The number of all possible strings of binary numbers of length 4 are found by raising 2^4=16. It is just a coincidence that 4^2=2^4=16 here. If the rows had been 5, the right answer would've been 2^5=32, not 5^2=25.
Great Video, you explained it super easy. I never got the connection between the codewords and the Generatormatrix. 1:33 is wrong, but another comment already explained why it is 2^Rowcount and not Rowcount^2
Just a question to be sure. Lets name G1 the 1st row of G, G2 the 2nd row, G3 the 3rd and G4 the 4th. We know that the codewords will be 16 so if we do these bit-by-bit additions: G1 G2 G3 G4 G5=G1+G2+G3+G4 G6=G1+G2+G3 G7=G1+G2+G4 C8=G1+G3+G4 G9=G2+G3+G4 G10=G1+G2 G11=G1+G3 G12=G1+G4 G13=G2+G3 G14=G2+G4 G15=G3+G4 G16=G1+G1=G2+G2=G3+G3=G4+G4=0000000 we basically have the same codewords as of your method, right?
Hello..thanks for the video. If the number of rows is 3..then how you proceed for step 2? because 3^2 is 9 and 9 divided by 2 gives you 4.5...how you deal with that? To ease things...its a [6,3] matrix Generator.. Thank you!
+Kowlessur Rajiv Hi, so I'm embarrassed to say that the reason it didn't work for you is cause I made a basic mistake in the video (which I have now annotated in the video, so thanks for that catch). I put 4^2, when it should have been 2^4 (in my case both 2^4 and 4^2 =16, so the example should still be okay). But in your case you would have 2^3, giving you 8 options, not 9. So now that should work for you. Same reasoning, just backwards. Binary code=>2, length 3==> 2^3. So sorry about the confusion.
Very helpful video thank you but i think you should state this generator matrix applies to using the parity equations p1 = d1 xor d2 xor d3, etc and not the parity equations p1 = d1 xor d2 xor d4 as with that generator matrix the 3rd and 4th row of the parity section(the non identity part) need to swap
Yeah, apparently the only way to find the nullspace of a matrix is the list EVERY VECTOR and check if it's in the kernel. Thank god the field was finite or we would be here for a long time.
The number of all possible strings of binary numbers of length 4 are found by raising 2^4=16. It is just a coincidence that 4^2=2^4=16 here. If the rows had been 5, the right answer would've been 2^5=32, not 5^2=25.
thats what I was about to say.
Choose k=3 instead, the 2^3 outcomes are more obvious.
Andratos95 thanks for this.. I was thinking the same thing
It says in the description as well, but thanks for pointing it out!
5:20 a.m., i've spent all the night studying without sleeping-> test in 3 hours-> didn't know where to get study material-> you just saved me.
Hope your test went well!
facts
omg same exact situation (including 5:20 am)!!!!! I am following you to greatness sir
@@ddxfraxinusdne A little late but I now have a BSc and a MSc. Thanks
Just remember addition is an XOR operation, so for example bit 5 in a given code word, if calculated as (1+0+1+0) the result = 0 since 1(XOR)1 = 0.
thank you
Great Video, you explained it super easy. I never got the connection between the codewords and the Generatormatrix.
1:33 is wrong, but another comment already explained why it is 2^Rowcount and not Rowcount^2
much easier explained than my prof was able to, thanks
Just a question to be sure. Lets name G1 the 1st row of G, G2 the 2nd row, G3 the 3rd and G4 the 4th.
We know that the codewords will be 16 so if we do these bit-by-bit additions:
G1
G2
G3
G4
G5=G1+G2+G3+G4
G6=G1+G2+G3
G7=G1+G2+G4
C8=G1+G3+G4
G9=G2+G3+G4
G10=G1+G2
G11=G1+G3
G12=G1+G4
G13=G2+G3
G14=G2+G4
G15=G3+G4
G16=G1+G1=G2+G2=G3+G3=G4+G4=0000000
we basically have the same codewords as of your method, right?
Hi I was wondering if you got 1+1+0+) =2 what would that be since its not binary?
what to do when it became [ 0+0+1+1] in last column and give bits product.
'
1+1 then 0 , if 1+1+1 then 1
Thanks! You made this super easy to understand
Hello..thanks for the video. If the number of rows is 3..then how you proceed for step 2? because 3^2 is 9 and 9 divided by 2 gives you 4.5...how you deal with that? To ease things...its a [6,3] matrix Generator..
Thank you!
+Kowlessur Rajiv Hi, so I'm embarrassed to say that the reason it didn't work for you is cause I made a basic mistake in the video (which I have now annotated in the video, so thanks for that catch).
I put 4^2, when it should have been 2^4 (in my case both 2^4 and 4^2 =16, so the example should still be okay). But in your case you would have 2^3, giving you 8 options, not 9. So now that should work for you. Same reasoning, just backwards. Binary code=>2, length 3==> 2^3. So sorry about the confusion.
+Theoretically Hello...thanks for the reply and welcome! Yes indeed it give you 8 options for my mentioned problem..
Best of luck for more videos..
thanks so much for explaining this complex theory in such a simple way. god bless you.
its really helped me in my maths test thanks a ton
what if we have 6,3 generator matrx .. then dividing by 2 wont work
Thank you very much.... It's helpful
Very helpful video thank you but i think you should state this generator matrix applies to using the parity equations p1 = d1 xor d2 xor d3, etc and not the parity equations p1 = d1 xor d2 xor d4 as with that generator matrix the 3rd and 4th row of the parity section(the non identity part) need to swap
I have been searching from 3 days how to get G matrix. everyone give example of 4 bit data . plz tell how to get G matrix for 8 bit
thank you very much for your video, it's really helpful!
thats great but where did the most important bit- the generating matrix come from?
Thank you so much....easy to understand
what if it has 5 rows isnt there any alternative sol
Yeah, apparently the only way to find the nullspace of a matrix is the list EVERY VECTOR and check if it's in the kernel. Thank god the field was finite or we would be here for a long time.
Thank you very much for explaining. Especially the binary row vector finding method was lifesaver.
what is meant by codeword?
Nice explanation, thank you!
This is so helpful! Liked and subscribed :)
which software she is using ?????
so in your case your making 16 different code words for 16 different message vectors
Thanks alot! i was so confused before
very good explanation!
Thanku girl,, you help me lot,, at this imp tym,, God bless yuu
thank u so much , i . having my test today and it will gonnaa help me ,, thank u thank u
Hello thanks to this video but please can tell us how get the generating matrix [G]
You could get it from the parity check matrix (recommend starting at 1min38sec of following link):
th-cam.com/video/oYONDEX2sh8/w-d-xo.html
+Theoretically thanks so much for your answer that kind from you 🌹
Your voice giving me a feeling like u are crying !
Thanks a lot!
not satisfied
Hi mam....how can i contact you for my queries
Nice Vid. Thanks!
life saver
Thank you
It's awesome! Thanks!
Thanks a lot
thanks so much
Why the hell are u sounding like this?😣
explaination is well but ur voice is spoiling it
too slow.. not good lecture at all..
it is 6 and a half minutes long.... it is officially the shortest "lecture" I've ever heard. And helpful!