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<Paper uid="P00-1047">
  <Title>A Polynomial-Time Fragment of Dominance Constraints</Title>
  <Section position="3" start_page="0" end_page="0" type="metho">
    <SectionTitle>
2 Dominance Constraints
</SectionTitle>
    <Paragraph position="0"> In this section, we de ne the syntax and semantics of dominance constraints. The variant of dominance constraints we employ describes constructor trees { ground terms over a signature of function symbols { rather than feature trees.</Paragraph>
    <Paragraph position="1">  So we assume a signature function symbols ranged over by f;g;:::, each of which is equipped with an arity ar(f)</Paragraph>
  </Section>
  <Section position="4" start_page="0" end_page="0" type="metho">
    <SectionTitle>
0. Constants { function
</SectionTitle>
    <Paragraph position="0"> symbols of arity 0 { are ranged over by a;b.</Paragraph>
    <Paragraph position="1"> We assume that contains at least one constant and one symbol of arity at least 2.</Paragraph>
    <Paragraph position="2"> Finally, let Vars be an in nite set of variables ranged over by X;Y;Z. The variables will denote nodes of a constructor tree. We will consider constructor trees as directed labeled graphs; for instance, the ground term f(g(a;a)) can be seen as the graph in Fig. 2.</Paragraph>
    <Paragraph position="3"> We de ne an (unlabeled) tree to be a nite directed graph (V;E). V is a nite set of nodes ranged over by u;v;w, and E V V is a set of edges denoted by e. The indegree of each node is at most 1; each tree has exactly one root, i.e. a node with indegree 0. We call the nodes with outdegree 0 the leaves of the tree.</Paragraph>
    <Paragraph position="4"> A ( nite) constructor tree is a pair (T;L) consisting of a tree T = (V;E), a node labeling L : V ! , and an edge labeling L : E ! N, such that for each node u 2 V and each 1 k ar(L(u)), there is exactly one edge (u;v) 2 E with L((u;v)) = k.1 We draw 1The symbol L is overloaded to serve both as a node and an edge labeling.</Paragraph>
    <Paragraph position="5"> constructor trees as in Fig. 2, by annotating nodes with their labels and ordering the edges along their labels from left to right. If = ((V;E);L), we write V = V , E = E, L = L. Now we are ready to de ne tree structures, the models of dominance constraints: De nition 2.1. The tree structure M of a constructor tree is a rst-order structure with domain V which provides the dominance relation and a labeling relation for each function symbol f 2 .</Paragraph>
    <Paragraph position="6"> Let u;v;v1;::: vn 2 V be nodes of . The dominance relationship u v holds i there is a path from u to v in E ; the labeling relationship u:f (v1;::: ;vn) holds i u is labeled by the n-ary symbol f and has the children v1;::: ;vn in this order; that is, L (u) = f,</Paragraph>
    <Paragraph position="8"> A dominance constraint ' is a conjunction of dominance, inequality, and labeling literals of the following form where ar(f) = n:</Paragraph>
    <Paragraph position="10"> is able constraint Let Var(') be the set of variables of '. A pair of a tree structure M and a variable assignment : Var(') ! V satis es ' i it satis es each literal in the obvious way. We say that (M ; ) is a solution of ' in this case; ' is satis able if it has a solution. We usually draw dominance constraints as constraint graphs. For instance, the constraint graph for X:f(X1;X2) ^ X1 Y ^ X2 Y is shown in Fig. 3. As for trees, we annotate node labels to nodes and order tree edges from left to right; dominance edges are drawn dotted. The example happens to be unsatis able because trees cannot branch upwards. null De nition 2.2. Let ' be a dominance constraint that does not contain two labeling constraints for the same variable.2 Then the constraint graph for ' is a directed labeled graph G(') = (Var(');E;L). It contains a (partial) node labeling L : Var(') and an edge labeling L : E ! N[f g.</Paragraph>
    <Paragraph position="11"> The sets of edges E and labels L of the graph G(') are de ned in dependence of the literals in ': The labeling literal X:f(X1;::: ;Xn) belongs to ' i L(X) = f and for each 1 i n, (X;Xi) 2 E and L((X;Xi)) = i. The dominance literal X Y is in ' i (X;Y ) 2 E and L((X;Y )) = .</Paragraph>
    <Paragraph position="12"> Note that inequalities in constraints are not represented by the corresponding constraint graph. We de ne (solid) fragments of a constraint graph to be maximal sets of nodes that are connected over tree edges.</Paragraph>
  </Section>
  <Section position="5" start_page="0" end_page="0" type="metho">
    <SectionTitle>
3 Normal Dominance Constraints
</SectionTitle>
