# Database Principles Assignment

**Topics:**Relational model, Florida Keys, Decomposition

**Pages:**6 (1441 words)

**Published:**February 24, 2013

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2. For your E/R diagram of exercise 1, (i) Select and specify keys, and (ii) Indicate appropriate referential integrity constraints. Solution:

Keys ssNo and number are appropriate for Customers and Accounts, respectively. Also, we think it does not make sense for an account to be related to zero customers, so we should round the edge connecting Owns to Customers. It does not seem inappropriate to have a customer with 0 accounts; they might be a borrower, for example, so we put no constraint on the connection from Owns to Accounts. The E/R diagram below showes underlined keys and the numerocity constraint. [pic]

3. The E/R diagram below represents ships. Ships are said to be sisters if they were designed from the same plans. Convert this diagram to a relational database schema. Solution:

Ships(name, yearLaunched)

SisterOf(name, sisterName)

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4. Consider a relation representing the present position of molecules in a closed container. The attributes are an ID for the molecule, the x, y, and z coordinates of the molecule, and its velocity in the x, y, and z dimensions. What FD’s would you expect to hold? What are the keys? Solution:

Surely ID is a key by itself. However, we think that the attributes x, y, and z together form another key. The reason is that at no time can two molecules occupy the same point.

Consider a relation with schema R(A,B,C,D) and FD’s AB->C, C->D, and D->A. What are all the nontrivial FD’s that follow from the given FD’s? You should restrict yourself to FD’s with single attributes on the right side. What are all the keys of R?

What are all the superkeys for R that are not keys?

Solutions:

(a)We could try inference rules to deduce new dependencies until we are satisfied we have them all. A more systematic way is to consider the closures of all 15 nonempty sets of attributes. For the single attributes we have A+ = A, B+ = B, C+ = ACD, and D+ = AD. Thus, the only new dependency we get with a single attribute on the left is C->A. Now consider pairs of attributes:

AB+ = ABCD, so we get new dependency AB->D. AC+ = ACD, and AC->D is nontrivial. AD+ = AD, so nothing new. BC+ = ABCD, so we get BC->A, and BC->D. BD+ = ABCD, giving us BD->A and BD->C. CD+ = ACD, giving CD->A. For the triples of attributes, ACD+ = ACD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABC->D, ABD->C, and BCD->A. Since ABCD+ = ABCD, we get no new dependencies.

The collection of 11 new dependencies mentioned above is: C->A, AB->D, AC->D, BC->A, BC->D, BD->A, BD->C, CD->A, ABC->D, ABD->C, and BCD->A. (b)

From the analysis of closures above, we find that AB, BC, and BD are keys. All other sets either do not have ABCD as the closure or contain one of these three sets. (c)

The superkeys are all those that contain one of those three keys. That is, a superkey that is not a key must contain B and more than one of A, C, and D. Thus, the (proper) superkeys are ABC, ABD, BCD, and ABCD.

5. For each of the following relation schemas and sets of FD’s: a) R(A,B,C,D) with FD’s AB->C,C->D, and D->A

b) R(A,B,C,D) with FD’s B->C and B->D

Do the following:

i) Indicate all the BCNF violations. Do not forget to consider FD’s that are not in the given set, but follow from them. However, it is not necessary to give violations that have more than one attribute on the right side. ii) Decompose the relations, as necessary, into collections of relations that are in BCNF. iii) Indicate all the 3NF violations.

iv)...

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