Normalization Questions and Answers
Normalization Questions and Answers
Database Systems, CSCI 4380-01 Database Systems, CSCI 4380-01
Sibel Adalı Sibel Adalı October 28, 2002 October 28, 2002
Question 1
Question 1 Suppose you are given a relationSuppose you are given a relation RR = = ((A , B , C , D , E A , B , C , D , E ) with the following functional) with the following functional dependencies:
dependencies: {{CE CE →→ D,D, DD →→B,B, C C →→AA}}.. a. Find all candidate keys.
a. Find all candidate keys.
b. Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). b. Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). c.
c. If the If the relrelatioation n is not is not in BCNF, decoin BCNF, decomposmpose e it until it it until it becobecomes BCNFmes BCNF. . At eacAt each h stestep, p, ideidentntify aify a new relation, decompose and re-compute the keys and the normal forms they satisfy.
new relation, decompose and re-compute the keys and the normal forms they satisfy.
Answer. Answer.
a. The only key is
a. The only key is {{C,C, E E }} b. The relation is in 1NF b. The relation is in 1NF c.
c. DecomDecompose into R1=(Apose into R1=(A,C) and R2=(B,C,D,C) and R2=(B,C,D,E). R1 is ,E). R1 is in BCNF, R2 is in BCNF, R2 is in 2NF. Decompose R2in 2NF. Decompose R2 into, R21=(C,D,E) and R22=(B,D). Both relations are in BCNF.
into, R21=(C,D,E) and R22=(B,D). Both relations are in BCNF.
Question 2
Question 2 Suppose you are given a relation R=(A,B,C,D,E) with the following functional de-Suppose you are given a relation R=(A,B,C,D,E) with the following functional de-pendencies:
pendencies: {{BC BC →→ADE,DADE,D →→ BB}}.. a. Find all candidate keys.
a. Find all candidate keys.
b. Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). b. Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). c.
c. If the If the relrelatioation n is not is not in BCNF, decoin BCNF, decomposmpose e it until it it until it becobecomes BCNFmes BCNF. . At eacAt each h stestep, p, ideidentntify aify a new relation, decompose and re-compute the keys and the normal forms they satisfy.
new relation, decompose and re-compute the keys and the normal forms they satisfy.
Answer. Answer.
a. The keys are
a. The keys are {{B,B, C C }} andand {{C,C, DD}} b. The relation is in 3NF
b. The relation is in 3NF
c. It cannot be put into BCNF, even if I remove D and put into a relation of the form (B,C,D) (I c. It cannot be put into BCNF, even if I remove D and put into a relation of the form (B,C,D) (I need C for the functional dependency), the resulting relation would not be in BCNF.
need C for the functional dependency), the resulting relation would not be in BCNF.
Question 3
Question 3 Suppose you are given a relation R=(A,B,C,D,E) with the following functional de-Suppose you are given a relation R=(A,B,C,D,E) with the following functional de-pendencies:
pendencies: BDBD →→ E,E, AA→→ C C ..
a. Show that the decomposition into R1=(A,B,C) and R2=(D,E) is lossy. You can show using any a. Show that the decomposition into R1=(A,B,C) and R2=(D,E) is lossy. You can show using any method.
method. My suggestioMy suggestion is n is to show how spurious tuples resuto show how spurious tuples result from this lt from this decomdecomposition with respectposition with respect to the table below:
to the table below:
A B C D E A B C D E 1 2 3 4 5 1 2 3 4 5 1 8 3 4 4 1 8 3 4 4
b. Find a single dependency from a single attribute X to another attribute Y such that when you b. Find a single dependency from a single attribute X to another attribute Y such that when you add the dependency
add the dependency X X →→ Y Y to the above dependencies, the decomposition in part a is no longerto the above dependencies, the decomposition in part a is no longer lossy.
lossy.
