Temple University – CIS Dept. CIS616– Principles of Data Management

Temple University – CIS Dept.CIS616– Principles of Data Management

V. Megalooikonomou

Database Design and Normalization

(based on notes by Silberchatz,Korth, and Sudarshan and notes by C. Faloutsos at CMU)

Overview Relational model

formal query languages commercial query languages (SQL)

Integrity constraints domain I.C., foreign keys functional dependencies

Functional Dependencies DB design and normalization

Overview - detailed

DB design and normalization pitfalls of bad design decomposition normal forms

Design ‘good’ tables sub-goal#1: define what ‘good’ means sub-goal#2: fix ‘bad’ tables

in short: “we want tables where the attributes

depend on the primary key, on the whole key, and nothing but the key”

Let’s see why, and how:

Goal

Pitfalls

takes1 (ssn, c-id, grade, name, address)

Ssn c-id Grade Name Address

123cs331 A smith Main

Pitfalls

‘Bad’ - why? because: ssn->address, name

Ssn c-id Grade Name Address

123 cs331 A smith Main

123 cs351 B smith Main

123 cs211 A smith Main

Pitfalls

Redundancy space (inconsistencies) insertion/deletion anomalies:

Pitfalls

insertion anomaly: “jones” registers, but takes no class -

no place to store his address!

Ssn c-id Grade Name Address

123 cs331 A smith Main

… … … … …

234 null null jones Forbes

Pitfalls

deletion anomaly: delete the last record of ‘smith’ (we

lose his address!)

Ssn c-id Grade Name Address

123 cs331 A smith Main

123 cs351 B smith Main

123 cs211 A smith Main

Solution: decomposition

split offending table in two (or more), e.g.:

Ssn c-id Grade Name Address123 cs331A smith Main123 cs351B smith Main123 cs211A smith Main

? ?

Overview - detailed

DB design and normalization pitfalls of bad design decomposition

lossless join dependency preserving

normal forms

Decompositions

there are ‘bad’ decompositions we want:

lossless and dependency preserving

Decompositions - lossy:

R1(ssn, grade, name, address) R2(c-id,grade)

Ssn c-id Grade Name Address123 cs331 A smith Main123 cs351 B smith Main

234 cs211 A jones Forbes

ssn->name, address

ssn, c-id -> grade

Ssn Grade Name Address123 A smith Main123 B smith Main

234 A jones Forbes

c-id Grade

cs331 Acs351 B

cs211 A

Decompositions - lossy:

can not recover original table with a join!

Ssn c-id Grade Name Address

123 cs331 A smith Main123 cs351 B smith Main

234 cs211 A jones Forbes

ssn->name, address

ssn, c-id -> grade

Ssn Grade Name Address123 A smith Main123 B smith Main

234 A jones Forbes

c-id Grade

cs331 Acs351 B

cs211 A

Decompositions – lossy: Another example

Decomposition of R = (A, B) into R1 = (A), R2 = (B)

A B

121

A

B

12

rA(r) B(r)

A (r) B (r)A B

1212

Decompositions

example of non-dependency preserving

S# address status123 London E125 Paris E

234 Pitts. A

S# -> address, status

address -> status

S# status123 E125 E

234 A

S# address123 London125 Paris

234 Pitts.

S# -> address S# -> status

Decompositions

is it lossless?

S# address status123 London E125 Paris E

234 Pitts. A

S# -> address, status

address -> status

S# status123 E125 E

234 A

S# address123 London125 Paris

234 Pitts.

S# -> address S# -> status

Decompositions - lossless

Definition: Consider schema R, with FD ‘F’. R1, R2 is a lossless join decomposition

of R if we always have:

An easier criterion?

rrr 21

Decomposition - lossless

Theorem: lossless join decomposition if the

joining attribute is a superkey in at least one of the new tables

Formally:

221

121

RRR

orRRR

Decomposition - losslessexample:

Ssn c-id Grade Name Address

123 cs331 A smith Main123 cs351 B smith Main

234 cs211 A jones Forbes

ssn->name, address

ssn, c-id -> grade

Ssn c-id Grade123 cs331 A123 cs351 B

234 cs211 A

Ssn Name Address123 smith Main234 jones Forbes

ssn->name, addressssn, c-id -> grade

R1R2

Overview - detailed

DB design and normalization pitfalls of bad design decomposition

lossless join decomp. dependency preserving

normal forms

Decomposition - depend. pres.

informally: we don’t want the original FDs to span two tables - counter-example:

S# address status123 London E125 Paris E

234 Pitts. A

S# -> address, status

address -> status

S# status123 E125 E

234 A

S# address123 London125 Paris

234 Pitts.

S# -> address S# -> status

Decomposition - depend. pres.

dependency preserving decomposition:

S# address status123 London E125 Paris E

234 Pitts. A

S# -> address, status

address -> status

S# address123 London125 Paris

234 Pitts.

S# -> address address -> status

address statusLondon EParis E

Pitts. A

(but: S#->status ?)

