<?xml version="1.0"?><!DOCTYPE article SYSTEM "/project/take/software/searchbench_offline_processing/paperxml_generator/aclextractor/src/python/../resource/dtd/paperxml.dtd"><article><header><firstpageheader><page local="1" global="78"/><title>Restrictions on Monadic Context-Free Tree Grammars</title><author surname="Fujiyoshi" givenname="Akio"><org  name="Ibaraki University" country="Japan" city="Ibaraki"/></author></firstpageheader><frontmatter><p><b>Restrictions on Monadic Context-Free Tree Grammars</b></p><p><b>Akio Fujiyoshi</b></p><p>Department of Computer and Information Sciences, Ibaraki University 4-12-1 Nakanarusawa, Hitachi, Ibaraki, 316-8511, Japan fujiyoshi@cis.ibaraki.ac.jp</p></frontmatter><abstract>In this paper, subclasses of monadic context-free tree grammars (CFTGs) are compared. Since linear, nondeleting, monadic CFTGs gen­erate the same class of string languages as tree adjoining grammars (TAGs), it is examined whether the restrictions of linearity and non-deletion on monadic CFTGs are necessary to generate the same class of languages. Epsilon-freeness on linear, nondeleting, monadic CFTG is also examined. </abstract></header><body><section number="1" title="Introduction"><p>The context-free tree grammars (CFTGs) were in­troduced by W. C. Rounds (1970) as tree gener­ating systems, the definition of which is a direct generalization of context-free grammars (CFGs) from strings to rooted, ordered, labeled trees. For the application of CFTGs to natural languages, many kinds of restrictions on CFTGs have been considered because the string languages generated by CFTGs are exactly indexed languages, whose emptiness problem and uniform membership prob­lem are exponential time complete, i.e., nonre-stricted CFTGs are formidable. One approach to define subclasses of CFTGs is to restrict the ranks of nonterminals. The rank of a nonterminal is a nat­ural number assigned to each nonterminal by which the number of children of the node labeled by the nonterminal is fixed. Through this approach, the simplest model of CFTGs is regular tree grammars (RTGs) (Brainerd, 1969), where the ranks of non­terminals are all 0. The string languages generated by RTGs are the languages generated by context-free grammars (CFGs). Since recent research on natural languages has suggested that formalisms for natural languages need to generate a slightly larger class of languages than CFGs, this paper focuses on monadic CFTGs, where the ranks of nonterminals are either 0 or 1.</p><p>Another formalism of tree generating systems is tree adjoining grammars (TAGs) (Joshi et al., 1975; Joshi and Schabes, 1996; Abeille and Rambow, 2000). TAGs have been widely studied relating them to natural languages, and it was shown that TAGs have the same generative power of string lan­guages as other formalisms for natural languages developed independently such as head grammars, combinatory categorial grammars and linear in­dexed grammars (Vijay-Shanker and Weir, 1994). It is also noteworthy that there are recognition algorithms for the string languages generated by TAGs that run in O(n<footnote anchor="6"/>) and O(M(n<footnote anchor="2"/>)) time (Ra-jasekaran, 1996; Rajasekaran and Yooseph, 1998). From the view point of CFTG, the languages gener­ated by TAGs were examined (Fujiyoshi and Kasai, 2000; Fujiyoshi, 2004; Moennich, 1997), and it was shown that linear, nondeleting, monadic CFTGs generate the same class ofstring languages as TAGs and a strictly larger class of tree languages than TAGs. Linearity is a restriction on CFTGs that requires the number of occurrences of every vari­able in the right-hand side of a rule be no more than 1, and nondeletion requires all variables in the left-hand side of a rule occur at least once in the right-hand side. In other words, linear, nondelet-ing, monadic CFTGs are those with nonterminals of rank 0 and 1 only and with exactly one occurrence of a variable in every right-hand side of a rule for a nonterminal of rank 1 .</p><p>In this paper, the subclasses of monadic CFTGs are compared to examine whether the restrictions of linearity and nondeletion on monadic CFTGs are necessary to generate the same class of string lan­guages as TAGs. It is shown that nondeletion is un­necessary since for any linear, monadic CFTG, there exists an equivalent linear, nondeleting, monadic CFTG. On the other hand, it is shown that linearity is necessary since there exists a non-linear, monadic CFTG which is not weakly equivalent to any linear, monadic CFTG.</p><p>For the development of parsing algorithm, the property of epsilon-freeness is very important, and in this paper, epsilon-freeness on linear, monadic CTFGs is also considered. Epsilon-freeness is a restriction on grammars that requires no use of epsilon-rules, that is, rules defined with the empty string.