Finding Compact Scheme Forests in. P. Thanisch. Department of Computer Science, University of Edinburgh,

Transcription

1 Finding Compact Scheme Forests in Nested Normal Form is NP-Hard P. Thanisch Department of Computer Science, University of Edinburgh, King's Buildings, Mayeld Road, Edinburgh EH9 3JZ, Scotland G. Loizou Department of Computer Science, Birkbeck College, University of London, Malet Street, London WC1E 7HX, England J. Nummenmaa Department of Computer Science, University of Edinburgh, King's Buildings, Mayeld Road, Edinburgh EH9 3JZ, Scotland August 29, 1997 On leave, funded by a Royal Society grant, from University of Tampere, Department of Computer Science, P.O. Box 607, Tampere, Finland 1

2 Running Head: Compact Nested Scheme Forests Proofs to be sent to: Dr. Peter Thanisch, Department of Computer Science, University of Edinburgh, King's Buildings, Mayeld Road, Edinburgh EH9 3JZ, Scotland. 2

3 Abstract In traditional relational databases, the data is stored in \at" tables. Query processing performance is dominated by the cost of joining such tables. By contrast, nested relational structures can avoid joins. If, however, such structures are decomposed into nested normal form (NNF) then the number of normal scheme trees in the resulting nested scheme forest may dominate query processing performance. Thus minimizing the number of such trees is an important design goal. We prove that the problem of nding a succinct NNF scheme forest is NP-hard even for the class of sets of unary multivalued dependencies, which is a subclass of the class of split-free sets of multivalued dependencies. 3

4 List of Symbols Used S set theory union T set theory intersection 2 set theory \element of" set theory proper subset set theory subset? Greek upper-case Gamma Greek upper-case Phi Greek lower-case pi Greek upper-case Psi Greek upper-case Theta P Boldface P R Boldface R E \calligraphic" E G \calligraphic" G T \calligraphic" T!! double arrow./ bow tie 4

5 1. INTRODUCTION Nested relational structures (Makinouchi, 1977; Van Gucht and Fischer, 1988) are a generalization of traditional relational databases in which the tuple components are allowed to be repeating group instances or even nested relations. Three normal forms have been proposed for the logical design of databases made up of such structures. Two of theses normal forms, \hierarchical" normal form (Benecke, 1987) and \partitioned" normal form (Roth et al., 1988), are with respect to special versions of the class of sets of functional dependencies that have been generalized to nested relations. In this paper, we examine Nested Normal Form (NNF) ( Ozsoyoglu and Yuan, 1987a), which takes into account both functional dependencies and multivalued dependencies (MVDs) ( Ozsoyoglu and Yuan, 1989). A functional dependency captures the semantics of a one-to-one, or many-to-one, relationship, whereas an MVD captures the semantics of a one-to-many relationship associated with repeating groups. We concentrate on MVDs, though the results generalize to the case where the set of data dependencies is a mixture of functional dependencies and MVDs. One of the advantages claimed for nested, as opposed to at, relational databases is that query processing can be more ecient, as computationally expensive join operations may be avoided (Abiteboul and Bidoit, 1986). The process of normalizing a nested relational structure involves nding a nested scheme forest composed of scheme trees. Algorithms for nding nested scheme forests work by splitting a scheme tree into two scheme trees in order to eliminate some redundancy. Given that the nested scheme forest must be in NNF, minimizing the number of scheme trees is important, as the number of join operations required in order to answer a query is related to the number of scheme trees involved in the expression of the query. Herein, we prove that the following problem is NP-hard even for the class of split-free sets of MVDs (Lien, 1982). Given a set of MVDs does there exist a NNF scheme forest with no more than k scheme trees? In fact, the problem is NP-complete, though we do not provide a proof of NPcompleteness in this paper. In Section 2, we introduce some necessary terminology from database theory. In Section 3, we give the necessary denitions from graph theory and we establish some properties of the class of graphical sets of MVDs. In Section 4, we describe the necessary concepts from the theory of nested relations. In Section 5, we prove various properties of scheme trees and forests. In Section 6, we give our NP-hardness proof for the problem SNNF (see Denition 6.2). The proofs of the technical lemmas are not particularly instructive. Thus they appear in an Appendix in order to improve the readability of the rest of the paper. Although we have included some examples in the text, the reader 5

6 is advised to refer to Ozsoyoglu and Yuan (1987a) for further background and examples. 6

7 2. TERMINOLOGY Let U = fa 1 ; A 2 ; : : : ; A n g be the universal set of attributes (abbreviated to universe). A relation scheme R is a subset of U. Corresponding to each attribute A i is a set DOM(A i ), i = 1; 2; : : : ; n, called the domain of A i. The domains are arbitrary, non-empty sets, nite or countably innite. Let DOM = S n DOM(A i=1 i). A relation I on the relation scheme R is a nite set of mappings (tuples) ft 1 ; t 2 ; : : : ; t p g from R to DOM with the restriction that for each mapping t 2 I, t(a i ) 2 DOM(A i ), i = 1; 2; : : : ; n. A database scheme for U is a collection of subsets of U, the union of which is U. Let I be a relation on the relation scheme, R, and let X R. The projection of I onto X, written X (I), is the relation obtained from I by striking out columns corresponding to attributes in R? X and removing duplicate tuples in what remains. In mapping notation, X (I) = ft(x) j t 2 Ig; t(x) is also known as the X-value of t. Let I and J be two relations over, respectively, the relation schemes R and S. The join of I and J, written I./ J, is the relation H over R S S of all tuples t over R S S such that there are tuples t I 2 I and t J 2 J with t I = t(r) and t J = t(s). Since R T S is a subset of both R and S, as a consequence of the denition t I (R T S) = t J (R T S). Thus every tuple in I./ J is a combination of a tuple from I and a tuple from J with equal (R T S)-values. Let H be a relation on R S S, with I = R (H) and J = S (H). Let H 0 = I./ J. If t is a tuple of H, then t(r) is in I and t(s) is in J, so t is also in H 0. Therefore, H H 0. If H = H 0, we say relation H decomposes losslessly onto relation schemes R and S. More generally, let R = fr 1 ; R 2 ; : : : ; R c g be a database scheme and let S c R i=1 i = U. A relation I, over U, has a lossless decomposition onto R if I = R1 (I)./ R2 (I)./ : : :./ Rc (I) : Let R = fr 1 ; R 2 ; : : : ; R c g be a database scheme. A relation I, over U, satises the join dependency (JD)./ [fr 1 ; R 2 ; : : : ; R c g] if I decomposes losslessly onto R 1 ; R 2 ; : : : ; R c. That is, I = R1 (I)./ R2 (I)./ : : :./ Rc (I). A multivalued dependency (MVD) is a special case of a JD, where c = 2. T Given./ [fr 1 ; R 2 g], let X = R 1 R2, Y = R 1? R 2 and Z = R 2? R 1. We use the notation X!! Y or, alternatively, X!! Y j Z, to denote this MVD. We refer to X as the left-hand side (l.h.s.) set of attributes and to Y as the right-hand side (r.h.s.) set of attributes. Let X S Y W U. We write X!! Y [W ] to mean X!! Y in the context W and, in general, we omit \[W ]" whenever the context is understood. In the sequel, where X and Y are sets of attributes, on occasion (as is now customary in database theory), we omit the union operator and write \XY " instead of \X S Y ". 7

