I read a statement "Relation R in BCNF with at-least one simple candidate key is also in 4NF"
I don't think that it is always true but I am not able to prove it.
Can someone please help ?
The statement is true and this is the sketch of the proof taken from the paper "Simple Conditions for Guaranteeing Higher Normal Forms in Relational Databases", by C.J.Date and R.Fagin, ACM TODS, Vol.17, No. 3, Sep. 1992.
A relation is in 4NF if, for every nontrivial multivalued dependency X →→ Y in F+, X is a superkey for R. So, if a relation is in BCNF, but not in 4NF, then there must exists a nontrivial multivalued dependency (MVD) X →→ Y such that X is not the key. We will show that this is in contradiction with the fact that the relation is in BCNF and has a candidate key K constituted by a unique attribute (simple candidate key).
Consider the fact that, in a relation R(T), when we have a nontrivial MVD X →→ Y, (assuming, without loss of generality that X and Y are disjoint), then also the MVD dependency X →→ Z must hold in the same relation, with Z = T - X - Y (that is Z are all the other attributes of the relation). We can now prove that each candidate key must contain at least an attribute of Z and an attribute of Y (so it must contain at least 2 attributes!).
Since we have X →→ Y and X →→ Z, and X is not a candidate key, assume that the hypothesis is false, that is that there is a candidate K which does not contain a member of Y (and for symmetry, neither a member of Z). But, since K is a key, we have that K → Y, with K and Y disjoint.
Now, there is an inference rule that says that, in general, if V →→ W and U → W, where U and W are disjoint, then V → W.
Applying this rule to our case, since X →→ Y, and K → Y, we can say that X → Y. But this is a contradiction, since we have said that R is in BCNF, and X is not a candidate key.
In other words, if a relation is not in 4NF, than each key must have at least 2 attributes.
And given the initial hypothesis, that we have a relation in BCNF with at least a simple candidate key, for the previous lemma, the relation must be in 4NF (otherwise every key should be constituted by at least 2 attributes!).
I am having some trouble understanding Multi-Valued Dependencies. The definition being: A multivalued dependency exists when there are at least 3 attributes (like X,Y and Z) in a relation and for value of X there is a well defined set of values of Y and a well defined set of values of Z. However, the set of values of Y is independent of set Z and vice versa.
Suppose we have a relation R(A,B,C,D,E) that satisfies the MVD's
A →→ B and B →→ D
How does MVD play into A->B and B->D here? Honestly I'm not sure I really understand the definition after looking at example problems.
If R contains the tuples (0,1,2,3,4) and (0,5,6,7,8), what other tuples must
necessarily be in R? Identify one such tuple in the list below.
a) (0,5,2,3,8)
b) (0,5,6,3,8)
c) (0,5,6,7,4)
d) (0,1,6,3,4)
I would have thought AB is 0,1 and 0,5 and BD is 1,3 and 5,7. None of the answers have 0,1,3,5,7.
MVDs (multi-valued dependencies) have nothing to do with "at least 3 attributes". (You seem to be quoting the Wikipedia article but that informal definition is partly wrong and partly unintelligible.) You need to read and think through a clear, precise and correct definition. (Which you were probably given.)
MVD X ↠ Y mentions two subsets of the set S of all attributes, X & Y. There are lots of ways to define when a MVD holds in a relation but the simplest to state & envisage is probably that the two projections XY and X(S-Y) join to the original relation. Which also mentions a third subset, S-Y. Which is what the (binary) JD (join dependency) {XY, X(S-Y)} says.
Wikipedia (although that article is a mess):
A decomposition of R into (X, Y) and (X, R−Y) is a lossless-join decomposition if and only if X ↠ Y holds in R.
From this answer:
MVDs always come in pairs. Suppose MVD X ↠ Y holds in a relation with attributes S, normalized to components XY & X(S-Y). Notice that S-XY is the set of non-X non-Y attributes, and X(S-Y) = X(S-XY). Then there is also an MVD X ↠ S-XY, normalized to components X(S-XY) & X(S-(S-XY)), ie X(S-XY) & XY, ie X(S-Y) & XY. Why? Notice that both MVDs give the same component pair. Ie both MVDs describe the same condition, that S = XY JOIN X(S-XY). So when an MVD holds, that partner holds too. We can write the condition expressed by each of the MVDs using the special explicit & symmetrical notation X ↠ Y | S-XY.
For R =
A,B,C,D,E
0,1,2,3,4
0,5,6,7,8
...
A ↠ B tells us that the following join to R:
A,B A,C,D,E
0,1 0,2,3,4
0,5 0,6,7,8
... ...
so R has at least
A,B,C,D,E
0,1,2,3,4
0,1,6,7,8
0,5,2,3,4
0,5,6,7,8
of which two are given and two are new but not choices.
B ↠ D tells us that the following join to R:
B,D A,B,C,E
1,3 0,1,2,4
5,7 0,5,6,8
... ...
so R has at least
A,B,C,D,E
0,1,2,3,4
0,5,6,7,8
which we already know.
So we don't yet know whether any of the choices are in R. But we do now know R is
A,B,C,D,E
0,1,2,3,4
0,1,6,7,8
0,5,2,3,4
0,5,6,7,8
...
Repeating, A ↠ B adds no new tuples but B ↠ D now gives this join:
B,D A,B,C,E
1,3 0,1,2,4
1,7 0,1,6,8
5,3 0,5,2,4
5,7 0,5,6,8
... ...
And one of the tuples in that join is choice b) (0,5,6,3,8).
