Equivalence of Regular Languages and FSMs

Transcription

1 Equivlence of Regulr Lnguges nd FSMs Red K & S 2.4 Red Supplementry Mterils: Regulr Lnguges nd Finite Stte Mchines: Generting Regulr Expressions from Finite Stte Mchines. Do Homework 8. Equivlence of Regulr Lnguges nd FSMs Theorem: The set of lnguges expressile using regulr expressions (the regulr lnguges) equls the clss of lnguges recognizle y finite stte mchines. Alterntively, lnguge is regulr if nd only if it is ccepted y finite stte mchine. Proof Strtegies Possile Proof Strtegies for showing tht two sets, nd re equl (lso for iff): 1. Strt with nd pply vlid trnsformtion opertors until is produced. Exmple: Prove: A (B C) = (A B) (A C) A (B C) = (B C) A commuttivity = (B A) (C A) distriutivity = (A B) (A C) commuttivity 2. Do two seprte proofs: (1), nd (2), possily using totlly different techniques. In this cse, we show first (y construction) tht for every regulr expression there is corresponding FSM. Then we show, y induction on the numer of sttes, tht for every FSM, there is corresponding regulr expression. We'll show this y construction. Exmple: For Every Regulr Expression There is Corresponding FSM *( )* Review - Regulr Expressions The regulr expressions over n lphet Σ* re ll strings over the lphet Σ {(, ),,, *} tht cn e otined s follows: 1. nd ech memer of Σ is regulr expression. 2. If α, β re regulr expressions, then so is αβ. 3. If α, β re regulr expressions, then so is α β. 4. If α is regulr expression, then so is α*. 5. If α is regulr expression, then so is (α). 6. Nothing else is regulr expression. We lso llow nd α +, etc. ut these re just shorthnds for * nd αα*, etc. so they do not need to e considered for completeness. Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 1

2 For Every Regulr Expression There is Corresponding FSM Formlizing the Construction: The clss of regulr lnguges is the smllest clss of lnguges tht contins nd ech of the singleton strings drwn from Σ, nd tht is closed under Union Conctention, nd Kleene str Clerly we cn construct n FSM for ny finite lnguge, nd thus for nd ll the singleton strings. If we could show tht the clss of lnguges ccepted y FSMs is lso closed under the opertions of union, conctention, nd Kleene str, then we could recursively construct, for ny regulr expression, the corresponding FSM, strting with the singleton strings nd uilding up the mchine s required y the opertions used to express the regulr expression. FSMs for Primitive Regulr Expressions An FSM for : An FSM for ( *): An FSM for single element of Σ: Closure of FSMs Under Union To crete FSM tht ccepts the union of the lnguges ccepted y mchines M1 nd M2: 1. Crete new strt stte, nd, from it, dd -trnsitions to the strt sttes of M1 nd M2. Closure of FSMs Under Conctention To crete FSM tht ccepts the conctention of the lnguges ccepted y mchines M1 nd M2: 1. Strt with M1. 2. From every finl stte of M1, crete n -trnsition to the strt stte of M2. 3. The finl sttes re the finl sttes of M2. Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 2

3 Closure of FSMs Under Kleene Str To crete n FSM tht ccepts the Kleene str of the lnguge ccepted y mchine M1: 1. Strt with M1. 2. Crete new strt stte S0 nd mke it finl stte (so tht we cn ccept ). 3. Crete n -trnsition from S0 to the strt stte of M1. 4. Crete -trnsitions from ll of M1's finl sttes ck to its strt stte. 5. Mke ll of M1's finl sttes finl. Note: we need new strt stte, S0, ecuse the strt stte of the new mchine must e finl stte, nd this my not e true of M1's strt stte. Closure of FSMs Under Complementtion To crete n FSM tht ccepts the complement of the lnguge ccepted y mchine M1: 1. Mke M1 deterministic. 2. Reverse finl nd nonfinl sttes. q1 q2 A Complementtion Exmple Closure of FSMs Under Intersection L1 L2 = Write this in terms of opertions we hve lredy proved closure for: L1 L2 Union Conctention Kleene str Complementtion An Exmple ( *)** Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 3

