Hey Everyone,

Post your solutions here for the Sep 25 In-Class Exercise.

Best,

Chris

R{ABCDEF}
FD: A->BE: R1(ABE) R2{ACDF) R1 is good. FD: C->D: R2 splits to: R3(C, D, B), R4(C, F)

R0(A, B, E) having FD(s): A → B; A → E. R1(C, D) having FD(s): C → D. R2(B, D) having FD(s): D → B. R4(A, C, F) having FD(s): (none).

R(A, B, C, D, E, F) can be split into R1(A, B, E) and R2(A, C, D, F) Now R1 is in BCNF. For R2, We can split it into R3(C, D, B) and R4(C ,F)

1. looking at A->BE, A+ = {A, B, E} != {A, B, C, D, E}, so R is not in BCNF.
2. A+ = {A, B, E}

so R1 = (A, B, E) and R1 = (C, D, F)

1. Not sure how to apply Monday's algorithm from there...

I will guess that FD1 = (A->B, A->E) and FD2 = {C->D}

1. Recursively check R1 and R1 and split if needed.

R(A,B,C,D,E, F) is decomposed into a set of relations: R1(A,B,E), R2(A,C,D,F). Here, R1 satisfies BCNF. Here, R2 can be split into R3(C,D,B) and R4(C,F)

Consider the relation R(A,B,C,D,E,F) with FD A→BE, C→D, D→B. R1(A,B,E) R2(A,C,D,F)-> R3(A,C,D)+R4(A,F)

R(A, B, C, D, E, F) is not in BCNF due to A->BE

R1 = A+ = {A, B, E}

R2= {A,C,D,F}

R1 now is in BCNF

for R2, it is split into R3 = {C,D,B} ABD R4={C,F}

both of them are in BCNF

so R1, R2, R3 sastify BCNF

R(A,B,C,D,E,F) with FD A→BE, C→D, D→B

1. A is not a super key
2. R1 = X+ = {A}+ R2 = X U R-X+

   FDs: A->B, A->E, C->D, D->B

   Let X:= {A}, Xold := empty Loop 1: Xold := {A}

	  X = {A, B, E}

Loop 2: Xold = {A, B, E} X = {A, B, E} {A}+ = {A, B, E}

1. R1={A,B,E} S1 = {A-> B, A-> E} this is in BCNF R2={A,C,D,F} S2 = {C -> D, D -> B} Not in BCNF
2. Decompose C -> D (to replace R2 = {A,C,D,F}, S2 = {C -> D, D->B}) R3 = {C}+

   Let X:= {C}, Xold := empty Loop 1: Xold := {C}

	X = {C, D}

Loop 2: Xold := {C,D} X = {C, D, B} Loop 3: Xold := {C, D, B} X = {C, D, B}

1. Answers R1 = {A,B,E} in BCNF S1 = {A -> B, A-> E}
   R3 = {C, D, B} in BCNF S3 = {C -> D, D -> B} R4 = {A, C, F} in BCNF S4 = {}

1. R(A,B,C,D,E,F) FD A->BE, C->D, D->B

A->BE is a violation of BCNF we compute A+ A+ = R1 = R(A,B,E) R2 = A and everything that is not in A+ then we have R(A,C,D,F) Now R1 is in BCNF C+ = R3 = C -> D we have R(C,D,B) now we take C and everything that is not in R3 for R4 we have R(C,F)