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DOI:   10.3217/jucs-001-01-0067

# Grammars based on the shuffle operation

Gheorghe Paun
Institute of Mathematics of the Romanian Academy of Sciences
PO Box 1 - 764, Bucuresti, Romania

Grzegorz Rozenberg
University of Leiden, Department of Computer Science
Niels Bohrweg 1, 2333 CA Leiden, The Netherlands

Arto Salomaa
Academy of Finland and University of Turku
Department of Mathematics, 20500 Turku, Finland

Abstract: We consider generative mechanisms producing languages by starting from a finite set of words and shuffling the current words with words in given sets, depending on certain conditions. Namely, regular and finite sets are given for controlling the shuffling: strings are shuffled only to strings in associated sets. Six classes of such grammars are considered, with the shuffling being done on a left most position, on a prefix, arbitrarily, globally, in parallel, or using a maximal selector. Most of the corresponding six families of languages, obtained for finite, respectively for regular selection, are found to be incomparable. The relations of these families with Chomsky language families are briefly investigated.

Categories: F4.2 [Mathematical Logic and Formal Languages]: Grammars and other Rewriting Systems: Grammar types, F4.3 [Mathematical Logic and Formal Languages]: Formal Languages: Operations on languages

Keywords: Shuffle operation, Chomsky grammars, L Systems

# 1. Introduction

In formal language theory, besides the basic two types of grammars, the Chomsky grammars and the Lindenmayer systems, there are many "exotic" classes of generative devices, based not on the process of rewriting symbols (or strings) by strings, but using various operations of adjoining strings. We quote here the string adjunct grammars in [7], the semi-contextual grammars in [3], the -grammars in [10], the pattern grammars in [2], and the contextual grammars [8]. The starting point of the present paper is this last class of grammars, via the paper [9], where a variant of contextual grammars was introduced based on the shuffle operation. Basically, one gives two finite sets of strings,and, over some alphabet, and one considers the set of strings obtained by starting from and iteratively shuffling strings from, without any restriction. This corresponds to the simple contextual grammars in [8].

We consider here a sort of couterpart of the contextual grammars with choice, where the adjoining is controlled by a selection mapping. In fact, we proceed in a way similar to that used in conditional contextual grammars [12], [13], [14] and in modular contextual grammars [15]: we start with several pairs of the form, a language (we consider here only the case when is regular or finite) and a finite set of strings, and allow the strings in to be shuffled only to strings in. Depending on the place of the string in in the current string generated by our grammar, we can distinguish several types of grammars: prefix (the string in appears in the left-hand of the processed string, as a prefix of it), leftmost (we look for the leftmost possible occurrence of a string in), arbitrary (no condition on the place where the string in appears), global (the whole current string is in), and parallel (the current string is portioned into strings in). An interesting variant is to use a substring as base for shuffling only when it is maximal. Twelve families of languages are obtained in this way. Their study is the subject of this paper.

It is worth noting that the shuffle operation appears in various contexts in algebra and in formal language theory; we quote only [1], [4], [5], [6] (and [11], for applications). In [1], [6] the operation is used in a generative-like way, for identifying families of languages of the form, the smallest family of languages containing the finite languages and closed under union, concatenation and shuffle. (From this point of view, the family investigated in [9] is a particular case,).

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The shuffle of the symbols of two words is also related to the concurrent execution of two processes described by these words, hence our models can be interpreted in terms of concurrent processes, too. For instance, the prefix mode of work corresponds to the concurrent execution of a process strictly at the beginning of another process, the 'beginning' being defined modulo a regular language; in the global case the elementary actions of the two processes can be freely intercalated.

As it is expected, the languages generated by shuffle grammars are mostly incomparable with Chomsky languages (due to the fact that we do not use nonterminals). Somewhat surprising is the fact that in the regular selection case the five modes of work described above, with only one exception, cannot simulate each other (the corresponding families of languages are incomparable).

