Consider a relation R(P,Q,R,S,T) with the following functional dependencies:…

🔹 To determine if the decomposition of R(P,Q,R,S,T) into R1(P,Q,S) and R2(R,S,T) is Lossless and/or Dependency Preserving, we must evaluate both properties systematically.

1. Lossless Join Property Test

A decomposition of R into R1 and R2 is lossless if and only if the intersection of their attributes functionally determines all attributes of at least one of the relations:

$$(R_1 \cap R_2) \rightarrow R_1 \quad \text{or} \quad (R_1 \cap R_2) \rightarrow R_2$$

1. Find the intersection:

   $$R_1 \cap R_2 = \{P, Q, S\} \cap \{R, S, T\} = \{S\}$$

2. Compute the attribute closure of the intersection under the original FDs (S+):

   • Start with S+ = {S}

   • Using S -> P, we get S+ = {P, S}

   • Using P -> QT, we get S+ = {P, Q, S, T}

   • Using T -> S (no new attributes)

   • Result: S+ = {P, Q, S, T}

3. Check containment:

   • Is R1 ⊆ S+? {P, Q, S} ⊆ {P, Q, S, T} -> Yes!

   • Is R2 ⊆ S+? {R, S, T} ⊄ {P, Q, S, T} (Missing R)

Since (R1 ∩ R2) -> R1, the decomposition is Lossless.

2. Dependency Preservation Test

A decomposition is dependency preserving if the union of the closures of FDs projected onto each individual relation contains all original FDs: (F1 ∪ F2)+ = F+.

Let's find the projected functional dependencies for R1 and R2:

FDs for R1(P,Q,S):

Compute closures using original FDs to find valid dependencies restricted only to {P, Q, S}:

• P+ = {P, Q, S, T} ∩ {P, Q, S} = {P, Q, S} => {P -> QS}

• S+ = {P, Q, S, T} ∩ {P, Q, S} = {P, Q, S} => {S -> PQ}

• Q+ = {Q}

FDs for R2(R,S,T):

Compute closures restricted only to {R, S, T}:

• T+ = {T, S} ∩ {R, S, T} = {S, T} => {T-> S}

• S+ = {P, Q, S, T}∩ {R, S, T} = {S, T} => {S -> T}

• (RS)+ = {P, Q, R, S, T} ∩ {R, S, T} = {R, S, T} => {RS -> T}

Check original FDs against (F1 ∪ F2):

• T -> S: Preserved directly in F2.

• S -> P: Preserved directly in F1.

• P -> QT: P -> Q is in F1. To see if P -> T is preserved, compute P+ under (F1 ∪ F2):

  $$P^+ = \{P\} \xrightarrow{P \rightarrow QS} \{P, Q, S\} \xrightarrow{S \rightarrow T} \{P, Q, S, T\}$$

  Since T is in the closure, P -> QT is preserved.

• RS -> Q: Compute (RS)+ under (F1 ∪ F2):

  $$({RS})^+ = \{R, S\} \xrightarrow{S \rightarrow PQ} \{P, Q, R, S\}$$

  Since Q is in the closure, RS -> Q is preserved.

All original functional dependencies are preserved, so the decomposition is Dependency Preserving.

Conclusion

The decomposition is both Lossless and Dependency Preserving.