 So, we are going to continue looking at non binary TCH4 and because we ended up last class I gave an example. There is a big window in the example, hopefully I didn't follow too much on the example, otherwise it would have been really really bad. So, let me go through and do this once again in co-opt detail. So, we said N equals 10, Q equals 9 and I kind of made a lot of statements which suggested that GF9, you know the 9 is not prime, so 9 is a composite number, so GF9 is not 0 to 8 modulo 9, that is not GF9. So, I made some statements which kind of maybe people felt it was like that, so how do you construct GF9? Yeah, GF3 square, so you need to find the primitive polynomial of degree 2 over GF3 and then define this as 8 plus 2 alpha, right AB and GF3, so alpha square plus alpha plus 2 equals 3. So, some primitive polynomial has to be set to be equal to beta, so that is GF9. So, after that things were okay, no I mean I think, so if you want to find the primitive 10th root of unity, GF81 will have it, yes GF81, so say some beta and GF81 was the primitive, we notice that 10 n equals 7 divides order of beta, which is what, 80, order of beta is 80 and 10 divides 80, so you notice that order of beta power 8 is going to be equal to, order of beta power 8 will be, order of beta is 80, so order of beta power 8 is going to be 10, so 10 which is what I want, so beta when you raise it to the power 80 you become 1, so if I have beta power 8, I have to raise it to the power 10 to get 1, so that is a simple rule and you can also use the complicated formula, but that is the rule you have to use to figure out this kind of order, so this is what I want, I want an element whose order is 8 in some extension field of 9, and 9 square gives you the answer. So, once you have that, that is the starting point and then what do you do, you form cyclotomic process from modulo of 10 multiplication by 9, here in that part of ok, we did that, there was 2 solution sets and then we have 1, 9, 2, 8, right, 3, 3 what, 3 times 9 times 27, 27 mod 10 is 7 right, so 3 7 and then 4, 4 6, so this part was ok, and then that, what means when we do x power 10 minus 1, of course we will have x minus 1 on x plus 1, I made some statements saying do x minus 1 is the same as x plus 8, so that is completely bogus, so that is not wrong, that is not correct, actually it is, so it keeps a plus 8, because in this case it is what is what we will get, let us not say that the spirit in which I said it was wrong, how do you convert x minus 1, ok, of course we have some many other factors, but what is x minus 1, mod, it is not n, it is mod, oh it is not mod 10, what should I, so minus 1 is in what field, GF 9 or ok, 9, so in GF 9 how do you convert minus 1, minus 1, 2 becomes what, let me see, characteristic is, see in GF 9 characteristic is actually 3, so minus 1 you have to do modulo of 3, so minus 1 will actually go to x plus 2, so that is the correct thing, actually it turns out plus 8 and plus 2 have the same thing modulo of 3, so it is not wrong, but where I said it, maybe it sounds like I have to do mod 9, mod 9 is meaningless, so it is completely wrong, should not do that, this G becomes x plus 2, that is simple, then you have x plus 1 and then you have other factors, so how do you find the term corresponding to 1, 9 for instance, so the 1, 9 would contribute x minus right, beta power 8 times what, x minus, x minus, this is the numbers in the cyclic on the cross set represents the power of the primitive n truth of unity, primitive n truth of unity is beta power 8, so if you have 1 it should be beta power 8 and if you have 9 what should it be, beta power 72, so that is it, if you multiply these things together you will get some polynomial in GF 9 and so that is also true, that will also happen, so far so on you can find all the polynomials, so keep that in mind, yesterday I was pointing out that this minus 1 does not become, it should not be modulo of 9 or anything, so minus 1 is not modulo of 9, modulo of 9 is meaningless, remember in GF q equal to 9 and all the characteristic is 3, so to do modulo of 3, so it is actually x plus 2, so that is the factorization, so once you have this factorization you can take generated polynomials, if you take any divisor of x part n minus 1 that will be a generated polynomial, it will have a 7 degree and that will tell you what to do with this, so that is something to keep in mind, so if you want to explicitly simplify this you need to construct GF 81, at least you should know some primitive polynomials and all that, it is a bit complicated so we are going to skip that but this is the way in which it is done, so let me do a slightly more practical sounding example, so we will take q to be equal to 4, so this is more practical we have a characteristic 2, so I want BCH codes over