4.15 How to Identify Normal Form Practice Question Part-3

[@KNOWLEDGEGATE_kg]

For complete DBMS, check out this video: kzbin.info/www/bejne/j4PRm3qbhsemfrM

[@shubhammandwe1528]

Had a nice experience, appreciate your way of teaching. Teachers like you are rare. Thank you for making even a non CS student (me) understand DBMS basics, especially normalisation.

[@KNOWLEDGEGATE_kg]

Hi Shubham Thank u for your kind word.We will upload many topic every week.Subscribe us to get all future updates on lectures.

[@saumyojitdas4212]

@@KNOWLEDGEGATE_kg please upload videos on os and computer architecture and computer organisation . These come in Amcat exams🇮🇳

[@lintaolu8202]

This unbelievable sir teaches me more than my professor!

[@KNOWLEDGEGATE_kg]

🔥Complete DBMS by Sanchit Sir: tiny.cc/DBMS_Sanchitsir_kg 🔥🔥All Computer Science Subjects by Sanchit Sir: tiny.cc/CSbundle_dbms_kg

[@shifasamreen8743]

I had been slogging with normalization concepts for a while.Your videos are a life saver! Simple, compact, clear and crisp! :D Eagerly looking forward for more lectures. Thank you so much Sir :)

[@abhisheklolage]

Very helpful lectures. Suggestion: Use LMR method to identify candidate keys. L = attrs appearing on left sides of FD (a must in a CK), M = attrs appearing on both sides of FD (may or may not be in CK -- go for combinations), R = attrs appearing on right sides of FD (cannot appear in CK -- this may reduce work). Just include attrs that are not in any FD by default in the CK.

[@RohIt_bhopali]

Thank you so much sir, words merely aren't enough for the tough concepts that you have made very clear for us! Kindly make videos on B,B+ trees and also upload video on ER to Relational table translation.

[@manohargupta7443]

i am able to solve any gate question all because of you sir. Thank you so much. :)

[@bharatbhushan2197]

Sir normal form aapki wajah se hi identify karna seekha hai....thanks a lot sir. God bless you always

[@sweetnishu005]

for the last question u said there is no clear cut case of partial dependency....as i can see for D-->B it is evident that D is a part of candidate key & B is a prime attribute as well & hence satisfies for 2NF. Please, have a look and reply :)

[@mohammadaazamchhipa3813]

saurabh sinha kehna kya chahte ho...he already said the table is in 2nf.

[@Ankit-we8ym]

sir will you please teach lossy and lossless decomposition and dependency preserving

[@shashikalaraju5769]

I will always remember I owe you one. Thank you so much for your work, generosity, love and hard-work. We love you too. Someday I could thank you personally. Now I just want to know, is it safe to say, pk---->pk nk---->nk pk = prime key nk = non prime key Can both be ignored while deciding the NF?

[@binodkumar-wy6th]

Nk=====Nk show transative property we can not ignore

[@apoorvaka4247]

Very Clear Explanation with more Practice problems, Thank you very much.

[@aditibose3486]

Sir, these videos are really very helpful and I find them easy to understand the concepts so please upload 4NF and 5NF videos too.

[@fifafreakk]

so so clear! amazing sir keep it up

[@nikhilnaudiyal7082]

Sir aap great ho

[@sheffalibarola4290]

and all we need is a teacher like u...thank u so much

[@nadianipa6402]

awesome tutorial... really a great work. u washed away all my confusion.Thank you sir :)

[@abhishekshivgan1884]

Thank you very much sir 🙏

[@KNOWLEDGEGATE_kg]

You're most welcome Abhishek ❤️❤️.. Keep learning and supporting !! Do visit our website www.knowledgegate.in for more courses & contents !!

[@preeti93100]

Superb explanation...

[@manigandankarthick]

Sir will u upload topics in COA ?? (pipelines ., etc )

[@shikhashah7722]

Thank You sir for the videos. They are helping me in this subject a lot. I have one question and i am not able to solve it. Request you to please help me in solving this question :- R (A,B,C,D,E,F) A --> BC BD --> E E --> F Convert this to highest normal form.

