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DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A BCNF Normalisation Algorithm
Input:I A specification containing:
1. a universal set of attributes U, and2. a set of functional dependencies (FDs) F over U.
I An entity-relationship diagram (ERD) conforming tothe specification.
Output: A database schema R which is in BCNF withrespect to F.
Notes:I R = {R1,R2, . . . ,Rn}, where each Ri is a subset of U.I The union of all schema(Ri ) is U.I R is called a decomposition of U.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A BCNF Normalisation Algorithm
Input:I A specification containing:
1. a universal set of attributes U, and2. a set of functional dependencies (FDs) F over U.
I An entity-relationship diagram (ERD) conforming tothe specification.
Output: A database schema R which is in BCNF withrespect to F.
Notes:I R = {R1,R2, . . . ,Rn}, where each Ri is a subset of U.I The union of all schema(Ri ) is U.I R is called a decomposition of U.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Strategy
Step 1. Convert ERD into a database schema S.Step 2. If any of the relation schemas in S are not in
BCNF with respect to F, then decomposethem further, using the DECOMPOSEalgorithm, given in this lecture.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Algorithm ERD-TO-BCNF(ERD, F)
1. Convert ERD into a database schema S = {S1, . . . ,Sm};2. let the output database schema R be empty;3. for each Si in S do4. if TEST-BCNF(Si , F) = YES5. add Si to R;6. else7. merge DECOMPOSE(Si , F) and R;8. end for9. return the decomposition R;
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Testing whether a relation schema is in BCNF
Algorithm TEST-BCNF(R, F)Assume F is a set of canonical FDs
1. for each (non-trivial) X → A in F+ do2. if X is not a superkey with respect to F3. return NO;4. end if5. end for6. return YES;
Note that, in general, we need to consider F+, theclosure of F , to check whether there are any FDs whichviolate BCNF.
But we can start trying to find violations in F , and onlyconsider F+ once we find no violations in F .
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Testing whether a relation schema is in BCNF
Algorithm TEST-BCNF(R, F)Assume F is a set of canonical FDs
1. for each (non-trivial) X → A in F+ do2. if X is not a superkey with respect to F3. return NO;4. end if5. end for6. return YES;
Note that, in general, we need to consider F+, theclosure of F , to check whether there are any FDs whichviolate BCNF.
But we can start trying to find violations in F , and onlyconsider F+ once we find no violations in F .
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example
I Let schema(PHONE) = {cust-name, phone-num,phone-network}.
I Let F = {cust-name → phone-num, phone-num →phone-network}.
Is PHONE in BCNF with respect to F ?
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Decomposition Condition used by AlgorithmSo phone-num → phone-network violates BCNF.
I phone-num is the left-hand side of the violating FDI phone-network is the right-hand side of the violating
FD
Split PHONE into two relation schemas:
1. R1 = NETWORK, with schema(NETWORK) ={phone-num, phone-network}, containing all theattributes in the violating FD, andF1 = { phone-num → phone-network }.
2. R2 = CUST, with schema(CUST) = {cust-name,phone-num}, containing the attributes inschema(PHONE) except those in the right-hand sideof the violating FD, andF2 = { cust-name → phone-num }.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Decomposition Condition used by AlgorithmSo phone-num → phone-network violates BCNF.
I phone-num is the left-hand side of the violating FDI phone-network is the right-hand side of the violating
FD
Split PHONE into two relation schemas:
1. R1 = NETWORK, with schema(NETWORK) ={phone-num, phone-network}, containing all theattributes in the violating FD, andF1 = { phone-num → phone-network }.
2. R2 = CUST, with schema(CUST) = {cust-name,phone-num}, containing the attributes inschema(PHONE) except those in the right-hand sideof the violating FD, andF2 = { cust-name → phone-num }.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Decomposition Condition used by AlgorithmSo phone-num → phone-network violates BCNF.
I phone-num is the left-hand side of the violating FDI phone-network is the right-hand side of the violating
FD
Split PHONE into two relation schemas:
1. R1 = NETWORK, with schema(NETWORK) ={phone-num, phone-network}, containing all theattributes in the violating FD, andF1 = { phone-num → phone-network }.
