Hardness of conjugacy, embedding and factorization of multidimensional subshifts

Transcription

1 Hrdness of conjugcy, embedding nd fctoriztion of multidimensionl subshifts Emmnuel Jendel, Pscl Vnier b LORIA Cmpus Scientifique - BP Vndoeuvre-les-Nncy Frnce b Lbortoire d Algorithmique Complexité et Logique Université de Pris Est, LACL, UPEC, Frnce Abstrct Subshifts of finite type re sets of colorings of the plne defined by locl constrints. They cn be seen s discretiztion of continuous dynmicl systems. We investigte here the hrdness of deciding fctoriztion, conjugcy nd embedding of subshifts in dimensions d > 1 for subshifts of finite type nd sofic shifts nd in dimensions d 1 for effective shifts. In prticulr, we prove tht the conjugcy, fctoriztion nd embedding problems re Σ 0 3-complete for sofic nd effective subshifts nd tht they re Σ 0 1-complete for SFTs, except for fctoriztion which is lso Σ 0 3-complete. Keywords: Subshifts, Computbility, Fctoriztion, Embedding, Conjugcy, Subshift of finite type, Arithmeticl Hierrchy, Tilings, SFTs. A d-dimensionl subshift is the set of colorings of Z d by finite set of colors in which fmily of forbidden ptterns never pper. These re shift-invrint spces, hence the nme. If the fmily of forbidden ptterns is finite, then it is subshift of finite type (SFT). If the fmily of forbidden ptterns is recursively enumerble, then the subshift is clled effective. Another clss of subshifts cn be defined by the help of locl mps, nmely the clss of sofic shifts: they re the letter by letter projections of SFTs. One cn lso see SFTs s tilings of Z d, nd in dimension 2 they re equivlent to the usul notion of tilings introduced by Wng [17]. Subshifts re wy to discretize continuous dynmicl systems: if X is compct spce nd φ : X X continuous mp, we cn prtition X in finite number of prts A = {1,..., n} nd trnsform the orbit of point x X into sequence (x n ) n N, where x i denotes the prt of X in which φ i (x) lies. Conjugcy is the right notion of isomorphism between subshifts, nd plys mjor role in their study: when two subshifts re conjugte they code ech other nd hence hve the sme dynmicl properties. Conjugcy is n equivlence reltion nd llows This work ws sponsored by grnts ANR-09-BLAN-0164, EQINOCS ANR 11 BS , TARMAC ANR 12 BS Emil ddresses: emmnuel.jendel@lori.fr (Emmnuel Jendel), pscl.vnier@lcl.fr (Pscl Vnier) Preprint submitted to Journl of Computer nd System Sciences My 22, 2015

2 to seprte SFTs into equivlence clsses. Deciding whether two SFTs re conjugte is clled the clssifiction problem. It is long stnding open problem in dimension one [5], lthough hs been proved decidble in the prticulr cse of one-sided SFTs on N, see [18]. It hs been known for long time tht in higher dimensions the problem is undecidble when given two SFTs, since it cn be reduced to the emptiness problem which is Σ 0 1-complete [2]. However, we prove here slightly stronger result: even by fixing the clss in dvnce, it is still undecidble to decide whether some given SFT belongs to it: Theorem 0.1. For ny fixed SFT X, given some SFT Y s n input, it is Σ 0 1-complete to decide whether X nd Y re conjugte (resp. equl). As for the clsses of sofic nd effective shifts, the complexity is higher: Theorem 0.2. Given two sofic/effective shifts X, Y, it is Π 0 2-complete to decide whether X nd Y re equl. Theorem 0.3. Given two sofic/effective subshifts X, Y, it is Σ 0 3-complete to decide whether X nd Y re conjugte. An interesting open question for higher dimension tht would probbly help solve the one dimensionl problem would be is conjugcy of subshifts decidble when provided n orcle nswering whether or not pttern is extensible?. A positive nswer to this question would solve the one dimensionl cse, even if the SFTs re considered on N 2 insted of Z 2. Fctoriztion is the notion of surjective morphism dpted to SFTs: when X fctors on Y, then Y is recoding of X, possibly with informtion loss: the dynmic of Y is simpler thn X s,i.e. it cn be deduced from X s. The problem of knowing if some SFT is fctor of nother one hs lso been much studied. In dimension one, it is only prtly solved for the cse when the entropies of the two SFTs X, Y verify h(x) > h(y ), see [4]. Fctor mps hve lso been studied with the hope of finding universl SFTs: SFTs tht cn fctor on ny other nd thus contin the dynmics of ll of them. However it hs been shown tht such SFTs do not exist, see [7, 3]. We prove here tht it is hrder to know if n SFT is fctor of nother thn to know if it is conjugte to it. Theorem 0.4. Given two SFTs/sofic/effective subshifts X, Y s input, it is Σ 0 3-complete to decide whether X fctors onto Y. The lst problem we will tckle is the embedding problem, tht is to sy: when cn n SFT be injected into some other SFT? If n SFT X cn be injected into nother SFT Y, tht mens tht there is n SFT Z Y such tht X nd Z re conjugte. In dimension 1, this problem is lso prtly solved when the two SFTs X, Y re irreducible nd their entropies verify h(x) > h(y ) [12]. We prove here tht the problem is Σ 0 1-complete for SFTs nd Σ 0 3-complete for effective nd sofic subshifts: Theorem 0.5. Given two SFTs X, Y s inputs, it is Σ 0 1-complete to decide whether X embeds into Y. Theorem 0.6. Given two sofic/effective subshifts X, Y s inputs, it is Σ 0 3-complete to decide whether X embeds into Y. 2