Answer. Answer.
a. If we were to decompose the relations into: a. If we were to decompose the relations into:
A A B B CC 1 1 2 2 33 1 1 8 8 33 D D EE 4 4 55 4 4 44
and then join the two (in this case with a cartesian product), we would get: and then join the two (in this case with a cartesian product), we would get:
A B C D E A B C D E 1 2 3 4 5 1 2 3 4 5 1 8 3 4 5 1 8 3 4 5 1 2 3 4 4 1 2 3 4 4 1 8 3 4 4 1 8 3 4 4
Tuples 2 and 3 are not in the original relation. Hence, this decomposition is lossy. Tuples 2 and 3 are not in the original relation. Hence, this decomposition is lossy.
b.
b. This decomThis decomposition cannoposition cannot be t be made losslemade lossless. ss. The probleThe problem is m is there is no longer a way to makethere is no longer a way to make sure
sure BDBD →→E E holds across two relationholds across two relations since s since they do they do not share any not share any attribattributes. utes. HowHoweveever, a r, a lossylossy decomposition of the form (A,B,C), (C,D,E) can be made lossless by adding an FD
decomposition of the form (A,B,C), (C,D,E) can be made lossless by adding an FD BB →→ C C ..
Question 4
Question 4 YYouou are giare givevenn the fthe folloollowinwingg setset ofof funfunctictionaonall depedependendencincies foes forr aa relrelatioationn R(AR(A,B,,B,C,DC,D,E,,E,F),F), F
F == {{ABAB →→ C,DC C,DC →→AE,E AE,E →→ F F }}.. a. What are the keys of this relation? a. What are the keys of this relation?
b. Is this relation in BCNF? If not, explain why by showing one violation. b. Is this relation in BCNF? If not, explain why by showing one violation. c.
c. Is the decompositioIs the decomposition (A,B,C,D) (B,C,Dn (A,B,C,D) (B,C,D,E,F) a dependenc,E,F) a dependency preservy preserving decomposing decomposition? ition? If not,If not, explain briefly.
explain briefly.
Answer. Answer.
a. What are the keys of this relation? a. What are the keys of this relation? {{A , B , DA , B , D}} andand {{B , C , DB , C , D}}..
b. Is this relation in BCNF? If not, explain why by showing one violation. b. Is this relation in BCNF? If not, explain why by showing one violation.
No, all functional dependencies are actually violating this. No dependency contains a superkey on No, all functional dependencies are actually violating this. No dependency contains a superkey on its left side.
its left side. c.
c. Is the decompositioIs the decomposition (A,B,C,D) (B,C,Dn (A,B,C,D) (B,C,D,E,F) a dependenc,E,F) a dependency preservy preserving decomposing decomposition? ition? If not,If not, explain briefly.
explain briefly. Yes,
Yes, ABAB →→ C C andandDC DC →→AA are preserved in the first relation.are preserved in the first relation. DC DC →→E E andandE E →→ F F are preservedare preserved in the second relation.
in the second relation.
Question 5
Question 5 You are given the below functional dependencies for relation R(A,B,C,D,E),You are given the below functional dependencies for relation R(A,B,C,D,E), F F == {{ABAB →→ C,ABC,AB →→ D,D, DD→→ A,BC A,BC →→D,BC D,BC →→ E E }}..
a. Is this relation is in BCNF? If not, show all dependencies that violate it. a. Is this relation is in BCNF? If not, show all dependencies that violate it. b. Is this relation in 3NF? If not, show all dependencies that violate it. b. Is this relation in 3NF? If not, show all dependencies that violate it.
c.
c. Is the Is the folfollolowinwing g depedependendency implincy implied by ed by the abovthe above e set of set of depedependendencincies? es? If so, If so, shoshow w hohow w usiusingng the Amstrong’s Axioms given in the book (p. 362-363):
the Amstrong’s Axioms given in the book (p. 362-363): ABC ABC →→ AE AE Answer.
Answer.
Keys for the relation:
Keys for the relation: {{A,A, BB}},, {{B,B, DD}},, {{B,B, C C }}.. a. Not in BCNF since
a. Not in BCNF since DD →→ AA does have a superkey on the left hand side.does have a superkey on the left hand side. b. In 3NF since in
b. In 3NF since in DD →→ AA, A is part of a key., A is part of a key. c.
c. BC BC →→ E E (given)(given)
ABC
ABC →→AE AE by the augmentation rule.by the augmentation rule.
Question 6
Question 6 You are given the table below for a relation R(A,B,C,D,E). You do not know theYou are given the table below for a relation R(A,B,C,D,E). You do not know the functional dependencies for this relation. This question is independent of Question 2 above.
functional dependencies for this relation. This question is independent of Question 2 above. A A B B C C D D EE ’’aa’ ’ 11222 2 1 1 ’’ss11’ ’ ’’aa’’ ’’ee’ ’ 22336 6 4 4 ’’ee22’ ’ ’’bb’’ ’’aa’ ’ 11999 9 1 1 ’’bb55’ ’ ’’cc’’ ’’bb’ ’ 22113 3 2 2 ’’zz88’ ’ ’’dd’’
Suppose this relation is decomposed into the following tw
Suppose this relation is decomposed into the following two o tablestables: : R1(A,BR1(A,B,C,D) and R2(A,C,E).,C,D) and R2(A,C,E). Is this decomposition lossless? Explain your reasoning.
