XPath Rewriting Using Multiple Views. Author. Published. Journal Title DOI. Copyright Statement. Downloaded from. Griffith Research Online

Transcription

1 XPth Rewriting Using Multiple Views Author Wng, Junhu, Yu, Jeffre Pulished 2008 Journl Title Leture Notes in Computer siene DOI Copright Sttement 2008 Springer Berlin / Heidelerg. This is the uthor-mnusript version of this pper. Reprodued in ordne with the opright poli of the pulisher. The originl pulition is ville t Downloded from Griffith Reserh Online

2 XPth rewriting using multiple views Junhu Wng 1 nd Jeffre Xu Yu 2 1 Griffith Universit, Gold Cost, Austrli J.Wng@griffith.edu.u 2 Chinese Universit of Hong Kong, Hong Kong, Chin u@se.uhk.edu.hk Astrt. We stud the prolem of tree pttern quer rewriting using multiple views for the lss of tree ptterns in P {//,[]}. Previous work hs onsidered the rewriting prolem using single view. We onsider two different ws of omining multiple views, define rewritings of tree pttern using these omintions, nd stud the reltionship etween them. We show tht when rewritings using single views do not eist, we m use suh omintions of multiple views to rewrite quer, nd even if rewritings using single views do eist, the rewritings using omintions of multiple views m provide more nswers thn those provided the union of the rewritings using the individul views. We lso stud properties of intersetions of tree ptterns, nd present lgorithms for finding rewritings using intersetions of views. 1 Introdution Quer rewriting using views hs mn pplitions inluding dt integrtion, quer optimiztion, quer hing, nd support of phsil dt independene [5]. Given quer, it studies finding nother quer using onl the views to produe orret nswers to the originl quer. In the literture, two tpes of quer rewritings re studied, nmel, equivlent rewritings nd ontined rewritings. Given view, V, nd quer, Q, n equivlent rewriting produes ll nswers to the originl quer Q using view V, wheres ontined rewriting m produe suset of the nswers to Q using V. Both tpes of rewritings hve een etensivel studied in the reltionl dtse ontet [5]. Reentl, ml quer rewriting hs ttrted ttention euse of the rising importne of ml dt [3, 14, 7, 6, 2, 9]. XPth lies in the enter of ll ml lnguges, where the mjor lsses of XPth epressions tht hve een studied re tree ptterns [1, 8]. Among previous work on rewriting XPth queries using views, Xu nd Ozsooglu [14] studied the ompleit of finding equivlent rewritings for four tpes of tree ptterns studied in [8] nd presented n pproh for finding nd minimizing suh rewritings. Mndhni nd Suiu [7] presented, in ddition to he orgniztion, n effiient ut inomplete method for finding equivlent rewritings of tree ptterns involving /, //, [] nd * when the ptterns re ssumed to e minimized nd m hve vlue-sed predites. In [6], Lkshmnn et l. studied miml ontined rewritings of tree ptterns where

3 pper setion setion n 1 referene pper susetion emple emple n 2 referene n 3 referene pper Fig. 1. Emple ml tree t pper oth the views nd queries re in P {//,[]}, oth in the sene nd in the presene of non-reursive, non-disjuntive dtds. As the nme implies, the miml ontined rewriting is rewriting tht ontins ll other ontined rewritings. All of the ove works fous on rewriting the quer using single view. In other words, no omintion of views is onsidered for the purpose of rewriting. In this pper we stud tree pttern rewriting using multiple views, nd our work is motivted the following oservtions. Suppose we hve set V of views nd the quer Q. Then 1. It is possile for Q to hve no ontined rewritings using n individul view in V, ut it n e (prtill) nswered using the omintion of some of the views. As simple emple, the quer Q = rtile[tle][figure]/uthor nnot e (prtill) nswered using either V 1 = rtile[tle]/uthor or V 2 = rtile[figure]/uthor, ut it n e nswered using V 1 V 2, whih is equivlent to Q. 2. Even if Q does hve ontined rewritings using V 1, nd/or Q does hve ontined rewritings using V 2, there m e ontined rewritings of Q using some omintion of V 1 nd V 2 (e.g., intersetion) whih provide stritl more nswers to Q thn the union of ll ontined rewritings of Q using the individul views. This is demonstrted Emple Even if quer Q does not hve n equivlent rewriting using V ording the onventionl definition of rewriting [6, 14, 7], it is still possile to find ll of the orret nswers to Q using V for ll ml trees. This is demonstrted in Emple 2. The emple elow demonstrtes oservtion 2. Emple 1. Consider the views V 1 = pper//susetion//referene, V 2 = pper//emple/referene nd the quer Q= pper//susetion//emple/referene//pper. It n e verified tht V 1 V 2 = pper//susetion//emple/referene, 2

