Resolution of The Linear-Bounded Automata Question

10/11/2021 ∙ by Tianrong Lin, et al. ∙ 0

This work resolve a longstanding open question in automata theory, i.e. the linear-bounded automata question ( shortly, LBA question), which can also be phrased succinctly in the language of computational complexity theory as NSPACE[n]?=DSPACE[n]. We prove that NSPACE[n]≠ DSPACE[n]. Our proof technique is based on diagonalization against all deterministic Turing machines working in O(n) space by an universal nondeterministic Turing machine running in O(n) space. Our proof also implies the following consequences: (1) There exists no deterministic Turing machine working in O(log n) space deciding the st-connectivity question (STCON); (2) L≠ NL; (3) L≠ P.

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1 Introduction

The automata theory, origin from Turing’s seminal work [Tur37], has witnessed two fundamental branches in the recent eight decades; The first of which is the study of notion of computability, see Rogers [Rog67], dealing with the main topics of what is computable by Turing machine. This kind of study has been extended to different type of automata such as finite automata, pushdown automata, linear-bounded automata and so on which including other important questions related to automata, for example, see Ginsburg [Gin66], Sipser [Sip13] and Kuroda [Kur64]. As observed by Stearns et al. [SHL65], an important goal of automata theory is a basic understanding of computational process for various classes of problems. Hence, along with the development, after more and more questions being answered, the theory of automata has gone beyond the notion of computability and included some measure of the difficulty of the computation and how that difficulty is related to the organization of the machine that performs the computation, which forms the second branch of automata theory, i.e. the complexity theory.

Perhaps, the most basic and fundamental measures of difficulty that appear particularly important are time and space (memory). The fundamental measure of time opened the door to the study of the extremely expressive time complexity class , and the rich theory of -completeness, see Arora and Barak [AB09]; While the basic measure of space also opened the door to the study of space complexity class and and the theory of -completeness [AB09]. These notions were important building block in theoretical computer science and complexity theory.

In 1964, Kuroda stated two research challenges in his seminal paper [Kur64], which subsequently became famously known as the LBA questions. See Hartmanis and Hunt III [HH74] for a discussion of the history and importance of this question. The first LBA question is whether the class of languages accepted by (nondeterministic) linear-bounded automata is equal to the class of languages accepted by deterministic linear-bounded automata. The second LBA question is whether the class of languages accepted by (nondeterministic) linear-bounded automata is closed under complement. These two questions can be phrased succinctly in the language of computational complexity theory as:

  1. ;

where we will define the notation and in Section 2 and is the complement of . Kuroda showed in [Kur64] that the class of languages accepted by (nondeterministic) linear-bounded automata (i.e. the complexity class ) is right the context-sensitive languages. The second LBA question has an affirmative answer, implied by the well-known Immerman-Sénizergues nondeterministic algorithm — proved 20 years after the question was raised — by Immerman and Sénizergues, see their famous result [Imm88, Sze88]. While the first LBA question still remains open and in fact becomes one of the most oldest open questions in the field of automata theory. From now on, we refer the LBA question to the first LBA question which is still open.

Savitch’s theorem [Sav70] gives us an initial insight on the LBA question: nondeterministic space Turing machines can be simulated efficiently by deterministic space Turing machines, with only quadratic loss in space usage. That is, for a space-constructible function. As we can see, the technique used in proof of Savitch’s theorem [Sav70] is an application of the reachability method of -connectivity in the configuration graph (which is a directed graph) of a nondeterministic space-bounded Turing machine.

Restricting our attention to the LBA question, only the restricted case, i.e. the Turing machines are limited to one-way case but not two-way, was shown by Hopcroft and Ullman [HU67] that the nondeterministic models are more power than deterministic models. However for two-way Turing machines of all other complexity classes it is an open question as to whether or not the deterministic and nondeterministic models are equivalent, i.e. the general LBA question. In this work, we will use the technique of diagonalization together the well-known Immerman-Sénizergues theorem to resolve this fundament open question. Specifically, we will use such technique which will be further used in author’s future and upcoming work [Lin21], i.e. of using an universal nondeterministic Turing machine to diagonalize over all single-tape deterministic Turing machine of space complexity . Our first main contribution is the following

Theorem 1.

There exists a language but , in other words, .

