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Riyadh Philanthropic Society For Science Prince Sultan College For Woman Dept. of Computer & Information Sciences CS 340 Introduction to Database Systems (Chapter 6 Tutorial)

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Chapter 6 Tutorial1 Exercise 1 Given the Following database schema: Specify the following queries in: Relation algebra Tuple relational calculus Domain relational calculus Only for a, b, c, f, i, j

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Chapter 6 Tutorial2 Exercise 1 - a Retrieve the names of all employees in department 5 who work more than 10 hours per week on the ‘Product x’ project. Relational Algebra: PROJ_X PNAME = ‘Product x’ (PROJECT) EMP_WORK_10 HOURS>10 (PROJ_X PNUMBER=PNO WORKS_ON) EMP_DEPT_5 DNO=5 (EMPLOYEE SSN=ESSN EMP_WORK_10 ) RESULT FNAME, LNAME (EMP_DEPT_5)

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Chapter 6 Tutorial3 Exercise 1 - a Tuple relational calculus: {e.FNAME, e.LNAME | EMPLOYEE(e) AND e.DNO=5 AND ( p) ( w) (PROJECT(p) AND WORKS_ON(w) AND p.PNAME=‘Product x’ AND w.HOURS>10 AND e.SSN=w.ESSN AND p.PNUMBER= w.PNO)} Domain relational calculus: {qs | ( z) ( a) ( b) ( e) ( f) ( g) (EMPLOYEE(qrstuvwxyz) AND PROJECT(abcd) AND WORKS_ON(efg) AND z=5 AND a=‘Product x’ AND g>10 AND t=e AND b=f)}

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Chapter 6 Tutorial4 Exercise 1 - b List the names of all employees who have a dependent with the same first name as themselves. Relational Algebra: EMP (EMPLOYEE SSN=ESSN AND FNAME=DEPENDENT_NAME DEPENDENT) RESULT FNAME, LNAME (EMP)

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Chapter 6 Tutorial5 Exercise 1 - b Tuple relational calculus: {e.FNAME, e.LNAME | EMPLOYEE(e) AND ( d) (DEPENDENT(d) AND e.SSN=d.ESSN AND e.FNAME=d.DEPENDENT_NAME)} Domain relational calculus: {qs | ( t) ( a) ( b) (EMPLOYEE(qrstuvwxyz) AND DEPENDENT(abcde) AND t=a AND q=b)}

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Chapter 6 Tutorial6 Exercise 1 - c Find the names of all employees who are directly supervised by ‘Franklin Wong’. Relational Algebra: FW FNAME=‘Franklin’ AND LNAME=‘Wong’ (EMPLOYEE ) FW_SSN SSN (FW) FW_EMP (EMPLOYEE SUPERSSN=SSN FW_SSN) RESULT FNAME, LNAME (FW_EMP)

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Chapter 6 Tutorial7 Exercise 1 - c Tuple relational calculus: {e.FNAME, e.LNAME | EMPLOYEE(e) AND ( s) (EMPLOYEE(s) AND s.FNAME=‘Franklin’ AND s.LNAME=‘Wong’ AND e.SUPERSSN=s.SSN)} Domain relational calculus: {qs | ( y) ( a) ( c) ( d) (EMPLOYEE(qrstuvwxyz) AND EMPLOYEE(abcdefghi) AND a=‘Franklin’ AND c=‘Wong’ AND y=d)}

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Chapter 6 Tutorial8 Exercise 1 - d For each project, list the project name and the total hours per week (by all employees) spent on that project. Relational Algebra: PROJ_HOURS1(PNO, T_HOURS) PNO SUM HOURS (WORKS_ON) PROJ_HOURS2 (PROJ_HOURS1 PNO=PNUMBER PROJECT) RESULT PNAME, T_HOURS (PROJ_HOURS2)

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Chapter 6 Tutorial9 Exercise 1 - e Retrieve the names of all employees who work on every project. Relational Algebra: PROJ_EMP PNO, ESSN (WORKS_ON) ALL_PROJ(PNO) PNUMBER (PROJECT) EMP_ALL_PROJ1 PROJ_EMP ALL_PROJ EMP_ALL_PROJ2 (EMPLOYEE SSN=ESSN EMP_ALL_PROJ1) RESULT FNAME, LNAME (EMP_ALL_PROJ2)..