3 The pper is orgnized s follows: first we give the necessry definitions nd fix the nottion in section 1, fter wht we give the proofs of the theorems bout conjugcy nd equlity in Section 2, bout fctoriztion in Section 3 nd bout embedding in Section 4. This rticle covers the results nnounced in [10] with the dditions of the results on sofic nd effective subshifts. 1. Preliminry definitions 1.1. SFTs nd effective subshifts We give here some stndrd definitions nd fcts bout multidimensionl subshifts, one my consult Lind [14] or Lind/Mrcus [13] for more detils. Let A be finite lphbet, its elements re clled symbols, the d-dimensionl full shift on A is the set A Zd of ll mps (colorings) from Z d to the A (the colors). For v Z d, the shift functions σ v : A Zd A Zd, re defined loclly by σ v (c x ) = c x+v. The full shift equipped with the distnce d(x, y) = 2 min{ v v Zd,x v y v} is compct metric spce on which the shift functions ct s homeomorphisms. An element of A Zd is clled configurtion. Every closed shift-invrint (invrint by ppliction of ny σ v ) subset X of A Zd is clled subshift, or shift. An element of subshift is clled point of this subshift. Alterntively, subshifts cn be defined with the help of forbidden ptterns. A pttern is function p : P A, where P, the support, is finite subset of Z d. Let F be collection of forbidden ptterns, the subset X F of A Zd contining the configurtions hving nowhere pttern of F. More formlly, X F is defined by { X F = x A Zd } z Z d, p F, x z+p p. In prticulr, subshift is sid to be subshift of finite type (SFT) when the collection of forbidden ptterns is finite. Usully, the ptterns used re blocks or r-blocks, tht is they re defined over finite subset P of Z d of the form B r = r, r d, r is clled its rdius. We my ssume tht ll ptterns of F re defined with blocks of the sme rdius r, nd sy the fmily F hs rdius r. We note r X the rdius of the SFT X, the smllest r for which there is fmily F of rdius r defining X. When the collection of forbidden ptterns is recursively enumerble (i.e. Σ 0 1), the subshift is n effective subshift. Given subshift X, pttern p is sid to be extensible if there exists x X in which p ppers, p is lso sid to be extensible to x. We lso sy tht pttern p 1 is extensible to pttern p 2 if p 1 ppers in p 2. A block or pttern is sid to be dmissible if it does not contin ny forbidden pttern. Note tht every extensible pttern is dmissible but tht the converse is not necessrily true. As mtter of fct, for SFTs, it is undecidble (in Π 0 1 to be precise) in generl to know whether pttern is extensible while it is lwys decidble efficiently (polynomil time) to know if pttern is dmissible, further detils bout tht will be introduced in Section 1.4. As we sid before, subshifts re compct spces, this gives link between dmissibility nd extensibility: if pttern ppers in n incresing sequence of dmissible ptterns, then it ppers in vlid configurtion nd is thus extensible. More generlly, if we hve n incresing sequence of dmissible ptterns, then we cn extrct from it sequence converging to some point of the subshift. 3

4 Note tht insted of using the formlism of SFTs for the theorems we could hve used the formlism of Wng tiles, in which numerous results hve been proved. In prticulr the undecidbility of knowing whether n SFT is empty. Since we will use construction bsed on Wng tiles, we review their definitions. Wng tiles re unit squres with colored edges which my not be flipped or rotted. A tileset T is finite set of Wng tiles. A coloring of the plne is mpping c : Z 2 T ssigning Wng tile to ech point of the plne. If ll djcent tiles of coloring of the plne hve mtching edges, it is clled tiling. The set of tilings of Wng tileset is SFT on the lphbet formed by the tiles. Conversely, ny SFT is isomorphic to Wng tileset. From recursivity point of view, one cn sy tht SFTs nd Wng tilesets re equivlent. In this pper, we will be using both terminologies indiscrimintely Conjugcy, Embedding nd Fctoriztion In the rest of the pper, we will use the nottion A X for the lphbet of the subshift X. Let X A Z2 X nd Y AZ2 Y be two subshifts, function F : X Y is block code if there exists finite set V = {v 1,..., v k } Z 2, the window, nd locl mp f : A V X A Y, such tht for ny point x X nd y = F (x), for ll z Z d, y z = f(x z+v1,..., x z+vk ). Tht is to sy F is defined loclly. Without loss of generlity, we my suppose tht the window is n r-block, r being then clled the rdius of F nd (2r + 1) its dimeter, we note r F the rdius of F. A fctoriztion or fctor mp is surjective block code F : X Y. When the function is injective insted of being surjective, it is clled n embedding, nd we sy tht X embeds into Y. By the Curtis/Lyndon/Hedlund Theorem [6], when block mp F is bijective then it is invertible nd its inverse is lso block code. Subshifts X nd Y for which there exist bijective block mp F : X Y re sid to be conjugte. In the rest of the pper, we will note with the sme symbol the locl nd globl functions, the context mking cler which one is being used. The entropy of subshift X is defined s log E n (X) h(x) = lim n n d where E n (X) is the number of extensible ptterns of X of support 0, n 1 d where d is the dimension. For instnce, the entropy of the full shift is h(a Zd ) = log A. The entropy is conjugcy invrint, tht is to sy, if X nd Y re conjugte, then h(x) = h(y ). It is in prticulr esy to see thnks to the entropy tht the full shift on n symbols is not conjugte to the full shift with n symbols when n n Sofic subshifts nd their reltion to effective subshifts The clss of subshifts tht re imges of SFTs by fctor mp is the clss of sofic subshifts. It is the smllest clss of subshifts tht is closed by fctoriztion. There is link between sofic subshifts of dimension d nd effective subshifts of dimension d

5 s s s q 0 q 0 s s s s s s s s s s s s q 0 s q 0 q 0 s () Tiles llowing to encode computtions of Turing mchine: the djcency rules re given by the mchine s trnsition tble δ(s, ) = (s,, d) with d determining to which side of the tile the rrow goes, s the new stte nd the letter written on the tpe. q 0 is the initil stte of the Turing mchine. Note tht there is no tile for the hlting stte (q 3, 5 2) q (q 4, 4 2) q (q 7, 3 3) time q (q 3, 2 2) q (q 1, 1 1) q (q 0, 0 0) spce (b) Spce-Time digrm of Turing mchine. q (c) Vlid tiling by the tileset corresponding to the Turing mchine. If the Turing mchine tkes no input, then the 0 i re ll blnks. Figure 1: How to encode Turing mchine computtions in tilings. Only the vlid infinite runs of the Turing mchine my tile the qurter plne without ny errors when the strting tile is present, since we forbid hlting sttes to pper in tile. 5

