Questions on Lossless Join

Questions on Lossless Join To Identify whether a decomposition is lossy or lossless, it must satisfy the following conditions :

1. R₁  ∪ R₂ = R
2. R₁  ∩ R₂ ≠ Φ and
3. R₁  ∩ R₂  → R₁  or R₁  ∩ R₂  → R₂

Question 1 :
R(ABC)
F = {A → B, A → C} decomposed into
D = R1(AB), R2(BC)
Find whether D is Lossless or Lossy ?

Solution :
D = {AB, BC}
Step 1: AB ∪ BC = ABC
Step 2: AB ∩ BC = B               //Intersection
Step 3: B+ = {B}                  //Not a super key of R1 or R2
⇒ Decomposition is lossy.

Question 2 :
R(ABCDEF)
F = {A → B, B → C, C → D, E → F} decomposed into
D = R1(AB), R2(BCD), R3(DEF).
Find whether D is Lossless or Lossy ?

Solution :
Step 1: AB ∪ BCD ∪ DEF = ABCDEF = R  // Condition 1 satisfies
step 2: AB ∩ BCD = B
        B+ = {BCD}        //superkey of R2
     ⇒ R12(ABCD)

        ABCD ∩ DEF = D
        D+ = {D}         // Not a superkey of R12 or R3
     ⇒ Decomposition is Lossy.

Question 3 :
R(ABCDEF)
F = {A → B, C → DE, AC → F} decomposed into
D = R1(BE), R2(ACDEF).
Find whether D is Lossless or Lossy ?

Solution :
Step 1: BE ∪ ACDEF = ABCDEF = R  // Condition 1 satisfies
step 2: BE ∩ ACDEF = E
        E+ = {E}        //Not a superkey of R1 or R2
     ⇒ Decomposition is Lossy.

Question 4 :
R(ABCDEG)
F = {AB → C, AC → B, AD → E, B → D, BC → A, E → G} decomposed into
(i)  D1 = R1(AB), R2(BC), R3(ABDE), R4(EG).
(ii) D2 = R1(ABC), R2(ACDE), R3(ADG).
Find whether D1 and D2 is Lossless or Lossy ?

Solution (i) :
Step 1: AB ∪ BC ∪ ABDE ∪ EG = ABCDEG = R  // Condition 1 satisfies
step 2: AB ∩ BC = B
        B+ = {BD}        //Not a superkey of R1 or R2
     ⇒ Decomposition is Lossy. No need to check further.

Solution (ii) :
Step 1: ABC ∪ ACDE ∪ ADG = ABCDEG = R  // Condition 1 satisfies
step 2: ABC ∩ ACDE = AC
        AC+ = {ACBDEG}        //superkey
     ⇒ R12(ABCDE)

        ABCDE ∩ ADG = AD
        AD+ = {ADEG}         //Superkey of R3
     ⇒ R123(ABCDEG)
     ⇒ Decomposition is LossLess.

Question 5 :
R(ABCDEFGHIJ)
F = {AB → C, B → F, D → IJ, A → DE, F → GH} decomposed into
(i)   D1 = R1(ABC), R2(ADE), R3(BF), R4(FGH),R5(DIJ).
(ii)  D2 = R1(ABCDE), R2(BFGH), R3(DIJ).
(iii) D3 = R1(ABCD), R2(DE), R3(BF), R4(FGH),R5(DIJ).
Find whether D1, D2 and D3 is Lossless or Lossy ?

Solution (i) :
Step 1: ABC ∪ ADE ∪ BF ∪ FGH ∪ DIJ = ABCDEFGHIJ = R  // Condition 1 satisfies
step 2: ABC ∩ ADE = A
        A+ = {ADEIJ}        //Superkey of R2
     ⇒ R12(ABCDE)

        ABCDE ∩ BF = B
        B+ = {BFGH}         //superkey of R3
     ⇒ R123(ABCDEF)

        ABCDEF ∩ FGH = F
        F+ = {FGH}          //Superkey of R4
     ⇒ R1234(ABCDEGH)

        ABCDEFGH ∩ DIJ = D
        D+ = {DIJ}          //Superkey of R5
     ⇒ R12345(ABCDEGHIJ)
     ⇒ Decomposition is LossLess.

Solution (ii) :
Step 1: ABCDE ∪ BFGH ∪ DIJ = R  // Condition 1 satisfies
step 2: ABCDE ∩ BFGH = B
        B+ = {BFGH}        //Superkey of R2
     ⇒ R12(ABCDEFGH)

        ABCDEFGH ∩ DIJ = D
        D+ = {DIJ}         //superkey of R3
     ⇒ R123(ABCDEFGHIJ)
     ⇒ Decomposition is LossLess.

Solution (iii) :
Step 1: ABCD ∪ DE ∪ BF ∪ FGH ∪ DIJ = ABCDEFGHIJ = R  // Condition 1 satisfies
step 2: ABCD ∩ DE = D
        D+ = {DIJ}        //Superkey of R2
     ⇒ R12(ABCDE)

        ABCDE ∩ BF = B
        B+ = {BFGH}         //superkey of R3
     ⇒ R123(ABCDEF)

        ABCDEF ∩ FGH = F
        F+ = {FGH}          //Superkey of R4
     ⇒ R1234(ABCDEGH)

        ABCDEFGH ∩ DIJ = D
        D+ = {DIJ}          //Superkey of R5
     ⇒ R12345(ABCDEGHIJ)
     ⇒ Decomposition is LossLess.

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