A note on the paper “disjunctive languages and compatible orders”

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I NFORMATIQUE THÉORIQUE ET APPLICATIONS

M. I TO

A note on the paper “disjunctive languages and compatible orders”

Informatique théorique et applications, tome 25, n^o5 (1991), p. 419-421

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Informatique théorique et Applications/Theoretical Informaties and Applications (vol 25, n°5, 1991, p 419 à 421)

A NOTE ON THE PAPER

"DISJUNCTIVE LANGUAGES AND COMPATIBLE ORDERS" (*)

by M. ITO C)

Communicated by J -E PIN

Abstract — In a previous paper we have dealt with m-disjunctwe languages, s-disjunctive Ianguages over an alphabet X and related compatible partial orders on X* However, in our former paper, the cardinahty of X has been assumed to be greater than one In this note, we will deal with the case X an alphabet with one letter We show that every disjunctive language over an alphabet with one letter is s-disjunctive

Resumé - Dans un article antérieur, nous avons étudié les langages m-disjonctifs et s-disjonctifs sur un alphabet X ainsi que les relations d'ordre partiel compatibles qui leur sont associées Cependant, nous n'avions considéré jusqu'à présent que des alphabets d'au moins deux lettres Dans cette note, nous traitons le cas d'un alphabet à une lettre et nous montrons que tout langage disjonctif sur un alphabet à une lettre est s-disjonctif

Let X be a imite nonempty alphabet and let L be a language over X. For ueX*, let L. .u—{(x,y) \ xuyeL}. The language L is called disjunctive if L, .u = L. . v implies u = v. If L is disjunctive, then the relation ^L defîned on X* by u ^Lv if and only if L. .u <= L. . v is a compatible partial order.

In the case that ^L is the identity relation, then L is said to be s-disjunctive;

otherwise L is called m-disjunctive. The purpose of this note is to prove that every disjunctive language over an alphabet with one letter is s-disjunctive.

One may think that this resuit has already been shown by Proposition 2.2 in [1]. The statement of Proposition 2.2 is as follows: If L <= X* is a semi- discrète language, then L is s-disjunctive, However, as the proof of Proposition 2.2 dépends on the fact \X\ ^ 2, we must show it independently for the case

(*) Received November 1989, revised July 1990.

l1) Faculty of Science, Kyoto Sangyo Umversity, Kyoto 603, Japan

Informatique théorique et Applications/Theoretical Informaties and Applications 098S-3754/91/05 419 03/S2 30/© AFCET-Gauthier-ViUars

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420 M. ITO

LEMMA 1: Let X be a finite nonempty alphabet^ let L <i X* be an m- disjunctive language and let ueX+, If L. .uL <•= L. .uj and L. .um <= L. .un for some U h m, n^.0, then (ƒ — i) (n~m) ^ 0.

Proof: S u p p o s e t h a t i<j a n d n < m. Let r, s be positive integers such that r(j-i) = s(m-n). Then L, ,u^ri ç L . ,u^rj and L . .u^{s m} ^L. .u^sn. There- fore L. .uri+sm^ L. ,urj+sm^L. .urj+sn. Since L is disjunctive, we have uri+sm = urj+sm, Le. i=j, a contradiction. This complètes the proof of the lemma.

LEMMA 2: Let yu y2, . . -, yfc be positive integers whose greatest common divisor equals y. Then there exists a positive integer s satisfying the following condition:

For every i ^ 0, there exist nonnegative integers a^l9 a², . ..,<x^fc such that

Proof: This follows from the définition of the greatest common divisor.

LEMMA 3: Let L <= X* be an m-disjunctive language, let ueX* and let L. . umi g L. ,uni with nt > mi and 1 ^ i g k. Let yt — ni — mi and let

? > m a x { m^{1 ?} m²⁹ . . ., m^k}. Ifu'eL, then^{Wt +} P m + P2ï2+- • >+Hu^eL for every Proof: Since ut = ut~miumi, t > mt and L. .umi g L . .w% we have ut~miuni = ut+yieL. The lemma follows by induction.

In the following lemma, L is assumed to be an m-disjunctive language satisfying the conditions in Lemma 3. Moreover, s, y, t and u are assumed to be the values and the element in X+ appearing in the preceding lemmas.

LEMMA 4: Let t < t' and let t = ^ ( m o d s y ) . Ifu^leL, then ü' U ul' us Y OY)* g ul U ü us y (wY)* g L.

Proof: This follows immediately from the preceding lemmas.

Now we are ready to prove that every disjunctive language over an alphabet with one letter is ^-disjunctive.

PROPOSITION 1: Let X be an alphabet with only one letter and let L <^ X* be a disjunctive language. Then L is s-disjunctive.

Proof: Let X={a). Suppose that L is not ^-disjunctive. Let H = {n — m | n # my L. ,am g L. .a"}. Since L is m-disjunctive, H is not empty. From Lemma 1, we have the following two cases.

Informatique théorique et Applications/Theoretical Informaties and Applications

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DISJUNCTIVE LANGUAGES AND COMPATIBLE ORDERS 4 2 1

Case 1: For every heH, h > 0. Let y be the greatest common divisor of the numbers from H. Then there exist y^{l 5} y², . . .⁵y^k such that Y = g.c.âf.{y^{1 )}y², . . . , y^{f c}} where y^l9 y², . . ., y^keH and y^l = n^l-m^t with L . .a™1 <=: L. .anK For every z, 1 ^ z ^ .sy, let f, be the least integer such that max {mu m2, . . ., mk} S t» tt = i (mod y s) and disL. Let

ƒ= {z | 1 ^ z ^ 5y and t^t exists}.

Then by Lemma 4, we have L = ( U ( ^ U ^ ^ ^ a7) * ) ) U F where

i e /

F S U aJ with

7"= max {mu m2, . . ., mfc}.

Clearly L is regular, a contradiction.

Case 2; For every heH, h < 0. Note that L. .a^m ^ L. .aⁿ implies (X*\L). . aⁿ g ( T * \ L ) . . a^m. As in Case 1, it can be shown that the language X*\L is regular. Hence L is regular, a contradiction. Therefore, L is 5-disjunctive.

ACKNOWLEDGEMENTS

This work was done during the author's stay at The University of Western Ontario The author would like to thank Prof G Thiernn for his invitation and usefui discussion The author would like also to thank the referee for his valuable comments

REFERENCES

1 M. ITO, H. J. SHYR, G. THIERRIN, Disjunctive Languages and Compatible Orders, R.A.LR.O. Inform. Théor. AppL, 1989, 23, pp 149-163

vol 25, n° 5, 1991