Decomposition - depend. pres.

informally: we don’t want the original FDs to span two tables

more specifically: … the FDs of the canonical cover

Let Fi be the set of dependencies F+ that include only attributes in Ri.

Preferably the decomposition should be dependency preserving, that is, (F1 F2 … Fn)

+ = F+

Otherwise, checking updates for violation of functional dependencies may require computing joins expensive

Decomposition - depend. pres.

why is dependency preservation good?

S# address123 London125 Paris

234 Pitts.

S# -> address address -> status

address statusLondon EParis E

Pitts. A

S# status123 E125 E

234 A

S# address123 London125 Paris

234 Pitts.

S# -> address S# -> status

(address->status: ‘lost’)

Decomposition - depend. pres.

A: eg., record that ‘Philly’ has status ‘A’

S# address123 London125 Paris

234 Pitts.

S# -> address address -> status

address statusLondon EParis E

Pitts. A

S# status123 E125 E

234 A

S# address123 London125 Paris

234 Pitts.

S# -> addressS# -> status

(address->status: ‘lost’)

Decomposition - depend. pres.

To check if a dependency is preserved in a decomposition of R into R1, R2, …, Rn we apply the following test (with attribute closure done w.r.t. F)

result = while (changes to result) do

for each Ri in the decompositiont = (result Ri)+ Ri

result = result t If result contains all attributes in , then functional dependency is

preserved

We apply the test on all dependencies in F to check if a decomposition is dependency preserving

The test takes polynomial time Computing F+ and (F1 F2 … Fn)

+ needs exponential time

Decomposition - conclusions

decompositions should always be lossless joining attribute -> superkey

whenever possible, we want them to be dependency preserving (occasionally, impossible - see ‘STJ’ example later…)

Normalization using FD When decomposing a relation schema R with a set of

functional dependencies F into R1, R2,…, Rn we want:

Lossless-join decomposition: otherwise … information loss

No redundancy: relations Ri preferably should be in either Boyce-Codd Normal Form or Third Normal Form

Dependency preservation: Let Fi be the set of dependencies in F+ that include only attributes in Ri.

Preferably the decomposition should be dependency preserving, i.e., (F1 F2 … Fn)

+ = F+

Otherwise, checking updates for violation of functional dependencies may require computing joins expensive

Normalization using FD - Example

R = (A, B, C)F = {A B, B C)

R1 = (A, B), R2 = (B, C) Lossless-join decomposition:

R1 R2 = {B} and B BC Dependency preserving

R1 = (A, B), R2 = (A, C) Lossless-join decomposition:

R1 R2 = {A} and A AB Not dependency preserving

(cannot check B C without computing R1 R2)

Overview - detailed

DB design and normalization pitfalls of bad design decomposition ( how to fix the problem) normal forms ( how to detect the

problem) BCNF, 3NF, (1NF, 2NF)

Normal forms - BCNF

We saw how to fix ‘bad’ schemas -but what is a ‘good’ schema?

Answer: ‘good’, if it obeys a ‘normal form’,i.e., a set of rules

Typically: Boyce-Codd Normal Form (BCNF)

Normal forms - BCNF

Defn.: Rel. R is in BCNF w.r.t. F, if

informally: everything depends on the full key, and nothing but the key

semi-formally: every determinant (of the cover) is a candidate key

Normal forms - BCNF

Example and counter-example:

Ssn Name Address123 smith Main123 smith Main

234 jones Forbes

ssn->name, address

Ssn c-id Grade Name Address123 cs331 A smith Main123 cs351 B smith Main

234 cs211 A jones Forbes

ssn->name, address

ssn, c-id -> grade

Normal forms - BCNF

Formally: for every FD a->b in F+ a->b is trivial (a is a superset of b) or a is a superkey (or both)

Normal forms - BCNF

Theorem: given a schema R and a set of FD ‘F’, we can always decompose it to schemas R1, … Rn, so that R1, … Rn are in BCNF and the decomposition is lossless

(…but, some decomp. might lose dependencies)

BCNF Decomposition

How? ….essentially, break off FDs of the cover

eg. TAKES1(ssn, c-id, grade, name, address)ssn -> name, addressssn, c-id -> grade

Normal forms - BCNF

eg. TAKES1(ssn, c-id, grade, name, address)ssn -> name, address ssn, c-id -> grade

name

addressgrade

c-id

ssn

Normal forms - BCNF

Ssn c-id Grade Name Address

123 cs331 A smith Main123 cs351 B smith Main

234 cs211 A jones Forbes

ssn->name, address

ssn, c-id -> grade

Ssn c-id Grade123 cs331 A123 cs351 B

234 cs211 A

Ssn Name Address123 smith Main123 smith Main

234 jones Forbes

ssn->name, addressssn, c-id -> grade

Normal forms - BCNF

pictorially: we want a ‘star’ shape

name

addressgrade

c-id

ssn

:not in BCNF

Normal forms - BCNF

pictorially: we want a ‘star’ shape

B

C

A G

E

Dor

F

H

Normal forms - BCNF

or a star-like: (e.g., 2 cand. keys):STUDENT(ssn, st#, name, address)

name

address

ssn

st#

=

name

address

ssn

st#

Normal forms - BCNF

but not:

or

B

C

A

D

G

E

D

F

H

BCNF Decomposition

result := {R};done := false;compute F+;while (not done) do

if (there is a schema Ri in result that is not in BCNF)then begin

let be a nontrivial functionaldependency that holds on Ri

such that Ri is not in F+, and = ;

result := (result – Ri) (Ri – ) (, ); end

else done := true;

