An Automaton Group with PSPACE-Complete Word Problem
aa r X i v : . [ c s . F L ] J un An Automaton Group with
PSPACE -Complete Word Problem
Jan Philipp Wächter and Armin Weiß
Institut für Formale Methoden der Informatik (FMI)Universität StuttgartUniversitätsstraße 3870569 Stuttgart, Germany
June 11, 2019
Abstract.
We construct an automaton group with a
PSPACE -completeword problem, proving a conjecture due to Steinberg. Additionally, the con-structed group has a provably more difficult, namely
EXPSPACE -complete,compressed word problem. Our construction directly simulates the computa-tion of a Turing machine in an automaton group and, therefore, seems to bequite versatile. It combines two ideas: the first one is a construction used byD’Angeli, Rodaro and the first author to obtain an inverse automaton semi-group with a
PSPACE -complete word problem and the second one is to utilizea construction used by Barrington to simulate circuits of bounded degree andlogarithmic depth in the group of even permutations over five elements.
The word problem is one of Dehn’s fundamental algorithmic problems in group theory[11]: given a word over the generators of a finitely generated group, the question iswhether this word represents the identity in the group. While, in general, the wordproblem is undecidable [20, 7], there are many classes of groups having a decidable wordproblem. Among them is the class of automaton group.In this context, the term automaton refers to finite state, letter-to-letter transducers.In such automata, every state q induces a length-preserving, prefix-compatible actionon the set of words, where an input word u is mapped to the output word obtained byreading u starting in q . The group or semigroup generated by the automaton is theclosure under composition of the actions of the different states and a (semi)group arisingin this way is called an automaton (semi)group .The interest in automaton groups was stirred by the observation that many groups withinteresting properties arise as automaton groups. Most prominently, the class contains1he famous Grigorchuk group (which is the first example of a group with sub-exponentialbut super-polynomial growth and admits other peculiar properties, see [15] for an ac-cessible introduction). There is also a quite extensive study of algorithmic problems inautomaton (semi)groups: the conjugacy problem and the isomorphism problem (here theautomaton is part of the input) – the other two of Dehn’s fundamental problems – areundecidable for automaton groups [25] . For automaton semigroups, the order problemcould be proved to be undecidable [13, Corollary 3.14]. Recently, this could be extendedto automaton groups [14] (see also [4]). On the other hand, the undecidability result forthe finiteness problem for automaton semigroups [13, Theorem 3.13] could not be liftedto automaton groups so far. Similarly, the freeness problem is known to be undecidablefor automaton semigroups [8] but nothing is known in the group case.The undecidability results show that the presentation of groups using automata is stillquite powerful. Nevertheless, it is not very difficult to see that the word problem forautomaton groups is decidable. One possible way is to show an upper bound on thelength of an input word on which a state sequence not representing the identity of thegroup acts non-trivially. In the most general setting, this bound is | Q | n where Q is thestate set of the automaton and n is the length of the state sequence. Another viewpointis that one can use a non-deterministic guess and check algorithm to solve the wordproblem. This algorithm uses linear space proving that the word problem for automaton(semi)groups is in PSPACE . This approach seems to be mentioned first by Steinberg [24,Section 3] (see also [10, Proposition 2 and 3]). In some special cases, better algorithmsor upper bounds are known: for example, for contracting automaton groups (and thisincludes the Grigorchuk group), the witness length is bounded logarithmically [19] andthe problem, thus, in
LOGSPACE ; other examples of classes with better upper bounds oralgorithms include automata with polynomial activity [5] or Hanoi Tower groups [6]. Onthe other hand, Steinberg conjectured that there is an automaton group with a
PSPACE -complete word problem [24, Question 5]. As a partial solution to his problem, an inverseautomaton semigroup with a
PSPACE -complete word problem has been constructed in[10, Proposition 6] . In this paper, our aim is to finally prove the conjecture for groups.In order to do so, we adopt the construction used by D’Angeli, Rodaro and the firstauthor from [10, Proposition 6]. This construction uses a master reduction and directlyencodes a Turing machine into an automaton. Already in [10, Proposition 6], it wasalso used to show that there is an automaton group whose word problem with a rationalconstraint (which depends on the input) is PSPACE -complete. To get rid of this rationalconstraint, we apply an idea used by Barrington [3] to transform NC -circuits (circuitsof bounded fan-in and logarithmic depth) into bounded-width polynomial-size branchingprograms. Similar ideas predating Barrington have been attributed to Gurevich (see[17]) and given by Mal’cev [18]. Nevertheless, this paper is fully self-contained and noprevious knowledge of either [10] or [3] is needed. In order to avoid ambiguities, we call a word over the states of the automaton a state sequence . So, inour case the input for the word problem is a state sequence. In fact, the semigroup is generated by a partial, invertible automaton. A priori, this seems to be astronger statement than that the semigroup is inverse and also an automaton semigroup. That iswhy the cited paper uses the term “automaton-inverse semigroup”. Only later, it was shown that thetwo concepts actually coincide [9, Theorem 25].
2n addition, we also investigate the compressed word problem for automaton groups.Here, the (input) state sequence is given as a so-called straight-line program (a context-free grammar which generates exactly one word). By uncompressing the input sequenceand applying the above mentioned non-deterministic linear-space algorithm, one cansee that the compressed word problem can be solved in
EXPSPACE . Thus, the moreinteresting part is to prove that this algorithm cannot be improved significantly: we showthat there is an automaton group with an
EXPSPACE -hard compressed word problem.This result is interesting because, by taking the direct product, we obtain a group whose(ordinary) word problem is
PSPACE -complete and whose compressed word problem is
EXPSPACE -complete and, thus, provably more difficult. To the best of our knowledge,this is the first example of a group for which this is possible.Explicit previous results on the compressed word problem for automaton groups donot seem to exist. However, it was observed by Gillibert [12] that the proof of [10, Propo-sition 6] also yields an automaton semigroup with an
EXPSPACE -complete compressedword problem in a rather straightforward manner. For the case of groups, it is possibleto adapt the construction used by Gillibert to prove the existence of an automaton groupwith an undecidable order problem [14] slightly to obtain an automaton group with a
PSPACE -hard compressed word problem [12].
Words and Alphabets with Involution.
We use common notations from formal lan-guage theory. In particular, we use Σ ∗ to denote the set of words over an alphabet Σ including the empty word. If we want to exclude the empty word, we write Σ + .For any alphabet Q , we define a natural involution between Q and a disjoint copy Q − = { q − | q ∈ Q } of Q : it maps q ∈ Q to q − ∈ Q − and vice versa. In partic-ular, we have ( q − ) − = q . The involution extends naturally to words over Q ∪ Q − :for q , . . . , q n ∈ Q ∪ Q − , we set ( q n . . . q ) − = q − . . . q − n . This way, the involution isequivalent to taking the group inverse if Q is a generating set of a group. Turing Machines and Complexity.