8 Given a set,, of MVDs, we let + denote the set of all MVDs that can be derived from by using a sound and complete set of MVD axioms (or inference rules). The following set of MVD axioms has been shown to be sound and complete (Ullman, 1988): MVD0 (complementation): X!! Y ` X!! Z, if XY Z = U and Y T Z X. MVD1 (reexivity): ` X!! Y, if Y X. MVD2 (augmentation): X!! Y ` (XW )!! (Y Z), if Z W. MVD3 (transitivity): X!! Y ; Y!! Z ` X!! (Z? Y ). MVD4 (disjoining): X!! Y ` X!! (Y? X). Although the above set of inference rules is sound and complete, the rule MVD4 is redundant in the sense that its validity follows from the other axioms. Let X U. We write DEP (X; ) to denote the dependency basis of X with respect to (Ullman, 1988), i.e. DEP (X; ) = fw U j X!! W 2 + and 8V W ; X!! V 62 + g: Let R U. We write P ROJ( + ; R) to denote the projection of + onto R. Let X R and X!! Y 2 +. Then X!! Y T R [R] 2 P ROJ( + ; R). The following denition describes four of the ways in which an MVD can be said to contain some redundant information ( Ozsoyoglu and Yuan, 1987a). DEFINITION 2.1. Given a set of MVDs, X!! W 2 + is said to be (1) trivial if XW = U or W X; (2) left-reducible if 9X 0, X 0 X, such that X 0!! W is in + ; (3) right-reducible if 9W 0, W 0 W, such that X!! W 0 is a nontrivial MVD in + ; (4) transferable if 9X 0, X 0 X, such that X 0!! W (X? X 0 ) is in +. An MVD X!! W is said to be reduced if it is nontrivial, left-reduced (that is, nonleft-reducible), right-reduced (that is, nonright-reducible) and nontransferable. We let REDUCED( ) denote the set of all reduced MVDs in +. DEFINITION 2.2. A set of MVDs is said to be a minimal cover if 8

9 (1) REDUCED( ), and (2) if? then? + +. The denition of NNF in Ozsoyoglu and Yuan (1987a) presupposes that the set of MVDs is a minimal cover. 3. GRAPHS AND GRAPHICAL SETS OF MVDS An undirected graph is a pair, G = (V; E), where V is a set of vertices and E is a set of edges such that if (x; y) 2 E then x 2 V, y 2 V and x 6= y. We assume that each edge in the graph has a unique label. EXAMPLE 3.1. Consider the following undirected graph, G. The numbers inside the nodes represent the vertex names and the letters represent the edge labels. Thus V = f1; 2; 3; 4; 5; 6; 7g and E = f(1; 2); (1; 3); (1; 4); (2; 5); (3; 6); (4; 7)g. The set of labels of the edges is L = fa; B; C; D; E; F g. In Fig. 1, the label A is associated with the edge (1; 2), B with the edge (1; 3), etc F E A 2 4 P PP C D B FIG. 1. The graph, G, with vertices and edges labelled. DEFINITION 3.1. A vertex cover, say W, is a subset of the set, V, of vertices with the property that for each edge, (x; y) 2 E, fx; yg T W 6= ;. EXAMPLE 3.2. Referring to the graph, G, in Fig. 1, the reader may verify that the set f2; 3; 4g is a vertex cover for G. 2 DEFINITION 3.2. dened as follows. The Vertex Cover problem, abbreviated to VC, is Instance: (1) A graph, G = (V; E), and (2) a bound, b, 0 b j V j. Question: Is there a vertex cover of size b or less for G? VC is known to be NP-complete (Garey and Johnson, 1979). 9

10 EXAMPLE 3.3. Referring, again, to the graph, G, in Fig. 1, if the instance of VC has the bound b = 3, then the answer to the decision problem is \yes", since the vertex cover f2; 3; 4g from Example 3.2 has 3 vertices. 2 We now dene the class of graphical sets of MVDs, so called because it is isomorphic to the class of undirected graphs. DEFINITION 3.3. Given a graph, G = (V; E), for each edge (x; y) 2 E, let l xy denote a unique label for the edge, L = fl xy j (x; y) 2 Eg. The graphical set of MVDs, say, corresponding to G has a universe, U = V S L, and for each edge (x; y) in E, where l labels (x; y), fxg!! flg 2 and fyg!! flg 2. EXAMPLE 3.4. Given the graph in Fig. 1, the corresponding universe, U, and graphical set of MVDs,, are given by U = f1; 2; : : : ; 7; A; B; : : : ; F g and = f f1g!! fag, f1g!! fbg, f1g!! fcg, f2g!! fag, f2g!! fdg, f3g!! fbg, f3g!! feg, f4g!! fcg, f4g!! ff g, f5g!! fdg, f6g!! feg, f7g!! ff g g. 2 Let = f X 1!! Y 1 ; X 2!! Y 2 ; : : : ; X k!! Y k g be a graphical set of MVDs for a universe, U. We let LEF T ( ) = k[ i=1 k[ X i and RIGHT ( ) = Y i : i=1 Furthermore, LHS( ) = fx j 9Y such that X!! Y 2 g and RHS( ) = fy j 9X such that X!! Y 2 g: We note that LHS( ) T RHS( ) = ; and that flef T ( ); RIGHT ( )g partitions U. EXAMPLE 3.5. Given U and as in Example 3.4, LEF T ( ) = f1; 2; : : : ; 7g, RIGHT () = fa; B; : : : ; F g, LHS() = ff1g; f2g; : : : ; f7gg and RHS() = ffag; fbg; : : : ; ff gg. 2 The following result, the proof of which can be found in the Appendix, gives a necessary and sucient condition for the lossless join property to hold in the presence of a graphical set of MVDs. 10

