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<Paper uid="W06-1504">
  <Title>The weak generative capacity of linear tree-adjoining grammars</Title>
  <Section position="4" start_page="25" end_page="31" type="intro">
    <SectionTitle>
3 Properties
</SectionTitle>
    <Paragraph position="0"> We now review several old results and prove a few new results relating the weak generative capacity of these formalisms to one another and to (linear) CFG and TAG. These results are summarized in</Paragraph>
    <Section position="1" start_page="25" end_page="25" type="sub_section">
      <SectionTitle>
3.1 Previous results
</SectionTitle>
      <Paragraph position="0"> that the higher formalism has strictly greater weak generative capacity than the lower.</Paragraph>
      <Paragraph position="1">  mura et al. 1999). SSL-TAG and ESL-TAG can be parsed in O(n5) time.</Paragraph>
    </Section>
    <Section position="2" start_page="25" end_page="27" type="sub_section">
      <SectionTitle>
3.2 Weak equivalence
</SectionTitle>
      <Paragraph position="0"> Proposition 5. The following formalisms are weakly equivalent: (i) ESL-TAG (ii) SL-TAG with substitution (iii) ESL-TAG with substitution (iv) SSL-TAG Proof. We prove this by proving four inclusions.</Paragraph>
      <Paragraph position="2"> deal rst with the left and right auxiliary trees, and then with off-spine adjunction.</Paragraph>
      <Paragraph position="3"> First, we eliminate the left and right auxiliary trees. Since these only insert material to the left or right of a node, just as in tree-insertion grammars (TIGs), we may apply the conversion from TIGs to tree-substitution grammars (Schabes and Waters, 1995), used in the proof of the context-freeness of  TIG.2 (Step 1a) For each active node X that is not the root of a left or right auxiliary tree, we create four copies of the containing elementary tree with X altered in the following ways: rst, leave X unchanged; then, add a copy of X above it, making both nodes no-adjunction nodes, and add a new left sister substitution node labeled LX or a new right sister substitution node labeled RX, or both. See Figure 2. (Step 1b) For each b that was originally a left (right) auxiliary tree with root/foot label X, relabel the root node as LX (RX) and delete the foot node, and create two copies of the containing elementary tree, one unchanged, and one with a new left (right) sister substitution node. See Figure 2. When the modi ed b substitutes at one of the new children of an e, the substitution clearly results in the same string that would have resulted from adjoining the original b to e.</Paragraph>
      <Paragraph position="4"> This construction might appear incorrect in two ways. First, the new grammar has trees with both an LX and an RX node corresponding to the same original node, which would correspond to adjunction of two auxiliary trees bL and bR at the same node X in the original grammar. But this new derivation generates a string that was generable in the original grammar, namely by adjoining bL at 2This corresponds to Steps 1 4 of that proof (Schabes and Waters, 1995, p. 486). Since that proof uses a more relaxed de nition of left and right auxiliary trees, it is probable that SSL-TAG could also be relaxed in the same way.</Paragraph>
      <Paragraph position="5"> X, then adjoining bR at the root of bL, which is allowed because the de nition of SSL-TAG prohibits adjunction constraints at the root of bL.</Paragraph>
      <Paragraph position="6"> Thus the rst apparent problem is really the solution to the second problem: in the original grammar, a left auxiliary tree bL could adjoin at the root of a right auxiliary tree bR, which in turn adjoined at a node e, whereas in the new grammar, bR does not have an LX substitution node to allow this possibility. But the same string can be generated by substituting both trees under e in the new grammar. In the case of a whole chain of adjunctions of left/right auxiliary trees at the root of left/right auxiliary trees, we can generate the same string by rearranging the chain into a chain of left auxiliary trees and a chain of right auxiliary trees (which is allowed because adjunction constraints are prohibited at all the roots), and substituting both at e. (Step 2) Next, we eliminate the case of a wrapping auxiliary tree b that can adjoin at an off-spine node e. (Step 2a) For each active off-spine node e, we relabel e with a unique identi er ^e and split the containing elementary tree at e:  (Step 2b) After step 2a has been completed for all nodes e, we revisit each e, and for every wrapping b that could adjoin at e, create a copy of b with root relabeled to T^e and foot relabeled to B^e.</Paragraph>
      <Paragraph position="8"> Then the original b is discarded. Substituting one of these copies of b at a T^e node and then substituting a B^e tree at the former foot node has the same effect as adjoining b at e. Finally, unless e had an obligatory-adjunction constraint, simulate the lack of adjunction at e by adding the initial</Paragraph>
      <Paragraph position="10"> construction is related to Lang's normal form which ensures binary-branching derivation trees (Lang, 1994), but guarantees that one adjunction site is on the spine and one is off the spine.</Paragraph>