6 Definition 1.1 (Lift). Let X be d-dimensionl subshift, then the lift X of X to dimension d + 1 is the subshift tht is formed of configurtions x X nd tht re identicl on the next component. Prticulrly, the projection of the lift of X long its d first components is X. Through lifting, one cn link sofic nd effective subshifts with the following theorem: Theorem 1.1 (Hochmn [8], Aubrun nd Sblik [1]). A subshift is effective if nd only if its lift is sofic. This theorem will be minly used to trnspose constructions of effective subshifts to the sofic cse, but in one more dimension Arithmeticl Hierrchy nd computbility We give now some bckground in computbility theory nd in prticulr bout the rithmeticl hierrchy. More detils cn be found in Rogers [16]. Given Turing mchine M, we will note M(n) when M hlts on input n nd M(n) when it does not. In computbility theory, the rithmeticl hierrchy is clssifiction of sets ccording to their logicl chrcteriztion. A set A N is Σ 0 n if there exists totl computble predicte R such tht x A y 1, y 2,..., Qy n R(x, y 1,..., y n ), where Q is or n depending on the prity of n. A set A is Π 0 n if there exists totl computble predicte R such tht x A y 1, y 2,..., Qy n R(x, y 1,..., y n ), where Q is o r n depending on the prity of n. Equivlently, set is Σ 0 n iff its complement is Π 0 n. We sy set A is mny-one reducible to set B, A m B if there exists computble function f such tht for ny x, f(x) A x B. Given n enumertion of Turing mchines M i with orcle X, the Turing jump X of set X is the set of integers i such tht M i hlts on input i. We note X (0) = X nd X (n+1) = (X (n) ). In prticulr 0 is the set of hlting Turing mchines. A set A is Σ 0 n-hrd (resp. Π 0 n) iff for ny Σ 0 n (resp. Π 0 n) set B, B m A. Furthermore, Σ 0 n-hrd (resp. Π 0 n) is Σ 0 n-complete (resp. Π 0 n-complete) if it is in Σ 0 n. An exmple of Σ 0 n-complete problem is 0 (n). The sets in Σ 0 1 re lso clled recursively enumerble nd the sets in Π 0 1 re clled the co-recursively enumerble or effectively closed sets. In this rticle, we will minly use two complete problem: TOTAL: this is the set of Turing mchines which hlt on ll inputs, see exmple 35.3 of [11]. COFIN: this is the set of Turing mchines which hlt on ll inputs but finite number. This problem is Σ 0 3-complete, see exmple 35.5 nd lemm 36.1 of [11]. Another fct tht will be used severl times is tht extensibility for the severl clsses of subshifts we consider is Π 0 1. Lemm 1.2. Extensibility for SFTs, sofic nd effective subshifts is Π 0 1. Proof. We my restrict ourselves to effective subshifts since SFTs nd sofic subshifts re effective. One my check if pttern M is extensible step by step: t the k-th step one enumertes k forbidden ptterns nd checks whether there exist pttern of 6

7 rdius r(m) + k contining M t its center contining none of the k forbidden ptterns enumerted so fr. If this lgorithm hlts, then M is not extensible. Otherwise we hve n infinite sequence of incresing ptterns M k from which we cn extrct converging subsequence whose limit is n extension of M contining no forbidden pttern. It is quite cler tht deciding whether pttern is dmissible for SFTs is decidble, while for effective shifts is Π 0 1 since one needs to enumerte ll forbidden ptterns to check. Admissibility for sofic shifts is lso Π 0 1 since they re effective. 2. Conjugcy nd equlity 2.1. SFTs We prove here the Σ 0 1-completeness of the conjugcy problem for SFTs in dimension d 2, even if we fix n SFT in dvnce. We first prove the following lemm, which is the first step to show tht conjugcy is Σ 0 1 nd lso proves tht equlity of SFTs is Σ 0 1. Lemm 2.1. Given F locl mp, X nd Y SFTs s inputs, deciding if F (X) Y is Σ 0 1. Proof. It is cler tht F (X) Y if nd only if F (X) does not contin ny configurtion where forbidden ptterns of Y ppers. We now show tht this is equivlent to the following Σ 0 1 sttement: there exists rdius r > mx(r F + r Y, r X ) such tht for ny dmissible r-block M of X, F (M) does not contin ny forbidden pttern in its center. We prove the result by contrposition, in both directions. Suppose there is configurtion x X such tht F (x) contins forbidden pttern. Then for ny rdius r > mx(r F + r Y, r X ), there exists n extensible, nd thus dmissible, pttern M of size r such tht F (M) contins forbidden pttern in its center. Conversely, if for ny rdius r > mx(r F + r Y, r X ), there exists n dmissible pttern M of X of size r such tht F (M) contins forbidden pttern in its center, then by compctness one cn extrct converging subsequence from these forbidden ptterns, its limit x is in X nd F (x) contins forbidden pttern in its center. In prticulr, if F is the identity we obtin: Corollry 2.2. Given two SFTs X, Y s n input, it is Σ 0 1 to decide whether X = Y. We my now prove one of the nnounced theorems: Theorem 2.3. Given two SFTs X, Y s n input, it is Σ 0 1 to decide whether X nd Y re conjugte. Proof. To decide whether two SFTs X nd Y re conjugte, we hve to check whether there exists two locl functions F : A Br F X A Y nd G : A Br G Y A X such tht the globl functions ssocited verify F X G Y = id Y nd G Y F X = id X. These functions being locl, we cn guess them with first order existentil quntifier. We prove tht X nd Y re conjugte if nd only if the following A 0 1 sttement is true : 7

8 There exist F, G nd k > mx(r X + r Y ) + r F + r G such tht F (X) Y nd G(Y ) X nd : for ny k-block b, if b is dmissible for X, then G F (b) 0 = b 0 for ny k-block b, if b is dmissible for Y, then F G(b) 0 = b 0 We only prove the sttement for G F the other one being identicl. The proof is by contrposition in both directions : Let x X be point such tht G F (x) x, we my suppose tht the difference is in 0 by shifting. For ll k, there exists n extensible pttern b of size k such tht G F (x) 0 b 0. Conversely, if there exists sequence b k of dmissible k-blocks such tht G F (b k ) 0 (b k ) 0, then by compctness we cn extrct subsequence converging to some point x X which by construction is different from its imge by G F in 0. As we hve seen in Lemm 2.1 tht checking whether F (X) Y is Σ 0 1, we hve the desired result. Theorem 2.4. For ny fixed SFT X of dimension d 2, the problem of deciding whether given Y n SFT of dimension d 2 s input, Y is conjugte (resp. equl) to X is Σ 0 1-hrd. Proof. We reduce the problem from 0, the hlting problem. Given Turing mchine M we construct SFT Y M such tht Y M is conjugte to X iff M hlts. Let R M be Robinson s SFT [15] encoding computtions of M: R M is empty iff M hlts 1. Now tke the full shift on one more symbol thn X, note it F. Let Y M be now the disjoint union of X nd R M F. If M hlts, Y M = X nd hence is conjugte to X. In the other direction, suppose M does not hlt, then R M F hs entropy strictly greter thn tht of X nd hence Y M is not conjugte to X. Corollry 2.5. Given two SFTs X, Y of dimension d 2 s n input, it is Σ 0 1-hrd to decide whether X = Y Sofic nd effective subshifts For effective nd sofic subshifts, the complexity becomes higher: checking whether pttern is dmissible or whether it is extensible is the sme complexity-wise. It is Π 0 1 in both cses, which disllows us from using the sme compctness tricks s in Lemm 2.1. Lemm 2.6. Given two effective subshifts X, Y nd locl function F, deciding if F (X) Y is Π Robinson s SFT is in dimension 2 of course, for higher dimensions, we tke the iterted lift: we tke the rules tht the symbol in x ± e i equls the symbol in x, for i > 2. 8

