Normalisation Lecture 3 Akhtar Ali 06/13/22 1
Jan 18, 2016
Normalisation
Lecture 3
Akhtar Ali
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Learning Objectives
1. To consider the process of Normalisation
2. To consider the definition and application of 1NF
3. To consider the definition and application of 2NF
4. To consider the definition and application of 3NF
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NORMALISATION PRINCIPLES
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Normalisation
• Definition : a systematic method that takes pre-existing relations and produces a canonical set of relations.– By canonical is meant well-designed, sound, or a
recognised and lawful form.
• It can be used both for :• designing canonical relations,• checking existing relations to ensure they are canonical.
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How Normalization Supports Database Design?
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Normal forms
1NF 3NF 2NF
Normalisation uses the concept of Normal Forms. They are organised in a sequence, each successive normal form being higher than the one before.
A normal form is higher because it applies more stringent constraints to a relation than a lower normal form.
A relation is said to a be in a certain “normal form” if it conforms to the constraints of that normal form.
Normalisation as a Relational Design Tool
• Sometimes, we need to use normalisation for designing relations. – For example, when ER modelling is not feasible or if we deal
with a small number of attributes.– So we need to learn normalisation.
• 1NF stands for First Normal Form, 2NF for Second Normal Form, and so on.
• The constraints of a particular normal form are those of the previous normal form– plus the additional constraint(s) peculiar to this particular
normal form.
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The Normalisation Procedure• The normalisation procedure starts with a set of
relations, for each of which, it is presumed to be un-normalised or in 0NF.
– DO FOR xNF = 1NF, ..... 5NF• DO FOR each relation that exists
– IF relation already conforms to xNF» THEN it is in xNF, so do nothing
– ELSE create 2 or more replacement relations from it that do conform to xNF.– END-LOOP
• END-LOOP
• 5NF is the highest possible normal form.• In practice, 3NF is the highest normal form usually
reached.
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What is a Normal Form?
• Each Normal Form has two parts – A definition that specifies exactly what constraints
apply to a relation in that normal form.• This is used to check whether any given relation is
already in that normal form or not.
– A method to be used to replace the relation with 2 or more that will be in that normal form.
• The method assumes that the relation-to-be-replaced is in the previous normal form.
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Normalising : Possibilities
The set of all un-normalised
relations
Relation
The set of all relations in 1NF
The set of all relations in 2NF
A given relation.
Already in 1NF. Nothing to do.
And so on.
Already in 2NF. Nothing to do.
Consequences of Normalisation• If new, replacement relations are created, then they
must be projections of the original.– New-Relation πset of attributes (Original-relation)
• The symbol π denotes projection of a set of attributes from a relation.
• Normalisation always creates new relations such that– Original-relation New-Rel-1 ⋈ New-Rel-2
• The symbol ⋈ denotes a join between two relations.
• This ensures that no information is ever lost.
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FIRST NORMAL FORM (1NF)
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Definition of 1NF
• A relation is in 1NFif and only if
every attribute value it can ever contain is an atomic value
• Question : What is an atomic value ?• Answer : A value that cannot meaningfully be
broken down into two or more constituent parts.
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Example : Purchase Order Relation
L6
315
Bloggs
D’ham
8 June
P3 Q7
Pump Motor
5 5
400
150 250
L5
127
Smith
N’cle
7 May
N8 B6 L4
Nut Bolt Nail
70 60 100
12
4 5 3
Ord Sno Sname Saddr Date Part Pname Qty Tot Price
The following relation holds data about purchase orders placed on suppliers for parts
Ord Order number that uniquely identifies every purchase order.Sno Supplier number that uniquely identifies any supplier.Sname The name of a supplier.Saddr The address of a supplier.Date The date on which the order was placed.Part Part number that uniquely identifies every kind of part used by the company.Pname The name of a particular kind of part.Qty The quantity of a particular kind of part ordered on a purchase order.Price The price of that quantity of that particular kind of part.Tot The total price to be paid for the whole order.