4 For Every FSM There is Corresponding Regulr Expression Proof: (1) There is trivil regulr expression tht descries the strings tht cn e recognized in going from one stte to itself ({} plus ny other single chrcters for which there re loops) or from one stte to nother directly (i.e., without pssing through ny other sttes), nmely ll the single chrcters for which there re trnsitions. (2) Using (1) s the se cse, we cn uild up regulr expression for n entire FSM y induction on the numer ssigned to possile intermedite sttes we cn pss through. By dding them in only one t time, we lwys get simple regulr expressions, which cn then e comined using union, conctention, nd Kleene str. Key Ides in the Proof Ide 1: Numer the sttes nd, t ech induction step, increse y one the sttes tht cn serve s intermedite sttes I K J Ide 2: To get from stte I to stte J without pssing through ny intermedite stte numered greter thn K, mchine my either: 1. Go from I to J without pssing through ny stte numered greter thn K-1 (which we'll tke s the induction hypothesis), or 2. Go from I to K, then from K to K ny numer of times, then from K to J, in ech cse without pssing through ny intermedite sttes numered greter thn K-1 (the induction hypothesis, gin). So we'll strt with no intermedite sttes llowed, then dd them in one t time, ech time uilding up the regulr expression with opertions under which regulr lnguges re closed. The Formul Adding in stte k s n intermedite stte we cn use to go from i to j, descried using pths tht don't use k: i k j R(i, j, k) = R(i, j, k - 1) R(i, k, k-1) R(k, k, k-1)* R(k, j, k-1) /* wht you could do without k /* go from i to the new intermedite stte without using k or higher /* then go from the new intermedite stte ck to itself s mny times s you wnt /* then go from the new intermedite stte to j without using k or higher Solution: R(s, q, N) q F Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 4

5 An Exmple of the Induction Going through no intermedite sttes: (1,1,0) = (1,2,0) = (1, 3, 0) = (2,3,0) = (3,3,0) = (3,4,0) = Allow 1 s n intermedite stte: Allow 2 s n intermedite stte: (1, 3, 2) = (1, 3, 1) (1, 2, 1)(2, 2, 1)*(2, 3, 1) = * = Allow 3 s n intermedite stte: (1, 3, 3) = (1, 3, 2) (1, 3, 2)(3, 3, 2)*(3, 3, 2) = ( )* ( ) = * (1, 4, 3) = (1, 4, 2) (1, 3, 2)(3, 3, 2)*(3, 4, 2) = ( )* = * An Esier Wy - See Pcket (1) Crete new initil stte nd new, unique finl stte, neither of which is prt of loop Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 5

6 (2) Remove sttes nd rcs nd replce with rcs lelled with lrger nd lrger regulr expressions. Sttes cn e removed in ny order, ut don t remove either the strt or finl stte * * 5 (Notice tht the removl of stte 3 resulted in two new pths ecuse there were two incoming pths to 3 from nother stte nd 1 outgoing pth to nother stte, so 2 1 = 2.) The two pths from 2 to 1 should e colesced y unioning their regulr expressions (not shown). 4 1 * * 5 ( * *)*( ) 4 5 Thus, the equivlent regulr expression is: ( * *)*( ) Mtching floting point numers: -? ([0-9]+(\.[0-9]*)? \.[0-9]+) Using Regulr Expressions in the Rel World (PERL) Mtching IP ddresses: ([0-9]+ (\. [0-9]+) {3}) Finding douled words: \< ([A-Z-z]+) \s+ \1 \> From Friedl, J., Mstering Regulr Expressions, O Reilly,1997. Note tht some of these constructs re more powerful thn regulr expressions. Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 6

7 Regulr Grmmrs nd Nondeterministic FSAs Any regulr lnguge cn e defined y regulr grmmr, in which ll rules hve left hnd side tht is single nonterminl hve right hnd side tht is, single terminl, single nonterminl, or single terminl followed y single nonterminl. Exmple: L={w {, }* : w is even} (() () () ())* S S T S T T T T S T S, S T, An Algorithm to Generte the NDFSM from Regulr Grmmr 1. Crete nonterminl for ech stte in the NDFSM. 2. s is the strt stte. 3. If there re ny rules of the form X w, for some w Σ, then crete n dditionl stte leled #. 4. For ech rule of the form X w Y, dd trnsition from X to Y leled w (w Σ ). 5. For ech rule of the form X w, dd trnsition from X to # leled w (w Σ). 6. For ech rule of the form X, mrk stte X finl. 7. Mrk stte # finl. Exmple 1 - Even Length Strings S S T S T T T T S T S S S B S C S A S C S ca S cb Exmple 2 - One Chrcter Missing A A A ca A B B B cb B C C C C C Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 7

8 An Algorithm to Generte Regulr Grmmr from n NDFSM 1. Crete nonterminl for ech stte in the NDFSM. 2. The strt stte ecomes the strting nonterminl 3. For ech trnsition δ(t, ) = U, mke rule of the form T U. 4. For ech finl stte T, mke rule of the form T. Exmple: X Y Conversion Algorithms etween Regulr Lnguge Formlisms Regulr Grmmr Regulr Expression NFSM (NFA) DFSM (DFA) Lecture Notes 7 Equivlence of Regulr Lnguges nd FSMs 8