9 Proof. F (X) Y if nd only if the imge of every extensible pttern of X is n extensible pttern of Y, which is equivlent to the following logicl sentence: For every pttern M of rdius r, M is extensible for X F (M) is extensible for Y. Which is clerly Π 0 2. Corollry 2.7. Given two effective subshifts X, Y it is Π 0 2 to decide whether X = Y. Let us now prove the Π 0 2-hrdness of the equlity problem in order to obtin the Π 0 2-completeness of it. Theorem 2.8. Given two effective subshifts X, Y, deciding whether X = Y is Π 0 2-hrd. Proof. To show tht the problem is Π 0 2-hrd, we strt from the TOTAL problem which is Π 0 2-complete. Tke the following two one-dimensionl effective subshifts: Let M be Turing mchine, X M is the subshift on two symbols {#, 0} where we forbid ll words #0 n # such tht M(n). Y is the subshift on the lphbet {#, 0} where we forbid the words #0 n # for ll n N such tht the subshift is composed only of the orbits the two following points: nd 000#000 These subshifts re effective, since for ny Turing mchine M one cn enumerte with Turing mchine ll n s such tht M(n). The two subshifts, X M nd Y re equl if nd only if the Turing mchine M hlts on ll inputs : if the mchine M does not hlt on n, then the subshift X M contins the periodic point #0 n #0 n # which is not in Y. Now using Theorem 1.1 llowing to lift n effective subshift of dimension 1 to sofic one of dimension 2, we obtin the following corollry: Corollry 2.9. Given two sofic subshifts X, Y of dimension d 2, knowing whether X = Y is Π 0 2-complete. We my now hed bck to the conjugcy problem: strightforwrd dpttion of the proof of Theorem 2.3 leds to the following upper bound: Theorem Given two effective subshifts X, Y s n input, it is Σ 0 3 to decide whether they re conjugte. Only remins the hrdness prt, which we prove by reducing to COFIN, the set of Turing mchines tht do not hlt on finite number of input only. Theorem Given two effective subshifts X, Y (resp. sofic of dimension d 2) s n input, deciding whether they re conjugte is Σ 0 3-hrd. Proof. We give construction for effective subshifts of dimension 1 which cn gin be lifted to sofic subshifts of dimension 2 or higher. Given Turing mchine M, we construct two subshifts X M nd Y M on the lphbet {#, 0, 1}: 9

10 X M : we forbid the words #1, 1#, 10, 0#, the words #0 k 1 when k is not of the form 2 i+1 with i N nd the words #0 2n+1 1 for ll n such tht M(n). The subshift is formed of the following biinfinite words: ###0 2n with M(n) ###### ### Y M : we forbid the words #1, 1#, 10, 0#, the words #0 k 1 when k is not of the form 2 i i with i N nd the words #0 2n+1 1 for ll n such tht M(n). The subshift Y M is formed of the following words: ###0 2n+1 +2 n 111 with M(n) ###### ### Let us now prove tht X M nd Y M re conjugte if nd only if the set H M = {n M(n) } is finite: If H M is finite, then there exists some integer N tht bounds ll its elements. Then there clerly exists conjugcy function F with rdius r F > 2 N N which consists only in shifting right-infinite sequence of ones by 2 n nd dding 2 n ones t the beginning. If H M is infinite, suppose there exists conjugcy function F : Y M X M. First note tht # r F, 0 r F, 1 r F respectively hve #, 0, 1 s imges. If this were not the cse, then the words ###0 k 111 would hve the sme imge imge for k > r F nd this would contrdict the injectivity of F. Now tke n H M such tht 2 n > 2r F +1. The point ###0 2n+1 +2 n 111 hs n imge tht does not belong to the subshift X M becuse it is of the form with w i, w i {#, 0, 1}. ###w 1 w 2rF 0 2n+1 +2 n 2r F w 1 w 2r F 111 Note In the previous proof, it ws only mde use of the injectivity of F for the reciprocl. This will be used in Corollry

11 3. Fctoriztion We will prove here tht fctoriztion is Σ 0 3-complete for SFTs, sofic nd effective subshifts. To do this we will prove tht the upper bound is Σ 0 3 in the effective cse nd tht the SFT cse is Σ 0 3-hrd, thus leding to the completeness result for ll clsses. We strt with two smll exmples to see why fctoriztion is more complex thn conjugcy in the SFT cse. Here the exmples re the simplest possible: we fix the SFT to which we fctor in very simple wy, thus mking the fctor mp known in dvnce. Theorem 3.1. Let Y be the SFT contining exctly one configurtion, uniform configurtion. Given n effective subshift X s n input, it is Π 0 1-complete to know whether X fctors onto Y. Proof. In this cse the fctor mp is forced: it hs to send everything to the only symbol of A Y. And the problem is hence equivlent to knowing whether SFT is not empty, which is Π 0 1-complete. Theorem 3.2. Let Y be the empty SFT. Given n effective subshift X s n input, it is Σ 0 1-complete to know whether X fctors onto Y. Proof. Here ny fctor mp is suitble, the problem is equivlent to knowing whether X is empty, which is Σ 0 1-complete. We study now the hrdness of fctoriztion in the generl cse, tht is to sy when two SFTs re given s inputs nd we wnt to know whether one is fctor of the other. We prove here with Theorems 3.3 nd 3.10 the Σ 0 3-completeness of the fctoriztion problem Fctoriztion is in Σ 0 3 Theorem 3.3. Given two effective subshifts X, Y s n input, deciding whether X fctors onto Y is Σ 0 3. Proof. The subshift X fctors onto Y iff there exists fctor mp F, locl function, such tht F (X) = Y. This forces one existentil quntifier, nd the result follows from the next lemm nd Lemm 2.6 which prove tht deciding whether F (X) = Y is Π 0 2. Lemm 3.4. Given two effective subshifts X, Y nd locl mp F s n input, deciding if Y F (X) is Π 0 2. Proof. We prove here tht the sttement Y F (X), tht is to sy, for every point y Y, there exists point x X such tht F (x) = y, is equivlent to the following Π 0 2 sttement: for ny pttern m, if m is extensible for Y, then F 1 (m) contins n extensible pttern for X. This sttement is Π 0 2 since checking tht m is extensible is Π 0 1. We now prove the equivlence. Suppose tht Y F (X), then ny extensible pttern m of Y ppers in configurtion y Y which hs preimge x X. Thus m hs n extensible preimge in X. This proves the first direction. Conversely, suppose ll extensible ptterns m of Y hve extensible preimges in X. Let y be point of Y, then we hve n incresing sequence m i of extensible ptterns converging to y. All of them hve t lest one extensible preimge m i. By compctness, we cn extrct from this sequence converging subsequence, note x its limit. By construction x is point of X nd preimge of y. 11

