File Information
File: 05-lr/acl_arc_1_sum/cleansed_text/xml_by_section/metho/93/e93-1032_metho.xml
Size: 27,330 bytes
Last Modified: 2025-10-06 14:13:17
<?xml version="1.0" standalone="yes"?> <Paper uid="E93-1032"> <Title>Towards efficient parsing with proof-nets</Title> <Section position="3" start_page="0" end_page="272" type="metho"> <SectionTitle> 2 Connection Graphs </SectionTitle> <Paragraph position="0"/> <Section position="1" start_page="0" end_page="269" type="sub_section"> <SectionTitle> 2.1 Links and Nodes </SectionTitle> <Paragraph position="0"> Definition 1: Let S be a set of signed vertices (i-e: labelled with letters belonging to an alphabet A and with a + or - sign). We define three types of links: * external links: +a ....... a or -a ...... +a, between complementary vertices (same letter and opposite signs) * internal links:</Paragraph> </Section> <Section position="2" start_page="269" end_page="269" type="sub_section"> <SectionTitle> 2.2 Connection Graphs (inductive definition) </SectionTitle> <Paragraph position="0"> We define by induction the class of connection graphs and the associated notions of unit and of linking a sequence of units.</Paragraph> <Paragraph position="2"> they will be noted respectively: o~ r 13 and c~ 1 \[~, and they link respectively: y, - (B/A), x, z and y, x, -(AkB), z where -(B/A) and -(AkB) are new units.</Paragraph> <Paragraph position="3"> * (II) if ~ is a connection graph which links -A, x and +B, then: and we obtain by (Ha):</Paragraph> <Paragraph position="5"> Proof: assume we have a connection graph \[3 which links y, -(AXB), z and a connection graph a which links x and +C, then, by (Ia) we obtain a connection graph which links y, -((AkB)/C), x and z. But since 13 links units having already a type 2 link, it necessarily comes from a \[y', -B, z\] and a Ix', +A\] such that y' x' = y.</Paragraph> <Paragraph position="6"> From \[y', -B, z\] and \[x, +C\] we obtain a connection graph which links y', -(B/C), x and z and from this graph and the graph which links Ix', +A\], we get a graph which links y', x', -(A~(B/C)), x and z, that means a graph which links y, -(A~(B/C)), x and z, which is identical to the graph which links y, -((AkB)/C), x and z. 0</Paragraph> </Section> <Section position="3" start_page="269" end_page="270" type="sub_section"> <SectionTitle> 2.3 Alternating trees </SectionTitle> <Paragraph position="0"> Definition 2: Let L1 and L2 respectively the sets of type 1 links and type 2 links. An alternating tree on L1uL2 is a tree in which all the nodes at a same level have the same sign, all the edges are type 1 or type 2 links and the sign of a node is alternatively + and along any path from the root to any leaf.</Paragraph> <Paragraph position="1"> the set of its external links. The set G-E is an ordered set of units, each of them consisting in an alternating tree, noted -A if the root is negative and +A if the root is positive.</Paragraph> <Paragraph position="2"> Proof. by induction on building a connection graph. 1) Let G consist in a single external link, if we remove the external link, we get two distinct vertices: +a and -a, which are alternating trees.</Paragraph> <Paragraph position="3"> 2) Let us assume the property true for ~ which links y, -B, z and for ct which links x and +A. -B, as a unit, is an alternating negative tree and +A is an alternating positive tree. By (Ia) and 0b), a type 2 link is added from the root of -B. We thus keep a tree the root of which is negative and the type of the added link is the same as that of the links at the same level. Moreover, no cycle is added because before this operation, the two graphs were not already connected. -(B/A) and -(AkB) are thus alternating Irees.</Paragraph> <Paragraph position="4"> 3) Let us assume now the property true for ct which links -A, x and +B, then it is also true for tl(C/0 and tr(cx) because a type 1 link is added from the positive root of +B. Obviously, no cycle is added when we exclude the external links. 0 2.4 Numbering the nodes of a connection graph Let F be the ordered set of alternating trees in a connection graph G.</Paragraph> <Paragraph position="5"> Proposition 3: F contains one and only one positive tree. It is the last tree of the set. Its root will be called the positive root of G.</Paragraph> <Paragraph position="6"> Proof.&quot; very easy, by induction on building a connection graph.0 Proposition 4: Let us assume that G contains 2n vertices. There is one and only one way of numbering these vertices in order that the following conditions are full filled: * ifX and Y are alternating trees and X< Y (X before Y in the order defined on 1-') the set I X of numbers associated to X and the set Iy are such that: I X < Iy - &quot;'&quot; ..