Jumping Pure Grammars

Jumping Pure Grammars Abstract This paper introduces and studies jumping pure grammars, which are conceptualized just like classical pure grammars except that during the applications of their productions, they can jump over symbols in either direction within the rewritten strings. The paper compares the generative power of jumping pure grammars with that of classical pure grammars while distinguishing between their versions with and without erasing productions. Apart from sequential versions, the paper makes an analogical study in terms of parallel versions of jumping pure grammars represented by 0L grammars. 1. INTRODUCTION Jumping versions of language-defining rewriting systems, such as grammars and automata, represent a brand new trend in formal language theory (see [1–9]). In essence, they act just like classical rewriting systems except that they work on strings discontinuously. That is, they apply a production so they erase an occurrence of its left-hand side in the rewritten string while placing the right-hand side anywhere in the string, so the position of the insertion may occur far away from the position of the erasure. The present paper contributes to this trend by investigating the generative power of jumping versions of pure grammars, whose original versions were introduced in [10], and their properties are still intensively investigated in language theory (see [11, 12]). Recently, regulated versions of these grammars have been discussed, too (see Chapter 5 in [13–15]). The notion of a pure grammar G represents a language-generating rewriting system based upon an alphabet of symbols and a finite set of productions (as opposed to the notion of a general grammar, its alphabet of symbols is not divided into the alphabet of terminals and the alphabet of nonterminals). Each production represents a pair of the form (x,y), where x and y are strings over the alphabet of G. Customarily, (x,y) is written as x→y, where x and y are referred to as the left-hand side and the right-hand side of x→y, respectively. Starting from a special start string, G repeatedly rewrites strings according to its productions, and the set of all strings obtained in this way represents the language generated by G. In a greater detail, G rewrites a string z according to x→y so it (i) selects an occurrence of x in z, (ii) erases it and (iii) inserts y precisely at the position of this erasure. More formally, let z=uxv, where u and v are strings. By using x→y, G rewrites uxv as uyv. The notion of a jumping pure grammar—that is, the key notion introduced in this paper—is conceptualized just like that of a classical pure grammar; however, it rewrites strings in a slightly different way. Let G, z, and x→y have the same meaning as above. G rewrites a string z according to x→y so it performs (i) and (ii) as described above, but during (iii), G can jump over a portion of the rewritten string in either direction and inserts y there. More formally, by using x→y, G rewrites ucv as udv, where u,v,w,c,d are strings such that either (a) c=xw and d=wy or (b) c=wx and d=yw. Otherwise, G works as described above. The present paper compares the generative power of classical and jumping versions of pure grammars. It distinguishes between these grammars with and without erasing productions. Apart from these sequential versions of pure grammars, it also considers parallel versions of classical and jumping pure grammars represented by 0L grammars (see [16]). As a result, the paper studies the mutual relations between eight language families corresponding to the following derivations modes (see Definition 2.1) performed by pure grammars both with and without erasing productions: classical sequential mode ( ⇒s); jumping sequential mode ( ⇒j); classical parallel mode ( ⇒p); jumping parallel mode ( ⇒jp). In essence, the paper demonstrates that any version of these grammars with erasing productions is stronger than the same version without them. Furthermore, it shows that almost all of the eight language families under considerations are pairwise incomparable—that is, any two families are not subfamilies of each other. The rest of the paper is organized as follows. Section 2 recalls all the terminology needed in this paper and introduces a variety of jumping pure grammars, illustrated by an example. Section 3 presents fundamental results achieved in this paper. Section 4 closes all the study by summing up ten open problems. 2. PRELIMINARIES AND DEFINITIONS This paper assumes that the reader is familiar with the basic notions of formal language theory (see [17–19]). Let A and B be two sets. By A⊆B, we denote that A is included in B and by A⊈B that A is not included in B. A⊂B denotes proper (or strict) inclusion. We say that A and B are incomparable iff A⊈B and B⊈A. The cardinality of A is expressed as card(A). For some n≥0, An denotes the n-fold Cartesian product of set A. By N, we denote the set of all positive integers. Let I⊂N be a finite nonempty set. Then, maxI denotes the maximum of I. For a (binary) relation ϱ over X, ϱi, ϱ+, and ϱ* denote the ith power of ϱ, for all i≥0, the transitive closure of ϱ, and the reflexive and transitive closure of ϱ, respectively. For x,y∈X, instead of (x,y)∈ϱ, we write xϱy throughout. Set dom(ϱ)={x∣xϱy}. Let Σ be an alphabet (finite nonempty set). Then, Σ* represents the free monoid generated by Σ under the operation of concatenation, with ε as the unit of Σ*. Set Σ+=Σ*−{ε}. For w∈Σ* and a∈Σ, #a(w) denotes the number of occurrences of a in w. By substr(w), we denote a set of all substrings of w, that is substr(w)={x∣w=uxv,u,x,v∈Σ*}. The length of w is denoted by ∣w∣. Let n≥0. A set J⊆Nn is said to be linear if there exist α,β1,β2,…,βm∈Nn,m≥0 such that J={x∣x=α+k1β1+k2β2+⋯+kmβm,ki∈N,1≤i≤m}. If J is the union of a finite number of linear sets, we say that J is semilinear. If Σ={a1,a2,…,an} is an alphabet, then for w∈Σ*, ϕ(w)=(#a1(w),#a2(w),…,#an(w)) denote the commutative (Parikh) image of w. For L⊆Σ*, ϕ(L)={ϕ(w)∣w∈L} denote the commutative (Parikh) map of L. We say that L is a semilinear language if and only if ϕ(L) is a semilinear set. A language family is semilinear if and only if it contains only semilinear languages. Let S be a finite set. Define a permutation in the terms of bijective mappings as follows: let I={1,2,…,card(S)} be a set of indices. The set of all permutations of elements of S, perm(S), is the set of all bijections from I to S. An unrestricted grammar is a quadruple G=(V,Σ,P,σ), where V is a total alphabet, Σ⊆V is an alphabet of terminal symbols, P⊆V+×V* is a finite relation and σ∈V+ is the start string of G, called axiom. Members of P are called productions. Instead of (x,y)∈P, we write x→y throughout. For brevity, we sometimes denote a production x→y with a unique label r as r:x→y, and instead of x→y∈P, we simply write r∈P. We say that x→y is a unit production if x,y∈V. A relation of direct derivation in G, denoted ⇒, is defined as follows: if u,v,x,y∈V* and x→y∈P, then uxv⇒uyv. The language generated by G, denoted L(G), is defined as L(G)={w∣σ⇒*w,w∈Σ*}. G is said to be context-free iff for every production x→y∈P, ∣x∣=1. Furthermore, G is said to be context-sensitive iff every x→y∈P satisfies ∣x∣≤∣y∣. A language is context-free iff it is generated by some context-free grammar, and a language is context-sensitive iff it is generated by some context-sensitive grammar. By CF and CS, we denote the families of context-free and context-sensitive languages, respectively. Next, we give the formal definition of pure grammars (see [20, 13]), together with six modes of derivations. Definition 2.1 Let G=(V,Σ,P,σ)be an unrestricted grammar. Gis a pure grammar (PG for short), if V=Σ. For brevity, we simplify G=(V,Σ,P,σ)to G=(Σ,P,σ). We say that Gis propagating or without erasing productions iff for every production x→y∈P, y≠ε. Next, we introduce six modes of direct derivation steps as derivation relations over Σ*. Let u,v∈Σ*. The six derivation relations are defined as follows: u⇒svin Giff there exists x→y∈Pand w,z∈Σ*such that u=wxzand v=wyz; u⇒ljvin Giff there exists x→y∈Pand w,t,z∈Σ*such that u=wtxzand v=wytz; u⇒rjvin Giff there exists x→y∈Pand w,t,z∈Σ*such that u=wxtzand v=wtyz; u⇒jvin Giff u⇒ljvor u⇒rjvin G; u⇒pvin Giff there exist x1→y1,x2→y2,…,xn→yn∈Psuch that u=x1x2…xnand v=y1y2…yn, where n≥0; u⇒jpvin Giff there exist x1→y1,x2→y2,…,xn→yn∈Psuch that u=x1x2…xnand v=yp(1)yp(2)…yp(n), where p∈perm({1,2,…,n}), n≥0.Let ⇒hbe one of the six derivation relations (i) through (vi) over Σ*. To express that Gapplies production rduring u⇒hv, we write u⇒hv[r], where r∈P. By u⇒h*v[π], where πis a sequence of productions from P, we express that Gmakes u⇒h*vby using π. The language that G generates by using ⇒h, L(G,⇒h), is defined as L(G,⇒h)={x∣σ⇒h*x,x∈Σ*}.The set of all PGs and the set of all PGs without erasing productions are denoted ΓPGand ΓPG−ε, respectively. Let G=(Σ,P,σ) be a PG. G is said to be a pure context-free grammar (PCFG for short) if every x→y∈P satisfies x∈Σ. The set of all PCFGs and the set of all PCFGs without erasing productions are denoted ΓPCFG and ΓPCFG−ε, respectively. Remark 1 The inclusions ΓPCFG⊆ΓPG, ΓPCFG−ε⊆ΓPCFG and ΓPG−ε⊆ΓPG are obvious. Definition 2.2 Set SP={L(G,⇒s)∣G∈ΓPG}; SP−ε={L(G,⇒s)∣G∈ΓPG−ε}; JSP={L(G,⇒j)∣G∈ΓPG}; JSP−ε={L(G,⇒j)∣G∈ΓPG−ε}; PP={L(G,⇒p)∣G∈ΓPG}; PP−ε={L(G,⇒p)∣G∈ΓPG−ε}; JPP={L(G,⇒jp)∣G∈ΓPG}; JPP−ε={L(G,⇒jp)∣G∈ΓPG−ε}; SPCF={L(G,⇒s)∣G∈ΓPCFG}; SPCF−ε={L(G,⇒s)∣G∈ΓPCFG−ε}; JSPCF={L(G,⇒j)∣G∈ΓPCFG}; JSPCF−ε={L(G,⇒j)∣G∈ΓPCFG−ε}; PPCF={L(G,⇒p)∣G∈ΓPCFG}; PPCF−ε={L(G,⇒p)∣G∈ΓPCFG−ε}; 0L={L(G,⇒p)∣G∈ΓPCFG,G=(Σ,P,σ),dom(P)=Σ} (see [16]); 0L−ε={L(G,⇒p)∣G∈ΓPCFG−ε,G=(Σ,P,σ),dom(P)=Σ} (see [16], where 0L−εis denoted byP0L); JPPCF={L(G,⇒jp)∣G∈ΓPCFG}; JPPCF−ε={L(G,⇒jp)∣G∈ΓPCFG−ε}. Example 1 Consider the following PCFG: G=(Σ={a,b,c,d},P,a) where P={a→abcd,a→a,b→b,c→c,d→d}. Observe that L(G,⇒s)=L(G,⇒p)={a}{bcd}* is a regular language, but L(G,⇒j)=L(G,⇒jp)={w∣#a(w)=1,#b(w)=#c(w)=#d(w),w∈Σ+} is a non-context-free language. 