    <Paragraph position="0"> Satis ability of dominance constraints can be decided easily in non-deterministic polynomial time; in fact, it is NP-complete (Koller et al., 1998).</Paragraph>
    <Paragraph position="2"> The NP-hardness proof relies on the fact that solid fragments can \overlap&amp;quot; properly. For illustration, consider the con-</Paragraph>
    <Paragraph position="4"> straint graph is shown in Fig. 4. In a solution of this constraint, either Y or Y1 must be mapped to the same node as X; if X = Y , the two fragments overlap properly. In the applications in computational linguistics, we typically don't want proper overlap; X should 2Every constraint can be brought into this form by introducing auxiliary variables and expressing X=Y as X Y ^ Y X.</Paragraph>
    <Paragraph position="5"> never be identi ed with Y , only with Y1. The subclass of dominance constraints that excludes proper overlap (and xes some minor inconveniences) is the class of normal dominance constraints.</Paragraph>
    <Paragraph position="6"> De nition 3.1. A dominance constraint ' is called normal i for all variables X;Y;Z 2 Var('),  1. X 6= Y in ' i both X:f(:::) and Y :g(:::) in ', where f and g may be equal (no overlap);3 2. X only appears once as a parent and once as a child in a labeling literal (treeshaped fragments); 3. if X Y in ', neither X:f(:::) nor Z:f(::: Y :::) are (dominances go from holes to roots); 4. if X Y in ', then there are Z;f such that Z:f(::: X :::) in ' (no empty frag- null ments).</Paragraph>
    <Paragraph position="7"> Fragments of normal constraints are treeshaped, so they have a unique root and leaves. We call unlabeled leaves holes. If X is a variable, we can de ne R'(X) to be the root of the fragment containing X. Note that by Condition 1 of the de nition, the constraint graph speci es all the inequality literals in a normal constraint. All constraint graphs in the rest of the paper will represent normal constraints.</Paragraph>
    <Paragraph position="8"> The main result of this paper, which we prove in Section 4, is that the restriction to normal constraints indeed makes satis ability polynomial: Theorem 3.2. Satis ability of normal dominance constraints is O((k+1)3n2 log n), where n is the number of variables in the constraint, and k is the maximum number of dominance edges into the same node in the constraint graph.</Paragraph>
    <Paragraph position="9"> In the applications, k will be small { in scope underspeci cation, for instance, it is 3Allowing more inequality literals does not make satis ability harder, but the pathological case X 6= X invalidates the simple graph-theoretical characterizations below.</Paragraph>
    <Paragraph position="10"> bounded by the maximum number of arguments a verb can take in the language if we disregard VP modi cation. So we can say that satis ability of the linguistically relevant dominance constraints is O(n2 log n).</Paragraph>
  </Section>
  <Section position="6" start_page="0" end_page="0" type="metho">
    <SectionTitle>
4 A Polynomial Satis ability Test
</SectionTitle>
    <Paragraph position="0"> Now we derive the satis ability algorithm that proves Theorem 3.2 and prove it correct.</Paragraph>
    <Paragraph position="1"> In Section 5, we embed it into an enumeration algorithm. An alternative proof of Theorem 3.2 is by reduction to a graph problem discussed in (Althaus et al., 2000); this more indirect approach is sketched in Section 6.</Paragraph>
    <Paragraph position="2"> Throughout this section and the next, we will employ the following non-deterministic choice rule (Distr), where X;Y are di erent variables.</Paragraph>
    <Paragraph position="4"> In each application, we can pick one of the disjuncts on the right-hand side. For instance, we get Fig. 5b by choosing the second disjunct in a rule application to Fig. 5a.</Paragraph>
    <Paragraph position="5"> The rule is sound if the left-hand side is normal: X Z ^Y Z entails X Y _Y X, which entails the right-hand side disjunction because of conditions 1, 2, 4 of normality and</Paragraph>
    <Paragraph position="7"> If the left-hand side is normal, so are both possible results.</Paragraph>
    <Paragraph position="8"> De nition 4.1. A normal dominance constraint ' is in solved form i (Distr) is not applicable to ' and G(') is cycle-free.</Paragraph>
    <Paragraph position="9"> Constraints in solved form are satis able.</Paragraph>
    <Section position="1" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
4.1 Characterizing Satis ability
</SectionTitle>
