number of reversals in instruction (2) is O(h). Since machine N works in time T (jxj)

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1 It is easy to see that N 0 recognizes L. Moreover from Lemma 4 it follows that the number of reversals in instruction (2) is O(h). Since machine N works in time T (jxj) instruction (2) is performed for h = 2; 4; : : : ; 2 a?1 ; 2 a, where (2 a?1 ) 2 < T (jxj). The reversal complexity of the algorithm is O( a i=12 i ) = O(2 a+1 ) = O( q T (jxj)): 2 3 Acknowledgment The author would like to thank Rudiger Reischuk and Andreas Jakoby for helpful discussions and comments on a draft of this paper. References [1] J. Chen, The dierence between one tape and two tapes: with respect to reversal complexity, Theoret. Comput. Sci. 73 (1990) [2] J. Chen and C.K. Yap, Reversal complexity, SIAM J. Comput. 20 (1991) [3] P.W. Dymond and S.A. Cook, Hardware complexity and parallel computetion, in: Proc. 21st Ann. IEEE Symp. on Foundations of Computer Science (1980) [4] P. Fischer, The reduction of tape reversal for o-line one-tape Turing machines, J. Comput. System Sci. 2 (1968) [5] J. Hartmanis, Tape-reversal bounded Turing machine computations, J. Comput. System Sci. 2 (1968) [6] J.W. Hong, On similarity and duality of computation, in: Proc. 21st Ann. IEEE Symp. on Foundations of Computer Science (1980) [7] J.E. Hopcroft, W. Paul and L. Valiant, On time versus space and related problems, J. ACM 24 (1975) [8] N. Pippenger, On simultaneous resource bounds, in: Proc. 20th Ann. IEEE Symp. on Foundations of Computer Science (1979)

2 of R j construct a string (p j m j?1p j m j p j m j +1$) ch : Next, using once again the machine from Lemma 2 construct the string (C$) ch, where C = p 1 m 1?1p 1 m 1 p 1 m 1 +1@ : : k m k?1p k m k p k m k +1 Note, that C is a start-segment conguration of N. Now, in string S (from Step 1) nd a substring (C ` D) ch and generate D ~ = (D$) ch. To nd (C ` D) ch in S within a constant number of reversals M writes down string (C$) ch on the second tape, and places the head on the rst tape at the beginning of the string S and the head on the second tape at the beginning of (C$) ch. Next moving simultaneously both heads to the right M searches the list S for the string (C ` D). Let ~D = (d 1?1d 1 0d 1 +1@ : : k?1d k 0d k +1$) ch : Using the machine from Lemma 3 tranform ~ D into (d 1?1) ch #(d 1 0) ch #(d 1 +1) ch : : : (d k?1) ch #(d k 0) ch #(d k +1) ch : end Then, for j = 1; : : : ; k and i =?1; 0; +1, replace each substring (p j m j +i$) ch R j by (d j i $) ch of After M has completed this algorithm, it examines the current conguration of N to decide if N accepts or rejects x in at most h 2 steps or if N makes more than h 2 steps on x. It is obvious that M works within O(h) reversals. 2 To prove Theorem 1 we use the above lemma and the well-known technique of estimating the value of T (jxj), for an input x. Proof of Theorem 1. Let L be accepted by a k-tape DTM N of time complexity T (n). We construct a 2-tape machine N 0 running within O( q T (n)) reversals that simulates N. Let x be an input for N. N 0 performs the following instructions. 1. h := 2; 2. check if N accepts or rejects x within t h 2 steps; to this end N 0 works like the machine M from Lemma 4; 3. if N makes more than h 2 steps on x then let h := 2h and goto 2; 4. if N accepts x then Accept else Reject. 5