11 LEMMA 1. Let be a graphical set of MVDs and let R = fr 1 ; : : : ; R c g be a collection of sets of attributes. Then j=./ [R] if and only if (1) 9R i 2 R such that LEF T ( ) R i and (2) 8Y 2 RHS( ), where X 1 ; X 2 2 LHS( ) such that X 1 6= X 2 and X 1!! Y ; X 2!! Y 2, 9R j 2 R such that X 1 Y R j or X 2 Y R j. 2 Next we dene the set of keys of. DEFINITION 3.4. Let be a set of MVDs. By using the denition of REDUCED( ), the set of keys of, denoted MKEY S( ), is dened by: MKEY S( ) = fx U j 9W such that X!! W 2 REDUCED( )g: The results and denitions concerning NNF in Ozsoyoglu and Yuan (1987a) are based on the assumption that is a minimal cover (see Denition 2.2). Part (a) of the the next lemma proves that this is not a problem where is graphical. LEMMA 2. Let be a graphical set of MVDs. (a) is a minimal cover. (b) For any X LEF T ( ), 9Y 2 DEP (X; ) such that LEF T ( )? X Y. (c) For each V 2 MKEY S( ); V LEF T ( ). 2 DEFINITION 3.5. Let be a set of MVDs that is a minimal cover and let V U. The set of fundamental multikeys ( Ozsoyoglu and Yuan 1987a) on V, denoted F K(V; ), is dened to be the set: F K(V; ) = f V \ X j X 2 LHS( ) ; V \ X 6= ; and 6 9Y 2 LHS( ) such that ; 6= Y \ V X \ V g : For graphical sets of MVDs, we can use a simpler denition, exploiting the fact that, for all X 2 LHS( ), j X j = 1. We note that if is a graphical set of MVDs, then F K(V; ) = fx 2 LHS( ) j X V g : 4. NESTED RELATIONS We shall restrict our attention to a special class of nested relations, namely hierarchical nested relations (Van Gucht and Fischer, 1988), where the nesting 11

12 structure is a kind of tree. Let U be a universe. A scheme tree, say T, is a rooted tree with vertices labelled by subsets of U. We use the following notations for scheme trees. AT T (n) is a label associated with the node, n, of T. AT T (n) U. ROOT (T ) is the root node of T. P ARENT (n) is the parent of the (non-root) node n in T. LEAV ES(n) is the set of leaf nodes in the subtree rooted at node n. KIDS(n) is the set of nodes fk j n = P ARENT (k)g in T. The nested relation scheme represented by T, denoted N RS(T ), is dened as follows: (1) If T comprises a single node, say r, then NRS(T ) = AT T (r). (2) Let r = ROOT (T ), AT T (r) = V and KIDS(r) = fv 1 ; v 2 ; : : : ; v h g. Let v i = ROOT (T i ), i = 1; 2; : : : ; h, where T 1 ; T 2 ; : : : ; T h denote the subtrees rooted at the nodes in KIDS(r). Then NRS(T ) = V (NRS(T 1 )) : : : (NRS(T h )) : In Section 2, we denoted the domain of an attribute, say A, by DOM(A). We used that denition in a set-of-mappings denition of a at relation. We now extend this notion in order to dene the domain of a nested relation scheme. This will be used in a set-of-tuples denition of a nested relation. The domain of NRS(T ), denoted DOM(NRS(T )), is dened recursively as follows: (1) If T comprises a single node, say r, such that AT T (r) = fa 1 ; : : : ; A s g, then DOM(NRS(T )) = DOM(A 1 ) : : : DOM(A s ). (2) Let r = ROOT (T ), AT T (r) = V and let T 1 ; T 2 ; : : : ; T h denote the subtrees rooted at the nodes in KIDS(r). Then DOM(NRS(T )) = DOM(V ) P(DOM(NRS(T 1 ))) P(DOM(NRS(T h ))); where P(D) denotes the power set of the set D. A nested relation, over a nested relation scheme NRS(T ), is a subset of DOM(NRS(T )). 12

13 Let U be a universe, T a scheme tree and u and v nodes in T. Throughout the paper, we use the following notations: ANC(v) = S fat T (u) j u is an ancestor of v in T g. DESC(v) = S fat T (u) j u is a descendant of v in T g. AT T (T ) = DESC(ROOT (T )). MV D(v) = ANC(P ARENT (v))!! DESC(v) [AT T (T )]; v 6= ROOT (T ). MV D(T ) = fmv D(v) j v 6= ROOT (T )g. We note that, for a leaf node, u, ANC(u) is the union of all the sets of attributes labelling scheme tree nodes in the path from the root node, r, to u in T. Let T be a scheme tree, r = ROOT (T ) and let LEAV ES(r) = fu 1 ; u 2 ; : : : ; u m g. Then the path set of T, denoted P AT HS(T ), is dened as follows: P AT HS(T ) = fanc(u 1 ); ANC(u 2 ); : : : ; ANC(u m )g: 5. SEMI-NORMAL AND NORMAL SCHEME TREES A scheme tree is semi-normal with respect to a set,, of MVDs if it does not contain any partial redundancies. We dene this notion, below, in Denition 5.1, in which we let be a set of MVDs for a universe, U, T a scheme tree, v a non-root node in T, u = P ARENT (v) and Z 2 MKEY S( ) such that Z!! DESC(v) [AT T (T )] 2 P ROJ( + ; AT T (T )): DEFINITION 5.1. A node v is partial redundant with respect to Z if Z ANC(u) and there exists Y 2 DEP (Z; ) such that DESC(v) = Y T AT T (T ). The MVD Z!! DESC(v) [AT T (T )] is called a partial dependency in T. We now dene a semi-normal scheme tree ( Ozsoyoglu and Yuan 1987a). DEFINITION 5.2. A scheme tree, T, is semi-normal with respect to if: (1) MV D(T ) P ROJ( + ; AT T (T )). (2) For each edge, (u; v), of T, ANC(u)!! DESC(v) [AT T (T )] is left- and right- reduced. (Thus there are no partial dependencies in T.) 13

14 (3) AT T (ROOT (T )) 2 MKEY S( ) and for each non-root node, v, in T, if F K(DESC(v); ) 6= ; then AT T (v) 2 F K(DESC(v); ). EXAMPLE 5.1. Consider the scheme tree T in Fig. 2 (a). Let U = f1; 2; Ag, and = f f1g!! f2g ; f2g!! fag g. Then node w is partial redundant in T with respect to f1g. 2 u v w AT T (u) = f1g AT T (v) = f2g AT T (w) = fag (a) AT T (y) = fbg AT T (v 1 ) = feg x % % % y u % % % v 1 v (b) AT T (x) = fag AT T (u) = fcg AT T (v) = fdg FIG. 2. Two example scheme trees. DEFINITION 5.3. Let T be a scheme tree that is semi-normal with respect to a graphical set, say, of MVDs. Let r = ROOT (T ). A simple path in T is a path of length one, starting at node r and ending at a node, w, such that w 2 LEAV ES(T ), w 2 KIDS(r) and AT T (w) 2 RHS( ). A complex path in T is a path from r to a leaf node of T that is not a simple path. A simple leaf node is a leaf node at the end of a simple path. A complex leaf node is a leaf node at the end of a complex path. The dierence between a semi-normal scheme tree and a normal scheme tree (with respect to a given set of MVDs) is that a semi-normal scheme tree may contain a node that is transitive redundant ( Ozsoyoglu and Yuan, 1987a). We now dene this notion. DEFINITION 5.4. Let T be a scheme tree and a set of MVDs for a universe, U. Let (u; v) be any edge of T with fv 1 ; v 2 ; : : : ; v s g KIDS(u)? fvg and W = S s i=1 i). The node v is transitive redundant in T with respect to Z if Z 2 MKEY S( ) such that (1) Z ANC(u)W. (2) ZW!! ANC(u) 62 P ROJ( + ; AT T (T )). (3) Z!! DESC(v) 2 P ROJ( + ; AT T (T )). 14

