I am confused with dependency preserving property of database relations (tables). Do we have to look at initial FD set or what else? I tried to solve some problems on this subject. The questions before this one all feed my initial estimation, which is 'look at the given FD set. If you don't lose any of them in your new relation set, then this is dependency preserving'.
But when I come to this question I am confused.
Consider the relation R = (A B C D E F G H) and the following FD set:
FD1 E -› D
FD2 B, E -› C G
FD3 D, G -› E
FD4 C -› A B
FD5 E, G -› C
FD6 A, E -› B D
FD7 C, E, D -› G
FD8 A, G -› E
These are the given relations
R1 (E F G H)
R2 (A B E G)
R3 (C D E G)
R4 (A B C)
Answer says that this decomposition is dependency preserving. According to my estimation we lose FD2 so, this must not be dependency preserving.
I need an expert to clarify this concept for me.
This question was a part of homework questions. I wasn't sure if I am thinking right when I do the homework.
In my answer I wrote:
This decomposition is not dependency preserving because in this decomposition we loose the FD DF--> BC .
And my database teacher accepted this answer as an right answer I wanted to clearify the subject here also.
Ferda
The decomposition is dependency preserving as FD2 BE->CG can be achieved by relations R2(BE->G) and R3(EG->C).
Closure of BE gives CG.
Related
I have to prove that the set of rational or regular languages is closed by morphism on their alphabet.
i.e. that the image of a rational language by a morphism is still rational.
h being a morphism from Σ to Σ', my idea is to start with an automaton A and to construct an automaton A' which recognizes the language h(L(A)).
I use the same initial and final states then, for any transition (q, a, q') in A, I consider 3 cases :
if h(a) = ε I add the states q, q' (if they do not already exist in A') and an ε transition (q, ε, q')
if h(a) = b ∈ Σ', I add the states q, q' (if they do not already exist in A') and a transition (q, b, q')
if h(a) = b_1b_2...b_n ∈ Σ'*, I add the states q, q' (if they do not already exist in A') plus n-1 new states and n transitions from (q, b_1, q_1) to (q_{n-1}, b_n, q')
Then it's "easy" to prove that h(L(A)) is included in L(A') following the construction steps, however I'm struggling to prove the converse, i.e. that L(A') is included in h(L(A))
I did create a dictionary with 2 languages(English/Persian) at the one file like this:
بگو B E G U
خزنده KH A Z A N D E
قدت GH A D E T
چنده CH A N D E
قد GH A D
من M A N
شب SH A B
hi H AA Y
hello H E L L O
how H O V
are AA R
you Y U
what V AA T
is I Z
your Y O R
name N E Y M
old O L D
where V E R
from F E R AA M
And used http://www.speech.cs.cmu.edu/tools/lmtool-new.html to build the language model. Then I tried to learn an acoustic model with that language model and test it.
It works good for Persian voices but doesn't work for English words. After some try&error I found that the problem is about my phoneset. I used my own phoneset as you can see above, but it seems pocketsphinx doesn't accept this phoneset for English words and it only accepts it's own phoneset for English!
So I want to know did I found the problem true? Should I use the pocketsphinx phoneset for my Persian words as well? Where should I find it's complete phoneset and a guide to learn how to use it for Persian words?
You have to build a new acoustic model with joined phoneset
I have been trying to solve these two questions, but haven't had much luck.
Question 1: Show that the decomposition rule:
A → BC implies A → B and A → C,
is a sound rule, namely, that the functional dependencies A → B and A → C
are logically implied by the functional dependency A → BC.
Question 2: Let F be the following collection of functional dependencies
for relation schema R = (A, B, C, D, E):
D → A
BA → C
C → E
E → DB .
a) Compute the closure F + of F .
b) What are the candidate keys for R? List all of them.
c) List the dependencies in the canonical cover of the above set of
dependencies F (in other words, compute F c , as we have seen in class).
Any input will be helpful.
I am trying to prove commutative property for agda. I tried to explore the standard library but there is lot of complex thing which i could not understand.
I tried in this way --
comm : (a b : Q) -> (a + b) === (b + a)
the problem here is + which is not defined over Q in library. Can't we proof this without defining + over Q.
Please guide me.
You cannot prove this without first defining +.
If you get confused exploring the standard library I suggest you try to prove something easier first, in order to become more acquainted with Agda, before tackling this.
Of course you can't prove commutativity of an undefined function _+_; as a stupid counter-example, would you expect to be able to prove (a - b) == (b - a)? If not, why not? _-_ is just as much of an undefined function as _+_ at this point; it just has a different name...
Note that you can define addition for ℚ using elementary school math:
n ÷ p + m ÷ q = (n * q + m * p) ÷ (p * q)
and simplifying it by dividing both n * q + m * p and p * q with their GCD. I have already explained the details of this last step in this answer.
What is the easiest way for Example1 to be converted to Example2 (I would be doing this with much longer lists)? Column C and D shall be associated to Col B for the output of Example2. This is not just to make Col B replicate Col A, although that is part of the solution. Thank you in advance!
Example1:
Col A Col B Col C Col D
a e d c
l l o a
e x g t
x a s s
Example2:
Col A Col B Col C Col D
a a s s
l l o a
e e d c
x x g t
It is not totally clear what you want to achieve and what the data qualities are, so a few assumptions:
all items in Col A are also in Col B
items in Col A are unique
Consider the following screenshot. Column A has been copied into column F. The formula in G1 is
=INDEX(B$1:B$4,MATCH($F1,$B$1:$B$4,0))
Copy the formula across to I1 and then copy G1 to I1 down.
If that does not do what you need, please edit your question, add a better data sample and more explanation.