We assume the reader to be familiar with basicnotions of complexity theory such as configurations for Turing machines, computationsand reductions in logarithmic space as well as complete and hard problems for
PSPACE and the class
EXPSPACE . See [21] or [2] for standard text books on complexity theory.We only consider deterministic, single-tape machines and write their configurations asword γ . . . γ i − pγ i . . . γ n where the γ j are symbols from the tape alphabet and p is astate. In this configuration, the machine is in state p and its head is over the symbol γ i .Using suitable normalizations, we can assume that every Turing machine admits asimple function which describes its transitions: Fact 1 (Folklore) . Consider a deterministic Turing machine with state set P and tapealphabet ∆ . After a straightforward transformation of the transition function and states, This follows immediately from the space hierarchy theorem [23, Theorem 6] (or e. g. [2, Theorem 4.8]). e can assume that the symbol γ ( t +1) i at position i of the configuration at time step t + 1 only depends on the symbols γ ( t ) i − , γ ( t ) i , γ ( t ) i +1 ∈ Γ at position i − , i and i + 1 at time step t . Thus, we may always assume that there is a function τ : Γ → Γ with Γ = P ⊎ ∆ mapping the symbols γ ( t ) i − , γ ( t ) i , γ ( t ) i +1 ∈ Γ to the uniquely determined symbol γ ( t +1) i for all i and t .Proof idea. The only problem appears if the machine moves to the left: if we have thesituation abpc or abpd and the machine moves to the left in state p when reading a c but does not move when reading a d , then the new value for the second symbol does notonly depend on the symbols right next to it; we can either be in the situation ap ′ bc ′ or abp ′ d ′ . To circumvent the problem, we can introduce intermediate states. Now, insteadof moving to the left, we go into an intermediate state (without movement). In the nextstep, we move to the left (but this time the movement only depends on the state and noton the current symbol). Group Theory and A . For elements h and g of a group G , we write g h for the con-jugation h − gh of g with h and [ h, g ] for the commutator h − g − hg . For the neutralelement of a group, we write . We write p = G q or p = q in G if two words p and q over the generators (and their inverses) of a group evaluate to the same group element.With A we denote the alternating group of degree five, i.e. the group of even permu-tations of five elements. It was used by Barrington to convert logical circuits of boundedfan-in and logarithmic depth (so-called NC -circuits ) to bounded-width, polynomial-sizebranching programs. We will not require this actual result (or knowledge of the involvedconcepts) in the following, but we will make heavy use of the next lemma and the ideato use iterated commutators, which we will outline below. Lemma 2 (see Lemma 1 and 3 of [3]) . There are σ, α, β ∈ A such that σ = [ σ β , σ α ] .Proof. Set σ = (13254) , α = (23)(45) and β = (245) .From now on, σ , α and β will refer to those mentioned in Lemma 2 (if not explicitlystated otherwise). Word Problem.
The word problem of a group G generated by a finite set Q is thedecision problem Constant: the group G Input: q ∈ ( Q ∪ Q − ) ∗ Question: is q = in G ?In addition, if C is a class of groups, we also consider the uniform word problem for C .Here, the group G ∈ C is part of the input (in a suitable representation). Automata.
We use the word automaton to denote what is more precisely called aletter-to-letter, finite state transducer. Formally, an automaton is a triple T = ( Q, Σ , δ ) consisting of a finite set of states Q , an input and output alphabet Σ and a set δ ⊆ × Σ × Σ × Q of transitions . For a transition ( p, a, b, q ) ∈ Q × Σ × Σ × Q , we usually usethe more graphical notation p q a/b and, additionally, the common way of depictingautomata p qa/b where a is the input and b is the output . We will usually work with deterministic and complete automata, i. e. automata where we have d p,a = (cid:12)(cid:12)(cid:12) { p q a/b ∈ δ | b ∈ Σ , q ∈ Q } (cid:12)(cid:12)(cid:12) = 1 for all p ∈ Q and a ∈ Σ . In other words, for every a ∈ Σ , every state has exactly onetransition with input a .A run of an automaton T = ( Q, Σ , δ ) is a sequence q q . . . q n a /b a /b a n /b n of transitions from δ . It starts in q and ends in q n . Its input is a . . . a n and its output is b . . . b n . If T is complete and deterministic, then, for every state q ∈ Q and every word u ∈ Σ ∗ , there is exactly one run starting in q with input u . We write q ◦ u for its outputand q · u for the state in which it ends. This notation can be extended to multiple states.To avoid confusion, we usually use the term state sequence instead of “word” (which wereserve for input or output words) for elements q ∈ Q ∗ . Now, for states q , q , . . . , q ℓ ∈ Q ,we set q ℓ . . . q q ◦ u = q ℓ . . . q ◦ ( q ◦ u ) inductively. If the state sequence q ∈ Q ∗ is empty,then q ◦ u is simply u .This way, every state q ∈ Q (and even every state sequence q ∈ Q ∗ ) induces a map Σ ∗ → Σ ∗ and every word u ∈ Σ ∗ induces a map Q → Q . If all states of an automatoninduce bijective functions, we say it is invertible and call it a G -automaton . For a G -automaton T , all bijections induced by the states generate a group (with composition asoperation), which we denote by G ( T ) . A group is called an automaton group if it arisesin this way. Clearly, G ( T ) is generated by the maps induced by the states of T and,thus, finitely generated. Example 3.
