Finding Number of CANDIDATE KEY | DBMS

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worldwide*

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Right

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Please teach the full lectures in english sir.

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Hi Abdullah, thank you so much for your kind and delightful words. We are glad that we could help you. :) :)

[@rohitsoni936]

CORRECTION: In the last problem ( question given as assignment), the very last F.D will be F --> EG not A --> EG.

[@abhinavsingh9175]

ok

[@abhinavsingh9175]

if this is correct then answer is 4 candidate keys

[@yashhingu7955]

If here a-->eg than answer is 2

[@debosmitadeb_]

@@yashhingu7955 yes

[@satyamgupta1446]

ED,AD,BD,FD are CK

[@Raghavcric]

Oh Bhai mere! Tera Lakh Lakh Shukar Hai! I am studying in Australia and using your lectures to understand what my 60 lecture slides couldn't explain. God bless India! and God Bless you my friend! You are an awesome teacher! Tooo Good! Come to Australia and you will make Millions hahah

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Hail India!

[@AdityaSingh-os6jf]

Raghav Sharma you

[@hareshparab7808]

wtf

[@wassaufkhalid8593]

Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F = {CH -> G, A -> BC, B -> CFH, E -> A, F -> EG} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R. How many candidate keys does the relation R have? (A) 3 (B) 4 (C) 5 (D) 6 here is the correct question

[@varun_639]

(B) = 4

[@hanusaran5076]

AD, ED, FD, BD

[@VisaliniKumaraswamy]

In 6th pbm - There is a mistake - Last FD is not A ->EG but it is F->EG The reason behind 4 candidate keys is that, if any attribute does not have incoming edge, then omit it as unreachable... Now by omitting D, we get, A+, B+, E+, F+.... No need of D attribute to be reached... So totally we get 4 Candidate keys...

[@renjithkurup9874]

right

[@satyakibose8402]

Did you solve the question? Could you please share the solution?

[@DivyaSharma-vb5ei]

Yes you are right! Four candidate keys are AD, BD, ED, FD

[@devanshmaurya9928]

Correct. According to the FDs given in the question, candidate keys will be only two.

[@kastriotblakaj168]

hei can we say that and A is an candidate key too ?

[@rajatsingh2094]

sir i think that there is only two candidate keys according to the question which is DE and DA because you cannot find D with the help of any other attribute and you cannot find E without A or vice-versa so any other combination apart from this must contain DE or DA in it,so no other combination will become the candidate keys in this relation

[@cutiepie6604]

Yes bro

[@sonorousgaming7202]

in last question, last relation is f>eg, not a>eg

[@vinayaksharma-ys3ip]

💯💯👍best Teacher......Sanchit Sir was born to be a teacher!!

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Thanks a lot dear Vinayak..Keep learning & supporting !! Do visit our website www.knowledgegate.in for more courses & contents !!

[@rishabhkumargupta1730]

You taught tough things in much easiest way.... I like your the way you teaching...

[@ramyamanmadhan5688]

truly blessed teacher..God bless you more and more

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WoW! Just Wow! Sir, We really need teachers like you! Normally watching you tube videos make me sleep ! But your efforts are so sound that I can't stop myself from focusing !!

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Hi Shrehal, Thank you so much for showing gratitude towards our efforts,it really means a lot to us :) Thanks for your support brother :) God bless you!

[@redrose5950]

Superb explanation Sir, thanks alot.

[@KNOWLEDGEGATE_kg]

You're welcome.. Keep learning and supporting !!

[@harivijaysinghchandel2542]

Hi Sanchit, the way you are teaching is awesome. I became fan of your teaching after watching these videos and also sharing this link to my friends and they are thanking me for the link. so I would like to thank you for this useful work. keep it up. Hope we will get more videos very soon with respect to gate 2017 exam.

[@harivijaysinghchandel2542]

:) thanks bhai

[@shubhamanurag9449]

Best video for dbms. Cleared my concepts.thanks sir ji!

[@digvijaygautam8369]

Seriously hats off to ur hard work. Amazing videos. keep it up.

