Cleveland State University CIS 611 Relational Databases...

Functional Dependencies

Source:

The Theory of Relational Databases

D. Maier, Ed. Computer Science Press

Available at: http://www.dbis.informatik.hu-berlin.de/~freytag/Maier/

Cleveland State UniversityCIS 611 – Relational DatabasesPrepared by Victor Matos

1

Functional Dependencies

• Two primary purposes of databases are to – attenuate data redundancy and – enhance data reliability.

• Any a priori knowledge of restrictions or constraints on permissible sets of data has considerable usefulness in reaching these goals.

• Data dependencies are one way to formulate such advance knowledge.

2

Example1

• Consider the relation

assign (Pilot, Flight, Date, Departs)PILOT FLIGHT DATE DEPARTS

CushingCushingClarkClarkClarkChinChinCopelyCopelyCopely

83116281301 83 83 116281281 412

9 AugIO Aug8 Aug12 Aug11 Aug13 Aug12 Aug9 Aug13 Aug15 Aug

10: 15a1:25p5:50a6:35p10: 15a10: 15a1:25p5:50a5:50a1:25p

3

Example1 - Observations

• The relation assign tells which pilot flies a given flight on a given day, and what time the flight leaves.

• Not every combination of pilots, flights, dates, and times is allowable in assign. • The following restrictions apply, among others:

1. For each flight there is exactly one time.2. For any given pilot, date, and time, there is only one flight.3. For a given flight and date, there is only one pilot.

• These restrictions are examples of functional dependencies.

• Informally, a functional dependency occurs when the values of a tuple on one set of attributes uniquely determine the values on another set of attributes.

• Our restrictions can be phrased as

1. TIME functionally depends on FLIGHT,2. FLIGHT functionally depends on {PILOT, DATE, TIME}, and3. PILOT functionally depends on {FLIGHT, DATE}.

4

FD DefinitionDef. Let r be a relation on scheme R, with X and Y subsets of R. Relation r satisfies the functional dependency (FD) X Y if for every X-value x,

y( X=x(r)) has at most one tuple.

One way to interpret this expression is to look at pairs of tuples, t1 and t2, in r.

If t1(X) = t2(X), then t1(Y) = t2(Y).

In the FD X Y the portion X is called the leftside and Y is called the right side.

5

FD Satisfies

Algorithm 4.1 SATISFIES

Input: A relation r and an FD X Y.

Output: true if T satisfies X Y, false otherwise.

SATISFIES(r, X Y);

1. Sort the relation r on its X columns to bring tuples with equal X-values together.

2. If each set of tuples with equal X-values has equal Y-values, return true. Otherwise, return false.

SATISFIES tests if a relation r satisfies an FD X Y.

6

Algorithm: Satisfies

Using algorithm satisfies to test if FLIGHT DEPARTS

PILOT FLIGHT DATE DEPART

Cushing 83 9-Aug 10: 15a

Clark 83 11-Aug 10: 15a

Chin 83 13-Aug 10: 15a

Cushing 116 IO Aug 1:25p

Chin 116 12-Aug 1:25p

Clark 281 8-Aug 5:50a

Copely 281 9-Aug 5:50a

Copely 281 13-Aug 5:50a

Clark 301 12-Aug 6:35p

Copely 412 15-Aug 1:25p

Question:DEPARTS FLIGHT ???

7

Inference Axioms

• The number of FDs that can apply to a relation r(R) is finite, since there is only a finite number of subsets of R.

• Thus it is always possible to find all the FDs that r satisfies, by trying all possibilities using the algorithm SATISFIES.

• This brute-force approach is time-consuming.

8

Inference Axioms

• Finding F requires semantic knowledge of the relation r.

• After knowing some members of F, it is often possible to infer other members of F.

• A set F of FDs implies the FD X Y, written

F X Y, if every relation that satisfies all the FDs in F also satisfies X Y.

• An inference axiom is a rule that states if a relation satisfies certain FDs, it must satisfy certain other FDs.

9

Inference Axioms

F = { . . . }

Set of functional dependencies

Set of all relations r(R)

satisfying FDs in

F

X Y

A set F of FDs implies the FD X Y, written F X Y, if every relation that

satisfies all the FDs in F also satisfies X Y.

