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<?xml version="1.0" standalone="yes"?> <Paper uid="C96-2215"> <Title>Most Probable Tree in Data-Oriented Parsing and Stochastic Tree Grammars. In Proceedings</Title> <Section position="4" start_page="1175" end_page="1178" type="metho"> <SectionTitle> 3 Complexity of MPPWG, MPS </SectionTitle> <Paragraph position="0"/> <Section position="1" start_page="1175" end_page="1178" type="sub_section"> <SectionTitle> and MPP 3.1 3SAT to MPPWG and MPS </SectionTitle> <Paragraph position="0"> The reduction from the 3SAT instance INS to an MPPWG problem must construct an STSG and a word-graph in deterministic polynomialtime. Moreover, the answers to the MPPWG instance must correspond exactly to the answers to INS. The presentation of the reduction shall be accompanied by an example of the following 3SAT instance (Barton et al., 1987): (ul V E2 V ua) A (~1 V l/,2 V U3). Note that a 3SAT instance is satisfiable iff at least one of the literals in each conjunct is assigned the value True. Implicit in this, but crucial, the different occurrences of the literals of the same variable must be assigned values consistently.</Paragraph> <Paragraph position="1"> Reduction: The reduction constructs an STSG and a word-graph. The STSG has start-symbol S, two terminals represented by T and F , non-terminals which include (beside S) all Ck, for mula does not contain repetition of conjuncts.</Paragraph> <Paragraph position="3"> 1 < k < rn, and both literals of each Boolean variable of the formula of INS. The set of elementary-trees and probability function and the word-graph are constructed as follows: 1. For each Boolean variable ui, 1 < i < n, construct two elementary-trees that correspond to assigning the values true and false to ui consistently through the whole formula.</Paragraph> <Paragraph position="4"> Each of these elementary-trees has root S, with children Ck, 1 5 k < rn, in the same order as these appear in the formula of INS; subsequently the children of Ck are the non-terminals that correspond to its three disjuncts dkl, dk2 and dk3. And finally, the assignment of true (false) to ui is modeled by creating a child terminal T (resp. F ) to each non-terminal ui and P (resp. T ) to each ul. The two elementary-trees for u~, of our example, are shown in the top left corner of figure 1.</Paragraph> <Paragraph position="5"> 2. The reduction constructs three elementary-trees for each conjunct Ck. The three elementary-trees for conjunct Ck have the same internal structure: root Ck, with three children that correspond to the disjuncts dkl, dk2 and dk3 In each of these 3.</Paragraph> <Paragraph position="6"> 4.</Paragraph> <Paragraph position="7"> .</Paragraph> <Paragraph position="8"> elementary-trees exactly one of the disjuncts has as a child the terminal T ; in each of them this is a different one. Each of these elementary-trees corresponds to the conjunct; where one of the three possible literals is assigned the value T . For the elementary-trees of our example see the top right corner of figure l.</Paragraph> <Paragraph position="9"> The reduction constructs for each of tile two literals of each variable ni two elementary-trees where the literal is assigned in one case T and in the other F . Figure 1 shows these elementary-trees for variable ul in the bottom left corner.</Paragraph> <Paragraph position="10"> The reduction constructs one elementary-tree that has root S with children Ck, 1 < k < rn, in the same order as these appear in the formula of INS (see the bottom right corner of figure 1).</Paragraph> <Paragraph position="11"> The probabilities of the elementary-trees that have the same root non-terminal sum up to 1. The probability of an elementary-tree with root S that was constructed in step 1 of this reduction is a value Pi, 1 _< i < n, where ui is the only variable of which the literals in the elementary-tree at hand are lexical- null ized (i.e. have terminal children). Let ni denote the number of occurrences of both literals of variable ui in the formula of INS. Then Pi = 0 (1/2)ni for some real 0 that has to fulfill some conditions which will be derived next. The probability of the tree rooted with S and constructed at step 4 of this reduction must then bep0 = \[1 - 2~i=lpl\]. The probability of the elementary-trees of root Ck (step 2) is (1), and of root ui or ul (step 3)is (1/2). For our example some suitable probabilities are shown in figure 1.