ɛ-closure, Kleene s Theorem,
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1 DEGefW5wiGH2XgYMEzUKjEmtCDUsRQ4d 1 A nice pper relevnt to this course is titled The Glory of the Pst 2 NICTA Resercher, Adjunct t the Austrlin Ntionl University nd Griffith University ɛ-closure, Kleene s Theorem, nd Kleene s Alger Chrles Gretton 2 27 Ferury 2014 NICTA Funding nd Supporting Memers nd Prtners Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 1/31
2 Regulr Expressions A regulr expression is well formed word. We use the following symols: Σ {ɛ ( ) +. } Tking Σ {ɛ, }, regulr expressions stisfy the following grmmr: regxp ::= (regxp) regxp + regxp regxp.regxp regxp The Kleene closure opertor, *, hs the highest precedence. Conctention,., is usully represented y juxtposition e.g. conctention. is written Conctention hs the next highest precedence. Union/choice hs the lest precedence. Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 2/31
3 Regulr Expressions Regulr expressions correspond to n lgeric tool to descrie lnguges. A lnguge you cn descrie is clled regulr lnguge. If E is regulr expression we write L(E) for the corresponding lnguge. The definition is recursive, nd proofs often proceed using structurl induction. Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 3/31
4 Regulr Expressions Exmples + is union,. is conctention, nd is closure. L() = {} L( + ) = {, } L(( + )) = {, } L((( + )) + ) = {,,,,...} L((( + )) ) = {ɛ,,,,,...} Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 4/31
5 Regulr Expressions Kleene s Theorem For every regulr expression there is DFA / NFA / ɛ-nfa which ccepts the corresponding regulr lnguge. AND For every DFA / NFA / ɛ-nfa there is regulr expression descriing tht mchine s lnguge. The first direction cn e shown y structurl induction. We will uild the utomton for the regulr lnguge. As usul, we tret the se cses first (constnts nd lphet symols), nd then ech of the opertors: Σ {ɛ ( ) +. } Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 5/31
6 Bse Cse Kleene s Theorem Σ ɛ Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 6/31
7 Union Step Cse Kleene s Theorem L 1 + L 2 ǫ L(L 1 ) ǫ ǫ L(L 2 ) ǫ Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 7/31
8 Conctention Step Cse Kleene s Theorem L 1.L 2 L(L 1 ) L(L 2 ) ǫ Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 8/31
9 Closure Step Cse Kleene s Theorem L 1 ǫ ǫ ǫ L(L 1 ) ǫ ǫ ǫ Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 9/31
10 A Regulr Expression ɛ-nfa QED Now, let us go in the other direction. Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 10/31
11 DFA A Regulr Expression Induction is on the sttes of the DFA. Ignore ny existing stte lels, nd numer them ritrrily 1, 2,.., n New concept: k-pth etween two sttes i nd j. Expression descriing lnguge if we cn only trverse sttes lelled k or less, in pth from i to j. Sttes i nd j re excluded from the k restriction. Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 11/31
12 k-pths Exmples Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. For 0-pths etween ny two nodes, note lelling strts t 1 y convention. 0-pths from 2 to 3 hve the expression lel 1-pths from 2 to 3 hve the expression lel + 2-pths from 2 to 3 hve the expression lel () + () Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 12/31
13 k-pths DFA A Regulr Expression Re-lelled DFA sttes with integers 1, 2,.., n The nottion R k ij is the regulr expression for ll k-pths from i to j BASIS: Tking i j, we hve R 0 ij = SUM δ(i,x)=j x R 0 ii = ɛ + SUM δ(i,x)=i x Note: SUM... = Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 13/31
14 k-pths Bsis Exmples Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. R1,2 0 = R1,1 0 = + ɛ you ll rememer the cncelltion lws right? Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 14/31
15 Alger nd Regulr Expressions Cncelltion lws: Identities Annihiltors L + = L + L = L ɛl = L Lɛ = L L = L = Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 15/31
