PQ={a bb bab abbbb bbbbbb babbbbb} (a+bb+bab)( Λ+bbbb)=a+bb+bab+ab4+b6+bab5 Example if M={Λ x xx} N={Λ y yy yyy yyyy …} Then MN={Λ y yy yyy yyyy … x xy xyy xyyy xyyyy … xx xxy xxyy xxyyy xxyyyy …} Using regular expression we could write: (Λ+x+xx)(y*)=y*+xy*+xxy*


Definition The following rules define the language associated with any regular expression. Rule1: the language associated with the regular expression that is just a single letter is that one-letter word alone and the language associated with Λ is just{Λ}, a one-word language. Rule2: if r1 is regular expression associated with the language L1 and r2 is regular expression associated with the language L2 then: i) The regular expression (r1)(r2) is associated with the language L1 times L2. Language(r1r2)=L1L2 ii) The regular expression r1+r2 is associated with the language formed by the union of the sets L1 and L2. Language(r1+r2)=L1+L2 iii) The language associated with the regular expression (r1)* is L1*, the kleene closure of the set L1 as a set of words. Language(r1)*=L1* Example L={baa abba bababa} The regular expression for this language is: (baa+abba+bababa) Example L={Λ x xx xxx xxxx xxxxx} The regular expression for this language is: (Λ+x+xx+xxx+xxxx+xxxxx) =(Λ+x)5

Example L= language((a+b)*(aa+bb)(a+b)*) =(arbitrary)(double letter)(arbitrary) {Λ a b ab ba aba bab abab baba …} these words are not included in L but they included by the regular expression: (Λ+b)(ab)*(Λ+a) Example E=(a+b)*a(a+b)*(a+Λ)(a+b)*a(a+b)* E=(a+b)*a(a+b)*a(a+b)*a(a+b)*+(a+b)*a(a+b)*Λ(a+b)*a(a+b)* We have: (a+b)*Λ(a+b)*=(a+b)* Then: E=(a+b)*a(a+b)*a(a+b)*a(a+b)*+(a+b)*a(a+b)*a(a+b)* The language associated with E is not different from the language associated with: (a+b)*a(a+b)*a(a+b)* Note: (a+b*)*=(a+b)* (a*)*=a* (aa+ab*)*≠ (aa+ab)* (a*b*)*=(a+b)* Example E=[aa+bb+(ab+ba)(aa+bb)*(ab+ba)]* Even-even={Λ aa bb aabb abab abba baab baba bbaa aaaabb aaabab…}


Finite Automata (FA) A finite automata is a collection of five things: 1. A finite set of states 2. An alphabet ∑ of possible input letters from which are formed strings that are to be read one letter at a time. 3. A finite set of transitions that tell for each state and for each letter of the input alphabet which state to go to next. 4. The initial state; and 5. The set of final states. Therefore formally a finite automata is a five-tuple: where:

Q is a set of states of the finite automata, Σ is a set of input symbols, and δ specifies the transitions in the automata. If from a state p there exists a transition going to state q on an input symbol a, then we write δ (p, a) = q. Hence, δ is a

function whose domain is a set of ordered pairs, (p, a), where p is a state and a is an input symbol, and the range is a set of states. Therefore δ defines a mapping from q0 is the initial state, and F is a set of final sates of the automata. For example: where

The transition diagram of this automata is:

Transition Diagram

Transition Table

For example:

where

0

Let x be 010. To find out if x is accepted by the automata or not, we proceed as follows: δ 1(q0, 0) = δ (q0, 0) = q1 Therefore, δ 1 (q0, 01 ) = δ {δ 1 (q0, 0), 1} = q0 δ 1 (q 0, 010) = δ {δ 1 (q 0, 0 1), 0} = q 1

In the finite automata discussed above, since δ defines mapping from Q × Σ to Q, there exists exactly one transition from a state on an input symbol; and therefore, this finite automata is considered a deterministic finite automata (DFA). Therefore, we define the DFA as the finite automata:

where: M = (Q, Σ, δ , q , F ), such that there exists exactly one transition from a state on a input symbol.

Example if ∑={a,b}, states={x,y,z} Rules of transition: 1. From state x and input a go to state y. 2. From state x and input b go to state z. 3. From state y and input a go to state x. 4. From state y and input b go to state z. 5. From state z and any input stay at the state z. Let x be the start state and z be the final state. a xy a b b z+ a,b Transition Diagram The FA above will accept all strings that have the letter b in them and no other strings. The language associated with(or accepted by) this FA is the one defined by the regular expression: a*b(a+b)* The set of all strings that do leave us in a final state is called the language defined by the FA. The word abb is accepted by this FA, but The word aaa is not. a b xy Z y x Z z+ z Z Transition Table


Example The following FA accept all strings from the alphabet {a,b} except Λ. a + b

a,b The regular expression is: (a+b)(a+b)*=(a+b)+ Example The following FA accept all words from the alphabet {a,b}. +a,b

The regular expression is: (a+b)*

Note: every language that can be accepted by an FA can be defined by a regular expression and every language that can be defined by a regular expression can be accepted by some FA. FA that accepts no language will be one of the two types: 1. FA that have no final states. Like the following FA: a b

a,b 2. FA in which the final states cannot be reached. Like the following FA: -

b

a,b

a

a,b a

+

b

a,b

Or Like the following FA: -

a,b

a,b

+

a,b Example The following FA accept all strings from the alphabet {a,b} that start with a. b

The regular expression is: a(a+b)*

a

a,b + a,b


Example The following FA accept all strings from the alphabet {a,b} with double letter. a

a b a

-

+

b

b

a,b

The regular expression is: (a+b)*(aa+bb) (a+b)* Example the following FA accepts the language defined by the regular expression: (a+b)(a+b)b(a+b)* a a,b

a,b

-

a,b b

+

a,b Example the following FA accepts only the word baa. b

-

a

a b

b

a

+

a,b

a,b Example the following FA accepts the words baa and ab. +

b

a

a,b

a

-

b

a b

a

b a,b

a,b

+


Example the following FA accepts the language defined by the regular expression: (a+ba*ba*b)+ b a a a

b

b b

+

a

Example the following FA accepts the language defined by the regular expression: (a+ba*ba*b)* a

+-

b

b

b

a

a Example the following FA accepts only the word Λ. a,b

+ -

a,b Example the following FA accepts all words from the alphabet {a,b} that end with a. b

a b

+ a

The regular expression for this language is: (a+b)*a


Example the following FA accepts all words from the alphabet {a,b} that do not end in b and accept Λ. b

+ -

a

a

b

The regular expression for this language is: (a+b)*a + Λ Example the following FA accepts all words from the alphabet {a,b} with an odd number of a's. a

-

+

a

b

b

The regular expression for this language is: b*a(b*ab*ab*)* Example the following FA accepts all words from the alphabet {a,b} that have different first and last letters. b

+

a

a

b

a

b

a

+

b b

a

The regular expression for this language is: a(a+b)*b + b(a+b)*a Example the following FA accepts the language defined by the regular expression (even-even): [aa+bb+(ab+ba)(aa+bb)*(ab+ba)]* b b