-c ~&quot; ..... &quot;&quot;: +b 5 ; 6 +a 1 -I-12 -a Definition 3: a connection graph G is said to be well numbered if and only if its nodes are numbered according to Proposition 4.</Paragraph> </Section> <Section position="4" start_page="270" end_page="270" type="sub_section"> <SectionTitle> 2.5 Completeness of Connection Graphs </SectionTitle> <Paragraph position="0"> with respect to the Associative Product-free</Paragraph> </Section> <Section position="5" start_page="270" end_page="270" type="sub_section"> <SectionTitle> Lambek Calculus </SectionTitle> <Paragraph position="0"> We show that every deduction d in the calculus A (for Associative Product-free Lambek calculus) may be translated into a connection graph %(d).</Paragraph> <Paragraph position="1"> axiom: a ---> a is translated into: -a ....... -+a or +a ........... a rules: \[L/q: if x ---> A translates into 13 and y B z --> C translates into y, y B/A x z translates into l\] ~r y \[L\\]: y x AkB z translates into 13 @1 7 \[R/\]: translates into tr(a) where ~ is the translation ofA x---> B \[R\\]: translates into tl(cx).0 Remark: this translation is not a one-to-one mapping, because several deductions can be translated into the same connection graph. We assume here that connection graphs provide a semantics for derivations. It is possible to show that this semantics is isomorphic to the associative directed lambda calculus (see Wansing 1990).</Paragraph> </Section> <Section position="6" start_page="270" end_page="271" type="sub_section"> <SectionTitle> 2.6 Soundness of Connection Graphs with </SectionTitle> <Paragraph position="0"> respect to A This paragraph is very similar to Roorda 1991, chap IIl, SS4.</Paragraph> <Paragraph position="1"> Lemma 1: If we remove a type 1 link from a connection graph G, we keep a connection graph. Proof: we may assume that this link has been added at the last stage of the construction.(> Definition 4: a type 2 link is called separable if it could have been added in the last stage of the conslruction.</Paragraph> <Paragraph position="2"> Lemma 2: If a connection graph, consisting of more than one link, has no terminal type 1 link, it has a separable type 2 link.</Paragraph> <Paragraph position="3"> Proof'. obvious.</Paragraph> <Paragraph position="4"> Proposition 5: To every connection graph G the units of which are: -A1. -A 2 ..... -A n, +B, there corresponds a deduction in A of the sequent: AI*. A2* ..... An* --~ B* (where X* is the formula associated with the alternating tree X) Proof: by induction on the structure of G. G has necessarily a last link, in the order of the construction. As seen in the previous lemma, it is necessarily either a type 1 link or a type 2 link. In the first case, when removing it, we still have a connection graph. In the second case, when removing it, we get two connection graphs ct and \[3 which correspond, by induction hypothesis, respectively to x ---> A and y B z---> C.</Paragraph> </Section> <Section position="7" start_page="271" end_page="271" type="sub_section"> <SectionTitle> 2.7 Non-Overlapping property </SectionTitle> <Paragraph position="0"> Definition 5: given a connection graph G, we call interval every set of integers \[i, j\] (ie: {x; i<x<j} such that i and j are indices associated with ending points of an external link (and i<j).</Paragraph> <Paragraph position="1"> Two intervals \[i, j\] and \[i', j'\] do not overlap if and only if: * \[i, j\] n \[i', j'\] = gl or * \[i, j\] D \[i', jq and i ~ i' and j ~: j' or * \[i', j'\] D \[i, j\] and i C/ i' and j C/ j' Given a family I of intervals, we say that it satisfies the Non Overlapping Condition (NOC) if it does not contain any pair of intervals which overlap.</Paragraph> <Paragraph position="2"> Theorem 1: in a well numbered connection graph G, the family of intervals associated with all the external links satisfies NOC.</Paragraph> <Paragraph position="3"> Proof: easy, by induction.<)</Paragraph> </Section> <Section position="8" start_page="271" end_page="271" type="sub_section"> <SectionTitle> 2.8 Linking the positive root </SectionTitle> <Paragraph position="0"> Theorem 2: in a connection graph G, the positive root is connected by an external link either to a negative vertex in the same tree (just below it) or to a negative root.