Regular Languages and Applications

Transcription

1 Regulr Lnguges nd Applictions Yo-Su Hn Deprtment of Computer Science Yonsei University 1-1 SNU 4/14

2 Regulr Lnguges An old nd well-known topic in CS Kleene Theorem in 1959 FA (finite-stte utomton) constructions: Thompson utomt, position utomt in 1960s Pttern Mtching Prolem in 1970s... in 1980s REVISIT Stte Complexity, Prime Decomposition, Pttern Mtching since mid 1990s 2-1 SNU 4/14

3 Regulr Lnguges An old nd well-known topic in CS Kleene Theorem in 1959 FA (finite-stte utomton) constructions: Thompson utomt, position utomt in 1960s Pttern Mtching Prolem in 1970s... in 1980s REVISIT Stte Complexity, Prime Decomposition, Pttern Mtching since mid 1990s XML, Bioinformtics - New Applictions 2-2 SNU 4/14

4 Overview Bsic Notions Position Construction nd XML DTD Regulr-Expression Pttern Mtching Stte Complexity Future Directions nd Conclusions 3-1 SNU 4/14

5 Regulr Expressions Regulr expressions re very convenient form tht represents (infinite) sets of strings clled regulr sets. Given finite lphet Σ, regulr expression over Σ is defined recursively s follows: 1., the empty-set symol, is regulr expression. 2. λ, the empty-string symol, is regulr expression. 3. Σ is regulr expression. 4. E +F (union), where E nd F re regulr expressions, is regulr expression. 5. E F (ctention), where E nd F re regulr expressions, is regulr expression. 6. E (Kleene str), where E is regulr expression, is regulr expression. 4-1 SNU 4/14

6 Finite-stte Automt (FAs) A finite-stte utomton A is specified y tuple (Q, Σ, δ, s, F ); Q finite set of sttes Σ finite lphet δ(p, ) = q set of trnsition rules s Q the strt stte F Q set of finl sttes 5-1 SNU 4/14

7 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } 6-1 SNU 4/14

8 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = 6-2 SNU 4/14

9 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 2 } 6-3 SNU 4/14

10 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 2 } 6-4 SNU 4/14

11 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = 6-5 SNU 4/14

12 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 3 } 6-6 SNU 4/14

13 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 4, q 6 } 6-7 SNU 4/14

14 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 7 } 6-8 SNU 4/14

15 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 8, q 9 } 6-9 SNU 4/14

16 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = Q = {q 9 } 6-10 SNU 4/14

17 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } T = ccepted!! Q = {q 9 } F 6-11 SNU 4/14

18 Finite-stte Automt - exmple q 4 q 5 q 1 q 2 q 3 q 9 q 6 q 7 q 8 q 10 s = q 1 F = {q 9, q 10 } L = L( ( + ( + ))) 6-12 SNU 4/14

19 REs into Finite-stte Automt The well-known Thompson construction y Ken Thompson in E = λ E = E = λ M(E 1 ) λ λ M(E 2 ) λ E 1 + E 2 λ M(E 1 ) M(E 2 ) E 1 E 2 λ M(E) λ λ E λ 7-1 SNU 4/14

20 REs into Finite-stte Automt The well-known Thompson construction y Ken Thompson in E = λ E = E = esy to understnd nd uild-up too mny λ trnsitions λ M(E 1 ) λ λ M(E 2 ) λ E 1 + E 2 λ M(E 1 ) M(E 2 ) E 1 E 2 λ M(E) λ λ E λ 7-2 SNU 4/14

21 Position Automt - nother utomton construction Proposed y Glushkov nd McNughton nd Ymd in 1960 independently. The construction is sed on the positions of chrcters of given regulr expression. 8-1 SNU 4/14

22 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) 9-1 SNU 4/14

23 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

24 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

25 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

26 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

27 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

28 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

29 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) SNU 4/14

30 Position Automt - n exmple E = ( + ) c( + ) E = (1 + 2) 3(4 + 5) c c c SNU 4/14

31 Position Automt The construction looks nice! All in-trnsitions of stte hve the sme lel. The numer of sttes = E + 1 Less sttes thn the Thompson utomt nd, thus usully fster! E = ( + ) c( + ) c c c SNU 4/14

32 Where do position utomt led us? 11-1 SNU 4/14

33 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic SNU 4/14

34 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-2 SNU 4/14

35 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-3 SNU 4/14

36 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-4 SNU 4/14

37 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-5 SNU 4/14

38 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-6 SNU 4/14

39 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-7 SNU 4/14

40 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-8 SNU 4/14

41 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N 12-9 SNU 4/14

42 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Given n one-unmiguous regulr expression E nd n input string w, we cn red w using one lookhed with respect to E. E = SEO(UL) N S E O U L U L N SNU 4/14

