A tutoril on sequentil functions Jen-Éric Pin LIAFA, CNRS nd University Pris 7 30 Jnury 2006, CWI, Amsterdm
Outline (1) Sequentil functions (2) A chrcteriztion of sequentil trnsducers (3) Miniml sequentil trnsducers (4) Minimiztion of sequentil trnsducers (5) Composition of sequentil trnsducers (6) An lgebric pproch (7) The wreth product principle
Prt I Sequentil functions
Informl definitions A trnsducer (or stte mchine) is n utomton equipped with n output function. A trnsducer computes reltion on A B. A sequentil trnsducer is trnsducer whose underlying utomton is deterministic (but not necessrily complete). A sequentil trnsducer computes prtil function from A into B. A pure sequentil trnsducer computes prtil function ϕ preserving prefixes: if u is prefix of v, then ϕ(u) is prefix of ϕ(v).
An exmple of pure sequentil trnsducer 01 1 2 b 0 ε On the input b, the output is 01001. 01 b 0 ε 01 1 2 2 1 2
Pure sequentil trnsducers A pure sequentil trnsducer is 6-tuple A = (Q, A, B, i,, ) where the input function (q, ) q Q nd the output function (q, ) q B re defined on the sme domin D Q A. q q q
Extensions of the trnsition nd output functions The trnsition function is extended to Q A Q. Set q ε = q nd, if q u nd (q u) re defined, q (u) = (q u). The output function is extended to Q A B. Set q ε = ε nd, if q u nd (q u) re defined, q (u) = (q u)((q u) ). u q u (q u) q q u q u
Pure sequentil functions The function ϕ: A B defined by ϕ(u) = i u is clled the function relized by A. A function is pure sequentil if it cn be relized by some pure sequentil trnsducer.
Exmples of pure sequentil functions Replcing consecutive white spces by single one: 1 2 ε Converting upper cse to lower cse letters: b b. 1 A B b.
Coding nd decoding Consider the coding 0 b 1010 c 100 d 1011 r 11 Decoding function 1 ε 1 r 1 ε 1 d 0 0 ε 0 c 0 b
Decoding 0 b 1010 c 100 d 1011 r 11 0 1 ε 1 2 1 r 0 b 1 d 0 c 0 ε 4 1 ε 3 010101101000101101010110 brcdbr
Sequentil trnsducers: informl definition A sequentil trnsducer is trnsducer whose underlying utomton is deterministic (but not necessrily complete). There is n initil prefix nd terminl function. 11 ε 01 00 1 2 b 0 ε On the input b, the output is 110100100.
Sequentil trnsducers A sequentil trnsducer is 8-tuple A = (Q, A, B, i,,, m, ρ) where (Q, A, B, i,, ) is pure sequentil trnsducer, m B is the initil prefix nd ρ: Q B is prtil function, clled the terminl function. 11 ε 01 00 1 2 b 0 ε
Sequentil functions The function ϕ: A B defined by ϕ(u) = m(i u)ρ(i u) is clled the function relized by A. A function is sequentil if it cn be relized by some sequentil trnsducer.
Some exmples of sequentil functions The function x x + 1 (in reverse binry) 1 0 ε 0 1 1 2 1 ε 0 0 1 1 The mp ϕ : A A defined by ϕ(x) = uxv. u 1 v
Addition (in reverse binry) (0, 0) 0 (0, 1) 1 (1, 0) 1 ε (1, 1) 0 0 1 ε (0, 0) 1 1 (0, 1) 0 (1, 0) 0 (1, 1) 1 In inverse binry nottion, 22 = 2 + 4 + 16 01101 nd 13 = 1 + 4 + 8 10110. Tking s input (0, 1)(1, 0)(1, 1)(0, 1)(1, 0), the output is 110001, the inverse binry representtion of 35 = 1 + 2 + 32.
