zone# Garbage_Truck ships_to Garbage_Structure structid max_age

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1 B561 Assignment 2 Solutions C.M. Wyss September 19, Questions a-c (below) concern the ER diagram in gure 1 (below). truckid driver zone# Garbage_Truck services Zone ships_to structid Garbage_Structure max_age type curr_age Figure 1: ER diagram for question 1. This ER diagram may be translated into the following tables: zone/ships to(zone#, structid) truck/services(truckid, zone#, driver) garbage structure(structid, type, curr age, max age) In addition, suppose that the functional dependencies F = f driver! truckid, type! max age, zone#! truckid, zone#! structid g hold. (a) Decompose the tables given above into a set of tables that are in BCNF with respect to the FDs in F. Make sure that you indicate primary keys and that these are minimal (given F ). 1

2 Solution: zone/ships to(zone#, structid) truck(truckid, driver) services(zone#, truckid) garbage structure(structid, type, curr age) structure max ages(type, max age) (b) Your BCNF decomposition (from part (a)) may contain remaining redundancies that indicate aws in the design of the original ER diagram. Use the techniques in your text to eliminate any remaining redundancy in your tabular representation. For example, if two tables have the same primary key, you may consider merging them. Make sure you indicate primary keys for your new tables. Solution: The tables zone/ships to and services have the same key, so we'll merge them. zone/ships to/services(zone#, structid, truckid) truck(truckid, driver) garbage structure(structid, type, curr age) structure max ages(type, max age) (c) Give an ER diagram that corresponds to the tables you obtained in part (b). Make sure you capture any new relationships that arise naturally in your new design so that, in particular, the information contained in the old design is completely recoverable. Solution: The merged table zone/ships to/services indicates we should use a ternary relationship instead of two binary relationships. Also, note that now that we have split garbage structures into two entities, we must capture the fact that structures are associated with a maximum age. Thus, we introduce a new relationship ("has max age") for this. The resulting modied diagram appears in gure 2. truckid driver zone# Garbage_Truck services/ ships_to Zone Garbage_Structure has_max_age Structure Max_Ages structid curr_age type max_age type Figure 2: Modied ER diagram for question 1, part (c). 2

3 2. Prove that the decomposition R(A; B; C; D), S(C; D; E; F), T (E; F; G; H) is in BCNF with respect to the set of FDs F = fg! B;AB! D; DE! C; FG! H; DB! E;AH! F; CE! A; AD! H; FG! Dg. Solution: We must show the BCNF property for each of R, S, and T. R: The only non-trivial FD that holds on R is AB! D. We must therefore show that AB is a key for R, i.e. that AB! C. We use Armstrong's axioms as follows. FD deduced 1 AB! D 2 F 2 AB! DB augmentation on 1 3 DB! E 2 F 4 AB! E transitivity on2 and 3 5 AB! DE union on 1 and 4 6 DE! C 2 F 7 AB! C transitivity on5 and 6 S: The only non-trivial FD that holds on S is DE! C. We must therefore show that DE is a key for S, i.e. that DE! F. We use Armstrong's axioms as follows. FD deduced 1 DE! C 2 F 2 AH! F 2 F 3 DE! CE augmentation on 1 4 CE! A 2 F 5 DE! A transitivity on3and4 6 DE! AD augmentation on 5 7 AD! H 2 F 8 DE! H transitivity on6and7 9 DE! AH union on 5and8 10 DE! F transitivity on2and9 T : The only non-trivial FD that holds on T is FG! H. We must therefore show that F G is a key for T, i.e. that F G! E. We use Armstrong's axioms as follows. FD deduced 1 FG! H 2 F 2 FG! D 2 F 3 G! B 2 F 4 FG! G decomposition on 2 5 FG! B transitivity on3 and 4 6 FG! DB union on 2 and 5 7 DB! E 2 F 8 FG! E transitivity on6 and 7 3

4 3. Consider the relation schema R(A; B; C; D; E; F; G; H) and set of FDs F = fa! D; DC! A; AH! F; FG! H; G! B;DE! Cg that hold in instances of R. (a) Find a decomposition of R that is in BCNF with respect to F. Solution: We use the decomposition method given in class (as part of the decomposition theorem). This is illustrated in gure 3. The nal decomposition is R 1 (A; D), R 2 (A; C), R 3 (F; G; H), R 4 (C; D; E), R 5 (B;G), and R 6 (D; E; F; G). R(A,B,C,D,E,F,G,H) DC > A R (A,C,D) * A > D R 1(A,D) R 2(A,C) R 3(F,G,H) R (B,C,D,E,F,G,H) FG > H R (B,C,D,E,F,G) DE > C R 4(C,D,E) R (B,D,E,F,G) G > B R 5(B,G) R (D,E,F,G) 6 Figure 3: BCNF decomposition. (b) Is your decomposition from part (a) lossless? Prove your answer. Solution: The decomposition is lossless due to the decomposition theorem. (c) Is your decomposition from part (a) dependency preserving? Prove your answer. Solution: The decomposition is not dependency preserving; for example, the dependency AH! F has been lost. To see this, note that AH! F is not derivable from the other FDs if F, and also it does not hold on any of the decomposed relations. (d) Find a decomposition of R that is in 3NF with respect to F, but is not in BCNF with respect to F. Solution: There are many such decompositions. One is R 1 (A; C; D), R 2 (F; G; H), R 3 (C; D; E), R 4 (B;G) and R 5 (D; E; F; G). Note, for example, that R 1 is not in BCNF since the FD A! D holds on R 1 but A is not a key for R 1 (since A 6! C). On the other hand, DC isakey for R 1, so D is part of the key DC for R 1. (e) Is your decomposition from part (c) lossless? Prove your answer. Solution: The decomposition is obtained by using the method for nding a BCNF decomposition for R (part (a)), however the step labelled * in gure 3 is 4