<page local="2" global="79"/> It is shown that for any linear, monadic CFTG, there exists an epsilon-free, linear, nondelet-ing, monadic CFTG that generate the same string language.</p></section><section number="2" title="Preliminaries"><p>In this section, some terms, definitions and former results which will be used in the rest of this paper are introduced.</p><subsection number="2.1" title="Ranked Alphabets, Trees and Substitution"><p>A <i>ranked alphabet </i>is a finite set of symbols in which each symbol is associated with a natural number, called the <i>rank </i>of a symbol. Let £ be a ranked al­phabet. For n <b>&gt; </b>0, it is defined that £<b>n </b>= {a <b>£ </b>£<b>| </b>the rank of a is n}.</p><p>The set Ts <i>(trees </i>over £) is the smallest set of strings over £, parentheses and commas such that (1) <b>£o </b><b>C </b>T<b>j </b>and (2) if a<b>i</b>, a <b>,.</b><b>..,«n </b><b>£ </b>T<b>j </b>and a <b>£ </b>£<b>n </b>for some n <b>&gt; </b>1, then a(a1, a2,..., an) <b>£ </b>T<b>j.</b></p><doubt alpha="63.9" length="230" tooSmall="False" monospace="0.0">Let A be the empty string. Let e be the special symbol that may be contained in £0. Theyieldof a tree is a function fromTsinto £* defined as fol­lows. Fora£Ts, (1) if a = a£(S0—{e}), yield (a) = a, (1') if a = e, yield(a) = A, and</doubt><p>(2) if a = a(a1 ,a2,...,an) for some a <b>£ </b>Snand a1, a2,..., an <b>£ </b>Ts, yield(a) = yield(a1) <b>• </b>yield (a<b>2</b>) <b>• • • </b>yield (a<b>n</b>).</p><p>Let X be the fixed countable set of variables x1, x2, — It is defined that X0 = <b>0 </b>and for n <b>&gt; </b>1, Xn = {x1, x2,..., xn}. x1 is situationally denoted by x. Ts(Xn) is defined to be TSuXn taking the ranks of elements in X are all 0. For a <b>£ </b>Ts(Xn) and ft,...,ftn <b>£ </b>T<b>j</b>(X), a[ft1 ,ft2,...,ftn] is defined to be the result of substituting each ftj (1 <b>&lt; </b>i <b>&lt; </b>n) for the occurences of the variable X<b>j </b>in a.</p><p>A tree a <b>£ </b>Ts (Xn) is <i>linear </i>if no variable occurs more than once in a, and <i>nondeleting </i>if all variables in X<b>n </b>occur at least once in a. The set of all linear trees and all nondeleting trees in Ts (Xn) are de­noted by Ts (|~Xn<b>]</b>) and Ts (|_Xn<b>J</b>), respectively.</p><p>In this papaer, the conventional way of illus­trating trees is used. See Figure 1. The tree A(b(a),a,B(E,d)) is illustrated as (1). An arbi­trary tree a <b>£ </b>Ts is illustrated as (2). When the variables of a tree ft <b>£ </b>Ts (X3) occur in the order of the tree is illustrated as (3).</p></subsection><subsection number="2.2" title="Context-Free Tree Grammars"><p>The context-free tree grammars (CFTGs) were in­troduced by W. C. Rounds (1970) as tree generating systems. The definition of CFTGs is a direct gener­alization of context-free grammars (CFGs).</p><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">A</doubt><doubt alpha="100.0" length="3" tooSmall="False" monospace="0.0">aEd</doubt><doubt alpha="36.4" length="11" tooSmall="False" monospace="0.0">X\ X2 X3 X\</doubt><doubt alpha="0.0" length="3" tooSmall="False" monospace="0.0">(3)</doubt><figure caption="Figure 1: Trees"></figure><p>A <i>context-free tree grammar </i>(CFTG) is a four-tuple <b>G </b>= (N, S, P, <i>S</i>), where:</p><p><b>• </b>N and S are disjoint ranked alphabets of <i>non­terminals </i>and <i>terminals, </i>respectively.</p><p><b>• </b>P is a finite set of <i>rules </i>of the form</p><doubt alpha="37.5" length="16" tooSmall="False" monospace="0.0">A(X1,X2,...,Xn)a</doubt><p>with n <b>&gt; </b>0, <i>A </i><b>£ </b>Nn and a <b>£ </b>TNu<b>s </b>(Xn). For <i>A </i><b>£ </b>N0, rules are written as A <b>— </b>a instead of</p><p>A() <b>- </b>a.</p><p><b>• </b>S, the <i>initial nonterminal, </i>is a distinguished symbol in N<b>0.</b></p><p>For a CFTG <b>G</b>, the <i>one-step derivation </i>is the relation on TNu<b>j </b><b>x </b>TNu<b>j </b>such that for a tree is an example of a one-step derivation where the rule A(x) <b>— </b>ft is applied to the tree a = a'[A(a'')] and the tree a'[ft[a'']] is obtained.</p><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">g</doubt><doubt alpha="48.5" length="169" tooSmall="False" monospace="0.0">a£TNuj, if a = a'[A(a1, a2,..., an)] for some a'£Tnue([X11)nTnue(LX1J),A£Nnanda1,a2,...,an£Tnus, and A(x1,X2,...,Xn)—ft is in P, then a==a'[ft[a1, a2,..., an]]. Figure 2</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">G</doubt><doubt alpha="66.4" length="131" tooSmall="False" monospace="0.0">An (n-step)derivationis a finite sequence of trees a0, a1,..., an£TNussuch thatn&gt;0 and a0=a1= • • • =an.When there exists a deriva-</doubt><doubt alpha="100.0" length="4" tooSmall="False" monospace="0.0">GGGn</doubt><doubt alpha="56.2" length="48" tooSmall="False" monospace="0.0">tion a0, a1,..., an, it is writen that a0=&gt; anor</doubt><p><b>a0 </b><b>= </b><b>an.</b></p><p>The <i>tree language generated by </i><b>G </b>is the set <i>generated by </i><b>G </b>is LS <b>(G) </b>= {yield(a) <b>| </b>a <b>£ </b>L<b>(G)}.