Following questions have been asked in GATE CS 2005 exam.

**1) Let r be a relation instance with schema R = (A, B, C, D). We define r1 = ‘select A,B,C from r’ and r2 = ‘select A, D from r’. Let s = r1 * r2 where * denotes natural join. Given that the decomposition of r into r1 and r2 is lossy, which one of the following is TRUE?**

(a) s is subset of r

(b) r U s = r

(c) r is a subset of s

(d) r * s = s

Answer (c)

Consider the following example with lossy decomposition of r into r1 and r2. We can see that r is a subset of s.

Table r A B C D --------------------------- 1 10 100 1000 1 20 200 1000 1 20 200 1001 Table r1 A B C ------------------ 1 10 100 1 20 200 Table r2 A D ----------- 1 1000 1 1001 Table s (natural join of r1 and r2) A B C D --------------------------- 1 10 100 1000 1 20 200 1000 1 10 100 1001 1 20 200 1001

**2) Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?**

(a) 2

(b) 3

(c) 4

(d) 5

Answer (b)

See http://geeksquiz.com/gate-gate-cs-2005-question-75/ for explanation.

**3) Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold: {A–>B, BC–> D, E–>C, D–>A}. What are the candidate keys of R?**

(a) AE, BE

(b) AE, BE, DE

(c) AEH, BEH, BCH

(d) AEH, BEH, DEH

Answer (d)

A set of attributes S is candidate key of relation R if the closure of S is all attributes of R and there is no subset of S whose closure is all attributes of R.

Closure of AEH, i.e. AEH+ = {ABCDEH}

Closure of BEH, i.e. BEH+ = {ABCDEH}

Closure of DEH, i.e. DEH+ = {ABCDEH}

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