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Chapter 6: The Relational Algebra and Relational Calculus 1

CHAPTER 6: THE RELATIONAL ALGEBRA AND RELATIONAL CALCULUS

Answers to Selected Exercises

6.15 Show the result of each of the example queries in Section 6.5 if they are applied to

the database of Figure 5.6.

Answer:

(QUERY 1) Find the name and address of all employees who work for the 'Research'

department.

Result: FNAME LNAME ADDRESS

John Smith 731 Fondren, Houston, TX

Franklin Wong 638 Voss, Houston, TX

Ramesh Narayan 975 Fire Oak, Humble, TX

2 Chapter 6: The Relational Algebra and Relational Calculus

Joyce English 5631 Rice, Houston, TX

(QUERY 2) For every project located in 'Stafford', list the project number, the

controlling department number, and the department manager's last name, address, and

birth date.

Result:

PNUMBER DNUM LNAME ADDRESS BDATE

10 4 Wallace 291 Berry, Bellaire, TX 20-JUN-31

30 4 Wallace 291 Berry, Bellaire, TX 20-JUN-31

(QUERY 3) Find the names of all employees who work on all the projects controlled by

department number 5.

Result: (empty because no tuples satisfy the result).

Chapter 6: The Relational Algebra and Relational Calculus 3

LNAME FNAME

(QUERY 4) Make a list of project numbers for projects that involve an employee whose

last name is 'Smith' as a worker or as a manager of the department that controls the

project.

Result: PNO

(QUERY 5) List the names of all employees with two or more dependents.

Result: LNAME FNAME

Smith John

Wong Franklin

4 Chapter 6: The Relational Algebra and Relational Calculus

(QUERY 6) List the names of employees who have no dependents.

Result: LNAME FNAME

Zelaya Alicia

Narayan Ramesh

English Joyce

Jabbar Ahmad

Borg James

(QUERY 7) List the names of managers who have at least one dependent.

Result: LNAME FNAME

Wallace Jennifer

Wong Franklin

6.16 Specify the following queries on the COMPANY relational database schema shown in Figure 5.5, using the relational operators discussed in this

Chapter 6: The Relational Algebra and Relational Calculus 5

chapter. Also show the result of each query as it would apply to the database state of Figure 5.6.

(a) Retrieve the names of employees in department 5 who work more than 10 hours per

week on the 'ProductX' project.

(b) List the names of employees who have a dependent with the same first name as

themselves.

(d) For each project, list the project name and the total hours per week (by all

employees) spent on that project.

(e) Retrieve the names of employees who work on every project.

(f) Retrieve the names of employees who do not work on any project.

6 Chapter 6: The Relational Algebra and Relational Calculus

(g) For each department, retrieve the department name, and the average salary of

employees working in that department.

(h) Retrieve the average salary of all female employees.

(i) Find the names and addresses of employees who work on at least one project located

in Houston but whose department has no location in Houston.

(j) List the last names of department managers who have no dependents.

Answers:

In the relational algebra, as in other languages, it is possible to specify the same query

in multiple ways. We give one possible solution for each query. We use the symbol s for

SELECT, P for PROJECT, J for EQUIJOIN, * for NATURAL JOIN, and f for FUNCTION.

Chapter 6: The Relational Algebra and Relational Calculus 7

(a) EMP_W_X <-- ( s PNAME='ProductX' (PROJECT)) J (PNUMBER),(PNO)

(WORKS_ON)

EMP_WORK_10 <-- (EMPLOYEE) J (SSN),(ESSN) ( s HOURS>10 (EMP_W_X))

RESULT <-- P LNAME,FNAME ( s DNO=5 (EMP_WORK_10))

Result:

LNAME FNAME

Smith John

English Joyce

(b) E <-- (EMPLOYEE) J (SSN,FNAME),(ESSN,DEPENDENT_NAME)

(DEPENDENT)

R <-- P LNAME,FNAME (E)

Result (empty):

LNAME FNAME

8 Chapter 6: The Relational Algebra and Relational Calculus

LNAME='Wong' (EMPLOYEE))

WONG_EMPS <-- (EMPLOYEE) J (SUPERSSN),(SSN) (WONG_SSN)

