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Database Management System | Dependency Preserving Decomposition

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  • Difficulty Level : Medium
  • Last Updated : 18 Jan, 2023
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Dependency Preservation: A Decomposition D = { R1, R2, R3…Rn } of R is dependency preserving wrt a set F of Functional dependency if

(F1 ∪ F2 ∪ … ∪ Fm)+ = F+.
Consider a relation R
R ---> F{...with some functional dependency(FD)....}

R is decomposed or divided into R1 with FD { f1 } and R2 with { f2 }, then
there can be three cases:

f1 U f2 = F -----> Decomposition is dependency preserving. 
f1 U f2 is a subset of F -----> Not Dependency preserving.
f1 U f2 is a super set of F -----> This case is not possible.


Let a relation R (A, B, C, D ) and functional dependency {AB –> C, C –> D, D –> A}. Relation R is decomposed into R1( A, B, C) and R2(C, D). Check whether decomposition is dependency preserving or not. 


R1(A, B, C) and R2(C, D)

Let us find closure of F1 and F2
To find closure of F1, consider all combination of
ABC. i.e., find closure of A, B, C, AB, BC and AC
Note ABC is not considered as it is always ABC 

closure(A) = { A }  // Trivial
closure(B) = { B }  // Trivial
closure(C) = {C, A, D} but D can't be in closure as D is not present R1.
           = {C, A}
C--> A   // Removing C from right side as it is trivial attribute

closure(AB) = {A, B, C, D}
            = {A, B, C}
AB --> C  // Removing AB from right side as these are trivial attributes

closure(BC) = {B, C, D, A}
            = {A, B, C}
BC --> A  // Removing BC from right side as these are trivial attributes

closure(AC) = {A, C, D}
AC --> D  // Removing AC from right side as these are trivial attributes             

F1 {C--> A, AB --> C, BC --> A}.
Similarly F2 { C--> D }

In the original Relation Dependency { AB --> C , C --> D , D --> A}.
AB --> C is present in F1.
C --> D is present in F2.
D --> A is not preserved.

F1 U F2 is a subset of F. So given decomposition is not dependency preserving.

Question 1: Let R (A, B, C, D) be a relational schema with the following functional dependencies:

A → B, B → C,
C → D and D → B. 

The decomposition of R into 
(A, B), (B, C), (B, D)

(A) gives a lossless join, and is dependency preserving 

(B) gives a lossless join, but is not dependency preserving 

(C) does not give a lossless join, but is dependency preserving 

(D) does not give a lossless join and is not dependency preserving 

Refer to this for a solution. 

Question 2 R(A, B, C, D) is a relation. Which of the following does not have a lossless join, dependency-preserving BCNF decomposition? 

(A) A->B, B->CD 

(B) A->B, B->C, C->D 

(C) AB->C, C->AD 

(D) A ->BCD

 Refer to this for a solution. Below is the Quiz of the previous year’s GATE Questions. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above

Imp: The 1NF, 2NF, and 3NF are valid for dependency-preserving decomposition.

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