4 d d e () Q d () V e () Q 1 d e (d) Q 1 V (e) Q 2 d (f) Q 2 V Fig. 2. Q hs no equivlent rewriting using V ording to onventionl definition, ut Q n e full nswered using V : Q 1 V Q 2 V = Q nd evluting the pttern referene//pper over the nswers of V 1 V 2 will produe the sme nswers s Q, tht is, referene//pper is n equivlent rewriting of Q using V 1 V 2. However, the miml ontined rewriting of Q using V 1 is referene//emple/referene//pper, nd the miml ontined rewriting of Q using V 2 is referene//susetion//emple/referene//pper. As we show net, the rewriting using V 1 V 2 produes stritl more nswers thn the union of the nswers produed the miml ontined rewritings using V 1 nd V 2 individull. For the ml tree t shown in Fig.1, evluting V 1 over t produes (the sutrees rooted t) n 1 nd n 2, nd evluting V 2 over t produes (the sutrees rooted t) n 2 nd n 3. Therefore, the miml ontined rewriting using V 1 or V 2 will produe no nswers for Q, ut the rewriting using V 1 V 2 will produe the pper node under n 2. The net emple demonstrtes oservtion 3. Emple 2. Consider the quer Q nd view V shown in Fig.2 () nd () respetivel. It n e verified tht Q hs no equivlent rewritings using V ording to the onventionl definition. But given n ml tree t, we n find Q(t) using the view s follows. We evlute the quer Q 1 = /[e]/ over the sutrees in V (t), nd denote the results s Q 1 (V (t)); we then evlute Q 2 = / over the sutrees in V (t), nd otining set denoted s Q 2 (V (t)). Finll, we tke the intersetion of Q 1 (V (t)) nd Q 2 (V (t)). It n e verified tht Q(t) = Q 1 (V (t)) Q 2 (V (t)). We will stud ontined nd equivlent rewritings of tree ptterns in P {//,[]} using multiple views in the sene nd presene of dtds, with the intersetion nd union opertions. Our min ontriutions re summrized elow. We define (ontined nd equivlent) rewritings of tree pttern using two different omintions of views, nd show the reltionship etween these rewritings. 3

5 We show the intersetion of some tree ptterns, if not empt, n e epressed s the union of tree ptterns, nd provide n effiient lgorithm to trnslte the intersetion into union. Bsed on the ove, we provide lgorithms for finding the miml ontined rewritings nd equivlent rewritings using the intersetion of views. We lso show the effet of n non-reursive dtd on the rewritings. The rest of the pper is orgnized s follows. Setion 2 provides the kground nd nottions. Setion 3 presents the lgorithm for reformulting intersetions of tree ptterns into union, nd studies rewritings using the intersetion of views. Setion 4 defines rewritings using different omintion views, nd ompre them with rewritings using the intersetion. Setion 5 onsiders the effets of non-reursive dtds. Setion 6 surves relted work. Finll Setion 7 onludes the pper. 2 Preliminries 2.1 XML trees nd tree ptterns Let Σ e n infinite set of tgs. An ml tree is tree suh tht ever node is leled with tg in Σ. A tree pttern (TP) is tree with unique distinguished node, nd with ever node leled with tg in Σ, nd ever edge leled with either / or //. Suh TP orresponds to n XPth epression involving the hild is, desendnt is, nd rnhing ondition []. A tree pttern hs tree representtion. Fig.2 show severl TPs, nd the TP in Fig.2 () represents the XPth epression //[]/[]/[d]. Here, single nd doule lines represent /- edges nd //-edges respetivel. The distinguished node is indited irle. Note: the TPs in our disussion orrespond to the frgment P {//,[]} defined in [8]. A suset of P {//,[]}, denoted P {//}, ontins ll TPs tht hs single root-to-lef pth. Let T e n ml tree or TP, nd v e node in T. We will use N(T ) to denote the set of ll nodes in T, rt(t ) to denote the root of T, nd lel(v) to denote the lel of v. For n TP P, DN P nd DP P denote, respetivel, the distinguished node nd distinguished pth of P (i.e., the pth from rt(p ) to DN P ). A mthing of TP, P, in n ml tree, t, is mpping δ from N(P ) to N(t) stisfing the following onditions: (1) root-preserving, i.e., δ(rt(p )) = rt(t), (2) lel-preserving, i.e., v N(P ), lel(v) = lel(δ(v)), nd (3) struturepreserving, i.e., for ever edge (, ) in P, if it is /-edge, then δ() is hild of δ(); if it is //-edge, then δ() is desendnt of δ(), i.e, there is pth from δ() to δ(). Eh mthing δ produes sutree of t rooted t δ(dn P ), denoted su t δ(dn P ), whih is lled n nswer to the TP. We use P (t) to denote the nswer set of P on t: P (t) = {su t δ(dn P ) δ is mthing of P in t}. Let T e set of ml trees. We use P (T ) to denote the union of nswer sets of Q on the trees in T. Tht is, P (T ) = t T P (t). 4