RESULT <-- P LNAME,FNAME (WONG_EMPS)

Result:

LNAME FNAME

Smith John

Narayan Ramesh

English Joyce

(d) PROJ_HOURS(PNO,TOT_HRS) <-- PNO f SUM HOURS (WORKS_ON)

RESULT <-- P PNAME,TOT_HRS ( (PROJ_HOURS) J (PNO),(PNUMBER)

(PROJECT) )

Chapter 6: The Relational Algebra and Relational Calculus 9

Result:

PNAME TOT_HRS

ProductX 52.5

ProductY 37.5

ProductZ 50.0

Computerization 55.0

Reorganization 25.0

Newbenefits 55.0

(e) PROJ_EMPS(PNO,SSN) <-- P PNO,ESSN (WORKS_ON)

ALL_PROJS(PNO) <-- P PNUMBER (PROJECT)

EMPS_ALL_PROJS <-- PROJ_EMPS -:- ALLPROJS (* DIVISION operation *)

RESULT <-- P LNAME,FNAME (EMPLOYEE * EMP_ALL_PROJS)

10 Chapter 6: The Relational Algebra and Relational Calculus

Result (empty):

LNAME FNAME

(f) ALL_EMPS <-- P SSN (EMPLOYEE)

WORKING_EMPS(SSN) <-- P ESSN (WORKS_ON)

NON_WORKING_EMPS <-- ALL_EMPS - WORKING_EMPS (* DIFFERENCE

RESULT <-- P LNAME,FNAME (EMPLOYEE * NON_WORKING_EMPS)

Result (empty):

LNAME FNAME

(g) DEPT_AVG_SALS(DNUMBER,AVG_SAL) <-- DNO f AVG SALARY

(EMPLOYEE)

RESULT <-- P DNUMBER,AVG_SAL ( DEPT_AVG_SALS * DEPARTMENT )

Chapter 6: The Relational Algebra and Relational Calculus 11

Result:

DNUMBER AVG_SAL

Research 33250

Administration 31000

Headquarters 55000

(h) RESULT(AVG_F_SAL) <-- f AVG SALARY ( s SEX='F' (EMPLOYEE) )

Result:

AVG_F_SAL

31000

(i) E_P_HOU(SSN) <--

P ESSN (WORKS_ON J(PNO),(PNUMBER) ( s PLOCATION='Houston' (PROJECT)))

D_NO_HOU <--

12 Chapter 6: The Relational Algebra and Relational Calculus

P DNUMBER (DEPARTMENT) - P DNUMBER ( s

DLOCATION='Houston' (DEPARTMENT))

E_D_NO_HOU <-- P SSN (EMPLOYEE J(PNO),(DNUMBER) (D_NO_HOU))

RESULT_EMPS <-- E_P_HOU - E_D_NO_HOU (* this is set DIFFERENCE *)

RESULT <-- P LNAME,FNAME,ADDRESS (EMPLOYEE * RESULT_EMPS)

Result:

LNAME FNAME ADDRESS

Wallace Jennifer 291 Berry, Bellaire, TX

(j) DEPT_MANAGERS(SSN)<-- P MGRSSN (DEPARTMENT)

EMPS_WITH_DEPENDENTS(SSN) <-- P ESSN (DEPENDENT)

RESULT_EMPS <-- DEPT_MANAGERS - EMPS_WITH_DEPENDENTS

RESULT <-- P LNAME,FNAME (EMPLOYEE * RESULT_EMPS)

Chapter 6: The Relational Algebra and Relational Calculus 13

Result:

LNAME FNAME

Borg James

6.18 Consider the LIBRARY relational schema shown in Figure 6.14, which is used to

keep track of books, borrowers, and book loans. Referential integrity constraints are

shown as directed arcs in Figure 6.14, as in the notation of Figure 6.7. Write down

relational expressions for the following queries on the LIBRARY database:

(a) How many copies of the book titled The Lost Tribe are owned by the library branch

whose name is \"Sharpstown\"?

(b) How many copies of the book titled The Lost Tribe are owned by each library

14 Chapter 6: The Relational Algebra and Relational Calculus

branch?

(d) For each book that is loaned out from the \"Sharpstown\" branch and whose DueDate is today, retrieve the book title, the borrower's name, and the borrower's address.

