1. CPSC-310 Database Systems
Professor Jianer Chen
Room 315C HRBB
Lecture #6

2. Computing Closure Computing the closure Y+ of a set Y of attributes.
Algorithm Closure(Y)
1. Y+ = Y;
2. While changes Do
2.1 look for an FD X → B where X is a subset of Y+ but B is not in Y+;
2.2 add B to Y+;
3. every A in Y+\Y satisfies Y → A Y+
new Y+ X A
It can be proved that the Algorithm Closure(Y) finds exactly those FDs Y → A implied by the given FDs.

3. Eliminating bad FD X → A (for all A)
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

4. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.

5. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;

6. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;
2. Record the superkeys and keys Y for R (Y+=Z);

7. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;
2. Record the superkeys and keys Y for R (Y+=Z);
3. If there is a subset Y of Z such that
Y+≠ Z and Y+≠ Y \\ Y → A is a bad FD for some A

8. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;
2. Record the superkeys and keys Y for R (Y+=Z);
3. If there is a subset Y of Z such that
Y+≠ Z and Y+≠ Y \\ Y → A is a bad FD for some A
Then
3.1 Call Decomposition(Y) to decompose the
relation R into two smaller relations R1 and R2;

9. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;
2. Record the superkeys and keys Y for R (Y+=Z);
3. If there is a subset Y of Z such that
Y+≠ Z and Y+≠ Y \\ Y → A is a bad FD for some A
Then
3.1 Call Decomposition(Y) to decompose the
relation R into two smaller relations R1 and R2;
3.2 Call Decomposed-FDs(Rk) for k = 1, 2 to
construct the FD sets T1 for R1 and T2 for R2;

10. Normalization Algorithm
Algorithm Normalization(R, T)
Input: A relation R with attribute set Z and FD set T.
1. For each subset Y of Z Do construct Y+;
2. Record the superkeys and keys Y for R (Y+=Z);
3. If there is a subset Y of Z such that
Y+≠ Z and Y+≠ Y \\ Y → A is a bad FD for some A
Then
3.1 Call Decomposition(Y) to decompose the
relation R into two smaller relations R1 and R2;
3.2 Call Decomposed-FDs(Rk) for k = 1, 2 to
construct the FD sets T1 for R1 and T2 for R2;
Example. R(A, B, C, D, E)
T = {AB → D, AC → E, BC → D, D → A, E → B}

11. Eliminating bad FD X → A (for all A)
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

12. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

13. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A:
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

14. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1;
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

15. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1; * X→A is not FD in R2=(X∪W) because A is not in X∪W.
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

16. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1; * X→A is not FD in R2=(X∪W) because A is not in X∪W.
Does Decomposition(X) change (i.e., lose or add extra) information for R?
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

17. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1; * X→A is not FD in R2=(X∪W) because A is not in X∪W.
Does Decomposition(X) change (i.e., lose or add extra) information for R: No.
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

18. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1; * X→A is not FD in R2=(X∪W) because A is not in X∪W.
Does Decomposition(X) change (i.e., lose or add extra) information for R: No.
Can R1 and R2 still have bad FDs?
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

19. Eliminating bad FD X → A (for all A)
Remarks.
Algorithm Decomposition(X) eliminates the bad FD X→A: * X→A is still an FD in R1=(X+) but now X is a superkey for R1; * X→A is not FD in R2=(X∪W) because A is not in X∪W.
Does Decomposition(X) change (i.e., lose or add extra) information for R: No.
Can R1 and R2 still have bad FDs: Yes.
Algorithm Decomposition(X)
1. Compute S1 = X+ using the FDs in R;
2. Let W be the set of attributes that
are not in S1;
3. Make a relation R1 of schema S1;
4. Make a relation R2 with schema
S2 = X ∪ W;
5. Compute the FDs for R1 and R2.
Algorithm Decomposed-FDs(R1)
1. T1 = ; \\ T1 is the FDs for R1
2. For each subset Y of S1 Do
2.1 compute Y+ using the FDs in R;
2.2 For each A in S1∩(Y+\Y)
add Y → A to T1;
3. While changes Do
3.1 Drop from T1 those FDs that are
derivable from the others;
3.2 For each XB → A in T1, if X → A is
implied by T1, then replace XB → A
in T1 by X → A.

20. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.

21. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)

22. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 1.
Likes(drinker, addr, zip, beer, manf)
FD’s: T = {drinker → addr, addr → zip, beer → manf}

23. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 1.
Likes(drinker, addr, zip, beer, manf)
FD’s: T = {drinker → addr, addr → zip, beer → manf}
The only key is {drinker, beer}.

24. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 1.
Likes(drinker, addr, zip, beer, manf)
FD’s: T = {drinker → addr, addr → zip, beer → manf}
The only key is {drinker, beer}.
In each FD, the left side is not a superkey.

25. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 1.
Likes(drinker, addr, zip, beer, manf)
FD’s: T = {drinker → addr, addr → zip, beer → manf}
The only key is {drinker, beer}.
In each FD, the left side is not a superkey.
Any one of these FD’s shows Likes is not in BCNF

26. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 2.
Beers(name, manf, manfAddr)
FD’s: T = {name → manf, manf → manfAddr}

27. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 2.
Beers(name, manf, manfAddr)
FD’s: T = {name → manf, manf → manfAddr}
The only key is {name}.

28. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 2.
Beers(name, manf, manfAddr)
FD’s: T = {name → manf, manf → manfAddr}
The only key is {name}.
name → manf does not violate BCNF.

29. Boyce-Codd Normal Form (BCNF)
Definition. A relation R is in Boyce-Codd Normal Form (BCNF) if every nontrivial FD X→A (i.e., AX) has its left side X a superkey.
(Thus, a relation in BCNF has no bad FDs.)
Example 2.
Beers(name, manf, manfAddr)
FD’s: T = {name → manf, manf → manfAddr}
The only key is {name}.
name → manf does not violate BCNF.
However, manf → manfAddr violates BCNF.