      <Paragraph position="0"> In a rst step, we characterize the unsatis ability of a normal constraint by the existence of certain cycles in the undirected version of its graph (Proposition 4.4). Recall that a cycle in a graph is simple if it does not contain the same node twice.</Paragraph>
      <Paragraph position="1"> De nition 4.2. A cycle in an undirected constraint graph is called hypernormal if it does not contain two adjacent dominance edges that emanate from the same node.</Paragraph>
      <Paragraph position="2">  For instance, the cycle in the left-hand graph in Fig. 5 is not hypernormal, whereas the cycle in the right-hand one is.</Paragraph>
      <Paragraph position="3"> Lemma 4.3. A normal dominance constraint whose undirected graph has a simple hypernormal cycle is unsatis able.</Paragraph>
      <Paragraph position="4"> Proof. Let ' be a normal dominance constraint whose undirected graph contains a simple hypernormal cycle. Assume rst that it contains a simple hypernormal cycle C that is also a cycle in the directed graph. There is at least one leaf of a fragment on C; let Y be such a leaf. Because ' is normal, Y has a mother X via a tree edge, and X is on C as well. That is, X must dominate Y but is properly dominated by Y in any solution of ', so ' is unsatis able.</Paragraph>
      <Paragraph position="5"> In particular, if an undirected constraint graph has a simple hypernormal cycle C with only one dominance edge, C is also a directed cycle, so the constraint is unsatis able. Now we can continue inductively. Let ' be a constraint with an undirected simple hypernormal cycle C of length l, and suppose we know that all constraints with cycles of length less than l are unsatis able. If C is a directed cycle, we are done (see above); otherwise, the edges in C must change directions somewhere. Because ' is normal, this means that there must be a node Z that has two incoming dominance edges (X;Z);(Y;Z) which are adjacent edges in C. If X and Y are in the same fragment, ' is trivially unsatis able. Otherwise, let '1 and '2 be the two constraints obtained from ' by one application of (Distr) to X;Y;Z. Let C1 be the sequence of edges we obtain from C by replacing the path from X to R'(Y ) via Z by the edge (X;R'(Y )). C is hypernormal and simple, so no two dominance edges in C emanate from the same node; hence, the new edge is the only dominance edge in C1 emanating from X, and C1 is a hypernormal cycle in the undirected graph of '1. C1 is still simple, as we have only removed nodes. But the length of C1 is strictly less than l, so '1 is unsatis able by induction hypothesis. An analogous argument shows unsatis ability of '2. But because (Distr) is sound, this means that ' is unsatis able too.</Paragraph>
      <Paragraph position="6"> Proposition 4.4. A normal dominance constraint is satis able i its undirected constraint graph has no simple hypernormal cycle. null Proof. The direction that a normal constraint with a simple hypernormal cycle is unsatis able is shown in Lemma 4.3.</Paragraph>
      <Paragraph position="7"> For the converse, we rst de ne an ordering '1 '2 on normal dominance constraints: it holds if both constraints have the same variables, labeling and inequality literals, and if the reachability relation of G('1) is a subset of that of G('2). If the subset inclusion is proper, we write '1 &lt; '2. We call a constraint ' irredundant if there is no normal constraint '0 with fewer dominance literals but ' '0. If ' is irredundant and G(') is acyclic, both results of applying (Distr) to ' are strictly greater than '.</Paragraph>
      <Paragraph position="8"> Now let ' be a constraint whose undirected graph has no simple hypernormal cycle. We can assume without loss of generality that ' is irredundant; otherwise we make it irredundant by removing dominance edges, which does not introduce new hypernormal cycles.</Paragraph>
      <Paragraph position="9"> If (Distr) is not applicable to ', ' is in solved form and hence satis able. Otherwise, we know that both results of applying the rule are strictly greater than '. It can be shown that one of the results of an application of the distribution rule contains no simple hypernormal cycle. We omit this argument for lack of space; details can be found in the proof of Theorem 3 in (Althaus et al., 2000). Furthermore, the maximal length of a &lt; increasing chain of constraints is bounded by n2, where n is the number of variables. Thus, applications of (Distr) can only be iterated a nite number of times on constraints without simple hypernormal cycles (given redundancy elimination), and it follows by induction that ' is satis able.</Paragraph>
    </Section>
    <Section position="2" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