3 1.3 Within O(h) reversals construct a string S = (v 1 ` z 1 ) ch #(v 2 ` z 2 ) ch # : : : #(v eh ` z eh ) ch ; where c is the constant as in 1.1, and write down S on the rst tape. Step 2 f Initial description g M will code the full contents of the tapes of N on input x at a step t by a sequence of strings R 1 ; R 2 ; : : : ; R k. Each R j has the form (b j?h$) ch # : : : #(b j?1$) ch #(b j 0$) ch #(b j 1$) ch # : : : #(b j h$) ch where b j i is the contents of the i-th block of tape T j at time t. Moreover, if the head on T j visits the m j -th block at time t then b j m j = u j 1hq; a j iu j 2, where u j 1a j u j 2 is the contents of the block and a j is the symbol read by this head in state q. Let [ denote the blank symbol and let the input be x = x 1 : : : x n. We assume that the input word is initially placed on the rst tape of N. To generate an initial description of N on x do 2.1 For j = 1; : : : ; k, within O(h) reversals, generate strings P j = p j?h# : : : #p j?1#p j 0#p j 1# : : : #p j h; where p 1 i = 8 >< >: x ih+1 x ih+2 : : : x ih+h if 0 i bn=hc? 1, x ih+1 : : : x n [ h?(n?ih) if i = bn=hc, [ h otherwise, and, for j = 2; : : : ; k, p j i = [ h. Next replace the rst symbol a j of each string p j 0 by the symbol hq 0 ; a j i, where q 0 is the inititial state of N. 2.2 Within O(h) reversals, for j = 1; : : : ; k, from string P j generate a string Step 3 f The simulation g Repeat h times begin Let R 1 ; R 2 ; : : : ; R k with R j = (p j?h$) ch # : : : #(p j 0$) ch # : : : #(p j h$) ch R j = (p j?h$) ch # : : : #(p j 0$) ch # : : : #(p j h$) ch describe the state and the contents of the tapes of N on x at time t. M will replace R 1 ; R 2 ; : : : ; R k by a new sequence that represents N at time t + h. Let m j be the index of a block on T j which is visited by the head. For each j = 1; : : : ; k using the machine from Lemma 2 from substring (p j m j?1$) ch #(p j m j $) ch #(p j m j +1$) ch 4

4 Proof. Let N be a k-tape DTM for some k 2 N. We will construct a 2-tape DTM M which given an integer h > 0 and a string x simulates h 2 steps of N on x. The integer h may be given in binary or unary form. Machine M performs the simulation in h stages. In each stage it simulates h consecutive steps of N on x within a constant number of reversals. To this end M groups the cells of each tape T j of N into blocks : : : B j?2b j?1b j 0B j 1B j 2 : : : labelled by the integers. Each block B j i consists of h consecutive cells of tape T j in the range from ih + 1 to ih + h. Since we want to simulate at most h 2 steps it is sucient to consider only blocks labelled by?h; : : : ;?1; 0; 1; : : : ; h. Let a segment denote three consecutive blocks of a tape and a segment conguration of N a word of the form w 1 1hq; a 1 iw 1 2@w 2 1hq; a 2 iw 2 2@ : : k 1hq; a k iw k 2; (2) where q is a state of N and for j = 1; : : : ; k, w1a j j w j 2 are words of length 3h over the alphabet of N. A segment conguration (2) represents the fact that the machine is in state q and the j-th head visits a segment containing w1a j j w j 2 and reads the symbol a j. Let a start-segment conguration be a segment conguration C such that each head in C visits the middle block of a segment. Now M works as follows: Step 1 f Generate an auxiliary string S g 1.1 Within O(h) reversals generate a list L = v 1 #v 2 # : : : #v eh of all start-segment congurations of machine N. The number of all startsegment congurations e h is bounded by c h, where c is a constant that depends only on N. M can construct L as follows. At the beginning it generates, as in [1] or [2], a list of all words of length k(3h + 1)? 1. Then M eliminates from this list all words which are incorrect start-segment congurations. 1.2 For each start-segment conguration v i nd a segment conguration z i such that N beginning its work in v i reaches z i in h steps. If N starting in v i cannot perform h steps then M nds the last conguration reached by N when starting in v i. M achieves this goal as follows. At the beginning it replaces the list L by a list L 0 = v 1`w 1 # v 2 `w 2 # : : : # v eh `w eh where w i = v i for each i = 1; : : : ; e h. Now M traverses the list L 0 h times. Every time, when scanning L 0, M replaces each w i by its immediate successor. Let the obtained list be of the form v 1 `z 1 # v 2 `z 2 # : : : # v eh `z eh 3