15 In this case, Z!! DESC(v) [AT T (T )] is a transitive dependency in T. EXAMPLE 5.2. Consider the scheme tree T in Fig. 2 (b). Let U = fa; B; C; D; Eg and = f fag!! fbg, fa; Cg!! feg ; feg!! fdg g. The reader may verify that T is semi-normal with respect to. We note that ANC(u) = fa; Cg and DESC(v) = fdg. The node v is transitive redundant with respect to the key feg. This is because feg 2 MKEY S( ), feg fa; C; Eg, feg!! fa; Cg 62 P ROJ( + ; AT T (T )) and feg!! fdg 2 P ROJ( + ; AT T (T )). 2 DEFINITION 5.5. Let T be a scheme tree and a set of MVDs. T is normal with respect to if: (1) T is semi-normal with respect to. (2) For each pair, say v and w, of distinct nodes in T, AT T (v) T AT T (w) = ;. (3) For each node, u, in T, there is no key X of such that u is transitive redundant with respect to X. (Thus there are no transitive dependencies in T.) The following lemma establishes some properties of a scheme tree that is normal with respect to a graphical set of MVDs. LEMMA 3. Let T be a scheme tree that is normal with respect to a graphical set of MVDs,. Then the following properties hold for T. (a) For each non-root node, say v, in T, if AT T (v) 62 LHS( ) then (1) AT T (v) RIGHT ( ) and (2) v 2 LEAV ES(T ) : (b) For each non-root node, v, in T, j KIDS(v) j 1. (c) There is at most one complex path in T. (d) If j LEAV ES(T ) j 2 then AT T (ROOT (T )) 2 LHS( ). (e) If LEF T ( ) AT T (T ) then T contains no transitive redundant nodes NESTED SCHEME FORESTS DEFINITION 6.1. ft 1 ; T 2 ; : : : ; T p g such that A nested scheme forest, say F, is a set of scheme trees (1) S p i=1 AT T (T i ) = U. (2) j=./ [fat T (T 1 ) ; AT T (T 2 ) ; : : : ; AT T (T p )g]. 15

16 We say that F is in NNF with respect to if it is a nested scheme forest in which each scheme tree is normal with respect to. LEMMA 4. Let F be a nested scheme forest that is in NNF with respect to a graphical set of MVDs,. Then there exists a nested scheme forest, say F 0, that is in NNF with respect to and such that j F 0 j = j F j, and for each T 2 F 0, AT T (ROOT (T )) 2 LHS( ). 2 We assume that the database designer will not be content with just any normalized database scheme. Consequently, we investigate the problem of determining whether or not there exists a succinct nested scheme forest that is in NNF with respect to a set,, of MVDs. We prove that the following problem is NP-hard for graphical sets of MVDs. DEFINITION 6.2. The problem Succinct Nested Normal Form, abbreviated to SNNF, is dened as follows. Instance: A universe, U, a set,, of MVDs and an integer bound, c. Question: Does there exist a nested scheme forest, say F, such that j F j c and F is in NNF with respect to? THEOREM 1. sets of MVDs. Succinct Nested Normal Form is NP-hard for graphical Proof. We prove that SNNF is NP-hard by showing that an arbitrary instance of VC, see Denition 3.2, can be transformed into an equivalent instance of SNNF. That is, we prove that, for any graph, G, with corresponding graphical set of MVDs,, there exists a vertex cover for G with no more than c nodes if and only if there exists a nested scheme forest, say F, such that j F j c and F is in NNF with respect to. [IF] Suppose that there is a vertex cover, say W, for G such that j W j c. We prove that there is a nested scheme forest, which is in NNF with respect to, and which contains no more than c normal scheme trees. Without loss of generality, we may assume that j W j = c, since, from Denition 3.1, a set of vertices that is a superset of a vertex cover is itself a vertex cover. Let W = fa 1 ; A 2 ; : : : ; A c g; obviously, W LEF T ( ). For each A i, i = 1; 2; : : : ; c? 1, we dene a scheme tree, T i, as follows. We let r i denote ROOT (T i ) and AT T (r i ) = fa i g. For each Y in the collection DEP (fa i g; ) T RHS( ), we have an element v 2 KIDS(r i ) with AT T (v) = Y. This completes T i, i.e. KIDS(r i ) = LEAV ES(r i ). 16

17 We dene T c as follows. ROOT (T c ) = r c, AT T (r c ) = fa c g and for each Y 2 DEP (fa c g; ) T RHS( ) 9v 2 KIDS(r c ) such that AT T (v) = Y and v 2 LEAV ES(r c ). Let LHS( )? fa c g = fx 1 ; X 2 ; : : : ; X g g. Then we dene the nodes v 1 ; v 2 ; : : : ; v g in T c, such that KIDS(v i ) = fv i+1 g; i = 1; 2; : : : ; g? 1, AT T (v i ) = X i ; i = 1; 2; : : : ; g and P ARENT (v 1 ) = r c. We prove that the nested scheme forest F = ft 1 ; T 2 ; : : : ; T c g is in NNF with respect to. We now refer to the two conditions in Denition 7.1. (1) Let S = fat T (T i ) j T i 2 Fg = fs 1 ; S 2 ; : : : ; S c g. Obviously, S S U. By the denition of a vertex cover, whereby each element of RIGHT ( ) corresponds to an edge label in the graph G, it follows that RIGHT ( ) S S. Now LEF T ( ) AT T (T c ) but flef T ( ); RIGHT ( )g partitions U, so U = S S. (2) By Lemma 1, j=./ [S]. Next, we prove that each scheme tree T i 2 F is semi-normal with respect to. We refer to the three conditions in Denition 5.2. (1) For i = 1; 2; : : : ; c? 1, AT T (ROOT (T i ))!! AT T (w) 2 +, for each w 2 KIDS(ROOT (T i )). Thus MV D(T i ) P ROJ( + ; AT T (T i )). Next, consider T c. By Lemma 2(b), for each X 2 LHS( ), 9Y 2 DEP (X; ) such that LEF T ( )?X Y. But LEF T ( ) AT T (T c ). Thus, for X = AT T (ROOT (T c )), X!! (LEF T ( )? X) 2 P ROJ( + ; AT T (T c )): By MVD2 and MVD4, for Z LEF T ( ), XZ!! (LEF T ( )? XZ) 2 P ROJ( + ; AT T (T c )): Let v 1 2 KIDS(r c ) such that AT T (v 1 ) LEF T ( ). For all v 2 KIDS(r c )? fv 1 g, by construction of T c we have v 2 LEAV ES(r c ) and AT T (v) 2 DEP (AT T (r c ); ). Thus AT T (r c )!! AT T (v) 2 P ROJ( + ; AT T (T )). It follows that MV D(T c ) P ROJ( + ; AT T (T c )). (2) Consider T i 2 F, i = 1; 2; : : : ; c. For each w 2 KIDS(ROOT (T i )), except for v 1 in T c, AT T (ROOT (T i ))!! DESC(w) 2. By Lemma 2(a), comprises only reduced MVDs. Next, consider T c. Let v be a non-root node in T c and let u = P ARENT (v). The MVD represented by the edge (u; v) is ANC(u)!! DESC(v), i.e. ANC(u)!! LEF T ( )?ANC(u). By Lemma 2(b), 9Y 2 DEP (ANC(u); ) such that LEF T ( )? ANC(u) Y. We have DESC(v) Y T AT T (T c ). Now let X ANC(u). Thus X!! DESC(v) 62 P ROJ( + ; AT T (T c )): Consequently, AN C(P AREN T (v))!! DESC(v) is left- and right- reduced 17