+a

a

a b b

a


Transition Graph A Transition Graph (TG) is a collection of three things: 1. A finite set of states, at least one of which is designed as the start state (-) and some (may be none) of which are designed as final states (+). 2. An alphabet ∑ of possible input letters from which input string are formed. 3. A finite set of transitions that show how to go from one state to another based on reading specified substrings of input letters (possibly even the null string Λ). Example the following TG accepts the word baa. baa

-

+

Example the following TG accepts the words with double letters. aa,bb

a,b

+ a,b

Example the following TG accepts the word baab in two different ways. ba

ab

-

+ baa

b

Note: in TG some words have several paths accept them while in FA there is only one Note: every FA is also a TG. Example the following TG accept nothing. Example the following TG accept Λ. +-


Example the following TG accept the words {Λ, baa, abba}. Λ abba + baa

-

Example the following TG accept all words that end with b. b

-

+

a,b Example the following TG accept all words that have different first and last letters. b

+

a a,b

b

a

+

a,b

Example the following TG accept all words in which a's occur in even clumps only and end in three or more b's. aa

b

b

b

aa b

Example the following TG for even-even. + aa,bb

ab,ba ab,ba aa,bb

+


Kleen's Theorem

Kleen's Theorem Any language that can be defined by: 1. Regular expression or 2. Finite automata or 3. Transition graph Can be defined by all three methods. Proof The three sections of our proof will be: Part1: every language that can be defined by a FA can also be defined by a TG. Part2: every language that can be defined by a TG can also be defined by a RE. Part3: every language that can be defined by a RE can also be defined by a FA. The proof of part1 This is the easiest part. Every FA is itself a TG. Therefore, any language that has been defined by a FA has already been defined by a TG. The proof of part2 The proof of this part will be by constructive algorithm. This means that we present a procedure that starts out with a TG and ends up with a RE that defines the same language. Let the start states be only one. b 2 1 b Λ 2 -1 ab 4 Λ 3 Becomes ab 4 -3 aa Λ aa 5 -5 Let the final states be only one. b 9+ aa ab

Becomes 12+

b aa ab

9

12

Λ Λ

+


Allow the edges to be labeled with regular expressions (reduce the number of edges or states in each time). r3 Becomes

r2

x

x r1+r2+r3

r1 r1

3

4

r2

2

r1

3

2

r1

3

r2

r3

4

4

Becomes

3

Becomes

r1+r2

4

r1r2

4

2

Becomes

2

r1r2*r3

r2

r1

1

2

4 3

r1r2*r3

3

r3 r4

4

r5

1

Becomes

4

r1r2*r5

5

r2

r1r2*r4

5 r4r2*r3

1

r1

2

r4

Becomes

3

1

2

r1r2*r3

r2

r4

3

r3 r2

Repeat the last step again and again until we eliminate all the states from TG except the unique start state and the unique final state. -

r1 r2 rn

+

Becomes

-

r1+r2+…+rn

+


Example Find the RE that defines the same language accepted by the following TG using Kleenes theorem. +

aa -

aa,bb bb +

a,b

aa -

Λ

aa,bb

+ bb

Λ

a,b aa -

Λ

aa+bb

+ bb

Λ

a+b

-

aa+bb

aa bb a+b

-

Λ

(aa+bb)(a+b)*aa

(aa+bb)(a+b)*bb

-

+

(aa+bb)(a+b)*aa (aa+bb)(a+b)*bb

RE=(aa+bb)(a+b)*(aa+bb)

+ Λ

+


H.W Find the RE that defines the same language accepted by the following TG using Kleenes theorem. ab,ba

+ -

ab,ba

aa,bb

aa,bb

The proof of part3 Rule1: there is an FA that accepts any particular letter of the alphabet. There is an FA that accepts only the word Λ. For example, if x is in ∑, then the FA will be:

All ∑

All ∑ axcept x All ∑

-

x

+

FA accepts only Λ will be: +-

a,b a,b

Rule2: if there is an FA called FA1, that accepts the language defined by the regular expression r1 and there is an FA called FA2, that accepts the language defined by the regular expression r2, then there is an FA called FA3 that accepts the language defined by the regular expression (r1+r2). We can describe the algorithm for forming FA3 as follows: Starting with two machines FA1, with states x1, x2, x3,…. And FA2 with states y1,y2,y3,…,build a new machine FA3 with states z1,z2,z3,… where each z is of the form "xsomething or ysomething". If either the x part or the y part is a final state, then the corresponding z is a final state. To go from one z to another by reading a letter from the input string, we see what happens to the x part and to the y part and go to the new z accordingly. We could write this as a formula: znew after letter p=[xnew after letter p]or[ynew after letter p]


Example We have FA1 accepts all words with a double a in them, and FA2 accepts all words ending in b. we need to build FA3 that accepts all words that have double a or that end in b. -x1 b

a b

x2

a

FA1

+x3

-y1

a,b

a

b a FA2

a b -x1 x2 x1 x2 x3 x1 +x3 x3 x3 The transition table for FA1 a b -y1 y1 y2 +y2 y1 y2 The transition table for FA2 z1=x1 or y1 z2= x2 or y1 z3=x1 or y2 z4=x3 or y1 z5=x3 or y2 a b -z1 z2 z3 z2 z4 z3 +z3 z2 z3 +z4 z4 z5 +z5 z4 z5 The transition table for FA3 -z1 b

a b

z2 a

+z3 b

a +z4

FA3

a

b a

+z5 b

+y2 b

H.W. Let FA1 accepts all words ending in a, and let FA2 accepts all words with an odd number of letters (odd length). Build FA3 that accepts all words with odd length or end in a, using Kleenes theorem. -x1

a b

+x2

b

-y1

a,b a,b

+y2

a FA1

FA2

Rule3: if there is an FA1 that accepts the language defined by the regular expression r1 and an FA2 that accepts the language defined by the regular expression r2, then there is an FA3 that accepts the language defined by the concatenation r1r2. We can describe the algorithm for forming FA3 as follows: We make a z state for each none final x state in FA1. And for each final state in FA1 we establish a z state that expresses the option that we are continuing on FA1 or are beginning on FA2. From there we establish z states for all situations of the form: Are in xsomething continuing on FA1 Or Have just started y1 about to continue on FA2 Or Are in ysomething continuing on FA2 Example We have FA1 accepts all words with a double a in them, and FA2 accepts all words ending in b. we need to build FA3 that accepts all words that have double a and end in b. -x1 b

a b

x2 FA1

a

+x3

-y1

a,b

a

a b -x1 x2 x1 x2 x3 x1 +x3 x3 x3 The transition table for FA1

b a FA2

+y2 b

-y1 +y2

a

b

y1 y1

y2 y2

The transition table for FA2 z1=x1 z2= x2 z3=x3 or y1 z4=x3 or y2 or y1 a z2 z3 z3 z3