43 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. Not ll regulr expressions re one-unmiguous. E = SEO(UL) UNI S E O U L U L N Not ll regulr lnguges re one-unmiguous. There re some regulr lnguges tht cnnot e defined y n one-miguous regulr lnguges. e.g. L(( + ) ( + ) k ), k SNU 4/14

44 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. <?xml version="1.0"?> <!DOCTYPE BOOK [ <!ELEMENT p (#PCDATA)> <!ELEMENT BOOK (OPENER,SUBTITLE?,INTRODUCTION?,(SECTION PART)+)> <!ELEMENT OPENER (TITLE_TEXT)*> <!ELEMENT TITLE_TEXT (#PCDATA)> <!ELEMENT SUBTITLE (#PCDATA)> <!ELEMENT INTRODUCTION (HEADER, p+)+> <!ELEMENT PART (HEADER, CHAPTER+)> <!ELEMENT SECTION (HEADER, p+)> <!ELEMENT HEADER (#PCDATA)> <!ELEMENT CHAPTER (CHAPTER_NUMBER, CHAPTER_TEXT)> <!ELEMENT CHAPTER_NUMBER (#PCDATA)> <!ELEMENT CHAPTER_TEXT (p)+> ]> SNU 4/14

45 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. <?xml version="1.0"?> <!DOCTYPE BOOK [ <!ELEMENT p (#PCDATA)> <!ELEMENT BOOK (OPENER,SUBTITLE?,INTRODUCTION?,(SECTION PART)+)> <!ELEMENT OPENER (TITLE_TEXT)*> <!ELEMENT TITLE_TEXT (#PCDATA)> <!ELEMENT SUBTITLE (#PCDATA)> <!ELEMENT INTRODUCTION (HEADER, p+)+> content models <!ELEMENT PART (HEADER, CHAPTER+)> <!ELEMENT SECTION (HEADER, p+)> <!ELEMENT HEADER (#PCDATA)> <!ELEMENT CHAPTER (CHAPTER_NUMBER, CHAPTER_TEXT)> <!ELEMENT CHAPTER_NUMBER (#PCDATA)> <!ELEMENT CHAPTER_TEXT (p)+> ]> SNU 4/14

46 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. <?xml version="1.0"?> <!DOCTYPE BOOK [ <!ELEMENT p (#PCDATA)> <!ELEMENT BOOK (OPENER,SUBTITLE?,INTRODUCTION?,(SECTION PART)+)> <!ELEMENT OPENER (TITLE_TEXT)*> <!ELEMENT TITLE_TEXT (#PCDATA)> <!ELEMENT SUBTITLE (#PCDATA)> BOOK <!ELEMENT ::= INTRODUCTION (HEADER, p+)+> OPENER <!ELEMENT (SUBTITLE+λ) PART (HEADER, (INTRODUCTION+λ) CHAPTER+)> (SECTION + PART)(SECTION + PART) <!ELEMENT SECTION (HEADER, p+)> <!ELEMENT HEADER (#PCDATA)> <!ELEMENT CHAPTER (CHAPTER_NUMBER, CHAPTER_TEXT)> <!ELEMENT CHAPTER_NUMBER (#PCDATA)> <!ELEMENT CHAPTER_TEXT (p)+> ]> SNU 4/14

47 One-Unmiguous Regulr Lnguges Proposed y Brüggemnn-Klein nd Wood. A regulr lnguge L is one-unmiguous if there is regulr expression E such tht L = L(E) nd the position utomton of E is deterministic. <?xml version="1.0"?> <!DOCTYPE BOOK [ <!ELEMENT p (#PCDATA)> <!ELEMENT BOOK (OPENER,SUBTITLE?,INTRODUCTION?,(SECTION PART)+)> <!ELEMENT OPENER (TITLE_TEXT)*> <!ELEMENT TITLE_TEXT (#PCDATA)> <!ELEMENT SUBTITLE (#PCDATA)> BOOK <!ELEMENT ::= INTRODUCTION (HEADER, p+)+> OPENER <!ELEMENT (SUBTITLE+λ) PART (HEADER, (INTRODUCTION+λ) CHAPTER+)> (SECTION + PART)(SECTION + PART) <!ELEMENT SECTION (HEADER, p+)> <!ELEMENT HEADER One-unmiguous (#PCDATA)> regulr expression!! <!ELEMENT CHAPTER (CHAPTER_NUMBER, CHAPTER_TEXT)> <!ELEMENT CHAPTER_NUMBER (#PCDATA)> <!ELEMENT CHAPTER_TEXT (p)+> ]> SNU 4/14