6 2.2 Intersetion nd etension of TPs Two tree ptterns, P nd Q, re sid to e omprle if lel(rt(p )) = lel(rt(q)) nd lel(dn P ) = lel(dn Q ). Let P 1,..., P n e omprle TPs. For n XML tree t, the intersetion of P 1,..., P n, denoted P 1 P n, returns P 1 (t) P n (t). The union of P 1,..., P n, denoted P 1 P n, returns P 1 (t) P n (t). Let P nd Q e TPs suh tht lel(dn P ) = lel(rt(q)). The etension of P using Q, denoted Q P, is the TP otined merging DN P with rt(q). The distinguished node of Q P is DN Q. Fig.2 (d) shows the etension of V (Fig.2 ()) using Q 1 (Fig.2 ()). It is es to see tht, for n ml tree t, (Q P )(t) = Q(P (t)), tht is, (Q P )(t) is equivlent to the union of nswer sets of Q on the sutrees in P (t). Let P 1,..., P n e omprle TPs, nd Q e TP suh tht rt(q) nd DN Vi hve identil lels. We denote Q (P 1 P n ) the etension of P 1 P n using Q, whih returns, for n t, Q(P 1 (t) P n (t)). 2.3 TP ontinment Let P nd Q e TPs. P is sid to e ontined in Q, denoted P Q, if for ever ml tree t, P (t) Q(t). It is shown in [1] tht P Q iff there is ontinment mpping from Q to P. Rell: A ontinment mpping (CM) from Q to P is mpping δ from N(Q) to N(P ) tht is lel-preserving, root-preserving s disussed in the lst setion, struture-preserving (whih now mens for n /-edge (,) in Q, (δ(), δ()) is /-edge in P, nd for n //-edge (, ), there is pth from δ() to δ() in P ) nd is output-preserving, whih mens δ(dn Q ) = DN P. A homomorphism from Q to P is similr to CM, eept there is no requirement of output-preserving. The following lemm is proved in [13]. Lemm 1. For TPs P 1,..., P n, P P {//,[]}, P P 1 P n i [1, n] suh tht P P i. iff there is 3 TP rewriting using intersetions of views A view is n eisting TP. We define rewritings using intersetions of views s follows. Definition 1. Let Q e quer, nd V 1,..., V n e omprle views. If V 1 V n is non-empt, nd there eists Q suh tht lel(rt(q )) = lel(dn V1 ), nd Q (V 1 V n ) Q, then we s Q (V 1 V n ) is ontined rewriting (CR) of Q using V 1 V n. If Q (V 1 V n ) = Q, we s Q is n equivlent rewriting (ER) using V. The miml ontined rewriting (MCR) of Q using V 1 V n, denoted MCR(Q, V 1 V n ), is the union of ll CRs of Q using V 1 V n. Note tht when n = 1, the ove definition redues to those in [6] (CR nd MCR) nd [14] (ER). 5

