KOBAYASHI, YUJI

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RIGHT:URL:CITATION:AUTHOR(S):ISSUE DATE:TITLE:Rewriting Systems with LowDerivational Complexity (Algebraicsystem, Logic, Language andRelated Areas in ComputerSciences II)KOBAYASHI, YUJIKOBAYASHI, YUJI. Rewriting Systems with Low Derivational Complexity (Algebraic system, Logic, Language and RelatedAreas in Computer Sciences II). 数理解析研究所講究録 2021, 2188: 81-862021-07http://hdl.handle.net/2433/26561981Rewriting Systems with Low Derivational Complexity * YUJI KOBAYASHI Laboratory of Mathematics and Games https//: math-game-labo.com 1 Rewriting systems and complexity Let A be an alphabet, a finite set of letters and let A* = { a 1 a 2 ... an I n ?: 0, ai E A} be the free monoid generated by A. The empty word in A* is denoted by 1. We denote by lxl the length n of a word x = a1a2 ... an EA*. A rewriting system R on A is a subset of A* x A*. An element r = ( u, v) of R is called a rule and is written as u ---+ v. R is finite if it is a finite set. For two words x and yin A*, if x = x 1ux2 , y = x 1 vx2 with x1 ,x2 EA*, we write as x ---+r y. If there are words x1, ... , Xk-1 EA* and rules r1, ... , rk E R such that (1) we write as x ➔i y or simply x ---+k y. We call (1) a derivation sequence in R of length k and say that y is derived from x for k steps. If there is no sequence of length larger thank starting with x, (1) is called maximal. For x E A* the derivational length JR(x) of x is the length of a maximal sequence starting with x, that is, JR(x) = max{k I :3y EA*, x ➔i y}. The ( derivational) complexity dR of R is defined by the function that relates the largest length of derivation sequences in R to the length of starting words; ( see [1] and [2]). If J R ( x) < oo for all x E A*, R is called terminating. If R is terminating, dR is a function from N to N. For two functions f, g: N---+ N, we write f = O(g) (resp. f = O(g)), if there is a constant C > 0 such that f(n) ::=; Cg(n) (resp. f(n) ?: Cg(n)) for sufficiently large n. We say f and g are equivalent, and write as f ~ g or f = 8(g) if f = O(g) and f = O(g). *this is a preliminary version and a full version will appear elsewhere 82Example 1.1. (1) The system R ={a ➔ 1} on {a} has linear complexity, in fact, oR(an) = n and dR(n) = n. (2) Any nonempty system R has at least linear complexity, that is, dR(n) = O(n). (3) The system R = { ab ➔ ba} on { a, b} has quadratic complexity. In fact, oR(anbn) = n 2 and dR(n) = ¼ n 2 = 8(n2 ). ( 4) The system R = { ab ➔ b2 a} on { a, b} has exponential complexity. In fact, oR(anbn) = n(2n - 1) and 0(2n) = dR(n) = 0(3n). Kobayashi [2] proved that for any real number a 2". 2 there is a finite rewriting system with complexity equivalent to na., if computational complexity of a is not very high (bounded by C2n for some C > 1), and posed the following problem. Question 1.2. For a real number a with 1 < a < 2, is there a finite rewriting system with complexity equivalent to na.? Recently, Talambutsa [3] has given a positive answer for any rational a with 1 < a < 2. That is, for any rational number a 2". 1 there is a finite rewriting system with complexity 8(na.). To his end he constructed a supplementary system which is length-preserving and has complexity 8(nlogn). In the next section we give a little different system with this complexity whose mechanism will appear in the system with complexity 8 ( n log log n) given in the last section. 2 System with complexity n log n Consider an alphabet A1 = {a,a,h,p,v,w} and a system Ro= {a2h ➔ ha, wh ➔ wp, pa ➔ ap} over A1 . Let x = wanhv with even number n 2". 0, then we have a derivation sequence in Ro. This is a maximal sequence starting with x, in which h travels for n/2 steps from right to left, and at the left end it changes to p and returns to the original position (the pair (h,p) shuttles once between v and w). Thus, OR0 (x) = n + 1. Adding a new rule r0 = (apv,ahv) to Ro, set R1 = Ro U {apv ➔ ahv}. Suppose n = 2i with i > 1 and let x = wanhv, then we have a maximal derivation sequence 83in R 1 . In this sequence the pair (h,p) shuttles i = log2 n times between v and w, and we have Next, let A2 = {b, b, f, q, v, w }, and consider a system R2 = {fb ➔ bf, fw ➔ qw, bq ➔ qb}. For a word x = vfbnw (n ~ 1) we have a maximal sequence in R2. In the sequence the pair (J, q) shuttles once between w and v, and we have (3) Now let A3 = A1 u A2 = {a,a, b, b, h,p, f, q,v,w}, and define a system R 3 by adding a rule r 1 = (apvq, ahvf) to the union of Ro and R 1 , that is, R3 = Ro U R2 U {ri} = {a2h ➔ ha, wh ➔ wp,pa ➔ ap, fb ➔ bf, fw ➔ qw, bq ➔ qb, apvq ➔ ahvf}. Let n = 2i (i ~ 1) and x = wanhvfbnw E A2. We have a maximal sequence n+l n 2n+l n n x -*Ro wa 2 pvfbnw ➔R2 wa 2 pvqbnw ---+r1 wa 2 hvfbnw ---+k:1 wa!1:pvfbnw ➔~:+1 wa!1:pvqbnw ---+r1 wa!1: hvfbnw (4) -*Ro ... ---+r1 wahvfbnw ➔~:+l wahvqbnw. in R 3 • In ( 4) the movements in the left side and in the right of v synchronize, one shuttle of (h,p) in the left corresponds to one shuttle of (J, q) in the right. The number of the shuttlings of ( h, p) is i = log2 n and the number of derivation steps in them is O(n) by (2) above. The number of applications of the rule r 1 is i, and the number of shuttlings of (J, q) in the right side is also i. Hence, the number of steps in the shuttling s of (J, g) is (2n + 1) log2 n by (3). The length of the sequence ( 4) is the sum of these numbers of steps and is dominated by the last number, and hence we see b R 3 ( x) = 8 ( n log n). Because ( 4) gives the maximum length relative to the length of the starting word among all sequences in R3 ( the details are omitted), we see Talambutsa asked about the existence of a finite system with complexity strictly between 8(n) and 8(nlogn). In the next section we give a system with complexity n log log n. 843 System with complexity n log log n Let A4 = {b,b,b,c,c,f,q,v,w}, and consider a system R4 over A4 similar to R2 : R4 = {fb---+ bf, Jc---+ cf, fw---+ qw, bq---+ qb, cq---+ qc}. For a word x = vfbmcnw (m, n ~ 0) we have a maximal sequence In (5) the pair (f, q) shuttles once between w and v, and we have Next, let A5 = {b,b,b,c,g,r,v,w}, and R 5 = {gb---+ bg, gb---+ bg, gc---+ rb2 , br---+ rb, br---+ rb}. Let x = vgbmcnw with m ~ 0, n ~ 1. Then, we have In this sequence the pair (g, r) shuttles once between v and c, and (5) Let A6 = A1 U A5 and let R 6 be the union of Ro and R5 adding a rule r2 = (apvrb, ahvg); R5 = Ro U R5 U {apvrb---+ ahvg}. Let i,j > m ~ 0 and n = 2i. For a word x = wanhvgbmdw E A6 we have In this situation we write x ====}(6) y. In (6) the pairs (h,p) and (g,r) both shuttle m + 1 times between v and w, and the number of steps in the shuttlings of (g,r) is 6 R 6 ( x) = 2 ( m + ( m + 1) + · · · + ( m + m)) + m + 1 = 8 ( m 2). (7) Finally, let A1 = A 1 UA4 UA5 = {a,a,b,b,b,c,c,h,p,f,q,g,r,v,w}, 85and let r 3 = (apvqb,ahvg) and r 4 = (apvrb,ahvfb). Define R1 RoUR4UR5U{r3,r4} { a2h _=--+ ha, wh ➔ wp, pa ➔ a_p, fb ➔ bf, f c ➔ cf, fw ➔ qw, bq ➔ qb, cq ➔ qc, gb ➔ bg, gb ➔ bg, gc ➔ rb2 , br ➔ rb, br ➔ rb, apvrb ➔ ahvg, apvqb ➔ ahvg, apvrb ➔ ahvfb }. Let n = 2i(i 2': 1) and x = wanhvfbcnw. We have a maximal sequence in R7 . Here, 0 ,S: k ,S: 2j-l, j is the number of the shuttlings of the pair (f, q), £ is the number of shuttlings of (g, r ), and s = q if k = 0 and s = r otherwise. Moreover, the pair (g, r) shuttles 2t-l times after the t-th shuttling of (f, q) for t < j and shuttles k times after the last j-th shuttling of (f, q). Thus we see £ = 1 + 2 + · · · + 2J-2 + k. Now, in the left side of the letter v in (8), the pair (h,p) shuttles i = log2 n times, and corresponding to it, in the right side the pairs (f, q) and (g, r) shuttle i + 1 times together. Hence, i + 1 = j + £ = j + 2j-l - 1 + k, (9) and so j = 8(logi) = 8(loglogn). Thus, the number of the steps in the shuttlings of (f, g) in (8) is (2n + 3)j = 8(nloglogn). On the other hand, the number of the steps in the shuttlings of (g, r) is 0(£2 ) by (7) and by (9) it equals 0(22J) = O(i2 ) = O(log2 n), and the number of the steps in the shuttlings of (h,p) is O(n) by (2). Further, the rules r 2 , r3 and r 4 are applied i = O(logn) times altogether. To estimate 0R7 (x), we can ignore these numbers and we may only take the shuttling of (g, r) into account. Thus, we see 0R7 (x) = 8(nloglogn). Because words of the form of x give the maximum derivation length relative to the length of the words, we finally have 86References [1] D. Hofbauer, C. Lautermann, Termination proofs and the length of deriva-tions, RTA 1989, LNCS 355, (1989), 167-177. [2] Y. Kobayashi, The derivational complexity of string rewriting systems, The-oret. Comp. Sci., 438 (2012), 1-12. [3] A. Talambutsa, On subquadratic derivational complexity of semi-Thue sys-tems, Proc. Computer Science in Russia-2020, LNCS, to appear. 

Rewriting Systems with Low Derivational Complexity (Algebraic system, Logic, Language and Related Areas in Computer Sciences II)

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Rewriting Systems with Low Derivational Complexity (Algebraic system, Logic, Language and Related Areas in Computer Sciences II)

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