The typical first example of an automaton generating a group is the addingmachine T = ( { q, id } , { , } , δ ) : q id1 / / / / It obviously is deterministic and complete and, therefore, we can consider the map in-duced by state q . We have q ◦
000 = q ◦
100 = q ◦
010 = 110 . From this example, itis easy to see that the action of q is to increment the input word (which is interpretedas a reverse/least significant bit first binary representation ←− bin( n ) of a number n ). Theinverse is accordingly to decrement the value. As the other state id acts like the identity,we obtain that the group G ( T ) generated by T is isomorphic to the infinite cyclic group.5 , . . . a ,m q , q , . . . q ,m − q ,m a , a ,m ... ... ... ... a n − , a n − ,m q n, q n, . . . q n,m − q n,m a n, . . . a n,m (a) Multiple combined cross diagrams u p q v (b) Abbreviated cross diagram Similar to extending the notation q ◦ u to state sequences, we can also extend thenotation q · u . For this, it is useful to introduce cross diagrams , another notation fortransitions of automata. For a transition p q a/b of an automaton, we write aq pb .Multiple cross diagrams can be combined into a larger one. For example, the crossdiagram in Figure 1a indicates that there is a transition q i,j − q i,ja i − ,j /a i,j for all ≤ i ≤ n and ≤ j ≤ m . Typically, we omit unneeded names for states and abbreviatecross diagrams. Such an abbreviated cross diagram is depicted in Figure 1b. If we set q n, . . . q , = p , u = a , . . . a ,m , v = a n, . . . a n,m and q = q n,m . . . q ,m , then it indicatesthe same transitions as the one in Figure 1a. It is important to note here, that theright-most state in p is actually the one to act first.If we have the cross diagram from Figure 1b, we set p · u = q . This is the same, assetting q n . . . q · u = q n . . . q · ( q ◦ u )( q · u ) inductively and, with the definition fromabove, we already have p ◦ u = v .Normally, we cannot simply re-order the rows of a cross diagram as the output interferesand we could get into different states. However, we can clearly re-order rows if they actlike the identity: Fact 4.
Let T = ( Q, Σ , δ ) be a deterministic automaton, w ∈ Σ ∗ and q , . . . , q ℓ ∈ Q ∗ such that q ◦ w = · · · = q ℓ ◦ w = w .Then, for any p = p k · · · p ∈ { q , . . . , q ℓ } ∗ , we have p · w = ( p k · w ) · · · ( p · w ) . We start by showing that the uniform word problem for automaton groups is
PSPACE -complete. Although this also follows from the non-uniform case proved below, it uses thesame ideas but allows for a simpler construction. This way, it serves as a good startingpoint and makes understanding the more complicated construction below easier.6 alanced Iterated Commutators.
We lift the notation g h for conjugation and [ h, g ] forthe commutator from groups to words over the generators: for an alphabet Q and words p , q ∈ Q ∗ , we write p q = q − pq and [ q , p ] = q − p − qp using the natural involution for Q ∪ Q − .We also need a balanced version of an iterated commutator; in fact, it will be crucialto our constructions. Definition 5.
Let Q be an alphabet and α, β ∈ ( Q ∪ Q − ) ∗ . For g t , . . . , g ∈ ( Q ∪ Q − ) ∗ ,we inductively define the word B β,α [ g t , . . . , g ] by B β,α [ g ] = g B β,α [ g t , . . . , g ] = h B [ g t , . . . , g ⌊ t ⌋ +1 ] β , B [ g ⌊ t ⌋ , . . . , g ] α i . Lemma 6.
On input of g t , . . . , g , one can compute B β,α [ g t , . . . , g ] in logarithmic space.Proof. We give a sketch for a (deterministic) algorithm which computes the symbol atposition i of B β,α [ g t , . . . , g ] in logarithmic space. For simplicity, we only describe thecase when t = 2 k for some k . Then, we have B β,α [ g t , . . . , g ] = β − B [ g t , . . . , g t +1 ] − β α − B [ g t , . . . , g ] − αβ − B [ g t , . . . , g t +1 ] β α − B [ g t , . . . , g ] α and the length ℓ ( t ) (as a word over α , β , the g i and their inverses) of B β,α [ g t , . . . , g ] isgiven by ℓ (1) = 1 and ℓ ( t ) = 8 + 4 ℓ ( t ) . This yields ℓ ( t ) = k − X i =0 i · ! + 4 k = 8 4 k −
13 + 4 k = 83 ( t −
1) + t = 113 t − and, thus, that the length of B β,α [ g t , . . . , g ] is polynomial in t . Therefore, we can iteratethe above algorithm for all positions ≤ i ≤ ℓ ( t ) to output B β,α [ g t , . . . , g ] entirely.To compute the symbol at position i , we first check whether i is the first or last position(notice that we need the exact value of ℓ ( t ) for testing the latter). In this case, we knowthat it is β − or α . Similarly, we can do this for the positions in the middle and at oneor three quarters. If the position falls into one of the four recursion blocks, we use twopointers into the input: left and right . Depending on the block, left and right eitherpoint to g and g t or to g t +1 and g t . Additionally, we also store whether we are in aninverse block or a non-inverse block. From now on, we disregard the input left of left and right of right (and do appropriate arithmetic on i ) and can proceed recursively.If we need to perform another recursive step, we update the variables left and right (instead of using new ones). Therefore, the whole recursion can be done in logarithmicspace. Here, we make an exception as α and β do not refer to the corresponding permutations from Lemma 2but are arbitrary words. Later on, however, we will apply the definition usually in such a way that α and β indeed are related to the ones from Lemma 2. emark . For readers familiar with the notions: using a more careful but tedious analy-sis (and appropriate padding symbols), one can see that the reduction form g , . . . , g t to B β,α [ g t , . . . , g ] can not only be done in logarithmic space but actually it is a DLOGTIME -uniform projection reduction (compare to the proof of Barrington’s result in [26, Theo-rem 4.52]).If we substitute σ , α and β by the actual elements from A , we can see (using a simpleinduction) that B β,α [ g t , . . . , g ] works as a t -ary logical conjunction: Lemma 8.
For all g t , . . . , g ∈ { id , σ } ⊆ A , we have B β,α [ g t , . . . , g ] = A ( σ if g = · · · = g t = σ otherwise.Proof. For simplicity, we write B instead of B β,α in the following. For ℓ = 1 , there isnothing to show. So, let ℓ > and, first, assume g = · · · = g t = σ . Then, we have B [ g t , . . . , g ] = (cid:2) B [ g t , . . . , g ⌊ t ⌋ +1 ] | {z } = A σ β , B [ g ⌊ t ⌋ , . . . , g ] | {z } = A σ α (cid:3) = A σ by induction and the choice of σ , α and β from Lemma 2. If there is some i ∈ { , . . . , ⌊ t ⌋} with g i = id , then, by induction, we have B [ g t , . . . , g ] = (cid:2) B [ g t , . . . , g ⌊ t ⌋ +1 ] | {z } = g ′ β , B [ g ⌊ t ⌋ , . . . , g ] | {z } = A α (cid:3) = A β − g ′− β α − α β − g ′ β α − α = A .The case i ∈ {⌊ t ⌋ + 1 , . . . , ℓ } is symmetric. Remark . Something similar can be done with the free group of rank two (instead of A ) (see [22]). This is interesting because A cannot be realized as an automaton groupover an alphabet with less than five elements but the free group of rank three can begenerated by an automaton with binary alphabet [1, 27]. Theorem 10.
