 Okay, so we've been looking at, so we can see in finite fields, just quickly recap so far. So we saw that a finite field has to have p part m elements for p being prime. And then we saw that it's essentially unique, okay. And we saw our construction for it. And for this construction, we need power effects, which is one, reducible polynomial. And as of dx, that's degree m. Then we saw that we could construct for p part m as polynomials, polynomials to coefficients in zp of degree less than or equal to m minus 1, okay. And when you do multiplication, modulo, power effects, you get a field. And then we saw that this field has a primitive element. Here's the same p part m, and we written as 0, 1, alpha, alpha squared, all the way to alpha part p part m minus 2, okay. There exists alpha such that this is true. So this is a primitive element. If you have a method for going back and forth between the alpha representation and the polynomial representation, you can do addition very easily in the polynomial representation. You can do multiplication very easily in the alpha power representation. So it works out, okay. So this is how you construct and work with the field of size p part m. So the only thing here is how do you find this p effect, p effects. And there are tables available for when you're using the BLM. But you can also write a program to find this kind of reducible or primitive polynomials even, okay. So I don't know if you're familiar with these things. There's something about how many primes are there in the first n natural numbers, roughly? How many are there? In terms of numbers, how many prime numbers are there from say 1 to n, roughly? Do you know of any result like that? Okay. Go back and look at it. So it turns out it's a fairly, it's just some number between n and 2n. There are some primes, etc. So there are some results like that you can use. Figure out that if you randomly come up with numbers in a certain way, then you can get to prime numbers. So similarly, there are methods to find the reducible polynomials of the VM. So if you keep picking random coefficients for the polynomials, of course making sure that you don't make obvious mistakes in the thing. For instance, if you keep the constant coefficient as 0 all the time, you'll never get the reducible polynomial. So you don't do some obvious mistakes everywhere. You pick every coefficient randomly to do something like that. It turns out you'll get an reducible polynomial soon enough. So you can write a program like that to find them. So since that is true, one can even write a program to quickly come up with this polynomial. So the construction and the operations of the finite field have used to be clear. And there are some, a lot of certain results that we saw about for what smaller fields will be contained in FB param. Those things are interesting just to give you a structure. And then the next thing we saw was about minimal polynomials. So that was an important kind of idea. So we saw that all the elements of the finite field are roots of X part P param minus X. So that is the result that we saw. And that was a crucial result which tied up everything together. And then we saw that it's in fact X part P param minus X is a product of all reducible polynomials in ZPX whose degree divides M. So that was an interesting result as well. And then we saw minimal polynomials for each of these elements. And so finally all that was basically tuned towards finding roots of a polynomial with coefficients from a finite field. What is the main result about roots of polynomials with coefficients from finite fields? There is an extension field of the finite field over which this polynomial factors into linear factors. So we know that there is a splitting field for every polynomial. How do you find that field to take its degree and then work can see which P param minus one little divide. And then once you see that you can figure out which field will have. So there's a way to easily do it. So one can do that for any polynomial. So that's a nice way of studying all that you want to know about the field. So usually in a field you're interested in finding roots for a polynomial. So that gives you some closure of the interesting. So you know how to find roots for every polynomial. You pretty much studied most of the algebraic properties of the field. If you're on that you're not too interested. So one crucial idea that I did emphasize in the class itself that in the problem set it showed up and I explained briefly how to do it. Is this notion of factoring x-par n minus one over bx. So let's spend some time on this. This is very important in something else that we're going to study. You'll see that this will show up as an important factor. So how do you factor x-par n minus one over z bx? So the first thing is one condition you need is to say n and p should be relatively prime. So this is a notation to