48 One-Unmiguous Regulr Lnguges vs XML DTD Regulr expressions for content models of DTD re one-unmiguous XML DTDs re LL(1) grmmrs [Wood 96] LL(k) grmmrs hve proper hierrchy [AU 72] k-unmiguous regulr lnguges?? 13-1 SNU 4/14

49 One-Unmiguous Regulr Lnguges vs XML DTD Regulr expressions for content models of DTD re one-unmiguous XML DTDs re LL(1) grmmrs [Wood 96] LL(k) grmmrs hve proper hierrchy [AU 72] k-unmiguous regulr lnguges?? We hve k-lookhed for processing n input string. X M L I 6-lookhed N S T A N C E current stte 13-2 SNU 4/14

50 k-lookhed Regulr Lnguges Two wys for defining k-lookhed regulr lnguges. The first is sed on lookhed of t most k 1 symols to determine the next, t most one, mtching position in given regulr expression: deterministic k-lookhed regulr expressions The second is similr except tht when we use lookhed of k symols, we must mtch the next k positions uniquely: k-lockdeterministic regulr expressions 14-1 SNU 4/14

51 Deterministic k-lookhed regulr lnguges fter reding i+1 t stte q i i i+1 i+2 i+k i+k+1 k-lookhed i i+1 i+2 t stte q i+1 k-lookhed i+k i+k SNU 4/14

52 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) 16-1 SNU 4/14

53 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) # 16-2 SNU 4/14

54 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) # 16-3 SNU 4/14

55 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed. 0? 1 2? 3 E = ( + ) # 16-4 SNU 4/14

56 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) # 16-5 SNU 4/14

57 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) # 16-6 SNU 4/14

58 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) # 16-7 SNU 4/14

59 Deterministic k-lookhed regulr lnguges A regulr lnguge L is deterministic k-lookhed if there is deterministic k-lookhed regulr expression for L. A regulr expression is deterministic k-lookhed if its position utomton is deterministic k-lookhed E = ( + ) E is deterministic 2-lookhed. # 16-8 SNU 4/14

60 Deterministic k-lookhed regulr lnguges Thm. L((+) (+) k ), for k 0, is deterministic (k+1)-lookhed SNU 4/14

61 Deterministic k-lookhed regulr lnguges Thm. L((+) (+) k ), for k 0, is deterministic (k+1)-lookhed. k = SNU 4/14

62 17-3 SNU 4/14 Deterministic k-lookhed regulr lnguges Thm. L((+) (+) k ), for k 0, is deterministic (k+1)-lookhed. k = 1 k = 1 k = 2

63 Deterministic k-lookhed regulr lnguges Thm. L((+) (+) k ), for k 0, is deterministic (k+1)-lookhed. There exists hierrchy for deterministic k-lookhed regulr lnguges k k 1 k 2 k SNU 4/14

64 k-lock-deterministic regulr lnguges i i+1 i+2 i+k i+k+1 t stte q i k-lookhed fter reding i+1 i+k i i+1 i+2 t stte q i i+k i+k+1 k-lookhed 19-1 SNU 4/14

65 k-lock-deterministic regulr lnguges We define regulr lnguge L to e k-lock-deterministic if there exists k-lock utomton A = (Q, Σ, Γ, δ, s, F ) tht stisfies the following conditions: 1. A is position utomton over Γ. 2. A is deterministic lock utomton. 3. L = L(A ). It is esy to verify tht position utomton A for n 1-deterministic regulr lnguge is 1-lock-deterministic SNU 4/14

66 k-lock-deterministic regulr lnguges Thm. There is proper hierrchy in k-lock-deterministic regulr lnguges. Sketch of Proof. A (k 1)-lock-deterministic regulr lnguge is k- lock-deterministic y definition. Thus, it is enough to show tht there is k-lock-deterministic regulr lnguge tht is not (k 1)-lockdeterministic. k k 1 k 2 k SNU 4/14

67 k 3 sttes q 1 q 2 q 3 q 4 q 5 A q 1 q 3 A q 4 q SNU 4/14

68 Two Wys... Thm. k-lock-deterministic regulr lnguges re proper sufmily of deterministic k-lookhed regulr lnguges. k-lookhed determinism k-lock determinism Generliztions of One-Deterministic Regulr Lnguges, Yo-Su Hn nd Derick Wood, Informtion nd Computtion, Vol. 206, , SNU 4/14

69 XML DTD vs XML Schem There s no vs XML Schem re much more flexile nd powerful Thus, there re lso much more difficult nd confusing 24-1 SNU 4/14

70 XML DTD vs XML Schem There s no vs XML Schem re much more flexile nd powerful Thus, there re lso much more difficult nd confusing XML DTD XML Schem 24-2 SNU 4/14

71 XML DTD vs XML Schem There s no vs XML Schem re much more flexile nd powerful Thus, there re lso much more difficult nd confusing XML DTD 1-lookhed determinism XML Schem k-lookhed determinism 24-3 SNU 4/14