18 for each non-root node v 2 AT T (T c ). (3) For each T i 2 F, AT T (ROOT (T )) 2 LHS( ) and, by Lemma 2(a), is a minimal cover. Thus AT T (ROOT (T )) 2 MKEY S( ). For i = 1; 2; : : : ; c, each non-root node, say v, in T i has AT T (v) RIGHT ( ), so F K(DESC(v); ) = ;. In T c, for each node, v, AT T (v) 2 LHS( ), thus AT T (v) 2 F K(DESC(v); ). Thus each T i 2 F is semi-normal with respect to. Next we prove that each T i 2 F is normal with respect to. The above argument establishes the rst of the three conditions in Denition 5.5. The second condition of Denition 5.5, i.e. pairwise-disjoint node labels in each T i 2 F, follows from the construction of F. We now establish the third condition of Denition 5.5. By Lemma 2(c), for each X 2 MKEY S( ), X LEF T ( ). For 1 i < c, each non-root node, say v, in T i is labelled so that AT T (v) T LEF T ( ) = ;. Thus there is no transitive redundant node in T i. By Lemma 3(e), there are no transitive redundant nodes in T c either. This completes the proof that each T i 2 F is normal with respect to. Thus if there is a vertex cover for G comprising no more than c vertices then there exists a nested scheme forest that is in NNF with respect to and that it comprises no more than c normal scheme trees. [ONLY IF] Suppose that there exists a nested scheme forest which is in NNF with respect to, and which has no more than c normal scheme trees. By Lemma 4, there exists a nested scheme forest, say F, that is in NNF with respect to which has no more than c normal scheme trees and such that for each T 2 F, AT T (ROOT (T )) 2 LHS( ). Let F = ft 1 ; T 2 ; : : : ; T g g, g c. Also, let W = fa 2 U j 9T i 2 F such that AT T (ROOT (T i )) = fagg: We shall prove that W is a vertex cover for G. Suppose that W is not a vertex cover for G. Then there is some Y 2 RHS( ) such that, where X 1!! Y ; X 2!! Y 2, X 1 6= X 2 ; X 1 6 W and X 2 6 W : Now S g i=1 AT T (T i) = U, so 9T 2 F such that Y AT T (T ). Let w be the node in T such that Y AT T (w). By Lemma 3(a) and Denition 5.3, AT T (w) RIGHT ( ), w 2 LEAV ES(ROOT (T )) and Y = AT T (w). By hypothesis, X 1 6 W and X 2 6 W, so AT T (ROOT (T )) 6= X 1 and AT T (ROOT (T )) 6= X 2. Suppose that there is a node, say n, in T such that X 1 AT T (n) or X 2 AT T (n). In either case, AT T (n) 6= AT T (ROOT (T )), so n 6= ROOT (T ), thus n is a proper descendant of ROOT (T ) and a proper ancestor of w. Thus, by Denition 5.1, X 1!! AT T (w) or X 2!! AT T (w) is a partial dependency in T, yet T is a 18

19 T normal scheme tree, which leads to a contradiction. Thus X 1 X 2 AT T (T ) = ; for each T 2 F such that Y AT T (T ). Let R = fat T (T ) j T 2 Fg. Thus, by Lemma 1, 6j=./ [R]. However, by hypothesis, F is in NNF, so, by Denition 7.1, j=./ [R], which leads to a contradiction. Thus W is a vertex cover for G. 2 The class of graphical sets of MVDs is a subclass of the class of split-free (Lien 1982) sets of MVDs. Thus we have the following Corollary to Theorem 1. COROLLARY 1. Succinct Nested Normal Form is NP-hard for the class of split-free sets of MVDs. 2 ACKNOWLEDGEMENTS We thank the referees for their helpful suggestions. APPENDIX This Appendix contains the proofs of the technical lemmas used for our main result, namely Theorems 1. The Appendix also contains some additional denitions and certain Propositions needed to obtain the proofs. The Chase algorithm (see Ullman (1988) for further details and a bibliography) is used for determining whether or not a set of MVDs logically implies a JD. The Chase uses a tableau which is a two-dimensional array with j U j columns. There is a bijection between the set of tableau columns and the set of attributes in U. We shall refer to the tableau column corresponding to the attribute A j as the A j -column. Let be a tableau and let./ [R] be a JD in which R = fr 1 ; R 2 ; : : : ; R p g. has p rows, with row i corresponding to R i. Tableau is initialized with \a" and \b" symbols as follows: [i; A j ] = a j if A j 2 R i ; otherwise [i; A j ] = b ij : We shall refer to the rst subscript of a \b" symbol as its row subscript and to the second as its column subscript. We let T ABLE(R) denote the state of the tableau initialized from the database scheme R. We note that, at this stage, all \b" symbols are unique and that all occurrences of a particular \a" symbol are in one column. The Chase algorithm proceeds by making changes to the tableau,, according to the following rule, which we shall henceforth refer to as the \Chase rule": 19

