@@GateSmashersl think you did not write the correct answer for dense index because for dense there is only linear search available for non order it's ans I think will be 200 + 1
for sparse indexing when we find the block, then we again need to search 4 records with in that block soo the worst case for sorted records is will be (log50)+log(4)=7.64~8. Not 7 as mentioned. Also in dense case we can't apply log because the data is unsorted in dense.
I think we don't need to search 4 times in sparse when coming to the data table in hardisk because data is ordered. A normal addition will help to find the data. but there's a problem in dense. let's assume the pointer table is ordered in dense but even then it doesn't make sense to add 1 because the main table's data isn't ordered so we'd need to add 4 instead of 1.
I have been learning from your channel since 2021 and I got through my master's because of you and came back to revise the concepts for a Job interview. Thanks sir. Truely legendary teaching.
The way you explain things, so far the best I've come across. Thank you for saving my time and I don't feel pressurised at all while learning rather it feels great!! Big shoutout to this channel!!
For those who are having doubt and think in dense it should be 200 + 1: Keys in the index table are arranged in order to make it easier to search the records, so that we can use binary search. Hence lg(200) + 1
But If we see acc. to you, we have to make index table sorted which takes extra time. Apart from these If you google search you would know that Dense Indexing is Faster than Sparse. Google: Dense indices are faster in general, but sparse indices require less space and impose less maintenance for insertions and deletions. @gatesmashers Sir please make a pin comment every one is clear for this lecture
Sir u r really doing great job. Ur smile starting from video and after giving so much effort you end video with smile . It's really spread positive vibes.....Thank u for teaching us...
Truly stupendous !!! Just completed all your 63 uploaded videos on DBMS. My DBMS fundamental have become more stronger. Thanks a lot Sir :) :) Just one request- As soon as you complete the DBMS course, kindly upload NET DBMS Solutions for past papers. Thanks again:)
I think there is a mistake in the end part of the video when I/O cost is being said for dense entry. Since it is unordered, there will be no concept of binary search, so it should be 200 searches in index table and 1 search for the corresponding hit in the index table. === (200 +1). And the average answer will be 200/2 = 100.
The key values will be inserted in sorted manner in the index table and they will be pointing to the corresponding block So, I believe we can do binary search in index table that will give us the block and then we can load the entire block in ram and then search for our key so it will be log2(200)+1
In dense we have made index table for all 10000 records not for 2500 blocks, So we can make it ordered and then can apply binary search. So sir is correct.
Sir,as you said dense is for unordered data so in that case time complexity should be of O(n) but why u did log200+1, according to concept it must be 200+1 due to linear search for unordered data.
Index is always sorted ...watch once again he clearly said that if index is nit sorted then there will be no diff in searching with or without indexing
I have a doubt about the case of Dense. In the case of Sparse, the data is ordered and there we can apply the binary search but in the case of unordered(Dence) the data in the Index table is also not sorted in such case we can't apply the binary search. I think in dense case it should be N ( which is 200 in the example case which is far less than 2500 earlier and far efficient). I might be wrong. Please correct me in case I'm wrong. Thanks
@@tejasghone5118 But we are binarily searching for value.... Not the index....We don't need to search the index as its already in sorted order and can be reached in o(1) time just like array element.
The keys which are copy of PK or CK from the records in hard disk are sorted in index table, so we perform binary search on keys then the pointer associated with a perticular key get us to the exact record
Please make lectures on Theory of Computation and Formal Languages i.e AUTOMATA. We need teachers like you to teach us. We want to understand concepts, not for preparing exams. Thank you for Computer Networks Lectures, I learnt from those videos and feel confident in it.
It's amazing sir.Your videos are always up to the mark and help me a lot .There is no other coaching classes required .All the doubts are always clear here .😀😃😃
Good Lecture Sir :) Having one doubt here : without indexing, we are not considering searching inside a page, only I/O operation is considerable.=>fine with indexing => when unsorted => not considering searching inside a index page => log200 + 1 =>fine=> because here we have to find the right page pointer. with indexing => when sorted => we have to consider searching inside an index page also because that's the only thing which gives us right DB page => so here time will be log50(searching right page among 50 pages) + log50(searching right record among 50 records) +1 Please correct me, if I made any mistake here.