3. RESULTS The organization of this section is divided into three parts. First, we give an overview about several elementary properties of pure grammars. Second, we investigate the mutual relations of SPCF, JSPCF, PPCF, JPPCF, CF and CS and summarize the results by Euler diagram in Fig. 1. Finally, we study the former without erasing productions and sum up the investigated relations in Table 1. Figure 1. View largeDownload slide Summary of hierarchy between SPCF, JSPCF, PPCF, JPPCF, CF and CS language families ( ? stands for an open problem of the existence of a witness language). Figure 1. View largeDownload slide Summary of hierarchy between SPCF, JSPCF, PPCF, JPPCF, CF and CS language families ( ? stands for an open problem of the existence of a witness language). Table 1. Mutual relations between investigated language families. A denotes the language family from the first column, B the language family from the table header. If the relation in the cell given by A and B is ⋆, then A⋆B. A∥B means that A and B are incomparable, but not disjoint, ? stands for an open problem, and the meaning of ⊂, =, and ⊃ is as usual. View Large Table 1. Mutual relations between investigated language families. A denotes the language family from the first column, B the language family from the table header. If the relation in the cell given by A and B is ⋆, then A⋆B. A∥B means that A and B are incomparable, but not disjoint, ? stands for an open problem, and the meaning of ⊂, =, and ⊃ is as usual. View Large 3.1. Elementary properties Many properties about pure grammars can be found in [18, 20]. Recall that1 SPCF⊂CF (see [18, 20]). As follows from (13) and (15) in Definition 2.2, 0L⊆PPCF. Furthermore, there exist languages that can be generated by parallel PCFG but cannot be generated by any 0L system (such a language is, for example, {a,aab}). Thus, 0L⊂PPCF. Lemma 3.1 Let X∈{SP, JSP, PP, JPP, SPCF, JSPCF, PPCF, JPPCF, 0L}. Then, X−ε⊆X. Proof Obvious.□ Theorem 3.1 SPCFand JSPCFare semilinear. Proof Since SPCF⊂CF and CF is semilinear (see [21]), SPCF must be also semilinear. Consider any PCFG G=(Σ,P,σ). From the definitions of ⇒s and ⇒j, it follows that ϕ(L(G,⇒s))=ϕ(L(G,⇒j)). Thus, JSPCF is semilinear as well.□ Theorem 3.2 SPCF⊂PPCF. Proof First, we prove the inclusion SPCF⊆PPCF. The proof is based on the proof of Theorem 4.2 in [16]. Let Σ be an alphabet. We claim that for every PCFG G=(Σ,P,σ), there is a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒p)=L(G,⇒s). Set P′=P∪{a→a∣a∈Σ}andσ′=σ Now, we prove the following two claims by induction on m≥0. Since both proofs are straightforward, we show only their induction steps. As the common hypothesis, assume that the claims hold for all 0≤m≤k, where k≥0. Claim 3.3 Let σ⇒smwin G, where w∈Σ*. Then σ′⇒p*win G′. Proof Let σ⇒sk+1w in G, where w∈Σ*. Express σ⇒sk+1w as σ⇒skuav⇒suxv, where u,v,x∈Σ*, a∈Σ, a→x∈P and uxv=w. By the induction hypothesis, there exists a derivation σ′⇒p*uav in G′. Since P⊆P′ and there are also unit productions b→b∈P′, for every b∈Σ, clearly uav⇒puxv in G′, which completes the induction step.□ Claim 3.4 Let σ′⇒pmwin G′, where w∈Σ*. Then σ⇒s*win G. Proof Let σ′⇒pk+1w in G′, where w∈Σ*. Express σ′⇒pk+1w as σ′⇒pkx⇒pw, where x∈Σ*. Set n=∣x∣. Express x and w as x=a1a2…an and w=y1y2…yn, respectively, where ai∈Σ, yi∈Σ* and ai→yi∈P′, 1≤i≤n. Observe that ai→yi∈P′ and ai≠yi implies ai→yi∈P, for all 1≤i≤n. Thus, x⇒s*w in G. By the induction hypothesis, we have that σ⇒s*x in G, which completes the induction step.□ By Claim 3.3 and Claim 3.4, σ⇒s*w in G iff σ′⇒p*w in G′, that is L(G,⇒s)=L(G′,⇒p) and therefore SPCF⊆PPCF. By Theorem 4.7 in [16], 0L⊈CF. Clearly, 0L⊈SPCF. Since 0L⊂PPCF, PPCF⊈SPCF and hence SPCF⊂PPCF.□ Corollary 3.1 SPCF⊂0L. Proof Observe that G′ from the proof of Theorem 3.2 is a correctly defined 0L system according to p. 304 in [16]. Theorem 3.5 SPCF⊂CF∩PPCF. Proof SPCF⊆CF∩PPCF is a consequence of recalled inclusion SPCF⊂CF and Theorem 3.2. Let Σ={a,b,c,d} be an alphabet and L={ab,ccdd} be a language over Σ. Clearly, L∈CF and also L∈PPCF since there is a PCFG G=(Σ,{a→cc,b→dd,c→c,d→d},ab) such that L=L(G,⇒p). We show by contradiction that there is no PCFG G′=(Σ,P′,σ) such that L(G′,⇒s)=L. Clearly, σ must be either ab or ccdd. If we take ccdd as the axiom, there must be c→ε or d→ε in P′ and hence cdd or ccd are contained in L, which is a contradiction. On the other hand, if we take ab, there is no possible way how to directly derive ccdd from ab by using ⇒s. Hence L∉SPCF, which completes the proof.□ Corollary 3.2 SPCF⊂CF∩0L. Theorem 3.6 For a unary alphabet, 0L=PPCF=JPPCF. Proof It follows directly from the definition of ⇒p and ⇒jp and from the definition of ⇒ in 0L systems (see [16]).□ Theorem 3.7 For a unary alphabet, SPCF=JSPCF. Proof It follows directly from the definition of ⇒s and ⇒j.□ Now, we recall Lemma 4.8 from p. 313 in [16]. Lemma 3.2 (Rozenberg, Doucet). Let Gbe a 0L system. Then there exists a number ksuch that for every string win L(G)there exists a derivation such that ∣u∣≤k∣w∣for every string uin that derivation. By analogy with the proof of Lemma 3.2, it is easy to prove the following lemma because PCFGs do not differ from 0L systems when only the longest sentential forms are considered during the derivation of any sentence w. Lemma 3.3 Let Gbe a PCFG. Let h∈{s,j,p,jp}. Then there exists a number ksuch that for every string win L(G,⇒h)there exists a derivation such that ∣u∣≤k∣w∣for every string uin that derivation. Lemma 3.4 CS−JPPCF≠∅. Proof The language X={ap∣pisaprime} over a unary alphabet {a} is a well-known context-sensitive non-context-free language (see [22]). By contradiction, we show that X∉JPPCF. Assume that there is a PCFG G=({a},P,σ) such that L(G,⇒jp)=X. Obviously a→ε∉P and σ=a2 since 2 is the smallest prime. As 3 is also prime, a2⇒jp*a3 and we have a→a∈P and a→a2∈P. Thus, a2⇒jp*a4. Since 4 is not a prime, we have a contradiction.□ Corollary 3.3 CS−JSPCF≠∅. Proof From Lemma 3.4, we have that X={ap∣pisaprime} is not contained in JPPCF. Since X is a unary language and for unary languages holds JSPCF=SPCF⊂PPCF=JPPCF (see Theorems 3.2, 3.6 and 3.7), we have that X∉JSPCF.□ Theorem 3.8 JPPCF⊂CS. Proof Let G=(Σ,P,σ) be a PCFG. As obvious, there is an unrestricted grammar H=(V,Σ,P′,S) such that L(H)=L(G,⇒jp). More precisely, we are able to construct H in the way that H simulates G. In this case, Lemma 3.3 also holds for H. Observe that Lemma 3.3 is the workspace theorem, and every language from JPPCF must be then context-sensitive. As CS−JPPCF≠∅ by Lemma 3.4, we have JPPCF⊂CS.□ Theorem 3.9 JSPCF⊂CS. Proof JSPCF⊆CS can be proved analogously as JPPCF⊆CS from Theorem 3.8. Together with Corollary 3.3, we have JSPCF⊂CS. □ 3.2. Mutual relations of SPCF, JSPCF, PPCF, JPPCF, CF and CS Now, we investigate all the mutual relations between SPCF, JSPCF, PPCF, JPPCF, CF and CS. We refer to them as language subfamilies A through T in Fig. 1, which presents them by using an Euler diagram. More precisely, in this diagram, JSPCF, PPCF, JPPCF and CF form Venn diagram with 16 subfamilies contained in CS; in addition, four more subfamilies are pictured by placing SPCF as a subset of CF∩PPCF (see Theorem 3.5). Hereafter, we study 20 subfamilies in the following 13 theorems and 7 open problems (Theorems and Open Problems 3.10–3.29). Theorem 3.10 (Subfamily A). PPCF−(CF∪JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b} be an alphabet. Let X={a2nb2n∣n≥0} be a language over Σ. Clearly, X∈PPCF, since there exists a PCFG, G=(Σ,{a→aa,b→bb},ab), such that L(G,⇒p)=X. X∉CF and X∉JSPCF is satisfied since X is not semilinear. By contradiction, we show that X∉JPPCF. Consider that there is a PCFG, G′=(Σ,P′,σ′), such that L(G′,⇒jp)=X. Observe that ab∈L(G′,⇒jp). Let a→x, b→y be productions from P′, x,y∈Σ*. Then, there exist two derivations, ab⇒jpxy and ab⇒jpyx, in G′. Now, consider the following cases: x=ε ( y=ε). If y∈X ( x∈X), then either ab is the only string derivable in G′ using ⇒jp or there is a derivation y⇒jp*z ( x⇒jp*z) in G′ such that ba∈substr(z), which is a contradiction. If y∉X, such as y∈{ε,a,b} ( x∉X, such as x∈{ε,a,b}), then ab⇒jpy ( ab⇒jpx), so y∈X ( x∈X), which is a contradiction as well. In the following, we assume that x≠ε and y≠ε. x=bx′ or y=by′, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpbz in G′, where z∈Σ*, and thus bz∈X, which is a contradiction. x=x′a or y=y′a, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpza in G′, where z∈Σ*, and thus za∈X, which is a contradiction. x=ax′b and y=ay′b, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpz in G′ such that ba∈substr(z), which is a contradiction. If a→x or b→y is missing in P′, then X is finite—a contradiction. No other cases are possible, which completes the proof.□ Several intersections of some language families are hard to investigate. Such an intersection is PPCF∩JSPCF. At this moment, we are not able to prove whether PPCF∩JSPCF⊆CF or not. For this reason, we leave the subfamilies B and C as open problems. Open Problem 3.11 (Subfamily B). Is it true that (PPCF∩JSPCF)−(CF∪JPPCF)≠∅? Open Problem 3.12 (Subfamily C). Is it true that (PPCF∩JSPCF∩JPPCF)−CF≠∅? Theorem 3.13 (Subfamily D). (PPCF∩JPPCF)−(CF∪JSPCF)≠∅ Proof For unary alphabet, 0L=PPCF=JPPCF (Theorem 3.6). Since CF and JSPCF are both semilinear, it is sufficient to find any non-semilinear language over unary alphabet which is also contained in PPCF. Such a language is indisputably {a2n∣n≥0}.