Formal Language & Automata Theory Shyamanta M Hazarika Computer Sc. & Engineering Tezpur University http://www.tezu.ernet.in/~smh

Regular Expressions A regular expression defines a language. Regular expressions are defined through union, concatenation, and closure. Union: L1  L2  {x | x in L1 or x in L2}. Concatenation: L1L2  {xy | x in L1 and y in L2}. (Kleene) Closure: L*  i=0.. Li, where L0 = {}, Li =LLi-1. Example: {01, 10}* = {, 01, 10, 0110, 0101, 1010, …}.

Formal Definition Definition: Regular expressions (over ). 1)  is a regular expression denoting  (empty set). 2)  is a regular expression denoting {}. 3) For each a  , a is a regular expression denoting {a}. 4) If E (F) is a reg. exp. denoting language L(E) (L(F)), then (E + F) is a reg. exp. denoting L(E)  L(F), (EF) ………………………….L(E)L(F), (E*) ………………………….L(E)*. We will omit parenthesis if no confusion.

Example 00(0 + 1)*01 +  = {x | x =  or x, |x|  4, begins with 00 and ends with 01} (An alphabet of {0,1} is assumed here, and in many other examples as well.)

Finite Automata and Reg. Exps. Regular expressions define the same class of languages as Finite Automaton. NFA DFA -NFA Reg. Exp.

Theorem Theorem: If R is a regular expression, then L(R) = L(E) for some NFA E with -moves. Proof: We construct E so that it has one final state and no transitions out of that state. We do this by induction on the number of operators in R.

Basis Only three cases: Case 1: R = . Case 2: R = . Case 3: R = a. q0 Start q0 qf Start q0 qf Start a

Inductive Step Case 1: R = R1 + R2. By the induction hypothesis, machines E1 and E2 exist for R1 and R2, respectively. Construct E as follows: q1 f1 E1  q0 Start f0 q2 f2 E2

Inductive Step (Continued) Case 2: R = R1R2. Start q1 f1 E1 q2 E2 f2  Case 3: R = R1*.  q1 f1 E1   q0 f0 Start 

Example 0(0 + 1)* 0: 1: 0 + 1: Start 1 Start    1  Start

Example (Continued) 1  0 + 1: (0 + 1)*: Start 1  Start

Example (Continued)      (0 + 1)*:   0(0 + 1)*:             (0 + 1)*: 0(0 + 1)*:  1  Start         1  Start

Example (Continued)  0(0 + 1)*:              1  Start Note: A much simpler machine exists.