SELF DOUBT gate 2002 normalisation

Question

PLEASE SOMEONE EXPLAIN ME THAT WHY 2ND OPTION IS TRUE ? I AM NOT GETTING THIS OPTION HELP ME

 

R(A,B,C,D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition?

1. A→B,B→CD
2. A→B,B→C,C→D
3. AB→C,C→AD 
4. A→BCD

Comments

Is the answer option 3?

---

yes but please explain me 2nd option

---

The BCNF decompostion of option 2 will be R1(A,B), R2(B,C) and R3(C,D).

We can see that these three relations are lossless (by drawing table).For dependency preserving, from R1 er get the fd A->B, from R2 we get B->C and from R3 we get C->D.Thus we can obtain all the FDs and hence the relation is dependency preserving

---

ok ok got it thanku bro