□ Theorem 3.14 (Subfamily E). SPCF−(JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b,c} be an alphabet. Let X={ancbn∣n≥0} be a language over Σ. Clearly, there exists a PCFG G=(Σ,{c→acb},c) such that L(G,⇒s)=X and hence X∈SPCF. We prove by contradiction that X is neither jumping sequential pure context-free nor jumping parallel pure context-free language. X∉JSPCF. Assume that there is a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒j)=X. Clearly, σ′=c must be the axiom since there must be no erasing productions in P′ (observe that ab,ac,cb∉X). Because acb∈X, we have that c→acb∈P′. But acb⇒jabacb and abacb∉X, which is a contradiction. X∉JPPCF. Assume that there is a PCFG H=(Σ,R,ω) such that L(H,⇒jp)=X. First, let k≥1 and assume that ω=akcbk is an axiom. Since ω⇒jp*c, there must be a productions a→ε, b→ε and c→c contained in R. Now, assume that dˆ→dx∈R, dˆ∈{a,b}, d∈Σ, x∈Σ*; then, ω⇒jp*udxcv and ω⇒jp*ucdxv and obviously for d=a holds ucdxv∉X and for d=b holds udxcv∉X, u,v∈Σ*; d=c is obvious; dˆ→xd∈R, dˆ∈{a,b}, d∈Σ, x∈Σ*; then, ω⇒jp*uxdcv and ω⇒jp*ucxdv and obviously for d=a holds ucxdv∉X and for d=b holds uxdcv∉X, u,v∈Σ*; d=c is obvious. Therefore, a→x,b→y∈R implies x=y=ε. Hence, only productions of the form c→z, where z∈X, can be considered. But the finiteness of R implies the finiteness of X, which is a contradiction. Clearly, the axiom must be ω=c, which implies that R contains productions of the form c→z, where z∈X. Obviously, there must be also productions a→x,b→y∈R, x,y∈Σ*. If x=y=ε, X must be finite. Thus, assume that x≠ε or y≠ε. Then, like before, we can derive a string which is not contained in X—a contradiction.□ Theorem 3.15 (Subfamily F) (SPCF∩JSPCF)−JPPCF≠∅ Proof Let Σ={a,b,c} be an alphabet and let X={aa,aab,aac,aabc} be a language over Σ. Consider a PCFG G=(Σ,{b→ε,c→ε},aabc). Clearly, L(G,⇒s)=L(G,⇒j)=X and hence X∈SPCF∩JSPCF. To show that X∉JPPCF, we use a contradiction. Assume that there exists a PCFG G′=(Σ,P′,σ) such that L(G′,⇒jp)=X. Since σ∈X and X⊆{aa}{b}*{c}*, there must be a production a→x in P′ with x∈Σ*. But this implies that there must be a derivation σ⇒jp*aa⇒jpxx in G′. The only string from X that has a form xx is aa so a→a is the only production with a on its left-hand side so a→a∈P′. Next, we choose σ. Clearly, σ≠aa. Furthermore, σ∉{aab,aac} since σ⇒jpaabc implies that σ⇒jp*abca, and abca∉X. Thus, the only possibility is to choose σ=aabc. But aabc⇒jpaab means that {b→b,c→ε}⊆P′ or {b→ε,c→b}⊆P′. In both cases, aabc⇒jpaba. As aba∉X, there is no PCFG G′ such that L(G′,⇒jp)=X, which is a contradiction.□ Theorem 3.16 (Subfamily G) SPCF∩JSPCF∩JPPCF≠∅ Proof Let G=({a},{a→a,a→aa},a) be a PCFG. It is easy to see that L(G,⇒s)=L(G,⇒j)=L(G,⇒jp)={a}+. □ Open Problem 3.17 (Subfamily H). Is it true that (SPCF∩JPPCF)−JSPCF≠∅? Theorem 3.18 (Subfamily I). (PPCF∩CF)−(SPCF∪JSPCF∪JPPCF)≠∅ Proof Let X={aabb,ccdd} be a language over an alphabet Σ={a,b,c,d}. Clearly, X∈CF. Since there exists a PCFG G=(Σ,{a→c,b→d},aabb) such that L(G,⇒p)=X, X∈PPCF. Furthermore, observe that derivations aabb⇒sccdd ( aabb⇒jccdd) or ccdd⇒saabb ( ccdd⇒jaabb) cannot be performed due to the definition of ⇒s ( ⇒j) and hence there is no PCFG G′ such that L(G′,⇒s)=X ( L(G′,⇒j)=X). Thus, X∉SPCF and X∉JSPCF. Now, suppose that there is a PCFG H=(Σ,P,σ) such that L(H,⇒jp)=X. For σ=aabb, we have aabb⇒jpccdd. If a→ε∈P or b→ε∈P, then aabb⇒jpx, where x∉X. Thus, a→y and b→z, where y,z∈{c,d}, are only possible productions in P. But aabb⇒jpcdcd and since cdcd∉X, there is no PCFG H such that L(H,⇒jp)=X. Analogously for σ=ccdd. We have a contradiction and therefore X∉JPPCF.□ Open Problem 3.19 (Subfamily J). Is it true that (PPCF∩CF∩JSPCF)−(SPCF∪JPPCF)≠∅? Open Problem 3.20 (Subfamily K). Is it true that (PPCF∩CF∩JSPCF∩JPPCF)−SPCF≠∅? Theorem 3.21 (Subfamily L). (PPCF∩CF∩JPPCF)−(SPCF∪JSPCF)≠∅ Proof Consider a language X={ab,cd,dc} over an alphabet Σ={a,b,c,d}. Clearly, X is neither classical sequential pure context-free nor jumping sequential pure context-free language since in some point during a derivation, we must rewrite two symbols simultaneously. As X is a finite language, X∈CF. As there exists a PCFG G=(Σ,{a→c,b→d,c→d,d→c},ab) such that L(G,⇒p)=L(G,⇒jp)=X, X∈PPCF∩JPPCF.□ Theorem 3.22 (Subfamily M). CF−(PPCF∪JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b} and let X={anbn∣n≥1} be a language over Σ. Indisputably, X is well-known context-free language. According to [16], X∉0L. Observe that every language Y that belongs to (PPCF−0L) can be generated by PCFG G=(Σ,P,σ) such that there exists c∈Σ such that for every x∈Σ*, c→x∉P. Thus, if X∈(PPCF−0L), then X must be a finite language (since either a or b blocks deriving of any string from axiom), which is a contradiction. Therefore, X∉(PPCF−0L) and clearly X∉PPCF. Next, we demonstrate that X∉JSPCF and X∉JPPCF. X∉JSPCF. Suppose that X∈JSPCF, so there exists a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒j)=X. As a,b∉X, there are no erasing productions in P′ and thus σ′=ab must be the axiom. Now consider a derivation ab⇒jaabb. There are exactly two possibilities how to get a string aabb directly from the axiom ab—either expand a to aab ( a→aab∈P′) or expand b to abb ( b→abb∈P′). Due to the definition of ⇒j, ab⇒jbaab in the first case, and ab⇒jabba in the second case. Since neither baab nor abba belongs to X, X∉JSPCF, which is a contradiction. X∉JPPCF. Suppose that X∈JPPCF, so there exists a PCFG H=(Σ,R,ω) such that L(H,⇒jp)=X. As for all i≥0, ai,bi∉X, there are no erasing productions in R and thus ω=ab must be the axiom. Clearly, ab⇒jpaabb. There are exactly three ways how to get aabb from ab: a→a∈R, b→abb∈R. In this case ab⇒jpaabb implies that ab⇒jpabba, but abba∉X. a→aa∈R, b→bb∈R. In this case ab⇒jpaabb implies that ab⇒jpbbaa, but bbaa∉X. a→aab∈R, b→b∈R. In this case ab⇒jpaabb implies that ab⇒jpbaab, but baab∉X. Thus, X∉JPPCF, which is a contradiction.□ Open Problem 3.23 (Subfamily N). Is it true that (CF∩JSPCF)−(PPCF∪JPPCF)≠∅? Theorem 3.24 (Subfamily O). (CF∩JSPCF∩JPPCF)−PPCF≠∅ Proof Let Σ={a,b} be an alphabet and let X={aabb,abab,abba,baab,baba,bbaa} be a language over Σ. Since X is finite, X is context-free. Given a PCFG G=(Σ,{a→a,b→b},aabb). Clearly, L(G,⇒j)=L(G,⇒jp)=X. Hence, X∈CF∩JSPCF∩JPPCF. By contradiction, we show that X∉PPCF. Assume that there is a PCFG H=(Σ,P,σ) such that L(H,⇒p)=X. First, we show that P contains no erasing productions: If a→ε∈P and b→ε∈P, we have ε∈X, which is a contradiction. If a→ε∈P, then b→x∈P implies that x∈{aa,bb,ab,ba} because for every w∈X, ∣w∣=4. Clearly, if b→aa∈P, then aaaa∈X, and if b→bb∈P, then bbbb∈X. As obvious, both cases represent a contradiction. On the other hand, if there are no productions in P starting from b apart from b→ab and/or b→ba, then aabb∉X, which is a contradiction. Similarly for b→ε∈P. Since all strings in X have the same length and there are no erasing productions in P, only unit productions can be contained in P. Because aaaa∉X and bbbb∉X, either P={a→a,b→b} or P={a→b,b→a}. In both cases, we never get X. Thus, there is no PCFG H such that L(H,⇒p)=X, and hence X∉PPCF.□ Theorem 3.25 (Subfamily P). (CF∩JPPCF)−(PPCF∪JSPCF)≠∅ Proof Consider a language Y={aabb, ccdd, cdcd, cddc, dccd, dcdc, ddcc} over an alphabet Σ={a,b,c,d}. Clearly, Y∈CF and also Y∈JPPCF because there is a PCFG G=(Σ,{a→c,b→d,c→c,d→d},aabb) such that L(G,⇒jp)=Y. The proof that Y∉PPCF is almost identical to the proof that X∉PPCF from Theorem 3.24, so it is omitted. Because it is not possible to rewrite two or more symbols simultaneously during direct derivation step by using ⇒j, we have Y∉JSPCF.□ Open Problem 3.26 (Subfamily Q). Is it true that JSPCF−(CF∪PPCF∪JPPCF)≠∅? Theorem 3.27 (Subfamily R). (JSPCF∩JPPCF)−(CF∪PPCF)≠∅ Proof Let Σ={a,b,c} be an alphabet and let X={w∣#a(w)−1=#b(w)=#c(w),w∈Σ+} be a language over Σ. X∈JSPCF∩JPPCF since there is a PCFG G=(Σ,{a→abca,a→a,b→b,c→c},a) such that L(G,⇒j)=L(G,⇒jp)=X. By pumping lemma for context-free languages, X∉CF. By contradiction, we show that X∉PPCF. Assume that there is a PCFG H=(Σ,P,σ) such that L(H,⇒p)=X. First, we show that σ=a. Assume that σ≠a. Then, σ⇒p*a implies that a→ε∈P and we have that ε∈X, which is a contradiction. Thus, a must be the axiom, and a→x∈P implies that x∈X. Let l=3max{∣β∣∣α→β∈P}. The smallest possible value of l is 3. Let ω=al+1blcl. Clearly, ω∈X. Then there is a direct derivation step θ⇒pω, where θ∈X. Next, we make the following observations about θ and P: θ≠a, since a→ω∉P. The choice of l excludes such situation. θ contains all three symbols a, b and c. a→a∈P is the only production with a on its left-hand side that is used during θ⇒pω. Observe that if a→x∈P is chosen to rewrite a during θ⇒pω, then x∈X and x must be a substring of ω. Only x=a meets these requirements. θ can be expressed as a+θ′, where θ′∈{b,c}*. This follows from the form of ω and the third observation. During θ⇒pω are used productions b→y,c→y′∈P such that each of y, y′ do not contain at least one symbol from Σ. This is secured by the choice of l. Every production with b on its left-hand side in P has the same commutative image of its right-hand side and every production with c on its left-hand side in P has the same commutative image of its right-hand side. To not break a number of occurrences of symbols a, b and c in ω during θ⇒pω, when b→y∈P is used, then the corresponding c→y′∈P must be also used simultaneously with it. To preserve the proper number of occurrences of a, b and c in ω, we have card({ψ(β)∣b→β∈P})=1 and card({ψ(γ)∣c→γ∈P})=1. Now, we inspect the ways how a+θ′⇒pω could be made. Suppose that the first symbol of θ′ is b: b→ε∈P was used. Then, c→bc∈P must be used ( c→cb is excluded since c is not before b in ω). As there are at least two c s in θ′, applying c→bc brings c before b which is in a contradiction with the form of ω. Let i≥1 and let b→ai∈P. Then, c→bi+1ci+1∈P. Since ∣bi+1ci+1∣ is at most l3, there are at least two occurrences of c in θ′ and then we obtain c before b in ω. Let i≥1 and let j be a non-negative integer such that j≤i+1. Let b→aibj∈P. Then c→bkcm∈P, where j+k=m=i+1. As in the previous case, when these productions are used during θ⇒pω, we get b before a or c before b in ω. No a s were added during θ⇒pω. In this case, the only productions with b and c on their left-hand sides in P can be either b→bc and c→ε, or b→b and c→c, or b→c and c→b. This implies that the only way how to get θ from a is to use a→θ production that is clearly not in P. For the case that c is the first symbol of θ′, we can proceed analogously. Therefore, ω∉L(H,⇒p), which implies that X∉PPCF.