4.2 Testing for Simple Hypernormal
Cycles
</SectionTitle>
      <Paragraph position="0"> We can test an undirected constraint graph for the presence of simple hypernormal cycles by solving a perfect weighted matching problem on an auxiliary graph A(G(')). Perfect weighted matching in an undirected graph G = (V;E) with edge weights is the problem of selecting a subset E0 of edges such that each node is adjacent to exactly one edge in E0, and the sum of the weights of the edges in E0 is maximal.</Paragraph>
      <Paragraph position="1"> The auxiliary graph A(G(')) we consider is an undirected graph with two types of edges.</Paragraph>
      <Paragraph position="2"> For every edge e = (v;w) 2 G(') we have two nodes ev;ew in A(G(')). The edges are as follows: (Type A) For every edge e in G(') we have the edge fev;ewg.</Paragraph>
      <Paragraph position="3"> (Type B) For every node v and distinct edges e;f which are both incident to v in G('), we have the edge fev;fvg if either v is not a leaf, or if v is a leaf and either e or f is a tree edge.</Paragraph>
      <Paragraph position="4"> We give type A edges weight zero and type B edges weight one. Now it can be shown (Althaus et al., 2000, Lemma 2) that A(G(')) has a perfect matching of positive weight i the undirected version of G(') contains a simple hypernormal cycle. The proof is by constructing positive matchings from cycles, and vice versa.</Paragraph>
      <Paragraph position="5"> Perfect weighted matching on a graph with n nodes and m edges can be done in time O(nmlog n) (Galil et al., 1986). The matching algorithm itself is beyond the scope of this paper; for an implementation (in C++) see e.g. (Mehlhorn and N aher, 1999). Now let's say that k is the maximum number of dominance edges into the same node in G('), then A(G(')) has O((k + 1)n) nodes and O((k + 1)2n) edges. This shows: Proposition 4.5. A constraint graph can be tested for simple hypernormal cycles in time O((k + 1)3n2 log n), where n is the number of variables and k is the maximum number of dominance edges into the same node.</Paragraph>
      <Paragraph position="6"> This completes the proof of Theorem 3.2: We can test satis ability of a normal constraint by rst constructing the auxiliary graph and then solving its weighted matching problem, in the time claimed.</Paragraph>
    </Section>
    <Section position="3" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
4.3 Hypernormal Constraints
</SectionTitle>
      <Paragraph position="0"> It is even easier to test the satis ability of a hypernormal dominance constraint { a normal dominance constraint in whose constraint graph no node has two outgoing dominance edges. A simple corollary of Prop. 4.4 for this special case is: Corollary 4.6. A hypernormal constraint is satis able i its undirected constraint graph is acyclic.</Paragraph>
      <Paragraph position="1"> This means that satis ability of hypernormal constraints can be tested in linear time by a simple depth- rst search.</Paragraph>
    </Section>
  </Section>
  <Section position="7" start_page="0" end_page="0" type="metho">
    <SectionTitle>
5 Enumerating Solutions
</SectionTitle>
    <Paragraph position="0"> Now we embed the satis ability algorithms from the previous section into an algorithm for enumerating the irredundant solved forms of constraints. A solved form of the normal constraint ' is a normal constraint '0 which is in solved form and ' '0, with respect to the order from the proof of Prop. 4.4.4 Irredundant solved forms of a constraint are very similar to its solutions: Their constraint graphs are tree-shaped, but may still 4In the literature, solved forms with respect to the NP saturation algorithms can contain additional labeling literals. Our notion of an irredundant solved form corresponds to a minimal solved form there.</Paragraph>
    <Paragraph position="1">  1. Check satis ability of '. If it is unsatisable, terminate with failure.</Paragraph>
    <Paragraph position="2"> 2. Make ' irredundant.</Paragraph>
    <Paragraph position="3"> 3. If ' is in solved form, terminate with success. null 4. Otherwise, apply the distribution rule  dundant solved forms of a normal constraint. contain dominance edges. Every solution of a constraint is a solution of one of its irredundant solved forms. However, the number of irredundant solved forms is always nite, whereas the number of solutions typically is not: X:a^Y :b is in solved form, but each solution must contain an additional node with arbitrary label that combines X and Y into a tree (e.g. f(a;b), g(a;b)). That is, we can extract a solution from a solved form by \adding material&amp;quot; if necessary.</Paragraph>
