3.5k views

Consider the following implications relating to functional and multivalued dependencies given below, which may or may not be correct.

1. if $A \rightarrow \rightarrow B$ and $A \rightarrow \rightarrow C$ then $A \rightarrow \rightarrow BC$
2. if $A \rightarrow B$ and $A \rightarrow C$ then $A \rightarrow \rightarrow BC$
3. if $A \rightarrow \rightarrow BC$ and $A \rightarrow B$ then $A \rightarrow C$
4. if $A \rightarrow BC$ and $A \rightarrow B$ then $A \rightarrow \rightarrow C$

1. If A→→ B and A → → C then A → BC
2. If A → B and A → C then A → → BC
3. If A → → BC and A → B then A → C
4. If A → BC and A → B then A → → C

Exactly how many of the above implications are valid?

1. $0$
2. $1$
3. $2$
4. $3$

edited | 3.5k views
–6
Here single arrow is functional dependency and doble arrow is multivalued dependency. But in axioms rule we only work with functional dependency . So ans. will be 0
+7
That's a very wrong way to put it :)
You should be able to solve by applying the definition of multi-valued dependencies.
+2
@Arjun, suggest best resource for learning MVD, 4NF. Also till which normal form one should study ? I studied till BCNF. Due to this questions I'll do 4NF. i!
+8
yes. 4NF is required. I had followed Korth for this. I guess any resource is fine- just understand the meaning of MVD.
+1
0
only B is true.
0

Please refer this video for better understanding of Multivalued dependencies.

Now according to multivalued dependency definition.

SSN -> -> CollegeName means

If we have the below tuples

 SSN CollegeName Hobby 100 Global Academy Of Technology Music 100 SJBIT Swimming

then we also need to have the below two tuples.

 SSN CollegeName Hobby 100 Global Academy Of Technology Swimming 100 SJBIT Music

Counter example of Option (i)If A→→ B and A → → C then A → BC

 A B C 100 X1 P1 100 X2 P2 100 X1 P2 100 X2 P1 200 Y Q 300 Z R

From this table we can see that A $\rightarrow \rightarrow$ B

and A $\rightarrow \rightarrow$ C.

But here A -> B and also A -> C do not hold true.

Counter example of Option (iii)If A → → BC and A → B then A → C

 A B C D(Rest) 100 X1 P1 D1 100 X1 P1 D2 200 Y1 Q1 D3 200 Y1 Q2 D4

For the same value of A,B,C the rest of the attributes(D here) can be interchanged.

Here A -> -> BC and A -> B hold true but A -> C is not true.

Please correct me if I'm wrong here.

+1
8 statements are given in question while there should be only 4.. Which 4 statements were there in original question

a. If A → → B and A  → →C then A → BC . So FALSE
b. If A → B and A → C then A→ BC.   So   A → →BC    TRUE..
c. If A → → BC and A → B  here B is Subset of AB and (A intersection BC) is phi so
A → B but not A → C so FALSE  (Coalescence rule )
d. If A → BC  then A → C   so  A → → C    TRUE
if A → B then A → → B  holds but reverse not true.

Correct Answer: $C$

by Veteran (60.5k points)
edited
0
I also think ii, iii statements are valid. But ans. given as 0
0
ya u r right..
edited..
0
a. This will be true if A->> B and A->>C then A->>BC .So it is false

b.  As functional dependency satisfying , it also supports multilevel dependency. So true

c.  proved by coalescence rule

d. It is proved by distributive rule of functional dependency and then multilevel dependency

am I right ?
0
b is true..  augmentation rule.
0
yes right
0
Is multivalued dependency 4nf there in syllabus?
0
@digvijay  IF A->->B  and A->->C then A->->BC is true ??. bcz  we prove it as  we do uinon of FD's  am i rt ?? and whatever rule we follow in FD  like union , decomposition  except transtive  are also applicable ?? in MVD
+2
4NF is part of GATE 2018 syllabus?
0
@digvijay pandey From where you have studied the theory that is required to solve this question?
+5

This is from Navathe, 6th ed.

Every FD is a MVD.

i.e suppose $x\rightarrow y$  $\Rightarrow$ $x\rightarrow \rightarrow y$

If y can be determined by x on y's single value then we can easily say x multi-determines y. as single value $\subseteq$ multiple value.

Now,

1. can't even possible.

2.  A -> BC which implies A -> -> BC. (true)

3. using given data we can't prove the then part.

4. given FDs are A->B & A->C, so using this we can say A->->C.(true)

here 2 implications are valid. i.e option C

by Active (3.1k points)