-z1 z2 z3 +z4

b z1 z1 z4 z4

The transition table for FA3

-z1

a

z2

b

a

z3

b

a

b a

+z4 b

FA3 H.W. Let FA1 accepts all words with a double a in them, and let FA2 accepts all words with an odd number of letters (odd length). Build FA3 that accepts all words with odd length and have double a in them using Kleen’s theorem. -x1 b

a b

x2 FA1

a

+x3 a,b

-y1

a,b a,b FA2

+y2

Rule4: if r is a regular expression and FA1 accepts exactly the language defined by r, then there is an FA2 that will accept exactly the language defined by r*. We can describe the algorithm for forming FA2 as follows: Each z state corresponds to some collection of x states. We must remember each time we reach a final state it is possible that we have to start over again at x1. Remember that the start state must be the final state also. Example If we have FA1 that accepts the language defined by the regular expression: r=a*+aa*b We want to build FA2 that accept the language defined by r*. a +x1 -

a

b

+x2 a,b

b

+x3 a,b

x4 FA1 a b -,+ x1 x2 x4 +x2 x2 x3 +x3 x4 x4 x4 x4 x4 The transition table for FA1 z1=x1 z2=x4 z3=x2 or x1 z4=x3 or x4 or x1 z5=x4 or x2 or x1 a b -,+ z1 z3 z2 z2 z2 z2 +z3 z3 z4 +z4 z5 z2 +z5 z5 z4 The transition table for FA2

a

a +z1 -

a

b

b

+z3 a,b

a

+z4

+z5

b

b

z2 FA2 H.W. Let FA1 accept the language defined by r1, find FA2 that accept the language defined by r1* using Kleene's theorem. r1= aa*bb* a -x1

a

b

x2 a,b x3 FA1

b b

+x4 a

NONDETERMINISM A nondeterministic finite automaton (NF) is a collection of three things: 1. A finite set of states with one start state (-) and some final states (+). 2. An alphabet ∑ of possible input letters. 3. A finite set of transitions that describe how to proceed from each state to other states along edges labeled with letters of the alphabet (but not the null string Λ), where we allow the possibility of more than one edge with the same label from any state and some states for which certain input letters have no edge. We can convert any NFA into a TG with no repeated labels from any single state as in the following: 2 a a

1

3

a 4

Is equivalent to 2 a 1

3 a

Λ

4 a

Λ

Λ

Any FA will satisfy the definition of an NFA. We have: 1. Every FA is an NFA. 2. Every NFA has an equivalent TG. 3. By Kleen's theorem, every TG has an equivalent FA. Therefore: Language of FA's language of NFA's language of TG's = language of FA's Theorem FA = NFA By which we mean that any language defined by a nondeterministic finite automaton is also definable by a deterministic (ordinary) finite automaton and vice versa.

NON-DETERMINISTIC FINITE AUTOMATA If the basic finite automata model is modified in such a way that from a state on an input symbol zero, one or more transitions are permitted, then the corresponding finite automata is called a "non-deterministic finite automata" (NFA).

Therefore, an NFA is a finite automata in which there may exist more than one paths corresponding to x in Σ* (because

zero, one, or more transitions are permitted from a state on an input symbol). Whereas in a DFA, there exists exactly one path corresponding to x in Σ*. Hence, an NFA is nothing more than a finite automata: in which δ defines mapping from Q × Σ to 2 (to take care of zero, one, or more transitions). For example, consider the

finite automata shown below:

The transition diagram of this automata is:

Q

Example Let FA1 be

a a a

a

b

-

b

b

+

a,b

b And let FA2 be

a b

b

+

a

-

a,b

a

b

Then NFA3= FA1 + FA2 is a

a a,b

a b

b a

+ b

a b

a

a

b a,b

b + b

It is sometimes easier to understand what a language is from the picture of an NFA that accepts it than from the picture of an FA as in the following example. Example The NFA and FA Below accepts the language of all words that contains either a triple a (the substring aaa) or a triple b (the substring bbb) or both.

a

a

+

a a,b

a,b

b

b

b

+ a,b

NFA

a a b a

-

a

b

b

+

b

b

a,b

a FA COMPARISON TABLE FOR AUTOMATA Start states Final states Edge labels Number of edges from each state Deterministic (every input string has one path) Every path represents one word

FA one Some or none Letter from ∑ One for each letter in ∑

TG One or more Some or none words from ∑*

NFA one Some or none Letter from ∑

Arbitrary

Arbitrary

Yes

Not necessarily

Not necessarily

Yes

Yes

Yes

Acceptance of Strings by Non-deterministic Finite Automata Since an NFA is a finite automata in which there may exist more than one path corresponding to x in Σ*, and if this is, indeed, the case, then we are required to test the multiple paths corresponding to x in order to decide whether or not x is accepted by the NFA, because, for the NFA to accept x, at least one path corresponding to x is required in the NFA. This path should start in the initial state and end in one of the final states. Whereas in a DFA, since there exists exactly one path corresponding to x in Σ*, it is enough to test whether or not that path starts in the initial state and ends in one of the final states in order to decide whether x is accepted by the DFA or not. Therefore, if x is a string made of symbols in Σ of the NFA (i.e., x is in Σ*), then x is accepted by the NFA if at least one path exists that corresponds to x in the NFA, which starts in an initial state and ends in one of the final states of the NFA. Since x is a member of Σ* and there may exist zero, one, or more transitions from a state on an input symbol, we Q

Q

define a new transition function, δ 1, which defines a mapping from 2 × Σ* to 2 ; and if δ 1 ({q0},x) = P, where P is a set For example, consider the finite automata shown below:

where:

If x = 0111, then to find out whether or not x is accepted by the NFA, we proceed as follows:

THE NFA WITH ∈-MOVES If a finite automata is modified to permit transitions without input symbols, along with zero, one, or more transitions on the input symbols, then we get an NFA with ‘∈ -moves,’ because the transitions made without symbols are called "∈ -transitions." Consider the NFA shown in.

Finite automata with ∈ -moves. This is an NFA with ∈ -moves because it is possible to transition from state q0 to q1 without consuming any of the input symbols. Similarly, we can also transition from state q1 to q2 without consuming any input symbols. Since it is a finite automata, an NFA with ∈ -moves will also be denoted as a five-tuple:

where Q, Σ, q0, and F have the usual meanings, and δ defines a mapping from

(to take care of the ∈ -transitions as well as the non ∈ -transitions).

