 Okay, so we're going to continue the problems and this is like the additional questions. I'm at the 6th question No, no, none of you have the printer I'm gonna, I thought I bought some last time It's all been It's okay, but at least the other sheets were there You have some? Yeah, maybe Okay, so gf16 This is defined there Okay, I don't know. I think this question is quite Seems to be quite easy So basically it just wants you to solve equations. So it says Find non-photo solution to x plus y plus y equal to zero Okay, so it's very easy just pick x and y arbitrarily non-zero It says not equal to y and then put z equals x plus y you'll get that Okay, the next thing you're supposed to solve Okay What do you do for this? Yeah, so I mean this is gf16. So if you have x plus y plus z is zero You'll also have x plus y plus y plus z plus zero. So you don't have to worry too much about it. The third part is a simultaneous Question, how do you do this? Okay, so you have to factor this So any factor that will happen You'll get x plus y plus z times x square square square square square square plus x y plus y z plus x z. You have to kind of know this factorization. If you didn't know this factorization I'm sorry This is going to be equal to I'm sorry Yeah, I mean no point in doing this factorization, right? So what do I do with this? It should be zero so then what do I do? I'm sorry Yeah, yeah, yeah, so this is wrong. No, this is the Yeah, I think something like that has to happen. So I don't know I'm sorry What is the solution? I don't know why I asked this question. So I'm just trying to think what is the So you can do it in another way. Okay, so maybe I should just talk about doing it in another way. Okay, so if you take I think I don't think you'll have a solution. Will you have a solution? Okay, so that's one way of solving it. So you put z is zero And then you pick x equals y. So what are you saying? I want z to be non-zero. Is it possible? Anyway, so I mean this is one solution, okay So I really don't know why you asked. There must be some Something I did. I can't remember Exactly what I did here Yeah, I think this is a valid solution. There's no doubt about it. That equals zero and x equals y Yeah, this is a valid solution. The thing is can you have a non-zero solution? It looks like it looks like you cannot Okay, so this will be equal to x plus z, no Yeah, so there's some relation like that. You can show this. I forgot how you do this But if x plus y plus y plus z is zero x cube plus y cube plus y cube is That is nothing but x cube plus y cube plus z cube plus x y z plus x y z, okay So that has to be equal to zero. So you get that to be equal to x y z Okay, so anyway, so that's the way you do it. So this has to be a solution. Anyway, it's not too critical for us I think that's Okay, so it's not a very critical Not so happy about this Okay, so let's go to the seventh question Okay, so seven So you put z equals x plus y Yeah, so you can do that also you put z equals x plus y here You'll get x 3 plus y 3 plus So you can write something so you can do it that well Okay, so there are various ways of doing it. Let me just not worry about it Seventh question is just a bch code question. It's very very similar to cross 2. So i'm going to skip this Okay, it is basically a two error correcting bch code and you don't have to do You have to find some dimension minimum distance etc. So i'm going to skip this very similar to your second question and Yeah, so the eighth question is also Very similar Okay, so here Okay, so eighth question also i'm going to skip. It's a simple bch code question Two error correcting bch code. You can see it. You'll see that it's it's not too hard So let me move on to the ninth question. It's a little bit interesting. So you want to look at bch codes n equals 6 63 and t being the error correcting capability Okay Okay, the first part wants you to find smallest t That's not k to the dimension of this code Is not equal to n minus 6. Hi, you already know the answer Okay, you did it, okay Okay, so that's when we're doing it you can also list it from t equals 1 and figure out the first one that We don't have length 6. Okay, it's not very hard. You can do it. So apparently the answer is t equals 5 Okay, so you can take this And the next question is n equals 255 In the same question k smallest p such that k not equal to n minus 6 Okay, so once again here, it's a little bit more twisted You have to do a few more things but apparently the answer is what? I'm sorry 8 p The answer is t equals 9 Okay, so so you have to keep finding the Cyclotomic concepts and stop and the length becomes smaller Okay, so that's the ninth question 10th question