GF 4, here that is what I want, so n has to be odd now, any GCD of n comma q to be 1 which means n has to be odd, it cannot be even, so we can take n to be anything, so let us say we take n to be 1, so the first thing is why will you have a primitive 15th root of unity and the characteristic you are looking at is 2, so you need 4, 4 squared, 4 part 3, etc. 4 squared itself has it, so you can take alpha along to GF 16, so if you do that you will get an element of order 15, so after that I have to now do cyclotonic process, modular 15 under multiplication by 4, remember that it is 4, it is not 2, because I have 2 equals 4, I want BCH codes over GF 4 not GF 2, if I wanted GF 2 then I have to of course do multiplication by 2, so remember that is the biggest change when you go to non-dual BCH codes, so 0 will be by itself, so after that you will have 1, 1, 4, that is it, 4 times 4 is 16, modular 15 it becomes 1, 2, 8, 32 becomes 2 again, and then you have 3, 12, 48 becomes 3 again, so 3, 12 and then 4 is covered, 5 will be I think by itself, 5 is a singleton guy, 6, 9, that is it, so one thing to remember is this alpha is in GF 16 and that is 4 squared, so any cyclotonic process will have either 2 guys or 1 girl, nothing else is possible, it is about something you can repeat of mine, and then 7, 13, what 13 right, I think that he means 12, and then 11, 4, okay, the only thing that missing is 10, 10 is a singleton by itself, okay, so this is how x part 15 minus 1, x part 15 plus 1 factors over GF 4, we know the factors are very well, but this is what GF 4, so the factorization changes, so you have 2, 5, x plus 1, I am sorry, it is okay, I am in characteristic 2, so Q is 4, once I have Q to be power of 2, I am in characteristic 2, so minus 1 plus 1 is the same, only when I have characteristic 3 or something like in the previous case, when you look at Q equals 9, characteristic is 3, so you have to worry about the minus sign, okay, so here everything is plus, there is no problem, okay, so x plus 1 will be there, and then you also have 5 and 10, okay, so both ways after the dense source, as it makes sense, alpha power 5 and alpha power 10, so they belong to GF 4, that is true, right, see GF 4 is a subfield of GF 4 squared, okay, and now do you find the primitive element of GF 4 and GF 4 squared, you have to find an element of order, order what, that is the order of the primitive element of GF 4, 3, okay, so you have to find an element of order 3 in GF 16, okay, and that will automatically generate the GF 4 for you, okay, that will be a primitive element in GF 4, okay, so because all these things work out, we want to go back and look at the subfield idea, so 0, 1, alpha power 5, alpha power 10 is actually GF 4, okay, since this is the copy of GF 4 that is sitting inside GF 16, okay, and then you have the other guys, so for instance here, explore alpha times, explore alpha power 4, okay, you can see the way the multiplication is going to work, if you multiply these things out, you get x squared, okay, we will get something in the middle, then there we will get alpha power 5, what will you get in the middle, okay, so let's say primitive, so let me say alpha power 4, alpha power 1 plus 0, okay, so what will you get in the middle, 1, okay, so 1 is basically x, so that is clearly normal in GF 4, and it will also be irreducible, we know it will be irreducible, right, it has to work out better, and then what else is there, x plus alpha squared times, x plus alpha power 8, so let's say it doesn't be like that, that's going to be x squared plus, again x plus alpha power 10, right, I would get alpha squared plus alpha power 8 is 1, see, you know, you just get alpha power 8 plus alpha squared plus 1 equals 0, there is no characteristic 2, so none of the cross terms will survive, it will not go away when you start, so clearly this is also a polynomial in GF 4, and you can quickly check that it's irreducible, it's very easy to check degree 2 polynomials, quadratic polynomials for irreducibility, right, but again anyway I know it has to be irreducible because it's coming from the factorization of x plus alpha power 15 plus 1, okay, and then you have x plus alpha power 3 times x plus alpha power 12, here the middle term is a little bit more complicated, but what will it be, okay, the last time is just 1, at what will the middle time be, alpha power 3 plus alpha power 12, what is that going to be, alpha squared plus alpha power 1, and that's going to be, has to be something, it has to be either 0 or 1 or alpha power 5 or alpha power 10, right, it cannot be anything else, so what is that, you should have some table in your previous notes of GF 16, yes, alpha power 3 plus alpha power 12, alpha power 10, okay, so it's going to be alpha power 10, x, it's very believable, okay, then you have the