(e) For each library branch, retrieve the branch name and the total number of books loaned out from that branch.

(f) Retrieve the names, addresses, and number of books checked out for all borrowers

who have more than five books checked out.

(g) For each book authored (or co-authored) by \"Stephen Kingitle and

the number of copies owned by the library branch whose name is \"Central\".

Answer: (Note: We will use S for SELECT, P for PROJECT, * for NATURAL JOIN, - for

Chapter 6: The Relational Algebra and Relational Calculus 15

SET DIFFERENCE, F for AGGREGATE FUNCTION)

(a) A <-- BOOKCOPIES * LIBRARY-BRANCH * BOOK

RESULT <-- P No_Of_Copies ( S BranchName='Sharpstown' and Title='The Lost

Tribe'

(A) )

Note: A better query would be to do the SELECTs before the JOIN as follows:

A <-- P No_Of_Copies ( ( S BranchName='Sharpstown' (LIBRARY-BRANCH) ) *

(BOOKCOPIES * ( S Title='The Lost Tribe'

(BOOK) ) ) )

(b) P BranchID,No_Of_Copies ( ( S Title='The Lost Tribe' (BOOK)) * BOOKCOPIES )

16 Chapter 6: The Relational Algebra and Relational Calculus

RESULT <-- P Name (BORROWER * NO_CHECKOUT_B)

(d) S <-- P BranchId ( S BranchName='Sharpstown' (LIBRARY-BRANCH) )

B_FROM_S <-- P BookId,CardNo ( ( S DueDate='today' (BOOKLOANS) ) * S )

RESULT <-- P Title,Name,Address ( BOOK * BORROWER * B_FROM_S )

(e) R(BranchId,Total) <-- BranchId FCOUNT(BookId,CardNo) (BOOK_LOANS)

RESULT <-- P BranchName,Total (R * LIBRARY_BRANCH)

(f) B(CardNo,TotalCheckout) <-- CardNo F COUNT(BookId) (BOOK_LOANS)

B5 <-- S TotalCheckout > 5 (B)

RESULT <-- P Name,Address,TotalCheckout ( B5 * BORROWER)

(g) SK(BookId,Title) <-- ( sAuthorName='Stephen King' ( BOOK_AUTHORS)) * BOOK

CENTRAL(BranchId) <-- sBranchName='Central' ( LIBRARY_BRANCH )

RESULT <-- P Title,NoOfCopies ( SK * BOOKCOPIES * CENTRAL )

Chapter 6: The Relational Algebra and Relational Calculus 17

6.22

(a)

P Q R A B C

10 a 5 10 b 6

10 a 5 10 b 5

25 a 6 25 c 3

(b)

P Q R A B C

15 b 8 10 b 6

15 b 8 10 b 5

(c)

P Q R A B C

18 Chapter 6: The Relational Algebra and Relational Calculus

10 a 5 10 b 6

10 a 5 10 b 5

15 b 8 null null null

25 a 6 25 c 3

(d)

P Q R A B C

15 b 8 10 b 6

null null null 25 c 3

15 b 8 10 b 5

(e)

P Q R

10a 5

Chapter 6: The Relational Algebra and Relational Calculus 19

15 b 8

25 a 6

10b 6

25 c 3

10b 5

(f)

P Q R A B C

10 a 5 10 b 5

6.24 Specify queries (a), (b), (c), (e), (f), (i), and (j) of Exercise 6.16 in both the

tuple relational calculus and the domain relational calculus.

Answer:

(a) Retrieve the names of employees in department 5 who work more than 10 hours per

20 Chapter 6: The Relational Algebra and Relational Calculus

week on the 'ProductX' project.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND e.DNO=5 AND (EXISTS p) (EXISTS w)

(

WORKS_ON(w) AND PROJECT(p) AND e.SSN=w.ESSN AND

w.PNO=p.PNUMBER AND

p.PNAME='ProductX' AND w.HOURS>10 ) }

Domain relational Calculus:

{ qs | EMPLOYEE(qrstuvwxyz) AND z=5 AND (EXISTS a) (EXISTS b) (EXISTS e)

(EXISTS f)