12 3.2. Fctoriztion is Σ 0 3-hrd We give two proofs here for the Σ 0 3-hrdness of fctoriztion : one for effective subshifts in dimension one nd one for SFTs in dimension d 2. The proof for SFTs gives us completeness for ll clsses, SFTs, sofic nd effective subshifts, but only for dimensions d 2. Also, the proof for effective subshifts in dimension one gives the ides tht will be refined to get the proof for SFTs Effective subshifts Theorem 3.5. Given two effective subshifts X, Y s n input, it is Σ 0 3-hrd to decide whether X fctors onto Y. Proof. We reduce the problem to COFIN, the set of Turing mchines tht do not hlt on finite set of inputs. Given Turing mchine M, we construct two effective subshifts X M nd Y M such tht X M fctors onto Y M if nd only if the set of inputs on which M does not hlt is finite: X M is defined on the lphbet 2 {#, W, R, 0, 1, B} nd is constituted of the following points nd their orbits: ###W n+1 0BBB with M(n) ###W n+1 1BBB with M(n) W W W bbbb with b {0, 1} ###W W W ###RRR ###### W W W W W W RRRRRR BBBBBB X M is effective since the only complex forbidden ptterns to enumerte re those of the form #W n bb with b {0, 1} with M(n) nd the other forbidden ptterns re the two symbol ones tht follow: 0#, 1#, W #, B#, R#, 0W, 1W, BW, RW, W R, 0R, 1R, BR, 00, #0, R0, 10, B0, 11, #1, R1, 01, B1, #B, W B, RB Y M is defined on the lphbet {#, W, 0, 1, B} nd consists of the following points 2 W stnds for white, B for blue nd R for red. 12

13 nd their orbits: ###0W n+1 BBB with M(n) ###1W n+1 BBB with M(n) ###bw W W with b {0, 1} W W W BBB ###### BBBBBB W W W W W W It is quite cler tht Y M is n effective subshift. We will cll the {0, 1} symbols decortions: they will pper t most once in configurtion nd only in some configurtions of X M. Hence when we will tlk bout the decortion of configurtion, we will men the only {0, 1} symbol of this configurtion. Now let us check tht X M fctors onto Y M if nd only if the set of inputs on which M does not hlt is finite: Suppose the number of inputs on which M does not hlt is finite, then there exists n upper bound N on ll of these inputs. We my tke fctor mp of rdius N + 3, which shifts the decortion tht is t the end of the W s to the beginning (i.e. just fter the # symbols). It lso mps the ###W W W configurtion to the ###0W W W configurtion nd the ###RRR configurtion to the ###1W W W configurtion, we cn deduce the other mppings esily from these. Note tht we hd to dd the configurtion ###RRR to X M in order to pllite to the missing decortion when the word W W becomes infinite, i.e. the limit cse. Suppose the number of inputs on which M does not hlt is infinite nd tht there exists fctor mp F from X M onto Y M of rdius r. Since there re points of the form ###bw n+1 BBB in Y M for n s tht re rbitrrily lrge, the r-blocks # #, W W nd B B must hve #, W nd B s imges. But since the fctor mp is of rdius r, then for ll n > r such tht M(n), the points ###W n+1 0BBB nd ###W n+1 1BBB of X M necessrily hve the sme imge: ###bw n+1 BBB. Thus, the points ###bw n+1 BBB of Y M hve no preimge in X M by F nd F cnnot be fctor mp Finite type The ide of the proof for SFTs of dimension d 2 is similr, but this time we hve to explicitly encode the computtions in the SFT, since there is no wy nymore to "hide" them in the forbidden ptterns. To do this, we use simpler version of the construction introduced in [9], which could ctully hve lredy been used in it 3. This construction introduced new wy to put 3 It would in prticulr hve led to lower constnt in Theorem

14 Figure 2: The tileset T. Turing mchine computtions in SFTs. In prticulr, the bse construction hs exctly one point (up to shift) in which computtions my be encoded. It will in prticulr behve nicely when submitted to locl mps. T-structures. We construct in this section n SFT in which exctly one configurtion up to shift forms sprse grid. It hs the property tht whenever two subshifts re bsed on it, nd one fctors onto or embeds into the other one, then the grid configurtions hve to be mpped to grid configurtions. This is chieved by using the tileset T of Figure 2. First, note tht the configurtion represented in figure 3, tht we will cll α, is in T. Configurtion α forms grid in which we cn encode computtions s we will see lter. We will lso see tht it is the only one contining n infinite grid nd how to encode computtions in it. Before proving nything, let us set the vocbulry tht will be used to describe the SFT X T. Tile 30 is the corner tile. Tiles 20 nd 27 re the end tiles. 14

