 Alright so we are going to look up the assignment and BCH and its element codes. Okay so let me begin with the first question, the first question says You have to find k and g of x, what is k? k is dimension And g of x is generator terminal All narrow sense binary BCH codes for any codes 31 So you might wonder what narrow sense is Narrow sense is exactly the kind of codes that we've been looking at So those are called narrow sense and I mean just ignore narrow sense Just to say it's specific BCH code So like I said the zeroes, so narrow sense implies Zeroes of t are a corrective code So t alpha of s squared is a little bit Alpha of 2p But alpha is a little bit more Out of the alphas And t alpha belongs to some Galois field Out of the alphas So this is the meaning of narrow sense So in fact it turns out you don't have to start at one You can start at any other power You can start with b plus 1, b plus 2 So on to b plus 2p for any b So if you put b equals zero it's called narrow sense, yes You can even find it And just set all alphas for b from here So I'll have five different g of x For the different alphas Five different what? I'll have six different Zeroes with four elements of degree So will I have to mention all of them Will I have to Each of them will give me a narrow sense So let me answer this question So dimension is easy to find without any knowledge Of minimal polynomial or construction Or something like that So it's okay to do that But if you want to generate a polynomial You need a specific construction with a finite field So that's what you're saying So which primitive polynomial Is used for the construction of the alphas So you can pick any one Doesn't matter It's enough if you do one Because any other field is isomorphic to the scale So you just do some change if you come to this So for dimension is very easy So first thing you'll do is Enumerate all the acetylpomic cosets Multiplication to modular 31 So if you do that you'll get zero And then you'll get one, two, four, eight, sixteen You'll get three, six, twelve, twenty four Seventeen One more Five, ten, twenty, nineteen Sorry, nine I'm sorry Nineteen what? Five, ten, twenty Nine, nine, nine and eight Nine right Nine and eight Nineteen Five, ten Five, ten, twenty, nine, twenty Oh it's not nineteen It's nine Okay All right And then seven, fourteen, twenty-eight, twenty-five Then nineteen Right The last one will be eleven, twenty-two, fourteen, twenty-six, twenty-one Okay So you have these minimal polynomials So what you can do So the best way to answer the GFX question Is not to find the polynomial explicitly So this, this one you know it's like What it is x plus one For this case you simply say m1, m1x Just call this m3x, m5x, m7x and m11x Okay These are all irreducible polynomials of degree five And in some order you can take it depending on how it works So then you can make a nice table of K, T, K and GFX Okay, you can fill out this table How will the first entry be? T equals one you'll have What? You know how to fill out this table, right? T, K, GFX For each T you find the corresponding K So it's better to start with T, GFX and then K So if you don't get confused a little bit It's easy to do with this We're okay All right So if you won't let me write down this table I'll see you next So I'll do that T, K, GFX T equals one The roots are alpha and alpha squared So GFX is the LCM of m alpha and m alpha squared And both of them come as m1x And K is what? N minus Degree of GFX Degree of GFX is 5 So 31 minus 5 is 26 For 2 again, what will happen? It will be m1 of x times m3 of x Maybe I call them 3 as m alpha or 3 of x or something It's all the same There will be 21 and so on You just fill it out Okay, so what's the first question? And then the second question is very similar The third question is a little bit different So the third question is also the same Except that n is not a Let's say we want the 1, 2, 7 It's n equals 65 So you have to first find the field in which you'll have it Or the 65 element How will you do that? It's the only non-trivial part So if you want n equals 65 You want alpha In T of 2 power m 6 times Alpha equals 65 Can someone find the smallest m for which this will be true? Well Yeah, so that's the idea So here this is a nice form So 2 power 6 plus 1, right? So 2 power 6 plus 1 times 2 power 6 minus 1 Will be 2 power 12 minus 1 So if you have an element of order 2 power 12 minus 1 In that C, you'll also have an element of order 2 power 6 plus 1 You see that? So what you do is you pick n equals 12 And you take an alpha belonging to G of 2 power 12 It's this limit So what does that mean? Alpha equals 2 power 12 minus 1 It will be actually 2 power 6 minus 1 times 2 power 6 plus 1 is 63 times 65 So what will be an element of order 65? No simplification, order of 63 equals 65 So you pick alpha power 63 to be an element So that's just to be sure of what to expect But you don't really need to do this So how do you figure out the dimension and all that Without doing these things? All you need is the cyclotomic forces And the multiplication by 2 modulo 65 So we'll simply start with that We'll start with 0 1, 2, 4, 8, 16, 32, 64 Is that correct? No That will end there, right? 