[@naillahgul2985]

Frm the edge diag its evident that d is not having any edge n also not any incoming edge,so d is required in any case to find all other attributes ,also d is not mentioned in any functional dependency,so we have to pair it with a,b,c,e,f,g,h ,,,,find the closure of all these u will get answers as ad,bd,de,df,,, any other combo other than this wud not be minimal.

[@phoolchandra8456]

There is one more dependency in question no. 6 which is F->EG then it has four candidate keys AD, BD, ED and FD. Otherwise, it has only two c.k. AD and ED.

[@dharinivijayan5101]

U said it has 2 but ad is not a key

[@phoolchandra3710]

@@dharinivijayan5101 Yup, you are right, I am sorry. there is only one c.k on a given condition ED because we can not derive E from AD.

[@srinadhj2638]

in example 4 @12:28 R(A,B,C,D,E) from B we can find C,D,E and from D we can find A so why is B not a candidate key ?????

[@ankitsharma8221]

how can you find c,d,e from b if the FDs are AB->CD,D->A,BC->DE? had it been any FD having the determinant only 'B' and not 'BC' your assumption would be right. Go through the timestamp once again :)

[@sheheryaryousaf3664]

I got AD and ED as candidate key.

[@pranjaljhala4305]

ad se e nahi niklega beta

[@abhinavsingh9175]

@@pranjaljhala4305 bhai AD se ni to A se EG to nikal skta hai na

[@ayushishukla4568]

So 2 is the correct answer?

[@rkrox99]

@@ayushishukla4568 AD BD ED FD is answer.(i.e 4 candidate keys). www.geeksforgeeks.org/gate-gate-cs-2013-question-54/

[@pavanthalla465]

@@rkrox99 wrong in the question of geeksforgeeks last fd is F->EG but the above sir given question is last fd is A->EG

[@shahvanshika254]

Awesome sir....I understand better than my college lecturer plz try to upload new video further chapter of dbms...

[@vanisameera9439]

Your smile is so sweet sir. Thank you for the lectures!

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Thank you Vani.. keep learning and supporting !!

[@aabidwani7441]

very diverse approach of pedagogical principle, hats off to u

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Sir you are best and you are every thing for b.tech students☺️☺️

[@NasweefKunjippa]

you are excellent Sir! ,, great message with in 20 minutes....thank you...

[@shahijakrishna6130]

Sir Ur way of taking class s very easy thank u

[@AjayJaiswal-gp6lf]

thanks alot sir , these is very helpfull for my GATE preparation and i really apreciate your hard work!!

[@ajitsafeway]

U r my saviour ... since last 3 days i m struggling to understand whole normalisation nd every-time i extract wrong candidate key .. after watching ths now i m confident to get correct candidate key ... many thnx bhai

[@deepakkumarsingh3484]

great respect for you. you are very good in explaining. Keep up the good work buddy.

[@Adarshsahni1]

thanks a lot sir , your lectures help me a lot...... you make it very easy to understand...

[@ayushmaansahejpal3025]

Sir in question 6 there is no path to reach D and E so it should only be the candidate key .. if we include DE in other attributes it will become Superkey... please explain how to get 4 keys!

[@MultiNaruto92]

thank you!! Because of you I got placed....

[@husainmadraswala4072]

paisa de usko

[@priyankabhagat8924]

Sir, You are awesome! :) You are the reason I get continuously motivated for GATE without coaching. Great work.

[@groovity5718]

huwa clear?

[@vrindavan3142]

good evening sir in 5th problum there is y=R,wx=R,wy=R,xy=R,xz=R, are candidate keys .Am i right sir

[@asharawat5908]

D has no incoming edge or outgoing edge.. how u r getting candidate key.. we cant find all other attribute.

[@abhinavsingh9175]

it is not candidate key but it will be part of candidate key like DA and DE

[@souravdutta7120]

Sir you are doing a wonderful job of giving education to all in the best explained way possible.hats off

[@geodesic616]

Very good lectures sir!