say that gcd. So this is basically gcd of n and p. So we assume that n and p are relatively prime. If n and p are not relatively prime what do you do? For instance if n is p times k what can you do? So you can, so it's very easy to, let's see you know that x plus y whole power p is the same as x power p plus y power p. So if it's n is a multiple of p you can write it as x power k minus one whole raise to the power p. It goes from minus one to the power p is when it's odd it's odd at the minus again. It is even and even minus and plus are the same so it doesn't matter. So it's always possible to break it down to a point where n and p are relatively prime. So only from there you have to worry about rules. Up to that point it's just multiplicities. It's just that the multiplicity keeps going. So you can always assume that n and p are relatively prime. So once you say n and p are relatively prime it turns out that there exists n in such that n divides p power n minus one. So this is a result which needs proof but it's quite easy to prove. It's not very hard because if you take all the numbers that are relatively prime to p less than p they will form a multiplicative group. You can show that. Once they form a multiplicative group you can use some properties from there and then you can show it. That's just so much little theorem or something like that. So there's a very standard itself. It says there will be an n such that n will divide p power n minus one. So it's not very hard to show but we're not going to prove it in this class but accept that. So once you accept that things become much nicer. So once you say n divides p power n minus one this implies that exists some beta on p power n such that other than p power n equals n. So that is the crucial idea. So you have an element in p power n. The order of the element is equal to n. How do you find that element? How do you find this element beta and p power n? Find the primitive element. Find the primitive element. So you know the primitive element and then what? What is that I am saying? Yeah p power n minus one by n. So you take the primitive element then raise it to the power p power n minus one divided by n. So that will be an element which is other n. So once you have this it's easy to see that x minus n minus one has to be equal to x minus beta times x minus beta square. So one three, x minus beta power n minus one times x minus beta by n divided by n minus beta by n. It's actually one. So how do you show that this is equal? How do you prove this result? Why? Yeah, so we argue as beta as a root of x bar n minus 1, so x minus beta divides x bar n minus 1 and then beta squared is also a root of x bar n minus 1 and beta squared is not equal to beta, why is that? Yeah, so x minus beta and x minus beta squared are mutually relatively prime factors of x bar n minus 1, they are no common factors. So it means x bar x minus beta times x minus beta squared will have to divide x bar n minus 1, so that way we show all the product of everything on the right hand side has to divide x bar n minus 1 and then it's a very simple observation about equating the degrees on both sides to show that this has to actually be an equality, cannot be just divided, x bar n minus 1 has to have degree n, this also has degree n, this is the only other factor you can have, there is a constant and a constant has to be a 1 by equating the coefficients of x bar n, so it's a very easy thing to show that these two have to be equal, once you make this argument of order. So what we have shown here is the splitting scene of x bar n minus 1 can be quite easily solved, it's a simple look at the smallest n for which n divides p bar n minus 1 and there it is going to split. The next step, so I am interested, so this is nice, it's very good to factor it, the only problem is this is a factorization over f p bar n x, while I was interested in a factorization over z p x, so that's the next step, for that you have to use ideas of minimal polynomials, you know that this element beta has some conjugates in f p bar m and the conjugates and elements beta, when you do x minus beta, x minus beta square, beta bar p, beta bar p square, etc, you actually get a polynomial which is the minimal polynomial of beta in f p bar m and that polynomial has coefficients only from z p, so that's what you want, so you want to prove the powers of beta as conjugates and then get to the minimal polynomials by multiplying them out and that way you would have factor x bar n minus 1 into irreducible factors over z p x, so you know each of the minimal polynomials are irreducible, so that's like trend factorization of x bar n minus 1, so that's the next step and for that we are going to introduce a slightly different kind of cyclotomic process, so when you want to list out conjugates, you don't have to carry the main term there, I am only going to find conjugates of beta with