63 And then? 61 It should go all the way to 12 It should add 12 So it will be 59 Is that okay? Did I make a mistake here? 128 mod 65 is what? 63 and then? 126 61, 57, no? 49 33 will stop So this will also give you a hint as to What M should be? What is the hint there? From here you can guess that M equals 12 is correct thing to do So this is enough The cyclotomic forces are enough You don't really need to know the explicit field in which it belongs Only if you want an explicit GFX you need that You can simply work with the minimal problem I mentioned That's fine Okay So then the fourth question The fourth question is a very typical question So let's see that So you have N equals 31, T equals 2 VCH code Okay Do you know what a binary symmetric channel or whatever? And you have to decode a receipt vector Artifact for this across every question's property And then what's the first part? Okay, so how do you do these questions? That's the first step Sorry Find GFX Do you really need GFX? I don't think you really need GFX here See So the decoding part You really don't need GFX The first step in the decoding is what? Syndrome computation Syndrome is what? RF alpha, RF alpha part 3 Those are the only two syndromes that will be relevant here And then you just follow the decoding method You know how to do this From the equations And you can go and ever locate a polynomial You find the ever locate a polynomial And then solve for the roots But the first one is quite easy to do In fact, you can do the inspection What is that? I can quickly find the code word Which is a distance 2 from this received word So what will be that code word? All 0s That's right So you can easily guess C naught of X starts to be 0 Okay, why? The distance between 0 and this X plus 7 plus X plus 30 Is actually 2 And I can claim that there will be no other code word Which will be closer than this to the RFX What will happen if there is some other code word Which is closer than this? Then there will be a violation of the minimum distance You cannot do it So that way you can do it But here in this step you don't have a very easy answer Unless you are really smart with these polynomials Or you know the polynomial You can't do it So here you will have to do the actual decoding So go back and make sure that If you have given an RFX like this You can compute the syndrome You can do the next step Find the... Parallocate the polynomial Go through the thing Make sure you can do that That's important Okay? Yes What is the click decoding? You can do it with the GFX and then what? And then you still have to find the syndrome What would be always a multiplicative effect? And you are saying remainder has to be the error So if the rate of the remainder is less than 2 Less than or equal to 2 Yes, you can conclude something like that Because you have a code word plus A remainder The rate of the remainder Need not be less than or equal to 2 But it's not less than or equal to 2 Then you can conclude that I don't know what the answer is for part b Is it okay? Okay, so the first question Gives you a special case of a BCH code That you can explicitly find the exact minimum distance And first one is 6 car like that Okay? And it's an interesting little Construction that I think The BCH code will violate the BCH bond I mean it will not be tight The BCH bond will not be tight If you design for a TRI correcting You expect the minimum distance equal to 2T plus 1 So for some values of T If you pick them peculiarly depending on N You can show for TRI correcting minimum distance will be Strictly greater than 2T plus 1 It's not a violation of the bound That's just that the bound is not tight Okay, so 5th and 6th are like that Okay, I'm going to skip that Okay, so 7th is for instance A question that showed up in the quiz It's a very typical question Okay, so you have 15,5 BCH code 3 Okay And you have a received factor Which is something like 0, 0, 0, 0 0, 1, 1, 