</b><b> </b>Note that <b>Ls</b><b>(G</b>) <b>C </b>(S<b>0 </b><b>— </b>{e})*.</p><doubt alpha="64.9" length="37" tooSmall="False" monospace="0.0">L(G)= {a£Ts|S==a}. Thestring language</doubt><page local="3" global="80"/><figure caption="Figure 2: One-step derivation"></figure><p>Let <b>Q </b>and <b>Q</b><i>' </i>be CFTGs. <b>Q </b>and <b>Q</b><i>' </i>are <i>equivalent </i>if L<b>(Q) </b>= L<b>(Q'). Q </b>and <b>Q</b>' are <i>weakly equivalent </i>if <b>Ls </b><b>(Q) </b>= <b>Ls </b><b>(Q').</b></p></subsection><subsection number="2.3" title="Restrictions on CFTGs"><p>A CFTG <b>Q </b>= (N, S,P,S) is <i>monadic </i>if the rank of any nonterminal is 0 or 1, i.e., N = N0 <b>U </b>N<b>i </b>and Nn = <b>0 </b>for n <b>&gt; </b>2. <b>Q </b>is <b>lin­ear </b>if for any rule A(x1; x2,..., xn) <b>— </b>a in P, a <b>G</b><b> </b>TNu<b>s</b><b>(r</b>X<b>n</b><b>l), </b>and <i>nondeleting </i>if for any rule , X<b>2,</b>..., xn) <b>— </b>a in P, a <b>G </b>Twu<b>s</b>(<b>L</b>X™<b>J</b>).</p><p>A CFTG <b>Q </b>= (N, S, P, S) is <i>epsilon-free </i>if for any rule A(x1; x2,..., xn) <b>— </b>a in P, the symbol e doesn't occur in a.</p><p>When <b>Q </b>is monadic, all rules are either of the form A(x) <b>— </b>a with A <b>G </b>N1 and a <b>G </b>TNu<b>s </b>(X1) or of the form B <b>— </b>ß with B <b>G </b>N0 and ß <b>G </b>TNu<b>s</b>. When <b>Q </b>is monadic, linear and nondeleting, for any rule A(x) <b>— </b>a with A <b>G </b>N1 in P, there exists exactly one occurrence of x in a.</p><p>For linear, nondeleting, monadic CFTGs, the fol­lowing results are known.</p><doubt alpha="64.3" length="42" tooSmall="False" monospace="0.0">Theorem 2.1(Fujiyoshi and Kasai,  2000)The</doubt><p><i>class ofstring languages generated by linear, non-deleting, monadic CFTGs coincides with the class ofstring languages generated by TAGs.</i></p><doubt alpha="66.7" length="45" tooSmall="False" monospace="0.0">Theorem 2.2(Fujiyoshi and Kasai, 2000)For any</doubt><p><i>linear, nondeleting, monadic CFTG, there exists a weakly equivalent linear, nondeleting, monadic CFTG </i><b>Q </b>= (N, S, P, S) <i>that satisfies the following conditions:</i></p><p><b>• </b>For any a <b>G </b>S, the rank of a is either 0 or 2.</p><p><b>• </b>For each A <b>G </b>N0, if A <b>— </b>a is in P, then either a = a with a <b>G </b>S0, or a = B(C) with B <b>G </b>N and C <b>G </b>N<b>0. </b>See (1) and (2) in Figure 3.</p><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">B</doubt><doubt alpha="30.0" length="10" tooSmall="False" monospace="0.0">B       AI</doubt><doubt alpha="21.7" length="23" tooSmall="False" monospace="0.0">A -&gt; a      A -&gt;II -&gt; &amp;</doubt><doubt alpha="66.7" length="3" tooSmall="False" monospace="0.0">&amp;xI</doubt><doubt alpha="8.3" length="24" tooSmall="False" monospace="0.0">(1)                (2)xx</doubt><doubt alpha="25.0" length="4" tooSmall="False" monospace="0.0">(3)x</doubt><doubt alpha="100.0" length="4" tooSmall="False" monospace="0.0">AEAE</doubt><doubt alpha="100.0" length="6" tooSmall="False" monospace="0.0">xcxxxc</doubt><doubt alpha="0.0" length="7" tooSmall="False" monospace="0.0">(4) (5)</doubt><figure caption="Figure S: Strong normal form"></figure><p>• For each A G N1, if A(x) —&gt; a is in P, then a is one of the following forms:</p><doubt alpha="54.5" length="22" tooSmall="False" monospace="0.0">a = B(C(x)) withB,CGN,</doubt><doubt alpha="52.8" length="36" tooSmall="False" monospace="0.0">a = b(C, x) with b G S2and C GN0, or</doubt><p>a = b(x, C) with b G S2 and C G N0.</p><p>See (S),(4) and (5) in Figure S.</p><p>If a linear, nondeleting, monadic CFTG satisfies the condition of Theorem 2.2, it is said that the grammar is in <i>strong normal form<footnote anchor="1"/> </i>.</p></subsection></section><section number="3" title="Linearity and Nondeletion on Monadic"><doubt alpha="100.0" length="5" tooSmall="False" monospace="0.0">CFTGs</doubt><p>Because linear, nondeleting, monadic CFTGs gen­erate the same class of string languages as TAGs, the question is whether the restrictions of linearity and nondeletion on monadic CFTGs are necessary to generate the same class of languages. First, it will be shown that nondeletion is unnecessary.</p><p><b>Theorem 3.1 </b><i>For any linear, monadic CFTG </i>G, <i>there exists an equivalent linear, nondeleting, monadic CFTG </i>G/.</p><p><i>Proof. </i>Let G = (N, S, P, S) be a linear, monadic CFTG. An equivalent linear, nondeleting, monadic follows.</p><doubt alpha="60.0" length="45" tooSmall="False" monospace="0.0">CFTG G/= (N/, S, P/, S) can be constructed as</doubt><p>The set of nonterminal is <i>N' = Nq </i>U <i>N[ </i>such that <i>Nd</i><i> </i><i>= N0 </i>U <i>{A\A </i>G <i>Ni}</i><i> </i>and <i>N[</i><i> = Nx. </i>For the preparation of the definition of P/, for a G Tvu<b>s</b>(X<b>l</b>) we define n(a) C TW'u<b>s</b>(X<b>l</b>) as the smallest set satisfying the following conditions:</p><doubt alpha="45.5" length="11" tooSmall="False" monospace="0.0">• a G n(a)D</doubt><footnote label="1">We say "strong" because a grammar in this normal form only preserves weak equivalence.