16 k-pths Bsis Exmples Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. R 0 1,2 = R 0 1,1 = + ɛ you ll rememer the cncelltion lws, right! R 0 1,1 = ɛ Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 16/31
17 k-pths Step Cse Re-lelled DFA sttes with integers 1, 2,.., n The nottion R k ij is the regulr expression for ll k-pths from i to j INDUCTION: We hve found n expression for ech R k 1 ij tret Rij k Cse 1: There is no pth from i to j vi k. R k 1 ij = R k ij Cse 2: Trversl vi k is possile. R k ij = R k 1 ij + R k 1 ik (R k 1 kk ) R k 1 kj nd now Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 17/31
18 k-pths DFA A Regulr Expression Re-lelled DFA sttes with integers 1, 2,.., n We will keep nottions δ 0 nd F for: Strting stte index in {1..n} Set of indices of ll finl sttes in {1..n} The nottion R k ij is the regulr expression for ll k-pths from i to j Conclusion.. equivlent Expression is: SUM i F R n δ 0 i Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 18/31
19 k-pths Conclusion Exmple Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. R23 3 = R R2 23 (R2 33 ) R33 2 R23 3 = R2 23 ɛ + R2 23 (R2 33 ) R33 2 cncelltion R23 3 = R2 23 (ɛ + (R2 33 ) R33 2 ) left distriutive R23 3 = R2 23 (R2 33 ) cncelltion nd power ssocitive Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 19/31
20 k-pths Induction Exmple Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. R23 3 = R2 23 (R2 33 ) R23 2 = () + () R33 2 = () ( + ) + () ( + ) Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 20/31
21 k-pths Induction Exmple Note exmple is symmetric. Let us lel sttes 1,2 nd 3 from left to right. R 3 23 = (() + () )(() ( + ) + () ( + )) Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 21/31
22 A Regulr Expression DFA Going from right to left ws rther cumersome. Also, we should e llowed to process ɛ-nfa directly. To void ny explosions moving from ɛ-nfa to DFA. How mny expressions does nive implementtion of the k-pth pproch generte? Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 22/31
23 A Regulr Expression DFA Going from right to left ws rther cumersome. Also, we should e llowed to process ɛ-nfa directly. To void ny explosions moving from ɛ-nfa to DFA. How mny expressions does nive implementtion of the k-pth pproch generte? For i, j, k 0..n, i 0 nd j 0, we need to get R k ij It s cuic in n. Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 23/31
24 A Regulr Expression DFA Going from right to left ws rther cumersome. Also, we should e llowed to process ɛ-nfa directly. To void ny explosions moving from ɛ-nfa to DFA. Is there more prgmtic wy of getting from n utomton to n equivlent regulr expression? Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 24/31
25 A Regulr Expression DFA Going from from right to left ws rther cumersome. Also, we should e llowed to process ɛ-nfa directly. To void ny explosions moving from ɛ-nfa to DFA Is there more prgmtic wy of getting from n utomton to n equivlent regulr expression? yes! Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 25/31
26 DFA Regulr Expression Process rc lels so there is only one rc etween ech pir of sttes., A,, B C D A B C D + A + + B C D + Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 26/31
27 DFA Regulr Expression Mke copy of the utomton for ech ccepting stte.,,, A B C D A B C D + A B + C + D A B C D Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 27/31
28 DFA Regulr Expression Itertively contrct sttes. + A ( + ) C + D + ( + ) + A C D Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 28/31
29 DFA Regulr Expression Itertively contrct sttes. + A ( + ) C + A ( + )( + ) D Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 29/31
30 DFA Regulr Expression Itertively contrct sttes. ( + ) ( + ) ( + ) ( + )( + ) + A ( + ) C + A ( + )( + ) D Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 30/31
31 DFA Regulr Expression Itertively contrct sttes. Lnguge of mchine is: (( + ) ( + )) + (( + ) ( + )( + )) Kleene s Theorem Copyright NICTA 2014 Chrles Gretton 2 31/31
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