Problem 5:

a) AB is the only key.  AB+ includes C because of AB->C, then we can add
E because of C->E, and we can add D because of B->D.
It is easy to check that A+ = A, and B+ = BD, so AB is minimal.

b) B->D or C->E are the obvious choices.

c) Using B->D, we get S = BD and T = ABCE.
   Using C->E, we get S = CE amd T = ABCD.

d) The key for BD is B; for CE it is C, and for ABCD and ABCE it is AB.

e and f) BD and CE are in BCNF.  ABCD and ABCE are not.
B->D is a problem for ABCD, causing us to decompose into BD and ABC.
C->E is a problem for ABCE, causing us to decompose into CE and ABC.
Either way, we wind up decomposing into the three BCNF relations
ABC, BD, and CE.


Problem 6:

a) 6*3 = 18

b) name ->-> color and name ->-> size are the only two possible answers.

c) The key is all three attributes: {name, color, size}.
Since there are no FD's, and an MVD does not affect keyness, there is no
other possibility.

d) It is not in 4NF.  Decompose into name-color and name-size.