正規表現に潜む対称性 〜等式公理による等価性判定〜

SLACS2016での招待講演に用いたスライドです．

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1. ਖ਼نදݱͷ౳Ձੑ൑ఆͷ ౳ࣜެཧΞϓϩʔν ৽԰ ྑຏ (౦ژେֶ) SLACS@໊ݹ԰େֶ (2016/8/25)
2. Prologue 2 (a + b)⇤ (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤ Q: Are these two expressions equivalent?
3. Prologue 2 (a + b)⇤ (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤ Q: Are these two expressions equivalent? A: Yes, you can check it by using automata.
4. Another A: Yes, you can check it as follows:
5. (S) (M) (T) (S) (M) (T) (M) (M) (T) (T) (M) (T) (a + b)⇤ ⌘ (a⇤b)⇤a⇤ ⌘ ((1 + aa⇤)b)⇤a⇤ ⌘ (b + aa⇤b)⇤a⇤ ⌘ b⇤(aa⇤bb⇤)⇤a⇤ ⌘ (bb⇤ + 1)(aa⇤bb⇤)⇤a⇤ ⌘ (bb⇤(aa⇤bb⇤)⇤ + (aa⇤bb⇤)⇤)a⇤ ⌘ (1 + bb⇤(aa⇤bb⇤)⇤ + aa⇤bb⇤(aa⇤bb⇤)⇤)a⇤ ⌘ (1 + (1 + aa⇤)bb⇤(aa⇤bb⇤)⇤)a⇤ ⌘ (1 + a⇤bb⇤(aa⇤bb⇤)⇤)a⇤ ⌘ a⇤ + a⇤bb⇤(aa⇤bb⇤)⇤a⇤ ⌘ a⇤ + a⇤bb⇤(aa⇤bb⇤)⇤(1 + aa⇤) ⌘ a⇤ + a⇤bb⇤(aa⇤bb⇤)⇤aa⇤ + a⇤bb⇤(aa⇤bb⇤)⇤ ⌘ a⇤(1 + bb⇤(aa⇤bb⇤)⇤aa⇤) + a⇤bb⇤(aa⇤bb⇤)⇤ ⌘ a⇤(bb⇤aa⇤)⇤ + a⇤bb⇤(aa⇤bb⇤)⇤ ⌘ (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤. (T) (M) (S) Another A: Yes, you can check it as follows:
6. Prologue 4 In the axiomatic approach, one can verify the equivalence of two regular expressions via a finite rewriting sequence with certain axioms. The previous example uses three axims, so called (T) (M) (S). , and
7. Prologue 5 These are the reasons why I chose the axiomatic approach to the equivalence of regular expressions as the topic of the talk:
8. Prologue 5 These are the reasons why I chose the axiomatic approach to the equivalence of regular expressions as the topic of the talk: It’s like a “mathematical puzzle”: fun, easy to understand.
9. Prologue 5 These are the reasons why I chose the axiomatic approach to the equivalence of regular expressions as the topic of the talk: It’s like a “mathematical puzzle”: fun, easy to understand. But it is still a challenging problem. There is a big open problem.
10. Prologue 5 These are the reasons why I chose the axiomatic approach to the equivalence of regular expressions as the topic of the talk: It’s like a “mathematical puzzle”: fun, easy to understand. But it is still a challenging problem. There is a big open problem. The symmetries in regular expressions can be found in this topic.
11. Resume 6 ´ ´ I. Automata and Regular Expressions II. History of Axiomatisation III.Group Identities IV. Cascade Decomposition V. Conway’s Last Conjecture
13. 7 શવ஌Βͳ͍ਓ޲͚ͷ࿩
14. ΦʔτϚτϯͱ ਖ਼نදݱ Automata and Regular Expressions I
15. Regular expressions 9 For an alphabet A, the set of regular expressions over A, denoted by , is defined as follows: The special symbol is a regular expression. The special symbol is a regular expression. Every letter is a regular expression. And the followings are regular for any RegA 0 1 E, F 2 RegA : (concatenation) E · F (or more simply EF ) , (sum) E + F, (star) E⇤. a 2 A
16. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: E L(E)
17. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: E L(E) L(0) = ; L(1) = {"} L(a) = {a}
18. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: E L(E) is the empty word (the word of zero length). " L(0) = ; L(1) = {"} L(a) = {a}
19. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: L(E + F) = L(E) [ L(F) E L(E) is the empty word (the word of zero length). " L(0) = ; L(1) = {"} L(a) = {a}
20. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: L(E + F) = L(E) [ L(F) E L(E) is the empty word (the word of zero length). " L(EF) = L(E)L(F) = {uv | u 2 L(E), v 2 L(F)} L(0) = ; L(1) = {"} L(a) = {a}
21. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: L(E + F) = L(E) [ L(F) L(E⇤) = L(E)⇤ = [ n 0 L(E)n E L(E) is the empty word (the word of zero length). " L(EF) = L(E)L(F) = {uv | u 2 L(E), v 2 L(F)} L(0) = ; L(1) = {"} L(a) = {a}
22. Interpretation 10 The interpretation of an expression over A, denoted by , is the set of words (or language) recursively defined as follows: L(E + F) = L(E) [ L(F) L(E⇤) = L(E)⇤ = [ n 0 L(E)n E L(E) Two expressions are equivalent if these interpretation coincide. is the empty word (the word of zero length). " L(EF) = L(E)L(F) = {uv | u 2 L(E), v 2 L(F)} L(0) = ; L(1) = {"} L(a) = {a}
23. Equivalence of expressions 11 Two expressions and are said to be equivalent, and denoted by , if these interpretations coincide E F E ⇡ F E ⇡ F , L(E) = L(F).