# 2. Classes of shuffle grammars

As usual,denotes the set of all words over the alphabet, the empty word is denoted by, the length ofbyand the set of non-empty words overis identified by. The number of occurrences of a symbolin a stringwill be denoted by. For basic notions in formal language theory (Chomsky grammars and L systems) we refer to [16], [17]. We only mention thatare the families of finite, regular, context-free, context-sensitive, and of 0L languages, respectively.

Forwe define the shuffle (product) of, denoted , as

Various properties of this operation, such as commutativity and associativity, will be implicitely used in the sequel. The operationis extended in the natural way to languages,

and iterated,

The grammars considered in [9] are triples of the formwhere is an alphabet,andare finite languages over. The language generated byis defined as the smallest languageovercontainingand having the property that ifand, then. (Therefore, this language is equal to.)

There is no restriction in [9] about the shuffling of elements ofto current strings. Such a control of the grammar work can be done in various ways of introducing a context-dependency. We use here the following natural idea:

Definition 1. A shuffle grammar is a construct

whereis an alphabet,is a finite language over,are languages overandare finite languages over,.

The parameteris called the degree of . Ifare languages in a given family,then we say thatis withchoice. Here we consider only the casesand

The idea is to allow the strings into be shuffled only to strings in the corresponding set. The setsare called selectors.

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Definition 2. For a shuffle grammaras above, a constant, and two strings, we define the following derivation relations:

These derivation relations are called arbitrary, prefix, leftmost, and global derivations, respectively. Moreover, we define the parallel derivation as

.

We denote.

Definition 3. The language generated by a shuffle grammarin the mode is defined as follows:

For the parallel mode of derivation we define

The corresponding families of languages generated by shuffle grammars withchoice,, are denoted by, respectively; bywe denote the family of languages generated by grammars as in [9], without choice. Subscriptscan be added towhen only languages generated by grammars of degree at most, are considered.

# 3 The generative capacity of shuffle grammars

From definition we have the inclusions for all ,and .

Every family contains each finite language. This is obvious, because for we have for all . In fact, we have

Theorem 1.

(i)
Every family includes strictly the family SL.
(ii)
The family SL is incomparable with each .
(iii)
.

Proof.

(i) If is a simple shuffle grammar, then for each and

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(The only point which needs some discussion is the fact that a parallel derivation in and , can be simulated by a -step derivation in , because the strings shuffled into do not overlap each other.) Consequently, .

The inclusion is proper in view of the following necessary condition for a language to be in (Lemma 10 in [9]): if and

then each string in is a subword of a string in , condition rejects languages such as ; this language can be generated by the shuffle grammar with finite choice

in all modes of derivation excepting the global one. For take

(ii) We have already seen that for Consider now the simple shuffle grammar

We have

Assume that for some shuffle grammar with finite and Take a string with arbitrarily large . A derivation must be possible in , with for some. We must have , and for . If , this derivation step can be omitted, hence we may assume that .

The sets are finite. Denote

Therefore , that is . For we have , hence either or . In both cases, the shuffling with modifies at most two of the three subwords , hence a parasitic string is obtained. (In the prefix derivation case we precisely know that only is modified.)

Consider now the parallel case. Take such that finite sets. For obtaining a string with large enough we need a derivation . This implies we have sets containing strings and with the corresponding sets containing strings (similarly for and ).

Assume that we find in the sets two strings and in the associated sets we find two strings. Similarly, we have pairs and . The string for is in and it can be rewritten using the same pairs of strings containing the symbols and and different pairs for :

Consequently, we must have

which implies

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For all such pairs we obtain the same value for . Denote it by . Rewriting some using such pairs we obtain

occurrences of . Continuing steps, we get occurrences of , hence a geometrical progression, starting at.

Symbols can be also introduced in a derivation by using pairs with containing occurrences of . At most one such pair can be used in a derivation step. Let be the largest number of symbols in strings as above (the number of such strings is finite). Thus, in a derivation , the number can be modified by such pairs in an interval .

We start from at most strings of the form , hence we have at most geometrical progressions . The difference can be arbitrarily large. When , the at most progressions can have an element between and ; each such element can have at most values. Consequently, at least one natural number between and is not reached, the corresponding string , although in , is not in . This contradiction concludes the proof of point (ii).