Acceptance of a String by the NFA with ∈-Moves A string x in Σ* will ∈ -moves will be accepted by the NFA, if at least one path exists that corresponds to x starts in an initial state and ends in one of the final states. But since this path may be formed by ∈ -transitions as well as non-∈ -transitions, to find out whether x is accepted or not by the NFA with ∈ -moves, we must define a function,

∈ -closure(q), where q is a state of the automata. The function ∈ -closure(q) is defined follows:

∈ -closure(q)= set of all those states of the automata that can be reached from q on a path labeled by ∈. For example, in the NFA with ∈ -moves given above:

∈ -closure(q0) = { q0, q1, q2} ∈ -closure(q1) = { q1, q2} ∈ -closure(q2) = { q2} The function

∈ -closure (q) will never be an empty set, because q is always reachable from itself, without dependence on any input symbol; that is, on a path labeled by ∈ , q will always exist in ∈ -closure(q) on that labeled path. If P is a set of states, then the ∈ -closure function can be extended to find ∈ -closure(P ), as follows:

Since x is a member of Σ*, and there may exist zero, one, or more transitions from a state on an input symbol, we Q

Q

define a new transition function, δ , which defines a mapping from 2 × Σ* to 2 . If x is written as wa, where a is the last symbol of x and w is a string made of remaining symbols of x then:

Q

Q

since δ 1 defines a mapping from 2 × Σ* to 2 .

such that P contains at least one member of F and:

For example, in the NFA with ∈ -moves, given above, if x = 01, then to find out whether x is accepted by the automata or not, we proceed as follows:

Therefore:

∈ -closure( δ 1 (∈ -closure (q0), 01) = ∈ -closure({q1}) = {q1, q 2} Since q2 is a final state, x = 01 is accepted by the automata.

Equivalence of NFA with ∈-Moves to NFA Without ∈-Moves For every NFA with ∈ -moves, there exists an equivalent NFA without ∈ -moves that accepts the same language. To obtain an equivalent NFA without 1 ∈ -moves, given an NFA with ∈ -moves, what is required is an elimination of

∈ -transitions from a given automata. But simply eliminating the ∈ -transitions from a given NFA with ∈ -moves will change the language accepted by the automata. Hence, for every ∈ -transition to be eliminated, we have to add some non-∈ -transitions as substitutes in order to maintain the language's acceptance by the automata. Therefore, transforming an NFA with ∈ -moves to and NFA without ∈ -moves involves finding the non-∈ -transitions that must be added to the automata for every ∈ -transition to be eliminated.

Consider the NFA with ∈ -moves shown in

Therefore, by adding these non-∈ -transitions, and by making the initial state one of the final states, we get the automata shown in.

Therefore, when transforming an NFA with ∈ -moves into an NFA without ∈ -moves, only the transitions are required to be changed; the states are not required to be changed. But if a given NFA with q0 and ∈ -moves accepts ∈ (i.e., if the ∈ -closure (q0) contains a member of F), then q0 is also required to be marked as one of the final states if it is not already a member of F. Hence: If M = (Q, Σ, δ , q0, F) is an NFA with ∈ -moves, then its equivalent NFA without ∈ -moves will be M1 = (Q, Σ, δ 1, q0, F1) where δ 1 (q, a) = ∈ -closure( δ ( ∈ -closure(q), a)) and F1 = F ∪ (q0) if ∈ -closure (q0) contains a member of F F1 = F otherwise For example, consider the following NFA with ∈ -moves:

M=({q0,q1,q2},{0,1}, δ, q0, {q2}) Where δ q0 q1 q2

0 {q0} Ф Ф

1 Ф {q1} {q2}

λ {q1} {q2} Ф

Its equivalent NFA without λ- moves will be: M=({q0,q1,q2},{0,1}, δ, q0, {q0,q2}) δ1 q0 q1 q2

0 {q0,q1,q2} Ф Ф

1 {q1,q2} {q1,q2} {q2}

FINITE AUTOMATA WITH OUTPUT We shall investigate two different models for FA's with output capabilities; these are Moore machine and Mealy machine. A Moore machine is a collection of five things: 1. A finite set of states q0,q1,q2,… where q0 is designed as the start state. 2. An alphabet of letters for forming the input string ∑= { a, b, c, …}. 3. An alphabet of possible output characters Г = { x, y, z, …}. 4. A transition table that shows for each state and each input letter what state is reached next. 5. An output table that shows what character from Г is printed by each state that is entered. A Moore machine does not define a language of accepted words, since every input string creates an output string and there is no such thing as a final state. The processing is terminated when the last input letter is read and the last output character is printed. Example Input alphabet: ∑ = {a, b} Output alphabet: Г = {0, 1} Names of states: q0, q1, q2, q3. (q0 = start state) Old state -q0 q1 q2 q3

Transition table New state After input a after input b q1 q3 q3 q1 q0 q3 q3 q2

The Moore machine is:

b q0/1 a q2/0

a

q1/0

b b b

a q3/1 a

Output table (the character printed in the old state) 1 0 0 1

Note: the two symbols inside the circle are separated by a slash "/", on the left side is the name of the state and on the right is the output from that state. If the input string is abab to the Moore machine then the output will be 10010. Example The following Moore machine will "count" how many times the substring aab occurs in a long input string. b

a

a q0/0

a

q1/0

q2/0

b

b

q3/1

a b

The number of substrings aab in the input string will be exactly the number of 1's in the output string. Input string State Output

q0 0

a q1 0

a q2 0

a q2 0

b q3 1

a q1 0

b q0 0

b q0 0

a q1 0

a q2 0

b q3 1

A Mealy machine is a collection of four things: 1. A finite set of states q0,q1,q2,… where q0 is designed as the start state. 2. An alphabet of letters for forming the input string ∑= { a, b, c, …}. 3. An alphabet of possible output characters Г = { x, y, z, …}. 4. A pictorial representation with states represented by small circles and directed edges indicating transitions between states. Each edge is labeled with a compound symbol of the form i/o where i is an input letter and o is an output character. Every state must have exactly one outgoing edge for each possible input letter. The edge we travel is determined by the input letter i; while traveling on the edge we must print the output character o.

b q0 0

Example The following Mealy machine prints out the 1's complement of an input bit string. 1/0,0/1 q0 If the input is 001010 the output is 110101. This is a case where the input alphabet and output alphabet are both {0,1}. Example The following Mealy machine called the increment machine. 0/0,1/1 0/1

no carry

0/1

start

1/0 carry

1/0 If the input is 1011 the output is 1100. Definition Given the Mealy machine Me and the Moore machine Mo, which prints the automatic start-state character x, we will say that these two machines are equivalent if for every input string the output string from Mo is exactly x concatenated with the output from Me. Note: we prove that for every Moore machine there is an equivalent Mealy machine and for every Mealy machine there is an equivalent Moore machine. We can then say that the two types of machine are completely equivalent. Theorem If Mo is a Moore machine, then there is a Mealy machine Me that is equivalent to it. Proof The proof will be by constructive algorithm.

