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Пікірлер: 79
@yon1623 Жыл бұрын
8 years letter i am watching this. thanks for making this , old man.
@Ihatenicknames19 жыл бұрын
Thank you so so much! That was very helpful! You are awesome, greetings from Norway :)
@eggfryrice3088 жыл бұрын
The way you determine if a decomposition is dependency preserving is entirely wrong, you need to compute the clousure of F which is F+ and compute F_x's projection's clousure F_x+, also F_y's projection's closure F_y+, and check if (F_x+ U F_y+) = F+
@sunoharachan38208 жыл бұрын
+yun sheng You realize there's more than one way of checking for dependency preservation right?
@lifeofdarshit8 жыл бұрын
Helpful video, but one thing to remember is that Dependencies can be inferred so first we need to take the closure or there is a better algorithm to do that : To check if a dependency is preserved in a decomposition of R into R1, R2, …, Rn we apply the following test (with attribute closure done with respect to F) result = alpha while (changes to result) do for each Ri in the decomposition t = (result intersection Ri)+ union Ri result = result t If result contains all attributes in Beta, then the functional dependency alpha to beta is preserved. We apply the test on all dependencies in F to check if a decomposition is dependency preserving This procedure takes polynomial time, instead of the exponential time required to compute F+ and (F1 union F2 union … union Fn)+
@muhammadrizki6012 жыл бұрын
Very Helpful Video! The material is clearer in this video than my lecturer!
@1995aditi17 жыл бұрын
Awesome video sir..This helped me a lot in preparation of my University examination and is still helpful for Gate exam also..:)
@sherwincho37345 жыл бұрын
Thanks, this is really helpful to understand the Chase test.
@ryanpramadya58796 жыл бұрын
Thanks brooo.. u are awesome :) Love u from indonesia
@rebekah73725 жыл бұрын
This was amazingly helpful, thank you so much!!
@zubainmulla53579 жыл бұрын
too low volume
@adithyashah10883 ай бұрын
Use headphones
@corpsansorgans5 жыл бұрын
There is a BIG MISTAKE here regarding the section about dependency preservation. When checking which functional dependencies are preserved for R[i], you shouldn't look at the group of functional dependencies F but at the group of all of the functional dependencies that could be deduced from F, that is F+. "The restriction of F to R[i] is the set F[i] of all functional dependencies in F+ that include only attributes of R[i]" (Silberschatz, Korth, Sudarshan: Database System Concepts, p. 347)
@corpsansorgans5 жыл бұрын
E.g.: if F={A->B, B->C} and we have a decomposition R[1] = (A,C), R[2] = (A, B), then F1 = {A->C} since it could be deduced from F using the transitivity rule. Therefore looking just a F could be misleading.
@elo50253 жыл бұрын
thank you sir , this is the best explanation so far!
@ivan-the-l4 жыл бұрын
Thanks a lot! You are the best teacher
@Colouredeye37 ай бұрын
Amazing explanation!❤
@tonyaltamura6 жыл бұрын
Sorry but both algorithms aren't correct. In dep. preserving decomposition check you have to take care of the closures of the functional dependencies. In the second, there is an algorithm for lossless join and it is a little different. In both cases I'm pretty sure you can find examples where you get wrong results (not completely sure yet). Correct me if I'm wrong.
@kshitijsrivastava61484 жыл бұрын
Yes you are right he explained wrong.
@Therabidmonkey4 жыл бұрын
Thank you so much, This was extremely helpful.
@akanshagupta73652 ай бұрын
Thankyouu so much sir
@ginzeng41352 жыл бұрын
F(A->C, B->C, C->D,DE->C,CE->A), Decomposition is AD, AB, BE, CDE. How to know if this Decomposition lossless or not? I use your method, and got confused.Could u please help me to figure it out, thank you.
@shivamshah65798 жыл бұрын
Where is the videos where the FD are decomposed??? Can't understand without that :-(
@m.design7 жыл бұрын
can't getting answer for this. R(ABCDEG) and FD sets {AB->C, AC->B, AD->E, B->D, BC->A, E->G} D=( ABC, ACDE, ADG) answer as 'Lossless'
@rishikeshkumar44277 жыл бұрын
bro,it is lossless decomposition. r2=ACDE row get filled by alpha..hence lossless moreover it is not dependency preserving.