</Paragraph> <Paragraph position="12"> 6. Let Q denote a threshold probability that shall be derived hereunder. The MPPWG (MPS) instance is: does the STSG generate a parse (resp. sentence) of probability > Q, for the word-graph WG = {T, F} 3m ? Deriving the probabilities: The parses generated by the constructed STSG differ only in the sentences on their frontiers. Therefore, if a sentence is generated by this STSG then it has exactly one parse. This justifies the choice to reduce reduction. This type of derivation corresponds to assigning values to all literals of some variable ui in a consistent manner. For all 1 < i < n the probability of a derivation of this type is 13m n pi( ) - ' = The second type of derivation corresponds to substituting the elementary-trees rooted with Ck in S -+ C1 ... C,~, and subsequently substituting in the open-trees that correspond to literals. This type of derivation corresponds to assigning to at least one literal in each conjunct the value true. The probability of any such derivation is</Paragraph> <Paragraph position="14"> Now we derive both the threshold Q and the parameter 0. Any parse (or sentence) that fulfills both the &quot;consistency of assignment&quot; requirements and the requirement that each conjunct has at least one literal with child T , must be generated by n derivations of the first type and at least one derivation of the second type. Note that a parse can never be generated by more than n derivations of the first type. Thus the threshold</Paragraph> <Paragraph position="16"> However, 0 must fulfill some requirements for our reduction to be acceptable: 1. For all i: 0<pi < 1. This means that for 1 < i <_ n: 0 < 0(1/2) '~' < 1, and 0 < P0 < 1. However, the last requirement on P0 implies that 0 < 20V&quot;~z--,i=l~2/(!~'~ < 1, which is a stronger requirement than the other n requirements. This requirement can also be stated as follows: 1 0 < 0 < ,.</Paragraph> <Paragraph position="17"> 2. Since we want to be able to know whether a parse is generated by a second type derivation only by looking at the probability of the parse, the probability of a second type derivation must be distinguishable from first type derivations. Moreover, if a parse is generated by more than one derivation of the second type, we do not want the sum of the probabilities of these derivations to be mistaken for one (or more) first type derivation(s). For any parse, there are at most 3 TM second type derivations (e.g. the sentence T ...T ). Therefore we require that: Which is equal to 0 > 2~=,(~C/__ 3. For the resulting STSG to be a probabilistic model, the &quot;probabilities&quot; of parses and sentences must be in the interval (0, 1\]. This is taken care of by demanding that the sum of the probabilities of elementary-trees that have the same root non-terminal is 1, and by the definition of the derivation's probability, the parse's probability, and the sentence's probability.</Paragraph> <Paragraph position="18"> There exists a 0 that fulfills all these requirements Polynomlality of the reduction: This reduction is deterministic polynomial-time in n because it constructs not more than 2n + 1 + 3m + 4n elementary-trees of maximum number of nodes 4 7m+ 1.</Paragraph> <Paragraph position="19"> The reduction preserves answers: The proof concerns the only two possible answers. Yes If INS's answer is Yes then there is an assignment to the variables that is consistent and where each conjunct has at least one literal assigned true. Any possible assignment is represented by one sentence in WG. A sentence which corresponds to a &quot;successful&quot; assigmnent must be generated by n derivations of the first type and at least one derivation of the second type; this is because the sentence w 3m fulfills n consistency requirements (one per Boolean variable) and has at least one W as Wak+l, Wak+2 or W3k+3 , for all 0 < k < m. Both this sentence and its corresponding parse have probability > Q.</Paragraph> <Paragraph position="20"> Thus MPPWG and MPS also answer Yes.