(iii) As only strings in the sets , of a grammar can be derived, we have . The inclusion has been pointed out at the beginning of this section.

Corollary. The families , contain non-context-free languages.

For and this assertion can be strenghtened.

Theorem 2. Every propagating unary 0L language belongs to the family .

Proof. For a unary 0L system with and

with , we construct the shuffle grammar

It is easy to see that .

This implies that contain one-letter non-regular languages. This is not true for the other modes of derivation.

Theorem 3. A one-letter language is in or in if and only if it is regular.

Proof. Take a shuffle grammar with regular sets . Clearly, . Because we work with one-letter strings, a string can be derived using a component if (and only if) the shortest string in is contained in , hence is of length at most . Therefore we can replace each , where

without modifying the generated language. Denote

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and construct the right-linear grammar with

At the first steps of a derivation in , the nonterminals count the number of symbols introduced at that stage, in order to ensure the correct simulation of components of when this number exceeds , then each component can be used, and this is encoded by the nonterminal . Thus we have the equalities

For the global derivation we start from an arbitrary shuffle grammar , with regular sets , take for each a deterministic finite automaton and construct the right-linear grammar

with containing the following rules:

New occurrences of , corresponding to sets , are added (it does not matter where) only when the current string belongs to , and this is checked by the simultaneous parsings of the current string in the deterministic automata recognizing . Consequently, is a regular language.

Conversely, each one-letter regular language is known to be equal with the disjoint union of a finite language and a finite set of languages of the form

.

(The constant is associated to , but is the same for all .) Assume we have languages; denote

Then , for the grammar

hence we have the theorem for .

For the global case, starting from regular, written as above

we consider the grammar

The equality is obvious and this concludes the proof.

In view of the previous result, it is somewhat surprising to obtain

Theorem 4. The family contains non-semilinear languages (even on the alphabet with two symbols only).

Proof. Consider the shuffle grammar with the following eight components:

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Examine the derivations in in the global mode. The whole current string must be in some in order to continue the derivation. Assume we start from a string (initially we have). This string is in only, hence we can add one more occurrence of . This can be done in all possible positions of and we obtain a string in, but only when we obtain we can continue the derivation, namely by using the second component of . Now a symbol can be added, and again the only case which does not block the derivation leads to. We can continue with the third component and either we block the derivation, or we get . From a string of the form we can continue only with the first component, hence the previous operations are iterated. When all symbols are doubled, we obtain the string .

Note that (more precisely, ), but for each intermediate string in this derivation at least one of the following inequalities holds:

.

To a string of the form of above only the fifth component of can be applied and again either the derivation must be finished, or it continues only in . The derivation proceeds deterministically through , inserting step by step new symbols between pairs of such symbols in order to obtain again triples . In this way we obtain either strings from which we cannot continue or we reach a string of the same form with , namely . The number of occurrences has been doubled again, whereas in the intermediate steps we have strings with different numbers of occurrences of at least two of .

The process can be iterated. From the previous discussion one can see that for all strings with we also have .

Consequently, denoting by the Parikh mapping with respect to the alphabet , we have

.

This set is not semilinear; the set of semilinear vectors of given dimension is closed under intersection, therefore is not a semilinear language.

Consider now the morphism defined by . Define the grammar

Shuffling a string with a string , in a way different from inserting as a block leads to strings with substrings of the forms or . Take, for example, and examine the possibilities to shuffle after . In order to not get a substring we must insert the first symbol of after the specified occurrence of in . If we continue with the symbol of , this is as starting after this occurrence of , if we continue with the symbol of , then we obtain . Starting after in , after introducing one from we either get (if we continue in ), or (if we alternate in , or (hence is inserted as a block). Similarly, take for example and shuffle the same. Introducing after we have to continue either with from or with from , in both cases obtaining the subword . The same result is obtained in all other cases. After producing a substring or , the derivation is blocked. Inserting as a compact block corresponds to inserting in. Consequently, the derivations in correspond to derivations in . When a derivation is blocked, it can terminate with a string with only when contains the same number of substrings , with the possible exception of such substrings destroyed by the last shuffling, that of when using the pair or of when using the pair .