Hrdwre pplictions (Wikipedi) The circuit digrm for 4 bit TTL counter
Other exmples Multipliction by 4 00 0 0 1 1 1 Replcing ech occurrence of 011 by 100. 1 100 1 1 0 0 1 0 ε 2 1 ε 3 ε 0 0 01 01
Multipliction by 10 1 1 0 11 0 0 0 1 1 0 1 3 0 0 ε 1 1 0 0 2 01 1 1 0 0 1 0 4 001 1 1
Prt II A chrcteriztion
The geodesic metric The distnce between bbb nd bbb is 7. b b b b b
The geodesic metric (2) Denote by u v the longest common prefix of the words u nd v. Then d(u, v) = u + v 2 u v Exmple: d(bbb, bbb) = 6 + 7 2 3 = 7. One cn show tht d is metric. (1) d(u, v) = 0 iff u = v, (2) d(u, v) = d(v, u), (3) d(u, v) d(u, w) + d(w, v).
A chrcteriztion of sequentil functions A function ϕ: : A B is Lipschitz if there exists some K > 0 such tht, for ll u, v A, d(ϕ(u), ϕ(v)) Kd(u, v) Theorem (Choffrut 1979) Let ϕ : A B be function whose domin is closed under tking prefixes. TFCAE: (1) ϕ is sequentil, (2) ϕ is Lipschitz, nd ϕ 1 preserves regulr sets.
A chrcteriztion of pure sequentil functions Theorem (Ginsburg-Rose 1966) Let ϕ : A B be function whose domin is closed under tking prefixes. TFCAE: (1) ϕ is pure sequentil function, (2) ϕ is Lipschitz nd preserves prefixes, nd ϕ 1 preserves regulr sets.
Prt III Miniml sequentil trnsducers
Residuls of lnguge Let L be lnguge over A. Let u A. The (left) residul of L by u is the set u 1 L = {x A ux L}. It is esy to see tht v 1 (u 1 L) = (uv) 1 L. Let A = {, b} nd L = A ba. Then 1 1 L = L 1 L = A ba ba b 1 L = L (b) 1 L = A ba A, etc.
Miniml utomton of lnguge The miniml utomton of lnguge L is equl to A(L) = (Q, A,, i, F) where Q = {u 1 L u A }, i = L nd F = {u 1 L u L}). The trnsition function is given by (u 1 L) = 1 (u 1 L) = (u) 1 L. u 1 L (u) 1 L
Exmple of miniml utomton Let A = {, b} nd L = A ba. Then 1 L = A ba ba = L 1 b 1 L = L b 1 L 1 = A ba A = L 2 1 L 1 = L 1 1 L 2 = A = L 3 b 1 L 2 = L 1 L 3 = b 1 L 3 = L 3 b, b L L b 1 L 2 L 3 b
Residuls of sequentil function Let ϕ : A B be function nd let u A. The residul of ϕ by u is the function u 1 ϕ : A B defined by (u 1 ϕ)(x) = (ϕ u) 1 ϕ(ux) where (ϕ u) is the longest common prefix of the words ϕ(ux), for ux Dom(ϕ). In other words, u 1 ϕ cn be obtined from the function x ϕ(ux) by deleting the prefix ϕ u of ϕ(ux).