other terms, so if you want I can do x plus alpha power 6, x plus alpha power 9, that's going to be x squared plus something here plus 1, what will that be, alpha power 6 plus alpha power 9, alpha power 5, okay, so that's quite easy to see why it has to be alpha power 5, right, you can multiply this by alpha power 5, and you'll get alpha power 9 plus alpha power 6 plus alpha power 5 equals 0, so alpha power 9 plus 6 has to be equal to alpha power 5, so these are some simple minor tricks you can use to get to the answer quickly, so I'm not done with the factorization, it's continuing on here, so you have x plus alpha power 7 times x plus alpha power 13, okay, and then you have x plus alpha power 11 times x plus alpha power 4, okay, so what will this give you, alpha power 2, alpha power 5 x plus 5, and this one will be alpha power 10, okay, so what about x squared plus x plus 1, why is that not showing up here, yeah, see exactly, keep that in mind, we are now dealing with gf4, right, x squared plus x plus 1 is not an irreducible polynomial of degree 2 over gf4, so you don't show up here in the factorization, you should not show up, then you're made a mistake, okay, because alpha power 5 and alpha power 10 are roots of that polynomial, okay, so this is why we factor x plus 1, and you can do a factorization over gf4 now, or just gf2, previously you could do a gf2, now you can do a factorization over gf4, so if you want gfx over gf, so if you want possible gfx, okay, so you can take possible gfx, okay, you can do a lot of things, for instance I might want to take x squared plus x plus alpha power 5 times alpha power 10, I mean, so any product of these polynomials, factors that we got here would be a possible gfx, but in case if you want to design a bph code, okay, so suppose I want to design bph code over gf4, okay, what n equals, so spinning, okay, so I have to do it carefully, right, so if I want to do bph codes, if I take say p equals 1, what should my gfx be? lcm of minimal polynomial of, only the, no, no, no, should be carefully, so if I say p equals 1, what is the formula for gfx, it's lcm of minimal polynomial of alpha, right, so just alpha will come, okay, so you have basically minimal polynomial of alpha, what about minimal polynomial of alpha? Minimal polynomial of alpha is this k, right, so it's x squared plus x plus alpha power 5, okay, remember I'm not looking at minimal polynomial of alpha over gf2, in which case the answer is easy, x power 4 plus x plus 1, now I want the minimal polynomial of alpha over gf4, okay, so gf4 will have x squared plus alpha power 5, okay, so this way here, you see that, okay, so very similar method, in gf2, if you start with alpha, you can't start, you can't stop with just 1 and 4, you'll go to 1, 2, 4 and 8, okay, so if you take these two ways and multiply them together, you will get x power 4 plus x plus 1, you have to work on that, okay, but now in gf4, I can stop with gf4, so I can have my polynomial over gf4, so this becomes my minimal polynomial, is it okay? No, no, no, see, that's why I said it's over gf4, you see that, it's not a binary dch code, it's a non-binary dch code, so I said b, so 4 array dch code, if dch code is over gf4, it's not a binary dch code, binary means I know the answer already, I mean I don't have to do all these things, we've done this several times, okay, so this is non-binary, so we're doing it in some detail, so what is k? 13, right, so you have 15, 13, 3 code, so in fact this will be an mds code, okay, so if you go to 3 equals 2, what do you get? Okay, so did I get this right? No, I think I'm wrong here, I'm wrong here, maybe I'm wrong here, am I wrong here? See, alpha and alpha squared don't have the same minimal polynomial, right, so I should be careful here with the LCM, I'm sorry, I'm sorry, I'm sorry, I'm sorry, it cannot be an mds code, something is wrong there, so it has to be actually, so let me get that back, I'm sorry, let me be careful with these examples, there's something wrong with that, so this has to be LCM of m alpha and m alpha squared, right, let me do that actually again, I'm sorry, remove non-binary, you'll pay some attention, it's very easy to go around, okay, gfx would be LCM of these two words, in the binary case, m alpha and m alpha squared effects are the same, okay, but notice when you go over gf4, this is your m alpha, right, this is m alpha, what is m alpha squared? This guy, right, so it becomes a different polynomial, okay, so I guess you're getting back your same old binary dch code which is a bit of a dummy thing here, so let me just make sure I'm doing this correctly, yeah, I think that's fine, okay, so you have to get alpha and alpha squared, okay, so otherwise it won't work, yeah, okay, so the next thing will change, okay, all right, so you have this, m alpha