The uniform word problem for automaton groups
Constant:
Σ = { a , . . . , a } Input: a G -automaton T = ( Q, Σ , δ ) , p ∈ Q ∗ Question: is p = in G ( T ) ?(even over a fixed alphabet with five elements) is PSPACE -complete.Proof.
It is known that the uniform word problem for automaton groups is in PSPACE (using a guess and check algorithm; see [24] or [10, Proposition 2]). For showing
PSPACE -hardness, we reduce the
DFA Intersection Problem In fact, even the corresponding problem for automaton semigroups is in
PSPACE . onstant: Γ = { a , . . . , a } Input: ℓ ∈ N and deterministic finite acceptors A = ( P , Γ , τ , p , , F i ) , . . . , A ℓ = ( P ℓ , Γ , τ ℓ , p ,ℓ , F ℓ ) Question: is T ℓi =1 L ( A i ) = ∅ ?to the uniform word problem for automaton groups in logarithmic space. Kozen [16,Lemma 3.2.3] showed that this problem is PSPACE -hard.For the reduction, we need to map the acceptors A = ( P , Γ , τ , p , , F i ) , . . . , A ℓ =( P ℓ , Γ , τ ℓ , p ,ℓ , F ℓ ) to an automaton T = ( Q, Σ , δ ) and a sequence of states p ∈ Q ∗ .We assume the state sets P i to be pairwise disjoint and set P = S ℓi =1 P i ⊎ { id , σ } and F = S ℓi =1 F i . Additionally, we set Σ = { a , . . . , a } ⊎ { $ } and assume that theelements σ, α, β ∈ A (from Lemma 2) act as the corresponding permutations on Σ . Forthe transitions, we set δ = ℓ [ i =1 n p q a/a | p q a ∈ τ i o ∪ n r id $ / $ | r ∈ P \ F o ∪ n f σ $ / $ | f ∈ F o ∪ n id id a/a | a ∈ Σ o ∪ n σ σ a/σ ( a ) | a ∈ Σ o .Thus, we take the union of the acceptors and extend it into an automaton by letting allstates act like the identity. With the new letter $ (the “end-of-word” symbol), we go to id for non-accepting states and to σ for accepting ones. Finally, we have the state id ,which acts like the identity, and the state σ , whose action is to apply the permutation σ to all letters of the input word, justifying the re-use of the name σ . Finally, we define T as the (disjoint) union of the just defined automaton ( P, Σ , δ ) with the automaton α α β βa /a . . .a /a $ / $ a /α ( a ) . . .a /α ( a )$ /α ($) a /a . . .a /a $ / $ a /β ( a ) . . .a /β ( a )$ /β ($) .Notice that T is deterministic, complete and invertible and that all states except σ, α and β act like the identity on words not containing $ . Also note that on input of A , . . . , A ℓ ,the automaton can clearly be computed in logarithmic space.For the state sequence, we set p = B [ p ,ℓ , . . . , p , ] where we use B as a short-handnotation for the balanced commutator B β ,α defined in Definition 5; for the commutator B β,α , from now on we will simply write B . Observe that, by Lemma 6, we can compute p in logarithmic space.This completes our description of the reduction and it remains to show its correctness.If there is some w ∈ T ℓi =1 L ( A i ) , we have to show p = G ( T ) . We have the cross diagram That the alphabet may be assumed to contain exactly four elements can be seen easily. In fact, wemay even assume it to be of size three or – with suitable encoding – size two (see Remark 11). $ p , q f, σw $ ... ... ... w $ p ,ℓ q f,ℓ σw $ where all q f,i ∈ F i are final states. Thus, by Fact 4, we also have the cross diagram w $ p = B [ p ,ℓ , . . . , p , ] B [ q f,ℓ , . . . , q f, ] B [ σ, . . . , σ | {z } ℓ times ] w $ .Without loss of generality, we may assume σ ( a ) = a and, since we have B [ σ, . . . , σ ] = σ in A and also in G ( T ) by Lemma 8, we obtain p ◦ w $ a = w $ σ ( a ) = w $ a .If, on the other hand, we have T ℓi =1 L ( A i ) = ∅ , we have to show p = G ( T ) . For this,let w ∈ Σ ∗ be arbitrary. If w does not contain any $ , we do not need to show anythingsince, by construction, only the states σ , α and β act non-trivially on these words andthey can only be reached after reading a $ . If w contains $ , we can write w = u $ v with u ∈ { a , . . . , a } ∗ . Since the intersection is empty, there is some i ∈ { , . . . , ℓ } with w L ( A i ) and we obtain the cross diagram u $ p , q g u $ ... ... ... u $ p ,i q i g i = id u $ ... ... ... u $ p ,ℓ q ℓ g ℓ u $ where g , . . . , g ℓ ∈ { id , σ } . Again by Fact 4, we also obtain the cross diagram u $ p = B [ p ,ℓ , . . . , p , ] B [ q ℓ , . . . , q i , . . . , q ] B [ g ℓ , . . . , g i , . . . , g ] u $ but, this time, we have B [ g ℓ , . . . , g i , . . . , g ] = in A and in G ( T ) by Lemma 8 since g i = id . Accordingly, we have p ◦ u $ v = u $ v .10 emark . With Remark 9, we obtain that the size of the fixed alphabet can be reducedfurther down to two. The crucial observation here is that we actually only check a regularlanguage and then collect information from the states reached at the end of the word.This is the case because all states except α , β and σ act trivially and store the necessaryinformation only in the state. Therefore, for all states except the non-trivially actingones α , β and σ , we can encode the letters from the alphabet Σ as blocks of length over { , } . Finally, we can replace α , β and σ with suitable elements from the free group (seeRemark 9). Unfortunately, these group elements are not fixed anymore: they depend onthe position in the iterated commutator. This makes the construction (and the overallproof) a bit more difficult and also more technical . In this section, we are going to lift the result from the previous section to the non-uniformcase. We show:
Theorem 12.