The function n 6n n x ϕ(x) 0 ε 0 1 1 011 2 01 0011 3 11 01001 4 001 00011 5 101 01111 6 011 001001 7 111 010101 8 0001 000011 n x ϕ(x) 9 1001 011011 10 0101 001111 11 1101 0100001 12 0011 0001001 13 1011 0111001 14 0111 0010101 15 1111 0101101 16 00001 0000011 17 10001 0110011
Let ε 1 ϕ = ϕ 0. Then ϕ 0 represents n 3n n x ϕ(x) 0 ε 0 1 1 011 2 01 0011 3 11 01001 4 001 00011 5 101 01111 6 011 001001 7 111 010101 8 0001 000011 n x ϕ(x) 9 1001 011011 10 0101 001111 11 1101 0100001 12 0011 0001001 13 1011 0111001 14 0111 0010101 15 1111 0101101 16 00001 0000011 17 10001 0110011
The function ϕ 0, representing n 3n n x ϕ 0 (x) 0 ε ε 1 1 11 2 01 011 3 11 1001 4 001 0011 5 101 1111 6 011 01001 7 111 10101 8 0001 00011 n x ϕ 0 (x) 9 1001 11011 10 0101 01111 11 1101 100001 12 0011 001001 13 1011 111001 14 0111 010101 15 1111 101101 16 00001 000011 17 10001 110011
Residuls of ϕ 0 Let ϕ 0, ϕ 1 nd ϕ 2 be the functions representing n 3n, n 3n + 1 nd n 3n + 2, respectively. ϕ 0 0 = 0 (0 1 ϕ 0 )(x) = 0 1 ϕ 0 (0x) = ϕ 0 (x) ϕ 0 1 = 1 (1 1 ϕ 0 )(x) = 1 1 ϕ 0 (1x) = ϕ 1 (x) Indeed, if x represents n, 1x represents 2n + 1, ϕ 0 (1x) represents 3(2n + 1) = 6n + 3 nd 1 1 ϕ 0 (1x) represents ((6n + 3) 1)/2 = 3n + 1.
The function ϕ 1, representing n 3n + 1 n x ϕ 1 (x) 0 ε ε 1 1 001 2 01 111 3 11 0101 4 001 1011 5 101 00001 6 011 11001 7 111 01101 8 0001 10011 n x ϕ 1 (x) 9 1001 00111 10 0101 100001 11 1101 010001 12 0011 101001 13 1011 000101 14 0111 110101 15 1111 011101 16 00001 100011 17 10001 001011
Residuls of ϕ 1 ϕ 1 0 = 1 (0 1 ϕ 1 )(x) = 1 1 ϕ 1 (0x) = ϕ 0 (x) Indeed, if x represents n, 0x represents 2n, ϕ 1 (0x) represents 3(2n) + 1 = 6n + 1 nd 1 1 ϕ 1 (0x) represents ((6n + 1) 1)/2 = 3n. ϕ 1 1 = 0 (1 1 ϕ 1 )(x) = 0 1 ϕ 1 (1x) = ϕ 2 (x) Indeed, if x represents n, 1x represents 2n + 1, ϕ 1 (1x) represents 3(2n + 1) + 1 = 6n + 4 nd 0 1 ϕ 1 (1x) represents (6n + 4)/2 = 3n + 2.
The function ϕ 2, representing n 3n + 2 n x ϕ 2 (x) 0 ε ε 1 1 101 2 01 0001 3 11 1101 4 001 0111 5 101 10001 6 011 00101 7 111 11101 8 0001 01011 n x ϕ 2 (x) 9 1001 10111 10 0101 000001 11 1101 110001 12 0011 011001 13 1011 100101 14 0111 001101 15 1111 111101 16 00001 010011 17 10001 101011
Residuls of ϕ 2 ϕ 2 0 = 0 (0 1 ϕ 2 )(x) = 0 1 ϕ 2 (0x) = ϕ 1 (x) Indeed, if x represents n, 0x represents 2n, ϕ 2 (0x) represents 3(2n) + 2 = 6n + 2 nd 0 1 ϕ 2 (0x) represents (6n + 2)/2 = 3n + 1. ϕ 2 1 = 1 (1 1 ϕ 2 )(x) = 1 1 ϕ 2 (1x) = ϕ 2 (x) Indeed, if x represents n, 1x represents 2n + 1, ϕ 2 (1x) represents 3(2n + 1) + 2 = 6n + 5 nd 1 1 ϕ 2 (1x) represents ((6n + 5) 1)/2 = 3n + 2.