is a little bit interesting Okay, I have one error correcting r s code over g f p So what I asked was Find the non-zero code word It was people save to come but the answer is instantly so Okay, so I mean let me let me tell you I would like to talk about if you want parity check matrix It's going to be 1 alpha alpha square Alpha 3 alpha 4 alpha 5 alpha 6 right Can be 1 alpha square Alpha 4 alpha 5 alpha 6 Alpha 5 alpha 8 alpha 4 I want a code word such that h times c transpose equals 0 Right, and I want c 4 c 5 c 6 c 7 equals 0 so h times c transpose equals 0 will apply 1 alpha alpha square 1 alpha square alpha 4 c 1 c 2 c 3 has to be equal to C Okay, so this is the equation you have to solve Okay, so this is very simple way of thinking about it You can also think about it in terms of gendered and matrix somehow Maybe there's a way to do that that this is uh interesting what I'm thinking about Okay, so you have to solve this equation. How will you solve this equation? You have to be there's an elimination Okay, so you do there's an elimination make it i e and something then pick c 3 arbitrarily then c 1 and c 2 will be fixed Okay, so that's the idea So if you want me to do this one more step, this is c 2 i 1 2 To take the second row and add it to the first row You get 0 alpha plus alpha square is something Okay, this is enough actually Okay, it's not too bad. So you put c 3 arbitrarily here and find c 2 from this equation from c 2 and c 3 Okay, I'm trying a little it's good enough You don't get the answer Okay, that's part a Part b is also very similar. You know, I mean part b basically says Find a non-glue of code word work c 2 equals c 4 equals c 7 equals c 6 So you do a very similar thing and you'll take a different equation. That's all you do But okay, let me move on to the 11th question Okay, and 11th question is too simple. I'm going to skip that Okay, so it asks you to just find that where the 8th element code question is a standard resolving code question One error correcting rate solvent code over 3 of 8 you have to find the characteristic gendered m matrices and then decode our code Yes, it's a very standard question. So I'm not going to do the detail Okay, well Okay, so 12 is a bch code question Okay, again, I'm going to skip it. It's very simple. I mean simple questions I'm going to skip because it just says that our bch codes with given n and t equals 1 2 3 4 find gendered m matrix Things like that. It's no thing said We've been also asked to find the exact minimum distance that will require some more work But it can be done in most cases gfx itself will end up being The minimum weight. I think in one case it doesn't end up being but I'm not sure when does it come? I don't know. Okay, right. I think most cases it will be like that Okay Okay, so the 13th question Okay, so the 13th question says You take n equals 15 Okay, okay and then you take So first code that I consider c1 will be binary bch Second code I consider will be You take a read Solomon code 3 equals 1 over cf8 How will I make n equals 15? Binary length I want it to be 15 Okay So I know each symbol is gf8 expands into 3 bits. So I need an n particle length more which is 5 over gf8 Right and I love length sustained binary equivalent code Excuse me So you have to find the dimension of the binary code in each method Yes, here. It's quite easy Okay, it will be 11 What would be k here? 3 over gf8 Okay, and then when I multiply by 3 I'll get 9 Okay, it will be the same right okay will reduce Yeah, but it will be 5 comma 3, you know if p is 1 See you shorten from 7 comma 5 And you shorten by 2 So it will be 3 comma 5 see read Solomon code always. It's n minus 2p k is n minus 2p So capital K will be 5 minus 2p which is 3 over gf8 And that when expanded will become K equals 9 So you see if you do a read Solomon route you get a lower rate. We knew that this will happen, right? This year just better than read Solomon as far as efficiency is concerned And then in part b you have to find probability of block error expressions And that bounded distance decoding of both codes. Okay, so let's do that So this is part a Part b if you take c1 Okay, it corrects only one error. So probability of block error will be 1 minus 1 minus p power 15 Right 15 p power times 1 minus p power 14, right? So this is a simple expression and the bounded distance decoding, right? How do you do it for c2? You also bounded distance but it is correct one symbol error So first you have to go from p to probability of symbol error probability of symbol error to 1 minus 1 minus p power 3 So the question asks you to find the dominant term as p tends to 0 Okay, so as p tends to 0 you have to just expand see if you