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A relation R(ABCDE) with FD set f = {A -> BC, C -> DE, D -> E} and the decomposition d = { R1(ABCD), R2(DE)}. Is this decomposition dependency preserving?
Shefali
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Jul 23, 2015
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The answer given is YES. But how is the dependency C -> DE preserved in this decomposition?
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database-normalization
Shefali
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Yes its dependency preserving ,
from R1 we get : A-> BCD and C->D
and R2 : D->E
now see , C->D amd D->E so we can say C->DE
So all dependency are preserved .
Pranay Datta 1
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Jul 24, 2015
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focus _GATE
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for dependency C--->DE --------------- 1 we have dependecy D--->E, now replace E with D in --------------- 1, we get C-------->DD means C---------->D. hence C------->D preserve in R1(ABCD) ?? am i right???
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Aug 3, 2015
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What has explained is pseudo transitivity rule , and yes you're right.
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