DMDB 7

Is A a candidate key given:

R(A,B,C,D,E,F)

A-->B

A-->EF

BE-->CD

• Yes

• No

Answer explanation

The closure of A includes the whole relation R:

A-->B thus: {A,B}

A-->EF thus: {A,**B,E**,F}

BE-->CD thus {A,B,C,D,E,F}

Is AB a candidate key given the same FDs:

R(A,B,C,D,E,F)

A-->B

A-->EF

BE-->CD

• Yes

• No

Answer explanation

No. Keys must be minimal. Since A is already a candidate key (= minimal set of attributes), AB is not minimal and thus not a candidate key!

Is AB a super key given this (different set of FD):

R(A,B,C,D,E,F)

A-->B

A-->F

BE-->CD

• Yes

• No

Answer explanation

AB is not a superkey, since the closure is not the whole relation!

Is ABC a super key given:

R(A,B,C)

A-->B

• Yes

• No

Answer explanation

Yes. AC would be a (minimal) candidate key, however as we are interested in a super key, and the whole relation is trivially a superkey, this holds.

The main goal(s) of the minimal basis algorithm is to/are to

• Reduce the number of functional dependencies

• Remove trivial dependencies

• Get to a normal form

• Ensure data integrity

• Retain an equivalent set of functional dependencies

Answer explanation

See the lecture slides! Otherwise: Removing trivial FDs is just a small step in the algorithm. Getting to normal forms needs more than just the minimal basis. Data integrity is not related to the algorithm, but rather to the concept of functional dependencies, ...

Why are redundancies problematic?

• They waste storage space

• They need to have application side logic

• They introduce UPDATE-Anomalies

• They introduce SELECT-Anomalies

Answer explanation

Also: Redundancy introduces duplication, which wastes sotrage space and needs application side logic to handle those copies.

Select all valid functional dependencies implied by this ER diagram.

• A-->B

• AC-->B

• BC-->A

• AB-->C

• C-->B

Answer explanation

Due to the cardinality constraints, given A,C we know B and given BC we know A. Thus, those are the valid functional dependencies.