24. Automata 12 An automaton over an alphabet is a finite directed edge -labelled graph, equipped with two kinds of special states “initial” and “final”. A A q0 q1 a a b b An automaton over A = {a, b}
25. Automata 12 An automaton over an alphabet is a finite directed edge -labelled graph, equipped with two kinds of special states “initial” and “final”. A A q0 q1 a a b b An automaton over A = {a, b} Initial states are indicated by source-less edges.
26. Automata 12 An automaton over an alphabet is a finite directed edge -labelled graph, equipped with two kinds of special states “initial” and “final”. A A q0 q1 a a b b An automaton over A = {a, b} Initial states are indicated by source-less edges. Final states are indicated by destination-less edges.
27. Automata 12 An automaton over an alphabet is a finite directed edge -labelled graph, equipped with two kinds of special states “initial” and “final”. A A q0 q1 a a b b An automaton over A = {a, b} ✓ 1 0 , ✓ b a a b ◆ , ✓ 1 0 ◆◆ and its matrix representation. Initial states are indicated by source-less edges. Final states are indicated by destination-less edges.
28. Behaviour 13 For an automaton , the set of all words that label a path from an initial state to a final state is called the behaviour of , denoted by A A L(A).
29. q0 q1 a a b b ✓ 1 0 , ✓ b a a b ◆ , ✓ 1 0 ◆◆
30. q0 q1 a a b b ⇡ (b + ab⇤a)⇤ ✓ 1 0 , ✓ b a a b ◆ , ✓ 1 0 ◆◆
31. q0 q1 a a b b ⇡ (b + ab⇤a)⇤ ✓ 1 0 , ✓ b a a b ◆ , ✓ 1 0 ◆◆ r0 r1 r2 r3 b a a b a b a, b 0 B B @ 1 0 0 0 , 0 B B @ a b 0 0 0 b a 0 a 0 0 b 0 0 0 a + b 1 C C A , 0 B B @ 0 0 0 1 1 C C A 1 C C A
32. q0 q1 a a b b ⇡ (b + ab⇤a)⇤ ✓ 1 0 , ✓ b a a b ◆ , ✓ 1 0 ◆◆ ⇡ (a + b)⇤bab(a + b)⇤ r0 r1 r2 r3 b a a b a b a, b 0 B B @ 1 0 0 0 , 0 B B @ a b 0 0 0 b a 0 a 0 0 b 0 0 0 a + b 1 C C A , 0 B B @ 0 0 0 1 1 C C A 1 C C A
33. Equivalence of automata 15 Two automata and are said to be equivalent, and denoted by , if these behaviours coincide A ⇡ B , L(A) = L(B). A ⇡ B A B
34. Fundamental theorem 16 In 1951, Stephen Kleene proved that equi- expressibility of automata and regular expressions (the birth of automata theory). Theorem For any language , the followings are equivalent. (1) is an interpretation of some regular expression. (2) is an behaviour of some automaton. L L L [Kleene51]
35. 17 [Kleene51]
36. 18 [Kleene51] lifetime by a relatively small mechanism. The questions or reducibility of other mechanisms to McCulloch-Pitts nerve nets (not always without increasing the size of the mechanisms) is significant on this basis, but trivial on the basis or explaining behavior over a fixed tiatt. time only. 7. Regular Events: 7.1 "Regi:lar events" defined: We shall presently des- cribe a class of events which we will call "regular events." •' (We would welcome any suggest1lll'US as to a more descriptive term. We assume for the purpose that the events refer to the inputs up through time£ on a set of! input neurons Ni, ••• ,Nk - the same for all events considered; but the definition applies equally well for any k > 1 or even ror k • o. - - - The events can refer to the value or £· Our objective is to show that all and only regular events oan be represented by nerve nets or finite automata. We have already seen in
37. From expressions to automata 19 Base cases: ((1), (0), (0)) 0 ⇡ ((1), (0), (1)) 1 ⇡ ✓ 1 0 , ✓ 0 a 0 0 ◆ , ✓ 0 1 ◆◆ a ⇡ a 2 A
38. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U)
39. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U) ✓ I J , ✓ M 0 0 N ◆ , ✓ T U ◆ , ◆ E + F ⇡ AE AF ! ! ! !
40. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U) ✓ I J , ✓ M 0 0 N ◆ , ✓ T U ◆ , ◆ E + F ⇡ AE AF ! ! ! ! EF ⇡ ✓ I 0 , ✓ M TJ 0 N ◆ , ✓ 0 U ◆ , ◆ AE ! AF! !
41. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U) ✓ I J , ✓ M 0 0 N ◆ , ✓ T U ◆ , ◆ E + F ⇡ AE AF ! ! ! ! EF ⇡ ✓ I 0 , ✓ M TJ 0 N ◆ , ✓ 0 U ◆ , ◆ AE ! AF! ! E⇤ ⇡ (I, M + TI, T) AE ! !
42. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U) ✓ I J , ✓ M 0 0 N ◆ , ✓ T U ◆ , ◆ E + F ⇡ AE AF ! ! ! ! EF ⇡ ✓ I 0 , ✓ M TJ 0 N ◆ , ✓ 0 U ◆ , ◆ AE ! AF! ! E⇤ ⇡ (I, M + TI, T) AE ! ! +
43. From expressions to automata 20 Let and be automata constructed from and . F E AE = (I, M, T) AF = (J, N, U) ✓ I J , ✓ M 0 0 N ◆ , ✓ T U ◆ , ◆ E + F ⇡ AE AF ! ! ! ! EF ⇡ ✓ I 0 , ✓ M TJ 0 N ◆ , ✓ 0 U ◆ , ◆ AE ! AF! ! E⇤ ⇡ (I, M + TI, T) AE ! ! We call this construction as the “basic construction”. +
44. From automata to expressions 21 0 1 E F H G A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆
45. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆
46. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ ✓ (E + FH⇤G)⇤ E⇤F(H + GE⇤F)⇤ H⇤G(E + FH⇤G)⇤ (H + GE⇤F)⇤ ◆ = A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆
47. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ ✓ (E + FH⇤G)⇤ E⇤F(H + GE⇤F)⇤ H⇤G(E + FH⇤G)⇤ (H + GE⇤F)⇤ ◆ = A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆ C(A2) = I · M⇤ · T = (E + FH⇤G)⇤
48. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ ✓ (E + FH⇤G)⇤ E⇤F(H + GE⇤F)⇤ H⇤G(E + FH⇤G)⇤ (H + GE⇤F)⇤ ◆ = A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆ C(A2) = I · M⇤ · T = (E + FH⇤G)⇤ C(A2) A2 and are equivalent.
49. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ ✓ (E + FH⇤G)⇤ E⇤F(H + GE⇤F)⇤ H⇤G(E + FH⇤G)⇤ (H + GE⇤F)⇤ ◆ = A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆ C(A2) = I · M⇤ · T = (E + FH⇤G)⇤ C(A2) A2 and are equivalent. We can compute an equivalent expression from any two-states automaton in the same way.
50. From automata to expressions 21 0 1 E F H G ✓ E F G H ◆⇤ ✓ (E + FH⇤G)⇤ E⇤F(H + GE⇤F)⇤ H⇤G(E + FH⇤G)⇤ (H + GE⇤F)⇤ ◆ = A2 = ✓ I = 1 0 , M = ✓ E F G H ◆ , T = ✓ 1 0 ◆◆ C(A2) = I · M⇤ · T = (E + FH⇤G)⇤ C(A2) A2 and are equivalent. We can compute an equivalent expression from any two-states automaton in the same way. Q: How it works for three-more-states automata?
51. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A
52. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A
53. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A = ✓ A B C D ◆
54. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A = ✓ A B C D ◆ M⇤ = ✓ A B C D ◆⇤ = ✓ (A + BD⇤C)⇤ A⇤B(D + CA⇤B)⇤ D⇤C(A + BD⇤C)⇤ (D + CA⇤B)⇤ ◆
55. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A = ✓ A B C D ◆ M⇤ = ✓ A B C D ◆⇤ = ✓ (A + BD⇤C)⇤ A⇤B(D + CA⇤B)⇤ D⇤C(A + BD⇤C)⇤ (D + CA⇤B)⇤ ◆
56. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A = ✓ A B C D ◆ M⇤ = ✓ A B C D ◆⇤ = ✓ (A + BD⇤C)⇤ A⇤B(D + CA⇤B)⇤ D⇤C(A + BD⇤C)⇤ (D + CA⇤B)⇤ ◆ These red-underlined parts are 2x2 matrices, hence we can construct its star, canonically. So we can define for three-states automata, and similarly (recursive division), for any automata. C(A)
57. From automata to expressions 22 M = 0 @ E F G H I J K L M 1 A = ✓ A B C D ◆ M⇤ = ✓ A B C D ◆⇤ = ✓ (A + BD⇤C)⇤ A⇤B(D + CA⇤B)⇤ D⇤C(A + BD⇤C)⇤ (D + CA⇤B)⇤ ◆ These red-underlined parts are 2x2 matrices, hence we can construct its star, canonically. So we can define for three-states automata, and similarly (recursive division), for any automata. C(A) This construction was introduced in [Conway71], thus we refer to this as “Conway’s method”.
58. Fundamental theorem 23 In 1951, Kleene proved that equi-expressibility of automata and regular expressions (the birth of automata theory). Theorem For any language , the followings are equivalent. (1) is an interpretation of some regular expression. (2) is an behaviour of some automaton. L L L [Kleene51]
59. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A
60. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A E F
61. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A E F basic construction AF AE
62. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A E F basic construction AF AE determinisation & minimisation A0 E A0 F
63. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A E F basic construction AF AE determinisation & minimisation A0 E A0 F compare
64. Equivalence is decidable 24 Fact For any automaton , there exists an equivalent unique canonical automaton . Moreover, is effectively constructable from . A0 A A0 A E F basic construction AF AE determinisation & minimisation A0 E A0 F compare E ⌘ F A0 E = A0 F ,
65. Summing up 25 Automata and regular expressions have a deep connection: these are convertible to each other. The proof of the decidability of the equivalence problem is easy. But from the complexity viewpoint, the equivalence problem is rather hard (PSPACE-complete).
