chap8 Homework 10-2024简答题3. Normalize schema :name，ssn，phone#,city1.R(name，ssn，phone#,city)，F={ssn® (name,City)} A.Please calculate all the candidate KEY.B.Is R in 2NF？if R in not in 2NF，Decompose to 2NF.C.Is R in 3NF？if R in not in 3NF，Decompose to 3NF.D.Is R in BCNF？if R in not in BCNF，Decompose to BCNF
Question
chap8 Homework 10-2024简答题3. Normalize schema :name，ssn，phone#,city1.R(name，ssn，phone#,city)，F={ssn® (name,City)} A.Please calculate all the candidate KEY.B.Is R in 2NF？if R in not in 2NF，Decompose to 2NF.C.Is R in 3NF？if R in not in 3NF，Decompose to 3NF.D.Is R in BCNF？if R in not in BCNF，Decompose to BCNF
Solution
A. Candidate Key Calculation: The given functional dependency is ssn -> (name, city). This means that 'ssn' uniquely determines both 'name' and 'city'. Therefore, the candidate key for this schema is 'ssn'.
B. Checking for 2NF: A relation is in 2NF if it is in 1NF and no non-prime attribute is dependent on any proper subset of any candidate key of the relation. Here, 'name' and 'city' are non-prime attributes and they are fully functionally dependent on 'ssn' which is a candidate key. So, the relation R is in 2NF.
C. Checking for 3NF: A relation is in 3NF if it is in 2NF and for each of its functional dependency X->Y, either Y is a subset of X, or X is a superkey. Here, for the functional dependency ssn -> (name, city), 'ssn' is a superkey. So, the relation R is in 3NF.
D. Checking for BCNF: A relation is in BCNF if for every one of its dependencies X -> Y, X is a superkey. Here, for the functional dependency ssn -> (name, city), 'ssn' is a superkey. So, the relation R is in BCNF.
Therefore, there is no need to decompose the relation R as it is already in BCNF.
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chap8 Homework 10-2024简答题3. Normalize schema :name，ssn，phone#,city1.R(name，ssn，phone#,city)，F={ssn® (name,City)} A.Please calculate all the candidate KEY.B.Is R in 2NF？if R in not in 2NF，Decompose to 2NF.C.Is R in 3NF？if R in not in 3NF，Decompose to 3NF.D.Is R in BCNF？if R in not in BCNF，Decompose to BCNF
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