72 XML DTD vs XML Schem There s no vs XML Schem re much more flexile nd powerful Thus, there re lso much more difficult nd confusing XML DTD 1-lookhed determinism XML Schem k-lookhed determinism 24-4 SNU 4/14

73 XML DTD vs XML Schem There s no vs XML Schem re much more flexile nd powerful Thus, there re lso much more difficult nd confusing XML DTD 1-lookhed determinism XML Schem? k-lookhed determinism 24-5 SNU 4/14

74 Pttern Mtching - n ppliction of regulr lnguges Given regulr expression pttern P nd text T, find ll sustrings of T tht re in L(P ). T = AGCT AAT CCCT GAGAGT CCAGT T AGT CCCAT P = T (AG + C) T 25-1 SNU 4/14

75 Pttern Mtching - n ppliction of regulr lnguges Given regulr expression pttern P nd text T, find ll sustrings of T tht re in L(P ). T = AGCT AAT CCCT GAGAGT CCAGT T AGT CCCAT P = T (AG + C) T 25-2 SNU 4/14

76 Pttern Mtching New Domins: WEB, Bioinformtics, Huge DB, Imges or Source Codes 26-1 SNU 4/14

77 Pttern Mtching - relted work Given text T nd regulr expression E, The recognition prolem: We cn report ll end positions of mtching sustrings of T in O(mn) time [Aho] or in O(mn/ log n) time [Myers]. The identifiction prolem: We cn report ll (strt, end) positions of mtching sustrings of T in O(mn 2 ) time [Aho] SNU 4/14

78 Pttern Mtching - recognition prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) T = 28-1 SNU 4/14

79 Pttern Mtching - recognition prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) T = Σ E 28-2 SNU 4/14

80 Pttern Mtching - recognition prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) T = 28-3 SNU 4/14

81 Pttern Mtching - recognition prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) T = Given E nd T, we cn find ll end positions of mtching sustrings of T in O(mn) time using O(m) spce, where E = m nd T = n [Aho] SNU 4/14

82 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) T = 29-1 SNU 4/14

83 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = 29-2 SNU 4/14

84 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = 29-3 SNU 4/14

85 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = 29-4 SNU 4/14

86 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = 29-5 SNU 4/14

87 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = 29-6 SNU 4/14

88 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = Running Time = No. of mtching end positions O(mn) = O(n) O(mn) = O(mn 2 ) SNU 4/14

89 Pttern Mtching - identifiction prolem Given E over Σ, we prepend Σ to E; this llows mtching to egin t ny position in T. E = ( + ) E R = ( + ) T = Running Time = No. of mtching end positions O(mn) = O(n) O(mn) = O(mn 2 ). We cn solve the idenftifiction prolem in O(mn 2 ) worst-cse time using O(m) spce [Aho] SNU 4/14

90 Prefix nd Infix Given two strings x nd y over Σ, we sy x is prefix of y if there exists z Σ such tht xz = y. x is n infix of y if there exists u, v Σ such tht uxv = y; we often cll x sustring of y SNU 4/14

91 Prefix nd Infix Given two strings x nd y over Σ, we sy x is prefix of y if there exists z Σ such tht xz = y. x is n infix of y if there exists u, v Σ such tht uxv = y; we often cll x sustring of y. y = seoul 30-2 SNU 4/14

92 Prefix nd Infix Given two strings x nd y over Σ, we sy x is prefix of y if there exists z Σ such tht xz = y. x is n infix of y if there exists u, v Σ such tht uxv = y; we often cll x sustring of y. y = seoul seo is prefix of y SNU 4/14

93 Prefix nd Infix Given two strings x nd y over Σ, we sy x is prefix of y if there exists z Σ such tht xz = y. x is n infix of y if there exists u, v Σ such tht uxv = y; we often cll x sustring of y. y = seoul eou is n infix of y SNU 4/14

94 Prefix nd Infix Given two strings x nd y over Σ, we sy x is prefix of y if there exists z Σ such tht xz = y. x is n infix of y if there exists u, v Σ such tht uxv = y; we often cll x sustring of y. We define pttern P to e prefix-free if no string in P is prefix of ny other strings in P. infix-free if no string in P is n infix of ny other strings in P SNU 4/14

95 Infix-free Regulr-Expression Mtching L IN L P RE L REG L REG L P RE L IN 31-1 SNU 4/14

96 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE T L IN 31-2 SNU 4/14

97 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE T = the recognition process L IN 31-3 SNU 4/14

98 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE T = the recognition process L IN 31-4 SNU 4/14

99 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE E R = T = the recognition process L IN 31-5 SNU 4/14

