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Database Systems &Applications

Lec 9Functional Dependencies

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Determinant

The attribute on the left-hand side ofthe arrow in a functionaldependency.

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Candidate (Primary) key

An attribute or combination of attributesthat uniquely identifies a row in a relation.It should satisfy

Unique identification. i.e., for every row, thevalue of the key must uniquely identify thatrow. This property implies that each nonkeyattribute is functionally dependent on the

candidate key. Nonredundancy : No attribute in the key (in the

case of multi-attribute key) can be deletedwithout destroying the property of uniqueidentification.

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Determinant vs Candidate key

Let us summarize the relationshipbetween determinants and candidatekeys. A candidate key is always a

determinant.

A determinant may or may not be a

candidate key.

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Examples

st_id st_name h_no r_no address

STUDENT1

STUDENT2

st_id c_no st_name h_no r_no address sem year grade

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Examples

Ex 1:In the example STUDENT1, theattribute st_id is a determinant

and also a candidate key.Becausest_id uniquely identifies theremaining (non key) attributes in

the relation STUDENT1.

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Examples (cont.)

From the above two examples, it isclear that

a candidate key is a determinant that

uniquely identifies the remaining(non key) attributes in a relation.

A determinant may be a candidate key (such as st_id in

STUDENT1)

Part of a composite candidate key (suchas st_id in STUDENT2).

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Examples (cont.)

Ex 2:

In the example STUDENT2, the

attribute st_id is a determinant butnot a candidate key. Becausest_id does not uniquely identify theremaining (non key) attributes in

the relation STUDENT2.

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Rules of FDs Assume W, X, Y, Z are sets of attributes of R.

If Y X then X Y (reflexivity) If X Y then ZX ZY (augmentation)

If X Y and Y Z then X Z (transitivity)

If X Y and X Z then X YZ (Union/Combining)

If X YZ then X Y and X Z(splitting/decomposition)

If W X and XY Z then WY Z

If XY ZY then XY Z

First three are known as Armstrongs Axioms

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Rules of FDs (cont.) Reflexivity:

If {B1,B2,.,Bj} {A1,A2,.,Ai} thenA1A2.Ai B1B2Bj.

Augmentation:

If A1A2.Ai B1B2Bj thenA1,A2,.,AiC1C2Ck B1B2BjC1C2Ck

for any set of attributes C1,C2,,Ck.

Transitivity:

If A1A2.Ai B1B2Bj and B1B2Bj C1C2Ck then A1A2.Ai C1C2Ck.

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Rules of FDs (cont.)

Union:

If A1A2.Ai B1B2Bj and

A1A2.Ai C1C2Ck then

A1A2.Ai B1B2BjC1C2Ck

Decomposition:

If A1A2.Ai B1B2BjC1C2Ck and

A1A2.Ai B1B2Bj then

A1A2.Ai B1B2BjC1C2Ck

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Rules of FDs (cont.)

If W X and XY Z then WY Z

If D1D2Dl A1A2.Ai and

A1A2Ai B1B2Bj C1C2Ck and

A1A2.Ai C1C2Ck

then D1D2Dl B1B2Bj C1C2Ck.

If XY ZY then XY Z

If A1A2Ai B1B2BjC1C2CkB1B2Bj

then A1A2.Ai B1B2Bj C1C2Ck.

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14

Trivial FunctionalDependencies

A1 A2 A3.A n B1 B2 B3 Bm is

Trivial if the Bs are a subset of the As.

Nontrivial if at least one of the Bs is notamong the As.

Completely nontrivial if none of the Bsalso one of the As.

