0 votes 0 votes How many Super keys Possible for R(A,B,C,D,E) with 1. {A,BC,DE} as the keys ? 2. {A,BC,CDE} as the keys ? Databases databases superkeys candidate-key + – Na462 asked Oct 13, 2018 Na462 829 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Na462 commented Oct 13, 2018 reply Follow Share For first part if i take A as a single bit BC and DE as two different bits Total possiblity = 2^3 = 8 i need to subtract 1 case where everybody will be absent = 8 - 1 = 7 is it correct ? For second i am confused 1 votes 1 votes Magma commented Oct 13, 2018 i edited by Magma Oct 13, 2018 reply Follow Share 1)23 2) 21 yes 1 votes 1 votes Na462 commented Oct 13, 2018 reply Follow Share 1. I don't know answer of first one 2. Answer given is 21 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes 1) R (A,B,C,D,E) Candidate key : {A,BC,DE} = super-key(A) + super-key(BC) + super-key(DE) - super-key(ABC) - super-key(ADE) - super-key(BCDE) + super-key(ABCDE) = 24 + 23 + 23 -22-22 -21 + 20 = 23 2) R (A,B,C,D,E) Candidate key : {A,BC,CDE} = super-key(A) + super-key(BC) + super-key(CDE) - super-key(ABC) - super-key(ACDE) - super-key(BCDE) + super-key(ABCDE) = 24 + 23 + 22 -22 - 21 -21 + 20 = 21 Magma answered Oct 13, 2018 Magma comment Share Follow See all 2 Comments See all 2 2 Comments reply Magma commented Oct 13, 2018 reply Follow Share See the basic concept is this ..if A is a candidate key then how many Super keys are possible ?? you can choose {B C D E} in 2 x 2 x 2 x 2 = 16 ways right ?? if BC is a candidate key then {A D E } in 2 x 2 x 2 = 8 ways 0 votes 0 votes Na462 commented Oct 18, 2018 reply Follow Share Magma in first one cant i consider A,BC and CDE as three bits so total cases are 8 i subtract 1 here so 7 (because where every bit is 0 so phie which is not super key ) Why this is wrong ? 0 votes 0 votes Please log in or register to add a comment.