Practice Exercise 3 - Question 5 - 4) Minimal cover

[Cindy Li]

Hi tutors, 

I'm a little stuck on this question, just want to know where i'm going wrong. 

The answer stated in the solution is 𝐹𝑚 = {𝐴𝐵 → 𝐶, 𝐷 → 𝐴,𝐷 → 𝐺, 𝐸 → 𝐵, 𝐴𝐵 → 𝐷, 𝐸 → 𝐴, 𝐶𝐷 → 𝐸}

Consider a relation 𝑅(𝐴,𝐵, 𝐶,𝐷, 𝐸, 𝐺, 𝐻) and its FD set 𝐹 = {𝐴𝐵 → 𝐶𝐷, 𝐸 → 𝐷, 𝐴𝐵𝐶 → 𝐷𝐸, 𝐸 → 𝐴𝐵,𝐷 → 𝐴𝐺, 𝐴𝐶𝐷 → 𝐵𝐸}

Step 1: Decompose into single attributes on RHS: 

F' = {AB -->C,  AB-->D,  E-->D,  ABC -->D,  ABC --> E,  E-->A ,  E-->B,  D-->A,  D-->G,  ACD --> B, ACD -->E}

Step 2: Reduce LHS attributes 

AB-->C and AB--D cannot be reduced 

ABC --> D can be replaced by AB-->D 

ABC --> E can be replaced by AB --> E 

ACD -->B can be replace by CD -->B 

ACD --> E can be replaced by CD --> E 

so we get F''' = {AB -->C,  AB-->D,  E-->D,  AB --> E,  E-->A ,  E-->B,  D-->A,  D-->G,  CD --> B, CD -->E}

From F'' I get that AB --> D is redundant via AB-->E and E-->D and also E--> A is redundant via E-->D and D --> A . So this shouldn't be in the minimal cover. 

Just wondering what steps i'm missing. 

Thanks! 

Kind regards, 

Cindy 

[Kamiyu]

Hey Cindy, your answer should be correct. I also spent forever trying to figure out the same thing before realising that there can be more than one possible minimal cover, which isn't made clear by the solution.