expand there we get 1 minus 15 p and then plus 15 choose to p squared and after that you'll have higher terms. They will not play around here. You have minus 15 p Okay, and then you'll have plus Yeah, 14 times 15, right? p squared so it will be 105 p squared minus 105 plus 2 can so you do that carefully you'll get the answer So if you come out to 105 p squared, what about here? You're going to do it for p as fast and then you just have to do this will go to 3p as p tends to 0 And then you put that here and then simplify you'll get an answer. Okay, so this tends to 145 square of p tends to 0 which will tend to 10 p squared and that will be 90 90 No, no, no, don't be asking Yeah, that time will be right that this term will depend so you will become 90 Okay, so that's the thing and the part c asks you to find one advantage of c1 over c2 and another advantage of c2 over c1 Okay, what do you think an advice an advantage of c1 over c2? Case what about c2 over c1? Yeah, that's one thing that it's not I mean it's just a constant multiplying p squared I mean p becomes 10 power minus 6 It's irrelevant to have the constant outside. Okay, so what is the real advantage? Because you're working over gf8 You don't have to work over gf16 Okay, so that's an advantage Okay Yeah, you're right. I mean probably the blocker is marginally smaller, but the order is the same. It doesn't really matter Okay, so next is the second question So it's a concatenation question So you have a 2 to 1 code Yes Which question? 10 10, huh? You're asking me Or else no bch code such kind of things are not guaranteed So each certain code you can show in any k positions out of n you will have a non-zero program You have seven right n equals seven you pick any any k k was for three right pick any three positions out of seven you will have a non-zero code that's supported That's the result you can very easily show it's not very hard from the mdf code that they become Okay, so bch code that's not guaranteed you can't it's not true that in every thing you input Okay, so 14 is 2n code c1 And a 64 code c2 Okay, that's systematic Okay, so what you do is you you're doing a concatenation. So what is concatenation? You have two bits You first encode it with c1 Okay, how many bits will you get Just four bits. Yes, so two code words will come out And then you encode it with c2 You'll get six bits Okay, so you do systematic encoding it Okay, the question asks you to find c2 such that The overall code becomes a six comma two comma three code many ideas Yeah, there are various possibilities. It's not very hard to do this Right these four bits can be 0000, 0011, 1100 and 1111, right Hmm Yeah, I think it seems like this works Is better Or should it be one Yeah, so anyway for any non-zero code word, you'll at least have two here Either these two are one which will give you this one in the moment. Okay, so there are various ways of doing it It's not very hard. It's a very simple question. Okay, so let me move on to the 15th question So here I will claim something you have to say if it if it can exist or not I'll describe an object you have to say if such an object can exist or not Okay, so in part A I'm saying that if a 64 code c binary code c intersects c per equals If you if you know that that's a thing exist, you have to provide an example It's not you have to prove why something like that cannot exist not exist If it decides to well then c intersects c per will be equal to c, right? Won't c intersect? Huh Yeah, I want to show see that's not what I want to show See I want to show that such a c cannot exist or give an example for such a c It's a slight logic difference. I mean there are of course a lot of codes will satisfy this That's not what I'm asking. It's better code which satisfies this but that's what I'm asking So you have to either find the code which satisfies this condition or That cannot happen. I don't think it will happen So when the c intersects c per we all zero let me see No in terms of generator matrix and particle check matrix, it's not very hard to come up with the condition See any code word is m times g any code word of the dual is m times h, right? Right, and you have g times h transpose being equal to zero So you don't want anything to overlap. So what should happen? Are you giving an answer I'm just saying under what conditions on g and h will you have c intersects c per being all zero? Is it possible? See what should happen think about it. Okay, mg even arbitrary code word f my code How do I test if a code belongs to the dual? How do I test the code? Ah, it's just for the original code for the dual code. What will I say? g transpose right so mg g transpose Equal to zero in