Non Trivial FD. Candidate Key FD’s that Hold on S.

Non Trivial FD

Candidate Key

FD’s that Hold on S

3NF

Consider R = {A, B, C, D, E, F, G, H} with a set of FDs

F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B}

The candidate keys are:

{BEFG, CEFG, EFGH}

F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B}

No, R w.r.t. F is NOT in 3NF, because CD→A violates the 3NF requirements.

i.e.

• CD→A is not trivial FD

• CD is not a superkey

• CD is not a key, but A is not part of any key of R either

Binary Decomposition Approach

Considering R:

Keys ={BEFG, CEFG, EFGH}

F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B} Decomposition #1:

CD→A is a violating FD

R is decomposed into R1 and R2:

R1 (A,C,D):

We need to project FDs F onto relation R1:

A+ = A

C+ = CDA (C →DA)

D+ = D

AC+ = ACD (AC→D)

AD+ = AD

CD+ = CDA (CD → A)

So, F1 = {C→DA, AC→D, CD→A}

R2 ( B,C,D,E,F,G,H):In general, we should project F onto R2. However, if we

look carefully, we can easily see that the only difference between R and R2 is attribute A. Attribute A has never appeared on LHS of any FD. So, removing it won’t make any change in F2.

So, F2={EC→H, GHB→B, C→D, H→B, BE→CD, EC→B} FDs that are lost in Decomposition #1 are:Lost = {GHB→A, EG→A}

Do we need further decomposition?

Consider R1(A,C,D):F1 = {C→DA, AC→D, CD→A}Since C+=ACD, C is a key.C (in C→DA), CD (in CD→A), and AC (in AC→D) are key/super keys.

Therefore, we have no violating FD. (So, we are done with this branch.)

Consider R2: F2={EC→H, C→D, H→B, BE→CD, EC→B}

Keys of R2 = Keys of R = {BEFG, CEFG, EFGH}EC→H is not a violating FD, since H is part of a key.C→D is a violating FD, since C is not a super key and D is not part of any

key.

So, further decomposition is needed.

Decomposition #2:C→D is a violating FDR2 is decomposed into R21 and R22:

R21 (C, D):We need to project F2 onto relation R21: C+=CD D+=DSo, F21={C→D}

R22 (B,C,E,F,G,H):In general, we should project F2 onto R22. However, if we look

carefully, we can easily see that the only difference between R2 and R22 is attribute D. Attribute D has never appeared on LHS of any FD. So, removing it won’t make any change in F22.

So, F22 = {EC→H, H→B, BE→C, EC→B}

FDs that are lost in Decomposition #2 is:Lost = {BE→D}So, overall, we’ve lost the following FDs:Lost = {GHB→A, EG→A, BE→D}Do we need further decomposition?Consider R21(C,D): F21={C→D}Since C+=CD, C is a key. Therefore, we have no

violating FD. (So, we are done with this branch.)Consider R22(B,C,E,F,G,H): Keys of R22 = Keys of R2 = {BEFG, CEFG, EFGH}

F22 = {EC→H, H→B, BE→C, EC→B}EC→H is not a violating FD since H is part of a key.H→B is not a violating FD since B is part of a key.BE→C is not a violating FD since C is part of a key.EC→B is not a violating FD since B is part of a key.

So, we are done with this branch.Overall, we have:R1 (A, C, D) F1 = {C→DA, AC→D,

CD→A}R21 (C, D) F21 = {C→D}R22 (B, C, E, F, G, H) F22 = {EC→H, H→B, BE→C,

EC→B}

Since R1 includes R21 we might want to remove R21. This is a loss-less join decomposition, but it is not

dependency preserving.To make the decomposition dependency preserving, we

need to add the lost FDs as new relations.The lost FDs are:Lost = {GHB→AB, EG→A, BE→D}

So, we add three relations:L1(A, B, G, H) FL1 = {GHB→AB}L2(A, E, G) FL2 = {EG→A}L3(B, D, E) FL3 = {BE→D}

Synthesis Approach

R = {A, B, C, D, E, F, G, H} with a set of FDs

F = {CD→A, EC→H, GHB→AB, C→D, EG→A, H→B, BE→CD, EC→B}

The candidate keys are {BEFG, CEFG, EFGH}

Canonical cover for F is:

FC = {C→AD, EC→H, GH→A, EG→A, H→B, BE→C}

FC = {C→AD, EC→H, GH→A, EG→A, H→B, BE→C}

Now, we create the relations:R1 = {A, C, D} F1 = {C→AD}R2 = {E, C, H} F2 = {EC→H}R3 = {A, G, H} F3 = {GH→A}R4 = {A, E, G} F4 = {EG→A}R5 = {B, H} F5 = {H→B}R6 = {B, C, E} F6 = {BE→C}Now, we need to check if at least one of the keys

exists in the above relations.The candidate keys are {BEFG, CEFG, EFGH}

Since none of these keys is in the relations, this decomposition is not lossless. So, we need to add an extra relation containing those attributes that form any key of R:

R7 = {B, E, F, G}F7 = { }

Check lossless join

Check if the decomposition of R(A, B, C, D, E, F, G) with the set of FDs F={C→AD, E→G, FG→A, EF→A, G→B, BE→C} into the following relations is lossless join.R1 = {A, C, D}R2 = {E, C, G} R3 = {A, F, G} R4 = {A, E, F}R5 = {B, G} R6 = {B, C, E}

Step 1- Table initialization

Round 1

Round 2