Is this decomposition lossless? Explain your reasoning.
Answer. Answer. R1 R1 A B A B C DC D ’’aa’ ’ 11222 2 1 1 ’’ss11’’ ’’ee’ ’ 22336 6 4 4 ’’ee22’’ ’’aa’ ’ 11999 9 1 1 ’’bb55’’ ’’bb’ ’ 22113 3 2 2 ’’zz88’’ R2 R2 A A C C EE ’’aa’ ’ 1 1 ’’aa’’ ’’ee’ ’ 4 4 ’’bb’’ ’’aa’ ’ 1 1 ’’cc’’ ’’bb’ ’ 2 2 ’’dd’’ R R11 RR22 A A B B C C D D EE ’’aa’ ’ 11222 2 1 1 ’’ss11’ ’ ’’aa’’ ’’ee’ ’ 22336 6 4 4 ’’ee22’ ’ ’’bb’’ ’’aa’ ’ 11999 9 1 1 ’’bb55’ ’ ’’cc’’ ’’bb’ ’ 22113 3 2 2 ’’zz88’ ’ ’’dd’’ ’’aa’ ’ 11222 2 1 1 ’’ss11’ ’ ’’aa’’ ’’aa’ ’ 11999 9 1 1 ’’bb55’ ’ ’’cc’’
Since the last two rows are not in the original relation, then this decomposition is lossy. Since the last two rows are not in the original relation, then this decomposition is lossy.
Question 7
Question 7 YYouou are gare givivenen the the belobeloww setset ofof funfunctioctionalnal depedependendencincies fes foror aa relrelatiationon R(AR(A,B,C,B,C,D,,D,E,FE,F,G),G),, F
F == {{ADAD →→ BF,CDBF,CD →→EGC,BDEGC,BD →→F,F, E E →→ D,D, F F →→ C,C, DD →→ F F }}..
a. Find the minimal cover for the above set of functional dependencies using the algorithm described a. Find the minimal cover for the above set of functional dependencies using the algorithm described in class. Give sufficient detail to show your reasoning, but be succinct. You do not have to list all in class. Give sufficient detail to show your reasoning, but be succinct. You do not have to list all the cases you test/cons
the cases you test/consider for the ider for the algoritalgorithm. hm. ShoShow all w all steps where you make chasteps where you make changes to nges to the abovethe above set in detail.
set in detail. b.
b. Using the functUsing the functional dependeional dependencies that you compuncies that you computed in step a, ted in step a, find the keys for this relation.find the keys for this relation. Is it in BCNF? Explain your reasoning.
Is it in BCNF? Explain your reasoning.
c. Suppose we decompose the above relation into the following two relations: c. Suppose we decompose the above relation into the following two relations: R1(A,B,C,D,E) R2(A,D,F,G)
R1(A,B,C,D,E) R2(A,D,F,G)
Use the functional dependencies in the minimal cover. For each relation, write down the functional Use the functional dependencies in the minimal cover. For each relation, write down the functional dependencies that fall within that relation (you can decompose a dependency of the form
dependencies that fall within that relation (you can decompose a dependency of the formADAD →→ BF BF
into two i.e.
Using these functional dependencies, determine if this decomposition is lossless and/or dependency Using these functional dependencies, determine if this decomposition is lossless and/or dependency preserving. Explain your reasoning.
preserving. Explain your reasoning.
Answers. Answers. a. a. Step 1. Step 1.
{{ADAD →→ B,ADB,AD →→F,CDF,CD →→ E , C DE , C D→→ G,CDG,CD →→C,BDC,BD→→ F,F, E E →→ D,D, F F →→C,C, DD→→ F F }} Step 2.