5 not included. Still, by the decomposition theorem, the decomposition is lossless, as for part (b). (f) Is your decomposition from part (c) dependency preserving? Prove your answer. Solution: The decomposition does not capture the dependency AH! F, as in part (c). 4. Do the following exercises from question 4.2 in your text: (d), (e), (f), (g). Solution: Given: two relations, R 1 and R 2 where jr 1 j = N 1 and jr 2 j = N 2 and N 1 >N 2 > 0. (a) Assume schema(r 1 ) =schema(r 2 ). Then N 1 jr 1 [ R 2 jn 1 + N 2. (b) Again assume schema(r 1 ) = schema(r 2 ). Then 0 jr 1 \ R 2 jn 2. (c) Again assume schema(r 1 ) = schema(r 2 ). Then N 1, N 2 jr 1, R 2 jn 1. (d) Assume schema(r 1 ) \ schema(r 2 )=;. Then jr 1 R 2 j = N 1 N 2. (e) Assume A 2 schema(r 1 ). Then 0 j A=5(R 1 )jn 1. (f) Assume A 2 schema(r 1 ). Then 1 j A (R 1 )jn 1. (g) Assume schema(r 1 )=fa; Bg and schema(r 2 )=fbg. Then 0 jr 1 =R 2 jn Given the relation schemas R(A; B; C), S(A; B; C), T (C; D; E) andu(c; D; E), prove (from rst principles) or disprove (using a counterexample) the following relational algebra (RA) identities. (a) (R./T)./ (S./ U) (R./S)./ (T./ U). Solution: True. Proof that ha; b; c; d; ei 2 (R./ T )./ (S./ U) implies ha; b; c; d; ei 2(R./S)./ (T./ U) follows. fact deduced 1 ha; b; c; d; ei 2(R./T)./ (S./ U) assumption 2 ha; b; c; d; ei 2R./T follows from 1 3 ha; b; c; d; ei 2S./ U follows from 1 4 ha; b; ci 2R follows from 2 5 hc; d; ei 2T follows from 2 6 ha; b; ci 2S follows from 3 7 hc; d; ei 2U follows from 3 8 ha; b; ci 2R./S follows from 4and6 9 hc; d; ei 2T./ U follows from 5and7 10 ha; b; c; d; ei 2(R./S)./ (T./ U) follows from 8and9 (b) (R./T)./ (S./ U) (R./S)./ (T./ U). Solution: True. Proof that ha; b; c; d; ei 2 (R./ S)./ (T./ U) implies ha; b; c; d; ei 2(R./T)./ (S./ U) follows. 5

6 fact deduced 1 ha; b; c; d; ei 2(R./S)./ (T./ U) assumption 2 ha; b; ci 2R./S follows from 1 3 hc; d; ei 2T./ U follows from 1 4 ha; b; ci 2R follows from 2 5 ha; b; ci 2S follows from 2 6 hc; d; ei 2T follows from 3 7 hc; d; ei 2U follows from 3 8 ha; b; c; d; ei 2R./T follows from 4and6 9 ha; b; c; d; ei 2S./ U follows from 5and7 10 ha; b; c; d; ei 2(R./T)./ (S./ U) follows from 8and9 (c) C (R./T)./ C (S./ U) C ((R./S)./ (T./ U)). Solution: False. Counterexample follows. R: a 1 b 1 c S: a 2 b 2 c T: c d 1 e 1 U: c d 2 e 2 For this counterexample, hci 2 C (R./ T )./ C (S./ U) but C ((R./ S)./ (T./ U)) = ;. (d) C (R./T)./ C (S./ U) C ((R./S)./ (T./ U)). Solution: True. Proof that hci 2 C ((R./S)./ (T./ U)) implies hci 2 C (R./ T )./ C (S./ U) follows. fact deduced 1 hci 2 C ((R./S)./ (T./ U)) assumption 2 hci 2 C ((R./T)./ (S./ U)) by part (b) 3 9a; b; d; e such that ha; b; c; d; ei 2(R./T)./ (S./ U) follows from 2 4 ha; b; c; d; ei 2R./T follows from 3 5 ha; b; c; d; ei 2S./ U follows from 3 6 hci 2 C (R./T) follows from 4 7 hci 2 C (S./ U) follows from 5 8 hci 2 C (R./T)./ C (S./ U) follows from 6 and 7 (e) (R, S)./ (T, U) (R./T), (S./ U). Solution: True. Proof that ha; b; c; d; ei 2(R,S)./ (T,U) implies ha; b; c; d; ei 2 (R./T), (S./ U) follows. 6

7 fact deduced 1 ha; b; c; d; ei 2(R, S)./ (T, U) assumption 2 ha; b; ci 2R, S follows from 1 3 hc; d; ei 2T, U follows from 1 4 ha; b; ci 2R follows from 2 5 ha; b; ci 62S follows from 2 6 hc; d; ei 2T follows from 3 7 hc; d; ei 62U follows from 3 8 ha; b; c; d; ei 2R./T follows from 4 and 6 9 ha; b; c; d; ei 62S./ U follows from 5 and 7 10 ha; b; c; d; ei 2(R./T),(S./ U) follows from 9 and 10 (f) (R, S)./ (T, U) (R./T), (S./ U). Solution: False. Counterexample follows. R: a b c S: a 0 b 0 c 0 T: c d e U: c d e For this counterexample, ha; b; c; d; ei 2 (R./ T ), (S./ U) but (R, S)./ (T, U) =;. 7