Not in 1NF
• Attributes Ord, Sno, Sname, Saddr, Date and Tot currently contain only atomic values, and in fact can only ever contain atomic values.
• Attributes Part, Pname, Qty and Price currently contain non-atomic values, and in fact may often contain non-atomic values.
• Therefore the relation is not in 1NF.
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Putting Purchase Order into 1NF
• Separate out the atomic and non-atomic attributes
• Put all the atomic attributes in a new replacement relation, which then by definition is in 1NF.
Date Saddr Sname Sno Ord Tot
8 June D’ham Bloggs 315 L6 400
7 May N’cle Smith 127 L5 12
The Non-Atomic Attributes
• We can’t just throw away this data because it is a nuisance to store!
• The values in all these attributes repeat together. – If a part is removed from an order, its values must
be removed from all 4 attributes. – If another part is placed on an order, there must
be a value for that part in all 4 attributes.
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Repeating Together
• Thus a set of values that repeat together should become a tuple in a new relation.
• Now the attributes in these tuples contain only atomic data !• Thus we form another new replacement relation to hold the tuples
of data that repeat together.• There is no intrinsic reason why all the non-atomic attributes in an
un-normalised relation should always repeat together.
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Part Pname Qty Price
N8 Nut 70 4
B6 Bolt 60 5
L4 Nail 100 3
P3 Pump 5 150
Q7 Motor 5 250
Foreign Keys• The problem with this relation is that the part data is no longer associated
with its order data. • We no longer know which part type was ordered on which purchase
order. • We can solve this problem by adding the (purchase) order number
attribute to this relation.• In general, we must add the attribute(s) which formed a candidate key in
the original relation, to this relation as a foreign key. This retains the relationship information.
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Ord Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
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Candidate Keys for Relations
Extend the candidate key to Ord and Part including the foreign key Ord*
The candidate key is Ord
Date Saddr Sname Sno Ord Tot
8 June D’ham Bloggs 315 L6 400
7 May N’cle Smith 127 L5 12
Ord* Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
SECOND NORMAL FORM (2NF)
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Definition of 2NF
A relation is in 2NF if and only if
it is in 1NF and
every non-key attribute is fully functionally dependent on the candidate key.
The extra constraint applied by 2NF
Note that 2NF is more strict than 1NF because it requires the relation to conform to the additional “full functional dependency” constraint.
Fully Functionally Dependent
• Question : What does fully functionally dependent mean?
• We will first consider the principle of
functional dependency, and then see– what full functional dependency means, – the application to achieve 2NF.
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Example of Functional Dependency
Account Number Payment Due
This type of arrow indicates a function dependency.
Assume some kind of loan account where payments of a certain amount have to be made on a regular basis to pay off the loan.
This means :
• A given account number determines what payment is due.
• In principle, given an account number, one can find out what regular payment is due. (May not always be easy or feasible in practice).
Terminology
• The Account Number is said to functionally determine the Payment Due.
• The Payment Due is said to be functionally dependent on the Account Number.
• Both are equally good means of expression, and convenience and emphasis usually determine which of the two is preferred in any particular situation.
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Further FD Examples Example 1:
Supplier Number
Supplier Name
Supplier Address
Supplier Telephone No.
Example 2:
Customer Name
Customer Address
Customer Telephone No.
A set containing one attribute determining a set of three attributes.
a set of two attributes determining a set containing one attribute.
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Full Functional Dependency & 2NF
• The definition of 2NF requires not merely functional dependency, but full functional dependency.
Definition of FULL Functional Dependency:
A set of attributes Y is fully functionally dependent on a set of attributes X
if and only if
Y is functionally dependent
on all the attributes of X and not just a subset of them.
Condition for 2NF
Thus, to be in 2NF means that:
all attributes not in the candidate key
are fully FD on
all those attributes that are in the candidate key.
