Greedy regular expression matching
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- Magnus Arthur Baker
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1 Alin Frisch INRIA Luc Crdelli MSRC ICALP
2 The mtching prolem The prolem Project the structure of regulr expression on flt sequence. R = ( ) w = v = [1 : [ 1 ; 2 ]; 2 : 1 ; 2 : 2 ; 1 : [ 3 ]] The result retins the structure of the regexp nd the content of the sequence. Result driven y the syntx of regexps utomt. Issues: efficiency dismigution.
3 Min motivtion Type-directed ntive representtion of vlues in XDuce-like lnguges: E.g.: [ int ] int [ int int* ] struct {int fst; int[] snd;} Advntges over uniform representtion: More compct representtion less oxing Fst rndom ccess Esier to interfce/integrte with other lnguge Requires coercion etween sutypes. Fltten sequences. Project the structure of the new regexp = mtching.
4 Other pplictions Regexp pckges with structured mtching semntics. Lexer-prser genertors. Opertion/representtion defined y induction on the structure of regexps (e.g.: Hosoy s filters).
5 Proof of concept A regexp itertor extension for C# oject[] = new oject[] {1234"c"45"xyz"67flse}; pplyregexp() ( ( ( int int )* string ) ( ( int )* ool ) )*;
6 Proof of concept A regexp itertor extension for C# oject[] = new oject[] {1234"c"45"xyz"67flse}; pplyregexp() ( ( { int sum = 0; } ( int x int y { sum += x*y; } )* string s { System.Console.WriteLine(s + ":" + sum); } ) ( { int sum = 0; } ( int x { sum += x; } )* ool { System.Console.WriteLine( + ":" + sum); } ) )*; c:14 xyz:20 Flse:13
7 Key issue: voiding cktrcking Consider the regexp: R = ( ) To void cktrcking nd still proceed y induction on the regexp we need to decide first which rnch to tke (left or right?) Unounded look-hed!
8 An exmple Astrct syntx tree of the regexp. R = ( ) ( ) w =
9 An exmple Astrct syntx tree of the regexp. Build n utomton on top of it. R = ( ) ( ) w = q f
10 An exmple Astrct syntx tree of the regexp. Build n utomton on top of it. Scn the input ckwrds ( suset construction ). R = ( ) ( ) w = q f
11 An exmple Astrct syntx tree of the regexp. Build n utomton on top of it. Scn the input ckwrds ( suset construction ). R = ( ) ( ) w = q f
12 An exmple Astrct syntx tree of the regexp. Build n utomton on top of it. Scn the input ckwrds ( suset construction ). R = ( ) ( ) w = q f
13 Second pss: the mtcher let rec loop = function ε -> () r 1 r 2 -> (loop r 1 loop r 2 ) r 1 r 2 -> if... then (1loop r 1 ) else (2loop r 2 ) r -> if... then (loop r)::(loop r) else [] c -> (* Consume the token *) Wht re the...? Given y the first pss. Dismigution: first-mtch for greedy semntics for
14 Non-termintion prolem The lgorithm lwys termintes except with suregexp R where R is nullle. Exmples: ( ) ( ) Sme prolem in the folklore syntx-directed recognizer: let rec loop r k w = mtch r with ε -> k w r 1 r 2 -> loop r 1 (loop r 2 k) w r 1 r 2 -> (loop r 1 k w) (loop r 2 k w) r -> (loop r (loop r k) w) (k w) c -> (w <> []) && (hd w = c) && (k (tl w)) let ccept r = loop r ( (=) [] ) r : regexp k : continution w : input sequence loop r k w = true w = w2 s.t. (r mtches w1) && (k w2 = true)
15 Non-termintion prolem: solutions Rewrite regexps to void the prolemtic sitution E.g.: ( ) (( +) +) The structure of the regexp is lost: not suitle for the mtching prolem. Interction with the dismigution policy? Prevent itertions in strs from ccepting empty sequences In the functionl recognizer replce (loop r (loop r k) w) (k w) with: (loop r (fun w -> (w <> w ) && (loop r k w ) w)) (k w) How to dpt our lgorithm?
16 An exmple R = ( )
17 An exmple R = ( ) Loop of ε-trnsitions q f
18 An exmple R = ( ) Loop of ε-trnsitions... now roken. Still finite stte utomton (sttes (q f )). q f f 0 f? = 1 f 1 f 1
19 Second pss: the mtcher let rec loop = function ε -> () r 1 r 2 -> (loop r 1 loop r 2 ) r 1 r 2 -> if... then (1loop r 1 ) else (2loop r 2 ) r -> if... then ( f := 0; (loop r)::(loop r)) else [] c -> f := 1; (* Consume the token *) The... re given y the first pss.
20 Conclusion Keep tight connection etween regexps nd utomt. Direct construction of the utomton Accept prolemtic regexps reject prolemtic mtchings. Result: liner time (two-psses) mtching lgorithm which cn e efficiently implemented. Astrct specifiction of the dismigution policy s n optimiztion prolem (not presented). Chrcteriztion nd study of prolemtic cses (not presented).
21 Future work Evlute the lterntive implementtion technique for XML lnguges. Optimiztions: the first pss is not lwys necessry. Use (ounded) look-hed s long s possile or lzy first pss. Non-locl dismigution policy e.g.: longest mtch semntics. Non-regulr lnguges.
22 Questions?
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