10

Example - Inference Axioms

F = { A B, B C }Set of functional

dependencies

Set of all relations r(R)

satisfying FDs in

F

A C

A set F of FDs implies the FD X Y, written F X Y, if every relation that

satisfies all the FDs in F also satisfies X Y.

11

Inference AxiomsThe Armstrong-Set of Inference Axioms • Axioms will implement the “intelligence” needed to

prove (or disprove) a sequence of derivations.

• Inference Machines are used to determine whether or not the application of the axioms on some ‘basic knowledge’ produces a ‘new’ valid piece of knowledge not there in the basic set.

• The first set we will consider is called the A-setproposed by W. Armstrong1.

1 William Armstrong: Dependency Structures of Data Base Relationships, page 580-583. IFIP Congress, 1974.

12

A-AxiomsA1. Reflexivity X X

A2. Augmentation If (Z W; X Y) then XW YZ

A3. Additivity If { (X Y) and (X Z)} then X YZ

A4. Projectivity If (X YZ) then X Y

A5. Transitivity If (X Y) and (Y Z) then (X Z)

A6. Pseudotransitivity If (X Y) and (YZ W) then XZ W

13

Inference Machine

INFERENCE MACHINE

A-AxiomsA1A2. . . A6

INPUT:Relation schema R

Set F of FDs on R

INPUT:

A “new” rule of the form X Y With X and Y in schema(R)

Output

YES

NO

Is the “new” rule XYderived from what is

known (R, F) by using the intelligence provided by the A-Axioms ?

If NO we must conclude that (F XY) is

not true

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Example1 - Using the A-AxiomsConsider R = (Street, Zip, City) ; and the dependencies

F = { City Street Zip, Zip City }

We want to show: Street Zip Street Zip City

Proof:

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Example1 - Using the A-AxiomsConsider R = (Street, Zip, City) ; and the dependencies

F = { City Street Zip, Zip City }

We want to show: Street Zip Street Zip City

Proof:

1. Zip City – Given

2. Street Zip Street City – Augmentation of (1) by Street

3. City Street Zip – Given

4. City Street City Street Zip – Augmentation of (3) by

City Street

5. Street Zip City Street Zip – Transitivity of (2) and (4)

16

Example2 – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies

F = { ABE AGJ BE I E G GI H }

Show that

AB GH

is derived by F.

17

If YES give a proof

If NO provide a counter-example

Example2A – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies


Show that

AB GH

is derived by F.

Step Statement Explanation

1 AB E Given

2 E G Given

3 AB G Transitivity on (1) and (2)

4 AB BE Augmentation (1) by B

5 BE I Given

6 AB I Transitivity on (4) and (5)

7 AB GI Additivity on (6) and (3)

8 GI H Given

9 AB H Transitivity on (7) and (8)

10 AB GH Additivity on (3) and (11)Q.E.D.

quod eratdemonstrandum

18

Example2B – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies


Show that

AB GH

is derived by F.


1 AB E Given

2 AB AB Reflexivity

3 AB B Projectivity on (2)

4 AB BE Additivity on (1) and (3)

5 BE I Given

6 AB I Transitivity on (4) and (5)

7 E G Given

8 AB G Transitivity on (1) and (7)

9 AB GI Additivity on (6) and (8)

10 GI H Given

11 AB H Transitivity on (9) and (10)

12 AB GH Additivity on (8) and (11)Q.E.D.quod eratdemonstrandum

19

again!

Example3 – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies


Show that

AEI H

is derived by F.


1

2

3

4

5

6

7

8

9

10

11

12 20

Your turn!

Reducing the A-Axioms

The set of A-Axioms is not minimal, therefore some of its rules could be eliminated.

Observations

• Rule A5 (transitivity) is a special case of rule A6 (pseudo-transitivity).

• Rules A3 (additivity) and A4 (projectivity) can be derived from A1 (reflexivity), A2 (augmentation), A6 (pseudo-transitivity).