Miniml sequentil trnsducer of function ϕ It is the sequentil trnsducer whose sttes re the residuls of ϕ nd trnsitions re of the form ψ ψ(ε) ψ ( 1 ψ)(ε) 1 ψ Recll tht ψ is the longest common prefix of the words ψ(x), for x Dom(ϕ). The initil stte is ε 1 ϕ nd the initil prefix is ϕ ε.
More formlly... It is the sequentil trnsducer A ϕ = (Q, A, B, i,,, m, ρ) defined by Q = {u 1 ϕ u A nd Dom(ϕ u) } i = ε 1 ϕ, m = ϕ ε nd, for q Q, ρ(q) = q(ε) A typicl trnsition of A ϕ : u 1 ϕ (u 1 ϕ) (u 1 ϕ)(ε) ((u) 1 ϕ)(ε) (u) 1 ϕ
The miniml sequentil function of n 6n 0 1 1 1 0 0 0 0 1 2 1 1 ε 0 1 1 0 0 01 185 = 1 + 8 + 16 + 32 + 128 nd 6 185 = 1110 = 2 + 4 + 16 + 64 + 1024. Thus ϕ(10011101) = 01101010001
Prt IV Minimizing sequentil trnsducers
The three steps of the lgorithm How to minimize sequentil trnsducer? (1) Obtin trim trnsducer (esy) (2) Normlise the trnsducer (tricky) (3) Merge equivlent sttes (stndrd)
Obtining trim trnsducer Let A = (Q, A, B, i,,, m, ρ) be sequentil trnsducer nd let F = Dom(ρ). The trnsducer A is trim if the utomton (Q, A,, q 0, F) is trim: ll sttes re ccessible from the initil stte nd one cn rech finl stte from ny stte. Algorithm: it suffices to remove the useless sttes.
Equivlent trnsducers ε b ε 2 1 3 b ε b ε 2 1 3 b ε ε b 2 1 3 b ε ε ε b 2 1 3 b ε These four sequentil trnsducers relize exctly the sme function ϕ : {, b} {, b}, with domin () b, defined, for ll n 0, by ϕ( 2n b) = (b) n.
Normlized trnsducer Let A = (Q, A, B, i,,, m, ρ) be sequentil trnsducer. For ech stte q, denote by m q the gretest common prefix of the words (q u)ρ(q u), where u rnges over the domin of the sequentil function. Equivlently, m q = ϕ q ε, where ϕ q is the sequentil function relized by the trnsducer derrived from A by tking q s initil stte nd the empty word s initil prefix. A sequentil trnsducer is normlized if, for ll sttes q, m q is the empty word.
m q = (q u)ρ(q u) ε ε 2 1 3 b ε ε 2 1 3 b ε 2 1 3 b ε ε 2 1 3 b ε (1) m 1 = ε m 2 = ε m 3 = ε (2) m 1 = ε m 2 = m 3 = ε (3) m 1 = ε m 2 = m 3 = ε (4) m 1 = m 2 = m 3 = ε
Normlising trnsducer Let A = (Q, A, B, i,,, m, ρ) be trim sequentil trnsducer. One obtins normlised trnsducer by chnging the initil prefix, the output function nd the terminl function s follows: q = m 1 q (q )m q m = mm i ρ (q) = m 1 q ρ(q)
Normlistion on n exmple ε ε b 2 1 3 b ε ε b ε 2 1 3 b ε One hs m 1 =, m 2 =, m 3 = ε. Thus m = mm 1 = ε = 1 = m 1 1 (1 )m 2 = 1 (b) = b 2 = m 1 2 (1 )m 1 = 1 (ε) = ε 1 b = m 1 1 (1 b)m 3 = 1 ()ε = ε ρ (3) = m 1 3 ρ(3) = ε 1 ε = ε
Computing the m q is not so esy... b b 1 2 ε ε b b b b 3 4 bb X 1 = bx 1 + bx 2 + bx 3 X 2 = X 1 + bx 4 X 3 = X 1 + bx 4 X 4 = bx 2 + bb
Solving the system X 1 = bx 1 + bx 2 + bx 3 X 2 = X 1 + bx 4 X 3 = X 1 + bx 4 X 4 = bx 2 + bb We work on k = A {0}. Addition is the lest common prefix opertor (u + 0 = 0 + u = u by convention). Observe tht u + u = u nd u(v 1 + v 2 ) = uv 1 + uv 2 (but (v 1 + v 2 )u = v 1 u + v 2 u does not hold in generl). Thus k is left semiring.