Bhai Unordered ya dense vale case me sir ne binary search kaise laga di kuki data sorted nahi hai to binary search kaise laga rahe hai...... yahi doubt hai mera!
Hello sir, in 8:59 how can we get a block in log2(50), when each block of index itself contains 50 blocks of hard disk ( in case of spars ) rather we get bulk of 50 blocks and to get the required block we have to again binary search in that block of index therefore to get the required block we have to search again log2(50) after this we will get successfully our required block, therefore total time complexity of searching in index is log2(50) + log2(50) which is log2(2500), and after this time we will again search our required record in that particular block. Correct me if I'm wrong, and thank you sir for this amazing series.
DOUBT - log(50) gives the corrosponding block and not the index for next block we need log(50) * log(50) ( or log(2500) better )to get index from index table
Bhai dekh dense me bhi tuze starting Ko sirf required index value hi milegi , par tuze main value find out karne ki liye us index value pe to Jana hi hoga na , so udhar jane ki liye tuze Aur 1 bar search karna padega isliye + 1 Liya hai ..
Sir, in case of mapping in dense indexing, all the 10000 records that get mapped into all the 200 index table blocks would still be UNORDERED. Thus, traversing them would require a time complexity of O(n) rather than O(log n), as in the index table we would need linear search, don't you think? Please comment.
Sir, Thankyou for the best lecture videos, Your teaching skills are above the level, from that elementary class students can also understand the same. Sir but i hv 1 doubt: At Duration(in Video): 9:10 u said that we have to search 6 + 1 . But when we approached to M.memory then why are adding 1 search only for it, as within the blocks there are no. of records(In our case its 4). Please tell me ??
Non ordered me toh linear search lgega na toh usme log200 kaise hoga log toh binary search ke case me aata tha na.........isme toh 200 aana chahiye worst case me 🤔🤔
I think there is some mistake in dense index table because you said we are scanning in each pages of page table after doing binary search but if we are not able to find the required page in that block then where we need to go because data is sored in unordered fasion.
sir can u please explain why have u added extra plus 1 to search a record in a block in indexing search where as in last lecture video in binary search of n =2500 (here n is number of blocks ) we got log2^n =12 where u haven't added plus 1 but in this case also we do search for block and after getting that particular block we should search of particular record only .
sir first you are saying dense indexing is used when data is unordered . so why at last you are doing binary search on record and giving ans as log(200) + 1, ans should be 200 in this case acc to me . Please help me to clear this doubt please.
No need of coaching classes your teaching style is superb ......thanku so much sir 🙏
Thank You so much for the appreciation. Bless you. keep learning and sharing
did u clear gate??
ua-cam.com/channels/S5cAoAFGB6-aUzbesg2svg.html
@@ayushibansal6635 yes 👍
Your analogy to real book made the concept clear ..sir..thank you
You are most welcome
Today I cracked my viva due to your all lectures on dbms , really i thankful to you 🙆♀️ sir
Excellent
@@GateSmashersl think you did not write the correct answer for dense index because for dense there is only linear search available for non order it's ans I think will be 200 + 1
for sparse indexing when we find the block, then we again need to search 4 records with in that block soo the worst case for sorted records is will be (log50)+log(4)=7.64~8. Not 7 as mentioned. Also in dense case we can't apply log because the data is unsorted in dense.
Yeah I too got same doubt... How could we apply log in case of dense..
Ya, for dense it won't be possible to use binary search
I think we don't need to search 4 times in sparse when coming to the data table in hardisk because data is ordered. A normal addition will help to find the data.
but there's a problem in dense.
let's assume the pointer table is ordered in dense but even then it doesn't make sense to add 1 because the main table's data isn't ordered so we'd need to add 4 instead of 1.
He said at start within block Indexing fast nai krti... Its all about I/O cost. Index table to Hard disk miss/hit ki bat hei... within block nai...