Proof

(a) First observation is trivial (just make Z= Ø)

(b) Axiom A3 (Additivity) states that two rules, say X Y and X Z, can be combined in one X YZ. Lets use A2 on X Y to produce XZ YZ. Repeat A2 this time on X Z to produce X XZ. Now apply A5 on X XZ and XZ YZ; we get X YZ. Therefore, we conclude that X YZ without using the rule A3 itself (see next page)

21

Reducing the A-Axioms

The set of A-Axioms is not minimal, therefore some of its rules could be eliminated.

Statement

Axiom A3 is redundant. Rule A3 (Additivity) states that two rules, say X Y and X Z, can be combined in one X YZ.

Proof

We can prove that this fact is true without using A31. X Y Given

2. XZ YZ (A2) Augmenting (1) by Z

3. X Z Given

4. X XZ (A2) Augmenting (3) by Z

5. X YZ (A5) Transitivity on (4) and (2)

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Characterizing the A-Axioms

• The set of A-Axioms is complete

Therefore every FD that is implied by a set F of FDs can be derived from the FDs in F and one or more applications of the A-Axioms ( FA XY )

• A-Axioms are correct

Applying the axioms to FDs in a set F can only produce FDs that are implied by F.

• The set of A-axioms is not minimal

Some rules are added for convenience but they can be removed without diminishing the expressive power of the A-axioms

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Correctness of the A-Axioms

The axioms can not be used to prove a false derivation. In such a case showing a counter-example is sufficient to establish the falsity of a statement.

Example

Assume schema R(XYZW). Does ( XY ZW ) A X Z ?

The correct answer is NO. To show support for our argument we produce a counter-example. For instance:

On the example table there are no violations to the fact that XY implies a unique ZW (12 34 and 15 67). However X=1 determines two different Z values, 3 and 6. Therefore X Z is not a valid dependency as shown in the counter-example.

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X Y Z W

1 2 3 4

1 5 6 7

F+

Closure F+

• Let F be a set of FDs for a relation r(R). The closure of F, denoted F+, is the smallest set containing F such that the A-axioms cannot be applied to the set to produce a new rule not included in the set already

• Since F+ must be finite, we can compute it by starting with F, applying A1, A2, and A6, and adding the derived FDs to F until no new FDs can be derived.

• The closure of F depends

on the scheme R.

• If R = (A B) then F+ will always

contain B B, but if R = (A C),

F+ never contains B B.

F

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Closure F+

• The set F derives an FD X Y if X Y is in F+.

• Since our inference axioms are correct, if F derives X Y, then F implies X Y ( F A X Y )

• Note that F+ = (F+)+

• It is desirable to determine whether F A X Y without computing F+

• Computing the entire set F+ is time-consuming and tedious

26

Closure F+

Example: Consider the relation schema <R,F>

where R = (A B C) and F = { AB C, C B }.

By the use of brute-force we produce all rules out of F.

F+ is the set of rules listed below

F+

F = { ABC, CB }

A A

B B

C C

C B

AB AB

AC AC

BC BC

AB C

AB A

AB B

…

BC C

ABC ABC

ABC A

ABC B

ABC C

ABC AB

…

ABC BC

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A A

B B

C C

C B

AB AB

AC AC

BC BC

AB C

AB A

AB B

…

BC C

ABC ABC

ABC A

ABC B

ABC C

ABC AB

…

ABC BC

Closure F+

Example: Consider the relation schema <R,F> where R = (ABC)

and F = { AB C, C B }.

Question: Does F B C ?

Answer: F+ is the set of rules listed below and B C is not in the set; therefore the rule B C is not implied by F.

F+

F = { ABC, CB }

B C

This rule is NOT reachable from F

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Closure F+

Aside: How many FDs are there in <R,F>

An upper bound is

Each sum term represents the possible combinations of r attributes made out of the total n domains for each of the m X Y rule in F.

for n=3 there are (23-1)2 = 49 possibilities, however for R

holding 10 attributes there are over a million possible FDs

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2

1 1

(2 1)n n

n

r r

n n

r r

Closure F+

Definition. An FD X Y is trivial if X Y.