</Paragraph> </Section> <Section position="9" start_page="271" end_page="272" type="sub_section"> <SectionTitle> 2.9 Connectivity and acyclicity by </SectionTitle> <Paragraph position="0"> external links and type 2 links Theorem 3: Let G be a connection graph. Let L1 be the set of its type 1 links. G-L1 is connected and acyclic (it is a tree).</Paragraph> <Paragraph position="1"> Proof: a type 2 link connects two connection graphs for the first time they meet and a type 1 link does neither connect two graphs, nor modify the topology of type 2 links and external links. 0 2.10 One-to-one mapping between nodes Theorem 4: for every i in a connection graph G, let (~(i) be the node linked to i by an external link, ~ is a one-to-one mapping from S onto S.</Paragraph> <Paragraph position="2"> Proof: trivial by induction. 0 2-11 Strong connectivity Definition 6: given a graph G, a spanning tree of G is defined as a tree on the complete set of nodes of G. A tree is said to be alternating on L2 u E, if each of its paths from the root to a leaf is alternatively composed by L2-edges and E-edges.</Paragraph> <Paragraph position="3"> Theorem 5: every connection graph G admits an alternating spanning tree with the positive root of G as the root.</Paragraph> <Paragraph position="4"> of +C to the root of -B which is alternating. Since it arrives at a negative vertex, its last link cannot be of type 2, then it is an external link.</Paragraph> <Paragraph position="5"> * There is also a path from the root of +A to any leaf of the spanning tree of ix, which is alternating. Since it comes from a positive vertex, it cannot begin with a type 2 link, hence it begins with an extrernal link. Thus, by inserting a type 2 link between the external link arriving at -B and the external link starting from +A, we get a path starting from the positive root of +C and arriving at any leaf of ct, inserted into 7, which is alternating.</Paragraph> <Paragraph position="6"> Therefore, there is an alternating path from the positive root of +C to any leaf of ct0)7.</Paragraph> <Paragraph position="7"> Let us assume now it is true for ct which links -A, x and +B. The transformation t r or t I does not modify the set of paths starting from the positive root of +B. 0 Definition 7: a node in a connection graph G will be said strongly connected to another node in the same graph if they are connected by an alternating path. Definition 8: a link will be said to be strong if its two ends are srongly connected.</Paragraph> <Paragraph position="8"> Theorem 6: in a connection graph G, every type 1 link is strong.</Paragraph> <Paragraph position="9"> Proof: this is shown when installing a new type 1 link. Such an installation does not modify the topology of G-L1. The previous graph (before applying t 1 or t r) was necessarily a connection graph. Thus by Theorem 5, it was scanned by an alternating spanning nee with as root the positive root of the graph. This tree is preserved by t I or t r, it contained an alternating path connecting the two vertices which are now linked by a type 1 link. 0 As a matter of recapitulation, we enumerate now the following properties, satisfied by any connection graph. Proof: By CG4, G-L1 is a tree on S, it is therefore a spanning tree of G. Let us consider a path ~ from the positive root +b (which is the root of the positive tree +B, and which is unique according to CG1) to a leaf a. We must notice that a cannot be positive, because if it was, it would necessarily be an end of a type 2 link and this type 2 link would be the last edge on the path a, but by CG0, it would be linked by an external link to another node and thus it would not be a leaf. Thus, a is necessarily negative, and we can write -a instead of a. If -a is isolated (as a negative root of a negative tree), we can remove the last external link and the type 2 link before the last, we are led to the same problem: a path c' arriving at a negative leaf, but or' is shorter than ~. If -a is not isolated, it is necessarily the end of a type 1 link, but by CG3, there is an alternating path joining -a and the positive node +c which is the other end.</Paragraph> <Paragraph position="10"> Removing this path and the type 2 link arriving at +c, we still get the same problem of a path c' arriving at a negative node, but again a' is shorter than g. We can proceed like that until we have a mere external link between the positive root +b and a vertex -b. In this case, the path is obviously alternate.