Your are right. This is the answer of the doubt.@@anamshahzadi8722
I have been learning from your channel since 2021 and I got through my master's because of you and came back to revise the concepts for a Job interview. Thanks sir. Truely legendary teaching.
The way you explain things, so far the best I've come across. Thank you for saving my time and I don't feel pressurised at all while learning rather it feels great!! Big shoutout to this channel!!
same! I only know alittle hindi from watching indian movie, but when I learn from him, he is such a good teacher!
For those who are having doubt and think in dense it should be 200 + 1: Keys in the index table are arranged in order to make it easier to search the records, so that we can use binary search. Hence lg(200) + 1
Yes you are right
But If we see acc. to you, we have to make index table sorted which takes extra time.
Apart from these If you google search you would know that Dense Indexing is Faster than Sparse.
Google:
Dense indices are faster in general, but sparse indices require less space and impose less maintenance for insertions and deletions.
@gatesmashers Sir please make a pin comment every one is clear for this lecture
@@abhishekjain-tc3vy no you are wrong sparse is faster then dense index
@@ZainAli-ih1mncheck in Google
And also in indexing we have to short the index table that is to be add while calculating the time complexity
Just amazing, how much effort this guy has put in, you have literally saved CSE engineers not going to lie sir...
The Book analogy was an eye opener. Great explanation!
The way of explanation is Awesome...शुक्रिया सर..:) Waiting for another video..
Sir u r really doing great job. Ur smile starting from video and after giving so much effort you end video with smile . It's really spread positive vibes.....Thank u for teaching us...
Truly stupendous !!! Just completed all your 63 uploaded videos on DBMS. My DBMS fundamental have become more stronger. Thanks a lot Sir :) :)
Just one request- As soon as you complete the DBMS course, kindly upload NET DBMS Solutions for past papers. Thanks again:)
Excellent Explanation... Hats Off to your efforts...
Sir you are doing best Please continue You will be the best You tuber in 2019
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I think there is a mistake in the end part of the video when I/O cost is being said for dense entry. Since it is unordered, there will be no concept of binary search, so it should be 200 searches in index table and 1 search for the corresponding hit in the index table. === (200 +1). And the average answer will be 200/2 = 100.
Bro You really helped me at 4:42AM...I'm also thinking about but you assured me that i'm not alone and make me feel right....Thanks for Sharing
True same doubt I also got
The key values will be inserted in sorted manner in the index table and they will be pointing to the corresponding block
So, I believe we can do binary search in index table that will give us the block and then we can load the entire block in ram and then search for our key
so it will be log2(200)+1
But after indexing we are searching in index file so it is sorted the main file was unsorted
In dense we have made index table for all 10000 records not for 2500 blocks, So we can make it ordered and then can apply binary search. So sir is correct.
Bhai ye underrated video hai, most important I can say
simplicity in one word "superb"
simply awesome !!!!! The way of explaining is superb, Thanks!
Sir,as you said dense is for unordered data so in that case time complexity should be of O(n) but why u did log200+1, according to concept it must be 200+1 due to linear search for unordered data.
index table is sorted.. I don't think sir explicitly told this but binary search is only applied on sorted list
@@syedalmastirmizi3477 Index table is not sorted. No where mentioned.
Index is always sorted ...watch once again he clearly said that if index is nit sorted then there will be no diff in searching with or without indexing
For confirmation
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He is a true genius. None can deny that
I have a doubt about the case of Dense. In the case of Sparse, the data is ordered and there we can apply the binary search but in the case of unordered(Dence) the data in the Index table is also not sorted in such case we can't apply the binary search.
I think in dense case it should be N ( which is 200 in the example case which is far less than 2500 earlier and far efficient).
I might be wrong. Please correct me in case I'm wrong.
Thanks
Index table is sorted but the data isnt hence logn
@@tejasghone5118 But we are binarily searching for value.... Not the index....We don't need to search the index as its already in sorted order and can be reached in o(1) time just like array element.
