The group of reversible Turing machines

Transcription

1 The group of reversible Turing machines Sebastián Barbieri, Jarkko Kari and Ville Salo LIP, ENS de Lyon CNRS INRIA UCBL Université de Lyon University of Turku Center for Mathematical Modeling, University of Chile AUTOMATA June, 2016

2 Motivation Recall that a Turing machine is defined by a rule : δ T : Σ Q Σ Q { 1, 0, 1}

3 Motivation Recall that a Turing machine is defined by a rule : δ T : Σ Q Σ Q { 1, 0, 1} q δ T (, q) = (, r, 1)

4 Motivation Recall that a Turing machine is defined by a rule : δ T : Σ Q Σ Q { 1, 0, 1} r δ T (, q) = (, r, 1)

5 Motivation This defines a natural action T : Σ Z Q Σ Z Q q T r

6 Motivation This defines a natural action T : Σ Z Q Σ Z Q Such that if (x, q) Σ Z Q and δ T (x 0, q) = (a, r, d) then : T (x, q) = (σ d ( x), q ) where σ : Σ Z Σ Z is the shift action given by σ d (x) z = x z d, x 0 = a and x Z\{0} = x Z\{0}.

7 Motivation The composition of two actions T T is not necessarily an action generated by a Turing machine. if the action T is a bijection then the inverse it not necessarily an action generated by a Turing machine.

8 Motivation The composition of two actions T T is not necessarily an action generated by a Turing machine. if the action T is a bijection then the inverse it not necessarily an action generated by a Turing machine. As in cellular automata, the class of CA with radius bounded by some k N is not closed under composition or inverses.

9 Definition Let s get rid of these constrains. Given F, F finite subsets of Z d, consider instead of δ T a function : f T : Σ F Q Σ F Q Z d,

10 Definition Let s get rid of these constrains. Given F, F finite subsets of Z d, consider instead of δ T a function : f T : Σ F Q Σ F Q Z d, Let F = F = {0, 1, 2} 2, then f T (p, q) = (p, q, d) means : p p Turn state q into state q Move head by d.

11 Moving head model f T defines naturally an action T Σ Zd Q Z d T q 2 q 1 f (, q 1 ) = (, q 2, (1, 1)) F = {(0, 0), (1, 0), (1, 1)}

12 Moving head model f T defines naturally an action T Σ Zd Q Z d T q 2 q 1 f (, q 1 ) = (, q 2, (1, 1)) F = {(0, 0), (1, 0), (1, 1)} Let Σ = n and Q = k. (TM(Z d, n, k), ) is the monoid of all such T with the composition operation ; (RTM(Z d, n, k), ) is the group of all such T which are bijective.

13 Moving head model : As cellular automata Let Q = {1,..., k} and Σ = {0,..., n 1}. Σ Zd = {x : Z d Σ} X k = {x {0, 1,..., k} Zd 0 / {x u, x v } = u = v} Let X n,k = Σ Zd X k and Y = Σ Zd {0 Zd }. Then :

14 Moving head model : As cellular automata Let Q = {1,..., k} and Σ = {0,..., n 1}. Σ Zd = {x : Z d Σ} X k = {x {0, 1,..., k} Zd 0 / {x u, x v } = u = v} Let X n,k = Σ Zd X k and Y = Σ Zd {0 Zd }. Then : TM(Z d, n, k) = {φ End(X n,k ) φ Y = id, φ 1 (Y ) = Y } RTM(Z d, n, k) = {φ Aut(X n,k ) φ Y = id}

15 Moving tape model f T defines naturally an action T Σ Zd Q T f q 1 q 2 f (, q 1 ) = (, q 2, (1, 1)) F = {(0, 0), (1, 0), (1, 1)}

16 Moving tape model f T defines naturally an action T Σ Zd Q T f q 1 q 2 f (, q 1 ) = (, q 2, (1, 1)) F = {(0, 0), (1, 0), (1, 1)} Let Σ = n and Q = k. (TM fix (Z d, n, k), ) is the monoid of all such T with the composition operation ; (RTM fix (Z d, n, k), ) is the group of all such T which are bijective.

17 Moving tape model : dynamical definition Let x, y Σ Zd. x and y are asymptotic, and write x y, if they differ in finitely many coordinates. We write x m y if x v = y v for all v m.

