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  1. Split the fd's such that rhs contains single attribute. 
  2. Find the redundant fd's and remove redundant ones. 
  3. Find the redundant attributes on lhs and remove AB->C ,A can be deleted if closure of B contains A
posted Jul 28, 2015 in Databases by Active (1,953 points) 1 8 24 | 267 views


Explanation with proper example would have been noble

X->YZ , Y->XZ , Z->X


So from decomposition rule:-

X->Y ,X->Z , Y->X ,Y->Z, Z->X 

if we remove Z->X still we get Z and X in closure

X->Y ,X->Z , Y->X ,Y->Z

if Y->X remove still we can Y and X

X->Y ,X->Z ,Y->Z

we can remove one more either X->z or Y->Z

X->Y ,X->Z 

we have still X,Y,Z but now we can remove dependecy

so minimal 2

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