</Paragraph> </Section> </Section> <Section position="4" start_page="272" end_page="273" type="metho"> <SectionTitle> 3 Well Linked Graphs (WLG) </SectionTitle> <Paragraph position="0"> (ie: every well linked graph could be obtained by the inductive construction of a connection graph, with the sequence of alternating trees as G-E).</Paragraph> <Paragraph position="1"> Proof: given a WLG on I-A1, -A2 ..... -An, +B\], it has a unique positive root +b (the root of +B). Thus it satisfies the property of uniqueness of the positive root. Let us assume there is a type 1 link from +b, then let</Paragraph> <Paragraph position="3"> Let us assume for instance that it is left-oriented: * The tree below this link may be moved towards the left end of the sequence of trees by the inverse of the construction rule (IIa). This move preserves the topological structure of EuL2, therefore, CG1, CG3 and CG4 are preserved. This move implies a renumbering but it does not destroy the non-overlapping property. Thus CG2 is preserved. CG0 is trivially preserved. The argument is similar for a right-oriented link. Thus after this removal, we keep a WLG.</Paragraph> <Paragraph position="4"> Let us assume now there is no type 1 link from +b.</Paragraph> <Paragraph position="5"> Then there is an external link which links +b to a vertex -b situated among the negative trees. If -b is not related to another node, we get an elementary WLG: -b .... +b, which is obviously a connection graph. If -b is related to another node, then by CG5, either -b is a leaf, or it is the starting point of a type 2 link. Let us assume -b is a leaf (of a non atomic tree), then -b is linked by a type 1 link to a vertex +a (and not to +b since we have assumed there is no longer type 1 link from +b).</Paragraph> <Paragraph position="6"> Because of CG3, -b and +a are connected by an alternating path on EuL2, thus -b is necessarily the starting point of a type 2 link, but in this case, -b is not a leaf. Therefore -b is not a leaf and it is the starting point of a type 2 link. Let +c the other end of this link. * Let us assume that this link is left-oriented: we remove the leftmost one if many. In this case, the scanning tree is broken into two parts and the connection graph is also separated into two pieces. One contains +b, the other contains +c.</Paragraph> <Paragraph position="7"> Let us consider the first one: * it keeps CG3 and CG4: for example CG3: - let us consider a type 1 link situated in this part. It does not come from +b since we have assumed there is no longer type 1 link from +b.</Paragraph> <Paragraph position="8"> - its ends are linked by an alternating path. Let us assume that the removed type 2 link belonged to this path. By removing it, we get either a single external link: -b ...... +b, but such a piece does not contain any type 1 link, or another kind of graph. If we want this graph has a type 1 link, it necessarily must contain another type 2 link starting from -b, and arriving, say, at +d, But an alternating path between two ends of a type 1 link can neither arrive by an external link at -b since -b is already connected by such a link to the positive root +b (and we have assumed there is no type 1 link attached to +b), nor pass through +d since, in this case, the path would have two consecutive type 2 links, which contradicts the definition of an alternating path. Therefore, the removed type 2 link cannot be on the alternating path linking the ends of a type 1 link in this part of the graph. Finally, no alternating path in the first component is destroyed by this removal, among all the alternating paths connecting ends of type 1 links.</Paragraph> <Paragraph position="9"> Let us consider the second one: - let us consider a type 1 link situated in this part and let us assume that its ends are linked by an alternating path passing through the removed type 2 link. The proof is the same as previously: the path can neither arrive at -b by an external link nor by a type 2 link. Moreover, it has one and only one positive root +c, because it does not contain +b, and +c is necessarily linked by an external link to either a negative root or a negative vertex just below it (if not, there would be a type 1 link +x -- -c, with -c externally linked to +c, the alternating path from -c to +x would thus necessarily pass through +c and -b, which is impossible according to the first part of the proof).</Paragraph> <Paragraph position="10"> When all the type 2 links attached to -b are removed, there remains only the external link -b .... +b which is a WLG, and we can perform this decomposition for each part resulting from a previous step.