3 Regular expressions


 Julianna Wilkins
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1 3 Regulr expressions Given n lphet Σ lnguge is set of words L Σ. So fr we were le to descrie lnguges either y using set theory (i.e. enumertion or comprehension) or y n utomton. In this section we shll introduce regulr expressions s n elegnt nd concise wy to descrie lnguges. We shll see tht the lnguges definle y regulr expressions re precisely the sme s those ccepted y deterministic or nondeterministic finite utomt. These lnguges re clled regulr lnguges or (ccording to the Chomsky hierrchy) Type 3 lnguges. As lredy mentioned in the introduction regulr expressions re used to define ptterns in progrms such s grep. grep gets s n rgument regulr expression nd then filters out ll those lines from file which mtch the regulr expression, where mtching mens tht the line contins sustring which is in the lnguge ssigned to the regulr expression. It is interesting to note tht even in the cse when we serch for specific word (this is specil cse of regulr expresion) progrms like grep re more efficient thn nive implementtion of word serch. To find out more out grep hve look t the UNIX mnul pge nd ply round with grep. Note tht the syntx grep uses is slightly different from the one we use here. grep lso use some convenient shorthnds which re not relevnt for theoreticl nlysis of regulr expressions ecuse they do not extend the clss of lnguges. 3.1 Wht re regulr expressions? We ssume s given n lphet Σ (e.g. Σ = {,, c,..., z}) nd define the syntx of regulr expressions (over Σ) 1. is regulr expression. 2. ɛ is regulr expression. 3. For ech x Σ, x is regulr expression. E.g. in the exmple ll smll letters re regulr expression. We use oldfce to emphsize the difference etween the symol nd the regulr expression. 4. If E nd F re regulr expressions then E + F is regulr expression. 5. If E nd F re regulr expressions then EF (i.e. just one fter the other) is regulr expression. 6. If E is regulr expression then E is regulr expression. 7. If E is regulr expression then (E) is regulr expression. These re ll regulr expressions. Here re some exmples for regulr expressions: ɛ hllo hllo + hello h( + e)llo (ɛ + )() (ɛ + ) As in rithmetic they re some conventions how to red regulr expressions: inds stronger then sequence nd +. E.g. we red s ( ). We hve to use prentheses to enforce the other reding (). Sequencing inds stronger thn +. E.g. we red + cd s () + (c). To enforce nother reding we hve to use prentheses s in ( + c)d. 3.2 The mening of regulr expressions We now know wht regulr expressions re ut wht do they men? For this purpose, we shll first define n opertion on lnguges clled the Kleene str. Given lnguge L Σ we define L = {w 0w 1... w n 1 n N i < n.w i L} Intuitively, L contins ll the words which cn e formed y conctenting n ritrry numer of words in L. This includes the empty word since the numer my e 0. As n exmple consider L = {, } {, } : L = {ɛ,,,,,,,... } You should notice tht we use the sme symol s in Σ ut there is sutle difference: Σ is set of symols ut L Σ is set of words. Alterntively (nd more strctly) one my descrie L s the lest lnguge (wrt ) which contins L nd the empty word nd is closed under conctention: w L v L = wv L We now define the semntics of regulr expressions: To ech regulr expression E over Σ we ssign lnguge L(E) Σ. We do this y induction over the definition of the syntx: 1. L( ) = 2. L(ɛ) = {ɛ} 3. L(x) = {x} where x Σ. 4. L(E + F ) = L(E) L(F ) 5. L(EF ) = {wv w L(E) v L(F )} 6. L(E ) = L(E) 7. L((E)) = L(E) 14 13