</footnote><page local="4" global="81"/><doubt alpha="0.0" length="3" tooSmall="False" monospace="0.0">(3)</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">g</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">G</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">a</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">S</doubt><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">G'</doubt><doubt alpha="0.0" length="3" tooSmall="False" monospace="0.0">(1)</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">i</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">x</doubt><doubt alpha="0.0" length="3" tooSmall="False" monospace="0.0">(2)</doubt><doubt alpha="66.7" length="3" tooSmall="False" monospace="0.0">LP)</doubt><p>a <b>== </b>ft.   For k <b>&gt; </b>1, assume that the statement</p><p><b>G' —</b>holds for any derivation of length less than k. If</p><p>a rule of the form A <b>— </b>7 with A <b>£ </b>N0 is used at the first step, the proof is rather simple, so we only prove the other case.   Suppose that the case (3), A(x) <b>— </b>77 is in P' and therefore, 5'[A(5'')] <b>£ </b>n(a) and 5'[A(5'')] <b>== </b>5'[7[5'']] <b>= </b>ft.</p><doubt alpha="63.2" length="38" tooSmall="False" monospace="0.0">a rule A( x)—7 with A£N1is used at the</doubt><doubt alpha="47.9" length="48" tooSmall="False" monospace="0.0">first step and a = a'[A(a'')]=a'[7[a'']]==ft for</doubt><doubt alpha="64.9" length="131" tooSmall="False" monospace="0.0">some a'£Twus^l) nTwus(LX1J)and a''£TNus. By the induction hypothesis, there exist 5£n(a'[7[a'']]) such that 5==ft. Here, we have to</doubt><doubt alpha="59.6" length="52" tooSmall="False" monospace="0.0">think of the three different cases: (1) 5£n(a'), (2)</doubt><doubt alpha="60.5" length="43" tooSmall="False" monospace="0.0">5can be written as5'[7]for some5'£n(a') and</doubt><doubt alpha="42.9" length="49" tooSmall="False" monospace="0.0">7£n(7), and (3) 5 can be written as 5'[7[5'']]for</doubt><doubt alpha="40.5" length="37" tooSmall="False" monospace="0.0">some5'£n(a'),7£n(7) and5''£n(a"). See</doubt><figure caption="Figure 4. In the case (1), 5£n(a) and 5==ft."></figure><doubt alpha="33.3" length="3" tooSmall="False" monospace="0.0">_ g</doubt><doubt alpha="64.3" length="42" tooSmall="False" monospace="0.0">In the case (2),A—&gt; 7is inP'and therefore,</doubt><doubt alpha="28.6" length="42" tooSmall="False" monospace="0.0">5'[ A }G11(a)and5'[ A } =&gt;6'[*/]%13.And in</doubt><doubt alpha="50.0" length="4" tooSmall="False" monospace="0.0">G'G'</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">k</doubt><p>The "if part is proved as follows. Let a <b>= </b>ft for some 7</p><doubt alpha="16.7" length="6" tooSmall="False" monospace="0.0">7 = t,</doubt><p>G n(a).   If k = a and a == ß.</p><p>O, then clearly For k &gt; l, as-</p><doubt alpha="41.4" length="70" tooSmall="False" monospace="0.0">{A—a|A£N0,A—a£P,a£n(a)}     5£Tnussuch that a' [7[a'' ]]£n(5), 5=ft.By</doubt><figure caption="Figure 4: The three different cases"></figure><p><b>• </b>If ft <b>£ </b>n(a) and ft = ft' [B(ft'')] for some B <b>£ </b><i>Nu </i><i>P&gt; </i><b>G </b>TN/uE(pril) <b>n </b>2V<b>uE(|*iJ) </b>and <i>P" </i>£%uE, then <i>/3'</i><i>[B</i><i> } e 11(a).</i></p><p>The set of rules is defined as follows.</p><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">P/</doubt><doubt alpha="52.2" length="23" tooSmall="False" monospace="0.0">U{A(x)—a|AGN1,A(x)—aGP,</doubt><doubt alpha="53.5" length="43" tooSmall="False" monospace="0.0">aeIl(a)nTN/uS([X1\)}U{A—&gt;a\AeNi,A(x)-&gt; aGP,</doubt><p>a <b>g </b>n(a) <b>n </b><b>tw</b>/<b>us</b><b>}</b></p><p>Because of the construction of N<b>' </b>and P<b>'</b>, <b>Q' </b>is monadic and nondeleting.</p><p>To show the equivalence of <b>Q </b>and <b>Q</b><b>'</b>, we prove the following statement holds for any a <b>G </b>TN<b>us</b>and ß <b>G </b>Ts by induction on the length of deriva­tions:</p><p>ß if and only if there exitsts a <b>G</b></p><p>n(a) such that a == ft.</p><p>We start with proving "only-if' part. Let a == ft.</p><doubt alpha="60.5" length="38" tooSmall="False" monospace="0.0">If k= 0,then clearly a=ft, a £n(a) and</doubt><p>the definition of IT, there exists a' <b>G </b>TN<b>us </b>(X1) and and a'[7[a"]] <b>G </b>TT(a'[Y[a"]]). Therefore, a ^ß.