</Paragraph> <Paragraph position="11"> It would then be possible to reconstruct the graph accordint to the induction schemes (I) and (II), starting only from axioms.</Paragraph> <Paragraph position="12"> Corollary: well linked graphs are sound and complete with respect to the calculus A.</Paragraph> </Section> <Section position="5" start_page="273" end_page="275" type="metho"> <SectionTitle> 4 Method of construction of a well </SectionTitle> <Paragraph position="0"> linked graph An alternating tree was defined by a set of signed vertices and a set of typed links which link them. We are now adding two new kinds of entity in order to facilitate tree-encoding.</Paragraph> <Section position="1" start_page="273" end_page="273" type="sub_section"> <SectionTitle> 4.1 Colours and anti-colours 4.1.1 Colours </SectionTitle> <Paragraph position="0"> Let us assign to each vertex in a sequence of trees \[-A1, -A2 ..... -An, +B\] a colour (originally unknown and represented by a free variable X) in order that: a) two nodes linked by a type 2 link have same colours b) two nodes which are not linked or which are linked by a type-1 link have not the same colours (X ~ Y).</Paragraph> <Paragraph position="1"> Proposition 8:for every connection graph G with set of type 1 links L1, the connectivity and acyclicity of G -L1 translates into: every external link links two nodes having differents colours. After linking by an external link, the two colours are equalized (X = Y).</Paragraph> <Paragraph position="2"> Anticolours are assigned to nodes in an alternating tree in order that: a) two nodes linked by a type 1 link have same anticolour, b) if a positive node receives an anticolour a, (by (a) or by an external link), the negated anticolour 9ct is transmitted to all other positive nodes having same colour.</Paragraph> <Paragraph position="3"> Rule: 1) When joining two nodes by an external link, which are associated with different (positive) anticolours tx and 13, C/t and ~ are said to be equalized, that means: put in a same multi-set.</Paragraph> <Paragraph position="4"> 2) When joining a node having a negated anticolour 913 to a node having a colour X by an external link, the anticolour --,13 is transmitted to the colour X as a label. 3) When linking two ends of a type 1 link by external links, the two occurrences of the same (positive) anticolour tx must meet only one colour, or two colours which have been already equalized and such that one of the two is not labelled by a negated anticolour 913 if 13 is an anticolour already equalized to ix.</Paragraph> <Paragraph position="5"> Proposition 9: in a connection graph G, the strong connectivity translates into: the anticolour proper to a type 1 link meets only one colour (or colours which have been equalized).</Paragraph> <Paragraph position="6"> Corollary: Every connection graph verifies: CGO, CGI, CG2, CG3', CG4', CG5', CG6 where: CG3' is the condition on unifying anti-colours, CG4' the conditions on colours, CG5' the fact that any connection graph is monocoloured.</Paragraph> </Section> <Section position="2" start_page="273" end_page="275" type="sub_section"> <SectionTitle> 4.2 Method </SectionTitle> <Paragraph position="0"> Definition 10: We call an ordered category a category where 6-tuples are ordered according to their index.</Paragraph> <Paragraph position="1"> Proposition 10: each alternating tree has one and only one encoding into an ordered category.</Paragraph> <Paragraph position="2"> not have a colour which has been labelled by a negated anticolour --,13 such that ~ and 13 have already been equalized, in a same multiset.</Paragraph> <Paragraph position="3"> * if one node is the positive root, the other is a negative root or a negative node just below it in the same tree (same anticolour).</Paragraph> <Paragraph position="4"> We scan the ordered list of nodes from left to right, creating links at each step, between the current node and all the possible mergeable nodes on its left or just shifting. When nodes are shifted, they are pushed onto a stack. Links are recorded on the chart in the following way. Each link is a node of the chart (in consequence, the chart has no more than n 2 nodes, where n is the number of nodes on the reading tape R). A link 1 is joined by arcs in the chart to all links already recorded 11 .... 