Step 2. removeCDremoveCD →→C C ,, ADAD→→ F F , and, and BDBD →→F F .. {{ADAD →→ B , C DB , C D→→ E , C DE , C D→→ G,G, F F →→C,C, DD →→F,F, E E →→ DD}} Step 3. remove D from
Step 3. remove D from CDCD →→ E E andand CDCD →→GG
{{ADAD →→ B,B, DD→→ E,E, DD →→G, F G,F →→ C,C, DD→→ F,F, E E →→DD}} Finally recombine
Finally recombine
{{ADAD →→ B,B, DD→→ EGF,F EGF,F →→ C,C, E E →→ DD}}.. b. Keys:
b. Keys: {{A,A, DD}},, {{A,A, E E }}. Not in BCNF since the last three functional dependencies do not have a. Not in BCNF since the last three functional dependencies do not have a superkey on the left hand side.
superkey on the left hand side. c.
c. R1(A,R1(A,B,C,D,B,C,D,E) DependenE) Dependencies:cies: ADAD →→ B,B, DD →→ E,E, E E →→ DD R2(A,D,F,G) Dependencies:R2(A,D,F,G) Dependencies: DD →→
GF GF ..
Not functional dependency preserving, the dependency
Not functional dependency preserving, the dependency F F →→ C C is is not not presepreserverved.d.
head
head((RR1)1)∩∩headhead((RR2) 2) =={{A,A, DD}} R1:
R1: ADAD→→ ABCDE ABCDE is not true since C is not implied by A,Dis not true since C is not implied by A,D R2:
R2: ADAD→→ ADFGADFG is true since this is implied byis true since this is implied by DD →→ GF GF as follows:as follows:
AD
AD →→ ADAD inclusion rule, sinceinclusion rule, since DD →→ GF GF , use set accumulation rule,, use set accumulation rule, ADAD →→ ADGF ADGF . . HenHence, thce, thisis is a lossless decomposition.
is a lossless decomposition.
Question 8
Question 8 YYouou are are givgivenen the the folfollowlowinging setset FF ofof funfunctioctionalnal depedependendencincieses forfor aa relrelatiationon R(AR(A,B,,B,C,DC,D,E,F,E,F):): F
F =={{ABC ABC →→ D,ABDD,ABD →→E , C DE , C D →→F,CDF F,CDF →→B,BF B,BF →→ DD}}.. a. Find all keys of R based on these functional dependencies. a. Find all keys of R based on these functional dependencies.
b. Is this relation in Boyce-Codd Normal Form? Is it 3NF? Explain your answers. b. Is this relation in Boyce-Codd Normal Form? Is it 3NF? Explain your answers. c.
c. Can the Can the set F set F be be simsimpliplified (by fied (by remremovoving functing functionional al depedependendencincies es or or by remoby removinving g attrattribuibutestes from the left hand side of functional dependencies) without changing the closure of F (i.e. F+)? from the left hand side of functional dependencies) without changing the closure of F (i.e. F+)? Hint. Consider the steps of the minimal cover algorithm. Do any of them apply to this functional Hint. Consider the steps of the minimal cover algorithm. Do any of them apply to this functional dependency?
dependency?
Answer. Answer.
a. Keys:
a. Keys: {{A , B , C A , B , C }} andand {{A,C,DA,C,D}} b. It is not in BCNF. Counterexample
b. It is not in BCNF. Counterexample ABDABD →→ E E and ABD is not a superkey.and ABD is not a superkey. It is not in 3NF. Counterexample
It is not in 3NF. Counterexample ABDABD →→ E E , and ABD is not a superkey and E is not prime, and ABD is not a superkey and E is not prime attribute (part of a key).
attribute (part of a key).
c. Let F’ be obtained by replacing
c. Let F’ be obtained by replacing CDCDF F →→ BB withwith CDCD →→ BB.. According to F and F’,
According to F and F’, CDCD+ + == {{C , D , B , F C , D , B , F }}. . HeHencnce, e, we can remowe can removve e F F frfrom this funom this functctionionalal dependency without changing the meaning of the system.
dependency without changing the meaning of the system.
Question 9
Question 9 ConsidConsider relation R(X, Y, Z). Relation R currener relation R(X, Y, Z). Relation R currently has three tuples: tly has three tuples: (6, 4, (6, 4, 2), (6, 2), (6, 6,6, 8)
8) and (6, and (6, 4, 4, 8). 8). WhiWhich of ch of the follothe followinwing g thrthree functee functionaional l depedependendencincies can es can yoyou u infinfer do er do not holdnot hold for relation R? Explain your answer.
for relation R? Explain your answer.
Y
Z Z →→ Y Y X
XY Y →→ Z Z Answer.