and m alpha squared, so in fact, when you multiply those two together, you will end up getting x power 4 plus x power 1, you can check this, okay, so you can multiply these two together and you will get x power 4 plus x power 1, you have to get it, right, so it's basically x plus alpha x plus alpha power 4, x plus alpha squared, x plus alpha power 8, I know how to do the multiplication, it has to be that, now if it equals 2, it will get gfx to be, so now something nice will happen, okay, so you have m alpha squared, m alpha squared effects, m alpha power 3 effects, and then m alpha power 4 effects, okay, so here you'll see these two days are the same, am I right, alpha and alpha power 4 have the same minimal polynomial over gf4, then alpha squared will come, alpha power 3 will come, okay, so you have the product plus these two days, x power plus x plus alpha power 5 times x power plus alpha power n, which is actually x power 4 plus x plus 1 times x power plus alpha power n x plus 1, is that okay, alright, so here for k equals p equals 1, we simply get k equals, k equals 11, okay, right, you don't get anything else, 15 minus 4, here you'll get what, k equals 2 will get k equals 9, okay, so that's a bit different from the binary case, so binary case for t equals 2, you already get 7, if you do them 15, 7, so here you get only k equals 9, okay, so for 2 and a correcting, if you go over gf4, you get a slight advantage in terms of dimension, okay, so this is understandable, we have this to go okay, so this is how we do BCH codes in non-binary, okay, so you have to pay some attention, we are so used to the binary case that very often we make mistakes assuming some things are the same, like I did several times, okay, so you have to pay a lot of attention, make sure, as a real BCH codes will not be MDS, okay, if you see some MDS kind of property then you're going wrong from there, okay, only the read-salomon kind of codes will be MDS, those will not be MDS, those are sanity checks you can use to make sure you don't make mistakes, okay, shall I do one more example, maybe one more length just to get you comfortable with this idea, okay, so let's do one more example, what characteristic shall we use, two or something else, you're okay with anything, okay, so let's pick the same q equals 4, it's not interesting enough, it's nice enough, let's pick a different one, okay, so let me say I pick n equals 21, okay, we did this before with the binary, so we'll do the same thing with the BCH, okay, so n equals 21, alright, is that a question? No, are you okay, alright, so this is how we do it, so basically once again remember, when I do this factor effect of 10 plus 1 and basically finding the minimal polynomials of all the powers of alpha, okay, but minimal polynomial is not in binary, but over GF4, some other intermediate field, so I won't get really larger degree, okay, so if you see, minimal polynomial of alpha are the binary coefficients at degree 4, if you have X4 plus X41, but if you say GF4 is good enough you only have a degree 2 polynomial, so it's something unusual that happens in finite fields, usually in real numbers complex you'll only have degree 2, but no different personality, okay, so let's go to this q equals 4 and n equals 21, okay, the first question is, I have to find n such that 21 divides plus 1 minus 1, okay, so this way of finding it is just try for n equals 1, 2, 3, etc, so 3 here will be good enough, if I take the small enough example so that we can do something, so 3 is good enough, so we'll get 64, GF64 to be of an element, okay, so if you say, so beta is GF64 is primitive, okay, so you select n equals 3, then if you set alpha to be equal to beta power over 3, right, then order of alpha equals n equals 21, so what I want, okay, so once you do this you have your primitive 21st group of unity in a suitable extension field, an extension field of q, now you can start doing cyclotomic process to do the minimal polynomial decomposition of x part 21 plus 1, okay, so let's do that we'll have 0, 1, remember multiplication by 4, modulo 21, okay, so 1 will go to 4 and 1, 4 will go to 16, 16 will go to 64 and that is 1, okay, so you have to stop there, okay, so nice thing about finding things like n equals 3 is, immediately you can figure out some things, okay, the only two sizes possible for cyclotomic process are 1 and 3, all the sizes possible, okay, so that's a useful rule to keep in mind if you're getting confused by this multiplication, okay, you can quickly use that rule to rule out a lot of cases, so 2, 8, 11, that's it, right, it has to stop there, I know it has to stop there, it cannot go beyond that, okay, is that correct, 11 or is it 11 correct? 