(EXISTS g) ( WORKS_ON(efg) AND PROJECT(abcd) AND t=e AND f=b AND

a='ProductX' AND

Chapter 6: The Relational Algebra and Relational Calculus 21

g>10 ) }

(b) List the names of employees who have a dependent with the same first name as

themselves.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND (EXISTS d) ( DEPENDENT(d) AND

e.SSN=d.ESSN

AND e.FNAME=d.DEPENDENT_NAME ) }

Domain relational Calculus:

{ qs | (EXISTS t) (EXISTS a) (EXISTS b) ( EMPLOYEE(qrstuvwxyz) AND

DEPENDENT(abcde)

AND a=t AND b=q ) }

22 Chapter 6: The Relational Algebra and Relational Calculus

Wong'.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND (EXISTS s) ( EMPLOYEE(s) AND

s.FNAME='Franklin' AND s.LNAME='Wong' AND e.SUPERSSN=s.SSN ) }

Domain relational Calculus:

{ qs | (EXISTS y) (EXISTS a) (EXISTS c) (EXISTS d) ( EMPLOYEE(qrstuvwxyz) AND

EMPLOYEE(abcdefghij) AND a='Franklin' AND c='Wong' AND y=d ) }

(e) Retrieve the names of employees who work on every project.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND (FORALL p) ( NOT(PROJECT(p)) OR

(EXISTS w) (

WORKS_ON(w) AND p.PNUMBER=w.PNO AND w.ESSN=e.SSN ) ) }

Chapter 6: The Relational Algebra and Relational Calculus 23

Domain relational Calculus:

{ qs | (EXISTS t) ( EMPLOYEE(qrstuvwxyz) AND (FORALL b) (

NOT(PROJECT(abcd)) OR

(EXISTS e) (EXISTS f) (WORKS_ON(efg) AND e=t AND f=b) ) }

(f) Retrieve the names of employees who do not work on any project.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND NOT(EXISTS w) ( WORKS_ON(w)

AND

w.ESSN=e.SSN ) }

Domain relational Calculus:

{ qs | (EXISTS t) ( EMPLOYEE(qrstuvwxyz) AND NOT(EXISTS a) (

WORKS_ON(abc) AND a=t )

24 Chapter 6: The Relational Algebra and Relational Calculus

) }

(i) Find the names and addresses of employees who work on at least one project located

in Houston but whose department has no location in Houston.

Tuple relational Calculus:

{ e.LNAME, e.FNAME, e.ADDRESS | EMPLOYEE(e) AND (EXISTS p) (EXISTS w) (

WORKS_ON(w) AND PROJECT(p) AND e.SSN=w.ESSN AND

w.PNO=p.PNUMBER AND

p.PLOCATION='Houston' AND NOT(EXISTS l) ( DEPT_LOCATIONS(l) AND

e.DNO=l.DNUMBER

AND l.DLOCATION='Houston' ) ) }

Domain relational Calculus:

{ qsv | (EXISTS t) (EXISTS z) ( EMPLOYEE(qrstuvwxyz) AND (EXISTS b) (EXISTS c)

Chapter 6: The Relational Algebra and Relational Calculus 25

(EXISTS e)

(EXISTS f) ( WORKS_ON(efg) AND PROJECT(abcd) AND t=e AND f=b AND

c='Houston' AND

NOT(EXISTS h) NOT(EXISTS i) ( DEPT_LOCATIONS(hi) AND z=h AND i='Houston'

) ) }

(j) List the last names of department managers who have no dependents.

Tuple relational Calculus:

{ e.LNAME | EMPLOYEE(e) AND (EXISTS d) ( DEPARTMENT(d) AND

e.SSN=d.MGRSSN AND

NOT(EXISTS x) (DEPENDENT(x) AND e.SSN=x.ESSN) ) }

Domain relational Calculus:

{ s | (EXISTS t) ( EMPLOYEE(qrstuvwxyz) AND (EXISTS c) ( DEPARTMENT(abcd)

26 Chapter 6: The Relational Algebra and Relational Calculus

AND t=c

AND NOT(EXISTS e) (DEPENDENT(efghi) AND e=t) ) }

6.25 No solution provided.

6.26 Specify queries of Exercise 6.18 in both the tuple relational calculus and the

domain relational calculus. Also specify these queries in the relational algebra.