[@akshatjain9762]

Thank u so much , now itz very easy 4 me 2 solve

[@shalinimaurya9662]

Thnku so much sir.....ur explainations method is very gud.

[@preetisaharan8907]

Thanks alot Sir .These videos have crystal cleared my concepts and developed my interest in the subject . Sir , your videos are really helping many confused minds. Looking forward to subsequent videos.

[@nadspro07]

Sir, I am really thankful and truly appreciate all the efforts you have put in to make these videos. Thank you, Sir. You are amazing. This video series really helped me to understand some concepts. Thank you.

[@vendetta3953]

Sir, is it necessary to break all the dependencies(using decomposition rule) which have 2 or more attributes on Right side and make them have a singleton attribute on Right side? Like you've done this in question 4 in this video while checking for 2NF for the dependency A->BC? couldn't it be done for BC being considered a single unit itself without checking for A->B and A->C separately? Plus, brilliant channel, keep it up sir! :)

[@ashutoshrai5342]

Awesome sir your lectures are very good.

[@raghuvanshmani4257]

very good explanation

[@shabanaansari2317]

sir, thanks for ur wonderful videos but pls... sir, pls recheck the question 6, i m getting confuse.... here BC->DEF, D is a prime so here prove that not in 2nd NF, after dat D->B , here also B is prime n also D is a part of candidate key... so 3 possibilities prove dat not is 2nd NF.... pls correct me if i'm wrong........

[@pray307]

Agr hm 2NF me check krte hai to hm sbse pehle check krte h ki functional dependencies jo question m hme di gyi h to vo khi partial dependency to nhi hai jisme hmara alpha prime atteibutes hota h or beta non-prime agr partial dependency exist krti h f.d. me to fir vo 2NF me nhi hoga isliye partial dependency exist n kr rhi to ye 2NF me hai

[@shubhamjaiswal3743]

A jornal uses (Volume,Number,StartPage,EndPage,Title,Year,Price) Primary Key is(Volume,Number,StartPage,EndPage) FD ARE- Volume,,Numaber,StartPage,EndPage-->Title Volume,Number-->Year Volume,,Numaber,StartPage,EndPage-->Price It is in--- INF or 2NF Sir Please Solve this Because according to me it is 1 NF

[@anshgyl]

It is in 1NF as the second dependency is a partial dependency because "Volume,Number" is a subset of the key.

[@bimalashrestha6761]

SIR , please post Tabular form examples.

[@chintanpatel3400]

Ur teaching is excellent sir..✌👌👍.. 100/100 ✔

[@chinthasrikar7212]

Wonderful way of presenting the information which makes us understand the concept very well!Thank you sir!

[@ranugupta673]

awesome video's......... sir plz make video's on sql query

[@dhruvkaith3160]

Thank you sir

[@KNOWLEDGEGATE_kg]

You're welcome Dhruv !! keep learning !!

[@mayankchadha5915]

sir pls solve the below ques: The best normal form of relaion schema R(A,B,C,D) with AB->C,AB->B,C->A,D->B what will be the candidate keys and in which normal form the above table is? Acc to me AB,CD,AD,BC are 4 keys and table is in 3NF correct me if I'm wrong

[@sgurjeet99]

AD,CD 1NF

[@st-yi4oi]

great job!! sir

[@harpreetv761]

Great sir

[@ankitsharma-bt8zr]

In 6th ques. c is non prime attribute so AB->C is partial dependency so not in 2nd NF it is still in 1st NF

[@Neetigya]

But AB is candidate key and to violate 2nf, a prime attribute must give non prime. Hence no error in that question.

[@ronitkumar1838]

thank you..

[@_iam_ami_]

Thank you sir.. dbms lectures are really helpful from exam point of view and for concept clearance.

[@pray307]

Sir myself Prabhat Yadav....... Sir is video series me aapne btaya 6th question me ki ye 1NF me nhi h jbki last video Part 4.12 m aapne iska same question no. 7 btaya tha ki vo 1NF m hai........ Jb sir last video me DC-> AE do me toot skta h jaise DC->A and DC-> E waise hi aapne is video series 4.13 m kyu nhi ise toda 2 parts me i.e., AB->CD ko AB->C and AB->D agr ye b do parts m toot ta hai to ye b fir 1NF m aayega

[@ahmedbaig5121]

your the man

[@mdashrafulhaque1247]

Thank you sir. Your lectures are awesome..thanks a lot

[@prerna3404]

Please upload lectures on computer organnisation as well

[@jaydeepgondaliya2853]

Hello sir, I have learned a lot from your videos and i like the way you are teaching point to point but why i am unable to find 15th video?