11 into 4 is 44, modulo 21 is 2, okay, 3, 12, 6, yeah, stops there, okay, then 4 is already there, 5, 20, 17, that's it, right, 0 to 1, so let's check for ourselves, 17 times 4 is what, 68, 68 modulo 21, 63 is the closest multiple, so it becomes 5, okay, and then you have 6 already there, then we'll go 7, 7 I think should be interesting, it should be itself, okay, right, 7 4s are 28, 28 mod 21 goes back to itself, okay, how would I know that 7 should be itself? Yeah, invariably if you have a factor of n, something will go wrong, okay, so that's something simple rule to remember, okay, usually if n is a composite number in your factor of n, something will go wrong, that's one rule to remember, there are other rules I think about, 8, 9, 15, that's it, yeah, I mean it's just a rule, and it's one rule of thumb, I mean it's sometimes, sometimes it works, sometimes it doesn't work, it also, yeah, I mean you can have several rules like you said, so whenever you have a multiple of divisor of 21, then pay attention, okay, it doesn't mean there will always be one, but something can happen, okay, so that's a good rule to keep in mind, 9, 36 is what, 15, 18, then that should be it, 18, 4 are 80, 72, 72 is 9, okay, and then you have 10, 10, 19, 19, 4 are 76, that should be me, 13, 13, 4 are 52, that takes me back to 10, okay, so that gives me some more guys, and then I'll be done yet, 11, 12, 13, 14, 14 is the only disengagement, 13 will be there itself, that's it, I think that's it, 15, 16, 17, 18, 19, 20, okay that's it, you can also comment it in another way, okay so now it's a little bit more painful to write it down, but anyway if I know the degrees for M alpha, M alpha, M alpha square, M alpha by 3, whatever that is everywhere, I know the degrees, so I don't have to really worry about the explicit form for these polynomials, but something I can find out, okay for the single term guys I can find out, this will be a plus 1, what will be this, X plus alpha by 7 or beta by 21, right, so I'll know that beta by 21 is actually GF4, right, so GF4, if you look for a copy of GF4 and GF64, it will be 0, 1, beta by 21, beta by 32, okay, so you have to find an element of order 3, right, order 3 is 21, okay, so that's why it works out that way, so beta by 21 is okay, so this will be a plus beta by 32, okay, so the other way is if you want you can find out it's a bit painful, yes I'm going to skip it, so let's just go to BCH codes, remember I'm saying over GF4, okay, so it's not binary, N equals 21, and if I say D equals 1, I would have worked, I need GFX to be, okay, LCMO, M alpha and M alpha squared, remember M alpha and alpha squared are not the same, okay, so we have M alpha, multiplying M alpha squared, and that has degree 6, so K becomes 15, okay, so as it turns out you won't gain much, so go to D equals 2, what will happen? GFX, M alpha and M alpha squared, then you have to have M alpha by 3, but M alpha by 4 you won't have to help, right, because it's already come, so degree is 9, so K becomes 12, okay, so that's a bit of an advantage, yeah, so if you go to D equals 3, let's go to D equals 3, now M alpha, I'll skip the alpha squared, M alpha by 3, M alpha by 4 you can skip, side you have to have, right, what else do you have to have, both 5, 6 is already there, okay, so only 5 you have to have, so K becomes 9, right, you just have to degree 12, only D squared, I don't know, so that's why we do non-binary DCH code, it's not very hard, yes, so of course the most interesting non-binary DCH code is the Reed Solomon code, yes, go ahead, you can't start with 7, it needs to be a primitive M through the unity, yeah, but consecutive roots, the powers of roots have to be, okay, so you're saying D equals 7, right, okay, so you want to start at 7 instead of starting at 1, okay, so yeah, even you can do that, you can do that, in which case you will have just 4 is what you're saying, okay, that's correct, okay, so those are smart things you can do, okay, so when you look at non-binary DCH codes, crazy things like that can happen, okay, so I think that is correct, no, no, the common difference has to be relatively prime to 21, you can't take sound input, so that is one thing you have to pay attention to, but you can try these tricks, you will get some minor gains, but it will not be that big, okay, so it's the same as saying, instead of starting at 1 and 2, I can start at 0 and 1, okay, so instead of starting at 7 and 8, I can start at 0 and 1, that will give you one advantage, in GF2 and all it's useful, start including 0, GF2 you usually don't have to, because 1 itself will include 2, so you can't do anything better than that, so if you put the X plus 1 also you get 4 minimum distance, it doesn't help you, but in GF4 and all, because 1 does not include 2, you can include 0 and gain something, that's correct, so that's a good point, okay, so once you go to these non-binary codes, all these narrow