18 Moving tape model : dynamical definition Let x, y Σ Zd. x and y are asymptotic, and write x y, if they differ in finitely many coordinates. We write x m y if x v = y v for all v m. Let T : Σ Zd Q Σ Zd Q be a function. T is a moving tape Turing machine T is continuous, and for a continuous function s : Σ Zd Q Z d and a N we have T (x, q) 1 a σ s(x,q) (x) for all (x, q) Σ Zd Q.

19 Moving tape model : dynamical definition Let x, y Σ Zd. x and y are asymptotic, and write x y, if they differ in finitely many coordinates. We write x m y if x v = y v for all v m. Let T : Σ Zd Q Σ Zd Q be a function. T is a moving tape Turing machine T is continuous, and for a continuous function s : Σ Zd Q Z d and a N we have T (x, q) 1 a σ s(x,q) (x) for all (x, q) Σ Zd Q. s : Σ Zd Q Z d is the shift indicator function

20 Equivalence of the models RTM fix (Z d, 1, k) = S k and Z d RTM(Z d, 1, k).

21 Equivalence of the models RTM fix (Z d, 1, k) = S k and Z d RTM(Z d, 1, k). Proposition If n 2 then : TM fix (Z d, n, k) = TM(Z d, n, k) RTM fix (Z d, n, k) = RTM(Z d, n, k).

22 Properties of RTM Proposition Let T TM fix (Z d, n, k). Then the following are equivalent : 1 T is injective. 2 T is surjective. 3 T RTM fix (Z d, n, k). 4 T preserves the uniform measure (µ(t 1 (A)) = µ(a) for all Borel sets A). 5 µ(t (A)) = µ(a) for all Borel sets A.

23 Properties of RTM Proposition If n 2 RTM(Z d, n, k) is not finitely generated.

24 Properties of RTM Proposition If n 2 RTM(Z d, n, k) is not finitely generated. Proof : We find an epimorphism from RTM to a non-finitely generated group. Let T RTM fix (Z d, n, k), therefore, it has a shift indicator s : Σ Zd Q Z d. Define α(t ) := E µ (s) = s(x, q)dµ, Σ Zd Q One can check that α(t 1 T 2 ) = α(t 1 ) + α(t 2 ). Therefore α : RTM(Z d, n, k) Q d is an homomorphism

25 Properties of RTM Now consider the machine T SURF,m where for all a Σ and q Q : q a q T SURF,m a f (0 m a, q) = (a0 m, q, 1). Otherwise f (u, q) = (u, q, 0).

26 Properties of RTM Now consider the machine T SURF,m where for all a Σ and q Q : q a q T SURF,m a f (0 m a, q) = (a0 m, q, 1). Otherwise f (u, q) = (u, q, 0). T SURF,m RTM(Z, n, k) and α(t SURF,m ) = 1/n m

27 Properties of RTM Now consider the machine T SURF,m where for all a Σ and q Q : q a q T SURF,m a f (0 m a, q) = (a0 m, q, 1). Otherwise f (u, q) = (u, q, 0). T SURF,m RTM(Z, n, k) and α(t SURF,m ) = 1/n m (1/n m ) m N α(rtm(z, n, k)) which is thus a non-finitely generated subgroup of Q.

28 Interesting subgroups of RTM LP(Z d, n, k) Local permutations. q T π r

29 Interesting subgroups of RTM LP(Z d, n, k) Local permutations. RFA(Z d, n, k) Reversible finite-state automata. q T r

30 Interesting subgroups of RTM LP(Z d, n, k) Local permutations. RFA(Z d, n, k) Reversible finite-state automata. OB(Z d, n, k) Oblivous machines LP, Shift.

31 Interesting subgroups of RTM LP(Z d, n, k) Local permutations. RFA(Z d, n, k) Reversible finite-state automata. OB(Z d, n, k) Oblivous machines LP, Shift. EL(Z d, n, k) Elementary machines LP, RFA.

32 Small group theory roadmap Residually Finite LEF Abelian Sofic Surjunctive Amenable LEA Res. finite groups are those where every non-identity element can be mapped to a non-identity element by a homomorphism to a finite group Amenable groups admit left invariant finitely additive measures. LEF and LEA stand for locally embeddable into (finite/amenable) groups. Sofic groups are generalizations of LEF and LEA. Surjunctive groups satisfy that all injective CA are surjective.