20 Let be a graphical set of MVDs and let X!! Y 2. If there is a pair of rows in, say h and i, such that (1) [h; X] = [i; X], (2) [h; Y ] 6= [i; Y ] and (3) (i) one of [h; Y ] and [i; Y ], say [i; Y ] contains an \a" symbol, say a j, then assign the symbol a j to [h; Y ]; (ii) if [h; Y ] and [i; Y ] both contain \b" symbols, then assign to both the \b" symbol with the smaller row subscript. We can summarize a Chase rule as an ordered triple: h h ; i ; X!! Y i ; where h and i are the matching rows and X!! Y 2 is the MVD used. This version of the Chase rule is simpler than the one given in Ullman (1988). Its correctness is proved in Thanisch and Loizou (1986). This simplication is possible because of the restricted nature of graphical sets of MVDs. The Chase terminates when no more changes are possible. We refer to the state of the tableau on termination of the Chase by using the notation CHASE(; ). The following proposition is from Maier et al. (1979). PROPOSITION 1. Let be a set of MVDs and./ [R] be a JD. Then CHASE(T ABLE(R); ) contains a row comprising all \a" symbols if and only if j=./ [R]. 2 LEMMA 1. Let be a graphical set of MVDs and let R = fr 1 ; : : : ; R c g be a collection of sets of attributes. j=./ [R] if and only if (1) 9R i 2 R such that LEF T ( ) R i and (2) 8Y 2 RHS( ), and X 1 ; X 2 2 LHS( ) such that X 1 6= X 2, and X 1!! Y ; X 2!! Y 2, 9R j 2 R such that X 1 Y R j or X 2 Y R j. Proof. Let = T ABLE(R). 20

21 [ IF ] Without loss of generality, let LEF T ( ) R c. Let LEF T ( ) = fa 1 ; A 2 ; : : : ; A m g and let RIGHT ( ) = fb 1 ; B 2 ; : : : ; B n g. Let f and g be functions from f1; 2; : : : ; ng to, respectively, f1; 2; : : : ; mg and f1; 2; : : : ; c? 1g such that, for j = 1; 2; : : : ; n, fa f (j) g!! fb j g 2 and fa f (j) ; B j g R g(j) : Let = h 1 ; 2 ; : : : ; h i be an ordered sequence of valid Chase rules, where j = hc ; g(j) ; fa f (j) g!! fb j g i ; j = 1; 2; : : : ; h : From the denitions of the functions f and g, such a sequence of Chase rules describes a valid Chase computation sequence when applied to. Rule j adds an \a" symbol to the B j -column of row c of the tableau. Thus row c must contain a row of all \a" symbols after these Chase rule applications. Thus, by Proposition 1, j=./ [R]. [ ONLY IF ] j=./ [R], so, by Proposition 1, some row in CHASE( ; ) contains a row of all \a" symbols. CASE 1: Suppose, for contradiction, that 8R i 2 R, LEF T ( ) 6 R i. Let 0 denote the state of the tableau immediately before the Chase rule application, say h h; j; V!! W i, that produces the rst tableau state in which some tableau row contains \a" symbols in all of the columns corresponding to the attributes in LEF T ( ). By hypothesis, neither 0 [h; LEF T ( )] nor 0 [j; LEF T ( )] contains only \a" symbols. Thus W 2 LHS( ), which leads to a contradiction. CASE 2: Suppose, for contradiction, that 9Y 2 RHS( ) such that, where X 1 ; X 2 2 LHS( ), X 1 6= X 2, X 1!! Y ; X 2!! Y 2 and 8R i 2 R, X 1 Y 6 R i and X 2 Y 6 R i. Let 0 be the state of the tableau immediately prior to the Chase rule application that produces the rst tableau state in which some row has \a" symbols in either the X 1 Y -columns or the X 2 Y -columns. Without loss of generality, let it be the former. Let the said Chase rule be h h; j; V!! W i. We subdivide the proof into two further cases. Case 2.1: W = X 1. Thus X 1 2 RHS( ), which leads to a contradiction. Case 2.2: W = Y. But V!! W 2, so, from Denition 3.3, either V = X 1 or V = X 2. In our case V = X 1. Thus either 0 [h; Y ] or 0 [j; Y ] contains an \a" symbol. But 0 [h; V ] is identical to 0 [i; V ]. Thus either 0 [h; X 1 Y ] or 0 [j; X 1 Y ] must comprise all \a" symbols, which leads to a contradiction. 2 21

22 Let X U. The dependency basis of X with respect to, denoted DEP (X; ), partitions the set U? X into the collection of sets of attributes fy 1 ; Y 2 ; : : : ; Y h g such that if Z U? X then X!! Z 2 + if and only if Z is the union of some of the Y i 's. This is expressed as follows: DEP (X; ) = fy j j= X!! Z and Z Y implies that Z = Y g : The uniqueness of this partition for sets of MVDs with identical closures was proved by Fagin (1977). The elements of the partition are called dependents. It will be useful for us to describe, briey, Beeri's algorithm (Beeri, 1980) for computing the dependency basis. (A more ecient version can be found in Galil (1982).) The algorithm starts with the partition fu? Xg and then uses the MVDs in to \rene" this partition, thereby creating a ner partition of U? X. Each successive partition in this sequence is called a candidate partition. The renement rule works as follows. Let V!! Z 2. V!! Z can rene a partition, P, if there exists Y 2 P such that V T Y = ;, Z T Y 6= ; and Y? Z 6= ;. In this case, generate the new partition by replacing Y in P with the pair of sets Z T Y and Y? Z. Repeated application of the renement rule eventually yields the partition DEP (X; ). A dependency basis computation can be characterised by the sequence h P 1 ; P 2 ; : : : ; P h i of candidate partitions generated by successive applications of the renement rule, where P 1 = fu? Xg, P h = DEP (X; ) and for j = 1; 2; : : : ; h? 1, P j+1 is a ner partition of U? X than P j. PROPOSITION 2. Let be a graphical set of MVDs. For any V U, each renement rule application in the computation of DEP (V; ) uses an MVD, X!! Y, such that X V. Proof. This result follows from Lemma 8.4 in Beeri et al. (1983). 2 LEMMA 2. Let be a graphical set of MVDs. (a) is a minimal cover. (b) For any X LEF T ( ), 9Y 2 DEP (X; ) such that LEF T ( )? X Y. (c) For each V 2 MKEY S( ); V LEF T ( ). Proof. (a) If is a graphical set of MVDs then for each X!! Y 2, X T Y = ;. If X!! Y 2 then, from Denition 3.3, 9Z U such that Z!! Y 2 and Z T XY = ;. Thus XY U. Consequently, X!! Y is non-trivial. 22