[@sshinge0777]

given R {ABCD} and a set F of functional dependencies on R given as F{AB-C,AB-D,C-A,D-B}.find ant two candidate keys of R given as F={AB-C,AB-D,C-A,D-B}. find any two candidate keys of R show each step.in what normal from is? justify plz tell me sir what is answer ths question

[@relax.moods.sounds]

last FD must be F->EG to get the answer 4 which are (AD,BD,ED,FD)

[@abhishekahlawat8772]

you `are very intelligent bro, i'm amazed

[@relax.moods.sounds]

well u must be

[@udaysagar9017]

Could you explain how BD could be a candidate key?

[@CProgrammingTutorialManish]

Why BD AND FD

[@paramasarbajna7602]

from BD-> we get BDCFH ,from CH-> we get G, from F-> we get EG and from E we get A; so closer of BD=ABCDEFGH

[@siddharthsingh371]

sir, to be honest i would say if we can find teachers like you in our college too, then there would be no doubts in any of the student's mind. you teach so well. your way of explaining and teaching the topic is amazing that even the tough topic appears to be easy when you teach. i truly appreciate your teaching skills.

[@nareshdhakad5883]

Sir love you 💝💝 your teaching is wonderful 👍👍 and ans is 2 ❤❤❤

[@DipPodderOfficial]

you are a genius in teaching 😀

[@utpalkumar1421]

Sir, Please provide solution of 6th Que

[@lalitverma5248]

Q6 giving only have one Candidate key DE, the 2013 Gate question includes F->EG dependency besides given ones which have 4 candidate key.

[@raghvendrarajak4839]

Lot of the thanks sir for explaining with so pretty way i compeletly understand what you explained .

[@ckumar2028]

Sanchit sir q no (1) jab aap question no one me B->For then F->GH same think in question no three clouser (CE) jab C->ADE then why not D->B in also fourth question Clouser (BE) B-> CDE then why not D->A also in five question Clouser (W)->Y then Y -> XZ nahi le rahe h ,then Y->XZ and Z->W kyun le rahe hain aur phir Clouser(WX) W->Y then Y-> Z kyun aise me bhi Clouser (WZ) bhi consider Kar sakte the. Finally first question is wrong or other lots of confusion So last question can't be solve by your lecture ..mughe ummid Hai sir reply jarur milna chahia....

[@arindamsarkar672]

can you please give a video about foreign key and every types of joins? that would so helpful to us

[@ridakhan7936]

sir your videos was too good. and i m very inspire from your videos.your way of lecture is very clearly for students.plzz upload sir relational algebra,and sql ...i will wait thanku.from rida khan.

[@shambhawimnit1611]

LAST FD is F->EG and i am getting the following keys AD,ED,FD,BD

[@vaishnavidongare4408]

Me too

[@archanabisht1457]

Superb explanation

[@KNOWLEDGEGATE_kg]

Thank you 🙂

[@INNOVATIVEAPPROACH]

sir after to much scratching head solving again again i found also 2 candidiate keys AD and ED thats its . please let me know your view too ...waiting

[@sreeparnadas2929]

Sir your videos are very good.. you teach very well.. aap ke bharose pe hi mai apni end sem exam dene jaa rhi hoon becoz I dont have the class notes properly :P But I wish there were videos on ER diagram, Indexing, B-Tree & B+ Tree, SQL and Relational Algebra too :( Hopefully you'll be updating that by mid of next year :)

[@utkarsh22smart]

Sir in the part 6 of the video series, you calculate A as a super key and candidate key, there you didn't talked about A+ , but here you say that for finding a candidate key you need a closure set, please correct the argument.

[@priyeshkumar4814]

Small question: On R(A,B,C,D) If we are able to find Keys as A, BD, CD, as per definition CK should be minimal set then why BD and CD are considered as candidate key?

[@sauhardjain9847]

Hi Sanchit, thanks for the superb lectures.. but i have one question here. In previous lecture while finding a candidate key you explained that we should not consider other FD or closure will not work. But here you are directly using closures to find a CK. Also here many keys are not super key because they are not covering all attributes so how they can become a Candidate key. Please clear this..

