Exercise 9: Normal Forms
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1 Data Modelling and Databases (DMDB) ETH Zurich Spring Semester 2017 Systems Group Lecturer(s): Gustavo Alonso, Ce Zhang Date: Assistant(s): Claude Barthels, Eleftherios Sidirourgos, Eliza Last update: August 16, 2017 Wszola, Ingo Müller, Kaan Kara, Renato Marroquín, Zsolt István Exercise 9: Normal Forms Solution The exercises marked with * will be discussed in the exercise session. You can solve the other exercises as practice, ask questions about them in the session, and hand them in for feedback. All exercises may be relevant for the exam. Ask Claude (claude.barthels@inf.ethz.ch) for feedback on this week s exercise sheet or give it to the TA of your session (preferably stapled and with your address). 1 LineItem Relation * Consider the following relational schema: LINEITEM (OrderNumber, ItemNumber, Description, Price, Quantity) 1. Find the functional dependencies and a candidate key of the relation above. Solution: ItemN umber Description, P rice OrderN umber, ItemN umber Quantity (OrderN umber, ItemN umber) is a candidate key of this relation. 2. What normal form is the above LINEITEM relation in? Solution: 1NF: All attributes have atomic domains, so LINEITEM is in 1NF. 2NF: Description and P rice depend on a subset of a key (namely the ItemNumber) and not on the whole key. Therefore, the relation is not in 2NF. Higher NFs: Since the relation is not in 2NF, it is also not in any higher normal form.
2 3. What are some disadvantages of this choice of schema? Solution: The item description and price are stored unnecessarily for each instance of the particular item in the LINEITEM relation, which might lead to anomalies. On the other hand, a positive effect of this is that, in order to find the total price of items, no join is needed. 2 Concert Relation * Consider the following relational schema describing musical events in Switzerland: Assumptions: Each event has at least one singer. Singers stick to one genre of music and they do not visit the same venue twice in the same year. Venue Year Singer Genre X 1999 Cher pop Z 1999 Cher pop Y 2001 Cher pop Y 2001 Porcupine Tree rock 1. Is this relation in 2NF? 3NF? Determine the functional dependencies and keys and justify your answer. Solution: The key is (V enue, Y ear, Singer). Genre depends on Singer, which is a strict subset of a key. Therefore, the relation is not in 2NF and consequently also not in 3NF. We now modify the schema to include the number of attendees: Venue Year Singer Attendees X 1999 Cher Z 1999 Cher Y 2001 Cher Y 2001 Porcupine Tree Is the new schema in 2NF? 3NF? Solution: 2NF: All non-key attributes (namely Attendees) depend on the whole key and nothing else. 3NF: There is only one non-key attribute, which can hence not depend on another non-key attribute. The point of this exercise is to show the difference between storing attributes that all belong to the same concept (Concert, including Attendees) and mixing attributes of different concepts (Concert and Singer).
3 3 Synthesis Algorithm * Consider the following relation: with the following functional dependencies: R(A, B, C, D) A D A, B C A, C B 1. Decide whether the following statements are true or false. If they are true, give a reason why. ID Statement True False 1 AB is a candidate key of R. X 2 A is a candidate key of R. X 3 B is a candidate key of R. X 4 ABC is a candidate key of R. X 5 AC is a candidate key of R. X 6 BC is a candidate key of R. X 7 R is in 3NF. X Explanation: Statement 1: Closure(F, AB) = (ABCD) AB is a super-key Closure(F, A) = (AD) Closure(F, B) = (B) AB is a candidate key Statement 5: Closure(F, AC) = (AC) AC is a super-key Closure(F, C) = (C) AC is a candidate key 2. Apply the synthesis algorithm to transform the schema into 3NF (which is lossless and preserves dependencies). Solution: 1. Minimal basis: F c = F 2. Create relations for FDs:
4 A D R1(A,D) AB C R2(A,B,C) AC B R3(A,C,B) 3. Relation for keys: not necessary 4. Eliminate redundant relations: R3 R2 drop R3 Thus, the schema in 3NF is R1(A,D) and R2(A,B,C). 4 Decomposition Algorithm Consider the following relations and functional dependencies: S1(A, B, C, D) with functional dependencies: AC D A B R(A, B, C, D, E, F ) and its decomposition into R1(A, B, D, E) and R2(A, C, F ) with the following functional dependencies: A BE A D F A AC F BC E C A S2(A, B, C) with functional dependencies: AB C C A 1. Fill in the valid candidate keys for each of the relations. Mark the candidate keys that you think are correct. Reason why you decided that a candidate key is correct for a relation. ID Statement R1 R2 S1 S2 1 A is a candidate key. 2 C is a candidate key. 3 AC is a candidate key. 4 AB is a candidate key. 5 AE is a candidate key. 6 BC is a candidate key. 7 ACF is a candidate key. 8 BCE is a candidate key.
