Consider a relation S(G,H,J,K). The following FD's are given:

• G H
• JK G
• H K

Is S in BCNF? Why or why not?

If it is not in BCNF:

• Decompose into BCNF as covered in class.
• Circle the relations in your final answer.
• Underline the keys in your final answer.

Show your work. Write your answer here:

• G {G, H, K}
• H {H, K}
• J {J}
• K {K}
• JK {J, K, G, H}

G H violates BCNF as H is not a trivial functional dependence of G and G is not a super key so we start by decomposing using G H

• S(G, H)
  • G H
• S(G, J, K)
  • JK G
  • G K (via G H and H K)

G K violates BCNF as K is not a trivial functional dependence of G and G is not a super key of S so now we decompose S using G K

• S(G, K)
  • G K
• S(G, J)

S is not in BCNF because G H and H K both violate BCNF. We can decompose S into

• S(<u>G</u>, H)
• S(<u>G</u>, K)
• S(<u>G</u>, <u>J</u>)