66. Summing up 25 Automata and regular expressions have a deep connection: these are convertible to each other. The proof of the decidability of the equivalence problem is easy. But from the complexity viewpoint, the equivalence problem is rather hard (PSPACE-complete). In this year, Nakamura proved that the almost equivalence problem is also hard (PSPACE-complete).
67. Almost equivalence problem (༨ஊ) 26 K L lim n ! 1 |(L4K) \ An| |An| = 0. Two languages and over are said to be almost equivalent, if their symmetric difference is “very small” (zero measure): A Two expressions are said to be almost equivalent if these interpretations are almost equivalence. In this year, Nakamura proved that the almost equivalence problem is also hard (PSPACE-complete).
68. 27 The Almost Equivalence by Asymptotic Probabilities for Regular Languages and Its Computational Complexities Yoshiki Nakamura Tokyo Institute of Technology Tokyo, Japan nakamura.y.ay@m.titech.ac.jp We introduce p-equivalence by asymptotic probabilities, which is a weak almost-equivalence based on zero-one laws in finite model theory. In this paper, we consider the computational complexities of p-equivalence problems for regular languages and provide the following details; First, we give some fundamental results and a logical characterization for p-equivalence, which can generate some algorithms by descriptive complexity. Second, we give the computational complexities of the p- equivalence problems by the logical characterization. Finally, we apply the proofs in this paper for p-equivalence to some generalized equivalences. 1 Introduction The study of the equivalence problem of regular languages dates back to the beginning of formal language theory. This problem is a fundamental problem and regular languages have many applications (see e.g., [2]). Regular expressions (REG), nondeterministic finite state automaton (NFA), and deterministic finite state automaton (DFA) are normally used to represent regular languages. Both the equivalence problem for NFAs and REGs are known as PSPACE-complete [17] and the equivalence problem for DFAs is known as NL-complete [14]. The paper will be appeared in GandALF 2016.
69. ެཧԽͱͦͷྺ࢙ History of Axiomatisation II
70. 29 The origin of the axiomatic approach goes back to [Kleene51]. Kleene observed several laws (axioms) in regular expressions.
71. 30 After the work of Kleene, several authors tackled this problem. A first complete axiomatisation is due to Arto Salomaa in 1966 [Salomaa66]. An complete equational axiomatisation for one-letter regular expressions was given by Redko [Redko64]. An elegant complete equational axiomatisation was conjecture by John Conway in 1971 [Conway71]. The Conway’s conjecture was finally proved by Daniel Krob [Krob91].
72. The “Bible” This is the first book written by British mathematician John Conway. [Conway71]
73. Trivial identities f 32 (T)
74. Trivial identities f 32 (T) (E + F) + G ⌘ E + (F + G) (E · F) · G ⌘ E · (F · G) (associativity)
75. Trivial identities f 32 (T) (E + F) + G ⌘ E + (F + G) (E · F) · G ⌘ E · (F · G) (associativity) E + F ⌘ F + E (commutativity)
76. Trivial identities f 32 (T) (E + F) + G ⌘ E + (F + G) (E · F) · G ⌘ E · (F · G) (associativity) E + F ⌘ F + E (commutativity) (distributivity) E · (F + G) ⌘ E · F + E · G (E + (F) · G) ⌘ E · G + F · G
77. Trivial identities f 32 (T) (E + F) + G ⌘ E + (F + G) (E · F) · G ⌘ E · (F · G) (associativity) E + F ⌘ F + E (commutativity) (unit elements) E + 0 ⌘ 0 + E ⌘ E E · 1 ⌘ 1 · E ⌘ E E · 0 ⌘ 0 · E ⌘ 0 (distributivity) E · (F + G) ⌘ E · F + E · G (E + (F) · G) ⌘ E · G + F · G
78. Trivial identities f 32 (T) (E + F) + G ⌘ E + (F + G) (E · F) · G ⌘ E · (F · G) (associativity) E + F ⌘ F + E (commutativity) (unit elements) E + 0 ⌘ 0 + E ⌘ E E · 1 ⌘ 1 · E ⌘ E E · 0 ⌘ 0 · E ⌘ 0 (distributivity) E · (F + G) ⌘ E · F + E · G (E + (F) · G) ⌘ E · G + F · G Trivial identities states exactly that the regular expressions over A admits a ring structure (T) (RegA, 0, 1, +, ·).