The keys which are copy of PK or CK from the records in hard disk are sorted in index table, so we perform binary search on keys then the pointer associated with a perticular key get us to the exact record
Yaa exactly the same doubt he is wrong there
I have never ever ever ever ever...seen a techer like you...🙏☺️ Sir ji
Please make lectures on Theory of Computation and Formal Languages i.e AUTOMATA. We need teachers like you to teach us. We want to understand concepts, not for preparing exams. Thank you for Computer Networks Lectures, I learnt from those videos and feel confident in it.
I would suggest video lectures from sipser, lol you prolly has been graduated by now
This man deserve more view and more likes ... thank you so much sir for saving our future
Thank you Sirji, very well explained, and understood, I have subscribed and have recommended to friends too.
Sir aapka bahut bahut shukriya mere DBMS pakka krne ke liye 🙏🔥
thanks for sharing your tremendous knowledge with us
It's amazing sir.Your videos are always up to the mark and help me a lot .There is no other coaching classes required .All the doubts are always clear here .😀😃😃
Beautiful teaching style...
sir you are the best teacher for computer science
Good Lecture Sir :)
Having one doubt here :
without indexing, we are not considering searching inside a page, only I/O operation is considerable.=>fine
with indexing => when unsorted => not considering searching inside a index page => log200 + 1 =>fine=>
because here we have to find the right page pointer.
with indexing => when sorted => we have to consider searching inside an index page also because that's the only thing which gives us right DB page =>
so here time will be log50(searching right page among 50 pages) + log50(searching right record among 50 records) +1
Please correct me, if I made any mistake here.
Bhai Unordered ya dense vale case me sir ne binary search kaise laga di kuki data sorted nahi hai to binary search kaise laga rahe hai...... yahi doubt hai mera!
same doubt
you do a great job, thanks a lot sir
In dense indexing the records are un ordered then how can we use binary search
really very good explanation.. I saw both videos part1 and part 2
Sir, your videos are just awesome. Please keep it up.
Thank You So Much Sir
Awesome videos. Thank you so much :)
Hello, Bang Saab
Hello sir, in 8:59 how can we get a block in log2(50), when each block of index itself contains 50 blocks of hard disk ( in case of spars ) rather we get bulk of 50 blocks and to get the required block we have to again binary search in that block of index therefore to get the required block we have to search again log2(50) after this we will get successfully our required block, therefore total time complexity of searching in index is log2(50) + log2(50) which is log2(2500), and after this time we will again search our required record in that particular block.
Correct me if I'm wrong, and thank you sir for this amazing series.
exactly same doubt
Sir, thank you so much for the video. It explains indexing clearly and very informative.
dense main +1 nahi aye ga aor log 2 200 nahi aye ga keun k data sorted nahi he. baki you did well. keep doing
explain your points please
He is right dense is for unsorted so we would use l linear search(O(n)) and binary for sorted(log n )
Excellent presentation...👏👏 you deserve more subscribers and views than what they are today...
DBMS my favourite subject because of You sir
Thank you
Masallah your explanations is excellent ❤❤❤
explanation mast h sir, Sir plz computer organisation k videos upload kardiye plz sir
Ur teaching style is fantastic.....i loved it.....plz make videos on Ugc net unit 1 (discreet structure and optimization)
* In dense , why 1 is added * as in sparse it require further search in page but in dense it have all records.
Please add dense and sparse indexing also in title. This will help students who will come in future.
DOUBT - log(50) gives the corrosponding block and not the index for next block
we need log(50) * log(50) ( or log(2500) better )to get index from index table
i think searching time in DENSE index should be O(200)+1 rather than log(200)+1 ?
Brilliant lecture.
Sir, at 9:09 it should be [log50] + [log4] = 8, as there are 4 records(sorted) in each block.
he only considered the no. of blocks needs for searching.
For Dense shouldn't it be O(n) because of linear search? We can't apply binary search here.
Dense me log +1 kuyn, + 1 to nahi chahiye na? ptr per record to hain hi, phir hard disk me kuyn search kare?