23 We have j X j = j Y j = 1 and ; 62 LHS( ), thus X!! Y is neither left- nor right- reducible. The fact that X!! Y is non-transferable also follows from the fact that j X j = 1. The only proper subset of X is ; and DEP (;; ) = U. Thus we have proved that an arbitrary MVD in a graphical set of MVDs is reduced. In order to establish that is a minimal cover, we must also prove that is non-redundant. Let X!! Y 2 and? =? f X!! Y g. is graphical, so 9Z 2 LHS( )? fxg such that Z!! Y 2. Furthermore, for each element, V 2 (LHS( )? fx; Zg), V!! Y Y 62 DEP (X;?), since Z 62 RIGHT ( ) and Z!! Y is the only MVD that can be used in a renement rule to generate the singleton Y in a candidate partition. Thus? + 6= +. We have thus proved that removing an arbitrary MVD from aects the closure. Thus is non-redundant. It follows that a graphical set of MVDs is a minimal cover. (b) This follows from the fact that does not split LEF T ( ). Consider the computation of DEP (X; ). The initial candidate partition is fu? Xg. LEF T ( )? X U? X. For each Z!! W 2 used in renement rule applications, W T LEF T ( ) = ;. Thus LEF T ( )? X continues to be a subset of an element of the resulting candidate partition. Thus 9Y 2 DEP (X; ) such that LEF T ( )? X Y. (c) If V 6 LEF T ( ), we obtain a contradiction since, from Proposition 4.1 in Ozsoyoglu and Yuan (1987b), this would imply that is not split-free. 2 PROPOSITION 3. Let be a graphical set of MVDs. For any V U, j DEP (V; )? RHS( ) j 1. Proof. Consider the dependency basis computation for DEP (V; ). Each renement rule application uses an MVD, say X!! Y 2. For some element, say P, of the current candidate partition, P T Y 6= ; and P? Y 6= ;. In the resulting candidate partition, P is replaced by the pair of elements P T Y and P? Y. X!! Y 2, so j Y j = 1 and Y 2 RHS( ). Thus P T Y = Y, i.e. P T Y 2 RHS( ). Thus, after each renement rule application there is at most one element of the resulting candidate partition, P, which is not an element of RHS( ), i.e. j P? RHS( ) j 1. Thus j DEP (V; )? RHS( ) j

24 LEMMA 3. Let T be a scheme tree that is normal with respect to a graphical set of MVDs,. Then the following properties hold for T. (a) For each non-root node, say v, in T, if AT T (v) 62 LHS( ) then (1) AT T (v) RIGHT ( ) and (2) v 2 LEAV ES(T ) : (b) For each non-root node, v, in T, j KIDS(v) j 1. (c) There is at most one complex path in T. (d) If j LEAV ES(T ) j 2 then AT T (ROOT (T )) 2 LHS( ). (e) If LEF T ( ) AT T (T ) then T contains no transitive redundant nodes. Proof. (a) Let v be a non-root node in T such that AT T (v) 62 LHS( ). We rst prove property (1). Suppose that AT T (v) 6 RIGHT ( ). The universe U is partitioned by flef T ( ); RIGHT ( )g. Thus AT T (v) T LEF T ( ) 6= ;. We have AT T (v) DESC(v), so, from Denition 3.5, F K(DESC(v); ) 6= ;. Thus, from Denition 5.2(3), AT T (v) 2 F K(DESC(v); ). However, F K(DESC(v); ) LHS( ), so AT T (v) 2 LHS( ), which leads to a contradiction. Thus AT T (v) RIGHT ( ). Next we prove property (2). Suppose that v 62 LEAV ES(T ). Let w 2 KIDS(v). AT T (v) ANC(v) and ANC(v)!! DESC(w) [AT T (T )] is leftand right- reduced. From property (1) of this part of the lemma, AT T (v) RIGHT ( ), so ANC(v)? AT T (v)!! DESC(w) 2 P ROJ( + ; AT T (T )); which leads to a contradiction. Thus v 2 LEAV ES(T ). (b) Suppose that there is a node, say v, in T such that v 6= ROOT (T ) and j KIDS(v) j > 1. Let u = P ARENT (v). Obviously, v 62 LEAV ES(T ). Thus, by Lemma 3(a), AT (v) 2 LHS( ). Let V = AT T (v) and X = ANC(u). By Denition 5.5(2), X T V = ; and both X and V are non-empty. Let w 1 ; w 2 2 KIDS(v), w 1 6= w 2. Let W 1 = DESC(w 1 ) and W 2 = DESC(w 2 ). T Again, by Denition 5.5(2), W 1 W2 = ;. Thus, we have: XV!! W 1 ; XV!! W 2 2 P ROJ( + ; AT T (T )): T is semi-normal with respect to, so from part (2) of Denition 5.2, both of these MVDs are left- and right-reduced. Thus there exist Y 1 ; Y 2 2 DEP (XV; ) such that Y 1 6= Y 2, W 1 Y 1 and W 2 Y 2. By Proposition 3, either Y 1 2 RHS( ) or Y 2 2 RHS( ). Without loss of generality, let Y 1 2 RHS( ). is graphical, so j Y 1 j = 1. Furthermore ; W 1 Y 1, so W 1 = Y 1. 24

25 We have Y 1 6= Y 2, so j DEP (XV; ) j 2. Consequently, in the computation of DEP (XV; ), there is at least one renement rule application. Further, W 1 = Y 1, so let Z!! W 1 be the MVD associated with the renement rule application that results in a candidate partition with W 1 as an element. By Proposition 2, Z XV. is graphical and Z 2 LHS( ), so j Z j = 1. Furthermore, v 6= ROOT (T ), so V XV. Thus j XV j 2. Consequently Z XV, so XV!! W 1 is a partial dependency in T, yet T is semi-normal with respect to. By part (2) of Denition 5.2, this is a contradiction. Thus, for each non-root node, v, in T, j KIDS(v) j 1. (c) Suppose that there are two or more complex paths in T. Let w 1 ; w 2 2 LEAV ES(T ) be two dierent complex leaf nodes in T. Let r = ROOT (T ) and Z = AT T (r). By Lemma 3(b), r is the lowest common ancestor of w 1 and w 2. Let v 1 ; v 2 2 KIDS(r) be such that v 1 is an ancestor of w 1 and v 2 is an ancestor of w 2. Then Z!! DESC(v 1 ) ; Z!! DESC(v 2 ) 2 P ROJ( + ; AT T (T )): T is a normal scheme tree, so, by Denition 5.5(2), DESC(v 1 ) T DESC(v 2 ) = ;. By Proposition 3, either DESC(v 1 ) 2 RHS( ) or DESC(v 2 ) 2 RHS( ). Without loss of generality, let DESC(v 1 ) 2 RHS( ). Thus j DESC(v 1 ) j = 1, so v 1 2 LEAV ES(T ) and AT T (v 1 ) 2 RHS( ). Thus, by Denition 5.3, the path from r to v 1 in T is simple, which leads to a contradiction. Thus there exists at most one complex path in T. (d) Let ROOT (T ) = v and AT T (v) = V. Suppose that V 62 LHS( ). By Denition 5.2, V 2 M KEY S( ), so, by Lemma 2(c), V LEF T ( ). Thus V is the union of two or more dierent elements of LHS( ). Consequently, j V j 2. By hypothesis, j LEAV ES(T ) j 2. Thus, by Lemma 3(b), j KIDS(v) j 2. Let w 1 ; w 2 2 KIDS(v) such that w 1 6= w 2. Let W 1 = DESC(w 1 ) and W 2 = T DESC(w 2 ). From Denition 5.5(2), it follows that W 1 W2 = ;. Furthermore, V!! W 1 and V!! W 2 are left- and right-reduced, since the scheme tree T is semi-normal. Let Y 1 ; Y 2 2 DEP (V; ) such that W 1 Y 1 and W 2 Y 2. By Proposition 3, at least one of Y 1 and Y 2 is an element of RHS( ). Without loss of generality, let Y 1 2 RHS( ). j Y 1 j = 1 and ; W 1 Y 1. Thus W 1 = Y 1. Consider the dependency basis computation for DEP (V; ). Let X!! Y 1 be the MVD used in the renement rule that generates the rst candidate partition in which Y 1 is an element. By Proposition 2, X V. X 2 LHS( ), so j X j = 1. j V j 2, so X V. But T is semi-normal with respect to, yet V!! W 1 is not left-reduced, which leads to a contradiction. Thus if j LEAV ES(T ) j 2 then AT T (ROOT (T )) 2 LHS( ). 25