There is an automaton group with a
PSPACE -complete word problem:
Constant: a G -automaton T = ( Q, Σ , δ ) Input: q ∈ Q ∗ Question: is q = in G ( T ) ? In order to prove this theorem, we are going to adapt the construction used in [10,Proposition 6] to show that there is an inverse automaton semigroup with a
PSPACE -complete word problem and that there is an automaton group whose word problem witha single rational constraint is
PSPACE -complete. The main idea is to use a masterreduction. Our automaton operates in two modes. In the first mode, which we willcall “TM-mode”, it will interpret its input word as a sequence of configurations of a(suitable)
PSPACE -machine and verifies that the configuration sequence constitutes avalid computation of the Turing machine. This verification is done by multiple states(where each state is responsible for a different verification part) and the informationwhether the verification was successful is stored in the state, not by manipulating theinput word. So we have successful states and fail states . Upon reading a special inputsymbol, the automaton will switch into a second mode, the “ A -mode”. More precisely,successful states go into a state which acts like σ from Lemma 2 and the fail states goesinto an identity state id . Finally, to extract the information from the states, we use theiterated commutator from Definition 5.The idea for the TM-mode is similar to the approach taken by Kozen to show PSPACE -completeness of the
DFA Intersection Problem where the input word is interpreted asa sequence of configurations of a
PSPACE
Turing machine where each configuration is oflength s ( n ) : γ (1)1 γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . . which is why we chose to present the easier (but a bit weaker) proof. Another advantage of theeasier proof is that it is self-contained (as it does not require a presentation of the free group by anautomaton).
11n Kozen’s proof, there is an acceptor for each position i of the configurations with ≤ i ≤ s ( n ) which checks for all t whether the transition form γ ( t ) i to γ ( t +1) i is valid.In our case, however, the automaton must not be dependent on the input (or its length n ) and we have to handle this a bit differently. The first idea is to use a “check-markapproach”. First, we check all first positions for valid transitions. Then, we put a check-mark on all first positions, which tells us that we now have to check all second positions(i. e. the first ones without a check-mark). Again, we put a check-mark on all these,continue with checking all third positions and so on (see Figure 1).The problem with this approach is that the check-marking leads to an intrinsically non-invertible automaton (see Figure 2). To circumvent this, we generalize the check-markapproach: before each symbol γ ( t ) i of a configuration, we add a k block (of sufficientlength k ). In the spirit of Example 3, we interpret this block as representing a binarynumber. We consider the symbol following the block as “unchecked” if the number iszero; for all other numbers, it is considered as “checked”. Now, checking the next symbolboils down to incrementing each block until we have encountered a block whose value waspreviously zero (and this can be detected while doing the addition). This idea is depictedin Figure 3. I would also be possible to have the check-mark block after each symbolinstead of before (which might be more intuitive) but it turns out that our ordering hassome technical advantages. Proof of Theorem 12.
Since the uniform word problem for automaton groups is in
PSPACE (see Theorem 10), so is the word problem of any (fixed) automaton group.Therefore, we only have to show the hardness part of the result.Consider an arbitrary
PSPACE -complete problem and let M be a deterministic, polyno-mially space-bounded Turing machine deciding it with input alphabet Λ , tape alphabet ∆ , blank symbol , state set P , initial state p and accepting states F ⊆ P . Thus,for any input word of length n , all configurations of M are of the form ∆ ℓ P ∆ m with ℓ + 1 + m = s ( n ) for some polynomial s . This makes the problem Constant: the
PSPACE machine M Input: w ∈ Λ ∗ Question: does M reach a configuration with a state from F from the initialconfiguration p w s ( n ) − n − ? Alternatively, we could also use a
PSPACE -universal Turing machine for our construction.
Figure 1: Illustration of the checkmark approach. γ (1)1 γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . .γ (1)1 X γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 X γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . .γ (1)1 X γ (1)2 X γ (1)3 . . . γ (1) s ( n ) γ (2)1 X γ (2)2 X γ (2)3 . . . γ (2) s ( n ) . . . checkcheck 12igure 2: Adding a check-mark yields a non-invertible automaton γ X / γ X γ/ γ X γ/γ γ X / γ X / Figure 3: The idea of our generalized check-marking approach. γ (1)1 γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . . γ (1)1 γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . . γ (1)1 γ (1)2 γ (1)3 . . . γ (1) s ( n ) γ (2)1 γ (2)2 γ (2)3 . . . γ (2) s ( n ) . . . g e n . c h ec k g e n . c h ec k PSPACE -complete. From the machine M , we construct the G -automaton T and, fromthe input w , we construct the state sequence q . Let Γ = ∆ ⊎ P . From now on, we will notwork with M anymore but rather only with its corresponding τ : Γ → Γ from Fact 1. Construction of the Automaton.
The automaton T works in the way described aboveand is the union of several simpler automata. For the alphabet, we use Σ = Γ ⊎ { , } ⊎{ , $ } with new letters , , and $ . The letters and will be used for the gener-alized check-mark approach described above, the letter is used to separate individualconfigurations and $ acts as an “end-of-computation” symbol switching the automatonfrom the TM-mode to the A -mode (mentioned above). For the A -mode, we write Σ = { a , . . . , a }⊎ B and assume that σ , α and β (from Lemma 2) operate on { a , . . . , a } such that σ ( a ) = a . With this, the first part of the automaton T used for the A -modeis σ id α βa i /σ ( a i )id B a i /a i id B a i /α ( a i )id B a i /β ( a i )id B where the a i -transitions exist for all i ∈ { , . . . , } and we use the convention that id X indicates x/x -transitions for all x ∈ X ⊆ Σ . Obviously, the state id acts as the identityand the action of the state σ on a word is to apply the permutation σ letter-wise (and toignore letters from B ), which justifies the dual use in notation as we can identify σ in theautomaton group with σ in A . For the intuition, it helps to see id as a “fail” state and σ as an “okay” state in the following. In the end, we will implement this intuition basicallyusing the iterated commutator from Definition 5. For this commutator, we also need theconjugating elements α and β , which work in the same way as σ . However, they do nothave an intuitive semantic and are mostly there for technical reasons.Now, let us describe the part of the automaton used for the TM-mode. First, we needtwo states which ignore everything in the TM-mode and then go to α or β :13 α β β id Γ ∪{ , , } $ / $ id Γ ∪{ , , } $ / $ where dotted states refer to the states defined above.The next part of our automaton is used to check that the input word (for the TM-mode)is of the form (0 ∗ Γ) + ( ∗ Γ) + ) ∗ : r σ / Γ / Γ / / $ Here, we use the convention that, whenever a transition is missing for some x ∈ Σ , thereis an implicit x/x -transition to the state id (as defined above). Note that we do notcheck that the factors in (0 ∗ Γ) + correspond to well-formed configurations fo the Turingmachine. This will be done implicitly by checking that the input word belongs to a validcomputation of the Turing machine, which we describe below.Next, we need a part which checks whether the input word contains a final state (ifthis is not the case, we want to “reject” the word): id f σ $ / $ id { , , }∪ Γ \ F id F id { , , }∪ Γ $ / $ Finally, we come to the more complicated parts of T . The first one is for the generalizedcheck-marking as described above and is depicted in Figure 4. In fact, we need this part twice : once for g = σ and once for g = id . Notice that, during the TM-mode phase (i. e.before the first $ ), the two versions behave exactly the same way; the only differenceis after switching to the A -mode: while X id still always acts like the identity, X σ actsnon-trivially on suitable input words.Additionally, we also need an automaton part verifying that every configuration symbolhas been check-marked (in the generalized sense): id c σ / / $ 1 / / / Γ ∪{ } $ / $ The last part is for checking the validity of the transition at all first so-far uncheckedpositions. While it is not really difficult, this part is a bit technical. Intuitively, for14 g g “So far, the original input digitblock of this symbol did notcontain a .”“The digit block of the lastsymbol contained at least one .”“Skip everything up to thenext configuration” / / / / Γ / / Γ / / Γ ∪{ } / / $ Figure 4: The automaton part used for generalized check-markingchecking the transition form time step t − to time step t at position i , we need tocompute γ ( t ) i = τ ( γ ( t − i − , γ ( t − i , γ ( t − i +1 ) from the configuration symbol at positions i − , i and i + 1 for time step t − . We store γ ( t ) i in the state (to compare to the actualvalue). Additionally, we need to store the last two symbols of configuration t we haveencountered so far (for computing what we expect in the next time step later on) andwhether we have seen a or only s in the check-mark digit block.For all this, we use the states γ , γ − γ , γ − γ − , γ γ ′ with γ − , γ , γ ′ ∈ Γ . The idea is the following. In the and states, we store the valuewe expect for the first unchecked symbol ( γ ) and the last symbol we have seen in thecurrent configuration ( γ − ). We are in the -state if we have not seen any in the digitblock yet and in the if we did. The latter two are used to skip the rest of the currentconfiguration and to compute the symbol we expect for the first unchecked position inthe next configuration ( γ ′ ).We use these states in the transitions schematically depicted in Figure 5. Here, thedashed transitions exist for all γ ′− and γ in Γ but go to different states, respectively,and the dotted states correspond to the respective non-dotted states with different valuesfor γ and γ − (with the exception of σ , which corresponds to the state defined above).We also define q γ ′ as the state on the bottom right (for every γ ′ ∈ Γ , respectively).The automaton parts depicted in Figure 4 and Figure 5 are best understood with an15 γ γ − γ − , γ γ ′ = τ ( γ − , γ , γ ) 1 γ γ − γ γ ′− τ ( γ − , γ , )0 γ ′ = q γ ′ σ / / γ /γ / / γ ′− /γ ′− / / / $ γ /γ id Γ ∪{ } $ / $ / Figure 5: Schematic representation of the transitions used for checking Turing machinetransitions and definition of q ′ γ ; the dashed transitions exist for all γ ′− and γ in Γ but go to different states, respectivelyexample. Consider the input word γ (0)1 γ (0)2 γ (0)3 γ (1)1 γ (1)2 γ (1)3 $ where we consider the γ ( t ) i to form a valid computation. If we start in state q γ (0)2 and readthe above word, we immediately take the / -transition and go into the corresponding state where we skip the rest of the digit block. Using the dashed transition, the nextsymbol γ (0)1 takes us back into a -state where the upper entry is still γ (0)2 but the lowerentry is now γ (0)1 (i. e. the last configuration symbol we just read). We loop at this statewhile reading the next three s and, since the next symbol γ (0)2 matches with the onestored in the state, we get into the states with entries γ (0)1 , γ (0)2 where we skip the nextthree s again. Reading γ (0)3 now gets us into the state with entry γ (1)2 since we have τ ( γ (0)1 , γ (0)2 , γ (0)3 ) = γ (1)2 by assumption that the γ ( t ) i form a valid computation. Here, weread / and the process repeats for the second configuration, this time starting in q γ (2)2 .When reading the final $ , we are in the state with entry τ ( γ (1)1 , γ (1)2 , γ (1)3 ) and finally goto σ . Notice that during the whole process, we have not changed the input word at all!If we now start reading the input word again in state X σ (see Figure 4 and also referto Figure 3), we turn the first into a , go to the state at the bottom, turn the next into a and go to the state on the right, where we ignore the next . When reading γ (0)1 , we go back to X σ . Next, we take the upper exit and turn the next into a . Theremaining s are ignored and we remain in the state at the top right until we read γ (0)2 , which gets usback into X σ . The second part works in the same way with the difference that we go to σ at the end since we encounter the $ instead of . The output word, thus, is γ (1)1 γ (1)2 γ (1)3 γ (2)1 γ (2)2 γ (2)3 $ and we have check-marked the next position in both configurations.This concludes the definition of the automaton and the reader may verify that T isindeed a G -automaton since all individual parts are G -automata. Furthermore, apartfrom the check-marking, all states except σ , α and β (which belong to the A -mode andare only entered when reading $ ) act like the identity. Definition of the State Sequence.
