A Graph Based Parsing Algorithm for Context-free Languages
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1 A Graph Based Parsing Algorithm for Context-free Languages Giinter Hot> Technical Report A 01/99 June hotzocs.uni-sb.de VVVVVV: de Abstract We present a simple algorithm deciding the word problem of c. f. languages in O(n 3 ). It decides this problem in time O(n') for unambiguous grammars and in time O(n) in the case of LR(k) grammars. Fachbereich 14 Informatik Universitat des Saarlandes P08tfach Saarbriicken Germany
2 1 Introduction There are several algorithms known deciding the word problem of general context-free languages in time O(n 3 ). The algorithm of Younger [You67] is very simple and it solves the problem in time O(n 3 ), but it takes no advantage of special cases. Kasami in [KT69] describes an algorithm, which decides this problem for unambiguous context-free grammars in time O(n 2 log n). Early [Ear70] developed an algorithm which decides the general word problem in time O(n 3 ) but does it for unambiguous grammars in time O(n 2 ) and for a wide class of grammars as LR(k) grammars [Knu65] in time O(n). His algorithm takes no advantage of grammars in a normal form. The proofs are hard to read. We present here a simple algorithm with the same runtime eciency as Early's algorithm. 2 Notations and Denitions Let V T be nite alphabets, V \ T = S 2 V and P (V V 2 ) [ (V T ) a c. f. production system in Chomsky normal form (Ch-NF). We assume that the grammar G := (V T P S) does not contain superuous variables. That means for each x 2 V we nd u1 u2 u 2 T such that x ;! u and s ;! u1xu2 holds. We dene linear forms with variables from V and coecients from the boolean algebra B. These are mappings : V ;! B and we write B hv i := f j : V ;! B g: We use the equivalent notation :=X (v) v We dene the sum and a product in B hv i : As usual we put v2v ( + )(v) :=(v) +(v) for v 2 V: The product x y for x y 2 V gives all possible reductions of xy relative to P. More formally we dene Now we put x y :=X z2v (z) z () ((z) =1() (z xy) 2 P: := X x y2v (x) (y) (x y) 1
3 we use in this denition for 2 B and 2 B hv i the operation ()(v) = (v) for v 2 V: The product is not associative. (B hv i + ) is distributive. We use furthermore the notation P ;1 (t) =X t z z t z =1() (z t) 2 P: z2v If the operation is associative then for u = t1 ::: t n and (u) := P ;1 (t1) ::: P ;1 (t n ) we have u 2 L(G) () (u)(s) =1: In this case (B hv i ) is a nite monoid and P ;1 : T homomorphism and therefore L(G) is regular. ;! (B (V ) ) is a 3 The Graph ;(G u) We assign to the grammar G and u 2 T an oriented graph ;=(K E) K is the set of vertices and E the set of edges and n := juj the length of u. K [ f(v i 0) j v 2 V 1 <i ng [ f(v i 1) j v 2 V 1 i<n+1g E [ f((v i 1) (v j 0)) j V ;! t i ::: t j;1 1 i<j n +1g Obviously it holds u 2 L(G) () ((s 1 1) (s n 0)) 2 E: The graph ; is closed under the following operation: Let be i<j<m edges of ; and If (z) =1 then the edge (x i 1) s 1 ;! (x j 0) (y j 1) s 2 ;! (y m 0) := x y: (z i 1) s 3 ;! (z m 0) 2
4 is in ;. We write in this case s3 := s1 s2 in general there may be several edges s 0 3 in the relation s 0 3 := s 1 s2. This closure property corresponds to and x ;! t i ::: t j;1 y ;! t j ::: t m;1 z ;! xy: Therefore we have z ;! t1 ::: t m;1 and from this follows by denition of ;, that s3 is in E. Lemma 1. If there are two dierent operations producing the same edge s3, then G is ambiguous. Proof 1. Let s1 s2 and s 0 1 s0 2 two pairs of edges from ; producing under the explained operation the edge s3, then we have the two dierent derivations z ;! xy x ;! u1 y ;! u2 u3 = u1 u2 z ;! x 0 y 0 x 0 ;! u 0 1 y0 ;! u 0 2 u 3 = u 0 1 u 0 2 : Now we assume G not containing superuous variables. Therefore exist the derivations s ;! ~uzu ;! ~uu1 u2u = ~uu 0 1 u 0 2 u 2 T : So we have more than one derivation of ~uu3u from S, i.e. G is ambiguous. 4 The algorithm We now construct a sequence ;1 ;2 ::: ; n of subgraphs of ; such that ;1 depends only on t1 and with ; n =;: We give an operation which constructs ; i+1 from ; i and estimate the complexity ofthisoperation. Let ; i := (K i E i ) for i =1 ::: n and K i := (v l ") 2 K j 1 l i " 2f0 1gg [ f(x i +1 0) j x 2 V g E i := fs 2 E j source(s), sink(s) 2 K i g: The construction of ;1 can be done in time O(1). 3