26 (e) Let AT T (ROOT (T )) = V. If a node is transitive redundant in T then we observe from Denition 5.4 that there must be at least two leaf nodes in T. Thus, by Lemma 3(d), V 2 LHS( ). It is given that LEF T ( ) AT T (T ). Let Z = LEF T ( )? V. From Denition 3.3, Z 6= ;. Thus, from Denition 5.3, Z is the union of the sets of attributes labelling non-root nodes on T 's complex path which, by Lemma 3(c), is unique. Suppose, for contradiction, that T contains some node, say v, that is transitive redundant in T. From Denition 5.4, for any transitive redundant node to exist, such a node must have some sibling nodes. From Lemma 3(b), this may only occur at depth one in T. We divide the proof into two cases, deriving a contradiction in each case. CASE 1: v is a node on a complex path. By Denition 5.3, v's sibling nodes are simple leaf nodes labelled by subsets of RIGHT ( ), i.e. non-key attributes. Thus, from Denition 5.4, v must be transitive redundant with respect to V. However, by MVD1, V!! ANC(P ARENT (v)) 2 P ROJ( + ; AT T (T )), which leads to a contradiction of Denition 5.4. Thus a node on the complex path cannot be transitive redundant. CASE 2: v is a node on a simple path. By Denition 5.3, v is a simple leaf node and AT T (v) 2 RHS( ). Furthermore, v must be a sibling to the node at depth one on the complex path. is graphical so 9X, X Z such that X!! AT T (v) 2. We shall derive a contradiction by showing that v is not transitive redundant with respect to X. If there exist some sibling nodes v 1 ; v 2 ; : : : ; v s of v in T such that W = s[ DESC(v i ) ; X ANC(P ARENT (v)) [ W and i=1 XW!! ANC(P ARENT (v)) 62 P ROJ( + ; AT T (T )) ; then v is transitive redundant in T with respect to X. In this case ANC(P ARENT (v)) = V. We have X V W and X 6= V, so Z W. Consider the computation of DEP (Z; ). For every Y 2 RHS( ) there exists an MVD X 0!! Y 2 such that X 0 Z, which is used in a renement rule. Thus U? Z is partitioned into RHS( ) and V (the remainder of U? Z). Clearly Z!! V 2 +. By MVD2 we can conclude that W!! V 2 P ROJ( + ; AT T (T )), which leads to a contradiction. Thus v is not transitive redundant with respect to X. 2 LEMMA 4. Let F be a nested scheme forest that is in NNF with respect to a graphical set of MVDs,. Then there exists a nested scheme forest, say F 0, that is in NNF with respect to and such that j F 0 j = j F j, and for each T 2 F 0, AT T (ROOT (T )) 2 LHS( ). 26

27 Proof. If for each T 2 F, AT T (ROOT (T )) 2 LHS( ), then the result follows immediately. So, suppose that 9T 2 F such that, where ROOT (T ) = r and AT T (r) = R, R 62 LHS( ). Now F is in NNF with respect to, so T is a scheme tree that is normal with respect to. Thus, from part (3) of Denition 5.2, R 2 MKEY S( ), so, by Lemma 2(c), R LEF T ( ). If j LEAV ES(T ) j 2 then, by Lemma 3(d), AT T (ROOT (T )) 2 LHS( ). Thus we assume that j LEAV ES(T ) j = 1, with KIDS(r) = fwg. Now let fx 1 ; X 2 ; : : : ; X g g be a partition of R, with g = j R j, and let v 1 ; v 2 ; : : : ; v g be node names that do not occur in T. We now form a new tree, T 0, that is identical to T except that the root node, r, is replaced by the path v 1 ; v 2 ; : : : ; v g and ROOT (T 0 ) = v 1. For i = 1; 2; : : : ; g, KIDS(v i ) = fv i+1 g, with v g+1 = w, and AT T (v i ) = X i. We prove that T 0 is a normal scheme tree. AT T (T ) = AT T (T 0 ), since fx 1 ; X 2 ; : : : ; X g g partitions R. ANC(v g ) in T 0 is identical to ANC(r) in T. Next we prove that T 0 satisifes the three conditions in Denition 5.2. (1) T comprises a single path, so each MVD in MV D(T 0 ) is trivial in the context of AT T (T 0 ). Thus MV D(T 0 ) P ROJ( + ; AT T (T 0 )). (2) If R!! DESC(w) [AT T (T 0 )] is left- and right-reduced then so is j[ X i!! i=1 g [ i=j+1 X i 1 A DESC(w) [AT T (T 0 )] ; j = 1; 2; : : : ; g? 1: (3) X 1 2 LHS( ), so X 1 2 MKEY S( ). For the nodes, v i, i = 2; 3; : : : ; g, AT T (v i ) 2 LHS( ). Thus AT T (v i ) 2 F K(DESC(v i ); ). Thus T 0 is semi-normal with respect to. In order to prove that T 0 is normal, we refer to Denition 5.5. The above argument establishes that the rst condition of Denition 5.5 is satisifed. The second condition of Denition 5.5 follows from the construction of T 0. Turning to the third condition of Denition 5.5, we observe that T 0 comprises a single path, so, by Denition 5.4, it is not possible for a node to be transitive redundant in T 0. From the above, it follows that the scheme tree T 0 is normal with respect to and that AT T (ROOT (T 0 )) 2 MKEY S( ). Next, we prove that F 0 is in NNF. We refer to Denition 7.1. Let F 0 = (F? T ) S ft 0 g. As already stated, AT T (T ) = AT T (T 0 ). Thus fat T (T i ) j T i 2 Fg = fat T (T j ) j T j 2 F 0 g: Consequently, S fat T (T j ) j T j 2 F 0 g = U and j=./ [fat T (T j ) j T j 2 F 0 g]. Thus F 0 is in NNF. Furthermore, j F 0 j = j F j. 2 REFERENCES 27

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