a

a/t

b

Becomes

q4/t

b/t

b

q4

b/t

Example Below, a Moore machine is converted into a Mealy machine: q1/1

a q0/0

a

b b

q2/0

a,b q3/1

q1

a/1 q0

becomes

b

a/1

b/0 b/0

a

a/1,b/1 q3

b/1

q2 a/0

Theorem For every Mealy machine Me there is a Moore machine Mo that is equivalent to it. Proof The proof will be by constructive algorithm. a

a/1 b/1

q4

becomes

b

q4/1 b

b/1

If there is more than one possibility for printing as we enter the state, then we need a copy of the state for each character we might have to print. (we may need as many copies as there are character in Г). a/0 b/1 b/0

a

b/1

q4

becomes a/1

b

b/1

b/1 b

1

q4 /0 a/1

q42/1 a/1

a a/0

q41/0

q6 a/0

Becomes

q4

a/0

b

q6

q42/1

b/1

a/0

b

Example Convert the following Mealy machine to Moore machine: b/1

q1

a/1

a/0

q0

a/1 b/0

b/0

q2 a/0

b/1

q3 Mealy machine b

q1/1

a

a

q01/0 q02/1

a b b

b

q21/0 a

a

b

a

b

q22/1

q3/0 Moore machine

Example Draw the Mealy machine for the following sequential circuit: input

NAND

A

OR

DELAY

B

OR

output

First we identify four states: q0 is A= 0 B= 0 q1 is A= 0 B= 1 q2 is A= 1 B= 0 q3 is A= 1 B= 1 the operation of this circuit is such that after an input of 0 or 1 the state changes according to the following rules: new B= old A new A = (input) NAND (old A OR old B) output = (input) OR (old B) Suppose we are in q0 and we receive the input 0. new B = old A = 0 new A = 0 NAND (0 OR 0) = 0 NAND 0 =1 output = 0 OR 0 = 0 the new state is q2 (since new A=1, new B=0) if we are in state q0 and we receive the input 1: new B= old A = 0 new A = 1 NAND (0 OR 0) =1 output = 1 OR 0 =1 the new state is q2. We repeat this process for every state and for each input to produce the following table: Old state After input 0 After input 1 New state Output New state Output q0 q2 0 q2 1 q1 q2 1 q0 1 q2 q3 0 q1 1 q3 q3 1 q1 1 q0

1/1

0/1

q1

1/1 1/1

q3

0/0,1/1 q2 0/0

0/1 Mealy machine

Comparison table for automata FA Start states

one

Final states

Some or none

Edge labels

Letters from ∑

Number of edges from each state deterministic output

One for each letter in ∑ yes no

NFA

NFA- Λ

Moore

Mealy

one

one

one

one

Some or none

Some or none

none

none

words from ∑*

Letters from ∑

Letters from ∑ or Λ

Letter from ∑

arbitrary

arbitrary

arbitrary

no no

no no

no no

TG One or more Some or none

One for each letter in ∑ yes yes

i/o i from ∑ o from Г One for each letter in ∑ yes yes

A phrase-Structure Grammar A phrase-Structure Grammar, called PSG, is a collection of three things: 1. An alphabet ∑ of letters called terminals. 2. A set of symbols called nonterminals that includes the start symbol S. 3. A finite set of productions of the form: String 1 → String 2 Where string 1 can be any string of terminals and nonterminals that contains at least one nonterminals and where string 2 is any string of terminals and nonterminals whatsoever. Definition: The language generated by the PSG is the set of all strings of terminals that can be derived starting at S. Example: the following is a phrase-structure grammar over Σ={a,b} with nonterminals X and S: S XS X aX | a aaaX ba In this language we can have the following derivation: S XS XXS XXXS XXX aXXX aaXXX aaaXXX baXX baaXX baaaX bba Example: S aSBA S abA AB BA bB bb bA ba aA aa

A context-sensitive grammar (CSG) A context-sensitive grammar (CSG) is a formal grammar in which the left-hand sides and right-hand sides of any production rules may be surrounded by a context of terminal and nonterminal symbols. Definition A formal grammar G = (N, Σ, P, S) Where N the Non - Terminal Σ the terminal P is context-sensitive if all rules in P are of the form αAβ → αγβ where A Є N (i.e., A is a single nonterminal), α,β Є (N U Σ)* ( α and β are strings of nonterminals and terminals) and γ Є (N U Σ)+ ( γ is a nonempty string of nonterminals and terminals). A rule of the form S → λ provided S does not appear on the right side of any rule where λ represents the empty string is permitted. The addition of the empty string allows the statement that the context sensitive languages are a proper superset of the context free languages, rather than having to make the weaker statement that all context free grammars with no →λ productions are also context sensitive grammars. Example: This grammar generates the context sensitive language: 1. 2. 3. 4. 5. 6. 7. 8. 9.

CONTEXT FREE GRAMMAR A context free grammar, called CFG, is a collection of three things: 1. An alphabet ∑ of letters called terminals from which we are going to make strings that will be the words of a language. 2. A set of symbols called nonterminals, one of which is the symbol S, standing for "start here". 3. A finite set of production of the form: One nonterminal → finite string of terminals and/ or nonterminals Where the strings of terminals and nonterminals can consist of only terminals or of any nonterminals, or any mixture of terminals and nonterminals or even the empty string. We require that at least one production has the nonterminal S as its left side. Definition The language generated by the CFG is the set of all strings of terminals that can be produced from the start symbol S using the production as substitutions. A language generated by the CFG is called a context free language (CFL). Example Let the only terminal be a. Let the only nonterminal be S. Let the production be: S → aS S→Λ The language generated by this CFG is exactly a*. In this language we can have the following derivation: S → aS → aaS → aaaS → aaaaS → aaaaaS → aaaaaΛ = aaaaa

Example Let the only terminal be a. Let the only nonterminal be S. Let the production be: S → SS S→a S→Λ The language generated by this CFG is also just the language a*. In this language we can have the following derivation: S → SS → SSS → SaS → SaSS → ΛaSS → ΛaaS → ΛaaΛ = aa Example Let the terminals be a, b. And the only nonterminal be S. Let the production be: S → aS S → bS S→a S→b The language generated by this CFG is (a+b)+. In this language we can have the following derivation: S → bS → baS → baaS → baab Example Let the terminals be a, b. And the only nonterminal be S. Let the production be: S → aS S → bS S→Λ The language generated by this CFG is (a+b)*. In this language we can have the following derivation: S → bS → baS → baaS → baaΛ=baa

Example Let the terminals be a, b,Λ. And the nonterminals be S,X,Y. Let the production be: S→X S→Y X→Λ Y → aY Y → bY Y→a Y→b The language generated by this CFG is (a+b)*. Example Let the terminals be a, b,Λ. And the nonterminals be S,X,Y. Let the production be: S → XY X→Λ Y → aY Y → bY Y→a Y→b The language generated by this CFG is (a+b)+. Example Let the terminals be a, b. Let the nonterminals be S,X. Let the production be: S → XaaX X → aX X → bX X→Λ The language generated by this CFG is (a+b)* aa(a+b)*. To generate baabaab we can proceed as follows: S→XaaX→bXaaX→baXaaX→baaXaaX→baabXaaX→baabΛaaX=baabaaX →baabaabX→baabaabΛ=baabaab

Example Let the terminals be a, b. Let the nonterminals be S,X,Y. Let the production be: S → XY X → aX X → bX X→a Y → Ya Y → Yb Y→a The language generated by this CFG is (a+b)* aa(a+b)*. To drive babaabb: S→XY→bXY→baXY→babXY→babaY→babaYb→babaYbb→babaabb Example Let the terminals be a, b. Let the nonterminals be S,BALANCED,UNBALANCED. Let the production be: S → SS S → BALANCED S S → S BALANCED S→Λ S → UNBALANCED S UNBALANCED BALANCED→ aa BALANCED→ bb UNBALANCED → ab UNBALANCED → ba The language generated by this CFG is even-even.