To describe the actual reduction, we have to definethe state sequence q such that q depends only on the input word w for the Turingmachine. Similar to the automaton T , this sequence consists of multiple parts. Each partwill verify a certain aspect of the input word and the general idea is that, after readinga word u $ , we are either in σ (if u satisfies the criterion we are currently checking) orin id (if it does not). Finally, using the balanced commutator from Definition 5, we canfind whether any of the criteria was not satisfied. For this to work easily, we define theindividual parts in such a way that the output will be u $ again, which will allow us toapply Fact 4.First, we simply use the state r to verify that u is from (0 ∗ Γ) + ( ∗ Γ) + ) ∗ . Thus, weonly need to consider the case that u is of the form ℓ (0)1 γ (0)1 ℓ (0)2 γ (0)2 . . . ℓ (0) I γ (0) I . . . ℓ ( T )1 γ ( T )1 ℓ ( T )2 γ ( T )2 . . . ℓ ( T ) IT γ ( T ) I T ( † )with γ ( t ) i ∈ Γ any further.Next, we need to verify that, for every ≤ i ≤ s ( n ) , we can check-mark the first i positions. For this, we use c i = X − i id X σ X i − as we have the cross diagram ←− bin(0) γ ( t )1 . . . ←− bin(0) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X i − X i − / id i − ←− bin( i − γ ( t )1 . . . ←− bin(1) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X σ X σ /σ ←− bin( i ) γ ( t )1 . . . ←− bin(2) γ ( t ) i − ←− bin(1) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X − X − / id − ←− bin( i − γ ( t )1 . . . ←− bin(1) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X − ( i − X − ( i − / id − ( i − ←− bin(0) γ ( t )1 . . . ←− bin(0) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ ←− bin( z ) denotes the reverse/least significant bit first binary representation of z (ofsufficient length). Here, it is useful to observe that, if j th block with j ≤ i is not longenough to count to its required value, then we will always end up in id after reading a $ . The same happens if I t < i (i. e. is one of the configurations is “too short”). So thisguarantees, I t < s ( n ) for all t .On the other hand, we use c ′ = X − s ( n )id c X s ( n )id to ensure that, after check-marking the first s ( n ) positions in every configurations, allsymbols have been check-marked (i. e. that no configuration is “too long”), which guar-antees I t = s ( n ) for all t .Now that we have ensured that the word is of the correct form and we can counthigh enough for our check-marking, we need to actually verify that the γ ( t ) i constitutea valid computation of the Turing machine with the initial configuration γ ′ . . . γ ′ s ( n ) = p w s ( n ) − n − for the input word w . To do this, we define q i = X − ( i − q γ ′ i X i − for every ≤ i ≤ s ( n ) as we have the cross diagram ←− bin(0) γ ( t )1 . . . ←− bin(0) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X i − X i − / id i − ←− bin( i − γ ( t )1 . . . ←− bin(1) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ q γ ′ i q τ ( γ ( t ) i − ,γ ( t ) i ,γ ( t ) i +1 ) /σ ←− bin( i − γ ( t )1 . . . ←− bin(1) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ X − ( i − X − ( i − / id − ( i − ←− bin(0) γ ( t )1 . . . ←− bin(0) γ ( t ) i − ←− bin(0) γ ( t ) i ←− bin(0) γ ( t ) i +1 . . . ←− bin(0) γ ( t ) I t / $ if γ ( t ) i is the expected γ ′ i . Otherwise (if γ ( t ) i = γ ′ i ), we always end in state id after readingthe $ . Finally, to ensure that the computation is not only valid but also accepting, weuse the state f .Summing this up, we define q = B [ f, q s ( n ) , . . . , q , c ′ , c s ( n ) , . . . , c , r ] where we use the short-hand notation B for the balanced commutator B β ,α fromDefinition 5. Observe that the individual parts of q can indeed be computed in logarith-mic space and that, thus, this is also true for q itself by Lemma 6.18 $ r σu $ c σu $ ... ... u $ c s ( n ) σu $ c ′ σu $ q σu $ ... ... u $ q s ( n ) σu $ f σu $ B [] ,,,,,,,, B [] ,,,,,,,, Figure 6: Crossdiagram
Correctness.
We need to prove that the action of q is equal tothe identity if and only if the Turing machine does not accept theinput word w . The easier direction is to assume that the Turingmachine accepts on the initial configuration p w s ( n ) − n − . Let γ (0)1 . . . γ (0) s ( n ) ⊢ γ (1)1 . . . γ (1) s ( n ) ⊢ · · · ⊢ γ ( T )1 . . . γ ( T ) s ( n ) be the correspondingcomputation with γ (0)1 = p , γ (0)2 . . . γ (0) n +1 = w and γ ( T ) i ∈ F for some ≤ i ≤ s ( n ) . We choose ℓ = ⌈ log( s ( n )) ⌉ + 1 and define u = 0 ℓ γ (0)1 . . . ℓ γ (0) s ( n ) ℓ γ (1)1 . . . ℓ γ (1) s ( n ) . . . ℓ γ ( T )1 . . . ℓ γ ( T ) s ( n ) .We now let q act on the word u $ a . Recall that we assume σ tooperate non-trivially on a . The reader may verify that we have theblack part of the cross diagram depicted in Figure 6. From Fact 4,we immediately also obtain the gray additions to the cross diagramwhere we use B instead of B β,α for the balanced commutator fromDefinition 5. By Lemma 8, we obtain B [ σ, . . . , σ ] = σ in A and,thus, in G ( T ) . Therefore, q acts non-trivially on u $ a .For the other direction, assume that no valid computation of M onthe initial configuration p w s ( n ) − n − contains an accepting statefrom F . We have to show that q acts like the identity on all wordsfrom Σ ∗ . If the word does not contain a $ , then all individual partsof q act on it like the identity by construction. This is clearly thecase for r , c ′ , the q i and f . For the c i , the only point to note is that X σ acts in the same way as X id on such words.Thus, we may assume that the word is of the form u $ v . If u is notof the form (0 ∗ Γ) + ( ∗ Γ) + ) ∗ , we have the cross diagram u $ r id u $ and, thus, u $ q B [ g t , . . . , g , id] u $ with g , . . . , g t ∈ { σ, id } . As we have, B [ g t , . . . , g , id] = A byLemma 8, we obtain that q acts like the identity on u $ v .Therefore, we assume u to be of the form mentioned in Equation † and use a similarargumentation for the remaining cases. If u does not contain a state from F , then weend up in state id after reading $ for f . As w is not accepted by the machine, thisincludes in particular all valid computations on the initial configuration p w s ( n ) − n − .If one of the blocks in u is too short to count to a valued required for the check-marking (i. e. one ℓ ( t ) i is too small), then the corresponding c i will go to (a state sequenceequivalent to) id . This is also true if one configuration is too short (i. e. I t < s ( n ) forsome t ). If one configuration is too long (i. e. I t > s ( n ) ), then this will be detected by c ′ as not all positions will be check-marked after check-marking all first s ( n ) positionsin every configuration. Finally, q i yields an id if γ (0) i is not the correct symbol from theinitial configuration or if we have γ ( t +1) i = τ ( γ ( t ) i − , γ ( t ) i , γ ( t ) i +1 ) for some t (where we let γ ( t ) − = = γ ( t ) s ( n )+1 ). 19 emark . Similar to what we described in Remark 11, we can also reduce the size ofthe alphabet Σ to two in the non-uniform case. However, this time, we do not only havestates acting like the identity in the TM-mode anymore (as we did in the uniform case).Therefore, we have to realize the block encoding of the letters from Σ over a binaryalphabet { a, b } in such a way that it remains compatible with the check-marking/binaryincrement. This is, for example, possible by encoding the two letters and as blocksstarting with a and all other letters as blocks starting with b . In this section, we re-apply our previous construction to show that there is an automatongroup with an
EXPSPACE -complete compressed word problem. The compressed wordproblem of a group is similar to the normal word problem. However, the input element(to be compared to the neutral element) is not given directly but as a straight-lineprogram. A straight-line program is a context-free grammar which generates exactly oneword.
Theorem 14.