5 Solution: To determine the key candidates for R1 and R2, let us first compute the minimal basis (of all attributes and FDs of R): Left reduction: Reduce AC F to C F because {F } Closure(F D, C) = (A, B, C, D, E, F ) Reduce BC E to C E because {E} Closure(F D, C) = (A, B, C, D, E, F ) Right reduction: Reduce C E to C because {E} Closure(F D\{C E}, C) = (A, B, C, D, E, F ) Reduce C A to C because {E} Closure(F D\{C A}, C) = (A, B, C, D, E, F ) After applying the union rule to FDs with the same left side, the following remains as minimal basis: (1)A BDE (2)F A (3)C F Note that the decomposition of R into R1 and R2 is in fact dependency-preserving: (1) is the only FD of R1, (2) and (3) are the FDs of R2, and toghether, they are a minimal basis of the dependencies of R. A is a candidate key of R1 because of (1); C is a candidate key of R2 because of (2) and (3). A candidate key of S1 is AC because it defines all other attributes. The candidate keys of S2 are AB and BC. AB is trivially a key and BC is a candidate key because C A holds. 2. Given the keys that you have identified, in which normal forms are the relations? Mark all correct forms and derive the highest normal form that is applicable. Normal Form R1 R2 S1 S2 1NF 2NF 3NF BCNF Explanation: R1: 2NF yes - BDE is dependent on A. 3NF yes - A is superkey. BCNF yes - A is superkey. R2: 2NF yes - AF is dependent on C.
6 3NF no - F A is not trivial. superkey. S1: 2NF no - B is not entirely dependent on (A, C). A is not part of the candidate key and F is not a S2: 2NF yes - there are no non-key attributes. 3NF yes - AB is a superkey, and A is part of a candidate key. BCNF no - because in C A, C is not a superkey. 3. Transform those relations that are lower than 3NF to conform to 3NF. Are the new relations in BCNF? Solution: The decomposition of R2(A, C, F ) into R21(C, F ) and R22(A, F ) is in 3NF and BCNF. Note: The synthesis algorithm is not defined for two or more relations. An alternative solution would be to join the relations R1 and R2 and then apply the algorithm. The result would be identical. The decomposition of S1(A, B, C, D) into S11(A, B) and S12(A, C, D) is in 2NF, 3NF, and BCNF. 5 Normalization up to BCNF * Consider the following relation: Shipping (ShipName, ShipT ype, T ripid, Cargo, P ort, Date) and the following functional dependencies: 1. Find the candidate key(s). ShipName ShipT ype T ripid ShipName, Cargo ShipName, Date T ripid, P ort. Solution: ShipN ame, Date and T ripid, Date.
7 2. Apply the synthesis algorithm. Solution: 1. Minimal basis: Given FDs are a minimal basis. 2. Create relations for FDs: R1(ShipName, ShipType) with ShipN ame ShipT ype - 3NF R2(TripId, ShipName, Cargo) with T ripid ShipName, Cargo - 3NF R3(ShipName, Date, TripId, Port) with ShipName, Date T ripid, P ort and T ripid ShipName - 3NF Note that R3 is indeed in 3NF. R3 has two candidate keys: (ShipName, Date) and (T ripid, Date). Both T ripid and ShipN ame are not non-key attributes. Thus, there is no 3NF violation. However, the relation is not in BCNF. 3. Relation for keys: not necessary 4. Eliminate redundant relations: none possible 3. Apply the decomposition algorithm. Solution: We take a shortcut and apply the decomposition algorithm to each of the resulting relations from the previous exercise. R1 and R2 are already in BCNF, so nothing needs to be done. For R3, we find FDs violating BCNF to decompose it: T ripid ShipName violates BCNF because the left-hand side is not a key of R3. We thus define: R31(TripId, ShipName) = {TripId} {ShipName} R32(Date, TripId, Port) = R3 - {ShipName} We can eliminate R32 from the final results since it is contained in R2. If we give the remaining relations meaningfull names, we get the following final result: Ships(ShipName, ShipType) with ShipN ame ShipT ype for R1 Trips(TripId, ShipName, Cargo) with T ripid ShipN ame, Cargo for R2 TripPort(Date, TripId, Port) with T ripid, Date P ort for R32 Note that the decomposition does not preserve the functional dependencies ShipN ame, Date T ripid and ShipName, Date P ort
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