highlights CSE 311: Foundations of Computing highlights 1 in third position from end Fall 2013 Lecture 25: Non-regularity and limits of FSMs

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1 CSE 3: Foundations of Computing Fall 23 Lecture 25: Non-regularity and limits of FSMs highlights NFAs from Regular Epressions ( )* highlights in third position from end Subset construction : NFA to DFA a a,b c c b, b,c NFA DFA, b a,b,c A,, B C D, {A, B, C} {A} {A, B} {A, C} {A, B, C, D} {A, C, D} {A, B, D} {A, D}

2 redrawing DFAs regular epressions A,, B C D, We have shown how to build an optimal DFA for every regular epression Build NFA Convert NFA to DFA using subset construction Minimize resulting DFA {A} {A,B} {A,C} {A,B,D} {A,B,C} {A,B,C,D} Theorem: A language is recognized by a DFA if and only if it has a regular epression {A,D} {A,C,D} generalized NFAs Like NFAs but allow Parallel edges Regular Epressions as edge labels NFAs already have edges labeled or a An edge labeled by Acan be followed by reading a string of input chars that is in the language represented by A A string is accepted iffthere is a path from start to final state labeled by a regular epression whose language contains starting from an NFA Add new start state and final state Then eliminate original states one by one, keeping the same language, until it looks like: Final regular epression will be A A

3 only two simplification rules Rule : For any two states q and q 2 with parallel edges (possibly q =q 2 ), replace A A B q q 2 by q q 2 B converting an NFA to a regular epression Considerthe DFA for the mod 3 sum Accept strings from {,,2}* where the digits mod 3 sum of the digits is Rule 2: Eliminate non-start/final state q 3 by replacing all B q A C AB*C q 3 q 2 by q q 2 t t t 2 for everypair of states q, q 2 (even if q =q 2 ) splicing out a node finite automaton without t Label edges with regular epressions t t t : *2 t t t 2 : * t 2 t t : 2*2 t 2 t t 2 : 2* s 2 t 2 t t 2 2 R : *2 R 2 : 2 * R 3 : 2*2 R 4 : 2* R 5 : R R 2 R 4 *R 3 s R R 2 t t 2 f s R 5 t R 3 f R 4 f Final regular epression: ( *2 (2 *)( 2*)*( 2*2))*

4 what can finite state machines do? We ve seen how we can get DFAs to recognize all regular languages What about some other languages we can generate with CFGs? { }? binary palindromes? strings of balanced parentheses? A={ } cannot be recognized by any DFA Consider the infinite set of strings S={,,,,,...} Claim: No two strings in S can end at the same state of any DFA for A Proof: Suppose n mand n and m end at the same state p of some DFA for A. Since n n is in A, following n after state p must lead to a final state. But then the DFA would also accept m n which is a contradiction to the DFA recognizing A. Given claim, the # of states of any DFA for A must be S which is not finite, which is impossible for a DFA. B = {binary palindromes} can t be recognized by any DFA Consider the infinite set of strings S={,,,,,...}={ n : n } Claim: No two strings in S can end at the same state of any DFA for B Proof: Suppose n mand n and m end at the same state p of some DFA for B. Since n n is in B, following n after state p must lead to a final state. But then the DFA would also accept m n which is not in B and is a contradiction since the DFA recognizes B. Given claim, the # of states of any DFA for B must be S which is not finite, which is impossible for a DFA. general: how to show language L has no DFA Find a hard infinite set S={s,s,...,s n,...} of strings that might be prefies of strings in L Show that S is hard by showing that no two strings s n s m in S can end at the same state of any DFA recognizing L For each pair s n s m find an etender string t depending on n,mso that eactly one of s n tand s m t is in L Conclude that any DFA for L would need S stateswhich is not finite, and so impossible

5 P = {strings of balanced parentheses} pattern matching Given a string, s, of ncharacters a pattern, p, of mcharacters usually m<<n Find all occurrences of the pattern pin the string s Obvious algorithm: try to see if pmatches at each of the positions in s stop at a failed match and try the net position 8 string s = y y y y y y y y y y y pattern p = y y y y y string s = y y y y y y y y y y y y y y y y 9 2

6 string s = y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y 2 22 string s = y y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y y y y y 23 24

7 string s = y y y y y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y y y y y y y y y y String s = y y y y y y y y y y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y y y y y y y y y y y 27 28

8 string s = y y y y y y y y y y y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y y y y y y y y y y y 29 3 string s = y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y string s = y y y y y y y y y y y y y y y y y y y y y y Worst-case time O(mn) y y y y y y y y y 3 32

9 string s = y y y y y y y y y y y y y Lots of wasted work better pattern matching via finite automata Build a DFA for the pattern (preprocessing) of size O(m) Keep track of the longest match currently active The DFA will have only m+ states Run the DFA on the string nsteps y y y y y Obvious construction method for DFA will be O(m 2 ) but can be done in O(m) time. Total O(m+n m+n) time building a DFA for the pattern preprocessing the pattern pattern p= y y y y y pattern p= y y y y y 35 36

10 preprocessing the pattern preprocessing the pattern pattern p= y y y y y pattern p= y y y y y preprocessing the pattern generalizing pattern p= y y y y y Can search for arbitrary combinations of patterns Not just a single pattern Build NFA for pattern then convert to DFA on the fly. Compare DFA constructed above with subset construction for the obvious NFA. 39 4