sense primitive and all make a difference, so if it's not narrow sense, you'll get some of that minimum distance expressed, so you can pay attention to that, that's a good point anyway, anything else, okay, so let's move on to read Solomon codes, which are the primitive codes, so for read Solomon codes, what you do is you make the following choice, okay, you first of all pick Q to be some prime power, so let's go some prime power, okay, and then you take N to be Q minus 1, you have to do the choice you make for read Solomon codes, okay, so when you do that, basically read Solomon codes are Q of E codes of length Q minus 1, okay, so in terms of BCH codes definition, non-binary BCH codes we will do, okay, so if you do that, the first thing we do notice is, if we pick alpha to be G of Q prime power then what do we have? all of alpha equals okay, so that's the first observation, so I don't have to go to any field greater than G of Q I immediately have my element of order n, okay, so we'll have 0, okay, and then we'll have 1, so if you click on the codes, let's remember under multiplication by Q modulo Q minus 1, okay, so what will happen is you multiply 1 by Q, I get 1 by itself okay, so if I take 2, what will happen? in i minus B2, so in fact, I go up to Q minus 1 Q minus 2, I will get similar elements okay, so remember i is that i times Q modulo Q minus 1 is what? same as i right? i modulo Q minus 1 if you like but i, okay so it's a very similar thing to have so basically what happens is all the cyclotomic process becomes single terms, okay, and alpha itself is the primitive element of G of Q okay, so if I want a non-binary BCH construction with P equals 1 I should take G of X to be what? okay X minus alpha times X minus alpha, that's it so I will always have degree of G to be 2 times T and K will be 1 minus T all that okay, so this will be for arbitrary T take an arbitrary T G of X is going to be X minus alpha X minus alpha squared all the way down to X minus alpha times 2T okay, because there's no question of having any common factor everything is a single term okay, so you get the same so essentially what you're using is you're using this factorization for X minus alpha 1 okay, I know that this is X minus 1, X minus alpha X minus alpha squared X minus alpha 2 minus you're using the basic factorization of X minus alpha 1 since my code is anyway G of Q right, so it doesn't matter if they really combine anything so I get everything and here I have K to be N minus 2T and my code becomes NDS so this is one MDS okay, so this is the way to bring in Reed-Salomon codes from the cyclic BCH codes point of view so now if you want a different N, if you want N to be less than Q minus 1, what do I do? it's simply shorter yeah, it doesn't lose any MDF property the same distance goes to and I go to the other direction right, we look at the 0's first okay, so what are the 0's of the PRE, PRE correcting QRE, BCH, Reed-Salomon code of length Q minus 1, it's alpha alpha squared all the way to alpha power 2T okay, I put down the parity check matrix which essentially enforces the same condition as this GFX okay, every code word is a multiple of GFX which means every code word has 0's alpha, alpha squared, alpha power 2T which means it satisfies the parity check matrix that I had said in terms of the loop in terms of these two different ways of doing okay, so in the last 10 minutes I'm going to give you, there was a problem in the assignment which talked about another way of constructing Reed-Salomon codes I don't know if you saw it you might have seen it, so I'm going to talk about that a little bit, it's a very interesting kind of construction, so we'll sign off with that a lot of people use that construction for proving some results particularly in the computer science area okay, so what is the what is the construction here, so this is the original construction according to Reed and Salomon okay, so I'll call it the Palomomol Evaluation Construction okay so what you do is uh... let's say you pick an so let's say you pick an element alpha in some GFQ out of alpha out of beta than I equal to 1 okay, so these are similar to what I had before and specifically if you want to you can pick n to be equal to Q minus 1, take alpha in GFQ as primitive, then out of alpha will be equal to Q minus 1 you can do that if you like, so that's one charge put in the envelope you can do so now what I'll do is I will consider the following set, so let's say what's the sex, some polynomial we've got 0 plus F1 X plus 1 S K minus 1 X by the K minus 1 okay and uh... yeah, okay, so each of these FIs are from GFQ arbitrary, an arbitrary polynomial so how many such polynomials will that be Q power K that's the number of such polynomials right, some that many polynomials now to get a code word of the weak polynomial