[@bittupaul100]

great lecture...really helpful

[@gulrezeqbal3048]

I have no words to describe to yourself !! Because you r best !

[@vikassuryawanshi5909]

thanks very much sir, all these videos are very heplfull

[@sonu_artist743]

upload more sir... this all video is very helpfull

[@sunilchaudhary3195]

sir when you going to upload more viedos on DBMS please uploads more viedos.. great teaching ..

[@KNOWLEDGEGATE_kg]

thanks sunil bhai for showing your interest, will be uploading more videos soon...

[@abhishekpatel6126]

T

[@varinderjitbindra7220]

sir your videos are very helpful. thanks for the quality lectures. upload more on algorithms

[@becomingpure2687]

Very nice Better understood Thank u

[@poorviasthana1587]

Sir, on what basis do we try random combinations... like in 5th, you tried wx.. etc..

[@saurabh2920]

sir in q? 6 A--->EG is wrong bcz i get 2 candidate key if i put F instead of A [F--->EG] then i get 4 candidate key

[@himanshugta1724]

In the last question given as an assignment, there are no incoming or outgoing arrows to D. So how can there be a candidate key at all? Because any closure will not be able to reach D. Please someone explain!

[@avijitbhowmik8545]

Thank you sir, 18.o8 ,It is F-->EG in Q.No.-6,then only we get 4 candidate key.

[@kpodjiemmanuel4319]

you are doing very good. i wish you have a completely different session in English only. combination of two languages makes it difficult to follow.

[@PhoneGeeks]

the last question you gave as a homework is work as the last value you wrote is "A implies EG" which gives only two candidate keys and their should be "F implies EG" which gives 4 candidate keys. Sir try to fix it

[@arpitgupta4013]

tnx bhai

[@muditmehrotra4517]

In last question you are missing one FD ie. f->EG , plz correct and reupload sir , and you are an amazing teacher.

[@tourwithkausik7730]

sir, E has no incoming edge and D has no dependency so,E,D is essential key..closure of ED:{A,B,C,D,E,F,G,H},.we found all the attribute from this..so ED is candidate key.how can you find 4 candidate key? help me sir

[@KNOWLEDGEGATE_kg]

hi, go through the video again, you will understand..

[@pallabsarkar5765]

The last correct question is: CH>G, A>BC, B>CFH , E>A , F>EG answer: Here I found essential attribute is D,but it is not a key.so i check closure for AD,BD,CD,ED,FD,GD,HD and among these AD , BD ,ED and FD i found candidate keys.so answer is 4.

[@DHANANJAYGROVER]

can u please explain why ad and bd are candidate keys as e and a are super keys as all other attributes can be find by these e and d so.....please tell

[@tech_Animus]

Only 2 AD,ED

[@tech_Animus]

BD determine BDCFHG so it's not a candidate key

[@tech_Animus]

And FD never possible.

[@tech_Animus]

​@@DHANANJAYGROVERyes, if Functional dependency is F determine EG. Not A determine EG

[@yash6680]

sir.. you're doing a great job 😇😇🤗🤗😍😍😘😘 god bless you.. keep it up..

[@abhishekgupta545]

In video find no of cndidate keys sir u wrote A-EG as last dependency in assignmnt Q and for this 2 candidate keys are there i solved and when u were solving tht Q in Practise problem on normalisation part 14 u wrote F-EG as last dependency and for this i solved answer is 4 candidate keys are there Many Students got confused sir because of this confusion

[@sujaysaha4308]

Sir Thanks for fantastic videos and clearing Concepts#Sir where is the Video of the Answer of the 6th Question?