      <Paragraph position="11"> (Step 0a) Ensure that the elementary trees are binary-branching. (Step 0b) Add a new root and foot node to every elementary tree:</Paragraph>
      <Paragraph position="13"> auxiliary tree has more than one substitution node.</Paragraph>
      <Paragraph position="14"> For any auxiliary tree with spine longer than four nodes, we apply the following transformation: target either the active node or its parent, and call it Y . Let Z1 be the child that dominates the foot node; let V1 be a fresh nonterminal symbol and insert V1 nodes above Y and below Z1, and excise the segment between the two V nodes, leaving behind an active obligatory-adjunction node.</Paragraph>
      <Paragraph position="15"> If Y has another child, call it Z2; let V2 be a fresh nonterminal symbol and insert a V2 node above Z2, and break off the subtree rooted in V2, leaving behind a substitution node. See Figure 3. This transformation reduces the spine of the auxiliary tree by one node, and creates two new trees that satisfy the desired form. We repeat this until the entire grammar is in the desired form.</Paragraph>
      <Paragraph position="16"> (Step 2) Next, we transform the grammar so that no initial tree has more than one substitution node, while maintaining the form acquired in step 1. For any initial tree with height greater than three nodes, we apply the same transformation as in step 1, except that Y is the child of the root node, Z1 is its left child, and Z2 is its other child if it exists and is not already a substitution node. See Figure 3. This transformation replaces an initial tree with at most two shorter initial trees, and one auxiliary tree in the desired form. Again we repeat this until the entire grammar is in the desired form.</Paragraph>
      <Paragraph position="17"> (Step 3) Finally, we convert each substitution node into an adjunction node (Schabes, 1990). For each substitution node e, let X be the label of e.</Paragraph>
      <Paragraph position="18"> Relabel e to SX with obligatory adjunction and place an empty terminal beneath e.</Paragraph>
      <Paragraph position="19">  For each initial tree with root label X, convert it into an auxiliary tree by adding a new root node labeled SX whose children are the old root node and a new foot node.</Paragraph>
      <Paragraph position="21"/>
    </Section>
    <Section position="3" start_page="27" end_page="31" type="sub_section">
      <SectionTitle>
3.3 Relation to tree-adjoining languages
</SectionTitle>
      <Paragraph position="0"> Our second result, also conjectured by Kato et al., is that the weak equivalence class established above is a proper subset of TAL.</Paragraph>
      <Paragraph position="1"> Proposition 6. The language  Before proceeding to the other half of the proof, we de ne a few useful notions. A marked string (as in Ogden's Lemma) over an alphabet S is a string over S x {0,1}, where a symbol &lt;s,1&gt; is marked and a symbol &lt;s,0&gt; is not. Marked strings over S can be projected into S[?] in the obvious way and we will talk about marked strings and their projections interchangeably.</Paragraph>
      <Paragraph position="2"> A decomposed string over S is a sequence of strings over S, which can be projected into S[?] by concatenating their members in order, and again we will talk about decomposed strings and their projections interchangeably. In particular, we will often simply write a decomposed string &lt;w1,... ,wn&gt; as w1 ***wn. Moreover, we may use the symbol wi to refer to the occurrence of the ith member of the decomposition in w; for example, if w is a marked string, we may say that a symbol in wi is marked, or if w is generated by a TAG derivation, we may say that wi is generated by some set of nodes in the derivation tree.</Paragraph>
      <Paragraph position="3"> The second half of the proof requires a doubledecker pumping lemma.</Paragraph>
      <Paragraph position="4"> Condition 1 (cf. Vijay-Shanker (1987), Theorem 4.7). Given a language L and a decomposed string x1zx2 [?] L with some symbols in z marked, there exists a decomposition of z into u1v1w1v2u2v3w2v4u3 such that one of the vi contains a mark, and L contains, for all k [?] 1, x1(u1vk1w1vk2u2vk3w2vk4u3)x2 Condition 2 (cf. Uemura et al. (1999), Lemma  1). Given a language L and a decomposed string x1z1z2x2z3z4x3 [?] L with some symbols in one of the zi marked, there exist decompositions of the zi into uiviwi such that one of the vi contains a mark, and L contains, for all k [?] 1, x1(u1vk1w1)(u2vk2w2)x2(u3vk3w3)(u4vk4w4)x3 Lemma 7. If L is an ESL-TAL, then there exists a constant n such that for any z [?] L with n symbols marked, Condition 1 holds of epsilon1 * z * epsilon1. Moreover, it holds such that the w1 and w2 it provides can be further decomposed into z1z2 and z3z4, respectively, such that for any marking of n symbols of any of the zj, either Condition 1 holds of z = x1zjx2 (where x1 and x2 are the surrounding context of zj) or Condition 2 holds of z = x1z1z2x2z3z4x3 (where x1, x2, and x3 are the surrounding context of z1z2 and z3z4).</Paragraph>