Moreover,

.

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Indeed, from a string containing occurrences of every symbol we get a string with

Conversely, if a string contains the same number of and occurrences, assume that it contains substrings substrings and substrings . Then it contains occurrences of and occurrences of . Consequently,

which implies . As we have seen in the first part of the proof, when , then also , hence , too. Therefore the obtained string corresponds to a string (in the sense ) such that . This implies and

In conclusion, also is not semi-linear.

For the case of finite selection we have

Theorem 5. .

Proof. Let be a shuffle grammar with all sets , finite. We construct a left-linear grammar as follows.

Let

.

Then

The derivation proceeds from right to left; the nonterminals in the left-hand side of sentential forms memorize the prefix of large enough length to control the derivation in in the prefix mode. Therefore,

Theorem 6. contains non-linear languages.

Proof. Take . We obviously have = the Dyck language over , known to be a (context-free) non-linear language.

Open problems. Are there non-semilinear languages in families , or in ? Are there non-context-free languages in families ?

We proceed now to investigating the relations among families for as above. We present the results in the next theorem, whose proof will consist of the series of lemmas following it.

Theorem 7.

(i)
Each two of the families are incomparable, excepting the case of for which we have the proper inclusion .

(ii)
Each pair

consists of incomparable families; the following inclusions are proper: for all , and .

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Lemma 1. For each we have

Proof. Follows from the fact that contains one-letter non-regular languages, but the one-letter languages in the other families are regular (Theorem 3).

Lemma 2.

Proof. For a shuffle grammar used in the arbitrary mode of derivation, construct .

For in , with , for some , we have , hence in .

Conversely, a derivation in corresponds to a derivation in , with the steps omitted when .

In conclusion,

Lemma 3. The language , for is not in , or .

Proof. We have

Assume that this language can be generated in one of the modes by a grammar . Each string in contains the same number of occurrences of and of , hence for each string in which is used in a derivation we must have the same property. In the mentioned modes of derivation, if we have a derivation (for we have) using a string , then we can obtain, too, hence the use of can be iterated, thus producing strings with arbitrary. For this is contradictory .

Lemma 4. The language is in but not in.

Proof. The language can be generated by the grammar in both modes of derivation and , but it cannot be generated by any shuffle grammar in modes or: in order to arbitrarily increase the number of occurrences we have to use a string, which is shuffled to a string containing the leftmost occurrence of the symbol in the strings in . In this way, strings beginning with can be produced, a contradiction.

Lemma 5. The language is in but not in .

Proof. For we have but cannot be generated in the mode by any grammar: we need a string u which introduces occurrences of and in the global mode this can be adjoined in the right hand of the rightmost occurrence of in the strings of .

Lemma 6. The language , for the grammar , is not in.

Proof. We have

(When the leftmost two symbols are not , the derivation is blocked.)

Assume that for some . For every we have , hence the same property holds for all strings in sets effectively used in derivations. Take such a nonempty string and examine the form of strings . Such strings

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cannot be prefixes of , because otherwise we can obtain parasitic strings by introducing in the left of the pair in strings . Suppose that is a prefix of different from . Then we can shuffle the string in such a way to obtain a pair in the right of the pair already existing in strings of the form . Again a contradiction, hence the strings of the form cannot be derived. This implies that they are obtained by derivations of the form . Consequently, also is possible (we have found that the prefix cannot be rewritten, hence the derivation is leftmost). Such strings are not in .

Lemma 7. The language in the previous lemma is not in the family .

Proof. Assume that for some . If a derivation is possible in , then also is possible, a contradiction. Consequently, the strings can be produced only by derivations of the form .

Assume that there is a derivation of the form .

We cannot have pairs with . Indeed, if such a pair exists, then from a derivation we can produce also, for each .

Examine the pairs , used in the derivation . We know that . If contains a symbol and also contains an occurrence of , then we can produce strings having the subword ; this is forbidden ( cannot generate strings of a form different from . Similarly, we cannot have occurrences of both in and in . Therefore, . Clearly, we must have (no other subwords consisting of or of only appear in the derived string) and (no other subwords consisting of or of only appear in the produced string). Such pairs can be clearly used for producing a parasitic string, a contradiction.