7 3.1 Properties of intersetions In this setion we identif onditions under whih the intersetion of TPs re not empt, nd present lgorithms to trnslte the intersetion into the union of TPs. We present these results using two TPs. Generliztion to more TPs is simple. In mn ses the intersetion of two TPs P nd Q is n empt quer, tht is, it lws returns the empt set. For emple, when P = // nd Q = ///. The question rises: when is P Q non-empt quer? We hve the follow nswer to this question. Lemm 2. P Q is non-empt iff there is pth P in P {//} suh tht P DP P, nd P DP Q. The lemm is true euse, when there is no dtd, ever TP is non-empt, nd P Q is non-empt iff DP P DP Q is non-empt. Let s ll the pth P in lemm 2 ommon distinguished pth of P nd Q. If DP P nd DP Q do not hve //-edges, then onl when DP P = DP Q there is CDP of P nd Q, whih is DP P itself. However, if DP P nd DP Q do hve //-edges, then there m e (infinitel) mn CDPs of P nd Q. For emple, if P = //[]//, Q = //[]//, then P nd Q hve the CDPs ////, //////, //////// nd so on. But there re onl finite numer of CDPs tht re of interest to us. These CDPs re nnotted to form nnotted CDPs. Definition 2. Let P nd Q e omprle TPs in P {//,[]}. An nnotted CDP (ACDP) of P nd Q is CDP P of P nd Q suh tht (1) ever node in P is nnotted with either 1, or 2, or oth, (2) there is (unique) CM m P from DP P to P suh tht ever node in DP P is mpped node nnotted with 1 (or oth 1 nd 2), nd for ever node v in P nnotted with 1 (or oth 1 nd 2), there is unique node u in DP P suh tht m P (u) = v, (3) there is (unique) CM m Q from DP Q to P suh tht ever node in DP Q is mpped node nnotted with 2 (or oth 1 nd 2), nd for ever node v in P nnotted with 2 (or oth 1 nd 2), there is unique node u in DP Q suh tht m Q (u) = v. Intuitivel, n ACDP of P nd Q is pth in P {//} whih ontins etl one op of DP P nd one op of DP Q (the nodes nnotted with 1 form op of DP P, nd the nodes nnotted with 2 form op of DP Q ) nd ever node ppers in t lest one of the opies. Fig.3 shows ll ACDPs of P = //[]// nd Q = //[]//. Net we present n lgorithm tht finds ll ACDPs of P nd Q. Let P 1 = DP P nd P 2 = DP Q. We n find ll ACDPs of P nd Q lling the funtion merge(p 1, P 2 ). In the funtion, position in P j refers to either node or //-edge in P j. Eh position in P j is given unique position numer, with the position numer of rt(p j ) eing 0, nd eh susequent position s position numer inreses 1. The funtion merge(p i, P j ) finds the ACDPs of P i nd P j inserting the nodes of P i into P j. To do so, it needs to find position, 6

8 1,2 1,2 1,2 1, ,2 1, ,2 Fig. 3. ACDPs of //[]// nd //[]// in P j, for ever node in P i. There re three stges in the proess. The first stge (lines 1-6) is to sn P i top-down, mrking eh node v in P i with set of mrkings. Eh mrking is of the form {s : A(v, s)} (eept for rt(p i ), whih is mrked {0}), where s is position, nd A(v, s) is set of positions. The mening of this mrking is tht if prent(v) is merged into P j in position s, then the possile positions where v n e inserted in P j re those in A(v, s). The riterion to hoose the positions in A(v, s) is lel-preservtion nd struturepreservtion. The seond stge (lines 8-10) removes the impossile positions from the mrkings, going from ottom-up. The onl possile position for DN Pi is the lst position in P j. Bsed on struture-preservtion, if impossile positions for v re found nd deleted, then impossile positions for prent(v) m e found nd deleted. In the lst stge (lines 13-29), if ever node hs possile position, we pik position s v for eh node v in P i ording to the mrkings, nd onstrut ACDP. For eh omintion of positions (eh omintion hs one position for ever node in P i ) there is ACDP onstruted. Finll the funtion returns ll ACDP onstruted this w. Emple 3. Let P 1 = ///// nd P 2 = //////. Fig.4 shows the proess of running merge(p 1, P 2 ). First, we identif nd lel eh position in P 2. There re 7 positions leled 0, 1,..., 6, s shown in the figure. In stge 1, we mrk eh node in P 2 with its possile positions. The -node is mrked {0}. If the -node is put in position 0, then the possile positions of the net node re 1,2,3,4,5. Therefore, the first -node is mrked with (0 : {1, 2, 3, 4, 5}). If the -node is put to position 1, then the net node m e put to positions 1,3 or 5, to preserve lel nd struture of P i. If the -node is put to position 2, then the possile positions for its hild re 3 nd 5,..., if the -node is put to position 5, then its hild must e put to position 5 (fter the -node). Therefore, the hild of the -node is mrked with the mrkings s shown in the figure. Finll, the lst node of P 1 must e put to the lst position, position 6, in P 2. Thus if its prent is put to position 1 or 3, then there re no possile positions for it. If its prent is put to position 5, the it n go to position 6. This eplins the mrkings of the lst node of P 1. In stge 2, we remove the impossile positions in the mrkings. Beuse position 1 nd 3 of the first -node prohiits the seond -node to e put in position 6 (s indited the mrkings (1, {}) nd (3, {}), we know 7