Example Let the terminals be a, b. Let the nonterminals be S,A,B. Let the production be: S → aB S → bA A→a A → aS A → bAA B→b B → bS B → aBB The language generated by this CFG is the language EQUAL of all strings that have an equal number of a's and b's.

H.W Find the RE for the following CFG: S → XS| Λ X → ZY Z → abZ| baZ| ab| ba Y → aa| bb

The empty string in Context Free Grammar’s Let G = (N,T,P,S) be any CFG with Λ- productions. Then G’ =(N,T,P’,S), where P’ is constructed from P as follows: 1. Put all the Λ-free productions of P into P’. Λ. 2. Find all the nonterminals AЄN such that A Theorem: For any CFL,L, there exists an Λ-free CFG,G, such that L(G) = L\{Λ} Example: S [E] | E E T| E+T | E-T T F| T*F | T/F F a| b| c| Λ The Λ-free grammar constructed from G has productions S [E] | E |[ ] E T| E+T | E-T | E+ | E- | +T | -T | + | T F| T*F | T/F | T* | T/ | *T | /T | * | / F a| b| c

Trees Example S → AA A → AAA | bA | Ab | a If we want to produce the word bbaaaab, the tree will be: S A

A

A A A

A

b

a a A

b

A

b

a

a

This diagram is called syntax tree or parse tree or generation tree or production tree or derivation tree. Example S → (S) | S&S | ~S | p| q The derivation tree for the word (~ ~ p & ( p & ~ ~ q )) will be: S (

S

)

S

&

S

~

S

(

S

)

~

S

S

&

S

p

p

~

S

~

q

Example S → S + S | S* S | number Does the expression 3+4*5 mean (3+4)*5 which is 35 or does it mean 3+(4*5) which is 23?

We can distinguish between these two possible meaning for the expression 3+4*5 by looking at the two possible derivation trees that might have produced it. S

S S

+

S

3

S

*

or S

S

5

3

4 S

S

S

+

S

3

S

*

S

S

5

4

*

S

+

S

5

4 S

3 + S

S

3

20

+

23

S

*

5

4

Or S

S 3

S

S

*

S

+

S

5

3

S

S

*

S

+

4

5

7

*

4 Example S → AB A→a B→b S → AB → aB → ab or S → AB → Ab → ab S

S A

B

A

B

a

b

a

b

There is no ambiguity of interpretation.

5

35

Definition A CFG is called ambiguous if for at least one word in the language that it generates there are two possible derivations of the word that correspond to different syntax trees. Example The CFG for palindrome S → aSa | bSb |a |b | Λ S → aSa → aaSaa → aabaa S a

S

a

a

S

a

b The CFG is unambiguous. Example S → aS |Sa | a In this case the word a3 can be generated by four different trees: S S S S a

S

a

S

S

a

a

a

S a

a

a

S

a

S

S

a

a

a

S

The CFG is therefore ambiguous. The same language can be defined by the CFG: S → aS | a For which the word a3 has only one production tree: S

This CFG is not ambiguous.

a

S

a

S a

Definition For a given CFG we define a tree with the start symbol S as its root and whose nodes are working strings of terminals and nonterminals. The descendants of each node are all possible results of applying every production to the working string, one at a time. A string of all terminals is a terminal node in the tree. The resultant tree is called the total language tree of the CFG. Example For the CFG S → aa | bX | aXX X → ab | b The total language tree is: S bX aXX

aa

bab bb

aabX abX

aabab aabb

aXab

aXb

aabab abab aabb abb abb abab

The total language has only seven different words. Four of it's words (abb, aabb, abab, aabab) have two different possible derivation because they appear as terminal nodes in this tree in two different places. However the words are not generated by two different derivation trees and the grammar is unambiguous. For example: S

Example

a

X X

a

b

b

Consider the CFG: S → aSb | bS | a The total tree of this language begins:

S aSb bS a

aaSbb abSb

aab baSb

bbS

aaaSbbb aabSbb aaabb The trees may get arbitrary wide as well as infinitely long.

ba

linear grammar A grammar is linear if it is context-free and all of its productions' right hand sides have at most one nonterminal. A linear language is a language generated by some linear grammar. Example A simple linear grammar is G with N = {S}, Σ = {a, b}, P with start symbol S and rules S → aSb S→Λ It generates the language

Relationship with regular grammars: Two special types of linear grammars are the following: 1. the left-linear or left regular grammars, in which all nonterminals in right hand sides are at the left ends; 2. the right-linear or right regular grammars, in which all nonterminals in right hand sides are at the right ends. These two special types of linear grammars are known as the regular grammars; both can describe exactly the regular languages. Another special type of linear grammar is the linear grammars in which all nonterminals in right hand sides are at the left or right ends, but not necessarily all at the same end. By inserting new nonterminals, every linear grammar can be brought into this form without affecting the language generated. For instance, the rules of G above can be replaced with S → aA A → Sb S→Λ

REGULAR GRAMMARS

Note : all regular languages can be generated by CFG's, and so can some non-regular languages but not all possible languages. Example Consider the FA below, which accepts the language of all words with a double a: a

-S

b

M

a

+F

b

a,b

All the necessary to convert it to CFG is that: 1. every edge between states be a production: X

c

Y

becomes

X→ cY

and 2. every production correspond to an edge between states: X→ cY

comes from

X

c

Y

or to the possible termination at a final state: X→ Λ only when X is a final state. So the production rules of our example will be: S → aM | bS M → aF | bS F →aF | bF | Λ Definition For a given CFG a semiword is a string of terminals (maybe none) concatenated with exactly one nonterminal (on the right), for example: (terminal) (terminal) . . . (terminal) (Nonterminal)

Example Consider the following FA with two final states: a

S±

M+

b

a

F

b

a,b

So the production rules of our example will be: S → aM | bS | Λ M → aF | bS | Λ F →aF | bF Theorem All regular languages can be generated by CFG's. This can be stated as: All regular languages are CFL's. Example The language of all words with an even number of a's (with at least some a's) is regular since it can be accepted by this FA: Sb

a

a M b

a

F+ b

We have the following set of productions: S → aM | bS M → aF | bM F →aM | bF | Λ Theorem If all the productions in a given CFG fit one of the two forms: Nonterminal→ semiword Or Nonterminal → word (where the word may be Λ) then the language generated by this CFG is regular. Proof We shall prove that the language generated by such a CFG is regular by showing that there is a TG that accepts the same language. We shall build this TG by constructive algorithm.

Let us consider a general CFG in the form: N1 → w1N2 N40 → w10 N41 → w23 N1 → w2N3 N2 → w3N4 ... ... Where N's are the nonterminals and w's are strings of terminals, and the parts wyNz are the semiwords used in productions. One of these N's must be S. Let N1 = S. Draw a small circle for each N and one extra circle labeled +. The circle for S we label - . S-

N2

N3 . . . Nx

...

