Question

5. Are the following relations in BCNF? 3NF? (if R is not in BCNF, decompose it to BCNF; if R is not in 3NF, decompose it to 3NF)
   1. R(X,Y, Z,T,V): XY->Z, Y->T, Z->V
   2. R(X,Y,Z,T): X->Y, Y->Z, Z->T
   3. R(A,B,C): AB->C, B->A, C->B
   4. R(ABCD): BD->C, AB->D, AC->B, BD->A
   5. R(ABCD): AD->C, CD->B, BD->C
   6. R(ABCD): A->C, B->A, A->D, AD->C
   7. R(ABCD): A->D, C->A, D->B, AC->B
   8. R(XYZT): XYT->Z, ZT->X, XZ->Y, XZ->T
   9. R(XYZT): XY->Z, XYT->Z, XYZ->T, XZ->T
   10. R(XYZT): YT->Z, XY->T, XZ->Y, YT->X
   11. R(XYZT): YZ->X, XT->Z, ZT->Y, YT->Z

Answer 1

Please find the solutions below in the screenshots,

Answer 2

R(X,Y, Z,T,V): XY->Z, Y->T, Z->V

This relation is not in BCNF, as the determinant Y does not contain all the attributes of the relation. Therefore, we need to decompose it into two relations: R1(X,Y,Z) and R2(Y,T,V). Both relations are in BCNF.

R(X,Y,Z,T): X->Y, Y->Z, Z->T

This relation is already in BCNF, as all determinants are candidate keys.

R(A,B,C): AB->C, B->A, C->B

This relation is already in BCNF, as all determinants are candidate keys.

R(ABCD): BD->C, AB->D, AC->B, BD->A

This relation is not in 3NF, as there is a transitive dependency between AC and D. Therefore, we need to decompose it into three relations: R1(ABD), R2(ACB), and R3(BCD). All three relations are in 3NF.

R(ABCD): AD->C, CD->B, BD->C

This relation is not in BCNF, as the determinant BD does not contain all the attributes of the relation. Therefore, we need to decompose it into two relations: R1(ABD) and R2(BCD). Both relations are in BCNF.

R(ABCD): A->C, B->A, A->D, AD->C

This relation is not in BCNF, as the determinant A does not contain all the attributes of the relation. Therefore, we need to decompose it into two relations: R1(ACD) and R2(ABD). Both relations are in BCNF.

R(ABCD): A->D, C->A, D->B, AC->B

This relation is not in BCNF, as the determinant AC does not contain all the attributes of the relation. Therefore, we need to decompose it into two relations: R1(ABCD) and R2(BCD). Both relations are in BCNF.

R(XYZT): XYT->Z, ZT->X, XZ->Y, XZ->T

This relation is not in BCNF, as the determinant XYT does not contain all the attributes of the relation. Therefore, we need to decompose it into two relations: R1(XYZ) and R2(XZT). Both relations are in BCNF.

R(XYZT): XY->Z, XYT->Z, XYZ->T, XZ->T

This relation is already in BCNF, as all determinants are candidate keys.

R(XYZT): YT->Z, XY->T, XZ->Y, YT->X

This relation is already in BCNF, as all determinants are candidate keys.

R(XYZT): YZ->X, XT->Z, ZT->Y, YT->Z

This relation is already in BCNF, as all determinants are candidate keys.