CS 430
Database Theory
Winter 2005
Lecture 9: Fourth and Fifth Normal Forms
1
Decompositions

Given a relation R = {A1, … , An} (all of the Ai
are unique), then a set of relation schemas
D = {R1, … , Rm} is a decomposition of R if R
is the union of the Ri, or
R
m
Ri
i =1

That is, all the attributes of R appear in the Ri
2
Goodness of Decomposition


When is a decomposition “good”?
Two standards:


Dependency Preservation
Lossless (Nonadditive) Join
3
Dependency Preservation

Suppose we have a set of FDs F on R and a
decomposition D = {R1, … , Rm}, the
projection of F on R is the set
Ri(F) = {X  Y  F+ | X  Y  Ri}
That is, Ri(F) consists of all the FDs in the
closure of F which are FDs on Ri
4
Dependency Preservation

D is Dependency Preserving with respect to
F if the closure of the union of the projections
of F onto the Ri is the closure of F. Or,
(R1(F)  …  Rm(F))+ = F+
Or, if we project F onto the individual Ri,
union the projections together, and compute
the closure, we get the original closure of F.

Or, no information contained in F is lost by
projecting F onto the individual Ri
5
Dependency Preservation Notes

Claim: It is possible to find a 3NF
decomposition of R (each of Ri is 3NF) which
is dependency preserving


See Algorithm 11.2, page 340. (No proof.)
Why do we want this?


When we update the database, we want to be
able verify FDs by verifying them on the individual
relations
The alternative is having to do joins to verify that
our update is good, slowing system.
6
Lossless (Nonadditive) Join Property



D has the Lossless (Nonadditive) Join
property with respect to a set of FDs F if for
every relation state r of R that satisfies F:
R1(r)  …  Rm(r) = r
( is the natural join)
Lossless means no loss of information
Nonadditive means that natural join doesn’t
add any information
7
Lossless (Nonadditive) Join Notes


Algorithm 11.1, page 337, provides a way to
test for this property
If D is a binary decomposition, D = {R1 ,
R2}, D is nonadditive if and only if:
(R1  R2)  (R1 - R2) is in F+, or
(R1  R2)  (R2 - R1) is in F+
That is, R1  R2 is a key for (at least) one
of R1 or R2
8
Aside: Null Problems with Nulls


See Figures 11.2, 11.3, Text Book
Bottom line: If nulls are present, especially
nulls in foreign keys then


May have to use outer joins instead of ordinary
(inner) joins
Have to be careful if using aggregation (e.g. sum
or average)
9
Multi-Value Dependencies


If X ,Y attributes of R there is a Multi-Valued
Dependency (MVD) X > Y, (we let Z = R - (X Y ))
if for all states r of R, and t1, t2 tuples of r such that
t1[X ] = t2[X ], then there exist tuples t3, t4 of r such
that:
t3[X ] = t4[X ] = t1[X ] = t2[X ]
t3[Y ] = t1[Y ], t4[Y ] = t2[Y ]
t4[Z ] = t1[Z ], t3[Z ] = t2[Z ]
An MVD X > Y, is trivial if Y  X , or X Y = R
10
Fourth Normal Form

R is 4NF with respect to a set of FDs and
MVDs F if for every non-trivial MVD X > Y,
X is a superkey of R.

See Figure 11.4(a, b) in Text Book.
11
Fourth Normal Form Notes

If a relation is not 4NF then there are update
anomalies:


If you add a relation you must also add the corresponding
relations
D is a lossless (nonadditive) decomposition of R,
D = {R1 , R2}, with respect to a set of FDs and
MVDs F if and only if:
(R1  R2) > (R1 - R2), which is the same as
(R1  R2) > (R2 - R1)
12
Fifth Normal Form
• JD(R1, … , Rm) is a Join Dependency (JD) for
a decomposition {R1, … , Rm} of R if for every
legal state r of R:
R1(r)  …  Rm(r) = r
• A JD is trivial if some Ri = R
• A relation R is in Fifth Normal Form (5NF) if
for every non-trivial JD of R, every Ri is a
superkey of R
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Notes on Fifth Normal Form