powers of beta, so I can simply drop beta and only use the powers, so I am going to define cyclotomic process modulo n under multiplication by p, so what do I do, first of all I try to find the cyclotomic process of 0, that's going to be 0 by itself, so this corresponds to this, so this conjugates which are beta bar 0 and that's what, basically 1, so x minus 1, so x minus 1 is a factor and you know and that occurs by itself, that's the minimal polynomial of 1, this is beta bar c and then what do you do for c 1, you have to do 1 p p square x bar 1, so it might repeat or it might stop, who knows, so you are doing modulo n and this will correspond to beta, beta bar p, beta bar p square, so on, this will be the minimal polynomial of, so basically this corresponds to the polynomial x minus 1, that corresponds to the minimal polynomial m dot by x, the next step what do you do, you look at a number that does not show up in the previous things and find the minimal polynomial of that, find the cyclotomic process of that, so what you can show likewise is this result that we had before, the cyclotomic process, remember this will be modulo n, the cyclotomic process at the former partition of 0, 1, 2 and minus 1, so you can show the similar result as we did before, so you understand what I am saying, so when I say partition what do I mean, you know that all those things is equal to 0, 1 and minus 1 and then what is the next thing that is implied by the partition, they are either equal or distant, so if you take the cyclotomic process, they will be either fully same or they will be distant, so you can throw away, so you can throw away repetitions and only take one representative and you can get it, this joint union of cyclotomic process which equals 0, 0 to n minus 1, so the best thing to illustrate this is an example, let us take a simple example, I will take n equals 9, so let us look at the cyclotomic process for 0 or we can do just 0 and then 1 is going to be, so let us say p equals 2, 1 is going to be what, 1, 2, 4, 8, then are you going to get everything, that seems a bit odd, you will get 7 and then 5, so you get 6 elements and then what is remaining, see 3 is remaining, so that will be 3, 6 that is it, so here is the disjoint partition of 0 to 8 through cyclotomic process, you have 0, you have 1, 2, 1, 2, 4, 5, 7, 8 and then 3, 6, so this is the cyclotomic process of the condition, so what it means is, what is it mean, it means that if you find an element beta in some f2000, so that order of beta equals 9, then what can you do, x bar n plus 1 in cyclotomic process, x plus 1 corresponding to 3, 0, then there will be a factor which is x plus beta power 3 times beta power 6 and that will be corresponding to see 3, but you know what will this product be, it has to be x square plus x plus 1, why should it be x square plus x plus 1, that is the only irreducible polynomial of degree 2L in x square, so this will be x square plus x plus 1, that is just something that you can find, then what will be the third term, so x plus beta times x plus beta square plus times x plus beta power 4 times x plus beta power 5 times x plus beta power 7 times x plus beta power 8, so this guy is going to be the minimal polynomial of, because he corresponds to C1 and it is the minimal polynomial of beta and what will be its degree, 6, so you know its degree is going to be 6, so this is how it works, so I think I did this in last class also, in Monday also I did it with explicit beta, what was the explicit beta that we found, we found beta to be equal to alpha plus 7, where alpha is the primitive element of what, C of 64, so we took that field and we found the primitive element and we found beta exclusively and then I found the minimal polynomial of alpha plus 7, so this cyclotomic process multiplication by 2 modulo n simplifies all that, so you can, before even finding out which field contains beta, you can quickly find out the degrees of the minimal polynomials for each element, so you don't have to do that next step, so anyway this is what it is going to work out, anyway you are going to find only conjugates for beta, you are all going to find conjugates for some other, only for beta, but also beta, so you can deal with beta itself directly and you can drop the beta from your notation and you get, so maybe I will do one more example, just to try from the point, what n shall we take, let's take something like, let's say n, we will take p equals 3, so how do you find cyclotomic process, so c0 which is 0 by 4, I hope this is not too trivial, c1 is going to be 1, then what, 3, 9, 6, 7, is that okay, remember modulo 10, so 7 times 3 is 21, that modulo 10 brings you back to 1, what about c2, 6, 2, 4, 2, 6, no, 6, 8, so am I right, that's it and then psi is the letter by itself, am I right, okay, psi times 3 is 15, both that is okay, so if you factor x bar n minus 1 over z3x, you are