15 Figure 3: Configurtion α: the unique vlid tiling by T in which there re 2 or more verticl lines. Tiles 30, 32, 33 nd 34 re the strt tiles. A horizontl line is connex horizontl lignment of tiles contining verticl blck line (tiles 5, 6, 7, 17, 21, 24, 25, 26, 31, 35, 36, 37). It my be terminted by strt tiles on its left nd by end tiles on its right. A verticl line is connex verticl lignment of tiles contining blck or blue verticl line (tiles 13, 14, 15, 16, 18, 19, 22, 23, 28, 32, 33, 34, 38). It my be terminted on top by tiles 5, 21, 26 nd on bottom by tiles 6, 20, 25, 27, 30, 36, 37. A digonl is connex digonl lignment (positions (i, j), (i + 1, j + 1),... ) of tiles mong 4, 11, 12. A squre of size k is n extensible pttern of support 0, k +1 2 such tht {0} 1, k nd {k + 1} 1, k re verticl lines nd 1, k {0} nd 1, k {k + 1} horizontl lines. A squre does not contin horizontl/verticl line in the subpttern of support 1, k 2. Remrk tht the colors bove nd below horizontl line differ nd tht this forces squre to contin digonl t positions (i, i), for 0 < i < k+1 nd counting signl (see next bullet point) somewhere in between the two horizontl lines. A row of squres is horizontl lignment of squres. A counting signl is connex pth of tiles mong 3, 7, 10, 12, 14, 19, 22, 32, 33, 38, such tht the red signl is connected. It my be strted (on the left) only by tiles 30, 32 nd ended (on the right) by tiles 7, 21. The counting signl counts the 15

16 number of squres on ech row nd forces this number to be exctly the height of these squres. An increse signl is formed by pth of tiles mong 7, 14, 22, 23, 24, 25, 26, 27, 30, 31, 32, 36, 38, such tht the blue signl is connected. This signl forces squres to increse their size by exctly one on the row right bove. And thus to increse their number by one lso. Let us first notice tht whenever the corner tile ppers in point, this point is necessrily shifted copy of α, the point of figure 3: the corner tile forces tile 33 to pper bove it nd tiles 31 nd then 27 to pper on its right. These tiles enforce the existence of the first squre of size 1, i.e. the first row. The increse signl forces the first squre of the row bove to be of size 2, nd so on... Lemm 3.6. SFT X T dmits t most one point, up to trnsltion, with two or more verticl lines. This is the α configurtion represented in figure 3. Proof. Let x be point contining two horizontl lines, these two lines necessrily fce ech other: either they re infinite, or they end on the left, in which cse they end on the sme column, strt tiles being ll in the sme column becuse of the color on their left. We my suppose tht there is no other horizontl line in between nd tht they re t distnce k + 1. Since the sides bove nd below horizontl line do not hve the sme color, there must necessrily be one or more digonls in between. Ech digonl forces verticl lines thus forming squres of size k. Furthermore, these squres re necessrily cut horizontlly by counting signl, which moves bove exctly once ech time it crosses verticl line. This gurntees tht there re exctly k squres in this row nd thus tht it is not infinite. The increse signl necessrily ppers on the verticl line formed by the right side of the bottommost squre nd forces the existence of squres of size k + 1 in the row bove nd of size k 1 1 in the row below. The increse signl lso forces squres to pper. The corner tile will pper t the bottom left corner of the only squre of size 1 of the bottommost row. One my notice tht if the corner tile ppers t position ( (0, 1) ) then there re horizontl lines of length (k + 1)k + 1 which strt t positions : theses lines form the 0, k(k+1) 2 bottom border of row of k squres, one my lso see tht the increse signl drws prbol. Lemm 3.7. Besides the α configurtion, the SFT X T contins points formed of uniform zones (one tile only) except for three infinite strips of finite width : verticl strip, horizontl strip nd digonl SW-NE strip. See figure 5. 16

17 k,j bi A ZZ c k,i,j f i,j Rnk 1 d i Rnk 2 e i g i,j h k,i B l k C D m k,i n i,j o i Rnk 3 p k E r i s i t i u i v i,j Rnk 4 x i F G H y i z i Rnk 5 I J K K L M N O P Q R S Rnk 6 T U V X Y Z Rnk 17 7 Figure 4: The configurtions of X T \ {α}, the subscripts indicte tht verticl lines, horizontl lines nd signls my be t different distnces.

18 Figure 5: Uniform qurter- nd eighth-plnes in non-α-configurtions. Proof. Lemm 3.6 sttes tht there is one configurtion t most (up to shift) tht hs two horizontl lines or more. The other configurtions necessrily hve one of the following shpes: There is horizontl line, in which cse there cn be t most one verticl line bove nd/or one verticl line below, otherwise there would necessrily be squres nd hence two horizontl lines. There cn lso be counting signl rbitrrily fr bove nd/or below. There is no horizontl line, then there is t most one verticl line. An increse signl my gin pper on the left nd/or on the right. All the points of X T re shown in figures 3 nd 4. Corollry 3.8. SFT X T single Turing mchine. is countble nd ll its configurtions re computble from Let s see now how to encode computtions inside the α configurtion. First note tht on ech squre s bottom line, there is t most one verticl line ending: suppose tht the corner tile is t bsciss 0, then on the k th row, contining exctly k squres of size k, the verticl lines re t bsciss i(k + 1), for 0 i k nd in the row bove the verticl lines re t bsciss i(k + 2) with 0 i k + 1. Since (i 1)(k + 2) < i(k + 1) < i(k + 2) for 0 < i k + 1, this is true for ll squres, except for the leftmost ones for which the first verticl line is the sme. So one my see α s grid, for which the number of intersections increses of 1 for ech row, see Figure 6. Now, if we wnt to encode computtions in T, we cn use clssicl encoding of Turing mchines s Wng tiles, with the tile strting the computtion on the corner. Since the grid grows, the Turing mchine will never run out of spce. Our reductions will use SFTs bsed on this construction, they will be feture different tilings on its grid. We qulify n SFT which is bsiclly T with tiling on its grid s hving T -structure. Definition 3.1 (T -structure). We sy n SFT X hs T -structure if it is copy of T to which we superimposed new symbols only on the symbols representing the horizontl/verticl lines nd their crossings. 18