□ Theorem 3.28 (Subfamily S). JPPCF−(CF∪PPCF∪JSPCF)≠∅ Proof Let Σ={a,b,c,aˆ,bˆ,cˆ} be an alphabet and let X={aˆbˆcˆ}∪{x∣#a(x)−1=#b(x)=#c(x),x∈{a,b,c}+} be a language over Σ. Following the pumping lemma for context-free languages, X∉CF. Since there is a PCFG G=(Σ,{aˆ→a,bˆ→ε,cˆ→ε,a→abca,a→a,b→b,c→c},aˆbˆcˆ) such that L(G,⇒jp)=X, X∈JPPCF. By contradiction, we show that X∉JSPCF and X∉PPCF. Suppose that X∈JSPCF. Then, there is a PCFG H=(Σ,P,σ) such that L(H,⇒j)=X. First, we choose σ. From the definition of X, a∈X and for every string x∈X−{a} holds ∣x∣≥3. Since we are able to erase only one symbol during direct derivation step by ⇒j and there is no string of length 2 contained in X, we must choose σ=a as the axiom. Because abca∈X and aˆbˆcˆ∈X, there must be two derivations, a⇒j*abca and a⇒j*aˆbˆcˆ, and this implies that there exists also a derivation a⇒j*aˆbˆcˆbca. Since aˆbˆcˆbca∉X, we have a contradiction. Next, suppose that X∈PPCF, so there exists a PCFG H′=(Σ,P′,σ′) such that L(H′,⇒p)=X. In this case, we must choose σ′=aˆbˆcˆ as the axiom. If we choose a, then a⇒p*abca and a⇒p*aˆbˆcˆ implies that a⇒p*u1au2aˆu3, u1,u2,u3∈Σ* and u1au2aˆu3∉X. If we choose abca or similar, then abca⇒p*a implies that a⇒p*ε, and ε∉X. Without loss of generality, assume that for every α→β∈P′, β∈{a,b,c}* (this can be assumed since aˆbˆcˆ is the only string over {aˆ,bˆ,cˆ} in X). As a∈X, a→ε, a→b and a→c are not contained in P′. The observations (1) to (3) from the proof of Theorem 3.27 hold also for H′. The rest of proof is similar to the proof of Theorem 3.27.□ Theorem 3.29 (Subfamily T). CS−(CF∪JSPCF∪PPCF∪JPPCF)≠∅ Proof Let X={ap∣pisaprime} be a language over unary alphabet {a}. X∈CS and X∉CF are a well-known containments (see [22]). By Lemma 3.4 and Corollary 3.3, X∉JPPCF and X∉JSPCF. As for unary languages PPCF=JPPCF, X∉PPCF.□ The summary of Theorems 3.10–3.29 is visualized in Fig. 1. 3.3. Absence of erasing productions As stated in Lemma 3.1, it is natural that the family of languages generated by pure grammars without erasing productions is included in the family of languages generated by pure grammars in which the presence of erasing productions is allowed. As we show further, for PCFG, the inclusions stated in Lemma 3.1 are proper. The PG case is left as an open problem. Theorem 3.30 Let X∈{SPCF,JSPCF,PPCF,JPPCF,0L}. Then, X−ε⊂X. Proof Let K={a,ab} and Y={aa,aab} be two languages over Σ={a,b}. Furthermore, let G=(Σ,{a→a,b→ε},ab) and G′=(Σ,{a→a,b→ε},aab) be two PCFGs. SPCF−ε⊂SPCF. Since K=L(G,⇒s), K∈SPCF. Assume that K∈SPCF−ε; then, there is a PCFG H=(Σ,P,σ) with no erasing productions in P such that L(H,⇒s)=K. Obviously, σ=a, so a→ab∈P. We have a⇒s*abb and since abb∉K, K∉SPCF−ε. JSPCF−ε⊂JSPCF. K∈JSPCF and K∉JSPCF−ε are proved analogously as in (a). PPCF−ε⊂PPCF. Since Y=L(G′,⇒p), Y∈PPCF. Assume that Y∈PPCF−ε, so there is a PCFG H=(Σ,P,σ) with no erasing productions in P such that L(H,⇒p)=Y. Obviously, σ=aa and then a→ab∈P. We have aa⇒p*abab and since abab∉Y, Y∉PPCF−ε. JPPCF−ε⊂JPPCF. Y∈JPPCF and Y∉JPPCF−ε are proved analogously as in (c). 0L−ε⊂0L (see Theorem 2.8 in [23]).□ Open Problem 3.31 Let X∈{SP,JSP,PP,JPP}. Is the inclusion X−ε⊆X, in fact, proper? From Fig. 1 and from mentioned theorems, we are able to find out the most of relations between investigated language families (even for those which are generated by PCFGs without erasing productions—the most of languages used in Fig. 1 have this property), but not all. Following theorems fill this gap. Theorem 3.32 SPCFand PPCF−εare incomparable, but not disjoint. Proof Let X={aa,aab} be a language over alphabet Σ={a,b}. Obviously, there is a PCFG G=(Σ,{a→a,b→ε},aab) such that L(G,⇒s)=X, so X∈SPCF. By Theorem 3.30, X∉PPCF−ε. Conversely, there is a language Y={a2n∣n≥0} over {a} such that Y∉SPCF and Y∈PPCF−ε (see D in Fig. 1 and observe that to get Y we need no erasing productions). Finally, {a}+∈SPCF∩PPCF−ε.□ Theorem 3.33 SPCFand 0L−εare incomparable, but not disjoint. Proof Analogous to the proof of Theorem 3.32.□ The mutual relation between JSPCF−ε and JPPCF−ε is either incomparability or JSPCF−ε⊂JPPCF−ε, but we do not know the answer now. We also do not know either if JSPCF−ε and JPPCF are incomparable or JSPCF−ε⊂JPPCF. Open Problem 3.34 What is the relation between JSPCF−ε and JPPCF−ε? Open Problem 3.35 What is the relation between JSPCF−ε and JPPCF? Theorem 3.36 PPCF−εand 0Lare incomparable, but not disjoint. Proof Let X={aa,aab} and Y={a,aab} be two languages over {a,b}. X∉PPCF−ε, X∈0L, Y∈PPCF−ε and Y∉0L proves the incomparability, while {a}+∈PPCF−ε∩0L proves the disjointness.□ 3.4. Remark on unary alphabets We close this section by showing how the mutual relations between investigated language families change if we consider only alphabets containing only one symbol. From Theorems 3.2, 3.6 and 3.7, we can conclude that for every unary alphabet SPCF=JSPCF⊂PPCF=JPPCF=0L. Trivially, SPCF−ε=JSPCF−ε⊂PPCF−ε=JPPCF−ε=0L−ε. As the following theorem demonstrates that PPCF−ε and SPCF are incomparable, but not disjoint, we can summarize the results for the unary alphabet in Fig. 2. Theorem 3.37 In the case of unary alphabets, SPCFand PPCF−εare incomparable, but not disjoint. Proof Clearly, the language {a}+ is contained in both SPCF and PPCF−ε. Since the language {ε,a,aa} from SPCF is not contained in PPCF−ε, SPCF⊈PPCF−ε. Conversely, PPCF−ε⊈SPCF since PPCF−ε is not semilinear.□ Figure 2. View largeDownload slide A mutual relations between investigated language families in the case of unary alphabets. The straight line between two families means that these families are identical. The arrow from family A to family B denotes that A⊂B. Figure 2. View largeDownload slide A mutual relations between investigated language families in the case of unary alphabets. The straight line between two families means that these families are identical. The arrow from family A to family B denotes that A⊂B. 4. CONCLUSION Consider SPCF, JSPCF, PPCF, JPPCF, 0L, SPCF−ε, JSPCF−ε, PPCF−ε, JPPCF−ε and 0L−ε (see Section 2). The present paper has investigated mutual relations between these language families, which are summarized in Table 1 and Fig. 1. As a special case, this paper has also performed an analogical study in terms of unary alphabets (see Fig. 2). Although we have already pointed out several open problems earlier in the paper (see Open Problems 3.11, 3.12, 3.17, 3.19, 3.20, 3.23, 3.26, 3.31, 3.34 and 3.35), we repeat the questions of a particular significance next. Is it true that (PPCF∩JSPCF)−(CF∪JPPCF)≠∅ (Open Problem 3.11)? Is it true that (PPCF∩JSPCF∩JPPCF)−CF≠∅ (Open Problem 3.12)? Is it true that (SPCF∩JPPCF)−JSPCF≠∅ (Open Problem 3.17)? Is it true that (PPCF∩CF∩JSPCF)−(SPCF∪JPPCF)≠∅ (Open Problem 3.19)? Is it true that (PPCF∩CF∩JSPCF∩JPPCF)−SPCF≠∅ (Open Problem 3.20)? Is it true that (CF∩JSPCF)−(PPCF∪JPPCF)≠∅ (Open Problem 3.23)? Is it true that JSPCF−(CF∪PPCF∪JPPCF)≠∅ (Open Problem 3.26)? Let X∈{SP,JSP,PP,JPP}. Is the inclusion X−ε⊆X, in fact, proper (Open Problem 3.31)? What is the relation between JSPCF−ε and JPPCF−ε (Open Problem 3.34)? What is the relation between JSPCF−ε and JPPCF (Open Problem 3.35)? Recall that the present study has only considered pure grammars based on context-free productions. Of course, from a broader perspective, we might reconsider all the study in terms of grammars that allow non-context-free productions as well. FUNDING This work was supported by The Ministry of Education, Youth and Sports of the Czech Republic from the National Programme of Sustainability (NPU II), project IT4Innovations excellence in science—LQ1602. ACKNOWLEDGEMENTS The authors deeply thank both anonymous referees for their invaluable comments and suggestions. Footnotes 1 According to its definition, SPCF in this paper coincides with PCF in [20]. REFERENCES 1 Meduna , A. and Zemek , P. ( 2012 ) Jumping finite automata . Int. J. Found. Comput. Sci. , 23 , 1555 – 1578 . Google Scholar CrossRef Search ADS 2 Křivka , Z. and Meduna , A. ( 2015 ) Jumping grammars . Int. J. Found. Comput. Sci. , 26 , 709 – 732 . Google Scholar CrossRef Search ADS 3 Fernau , H. , Paramasivan , M. and Schmid , M.L. ( 2015 ) Jumping Finite Automata: Characterizations and complexity. In Drewes, F. (ed.), Proc. 20th Int. Conf. Implementation and Application of Automata (CIAA 2015), Umeå, Sweden, August 18–21, pp. 89–101. 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( 2001 ) Introduction to Automata Theory, Languages, and Computation ( 2nd edn ). Addison-Wesley-Longman , Boston . 23 Herman , G.T. and Rozenberg , G. ( 1975 ) Developmental Systems and Languages . North-Holland Publishing Company , Amsterdam . Author notes Handling editor: Fairouz Kamareddine © The British Computer Society 2018. All rights reserved. For permissions, please email: journals.permissions@oup.com http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png The Computer Journal Oxford University Press

Jumping Pure Grammars

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0010-4620
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1460-2067
D.O.I.