The way the question is phrased, they are probably expecting you to use a definition that they will have given you that is like another two in Wikipedia. One says that α ↠ β holds in R when
[...] if we denote by (x, y, z) the tuple having values for α, β, R − α − β collectively equal to x, y, z, then whenever the tuples (a, b, c) and (a, d, e) exist in r, the tuples (a, b, e) and (a, d, c) should also exist in r.
(The only sense in which this gives the "formal" definition "in more simple words" is that this is also a definition. Because this isn't actually paraphrasing it, because this uses R − α − β whereas it uses R − β.)
By applying this rule to repeatedly generate further tuples for R starting from the given ones, we end up generating b) (0,5,6,3,8) much as we did above.
PS I would normally suggest that you review the (unsound) reasoning that led you to expect "AB is 0,1 and 0,5 and BD is 1,3 and 5,7" (whatever that means) or "0,1,3,5,7". But the "definition" you give (from Wikipedia) doesn't make any sense. So I suggest that you consider what you were doing with it.
I mean if I have several X and several Y
and I do a match like this:
X -[ W ]-> Y
With X and Y related by several W ( there can be several W between same pairs (X,Y) )
I want top ten X for each Y with the property sum(W.property)
If I return
return Y , sum(W.property) , X order by sum(W.property) desc Limit 10
I Just get 10 but I need for every Y,
Is there a way to do that?
MATCH X -[ W ]-> Y
WITH Y, sum(W.property) AS total, X
ORDER BY total DESC
WITH Y, collect({sum: total, X: X})[0..10] AS values
UNWIND values AS value
RETURN Y, value.sum, value.X
You can actually skip the UNWIND and just change that second WITH to a RETURN if you're OK with it returning it as an array. It would be a bit more efficient because you're not repeating values of Y over and over. If you were going to do that you could even change the map structure into an array like this:
collect([total, X])[0..10]
The following code is from perm.v in the Ssreflect Coq library.
I want to know what this result is.
Lemma perm_invK s : cancel (fun x => iinv (perm_onto s x)) s.
Proof. by move=> x /=; rewrite f_iinv. Qed.
Definitions in Ssreflect can involve lots of concepts, and sometimes it is hard to understand what is actually going on. Let's analyze this by parts.
iinv (defined in fintype.v) has type
iinv : forall (T : finType) (T' : eqType) (f : T -> T')
(A : pred T) (y : T'),
y \in [seq f x | x in A] -> T
What this does is to invert any function f : T -> T' whose restriction to a subdomain A \subset T is surjective on T'. Put in other words, if you give me an y that is in the list of results of applying f to all elements of A, then I can find you an x \in A such that f x = y. Notice that this relies crucially on the fact that T is a finite type and that T' has decidable equality. The correctness of iinv is stated in lemma f_iinv, which is used above.
perm_onto has type codom s =i predT, where s is some permutation defined on a finite type T. This is saying, as its name implies, that s is surjective (which is obvious, since it is injective, by the definition of permutations in perm.v, and by the fact that the domain and codomain are the same). Thus, fun x => iinv (perm_onto s x) is a function that maps an element x to an element y such that s y = x. In other words, its the inverse of s. perm_invK is simply stating that this function is indeed the inverse (to be more precise, it is saying that it is the left inverse of s).
The definition that is actually useful, however, is perm_inv, which appears right below. What it does is that it packages fun x => iinv (perm_onto s x) with its proof of correctness perm_invK to define an element of perm_inv s of type {perm T} such that perm_inv s * s = s * perm_inv s = 1. Thus, you can view it as saying that the type {perm T} is closed under inverses, which allows you to use a lot of the ssr machinery for e.g. finite groups and monoids.
What are the non-trivial functional dependencies in the following table?
A B C
1 1 1
1 1 0
2 3 2
2 3 2
What the basic concept?
A functional dependency answers the question, "Given one value for X, do I find one and only one value for Y?" Both X and Y are sets; each one represents one or more attributes.
So we can ask ourselves, "Given one value for 'A', do I find one and only one value for 'B'?" And the answer is "Yes". (Assuming the sample data is representative.) That leads to the nontrivial functional dependency A->B.
And we continue with the question, "Given one value for 'A', do I find one and only one value for 'C'?" And the answer is "No". Given 1 for 'A', we find two different values for 'C': 1 and 0. No functional dependency there.
Repeat for every possible combination of attributes.
Trivial: If an FD X → Y holds where Y subset of X, then it is called a trivial FD. Trivial FDs are always hold.
Non-trivial: If an FD X → Y holds where Y is not subset of X, then it is called non-trivial FD.
Completely non-trivial: If an FD X → Y holds where x intersect Y = Φ, is said to be completely non-trivial FD.
For example:
X = { b, c } and Y = { b, a }. If X → Y, then the FD is non-trivial but not completely non-trivial.
See the examples here: http://en.wikipedia.org/wiki/Functional_dependency
Especially the lecture one. I think in this case (for the data set you show) for instance if A=1 B=2 and if A=2 B=3. That is probably the dependency you are talking about.
non trivial dependency means X-->Y that is if Y is not proper subset of X table or relation with X then it said to be non trivial functional dependency.
A FD (functional dependency) is trival, non-trivial or semitrivial.
Write what all attributes have functional dependency between them:
A->B, B->A, C->A, C->B
Using the augmentation inference rule we also get:
AC->B, BC->A
Augmentation says that if A -> B holds then AX -> BX holds.
So in total we have 5 non-trivial functional dependencies.
Trivial fd: x,y some attributes sets, if y is a subset of x then x->y implies is a trivial fd.
Non-trivial fd; x,y some attributes sets ,
if x intersection y goes to phi. then x->