• If F is a set of FDs over R and X is a subset of R, then there is a FD X Y in F+ such that Y is maximal: for any other FD X Z in F+, Y Z.

– This result follows from additivity. – The right side Y is called the closure of X and is

denoted by X+.

• The closure of X always contains X, by reflexivity.

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Derivations and DDAGs

• If F XY, then either XY is in F, or a series of applications of the inference A-axioms to F will yield X Y.

• This sequence of axiom applications and resulting FDs is called a derivation of X Y from F.

• More formally, let F be a set of FDs over scheme R . A sequence P of FDs over R is a derivation sequence on F if every FD in P either– is a member of F, or– follows from previous FDs in P by an application of one of the inference

axioms A1 to A6.

• P is a derivation sequence for XY if XY is one of the FDs in P.

• DefinitionLet P be a derivation sequence on F. The use set of P is the collection of all FDs (originally) in F that appear in P.

31

Derivations and DDAGsEXAMPLE

Consider schema r(ABCDEG) and functional dependencies F = { A BC, BD G, C ED }

A derivation sequence for A E is

Try…

Step Explanation

12345

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Derivations and DDAGsEXAMPLE

Consider schema r(ABCDEG) and functional dependencies F = { A BC, BD G, C ED }

A derivation sequence for A E is

The set P for AE (five rules written above) is a derivation sequence on F. The Use_Set_Of_P is = {A BC, C ED }

Step Explanation

12345

A BCA C C EDC EA E

(given)(Projectivity [A4] on 1)(given)(Projectivity[A4] on 3)(Transitivity[A5] on 2 and 4)

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Derivations and DDAGsExample. Consider schema <R, F> where R= { A B C D E G H I J }

and F = { ABE, AGJ, BE I, E G, GI H}The following sequence is a derivation sequence for A B G H.

This P sequence contains unneeded FDs, such as 12 and 13, and is also a derivation sequence for other FDs, such as A B G I.The Use_Set_Of_P is {C ED, BE I, E G, GI H }

Step Explanation

1. AB E2. AB AB3. AB B4. AB BE5. BE I6. AB I7. E G8. AB G9. AB GI

10. GI H11. AB H12. GI GI13. G I I14. ABG H

given)(reflexivity)(projectivity from 2)(additivity from 1 and 3)(given)(transitivity from 4 and 5)(given)(transitivity from 1 and 7)(additivity from 6 and 8)(given)(transitivity from 9 and 10)(reflexivity)(projectivity from 12)(additivity from 8 and 11)

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B-AxiomsDefinition:

The B-Axioms set is a small and complete collection of inference rules. It is not a subset of A1 to A6, however it is equally expressive. For a relation (R), with W, X, Y, and Z subsets of R, and C an attribute in R then:

B1. Reflexivity X X

B2. Accumulation If (X YZ) and (Z CW) then X YZC

B3. Projectivity If (X YZ) then X Y

Motivation:This is another approach to the problem of finding a sequence of derivations using a smaller set of axioms.

Significance:Since B-Axioms are complete, we can always find a derivation sequence using only the three B-axioms to assert whether or not FB XY.

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B-AxiomsExample: Let R = (ABCDEGHI) and F = {ABE

AG J BE I E G GI H}

ProblemFind a derivation sequence P showing FB AB GH using only B-axioms.

AnswerSee P sequence on the right

Comment(a) Use_Set_Of_P contains

rules on lines 2, 5, 9, 11.(b) Too many steps!

Step Explanation

1

2

3

4

5

6

7

8

9

10

11

12

13

14

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B-AxiomsExample: Let R = (ABCDEGHI) and F = {ABE

AG J BE I E G GI H}

ProblemFind a derivation sequence P showing FB AB GH using only B-axioms.

AnswerSee P sequence on the right

Comment(a) Use_Set_Of_P contains

rules on lines 2, 5, 9, 11.(b) Too many steps!

