Decoder Part 3 | GATE 2023

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[@sujoykrhaldar]

A Quick correction 11:18 : we realised ( n * 2^n ) or (logn * n) - (log base 2) decoder using (m * 2^m) or (log m * m) - (log base 2) so no of steps - celling fun(n/m) - step -1 2^n/2^m and so on to kth step - 2^n/2^km in log terms it will be - celling fun ( log [base m] n) as in step 1 -> n/m and so on to kth step - n/m^k correction: Sir mistakenly divide m/n so do check this - so you can swap the question as Realise (m * 2^m) or (log m * m) by ( n * 2^n ) or (logn * n) or do your own correction Its not a big deal you can spot the mistake and do the correction as well Thanks

[@huzaifam7060]

VIDEO NOTES 3x8 using 1x2's: 4 final, 2 before, 1 before = 4+2+1 =7 6x64 using 1x2's= 32 final, 16 before, 4 before, 2 before, 1 bfore = 32+16+4+2+1= 55 4x16 from 1x2= 8+4+2+1=15 6x64 from 2x4= 16+4+1=21

[@alokbaidya1247]

12:10 Ans should be sigma k =1 to floor(log base m n) floor(n/m^k)(i.e the inverse of what sir has written)

[@harshilshah8983]

it will be ceiling function and not floor

[@shivamjangid5097]

sir in 15:09 we are realizing lognxn from logmxm so the formula would be n/(m)^k right?

[@unkonw9621]

yes

[@adityaSingh-wk9jk]

Thank you sir loved it , like always ❤❤

[@pawanchoure1289]

great😊

[@foodmankey]

Sir 17:30 me 11th decoder and 13th line aayega

[@nishantpatil5859]

Hi. 14 th line hi aaega. Reason - In 11th decoder, o/p lines will from 160 to 175 (160 inclusive). So, 0000 will be 1st line on 11th decoder, 0001 will be 2nd and so on. Therefore, 1101 will appear in front of 14th line. I suggest you to draw 11th decoder with 16 o/p lines and check where 1101 appears. Hope this helps you.

[@pawanchoure1289]

correct✔✔☑​@@nishantpatil5859

[@J_BALVIN_007]

Sir is Discrete maths completed?

[@AmitKhuranaSir]

Yes

[@zeeshanyounas4947]

🥰🥰

[@Vaishnavidwivedi11]

💮💕💕💕✨