10.1093/comjnl/bxy027
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Abstract

Abstract This paper introduces and studies jumping pure grammars, which are conceptualized just like classical pure grammars except that during the applications of their productions, they can jump over symbols in either direction within the rewritten strings. The paper compares the generative power of jumping pure grammars with that of classical pure grammars while distinguishing between their versions with and without erasing productions. Apart from sequential versions, the paper makes an analogical study in terms of parallel versions of jumping pure grammars represented by 0L grammars. 1. INTRODUCTION Jumping versions of language-defining rewriting systems, such as grammars and automata, represent a brand new trend in formal language theory (see [1–9]). In essence, they act just like classical rewriting systems except that they work on strings discontinuously. That is, they apply a production so they erase an occurrence of its left-hand side in the rewritten string while placing the right-hand side anywhere in the string, so the position of the insertion may occur far away from the position of the erasure. The present paper contributes to this trend by investigating the generative power of jumping versions of pure grammars, whose original versions were introduced in [10], and their properties are still intensively investigated in language theory (see [11, 12]). Recently, regulated versions of these grammars have been discussed, too (see Chapter 5 in [13–15]). The notion of a pure grammar G represents a language-generating rewriting system based upon an alphabet of symbols and a finite set of productions (as opposed to the notion of a general grammar, its alphabet of symbols is not divided into the alphabet of terminals and the alphabet of nonterminals). Each production represents a pair of the form (x,y), where x and y are strings over the alphabet of G. Customarily, (x,y) is written as x→y, where x and y are referred to as the left-hand side and the right-hand side of x→y, respectively. Starting from a special start string, G repeatedly rewrites strings according to its productions, and the set of all strings obtained in this way represents the language generated by G. In a greater detail, G rewrites a string z according to x→y so it (i) selects an occurrence of x in z, (ii) erases it and (iii) inserts y precisely at the position of this erasure. More formally, let z=uxv, where u and v are strings. By using x→y, G rewrites uxv as uyv. The notion of a jumping pure grammar—that is, the key notion introduced in this paper—is conceptualized just like that of a classical pure grammar; however, it rewrites strings in a slightly different way. Let G, z, and x→y have the same meaning as above. G rewrites a string z according to x→y so it performs (i) and (ii) as described above, but during (iii), G can jump over a portion of the rewritten string in either direction and inserts y there. More formally, by using x→y, G rewrites ucv as udv, where u,v,w,c,d are strings such that either (a) c=xw and d=wy or (b) c=wx and d=yw. Otherwise, G works as described above. The present paper compares the generative power of classical and jumping versions of pure grammars. It distinguishes between these grammars with and without erasing productions. Apart from these sequential versions of pure grammars, it also considers parallel versions of classical and jumping pure grammars represented by 0L grammars (see [16]). As a result, the paper studies the mutual relations between eight language families corresponding to the following derivations modes (see Definition 2.1) performed by pure grammars both with and without erasing productions: classical sequential mode ( ⇒s); jumping sequential mode ( ⇒j); classical parallel mode ( ⇒p); jumping parallel mode ( ⇒jp). In essence, the paper demonstrates that any version of these grammars with erasing productions is stronger than the same version without them. Furthermore, it shows that almost all of the eight language families under considerations are pairwise incomparable—that is, any two families are not subfamilies of each other. The rest of the paper is organized as follows. Section 2 recalls all the terminology needed in this paper and introduces a variety of jumping pure grammars, illustrated by an example. Section 3 presents fundamental results achieved in this paper. Section 4 closes all the study by summing up ten open problems. 2. PRELIMINARIES AND DEFINITIONS This paper assumes that the reader is familiar with the basic notions of formal language theory (see [17–19]). Let A and B be two sets. By A⊆B, we denote that A is included in B and by A⊈B that A is not included in B. A⊂B denotes proper (or strict) inclusion. We say that A and B are incomparable iff A⊈B and B⊈A. The cardinality of A is expressed as card(A). For some n≥0, An denotes the n-fold Cartesian product of set A. By N, we denote the set of all positive integers. Let I⊂N be a finite nonempty set. Then, maxI denotes the maximum of I. For a (binary) relation ϱ over X, ϱi, ϱ+, and ϱ* denote the ith power of ϱ, for all i≥0, the transitive closure of ϱ, and the reflexive and transitive closure of ϱ, respectively. For x,y∈X, instead of (x,y)∈ϱ, we write xϱy throughout. Set dom(ϱ)={x∣xϱy}. Let Σ be an alphabet (finite nonempty set). Then, Σ* represents the free monoid generated by Σ under the operation of concatenation, with ε as the unit of Σ*. Set Σ+=Σ*−{ε}. For w∈Σ* and a∈Σ, #a(w) denotes the number of occurrences of a in w. By substr(w), we denote a set of all substrings of w, that is substr(w)={x∣w=uxv,u,x,v∈Σ*}. The length of w is denoted by ∣w∣. Let n≥0. A set J⊆Nn is said to be linear if there exist α,β1,β2,…,βm∈Nn,m≥0 such that J={x∣x=α+k1β1+k2β2+⋯+kmβm,ki∈N,1≤i≤m}. If J is the union of a finite number of linear sets, we say that J is semilinear. If Σ={a1,a2,…,an} is an alphabet, then for w∈Σ*, ϕ(w)=(#a1(w),#a2(w),…,#an(w)) denote the commutative (Parikh) image of w. For L⊆Σ*, ϕ(L)={ϕ(w)∣w∈L} denote the commutative (Parikh) map of L. We say that L is a semilinear language if and only if ϕ(L) is a semilinear set. A language family is semilinear if and only if it contains only semilinear languages. Let S be a finite set. Define a permutation in the terms of bijective mappings as follows: let I={1,2,…,card(S)} be a set of indices. The set of all permutations of elements of S, perm(S), is the set of all bijections from I to S. An unrestricted grammar is a quadruple G=(V,Σ,P,σ), where V is a total alphabet, Σ⊆V is an alphabet of terminal symbols, P⊆V+×V* is a finite relation and σ∈V+ is the start string of G, called axiom. Members of P are called productions. Instead of (x,y)∈P, we write x→y throughout. For brevity, we sometimes denote a production x→y with a unique label r as r:x→y, and instead of x→y∈P, we simply write r∈P. We say that x→y is a unit production if x,y∈V. A relation of direct derivation in G, denoted ⇒, is defined as follows: if u,v,x,y∈V* and x→y∈P, then uxv⇒uyv. The language generated by G, denoted L(G), is defined as L(G)={w∣σ⇒*w,w∈Σ*}. G is said to be context-free iff for every production x→y∈P, ∣x∣=1. Furthermore, G is said to be context-sensitive iff every x→y∈P satisfies ∣x∣≤∣y∣. A language is context-free iff it is generated by some context-free grammar, and a language is context-sensitive iff it is generated by some context-sensitive grammar. By CF and CS, we denote the families of context-free and context-sensitive languages, respectively. Next, we give the formal definition of pure grammars (see [20, 13]), together with six modes of derivations. Definition 2.1 Let G=(V,Σ,P,σ)be an unrestricted grammar. Gis a pure grammar (PG for short), if V=Σ. For brevity, we simplify G=(V,Σ,P,σ)to G=(Σ,P,σ). We say that Gis propagating or without erasing productions iff for every production x→y∈P, y≠ε. Next, we introduce six modes of direct derivation steps as derivation relations over Σ*. Let u,v∈Σ*. The six derivation relations are defined as follows: u⇒svin Giff there exists x→y∈Pand w,z∈Σ*such that u=wxzand v=wyz; u⇒ljvin Giff there exists x→y∈Pand w,t,z∈Σ*such that u=wtxzand v=wytz; u⇒rjvin Giff there exists x→y∈Pand w,t,z∈Σ*such that u=wxtzand v=wtyz; u⇒jvin Giff u⇒ljvor u⇒rjvin G; u⇒pvin Giff there exist x1→y1,x2→y2,…,xn→yn∈Psuch that u=x1x2…xnand v=y1y2…yn, where n≥0; u⇒jpvin Giff there exist x1→y1,x2→y2,…,xn→yn∈Psuch that u=x1x2…xnand v=yp(1)yp(2)…yp(n), where p∈perm({1,2,…,n}), n≥0.Let ⇒hbe one of the six derivation relations (i) through (vi) over Σ*. To express that Gapplies production rduring u⇒hv, we write u⇒hv[r], where r∈P. By u⇒h*v[π], where πis a sequence of productions from P, we express that Gmakes u⇒h*vby using π. The language that G generates by using ⇒h, L(G,⇒h), is defined as L(G,⇒h)={x∣σ⇒h*x,x∈Σ*}.The set of all PGs and the set of all PGs without erasing productions are denoted ΓPGand ΓPG−ε, respectively. Let G=(Σ,P,σ) be a PG. G is said to be a pure context-free grammar (PCFG for short) if every x→y∈P satisfies x∈Σ. The set of all PCFGs and the set of all PCFGs without erasing productions are denoted ΓPCFG and ΓPCFG−ε, respectively. Remark 1 The inclusions ΓPCFG⊆ΓPG, ΓPCFG−ε⊆ΓPCFG and ΓPG−ε⊆ΓPG are obvious. Definition 2.2 Set SP={L(G,⇒s)∣G∈ΓPG}; SP−ε={L(G,⇒s)∣G∈ΓPG−ε}; JSP={L(G,⇒j)∣G∈ΓPG}; JSP−ε={L(G,⇒j)∣G∈ΓPG−ε}; PP={L(G,⇒p)∣G∈ΓPG}; PP−ε={L(G,⇒p)∣G∈ΓPG−ε}; JPP={L(G,⇒jp)∣G∈ΓPG}; JPP−ε={L(G,⇒jp)∣G∈ΓPG−ε}; SPCF={L(G,⇒s)∣G∈ΓPCFG}; SPCF−ε={L(G,⇒s)∣G∈ΓPCFG−ε}; JSPCF={L(G,⇒j)∣G∈ΓPCFG}; JSPCF−ε={L(G,⇒j)∣G∈ΓPCFG−ε}; PPCF={L(G,⇒p)∣G∈ΓPCFG}; PPCF−ε={L(G,⇒p)∣G∈ΓPCFG−ε}; 0L={L(G,⇒p)∣G∈ΓPCFG,G=(Σ,P,σ),dom(P)=Σ} (see [16]); 0L−ε={L(G,⇒p)∣G∈ΓPCFG−ε,G=(Σ,P,σ),dom(P)=Σ} (see [16], where 0L−εis denoted byP0L); JPPCF={L(G,⇒jp)∣G∈ΓPCFG}; JPPCF−ε={L(G,⇒jp)∣G∈ΓPCFG−ε}. Example 1 Consider the following PCFG: G=(Σ={a,b,c,d},P,a) where P={a→abcd,a→a,b→b,c→c,d→d}. Observe that L(G,⇒s)=L(G,⇒p)={a}{bcd}* is a regular language, but L(G,⇒j)=L(G,⇒jp)={w∣#a(w)=1,#b(w)=#c(w)=#d(w),w∈Σ+} is a non-context-free language. 