Answer:

We will only do tuple and domain relational calculus here.

(a) Retrieve the names of all senior students majoring in 'COSC' (computer science).

Tuple relational Calculus:

{ s.Name | STUDENT(s) AND AND s.Class=5 AND s.Major='COSC' }

Domain relational Calculus:

Chapter 6: The Relational Algebra and Relational Calculus 27

{ a | (EXISTS c) (EXISTS d) ( STUDENT(abcd) AND c=5 AND d='COSC' ) }

(b) Retrieve the names of all courses taught by professor King in 85 and 86.

Tuple relational Calculus:

{ c.CourseName | COURSE(c) AND (EXISTS s) ( SECTION(s) AND

c.CourseNumber=s.CourseNumber AND s.Instructor='King' AND ( s.Year='85' OR

s.Year='86' ) ) }

Domain relational Calculus:

{ a | (EXISTS b) (EXISTS f) (EXISTS h) (EXISTS i) ( COURSE(abcd) AND

SECTION(efghi) AND

f=b AND i='King' AND ( h='85' OR h='86' ) ) }

28 Chapter 6: The Relational Algebra and Relational Calculus

year, and number of students who took the section.

This query cannot be done in basic relational calculus as it requires a COUNT function.

(d) Retrieve the name and transcript of each senior student (Class=5) majoring in

COSC. Transcript includes course name, course number, credit hours, semester, year,

and grade for each course completed by the student.

Tuple relational Calculus:

{s.Name, c.CourseName, c.CourseNumber, c.CreditHours, t.Semester, t.Year, g.Grade |

STUDENT(s)

AND COURSE(c) AND SECTION(t) AND GRADE_REPORT(g) AND s.Class=5 AND

s.Major='COSC' AND s.StudentNumber=g.StudentNumber AND

Chapter 6: The Relational Algebra and Relational Calculus 29

g.SectionIdentifier=t.SectionIdentifier

AND t.CourseNumber=c.CourseNumber}

Domain relational Calculus:

{aefgklp | (EXISTS b) (EXISTS c) (EXISTS d) (EXISTS n) (EXISTS o) (EXISTS j)

(EXISTS i)

(STUDENT(abcd) AND COURSE(efgh) AND SECTION(ijklm) AND

GRADE_REPORT(nop) AND

c=5 AND d='COSC' AND b=n AND i=o AND j=f)}

(e) Retrieve the names and major departments of all straight A students (students who

have a grade of A in all their courses).

Tuple relational Calculus:

{ s.Name, s.Major | STUDENT(s) AND (FORALL g) ( NOT(GRADE_REPORT(g))

30 Chapter 6: The Relational Algebra and Relational Calculus

NOT(s.StudentNumber=g.StudentNumber) OR g.Grade='A' ) }

Domain relational Calculus:

{ ad | (EXISTS b) ( STUDENT(abcd) AND (FORALL e) (FORALL g) (

NOT(GRADE_REPORT(efg)) OR NOT(b=e) OR g='A' ) ) }

(f) Retrieve the names and major departments of all students who do not have any grade

of A in any of their courses.

Tuple relational Calculus:

{ s.Name, s.Major | STUDENT(s) AND NOT(EXISTS g) ( GRADE_REPORT(g) AND

s.StudentNumber=g.StudentNumber AND g.Grade='A' ) }

Domain relational Calculus:

Chapter 6: The Relational Algebra and Relational Calculus 31

{ ad | (EXISTS b) ( STUDENT(abcd) AND NOT(EXISTS e) NOT(EXISTS g) (

GRADE_REPORT(efg)

AND b=e AND g='A' ) ) }

6.27 In a tuple relational calculus query with n tuple variables, what would be the

typical minimum number of join conditions? Why? What is the effect of having a

smaller number of join conditions?

Answer:

Typically, there should be at least (n-1) join conditions; otherwise, a cartesian product

with one of the range relations would be taken, which usually does not make sense.