Choffrut s lgorithm (2003) The prefix order is prtil order on k (with u 0 by convention). One cn extend this order to k n componentwise. Proposition For ll u, v k n, the function f(x) = ux + v is incresing. The sequence f n (0) is decresing nd converges to the gretest fixpoint of f. The gretest solution of our system is exctly (m 1, m 2, m 3, m 4 ).
Exmple of Choffrut s lgorithm X 1 = bx 1 + bx 2 + bx 3 X 2 = X 1 + bx 4 X 3 = X 1 + bx 4 X 4 = bx 2 + bb The sequence f n (0) is (0, 0, 0, 0), (0, 0, 0, bb), (0, bbb, bbb, bb), (bbbb, bbb, bbb, b), (bb, ε, bbb, b), (b, ε, bb, b), (b, ε, b, b), (b, ε, b, b). Thus m 1 = b, m 2 = ε, m 3 = b, m 4 = b.
Merging sttes Two sttes re equivlent if they re equivlent in the input utomton (Q, A, i, F, ), hve the sme output functions nd the sme terminl functions: p q p q p = q ρ(p) = ρ(q)
Prt V Composition of sequentil functions
Composition of two pure sequentil trnsducers Theorem Pure sequentil functions re closed under composition. Let σ nd τ be two pure sequentil functions relized by the trnsducers A = (Q, A, B, q 0,, ) nd B = (P, B, C, p 0,, ) The wreth product of B by A is obtined by tking s input for B the output of A. It relizes τ σ.
Wreth product of two pure sequentil trnsducers The wreth product is defined by B A = (P Q, A, C, (p 0, q 0 ),, ) (p, q) = (p (q ), q ) (p, q) = p (q ) (p, q) p (q ) (p (q ), q )
Composition of two sequentil trnsducers Theorem Sequentil functions re closed under composition. Let ϕ nd ψ be two sequentil functions relized by the trnsducers A (equipped with the initil word n nd the terminl function ρ) nd B (equipped with the initil word m nd the terminl function σ). The wreth product of B by A is obtined by tking m(p 0 n) s initil word nd, s terminl function, ω(p, q) = (p ρ(q))σ(p ρ(q)).
Iterting sequentil functions... Iterting sequentil functions cn led to difficult problems... { 3n + 1 if n is odd Let f(n) = n/2 if n is even It is conjectured tht for ech positive integer n, there exists k such tht f k (n) = 1. The problem is still open.
Miniml trnsducer of the 3n + 1 function Let f(n) = { 3n + 1 if n is odd n/2 if n is even 0 0 1 1 1 1 0 0 ε 0 0 0 1 0 ε 1 0 2 1 3 4 5 ε ε 01 1 0 1 1 1 ε
Iterting the 3n + 1 function... 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1.
A useful result Let ϕ : A B be pure sequentil function relized by A = (Q, A, B, q 0,, ). Let L be the regulr lnguge over B recognized by the DFA B = (P, B,, p 0, F). The wreth product of B by A is the DFA B A = (P Q, A, (p 0, q 0 ), ) defined by (p, q) = (p (q ), q ). Theorem The lnguge ϕ 1 (L) is recognized by B A.
Exmple 1 Let ϕ(u) = n, where n is the number of occurrences of b in u. This function is pure sequentil: b ε ε b ε ε 1 2 3 b ε Then ϕ 1 () is the set of words contining exctly one occurrence of b.