[@ridhusaini2365]

sir can you please help me for following question: R(A,B,C,D,E) Given FD'S AB->C BC->D CD->E What normal form is for this relation?

[@anshgyl]

Key: AB Prime Attributes: A, B Non-Prime Attributes: C, D, E BCNF AB --> C (LHS is Super-key) BC --> D (LHS is NOT Super-key) Hence, NOT in BCNF 3NF AB --> C (LHS is super-key) BC --> D (LHS is NOT super-key and RHS is NOT prime) Hence, NOT in BCNF 2NF AB --> C (AB is key, so no problem) BC --> D (BC is not a subset of any key, so no problem) CD --> E (CD is not a subset of any key, so no problem) Hence the above relation is in 2NF

[@shubhamtariyal439]

@@anshgyl that's some time consuming answer to type.

[@udarachinthaka2766]

thank you so much sir,respect lot

[@henafatma5101]

Thanks a lot for your help

[@srishtidhariwal8325]

I have a doubt. Third normal form is inadequate in situations where the relation a) Has multiple candidate keys b) Has candidate keys that are composite c) Has overlapped candidate keys d) None of the above What should be the answer?

[@preetyparamanick]

sir in last question there is one more candidate key which is ACD and if ACD is a candidate key the second dependency which is CD->EF is a case of partial dependency so that schema is also in 1NF not in 2NF. Please Correct me if i am wrong @KNOWLEDGE GATE .

[@psuthar08]

Sir .. Ur videos r superb .. Thnku.. N plz upload lect of serializable and view serializable numericals

[@livelife9229]

Really well explained !

[@mayankchadha5915]

Sir, pls solve sir pls solve the below ques: The best normal form of relaion schema R(A,B,C,D) with AB->C,AB->B,C->A,D->B what will be the candidate keys and in which normal form the above table is? Acc to me AB,CD,AD,BC are 4 keys and table is in 3NF correct me if I'm wrong

[@vendetta3953]

As no other attribute determines D so D will always be a part of Candidate key, the keys here are AD, CD.

[@ashieshhh]

Mayank Chadha the keys are AD and AC also the dependency D -> B , being Partial Dependency the scema is in 1NF

[@vedantkumar007]

seriously damn good, better than my American professors

[@narendraparmar1631]

Thanks sir ji😆

[@madhavsingla1792]

cleared ol my doubts regarding NFS :)

[@shivamgarg3069]

Sir! You are too good.

[@rfreedom6273]

Hello Sir your video is superb,,,,,,,,,,,,,,,,,,,,,,,,,,, But please set all Video by Sequential Series ,,,, i Can not find YOUR PART Number 15

[@egehandogan3421]

Harikasın kral

[@la55u]

In the 4th example why did we check for further candidate keys and didn"t stop when we found all 2 attribute keys? I mean how do we know when we have to stop searching?

[@vasanthraj5752]

at final qus how is it possible AD also a candidate key? there is only two AB and ABC.Am i right sir?

[@narendraparmar1631]

thanks for this video sir ji😆

[@AshishSingh-en1lu]

in question no 2 . we can have (ACEH) and (BCEH) as candidate key , i think sir u have missed them . !

[@pulkitagrawal1220]

no they will be super key not candidate key

[@shahidshaikh7438]

in 4th example . for 2NF it also should be that all non key attributes must depend on key attribute. now c was not depending on A so how can it be 2NF

[@PoojaVerma-mc3lh]

sir in Q 5. how is it in 2NF?? A->B here A is a prime attribute while B is a non prime. so it's a case of partial dependency.

[@kartikanand2701]

my also same doubt why sir it satifying 2nd noraml form condition sir

[@anshgyl]

It is not a case of partial dependency. In the case where a single attribute is the candidate key, partial dependency can never occur. Partial dependency means that a non prime attribute is derived from a part of a candidate key and not the whole key. But in this case "A" itself is the candidate key.