@m.design7 жыл бұрын
I got A,B,C,D,E filled with 'Alpha' but can't getting 'Alpha' in 'G'...help me!
@lethiminhngoc33914 жыл бұрын
Thank you, but I compute that: R3(A,B,D,E): B-->D, AB-->DE, AD-->E, BE-->D Pls help me to explain. Thank you!
@rahulkhandelwal87817 жыл бұрын
Dependency preserving algo might be wrong.
@Farahat12344 жыл бұрын
Why
@shatadruroychowdhury63196 жыл бұрын
Can you give me the solution if the BCNF decomposition of R(ABCDE) is Lossless or not, dependency preserving or not? FD: {AB-->C, C--D, D-->E, E-->A}
@watchmovie72887 жыл бұрын
Thanks, It is very helpful...
@ompatel72173 ай бұрын
what this algorithm is called
@pallavigaonkar17698 жыл бұрын
smart explanation about lossless property.. but there is an algorithm for lossless join better you explain with it..
@parameshchakraborty11467 жыл бұрын
Pallavi Gaonkar will u plz help me
@Unleashing_Creativity Жыл бұрын
@@parameshchakraborty1146Ab tk pallavi ke bacche ho gye honge 😂 daily help deti hogi
@kshitijsrivastava61484 жыл бұрын
I don't think that's a proper definition of dependency preserving. You said that if you see that if particular functional dependency does not belong to R1,R2,R3 and R4 then it is not Dependency preserving.But That's wrong we have to check closure for (R1 U R2 U R3 U R4) and if in that functional Dependency of R does not belong then it is not Dependency preserving.
@fatima-cq6qhАй бұрын
Mashallah
@nehairinirm78714 жыл бұрын
Thank u thank u..... Soooo much sir..
@nezmika29116 жыл бұрын
Thanks! Its helpful!
@anjanbasumatary15108 жыл бұрын
Sir why didn't you compute for----AC->B,AB->C,BC->A.......
@sumitkumar-rp6kn6 жыл бұрын
thanks...
@melikeb.bilgen20776 жыл бұрын
clearly explained. Thanks
@souviksarkar89757 жыл бұрын
thank you...that was very helpful
@priyanshpawar3322 жыл бұрын
thank you bro❣
@ginzeng41352 жыл бұрын
very well, dude
@jassikaur61809 жыл бұрын
thanks you so much for great video
@ssc__120212 жыл бұрын
Thank you❤️
@ohonaahona34046 жыл бұрын
please improve audio quality
@vishakhagautam14258 жыл бұрын
R(A,B,C,D) and have following fds A→B ,B→C, C→D,D→B. and the decomposition of R into (AB),(BC),(BD) is this lossless and dependency preserving. sir answer is not coming can u explain plz
@googlelife26967 жыл бұрын
Vishakha Gautam answer is lossy decomposition and not dependency preserving too which i have find.
@pallavichoudhary80807 жыл бұрын
Shelly Sood I think the answer is lossless
@harshvardhansrivastava57856 жыл бұрын
This decomposition is FD preserving and lossless. The method in the video is not 100% working.
@rbk09655 жыл бұрын
the answer is lossless because even if in 1 row we are getting all X then we can call it lossless so R1 is all X in this case
@shreyanalpet28584 ай бұрын
This video is 9 year old.... And now I am here 😮
@PriyanshuYadav-ct1lv Жыл бұрын
Thanks sir 🙏
@brilliantboy81475 жыл бұрын
thanks great knowledge
@syedarbaz37815 жыл бұрын
It is not always true for depndncy presvng
@tkdf136 жыл бұрын
Thank you!
@dazzykin3 жыл бұрын
What is alpha?
@mayursandbhor88247 жыл бұрын
Is loseless join and loseless decomposition same or its different?
@kevinaud64617 жыл бұрын
I believe a lossless decomposition is one in which you can perform a lossless join
@sahilsunny98857 жыл бұрын
Thank You
@parameshchakraborty11467 жыл бұрын
de -G why? assumption?
@Naveen-qz1rv8 жыл бұрын
thank you... helpful :)
@vishal48577 жыл бұрын
thnks
@nilanjanasarkar27557 жыл бұрын
please increase the sound level its not audible at alll!!!!!!!!