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Find Prime and Non-Prime Attribute:

Find Prime and Non – prime attribute from attribute closure:

Before seeing the chapter please follow the previous chapter: Concept of Attribute Closer, Find Super and Candidate Key from Attribute Closer.

Example 1:

A → R and A are minimal then attribute A is Super key and Candidate key also.

Calculating Candidate Keyand prime and non-prime attributes

PA (Prime Attribute) = {A,

CK (Candidate Key) = {A

Now we check from FDs where we will get prime attribute (PA) at the right-hand side of FD. If we get any PA to replace it with its left-hand side.

Why we do these because

We replace A with BC if A is key then BC also a key.

PA = {A, B, C}

CK = {A, BC}

So, A, B, C is Prime Attribute but there is no PA at left hand side of rest of FDs.

So, PA = {A, B, C} and CK = {A, BC}

Non-prime attribute (NPA) = {D, E, F}

Example 2:

CK = AB, PA = A, B

Now we search for PA at FDs on the right-hand side if present then replaces it by its left-hand side F → A (Here find PA – A at this FDs) i.e. we can replace A by F.

Check for Candidate Key (CK)

So, FB is SK and CK also.

CK = {AB, FB}

PA = {A, B, F}

Now again find PA at FDs.

We get

So, we can put E in place of F

Check for CK

So, EB is SK and CK also.

CK = {AB, FB, EB}

PA = {A, B, F, E}

Now again find PA at FDs 

Check for CK

So, DB is SK and CK also

CK = {AB, FB, EB, DB}

PA = {A, B, F, E, D}

Again we find Prime Attribute – D at C → D

Check for CK

So, DB is SK, CK

CK = {AB, FB, EB, DB}

PA = {A, B, F, E, D, C}

If we put again AB → C

CK and PA is final and NPA = ɸ = { }

If the only Relation is given with no FDs like R (ABCDE) we can make one trivial FD.

CK = {ABCDE}, PA = {A, B, C, D, E}, NPA = ɸ = { }