#Functional Depencency

Question

max number of FD in a relation with 'n' attributes=$2^{2n}$
how this formula is obtained?

Comments

LET R BE (AB)

FDS BE X->Y

X CAN BE (PHI,A,B,AB) Y CAN BE (PHI,A,B,AB)

CROSS PRODUCT WILL GIVE ALL THE FDS POSSIBLE (IN ABOVE CASE 16)

TAKE SMALL EXAMPLES ,YOU WILL GET THE FORMULA

Answer 1

n attributes each can be present or absent so for the left side we have 2$^{n}$ choice similarly for the right we have 2$^{n}$. Mapping 2$^{n}$ entries to another 2$^{n}$ entries we get 2$^{2n}$. This is an purely discrete math concept we just have to find how many pairs can we make including NULL

Comments

but shouldn't that be $(2^n)^{2^n}$ mappings then?

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No, 2$^{n}$ cross product with 2$^{n}$

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oh.. okay