[@psuthar08]

Sir , I have 1 question regarding normalization ... Y don't you check trivial dependency in BCNF and 3NF??plzz reply soon..

[@abhishekchaudhary453]

In the last Ques...ACD is also a Candidate key

[@chintanpatel3400]

Sir you taught to find a candidate key we have to see essential attribute or we can say the attribute which don't have any incoming edge but what suppose to do if I get all attribute with incoming edge.?

[@anshgyl]

In that scenario we need to look at all the combinations as explained by sir in previous videos.

[@abhishekmadaan8980]

sir ur videos r really helpful. plz upload a video fr lossless and lossy join decomposition thanx :)

[@pintusaini6838]

why are we checking for 3 attribute CK when we got 2 attribute CK, CK is minimum attribute key. Right na?

[@ARUNKUMARNATHGAU-C-]

please upload few video on decomposition of relation to bcnf & 3nf

[@pavankulkarni9756]

Sir when will you be upload video on decomposition of tables in desired normal form? Sir I am really in need of it, My BTech Exam starting in about 15 days from now , Sir it would be extremely helpful if you upload a video. I am really thankful for such a informative video provide by you till date. Thanks, Regards.

[@pavankulkarni9756]

OK sir . Humble request , one last video you upload sir . This will boost up-to 10 marks in my university exam . Thank you sir for your reply

[@lakhanmahto2651]

Sir I like ur all video and plz upload DDM DML and DCL

[@kuljitray2803]

Sir plz upload videos on Multivalued Dependency 4nf,5nf....

[@mrsumitraaz]

Can u tell me the maximum normal form in which this relation is in? R(ABCD) {AB->C, BC->D}.....

[@anshgyl]

Key: AB Prime Attributes: A,B Non-prime Attributes: C,D BCNF AB --> C (LHS is a super-key) BC --> D (LHS is NOT a super-key) Hence, NOT in BCNF 3NF AB --> C (LHS is a super-key) BC --> D (LHS is NOT a super-key, and RHS is NOT a prime attribute) Hence, NOT 3NF 2NF AB --> C (AB is a key) BC --> D (BC is not a subset of the key) Hence the maximum normal form is 2NF

[@vishalalhade2376]

Sir,In which normal form relation would if it is not in BCNF and it not contains transitive dependency but contains partial dependency

[@SumanSharma-zh6mv]

if it is not in bcnf and does not contain transitive dependency but contains partial dependency, it must be in 1nf

[@AnilGupta-iv1rz]

Sir, it would be very helpful if you could upload videos on relational algebra and ER model for DBMS.

[@c--rayashreemondal1913]

Sir, plz upload video on lossy and lossless join decomposition

[@KNOWLEDGEGATE_kg]

Hi Surjani, Thank you so much for showing your interest, We are working on LOSSY and LOSSLESS decomposition. Please subscribe our channel to follow all updates. :)

[@anmolgoyal6010]

There is a relation named Emp_Dept(SSN, Ename, Bdate, Address, Dnumber, Dname, Dmgrssn), with the set of functional dependencies as SSN -> Ename, Bdate, Address Dnumber -> Dname, Dmgrssn SSN is the key attribute for the relation In which normal form the relation is? Normalize it to the next higher normal form. sir this question is confusing me lot

[@JaskaranSingh-hw5jf]

at 9.20 if CEH is not a ck then shouldn't we make furthur combinations of CEH with A,B,D??

[@anuj4802]

in last question the FD is AB->CD the 2nd normal form is not exist because D is prime attribute and C is non prime attribute so it is follow the partial dependecy .

[@PrateekGupta26]

It is in 2NF because AB is not a partial key... here both C & D are depending on entire key not just part of key

[@salmananjum-thestoryteller1515]

for the 2nd case where BC -> DEF here E is non-prime and it is depending on BC where B is the part of key AB. a/c to my understanding this is partial dependency . please clarify

[@anubhavsingh9049]

Sir as you said that in 2NF, the condition is that it is partialy dependency but i question when you find partial dependency then tell to write 1NF why?