2. R2 = CUST, with schema(CUST) = {cust-name,phone-num}, containing the attributes inschema(PHONE) except those in the right-hand sideof the violating FD, andF2 = { cust-name → phone-num }.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Algorithm DECOMPOSE(R, F )Assume F is a set of canonical FDs
1. let the output database schema Out be empty;2. if TEST-BCNF(R, F) = YES then3. add R to Out;4. else5. let X → A in F+ be nontrivial (i.e. A is not in X)
such that X is not a superkey with respect to F;6. let R1 be a relation schema,
with schema(R1) = X merged with A;7. merge DECOMPOSE(R1, F ) and Out;8. let R2 be a relation schema,
with schema(R2) = schema(R) except A;9. merge DECOMPOSE(R2, F ) and Out;10. end if11. return Out;
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Result. DECOMPOSE(R, F) returns a decomposition ofschema(R).
Result. The natural join can be applied to all of therelations in DECOMPOSE(R, F) to recover precisely theinformation stored in any relation over schema(R); this isknown as the lossless join property.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Lossless join
Recall example: S is a relation schema, withschema(S) = {ENAME, CNAME, SAL} andsingle FD: ENAME → SAL
(Modified) relation s over S is given by
ENAME CNAME SALJack Diane 25Jack John 25
Donald Diane 30Donald David 30
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
If we decompose S into {ENAME,CNAME} and{CNAME,SAL} as follows:
ENAME CNAMEJack DianeJack John
Donald DianeDonald David
CNAME SALDiane 25John 25Diane 30David 30
and then perform the natural join, we get
ENAME CNAME SALJack Diane 25Jack Diane 30Jack John 25
Donald Diane 25Donald Diane 30Donald David 30
⇒ with two tuples that were not in the original relation
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
If we decompose S into {ENAME,CNAME} and{CNAME,SAL} as follows:
ENAME CNAMEJack DianeJack John
Donald DianeDonald David
CNAME SALDiane 25John 25Diane 30David 30
and then perform the natural join, we get
ENAME CNAME SALJack Diane 25Jack Diane 30Jack John 25
Donald Diane 25Donald Diane 30Donald David 30
⇒ with two tuples that were not in the original relation
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
I A decomposition such as that into {ENAME,CNAME}and {CNAME,SAL} is called lossy
I We started knowing Jack’s salary was 25I After decomposing, if we query Jack’s salary we get
both 25 and 30I The decomposition does not faithfully represent the
original information we had
I A decomposition which does faithfully represent theoriginal information is called lossless
I The lossless condition is guaranteed if we ensurethat the common attributes between a pair ofdecomposed relation schemas is a key for one ofthem
I The BCNF algorithm ensures losslessdecompositions
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
I A decomposition such as that into {ENAME,CNAME}and {CNAME,SAL} is called lossy
I We started knowing Jack’s salary was 25I After decomposing, if we query Jack’s salary we get
both 25 and 30I The decomposition does not faithfully represent the
original information we hadI A decomposition which does faithfully represent the
original information is called losslessI The lossless condition is guaranteed if we ensure
that the common attributes between a pair ofdecomposed relation schemas is a key for one ofthem
I The BCNF algorithm ensures losslessdecompositions
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of BCNF Decomposition
Recall example involving employee names (ENAME),salaries (SAL) and children (CNAME). Let’s call therelation schema EMP.
The assumed set of FDs was F2 = { ENAME → SAL }.