3. RESULTS The organization of this section is divided into three parts. First, we give an overview about several elementary properties of pure grammars. Second, we investigate the mutual relations of SPCF, JSPCF, PPCF, JPPCF, CF and CS and summarize the results by Euler diagram in Fig. 1. Finally, we study the former without erasing productions and sum up the investigated relations in Table 1. Figure 1. View largeDownload slide Summary of hierarchy between SPCF, JSPCF, PPCF, JPPCF, CF and CS language families ( ? stands for an open problem of the existence of a witness language). Figure 1. View largeDownload slide Summary of hierarchy between SPCF, JSPCF, PPCF, JPPCF, CF and CS language families ( ? stands for an open problem of the existence of a witness language). Table 1. Mutual relations between investigated language families. A denotes the language family from the first column, B the language family from the table header. If the relation in the cell given by A and B is ⋆, then A⋆B. A∥B means that A and B are incomparable, but not disjoint, ? stands for an open problem, and the meaning of ⊂, =, and ⊃ is as usual. View Large Table 1. Mutual relations between investigated language families. A denotes the language family from the first column, B the language family from the table header. If the relation in the cell given by A and B is ⋆, then A⋆B. A∥B means that A and B are incomparable, but not disjoint, ? stands for an open problem, and the meaning of ⊂, =, and ⊃ is as usual. View Large 3.1. Elementary properties Many properties about pure grammars can be found in [18, 20]. Recall that1 SPCF⊂CF (see [18, 20]). As follows from (13) and (15) in Definition 2.2, 0L⊆PPCF. Furthermore, there exist languages that can be generated by parallel PCFG but cannot be generated by any 0L system (such a language is, for example, {a,aab}). Thus, 0L⊂PPCF. Lemma 3.1 Let X∈{SP, JSP, PP, JPP, SPCF, JSPCF, PPCF, JPPCF, 0L}. Then, X−ε⊆X. Proof Obvious.□ Theorem 3.1 SPCFand JSPCFare semilinear. Proof Since SPCF⊂CF and CF is semilinear (see [21]), SPCF must be also semilinear. Consider any PCFG G=(Σ,P,σ). From the definitions of ⇒s and ⇒j, it follows that ϕ(L(G,⇒s))=ϕ(L(G,⇒j)). Thus, JSPCF is semilinear as well.□ Theorem 3.2 SPCF⊂PPCF. Proof First, we prove the inclusion SPCF⊆PPCF. The proof is based on the proof of Theorem 4.2 in [16]. Let Σ be an alphabet. We claim that for every PCFG G=(Σ,P,σ), there is a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒p)=L(G,⇒s). Set P′=P∪{a→a∣a∈Σ}andσ′=σ Now, we prove the following two claims by induction on m≥0. Since both proofs are straightforward, we show only their induction steps. As the common hypothesis, assume that the claims hold for all 0≤m≤k, where k≥0. Claim 3.3 Let σ⇒smwin G, where w∈Σ*. Then σ′⇒p*win G′. Proof Let σ⇒sk+1w in G, where w∈Σ*. Express σ⇒sk+1w as σ⇒skuav⇒suxv, where u,v,x∈Σ*, a∈Σ, a→x∈P and uxv=w. By the induction hypothesis, there exists a derivation σ′⇒p*uav in G′. Since P⊆P′ and there are also unit productions b→b∈P′, for every b∈Σ, clearly uav⇒puxv in G′, which completes the induction step.□ Claim 3.4 Let σ′⇒pmwin G′, where w∈Σ*. Then σ⇒s*win G. Proof Let σ′⇒pk+1w in G′, where w∈Σ*. Express σ′⇒pk+1w as σ′⇒pkx⇒pw, where x∈Σ*. Set n=∣x∣. Express x and w as x=a1a2…an and w=y1y2…yn, respectively, where ai∈Σ, yi∈Σ* and ai→yi∈P′, 1≤i≤n. Observe that ai→yi∈P′ and ai≠yi implies ai→yi∈P, for all 1≤i≤n. Thus, x⇒s*w in G. By the induction hypothesis, we have that σ⇒s*x in G, which completes the induction step.□ By Claim 3.3 and Claim 3.4, σ⇒s*w in G iff σ′⇒p*w in G′, that is L(G,⇒s)=L(G′,⇒p) and therefore SPCF⊆PPCF. By Theorem 4.7 in [16], 0L⊈CF. Clearly, 0L⊈SPCF. Since 0L⊂PPCF, PPCF⊈SPCF and hence SPCF⊂PPCF.□ Corollary 3.1 SPCF⊂0L. Proof Observe that G′ from the proof of Theorem 3.2 is a correctly defined 0L system according to p. 304 in [16]. Theorem 3.5 SPCF⊂CF∩PPCF. Proof SPCF⊆CF∩PPCF is a consequence of recalled inclusion SPCF⊂CF and Theorem 3.2. Let Σ={a,b,c,d} be an alphabet and L={ab,ccdd} be a language over Σ. Clearly, L∈CF and also L∈PPCF since there is a PCFG G=(Σ,{a→cc,b→dd,c→c,d→d},ab) such that L=L(G,⇒p). We show by contradiction that there is no PCFG G′=(Σ,P′,σ) such that L(G′,⇒s)=L. Clearly, σ must be either ab or ccdd. If we take ccdd as the axiom, there must be c→ε or d→ε in P′ and hence cdd or ccd are contained in L, which is a contradiction. On the other hand, if we take ab, there is no possible way how to directly derive ccdd from ab by using ⇒s. Hence L∉SPCF, which completes the proof.□ Corollary 3.2 SPCF⊂CF∩0L. Theorem 3.6 For a unary alphabet, 0L=PPCF=JPPCF. Proof It follows directly from the definition of ⇒p and ⇒jp and from the definition of ⇒ in 0L systems (see [16]).□ Theorem 3.7 For a unary alphabet, SPCF=JSPCF. Proof It follows directly from the definition of ⇒s and ⇒j.□ Now, we recall Lemma 4.8 from p. 313 in [16]. Lemma 3.2 (Rozenberg, Doucet). Let Gbe a 0L system. Then there exists a number ksuch that for every string win L(G)there exists a derivation such that ∣u∣≤k∣w∣for every string uin that derivation. By analogy with the proof of Lemma 3.2, it is easy to prove the following lemma because PCFGs do not differ from 0L systems when only the longest sentential forms are considered during the derivation of any sentence w. Lemma 3.3 Let Gbe a PCFG. Let h∈{s,j,p,jp}. Then there exists a number ksuch that for every string win L(G,⇒h)there exists a derivation such that ∣u∣≤k∣w∣for every string uin that derivation. Lemma 3.4 CS−JPPCF≠∅. Proof The language X={ap∣pisaprime} over a unary alphabet {a} is a well-known context-sensitive non-context-free language (see [22]). By contradiction, we show that X∉JPPCF. Assume that there is a PCFG G=({a},P,σ) such that L(G,⇒jp)=X. Obviously a→ε∉P and σ=a2 since 2 is the smallest prime. As 3 is also prime, a2⇒jp*a3 and we have a→a∈P and a→a2∈P. Thus, a2⇒jp*a4. Since 4 is not a prime, we have a contradiction.□ Corollary 3.3 CS−JSPCF≠∅. Proof From Lemma 3.4, we have that X={ap∣pisaprime} is not contained in JPPCF. Since X is a unary language and for unary languages holds JSPCF=SPCF⊂PPCF=JPPCF (see Theorems 3.2, 3.6 and 3.7), we have that X∉JSPCF.□ Theorem 3.8 JPPCF⊂CS. Proof Let G=(Σ,P,σ) be a PCFG. As obvious, there is an unrestricted grammar H=(V,Σ,P′,S) such that L(H)=L(G,⇒jp). More precisely, we are able to construct H in the way that H simulates G. In this case, Lemma 3.3 also holds for H. Observe that Lemma 3.3 is the workspace theorem, and every language from JPPCF must be then context-sensitive. As CS−JPPCF≠∅ by Lemma 3.4, we have JPPCF⊂CS.□ Theorem 3.9 JSPCF⊂CS. Proof JSPCF⊆CS can be proved analogously as JPPCF⊆CS from Theorem 3.8. Together with Corollary 3.3, we have JSPCF⊂CS. □ 3.2. Mutual relations of SPCF, JSPCF, PPCF, JPPCF, CF and CS Now, we investigate all the mutual relations between SPCF, JSPCF, PPCF, JPPCF, CF and CS. We refer to them as language subfamilies A through T in Fig. 1, which presents them by using an Euler diagram. More precisely, in this diagram, JSPCF, PPCF, JPPCF and CF form Venn diagram with 16 subfamilies contained in CS; in addition, four more subfamilies are pictured by placing SPCF as a subset of CF∩PPCF (see Theorem 3.5). Hereafter, we study 20 subfamilies in the following 13 theorems and 7 open problems (Theorems and Open Problems 3.10–3.29). Theorem 3.10 (Subfamily A). PPCF−(CF∪JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b} be an alphabet. Let X={a2nb2n∣n≥0} be a language over Σ. Clearly, X∈PPCF, since there exists a PCFG, G=(Σ,{a→aa,b→bb},ab), such that L(G,⇒p)=X. X∉CF and X∉JSPCF is satisfied since X is not semilinear. By contradiction, we show that X∉JPPCF. Consider that there is a PCFG, G′=(Σ,P′,σ′), such that L(G′,⇒jp)=X. Observe that ab∈L(G′,⇒jp). Let a→x, b→y be productions from P′, x,y∈Σ*. Then, there exist two derivations, ab⇒jpxy and ab⇒jpyx, in G′. Now, consider the following cases: x=ε ( y=ε). If y∈X ( x∈X), then either ab is the only string derivable in G′ using ⇒jp or there is a derivation y⇒jp*z ( x⇒jp*z) in G′ such that ba∈substr(z), which is a contradiction. If y∉X, such as y∈{ε,a,b} ( x∉X, such as x∈{ε,a,b}), then ab⇒jpy ( ab⇒jpx), so y∈X ( x∈X), which is a contradiction as well. In the following, we assume that x≠ε and y≠ε. x=bx′ or y=by′, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpbz in G′, where z∈Σ*, and thus bz∈X, which is a contradiction. x=x′a or y=y′a, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpza in G′, where z∈Σ*, and thus za∈X, which is a contradiction. x=ax′b and y=ay′b, where x′,y′∈Σ*. Then, there is a derivation ab⇒jpz in G′ such that ba∈substr(z), which is a contradiction. If a→x or b→y is missing in P′, then X is finite—a contradiction. No other cases are possible, which completes the proof.□ Several intersections of some language families are hard to investigate. Such an intersection is PPCF∩JSPCF. At this moment, we are not able to prove whether PPCF∩JSPCF⊆CF or not. For this reason, we leave the subfamilies B and C as open problems. Open Problem 3.11 (Subfamily B). Is it true that (PPCF∩JSPCF)−(CF∪JPPCF)≠∅? Open Problem 3.12 (Subfamily C). Is it true that (PPCF∩JSPCF∩JPPCF)−CF≠∅? Theorem 3.13 (Subfamily D). (PPCF∩JPPCF)−(CF∪JSPCF)≠∅ Proof For unary alphabet, 0L=PPCF=JPPCF (Theorem 3.6). Since CF and JSPCF are both semilinear, it is sufficient to find any non-semilinear language over unary alphabet which is also contained in PPCF. Such a language is indisputably {a2n∣n≥0}.