79. Substitution principle 33 E ⌘ F E + G ⌘ F + G GEH ⌘ GFH E ⌘ F E⇤ ⌘ F⇤ E ⌘ F
80. Idempotent identities 34 E + E ⌘ E 1⇤ ⌘ 1 E⇤E⇤ ⌘ E⇤ E⇤⇤ ⌘ E⇤ E + E⇤ ⌘ E⇤ E⇤ ⌘ (1 + E)⇤ 1 + 1 ⌘ 1
81. Star-free expressions are trivial 35 If we consider regular expressions with out star (thus denoting finite languages), the axiom consists of the trivial identities and an idempotent identity is enough to be complete. E + E ⌘ E (T)
82. Axiomatising the equivalence of regular expressions. Axiomatising the meaning of “star”. ;
83. Meaning of * 37
84. Meaning of * 37 E⇤ ⌘ 1 + E + EE + EEE + EEEE + EEEEE + · · ·
85. Meaning of * 37 E⇤ ⌘ 1 + EE⇤ E⇤ ⌘ 1 + E + EE + EEE + EEEE + EEEEE + · · ·
86. Meaning of * 37 Lemma Let and be two languages, where does not contain the empty word . Then is the unique solution of [Arden61] L K X = KX [ L. K⇤L " K E⇤ ⌘ 1 + EE⇤ E⇤ ⌘ 1 + E + EE + EEE + EEEE + EEEEE + · · ·
87. 38 Theorem [Salomaa66] Meaning of * Two identities and , and the following inference rule (solution of equatinos) make a complete axiomatisation. E⇤ ⌘ 1 + EE⇤ E⇤ ⌘ (1 + E)⇤ EF + G ⌘ F " / 2 L(E) F ⌘ E⇤G E⇤ ⌘ 1 + EE⇤ E⇤ ⌘ 1 + E + EE + EEE + EEEE + EEEEE + · · ·
88. 39 Discussion Salomaa’s axiomatisation is very simple, but it merely adopts the semantical meaning of star (Arden’s lemma). After the work of Salomaa, Kozen introduced a slightly modified axiomatisation [Kozen94]. But the syntactical meaning of star is still unclear. A purely equational axiomatisation is needed…
89. 40 Equational axiomatisation An axiomatisation is equiational if it consists of the set of identities (axioms) and only the substitution principle. An equational axiomatisation is finite if its contains only finite identities.
90. 40 Equational axiomatisation An axiomatisation is equiational if it consists of the set of identities (axioms) and only the substitution principle. An equational axiomatisation is finite if its contains only finite identities. Theorem [Redko64] RegA A complete equational axiomatisation is necessary to be infinite, even for over a one letter alphabet A = {a}. [Conway71]
91. The most important problem in this area is to construct a "good" system of rational identities that would permit us to obtain by a logical deductive process (i.e. by a rewriting process) every possible rational identity; such a system will be called complete. “ ” [Krob91]
92. Two standard identities of * 42 (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤
93. Two standard identities of * 42 (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (M)
94. Two standard identities of * 42 (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M)
95. 43 (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
96. 43 E F (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
97. 43 E F E F E F ⇡ (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
98. 43 E F E F E F ⇡ (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M) E F E F ⇡
99. 43 E F E F E F ⇡ (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M) E F E F ⇡
100. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. Answer: (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
101. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. E⇤ Answer: (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
102. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. E⇤ (E · 1)⇤ (T) Answer: (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
103. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. E⇤ (E · 1)⇤ (T) 1 + E(1 · E)⇤1 (M) Answer: (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
104. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. E⇤ (E · 1)⇤ (T) 1 + E(1 · E)⇤1 (M) Answer: (T) 1 + EE⇤. (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
105. Exercise 1 44 Prove that is a generalisation of i.e., E⇤ ⌘ 1 + EE⇤, (M) (M) ` E⇤ ⌘ 1 + EE⇤. E⇤ (E · 1)⇤ (T) 1 + E(1 · E)⇤1 (M) Answer: Corollary: (M) ` 0⇤ ⌘ 1. (T) 1 + EE⇤. (mult-star) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
106. 45 (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
107. 45 (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ E F (S)
108. 45 (sum-star) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ E F (S) ⇡ E 1 F 1 E
109. Elegant property of idempotency 46 E + E ⌘ E 1⇤ ⌘ 1 E⇤E⇤ ⌘ E⇤ E⇤⇤ ⌘ E⇤ E + E⇤ ⌘ E⇤ E⇤ ⌘ (1 + E)⇤ 1 + 1 ⌘ 1
110. Elegant property of idempotency 46 E + E ⌘ E 1⇤ ⌘ 1 E⇤E⇤ ⌘ E⇤ E⇤⇤ ⌘ E⇤ E + E⇤ ⌘ E⇤ All these identities are the consequence of (M), (S) and (I) (idempotency) E⇤ ⌘ (1 + E)⇤ 1 + 1 ⌘ 1
111. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
112. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: 1 + 1 (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
113. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: 1 + 1 1 + 1⇤ (I) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
114. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: 1 + 1 1 + 1⇤ (I) (T) 1 + 1 · 1⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
115. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: 1 + 1 1 + 1⇤ (I) (T) 1 + 1 · 1⇤ 1⇤ 1. (M) (I) (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
116. Exercise 2 47 Prove that the identity (I) ^ (M) ` 1 + 1 ⌘ 1. Answer: 1 + 1 1 + 1⇤ (I) (T) 1 + 1 · 1⇤ 1⇤ 1. (M) (I) Corollary: (I) ^ (M) ` E + E ⌘ E. (EF)⇤ ⌘ 1 + E(FE)⇤F (M)