Answer. The first functiThe first functional dependenonal dependency holds, but the rest do cy holds, but the rest do not hold. not hold. The second and thirdThe second and third tuples both have 8 for Z but different values of Y. The first and third tuples both have 6 and 4 for tuples both have 8 for Z but different values of Y. The first and third tuples both have 6 and 4 for X and Y but different values for Z.
X and Y but different values for Z.
Question 10
Question 10 Consider the relation R(V, W, X, Y, Z) with functional dependenciesConsider the relation R(V, W, X, Y, Z) with functional dependencies {{Z Z →→ Y,Y, Y Y →→
Z,
Z, X X →→ Y,Y, X X →→ V , V W V , V W →→X X }}..
a) List the possible keys for relation R based on the functional dependencies above. a) List the possible keys for relation R based on the functional dependencies above. b) Show the closure for attribute X given the functional dependencies above.
b) Show the closure for attribute X given the functional dependencies above.
c) Suppose that relation R is decomposed into two relations, R1(V, W, X) and R2(X, Y, Z). Is this c) Suppose that relation R is decomposed into two relations, R1(V, W, X) and R2(X, Y, Z). Is this decom
decomposition a position a losslelossless ss decomdecomposition? position? ExplaExplain in youyour r answanswer.er.
Answer. Answer. a. a. {{V,V, W W }},, {{X,X, W W }} b. b. X X ++ =={{X , V , Y , Z X , V , Y , Z }} c.
c. YYes it es it is losslis losslessess. . TTo o be losslebe lossless the attriss the attributbutes in es in comcommon betwmon between the tween the two o relrelatiations musons mustt funct
functionally deterionally determine all mine all the attributthe attributes in es in one of one of the two relationthe two relations. s. The only attribute in commonThe only attribute in common is X and it functionally determines all the attributes in R2.
is X and it functionally determines all the attributes in R2.
Question 11
Question 11 GivGiven en relrelatioation n R(WR(W, , X, X, Y, Y, Z) Z) and and set of set of funfunctictionaonal l depedependendencincieses F F == {{X X →→
W , W Z
W , W Z →→X Y , Y X Y , Y →→ WW XXZ Z }}. Compute the minimal cover for. Compute the minimal cover for F F ..
Answer. Answer.
Step 1:
Step 1: X X →→ W , W Z W , W Z →→X , W Z X , W Z →→Y,Y, Y Y →→ W,W, Y Y →→ X,X, Y Y →→ Z Z
Step 2: Don’t need
Step 2: Don’t need WW Z Z →→ X X , since, since WW Z Z →→Y Y andand Y Y →→X X
Don’t need
Don’t need Y Y →→ W W , since, since Y Y →→X X andand X X →→ W W
This leaves
This leaves {{X X →→ WW WW Z Z →→ Y,Y, Y Y →→ X,X, Y Y →→ Z Z }} Step 3: Only need to consider
Step 3: Only need to consider WW Z Z →→Y Y . Can’t eliminate W or Z. So nothing is eliminated.. Can’t eliminate W or Z. So nothing is eliminated. Step 4:
Step 4: {{X X →→ WW WW Z Z →→Y,Y, Y Y →→ XXZ Z }} is the minimal coveris the minimal cover
Question 12
Question 12 Given relation R(W, X, Y, Z) and set of functional dependenciesGiven relation R(W, X, Y, Z) and set of functional dependencies GG=={{Z Z →→W,W, Y Y →→
XZ,XW
XZ,XW →→Y Y }}, where G is a minimal cover:, where G is a minimal cover:
a) Decompose R into a set of relations in Third Normal Form. a) Decompose R into a set of relations in Third Normal Form.
b)Is your decomposition in part a) also in Boyce Codd Normal Form? Explain your answer. b)Is your decomposition in part a) also in Boyce Codd Normal Form? Explain your answer.
Answer. Answer.
a. Possible keys:
a. Possible keys: {{Y Y }},,{{X, Z X,Z }},,{{W,W, X X }} R1=(Z, W), R2=(X, Y, Z), R3=(X, Y, W) R1=(Z, W), R2=(X, Y, Z), R3=(X, Y, W) b.Y
b.Yes. es. In each of the In each of the three relatthree relations, the left side of ions, the left side of the funcatiothe funcational dependennal dependencies that apply arecies that apply are superkeys for the relation. Hence, all three relations satisfy the definition of BCNF.
superkeys for the relation. Hence, all three relations satisfy the definition of BCNF.