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Examples: Purchase Order RelationsP_ORDER_1: FD Diagram
The functional dependencies of the non-key attributes in P_ORDER_1 on its candidate key can be represented by the following FD diagram :-
As they are all fully FD on Ord, the relation is already in 2NF.
Date Saddr Sname Sno Ord Tot
8 June D’ham Bloggs 315 L6 400
7 May N’cle Smith 127 L5 12
P_ORDER_1
Ord
Sno
Sname
Saddr
Date
Tot
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Ord* Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
P-ITEM-1
P-ITEM-1: FD Diagram
Reason for non-2NF• Attributes Price and Qty depend on the full key.
– They depend not only on what kind of part they refer to, but also on the order itself
• the quantity of a part type ordered will vary with & depend on the order, as will the price since it depends on the quantity.
• However Pname depends solely on the type of part. – A particular kind of part will have the same name on
every order on which it appears.
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Three Problems of a Non-2NF Relation
• Redundant data may be stored.
• Update anomalies– there can be problems in inserting, deleting and
amending some of the data.
• Semantic problems. – relation does not reflect the real-world meaning
of the data, leading to problems in its use.
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Redundant Data
Example: Pname is unnecessarily repeated.
Every time a part type appears on an order (say Q7), its name (Motor) also appears.
N.B. The part number (say Q7) is enough to identify the part type.
Motor is repeated in orders L6 & L7. One order is sufficient to give us the name, so the Pname is redundant (either one).
P-ITEM-1Ord* Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
L7 Q7 Motor 2 100
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Ord* Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
L7 Q7 Engine 2 100
?? F5 Flange ? ??
Example: Part type details (Part and Pname) cannot always be updated.
Update AnomaliesP-ITEM-1
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Semantic Problems
Q7 now has two different names.
Ord* Part Pname Qty Price
L5 N8 Nut 70 4
L5 B6 Bolt 60 5
L5 L4 Nail 100 3
L6 P3 Pump 5 150
L6 Q7 Motor 5 250
L7 Q7 Engine 2 100
P-ITEM-1
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Putting P_ITEM_1 into 2NF (1) The problem is caused by ‘Pname’ being FD on just part, not the whole of the candidate key. The solution is to separate out each determinant and its dependents. Create 2 replacement relations based on these FDs.
Ord
Part Pname
Qty
Price
Ord
Qty
Price
Part
Part Pname
Satisfaction of 2NF
• A relation created with a determinant as its candidate key, and with non-key attributes that are fully functionally dependent on that candidate key, must be in 2NF by definition.
• Note that a determining attribute - Part in the above example - can appear in more than one complete determinant. – This is perfectly acceptable. It just depends what
attributes form determinants.04/21/23 37
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Putting P_ITEM_1 into 2NF (2)
Ord* Part Qty Price
L5 N8 70 4
L5 B6 60 5
L5 L4 100 3
L6 P3 5 150
L6 Q7 5 250
L7 Q7 2 100
P-ITEM-2
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Putting P_ITEM_1 into 2NF (3)
Part Pname
N8 Nut
B6 Bolt
L4 Nail
P3 Pump
Q7 Motor
PART_2
Benefits of 2NF
• No information has been lost. – A natural join of P_ITEM_2 and PART_2 on
attribute Part will re-create the original relation P_ITEM_1.
• Problems Solved:– Redundant data removed – each Pname in once– Update anomalies – no side effects in operations– Semantic problems – each part type has just one
name
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THIRD NORMAL FORM (3NF)
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Definition of 3NF
A relation is in 3NF if and only if
it is in 2NF and
every non-key attribute is non-transitively fully FD on the candidate key.
The extra constraint applied by 3NF
Question : What does non-transitively mean ?
Note that 3NF is more stringent than 2NF, as it requires that the relation not only have full functional dependencies on the candidate key, but that these dependencies must now additionally be “non-transitive”.