9 Algorithm 1 merge(p i, P j) 1: let pos(rt(p i )) = {0}, S = 2: for ever susequent node v in P i do 3: for ll s pos(prent(v)) do 4: let position(v, s) = Find Position(v, P i, s, P j ) 5: mrk v with (s : position(v, s)) 6: let pos(v) = position(v, s) s pos(prent(v)) 7: if pos(v) = then return 8: for eh node v in P i (strting from v = DN Pi ) do 9: while mrking (s : position(v, s)) for v suh tht position(v, s) = do 10: delete s from position(prent(v), s ) for ll s 11: if pos(v) = return 12: for eh node v in P i hoose position s v from its mrkings. Initill, v = rt(p i) nd s v = 0. For eh susequent node v, pik position from position(v, s prent(v) ) 13: for eh omintion of positions found ove do 14: let P = P j, nnotte ever node with j 15: for v in P i do 16: if position s v points to node u in P then 17: nnotte u with i. 18: if prt(u) is nnotted with i or i, j nd (prent(v), v) is /-edge then 19: hnge (prt(u), u) to /-edge 20: else if position s v represents //-edge (u 1, u 2) in P then 21: insert lel(v)-node u 0 etween u 1 nd u 2 22: if u 1 is nnotted with i or i, j then 23: let edge (u 1, u 0 ) e of the sme tpe s (prent(v), v) 24: else 25: let edge (u 1, u 0) e of tpe // 26: let edge (u 0, u 2 ) e of tpe //; let position s v point to this edge (u 0, u 2 ) 27: dd P to S 28: return S positions 1 nd 3 re impossile for the first -node. Thus we delete them from its mrkings. In stge 3, for eh omintion of the positions, we onstrut n ACDP. A omintion of position is mde of position for eh node in P 1. In this emple, the omintions of positions re (0,1,5,6), (0,2,5,6),..., (0,5,5,6). We use (0,5,5,6) to eplin the onstrution proess. Initill P = P 2 nd ever node in P is nnotted with 2. Position 0 points to the root of P 2, therefore we nnotte rt(p 1 ) with 1. Position 5 points to //-edge, therefore we insert n -node, 0, in this position nd nnotte this node with 1. The edge (, 0 ) nd ( 0, ) re to e of tpe // in this se. Now position 5 points to the edge ( 0, ). Sine the position for the net node is lso 5, we insert nother -node, 1, etween 0 nd, nd nnotte 1 with 1. The lst node of P 1 hs position 6, so we nnotte the -node in P 2 with 1. Sine 1 is nnotted with 1, we n hnge the the edge tpe of ( 1, ) to tht of the orresponding edge in P 1, in this se, /. The resulting P is ACDP: it is 1,2 // 2 // 2 // 1 // 1 / 1,2, where the supersripts indite the nnottion. The ACDPs onstruted using other omintions re: 1,2 // 1 // 2 // 2 // 1 / 1,2, 1,2 // 1,2 // 2 // 1 / 1,2, 1,2 // 2 // 1 // 2 // 1 / 1,2, nd 1,2 // 2 // 1,2 // 1 / 1,2. Numer of ACDPs. Let n nd m e the numer of edges in P i nd P j respetivel, nd f(m, n) e the worst-se numer of ACDPs of P i nd P j (whih ours when ll nodes in P i nd P j hve the sme lel, nd ll edges re //-edges). The following theorem n e proved using indution on m nd n. 8

10 Algorithm 2 Find Position(v, P i, s, P j) 1: if s is node u in P j then 2: if (prent(v), v) is //-edge then 3: let A onsist of ll //-edges fter u nd ll nodes leled lel(v) fter u. 4: else if (prent(v), v) is /-edge then 5: if u hs hild u then 6: if (u, u ) is //-edge then dd this //-edge to A 7: if lel(u ) = lel(v) then dd u to A 8: if s is //-edge (u 1, u 2 ) in P j then 9: dd s to A 10: if (prent(v), v) is //-edge then 11: dd ll //-edges fter u 2 nd ll nodes leled lel(v) fter u 1 to A 12: if (prent(v), v) is /-edge nd lel(u 2 ) = lel(v) then dd u 2 to A 13: if v is the lst node in P i then 14: delete from A ll positions eept tht of DN Pj 15: return A 0 1 {0} 2 (0:{1, 2, 3, 4, 5}) 3 4 (1:{1, 3, 5}), (2:{3, 5}), (3:{3, 5}), (4:{5}), (5:{5}) 5 6 (1:{} ), (3:{}), (5:{6}). p 2 p 1 Fig. 4. Finding ADCPs of P 1 nd P 2 Theorem 1. f(m, n) = f(n, m), whih n e lulted reursivel s follows: f(m, n) = f(m 1, n) + 2(f(m 1, n 1) + f(m 1, n 2) + + f(m 1, 1)). For emple, f(m, 1) = 1, f(2, 2) = 3, f(3, 2) = 5, f(3, 3) = 13, f(4, 3) = 25, f(4, 4) = 63 nd so on. Thus f(m, n) grows eponentill in generl. However, in most prtil ses, the numer of ACDPs is muh smller thn f(m, n). Compleit. Algorithm merge(p i, P j ) runs in O( P i P j 2 ), where P i is the numer of nodes in P i : funtion Find Position runs in O( P j ), the topdown sn visits eh node in P i one, nd for eh node in P i, the funtion Find Position is lled t most 2 P j times. The ottom-up sn nd the onstrution of ACDPs n e done in O( P i P j ). Let P e TP in P {//,[]}. For ever node v in P, we use P v to denote the supttern of P rooted t v. Let v e node in DP P, nd u e the hild of v on DP P (if u eists). We ll the supttern otined removing P u from P v the rnhing sutree t v. For emple, in Fig.5 (), the rnhing sutrees re indited the dotted ovl. Net we define merged TPs (MTPs) of P 1 nd P 2. Definition 3. Let P 1 nd P 2 e omprle TPs, nd P e n ACDP of P 1 nd P 2. Let δ i e the unique CM from P i to P tht mps ever node in P i to node in P nnotted with i. The merged TP (MTP) of P 1 nd P 2 wrt to P is the 9