Question 13
Question 13 ConsidConsider a er a relatirelation named on named EMP DEPT with attributesEMP DEPT with attributes: : ENAMENAME, SSN, E, SSN, BDABDATE,TE, ADDRESS, DNUMBER, DNAME, and DMGRSSN. Consider also the set G of functional ADDRESS, DNUMBER, DNAME, and DMGRSSN. Consider also the set G of functional depen-dencies for EMP DEPT:
G
G =={{SSSS N N →→ ENAENAMEME BDBDAATETE AADDDDRERESSSS DNUMBERDNUMBER,, DNUMBERDNUMBER→→ DNAME,DMGRSSM DNAME,DMGRSSM }}.. a) Calculate the closures SSN+ and DNAME+ with respect to G.
a) Calculate the closures SSN+ and DNAME+ with respect to G.
b) Is the set of functional dependences G minimal? If not, find a minimal set of functional b) Is the set of functional dependences G minimal? If not, find a minimal set of functional depen-dencies that is equivalent to G.
dencies that is equivalent to G.
c) List an update anomaly that can occur for relation EMP DEPT. c) List an update anomaly that can occur for relation EMP DEPT. d) List an insertion anomaly that can occur for relation EMP DEPT. d) List an insertion anomaly that can occur for relation EMP DEPT. e) List a deletion anomaly that can occur for relation EMP DEPT. e) List a deletion anomaly that can occur for relation EMP DEPT.
Answer. Answer.
a)
a) SSSS N N ++ == {{SSN,ENAME,BDATE,ADDRESS,DNUMBER,DNAME,DMGRSSN SSN,ENAME,BDATE,ADDRESS,DNUMBER,DNAME,DMGRSSN }}
DNAME
DNAME ++ =={{DNAME DNAME }}
b) It is minimal. b) It is minimal.
c) Since every member of a department has a reference to the manager of that department (i.e., c) Since every member of a department has a reference to the manager of that department (i.e., Dmgrssn), when the department manager changes this reference must be changed multiple places. Dmgrssn), when the department manager changes this reference must be changed multiple places. This leads to the possibility of an inconsistency in the database if they are not all changed.
This leads to the possibility of an inconsistency in the database if they are not all changed. d) You cannot enter data about a department until you have employees for the department. d) You cannot enter data about a department until you have employees for the department.
e) If you delete the last employee for a department, you lose all information about the department. e) If you delete the last employee for a department, you lose all information about the department.
Question 14
Question 14 You are given the following functional dependencies for the ”EMPLOYEE” relation.You are given the following functional dependencies for the ”EMPLOYEE” relation. Explain whether the relation ”EMPLOYEE” is BCNF and 3NF?
Explain whether the relation ”EMPLOYEE” is BCNF and 3NF? Database:
Database:
EMPLOYEE
EMPLOYEE(ssn, (ssn, first-nfirst-name, ame, last-namlast-name, e, addressaddress, , date-jodate-joined, ined, supervissupervisor-ssn)or-ssn) DEPARTME
DEPARTMENT(deptNT(dept-no, -no, name, name, manager-manager-ssn)ssn) WORKS-IN
WORKS-IN(employ(employee-ssn, ee-ssn, dept-nodept-no)) INVENTOR
INVENTORY(dept-Y(dept-no, no, item-iditem-id, , quantitquantity)y) ITEMS(it
ITEMS(item-id, em-id, item-namitem-name, e, type)type)
Foreign keys: Foreign keys:
1. EMPLOYEE.supervisor-ssn and WORKS-IN.employee-ssn point to EMPLOYEE.ssn. 1. EMPLOYEE.supervisor-ssn and WORKS-IN.employee-ssn point to EMPLOYEE.ssn. 2. WORKS-IN.dept-no and INVENTORY.dept-no point to DEPARTMENT.dept-no. 2. WORKS-IN.dept-no and INVENTORY.dept-no point to DEPARTMENT.dept-no. 3. INVENTORY.item-id points to ITEMS.item-id.
3. INVENTORY.item-id points to ITEMS.item-id.
{{ssnssn →→ supervisorsupervisor −−ssn,, ssnssn ssn →→ firstfirst−−namname,e, ssnssn→→ lastlast −−namname,e, ssnssn →→ datedate−− joined, joined, ssnssn →→
address
address,, addressaddress→→ ssnssn}}..
Answer.