In conclusion, the strings cannot be derived, hence such strings with arbitrarily long must be obtained by derivations . Then the symbols and in the prefix of must be separated by an occurrence of , respectively of . Irrespective whether this must be introduced after the first or before the second , we can introduce it before the first , respectively after the second , thus preserving the pair . If the pair in its left-hand has been correctly shuffled with , then we obtain a string of the form with , which is not in . The equality is impossible.

Lemma 8. There are languages in .

Proof. Consider the grammar

.

We have

,

which is not a parallel shuffle language. The argument is similar to that in the previous proof, hence it is left to the reader.

Lemma 9. There are languages in .

Proof. For the grammar we have

Assume that for some . For every effectively used, we must have . No derivation can use a pair , with containing a symbol , otherwise we can produce strings of the form . Consequently, all selectors can be supposed to contain only strings in . A pair can be used in a derivation if (and only if) the pair can be used, for the shortest string in ; for two pairs , only that with the shortest can be used. Denote

From a string in with we can produce no other string; when , we produce only strings of the form without modifying the number ; when and , then we

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can produce strings of the form . However, contains strings with simultaneously arbitrarily large , a contradiction.

Lemma 10. The language generated by in the leftmost mode is not in .

Proof. We have

.

Here is a derivation in in the leftmost mode:

Assume that for some

Examine the possibilities to obtain the strings with arbitrarily large . We cannot have a derivation for a string in , , because we must separate both the left occurrences of and the pairs appearing in the string; irrespective of the used strings to be shuffled we can do only part of these operations, letting for instance a pair unchanged and separating all pairs ; we obtain a parasitic string. Therefore we must have derivations . If in such a derivation we use a pair , such that occurs both in and in , then we can produce strings with substrings . Starting from a string we can then produce strings having not in the left-hand end, a contradiction. It follows that symbols are introduced only by pairs with (no other subword of consists of only occurrences of . In order to cover we must use also pairs with containing and not containing . This implies or . The use of introduces at least one new occurrence of for each , hence, in order to keep the number of and equal, we must have , which implies . Using such pairs we can produce again strings containing pairs not in the left-hand end, a contradiction which concludes the proof.

Let us note that in all previous lemmas the considered languages are generated by grammars with only one component . Consequently, we have

Corollary 1. For all families with which are incomparable, all are incomparable, too, for every .

Corollary 2. Each family such that is incomparable with each family .

Proof. The fact that , for each , has been already pointed out. On the other hand, the language in Lemma 4 is not in or in , and the language in Lemma 6 is not in or on , hence it is not in . These languages are regular, which concludes the proof.

By a straightforward simulation in terms of context-sensitive grammars of checking the conditions defining the correct derivation in a shuffle grammar and of performing such a derivation, we get

Theorem 8. All families are strictly included in CS.

The properness of these inclusions follows from the previous Corollary 2.

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# 4. The case of using maximal selectors

For a shuffle grammar as in the previous sections, we can also define the following mode of derivation: for and , write

(We use for shuffling only if no longer string can be used containing that occurrence of as a substring.)

We denote by the language generated in this way and by , the corresponding families of languages.

Lemma 11. .

Proof. For the assertion is proved by the language in Lemma 4: in general, if (one component, with a singleton selector), then the maximality has no effect, the maximal derivations in are arbitrary derivations. This is the case with the grammar in Lemma 4.

This is also true for the grammar in the proof of Lemma 9; that language covers the case .

Consider now the grammar

.

Starting from a string (initially we have ), we can shuffle the string in the prefix in five essentially different ways:

In cases (2) and (5) the derivation cannot continue (all selectors contain the subword ; in cases (1) and (3) we cannot use the selector because or are present; the latter selectors entail the shuffle of , which changes nothing, hence the derivation is blocked. Only case (4) can be continued. Consequently,

.