11 z z () P 1 () P 2 z z () MTP Fig. 5. TPs P 1, P 2 nd the MTP of P 1, P 2 TP otined s follows: for ever node v in P i, dd the rnhing sutree t v i under δ i (v i ). Fig.5 () shows the onl MTP of the TPs in Figures 5 () nd (). The following theorem is strightforwrd. Theorem 2. Let P 1 nd P 2 e TPs in P {//,[]}. The union of ll MTPs of P 1 nd P 2 is equivlent to P 1 P 2. Note lso tht the suptterns of the MTPs rooted t the distinguished nodes re ll identil. 3.2 Finding MCRs using V 1 V 2 Let Q e the quer nd V 1 nd V 2 e omprle views. We ssume V 1 V 2 is not empt, nd V 1 V 2, V 2 V 1. Suppose V 1 V 2 is equivlent to V 1 V k. Clerl Q (V 1 V 2 ) Q if nd onl if Q V i Q for ll i [1, k]. In other words, Q is CR of Q using V 1 V 2 iff it is CR of Q using V i for ll i [1, k]. Therefore, to find the MCR of Q using V 1 V 2, we n find the MCR of Q using eh V i nd interset them. Tht is, () V 1 MCR(Q, V 1 V 2 ) = () V 2 () V 1 k i=1 (d) V 2 MCR(Q, V i ). z (e) Q Fig. 6. Finding MCR using intersetion Emple 4. Consider the views V 1, V 2, V 1, V 2 nd the quer Q in Fig.6. V 1 V 2 = V 1 V 2. We find the MCR of Q using V 1, whih is /z, nd the MCR of Q using V 2, whih is lso /z. Thus /z is the MCR of Q using V 1 V 2. 10

12 Algorithm 3 Finding equivlent rewriting 1: for i = 1 to n do 2: if eists node v suh tht DP v Q is isomorphi to pi then 3: if su v Q V i = Q then 4: if j [1, k], su v Q V j Q then 5: return su v Q 3.3 Finding ERs using V 1 V 2 Suppose Q hs n ER Q using V 1 V 2, nd V 1 V 2 = V 1 V k, tht is, Q (V 1 V k ) = Q. B Lemm 1, there eists i [1, k], suh tht Q V i = Q. Hene for ll j [1, k], Q V j Q V i, nd thus there is CM from Q V i to Q V j, so the length of DP V i nnot e longer thn tht of DP V j. Furthermore, DP Q V i is isomorphi to DP Q, nd the supttern of Q rooted t the node tht orresponds to DN V i, is n ER of Q using V i ( Lemm 4.8 of [14]). Using the properties ove, we provide heuristi lgorithm, Algorithm 3, for finding ERs using V 1 V 2. In the lgorithm, we ssume p 1,, p n (n k) re the shortest ACDPs of V 1 nd V 2, nd the orresponding MTPs re V 1,, V n 3. For n node v on DP Q, the pth from rt(q) to v is denoted DP v Q, nd the supttern rooted t v is denoted su v Q. The si ide of the lgorithm is s follows. For eh shortest ACDP p i, we first hek whether it is isomorphi to some DP v Q, if not, there is no ER using V i ; if es, we further hek whether suv Q is n ER using V i. If es, we further hek whether the etension of ll other MTPs of V 1 nd V 2 with su v Q re ontined in V i. If es, su v Q is returned s n ER of Q using V 1 V 2. Let u e node on the distinguished pth of V. We use V (u) to denote the pttern otined from V removing ll desendnts of u nd mking u the distinguished node, nd V to denote V (DN V ). It is es to prove the following theorem. Theorem 3. (1) Suppose V 1 V 2 = V 1 V n. If Q hs n ER using V 1 V 2, then there is i [1, n] suh tht (i) V i hs the shortest distinguished pth mong V 1,..., V n, (ii) for ll j [1, n], there eists u j in DP V j suh tht V (uj) j V i. (2) Algorithm 3 finds the ER if it eists. Emple 5. (1) Consider the views in Fig.6. Sine V 1 V 2 = V 1 V 2, DP V 1 is 2 suh shorter thn DP V 2, nd there is no node u on the distinguished pth of V tht V (u) 2 V 1, we know there is no ER of Q using V 1 V 2 for n Q. (2) Consider the two views V 1 nd V 2 in Fig.7. There re two MTPs V 1 nd V 2, nd V 2 V 1. Therefore, V 1 V 2 = V 1. For n TP Q, Q is n ER of Q using V 1 V 2 iff Q is n ER using V 1. 3 Note tht lthough there m e mn ACDPs of V 1 nd V 2, usull there re few shortest ones. 11