code what I'll do is okay I will take my F0 to FK minus 1 to be the message vector so I'll take the message vector to be F0 on FK minus 1 so that's a very, a very important message to find the code word I'll do the following, I will evaluate F set let's say uh... alpha I'll prescribe right alpha, alpha, alpha 3 let's say I do that okay, so this is my code word okay, so my code word has 1, 1 okay, and it belongs to each code word uh... each code word, the position coordinate actually belongs to GFQ, so that's very easy and that's my, that's my concept okay, so let's look at minimum distance, okay, so dimension is easy dimension is what? K right, law of length is n dimension is K, what about minimum distance okay, for minimum distance you have to observe that a polynomial of degree K minus 1 over GFQ can have at most how many zeros K minus 1 zeros so if I evaluate this polynomial over n different values how many of these values can actually be zero maximum of K minus 1, right of these n parameters only a maximum of K minus 1 coordinates can be zero, the others have to evaluate the non-zero, why is that? all these alpha, alpha, prn are actually elements of GFQ right, and over a field I can have only at most K minus 1 zeros so that implies automatically the length of code word is greater than or equal to what? n minus K minus 1 which is n minus K plus 1 and I know from the situation that minimum length has to be equal to n minus K plus 1 so that's how you go to the MDS property as a polynomial, from a polynomial evaluation point of view so it turns out this is the same as the read Solomon code, it looks very drastically different that it is the exact same thing as the read Solomon code, you can go back and check the assignment that I gave you there is a problem which shows the relationship between this and the read Solomon code exactly the read Solomon code, if you pick n to be equal to Q minus 1 inside this will be cyclic, yes all those things you can show, it's not very hard so you can go back and work with but this is the basic idea and the read Solomon code construction where you are with MDS, it's because it's related to polynomial evaluation and polynomial has to be K minus 1 so it cannot be and minus K plus 1, so this conceivable roots idea for enforcing the minimum distance, it's kind of RTC shell so you have the random on matrix and all other things, doesn't seem very fundamental if you want to know where it really comes from it's because of these kind of properties it's a much more fundamental property where the polynomial cannot have more than K minus 1 roots, more fundamental than coming up with some random on matrix and saying it's becoming interesting, so this is a nice construction from that point of view but it's not, I mean it's useful particularly in computer science when people do a lot of scary proofs, they use these codes for some constructions but in practice it's not that useful, I mean it's used actually sometimes for decoding the latest power decoding that people have proposed for the Solomon codes, they use these things once again it came from the computer science side so it's really useful, mostly in communication side people will use the polynomial, the other generator polynomial parity check matrix, that kind of ideas are more popular so that's the so the other six is in the function you can check that this code is linear it's really easy to check that it's linear if you take two code words and add you get another polynomial so of course it's linear, for any q minus 1 you can check that this is cyclic, so that's a much more work you can do it, it's not very hard you can check that it's cyclic and then you can also check if you evaluate it like this and if you have K the corresponding powers of alpha will be zeros of this code word you can check that, so you can use some properties of the field and you can show that it's evaluated zero so all those things you can check and so this becomes exactly the redevelopment code and if you shorten it, it becomes a shortened redevelopment code all those things you can check did you have a question? yeah, exactly so that's the other connection it's similar to read miller codes also but there instead of evaluating it as a polynomial with coefficients from GFQ you view it as a Boolean primary coefficient polynomial with multiple variables and then you try and evaluate it so that's what makes it non-MDS so that those polynomials multivariable polynomials don't have such good I mean you don't have such good contrary to the zeros as you do with arbitrary polynomials that's not a field so you look at it becomes a vector space and then you have to worry about how that polynomial works and all that it's a bit more complicated to evaluate okay so I think this is where I want to stop on lectures so we'll stop here