□ Theorem 3.14 (Subfamily E). SPCF−(JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b,c} be an alphabet. Let X={ancbn∣n≥0} be a language over Σ. Clearly, there exists a PCFG G=(Σ,{c→acb},c) such that L(G,⇒s)=X and hence X∈SPCF. We prove by contradiction that X is neither jumping sequential pure context-free nor jumping parallel pure context-free language. X∉JSPCF. Assume that there is a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒j)=X. Clearly, σ′=c must be the axiom since there must be no erasing productions in P′ (observe that ab,ac,cb∉X). Because acb∈X, we have that c→acb∈P′. But acb⇒jabacb and abacb∉X, which is a contradiction. X∉JPPCF. Assume that there is a PCFG H=(Σ,R,ω) such that L(H,⇒jp)=X. First, let k≥1 and assume that ω=akcbk is an axiom. Since ω⇒jp*c, there must be a productions a→ε, b→ε and c→c contained in R. Now, assume that dˆ→dx∈R, dˆ∈{a,b}, d∈Σ, x∈Σ*; then, ω⇒jp*udxcv and ω⇒jp*ucdxv and obviously for d=a holds ucdxv∉X and for d=b holds udxcv∉X, u,v∈Σ*; d=c is obvious; dˆ→xd∈R, dˆ∈{a,b}, d∈Σ, x∈Σ*; then, ω⇒jp*uxdcv and ω⇒jp*ucxdv and obviously for d=a holds ucxdv∉X and for d=b holds uxdcv∉X, u,v∈Σ*; d=c is obvious. Therefore, a→x,b→y∈R implies x=y=ε. Hence, only productions of the form c→z, where z∈X, can be considered. But the finiteness of R implies the finiteness of X, which is a contradiction. Clearly, the axiom must be ω=c, which implies that R contains productions of the form c→z, where z∈X. Obviously, there must be also productions a→x,b→y∈R, x,y∈Σ*. If x=y=ε, X must be finite. Thus, assume that x≠ε or y≠ε. Then, like before, we can derive a string which is not contained in X—a contradiction.□ Theorem 3.15 (Subfamily F) (SPCF∩JSPCF)−JPPCF≠∅ Proof Let Σ={a,b,c} be an alphabet and let X={aa,aab,aac,aabc} be a language over Σ. Consider a PCFG G=(Σ,{b→ε,c→ε},aabc). Clearly, L(G,⇒s)=L(G,⇒j)=X and hence X∈SPCF∩JSPCF. To show that X∉JPPCF, we use a contradiction. Assume that there exists a PCFG G′=(Σ,P′,σ) such that L(G′,⇒jp)=X. Since σ∈X and X⊆{aa}{b}*{c}*, there must be a production a→x in P′ with x∈Σ*. But this implies that there must be a derivation σ⇒jp*aa⇒jpxx in G′. The only string from X that has a form xx is aa so a→a is the only production with a on its left-hand side so a→a∈P′. Next, we choose σ. Clearly, σ≠aa. Furthermore, σ∉{aab,aac} since σ⇒jpaabc implies that σ⇒jp*abca, and abca∉X. Thus, the only possibility is to choose σ=aabc. But aabc⇒jpaab means that {b→b,c→ε}⊆P′ or {b→ε,c→b}⊆P′. In both cases, aabc⇒jpaba. As aba∉X, there is no PCFG G′ such that L(G′,⇒jp)=X, which is a contradiction.□ Theorem 3.16 (Subfamily G) SPCF∩JSPCF∩JPPCF≠∅ Proof Let G=({a},{a→a,a→aa},a) be a PCFG. It is easy to see that L(G,⇒s)=L(G,⇒j)=L(G,⇒jp)={a}+. □ Open Problem 3.17 (Subfamily H). Is it true that (SPCF∩JPPCF)−JSPCF≠∅? Theorem 3.18 (Subfamily I). (PPCF∩CF)−(SPCF∪JSPCF∪JPPCF)≠∅ Proof Let X={aabb,ccdd} be a language over an alphabet Σ={a,b,c,d}. Clearly, X∈CF. Since there exists a PCFG G=(Σ,{a→c,b→d},aabb) such that L(G,⇒p)=X, X∈PPCF. Furthermore, observe that derivations aabb⇒sccdd ( aabb⇒jccdd) or ccdd⇒saabb ( ccdd⇒jaabb) cannot be performed due to the definition of ⇒s ( ⇒j) and hence there is no PCFG G′ such that L(G′,⇒s)=X ( L(G′,⇒j)=X). Thus, X∉SPCF and X∉JSPCF. Now, suppose that there is a PCFG H=(Σ,P,σ) such that L(H,⇒jp)=X. For σ=aabb, we have aabb⇒jpccdd. If a→ε∈P or b→ε∈P, then aabb⇒jpx, where x∉X. Thus, a→y and b→z, where y,z∈{c,d}, are only possible productions in P. But aabb⇒jpcdcd and since cdcd∉X, there is no PCFG H such that L(H,⇒jp)=X. Analogously for σ=ccdd. We have a contradiction and therefore X∉JPPCF.□ Open Problem 3.19 (Subfamily J). Is it true that (PPCF∩CF∩JSPCF)−(SPCF∪JPPCF)≠∅? Open Problem 3.20 (Subfamily K). Is it true that (PPCF∩CF∩JSPCF∩JPPCF)−SPCF≠∅? Theorem 3.21 (Subfamily L). (PPCF∩CF∩JPPCF)−(SPCF∪JSPCF)≠∅ Proof Consider a language X={ab,cd,dc} over an alphabet Σ={a,b,c,d}. Clearly, X is neither classical sequential pure context-free nor jumping sequential pure context-free language since in some point during a derivation, we must rewrite two symbols simultaneously. As X is a finite language, X∈CF. As there exists a PCFG G=(Σ,{a→c,b→d,c→d,d→c},ab) such that L(G,⇒p)=L(G,⇒jp)=X, X∈PPCF∩JPPCF.□ Theorem 3.22 (Subfamily M). CF−(PPCF∪JSPCF∪JPPCF)≠∅ Proof Let Σ={a,b} and let X={anbn∣n≥1} be a language over Σ. Indisputably, X is well-known context-free language. According to [16], X∉0L. Observe that every language Y that belongs to (PPCF−0L) can be generated by PCFG G=(Σ,P,σ) such that there exists c∈Σ such that for every x∈Σ*, c→x∉P. Thus, if X∈(PPCF−0L), then X must be a finite language (since either a or b blocks deriving of any string from axiom), which is a contradiction. Therefore, X∉(PPCF−0L) and clearly X∉PPCF. Next, we demonstrate that X∉JSPCF and X∉JPPCF. X∉JSPCF. Suppose that X∈JSPCF, so there exists a PCFG G′=(Σ,P′,σ′) such that L(G′,⇒j)=X. As a,b∉X, there are no erasing productions in P′ and thus σ′=ab must be the axiom. Now consider a derivation ab⇒jaabb. There are exactly two possibilities how to get a string aabb directly from the axiom ab—either expand a to aab ( a→aab∈P′) or expand b to abb ( b→abb∈P′). Due to the definition of ⇒j, ab⇒jbaab in the first case, and ab⇒jabba in the second case. Since neither baab nor abba belongs to X, X∉JSPCF, which is a contradiction. X∉JPPCF. Suppose that X∈JPPCF, so there exists a PCFG H=(Σ,R,ω) such that L(H,⇒jp)=X. As for all i≥0, ai,bi∉X, there are no erasing productions in R and thus ω=ab must be the axiom. Clearly, ab⇒jpaabb. There are exactly three ways how to get aabb from ab: a→a∈R, b→abb∈R. In this case ab⇒jpaabb implies that ab⇒jpabba, but abba∉X. a→aa∈R, b→bb∈R. In this case ab⇒jpaabb implies that ab⇒jpbbaa, but bbaa∉X. a→aab∈R, b→b∈R. In this case ab⇒jpaabb implies that ab⇒jpbaab, but baab∉X. Thus, X∉JPPCF, which is a contradiction.□ Open Problem 3.23 (Subfamily N). Is it true that (CF∩JSPCF)−(PPCF∪JPPCF)≠∅? Theorem 3.24 (Subfamily O). (CF∩JSPCF∩JPPCF)−PPCF≠∅ Proof Let Σ={a,b} be an alphabet and let X={aabb,abab,abba,baab,baba,bbaa} be a language over Σ. Since X is finite, X is context-free. Given a PCFG G=(Σ,{a→a,b→b},aabb). Clearly, L(G,⇒j)=L(G,⇒jp)=X. Hence, X∈CF∩JSPCF∩JPPCF. By contradiction, we show that X∉PPCF. Assume that there is a PCFG H=(Σ,P,σ) such that L(H,⇒p)=X. First, we show that P contains no erasing productions: If a→ε∈P and b→ε∈P, we have ε∈X, which is a contradiction. If a→ε∈P, then b→x∈P implies that x∈{aa,bb,ab,ba} because for every w∈X, ∣w∣=4. Clearly, if b→aa∈P, then aaaa∈X, and if b→bb∈P, then bbbb∈X. As obvious, both cases represent a contradiction. On the other hand, if there are no productions in P starting from b apart from b→ab and/or b→ba, then aabb∉X, which is a contradiction. Similarly for b→ε∈P. Since all strings in X have the same length and there are no erasing productions in P, only unit productions can be contained in P. Because aaaa∉X and bbbb∉X, either P={a→a,b→b} or P={a→b,b→a}. In both cases, we never get X. Thus, there is no PCFG H such that L(H,⇒p)=X, and hence X∉PPCF.□ Theorem 3.25 (Subfamily P). (CF∩JPPCF)−(PPCF∪JSPCF)≠∅ Proof Consider a language Y={aabb, ccdd, cdcd, cddc, dccd, dcdc, ddcc} over an alphabet Σ={a,b,c,d}. Clearly, Y∈CF and also Y∈JPPCF because there is a PCFG G=(Σ,{a→c,b→d,c→c,d→d},aabb) such that L(G,⇒jp)=Y. The proof that Y∉PPCF is almost identical to the proof that X∉PPCF from Theorem 3.24, so it is omitted. Because it is not possible to rewrite two or more symbols simultaneously during direct derivation step by using ⇒j, we have Y∉JSPCF.□ Open Problem 3.26 (Subfamily Q). Is it true that JSPCF−(CF∪PPCF∪JPPCF)≠∅? Theorem 3.27 (Subfamily R). (JSPCF∩JPPCF)−(CF∪PPCF)≠∅ Proof Let Σ={a,b,c} be an alphabet and let X={w∣#a(w)−1=#b(w)=#c(w),w∈Σ+} be a language over Σ. X∈JSPCF∩JPPCF since there is a PCFG G=(Σ,{a→abca,a→a,b→b,c→c},a) such that L(G,⇒j)=L(G,⇒jp)=X. By pumping lemma for context-free languages, X∉CF. By contradiction, we show that X∉PPCF. Assume that there is a PCFG H=(Σ,P,σ) such that L(H,⇒p)=X. First, we show that σ=a. Assume that σ≠a. Then, σ⇒p*a implies that a→ε∈P and we have that ε∈X, which is a contradiction. Thus, a must be the axiom, and a→x∈P implies that x∈X. Let l=3max{∣β∣∣α→β∈P}. The smallest possible value of l is 3. Let ω=al+1blcl. Clearly, ω∈X. Then there is a direct derivation step θ⇒pω, where θ∈X. Next, we make the following observations about θ and P: θ≠a, since a→ω∉P. The choice of l excludes such situation. θ contains all three symbols a, b and c. a→a∈P is the only production with a on its left-hand side that is used during θ⇒pω. Observe that if a→x∈P is chosen to rewrite a during θ⇒pω, then x∈X and x must be a substring of ω. Only x=a meets these requirements. θ can be expressed as a+θ′, where θ′∈{b,c}*. This follows from the form of ω and the third observation. During θ⇒pω are used productions b→y,c→y′∈P such that each of y, y′ do not contain at least one symbol from Σ. This is secured by the choice of l. Every production with b on its left-hand side in P has the same commutative image of its right-hand side and every production with c on its left-hand side in P has the same commutative image of its right-hand side. To not break a number of occurrences of symbols a, b and c in ω during θ⇒pω, when b→y∈P is used, then the corresponding c→y′∈P must be also used simultaneously with it. To preserve the proper number of occurrences of a, b and c in ω, we have card({ψ(β)∣b→β∈P})=1 and card({ψ(γ)∣c→γ∈P})=1. Now, we inspect the ways how a+θ′⇒pω could be made. Suppose that the first symbol of θ′ is b: b→ε∈P was used. Then, c→bc∈P must be used ( c→cb is excluded since c is not before b in ω). As there are at least two c s in θ′, applying c→bc brings c before b which is in a contradiction with the form of ω. Let i≥1 and let b→ai∈P. Then, c→bi+1ci+1∈P. Since ∣bi+1ci+1∣ is at most l3, there are at least two occurrences of c in θ′ and then we obtain c before b in ω. Let i≥1 and let j be a non-negative integer such that j≤i+1. Let b→aibj∈P. Then c→bkcm∈P, where j+k=m=i+1. As in the previous case, when these productions are used during θ⇒pω, we get b before a or c before b in ω. No a s were added during θ⇒pω. In this case, the only productions with b and c on their left-hand sides in P can be either b→bc and c→ε, or b→b and c→c, or b→c and c→b. This implies that the only way how to get θ from a is to use a→θ production that is clearly not in P. For the case that c is the first symbol of θ′, we can proceed analogously. Therefore, ω∉L(H,⇒p), which implies that X∉PPCF.