</p><doubt alpha="44.4" length="45" tooSmall="False" monospace="0.0">a''GTNuSsuch that a = a'[A(a'')]=^a'[7[a'']],</doubt><doubt alpha="50.0" length="36" tooSmall="False" monospace="0.0">And in the case (3),a=a'[A]^ä'[j]=^ß</doubt><doubt alpha="64.7" length="85" tooSmall="False" monospace="0.0">for some a'GTw'us([X1])nTW'us(LX1J).By the definition of P', A(x)—7 is in P such that</doubt></section><section number="7" title="G TT(y) . By the induction hypothesis, for any"><doubt alpha="60.0" length="40" tooSmall="False" monospace="0.0">8GTnussuch thata'[7]GTT(&lt;j),8^ ß. By the</doubt><p>definition of TT, there exists a' <b>G </b><b>Tn</b><b>us </b>(X1) and</p><doubt alpha="43.5" length="46" tooSmall="False" monospace="0.0">a''GTNussuch that a = a'[A(a'')]=&gt; a'[7[a'']],</doubt><doubt alpha="0.0" length="1" tooSmall="False" monospace="0.0">□</doubt><p>Next, consideration will be given to whether the restriction of linearity can be removed from sume that the statement holds for any derivation of length less than k. The rule used at the first step is one of the following forms: (1) A <b>— </b>77 with  and a'[7] <b>G </b>ü(a'[7[a"]]). Therefore, a <b>== </b>ß.</p><doubt alpha="0.0" length="1" tooSmall="False" monospace="0.0">1</doubt><doubt alpha="66.1" length="115" tooSmall="False" monospace="0.0">AGN0,(2)A(x)7 withAGNuor (3)A7 withAGN^ - N0.The proof of the case (1) is similar to the proofs of the other cases,</doubt><doubt alpha="45.3" length="95" tooSmall="False" monospace="0.0">so we start proving the case (2). In the case (2), 7 = 7'[A( 7'')]G'7'[77[7'']]G't for some 7'£</doubt><doubt alpha="62.1" length="87" tooSmall="False" monospace="0.0">Tn'us([X11)nTn'us(LX1J) and a''£Tw'us.By the definition of P', A(x)—7 is in P such that</doubt><p>Because <i>U(S)</i><i> </i><i>=</i><i> </i><i>{S},</i><i> </i><i>L(Q)</i><i> </i><i>=</i><i> </i><i>L(Q&lt;).</i></p><page local="5" global="82"/><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">G</doubt><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">G'</doubt><p>monadic CFTGs to generate the same class of lan­guages. The answer is negative. The following ex­ample is a non-linear, monadic CFTG thatgenerates a string language that no linear, monadic CFTG can generate.</p><doubt alpha="63.3" length="311" tooSmall="False" monospace="0.0">Example 3.2The following is an example of a monadic CFTG that generates the string language Lw4= {wwww|w£{a, 6}+}.G= (N, S, P, S) where N = {S, A}, the ranks ofS and A are 0 and 1, respectively, S = {a, 6, c, d}, the ranks of a, 6, c and d are 0, 0, 2 and 4, respectively, and P consists of the following rules:</doubt><p>S <b>— </b>A(a), S <b>— </b>A(6), A(x) <b>— </b>d(xxxx), A(x) <b>— </b>A(c(xa)), and A(x) <b>— </b>A(c(x6)).</p><p>Because <b>G </b>has the rule A(x) <b>— </b>d(xxxx), <b>G </b>is not linear.</p><p><b>Theorem 3.3 </b><i>There exists a monadic CFTG which is not weakly equivalent to any linear, monadic</i> <i>CFTG.</i><i></i></p><p><i>Proof </i>It is known that the string language L<b>w4 </b>in Example 3.2 cannot be generated by any TAG. It cannot be generated by any linear, monadic CFTG, neither. □</p></section><section number="4" title="Epsilon-Freeness on Linear, Monadic CFTGs"><p>According to our definition of CFTGs, they are al­lowed to generate trees with the special symbol e, which is treated as the empty string while taking the yields of trees. In this section, it will be seen that for any linear, monadic CFTG, there exists a weakly equivalent epsilon-free, linear, nondeleting, monadic CFTG. Because any epsilon-free CFTG cannot generate a tree with e, it is clear that for a CFTG with epsilon-rules, there generally doesn t exist an equivalent epsilon-free CFTG.</p><p><b>Theorem 4.1 </b><i>For any linear, monadic CFTG </i><b>G </b>= (N, S,P, S), <i>if </i>A <b>£ Ls(G), </b><i>then there exists a weakly equivalent epsilon-free, linear, nondeleting, monadic CFTG </i><b>G'</b><i>. If </i>A <b>£ Ls (G), </b><i>then there exists </i><b>G' </b><i>whose epsilon-rule is only </i>S <b>— </b>e.</p><p><i>Proof. </i>Since it is enough to show the existence of a weakly equivalent grammar, without loss of gen­erality, we may assume that <b>G </b>is in strong normal form. We may also assume that the initial nonter­minal S doesn t appear in the right-hand side ofany rule in P.</p><p>We first construct subsets of nonterminals E0 and E1 as follows. For initial values, we set E0 = {A <b>£</b></p><p>N0 |A <b>— </b>e <b>£ </b>P<b>} </b>and E1 <b>= 0. </b>We repeat the fol­lowing operations to E<b>0 </b>and E<b>1 </b>until no more oper­ations are possible:</p><doubt alpha="62.5" length="32" tooSmall="False" monospace="0.0">•If A—B(C)with B£E1and C£E0is in</doubt><p>P, then add A <b>£ </b>N<b>0 </b>to E<b>0.</b></p><doubt alpha="53.8" length="39" tooSmall="False" monospace="0.0">•If A(x)—6(C, x) with C£E0is in P, then</doubt><p>add A <b>£ </b>N<b>1 </b>to E<b>1.</b></p><doubt alpha="55.3" length="38" tooSmall="False" monospace="0.0">•If A(x)—6(x, C)with C£E0is in P, then</doubt><doubt alpha="52.8" length="36" tooSmall="False" monospace="0.0">•If A(x)—B(C(x)) with B,C£E1is in P,</doubt><p>then add A <b>£ </b>N<b>1 </b>to E<b>1.</b></p><p>In the result, E0 satisfies the following.