100 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE E R = T = L IN the recognition process Becuse of infix-freeness, ech pir of (, ) from left to right must e mtching sustring SNU 4/14

101 Infix-free Regulr-Expression Mtching L IN L P RE L REG Given n infix-free regulr expression E nd text T : y = seoul eou is n infix of y. L REG E L P RE E R = T = L IN the recognition process Becuse of infix-freeness, ech pir of (, ) from left to right must e mtching sustring. We cn find ll mtching sustrings in O(mn) time [HWW07]. Prefix-Free Regulr Lnguges nd Pttern Mtching, Yo-Su Hn, Yjun Wng nd Derick Wood Theoreticl Computer Science Vol. 389, , SNU 4/14

102 Prefix-free Regulr-Expression Mtching L IN L P RE L REG If E is infix-free, we hve n O(mn) running time lgorithm If E is (norml) regulr expression, we hve n O(mn 2 ) running time lgorithm If E is prefix-free, then there re t most n mtching sustrings of T tht elong to L(E), where n is the size of T SNU 4/14

103 Prefix-free Regulr-Expression Mtching L IN L P RE L REG If E is infix-free, we hve n O(mn) running time lgorithm If E is (norml) regulr expression, we hve n O(mn 2 ) running time lgorithm If E is prefix-free, then there re t most n mtching sustrings of T tht elong to L(E), where n is the size of T. c c c c T = SNU 4/14

104 Prefix-free Regulr-Expression Mtching L IN L P RE L REG If E is infix-free, we hve n O(mn) running time lgorithm If E is (norml) regulr expression, we hve n O(mn 2 ) running time lgorithm If E is prefix-free, then there re t most n mtching sustrings of T tht elong to L(E), where n is the size of T. c c c c T = 13 ccc is prefix of ccc. This contrdicts tht L(E) is prefix-free SNU 4/14

105 Prefix-free Regulr-Expression Mtching L IN L P RE L REG If E is infix-free, we hve n O(mn) running time lgorithm If E is (norml) regulr expression, we hve n O(mn 2 ) running time lgorithm If E is prefix-free, then there re t most n mtching sustrings of T tht elong to L(E), where n is the size of T. Cn we hve n O(mn) time lgorithm? c c c c T = 13 ccc is prefix of ccc. This contrdicts tht L(E) is prefix-free SNU 4/14

106 Prefix-free Regulr-Expression Mtching L IN L P RE L REG If E is infix-free, we hve n O(mn) running time lgorithm If E is (norml) regulr expression, we hve n O(mn 2 ) running time lgorithm If E is prefix-free, then there re t most n mtching sustrings of T tht elong to L(E), where n is the size of T. Cn we hve n O(mn) time lgorithm? c c c c T = 13 ccc is prefix of ccc. This contrdicts tht L(E) is prefix-free SNU 4/14

107 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E T SNU 4/14

108 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E T = the recognition process 33-2 SNU 4/14

109 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E T = the recognition process 33-3 SNU 4/14

110 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process 33-4 SNU 4/14

111 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process 33-5 SNU 4/14

112 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process prllel processing strts 33-6 SNU 4/14

113 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process T = Running Time = No. of mtching end positions O(mn) = O(n) O(mn) = O(mn 2 ). prllel processing strts 33-7 SNU 4/14

114 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process T = Running Time = No. of mtching end positions O(mn) = O(n) O(mn) = O(mn 2 ). prllel processing strts 33-8 SNU 4/14

115 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process T = Running Time = No. of mtching end positions O(mn) = O(n) O(mn) = O(mn 2 ). prllel processing strts 33-9 SNU 4/14

116 Prefix-free Regulr-Expression Mtching Sketch of our lgorithm: E E R = T = the recognition process Becuse of prefix-freeness, no two process cn hve the sme stte of E t the sme time. This implies tht single reverse scn is enough to find corresponding strt positions for ech end position. set of sttes for 12 set of sttes for SNU 4/14

117 Prefix-free Regulr-Expression Mtching Given prefix-free regulr expression E nd text T, we cn identify ll mtching sustrings of T tht elong to L(E) in O(mn) worst-cse time - [HWW07]. Prefix-Free Regulr Lnguges nd Pttern Mtching, Yo-Su Hn, Yjun Wng nd Derick Wood Theoreticl Computer Science Vol. 389, , SNU 4/14

118 Stte Complexity Wht is the stte complexity of regulr lnguge L? 35-1 SNU 4/14

119 Stte Complexity Wht is the stte complexity of regulr lnguge L? Stte complexity is descriptionl complexity of L L hs unique miniml DFA A We define the stte complexity of L to e the numer of sttes in A 35-2 SNU 4/14

