Lecture 1 String and Language String string is

Lecture 1 String and Language

String • string is a finite sequence of symbols. For example, string ( s, t, r, i, n, g) CS 4384 ( C, S, 4, 3, 8) 101001 (1, 0) • Symbols are given through alphabet. • An alphabet is a finite set of symbols.

Examples of Alphabet • {a, b, c, . . . , x, y, z} (Roman alphabet) • {0, 1, . . . , 9} • {0, 1} (binary alphabet)

Length of a String • The length of a string x is the number of symbols contained in the string x, denoted by |x|. • For example, | string | = 6, • |CS 5400| = 6, | 101001 | = 6. • The empty string is a string having no symbol, denoted by ε.

Equal • Two strings x 1 x 2···xn and y 1 y 2···ym are equal if and only if (1) n=m and (2) xi=yi for all i. • For example, 01 ≠ 010 and 1010 ≠ 1110.

Substring • s is a substring of x if there exist strings y and z such that x = ysz. • In particular, when x = sz (y=ε), s is called a prefix of x; when x = ys (z=ε), s is called a suffix of x. For example, CS is a prefix of CS 5400 • and 5400 is a surfix of CS 5400.

Concatenation • The concatenation of two strings x and y is a string xy, i. e. , x is followed by y. • For example, CS 5400 is a concatenation of CS and 5400. • In particular, we denote 4 2 3 xx = x, xxxx = x, . . . , and define 0 x =ε 3 0 • For example, 101010 = (10), (10) = ε

Solve equation 011 x=x 011 • • If x=ε, then ok. If |x|=1, then no solution. If |x|=2, then no solution. If |x|>3, then x=011 y. Hence, 011 x=011 y 011. So, x=y 011. Hence, 011 y=y 011. k • x=(011) for k > 0

Language • A language is a set of strings. 0 1 2 For example, {0, 1}, {all English words}, {0, 0, 0, . . . } are all languages. • The following are operations on sets and hence also on languages. Union: A U B Intersection: A ∩ B Difference: A _ B (A - B when B A) Complement: A = Σ* - A where Σ* is the set of all strings on alphabet Σ.

Concatenation of Languages • Concatenation: AB = {ab | a in A, b in B} • For example, {0, 1}{1, 2} = {01, 02, 11, 12}. • Especially, we denote A = A, A = AA, . . . , 1 2 and define A = {ε}. 0

If AB=B for any B, then A ={ε}. • Choose B = {ε }. Then A ≠ empty and A cannot contain a nonempty string.

Examples 2 • For Σ = {0, 1}, Σ = {00, 01, 10, 11}, k • (Σ is the set of all strings of length k on Σ. ) Therefore, 0 1 2 • Σ* = Σ U Σ U ···.

Kleene Closure • Kleene closure: 0 1 2 A* = A U A U ··· • Notation: + 1 2 3 A = A U A U ···

• A={grand, ε}, B={father, mother}. What is A*B? • A*B={father, mother, grandfather, grandmother, …}

What is • Where ? ? ? is the empty language.

+ A* = A if and only if ε is in A + + • If ε is in A, then ε is in A. Hence A* = A. + • If ε is not in A, then ε is not in A. Hence A* ≠ A. +

{0, 10}* is the language of strings not containing substring 11 and not ending with 1. • What is the language of strings not containing substring 11 and ending with 0? + • {0, 10}

Puzzle • How many strings of length at most 40 are in the following language ?

Lecture 2 Regular Language and Regular Expression.

Regular Languages • The concept of regular languages on an alphabet Σ is defined recursively as follows: (1) The empty language is regular. (2) For every symbol a Σ, {a} is regular. (3) If A and B are regular languages, then A U B, AB, and A* are regular. (4) Nothing else is a regular language.

{ε} is regular. • Because the empty language = {ε} is regular,

For Σ={0, 1}, {011} is regular. • Since {0} and {1} are regular, {011}={0}{1}{1} is regular • Remark: Every language containing only one string is regular.

{011, 100} is regular. • Because {011} and {100} are regular, {011, 100} = {011}U{100} is regular. • Remark: Every finite language is regular. • Remark: Every infinite regular language must be obtained with Kleene closure.

Operation Preference • ({0}*U{0}{1}{1}*){0}{0}{1}* • (1) Kleene closure has the higher preference over union and concatenation. • (2) Concatenation has the higher preference over union.

The language of all binary strings starting with 01 is regular. Proof. The string in this language is in form 01 x 1··· xn where x 1··· xn {0, 1}*. Therefore, the language can be written as {01} {0, 1}* = ({0}{1})({0} U {1})*, which is regular.

The language of all binary strings ending at 01 is regular. Proof. The string in this language is in form x 1··· xn 01 where x 1 ··· xn {0, 1}*. Therefore, the language can be written as {0, 1}*{01} = ({0} U {1})*({0}{1}), which is regular.

The language of all binary strings having substring 01 is regular. Proof. The string in this language is in form x 1 ··· xn 01 y 1 ··· ym where x 1 ··· xn, y 1 ··· ym {0, 1}*. Therefore, the language can be written as {0, 1}* {01} {0, 1}* =({0}U{1})*({0}{1})({0}U{1})*, which is regular.

Question: Do you fell that the expression of the regular set in the above example contains too many parentheses? • Here is a simple expression -- Regular Expression

Regular Expression • (1) is a regular expression of the empty language. • (2) ε is a regular expression of {ε}. • (3) For any symbol a, a is a regular expression of {a}. • (4) If r. A and r. B are regular expressions of languages A and B, then r. A+r. B is a regular expression of A U B, r. Ar. B is a regular expression of AB, and r. A* is a regular expression of A*.

Examples • • 011 is a regular expression of {0}{1}{1}. 0+1 is a regular expression of {0, 1}. (0+1)* is a regular expression of {0, 1}*. Remark: (0+1)+ is also considered to be + a regular expression of {0, 1}.

• The language of all binary strings starting with 01 has a regular expression 01(0+1)*. • The language of all binary strings ending at 01 has a regular expression (0+1)*01. • The language of all binary strings having substring 01 has a regular expression (0+1)*01(0+1)*.

Induction Proof • Because the regular language is defined recursively, • we can prove the property of regular languages by • proving the following: (1) has the property. (2) For any symbol a Σ, {a} has the property. (3) If A and B has the property, then all A U B, AB, and A* have the property. • Actually, this is an induction proof. (1), (2) serve the basis step and (3) is the induction step.

• For a string x=x 1 x 2…xn, x R =xn…x 2 x 1. R • For a language A, A = {x R| x A}. R • Show that if A is regular, so is A. Proof. (1) is regular. R (2) For any symbol a, {a} = {a} is regular. R (3) Suppose that for regular languages A and B, A and B Rare regular. Then R R R (A U B) = A U B is regular, R R R (AB) = B A is regular. R R (A*) = (A )* is regular.

Find a regular expression for R {xwx | x (0+1)*, w (0+1)*} • {xwx R | x (0+1)*, w (0+1)*} = (0+1)*

Find a regular expression for R + {xwx | x (0+1), w (0+1)*} + • x (0+1), w = 0(0+1)*0 + 1(0+1)*1 {xwx. R | (0+1)*}

Puzzle • How many regular expressions can a language have?