</p><doubt alpha="50.0" length="30" tooSmall="False" monospace="0.0">E0={A£N0|3a£Ts,A=a,yield(a)=A}</doubt><p>We construct <b>G</b><b>' </b><b>= </b>(N<b>'</b>, <b>S</b>',P',S) as follows. The set of nonterminals is <i>N' = Nq </i>U <i>N[ </i>such that <i>No"</i><i> </i><i>=</i><i> </i><i>N0 </i>U <i>(A\A </i>G <i>Nx}</i><i> </i>and <i>N[ = </i><i>Nx.</i><i> </i>The set of terminal is <b>S' = S U </b>{c}, where c is a new sym­bol of rank 1. The set of rules P' is the smallest set satisfying following conditions:</p><p><b>• </b>P' contains all rules in P except rules of the form A <b>— </b>e.</p><p><b>• </b>If S <b>£ </b>E<b>0</b>,then S <b>— </b>e is in P<b>'</b>.</p><p>• If A -»■ <i>B(C)</i><i> </i>is in <i>P</i><i> </i>and <i>C</i><i> </i>G E0, then <i>A -»■ </i><i>B </i>is in <i>P'.</i></p><p><i>•</i><i> </i><i>lfA(x)</i><i> </i>-»■ <i>B(C{x))</i><i> </i>is in <i>P,</i><i> </i>then <i>A^</i><i> </i><i>B(C)</i><i> </i>is in P'.</p><p><b>• </b>If A(x) <b>— </b>6(C, x) or A(x) <b>— </b>6(x, C<b>) </b>is in P and C <b>£ </b>E0, then A(x) <b>— </b>c(x) is in P<b>'</b>.</p><p><i>•</i><i> </i><i>UA{x)</i><i> </i>-»■ <i>b(C,</i><i> </i><i>x)</i><i> </i>or <i>A</i><i>(x)</i><i> </i>-»■ <i>b(x,</i><i> </i><i>C)</i><i> </i>is in <i>P, </i>then <i>A     </i><i>c(C)</i><i> </i>is in <i>P'.</i></p><p>To show <b>Ls(G</b><b>'</b><b>) = Ls(G), </b>we prove the follow­ing (i), (ii) and (iii) hold by induction on the length of derivations:</p><doubt alpha="58.6" length="29" tooSmall="False" monospace="0.0">(i) For A£N0,A''and'£Tsif and</doubt><p>only if A <b>G </b>for some <b>£ </b>T<b>s </b>such that yield( <b>) = </b>yield( <b>'</b><b>) = </b>A.</p><doubt alpha="51.4" length="74" tooSmall="False" monospace="0.0">(ii) For A£N1,A(x)==a' and a'£Ts(X^ if and only A(x)       for some£Ts(X1)</doubt><p>such that yield( <b>) = </b>yield( <b>'</b><b>)</b>.</p><p>(iii) For <i>A </i>G <i>- N0, </i><i>A </i>% <i>a' </i>and <i>a' </i>G Ts if and that yield( [e]) <b>= </b>yield( <b>'</b><b>) = </b>A.</p><footnote label="0">G ' only if A(x) == a for some a £ T s (X 1 ) such</footnote><page local="6" global="83"/><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">G</doubt><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">G'</doubt><doubt alpha="50.0" length="4" tooSmall="False" monospace="0.0">G'G'</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">k</doubt><p>We start with "only if part. For 0-step deriva­tions, (i), (ii) and (iii) clearly hold since there doesn't exists a' £ Ts nor a' £ Ts(X1) for each statement.</p><p>We consider the cases for 1-step derivations.</p><p><i>A=?B(C)p </i>ft'[7'] <i>= </i><i>a'</i><i> </i>for some <i>13'</i><i> </i><b>g </b>Ts(Xi) and V <b>g </b>Ts such that <i>B(x) </i><i>^13' </i>and <i>C =jfj'.</i></p><p>By the induction hypothesis of (ii), there exists ft £ T<b>s</b>(X<b>1</b>) such that B(x) = ft and yield(ft) = for some a £ S0 and the rule A — a in P<b>' </b>has been used. Therefore, A — a is in P and A a.</p><doubt alpha="66.3" length="86" tooSmall="False" monospace="0.0">[Proof of (i)] If A == a' and a' £Ts, then a' = ayield(ft').Bytheinductionhypothesisof</doubt><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">'s</doubt><doubt alpha="33.3" length="3" tooSmall="False" monospace="0.0">G''</doubt><doubt alpha="45.7" length="46" tooSmall="False" monospace="0.0">[Proof of (ii)] If A(x) == a' and a' £ Ts(X1),</doubt><doubt alpha="51.3" length="39" tooSmall="False" monospace="0.0">then'=  c(x) and the rule A(x)   — c(x)</doubt><p>in P' has been used.   By the definition of P<b>'</b>, some C £ E<b>0.</b><b>  </b>There exists 7 £ T<b>s </b>such that</p><doubt alpha="40.9" length="44" tooSmall="False" monospace="0.0">A(x) — 6(C, x) or A(x) — 6(x, C) is in P for</doubt><doubt alpha="51.1" length="45" tooSmall="False" monospace="0.0">C    7 and yield(7) = A.   Therefore, A(x)G G</doubt><doubt alpha="23.8" length="42" tooSmall="False" monospace="0.0">6(C,x) ==6(7,x) or A(x) ==6(x,C) ==6(x,7),</doubt><doubt alpha="60.0" length="5" tooSmall="False" monospace="0.0">G G G</doubt><doubt alpha="59.8" length="102" tooSmall="False" monospace="0.0">and yield(6(7, x)) = yield(6(x, 7)) = yield(c(x)). [Proof of (hi)] There doesn't existsa'gTEsuch thatA</doubt><p>For k &gt; 2, assume that (i), (ii) and (iii) holds for any derivation of length less than k.</p><doubt alpha="64.4" length="45" tooSmall="False" monospace="0.0">[Proof of (i)] If Ak'', then the rule used at</doubt><doubt alpha="54.1" length="122" tooSmall="False" monospace="0.0">the first step is one of the follwoing form: (1)A-»■B(C)or (2)A-»■B.In the case (1), A = B(C) ==ft'[7']= a' for some ft' £</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">C</doubt></section><section number="7" title="' . By the induction hypothesis of (ii), there"><doubt alpha="63.2" length="38" tooSmall="False" monospace="0.0">there exists 7  £ Ts(X1)such that C(x)</doubt><doubt alpha="55.8" length="52" tooSmall="False" monospace="0.0">and yield (7 [e])  = yield (7').   