Trick question about FD's? X  Y  Z
x1 y1 z1
x1 y1 z1
x1 y1 z2
x2 y1 z1


The questions  are :

a. List all the non-trivial original functional dependencies that this relation instance preserves

b. Change a single value to eliminate all FDs that the above instance preserves.



To confirm that there is no typo from my part (could be from the professor):
I see several non trivial dependencies (9 if I count right!) 

How to find them: Since, the relation has few attributes, we can examine each possible dependency separately, taking the respective projection. (The possible dependencies with just 3 attributes are not very few. They are 64 but see below for some ways to make that number smaller.)

Example: X -&gt; Y. Does Y depend on X? Yes.

X  Y
x1 y1
x2 y1


Example: Y -&gt; X. Does X depend on Y? No.

Y  X  
y1 x1 
y1 x2 


Example: XY -&gt; Z. Does Z depend on XY? No.

X  Y  Z
x1 y1 z1
x1 y1 z2
x2 y1 z1


Also we don't have to examine them all (64 possible dependencies).

First, we need to examine only the (37) non-trivial ones. The XY -&gt; X for example is trivial and of course holds, We don't have to examine the (27) trivial dependencies at all. 

Second, we can deduce several ones, in both directions:

Once we find that X -&gt; Y holds, we can deduce that XZ -&gt; Y holds, too.
Once we find that XY -&gt; Z does not hold, we can deduce that X -&gt; Z and Y -&gt; Z don't either.
We can also use Armstrong's axions to make further deductions. 

I won't go on further details but I think the exercise is good - and tricky.

If you are observant enough, you only need to find a single dependency and then you can prove the rest (8) that hold, using these deductions. 

And only find that 2 more dependencies do not hold and then you could prove that the above 9 are the only ones and no other holds.