I ENAME → SAL violates BCNF in EMP, sodecompose EMP intoES, with schema(ES) = { ENAME, SAL }, andEC, with schema(EC) = { ENAME, CNAME }
I Both ES and EC are in BCNF:I ENAME is a key in ESI the only key for EC is (ENAME, CNAME)
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example of BCNF DecompositionLet STUD be a relation schema, with schema(STUD) ={SNUM, POSTCODE, CITY, COUNTRY}, with FDs{SNUM → POSTCODE, POSTCODE → CITY,CITY → COUNTRY}
I CITY → COUNTRY violates BCNF in STUD, sodecompose STUD intoCC, with schema(CC) = {CITY, COUNTRY}, andSTUD1, with schema(STUD1) = {SNUM,POSTCODE, CITY}
I CC is in BCNF while POSTCODE → CITY violatesBCNF in STUD1, so decompose STUD1 intoPC, with schema(PC) = {POSTCODE, CITY}, andSINFO = {SNUM, POSTCODE}.
I All the relation schemas in the database schema{CC, PC, SINFO} are now in BCNF
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example of BCNF DecompositionLet STUD be a relation schema, with schema(STUD) ={SNUM, POSTCODE, CITY, COUNTRY}, with FDs{SNUM → POSTCODE, POSTCODE → CITY,CITY → COUNTRY}
I CITY → COUNTRY violates BCNF in STUD, sodecompose STUD intoCC, with schema(CC) = {CITY, COUNTRY}, andSTUD1, with schema(STUD1) = {SNUM,POSTCODE, CITY}
I CC is in BCNF while POSTCODE → CITY violatesBCNF in STUD1, so decompose STUD1 intoPC, with schema(PC) = {POSTCODE, CITY}, andSINFO = {SNUM, POSTCODE}.
I All the relation schemas in the database schema{CC, PC, SINFO} are now in BCNF
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example of BCNF DecompositionLet STUD be a relation schema, with schema(STUD) ={SNUM, POSTCODE, CITY, COUNTRY}, with FDs{SNUM → POSTCODE, POSTCODE → CITY,CITY → COUNTRY}
I CITY → COUNTRY violates BCNF in STUD, sodecompose STUD intoCC, with schema(CC) = {CITY, COUNTRY}, andSTUD1, with schema(STUD1) = {SNUM,POSTCODE, CITY}
I CC is in BCNF while POSTCODE → CITY violatesBCNF in STUD1, so decompose STUD1 intoPC, with schema(PC) = {POSTCODE, CITY}, andSINFO = {SNUM, POSTCODE}.
I All the relation schemas in the database schema{CC, PC, SINFO} are now in BCNF
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example of BCNF DecompositionLet STUD be a relation schema, with schema(STUD) ={SNUM, POSTCODE, CITY, COUNTRY}, with FDs{SNUM → POSTCODE, POSTCODE → CITY,CITY → COUNTRY}
I CITY → COUNTRY violates BCNF in STUD, sodecompose STUD intoCC, with schema(CC) = {CITY, COUNTRY}, andSTUD1, with schema(STUD1) = {SNUM,POSTCODE, CITY}
I CC is in BCNF while POSTCODE → CITY violatesBCNF in STUD1, so decompose STUD1 intoPC, with schema(PC) = {POSTCODE, CITY}, andSINFO = {SNUM, POSTCODE}.
I All the relation schemas in the database schema{CC, PC, SINFO} are now in BCNF
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
I Consider a modified relation schema EMP, withattributes ENAME, CNAME (child name), DNAME(department name) and MNAME (manager name).
I The set of FDs is F = {E → D, D → M, M → D}, whereE stands for ENAME, D stands for DNAME and Mstands for MNAME (and C stands for child name).
I All three FDs violate BCNF since EC is the only key.I We can choose any one of them as the basis for the
first decomposition step.I We will consider all three decompositions in turn.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Third Example: Decomposition 1
I If we first decompose using D → M, we get twoschemas with attributes {D, M} and {E, C, D}.
I FDs D → M and M → D are applicable to {D, M}, butboth D and M are keys.
I FD E → D is applicable to {E, C, D} and E is not asuperkey.
I So we decompose {E, C, D} into {E, D} and {E, C}.I E is a key for {E, D} and EC is the key for {E, C}.I So the final database schema comprises
{D, M}, {E, D} and {E, C}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Third Example: Decomposition 2
I If we first decompose using E → D, we get twoschemas with attributes {E, D} and {E, C, M}.