5 We assume ; i, i<nhas been constructed. We add t i+1 and f(v i +1 1) j v 2 V [fv i +2 0) j v 2 V g to K i. We in the rst step add the following edges of E to E i : (v i +1 1) ;! (v i +2 0) for v ;! t i+1: Let ; 0 i the result of this construction. Now we apply the closure operations s1 s2 ;! s3 to edges s1 s2 from ; 0 i : ; i being closed under these operations we have to begin with the new edges in ; 0 i. We have the following situation (x j 1) s 1 ;! (x i +1 0) (y i +1 1) s 2 ;! (y i +2 0): We built from s1 s2 (z j 1) s 3 ;! (z i +2 0) if (z xy) 2 P. Iterating this construction in the worst case we need O(n 2 ) elementary operations to construct ; i+1 from ; i, because each edge of ; 0 i we have to consider only once. To construct ; n by this procedure therefore needs in the worst case O(n 3 ) -operations. If the grammar is unambiguous we construct each edge only one time. Operations s1 s2 which do not produce a new edge we are able to exclude by only once inspecting the pairs of vertices (x l 0) (y l 1). If x y =0, then none of the pairs s 1 ;! (x l 0) (y l 1) s 2 ;! has to be considered. Therefore in this case we need only O(n 2 ) steps because this is the bound for the number of edges in ;. So we proved the Theorem 1. The algorithm dened here solves the word problem for c. f. languages in time O(n 3 ). In the case of unambiguous grammars the running time of the algorithm is O(n 2 ). 4
6 Corollar 1. The algorithm solves the word problem in the case of grammars with m-bound ambiguity in time O(n 2 m). Proof 2. From the m-bound ambiguity it follows that the algorithm draws each new edge maximal m times. Now we study the case G is a LR(k) grammar. LR(k) grammars are characterized by the following property: For uvu 0 2 L(G) and jvj = k let w1 ::: w l be the reduced words of u v relative to G. Then the set of this words has a common prex u, where u is a reduced word of u, such that we can write w1 + :::+ w l = u (v1 + :::+ v l ) jv i jk for i =1 ::: l: This property enables us to compute an upper bound for the number j; i j of edges in ; i. Obviously we have j;1j m for m := #V: We assume ; i being constructed. We then get ; i+1 by the following steps: 1. We compute P ;1 (t i+1), which produces not more than m new edges. 2. We match the new edges with the existing edges. This leads to new edges connecting vertices belonging to (v1 + :::+ v l ) P ;1 (t i+1)g and edges connecting vertices belonging to vt i+1 with edges belonging to u. The number of edges belonging to the rst class is bound by a constant c depending on m = #V and k. The number of the edges belonging to the second class is 0 if u i is prex of u i+1. It is 1 if ju i+1j = ju i j and it is ju i j;ju i+1j if reductions of the reduced word u i take place. So we have From this we get From this follows j; i+1j j; i j + C + ju i j;ju i+1j +1: j; n j = O(n): Theorem 2. The given graph algorithm solves the word problem for LR(k) grammars G := (V T P S) and words w 2 T with = O(n) -operations. It is obvious that the -operations can be performed on a computer in time only depending on G. This means that it can be done in constant time relative tojwj. 5
7 Literatur [Ear70] J. Early. An ecient context-free parsing algorithm. Com. ACM, 13, [Knu65] D. E. Knuth. On the translation of languages from left to right. Information and Control, 8, [KT69] T. Kasami and K. Tori. A syntax-analysis procedure for unambiguous context-free grammars. ACM, 16, [You67] D. H. Younger. Recognition and parsing of context-free languages in time n 3. Information and Control, 10,
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