2 Sutle points: in 1. the symol my e used s regulr expression (s in L( )) or the empty set ( = {}). Similrily, ɛ in 2. my e regulr expression or word, in 6. my e used to construct regulr expressions or it is n opertion on lnguges. Which lterntive we men ecomes only cler from the context, there is no generlly greed mthemticl nottion 1 to mke this difference explicit. Let us now clculte wht the exmples of regulr expressions from the previous section men, i.e. wht re the lnguges they define: ɛ hllo By 2. L(ɛ) = {ɛ} Let s just look t L(h). We know from 3: Hence y 5: L(h) = {h} L() = {} L(h) = {wv w {h} v {}} = {h} Continuing the sme resoning we otin: hllo + hello h( + e)llo L(hllo) = {hllo} From the previous point we know tht: Hence y using 4 we get: Using 3 nd 4 we know Hence using 5 we otin: L(hllo) = {hllo} L(hello) = {hello} L(hllo + hello) = {hllo} {hello}} = {hllo, hello} L( + e) = {, e} L(h( + e)llo) = {uvw u L(h) v L( + e) w L(llo)} = {uvw u {h} v {, e} w {(llo}} = {hllo, hello} 1 This is different in progrmming, e.g. in JAVA we use "... " to signl tht we men things literlly. Let us introduce the following nottion: Now using 6 we know tht w i = ww... w }{{} i times L( ) = {w 0w 1... w n 1 n N i < n.w i L()} = {w 0w 1... w n 1 n N i < n.w i {}} = { n n N} nd hence using 5 we conclude L( ) = {uv u L( ) v L( )} = {uv u { n n N} v { m m N}} = { n m m, n N} I.e. L( ) is the set of ll words which strt with (possily empty) sequence of s followed y (possily empty) sequence of s. (ɛ + )() (ɛ + ) Let s nlyze the prts: Hence, we hve L(ɛ + ) = {ɛ, } L(() ) = { i i N} L(ɛ + ) = {ɛ, } L((ɛ + )() (ɛ + )) = {u() i v u {ɛ, } i N v {ɛ, } In english: L((ɛ + )() (ɛ + )) is the set of (possily empty) sequences of interchnging s nd s. 3.3 Trnslting regulr expressions to NFAs Theorem 3.1 For ech regulr expression E we cn construct NFA N(E) s.t. L(N(E)) = L(E), i.e. the utomton ccepts the lnguge descried y the regulr expression. Proof: We do this gin y induction on the syntx of regulr expressions: 1. N( ): 16 15
3 x N(0) N(x) which will reject everything (it hs got no finl sttes) nd hence L(N( )) = = L( ) 2. N(ɛ): This utomton only ccepts the word x, hence: L(N(x)) = {x} = L(x) 4. N(E + F ): We merge the digrms for N(E) nd N(F ) into one: N(E) N(F) N( ε) I.e. given N(E+F) N(E) = (Q E, Σ, δ E, S E, F E) N(F ) = (Q F, Σ, δ F, S F, F F ) This utomton ccepts the empty word ut rejects everything else, hence: Now we use the disjoint union opertion on sets (see the MCS lecture L(N(ɛ)) = {ɛ} = L(ɛ) 3. N(x): 18 17
4 notes [Alt01], section 4.1) Q E+F = Q E + Q F δ E+F ((0, q), x) = {(0, q ) q δ E (q, x)} δ E+F ((1, q)), x = {(1, q ) q δ F (q, x)} S E+F = S E + S F F E+F = F E + F F N(E + F ) = (Q E+F, Σ, δ E+F, S E+F, F E+F ) The disjoint union just signls tht we re not going to identify sttes, even if they ccidently hppen to hve the sme nme. Just thinking of the gme with mrkers you should e le to convince yourself tht L(N(E + F )) = L(N(E)) L(N(F )) Moreover to show tht we re llowed to ssume tht L(N(E + F )) = L(E + F ) L(N(E)) = L(E) L(N(F )) = L(F ) tht s wht is ment y induction over the syntx of regulr expressions. Now putting everything together: 5. N(EF ): L(N(E + F )) = L(N(E)) L(N(F )) = L(E) L(F ) = L(E + F ) We wnt to put the two utomt N(E) nd N(F ) in series. We do this y connecting the finl sttes of N(E) with the initil sttes of N(F ) in wy explined