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CS6400 - Exam Questions Solved 100% Correct | Verified Answers
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CS6400 - Exam Questions Solved 100% Correct |Verified Answers
First Normal Form (1NF) - ✔✔A relation that has a primary key and in which there are no
repeating groups. 1NF disallows relations within relations or relations as attribute values within
tuples. all domain values are atomic.
Second Normal Form (2NF) - ✔✔is based on the concept of full functional dependency. a
relation is in 1NF and there are no partial dependencies. every nonkey attribute is fully
dependent on the key.
full functional dependency - ✔✔removal of any attribute A from X means that the
dependency does not hold anymore
partial dependency - ✔✔if some attribute A ε X can be removed from X and the dependency
still holds
Third Normal Form (3NF) - ✔✔is based on the concept of transitive dependency. R is 2NF
and every nonkey attribute is non-transitively dependent on the key.
transitive dependency - ✔✔if X → Y and Y → Z, then X → Z
general alternative definition of 3NF - ✔✔A relation schema R is in 3NF if every nonprime
attribute of R meets both of the following conditions: It is fully functionally dependent on every
key of R.
all-key relation - ✔✔obvious redundancy in the relation
X → → Y - ✔✔X multi-determines Y
,Fourth Normal Form (4NF) - ✔✔A table is in 4NF if it is in 3NF and contains no multiple
independent sets of multivalued dependencies.
join dependency (JD) - ✔✔specifies a constraint on the states r of R.
Fifth Normal Form (5NF) - ✔✔A normal form necessary to eliminate an anomaly where a
table can be split apart but not correctly joined back together.
Also known as Project-Join Normal Form (PJ/NF).
relational design by synthesis - ✔✔bottom-up approach to design that presupposes that the
known functional dependencies among sets of attributes in the Universe of Discourse (UoD)
have been given as input.
properties of decompositions - ✔✔the dependency preservation property and the
nonadditive (or lossless) join property
universal relation - ✔✔hypothetical relation containing all the attributes
Armstrong's axioms - ✔✔IR1 (reflexive rule)2: If X ⊇ Y, then X →Y. IR2 (augmentation rule)3:
{X → Y} |=XZ → YZ. IR3 (transitive rule): {X → Y, Y → Z} |=X → Z.
Proof of IR2 (contradiction) - ✔✔the rule does not hold and shows that this is not possible
Proof of IR1 - ✔✔Suppose that X ⊇ Y and that two tuples t1 and t2 exist in some relation
instance r of R such that t1 [X] = t2 [X]. Then t1[Y] = t2[Y] because X ⊇ Y; hence, X → Y must
hold in r.
Proof of IR3 - ✔✔Assume that (1) X → Y and (2) Y → Z both hold in a relation r., X → Z must
hold in r.
, Minimal Cover (of a set of FDs) - ✔✔A minimal cover of a set of FDs G is a minimal set of FDs
that is equivalent to G
extraneous attribute - ✔✔if we can remove it without changing the closure of the set of
dependencies.
minimal cover - ✔✔E is a minimal set of dependencies that is equivalent to E . a set of
dependencies in a standard or canonical form and with no redundancies.
universal relation assumption - ✔✔states that every attribute name is unique
D (decomposition ) - ✔✔is called a decomposition of R. decompose the universal relation
schema R into a set of relation schemas D = {R1, R2, ... , Rm} that will become the relational
database schema
attribute preservation - ✔✔make sure that each attribute in R will appear in at least one
relation schema Ri in the decomposition so that no attributes are lost
dependency preserving - ✔✔union of the projections of F on each Ri in D is equivalent to F;
that is, ((πR1(F)) ∪ K ∪ (πRm(F)))+ = F+.
projection of F on Ri, - ✔✔πRi(F) where Ri is a subset of R, is the set of dependencies X → Y
in F+ such that the attributes in X ∪ Y are all contained in Ri
lossless joins guarantee - ✔✔the join filed must be a key in at least one of the relations
lossless - ✔✔refers to loss of information, not to loss of tuples
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