bcnf

问题 I have the following functional dependencies (they represent all the functional dependencies on my relation): (1) BrokerName -> Office (2) StockName -> Dividend (3) InvestorId -> BrokerName (4) InvestorId, Stockname -> Quantity (5) InvestorId, Stockname -> Office I know from using the techniques in this YouTube video that (InvestorId, Stockname) is my one and only candidate key. According to @nvogel's solution in this SO thread: A relation, R, is in BCNF iff for every nontrivial FD (X->A)

问题 I looked in Decomposing a relation into BCNF answers and tried it on my homework, but i don't get the correct answers, so i ask for help in BCNF decomposition Consider R=(ABCDEG) & F={BG->CD, G->A, CD->AE, C->AG, A->D} . I start pick A->D . Now i got S=(AD), R'=(ABCEG). I pick G->A . Now i got S=(AD,AG) R'=(BCEG) . I pick C->G . Now i think i need to get S=(AD,AG,CG) and R'=(BCE) , But the answer in the end is (AD,AG,CGE,BC) .what went wrong? or perhaps, a better algorithm? 回答1: To convert a