0 1, 0, 0, 0 1, 0, 0 Yeah, and you have to find Sticker So that's it Again a very typical question So here you have T equals 3 So you might be able to type up to 3 errors And it's just GS16 So you have to work it out Find out the Syndrome Go through the Exercise And you get the answer So that's 7 And let's look at the 8th problem Okay, so here you have to look at N equals 7, RS code Over GS8 But T equals 2 So this gives you a lot of information From here you can Look at what GSx will do RS code, GSx itself is quite easy It will do GSx X-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x 3 times 6 times 4 Okay, that will be GSx So let's impact one of the questions And they're asked to encode In fact, what is k? k is 3. So, the message is of length 3, you have to encode this systematically. So, you know what the procedure is for systematic encoding. So, multiply by h bar n minus k, then divide by g of x. Whatever reminder you get, you plug it into the beginning, you get to encode. So, that's that. And then part b, this is part a, part b asks you to list the parameters of the binary expanded code. So, over g of 8, parameters are 7, 3, 7. So, when you do binary expansion, what will happen? Each symbol in g of 8 is replaced by 3 bits. So, n will become 21, k will become 9, what about d? I can say greater than or equal to 5. You don't know whether it will be equal to 5 or not. So, you have to check this, but greater than or equal to 5. So, if you want this, it is difficult to say. So, then you are asked to provide the binary equivalent of the code word corresponding to this. So, you look at the code word, go to the vector notation, replace each symbol with the vector notation, you will get it when you do binary expansion. And then part c asks you to decode 0, 1, alpha, alpha squared, alpha, part 3, 1, 0. So, again, what's the method? You have the standard Peterson method. So, it's just equals to g of 8. It's not very hard. It's quick to do. Find the syndromes, form the equations, check the rank. At best, you'll have 2 by 2 matrices. It's very quick to find the determinant of 0. So, it's a typical question. So, let's see, the ninth question. Also, very, very similar culture showed up in the final exam. So, you have to look at n equals 15. So, in case somebody says, reach the element code over g of 16 and does not specify the block length, you have to naturally assume the block length is 15. So, that's the maximum block length possible over g of 16. So, you just take it out. So, n equals 15, rs code over g of 16 and t equals 2. So, this seems to be a... So, you're asked to write down, generate a polynomial and what's written down as a check polynomial. So, what is a check polynomial? So, g of x is going to be the example. It's going to be alpha squared, x plus alpha power 3, x plus alpha power 4. The check polynomial is basically... I think I defined this, you may have noticed, it's basically x power n plus 1 developed by g of x. So, in this case, if you have a form like x plus alpha power 5, x plus alpha power 6 all the way to x plus alpha power 3 and it will also have x plus 1. So, it will be all this. And then you're asked to provide a parity check matrix for the code. So, it's not very hard, you know, the standard parity check matrix. So, it will be a parity check matrix. 1 alpha, alpha squared all the way to alpha 14, 1 alpha squared, alpha 4, so on. I mean, you know how to do this. So, it will be 1 alpha 4, 4 alpha power 8. So, it will have 4 rows and 15 columns. And then the final part asks you to decode or receive vector. So, the receive vector has been given, you have to decode it as far as it is. So, the third question asks you to show that the dual of an RS code, if you take an RS code, take its dual, you get another RS code. So, how do you go about doing it? So, what's the, see, remember, RS code is a click code. You know the GFX. How will you find the gendered polynomial for the dual? So, you have to find the HFX. Yes, the GFX is going to be, suppose it's a T error correcting code. GFX is going to be x plus alpha, x plus alpha squared all the way to x plus alpha power 20, right. So, I'm sorry. Yes, you mean that it's a... It's just a click code. Yeah, so n equals the maximum n that is possible. So, r of alpha equal to 1, that's assumed. Well, for any generic RS code, you can say it's a click code. Or it needs some voting mechanism. Yeah, yeah. Let's see how it works. Okay, so I don't know. Yeah, it may not be starting from 1. So, I think that's the format you picked. So, the first step is to find the HFX. And