33 Small group theory roadmap Residually Finite LEF Abelian Sofic Surjunctive Amenable LEA Theorem n 2, RTM(Z d, n, k) is LEF but neither amenable nor residually finite.

34 Some properties : LP(Z d, n, k) For n 2, we have S LP(Z d, n, k).

35 Some properties : LP(Z d, n, k) For n 2, we have S LP(Z d, n, k). This means that RTM is not residually finite, and that it contains all finite groups.

36 Some properties : LP(Z d, n, k) For n 2, we have S LP(Z d, n, k). This means that RTM is not residually finite, and that it contains all finite groups. LP(Z d, n, k) is locally finite and amenable.

37 Some properties : LP(Z d, n, k) For n 2, we have S LP(Z d, n, k). This means that RTM is not residually finite, and that it contains all finite groups. LP(Z d, n, k) is locally finite and amenable. In particular, for n 2 LP(Z d, n, k) is not finitely generated.

38 Some properties : OB(Z d, n, k) Now let s add the shift. Recall that OB(Z d, n, k) = LP, Shift.

39 Some properties : OB(Z d, n, k) Now let s add the shift. Recall that OB(Z d, n, k) = LP, Shift. OB(Z d, n, k) is amenable.

40 Some properties : OB(Z d, n, k) Now let s add the shift. Recall that OB(Z d, n, k) = LP, Shift. OB(Z d, n, k) is amenable. Proof : α gives a short exact sequence 1 LP(Z d, n, k) OB(Z d, n, k) Z d 1.

41 Some properties : OB(Z d, n, k) Now let s add the shift. Recall that OB(Z d, n, k) = LP, Shift. OB(Z d, n, k) is amenable. Proof : α gives a short exact sequence 1 LP(Z d, n, k) OB(Z d, n, k) Z d 1. OB(Z d, n, k) contains all generalized lamplighter groups G Z d.

42 Some properties : OB(Z d, n, k) Now let s add the shift. Recall that OB(Z d, n, k) = LP, Shift. OB(Z d, n, k) is amenable. Proof : α gives a short exact sequence 1 LP(Z d, n, k) OB(Z d, n, k) Z d 1. OB(Z d, n, k) contains all generalized lamplighter groups G Z d. Theorem OB(Z d, n, k) is finitely generated.

43 OB(Z d, n, k) is finitely generated. This proof is based on the existence of strongly universal reversible gates for permutations of Σ m. A controlled swap is a transposition (s, t) where s, t have Hamming distance 1 in Q Σ m. Theorem The group generated by the applications of controlled swaps of Q Σ 4 at arbitrary positions generates Sym(Q Σ m ) if Σ is odd and Alt(Q Σ m ) if it s even. Corollary : [Sym(Q Σ m )] m+1 [Sym(Q Σ 4 )] m+1.

44 OB(Z d, n, k) is finitely generated. Using this result, a generating set can be constructed : A 1 = Shifts T ei for {e i } i d a base of Z d. A 2 =All T π LP(Z d, n, k) of fixed support E Z d of size 4. A 3 = The swaps of symbols in positions ( 0, e i ).

45 Some properties : RFA(Z d, n, k) Recall that RFA(Z d, n, k) is the subgroup of machines which do not modify the tape. For n 2, RFA(Z d, n, k) contains all countable free groups.

46 Some properties : RFA(Z d, n, k) Recall that RFA(Z d, n, k) is the subgroup of machines which do not modify the tape. For n 2, RFA(Z d, n, k) contains all countable free groups. In particular, this means that RFA and RTM are not amenable.

47 Some properties : RFA(Z d, n, k) Recall that RFA(Z d, n, k) is the subgroup of machines which do not modify the tape. For n 2, RFA(Z d, n, k) contains all countable free groups. In particular, this means that RFA and RTM are not amenable. RFA(Z d, n, k) is residually finite but not finitely generated.

48 Some properties : EL(Z d, n, k) and RTM(Z d, n, k) Recall that EL(Z d, n, k) = LP(Z d, n, k), RFA(Z d, n, k) is the subgroup of elementary Turing machines. Question : Is EL(Z d, n, k) = RTM(Z d, n, k)?