13 () V 1 () V 2 () V 1 (d) V 2 Fig. 7. Emple for illustrting ER using intersetion 4 Rewriting using other omintions of views We now define seond tpe of rewritings of Q using multiple views tht does not require the views to e omprle. Definition 4. Let V 1,..., V n e some views with identil root lel (possil V i = V j for some i, j), nd Q e quer. If there re TPs Q 1,..., Q n suh tht n i=1 (Q i V i ) Q, nd n i=1 (Q i V i ) is non-empt, then we s Q 1,..., Q n is ontined rewriting of Q using V 1,..., V n. If n i=1 (Q i V i ) Q lso holds, we ll the CR n equivlent rewriting. Intuitivel, if there is ontined rewriting n i=1 (Q i V i ), then to prtill nswer Q over t, we n evlute Q i over V i (t) nd then find the intersetion n i=1 Q i(v i (t)). Note tht, when n = 1, the definition redues to tht in [6] (CR nd MCR) nd [14]. 4.1 Reltionship etween rewritings using V 1, V 2 nd rewritings using V 1 V 2 First, it is es to prove the following lemm. Lemm 3. Let V 1 nd V 2 e omprle views. Then Q (V 1 V 2 ) (Q V 1 ) (Q V 2 ). However, generll Q (V 1 V 2 ) (Q V 1 ) (Q V 2 ). Consider V 1 = /// nd V 2 = ///, nd Q = //z. Clerl V 1 V 2 =, hene Q (V 1 V 2 ) =. However, (Q V 1 ) (Q V 1 ). This emple lso shows tht sometimes even if V 1 V 2 is empt, it is still possile to hve CRs of Q using V 1, V 2, lthough there re lerl no CRs of Q using V 1 V 2. The net lemm identifies some speil ses where Q (V 1 V 2 ) = (Q V 1 ) (Q V 2 ). Lemm 4. Let V 1 nd V 2 e omprle views. If one of the following onditions holds, then Q (V 1 V 2 ) = (Q V 1 ) (Q V 2 ). (1) DP V1 = DP V2, nd ll edges in DP V1 re /-edges. (2) Ever edge in DP Q is /-edge. (3) V 1 V 2 or V 2 V 1. 12

14 Using the ove lemms, we n prove the following theorem: Theorem 4. Let V 1, V 2 e omprle views. For n quer Q, if there is CR of Q using V 1 V 2, then there is CR of Q using V 1, V 2. Note tht the ove theorem does not s there is CR using V 1, V 2 whih ontins the CR using V 1 V 2. The net emple shows tht it is possile for Q to hve CR using V 1 V 2, nd this rewriting is not ontined in n CRs of Q using V 1, V 2. Emple 6. Let V 1 = ////z, V 2 = ///z, nd Q = /////z//. It n e verified tht V 1 V 2 = /////z. Now let Q = z//, then Q (V 1 V 2 ) = Q. Thus Q is n ER of Q using V 1 V 2. If there re Q 1 nd Q 2 suh tht (Q 1 V 1 ) (Q 2 V 2 ) Q, then DP Q1 must e z/ or z//, nd DP Q2 must e z/ or z//, euse rt(q i ) must e leled z nd DN Qi must e leled. One n verif tht if DP Q2 = z//, then (Q 1 V 1 ) (Q 2 V 2 ) is not ontined in Q, nd if DP Q2 = z/, then (Q 1 V 1 ) (Q 2 V 2 ) is not equivlent to Q either. 5 The presene of non-reursive dtds In the following, we ssume ever TP P is stisfile under non-reursive dtd G, tht is, there is n ml tree t whih onforms to G, nd P (t). In the presene of G, no lel in n XML tree n pper in pth more thn one. Thus n TP tht is stisfile under G nnot hve pth tht ontins two or more nodes with the sme lel. Therefore, for n omprle views V 1 nd V 2, there is t most one ACDP of V 1 nd V 2, nd t most one MTP of V 1 nd V 2. In other words, V 1 V 2 is equivlent to single TP V under G. Therefore, to find the MCR or ER of Q using V 1 V 2, we onl need find the MCR or ER of Q using V, nd this n e done using the method of [6]. Furthermore, we n prove (see full version of this pper) the following theorem, whih implies tht, in the presene of G, Q is CR using V 1 V 2 iff Q, Q is CR of Q using V 1, V 2. Theorem 5. Let V 1 nd V 2 e omprle views, nd lel(rt(q )) = lel(dn V1 ). In the presene of G, Q (V 1 V 2 ) is equivlent to (Q V 1 ) (Q V 2 ). 6 More relted work Besides the reent works disussed in Setion 1, severl other ppers delt with tree pttern quer rewriting. In prtiulr, [10] studied the prolem of quer nswerilit using views for generl XPth queries, tht is, given Q nd V 1,, V n, whether there re Q 1,, Q n suh tht Q 1 V 1 Q n V n = Q. [3] ddressed the prolem of nswering XPth queries using single mterilized view where, for the view, omintion of node referenes, tped dt vlues, nd full pths m e stored. However, the w in whih quer is nswered using the view 13