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Transitivity
Let
and
Then
A C
A B
B C
These FDs are non-transitive i.e. direct, because they do not go via any other sets of attributes.
This FD is transitive, because it is via another set of attributes, in this case ‘B’.
Assume there are three sets of attributes, ‘A’, ‘B’ and ‘C’.
If A determines B, and B determines C, then logically A determines C, but transitively via B.
Example of Transitive FD
• Suppose pilots always fly the same aircraft – then if we know the pilot, we know the aircraft; so pilot
functionally determines aircraft.
• If we know the aircraft, then we know the airline that owns it – so aircraft functionally determines airline.
• Putting these two dependencies together– then pilot functionally determines airline.
• But the functional dependency of airline on pilot is transitive, because it goes via aircraft.
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Non-Transitive Full FD & 3NF
So, to be in 3NF means that
all attributes not in the candidate key are non-transitively - i.e. directly - fully FD on all those attributes that are in candidate key,
and not FD on the candidate key via some other non-key attribute.04/21/23 45
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Reviewing the Definition of 3NF 1.
Key NK1 NK2 NK3
R1( Key, NK1, NK2, NK3 ) R1’s FD diagram shows a “chain of dependencies”. It is not in 3NF.
2.
R2( Key, NK1, NK2, NK3 )
Key
NK1
NK2
NK3
R2’s FD diagram shows no “chain of dependencies”. It is in 3NF.
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Example: P_ITEM_2
Neither ‘Price’ nor ‘Qty’ is FD on the candidate key via the other, but non-transitively FD on the key.
Thus P_ITEM_2 is already in 3NF.
Ord* Part Qty Price
L5 N8 70 4
L5 B6 60 5
L5 L4 100 3
L6 P3 5 150
L6 Q7 5 250
L7 Q7 2 100
P-ITEM-2
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Example : PART_2
Thus PART_2 is already in 3NF.
If a 2NF relation only has one non-key attribute, then it must already be in 3NF, as there is no other non-key attribute via which a transitive dependency can occur.
Part Pname
N8 Nut
B6 Bolt
L4 Nail
P3 Pump
Q7 Motor
PART_2
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Example : P_ORDER_1
As we have already seen, its FD diagram is :- However, not all of these FDs are non-transitive FDs (= NTFDs).
P_ORDER_1
Date Saddr Sname Sno Ord Tot
8 June D’ham Bloggs 315 L6 400
7 May N’cle Smith 127 L5 12
Ord
Sno
Sname
Saddr
Date
Tot
Taking account now of transitivity, the FD diagram can be re-drawn as:-
Ord
Sname
Date
Sno
Tot
Saddr
Hence P_ORDER_1 is not in 3NF.
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Putting P_ORDER_1 into 3NF (1) The problem is caused by ‘Sname’ and ‘Saddr’ being only transitively FD on the candidate key.
Solution: separate out each determinant and its NTFD dependents, & create 2
replacement relations based on them.
Sname
Date
Sno
Tot
Saddr
Ord
Date
Tot
Sno
Ord
Sname
Sno Saddr
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Putting P_ORDER_1 into 3NF (2)
The corresponding relation is:-
Date
Tot
Sno
Ord
P_ORDER_3
Sno Ord
315 L6
L5
Date Tot
8 June 400
7 May 12 127
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Putting P_ORDER_1 into 3NF (3)
The corresponding relation is:-
Sname
Sno Saddr
SUPPLIER_3
Saddr Sname Sno
D’ham Bloggs 315
N’cle Smith 127
Benefits
• No information has been lost. – A natural join of P_ORDER_3 and SUPPLIER_3 on
attribute Sno will re-create the original relation P_ORDER_1.
• Problems Solved:– Redundant data removed – each Sname in once– Update anomalies – no side effects in operations– Semantic problems – each supplier has just one
name
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