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Examples

a, b. b. a, b. a, b. a, b. 0. 1. 2. 3. 4. . . 2. 4. . a. 0. 1. . 5. 3. b. . Examples. Definition of regular sets Basis: , { } and {a}, for every a   are regular sets over .

daryl-wilcox
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Examples

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  1. a, b b a, b a, b a, b 0 1 2 3 4   2 4  a 0 1  5 3 b  Examples

  2. Definition of regular sets • Basis: , {} and {a}, for every a   are regular sets over . • Recursive step: Let X and Y be regular sets over . Then the sets XY, XY, and X* are regular sets over . • Closure: X is a regular set over  only if it can be obtained from the basis elements by a finite number of applications of the recursive step • Examples: Let  = {0, 1} • Following are some regular sets over : • , {}, {0}, {1}, {00, 11}, {0}*, {0, 1}*

  3. Definition of regular expressions • Let  be an alphabet. The regular expressions over  are defined recursively as follows: • Basis:  is a regular expression denoting the regular set .  is a regular expression denoting {}. For every a  , a is a regular expression denoting {a}. • Recursive step: Let x and y be regular expressions over . Then the expressions x+y , xy , x* are regular expressions over . • Closure: x is a regular expression over  only if it can be obtained from the basis elements by a finite number of applications of the recursive step. • If x represents X and y represents Y, x+y represents XY, xy represents XY, and x* represents X*. (x) represents X.

  4. Examples: Let  = {0, 1} Regular expression Regular set 0 {0} 00 {00} 0 + 0 {0} 0* {0}* (0+1) {0, 1} 0(0+1) {00, 01} (00)* {00}*

  5. Regular expression identities: • x = x =  • x = x= x • * =  • * =  • x + y = y + x • x +  = x • x + x = x • (x*)* = x* • x(y+z) = xy + xz • (x+y)z = xz + yz • (xy)*x = x(yx)* • (x+y)* = (x*+y)* = x*(x+y)* = (x+yx*)* = (x*y*)* =x*(yx*)*=(x*y)*x*

  6. Example (construction of regular sets): Construct a regular expression to represent the set of strings over {a, b} that contain the substring ba and the substring ab. ---ba---ab--- ---bab--- ---ab---ba-- ---aba---

  7. -NFA NFA RE DFA Theorem: NFA, DFA, and Regular Expressions are Equivalent

  8. Example: Construct an NFA equivalent to the regular expression 0(1+0)*

  9. 2 a 1 Example: Construct equivalent DFA 4 a b 0 b a 3

  10. Rij(k) – regular expression representing the set of labels of all paths from state i to state j going through intermediate states {1, 2, 3, …, k} only (defined recursively) R1f(n) – regular expression representing strings accepted by f, f is in F Equivalent regular expression is the union of all R1f(n) Construction: BASE: Rij(0) = a1 + a2 + … + ak where i  j and am are labels of arcs from state i to state j. Rij(0) =  + a1 + a2 + … + ak where i = j and am are labels of arcs from state i to state j. INDUCTON: Rij(k) = Rij(k-1) + Rik(k-1)(Rkk(k-1))*Rkj(k-1) Construction of regular expression from FA

  11. a,b Example: Find equivalent regular expression for the following NFA: a 3 start 2 1 b b

  12. 0.1 start A B C D 1 0,1 0,1 State elimination method - example

  13. Applications of Finite Automata: • Software for digital circuit design • Lexical Analyzer • Keyword search • Software for verifying finite state systems

  14. Key word search  e n d b e g i n

  15. Pattern matching (KMP Algorithm) procedure fsasearch; begin state := 0; for i := 1 to N do begin state := next(state, a[i]); if (state = final) then return true; end; return false; end; e.g. search for 10100110

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