1. Schema Refinement and Normalization
Zachary G. Ives
University of Pennsylvania
CIS 550 – Database & Information Systems
September 25, 2003
Some slide content courtesy of Susan Davidson & Raghu Ramakrishnan

2. Administrivia HW2 due now HW3 handed out
Readings on XML: papers rather than textbook
Will shortly have readings for each project group
Q: Dinkar’s office hours conflict with 505 on Monday. Would you prefer them on Friday?

3. Schema Refinement & Design Theory
ER Diagrams give us a start in logical schema design
Sometimes need to refine our designs further
There’s a system and theory for this
Focus is on redundancy of data
Causes update, insertion, deletion anomalies

4. Not All Designs are Equally Good
Why is this a poor schema design?
And why is this one better?
Stuff(sid, name, serno, subj, cid, exp-grade)
Student(sid, name)
Course(serno, cid)
Subject(cid, subj)
Takes(sid, serno, exp-grade)

5. Focus on the Bad Design Certain items (e.g., name) get repeated
Some information requires that a student be enrolled (e.g., courses) due to the key
sid
name
serno
subj
cid
exp-grade 1
Sam
570103 AI
570 B 23
Nitin
550103 DB
550 A 45
Jill
505103 OS
505 C

6. Functional Dependencies Describe “Key-Like” Relationships
A key is a set of attributes where:
If keys match, then the tuples match
A functional dependency (FD) is a generalization:
If an attribute set determines another, written A ! B then if two tuples agree on attribute set A, they must agree on B: sid ! name
What other FDs are there in this data?
FDs are independent of our schema design choice

7. Formal Definition of FD’s
Def. Given a relation schema R and subsets X, Y of R:
An instance r of R satisfies FD X  Y if, for any two tuples t1, t2 2 r, t1[X ] = t2[X] implies t1[Y] = t2[Y]
For an FD to hold for schema R, it must hold for every possible instance of r
(Can a DBMS verify this? Can we determine this by looking at an instance?)

8. General Thoughts on Good Schemas
We want all attributes in every tuple to be determined by the tuple’s key attributes, i.e. part of a superkey (for key X  Y, a superkey is a “non-minimal” X)
What does this say about redundancy?
But:
What about tuples that don’t have keys (other than the entire value)?
What about the fact that every attribute determines itself?

9. Armstrong’s Axioms: Inferring FDs
Some FDs exist due to others; can compute using Armstrong’s axioms:
Reflexivity: If Y  X then X  Y (trivial dependencies)
name, sid  name
Augmentation: If X  Y then XW  YW
serno  subj so serno, exp-grade  subj, exp-grade
Transitivity: If X  Y and Y  Z then X  Z
serno  cid and cid  subj
so serno  subj

10. Armstrong’s Axioms Lead to…
Union: If X  Y and X  Z then X  YZ
Pseudotransitivity: If X  Y and WY  Z then XW  Z
Decomposition: If X  Y and Z  Y then X  Z
Let’s prove these from Armstrong’s Axioms

11. {X  Y | X  Y is derivable from F by Armstrong’s Axioms}
Closure of a Set of FD’s
Defn. Let F be a set of FD’s. Its closure, F+, is the set of all FD’s:
{X  Y | X  Y is derivable from F by Armstrong’s Axioms}
Which of the following are in the closure of our Student-Course FD’s?
name  name
cid  subj
serno  subj
cid, sid  subj
cid  sid

12. Attribute Closures: Is Something Dependent on X?
Defn. The closure of an attribute set X, X+, is:
X+ =  {Y | X  Y  F +}
This answers the question “is Y determined (transitively) by X?”; compute X+ by:
Does sid, serno  subj, exp-grade?
closure := X;
repeat until no change {
if there is an FD U  V in F
such that U is in closure
then add V to closure}

13. Equivalence of FD sets
Defn. Two sets of FD’s, F and G, are equivalent if their closures are equivalent, F + = G +
e.g., these two sets are equivalent:
{XY  Z, X  Y} and
{X  Z, X  Y}
F + contains a huge number of FD’s (exponential in the size of the scheme)
Would like to have smallest “representative” FD set

14. Minimal Cover we express each FD in simplest form
Defn. A FD set F is minimal if:
1. Every FD in F is of the form X  A, where A is a single attribute
2. For no X  A in F is: F – {X  A } equivalent to F
3. For no X  A in F and Z  X is:
F – {X  A }  {Z  A } equivalent to F
Defn. F is a minimum cover for G if F is minimal and is equivalent to G.
e.g.,
{X  Z, X  Y} is a minimal cover for
{XY  Z, X  Y}
in a sense,
each FD is
“essential”
to the cover

15. More on Closures
If F is a set of FD’s and X  Y  F + then for some attribute A  Y, X  A  F +
Proof by counterexample.
Assume otherwise and let Y = {A1,..., An} Since we assume X  A1, ..., X  An are in F +
then X  A1 ... An is in F + by union rule,
hence, X  Y is in F + which is a contradiction

16. Why Armstrong’s Axioms?
Why are Armstrong’s axioms (or an equivalent rule set) appropriate for FD’s? They are:
Consistent: any relation satisfying FD’s in F will satisfy those in F +
Complete: if an FD X  Y cannot be derived by Armstrong’s axioms from F, then there exists some relational instance satisfying F but not X  Y
In other words, Armstrong’s axioms derive all the FD’s that should hold

