functional-dependencies

So I am looking over my database notes and material trying to refresh myself on the general concepts and terminology for upcoming interviews. I have gotten stuck at dependency however and lossless-join decompositions though. I have searched all over and see lots of mathy equations but I am looking for a plain and simple English response or example. I have a found a powerpoint from http://www.cs.kent.edu/~jin/DM09Fall/lecture6.ppt which illustrates an example that I cannot fully understand. It is posted below. R = (A, B, C)F = {A → B, B → C) Can be decomposed in two different ways R1 = (A, B),

Apologies if this is a ridiculous question; I've searched high and low for an answer with no avail. I know how to calculate minimal cover; ie ensure each functional dependency only has one attribute on the the RHS, remove extratraneous/redudant lhs attributes by calculating closure of each examining all FD's, seeing if any can be removed (again by calculating closure) Is 'canonical' cover just another word for the same thing? A canonical cover is "allowed" to have more than one attribute on the right hand side. A minimal cover cannot. As an example, the canonical cover may be "A -> BC" where

Consider the code below : {-# LANGUAGE MultiParamTypeClasses,FlexibleInstances,FunctionalDependencies,UndecidableInstances,FlexibleContexts #-} class Foo a c | a -> c instance Foo Int Float f :: (Foo Int a) => Int -> a f = undefined Now when I see the inferred type of f in ghci > :t f > f :: Int -> Float Now If I add the following code g :: Int -> Float g = undefined h :: (Foo Int a) => Int -> a h = g I get the error Could not deduce (a ~ Float) I am not able to understand what has happened here? The restriction Foo Int a should have restricted the type of h to Int -> Float as shown in the

Suppose the relation R( K, L, M, N, P) , and the functional dependencies that hold on R are: - L -> P - MP -> K - KM -> P - LM -> N Suppose we decompose it into 3 relations as follows: - R1(K, L, M) - R2(L, M, N) - R3(K, M, P) How can we tell whether this decomposition is lossless? I used this example R1 ∩ R2 = {L, M}, R2 ∩ R3 = {M}, R1 ∩ R3 = {K,M} we use functional dependencies, and this is not lossless in my opinion, but a little bit confused. It helps if we demystify the concept of lossless decomposition a bit: it really just means that joining R1, R2 and R3 should yield the original R. Do

Just curious: is there some reason why one cannot do all necessary normalizations in a single step? Isnt normalization ultimately the redrawing of the Functional Dependency (FD) graph? We start out with an FD diagram/graph and we want to end up with a graph (vertices are attributes, there is an edge between attributes a,b if b is FD on a ) representing a relation in (Edit) BCNF ? EDIT: What I mean is : we start with a FD graph , which is a graph pairing attributes a,b iff b is FD on A, i.e., we join a and b with an edge iff b=f(a). From this graph we want to obtain a graph (FD)_2 with certain

I have a relation A,B,C,D,E with functional dependencies 1) A->BC 2) CD->E 3) B->D 4) E->A Using 1 gives A,D,E and then using 4 will make it D,E Using 2 gives A,B,C,D and then using 3 gives A,B,C and using 1 gives A Using 2 gives A,B,C,D and using 1 gives A,D Using 4 gives B,C,D,E and using 2 gives B,C,D and using 3 gives B,C Using 3 gives A,B,C,E and using 1 gives A,E and using 4 gives E So I would have 5 super keys? (A, E, AD, BC, DE). And from my super keys I would pick the unique ones. Since I can get A from E, I can remove A and AD(since DE is the same) and since I can get BC from A I can