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Chapter 6 Tutorial Exercise 1 - f Retrieve the names of all employees who do not work on any project. Relational Algebra: ALL_EMP SSN (EMPLOYEE) W_EMP(SSN) ESSN (WORKS_ON) N_W_EMP ALL_EMP - W_EMP RESULT FNAME, LNAME (EMPLOYEE * N_W_EMP) 10

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Chapter 6 Tutorial Exercise 1 - f Tuple relational calculus: {e.FNAME, e.LNAME | EMPLOYEE(e) AND NOT( w) (WORKS_ON(w) AND e.SSN=w.ESSN)} Domain relational calculus: {qs | ( t) (EMPLOYEE(qrstuvwxyz) AND NOT( a) (WORKS_ON(abc) AND t=a))} 11

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Chapter 6 Tutorial12 Exercise 1 - g For each department, retrieve the department name and the average salary of all employees working in that department. Relational Algebra: D_AVG_S(DNUMBER, AVG) DNO AVERAGE SALARY (EMPLOYEE) RESULT DNAME, AVG (D_AVG_S * DEPARTMENT)

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Relational Algebra: F_EMP SEX=‘F’ (EMPLOYEE) RESULT(AVG_F_S) AVERAGE SALARY (F_EMP) Chapter 6 Tutorial13 Exercise 1 - h Retrieve the average salary of all female employees.

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Chapter 6 Tutorial14 Exercise 1 - i Find the names and addresses of all employees who work on at least one project located in Houston but whose department has no location in Houston. Relational Algebra: H_PROJ PLOCATION=‘Houston’ (PROJECT) E_H_PROJ(SSN) ESSN (WORKS_ON PNO=PNUMBER H_PROJ) H_DEPT DNUMBER ( DLOCATION=‘Houston’ (DEPT_LOCATION)) ALL_DEPT DNUMBER (DEPARTMENT) N_H_DEPT ALL_DEPT - H_DEPT E_N_H_DEPT SSN (EMPLOYEE DNO=DNUMBER N_H_DEPT) EMP E_H_PROJ E_N_H_DEPT RESULT FNAME, LNAME, ADDRESS (EMPLOYEE * EMP)

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Chapter 6 Tutorial Exercise 1 - i Tuple relational calculus: {e.FNAME, e.LNAME, e.ADDRESS | EMPLOYEE(e) AND ( p) ( w) (PROJECT(p) AND WORKS_ON(W) AND p.PLOCATION=‘Houston’ AND e.SSN=w.ESSN AND w.PNO=p.PNUMBER AND NOT( l) (DEPT_LOCATION(l) AND l.DLOCATION=‘Houston’ AND e.DNO=l.DNUMBER))} Domain relational calculus: 15 {qsv | ( t) ( z) (EMPLOYEE(qrstuvwxyz) AND ( b) ( c) ( e) ( f) (PROJECT(abcd) AND WORKS_ON(efg) AND c=‘Houston’ AND t=e AND b=f AND NOT( h) NOT( i) (DEPT_LOCATIONS(hi) AND i=’Houston’ AND z=h))}

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Chapter 6 Tutorial16 Exercise 1 - j List the last names of all department managers who have no dependents. Relational Algebra: D_MGR(SSN) MGRSSN (DEPARTMENT) E_W_DEPENDENT(SSN) ESSN (DEPENDENT) EMP D_MGR - E_W_DEPENDENT RESULT LNAME (EMPLOYEE * EMP)

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Chapter 6 Tutorial Exercise 1 - j Tuple relational calculus: {e.LNAME | EMPLOYEE(e) AND ( d) (DEPARTMENT(d) AND e.SSN=d.MGRSSN AND NOT( x) (DEPENDENT(x) AND e.SSN=x.ESSN))} Domain relational calculus: {s | ( t) (EMPLOYEE(qrstuvwxyz) AND ( c) (DEPARTMENT(abcd) AND t=c AND NOT( e) (DEPENDENT(efghi) AND t=e))} 17

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