Bhai dekh dense me bhi tuze starting Ko sirf required index value hi milegi , par tuze main value find out karne ki liye us index value pe to Jana hi hoga na , so udhar jane ki liye tuze Aur 1 bar search karna padega isliye + 1 Liya hai ..
Thnx u paji bohat badiya
superb
Sir, in case of mapping in dense indexing, all the 10000 records that get mapped into all the 200 index table blocks would still be UNORDERED. Thus, traversing them would require a time complexity of O(n) rather than O(log n), as in the index table we would need linear search, don't you think? Please comment.
Index table is always ordered..it can not be unordered
@@GateSmashers Thank you, sir.
@@srijitanitamajumdar5904 I have same doubt also
@@GateSmashers ok sir..thank you sir 👍
Thanks sir!!
log(200) cant be done in unordered it should be 200/2=100 for unordered i.e. dense
thank you sir
In dense indexing we can directly go the record why it will be log200 if data is unordered ?? Please verify last part of the video
Best........Thanks a lot🙏
Thanks Sir.
Nice explanation sir
Sir, why we use log based time complexity in unordered list (Dense)
Sir mein sirf apki videos follow kar Raha hai ye videos comptetive exams ke liye sufficient hai .
i am bit confused...
for dense will the unordered data be stored in ordered form in index pages?
10:27 Why we are calculating the log2 (200) as we are using dense and records are unordered and cant use binary search.
How can dense searching take log time when data is unordered?
Isn't binary search applicable for ordered data only?
Excellent sir
can we store the index as a hashmap? then the searching would be constant and not logn.
can anyone explain me that why we take the size of index block same as that of hard disk block ?? why not different ??
how come sir, we are using binary in case of Dense as it is unordered it will have normal liner search?
Sir, Thankyou for the best lecture videos, Your teaching skills are above the level, from that elementary class students can also understand the same.
Sir but i hv 1 doubt: At Duration(in Video): 9:10
u said that we have to search 6 + 1 . But when we approached to M.memory then why are adding 1 search only for it, as within the blocks there are no. of records(In our case its 4). Please tell me ??
Same doubt
it should be:
[log50] + [log4] = 8
why log(200)+1 in case of dense searching as data is unordered in index table..it should be avg of 200+1, right?
Thank you for the videos sirrr..
Dear sir indexing in your 3 lectuers are 101% clear but plzzzzzzzzzz explain b tree and b+tree and as well as hashing topic......
excellent explaination
Non ordered me toh linear search lgega na toh usme log200 kaise hoga log toh binary search ke case me aata tha na.........isme toh 200 aana chahiye worst case me 🤔🤔
Bhaiya just wanted to know, are entries in Index Table always sorted?
Too good😍
Hello ! Can anyone tell me whether this playlist would be enough for doing a course which has hands on project ? Please tell me. Thank You.
I think there is some mistake in dense index table because you said we are scanning in each pages of page table after doing binary search but if we are not able to find the required page in that block then where we need to go because data is sored in unordered fasion.
Even in sparse we have to search in all 50 blocks then again for 50 entries
please make videos on algorithms ur explanation is good
10:44 whty are we taking log(200) the data is unordered so avg time should be 200/2 = 100 or simply O(no. of index blocks) ?
Thank you sir👍
Index table me Binary search kaise apply ho raha hai. Please explain
sir....thanks a lot !
yeh ek accha video hai
Superbbbb 👍
sir can u please explain why have u added extra plus 1 to search a record in a block in indexing search where as in last lecture video in binary search of n =2500 (here n is number of blocks ) we got log2^n =12 where u haven't added plus 1 but in this case also we do search for block and after getting that particular block we should search of particular record only .
sir first you are saying dense indexing is used when data is unordered . so why at last you are doing binary search on record and giving ans as log(200) + 1, ans should be 200 in this case acc to me . Please help me to clear this doubt please.
Sir at 10:27 log200 base 2 value is not 8 but it is 7.64
nice explanation
Sir, please give the serial no means sequence no of all videos.....
sir pls a make on a vedio of past year ugc net exam..
Thank you!!!
tussi great ho paajiiiiiiii😁😁
Sir Plz take a class about RAID