Step Explanation

1 EI EI Reflexivity (B1)

2 E G Given

3 EI EIG Accumulation (B2)

4 EI GI Projectivity (B3) from (3)

5 GI H Given

6 EI GHI Accumulation from (4) and (5)

7 EI GH Projectivity from (6)

8 AB AB Reflexivity

9 AB E Given

10 AB ABE Accumulation from (8) and (9)

11 BE I Given

12 AB ABEI Accumulation from (10) and (11)

13 AB ABEIG Accumulation from (4) and (12)

14 AB ABEGHI Accumulation from (7) and (13)

15 AB GH Projectivity from (14) 37

Ok, but useless

RAP-Derivation SequenceRAP: Stands for: Reflexivity, Augmentation, Projectivity

Definition: Consider derivation sequences for X Y on a set F of FDs using the B-axioms that satisfy the following constraints:1. The first FD is X X2. The last FD is X Y3. Every FD other than the first and last is either an FD in F (given) or

and FD of the form X Z that was derived using axiom B2 (Accumulation).

Such a derivation is called a RAP-derivation sequence

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RAP-Derivation SequenceExample:

Let R = (ABCDEGHI) and


Find a RAP-sequence for AB GH

Comments

1. The table contains a RAP sequence

for ABGH.

2. Each rule in P is either given in F

or the result of applying B2 on

previous rules in P.

3. First and Last lines agree with the

definition of RAP sequence.

4. Use_Set_Of_P contains rules in

lines 2, 4, 6, 8.

Step Explanation1 AB AB B1

2 AB E Given

3 AB ABE B2

4 BE I Given

5 AB ABEI B2

6 E G Given

7 AB ABEIG B2

8 GI H Given

9 AB ABIGH B2

10 AB GH B3

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Let R = (ABCDEGHI) and


Find a RAP-sequence for BHE GI

Your turn…

Step Explanation1

2

3

4

5

6

7

8

9

10

40


Let R = (ABCDEG) and

F = { A BC, BD G, C ED }

Find a RAP-sequence for AD GE

Your turn…

Step Explanation1

2

3

4

5

6

7

8

9

10

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Let R = (ABCDEI) and

F = { A D, AB E, BI E, CD I, E C}

Find a RAP-sequence for AE DCI

Your turn…

Step Explanation1

2

3

4

5

6

7

8

9

10

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Derivation DAGsA directed acyclic graph (DAG) is a directed graph with no directed paths from any node to itself.

A labeled DAG is a DAG with an element from some labeling set L associated with each node.

Valid DAG (disconnected but OK)

Not a Valid DAG (Path: A-D-A makes a cycle)

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Derivation DAGs• DAGS are a convenient way of graphically representing a

derivation sequence of the form F B X Y

• Whenever there is a RAP derivation sequence there is an equivalent DDAG (and conversely)

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Derivation DAGsEXAMPLE

Consider schema r(ABCDEG) and functional dependencies

F = { A BC, BD G, C ED }. Show a DDAG for AD GE

NOTE:

The Use_Set of the derivation sequence is { A BC, BD G, C E }

A

D

B

C

G

E

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Derivation DAGsRules for Constructing a DDAG

Rule 1. Any set of unconnected nodes with labels from r(R) is an F-based DDAG

Rule 2. Let H be a DDAG including nodes labeled A1 … Ak. Let rule A1…Ak B be part of F. Form graph H’ by adding a new node labeled “B” and new edges <A1,B>,…,<AK,B>.

Rule 3. Nothing else is an F-based derivation DDAG.

A1 A2

…

A1

AK

AN

B

New edges

New node

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Derivation DAGsExample

Consider the relation schema r(ABCDEGHIJ) subject to the dependencies

in F = { AB E, AG J, BE I, E G, GI H }.

Draw a DDAG for rule AB GH

Note: The Use_Set of the derivation sequence is { AB E, BE I, E G, GI H }

A

B

E

I

G

H

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Consider the relation schema r(ABCDEGHIJ) subject to the dependencies


Draw a DDAG for the new rule BIG JA

NOTE: No path from source to destination is possible, therefore the new rule BIG AJ is not derivable from F.

B

I H A

G

J

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Consider the relation schema R(ABCDEGHIJ) subject to the dependencies


Draw a derivation DDAG for the new rule AB HC

NOTE: Node C is not reachable from the source. Therefore the rule AB CH cannot be deduced from F.

