Grammars based on the shuffle operation
Institute of Mathematics of the Romanian Academy of Sciences
PO Box 1 - 764, Bucuresti, Romania
University of Leiden, Department of Computer Science
Niels Bohrweg 1, 2333 CA Leiden, The Netherlands
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
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
Keywords: Shuffle operation, Chomsky grammars, L Systems
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 , the semi-contextual grammars in , the
-grammars in , the pattern grammars in , and the
contextual grammars . The starting point of the present paper is
this last class of grammars, via the paper , 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
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 , ,
 and in
modular contextual grammars : 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 , , ,  (and , for applications). In
,  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  is a
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
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 , .
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
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,
The grammars considered in  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  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:
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 idea is to allow the strings into be shuffled only to strings in the
corresponding set. The setsare called selectors.
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
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,
bywe denote the family of languages generated by grammars as in , 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
- Every family includes strictly the family SL.
- The family SL is incomparable with each .
(i) If is a simple shuffle grammar, then for each and
(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 ): 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
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
. 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
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
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
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.
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
, 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
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.
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
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:
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
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 .
Indeed, from a string containing occurrences of every symbol
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
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)
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.
- Each two of the families
are incomparable, excepting the case of
for which we have the proper inclusion .
consists of incomparable families; the following
inclusions are proper: for all ,
Lemma 1. For each we have
Proof. Follows from the fact that contains one-letter
languages, but the one-letter languages in the other families are regular
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 .
Lemma 3. The language , for
is not in , or .
Proof. We have
Assume that this language can be generated in one of the modes
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,
Lemma 5. The language is in but not
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 .
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
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
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
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
, then we
can produce strings of the form
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.
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
(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
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
- ( )
- 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 .
Proof. Consider the grammar
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
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 .
Conversely, take a grammar and write
each , in the form
for finite languages, and associated with .
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
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.
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
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