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# Assignment1 solution

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2010-1 Database. assignment #1 (relation algebra)

Taemin Lee.
Dataknow Lab.
Computer Science Education Dept.
Korea University.

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### Assignment1 solution

1. 1. Database System Conceps : CH2 Exercises and solutions
2. 2. Relation Algebra <ul><li>Six basic operators </li></ul><ul><ul><li>select:  </li></ul></ul><ul><ul><li>project:  </li></ul></ul><ul><ul><li>union:  </li></ul></ul><ul><ul><li>set difference: – </li></ul></ul><ul><ul><li>cartesian product: x </li></ul></ul><ul><ul><li>rename:  </li></ul></ul><ul><li>Additional Operations </li></ul><ul><ul><li>Natural join: </li></ul></ul><ul><ul><li>Aggregation: </li></ul></ul><ul><ul><li>Outer Join: </li></ul></ul><ul><ul><li>Division:  </li></ul></ul>
3. 3. Problem 2.1.
4. 4. Problem 2.1.a. σ (Street = M_street ∧ City = M_city) ( 위 relation) t1 ← employee manages t2 ← ρ manager (manager_name, m_street, m_city) (employee) Person_name Street City Manager_name M_street M_city Person_name Street City Manager_name Manager_name M_street M_city
5. 5. Problem 2.1.a. t1 ← employee manages t2 ← ρ manager (manager_name, m_street, m_city) (employee) Π (person_name) σ (Street = M_street ∧ City = M_city) (t1 t2) Person_name Street City Manager_name M_street M_city Person_name Street City Manager_name Manager_name M_street M_city
6. 6. Problem 2.1.a. t1 ← employee manages t2 ← ρ manager (manager_name, m_street, m_city) (employee) Π (person_name) σ (Street = M_street ∧ City = M_city) (t1 t2)
7. 7. Problem 2.1.b.
8. 8. Problem 2.1.b. wrong answer! Set different operation 할 때에 attribute 의 수가 동일해야 함 ! person_name company_name salary person_name
9. 9. Problem 2.1.c.
10. 10. Problem 2.1.c.
11. 11. Problem 2.1.c.
12. 12. Problem 2.3.a.
13. 13. Problem 2.3.a. wrong answer! Update 할 때에 정보의 누수를 조심 !
14. 14. Problem 2.3.b.
15. 15. Problem 2.5.a.
16. 16. Problem 2.5.b. & 2.5.c
17. 17. Problem 2.5.d. & 2.5.e
18. 18. Problem 2.6.
19. 19. Problem 2.6.a.
20. 20. Problem 2.6.a.
21. 21. Problem 2.6.b. & 2.6.c
22. 22. Problem 2.7.a.
23. 23. Problem 2.7.b.
24. 24. Problem 2.7.b.
25. 25. Problem 2.7.c.
26. 26. Problem 2.8.a. t1 account_number count(customer_name)
27. 27. Problem 2.8.b. account customer 1 customer 2 customer 3
28. 28. Problem 2.8.b. account customer 1 customer 2 customer 3 depositor 1 account 1 customer 1 depositor 2 account 2 customer 2 depositor 3 account 3 customer 3
29. 29. Problem 2.8.b. account customer 1 customer 2 customer 3 customer 1 customer 2 customer 3 account 1 account 2 account 3 depositor 1 account 1 customer 1 depositor 2 account 2 customer 2 depositor 3 account 3 customer 3
30. 30. Problem 2.8.b. account customer 1 customer 2 customer 3 customer 1 customer 2 customer 3 account depositor 1 account 1 customer 1 depositor 2 account 2 customer 2 depositor 3 account 3 customer 3
31. 31. Problem 2.8.b.
32. 32. Problem 2.9.a. t1 t2 company_name count-distinct(person_name) max(count-tistinct(person_name)
33. 33. Problem 2.9.b.
34. 34. Problem 2.9.b.
35. 35. Problem 2.9.c.
36. 36. Problem 2.11.a
37. 37. Problem 2.11.b.
38. 38. Problem 2.11.c. common mistake! t1 name 이 존재하지 않음 ! empno count_distinct(isbn) empno name office age isbn title authors publisher date
39. 39. Problem 2.11.c. & 2.11.d