The directed -connectivity, denoted STCON, is one of the most widely studied questions in theoretical computer science and simple to state: Given a directed graph together with vertices and , the -connectivity question is to determine if there is a directed path from to . STCON plays an important role in complexity theory as it is complete for the complexity class Nondeterministic Logspace under Logspace reductions [AB09, Mic92, Sip13]. Currently, the best known space upper bound is 222 We write to mean in this work. using Savitch’s algorithm [Sav70]. In a breakthrough result, Reingold [Rei08] showed that, by using some tools in [RVW02], the undirected -Connectivity question can be solved in space which renews the enthusiasm to improve Savitch’s bound for STCON, since we are aware that one of the obvious direction is to extend Reingold’s algorithm to the directed case. In addition, it is considered that proving any nontrivial space lower bound on a general Turing machine is beyond the reach of current techniques. However,to our best of knowledge, the proof techniques of Theorem 1 also obviously imply the following important consequences:

Theorem 2.

There exists no deterministic Turing machine working in space deciding the STCON.

For the classes , , and the deterministic polynomial-time class , there is a well-known tower of inclusions [AB09, Sip13] :

It is unknown whether and , see open question list in Computer Science [A1]. In fact, and STCON are closely correlated since STCON is Log-space complete for , see Arora and Barak [AB09] and Sipser [Sip13]. Thus, if STCON can be decided deterministically in space , then will follow, and vice versa. By Theorem 2, we resolve the famous open questions by the following:

Corollary 1.

, that is,

from which it immediately follows that :

Corollary 2.

, that is, .

However, currently it is unknown whether , and to the best of our knowledge, it is also an important open question in Theoretical Computer Science.

The remainder of this work is organized as follows: For convenience of the reader, in the next Section, we review notation and some notions which are closely associated with our discussions appearing in Section 1. In Section 3 we will prove our first main result of Theorem 1. The proofs of Theorem 2 is put in to Section 4. Finally, a brief conclusion is drawn and some open question are raised in the last Section.

2 Preliminaries

In this Section, we describe some notions and notation required in the following context. Our style of writing after the previous Section is fully influenced by that in Aho, Hopcroft and Ullman’s book [HU69, AHU74, HU79].

The computational modes discussed in this work are Turing machine and linear-bounded automata. We first give the formal description of Turing machine in the following. For other standard description of on-line Turing machines we refer the readers to Hopcroft and Ullman’s book [HU69, AHU74, HU79].

Definition 1.

(-tape deterministic Turing machine,[AHU74]) A -tape deterministic Turing machine (shortly, DTM) is a seven-tuple where :

  1. is the set of states.

  2. is the set of tape symbols.

  3. is the set of input symbols; .

  4. , is the blank.

  5. is the initial state.

  6. is the final (or accepting) state.

  7. is the next-move function, maps a subset of to .
    That is, for some -tuples consisting of a state and tape symbols, it gives a new state and pairs, each pair consisting of a new tape symbol and a direction for the tape head. Suppose , and the deterministic Turing machine is in state with the th tape head scanning tape symbol for . Then in one move the deterministic Turing machine enters state , changes symbol to , and moves the th tape head in the direction for .

The definition of a nondeterministic Turing machine is similar to that of deterministic Turing machine, except that the next-move function is a mapping from to subsets of , stated as follows :

Definition 2.

(-tape nondeterministic Turing machine,[AHU74]) A -tape nondeterministic Turing machine (shortly, NTM) is a seven-tuple where all components have the same meaning as for the ordinary deterministic Turing machine, except that here the next-move function is a mapping from to subsets of .

The language accepted by the Turing machine is the set of in such that enters an accepting state when started with on its input.

A Turing machine is of space complexity if for each input of length , uses at most cells of its storage tape. A language is of space complexity for some machine model if it is defined by a Turing machine of that model which is of space complexity . Similar to [Sav70], we are considering only the growth rates of storage functions. That is, for our purposes, the storage function and are the same, for any constant . Thus, we have the following accurate definition:

Definition 3.

(Space-Bounded Computation. [AB09]) Let and . We say that (respective, if there is a constant and deterministic (respectively, nondeterministic) Turing machine deciding such that on every input , the total number of tape-cells that uses is at most for some constant .

In particular, is the class and is the class , i.e. the class of languages accepted by deterministic and nondeterministic Turing machine of space complexity for some constant ., respectively.