that's going to have x plus alpha power 20 plus 1 all the way to x plus alpha power n minus 2. And it will also have x plus alpha power n minus 1, which in this case, should I have x plus 1, right? This should be n minus 1. Okay, and then you'll also have x plus 1, okay? You'll also have the additional x plus 1. So, that's how it works out. So, the duals generated for the normal basically work. Yeah, so you have to do it carefully. So, it's going to be x bar k times which of x inverse, right? k is what? k is n minus 2t, right? If we have the generated polynomial here as 2t, k is n minus 2t. Degree of HFX will be equal to k, right? Is that okay? And then you have x bar k into HFX inverse. So, what you do then is you distribute the x bar k throughout and rearrange it. You can find the zeros of the dual. That's the idea. So, what happens is this will be like x bar k times, those things, but I have to put x inverse, right? But then there are exactly k terms in this product. So, you take 1xH and distribute it to each of these terms. So, you'll get something like 1 plus alpha power 2t plus 1x, 1 plus alpha power n minus 1x and 1 plus x, okay? So, essentially the zeros of the dual, it's in some form. The zeros of the dual are basically what? We'll have x plus 1 is there. So, you have 1 and then you have, sorry? Alpha power 1 minus n. So, alpha power 1 minus n would be what? Alpha. Yeah. So, you'll have alpha and then you'll have alpha squared all the way to n minus 2t minus 1. So, okay? Okay, the last thing would be alpha power n minus 2t. So, alpha power minus 2t minus 1, then you put an n in the front. I mean, alpha power n minus 1. So, you just put it in front so that you get a positive, right? And it would be in a proper sequence. So, you go from alpha power 0, alpha power 1, alpha squared all the way to alpha power n minus 2t minus 1. So, basically, there are n minus 2t consecutive zeros in this, in the dual. So, the minimum distance will become n minus 2t plus 1. And the dimension of the dual is basically also 2t, okay? So, the dual will end up having parameters n, 2t, n minus 2t. Which means, and it's also a reach element code the way we defined. It's also MDS and all that. So, in general what you can show is if you have an MDS code, some code that meets the simpleton bound, its dual will also meet the simpleton bound. Okay, so that's the general result. You don't have to be so specific about reach element and all that. At the MDS level, dual as an MDS code is also an MDS code. Okay, so you can't avoid that. Okay, so the element question I'll skip. It's not the dual because it's reasonably important. Okay? So, let's say we look at... Okay, so I'll slightly reword the question. n equals 2 power n minus 1 out of 4 over 2 power n. Okay? And let's say t error corrected. Yes, I suppose c is the script. Okay? The question asks you to show... So, let's say this is CRS. Then you can also look at a CBCH. It's again n equals 2 power n minus 1. The primary BCH. T error corrected. The question asks you to show that CBCH is contained in CRS. Okay, so you have to show this. Okay, it's quite easy. It's not hard at all. Okay? So, look at the zeroes of this code. What are the zeroes of this code? Okay, so you have one... Alpha squared over 2. Alpha squared over 2t. What are the zeroes here? Yeah, so you have to include bicyclists on the crosshairs. But definitely, zeroes are contained. So if you have a GRS of x, which is x plus alpha, but that's x plus alpha for 2t. Right? What is GBCH here? Yeah, so GBCH is going to be LCM of m alpha x m alpha squared x m alpha power 2t of x. So clearly, you can argue from this that that implies GRS of x divides GBCH of x. Okay, so this is a very easy statement to prove. You show first that x plus alpha has to divide GBCH of x. x plus alpha squared has to divide GBCH of x. All the way to x plus alpha power 2t has to divide GBCH of x. And all these guys are relatively trained. They don't have any common factor. So the product of all of them has to divide the GBCH of x. And that shows GRS of x divides GBCH of x. So what does that mean? The code word represented by GBCH of x, GBCH of x is the polynomial. The code word corresponding to it is also a code word of the reach element code. But interestingly, what is happening there? What is so nice about that code word? It has only binary coefficients. There's no non-binary code. So this GBCH code is also referred to as the subfield subcode of the reach element code. So we also knew this before. There are so many other ways of proving it. Just by definition, this