1 From NFA to regular expression

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1 Note 1: How to convert DFA/NFA to regulr expression Version: 1.0 S/EE 374, Fll 2017 Septemer 11, 2017 In this note, we show tht ny DFA cn e converted into regulr expression. Our construction would work y llowing regulr expressions to e written on the edges of the DFA, nd then showing how one cn remove sttes from this generlized utomt (getting new equivlent utomt with the fewer sttes). In the end of this stte removl process, we will remin with generlized utomt with single il stte nd single ccepting stte, nd it would e then esy to convert it into single regulr expression. 1 From NFA to regulr expression 1.1 NFA A Generlized NFA onsider n NFA N where we llowed to write ny regulr expression on the edges, nd not only just symols. The utomt is llowed to trvel on n edge, if it cn mtches prefix of the unred input, to the regulr expression written on the edge. We will refer to such n utomt s NFA (generlized non-deterministic fe utomt [Don t you just love ll these shortcuts?]). Thus, the NFA on the right, ccepts the string, since A B B E. To simplify the discussion, we would enforce the following conditions: A () B E (1) There re trnsitions going from the il stte to ll other sttes, nd there re no trnsitions into the il stte. (2) There is single ccept stte tht hs only trnsitions coming into it (nd no outgoing trnsitions). (3) The ccept stte is distinct from the il stte. (4) Except for the il nd ccepting sttes, ll other sttes re connected to ll other sttes vi trnsition. In prticulr, ech stte hs trnsition to itself. When you cn not ctully go etween two sttes, NFA hs trnsitions lelled with, which will not mtch ny string of input chrcters. We do not hve to drw these trnsitions explicitly in the stte digrms. 1.2 Top-level outline of conversion We will convert DFA to regulr expression s follows: (A) onvert DFA to NFA, dding new il nd finl sttes. (B) emove ll sttes one-y-one, until we hve only the il nd finl sttes. 1

2 () Output regex is the lel on the (single) trnsition left in the NFA. (The word regex is just shortcut for regulr expression.) Lemm 1.1. A DFA M cn e converted into n equivlent NFA G. Proof: We cn consider M to e n NFA. Next, we dd specil il stte q tht is connected to the old il stte vi ε-trnsition. Next, we dd specil finl stte q fin, such tht ll the finl sttes of M re connected to q fin vi n ε-trnsition. The modified NFA M hs n il stte nd single finl stte, such tht no trnsition enters the il stte, nd no trnsition leves the finl stte, thus M comply with conditions (1 3) ove. Next, we consider ll pir of sttes x, y Q(M ), nd if x y there is no trnsition etween them, we introduce the trnsition. The resulting NFA G from M is now complint lso with condition (4). It is esy now to verify tht G is equivlent to the originl DFA M. We will remove ll the intermedite sttes from the GNFA, leving NFA with only il nd finl sttes, connected y one trnsition with (typiclly complex) lel on it. The equivlent regulr expression is ovious: the lel on the trnsition. some regex q S q F Lemm 1.2. Given NFA N with k = 2 sttes, one cn generte n equivlent regulr expression. Proof: A NFA with only two sttes (tht comply with conditions (1)-(4)) hve the following form. q S some regex q F The NFA hs single trnsition from the il stte to the ccepting stte, nd this trnsition hs the regulr expression ssocited with it. Since the il stte nd the ccepting stte do not hve self loops, we conclude tht N ccepts ll words tht mtches the regulr expression. Nmely, L(N) = L(). 1.3 Detils of ripping out stte We first descrie the construction. Since k > 2, there is t lest one stte in N which is not il or ccepting, nd let q rip denote this stte. We will rip this stte out of N nd fix the NFA, so tht we get NFA with one less stte. Trnsition pths going through q rip might come from ny of vriety of sttes q 1, q 2, etc. They might go from q rip to ny of nother set of sttes r 1, r 2, etc. For ech pir of sttes q i nd r i, we need to convert the trnsition through q rip into direct trnsition from q i to r i. q 1 q 2 q 3 q rip r 1 r 2 r 3 2