120 Stte Complexity Wht is the stte complexity of regulr lnguge L? Stte complexity is descriptionl complexity of L L hs unique miniml DFA A We define the stte complexity of L to e the numer of sttes in A We cn estimte needed resource SNU 4/14

121 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? 36-1 SNU 4/14

122 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? Upper ound m 1 m SNU 4/14

123 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? Upper ound t most m 1 m 2 f(m 1, m 2 ) 36-3 SNU 4/14

124 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? Upper ound t most m 1 m 2 f(m 1, m 2 ) Lower ound 36-4 SNU 4/14

125 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? Upper ound t most m 1 m 2 f(m 1, m 2 ) Lower ound Present two (generl) L 1 nd L 2 such tht the stte complexity of L 1 L 2 lwys reches the upper ound SNU 4/14

126 Stte Complexity Prolem Given two (ritrry) regulr lnguges L 1 nd L 2, wht is the stte complexity of L 1 L 2? Upper ound t most m 1 m 2 f(m 1, m 2 ) Lower ound Present two (generl) L 1 nd L 2 such tht the stte complexity of L 1 L 2 lwys reches the upper ound. Tight ound: UB = LB, the stte complexity of the intersection of two regulr lnguges is f(m 1, m 2 ) 36-6 SNU 4/14

127 Stte Complexity - Motivtion 1970s 2011 In recent yers, there hve een mny new pplictions of FAs, such s in nturl lnguge nd speech processing, softwre engineering, nd imge genertion nd encoding tht need lrge numer of sttes. the Bell Ls multilingul TTS system: 26.6MB for Germn, 30.0MB for French nd 39.0MB for Chinese SNU 4/14

128 Stte Complexity - motivtion New Helper: FA mnipultion softwre systems such s Gril+, Automte nd FireLite 38-1 SNU 4/14

129 Stte Complexity - motivtion New Helper: FA mnipultion softwre systems such s Gril+, Automte nd FireLite We clculte the upper ound SNU 4/14

130 Stte Complexity - motivtion New Helper: FA mnipultion softwre systems such s Gril+, Automte nd FireLite We clculte the upper ound. We guess lower ound nd verify it, nd repet this step until we find mtching lower ound SNU 4/14

131 Stte Complexity - motivtion New Helper: FA mnipultion softwre systems such s Gril+, Automte nd FireLite We clculte the upper ound. We guess lower ound nd verify it, nd repet this step until we find mtching lower ound. helper 38-4 SNU 4/14

132 Stte Complexity opertion finite lnguges regulr lnguges L 1 L 2 O(mn) mn L 1 L 2 O(mn) mn Σ \ L 1 m m L 1 L 2 (m n + 3)2 n 2 1 (2m 1)2 n 1 L 1 2 m m 4, for m 4 2 m m 2 L R p 1 1 if m = 2p 2 p 1 if m = 2p 1 2 m 39-1 SNU 4/14

133 Union of Finite Lnguges Given two miniml DFAs A nd B for non-empty finite lnguges L 1 nd L 2, we cn construct DFA for L(A) L(B) sed on the Crtesin product of sttes s follows: Let A = (Q 1, Σ, δ 1, s 1, F 1 ) nd B = (Q 2, Σ, δ 2, s 2, F 2 ). M = (Q 1 Q 2, Σ, δ, (s 1, s 2 ), F ), where for ll p Q 1 nd q Q 2 nd Σ, δ((p, q), ) = (δ(p, ), δ(q, )) nd F = (F 1 Q 2 ) (Q 1 F 2 ). M is deterministic SNU 4/14

134 Union of Finite Lnguges - Crtesin Product of Sttes 1,1 1,2 1,3 1,n-1 1,n 2,1 The m 1th stte in A is the finl stte whose outtrnsitions go to the sink stte, the mth stte. m-1,1 m,1 m-1,n-1 m,n-1 m-1,n m,n 41-1 SNU 4/14

135 Union of Finite Lnguges - Crtesin Product of Sttes 1,1 1,2 1,3 1,n-1 1,n 2,1 The m 1th stte in A is the finl stte whose outtrnsitions go to the sink stte, the mth stte. For stte (i, j) in M, L i,j (M) = L i (A) L j (B). m-1,1 m,1 m-1,n-1 m,n-1 m-1,n m,n 41-2 SNU 4/14

136 Union of Finite Lnguges - Crtesin Product of Sttes 1,1 1,2 1,3 1,n-1 1,n 2,1 The m 1th stte in A is the finl stte whose outtrnsitions go to the sink stte, the mth stte. For stte (i, j) in M, L i,j (M) = L i (A) L j (B). m-1,1 m,1 m-1,n-1 m,n-1 m-1,n m,n ll sttes re unrechle from stte (1,1) since A nd B re non-returning SNU 4/14