is true, right? If you look at how we define it, we said they have the same particle number. I'm saying I'll restrict the code words over binary. So clearly, it has to be contained in the code word. Nothing so great about it. But usually, people define these codes from the zeroes point of view. We defined it from the paradigm matrix that is obvious to us. But usually, people define it from the zeroes and you have to make these arguments to come to the conclusion. So that's 12. So let's go to 13. It's a question that showed up in the finals. So you have ASA in G of 16. Then it takes very complicated questions in the quizzes. So the quizzes will be just simple, straightforward application of what was done in class. I don't expect these kind of questions. I'm doing this just for completion. But these kind of questions may not come up so important for those quizzes. So the question gives you some parity check matrices and you have to find the minimum list. The first one, let's do one. This one is 0, 0, 0, 1. So you have the identity matrix. And then you have the alpha squared, alpha part 3, I'll go with the alpha part 14. Then you have alpha squared, alpha squared squared, alpha squared, alpha squared, alpha squared, alpha squared. So you have to find the minimum distance of this quote. And part B, it's actually more complicated. So you have a similar looking problem but instead of 2, you have 3. So how do I argue for the minimum distance? First one is 3. What's the elimination? Second problem is? 0, 0, 0, 1. So remember, the argument again, you have to be very careful when you argue about the basic matrix for minimum distance. What is the definition? Minimum number of linearly dependent columns. This part nothing will do the rank. So don't think of personal elimination, rank and all that. There's no connection. There's a connection in terms of the bound. Be very careful if you're using the same elimination. So you start with weight 1. Can there be a code word of weight 1? Clearly rolled out, no. What about weight 2? It's going to be a close and not be a close. From 1, 1 to alpha for 14, alpha squared for 14, we know there's nothing on weight 2. Now I think I take 1, 1, 1, 1, 1. So that's the idea. So you have to be careful in that. You have to show that there's no code word of weight 2. So if the code word is supported only here, if these are 0, then I have a code word of the DCH code and error correcting capability 2. So from the random argument, I know there's no code word of weight 2. And weight will be 3. It's a 1 error correcting DCH code or whatever. 1 error correcting DCH code. It's got 2 successive roots, so it's like 3. I think it's a code over GF16. I'm sorry, code over GF16. It's not binary DCH. But anyway, I have 2 consecutive roots. Minimum distance will be 3. Because these gates are not 0. So the first one is non-zero. Then what happens? Can I have a code word of weight 2? Is it possible? Yeah, it's not going to work out. So if this one is, suppose you have 0 and something non-zero here. How do you argue? Sorry? Yeah, you can try a lot of things, but try to use what you know already. We're reworking it. If that is non-zero, then my second parity check matrix, second condition is useless. I mean, second condition is useless. But what about the first condition? It has to be fully satisfied, unless there's no way you can have just one more guy being giving you that. So the first one will clearly evaluate that. You have to at least have 2 non-zero gates there, so minimum distance will be greater than 3. And same thing works for the other way around also. The first thing is non-zero, non-zero, zero also, same way you can argue. So if both are non-zero, there should be at least one guy there, right? So again, you argue that way and show that minimum weight has to be at least greater than negative 3. I mean, if it sounds a little silly at this point, be careful about that. All these steps are required. And you can't just say there is no code word I'm going to. People will make that statement. It's true. I'm not saying no, but it means checking. And you have a word convincing me that you know what the arguments are, not just saying that I know what the answers. So if you have the minimum distance, you have to show 2, 3, sorry. How do you show z equal to 3? Completely find the, this thing, 111 fell out by high point of zero. It's a code word. So minimum distance will be greater than 3. What about next step? No. So, what? It's for 1 to power 14, and for 2 power 14, has an equal distance of 5. 