I E → D is applicable to {E, D}, but E is a key.I What FDs are applicable to {E, C, M}?I None of E → D, D → M or M → D apply because D is
not in {E, C, M}.I We have to consider all FDs in F+.I Recall that E → M follows from E → D and D → M.I E → M violates BCNF in {E, C, M} because E is not a
key.I So we decompose {E, C, M} into {E, M} and {E, C}.I So the final database schema comprises
{E, D}, {E, M} and {E, C}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Third Example: Decomposition 3
I If we first decompose using M → D, we get twoschemas with attributes {M, D} and {E, C, M}.
I FDs D → M and M → D are applicable to {M, D}, butboth D and M are keys.
I Once again we have {E, C, M}, so it is decomposedas before into {E, M} and {E, C}.
I So the final database schema comprises{M, D}, {E, M} and {E, C}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
An example for you to tryLet R be a relation schema, with schema(R) ={C,T,H,R,S,G}.
I C stands for a course,I T stands for a teacher,I H stands for hour,I R stands for room,I S stands for student andI G stands for grade.
An example set of FDs F over R :
1. C → T,2. HR → C,3. HT → R,4. CS → G and5. HS → R.
Decompose R into BCNF.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Dependency Preservation
Recall example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
{S,C,P} is not in BCNF
Decompose {S,C,P} into {P,C} and {P,S}
P Cp1 cp2 c
P Sp1 sp2 s
Only FD that can be tested in the decomposition is P → C
When we join the two relations, we see that SC → P isviolated.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Dependency Preservation
A decomposition is dependency preserving if the FDswhich hold on the original relation schema can be testedon the decomposed schemas, without using joins.
We cannot always find a BCNF decomposition that isdependency preserving.
To test that no FDs are violated, we may need to joinrelations (expensive).
We can always find a 3NF dependency-preservingdecomposition.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Dependency Preservation
For a starting set of attributes and FDs, some BCNFdecompositions may be dependency preserving andsome not.
Consider the example with attributes { E, C, D, M } andFDs F = {E → D, D → M, M → D}.
We had three possible decompositions1. {D, M}, {E, D} and {E, C}.2. {E, D}, {E, M} and {E, C}.3. {M, D}, {E, M} and {E, C}.
Which of them is dependency-preserving?
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
3NF Synthesis Algorithm
Given a relation schema R and a set of FDs F, thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:
1. Find a minimal cover for F, say G.2. For each FD X → A in G, use XA as the schema of
one of the relations in the decomposition.3. If none of the schemas from Step 2 includes a
superkey for R, add another relation schema that is akey for R.
4. Delete any of the schemas from Step 2 that iscontained in another.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
3NF Synthesis Algorithm
Given a relation schema R and a set of FDs F, thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:
1. Find a minimal cover for F, say G.
2. For each FD X → A in G, use XA as the schema ofone of the relations in the decomposition.
3. If none of the schemas from Step 2 includes asuperkey for R, add another relation schema that is akey for R.
4. Delete any of the schemas from Step 2 that iscontained in another.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
3NF Synthesis Algorithm
Given a relation schema R and a set of FDs F, thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:
1. Find a minimal cover for F, say G.2. For each FD X → A in G, use XA as the schema of
one of the relations in the decomposition.
3. If none of the schemas from Step 2 includes asuperkey for R, add another relation schema that is akey for R.
4. Delete any of the schemas from Step 2 that iscontained in another.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
3NF Synthesis Algorithm
Given a relation schema R and a set of FDs F, thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:
1. Find a minimal cover for F, say G.2. For each FD X → A in G, use XA as the schema of
one of the relations in the decomposition.3. If none of the schemas from Step 2 includes a
superkey for R, add another relation schema that is akey for R.
4. Delete any of the schemas from Step 2 that iscontained in another.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
3NF Synthesis Algorithm
Given a relation schema R and a set of FDs F, thefollowing steps produce a 3NF decomposition of R thatsatisfies the lossless join condition and is dependencypreserving:
1. Find a minimal cover for F, say G.2. For each FD X → A in G, use XA as the schema of
one of the relations in the decomposition.3. If none of the schemas from Step 2 includes a
superkey for R, add another relation schema that is akey for R.