Wreth product of the two utomt 1 2 1, 2 b 1, 3 2, 2 b 2, 3 1, 1 b 2, 1 b b b
The opertion L LA Let A = (Q, A, B, q 0, F, ) be DFA. Let B = Q A nd let σ: A B be the pure sequentil function defined by σ( 1 n ) = (q 0, 1 )(q 0 1, 2 ) (q 0 1 n 1, n ) q (q, ) q Let A nd let C = F {} B. Then σ 1 (B CB ) = LA.[Proof on blckbord!]
Exmple 2 Therefore B A recognizes LA, where B is the miniml utomton of B CB. B \ C B C 1 2 Note tht if ϕ is formul of liner temporl logic, then L(F(p Xϕ)) = A L(ϕ)
Prt VI The lgebric pproch Ide: replce utomt by monoids.
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b, c b, c b 1 2 3 Reltions:
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b, c b, c b 1 2 3 Reltions: =
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 b, c b, c b 1 2 3 Reltions: =
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 b, c b, c b 1 2 3 Reltions: = c =
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 b, c b, c b 1 2 3 Reltions: = c = b =
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 b, c b, c b 1 2 3 Reltions: = c = b = bb = b
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc cc = c
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc cc = c bc = b
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc cc = c bc = b bc = c
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc cc = c bc = b bc = c cb = bc
Trnsformtion monoid of n utomton 1 1 2 3 2 2 2 b 1 3 3 c - 2 3 b 3 3 3 bc - 3 2 c - 2 2 b, c b, c b 1 2 3 Reltions: = c = b = bb = b cb = bc cc = c bc = b bc = c cb = bc The end!
Recognizing by morphism Definition Let M be monoid nd let L be lnguge of A. Then M recognizes L if there exists monoid morphism ϕ : A M nd subset P of M such tht L = ϕ 1 (P). Proposition A lnguge is recognized by finite monoid iff it is recognized by finite deterministic utomton.
Syntctic monoid Definition (lgorithmic) The syntctic monoid of lnguge is the trnsition monoid of its miniml utomton. Definition (lgebric) The syntctic monoid of lnguge L A is the quotient monoid of A by the syntctic congruence of L: u L v iff, for ech x, y A, xvy L xuy L
Prt VII The wreth product principle The wreth product N K of two monoids N nd K is defined on the set N K K by the following product: (f 1, k 1 )(f 2, k 2 ) = (f, k 1 k 2 ) with f(k) = f 1 (k)f 2 (kk 1 ) Strubing s wreth product principle provides description of the lnguges recognized by the wreth product of two utomt (or monoids).
The wreth product principle Proposition Let M nd N be monoids. Every lnguge of A recognized by M N is finite union of lnguges of the form U σϕ 1 (V ), where ϕ : A N is monoid morphism, U is lnguge of A recognized by ϕ nd V is lnguge of BN recognized by M.
The wreth product principle 2 Theorem Let L A be lnguge recognized by n wreth product of the form (P, Q A) (Q, A). Then L is finite union of lnguges of the form W σ 1 (V ), where W A is recognized by (Q, A), σ is C-sequentil function ssocited with the ction (Q, A) nd V (Q A) is recognized by (P, Q A).
References I C. Choffrut, Contributions à l étude de quelques fmilles remrqubles de fonctions rtionnelles, Thèse de doctort, Université Pris 7, Pris (LITP), 1978. C. Choffrut, A generliztion of Ginsburg nd Rose s chrcteriztion of gsm mppings, in Automt, Lnguges nd Progrmming, H. A. Murer (éd.), pp. 88 103, Lecture Notes in Comput. Sci. vol. 71, Springer, 1979.
References II C. Choffrut, Minimizing subsequentil trnsducers: survey, Theoret. Comp. Sci. 292 (2003), 131 143. S. Ginsburg nd G. F. Rose, A chrcteriztion of mchine mppings, Cnd. J. Mth. 18 (1966), 381 388.