going to get x minus 1 as one factor, what's going to be x minus beta part 5, it's a irreducible polynomial of degree 1, what is the other irreducible polynomial of degree 1, x minus 2, so it's going to be x minus 1 times x minus 2, that is a minimal polynomial of degree 4, another minimal polynomial of degree 4, irreducible polynomial of degree 4, irreducible polynomial of degree 4, it's a bit difficult to figure out what they are directly, if you know the list, you can maybe figure out what they are, okay, so let's see what will be the extension field over 3 in which you will have other 10 elements, 4 right, so 3 power 4 minus 1 is 80, 10 divided by 80, so you can go to an explicit g of 3 power 4, 30, 81, then work with the minimal polynomials there and find these polynomials explicitly, if you want to find them all, you can do computations in g of 81, configure it, is that okay, so this is where we work it, but you cannot, you don't have to do that, okay, the main thing about the emphasis is you can find the degrees of the minimal polynomials without knowing anything about the field, right, only with this simple cyclinomic process algebra, this computations, you can find the degrees and it will turn out that is important for us for some other reason, okay, explicit polynomials also have to be found, if they have to be found, you go to the field extension field and do the computation, but if you are only interested in the degree, you can find any problem, is that okay, perfect, so that's the, that's kind of the point where we are going to start looking at finite fields, then move on to codes once again, does any question please go ahead and ask it, is that correct, wait, the degree of, okay, so the claim is you are saying the degree of m beta of x is always equal to the number of elements less than the other and relatively prime to what, to the other, is that correct, it's, is that correct, so here if you take, yeah, 1, 3, 7 and 9, so I don't know if you should get all the numbers, that's the only thing I'm thinking of, will that come, so sometimes it may not, so it depends on whether or not 2 or 3, the prime p is the primitive element in the, in what is that, the sudden star, you know what the sudden star, the sudden star basically, so if you take all the numbers from 1 to n minus 1, which are relatively prime to n, you put them in a multiplicative group, modulo n, you can do that, that is the proper multiplicative group here, in that element it depends on whether or not p is primitive, if, if your prime p is primitive in that, then you will get everything in that thing, but if in case p is not primitive in that, you may not get and I don't know if in general every prime p is a primitive element in every sudden star, that may not be, so you might have exceptions, I mean, you might be able to quickly find the p equals 2 itself, in some cases. Is it necessary, so for instance if you take, if you take p equals 2, n equals 15, right, if you take p equals 2, n equals 15, you will have only 4, that's the degree, okay, right, and I think there are more than, there are in fact 8 numbers which are less than 15 and relatively prime to n. So the only thing you can say is it will divide what's called c of n, c of n is the number of numbers less than n relatively prime to it, but all the c functions, it will divide the all the c functions, it's the only thing you can say, it need not be equal, depends on whether or not it's primitive, so for instance the example I gave you, if you take p equals 2 and n equals 15, you will only 4, on the other hand you have 8, if you have 15 and 8, 8 numbers, okay, any other question? Okay, so we're going to go back to codes once again, so let me see if you remember, I found a departure and getting into finite fields, why did we get into finite fields, where did I depart from binary linear block codes, then I gave you a reason, we were constructing some codes and I gave you a reason for code constriction which will take us towards finite fields, what was that reason? Minimum this, yeah, d equals 5 and greater, right, so up to d equals 4, we saw that we could optimally construct some nice codes in binary field, but we could never construct a code with d equals 5, we never had a nice algorithm for doing it, it was not obvious that how you guaranteed d equals 5, so that was kind of a point of departure and we said the one reason why you study finite fields, there are several reasons, one reason is to see whether we can come up with constructions for parity checkmate, this is our finite fields, where extension fields are the binary fields, where d equals 5 and all can be guaranteed, so it turns out it's true and that was kind of like the starting point of everything, I mean that's why finite fields became important and all that, the construction of that code goes back to those Chaudhary and Huck and Hamas, I don't know how to