+

Draw a directed edge from state Nx to Nz and label it with the word wy. wy

Nx

Nz

If the two nonterminals above are the same the path is a loop. For every production rule of the form: Np → wq Draw a directed edge from Np to + and label it with the word wq. Np

wq

+

We have now constructed a transition graph. Definition A CFG is called a regular grammar if each of its productions is of one of the two forms: Nonterminal → semiword Nonterminal → word Example Consider the CFG: S → aaS | bbS | Λ

It is a regular grammar and the whole TG is: -

Λ

+

aa,bb It is corresponds to the regular expression:

(aa+bb)*

Example Consider the CFG: S → aaS | bbS | abX | baX |Λ X → aaX | bbX | abS | baS The TG for even-even is: ab,ba

±

ab,ba

aa,bb

X aa,bb

Example Consider the CFG: S → aA | bB A → aS | a B → bS | b The corresponding TG is:

a -

A a

a +

b b B

b

This language can be defined by the regular expression: (aa+bb)+ H.W Find CFG that generate the regular language over the alphabet ∑={a,b} of all strings without the substring aaa.

CHOMSKY NORMAL FORM (CNF) Theorem If L is a context-free language generated by CFG that includes Λproductions, then there is a different CFG that has no Λ-production that generates either the whole language L(if L does not include the word Λ) or else generates the language of all the words in L that are not Λ. Definition In a given CFG, we call a nonterminal N nullable if: • There is a production: N→ Λ Or • There is a derivation that start at N and leads to Λ: N→…→ Λ Example Consider the CFG: S → a| Xb| aYa X→Y|Λ Y→b|X X and Y are nullable. The new CFG is: S → a| Xb| aYa| b| aa X→Y Y→b|X Example Consider the CFG: S → Xa X → aX| bX | Λ

X is the only nullable nonterminal. The new CFG is: S → Xa| a X → aX| bX |a| b Example Consider the CFG: S → XY X → Zb Y→ bW Z → AB W→ Z A → aA| bA| Λ B → Ba| Bb| Λ A, B, W and Z are nullable. The new CFG is: S → XY X → Zb| b Y→ bW| b Z → AB| A| B W→ Z A → aA| bA| a| b B → Ba| Bb| a| b Definition A production of the form: One Nonterminal → One Nonterminal is called a unit production.

Theorem If there is a CFG for the language L that has no Λ-production, then there is also a CFG for L with no Λ-production and no unit production. Example Consider the CFG: S → A| bb A → B| b B → S| a S→A

gives S → b

S→A→B

gives S → a

A→B

gives A → a

A→B→S

gives A → bb

B→S

gives B → bb

B→S→A

gives B → b

The new CFG for this language is: S → bb| b| a A → b| a| bb B → a| bb| b Theorem If L is a language generated by some CFG then there is another CFG that generates all the non- Λ words of L, all of these productions are of one of two basic forms: Nonterminal → string of only Nonterminals Or Nonterminal → One Terminal

Example Consider the CFG: S → X1| X2aX2| aSb| b X1 → X2X2| b X2 → aX2| aaX1 Becomes: S →X1 S →X2AX2 S →ASB S →B X1 → X2X2 X1 → B X2 → AX2 X2 → AAX1 A→ a B→b Example Consider the CFG: S → Na N → a| b Becomes: S → NA N → a| b A→a Theorem For any CFL the non- Λ words of L can be generated by a grammar in which all productions are of one of two forms: Nonterminal → string of exactly two Nonterminals Or Nonterminal → One Terminal

Definition If a CFG has only productions of the form: Nonterminal → string of two Nonterminals Or of the form: Nonterminal → One Terminal It is said to be in Chomsky Normal Form (CNF). Example Convert the following CFG into CNF: S→ASA S→ BSB S→AA S→BB S→a S→b A→a B→b The CNF: S→AR1 R1→SA S→ BR2 R2→ SB S→AA S→BB S→a S→b A→a B→b

S→ aSa| bSb| a| b| aa| bb

Example Convert the following CFG into CNF: S→bA| aB A→ bAA| aS| a B→aBB| bS| b The CNF: S→YA| XB A→ YR1| XS| a B→XR2| YS| b X→ a Y→ b R1→ AA R2→ BB Example Convert the following CFG into CNF: S→ AAAAS S→ AAAA A→ a The CNF: S→ AR1 R1→ AR2 R2 → AR3 R3→ AS S → AR4 R4→ AR5 R5→ AA A→ a

Greibach normal form a context-free grammar is in Greibach normal form (GNF) if the right-hand sides of all productions start with a terminal symbol. A context-free grammar is in Greibach normal form, if all production rules are of the form: where A is a nonterminal symbol, a is a terminal symbol, X is a (possibly empty) sequence of nonterminal symbols not including the start symbol.

Example1:convert the following CFG to GNF: S A B

AB BS| b SA| a 1. Convert the CFG to CNF 2. Rename nonterminals (A,S,B) S becomes A1 A becomes A2 B becomes A3 The grammar becomes: A1 A2A3 A2 A3A1| b A3 A1A2| a 3. Compare the value of i,j as Ai A1 A2A3 2>1 A2 A3A1| b 3>2 A3 A1A2| a 1i

A3 A3 B3

A2A3A2 | a A3A1A3A2 | bA3A2 | a A1A3A2 | A1A3A2B3

21 1< 2

A2

A3A2A4 | A3A4A4

The GNF: A2 aA2A4 | aA4A4 A1 aA2 | aA4 A3 a A4 b

Chomsky hierarchy The Chomsky hierarchy consists of the following levels: Type-0: grammars (unrestricted grammars) include all formal grammars. Type-1: grammars (context-sensitive grammars) generate the contextsensitive languages. Type-2: grammars (context-free grammars) generate the context-free languages. Type-3 grammars (regular grammars) generate the regular languages.

Every regular language is context-free, every context-free language, not containing the empty string, is context-sensitive and every context-sensitive language is recursive and every recursive language is recursively enumerable. The following table summarizes each of Chomsky's four types of grammars, the class of language it generates, the type of automaton that recognizes it, and the form its rules must have. Grammar Languages Type-0

Recursively enumerable

Type-1

Context-sensitive

Type-2

Context-free

Type-3

Regular

Automaton Turing machine

Production rules (constraints) α βγ (no restrictions)

Linear-bounded non-deterministic αAβ Turing machine Non-deterministic pushdown A automaton A Finite state automaton

and A

αγβ γ a aB

PUSHDOWN AUTOMATA (PDA) Definition A PDA is a collection of eight things: 1. An alphabet ∑ of input letters. 2. An input TAPE (infinite in one direction). Initially the string of input letters is placed on the TAPE starting in cell i. The rest of the TAPE is blanks. 3. An alphabet Гof STACK characters. 4. A pushdown STACK (infinite in one direction). Initially the STACK is empty (contains all blanks) 5. One START state that has only out_adges, no in-edges. START

6. HALT states of two kinds: some ACCEPT and some REJECT they have in-edges and no out-edges. REJECT

ACCEPT

7. Finitely many non branching PUSH states that introduce characters onto the top of the STACK. they are of the form: PUSH X

Where X is any letter in Γ. 8. Finitely many branching states of two kinds: i. States that read the next unused letter from the TAPE. READ

Which may have out-edges labeled with letters from ∑ and the blank character ∆, with no restrictions on duplication of labels and no insistence that there be a label for each letter of ∑, or ∆. ii. States that read the top character of STACK. POP

Which may have out-edges labeled with letters from Гand the blank character ∆, again with no restrictions.

