made easy cbt2

Question

consider the following relational schema R(A,B,C,D,E,F,G)

AB->C,BC->A,AC->B,B->D,D->E

how many minimum relations required to decompose R into BCNF which satisfy lossless and FDP decomposition?

 

initially i got 4 relation as

R1(A,B,C)

R2(D,E)

R3(B,D)

R4(A,B,F,G)

but in R1 and R2 there is no common attribute, hence it will not be lossless??am i correct

Comments

I think the decomposed Relations are:

R1(A,B,C)
R2(D,E)
R3(B,D)
R4(F)   -- trivial

---

no no

even they have made the same relation

---

which 3 relations would suffice according to you?

---

R1(A,B,C)

R2(B,D,E)

R3(A,B,F,G)

Although am not sure!!

but i just wanna know how it can be 4 when R1(A,B,C) AND R2(D,E) dont have anything common

---

In R2(B,D,E), D is neither candidate key nor super key, so the relation won't be in BCNF.

R1(A,B,C),R2(D,E),R3(B,D)

As R2,R3 has D in common and R3,R1 has B in common, i think it satisfies lossless decomposition. Moreover i check Lossless decomposition in some other way, so not entirely sure about whether there must be a common key in both while decomposition.

---

or think of it like this:

try decomposing into R1(A,B,C) AND R2(B,D,E) since B is the key and then try to decompose R2 into R2(B,D) and R3(D,E) as D will be the key during this decomposition.

---

R1(A,B,C), R2(D,E), R3(B,D) and R4(A,B,F,G)

but in R1 and R2 there is no common attribute, hence it will not be lossless

  " it will not be lossless " -------- is it question or answer ?

---

i just wanna ask

will this decomposition be lossless or not??

as there is no common attribute between R1 and R2

---

then what this is mean " so shouldn’t the answer be 3?  "

 

will this decomposition be lossless or not??

it is loss-less !

R1(A,B,C), R2(D,E), R3(B,D) and R4(A,B,F,G)

step 1:- combine R2 and R3 ==> R$_x$

step 2 :- combine R1 and R$_x$ ===> R$_y$

step 3 :- combine R4 and R$_y$ ===> R

---

ok ,

actually we say that to be lossless,if R is decomposed into R1 and R2; then there must be a common attribute in them.

so i thought there should be a common attribute in all relations.

---

" so shouldn’t the answer be 3?  "

what it means ?

---

i was confused at that time.

hiding it, sorry