4. Delete any of the schemas from Step 2 that iscontained in another.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of 3NF Synthesis
Recall the example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
Step 1 of the algorithm finds that F3 is a minimal cover.
Step 2 of the algorithm would produce {P,C} and {S,C,P}.
Step 3 finds that SC is a superkey.
Step 4 deletes {P,C} to leave just {S,C,P}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of 3NF Synthesis
Recall the example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
Step 1 of the algorithm finds that F3 is a minimal cover.
Step 2 of the algorithm would produce {P,C} and {S,C,P}.
Step 3 finds that SC is a superkey.
Step 4 deletes {P,C} to leave just {S,C,P}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of 3NF Synthesis
Recall the example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
Step 1 of the algorithm finds that F3 is a minimal cover.
Step 2 of the algorithm would produce {P,C} and {S,C,P}.
Step 3 finds that SC is a superkey.
Step 4 deletes {P,C} to leave just {S,C,P}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of 3NF Synthesis
Recall the example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
Step 1 of the algorithm finds that F3 is a minimal cover.
Step 2 of the algorithm would produce {P,C} and {S,C,P}.
Step 3 finds that SC is a superkey.
Step 4 deletes {P,C} to leave just {S,C,P}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Example of 3NF Synthesis
Recall the example: F3 = {SC → P, P → C}.S stands for Street, C stands for City and P stands forPostcode.
Step 1 of the algorithm finds that F3 is a minimal cover.
Step 2 of the algorithm would produce {P,C} and {S,C,P}.
Step 3 finds that SC is a superkey.
Step 4 deletes {P,C} to leave just {S,C,P}.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example
Recall the example: F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.
Step 1 of the algorithm finds that F2 is a minimal cover.
Step 2 of the algorithm would produce {E,S}.
Step 3 finds no superkey, so adds relation schema {E,C}.
Step 4 finds nothing to delete.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example
Recall the example: F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.
Step 1 of the algorithm finds that F2 is a minimal cover.
Step 2 of the algorithm would produce {E,S}.
Step 3 finds no superkey, so adds relation schema {E,C}.
Step 4 finds nothing to delete.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example
Recall the example: F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.
Step 1 of the algorithm finds that F2 is a minimal cover.
Step 2 of the algorithm would produce {E,S}.
Step 3 finds no superkey, so adds relation schema {E,C}.
Step 4 finds nothing to delete.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example
Recall the example: F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.
Step 1 of the algorithm finds that F2 is a minimal cover.
Step 2 of the algorithm would produce {E,S}.
Step 3 finds no superkey, so adds relation schema {E,C}.
Step 4 finds nothing to delete.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
Another Example
Recall the example: F2 = {E → S}.E stands for ENAME, S stands for SAL and C stands forCNAME.
Step 1 of the algorithm finds that F2 is a minimal cover.
Step 2 of the algorithm would produce {E,S}.
Step 3 finds no superkey, so adds relation schema {E,C}.
Step 4 finds nothing to delete.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
A Third Example
Consider the example: F = {AB → CD, C → AD, D → A }.
Step 1 of the algorithm finds that F is not a minimal cover.
First we form a canonical set of FDs:{AB → C, AB → D, C → A, C → D, D → A }.
Then we find that AB → D and C → A are redundant.
So we are left with minimal coverG = {AB → C, C → D, D → A }.
The rest is easy.
DatabaseManagement
Peter Wood
NormalisationAlgorithmsBCNF Algorithm
Lossless Join
BCNF Examples
Dependency Preservation
3NF Algorithm
An example for you to try
Consider the set of attributes { Drinker, Address, Pub,Location, Beer, Cost }, along with the following set ofFDs:
I Drinker → AddressI Pub → LocationI Pub, Beer → Cost, Location
Produce a set of 3NF relation schemas for the above.