# Dr. D.M. Akbar Hussain

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1 1 2 Compiler Construction F6S Lecture - 2 1

2 3 4 Compiler Construction F6S Lecture - 2 2

3 5 #include.. #include main() { char in; in = getch ( ); if ( isalpha (in) ) in = getch ( ); else error (); while ( isalpha (in) isdigit (in) ) in = getch ( ); } For an identifier 6 Compiler Construction F6S Lecture - 2 3

4 REGULAR EXPRESSIONS (RE) 7 REGULAR EXPRESSIONS (RE) 8 Compiler Construction F6S Lecture - 2 4

5 RE OPERATIONS Choice (alternation) Concatenation Repetition Choice: Given two RE M and N, it is written with the alternation operator (vertical bar) as M N. A string is in the language M N if it is in the language M or in the language N. It is a union, e.g., for a language a b it contains strings a and b. L(a b) =L(a) UL(b) ={a,b} 9 RE OPERATIONS 10 Compiler Construction F6S Lecture - 2 5

6 RE OPERATIONS U a* = a a n= 0 n 0 = 11 RE OPERATIONS (a b)*, a, b, aa, bb, ab, abb, a (b)*, a, b, bb, bbb,. * Has the highest precedence then concatenation and alternation Some short cuts (abbreviations): ab c means (a.b a.b) c (a ) means (a ) [b - g] means [bcdefg bcdefg] [abcd] mean (a b c d a b c d) 12 Compiler Construction F6S Lecture - 2 6

7 RE OPERATIONS EXTENSION Why it is required: Example: a* repetition zero or more, this could be problem for natural numbers, we need to have one instance in which it guarantees that we have a match, disallowing the empty string. Binary number example: (0 1)* Although, we can solve it by: (0 1) (0 1)*, 0, 00, 11,.. But this is a simple situation and it is easy, but things may not be that simple in a language. So the solution is M + (M.M*). 13 EXTENSION XTENSION RE O RE OPERATIONS Another Situation Demands: any character Common requirement is to have a match of any character in the alphabet. Which means we have to literally write each character with alternation. The extension is to use meta character. period, which gives the option not to put every character alternative. Example:.*b.* True for all strings containing at least one b. 14 Compiler Construction F6S Lecture - 2 7

8 RANGE OF CHARACTERS Usually we do like this for range of characters: a b c d.. z So the extension is [a - z] [0-9] Any Character not in a given set: It is not possible to get to a situation, where we have RE for all characters except one character. Tilde or Carat can be used. For example: ~a a character which is not a ~(a b c) a character which is neither a,b or c. In Lex: ^a, ^(a b c) Optional Occurrence: M? (M ) 15 RE NOTATIONS 16 Compiler Construction F6S Lecture - 2 8

9 FINITE AUTOMATA RE are very convenient for specifying token, but some formalism is required to implement it. Finite automata is a mathematical way of describing machines. In particular for the process of recognizing input tokens FA is the proper formalism. FA are the best construct to implement scanners. Formal Definition: FA is a finite automaton with finite set of states. Edges lead from one state to other states. Each edge is labeled with a symbol. One is unique start state and some are final states. 17 letter letter 1 2 O l e digit O l e Compiler Construction F6S Lecture - 2 9

10 19 20 Compiler Construction F6S Lecture

11 S K A K Compiler Construction F6S Lecture

12 23 24 Compiler Construction F6S Lecture

13 25 26 Compiler Construction F6S Lecture

14 27 28 Compiler Construction F6S Lecture

15 Suppose we have three tokens :=, <=, = : : = := < = <= = = 29 : = := < = = <= Suppose we have tokens <=, <>, < can we do that < = = <= < > < <> < 30 Compiler Construction F6S Lecture

16 = <= < > <> others < 31 : = := < = <= = = 32 Compiler Construction F6S Lecture

17 a 2 b 1 a 3 4 a b b a b b a c b a 34 Compiler Construction F6S Lecture

18 35 36 Compiler Construction F6S Lecture

19 Concatenation: M M N RE = MN so Thomson construction follows: Alternation/Choice: RE = M N so Thomson construction follows: M N 38 Compiler Construction F6S Lecture

20 Repetition: RE = M* so Thomson construction follows: M 39 Thomson construction is not unique, there are other possibilities e.g., for MN: M N However, it is only possible if the accepting state has no transitions to other states. 40 Compiler Construction F6S Lecture

21 Obviously, first thing to remove is transitions and second the multiple transitions from a state on a single character input. 1 For example: RE = M* a closure of a single state S is a set of states reachable by a series of zero or more transitions and denoted by S. 1 = {1, 2, 4}, 2 = {2}, 3 = {2, 3, 4}, 4 = {4} a {1, 2, 4} {2, 3, 4} a 41 RE = ab a* 1 = {2, 6, 1}, 2 = {2}, 3 = {3, 4}, 4 = {4}, 5 = {8, 5}, 6 = {6} 7 = {8, 7}, 8 = {8} 1 2 a 3 4 b 5 a {2, 6, 1} {3, 4, 7, 8} {5, 8} a b 42 Compiler Construction F6S Lecture

22 43 44 Compiler Construction F6S Lecture

23 45 (a a )b* Actual DFA a 1 2 b b a b {1} {2, 3} b 3 b 46 Compiler Construction F6S Lecture

24 47 48 Compiler Construction F6S Lecture

25 49 Compiler Construction F6S Lecture

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