6.28 Rewrite the domain relational calculus queries that followed Q0 in Section 6.7 in

32 Chapter 6: The Relational Algebra and Relational Calculus

the style of the abbreviated notation of Q0A, where the objective is to minimize

the number of domain variables by writing constants in place of variables

wherever possible.

Answer:

Q1A: { qsv | (EXISTS z) (EXISTS m) ( EMPLOYEE(q,r,s,t,u,v,w,x,y,z) AND

DEPARTMENT('Research',m,n,o) AND m=z ) }

Q2A: { iksuv | (EXISTS m) (EXISTS n) (EXISTS t) ( PROJECT(h,i,'Stafford',k) AND

EMPLOYEE(q,r,s,t,u,v,w,x,y,z) AND DEPARTMENT(l,m,n,o) ) }

The other queries will not be different since they have no constants (no selection

conditions; only join conditions)

6.30 Show how you may specify the following relational algebra operations in both

Chapter 6: The Relational Algebra and Relational Calculus 33

tuple and domain relational calculus.

(a) SELECT A=c (R(A, B, C)):

(b) PROJECT (R(A, B, C)):

(d) R(A, B, C) UNION S(A, B, C):

(e) R(A, B, C) INTERSECT S(A, B, C):

(f) R(A, B, C) MINUS S(A, B, C):

(g) R(A, B, C) CARTESIAN PRODUCT S(D, E, F):

(h) R(A, B) DIVIDE S(A):

Answer:

For each operation, we give the tuple calculus expression followed by the domain

calculus expression.

34 Chapter 6: The Relational Algebra and Relational Calculus

(a) { t | R(t) AND t.A=c}, { xyz | R(xyz) AND x=c }

(b) { t.A, t.B | R(t) }, { xy | R(xyz) }

{ xyzvw | R(xyz) AND (EXISTS u) ( S(uvw) AND z=u ) }

(d) { t | R(t) OR S(t) },

{ xyz | R(xyz) OR S(xyz) }

(e) { t | R(t) AND S(t) },

{ xyz | R(xyz) AND S(xyz) }

(f) { t | R(t) AND NOT(S(t)) },

{ xyz | R(xyz) AND NOT(S(xyz)) }

(g) { t.A, t.B, t.C, q.D, q.E, q.F | R(t) AND S(q) },

( xyzuvw | R(xyz) AND S(uvw) }

Chapter 6: The Relational Algebra and Relational Calculus 35

(h) { t.B | R(t) AND (FORALL s)

( NOT(S(s)) OR (EXISTS q) ( R(q) AND s.A=q.A AND q.B=t.B ) ) },

{ y | R(xy) AND (FORALL z) ( NOT(S(z)) OR (EXISTS u) ( R(uy) AND z=u ) }

6.32 A nested query is query within a query. More specifically, a nested query is a parenthesized query that can be used as a value in a number of places, such as instead of a relation or a selection condition. Specify the following queries on the database specified in Figure 5.5 using the concept of nested queries and the relational operators discussed in this chapter. Also show the result of each query as it would apply to the database state of Figure 5.6.

a. List the names of all employees who work in the department that has the employee with the highest salary among all employees.

b. List the names of all employees whose supervisor’s supervisor has '888665555' for SSN.

c. List the names of employees who make at least $10,000 more than the employee who is paid the least in the company.

6.33 State whether the following conclusion are true or false:

36 Chapter 6: The Relational Algebra and Relational Calculus

a. NOT (P(x) OR Q(x))  (NOT (P(x)) AND (NOT (Q(x)))

b. NOT ( x) (P(x))   x (NOT (P(x))

b. ( x) (P(x))   x ((P(x))

Answers

6.31

a) RESULT <-- P LNAME ( S DNO = ( P DNO ( S SALARY = MAX(SALARY)

EMPLOYEE ) ) EMPLOYEE )

b) RESULT <-- P LNAME ( S SUPERSSN = (P SSN ( S SUPERSSN =

‘88866555’ EMPLOYEE ) ) EMPLOYEE )

c) RESULT <-- P LNAME ( S SALARY >= 10000 + P SALARY ( S SALARY =

MIN(SALARY) EMPLOYEE ) EMPLOYEE )

6.32

a. TRUE

b. TRUE

Chapter 6: The Relational Algebra and Relational Calculus 37

c. FALSE

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