1. Relational Database Design (Discussion Session)
Jianguo Wang
Office hour: Friday 5pm – 6pm at CSE 3217

2. Functional dependency
A functional dependency (FD) has the form of X → Y (X implies Y), where X and Y are sets of attributes
Whenever two tuples are identical on all the attributes in X, they must also be identical in all the attributes in Y .
Examples
sid → sname
{sid, cid} → gpa

3. Closure of an attribute set
The closure of a set X of attributes is the set of all such attributes A that X → A can be deduced from F
R = ABCDEF F = {AC, BCD, ADE} X = AB
Computing X+ :
Initially: X(0) = AB
Applying AC: X(1) = ABC
Applying BCD: X(2) = ABCD
Applying ADE: X(3) = ABCDE
No more changes: X(4) = X(3)
X+ = ABCDE

4. Candidate key A candidate key is a set X of attributes in R such that
X+ includes all the attributes in R.
There is no proper subset Y of X such that Y+ includes all the attributes in R.
A super key is a super set of key
Example: R(A, B, C, D), F = {A → B, B → C}
A is not a candidate key, because A+ = {A, B, C}
AD is a candidate key, because AD+ = {A, B, C, D}
ABD is a super key

5. Normal forms – BCNF For every functional dependency X→Y (YX) Example:
X has to be a super key, i.e., X has to contain a candidate key
Example:
4 attributes A, B, C, D, and F = {A → B, B → C}
R(A, B, C) does not satisfy BCNF
A is the key
B → C does not include A
R(A, B) satisfies BCNF

6. BCNF decomposition with lossless join
For each FD XA, if it does not satisfy BCNF (i.e., X is not a super key), decompose it
S1 = XA
S2 = X(S  A)

7. FDs: CT, CS G, HR C, HS R, TH R
Example
FDs: CT, CS G, HR C, HS R, TH R
CTHRSG
violation: CS  G (CS)+ = CSTG, i.e., CS is not a super key
CSG
CTHRS
no violation, (CS)+ = CSG
violation: C  T C+ = CT CT
CHRS
no violation: 2 attributes
violation: CH  R (CH)+ = CHTR
CHR
CHS
no violation: C+ = CT, H+ = H, R+ = R, S+ = S
Resulting decomposition: CSG, CT, CHR, CHS

8. Normal forms – 3NF
For every functional dependency X→Y (YX), at least one of the following holds:
X is a super key, i.e., X contains a candidate key
Y belongs to a candidate key
No, A and D are candidate key, B → C violates

9. 3NF decomposition with dependency preserving
Step 1: Simplify the set of FDs (eliminate redundancies) , called the minimal cover
Rewrite the FDs with single attributes on RHS
Replace AB  CD with AB  C and AB  D
Eliminate redundant FDs
F = {A  C, A  B, B  C}
A  C is redundant (it is implied by A  B and B  C)
Eliminate redundant attributes from LHS of FDs
F = {A  B, AB  C}
B is redundant in AB  C because A by itself determines C (A  C is implied by F)

10. 3NF decomposition with dependency preserving
Step 2: the decomposition is defined as  = {XA1 … Am | XAiF}{BR | B does not occur in F}
Example
R = CTHRSG
F =
CT
CS G
HR C
HS R
HT R
Then  = {CT, CSG, HRC, HSR, HTR}

11. Decomposition with lossless join and dependency preserving
Step 1: compute 3NF decomposition with dependency preserving
Step 2: examine each set to see whether they have already contained a candidate key
if not, add it

12. Example R: ABCDE F = {AC, BCD ADE}  = {AC, BCD, ADE}
None of AC, BCD, ADE is a superkey
AC+ = AC, BCD+ = BCD, ADE+ = ADEC
AB is a key, add it to 
 = {AB, AC, BCD, ADE}

13. Testing lossless join Consider 5 attributes: A, B, C, D, E
FD = {ABC, CD, DE}
Consider a decomposition is R1(A,B,C); R2(C,D); and R3(D,E)
Is this satisfying lossless join decomposition? A B C D E a1 a2 a3 - a4 a5 R1 a4 a5 R2 R3