5? Why 4 at a time? It's called t, right? It's called t, right? 3 is considered to be roots. It's for 4. 5, how can it be 5, man? You have 3 rows. Any 4 columns will be linearly dependent. 4. So it's 4. Minimum distance cannot be greater than 4. And I will, it's okay. So you have 4 for that. And then you have to argue the remaining things. It's a bit more complicated, you know? It's not so direct. You have to argue for 3 weight. So what I can do? How do you argue that? Assume I take 1 of the vectors from 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, and 3 of the vectors from there. Okay. So once again I'll say, try to use the constitutive roots argument as much as possible. It's clean and simple and convincing. Okay. You try to do your own reinventing effect. It will just go around and around and around and you'll get confused. Okay. So try to use to the best possible situation try to use the constitutive roots argument. Okay? It's a very standard argument. Right? If all these guys are 0, then clearly it's 4. If any one of them is non-zero, what will happen? You have constitutive roots except if this guy is non-zero. What happens when this guy is non-zero? Then you don't have constitutive roots. Okay? So it's okay when that guy is non-zero here because in this case you have alpha and alpha power 3 and the minimum, the difference between those two powers is 2 and 2 is relatively prime to 15. So even if this is 0, you will have a minimum weight 3 there. You can also show that as in ways you can think about it so you will need a minimum weight of 3 for that part. So 4 will be fine. And then you have to worry about two guys being non-zero here. Two guys being non-zero is relatively easier. You'll have one condition but it will be 4 and all three are non-zero. I mean one more has to be non-zero. So you'll get it. Okay? So you have to do all these steps. Okay? Right? So the next thing I would urge you to try it's not given as a question here. Try the 4 case. Okay? Put an i4 here. Identity matrix 4. Then go all the way down to alpha power 4. There it will not turn out to be 5. So remember correctly. You can find I think the weight 4 code over there. Okay? Go back and look at that. So this is not a way to simply extend your minimum distance. Don't extend like that. It's a place where it stops. Okay? So 14th and 15th are the ways of adding an overall parity check matrix. Again the similar method will help. The way to prove the minimum distance for these constructions is by using a similar method. Okay? So you go, if you've given a parity check matrix which has the reach element structure as part of it and then something else is added to it. You look at particular cases for those things and argue it out like this. Okay? So this is a general thing you can do. And 14 and 15 are about that. And 16 is about this polynomial elevation idea. Okay? So 14, 15 and 16 are really not crucial. So maybe I'll discuss just 15. 15 is a nice, a little problem. Okay? So you have HB, parity check matrix. Let's say Nk, N minus 4 plus 1 are a score. Okay? And this is like a cyclic code. You have the parity check matrix. So it allows its form. Okay? So it will be 1 alpha, alpha squared, alpha over 2, alpha bar, N minus 1, 1 alpha squared, alpha square, alpha bar, N minus 1, all the way down to 1. This will be the last part. N minus k, right? N minus k squared over 2, alpha squared. Okay? So it's an Nk code. It should have N minus k rows. So from there, it's simply going to be N minus k. Okay? Okay? This is H. And then you're asked to consider a parity check matrix, which is not exactly H, but you take to H, and then you do this. You take a 1 here, follow the R0s, and then you put 0 here in the last one of them. Okay? So your graph length has become N plus 2. Okay? The rank will remain the same. Okay? So this will still be N minus k. Okay? So the dimension will become, will still remain k. Oh no, dimension will go to k plus 2. K plus 2. Okay? Is that okay? So you have an N plus 2 comma k plus 2 code. What is the minimum distance is the question? Okay? So here, D equals N minus k plus 1. This path has N minus k plus 1. There's no problem. Okay? So if you have 0 root of here, it's easy to argue that every code word has to have at least the D. Okay? What is one of these is non-zero? Okay? Then you say that the rest of my code word, leaving this out, has to satisfy D minus 1, D minus 2 consecutive 0s. Right? So it's the same from alpha all the way to alpha. N minus k minus 1. Right? If this game is non-zero, the last characteristic matrix may not be satisfied. The last characteristic