partial functional dependency
why is fd2 and fd3 here a partial dependency? my understanding is that in a partial dependency if I remove an attribute from the left hand side of the dependency then the dependency would still exist .. doesn't removing clientNo from fd2 deletes the dependency?
fd1 clientNo, propertyNo -> rentStart, rentFinish
fd2 clientNo -> cName
fd3 propertyNo -> pAddress, rent, ownerNo oName
fd4 ownerNo -> oName
fd5 clientNo, rentStart -> propertyNo, pAddress, rentFinish, rent ownerNo, oName
fd6 propertyNo, rentStart -> clientNo, cName, rentFinish
Partial FD normally means that the determinant (left hand side) is not a candidate key but is only part of a candidate key. You haven't specified any keys in your example so we can't tell.
Hi,
You cannot talk about partial dependency when you have only one attribute as the determinant.
For example: cName is determined by clientNo, and you need full clientNo to determine cName so this is not a partial dependency.
Regards,
Franck.
You cannot talk about partial dependency when you have only one attribute as the determinant.
For example: cName is determined by clientNo, and you need full clientNo to determine cName so this is not a partial dependency.
Regards,
Franck.
ASKER
here's my real example if you want to see, I don't quite get the explanation there from 1NF to 2NF
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ASKER
shouldn't that be a fully functionally dependent then? as if you remove the NIN then the dependency is gone... if it's partial then even if you remove NIN then that dependency would still exist
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ASKER
Here's what my text book explains:
A functional dependency A -> B if there is some attribute that can be removed from A and yet the dependency still holds.
In the NIN example above, I can't remove NIN right?
A functional dependency A -> B if there is some attribute that can be removed from A and yet the dependency still holds.
In the NIN example above, I can't remove NIN right?
ASKER
Just explain to me why NIN -> eName is partial dependency and I'll be happy
>> In the NIN example above, I can't remove NIN right?
Right. I think you left out part of the sentence you quoted from your book. What it describes is a reducible dependency, nothing to do with partial key depdendency.
>> Just explain to me why NIN -> eName is partial dependency
Because NIN is only part of a key, not the whole key in the table in which eName appears.
Right. I think you left out part of the sentence you quoted from your book. What it describes is a reducible dependency, nothing to do with partial key depdendency.
>> Just explain to me why NIN -> eName is partial dependency
Because NIN is only part of a key, not the whole key in the table in which eName appears.
ASKER
what is the whole key?
ASKER
and I did not copy the definition wrong
In the diagram, the key is (NIN,contractNo) and it has to be because of FD(1)
ASKER
oh I get it, so I guess if it were NIN, contractNo -> eName then it would be full? right?
Correct
ASKER
how about transitively dependent? why is the following transitively dependent
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How about looking up Transitive Dependency yourself? You can't expect others to do your homework for you.
ASKER
It's a condition where A, B, and C are attributes of a relation such that if A -> B and B->C is transitively dependent on A via B. I just don't understand why in the example above FD4 and FD3 satisfies this definition
Because A -> B B -> C is what is in your example:
contractNo -> hNo hNo -> hLoc
contractNo -> hNo hNo -> hLoc
ASKER
how do I know how to split up the table once I know the problem