exactly to pronounce the third name, the BCH, okay, so that construction really started everything, so what I'm going to do is to provide that construction first in a kind of a high level and then tell you why the minimum distance property comes and then we'll go back and we'll get some more fundamental results about these codes, okay, so what we're going to study next is basically BCH codes and their construction, mostly from a constructive point of view and then we'll see more properties, in fact concept BCH codes are a special class of what are called cyclic codes, we'll study cyclic codes more generally and see how it works out, okay, so we'll see that later for now we'll begin with the BCH codes construction, yes it's also historically accurate and on top of that it's also a very nice simple and elegant construction which is very powerful, there's really no other construction which beats the BCH codes construction in many ways, so it's a very nice construction to know, okay so remember what the definition was, let's go back to our linear course, you have a linear, so some of the different binary BCH codes for now, we'll go back to non-binary like maybe even something else, okay so we'll do that maybe even non-binary, so I don't know, so let's think about codes, so if you have a linear block code with the binary BCH matrix, okay, so there's a way to relate minimum distance to edge, so what is the definition of minimum distance, so minimum distance from its definition equals what, minimum number of problems that are linearly dependent, so you had a problem in your exam for instance that the BCH was defined over GF16 and the way to go about doing that is to find the minimum number of columns which are linearly dependent, that problem was really really simple, there are only two rows, so once you do elimination and get it to I, it's easy to see that there should be three columns that are linearly dependent and no two columns will be linearly dependent, that will require some proof, you have to look at it and it's easy to check about, so you can do that, for simple cases you can do that, okay, but in general we want to be able to use this definition and construct a suitable edge in which something like this can be guaranteed, you can find based on the construction of edge a number of B's such that lesser than or equal to less than D columns of H are always linearly independent, that's what I want to guarantee in my construction, I'm going to use finite fields for that, okay, so what I'll do is I'll construct a parity check matrix from some finite field, GFP param, okay, so let's not worry about the code for now, we'll simply construct a parity check matrix from some finite field GFP param or for simplicity maybe even GFP param, it doesn't matter why we construct it from some GFP param, so which I will be able to give such a guarantee, I will be able to guarantee that there is a number of B, okay, such that less than D columns of H are always linearly independent, okay, so that's my first starting point, I'll just give you the construction, there's no way to derive the construction, I'll just give you the construction and we'll see why that is true, okay, it's a very simple result, we'll see that, okay, so for that, we need some starting points, so let's say I'll move the block line, push the block line to this key which determines the number of columns of H, okay, so that I need, okay, so I need to know how many columns will be there in H, okay, total number of columns and now I want to come up with the construction, so for that, first thing I'll need is, I'll need an element beta in some finite field, it's a fb param, it doesn't matter, so we need there, so this is always possible, this is not very hard, such that order of beta is greater than n, okay, so that's just some number, so it's always possible, so for any n, you can always find a finite field large enough, such that there will be an element of either, it's greater than n, in fact equal than itself is possible and why not, greater than n is possible, every p is possible, so you just pick n large enough so that beta n is greater than n, that's the simplest thing you can do, then take the primitive element of the p, that will be, that will satisfy this, you don't have to necessarily take primitive, so many other elements which are greater than n, okay, so I'll start with this beta and the construction will be using this beta, okay, so the order of beta is greater than n, the next thing I'm going to do is the following, okay, so I'm going to say my edge to look like this, okay, first row is going to be 1, beta, beta squared, so for example beta squared, n minus 1, okay, so next row is going to be 1, beta squared, beta squared squared, for example, beta squared, h squared, n minus 1, okay, so all the way down to 1, beta squared, b minus 1, beta squared, b minus 1, whole squared, so on to beta