19 Figure 6: How the grid of α my be seen s regulr grid. In prticulr, one cn see how informtion my be trnsmitted from one intersection to its neighbors. Note tht n SFT my hve T -structure while hving no α-configurtion: for instnce if you put encode computtions of Turing mchine tht lwys hlts nd produce n error when it hlts. The reduction. The next lemm sttes very intuitive result, tht will be used lter, nmely tht if n SFT with T -structure fctors to nother one, then the structure of ech point is preserved by fctoriztion. Furthermore, it shows tht the fctor mp cn only send cell to its corresponding one, tht is to sy cell of the preimge hs to be in the window of the imge. Lemm 3.9. Let X, Y be two SFTs with T -structure, such tht X fctors onto Y. Let r be the rdius of the fctor mp, then ny α-configurtion of Y is fctored on by n α-configurtion of X shifted by v, with v r. Proof. By Lemm 3.7 we know tht non-α configurtions hve two uniform (sme symbols everywhere) qurter-plnes nd four uniform eighth-plnes, s seen on Figure 7. The two north est eighth-plnes re not uniform in configurtion α. Thus these configurtions cnnot be fctored on α. It remins to prove the second prt: tht in the fctoring process the α-structure is t most shifted by the rdius of the fctoriztion. We do tht by reductio d bsurdum, suppose tht n α-configurtion x of X is mpped to n α-configurtion y of Y nd shifts it by v = (v x, v y ), with v > r. Without loss of generlity we my suppose tht v x > r nd v y > 0 nd tht the corner tile of the preimge is t position (0, 1). We re now going to show tht this is not possible. For ll k N there is squre with lower left corner t (2k 2 +k, 2k 2 +k), see Figure 8 on the left. Inside this squre, there re two (k 1) (k 1) uniform smller squre subptterns, see Figure 8 on the right. Now tke k such tht k > ( v + 2r + 1). By hypothesis, there is verticl line symbol t t z p = (2k 2 + 2k + 1, 2k 2 + k) on x, nd thus 19

20 r h r h r v v r () The uniform qurters nd eights of plnes of the non-α configurtions. The horizontl nd verticl distnces h nd v between these plnes correspond to the distnces between lines nd signls. (b) The imge by F of these uniform subplnes still is uniform, note tht it hs been mputted of strips of width r, the rdius of F. Figure 7: The imges of non-α configurtions re necessrily non-α configurtions. The white zones represent the zones which re not uniform. t z i = (2k 2 + 2k v x, 2k 2 + k + v y ) on y. We know x zi+b r hs imge t, nd by wht precedes tht x zi+b r = x zi+(1,0)+b r since they re both uniform, therefore, there should be two t symbols next to ech other in y t z i nd z i + (1, 0). This is impossible. Theorem Given two SFTs X, Y s n input, deciding whether X fctors onto Y is Σ 0 3-hrd. For this proof, we will reduce from the problem COFIN, which is known to be Σ 0 3- complete, see Kozen [11]. COFIN is the set of Turing mchines which run infinitely only on finite set of inputs, s stted erlier. Proof. Given Turing mchine M, we construct two SFTs X M nd Y M such tht X M fctors on Y M iff the set of inputs on which M does not hlt is finite. We first introduce n SFT Z M on which both will be bsed. It will hve T structure. Above the T bse, we llow the cells of the grid to be either white or blue ccording to the following rules: All cells on sme horizontl line re of the sme color. A blue horizontl line my be bove white horizontl line, but not the contrry. We now llow computtion on blue cells only. The Turing mchine M is lunched on the input formed by the size of the first blue line (in number of cells). We forbid the mchine to hlt. 20

21 k 1 k 1 2k k (2k 2 + k, 2k 2 + k) k 1 Figure 8: For every k N, the squre strting t position (2k 2 + k, 2k 2 + k) is of the form on the right. We cn see tht there re two uniform (k 1) (k 1) squre subptterns t (2k 2 + 2k + 2, 2k 2 + k + 1) nd (2k 2 + k + 1, 2k 2 + 2k + 2) respectively. n Figure 9: Computtion on input n in the SFT Z, the blue zone contins the computtion, nd its distnce from the corner tile corresponds to the input. So for ech n on which M does not hlt, there is configurtion with white cells until the first blue digonl ppers, then computtion occurs inside the blue cone, see Figure 9 for schemtic view. If M hlts on n, then there is no configurtion where the first blue line codes n. By compctness, there is of course configurtion with only white lines. If M is totl, then the only α-configurtion in Z M is the one with no blue horizontl lines. Now from Z M, we cn give X M nd Y M : X M : Let Z M be copy of Z M to which we dd two decortions 0 nd 1 on the blue cells only, nd ll blue cells in configurtion must hve the sme decortion. Now X M is Z M to which we dd third color, red, tht my only pper lone, insted of white nd blue (one cn see this s dding copy of the configurtions with only white horizontl lines). No computtion is superimposed on red. Y M is copy of Z M where we decorted only the corner tile with two symbols 0 nd 1. We now check tht X M fctors onto Y M iff M does not hlt on finite set of inputs: Suppose the set of inputs on which M does not hlt is finite: there exists N such tht M hlts on every input greter thn N. The following fctor mp F works: 21

22 F is the identity on Z M. Note tht the dditionl copy of T is lso sent to the component Z M. F hs rdius big enough so tht if its window is centered on the corner tile, it would cover the beginning of computtion on input N, tht is r F > N 2 + N. An α-configurtion x of X M is sent on the sme α-configurtion y in Y M. For the decortions, when there is computtion on x, the fctor mp cn see it nd gives the sme decortion to the corner tile of y. When there is no computtion, the fctor mp doesn t see computtion zone nd gives decortion 0 to the corner tile. The configurtion with only white digonls nd decortion 1 of Y M is fctored on by the α-configurtion colored in red contined in X M. Note tht this lso works when M is totl. Conversely, suppose M does not hlt on n infinite set of inputs, nd tht there exists fctor mp F with rdius r: Lemm 3.9 sttes tht ll α-configurtions of Y M re fctored on by α-configurtions of X M. Now, there is n infinite number of α-configurtions with corner tile decorted with 0 (resp. 1) in Y M, they ll must be fctored on by some α-configurtion of X M. Still by Lemm 3.9, the corner tile of the preimge must be in the window of the corner tile of the imge. However, there cn only be finite number of configurtions in which the symbols in this window differ. So the α-configurtions of X M fctor to finite number of α-configurtions of Y M with one of the decortions. This is impossible. Note tht the construction of X M nd Y M from the description of M is computble nd uniform. The reduction is thus mny-one. 4. Embedding 4.1. SFTs We prove now Theorem 0.5 stting tht the embedding problem for SFTs is Σ 0 1- complete for SFTs. We strt with n nlogue of Lemm 3.9 : Lemm 4.1. Let X, Y be two SFTs with T -structure, such tht X embeds into Y. Let r be the rdius of the embedding, then ny α-configurtion of X is mpped to n α-configurtion of Y shifted by v, with v r. Proof. First note tht the uniform points of X must be mpped to uniform points of Y. So ll different uniform points, nd thus ll uniform ptterns of support B r, hve different imges. Now n α-configurtion of X hs rbitrrily lrge uniform res, s seen in Lemm 3.9, see lso Figure 8. These uniform res lternte, so their imge lso lterntes when they re sufficiently lrge. The only configurtions tht hve incresingly lrge lternting uniform res re α-configurtions. So α-configurtions of X re mpped to α-configurtions of Y. The proof tht these mppings do not shift the T -structure by more thn r is exctly the sme s in Lemm 3.9. Lemm 4.2. Let X nd Y be two SFTs, it is Σ 0 1 to check whether X embeds into Y. 22