137 Union of Finite Lnguges - Crtesin Product of Sttes 1,1 1,2 1,3 1,n-1 1,n 2,1 The m 1th stte in A is the finl stte whose outtrnsitions go to the sink stte, the mth stte. For stte (i, j) in M, L i,j (M) = L i (A) L j (B). m-1,1 m,1 m-1,n-1 m,n-1 m-1,n m,n ll sttes equivlent since L m 1,n 1 = L m 1,n = L m,n 1 = {λ}. ll sttes re unrechle from stte (1,1) since A nd B re non-returning SNU 4/14

138 Union of Finite Lnguges - Crtesin Product of Sttes 1,1 1,2 1,3 1,n-1 1,n 2,1 The m 1th stte in A is the finl stte whose outtrnsitions go to the sink stte, the mth stte. For stte (i, j) in M, L i,j (M) = L i (A) L j (B). m-1,1 m,1 m-1,n-1 m,n-1 m-1,n m,n ll sttes equivlent since L m 1,n 1 = L m 1,n = L m,n 1 = {λ}. Lemm ll sttes 1. mn (m+n 2) 2 re unrechle from= stte mn(1,1) (m + n) sttes re sufficient forsince L(A) A nd L(B). B re non-returning SNU 4/14

139 Union of Finite Lnguges Lemm 1. mn (m + n) sttes re sufficient for L(A) L(B). The next question is whether or not the ound is rechle in generl SNU 4/14

140 Union of Finite Lnguges Lemm 1. mn (m + n) sttes re sufficient for L(A) L(B). The next question is whether or not the ound is rechle in generl. The nswer is YES nd NO SNU 4/14

141 Union of Finite Lnguges Lemm 2. The upper ound mn (m + n) cnnot e reched with fixed lphet when m nd n re ritrrily lrge. Proof. Let A hve {p 0, p 1,..., p m 1 } nd B hve {q 0, q 1,..., q n 1 }. We order the sttes such tht if p j is rechle from p i, then i < j. Let i {1,..., m 1}. Any string tht reches p i from p 0 cn go through only the sttes p 1,..., p i 1 in etween nd cnnot visit the sme stte twice. Hence, there re t most t + t t i = t(ti 1) t 1 = def D(i) strings tht cn rech p i from p SNU 4/14

142 Union of Finite Lnguges Lemm 2. The upper ound mn (m + n) cnnot e reched with fixed lphet when m nd n re ritrrily lrge. Proof. Let A hve {p 0, p 1,..., p m 1 } nd B hve {q 0, q 1,..., q n 1 }. We order the sttes such tht if p j is rechle from p i, then i < j. Let i {1,..., m 1}. Any string tht reches p i from p 0 cn go through only the sttes p 1,..., p i 1 in etween nd cnnot visit the sme stte twice. Hence, there re t most t + t t i = t(ti 1) t 1 = def D(i) strings tht cn rech p i from p 0. Since M is deterministic, for ny fixed i for 1 i < m 1, t most D(i) of the pir-sttes (p i, q j ) re rechle from (p 0, q 0 ) in M. Thus, if n 2 > D(i), then some pir-sttes with p i s the first component re not rechle. Therefore, the ound mn (m + n) is not rechle SNU 4/14

143 Union of Finite Lnguges Lemm 2. The upper ound mn (m + n) cnnot e reched with fixed lphet when m nd n re ritrrily lrge. Wht if the size of n lphet is NOT fixed? 44-1 SNU 4/14

144 Union of Finite Lnguges Lemm 2. The upper ound mn (m + n) cnnot e reched with fixed lphet when m nd n re ritrrily lrge. Wht if the size of n lphet is NOT fixed? Lemm 3. The upper ound mn (m + n) is rechle if the size of the lphet cn depend on m nd n SNU 4/14

145 Union of Finite Lnguges Lemm 3. The upper ound mn (m + n) is rechle if the size of the lphet cn depend on m nd n. We prove the lemm y presenting two finite lnguges whose union reches the ound. Let Σ = {, c} { i,j 1 i m 2, 1 j n 2 nd (i, j) (m 2, n 2)} Let A = (Q 1, Σ, δ 1, p 0, {p m 2 }), where Q 1 = {p 0, p 1,..., p m 1 } nd δ 1 is defined s follows: δ 1 (p i, ) = p i+1, for 0 i m 2. δ 1 (p 0, i,j ) = p i, for 1 i m 2 nd 1 j n 2, (i, j) (m 2, n 2). Let B = (Q 2, Σ, δ 2, q 0, {q n 2 }), where Q 2 = {q 0, q 1,..., q n 1 } nd δ 2 is defined s follows: δ 2 (q i, c) = q i+1, for 0 i n 2. δ 2 (q 0, i,j ) = q j, for 1 j n 2 nd 1 i m 2, (i, j) (m 2, n 2) SNU 4/14

