The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
  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 (2,011 points) | 320 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

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

33,620 questions
40,170 answers
38,552 users