17. Proving Consistency
We prove that the axioms’ definitions must be true for any instance, e.g.:
For augmentation (if X  Y then XW  YW): If an instance satisfies X  Y, then:
For any tuples t1, t2 r, if t1[X] = t2[X] then t1[Y] = t2[Y] by defn.
If, additionally, it is given that t1[W] = t2[W], then t1[YW] = t2[YW]

18. Proving Completeness
Suppose X  Y  F + and define a relational instance r that satisfies F + but not X  Y:
Then for some attribute A  Y, X  A  F +
Let some pair of tuples in r agree on X+ but disagree everywhere else:
x1 x xn a1,1 v1 v vm w1,1 w2,1...
x1 x xn a1,2 v1 v vm w1,2 w2,2... X A
X+ – X
R – X+ – {A}

19. Proof of Completeness cont’d
Clearly this relation fails to satisfy X  A and X  Y. We also have to check that it satisfies any FD in F + .
The tuples agree on only X Thus the only FD’s that might be violated are of the form X’  Y’ where X’  X+ and Y’ contains attributes in R – X+ – {A}.
But if X’  Y’ F+ and X’  X+ then Y’  X+ (reflexivity and augmentation). Therefore X’  Y’ is satisfied.

20. Stuff(sid, name, serno, subj, cid, exp-grade)
Decomposition
Consider our original “bad” attribute set
We could decompose it into
But this decomposition loses information about the relationship between students and courses. Why?
Stuff(sid, name, serno, subj, cid, exp-grade)
Student(sid, name)
Course(serno, cid)
Subject(cid, subj)

21. Lossless Join Decomposition
R1, … Rk is a lossless join decomposition of R w.r.t. an FD set F if for every instance r of R that satisfies F,
ÕR1(r) ⋈ ... ⋈ ÕRk(r) = r
Consider:
What if we decompose on (sid, name) and (serno, subj, cid, exp-grade)?
sid
name
serno
subj
cid
exp-grade 1
Sam
570103 AI
570 B 23
Nitin
550103 DB
550 A

22. Testing for Lossless Join
R1, R2 is a lossless join decomposition of R with respect to F iff at least one of the following dependencies is in F+
(R1  R2)  R1 – R2
(R1  R2)  R2 – R1
So for the FD set:
sid  name
serno  cid, exp-grade
cid  subj
Is (sid, name) and (serno, subj, cid, exp-grade) a lossless decomposition?

23. Dependency Preservation
Ensures we can “easily” check whether a FD X  Y is violated during an update to a database:
The projection of an FD set F onto a set of attributes Z, FZ is
{X  Y | X  Y  F +, X  Y Í Z}
i.e., it is those FDs local to Z’s attributes
A decomposition R1, …, Rk is dependency preserving if F + = (FR1 ... FRk)+
The decomposition hasn’t “lost” any essential FD’s, so we can check without doing a join

24. Example of Lossless and Dependency-Preserving Decompositions
Given relation scheme
R(name, street, city, st, zip, item, price)
And FD set name  street, city
street, city  st
street, city  zip
name, item  price
Consider the decomposition
R1(name, street, city, st, zip) and R2(name, item, price)
Is it lossless?
Is it dependency preserving?
What if we replaced the first FD by name, street  city?

25. Another Example Given scheme: R(sid, fid, subj) and FD set: fid  subj
sid, subj  fid
Consider the decomposition
R1(sid, fid) and R2(fid, subj)
Is it lossless?
Is it dependency preserving?

26. FD’s and Keys
Ideally, we want a design s.t. for each nontrivial dependency X  Y, X is a superkey for some relation schema in R
We just saw that this isn’t always possible
Hence we have two kinds of normal forms

27. Two Important Normal Forms
Boyce-Codd Normal Form (BCNF). For every relation scheme R and for every X  A that holds over R,
either A  X (it is trivial) ,or
or X is a superkey for R
Third Normal Form (3NF). For every relation scheme R and for every X  A that holds over R,
either A  X (it is trivial), or
X is a superkey for R, or
A is a member of some key for R

28. Normal Forms Compared
BCNF is preferable, but sometimes in conflict with the goal of dependency preservation
It’s strictly stronger than 3NF
Let’s see algorithms to obtain:
A BCNF lossless join decomposition
A 3NF lossless join, dependency preserving decomposition

29. BCNF Decomposition Algorithm (from Korth et al
BCNF Decomposition Algorithm (from Korth et al.; our book gives recursive version)
result := {R}
compute F+
while there is a schema Ri in result that is not in BCNF {
let A  B be a nontrivial FD on Ri s.t. A  Ri is not in F+ and A and B are disjoint
result:= (result – Ri)  {(Ri - B), (A,B)} }

30. 3NF Decomposition Algorithm
Let F be a minimal cover
i:=0
for each FD A  B in F {
if none of the schemas Rj, 1 j  i, contains AB {
increment i
Ri := (A, B) }
if no schema Rj, 1  j  i contains a candidate key for R {
Ri := any candidate key for R
return (R1, …, Ri)
Build dep.- preserving
decomp.
Ensure lossless decomp.

31. Summary We can always decompose into 3NF and get:
Lossless join
Dependency preservation
But with BCNF we are only guaranteed lossless joins
BCNF is stronger than 3NF: every BCNF schema is also in 3NF
The BCNF algorithm is nondeterministic, so there is not a unique decomposition for a given schema R