Decomposition Liangsheng Deng Section 2 11/15/2005

Recall Functional Dependencies  A->B holds iff for any pairs of tuples t1 and t2 such that if t1[A] = t2[A], then t1[B] = t2[B].  Armstong’s axioms:  Reflexivity rule. If a is a set of attributes and b is a subset of a, then a->b holds.  Augmentation rule. If a->b holds and c is a set of attributes, then ac->bc holds.  Transitivity rule. If a->b holds and b->c holds, then a->c holds.

Closure of a Set of FD  Given f, a set of FD of a relational schema, we can use Armstrong axioms to derive other possible FD implied by f.  The closure of f is the set of all FD implied by f.  Determine whether S, a set of attributes, is a superkey by computing the closure of S.

Closure Computing Example  Given R(A, B, C, D, E, H) with FD1. BD->A, FD2. AC->H, FD3. D->C, FD4. E->D. Is BE a superkey of R? {BE}+ = {BDE} because E->D {BE}+ = {BCDE} because D->C {BE}+ = {ABCDE} because BD->A {BE}+ = {ABCDEH} because AC->H BE is a superkey of R.

Lossless Decomposition The purpose of using lossless decomposition:  Reduce unnecessary redundancy.  Retrieve information efficiently How do we decide a decomposition is a lossless decomposition?  Let R be a relation schema.  Let F be a set of functional dependencies on R.  Let R1 and R2 form a decomposition of R. The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies holds. R1 ∩ R2 -> R1 or R2 ∩ R1 -> R2

Lossless Decomposition Testing (1)  Given R1(A, B), R2(A, C, D) and F = {A->C; D->B; AC->B}. Is the join of R1 and R2 lossless?   R1 ∩ R2 = {A} If we can prove A->B, then the answer is yes. If we can prove A->B, then the answer is yes.Proof: A->AC because A->C A->B because A->AC and AC->B So, it is a lossless cecomposition.

Lossless Decomposition Testing (2)  To test whether  To test whether R1 ∩ R2 -> R1 or R2, we can test whether R1 ∩ R2 is a superkey of R1 or R2 by computing the closure of R1 ∩ R2.   Same question in the previous slide.   {R1 ∩ R2}+ = {A}+, {A}+ = {AC} because A->C {A}+ = {ABC} because AC->B Conclusion: A is a superkey of R1. Therefore, it is a lossless decomposition.

Lossless Decomposition Testing (3)  For the case of decomposition of a schema into more than 2 schemas, we can use the chase matrix algorithm.  For example, F = {A->C, B->C, C->D, DE->C, CE->A}, R1(AD), R2(AB), R3(BE), R4(CDE) and R5(AE). ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b23b24b25 R3(BE)b31a2b33b34a5 R4(CDE)b41b42a3a4a5 R5(AE)a1b52b53b54a5

Lossless Decomposition Testing (3) Cont. A->C, we get ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b13b24b25 R3(BE)b31a2b33b34a5 R4(CDE)b41b42a3a4a5 R5(AE)a1b52b13b54a5 B->C, we get ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b13b24b25 R3(BE)b31a2b13b34a5 R4(CDE)b41b42a3a4a5 R5(AE)a1b52b13b54a5

Lossless Decomposition Testing (3) Cont. C->D, we get ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b13a4b25 R3(BE)b31a2b13a4a5 R4(CDE)b41b42a3a4a5 R5(AE)a1b52b13a4a5 DE->C ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b13a4b25 R3(BE)b31a2a3a4a5 R4(CDE)b41b42a3a4a5 R5(AE)a1b52a3a4a5

Lossless Decomposition Testing (3) Cont. CE->A ABCDE R1(AD)a1b12b13a4b15 R2(AB)a1a2b23a4b25 R3(BE)a1a2a3a4a5 R4(CDE)a1b42a3a4a5 R5(AE)a1b52a3a4a5 The third tuple has a1, a2, a3,a4 and a5, so it is a lossless decomposition. it is a lossless decomposition.

Dependency Preservation Let F be a set of functional dependencies on schema R. Let F be a set of functional dependencies on schema R. Let {R1,R2,…Rn} be a decomposition of R. Let {R1,R2,…Rn} be a decomposition of R. The restriction of F to Ri is the set of all functional dependencies in F+ that include only attributes of Ri. The restriction of F to Ri is the set of all functional dependencies in F+ that include only attributes of Ri. Functional dependencies in a restriction can be tested in one relation, as they involve attributes in one relation schema. Functional dependencies in a restriction can be tested in one relation, as they involve attributes in one relation schema.

Dependency Preservation Cont.  The set of restrictions F1,F2,…Fn is the set of dependencies that can be checked efficiently.  We need to know whether testing only the restrictions is sufficient.  Let F’ = F1 U F2 U….Fn.  F’ is a set of functional dependencies on schema R, but in general, F’ != F.

Dependency Preservation Cont.  A decomposition having the property that is a dependency-preserving decomposition.

Dependency Preservation Cont. An Easier Way To Test For Dependency Preservation. Rather than compute F+ and F’+, and see whether they are equal, we can do this: Rather than compute F+ and F’+, and see whether they are equal, we can do this: Find F – F’, the functional dependencies not checkable in one relation. See whether this set is obtainable from F' by using Armstrong's Axioms. This should take a great deal less work, as we have (usually) just a few functional dependencies to work on.

Dependency Preservation Cont.  For example, given F={AB->C, C->A}, R1(AC) and R2(BC).  F’ = {C->A}. F – F’ = {AB->C}. AB->C can not be derived from F’. So, it is not a dependency-preserving decomposition.

Reference al/notes/Chapter7/node1.html al/notes/Chapter7/node1.html