□ Theorem 3.28 (Subfamily S). JPPCF−(CF∪PPCF∪JSPCF)≠∅ Proof Let Σ={a,b,c,aˆ,bˆ,cˆ} be an alphabet and let X={aˆbˆcˆ}∪{x∣#a(x)−1=#b(x)=#c(x),x∈{a,b,c}+} be a language over Σ. Following the pumping lemma for context-free languages, X∉CF. Since there is a PCFG G=(Σ,{aˆ→a,bˆ→ε,cˆ→ε,a→abca,a→a,b→b,c→c},aˆbˆcˆ) such that L(G,⇒jp)=X, X∈JPPCF. By contradiction, we show that X∉JSPCF and X∉PPCF. Suppose that X∈JSPCF. Then, there is a PCFG H=(Σ,P,σ) such that L(H,⇒j)=X. First, we choose σ. From the definition of X, a∈X and for every string x∈X−{a} holds ∣x∣≥3. Since we are able to erase only one symbol during direct derivation step by ⇒j and there is no string of length 2 contained in X, we must choose σ=a as the axiom. Because abca∈X and aˆbˆcˆ∈X, there must be two derivations, a⇒j*abca and a⇒j*aˆbˆcˆ, and this implies that there exists also a derivation a⇒j*aˆbˆcˆbca. Since aˆbˆcˆbca∉X, we have a contradiction. Next, suppose that X∈PPCF, so there exists a PCFG H′=(Σ,P′,σ′) such that L(H′,⇒p)=X. In this case, we must choose σ′=aˆbˆcˆ as the axiom. If we choose a, then a⇒p*abca and a⇒p*aˆbˆcˆ implies that a⇒p*u1au2aˆu3, u1,u2,u3∈Σ* and u1au2aˆu3∉X. If we choose abca or similar, then abca⇒p*a implies that a⇒p*ε, and ε∉X. Without loss of generality, assume that for every α→β∈P′, β∈{a,b,c}* (this can be assumed since aˆbˆcˆ is the only string over {aˆ,bˆ,cˆ} in X). As a∈X, a→ε, a→b and a→c are not contained in P′. The observations (1) to (3) from the proof of Theorem 3.27 hold also for H′. The rest of proof is similar to the proof of Theorem 3.27.□ Theorem 3.29 (Subfamily T). CS−(CF∪JSPCF∪PPCF∪JPPCF)≠∅ Proof Let X={ap∣pisaprime} be a language over unary alphabet {a}. X∈CS and X∉CF are a well-known containments (see [22]). By Lemma 3.4 and Corollary 3.3, X∉JPPCF and X∉JSPCF. As for unary languages PPCF=JPPCF, X∉PPCF.□ The summary of Theorems 3.10–3.29 is visualized in Fig. 1. 3.3. Absence of erasing productions As stated in Lemma 3.1, it is natural that the family of languages generated by pure grammars without erasing productions is included in the family of languages generated by pure grammars in which the presence of erasing productions is allowed. As we show further, for PCFG, the inclusions stated in Lemma 3.1 are proper. The PG case is left as an open problem. Theorem 3.30 Let X∈{SPCF,JSPCF,PPCF,JPPCF,0L}. Then, X−ε⊂X. Proof Let K={a,ab} and Y={aa,aab} be two languages over Σ={a,b}. Furthermore, let G=(Σ,{a→a,b→ε},ab) and G′=(Σ,{a→a,b→ε},aab) be two PCFGs. SPCF−ε⊂SPCF. Since K=L(G,⇒s), K∈SPCF. Assume that K∈SPCF−ε; then, there is a PCFG H=(Σ,P,σ) with no erasing productions in P such that L(H,⇒s)=K. Obviously, σ=a, so a→ab∈P. We have a⇒s*abb and since abb∉K, K∉SPCF−ε. JSPCF−ε⊂JSPCF. K∈JSPCF and K∉JSPCF−ε are proved analogously as in (a). PPCF−ε⊂PPCF. Since Y=L(G′,⇒p), Y∈PPCF. Assume that Y∈PPCF−ε, so there is a PCFG H=(Σ,P,σ) with no erasing productions in P such that L(H,⇒p)=Y. Obviously, σ=aa and then a→ab∈P. We have aa⇒p*abab and since abab∉Y, Y∉PPCF−ε. JPPCF−ε⊂JPPCF. Y∈JPPCF and Y∉JPPCF−ε are proved analogously as in (c). 0L−ε⊂0L (see Theorem 2.8 in [23]).□ Open Problem 3.31 Let X∈{SP,JSP,PP,JPP}. Is the inclusion X−ε⊆X, in fact, proper? From Fig. 1 and from mentioned theorems, we are able to find out the most of relations between investigated language families (even for those which are generated by PCFGs without erasing productions—the most of languages used in Fig. 1 have this property), but not all. Following theorems fill this gap. Theorem 3.32 SPCFand PPCF−εare incomparable, but not disjoint. Proof Let X={aa,aab} be a language over alphabet Σ={a,b}. Obviously, there is a PCFG G=(Σ,{a→a,b→ε},aab) such that L(G,⇒s)=X, so X∈SPCF. By Theorem 3.30, X∉PPCF−ε. Conversely, there is a language Y={a2n∣n≥0} over {a} such that Y∉SPCF and Y∈PPCF−ε (see D in Fig. 1 and observe that to get Y we need no erasing productions). Finally, {a}+∈SPCF∩PPCF−ε.□ Theorem 3.33 SPCFand 0L−εare incomparable, but not disjoint. Proof Analogous to the proof of Theorem 3.32.□ The mutual relation between JSPCF−ε and JPPCF−ε is either incomparability or JSPCF−ε⊂JPPCF−ε, but we do not know the answer now. We also do not know either if JSPCF−ε and JPPCF are incomparable or JSPCF−ε⊂JPPCF. Open Problem 3.34 What is the relation between JSPCF−ε and JPPCF−ε? Open Problem 3.35 What is the relation between JSPCF−ε and JPPCF? Theorem 3.36 PPCF−εand 0Lare incomparable, but not disjoint. Proof Let X={aa,aab} and Y={a,aab} be two languages over {a,b}. X∉PPCF−ε, X∈0L, Y∈PPCF−ε and Y∉0L proves the incomparability, while {a}+∈PPCF−ε∩0L proves the disjointness.□ 3.4. Remark on unary alphabets We close this section by showing how the mutual relations between investigated language families change if we consider only alphabets containing only one symbol. From Theorems 3.2, 3.6 and 3.7, we can conclude that for every unary alphabet SPCF=JSPCF⊂PPCF=JPPCF=0L. Trivially, SPCF−ε=JSPCF−ε⊂PPCF−ε=JPPCF−ε=0L−ε. As the following theorem demonstrates that PPCF−ε and SPCF are incomparable, but not disjoint, we can summarize the results for the unary alphabet in Fig. 2. Theorem 3.37 In the case of unary alphabets, SPCFand PPCF−εare incomparable, but not disjoint. Proof Clearly, the language {a}+ is contained in both SPCF and PPCF−ε. Since the language {ε,a,aa} from SPCF is not contained in PPCF−ε, SPCF⊈PPCF−ε. Conversely, PPCF−ε⊈SPCF since PPCF−ε is not semilinear.□ Figure 2. View largeDownload slide A mutual relations between investigated language families in the case of unary alphabets. The straight line between two families means that these families are identical. The arrow from family A to family B denotes that A⊂B. Figure 2. View largeDownload slide A mutual relations between investigated language families in the case of unary alphabets. The straight line between two families means that these families are identical. The arrow from family A to family B denotes that A⊂B. 4. CONCLUSION Consider SPCF, JSPCF, PPCF, JPPCF, 0L, SPCF−ε, JSPCF−ε, PPCF−ε, JPPCF−ε and 0L−ε (see Section 2). The present paper has investigated mutual relations between these language families, which are summarized in Table 1 and Fig. 1. As a special case, this paper has also performed an analogical study in terms of unary alphabets (see Fig. 2). Although we have already pointed out several open problems earlier in the paper (see Open Problems 3.11, 3.12, 3.17, 3.19, 3.20, 3.23, 3.26, 3.31, 3.34 and 3.35), we repeat the questions of a particular significance next. Is it true that (PPCF∩JSPCF)−(CF∪JPPCF)≠∅ (Open Problem 3.11)? Is it true that (PPCF∩JSPCF∩JPPCF)−CF≠∅ (Open Problem 3.12)? Is it true that (SPCF∩JPPCF)−JSPCF≠∅ (Open Problem 3.17)? Is it true that (PPCF∩CF∩JSPCF)−(SPCF∪JPPCF)≠∅ (Open Problem 3.19)? Is it true that (PPCF∩CF∩JSPCF∩JPPCF)−SPCF≠∅ (Open Problem 3.20)? Is it true that (CF∩JSPCF)−(PPCF∪JPPCF)≠∅ (Open Problem 3.23)? Is it true that JSPCF−(CF∪PPCF∪JPPCF)≠∅ (Open Problem 3.26)? Let X∈{SP,JSP,PP,JPP}. Is the inclusion X−ε⊆X, in fact, proper (Open Problem 3.31)? What is the relation between JSPCF−ε and JPPCF−ε (Open Problem 3.34)? What is the relation between JSPCF−ε and JPPCF (Open Problem 3.35)? Recall that the present study has only considered pure grammars based on context-free productions. Of course, from a broader perspective, we might reconsider all the study in terms of grammars that allow non-context-free productions as well. FUNDING This work was supported by The Ministry of Education, Youth and Sports of the Czech Republic from the National Programme of Sustainability (NPU II), project IT4Innovations excellence in science—LQ1602. ACKNOWLEDGEMENTS The authors deeply thank both anonymous referees for their invaluable comments and suggestions. Footnotes 1 According to its definition, SPCF in this paper coincides with PCF in [20]. REFERENCES 1 Meduna , A. and Zemek , P. ( 2012 ) Jumping finite automata . Int. J. Found. Comput. Sci. , 23 , 1555 – 1578 . Google Scholar CrossRef Search ADS 2 Křivka , Z. and Meduna , A. ( 2015 ) Jumping grammars . Int. J. Found. Comput. Sci. , 26 , 709 – 732 . Google Scholar CrossRef Search ADS 3 Fernau , H. , Paramasivan , M. and Schmid , M.L. ( 2015 ) Jumping Finite Automata: Characterizations and complexity. In Drewes, F. (ed.), Proc. 20th Int. Conf. Implementation and Application of Automata (CIAA 2015), Umeå, Sweden, August 18–21, pp. 89–101. 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