There is an automaton group with an
EXPSPACE -complete compressedword problem:
Constant: a G -automaton T = ( Q, Σ , δ ) Input: a straight-line program generating a state sequence q ∈ Q ∗ Question: is q = in G ( T ) ?Proof. Before starting, we observe that, whenever we have a variable X in a straight-lineprogram generating a word representing a group element g , we can easily obtain a variable X − generating a word representing g − by mirroring all rules for X , replacing all letters a by a − and all variables A by A − (where we have to continue recursively). Hence, wewill always assume that we also have A − if we have described A in the following.For the actual proof, we use the same construction as in the proof of Theorem 12,but we start with a Turing machine M for an EXPSPACE -complete problem. Now,all configurations on input of a word w of length n are of the form ∆ ℓ P ∆ m with ℓ + 1 + m = s ( n ) where s ( n ) is of the form n e for some constant e ∈ N .Recall that, for the (normal) word problem, we used q = B [ f, q s ( n ) , . . . , q , c ′ , c s ( n ) , . . . , c , r ] (1)for the reduction where c i = X − i id X σ X i − , c ′ = X − s ( n )id c X s ( n )id and q i = X − ( i − q γ ′ i X i − for ≤ i ≤ s ( n ) . The problem now is that we have exponentially many c i and q i . Thus,we cannot output them in logarithmic space (or polynomial time), not even if we use astraight-line program. So, we will have to accommodate for this. The good news is that,for c ′ , this is not a problem: we can output a straight-line program with starting variable C ′ generating c ′ : C ′ → M ne cM − ne , M ne → M ne − M ne − , . . . , M → M M , M → X id n and, clearly, C ′ evaluates to c ′ .To circumvent the problem with the c i , we use the fact that the behavior of c i is struc-turally the same as the behavior of c j . This can be exploited by defining a slightly modi-fied version of the balanced commutator B (where B is as in the proof of Theorem 12):let B c [1] = X − X σ and B c [ k ] = (cid:20) B c [ ⌈ k ⌉ ] β X ⌊ k ⌋ id , B c [ ⌊ k ⌋ ] α (cid:21) = X −⌊ k ⌋ id β − B c [ ⌈ k ⌉ ] − β X ⌊ k ⌋ id α − B c [ ⌊ k ⌋ ] − α X −⌊ k ⌋ id β − B c [ ⌈ k ⌉ ] β X ⌊ k ⌋ id α − B c [ ⌊ k ⌋ ] α .The intuitive idea behind this definition is that we start with all check-mark countersat value zero (“nothing is check-marked”). Then the bottom right part checks that wecan check the first ⌊ k ⌋ positions. However, the counters are reset to zero during thischecking (by induction), so we check-mark the positions again (without modifying thevalue in the A -mode) using a suitable power of X id . Then the part on the bottom leftchecks that we can check-mark the next ⌈ k ⌉ positions. After this, the counters are resetto zero. Then the same happens again only using the inverse group elements this timeto realize the commutator. Here, it is important to note that the top left part sees thesame situation as the bottom left part, so it behaves in the same way except for going to σ − instead of σ in the A -mode.Formally, we can show B c [ k ] = G ( T ) B [ c k , . . . , c ] by induction. The base case holdsby definition and, for the induction step, we observe that X id commutes with α and β and that we have c i + d = X − d id c i X d id in G ( T ) . Since we have [ y, x ] z = [ y z , x z ] in any group,we obtain in G ( T ) B c [ k ] = (cid:20) B [ c ⌈ k ⌉ , . . . , c ] β X ⌊ k ⌋ id , B [ c ⌊ k ⌋ , . . . , c ] α (cid:21) (by induction) = " B [ c X ⌊ k ⌋ id ⌈ k ⌉ , . . . , c X ⌊ k ⌋ id ] β , B [ c ⌊ k ⌋ , . . . , c ] α = h B [ c k , . . . , c ⌊ k ⌋ +1 ] β , B [ c ⌊ k ⌋ , . . . , c ] α i = B [ c k , . . . , c ] . In particular, this shows that no B c [ k ] modifies a word not containing $ (i. e. all B c [ k ] act like the identity in the TM-mode). This is important as we will be using B c [ s ( n )] = B c [2 n e ] for checking that we can check-mark all positions in the end.The straight-line program for B c [ s ( n )] is similar to the one of c ′ (and uses the variables M i defined above). We let B c, ℓ → M − ℓ − β − B − c, ℓ − β M ℓ − α − B − c, ℓ − α M − ℓ − β − B c, ℓ − β M ℓ − α − B c, ℓ − α ℓ > and B c, → X − X σ . This way, we can compute B c, ne as a variable for B c [ s ( n )] in logarithmic space.At first, it seems that we cannot use a similar idea for the q i as they are responsi-ble for different symbols of the configuration. However, as the initial configuration is p w ne − n , all q i except of the first n + 1 ones are responsible for the same tape sym-bol (the blank symbol ). We define B q [1] = q ; the inductive step is the same as in thedefinition of B c [ k ] and the same ideas apply. This time, however, we want to start withthe first n + 1 positions check-marked (this is the p w part) and, thus, we will be using X − n − B q [ s ( n ) − n − X n +1id and the original q i for the first n + 1 positions. With thesame arguments as for B c [ s ( n )] , one can show X − n − B q [ s ( n ) − n − X n +1id = G ( T ) B [ q s ( n ) , . . . , q n +2 ] .A straight-line program for B q [ s ( n ) − n − can be obtained in a similar way to theone for B c [ s ( n )] and the X id blocks are short enough to output them directly.Summing this up, we set Q → B [ f, X − n − B q X n +1id , q n +1 , . . . , q , C ′ , B c,s ( n ) , r ] where B q is the starting variable of the straight-line program for B q [ s ( n ) − n − . ByLemma 6, we can compute the right-hand side of this rule in logarithmic space. Bythe construction of the straight-line programs for B q , B c,s ( n ) and C ′ (and the provedequalities), it follows that Q evaluates to B [ f, B [ q s ( n ) , . . . , q n +2 ] , q n +1 , . . . , q , c ′ , B [ c s ( n ) , . . . , c ] , r ] . Now, using Lemma 8, it is easy to see that this is equal to q from (1) in G ( T ) : beforereading the first letter $ , it operates as the identity and, after the first letter $ , it operateseither as the identity or as σ depending on whether all components of the commutatoract like σ . Corollary 15.
There is an automaton group with a
PSPACE -complete word problem andan
EXPSPACE -complete compressed word problem.Proof.
The result follows from the closure of the class of automaton groups under directproduct (which is well-known and easy to see) in combination with Theorem 12 andTheorem 14. Be aware that this is a bit more technical, since we are not only working with powers of two anymoreand need to take the rounding into consideration. eferences [1] S. V. Aleshin. A free group of finite automata. Vestnik Moskovskogo Universiteta.Seriya I. Matematika, Mekhanika , (4):12–14, 1983.[2] Sanjeev Arora and Boaz Barak.
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