3 1.3.1 eworking connections for specific triple of sttes To understnd how this works, let us focus on the connections etween q rip nd two other specific sttes q in nd q out. Notice tht q in nd q out might e the sme stte, ut they oth hve to e different from q rip. The stte q rip hs self loop with regulr expression rip ssocited with it. So, consider frgment of n ccepting trce tht goes through q rip. It trnsition into q rip from stte q in with regulr expression in nd trvels out of q rip into stte q out on n edge with the ssocited regulr expression eing out. This trce, corresponds to the regulr expression in followed y 0 or more times of trveling on the self loop ( rip is used ech time we trverse the loop), nd then trnsition out to q out using the regulr expression out. As such, we cn introduce direct trnsition from q in to q out with the regulr expression = in ( rip ) out. rip lerly, ny frgment of trce trveling q in q in q rip q out cn e replced y the direct q rip q out trnsition q in q out. So, let us do this replcement for ny two such stges, we connect them directly vi new trnsition, so tht they no longer need to trvel through q rip. in ( rip ) out lerly, if we do tht for ll such pirs, the new utomt ccepts the sme lnguge, ut no longer need to use q rip. As such, we cn just remove q rip from the resulting utomt. And let M denote the resulting utomt. The utomt M is not quite legl, yet. Indeed, we will hve now prllel trnsitions ecuse of the ove process (we might even hve prllel self loops). But this is esy to fix: We replce two such prllel trnsitions q 1 i qj nd q 2 i qj, y single trnsition q i qj. As such, for the triple q in, q rip, q out, if the originl lel on the direct trnsition from q in to q out ws originlly dir, then the output lel for the new trnsition (tht skips q rip ) will e in out dir + in ( rip ) out. (1) lerly the new trnsition, is equivlent to the two trnsitions it replces. If we repet this process for ll the prllel trnsitions, we get new NFA M which hs k 1 sttes, nd furthermore it ccepts exctly the sme lnguge s N. 1.4 Proof of correctness of the ripping process Lemm 1.3. Given NFA N with k > 2 sttes, one cn generte n equivlent NFA M with k 1 sttes. Proof: Since k > 2, N contins lest one stte in N which is not ccepting, nd let q rip denote this stte. We will rip this stte out of N nd fix the NFA, so tht we get NFA with one less stte. For every pir of sttes q in nd q out, oth distinct from q rip, we replce the trnsitions tht go through q rip with direct trnsitions from q in to q out, s descried in the previous section. 3

4 orrectness. onsider n ccepting trce T for N for word w. If T does not use the stte q rip thn the sme trce exctly is n ccepting trce for M. So, ssume tht it uses q rip, in prticulr, the trce looks like T =... q i S i qrip 0 or more times {}}{ S i+1 S j 1 qrip... qrip S j 1 qj.... Where S i S i+1..., S j is sustring of w. lerly, S i in, where in is the regulr expression ssocited with the trnsition q i q rip. Similrly, S j 1 out, where out is the regulr expression ssocited with the trnsition q rip q j. Finlly, S i+1 S i+2 S j 1 ( rip ), where rip is the regulr expression ssocited with the self loop of q rip. Now, clerly, the string S i S i+1... S j mtches the regulr expression in ( out ) out. in prticulr, we cn replce this portion of the trce of T y S i S i+1... S j 1 S j T =... q i qj.... This trnsition is using the new trnsition etween q i nd q j introduced y our construction. epeting this replcement process in T till ll the ppernces of q rip re removed, results in n ccepting trce T of M. Nmely, we proved tht ny string ccepted y N is lso ccepted y M. We need lso to prove the other direction. Nmely, given n ccepting trce for M, we cn rewrite it into n equivlent trce of N which is ccepting. This is esy, nd done in similr wy to wht we did ove. Indeed, if portion of the trce uses new trnsition of M (tht does not pper in N), we cn plce it y frgment of trnsitions going through q rip. In light of the ove proof, it is esy nd we omit the strightforwrd ut tedious detils. Theorem 1.4. Any DFA cn e trnslted into n equivlent regulr expression. Proof: Indeed, convert the DFA into NFA N. As long s N hs more thn two sttes, reduce its numer of sttes y removing one of its sttes using Lemm 1.3. epet this process till N hs only two sttes. Now, we convert this NFA into n equivlent regulr expression using Lemm Exmples 2.1 Exmple: From NFA to regex in 8 esy figures 1: The originl NFA. A B 2: Normlizing it. A B A, + 4

5 3: emove stte A. A B A + 4: edrwn without old edges. B A + 5: emoving B. 6: edrwn. B + A + + A 7: emoving. ( +)( +) + 8: edrwn. ( +)( +) A + A Thus, this utomt is equivlent to the regulr expression ( + )( + ). 5

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