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Unit 2: Relational Query Languages

20247m

Find the highest normal form of a relation R(ABCD) with FD set
(B→A, A→C, BC→D, AC→BE).

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Find highest normal form of R(ABCD) with FDs B->A, A->C, BC->D, AC->BE. Note E not in R - likely typo, assume R is ABCDE or FD is AC->B. Proceed with typical interpretation R(ABCD) with B->A, A->C, BC->D. Compute candidate keys:

Closure: B+ = B -> A -> C, BC -> D, so B+ = ABCD. So B is candidate key (single attribute). Also BC is superkey etc. Only B is CK.

Check 1NF: assume atomic, so 1NF.
2NF: No partial dependency because CK is single attribute B, no proper subset, so automatically 2NF.
3NF: Check each FD: B->A where B is superkey so OK. A->C where A not superkey and C non-prime -> transitive violation, so not 3NF.
Hence highest is 2NF.

If considering FD AC->BE with E present and R as ABCDE: CK still B (B+= ABCDE via AC->BE gives B->AC->BE then BC->D), same violation A->C makes not 3NF, highest 2NF. If strict, R(ABCD) typo, answer remains 2NF. Mention anomaly if E included needs BCNF check fails.

Provide steps for closure, CK, normal form tests.

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