DBMS - Problem-2 To check Whether a Decomposition is Lossless

[@TutorialsPoint_]

To Get Certification, Click Here: bitly.ws/PXXy Use coupon "KZbin12" to get ‘’FLAT 12%’’ OFF at Checkout.

[@jugjyotiborgohain3212]

THANK YOU SO MUCH SIR....(IF YOU ARE WATCHING AT THE LAST MOMENT PLAY IT IN 1.25 OR 1.5X)

[@prasadreddy5824]

2x

[@fahansheikh]

Sir your accent is so sweet ....for real

[@varunbabbar1478]

Sir, I have a doubt. In the first row, there should be alpha E because in the dependencies there was E dependent on A so since A is alpha so E should also be alpha. If I am wrong then please correct me.

[@axelr7908]

Incorrect. R2 and R3 have all beta for A which implies that you can't know what E for R1 you have. If R2 would instead have decomposition A,C,E for example, your answer would have been correct.

[@DaLastOne]

Thank you very much

[@shubhamshrivastava296]

I have a same ques in my book but the answer there is lossless and your answer is lossy I’m confused

[@rohandevaki4349]

very simple and accurate explainaiton sir, thankyou very much

[@akshyananda7219]

Good explanation of solution. Thank you Sir.

[@lakshyachaturvedi2712]

First row should contain alpha only to be lossy?

[@somshekharmukherjee1760]

SIR, one doubt while checking FDs all the rows should have the same value or just two rows having same value is sufficient for making a change in RHS of FD ??

[@rejusubash5297]

I hope you passed your exam

[@simranrastogi4926]

You don't tell us the other two options such as if 1) dependency preserving and lossy 2) not dependency preserving and lossless

[@axelr7908]

Not dependency preserving because A-->E is not contained in R1(a,b,c), R2(b,c,d) and R3(c,d,e). Other way of saying it: No decomposition has A and E in them --> Not dependency preserving

[@mr.mentpolynopeace]

Thx

[@jayasreed2724]

Great

[@abhisheksingh2520]

Thanks sir!

[@M_Rimi]

Thank you sir

[@animraj99]

good tutorial. thanks

[@hackeroromis]

Great tutorial! Thank you

[@hardhishringeri5448]

that suit is gas

[@jkssbjobtutorial.5061]

thanku sir

[@nelsonmd]

This decomposition is NOT dependency-preserving because not all functional dependencies are subsets in all the relations, but definitively, this decomposition is Lossless. You should correct this example because many students are inferring something different. According to your method, there is an alfa in R1/D just because a functional dependency allows that. So, all R1 has alfas in all attributes. Even without applying your method, the natural join of the decomposed relations R1∪R2∪R3=(a,b,c)∪(b,c,d)∪(c,d,e) results in the R(a,b,c,d,e), the original relation, indicates losslessness. With all respect, please correct this - To err is human -.

[@Ayush-_-007]

don't watch wrong solution

[@amartyamukherjee8137]

your process is WRONG...plz dont post this kind OF HARMFUL VIDEOS...

[@axelr7908]

It's correct

[@dhanarajs2707]

how slow is his accent

[@AdamRubiks]

thanks sir