49 Some properties : EL(Z d, n, k) and RTM(Z d, n, k) Recall that EL(Z d, n, k) = LP(Z d, n, k), RFA(Z d, n, k) is the subgroup of elementary Turing machines. Question : Is EL(Z d, n, k) = RTM(Z d, n, k)? For n 2, α(el(z d, n, k)) = α(rfa(z d, n, k)) has bounded denominator.

50 Some properties : EL(Z d, n, k) and RTM(Z d, n, k) Recall that EL(Z d, n, k) = LP(Z d, n, k), RFA(Z d, n, k) is the subgroup of elementary Turing machines. Question : Is EL(Z d, n, k) = RTM(Z d, n, k)? For n 2, α(el(z d, n, k)) = α(rfa(z d, n, k)) has bounded denominator. In particular, this means that EL RTM.

51 Some properties : EL(Z d, n, k) and RTM(Z d, n, k) Recall that EL(Z d, n, k) = LP(Z d, n, k), RFA(Z d, n, k) is the subgroup of elementary Turing machines. Question : Is EL(Z d, n, k) = RTM(Z d, n, k)? For n 2, α(el(z d, n, k)) = α(rfa(z d, n, k)) has bounded denominator. In particular, this means that EL RTM. Open : Is EL = Ker α (RTM), Shift?

52 Some properties : EL(Z d, n, k) and RTM(Z d, n, k) Recall that EL(Z d, n, k) = LP(Z d, n, k), RFA(Z d, n, k) is the subgroup of elementary Turing machines. Question : Is EL(Z d, n, k) = RTM(Z d, n, k)? For n 2, α(el(z d, n, k)) = α(rfa(z d, n, k)) has bounded denominator. In particular, this means that EL RTM. Open : Is EL = Ker α (RTM), Shift? RTM is a LEF group, in particular, it is sofic.

53 Computability properties Given a finite rules : f, f : It is decidable (in any model) whether T f = T f. We can effectively calculate a rule for T f T f. It is decidable whether T f is reversible. If it is, we can effectively compute a rule for T 1 f.

54 Computability properties Given a finite rules : f, f : It is decidable (in any model) whether T f = T f. We can effectively calculate a rule for T f T f. It is decidable whether T f is reversible. If it is, we can effectively compute a rule for T 1 f. RTM(Z d, n, k) is a recursively presented group with decidable word problem.

55 Computability properties Given a finite rules : f, f : It is decidable (in any model) whether T f = T f. We can effectively calculate a rule for T f T f. It is decidable whether T f is reversible. If it is, we can effectively compute a rule for T 1 f. RTM(Z d, n, k) is a recursively presented group with decidable word problem. What can we say about the torsion ( n such that T n = 1) problem?

56 Computability properties Given a finite rules : f, f : It is decidable (in any model) whether T f = T f. We can effectively calculate a rule for T f T f. It is decidable whether T f is reversible. If it is, we can effectively compute a rule for T 1 f. RTM(Z d, n, k) is a recursively presented group with decidable word problem. What can we say about the torsion ( n such that T n = 1) problem? It is undecidable in RTM(Z d, n, k) if n 2. What about RFA?

57 The torsion problem for RFA RFA(Z, n, k) has decidable torsion problem. Proof : As Z is two-ended, any non-torsion machine must shift to the left or right in at least a periodic configuration.

58 The torsion problem for RFA RFA(Z, n, k) has decidable torsion problem. Proof : As Z is two-ended, any non-torsion machine must shift to the left or right in at least a periodic configuration. Theorem RFA(Z d, n, k) has undecidable torsion problem for d, n 2.

59 The torsion problem for RFA RFA(Z, n, k) has decidable torsion problem. Proof : As Z is two-ended, any non-torsion machine must shift to the left or right in at least a periodic configuration. Theorem RFA(Z d, n, k) has undecidable torsion problem for d, n 2. Proof : Reduction to the snake tiling problem, which reduces to the domino problem for Z d.

60 The snake problem ɛ Can we tile the plane in a way which produces a bi-infinite path?

61 The snake problem Theorem (Kari) The snake tiling problem is undecidable. The proof uses a plane filling curve generated by a substitution.

62 The snake problem Theorem (Kari) The snake tiling problem is undecidable. The proof uses a plane filling curve generated by a substitution. For every instance of the snake tiling problem, one can construct T RFA which walks the path of the snake, and turns back if it encounters a problem.

63 Thank you for your attention!