23 Proof. To decide whether X embeds into Y, we hve to check if there exists n injective locl function F : X Y. Such function being locl, it cn be guessed with first order existentil quntifier. To check tht it is n embedding, we hve to check tht F (X) Y nd tht for ll x 1, x 2 X, x 1 x 2 F (x 1 ) F (x 2 ). We know from Lemm 2.1 tht checking F (X) Y is Σ 0 1. We now show tht the second prt is lso Σ 0 1 by showing tht the two following sttements re equivlent. There exist x 1, x 2 X such tht x 1 x 2 nd F (x 1 ) = F (x 2 ). For ll r > mx(r F, r X ), there exist two r-blocks M 1 nd M 2 such tht M 1, M 2 re extensible nd (M 1 ) 0 (M 2 ) 0 nd F (M 1 ) = F (M 2 ). It is cler tht the second sttement is Π 0 1 nd tht the first sttement is the negtion of the definition of injectivity. Now to the proof of their equivlence: Suppose there exist two different points x 1, x 2 X such tht F (x 1 ) F (x 2 ), we my ssume x 1 nd x 2 differ in 0 by shifting. For ll r > mx(r F, r X ), the centrl r-blocks M 1, M 2 of x 1, x 2 re extensible nd differ in 0 Suppose now tht for ll r > mx(r F, r X ) there exist two extensible r-blocks M r 1, M r 2 differing in 0 nd such tht F (M r 1 ) = F (M r 2 ). By the pigeonhole principle, there is n infinity of M r 1 which hve the sme symbol in 0 nd thus of M r 2 without this symbol in 0. Tke these subsequences of M r 1 nd M r 2, by compctness we cn extrct converging subsequences from them which converge to two points x 1, x 2 X with different symbols in 0. These two points hve the sme imge, by construction. Lemm 4.3. Given two SFTs X, Y s n input, deciding whether X embeds into Y is Σ 0 1-hrd. We will use reduction from the hlting problem, the set of Turing mchines tht hlt on blnk input, nd construction bsed on T -structure, s before. Proof. Given Turing mchine M, we construct two SFTs X M nd Y M such tht X M embeds into Y M iff the Turing mchine M hlts. Both SFTs hve s bse n SFT Z M with T -structure, in which we encode computtions of M. Let us describe Z M : Z M is only T on which we directly encode the computtion of M, it my eventully rech hlting stte in which cse the remining spce is given new color, sy blue. So our SFT Z M cn tke two different forms : if the mchine M hlts, then blue zone ppers, if it does not hlt, then this zone does not pper. Now X M is Z M for which we dd decortion to the corner tile, 0 or 1, so there re two different grid points in ny cse, whether the mchine M hlts or not. Y M is Z M for which we dd decortion to the hlting stte only (it ppers t most once), there re two different grid points only when the mchine M hlts. Let us check now tht X M embeds into Y M if nd only if M hlts. 23

24 When the mchine M hlts, X M embeds into Y M : the rdius of the embedding r is the distnce between the hlting stte nd the corner, the decortion of the corner is just trnslted to the hlting stte. All the rest remins unchnged. Note tht there re less non α-configurtions in X M thn in Y M : they re the sme except for the configurtions contining exctly one horizontl line nd two verticl lines with hlting stte t their crossing. They hve different decortions in Y M but not in X M. When the mchine M does not hlt, there re two different α-configurtions in X M up to shift, while there is only one in Y M, so they must hve the sme imge Effective nd sofic shifts Lstly, we prove Theorem 0.6: Lemm 4.4. It is Σ 0 3 to decide given two effective subshifts X, Y whether X embeds in Y. Proof. To decide whether X embeds into Y one needs to decide whether there exists F such tht F (X) Y nd for ll x 1, x 2 X, x 1 x 2 F (x 1 ) F (x 2 ). Guessing F nd checking whether F (X) Y is Σ 0 3, s consequence of Lemm 2.6, while checking the injectivity prt remins Σ 0 1: s for the SFT cse, this is equivlent to negting the following sttement which remins Σ 0 1 in the effective cse: For ll r > mx(r F, r X ), there exist two r-blocks M 1 nd M 2 such tht M 1, M 2 re extensible nd (M 1 ) 0 (M 2 ) 0 nd F (M 1 ) = F (M 2 ). And corollry of the proof of Theorem 2.11, we obtin the Σ 0 3-hrdness for effective subshifts. Corollry 4.5. Given X, Y two effective subshifts, it is Σ 0 3-hrd to decide whether X embeds in Y. [1] Aubrun, N., Sblik, M., Simultion of effective subshifts by two-dimensionl subshifts of finite type. Act Applicnde Mthemtice. [2] Berger, R., The Undecidbility of the Domino Problem. Ph.D. thesis, Hrvrd University. [3] Boyer, L., Theyssier, G., On fctor universlity in symbolic spces. In: MFCS. pp [4] Boyle, M., Lower entropy fctors of sofic systems. Ergodic Theory nd Dynmicl Systems 3, [5] Boyle, M., Open Problems in Symbolic Dynmics. Contemporry Mthemtics 469, [6] Hedlund, G. A., Endomorphisms nd utomorphisms of the shift dynmicl system. Theory of Computing Systems 3 (4), [7] Hochmn, M., A note on universlity in multidimensionl symbolic dynmics. Discrete nd Continuous Dynmicl Systems S 2 (2). [8] Hochmn, M., Apr On the dynmics nd recursive properties of multidimensionl symbolic systems. Inventiones Mthemtice 176 (1), [9] Jendel, E., Vnier, P., Π 0 1 sets nd tilings. In: Theory nd Applictions of Models of Computtion (TAMC). Vol of Lecture Notes in Computer Science. pp