5 where DREV ERSAL 2 (S) denote the class of languages recognized by 2-tape DTMs making at most O(S(n)) reversals. By a 2-tape TM we mean a machine with 2 two-way read-write tapes. We assume that the input word is placed initially on the rst tape. It is not dicult to prove that (1) holds also for functions S(n) log n which are not reversal constructible (in [2], Chen and Yap show that, for such functions, DSP ACE(S) is included in DREV ERSAL 2 (S 2 )). Using (1) and the well known result of Hopcroft, Paul and Valiant ([7]) relating time and space one can show that any multitape DTM of time complexity T (n) can be simulated by a deterministic TM making at most R(n) = T (n)= log T (n) reversals. In this paper it is shown that the function R(n) can be reduced to q T (n). Theorem 1 Let T (n) n be an arbitrary function. Then any language L accepted by a multitape DTM in time T (n) can be accepted by a 2-tape DTM within O( q T (n)) reversals. The problem if the inclusion (1) is proper remains open. Our result relating time to reversal seems to conrm a conjecture that R(n) reversal bounded machines are more powerful than the machines working in space R(n). 2 A reversal ecient simulation of time bounded DTM To prove that for all functions T (n), DT IME(T ) DREV ERSAL 2 ( p T) methods developed by Chen and Yap in [2] will be used. Let us begin with two technical lemmata. In these lemmata it will be assumed that s is a xed constant, is a xed alphabet and that $; # 62. Lemma 2 There is a 2-tape DTM M such that for any positive integer r and n, given a string (x 1 $) r #(x 2 $) r # : : : #(x s $) r ; where x 1 ; x 2 ; : : : ; x s 2 n, M constructs the string (x 1 x 2 : : : x s $) r within O(1) reversals. Lemma 3 There is a 2-tape DTM M such that for any positive integer r and n, given a string (x 1 x 2 : : : x s $) r ; where x 1 ; x 2 ; : : : ; x s 2 n, M constructs the string (x 1 $) r #(x 2 $) r # : : : #(x s $) r within O(1) reversals. The above lemmata have straightforward proofs and we omit them. The theorem above will follow from the following lemma. Lemma 4 For any multitape DTM N there is a 2-tape DTM M that given an integer h > 0 and a string x checks within O(h) reversals if N accepts x in at most h 2 steps. 2

6 On the Relationship Between Deterministic Time and Deterministic Reversal Maciej Liskiewicz Institut fur Theoretische Informatik, TH Darmstadt, Darmstadt, Germany and Instytut Informatyki, Uniwersytet Wroc lawski, Wroc law, Poland Correspondence to: Maciej Liskiewicz, Institut fur Theoretische Informatik, TH Darmstadt, Alexanderstr. 10, 6100 Darmstadt, Germany Abstract We investigate the relationship between the time and the reversal complexity measure for deterministic Turing machines. It is shown that for any function T DT IM E(T ) DREV ERSAL( p T ): Keywords: Computational complexity, Turing machine, reversal, time complexity 1 Introduction Besides time and space, the number of reversals made by the tape heads during a computation is an important and widely used complexity measure for Turing machines (TM). The reversal complexity was introduced by Hartmanis ([5]) and Fischer ([4]) and intensively studied by many authors. For example, it has been shown that the reversal complexity on multitape deterministic TMs is intimately connected with parallel time ([3],[6],[8]). In this paper we investigate the relationship between reversals and deterministic time. It is obvious that the heads of a multitape deterministic TM (DTM) working in time T (n) make at most O(T (n)) reversals. Moreover, it is easy to see that a multitape DTM of time complexity T (n) can be simulated by a 1-tape DTM making at most T (n) reversals. An interesting open problem is for each T (n) to nd a minimal function R(n) such that every language accepted in time T (n) can be accepted by a DTM of reversal complexity R(n). Recently Chen and Yap ([2]) have shown that for any reversal constructible function S(n) log n DSP ACE(S) DREV ERSAL 2 (S); (1) This research was supported by the Alexander-von-Humboldt-Stiftung 1