By the definition</doubt><doubt alpha="56.4" length="39" tooSmall="False" monospace="0.0">of P', A(x) — B(C(x)) is inP.Therefore,</doubt><doubt alpha="55.4" length="410" tooSmall="False" monospace="0.0">A(x)fB(C(x))f/3[i]and yield (ft[7[e]]) = yield(ft/[7/]). In the case (2),Afc(C)fc(7') ='for some7'£ Tssuch that CG'7'. By the in­duction hypothesis of (i), there exists 7 £ Tssuch that CG7 and yield(7) = yield(7'). By the def­inition of P', A(x) — 6(C,x) or A(x) — 6(x,C) is inP.Without loss of generality, we may as­sume that A(x) — 6(C, x) is in P. Therefore, A(x) ==6(C, x) ==6(7, x) and yield(6(7, x)[e]) =</doubt><p>yield(c(7')).</p><p>The "if part is similarly proved as follows. For 0-step derivations, (i), (ii) and (iii) clearly hold since there doesn t exists £ T<b>s </b>nor £ T<b>s</b>(X<b>1) </b>for each statement.</p><p>The cases for 1-step derivations are proved.</p><doubt alpha="42.9" length="49" tooSmall="False" monospace="0.0">[Proof of (i)] If A       and    £ Ts,then    = a</doubt><p>for some a £ S0 and the rule A — a in P has been used. Therefore, A — a is in P' and A a.</p><doubt alpha="62.2" length="98" tooSmall="False" monospace="0.0">exists ft  £  Ts(X1) such that B (x)==ftand     [Proof of (ii) and (iii)] There doesn't existsa£Ts</doubt><doubt alpha="62.0" length="50" tooSmall="False" monospace="0.0">and yield(7) = yield(7'). By the definition of P',</doubt><doubt alpha="47.8" length="46" tooSmall="False" monospace="0.0">A — B(C) is in P. Therefore, A = B(C) == ft[7]</doubt><doubt alpha="57.3" length="89" tooSmall="False" monospace="0.0">and yield(ft[7]) = yield(ft/[Y]). In the case (2),A=&gt;B^a'.By the definition ofP',A-»■B(C)</doubt><doubt alpha="62.2" length="98" tooSmall="False" monospace="0.0">is in P for some C £ E0.There exists 7 £ Tssuch that C    7 and yield(7) = A. By the induction hy-</doubt><doubt alpha="65.0" length="103" tooSmall="False" monospace="0.0">pothesis of (iii), there existsft£Ts(X1) such that B (x)==ft and yield (ft [e]) = yield(a'). Therefore,</doubt><p>A =B(C) =ft[7] andyield(ft[7]) = yield(a').</p><doubt alpha="100.0" length="2" tooSmall="False" monospace="0.0">GG</doubt><doubt alpha="63.0" length="46" tooSmall="False" monospace="0.0">[Proof of (ii)] If A(x)k'', then the rule used</doubt><p>at the first step is one of the follwoing form: (1) A(x) — B(C(x)), (2) A(x) — 6(C,x) or (3) A(x) — 6(x, C). Becasue these rule are in P, the proofs are direct from the induction hypothesis like the proof of the case (1) of(i).</p><doubt alpha="65.2" length="46" tooSmall="False" monospace="0.0">[Proof of (iii)] If A'', then the rule used at</doubt><p>the first step is one of the follwoing form: (1) such that A .</p><doubt alpha="43.6" length="39" tooSmall="False" monospace="0.0">A-»■B(C)or (2)A-»■c(C).In the case (1),</doubt><doubt alpha="59.6" length="52" tooSmall="False" monospace="0.0">[Proof of (i)] If A      , then the rule used at the</doubt><doubt alpha="66.0" length="47" tooSmall="False" monospace="0.0">first step must be of the form A — B (C). Thus,</doubt><doubt alpha="48.6" length="37" tooSmall="False" monospace="0.0">A = B(C)=ft[7]= a for some ft £ Ts(X^</doubt><p>and 7 £ T<b>s </b>such that B(x)    t and C 7.</p><doubt alpha="62.1" length="140" tooSmall="False" monospace="0.0">Here, we have to think of the two cases: (1) yield( 7) = A and (2) yield(7) = A. In the case (1), by the induction hypothesis of (ii), there</doubt><doubt alpha="64.9" length="37" tooSmall="False" monospace="0.0">exists t'£ Ts(X1)such that B(x)'t'and</doubt><p>yield(t') = yield(t), and bythe inductionhypothe-sis of (i), there exists 7' £ T<b>s </b>such that C ' 7' and and yield(ft/[Y]) = yield(/%]). In the case (2), <i>C </i><b>g </b><i>Eq. </i>Thus, <i>A —&gt; B </i>is in <i>P'. </i>By the induction hypothesis of (iii), there exists ft' <b>g </b>TE <i>(X\)</i><i> </i>such that <i>B </i>ft' and yield (ft') = yield (ft [e]). There­fore, <i>A^B</i><i> </i><i>^13' </i>andyield(ft') = yield(/%]).</p><doubt alpha="58.0" length="50" tooSmall="False" monospace="0.0">yield(7') = yield(7). By the definition of P', A —</doubt><doubt alpha="59.5" length="37" tooSmall="False" monospace="0.0">B(C) is inP.Therefore, AG'B(C)Gt'[7']</doubt><doubt alpha="62.1" length="29" tooSmall="False" monospace="0.0">Ts(X1)and7'£ Tssuch that B(x)</doubt><doubt alpha="80.0" length="5" tooSmall="False" monospace="0.0">t'and</doubt><p>yield(t) = yield(t'). By the induction hypothesis of (i), there exists 7 £ T<b>s </b>such that C the first step is one of the follwoing form:<page local="7" global="84"/> (1) [Proof of (ii)] If A(x) k   , then the rule used at form.