</Paragraph> <Paragraph position="21"> No IfINS's answer is No, then all possible assignmcnts are either not consistent or result in at least one conjunct with three false disjuncts, or both. The sentences (parses) that correspond to non-consistent assignments each have a probability that cannot result in a Yes answer. This is the case because such sentences have fewer than n derivations of the first type, and the derivations of the second type can never compensate for that (the requirements on 0 take care of this). For the sentences (parses) that correspond to consistent assignments, there is at least some 0 < k < m such that wak+l , wak+2 and Wak+3 are all F . These sentences do not have second type derivations. Thus, there is no sentence (parse) that has a probability that can result in a Yes answer; the answer of MPPWG and MPS is NO.</Paragraph> <Paragraph position="22"> We conclude that MPPWG and MPS are both NP-hard problems.</Paragraph> <Paragraph position="23"> Now we show that MPPWG and MPS are in NP. A problem is in NP if it is decidable by a non-deterministic Turing machine. The proof here is informah we show a non-deterministic algorithm that keeps proposing solutions and then checking each of them in deterministic polynomial time cf. (Barton et al., 1987). If one solution is successful then the answer is Yes. One possible non-deterministic algorithm for the MPPWG and MPS, constructs firstly a parse-forest for WG in deterministic polynomial time based on the algorithms in (Schabes and Waters, 1993; Sima'an, 1996), and subsequently traverses this parse-forest (bottom-up for example) deciding at each point what path to take. Upon reaching the start non-terminal S, it retrieves the sentence (parse) and evaluates it in deterministic polynomial-time (Sima'an et al., 1994), thereby answering the decision problem.</Paragraph> <Paragraph position="24"> This concludes the proof that MPPWG and MPS are both NP-complete.</Paragraph> </Section> <Section position="2" start_page="1178" end_page="1178" type="sub_section"> <SectionTitle> 3.2 NP-eompletetless of MPP </SectionTitle> <Paragraph position="0"> The NP-completeness of MPP can be easily deduced from the previous section. In the reduction the terminals of the constructed STSG are new symbols vii, 1 < i < m and 1 < j < 3, instead of T and F that becomc non-terminals.</Paragraph> <Paragraph position="1"> Each of the elementary-trees with root S or Ck is also represented here but each T and F on the frontier has a child vkj wherever the T or F appears as the child of the jth child (a literal) of Ck. For each elementary-tree with root ui or ui, there are 3m elementary-trees in tile new STSG that correspond each to creating a child ~)ij for the T or F on its frontier. The probability of an elementary-tree rooted by a literal is 1 The probabilities of elementary-trees rooted gm&quot; with Ck do not change. And the probabilities of the elementary-trees rooted with S are adapted from the previous reduction by substituting for 1 every (1/2) the value ~. The threshold Q and the requirements on 0 are also updated accordingly.</Paragraph> <Paragraph position="2"> The input sentence which the reduction constructs is simply v11.., v3,~. The decision problem is whether there is a parse generated by the resulting STSG for this sentence that has probability larger than or equal to Q.</Paragraph> <Paragraph position="3"> The rest of the proof is very similar to that in section 3. Therefore the decision problem MPP is NP-complete.</Paragraph> </Section> <Section position="3" start_page="1178" end_page="1178" type="sub_section"> <SectionTitle> 3.3 MPS under SCFG </SectionTitle> <Paragraph position="0"> The decision problem MPS is NP-complete also under SCFG. The proof is easily deducible from the proof concerning MPS for STSGs. The reduction simply takes the elementary-trees of the MPS for STSGs and removes their internal structure, thereby obtaining simple CFG productions. Crucially, each elementary-tree results in one unique CI&quot;G production. The probabilities are kept the same. The word-graph is also the same word-graph as in MPS for STSGs. The problem is: does the SCFG generate a sentence with probability _> Q, for the word-graph W G -- {T, F} 3m. The rest of the proof follows directly from section 3.</Paragraph> </Section> </Section> class="xml-element"></Paper>