    <Paragraph position="4"> The main workhorse of the enumeration algorithm, shown in Fig. 6, is the distribution rule (Distr) we have introduced in Section 4. As we have already argued, (Distr) can be applied at most n2 times. Each end result is in solved form and irredundant. On the other hand, distribution is an equivalence transformation, which preserves the total set of solved forms of the constraints after the same iteration. Finally, the redundancy elimination in Step 2 can be done in time O((k+1)n2) (Aho et al., 1972). This proves: Theorem 5.1. The algorithm in Fig. 6 enumerates exactly the irredundant solved forms of a normal dominance constraint ' in time O((k +1)4n4N log n), where N is the number of irredundant solved forms, n is the number of variables, and k is the maximum number of dominance edges into the same node.</Paragraph>
    <Paragraph position="5"> Of course, the number of irredundant solved forms can still be exponential in the size of the constraint. Note that for hypernormal constraints, we can replace the quadratic satis ability test by the linear one, and we can skip Step 2 of the enumeration algorithm because hypernormal constraints are always irredundant. This improves the runtime of enumeration to O((k + 1)n3N).</Paragraph>
  </Section>
  <Section position="8" start_page="0" end_page="0" type="metho">
    <SectionTitle>
6 Reductions
</SectionTitle>
    <Paragraph position="0"> Instead of proving Theorem 4.4 directly as we have done above, we can also reduce it to a con guration problem of dominance graphs (Althaus et al., 2000), which provides a more general perspective on related problems as well. Dominance graphs are unlabeled, directed graphs G = (V;E ] D) with tree edges E and dominance edges D. Nodes with no incoming tree edges are called roots, and nodes with no outgoing ones are called leaves; dominance edges only go from leaves to roots. A con guration of G is a graph G0 = (V;E ] E0) such that every edge in D is realized by a path in G0. The following results are proved in (Althaus et al., 2000): 1. Con gurability of dominance graphs is in O((k + 1)3n2 log n), where k is the maximum number of dominance edges into the same node.</Paragraph>
    <Paragraph position="1"> 2. If we specify a subset V 0 V of closed leaves (we call the others open) and require that only open leaves can have outgoing edges in E0, the con gurability problem becomes NP-complete. (This is shown by encoding a strongly NP-complete partitioning problem.) 3. If we require in addition that every open leaf has an outgoing edge in E0, the problem stays NP-complete.</Paragraph>
    <Paragraph position="2"> Satis ability of normal dominance constraints can be reduced to the rst problem in the list by deleting all labels from the constraint graph. The reduction can be shown to be correct by encoding models as con gurations and vice versa.</Paragraph>
    <Paragraph position="3"> On the other hand, the third problem can be reduced to the problems of whether there is a plugging for a description in Hole Semantics (Bos, 1996), or whether a given MRS description can be resolved (Copestake et al., 1997), or whether a given normal dominance constraints has a constructive solution.5 This reduction is by deleting all labels and making leaves that had nullary labels closed. This means that (the equivalent of) deciding satisability in these approaches is NP-hard.</Paragraph>
    <Paragraph position="4"> The crucial di erence between e.g. satis ability and constructive satis ability of normal dominance constraints is that it is possible that a solved form has no constructive solutions. This happens e.g. in the example from Section 5, X:a ^ Y :b. The constraint, which is in solved form, is satis able e.g. by the tree f(a;b); but every solution must contain an additional node with a binary label, and hence cannot be constructive.</Paragraph>
    <Paragraph position="5"> For practical purposes, however, it can still make sense to enumerate the irredundant solved forms of a normal constraint even if we are interested only in constructive solution: It is certainly cheaper to try to nd constructive solutions of solved forms than of arbitrary constraints. In fact, experience indicates that for those constraints we really need in scope underspeci cation, all solved forms do have constructive solutions { although it is not yet known why. This means that our enumeration algorithm can in practice be used without change to enumerate constructive solutions, and it is straightforward to adapt it e.g. to an enumeration algorithm for Hole Semantics.</Paragraph>
  </Section>
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