Application that the general definitions of 2NF and also 3NF may identify additional redundancy resulted in by dependencies that violate one or much more candidate keys. However, despite these additional constraints, dependencies have the right to still exist that will reason redundancy come be present in 3NF relations. This weakness in 3NF, caused the presentation the a more powerful normal form called Boyce–Codd Normal form (Codd, 1974).Although, 3NF is sufficient normal form for relational database, still, this (3NF) normal kind may not remove 100% redundancy since of X?Y sensible dependency, if X is not a candidate vital of offered relation. This deserve to be solve by Boyce-Codd Normal type (BCNF).

You are watching: A table where every determinant is a candidate key is said to be in _____.


Attention reader! Don’t stop discovering now. Practice GATE exam well before the really exam v the subject-wise and overall quizzes easily accessible in GATE Test collection Course.Learn all GATE CS concepts with totally free Live Classes on our youtube channel.
Boyce-Codd Normal type (BCNF):Boyce–Codd Normal form (BCNF) is based on functional dependencies the take right into account all candidate tricks in a relation; however, BCNF likewise has added constraints contrasted with the general meaning of 3NF.


A relationship is in BCNF iff, X is superkey because that every useful dependency (FD) X?Y in given relation.In various other words,A relationship is in BCNF, if and also only if, every determinant is a type (BCNF) candidate key.Note – To test whether a relation is in BCNF, we recognize all the determinants and also make sure that they space candidate keys.
The Normal type Hierarchy
You came across a comparable hierarchy recognized as Chomsky typical Form in concept of Computation. Now, very closely study the pecking order above. It deserve to be inferred the every relationship in BCNF is additionally in 3NF. To placed it another way, a relationship in 3NF need not to it is in in BCNF. Ponder over this statement because that a while.To identify the highest possible normal type of a offered relation R with useful dependencies, the first step is to check whether the BCNF condition holds. If R is uncovered to it is in in BCNF, it can be safely deduced that the relationship is additionally in 3NF, 2NF and 1NF together the power structure shows. The 1NF has actually the the very least restrictive constraint – it just requires a relation R to have actually atomic worths in every tuple. The 2NF has actually a slightly an ext restrictive constraint.Read this because that a clear knowledge of 2NFThe 3NF has an ext restrictive constraint 보다 the very first two typical forms but is much less restrictive than the BCNF. In this manner, the restriction increases as we traverse down the hierarchy.Example-1:
Find the highest possible normal form of a relation R(A, B, C, D, E) with FD set as: BC->D, AC->BE, B->E Explanation:Step-1: as we have the right to see, (AC)+ =A, C, B, E, D but none that its subset deserve to determine all attribute the relation, so AC will certainly be candidate key. A or C can’t be obtained from any type of other attribute that the relation, therefore there will certainly be only 1 candidate crucial AC.Step-2: Prime features are those attribute which are component of candidate vital A, C in this example and also others will be non-prime B, D, E in this example.Step-3: The relationship R is in first normal form as a relational DBMS go not enable multi-valued or composite attribute.The relation is in 2nd normal kind because BC->D is in 2nd normal form (BC is not a ideal subset of candidate vital AC) and also AC->BE is in second normal kind (AC is candidate key) and also B->E is in second normal type (B is not a proper subset of candidate vital AC).


The relation is no in 3rd normal type because in BC->D (neither BC is a super an essential nor D is a element attribute) and in B->E (neither B is a super crucial nor E is a element attribute) but to satisfy 3rd normal for, either LHS of an FD should be super an essential or RHS must be prime attribute. So the highest possible normal kind of relation will be 2nd Normal form.Note –
A element attribute cannot be transitively dependence on a vital in BCNF relation.Consider these useful dependencies of some relation R,AB ->CC ->BAB ->BSuppose, the is known that the only candidate vital of R is AB. A careful observation is compelled to conclude that the over dependency is a leg Dependency together the prime attribute B transitively depends on the vital AB v C. Now, the very first and the third FD room in BCNF together they both contain the candidate crucial (or simply KEY) on your left sides. The second dependency, however, is no in BCNF however is certainly in 3NF due to the presence of the element attribute on the appropriate side. So, the highest possible normal type of R is 3NF together all three FD’s meet the necessary conditions to it is in in 3NF.

See more: How Many Calories In 1/4 Cup Brown Sugar ? Brown Sugar Nutrition Facts

Example-2:For example consider relation R(A, B, C)A -> BC, B -> AA and also B both room super secrets so over relation is in BCNF.Note –BCNF decomposition may constantly not feasible with suspended preserving, however, it constantly satisfies lossless sign up with condition. Because that example, relation R (V, W, X, Y, Z), with practical dependencies:V, W -> XY, Z -> X W -> Y It would not accomplish dependency maintaining BCNF decomposition.Note -:Redundancies are periodically still current in a BCNF relation as it is no always feasible to eliminate them completely.Refer for:4th and 5th Normal type and recognize the highest normal form of a provided relation.