1 n such that 1 makes a correct partial linking by insertion into the linkings represented by the paths arriving at 11 .... In. and 1 has a left extremity which coincides with either the rightmost right extremity of a link already recorded on such a path, or with a top of stack attached to such a previously recorded link. Thus, a link 1 may be an arriving point for several paths. In this case, we will consider 1 as a new starting point. That means that when joining a new link 1' to links above 1 in the chart, we only test the correctness of a partial linking down to the link I. We consider here that if 1 is in the chart at this place, there is necessarily a correct path up to it, and all the partial paths from I to the current node are, by definition, correct. Thus, when adding a link above 1 (and before a possible new &quot;crossroads&quot;), even if there are many paths joining I to it, there is at least one correct path from the bottom of the chart to the current node. Each time a link is recorded and joined to other ones, we record for each arc arriving at it, the possible tops of stack, the possible</Paragraph> <Paragraph position="6"> rightmost right extremities, the list of nodes through which the path has passed since the previous embranchment, the list of equalized colours (possibly labelled with negated anticolours) and the list of equalized anticolours (for the piece of path coming from the previous embranchment). When joining a new link, we have to retrieve a new top of the stack, if added by consuming a previous one or a previous rightmost fight extremity, and to test the correctness of the path. This necessitates a descent along paths down to the bottom of the chart. This descent is made deterministic because of the informations stored on arcs. If n is the number of nodes in the original sequent, a maximum of n 2 links may be created, and there can be a maximum of n 4 arcs in the chart. At step i, there can be a maximum of i 4 arcs. We add new links on the basis of stack informations stored on arriving arcs at each previously recorded link. Each checking does not take more than i steps, and there are at most i 2 nodes to check at step i.</Paragraph> <Paragraph position="7"> For a given link to add, when looking for new tops of the stack and checking the correctness of the new linking, we explore the current state of the chart by scanning no more than twice (one in one direction, one in the other) each arc it contains. Thus joining a new link to previous ones entails a maximum of 2i 7 steps.</Paragraph> <Paragraph position="8"> At step i, i new links can be added. Thus step i entails a maximum of 2i 8 steps. Thus, when reaching step n, we have done a maximum of 2Y.i 8 steps (i=l to n), that is O(n9). This is obviously a too big order. Nevertheless, the method is time-polynomial and more improvements can be expected in the future.</Paragraph> <Paragraph position="9"> Example: Suppose we have to demonstrate the sequent: (a/a)/(a/a) a/a a/a a a\a --) a (cf fig 1) At beginning steps 1, 2, 3, 4, nodes are pushed onto the top of the stack. At step 5, the link (4 5) is created and recorded in the chart. The new top of stack 3 is attached to it. At step 6, (3 6) is added (with new top of stack 2), on the top of the previous link. At step 7, (6 7) is created and joined to (4 5) (with top of stack 3) and not to (3 6) (because they have the node 6 in common). (2 7) is joined to (3 6) (with top of stack 1). At step 8, (3 8) is created and joined to (6 7) (with top of stack 2), but not to (2 7) because of anticolours (7 receives 913 and 8 receives ~ and they have same colour). (1 8) is not created because they have same colour. At step 9, (2 9) is created and joined to (3 8), and (8 9) is created too, but joined to (6 7) and (2 7). At step 10, (1 10) is joined to (8 9), (3 10) is joined to (8 9) and (6 7), (9 10) to (3 8). (7 10) is also joined to (8 9) and (3 8) because 7 is a rightmost right extremity in paths leading to these nodes. In such a circumstance, the node previously linked to the released right extremity, here 2 or 6 is pushed onto the stack. After that, (2 11) may be added to '(7 10) and (9 10) but not to (2 11) because of anticolours. And finally, (1 12) may be joinedto (2 11) and (11 12) to (1 10). By looking at the list of nodes attached to links installed in the chart, we see that these last moves lead to complete linkings. By going down to the bottom of the chart, we find the three solutions: \[1 \[2 \[3 \[4 5\] 6\] 7\] \[8 9\] 10\]\[11 12\], \[1 \[2 \[3 \[4 5\] 6\] \[7 \[8 9\] 10\] 11\] 12\] and \[1 \[2 \[3 \[4 5\] \[6 7\] 8\] \[9 10\] 11\] 12\] (cf fig 2).</Paragraph> <Paragraph position="10"> fig2: the final chart.</Paragraph> <Paragraph position="12"/> </Section> </Section> class="xml-element"></Paper>