The original notion of a (nondeterministic) linear-bounded automaton is an on-line (nondeterministic) Turing machine with one-tape whose read/write head can not move beyond the left and right boundary symbols and , see Landweber [Lan63] and Kuroda [Kur64]. But in this work, we shall give another description in accord with the Turing machine stated above. They are essentially the same in nature:

Formally, a (nondeterministic) linear-bounded automaton is a (nondeterministic) Turing machine of space complexity .

The following Lemmas are important premise tools for our work whose proofs are omitted here :

Lemma 1.

(Corollary 3, [AHU74], p. 372) If is accepted by a -tape deterministic Turing machine of space complexity , then is accepted by a single-tape deterministic Turing machine of space complexity .

Lemma 2.

(Corollary 2, [AHU74], p. 372) If is accepted by a -tape nondeterministic Turing machine of space complexity , then is accepted by a single-tape nondeterministic Turing machine of space complexity .

Lemma 3.

(Immerman-Sénizergues Theorem, [Imm88, Sze88]) Let be a space-constructible function. Then the nondeterministic space of is closed under complement. That is .

Other background information and notions will be given along the way in proving our main results stated in Section 1.

3 Proof of Theorem 1

To obtain the result of Theorem 1 we need to enumerate the deterministic Turing machines, that is, assign an ordering to DTMs so that for each nonnegative integer there is a unique DTM associated with .

We next use the method presented in [AHU74], p. 407, to encode a deterministic Turing machine into an integer.

Without loss of generality, see [AHU74], we can make the following assumptions about the representation of a single-tape deterministic Turing machine :

  1. The states are named for some , with the initial state and the accepting state.

  2. The input alphabet is .

  3. The tape alphabet is for some , where , , and .

  4. The next-move function is a list of quintuples of the form ,meaning that , and is the direction, , , or , if , or , respectively. We assume this quintuple is encoded by the string .

  5. The Turing machine itself is encoded by concatenating in any order the codes for each of the quintuples in its next-move function. Additional ’s may be prefixed to the string if desired. The result will be some string of ’s and ’s, beginning with , which we can interpret as an integer.

By this encoding, any integer which can not be decoded is deemed to represent the trivial Turing machine with an empty next-move function. Every single-tape deterministic Turing machine will appear infinitely often in the enumeration, since given a deterministic Turing machine, we may prefix ’s at will to find larger and larger integers representing the same set of quintuples.

We can now design a four-tape NTM which treats its input string both as an encoding of a single-tape DTM and also as the input to . One of the capabilities possessed by is the ability to simulate a Turing machine, given its specification. We shall have determine whether the Turing machine accepts the input without using more than tape-cells for some function . If accepts in space , then does not. Otherwise, accepts . Thus, for all , either disagrees with the behavior of the th DTM on that input which is the binary representation of , or the th DTM uses more than tape-cells on input . We first show the following more general theorem comparing to Theorem 1:

Theorem 3.

Let be a space-constructible function. Then there exists a language accepted by a NTM by using space but by no DTM of space complexity . That is, but .

Proof. Let be a four-tape NTM which operates as follows on an input string of length of .

  1. By using space, decodes the Turing machine encoded by . If is not the encoding of some single-tape DTM then GOTO , else determines , the number of tape symbols used by this Turing, and , its number of states. The third tape of can be used as “scratch” memory to calculate .

  2. marks off cells on each tape. After doing so, if any tape head of attempts to move off the marked cells, halts without accepting.

  3. Let be the Turing machine encoded by . Then lays off on its second tape blocks of cells each, the blocks being separated by single cells holding a marker , i.e. there are cells in all. Each tape symbol occurring in a cell of ’s tape will be encoded as a binary number in the corresponding block of the second tape of . Initially, places its input, in binary coded form, in the blocks of tape , filling the unused blocks with the code for the blank.

  4. On tape , sets up a block of cells, initialized to all ’s, provided again that this number of cells does not exceed . Tape is used as a counter to count up to .

  5. By using nondeterminism 333 When simulating a DTM, the behavior of the universal NTM is somewhat deterministic, since there is no nondeterministic choices in a DTM. The author receives some feedbacks that the construction of this universal NTM is almost the same as an universal DTM, so some reader confuses that is not accepted by any DTMs of space complexity but accepted by the “DTM” constructed here of space complexity . We modify the as above and clarify that we constructed here is a NTM of space complexity , that is, it must accepts some language in . We hope this explanation will clear the confusion of some reader. in cells, simulates , using tape , its input tape, to determine the moves of and using tape to simulate the tape of . The moves of are counted in binary in the block of tape , and tape is used to hold the state of . If accepts, then halts without accepting. accepts if halts without accepting, if the simulation of attempts to use more than the allotted cells on tape , or if the counter on tape overflows, i.e. the number of moves made by exceeds .