Note: each Ri is in BCNF, and decomposition is lossless-join

Normal forms - 3NF

consider the ‘classic’ case:STJ( Student, Teacher, subJect)

T-> JS,J -> T

is it BCNF? S

T

J

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

How to decompose it to BCNF?

S

T

J

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

1) R1(T,J) R2(S,J) (BCNF? - lossless? - dep. pres.? )

2) R1(T,J) R2(S,T) (BCNF? - lossless? - dep. pres.? )

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

1) R1(T,J) R2(S,J) (BCNF? Y+Y - lossless? N - dep. pres.? N )

2) R1(T,J) R2(S,T) (BCNF? Y+Y - lossless? Y - dep. pres.? N )

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

in this case: impossible to have both BCNF and dependency preservation

Welcome 3NF (…a weaker normal form)!

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

S

J

T

informally, 3NF ‘forgives’ the red arrow in the can. cover

Normal forms - 3NF

STJ( Student, Teacher, subJect)T-> J S,J -> T

S

J

T

Formally, a rel. R with FDs ‘F’ is in 3NF if:

for every a->b in F+:

•it is trivial or

•a is a superkey or

•each b-a attr.: part of a cand. key

Normal forms - 3NF

R = (J, K, L)F = {JK L, L K}Two candidate keys = JK and JL

R is not in BCNF Any decomposition of R will fail to preserve

JK L BCNF decomposition has (JL) and (LK)

Testing for JK L requires a join

R is in 3NFJK L JK is a superkeyL K K is contained in a candidate key

There is some redundancy in this schema…

Normal forms - 3NF TESTING FOR 3NF

Optimization: Need to check only FDs in F, need not check all FDs in F+

Use attribute closure to check, for each dependency , if is a superkey

If is not a superkey, we have to verify if each attribute in is contained in a candidate key of R this test is more expensive; it involves finding

candidate keys testing for 3NF is NP-hard Interestingly, decomposition into 3NF (described

shortly) can be done in polynomial time

Decomposition into 3NF

Let Fc be a canonical cover for F;i := 0;for each functional dependency in Fc do

if none of the schemas Rj, 1 j i contains then begin

i := i + 1;Ri :=

endif none of the schemas Rj, 1 j i contains a candidate key for R

then begini := i + 1;Ri := any candidate key for R;

end return (R1, R2, ..., Ri)

The dependencies are preserved by building explicitly a schema for each given dependency

Guarantees a lossless-join decomposition by having at least one schema containing a candidate key for the schema being decomposed

Normal forms - 3NF

how to bring a schema to 3NF?In short ….for each FD in the cover, put it in a

table

Normal forms - 3NF vs BCNF

If ‘R’ is in BCNF, it is always in 3NF (but not the reverse)

In practice, aim for BCNF; lossless join; and dep. preservation

if impossible, we accept 3NF; but insist on lossless join and dep.

preservation 3NF has problems with transitive

dependecies

3NF vs BCNF (cont.) Example of problems due to redundancy in 3NF

R = (J, K, L)F = {JK L, L K}

A schema that is in 3NF but not in BCNF has the problems of repetition of information (e.g., the relationship l1, k1) need to use null values (e.g., to represent the relationship

l2, k2 where there is no corresponding value for J).

J L Kj1

j2

j3

null

l1

l1

l1

l2

k1

k1

k1

k2

Normal forms - more details

why ‘3’NF? what is 2NF? 1NF? 1NF: attributes are atomic (i.e., no

set-valued attr., a.k.a. ‘repeating groups’)Ssn Name Dependents123 Smith Peter

MaryJohn

234 Jones AnnMichael

not 1NF

Normal forms - more details

2NF: 1NF and non-key attr. fully depend on the keycounter-example: TAKES1(ssn, c-id, grade, name, address)ssn -> name, address ssn, c-id -> grade

name

addressgradec-id

ssn

Normal forms - more details

3NF: 2NF and no transitive dependencies counter-example:

B

C

A

Din 2NF, but not in 3NF

Normal forms - more details

4NF, multivalued dependencies etc:

later… in practice, E-R diagrams usually

lead to tables in BCNF

Overview - conclusions

DB design and normalization pitfalls of bad design decompositions (lossless, dep.

preserving) normal forms (BCNF or 3NF)

“everything should depend on the key, the whole key, and nothing but the key”