Question

The next best alternate after BCNF is 3NF. We can say that this schema is 3NF but not BCNF. Match the following statements that make this 3NF and not BCNF

dept_advisor(s_ID, i_ID, dept_name)

§ With function dependencies:

             i_ID → dept_name

             s_ID, dept_name → i_ID

§ Two candidate keys =  { s_ID ,  dept_name }, { s_ID ,   i_ID   }

Match each statement below with the matching letter to make all statements 3NF:

i_ID is not a superkey { dept_name} – {i_ID }    dept_name   is contained in a candidate key   s_ID, dept_name

A. {dept_name } B. Since each attribute A in β – α is contained in a candidate key for R, dept_advisors is 3NF C. dept_advisor is not BCNF D. superkey

Answer 1

Consputer science Match the following to make 3 NF i-ID dept-name, s-10, dept_name in ID i_ID is not a super key super key {deft name - {1- 10} {dept-name dept.name is contained Since each athibute A un B - & is contained candidat dey for R, dept-advisors in candidate key. är Ừ 3 NF S-10, de pta name deft- advisa is not BC NF Candidate keys={s. ID, dept name}, {s. 10, 1-11)