This language cannot be generated by a shuffle grammar in the arbitrary mode. Assume the contrary. Every string in , in a grammar as above, must contain the symbol and every string in must contain the same number of occurrences of and of . If , then this string can be ignored without modifying the language of . In order to generate strings with arbitrarily large we must have derivation steps with or (the other strings in contain symbols or in front of ). Starting from we have to introduce one occurrence of between symbols in the substring and one occurrence of between symbols in , that is we need a pair with , and . Then from we can also produce , a parasitic string. If uses a pair with , then can be produced, for arbitrary (we introduce in the left of , thus leaving the occurrence of in unchanged). Consequently, we have to use a pair with . More exactly, is a substring

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of . As contains occurrences of both and , we can derive with arbitrary (all such strings are in in such a way to obtain with containing a substring or a substring . Such a string is not in , hence the grammar cannot generate strings with arbitrarily large . The equality is impossible.

Lemma 12. .

Proof. As and, clearly, , the case is obvious.

Take now a grammar with regular and construct with

.

We have

( )
If , that is then the derivation is maximal, we also have . Consequently, if derives in in the global mode, then derives in in the maximal mode.
( )
Take a derivation . If , then . If , then we have , otherwise the derivation is not allowed: for we have ,hence the derivation in is not allowed. By induction on the length of derivations, we can now show that for every maximal derivation in there is a global derivation in producing the same string, hence .

Lemma 13

Proof. Consider the grammar

We obtain

.

This language is not in . Assume the contrary, that is for some .

If we have a derivation of the form , then we have to separate the symbols in the subwords and by or some , and by or some , respectively. To this aim, a string of the form must be used as selector and a string must be shuffled. But then also is possible, and this is not in the language , a contradiction.

Assume that we have a derivation . Then we must have for some such that . Clearly, . Then from we can derive . If and starts by an occurrence of , then we can produce , for some ; if starts by , then we can produce , for some . No one of these strings is in , a contradiction. Therefore we must have . If contains the symbol , then we can obtain , for some , by appropriate shuffling of and . If contains the letter , then we can obtain , for some , by appropriate shuffling. In both cases we have obtained parasitic strings.

Consequently, the strings cannot be generated, they must be introduced as axioms; this is impossible, because the set is finite, hence .

Lemma 14. A one-letter language is regular if and only if it is in .

Proof. The proof of Theorem 3 shows the fact that every one-letter regular language is in .

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Conversely, take a grammar and write each , in the form

for finite languages, and associated with . Denote

Take for each

a deterministic finite automaton

.

We construct the right-linear grammar with

and contains the following rules: For every language , the difference between the length of two consecutive strings in is bounded by . Therefore the difference between two strings in any of these languages is bounded by . In the nonterminals of we memorize the last states used by the automata when recognizing the current string. A prolongation to right is possible (by rules of type (3)) only according to that language which contains the largest subword of the current string (thus we check the maximality). The derivation can stop in any moment by rules of type (4). Consequently, , hence , which concludes the proof.

Summarizing these lemmas and the fact that contains one-letter non-regular languages, we obtain

Theorem 9. includes strictly and is incomparable with , .

Open problems. Are the differences , for , non-empty ?

The diagram in figure 1 summarizes the results in the previous sections (an arrow from to indicates the strict inclusion of the family .

The maximal derivation discussed above can be considered as arbitrary-maximal, since we are not concerned with the place of the selector in the rewritten string, but only with its maximality. Similarly, we can consider prefix-maximal and leftmost -maximal derivations, when a maximal prefix or a substring which is both maximal and leftmost is used for derivation, respectively. We hope to return to these cases. At least the prefix-maximal case looks quite interesting. For instance, if the sets are disjoint, then in every step of a derivation only one of them can be used, which decreases the degree of nondeterminism of the derivations.

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# 5. Final remarks

We have not investigated here a series of issues which are natural for every new considered generative mechanisms, such as closure and decidability problems, syntactic complexity, or the relations with other grammars which do not use the rewriting in generating languages. Another important question is whether or not the degree of shuffle grammars induces an infinite hierarchy of languages. We close this discussion by emphasizing the variety of places where the shuffle operation proves to be useful and interesting, as well as the richness of the area of grammars based on adjoining strings.

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