elow. In this digrm I only depicted one initil nd one finl stte of ech of the utomt lthough they my e severl of them. Here is how we construct N(EF ) from N(E) nd N(F ): N(E) = (Q E, Σ, δ E, S E, F E) N(F ) = (Q F, Σ, δ F, S F, F F ) The sttes of N(EF ) re the disjoint union of the sttes of N(E) nd N(F ): Q EF = Q E + Q F The trnsition function of N(EF ) contins ll the trnsitions of N(E) nd N(F ) (s for N(E + F )) nd for ech stte q of N(E) which hs trnsition to finl stte of N(E) we dd trnsition with the sme lel to ll the initil sttes of N(F ). δ EF ((0, q), x) = {(0, q ) q δ E(q, x)} {(1, q ) q.q δ E(q, x) q S E} δ EF ((1, q)) = {(1, q ) q δ F (q))} The initil sttes of N(EF ) re the initil sttes of N(E), nd the initil sttes of N(F ) if there is n initil stte of N(E) which is lso finl stte. S EF = {(0, q) q S E} {(1, q) q S F S E F E } The finl sttes of N(EF ) re the finl sttes of N(F ). We now set F EF = {(1, q) q F F } N(EF ) = (Q EF, Σ, δ EF, S EF, Z EF ) I hope tht you re le to convince yourself tht nd hence we cn reson L(N(EF )) = {uv u L(N(E)) v L(N(F )) N(E) N(F) L(N(EF )) = {uv u L(N(E)) v L(N(F )) = {uv u L(E) v L(F ) 6. N(E ): = L(EF ) We construct N(E ) from N(E) y merging initil nd finl sttes of N(E) in wy similr to the previous construction nd we dd new stte which is initil nd finl. N(EF) 20 19
5 * since we cn run through the utomton n ritrry numer of times. The new stte llows us lso to ccept the empty sequence. Hence: L(N(E )) = {w 0w 1... w n 1 n N i < n.w i L(N(E))} = L(N(E)) = L(E) = L(E ) N(E) N(E*) 7. N((E)) = N(E) I.e. using rckets does not chnge nything. As n exmple we construct N( ). First we construct N(): Given we construct N(E ). N(E) = (Q E, Σ, δ E, S E, F E ) We dd one extr stte : Q E = Q E + { } N E inherits ll trnsitions form N E nd for ech stte which hs n rrow to the finl stte lelled x we lso dd n rrow to ll the initil sttes lelled x. N() Now we hve to pply the construction nd we otin: δ E ((0, q), x) ={(0, q ) q δ E (q, x)} {(0, q ) δ E (q, x) F E q S E } The initil sttes of N(E ) re the initil sttes of N(E) nd : S E = {(0, q) q S E } {(1, )} The finl sttes of N E re the finl sttes of N E nd : F E = {(0, q) q F E } {(1, )} We define We clim tht N(E ) = (Q E, Σ, δ E, S E, F E ) N(*) L(N(E )) = {w 0 w 1... w n 1 n N i < n.w i L(N(E))} N( ) is just the sme nd we get 22 21
6 N(*) N(*) nd now we hve to serilize the two utomt nd we get: 1. L is given y regulr expression. 2. L is the lnguge ccepted y n NFA. 3. L is the lnguge cceped y DFA. Proof: We hve tht 1. = 2 y theorem 3.1. We know tht 2. = 3. y2.2 nd 3. = 1. y 3.2. As indicted in the introduction: the lnguges which re chrcterized y ny of the three equivlent conditions re clled regulr lnguges or type3 lnguges. N(**) Now, you my oserve tht this utomton, though correct, is unnecessry complicted, since we could hve just used However, we shll not e concerned with minimlity t the moment. 3.4 Summing up... From the previous section we know tht lnguge given y regulr expression is lso recognized y NFA. Wht out the other wy: Cn lnguge recognized y finite utomton (DFA or NFA) lso e descried y regulr expression? The nswer is yes: Theorem 3.2 (Theorem 3.4, pge 91) Given DFA A there is regulr expression R(A) which recognizes the sme lnguge L(A) = L(R(A)). We omit the proof (which cn e found in the [HMU01] on pp.9193). However, we conclude: Corollry 3.3 Given lnguge L Σ the following is equivlent: 24 23
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