Note: we require that the states be connected so as to become a connected directed graph. Theorem For every regular language L there is some PDA that accepts it. Poof Since L is regular, so it is accepted by some FA, then we can convert FA to PDA (as in the following example). Example a

-

+

b

b

Becomes:

a START

b b

READ

a

READ

∆

ACCEPT

a

∆ REJECT

Example -

a

a

+

b

a,b

b Becomes: START

a,b b

b

READ

∆ REJECT

a

READ

a

READ

∆

ACCEPT

∆ REJECT

Note: we can find PDA accepts some non regular languages(as in the following example).

Example The language accepted by this PDA is exactly: {anbn,n=0,1,2,…} START

PUSH a

a

∆

READ

POP

b REJECT

b,∆

b a

POP

a,b

∆ ∆ ACCEPT

READ

a REJECT

Or START

PUSH X

a

∆

READ

POP

b

b X

POP

READ

a

∆

REJECT

Language accepted by nondeterministic PDA Language accepted by deterministic PDA Language accepted by FA or NFA or TG

∆ ∆ ACCEPT

X

REJECT

Example Consider the palindrome X, language of all words of the form: sXreverese(s), where s is any string in (a+b)*, such as {X aXa bXb aaXaa abXba aabXbaa …} START

a PUSH a

a

X

READ

a

READ

b

∆

b PUSH b

∆

ACCEPT

POP

b

POP POP

Example odd palindrome ={a b aaa aba bab bbb …} START

a PUSH a

a

READ

a,b

READ

∆

b PUSH b ACCEPT

∆

a

POP

b b

POP POP

Nondeterministic PDA Example Consider the language generated by CFG: 4

S→S+S|S*S|4

+

READ1

*

READ2

READ3

START

S

+

*

∆ ∆

POP

PUSH1 S PUSH2 S

S

ACCEPT

S

READ4 PUSH5 S

PUSH3 +

PUSH6 *

PUSH4 S

PUSH7 S

Now we trace the acceptance of the string: 4+4*4 State start push1 S pop push2 S push3 + push4 S pop read1 pop read2 pop push5 S push6 * push7 S pop read1 pop read3 pop read1 pop read4 accept

Stack ∆ S ∆ S +S S+ S +S +S S S ∆ S *S S*S *S *S S S ∆ ∆ ∆ ∆ ∆

Tape 4+4*4 4+4*4 4+4*4 4+4*4 4+4*4 4+4*4 4+4*4 +4*4 +4*4 4*4 4*4 4*4 4*4 4*4 4*4 *4 *4 4 4 ∆ ∆ ∆ ∆

H.W Find a PDA that accepts the language: {ambnan, m=1,2,3,…,n=1,2,3,…}

TURING MACHINE

Definition A Turing machine (TM) is a collection of six things: 1. An alphabet ∑ of input letters. 2. A TAPE divided into a sequence of numbered cells each containing one character or a blank. 3. A TAPE HEAD that can in one step read the contains of a cell on the TAPE, replace it with some other character, and reposition itself to the next cell to the right or to the left of the one it has just read. 4. An alphabet Гof character that can be printed on the TAPE by the TAPE HEAD. 5. A finite set of states including exactly one START state from which we begin execution, and some (may be none) HALT states that cause execution to terminate when we enter them. The other states have no functions, only names: q1, q2, … or 1, 2, 3, … 6. A program, which is a set of rules that tell us on the basis of the letter the TAPE HEAD has just read, how to change states, what to print and where to move the TAPE HEAD. We depict the program as a collection of directed edge connecting the states. Each edge is labeled with a triplet of information: (letter, letter, direction). The first letter (either ∆ or from ∑ or Γ) is the character that the TAPE HEAD reads from the cell to which it is pointing, the second letter (also ∆ or from Γ) is what the TAPE HEAD prints in the cell before it leaves, the third component, the direction, tells the TAPE HEAD whether to move one cell to the right(R) or to the left (L).

Note: TM is deterministic. This means that there is no state q that has two or more edges leaving it labeled with the same first letter. For example, the following TM is not allowed: q2

(a,a,R) q1 (a,b,L)

q3

Example Find TM that can accepts the language defined by the regular expression: (a+b)b(a+b)* (a,a,R)

(b,b,R)

START 1

(∆,∆,R)

2

3

HALT 4

(b,b,R) (a,a,R), (b,b,R)

Now we trace the acceptance of the string: aba 1→2→3 → 3 → 4 aba aba aba∆ aba∆ aba∆∆ Example Find TM that can accepts the language {anbn} (a,a,R), (B,B,R)

START 1

(a,A,R)

2

(B,B,L)

(b,B,L)

3

(B,B,R) (A,A,R)

5 (∆,∆,R)

(A,A,R)

4 (a,a,L)

(a,a,L)

HALT

Now we trace the acceptance of the string: aaabbb aaabbb Aaabbb Aaabbb Aaabbb AaaBbb AaaBbb AaaBbb AaaBbb AaaBbb AaaBbb AaaBbb AAaBBb AAaBBb AAaBBb AAaBBb AAABBb AAABBb AAABBb AAABBB AAABBB AAABBB AAABBB AAABBB AAABBB AAABBB∆ HALT Example Find TM that can accepts the language palindrome.

(a,a,R), (b,b,R) (a,∆,R)

2

(b,b,L), (a,a,L)

(∆,∆,L)

(a,∆,L)

3

(∆,∆,R)

4

(∆,∆,R) (∆,∆,R)

START 1

HALT 8

(∆,∆,R)

(a,a,R), (b,b,R) (b,∆,R)

5

(∆,∆,L)

6

(b,b,L), (a,a,L)

(b,∆,L)

(∆,∆,R)

7

Let us trace the running of this TM on the input string: ababa 1

→ →

4

→

∆bab∆

∆bab∆ 5

→

∆baba

ababa 4

2

→ 6

∆∆ab∆

→

∆∆ab∆

2

→

∆baba 4

→

Find TM for even-even.

4

→

∆bab∆ →

∆∆a∆∆

H.W

→

∆baba

∆bab∆ 7

2

7

→

∆∆a∆∆

2

→

∆baba 1

→

∆baba∆ →

∆bab∆ 1

2 5

→

∆∆ab∆ →

∆∆a∆∆

2

3

→

∆baba 5

→

∆∆ab∆ →

∆∆∆∆∆

3 → ∆∆∆∆∆

8 HALT