117. Exercise 3 48 Prove that the identity (I) ^ (S) ` E⇤ ⌘ (1 + E)⇤. Answer: (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
118. Exercise 3 48 Prove that the identity (I) ^ (S) ` E⇤ ⌘ (1 + E)⇤. Answer: (1 + E)⇤ (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
119. Exercise 3 48 Prove that the identity (I) ^ (S) ` E⇤ ⌘ (1 + E)⇤. Answer: (1 + E)⇤ 1⇤(E1⇤)⇤ (S) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
120. Exercise 3 48 Prove that the identity (I) ^ (S) ` E⇤ ⌘ (1 + E)⇤. Answer: (1 + E)⇤ (I) 1(E1)⇤ 1⇤(E1⇤)⇤ (S) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
121. Exercise 3 48 Prove that the identity (I) ^ (S) ` E⇤ ⌘ (1 + E)⇤. Answer: (1 + E)⇤ (I) 1(E1)⇤ (T) E⇤. 1⇤(E1⇤)⇤ (S) (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S)
122. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4
123. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4
124. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4
125. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
126. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (S) 1 + (E + 1)⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
127. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (S) 1 + (E + 1)⇤ (I) ^ (S) 1 + E⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
128. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (S) 1 + (E + 1)⇤ (I) ^ (S) 1 + E⇤ (M) 1 + 1 + EE⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
129. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (S) 1 + (E + 1)⇤ (I) ^ (S) 1 + E⇤ (M) 1 + 1 + EE⇤ (I) 1 + EE⇤ (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
130. 49 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤⇤ ⌘ E⇤. E⇤⇤ (M) 1 + E⇤E⇤⇤ (S) 1 + (E + 1)⇤ (I) ^ (S) 1 + E⇤ (M) 1 + 1 + EE⇤ (I) 1 + EE⇤ E⇤. (M) (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M) Exercise 4 (T) 1 + E⇤(1E⇤)⇤
131. Exercise 5 50 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤E⇤ ⌘ E⇤. (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M)
132. Exercise 5 50 Prove that the identity Answer: (I) ^ (M) ^ (S) ` E⇤E⇤ ⌘ E⇤. (S) E⇤ (I) ^ (S) (E + 1)⇤ E⇤(1E⇤)⇤ (T) E⇤E⇤⇤ (I) ^ (M) ^ (S) E⇤E⇤. (EF)⇤ ⌘ 1 + E(FE)⇤F (E + F)⇤ ⌘ E⇤(FE⇤)⇤ (S) (M)
133. Implicational hierarchy 51 E⇤⇤ ⌘ E⇤ , E + E ⌘ E 1⇤ ⌘ 1 E⇤E⇤ ⌘ E⇤ 1 + 1 ⌘ 1 ) ) 1⇤1⇤ ⌘ 1⇤ , ) (cf. Fig.12.2 of [Conway71])
134. Three fundamental identities 52 (E + F)⇤ ⌘ (E⇤F)⇤E⇤ (S) (M) (EF)⇤ ⌘ 1 + E(FE)⇤F (I) 1⇤ ⌘ 1 (sum-star) (mult-star) (idempotency)
135. Three fundamental identities 52 (E + F)⇤ ⌘ (E⇤F)⇤E⇤ (S) (M) (EF)⇤ ⌘ 1 + E(FE)⇤F (I) 1⇤ ⌘ 1 (sum-star) (mult-star) (idempotency) Q: Is complete? (I) ^ (M) ^ (S)
136. Three fundamental identities 52 (E + F)⇤ ⌘ (E⇤F)⇤E⇤ (S) (M) (EF)⇤ ⌘ 1 + E(FE)⇤F (I) 1⇤ ⌘ 1 (sum-star) (mult-star) (idempotency) A: NO! Q: Is complete? (I) ^ (M) ^ (S)
137. Cyclic identities 53 (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤
138. Cyclic identities 53 E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) Example: (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤
139. Cyclic identities 53 E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) Example: E (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤
140. Cyclic identities 53 E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) Example: E (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ ⇡ E E E
141. Cyclic identities 53 E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) Example: E E E ⇡ (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ ⇡ E E E
142. Cyclic identities 53 E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) Example: E E E ⇡ is independent from for every (P(n)) (I) ^ (M) ^ (S) n 2. (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ ⇡ E E E
143. (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ 54 Q: Is complete? (I) ^ (M) ^ (S) ^ (P(n))n 2 Cyclic identities
144. (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ 54 A: No, but not too bad. Q: Is complete? (I) ^ (M) ^ (S) ^ (P(n))n 2 Cyclic identities
145. (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ 54 A: No, but not too bad. Q: Is complete? (I) ^ (M) ^ (S) ^ (P(n))n 2 Cyclic identities Theorem (I) ^ (M) ^ (S) ^ (P(n))n 2 is complete for over a one letter alphabet A = {a}. RegA [Redko64][Conway71]
146. ܈౳ࣜ Group Identities III
147. Observation 56 ⇡ E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) a a a
148. Observation 56 ⇡ E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) The transition monoid of the righthand-side automaton is Z/2Z. a a a
149. Observation 56 ⇡ E⇤ ⌘ (1 + E)(EE)⇤ (P(2)) (P(n)) E⇤ ⌘ (1 + E + · · · + En 1)(En)⇤ Z/nZ The transition monoid of the righthand-side automaton is Z/2Z. a a a
150. Observation 57 For a one letter alphabet the concatenation operation is commutative (up to equivalence) A = {a}, EF ⇡ FE for any E, F 2 Reg A. The fundamental theorem of finite abelian group: every finite abelian group is isomorphic to a finite product of cyclic groups Z/nZ.
151. Groups as automata 58 From a finite group G, we can construct a (deterministic) automaton where the transition the natural group action on itself AG = (G, G, ·, 1, G) · : G ⇥ G ! G g · h = gh. L(AG) = G⇤. AG The initial state is the unit, and all states are final hence is called the group automaton of G.
