Exam 1 Solutions Spring 2016

Transcription

1 Exam 1 Solutions Spring 2016

2 Problem 1 1. R 1 := σ color= red OR color= green (P arts) Result := Π sid (R 1 Catalog) 2. R 1 := σ sname= Y osemitesham (Suppliers) R 2 := Π pid,cost (R 1 Catalog) R 3 (pid1, cost1) := R 2 R 4 (pid2, cost2) := R 2 R 5 := Π pid1,cost1 (R 3 pid1=pid2 AND cost1<cost2 R 4 ) R 6 := R 3 R 5 Result := Π pid (R 6 ) 3. R 1 := Π sid,pid (Catalog) R 2 (sid1, pid1) := R 1 R 3 (sid2, pid2) := R 1 Result := Π pid1 (R 2 sid1<>sid2 AND pid1=pid2 R 3 ) Page 2 of 11

3 Problem 2 1. A B E F A B Count(D) AVG(C) We don t count NULL in COUNT and AVG. Page 3 of 11

4 Problem 3 According to FD s the key is ID, AdvisorID, so all the given FD s violates BCNF because on the left side of them, we don t have the superkey. In order to decompose, we take one of them that violate BCNF. For example ID Name. {ID} + = {ID, Name, FavoriteAdvisorID} So we compose the table to: Students1 (ID,Name,FavoriteAdvisorID) Students2 (ID,AdvisorID,AdvisorName) But we are not done here! We should consider all relevant FD s for each table to see that there is still any violation of BCNF or not. For Studens1 there is no violation of BCNF because the key is ID and all relevant FD s have the key on their left side. For Students2 the key is ID, AdvisorID but the relevant FD that is {AdvisorID AdvisorName} don t have supper key on the left side. So we have BCNF violation and we should decompose this table. {AdvisorID} + = {AdvisorID, AdvisorName} So we have: Students3 (AdvisorID,AdvisorName) Students4 (AdvisorID,ID) Because they are two-attribute relations so there are in BCNF. Therefore at the end we will have three relations that are the final decomposition: Students1 (ID,Name,FavoriteAdvisorID) Students3 (AdvisorID,AdvisorName) Students4 (AdvisorID,ID) Page 4 of 11

5 Problem 4 a. Π name ssn=buyer ssn Person Purchase maker sid=sid Π pid,maker cid Π cid Product σ country= USA Company b. 1) The correct answer is: R := P erson2 P urchase P erson1 Answer := Π P 2.ssn (σ P 1.ssn=buyers ssn AND P 2.ssn=seller ssn AND P 2.phone number=p 1.phone number (R)) The common mistake was: R 1 := P urchase buyer ssn=ssn P erson R 2 := P urchase seller ssn=ssn P erson Result := R 1 R1.phone number=r2.phone number R 2 The reason why that is not correct because the result contains inappropriate tuples and does not cover all correct answers. Page 5 of 11

6 2) Π seller ssn (σ buyer ssn=seller ssn (P urchase)) 3) R = (P urchase1 P urchase2) Π buyer ssn (σ P 1.buyer ssn=p 2.seller ssn AND P 1.store=P 2.store (R)) Page 6 of 11

7 Problem 5 Property 3NF BCNF 4NF Eliminates redundancy due to FD s No Yes Yes Eliminates redundancy due to MVD s No No Yes Preserve FD s Yes No no Preserve MVD s No No no 3NF BCNF 4NF Page 7 of 11

8 Problem 6 [464] a) A D does not hold. b) Step 1. Initialize a b 1 c 1 d e 1 a b c d 2 e Step 2. Apply A BC a b 1 c 1 d e 1 a b 1 c 1 d 2 e a b c d 2 e a b c d e 1 Step 3. Apply B D a b 1 c 1 d e 1 a b 1 c 1 d e a b c d e a b c d e 1 Step 4. Apply C E a b 1 c 1 d e 1 a b 1 c 1 d e a b 1 c 1 d e a b 1 c 1 d e 1 a b c d e a b c d e 1 a b c d e 1 a b c d e Page 8 of 11

9 A D hold in that relation. [564] Step 1. Initialize a b c d 1 e 1 a 2 b c d e 2 a b 3 c d 3 e Step 2. Apply A D a b c d 1 e 1 a 2 b c d e 2 a b 3 c d 1 e Step 3. Apply D E a b c d 1 e a 2 b c d e 2 a b 3 c d 1 e Step 4. Apply B D a b c d e a 2 b c d e 2 a b 3 c d 1 e Decomposition is lossless. Page 9 of 11

10 Problem 7 [464] Studios (name, addr) Crews (number, studioname, crewchief) When weak entity relationship does not have attributes, there is no need to convert it to a relation. [564] Flights (FlightNumber, Day, Aircraft) Customers (SSN, Name, Address) Booking (SSN, FlightNumber, Day, Row, Seat) FD s for Flights is: F lightnumber, Day Aricraft It does not violate BCNF because left side has the keys (super key). FD s for Customers is: SSN Name SSN Address It does not violate BCNF because left side has the keys (super key). FD s for Booking is: SSN, F lightnumbers, Day Row SSN, F lightnumbers, Day Seats It does not violate BCNF because left side has the keys (super key). Page 10 of 11

11 Problem 8 We can use contrapositive in order to prove this. So, we are going to show that if X + is not a subset of Y +, then it must be that X is not a subset of Y. Imagine we have A 1 A 1...A n attributes in X + that is not in Y + (Because we said that (X + Y + ) ). If any of these attributes were originally in X then we are done because Y doesn t have any of this attributes but if these attributes were added by the closure, then we must see the case further. Assume that there was sum FD s like C 1 C 2...C m A 1 A 1...A i where A 1 A 1...A i is some subset of A 1 A 1...A n. So, it must be C 1 C 2...C m or some subset of C 1 C 2...C m is in X. But because the attribute C 1 C 2...C m cannot be in Y because we assumed that attributes A 1 A 1...A n are only in X + therefore X is not a subset of Y. By proving the contrapositive we have also proved if X Y then X + Y +. Page 11 of 11