15 is different from ours (nd those in Setion 1): one n follow node referenes to go to the originl doument, so the originl ml tree nnot e disrded. [2] studied different tpe of equivlent rewriting using multiple views in the presene of struturl summries nd integrit onstrints: the nswer sets of the views re nodes rther thn sutrees, nd the nswers to the new quer re otined omining nswers to the views through numer of lgeri opertions. [11] studied orret rewritings of TPs, using single view, whih n e seen s speil form of ontined rewritings. [4] ttempted to speed-up the finding of MCRs using single views omining the views into single tree. [12] studied equivlentl nswering XPth queries using multiple views sed on the ssumption tht the Dewe odes re stored in the mterilized views so tht the ommon nestors of nodes in different views n e found. Our work is lerl different from ll of the ove. 7 Conlusion We studied the prolem of rewriting TP queries using multiple views for the lss P {//,[]}, nd defined rewritings using two different omintions of views. We studied the reltionship etween the two tpes of rewritings nd presented effiient lgorithms to reformulte the intersetion of TPs into union of TPs, s well s lgorithms for finding the MCRs nd ERs using intersetions of views. Our definitions nd lgorithms enle us to mke etter use of the views in order to nswer quer. Aknowledgement This work is prtill supported Griffith Universit New Reserher s Grnt (GUNRG36621) nd grnt from the Reserh Grnt Counil of the Hong Kong Speil Administrtive Region, Chin (CUHK418205). The uthors re grteful for helpful omments Professor Rodne Topor. Referenes 1. S. Amer-Yhi, S. Cho, L. V. S. Lkshmnn, nd D. Srivstv. Minimiztion of tree pttern queries. In SIGMOD, A. Arion, V. Benzken, I. Mnolesu, nd Y. Ppkonstntinou. Strutured mterilized views for XML queries. In VLDB, A. Blmin, F. Özn, K. S. Beer, R. Cohrne, nd H. Pirhesh. A frmework for using mterilized XPth views in XML quer proessing. In VLDB, J. Go, T. Wng, nd D. Yng. MQTree sed quer rewriting over multiple XML views. In DEXA, A. Y. Hlev. Answering queries using views: A surve. VLDB J., 10(4), L. V. S. Lkshmnn, H. Wng, nd Z. J. Zho. Answering tree pttern queries using views. In VLDB, B. Mndhni nd D. Suiu. Quer hing nd view seletion for XML dtses. In VLDB, G. Miklu nd D. Suiu. Continment nd equivlene for n XPth frgment. In PODS,

16 9. N. Onose, A. Deutsh, Y. Ppkonstntinou, nd E. Curtmol. Rewriting nested XML queries using nested views. In SIGMOD, K. Tjim nd Y. Fukui. Answering XPth queries over networks sending miniml views. In VLDB, J. Tng nd S. Zhou. A theoreti frmework for nswering XPth queries using views. In XSm, N. Tng, J. X. Yu, M. T. Özsu, B. Choi, nd K.-F. Wong. Multiple mterilized view seletion for pth quer rewriting. In ICDE, J. Wng, J. X. Yu, nd C. Liu. On tree pttern rewriting using views. In WISE, W. Xu nd Z. M. Özsooglu. Rewriting XPth queries using mterilized views. In VLDB,