      <Paragraph position="5"> Proof. Since L is an ESL-TAL, it is generated by some ESL-TAG G. Let k be the number of elementary trees in G and t be the maximum number of terminal symbols in any elementary tree of G.</Paragraph>
      <Paragraph position="6"> Then set n = 2k+1t.</Paragraph>
      <Paragraph position="7"> The rst invocation of Condition 1 is the TAG version of Ogden's lemma (Hopcroft and Ullman, 1979). To show that it holds, we need to nd a path P in the derivation tree of z that has a cycle that generates at least one marked symbol. Dene a branch point to be a node h in the derivation tree such that the marked nodes generated by the subderivation of h are not all generated by the sub-derivation of a single child of h. We seek a P that has at least k + 1 branch points. Start by adding the root of the derivation tree to P. Thereafter let h be the last node in P. If h is a leaf, stop; otherwise, add to P the child of h whose subderivation generates the most marked symbols. Note that if a branch point in P generates m marked symbols, the next branch point generates at least m[?]t2 . Our choice of n then guarantees that P has at least k+1 branch points, at least two of which must correspond to the same auxiliary tree. Call these nodes h1 and h2.</Paragraph>
      <Paragraph position="8"> These two nodes divide the derivation up into three phases: rst, the derivation segment from the root to h1, which we call a (because it can be thought of as the derived initial tree it generates); then the segment from h1 to h2, which we call b1 (because it can be thought of as the derived auxiliary tree it generates); then subderivation of h2, which we call b2. Note that we can form new valid derivations of G by repeating b2: that is, in terms of derivation trees, stacking a on top of one or more copies of b1, on top of b2 or in terms of derived trees, repeatedly adjoining b1 into a and then adjoining b2.</Paragraph>
      <Paragraph position="9"> If b2 adjoins into the spine of b1, then let &lt;u1,u2,u3&gt; be the parts of z generated by a, &lt;v1,v2,v3,v4&gt; the parts generated by b1, and &lt;w1,w2&gt; the parts generated by b2 (see Figure 4a). Then these new derivations generate the strings u1vk1w1vk2u2vk3w2vk4u3.</Paragraph>
      <Paragraph position="10"> But if b2 adjoins at a node to the left of the spine of b1, then let &lt;u1,v42,u3&gt; be the parts of the z generated by a, &lt;v1,u2,v41,v43&gt; the parts generated by b1, and &lt;w1,w2&gt; the parts generated by b2 (see Figure 4b). Then let v2 = v3 = epsilon1 and v4 = v41v42v43; the new derivations will generate the strings u1vk1w1vk2u2vk3w2vk4u3. The case where b2 adjoins to the right of the spine.</Paragraph>
      <Paragraph position="11"> Now we focus attention on b2. Let S be the longest path of the derivation of b2 containing the root of the derivation and auxiliary trees adjoined at spine nodes. This S is unique because each spine can only have one active node. Let h3 be the last node in S, which divides the derivation of b2 into two phases: the segment from the root to h3, which we call b21, and the subderivation of h3, which we call b22. This gives a decomposition &lt;w1,w2&gt; = &lt;z1z21z22,z31z32z4&gt; , where b22 generates z21 and z32 (see Figure 5). Note that the derivation nodes in S are the only ones that can generate symbols in z1,z22,z31, and z4 at once; the other derivation nodes only generate symbols in a single zi. We let z2 = z21z22 and z3 = z31z32 and hand off the decomposition &lt;w1,w2&gt; = &lt;z1z2,z3z4&gt; to our adversary, who may choose a zj and mark n symbols in it.</Paragraph>
      <Paragraph position="12"> Then we recapitulate the reasoning above to get a path Pprime starting from the root of the derivation of b2 and containing at least k + 1 branch points, two of which correspond to the same auxiliary tree. Call these nodes h4 and h5 and the segment between them b3, and let &lt;v1,v2,v3,v4&gt; now stand for the parts of &lt;w1,w2&gt; generated by b3.</Paragraph>
      <Paragraph position="13"> Once again, we are going to repeat b3 to generate new derivations, pumping copies of the vi into &lt;w1,w2&gt; . But the location of the vi depends on h5: if h5 is in S, then the vi will appear inside each of the zi, satisfying Condition 2. Otherwise, they will all appear inside zj.</Paragraph>
      <Paragraph position="14">  Finally we complete the proof of Proposition 6.</Paragraph>
      <Paragraph position="15"> Proof of Proposition 6 (L /[?] ESL-TAL). Suppose L is an ESL-TAL. Let z be the string obtained by setting p = q = r = n, and mark the a1s. Then Lemma 7 must hold. The rst invocation of Condition 1 must give a w1 of the form a[?]1bn1bn2cn1cn2a[?]2 and a w2 of the form a[?]3cn3 cn4bn3bn4a[?]4. Lemma 7 must further decompose w1 into z1z2. Obviously, either z1 contains all the bjs or z2 contains all the cjs. Supposing the former, we can obtain a contradiction by marking the b1s: Condition 2 is impossible because it would give unequal numbers of b1s and b2s; Condition 1 is impossible because it would give unequal numbers of b1s and b3s. On the other hand, if z2 contains all the cjs, we mark the c1s, and both Conditions are again rendered impossible.</Paragraph>
    </Section>
  </Section>
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