[@anshgyl]

Because we can see that the relation doesn't have any multi-valued attribute hence it is always in 1NF even if any other normal form fails.

[@Epic_priya]

in ques 6 y to work with the ac+,ae+,af+..................we hv not done for d earlier questions 1,2,3,4,5

[@KNOWLEDGEGATE_kg]

because here a is essential attribute

[@Epic_priya]

thank u sir ........plz video in 4NF and 5NF

[@gunamoniphukan5074]

please include Relation algebra and SQL part also

[@gunamoniphukan5074]

thanks sir!!!

[@salmananjum-thestoryteller1515]

In last question, 3rd functional dependency " BC -> DEF" let me decompose it for simplicity BC - > E here E is non prime and it is depending on BC. where B is part of a key (AB) so it is the case of functional dependency! pls clarify!

[@anshgyl]

This is not the case of partial dependency as the functional dependency says "BC --> E" and NOT "B --> E". "B" is a subset of a key but "BC" is not a part of any key. For partial dependency the whole left side should be a subset of a key, hence this is not a case of partial dependency, therefore, the table is in 2NF.

[@salmananjum-thestoryteller1515]

Thanks for replying! ok I've got your point but the problem with Partial Dependencies due to which they are prohibited in RDBs is that "It is allowed for one attribute of Primary Key/Candidate Key to become Null both both can't become Null at the same point of time & other than Primary Key every member can become Null " so in this case suppose if B = null and C = null " how would u find the value of E?

[@anshgyl]

AB --> CD CD --> EF

[@parvathymenon9922]

I cannot thank you enough

[@mansiagnihotri7672]

sir plz post a video regarding wid writing relational expression for give db queries

[@ramshah7853]

Sir question no.2 has one more c.k. which is DCEH so question 2 is in 3nf.

[@RohIt_bhopali]

DCEH can't be a CK since you see, DEH already a ck and if DCEH a ck then no proper subset of it can be a key, but here DEH which is a proper subset contradicting our fact that DCEH a key so here DCEH not a CK. Hope it clears your doubt.

[@rubyyadav471]

Sir, Please provide video of 4NF and 5NF

[@relax.moods.sounds]

please do R(ABCD) A->BCD, B->ACD, D->B

[@anshgyl]

Key: A, B, D Prime: A, B, D Non-Prime: C BCNF A --> BCD (LHS is super-key, hence valid) B --> ACD (LHS is super-key, hence valid) D --> B (LHS is super-key, hence valid) Therefore the above relation is in BCNF

[@premshankar5967]

please upload videos on computer networks and algorithms sir. I am very confused

[@hinasingh1728]

Hello sir in this series of 1st question u said it will hv 1 candidate key bt according to me it will have two plz clear d confusion.

[@faheemsajaad3361]

Hina Singh there is only 1 key CE as no edges lead to them.

[@foodandnaturelover]

Sir plz upload a video of relational calculus and it's queries ...this topic is very confusing.🙏🙏

[@chintanpatel3400]

and in video 14.. question 4 why we check the other possibilities with length of 3 key even (from failure of lenght 2 candidate key) bcz the candidate key is minimul super key so we got length 2 candidate key why we go for lenght 3 from failures from them..??

[@anshgyl]

Minimul means irreducible. Candidate key is that key from which if we remove 1 attribute it no more remains a candidate key. Hence it doesn't matter if another key has a smaller length. The only thing that matters is that the set attributes should be able to derive all other attributes and should be irreducible.

[@SHUBHAMGOELBCE]

Sir, thanks for the awesome videos. Sir, can u Plzz upload videos on conversions of normal forms as soon as possible. It's urgent

[@neerajchauhan3790]

Why it was urgent?🤣😅

[@vidyush96]

if a non prime attribute is finding a prime attribute will it be in 3nf? ???

[@aneekroy4025]

Sir would you please add videos on Query Processing

[@KNOWLEDGEGATE_kg]

we shall :)

[@aneekroy4025]

KNOWLEDGE GATE Thanks a lot Sir Query Processing and Optimization really bugs me, it would be great if you would add videos on them.

[@Anytime4yo]

sir please posted our results too..so that i am able to find my mistake.. and thanks sir for your response