146 Union of Finite Lnguges Lemm 3. The upper ound mn (m + n) is rechle if the size of the lphet cn depend on m nd n. A, 11, 12, , 22, 23 31, 32, 33 41, 42 B c, 11, 21, 31, 41 c c c , 22, 32, 42 13, 23, 33 An exmple of two miniml DFAs for finite lnguges whose sizes re 6 nd 5, respectively, where stte 5 ove nd stte 4 elow re sink sttes 46-1 SNU 4/14

147 Union of Finite Lnguges Lemm 3. The upper ound mn (m + n) is rechle if the size of the lphet cn depend on m nd n. Let L = L(A) L(B). We shows tht there exists set R consisting of mn (m+n) strings over Σ tht re pirwise inequivlent modulo the right invrint congruence of L. Let R = R 1 R 2 R 3, where R 1 = { i 0 i m 1}. R 2 = {c j 1 j n 3}. (Note tht R 2 does not include strings c 0, c n 2 nd c n 1.) R 3 = { i,j 1 i m 2 nd 1 j n 2 nd (i, j) (m 2, n 2)}. It is esy to verify tht ll strings in R re pirwise inequivlent. complete proof is given in the proceedings.) Then, R = mn (m + n). (The 47-1 SNU 4/14

148 Union of Finite Lnguges Theorem 1. Given two miniml DFAs A nd B for finite lnguges, mn (m + n) sttes re necessry nd sufficient in the worst-cse for the miniml DFA of L(A) L(B), where m = A nd n = B SNU 4/14

149 Union of Finite Lnguges Lemm 2 shows tht the upper ound is unrechle if Σ is fixed wheres Lemm 3 shows tht the upper ound is rechle if Σ depends on m nd n. Then, wht is the stte complexity of union with fixed sized lphet? 49-1 SNU 4/14

150 Union of Finite Lnguges Lemm 4. There exist DFAs A nd B, with m nd n sttes respectively, tht recognize finite lnguges over Σ such tht the miniml DFA for L(A) L(B) requires c(min{m, n}) 2 sttes. Proof. Let s 1 e ritrry nd r = log s. We define the finite lnguge L 1 = {w 1 w 2 w 1 = 2r, w 2 = odd(w 1 ) {, }, even(w 1 ) {c, d} }. L 1 cn e recognized y DFA A with t most 10s sttes SNU 4/14

151 Union of Finite Lnguges L 1 = {w 1 w 2 w 1 = 2r, w 2 = odd(w 1 ) {, }, even(w 1 ) {c, d} }. c, d c, d c, d c, d c, d c, d expnding tree for w 1 prt c, d c, d c, d c, d c, d c, d c, d c, d merging tree for w 2 prt A DFA A tht recognizes L 1 when r = 3. We omit the sink stte nd its in-trnsitions SNU 4/14

152 Union of Finite Lnguges Symmetriclly, we define L 2 = {w 1 w 2 w 1 = 2r, odd(w 1 ) {, }, w 2 = even(w 1 ) {c, d} }. The lnguge L 2 consists of strings uv, where u = 2r, odd chrcters of u re in {, }, even chrcters of u re in {c, d} nd even(u) coincides with v. By similr rgument, L 2 cn e recognized y DFA B with t most 10s sttes SNU 4/14

153 Union of Finite Lnguges Now let L = L 1 L 2. Let u 1 nd u 2 e distinct strings of length 2r such tht odd(u i ) {, } nd even(u i ) {c, d} for i = 1, 2. If odd(u 1 ) odd(u 2 ): u 1 odd(u 1 ) L 1 L ut u 2 odd(u 1 ) / L. Hence, u 1 nd u 2 re not equivlent modulo the right invrint congruence of L. If even(u 1 ) even(u 2 ): u 1 even(u 1 ) L 2 L ut u 2 even(u 1 ) L. The ove implies tht the right invrint congruence of L hs t lest 2 r 2 r s 2 different clsses. Therefore, if m = n = 10s is the size of the miniml DFAs for the finite lnguges L 1 nd L 2, then we know tht the miniml DFA for L = L 1 L 2 needs t lest n2 sttes SNU 4/14

154 RECAP Structurl properties of the k-lookhed determinism tht might led to n efficient XML Schem prser Fst regulr-expression pttern mtching lgorithms Stte Complexity 54-1 SNU 4/14

155 Future Directions nd Conclusions Hierrchy of k-lookhed determinism XML Schem prser regulr-expression pttern mtching system for source codes pttern mtching + indexing pure theory prcticl ppliction stte complexity 55-1 SNU 4/14

156 THANK YOU ANY QUESTIONS?? 56-1 SNU 4/14