</p><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">G</doubt><doubt alpha="50.0" length="2" tooSmall="False" monospace="0.0">G'</doubt><doubt alpha="33.3" length="3" tooSmall="False" monospace="0.0">_ g</doubt><doubt alpha="50.0" length="4" tooSmall="False" monospace="0.0">G'G'</doubt><doubt alpha="100.0" length="1" tooSmall="False" monospace="0.0">k</doubt><doubt alpha="49.7" length="183" tooSmall="False" monospace="0.0">A(x) — B(C(x)), (2) A(x) — 6(C,x) or (3) A(x) — 6(x, C). The proof of the case (1) is di­rect from the induction hypothesis. In the case (2), A(x)G6(C, x)G6(7, x) =    for some 7 £ Ts</doubt><doubt alpha="66.7" length="51" tooSmall="False" monospace="0.0">such that C    7. Here, we have to think of the two</doubt><doubt alpha="60.1" length="138" tooSmall="False" monospace="0.0">cases: (a) yield(7) = A and (b) yield(7) = A. (a) If yield( 7) = A, then by the induction hy­pothesis of (i), there exists 7'£ Tssuch that</doubt><doubt alpha="52.1" length="48" tooSmall="False" monospace="0.0">C == 7' and yield(7') = yield(7). By the defini-</doubt><doubt alpha="42.6" length="94" tooSmall="False" monospace="0.0">tion of P', A(x) — 6(C, x) is in P'. Therefore, A(x) = b(C,x) == 6(7', x) and yield(6(7',x)) =</doubt><doubt alpha="49.0" length="96" tooSmall="False" monospace="0.0">yield(6(7,x)). (b) If yield(7) = A, then C £ E0and A( x) — c( x) is in P'. Therefore, A(x)G'c(x)</doubt><p>and yield(c(x)) = yield(6(7, x)). The proofofthe case (3) is similar to that ofthe case (2).</p><doubt alpha="58.0" length="50" tooSmall="False" monospace="0.0">[Proof of (iii)] If A(x)      , then the rule used</doubt><p>at the first step is one of the follwoing form:</p><doubt alpha="22.5" length="40" tooSmall="False" monospace="0.0">(1) A(x)  —  B(C(x)), (2) A(x)  — 6(C,x)</doubt><doubt alpha="35.7" length="42" tooSmall="False" monospace="0.0">or (3) A(x)  — 6(x, C).   In the case (1),</doubt><doubt alpha="42.1" length="38" tooSmall="False" monospace="0.0">A(x)GB(C(x))Gt[7] =    for some t, 7 £</doubt><doubt alpha="62.2" length="37" tooSmall="False" monospace="0.0">Ts(X1)such that B(x)Gt and C(x)G7. By</doubt><p>the definition of <i>P', A </i>-»■ <i>B(C)</i><i> </i>is in <i>P'. </i>By the induction hypothesis of (ii), there exists t' £ yield (ft). By the induction hypothesis of (iii), there exists 7' € Ts such that <i>C </i>^7' and yield(7') = the definition of <i>P', A </i>-»■ c(C) is in <i>P'.</i><i> </i>By the induction hypothesis of (i), there exists 7' £</p><doubt alpha="60.5" length="38" tooSmall="False" monospace="0.0">Ts(X1)such that B(x)'t'and yield(t') =</doubt><doubt alpha="53.8" length="39" tooSmall="False" monospace="0.0">yield(7[e]).Therefore,A^B{C)|&gt; /%'] and</doubt><doubt alpha="45.8" length="48" tooSmall="False" monospace="0.0">yield(t'[7']) = yield(t[7[e]]). In the case (2),</doubt><doubt alpha="35.0" length="40" tooSmall="False" monospace="0.0">A(x)G6(C, x)G6(7, x)  =     for some 7 £</doubt><doubt alpha="63.9" length="36" tooSmall="False" monospace="0.0">Tssuch that CG7 and yield(7) = A. By</doubt><p>T<b>s </b>such that C  ' 7' and yield(7') = yield(7).</p><doubt alpha="52.3" length="44" tooSmall="False" monospace="0.0">Therefore,A=&gt;c(C)=r&gt;c(7;) and yield(c(7')) =</doubt><p>yield(6(7, x)[e]). The proofofthe case (3) is simi­lar to that of the case (2).</p><doubt alpha="61.0" length="41" tooSmall="False" monospace="0.0">By (i), we have the resultLs(Qr) =Ls{Q).□</doubt></section><section number="5" title="Conclusions"><p>In this paper, the desirable features of linear monadic CFTGs have been discovered: the re­striction of nondeletion doesn't affect their gener­ative power of tree languages, and the restriction of epsilon-freeness can be assumed when their gener­ation of string languages is considered. The key to the proofs of this paper was the simplicity of the def­inition of linear, monadic CFTGs and their normal</p><p>Recently, the class of grammars called mildly context-sensitive grammars has been studied very actively, to which TAGs and other well-established formalisms for natural languages belong. Since it is not difficult to study formal properties of linear, monadic CFTGs, they are helpful tools for the study of mildly context-sensitive grammars.</p></section><references><p>Anne Abeille and Owen Rambow, editors. 2000. <i>Tree adjoining grammars: formalisms, linguis­tic analysis and processing. </i>CSLI Publications, Stanford, California.</p><p>Walter S. Brainerd. 1969. Tree generating regular systems. <i>Information &amp; Control, </i>14(2):217-231.</p><p>Akio Fujiyoshi and Takumi Kasai. 2000. Spinal-formed context-free tree grammars. <i>Theory of Computing Systems, </i>33(1):59-83.</p><p>Akio Fujiyoshi. 2004. Epsilon-free grammars and lexicalized grammars that generate the class of the mildly context-sensitive languages. In <i>7th In­ternational Workshop on Tree Adjoining Gram­mar and Related Formalisms: Proceedings ofthe Workshop, </i>Vancouver, pages 16-23.</p><p>Aravind K. 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