place. What should happen? m will be zero. So what should happen? G g transpose should have full Rank should be an invertible matrix That g g transpose will be a k cross k matrix It should be an invertible matrix. So that's the condition So you have to find that three by six matrix such that g g transpose equals. Let's say i is it possible Okay, so that this is the condition, okay G g transpose should have full rank and there are words to satisfy. I picked about how you would do it It's not very hard. I think there are codes like this, okay So let's go on to part b. It is a little bit more interesting From two to seven By what you call Okay, so these kind of questions is easier to think in terms of gender dimensions, okay So there are four code words. There are two rows in the gender dimensions. That's what we want we Okay, so this one has weight greater than or equal to seven It's also has weight greater than or equal to seven Yeah, so what will happen if this is weight seven and this is weight seven In how many places will they have to definitely overlap? Can they be without overlapping? It's not possible. Well, it's total volume is only 10. Okay, so if I can't do seven plus seven Okay, and how many places at least four places we have to intersect so when you add these two what will happen to the weight? It'll be them six set less So 10 to seven cannot Okay So this is uh This is a logic you can use this cannot exist Okay, so 16th question We have given that there was a seven three four in your code And you have two code words zero zero one one one zero one Then you have zero one zero zero So basically you have to find the code okay find possible value check matrices So I think it's these kind of questions are best done with the value check matrix Okay, you know the seven three Okay, so each column is going to have Uh weight four So So let's let's think of the value check matrix in systematic form Okay, and you have four three missing columns, right? So let me call the three missing columns with ab okay And then you have equations now see this code word gives you one equation Right, and this code word also gives you another equation. Okay, the only remaining piece of information you'll use is four Okay So what happens then is if the minimum distance is four the weight of each column should be at least three Okay, you cannot go less than three Right the weight of each column goes less than three what will happen to the minimum distance? Suppose a has weight say it Okay, then You take the first column second column. So maybe the two is in the first two positions Okay, we will get a code word of weight three. Okay, so you cannot have any column have weight less than two So only weight three is possible and you can quickly solve those equations assuming weight three or weight four So you quickly get the answer. It's not very hard Okay, so because I'm more solving after this I'm not going to get I'm not going to grade details here, but you write those equations and then use the fact that weight of abc is at least three Okay, we're only like type possibilities and you'll see the equations are very easy to solve Okay, so you have just problem error with one of them. We'll quickly get it Okay Okay, so 17th is just a standard dc or sweet salomon question. I'm gonna skip it and then Okay, I'm gonna skip it And 18 I think we did this problem before Okay, so but anyway We did this before right in class Some concatenation and I showed how you'll have to count risky So the critical thing here is We have four errors that you split split into Blocks in a certain way And the symbols in a certain way to consider all possible splits and show that you can always Okay, I'm going to skip 18 19 is about 18 has been done in class 19 is about burst duration correcting capability. Okay burst weather correcting capability. Okay, so I think I did this in class Okay, so I'm not going to go into detail here. So you know about burst other type of capability The tip here is Figure out what length burst will pass prs Okay, so it's not very hard So if you want to find it exactly you have to do some numerals, but it's not too hard Okay Okay, so 28th question and again, uh, sweet salomon question So you take p equals to n equals 8 over n equals 7 right Oh Oh There's something wrong with question 20 It says 8 4 read salomon call over g of 8. Yes. I'm doing some extension here. So let's just keep 20 20. There is some dug Take a look at it. That's not very interesting. So 21 is about Again puncturing that again, there is a problem that I said 8 comma 4 RS code over g of 8 is for deliver 4 bug here. So we can skip these two questions All right, so that takes me to the end Salomon