152. 59 AZ/3Z AZ/2Z a0 a1 a1 a0 a1 a0 a0 a1 a2 a1 a2 a0 a1 a2 a0 a1 a2 a0
153. Group identities 60 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤
154. Group identities 60 The identity of a group , denote by , is the identity between and AZ/2Z a0 a1 a1 a0 a1 a0 G C(AG) (P(G)) G⇤
155. Group identities 60 The identity of a group , denote by , is the identity between and AZ/2Z a0 a1 a1 a0 a1 a0 G C(AG) (P(G)) G⇤ (a0 + a1)⇤ ⌘ C(AZ/2Z) (P(AZ/2Z))
156. Group identities 60 The identity of a group , denote by , is the identity between and AZ/2Z a0 a1 a1 a0 a1 a0 G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) (P(AZ/2Z))
157. Group identities 61 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) (P(AZ/2Z))
158. Group identities 61 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) When we put and , we obtain a0 = 0 a1 = a (P(AZ/2Z))
159. Group identities 61 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) When we put and , we obtain a0 = 0 a1 = a a⇤ ⌘ (aa)⇤ + a(aa) (P(AZ/2Z))
160. Group identities 61 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) When we put and , we obtain a0 = 0 a1 = a a⇤ ⌘ (aa)⇤ + a(aa) ⌘ (1 + a)(aa)⇤ (P(AZ/2Z))
161. Group identities 61 The identity of a group , denote by , is the identity between and G C(AG) (P(G)) G⇤ = (a0 + a1a⇤ 0 a1)⇤ + a⇤ 0 a1(a0 + a1a⇤ 0 a1)⇤. (a0 + a1)⇤ ⌘ C(AZ/2Z) When we put and , we obtain a0 = 0 a1 = a a⇤ ⌘ (aa)⇤ + a(aa) ⌘ (1 + a)(aa)⇤ and thus implies (P(AZ/2Z)) (P(2)). (P(AZ/2Z))
162. 62 73 Tr(1, 1)= (T1,1 + T1,3T;,3T3,1 + T r (1, 2 ) = T r (1, 1)(TI,~ + T2,3 T~,3 Ta a ) ( T2,2 + T2,3 T~,3 T3,2 ) * Tr(1, 3) = Tr(1, 1)T1,3T~, 3 + Tr(1,2)T2,3T;, 3 Note that the interpretations of these expressions are quite simple. Indeed, Tr(1, i) is exactly the set of the words which corresponds to the permutations mapping 1 on i. COROLLARY V.6 : Let A be an alphabet. Then, the following system : (M), (S), (P(~,,)),,>__2 is a complete system of B-rational identities for A. Example : In order to understand the complexity which is hidden under the identities P(6,~), let us precise the group identity associated with the symmetric group of order 3 : e#3 = {i-- Id, Pl = (123), P2 = (132), al = (23), a2 = (13), a3-- (12)} Then let us consider the following rational expressions constructed over the alphabet which is naturally associated with 63 : T u=a~+a~l T m=ap,+a~ 3 T1,3=ap2+a~ T~,, = ap~ + a~ T2,2 = ai + ao2 T2,3 = ap, + a~, T3,1 = ap~ + ao~ T3,2 = a~ + a~, T3,3 = ai + aa~ Observe that the interpretation of the expression T/j is simply the set of the permutations of 63 that send i on j. With these denotations, we can now give an identity which is equivalent modulo (M), (S) and P(2) to P(e~a) : (hi + ca, + ap, + a~,~ + a~ + co2)* .~ Tr(1,1) + Tr(1,2) + Tr(1,3) where the three expressions Tr(1,1), Tr(1, 2) and Tr(1, 3) stand for : Example from [Krob90]
163. Conway’s conjecture 63 First Conjecture where is the set of all finite groups, is complete. G [Conway71] (I) ^ (M) ^ (S) ^ (P(G))G2G
164. Conway’s conjecture 63 First Conjecture where is the set of all finite groups, is complete. G [Conway71] (I) ^ (M) ^ (S) ^ (P(G))G2G where is the symmetric group of order , is complete. Sn n (I) ^ (M) ^ (S) ^ (P(Sn))n 2 Second Conjecture
165. Krob’s answer 64 In 1991, french mathematician Daniel Krob proved Conway’s first & second conjecture, affirmatively.
166. Krob’s answer 64 In 1991, french mathematician Daniel Krob proved Conway’s first & second conjecture, affirmatively. Krob’s paper [Krob91] consists of about 140 pages!