[@afeefarafeeque6365]

awesome videos seriously jz enjoying dbms :) i jz wanna say instead of drawing edge diagram we can also jz have a look on RHS , which attribute is missing.... more time will be saved :)

[@MMadni-ke9nw]

Excellent sir and Thanks a lot #From Pakistan

[@sahid197]

in Q3 how can CD be a candidate key since C alone cannot determine any attribute other than C itself. It requires BC together

[@aditiranjan303]

Nice explanation sir

[@7guitarlover]

You Are Awesome !!! Eagerly Waiting for normalization vids !!! =)

[@gopesh97]

for 4 candidate keys , i think the last fd should be either H-> EG or F->EG

[@pradeepkumarsingh6964]

thank u sir, u really made us have a clear vision regarding such a important topic

[@Anytime4yo]

easy way to find the candidate key .thanq very much sir

[@prakashkrsingh0]

Sir,aap offline class Lete ho kya????

[@changlife7581]

Great video!! Thanks a lot!!!

[@harleenkaur4048]

In 5th Question , WZX is also a candidate key . Its proper subsets are not candidate keys and its closure is entire R

[@ankitsharma8221]

WX is a proper subset of WZX and it is a candidate key.

[@bunalucky8464]

sir thank you....i got 2 candidate keys------->AD,DE..... BUT sir you're saying 4 C.K,,,,,,,YES,it would be true if F----->EG insted of A----->EG.....

[@pranshibhardwaj9197]

R(A,B,C,D,E,F,G,H) CH->G A->BC B->CFH E->A first candidate key is DE.after that I make closure to other attribute than I found ED candidate key. so my answer is 2 candidate keys.

[@kahmad3428]

agree, but dnt know why he said 4 :@

[@sukhbirsingh8053]

GATE | GATE CS 2013 | Question 54 Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F = {CH -> G, A -> BC, B -> CFH, E -> A, F -> EG} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R. sir gave wrong question. he wrote the last Functional dependency as A->EG, but the actual function was F->EG..now solve, u will get 4 candidate keys

[@deepakkumarprajapati8823]

no DE is already a candidate key so DEB is a super key but not a candidate key

[@arindamsarkar672]

thank you for clarifying this mistake.,it helps me out to right answer😋

[@avkaransingh1015]

right

[@dhavisachan2672]

sir , in previous video c-->ad is not a super key coz c+ doesn't enclose all elements but here in 1st eg u made ab a candidate key i.e. a super key though it's closure do not enclose all elements as explained in previous video . plz explain it sir

[@laahiridhavala86]

Is there any vdo describing how to find a candidate key

[@vishalvariyani4428]

Sir i have a problem in a question that was asked in my clg's weekly test. Here is the question: R=(A,,B,C,D,E) and functional dependencies F = {A->BC, CD->E, B->D, E->A} prove that AE is not a candidate key. I found that AE is the candidate key but in test i got no marks for this question....

[@vishalvariyani4428]

Ok thnx😊

[@rizaviasayeed2632]

Vishal Variyani: AE is not a candidate key bcuz A is a candidate key and hence AE is super key.

[@vishalvariyani4428]

thnx bro..

[@isaitech4602]

Edge digram is great sir!!!!🎀🎀🎀🎀🎀🚶❤️❤️❤️

[@naheedakhter3155]

E ,AB,ABC,ACH ARE the C.Keys. and in question D, is not determine by the other keys it also don't determine the others key.

[@sujatabiswas4158]

there is no path to reach D in the last problem... so how can we have any keys at all ?

[@vidyush96]

okiee , ty sir .. coz i think i hv got some problems in making er models..

[@UTKARSHKRISHAKBCE]

Sir, The answer for my last question is coming (b) ie 2 candidate keys namely AD and ED

[@sauravkumar7508]

i have only one ck i.e ED

[@sauravkumar7508]

sorry u are correct AD and ED

[@v.j.minimilitiaking5694]

@@sauravkumar7508 AD kese c.k. Hoti hay?

[@sqlearth7045]

for the given question ans. is AD & DE. but question) R(A,B,C,D,E,F,G,H) FD'S: CH-->G A-->BC B--->CFH E-->A F-->EG FOR this question ans is 4.

[@madhulikamittal]

at 16:53 you made combinations wx xz wz but instead of xz u taught yz...but we could not include y attribute bcaz y is a candidate key...pls correct it sir

[@supreetsingh1368]

In last question D is never achieved so how can we find any candidate key??

[@vaibhavjain2826]

This lecture series is love.