A

B

E

I

C

H

G

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X + Closure of a Set of Attributes

• In order to simplify the asserting of whether or not a rule X Y follows from a set F of FDs, we will compute X+ the closure of a set of attributes X

• The set X+ is the maximal set of attributes which can be derived from X using a RAP derivation sequence starting on X

• We will say that X Y is in F+ whenever Y is in X+

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Computing X +

The following algorithm to compute X+ has poor performance but is easy to understand

Algorithm: X-ClosureInput: A set of attributes X and a set of FDs FOutput: The closure of X under F denoted X+

function X-CLOSURE (X, F)begin

OldDep = ; NewDep = X;while ( NewDep OldDep ) do begin

OldDep = NewDepfor every FD A B in F do

if ( NewDep A ) then NewDep = NewDep B;end while;return ( NewDep )

end function;

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Computing X +

EXAMPLES

Consider the relation schema r(ABCDEGHIJ) subject to the dependencies in

F = { AB E, AG J, BE I, E G, GI H }.

Compute closure of AB

(AB) + = A B reflexivity

A B E using AB E

A B E I BE I

A B E I G E G

A B E I G HJ GI H

nothing else could be added to AB+

Note: Observe that AB ABEIHG. This rule is a compact notation for the 27 FDs having AB as LHS.52

Computing X +

EXAMPLES

Consider the relation schema r(ABCDEGHIJ) subject to the dependencies in

F = { AB E, AG J, BE I, E G, GI H }.

Compute closure DEC

(DEC) + = D E C

D E C G using E G

nothing else could be added

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Member Algorithm

Checking Membership

In order to verify whether or not a functional dependency X Y could be derived from a set F of FDs the following simple test could be applied

F B X Y if Y is part of X+

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Member AlgorithmMember Algorithm

Input: Rule X Y and functional dependencies F

Output: TRUE whenever the rule is derived from F

Method:

begin

if ( Xclosure (X, F) Y ) then

return( True )

else

return( False );

end;

Example

Question: Does rule AB EH follow from F = { AB E, AG J, BE I, E G, GI H }

Answer: YES. Observe that (AB)+ = ABEIGHJ EH.

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Linear Closure – XF+

Input: A set of attributes X and a set of functional dependencies F

Output: The closure of X under F demote XF+

Procedure LINCLOSURE ( Attribute X, SetOfFDs F)

BEGIN

/* Initialization */

for each FD W Z in F do begin

COUNT[ W Z ] = lenghtOf(w);

for each attribute A in W do

add rule W Z into LIST[ A ];

end;

NEWDEP = X; UPDATE = X;

/* Computation */

while ( UPDATE Ø) do begin

Choose an attribute A in UPDATE;

UPDATE = UPDATE - A;

for each FD W Z in LIST[A] do begin

COUNT[W Z] = COUNT[W Z] - 1;

if ( COUNT[w Z] = 0 ) then

ADD = Z - NEWDEP;

NEWDEP = NEWDEP ADD;

UPDATE = UPDATE ADD;

end if;

end for;

end while;

END 56


Example

Consider the schema r(ABCDEI) subject to the dependencies

F = { A D, AB E, BI E, CD I, E C}

Find the closure of AE using the Linear Closure algorithm

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Example (continuation…) Tracing the execution of the linear time closure algorithm

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UPDATE= AE NEWDEP= AE

List[A] Rule 1 A D Count[1] = 0therefore add D to both stringsRule 2 AB E Count[2] = 1

UPDATE= ED NEWDEP= AED

List[E] Rule 5 E C Count[5] = 0therefore add C to both strings

UPDATE= DC NEWDEP= AEDC

List[D] Rule 4 CD I Count[5] = 1

UPDATE= CNEWDEP= AEDC

List[C] Rule 4 CD I Count[4] = 0therefore add I to both strings

UPDATE= INEWDEP= AEDCI

List[I] Rule 3 BI E Count[3] = 1

UPDATE= ØNEWDEP= AEDCI

therefore (AE) + = AEDCI


Homework

You will create a CASE tool for designing ‘good’ databases.

The first step involves the implementation of the XLinearClosure() algorithm.

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