may not be satisfied. But from the first to N minus k minus 1 has to be satisfied. So 1 to D minus 1. So the rest of this code word has to satisfy D minus 1 consecutive 0s. So which means its weight should be at least greater than or equal to B. Yeah, D minus 2. So D minus 1 plus 1 will become D again. Okay? Same way you can argue for X 0 also. What we do then? Again you have D minus 2 consecutive 0s. So that part has to have weight D minus 1. And already we have 1 here. So D minus 2 plus 1 becomes D minus 1. No, that's also a plus 1. Yeah, okay. Okay, so both sides. Anyway, I mean think about it. It has to be. So the next argument is what if both of these are non-zero? Then you have a D minus 3 consecutive 0s. So the minimum weight there will be D minus 2 plus these two guys will make it Dn. Okay? So we'll put that one strategically so that whatever happens here you have consecutive 0s. Okay? In the previous thing there will be a problem. When I do this, there is a middle thing I have to account for. Okay? If I ask, the common difference relatively prime to N is okay. But if I put 4, then you can have situations where the common difference will be 3 and it will not be relatively prime to 15 and that will not give you minimum distance guarantees. Yes, some crazy things can go wrong here. Okay? So you have to be careful here. So this is a regular extension method. You can extend that to 2 and get the same minimum distance. Okay? So this is like an extension, right? So typically when we extend, we put all ones here. You can also do some all one kind of extension. It's also done. This is not all one type of extension. Okay? So for instance, 14 is the all one type extension. 14 source have C being a RS code with 0s, R for R, R for R, D minus 1. Okay? So it will have D minus 1 consecutive 0s, minimum distance is D. So what we do is, we make a C cap which is also C cap, also C E an extended version which has U of U2, U2 all the way to UN. And then you have UN plus 1. Okay? That's what U1, U2, U1 belongs to C. And then what? Basically, X cap. So UN plus 1 is U1 plus U2 plus 12 UN. Okay? So we have to sign the parameters of C. Okay? All right? Two parameters would be easier. UN plus 1 and K. Right? These two are easy. Why is it easy? UN plus 1 is very easy. Why is it still K? It's just UN plus 1 is another parity. It's not a message. But given the message, message bits of C, we'll find all the U1 through UN. And then somewhere, you can find UN plus 1 also. It's not a new message bit. It's a parity bit. So this will be K. What about minimum distance? We're not going to equal to T. Okay? See, in the binary case, we can't say too much. Right? This is a binary code. The weight is odd. It becomes D plus 1. The weight is even. It is still. It could remain D. But it turns out, for the Reap Salomon code, you can say whether it's even or odd, minimum distance will be D plus 1. Even odd doesn't matter. Okay? The reason is, let's make a parity check matrix for the extension. Okay? So the crucial argument is always the parity check matrix. Okay? So we have one part, which is 1 alpha all the way to alpha part. N minus 1, 1 alpha square all the way to alpha square part, N minus 1. The usual Reap Salomon part goes all the way to D minus 1. And then you have the extension. What is the extension there? I'll put the U N plus 1 in the beginning. It can also be pulled at the end. It doesn't matter. I'm putting it in the beginning. We'll have an old bunch of 0s and then you have a 1. And then this will be... Okay? Or how do I argue the D plus 1? So the first is 0. See, if this location is 0, what happens? I have D constitutive roots. Why do I have D constitutive roots? Alpha part 0 just got added as a root. And this is 0. This gate corresponds to 1 alpha part 0, alpha part 0, square, so on. So from alpha part 0 to alpha part D minus 1, I have D constitutive roots. The way it will be greater than or equal to D plus 1. Now if this is non-zero, what happens? I have D minus 1 constitutive roots. That part will be D at least. And then you also have the first part will be non-zero, so it becomes D plus 1. The same extension argument you can use to go to D plus 1. Okay? So even as D is even, this extension, overall penalty extension for a non-binary DCH or for a reach-aliment code, will increase minimum distance by 1. Okay? So this can be used sometimes. Sometimes people use this. They like this extension a little bit for various reasons. Is that all right? Okay, so I think these are all the questions that I wanted to do for today. We will stop here.