167. ΦʔτϚτϯͷ ෼ղఆཧ Krohn-Rhodes Decomposition IV
168. Cascade product 66 A B C A B C Cascade product is a hierarchical composition of autoamta.
169. 67 Cascade product
170. 68 Theorem [KR65] For any automaton , there exists a cascade decomposition where is either reset automaton or group automaton for each A 1  i  n. A  B1 B2 · · · Bn Bi Krohn-Rhodes theorem
171. The reset automaton f 69 0 1 b a a b ✓ a b a b ◆ R R
172. The reset automaton f 69 0 1 b a a b ✓ a b a b ◆ R R ✓ a b a b ◆⇤ = ✓ (a + bb⇤a)⇤ a⇤b(b + aa⇤b)⇤ b⇤a(a + bb⇤a)⇤ (b + aa⇤b)⇤ ◆
173. The reset automaton f 69 0 1 b a a b ✓ a b a b ◆ R R C(R) = 1 0 ✓ a b a b ◆⇤ ✓ 1 1 ◆ = (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤ ✓ a b a b ◆⇤ = ✓ (a + bb⇤a)⇤ a⇤b(b + aa⇤b)⇤ b⇤a(a + bb⇤a)⇤ (b + aa⇤b)⇤ ◆
174. Identity of 70 0 1 b a a b R R (a + b)⇤ ⌘ (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤
175. Exercise 6 71 Prove that the identity (M) ^ (S) ` (a + b)⇤ ⌘ (a + bb⇤a)⇤ + a⇤b(b + aa⇤b)⇤.
176. 72 Answer: (M) (T) (S) (M) (M) (a + b)⇤ ⌘ 1 + (a + b)⇤(a + b) ⌘ 1 + (a + b)⇤a + (a + b)⇤b ⌘ 1 + b⇤(ab⇤)⇤a + a⇤(ba⇤)⇤b ⌘ b⇤(ab⇤)⇤a + (a⇤b)⇤ ⌘ b⇤(ab⇤)⇤a + 1 + a⇤b(a⇤b)⇤ ⌘ (b⇤a)⇤ + a⇤b⇤(a⇤b)⇤ ⌘ ((1 + bb⇤)a)⇤ + a⇤b((1 + aa⇤)b)⇤ ⌘ (a + bb⇤a)⇤a⇤b(b + aa⇤b)⇤. (M) (T) (T) Exercise 6
177. 73 Theorem [KR65] For any automaton , there exists a cascade decomposition where is either reset automaton or group automaton for each A 1  i  n. A  B1 B2 · · · Bn Bi Krohn-Rhodes theorem
178. 73 Theorem [KR65] For any automaton , there exists a cascade decomposition where is either reset automaton or group automaton for each A 1  i  n. A  B1 B2 · · · Bn Bi Krohn-Rhodes theorem Every automaton can be decomposed into two parts: (1) aperiodic part and (2) symmetric part.
179. Conwayͷ ࠷ऴ༧૝ Conway Last Conjecture ’s V
180. 75 AZ/3Z AZ/2Z a0 a1 a1 a0 a1 a0 a0 a1 a2 a1 a2 a0 a1 a2 a0 a1 a2 a0 Every cyclic group is generated by one element . a1 So cyclic groups are also called monogenic group.
181. 76 a0 a1 a1 a1 a0 a1 a2 a1 a1 a1 A0 Z/2Z A0 Z/3Z Can we simplify group automata by restricting labels to only generators?
182. 77 Fact The symmetric group of order n is generated by two elements and for any n. Sn a = (1 2 · · · n) b = (1 2)
183. 77 Fact The symmetric group of order n is generated by two elements and for any n. Sn a = (1 2 · · · n) b = (1 2) 3 2 1 5 4 a + b a b a b a b a b A0 S5
184. (R(n)) Symmetric identities (E + F)⇤ ⌘ ⇣ (E + F)(F + (EF⇤)n 2E) ⌘⇤ ⇣ 1 + (E + F) n 2 X k=0 (EF⇤)k ⌘
185. Symmetric identities 3 2 1 5 4 a + b a b a b a b a b A0 S5 (R(5)) (a + b)⇤ ⇣ (a + b)(b + (ab⇤)3a) ⌘⇤ ⇣ 1 + (a + b)(1 + ab⇤ + (ab⇤)2 + (ab⇤)3) ⌘ ⌘
186. Symmetric identities 3 2 1 5 4 a + b a b a b a b a b A0 S5 (R(5)) (a + b)⇤ ⇣ (a + b)(b + (ab⇤)3a) ⌘⇤ ⇣ 1 + (a + b)(1 + ab⇤ + (ab⇤)2 + (ab⇤)3) ⌘ ⌘ One can easily observe that this red-underlined part fixes state 0.
187. where is the symmetric group of order , is complete. Sn n Last Conjecture (I) ^ (M) ^ (S) ^ (R(n))n 2 Conway’s last conjecture [Conway71]
188. Epilogue 81 These are the reasons why I chose the axiomatic approach to the equivalence of regular expressions as the topic of the talk: It’s like a “mathematical puzzle”: fun, easy to understand. But it is still a challenging problem. There is a big open problem. The symmetries in regular expressions can be found in this topic.
189. 82 Fin
191. ʙ౳ࣜެཧʹΑΔ౳Ձੑ൑ఆʙ ਖ਼نදݱʹજΉରশੑ
192. Author Ryoma Sin’ya  ʮΞϓϦέʔγϣϯ ʪόοδʫ ʯ Λ͝ཡ͍ͩ͘͞ʫ ۜҍϚʔΫʹۚͱۜΛ࢖༻ɹେֶ໊ʹۚΛ࢖༻ ۜҍϚʔΫʹۚͱۜΛ࢖༻ɹେֶ໊ʹۜΛ࢖༻ ۜҍϚʔΫʹۚͱۜΛ࢖༻ɹେֶ໊ʹϒϥοΫΛ࢖༻ ϒϥοΫ (Posdoc)