  6. Since is not encoding of some single-tape DTM. Then marks off cells on each tape. After doing so, if any tape head of attempts to move off the marked cells, halts without accepting. Then, on tape , sets up a block of cells, initialized to all ’s. Tape is used as a counter to count up to . By using its nondeterministic choices, moves as per the path given by . The moves of are counted in binary in the block of tape . rejects if the number of moves made by exceeds or reaches a reject state before reaching , otherwise accepts. Note that the number of and in is fixed, i.e. it is default.

The NTM described above is of space complexity , specifically 444 Note that for a fixed DTM, its tape-symbols and states are fixed, so and are fixed constants.. By Lemma 2 it is equivalent to a single-tape NTM of space complexity , and it of course accepts some language .

Suppose now were accepted by some DTM of space complexity . By Lemma 1 we may assume that is a single-tape Turing machine. Let have states and tape symbols. Since 555 denotes the set of binary strings which encodes DTM . We know that by stringing together the quintuples of , and we may prefix ’s at will to find larger and larger integers representing the same set of quintuples of the same DTM , thus there are infinitely binary strings of sufficiently long which represents . appears infinitely often in the enumeration, and

so, there exists a such that for any ,

which implies that for a sufficiently long , say , and denoted by such is , we have that

Thus, on input , has sufficient space 666 By our construction of , requires space to simulate . to simulate and accepts if and only if rejects 777 The simulating by an universal nondeterministic Turing machine is using only a constant factor of space overhead, see [AB09], p. 83. But we did not find its formal proof in [AB09] or elsewhere.. But we assumed that accepted , i.e. agreed with on all inputs. We thus conclude that does not exist, i.e. .

In the above, we need say more about (3). Since is a NTM, so it is not obvious for it to flip the answer, i.e. . However, by Lemma 3, i.e. the Immerman-Sénizergues Theorem, this can be done for in space to flip the answer 888 In fact, Lemma 3 is necessary only for diagonalizing against a NTM. When diagonalizing against a DTM of space complexity , since for any space-constructible function , the universal NTM can immediately flip the answer in space . In time complexity, a NTM of time complexity can have different branches, it is unknown for an universal NTM flips the answer in time , see [AB09], p. 70, proof of Theorem 3.2. But for space complexity, an universal NTM can flip the answer in space when diagonalizing against other NTMs of space complexity because of Lemma 3..

We thus conclude that

 

Now the Theorem 1 is a special case when making for some constant .

Proof of Theorem 1. Making for some constant and repeating the proving process of Theorem 3.

4 Proof of Theorem 2

At this point, the proof of Theorem 2 can be made naturally by a contradiction to Theorem 1.

Proof of Theorem 2. If STCON has deterministic algorithms with space complexity , we can use it in Savitch’s algorithm [Sav70] to deterministically simulate an NTM, which further implies that , a contradiction to Theorem 1.

Indeed, Theorem 2 is a special case when making for some constant . Hence, taking in the proof of Theorem 3 we can obtain the proof of Theorem 2.

Proof of Corollary 1. Since STCON is –complete for the complexity class (See complexity book by Arora and Barak [AB09] and Michel’s book [Mic92]), we know that if and only if STCON , i.e. there is a deterministic algorithm deciding STCON in space .

In fact, the later proof of Theorem 2 already implies that

5 Conclusions

To summarize, we have shown that the class of languages accepted by (nondeterministic) linear-bounded automata (i.e. the context-sensitive languages) is not equal to the class of languages accepted by deterministic linear-bounded automata. Thus we resolve the longstanding LBA question.

In fact, we have shown a more general result, i.e. for any space-constructible function , . We achieved this by enumerating all deterministic Turing machine of space complexity , and then diagonalizing over all deterministic Turing machine in the enumeration.

As two special cases of our result, the STCON has been resolved. That is there exists deterministic Turing machine working in deciding the -connectivity question where is the number of vertices of directed graph in question, and , which also resolves a famous open question from Theoretical Computer Science, see [A1], from which it follows that .

Finally, there are many important questions in Computer Science we did not touch in this work, see [A1].

References

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