An MVD is a JD with m = 2
Finding all the JDs of a database of any size
is probably not feasible
Example: See Figure 11.4 (c, d) of Text Book
14
Products, Salesmen, Territories
A Data Design Problem

Salesman




Product



Sells specific products
Has specific territories
Has a quota: How much he is supposed to sell
Sold by salesmen
Has a price
Territory

Worked by salesmen
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ER Model Version 1
Salesman
Sells
Product
Product
Price
Quota
Works
Territory
Territory
A Salesman can sell any Product he sells in any Territory he works.
A Product has one Price for all Salesmen and all Territories.
A Salesman has one Quota for all his sales.
Note: Each Entity and Relation becomes a relation in our database.
16
ER Model Version 2
Salesman
Sells
Product
Quota
Works
Territory
Product
Price
Territory
A Salesman has a Quota for each product he sells.
17
ER Model Version 3
Salesman
Sells
Product
Quota
Sold
In
Product
Works
Territory
Territory
Price
Products are only sold in specific Territories.
A Product has a Price set for each Territory where it is sold.
A Salesman can sell any Product he sells in any Territory he works
where that Product is sold.
Note JD between “Sells Product”, “Sold In”, and “Works Territory”.
18
ER Model Version 4
Salesman
Quota
Sells
Product
in Territory
Sells
Product
Product
Sold
In
Territory
Price
A Salesman is assigned to sell specific Products in specific Territories.
A Salesman has a Quota for each Product he sells in each Territory.
Possible Integrity Constraint: Keys of “Sells Product” and “Sold In” are
projections of “Sells Product in Territory”.
19
ER Model Version 4A
Salesman
Sells
Product
in Territory
Product
Sold
In
Quota
Territory
Price
Possible Integrity Constraint: Key of “Sold In” is projection of “Sells
Product in Territory”. (But I might want to assign a Price even though
no Salemen have yet been assigned that Product in that Territory.)
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Sample Fields

Employee





Manager ID Number
Manager Name


Territory



Product
Employee ID Number
Employee Name
Work Location
Manager



Territory Number
Territory Name
Territory Bonus




Product Number
Product Name
Price
Actual_Sales
Target_Sales
Other




Quota
Commission Rate
Commission
Manager Commission
21
Possible Functional Dependencies

{Employee ID Number} 


{Manage ID Number} 


{Manager Name, Manager Commission(?)}
{Territory Number} 


{Employee Name, Work Location, Manager ID
Number, Manager Commission(?)}
{Territory Name, Territory Bonus(?)}
{Product Number} 

{Product Name, Price(?), Actual Sales(?), Target
Sales (?)}
22
More Possible FDs

{Employee ID Number, Territory Number} 


{Employee ID Number, Product Number} 


{Territory Bonus(?), Quota(?), Commission
Rate(?)}
{Quota(?), Commission Rate(?)}
{Territory Number, Product Number} 

{Price(?), Actual Sales(?), Target Sales(?),
Territory Bonus(?), Commission Rate(?),
Commission(?), Manager Commission(?)}
23
More Possible FDs

{Employee ID Number, Product Number,
Territory Number} 


{Actual Sales, Commission Rate} 


{Quota(?), Actual Sales(?), Target Sales(?),
Commission Rate(?), Commission(?) , Manager
Commission(?)}
{Commission}
{Actual Sales, Manager Commission Rate} 

{Manager Commission}
24
Proposed Solution




Employee(Employee ID Number, Employee
Name, Work Location, Manager ID Number)
Manager(Manager ID Number, Manager
Name, Manager Commission)
Territory(Territory Number, Territory Name)
Product(Product Number, Product Name)
25
More Proposed Solution




Product_Territory(Product Number, Territory
Number, Price)
Employee_Territory(Employee ID Number,
Territory Number, Territory Bonus)
Employee_Product(Employee ID Number,
Product Number, Commission Rate)
Employee_Product_Territory(Employee ID
Number, Product Number, Territory Number,
Quota, Actual Sales, Target Sales,
Commission, Manager Commission)
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