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PILOT FLIGHT DATE DEPARTS

Soumini 83 9 Aug 10:15am

Soumini 116 10 Aug 1:25pm

Samulya 281 8 Aug 5:50pm

Samulya 301 12 Aug 6:35pm

Samulya 83 11 Aug 10:15am

Sekhar 83 13 Aug 10:15amSekhar 116 12 Aug 1:25pm

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PILOT FLIGHT DATE DEPARTS

Soumini 83 9 Aug 10:15am

Soumini 116 10 Aug 1:25pm

Samulya 281 8 Aug 5:50pm

Samulya

301 12 Aug 6:35pm

Samulya

83 11 Aug 10:15am

Sekhar 83 13 Aug 10:15a

mDEPARTS is determined by FLIGHT

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PILOT FLIGHT DATE DEPARTS

Soumini 83 9 Aug 10:15am

Soumini 116 10 Aug 1:25pm

Samulya 281 8 Aug 5:50pm

Samulya

301 12 Aug 6:35pm

Samulya

83 11 Aug 10:15am

Sekhar 83 13 Aug 10:15a

mPILOT is determined by FLIGHT, DATE

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FLIGHT DATE FLIGHT

FLIGHT DATE DATE

FLIGHT DATE

DEPARTS DATE

FLIGHT DEPARTS

FLIGHT DATE PILOT

Trivial

NonTrivial

Completely

Non Trivial

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Attribute closure

Suppose {A1,A2,.,An} is a set ofattributes and S is a set of FDs. The closureof {A1,A2,.,An} under the FDs in S is the

set of all attributes B such that everyrelation that satisfies all the FDs in set Salso satisfies A1,A2,.,An B. i.e., A1,A2,.,An follows from the FDs of S. It isdenoted as A1A2An by {A1,A2,,An}+.

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Attribute closure (Contd)

Starting with the given set ofattributes, we repeatedly expand the

set by adding the right sides of FDsas soon as we have included their leftsides. Eventually, we can not expandthe set any more and the resulting

set is the closure.

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Attribute closure (cont.) The following steps explain in more detail about the above:

step1. Let X be a set of attributes that eventually will

become the closure. First initialize X to {A1,A2,,An}.step2. Search for some FD B1B2Bm C such that all of

B1,B2,,Bm are in the set of attributes X, but not C.

Then add C to the set X.

step3. Repeat step 2 as many times as necessary until no

more attributes can be added to X. Since X can only grow

and number of attributes of any relation schema are finite,eventually nothing can be added to X.

step4. The set X, after no more attributes can be added to it is the

correct value of {A1,A2,,An}+.

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Attribute closure (Contd)

Let F be a set of FDs holding on a relationR. Let X, Y be sets of attributes of R. ThenY is said to be attribute closure of X,denoted by X+ = Y, if X Y follows from F

Algorithm

Ex: Find attribute closure of {A,B} w.r.tthe FDs:

AB C, BC AD, D E Useful to check whether a given FD follows from

given set of FDs: example AB D, D A Useful to check whether a given set of attributes

forms key w.r.t the FDs

Useful to find all FDs that hold on a relation

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Closure Test

An easier way to test is to computethe closure ofY, denoted Y+.

Basis: Y+ = Y.

Induction: Look for an FDs left sideXthat is a subset of the current Y+. Ifthe FD isX->A, addA to Y+.

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Attribute closure(example)

A relation has attributes

st_id, st_name, gender, dob, fname, hostel_no,room_no,

c_no,grade,sec_no,class_room

A = st_id, B=st_name,C=gender,

D=dob, E=fname, F=hostel_no,G=room_no, H=c_no, I=grade,J=sec_no,

K=class_room

The above relation has the FDs

AB CD, AH IJ, HJ K , A BE ,A F, AF G

Att ib t l ( l [ t ])

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Attribute closure (example[cont.]) The closure of {A,H} is {A,H}+.

Let X = {A,H}

These two attributes are on the left side of FD, AH IJ are in X, so add I,J to X,i.e., X= {A,H,I,J}.

In FD HJ K, the left side attributes H , J, form the subset of X, so add K to X,i.e., X={A,H,I,J,K}.

In FD A BE , the left side attributes A form subset of X, so add B,E to X, i.e.,X={A,B,E,H,I,J,K}.

In FD AB CD, the left side attributes A B form subset of X, so add C,D to X ,i.e.,X = {A,B,E,H,I,J,K,C,D}

In FD A F , the left side attributes A form subset of X, so add F to X, i.e.,X={A,B,C,D,E,F,H,I,J,K}.

In FD AF G , the left side attributes A form subset of X, so add G to X, i.e.,X={A,B,C,D,E,F,G,H,I,J,K}.

Therefore AH is the candidate key for the given relation.

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Is AH K ?