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lubasha [3.4K]
3 years ago
10

Consider the relation Courses(C, T, H, R, S, G), whose attributes can be thought of informally as course (C), teacher (T), hour

(H), room (R), student (S), and grade (G). Let the set of FD’s for Courses be C →T, HR →C, HT →R, HS →R, and CS →G.
Intuitively, the first says that a course has a unique teacher, and the second says that only one course can meet in a given room at a given hour. The third says that a teacher can be in only one room at a given hour, and the fourth says the same about students. The last says that students get only one grade in a course.
(a) What are all the keys for Courses?
(b) Verify that the given FD’s are their own minimal basis.
(c) Use the 3NF synthesis algorithm to find a lossless-join, dependency preserving decomposition of R into 3NF relations. Are any of the relations not in BCNF?
Computers and Technology
1 answer:
k0ka [10]3 years ago
3 0

Answer and Explanation:

a) Key for the relation courses is HS.

b) FD's are already minimal basis, as no attributes can be removed from the left side, right side is having only single attributes and none of the FD can be removed as given in the question.

c) By using the 3NF synthesis algorithm, the final set of relations obtained will be CT, HRC, HTR, HSR, CSG (FD'S are already proved to be minimal basis in part (b)). All the relations are in BCNF.

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#include <iostream>

using namespace std;

int main(){

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