squared, b minus 1, whole to the power n minus 1, this is my construction, okay, so this is, this you can call as a dch construction for a parity check matrix, okay, so you can call it dch construction, and it turns out in this matrix, the claim is, b minus 1 or fewer columns are linearly independent, okay, so what's the punchline and I'm giving it to you at the very beginning and we'll prove it, which is also quite easy, and the construction is also quite simple, I mean, if you think about it, it's, it's, it's generic, you can use it for any, for any d you want, right, so it's a very simple generic construction, very easy to come up with. So for instance in your quiz, the construction follow exactly this method, right, you can see that the question had the first word was 1 alpha, alpha square, alpha power 3, alpha power 4, and the next word was 1 alpha square, alpha power 4, alpha power 6, alpha power 8, alpha power 3, I just use this construction, okay, so it's a generic dch construction for a parity check matrix, and you can show that if you have beta, beta square all the way to beta power d minus 1, and you do this repeatedly, repeatedly for all these things, d minus 1 or fewer columns will be linearly independent. Okay, so the way to prove this is to assume the opposite, okay, and then we will show that there is a contradiction, okay, so if you say, well, not necessarily, we can also directly prove it, we'll directly prove it, okay, so suppose I take some, I don't know some d columns, okay, which are with d less than or equal to d minus 1, so let's say v is less than or equal to minus 1, and you take columns, okay, so I'll number the columns from 0, 1, 2 all the way to n minus 1, okay, so these are basically numbers for the columns, just so that I can refer it, refer to it, so I'll consider d columns, let's say i1, i2, all the way to ib, okay, so these are basically the numbers here, okay, so if I want to consider the first, first d columns I'll be looking at 0, 1, 2, d minus 1, okay, so for any other set of d elements here, those are the columns that I'm looking at, okay, so I'm going to pick up the b columns, so let's look at and see if these b columns have to be linearly independent or not, so what are these b columns, okay, so the b columns are actually what, okay, so we make a sublet of grids, what is the i1, ib, right, what is the i1 column, beta column, i1, beta column, 2, i1, all the way down to beta column, d minus 1 times i1, what about i2 column, beta column, i2, beta column, 2, i2, all the way down to beta column, d minus 1, i2, what about the last one, beta column, ib, beta column, 2, ib, down to beta power b minus 1 that is my sub matrix. Now, I want to argue that this matrix has full column rank. So, once I do that I know that the columns of this matrix are linearly independent and that is done. So, for that what I will do is I will only take the first b rows. So, I will take only the first b rows. So, why would I be doing only the first b rows? So, then I get a square sub matrix and I will show that the determinant of that first b rows is actually non-zero. Square matrices I have determinant and determinant we can show as non-zero. Once you show determinant as non-zero what is nice? The columns are linearly independent. So, once the first b rows itself has linearly independent columns then if you add more linearly independent it is not going to change because b itself is linearly independent. So, you are over. So, that is the strategy in showing that. So, let us look at the first b rows. So, if you do the first b rows what do you get? You get what is known as a Van der Beijn matrix. So, it is a very standard matrix. So, you will get something which looks like this. So, let me have this a. So, basically you will get beta power i1. I will write it a little bit differently. Beta power i1 square all the way down to beta power i1 raise to the power b. So, I will write it and I will take that i1 inside and take this outside. You get beta power i2, beta power i2 square all the way down to beta power i2 raise to the power b. This is a simple thing. So, I have beta times i1. I am pulling i1 inside and taking b outside. So, all the way up to the last one which is beta power ib, beta power ib square all the way down to beta power ib power b. So, such a structure is called the Van der Beijn matrix and it turns out the determinant of the Van der Beijn matrix has an explicit expression. And what is their explicit expression? So, it turns out the determinant here is equal to product beta power ij minus beta power ij prime j not equal to j prime. And I think you have to say j should be smaller than j prime or something like that. So, you should have some convention so that the sign is accurate. But the sign I do not care about. So, it does not matter. So, what does this give me? Once I know that this product is true, since the order of beta was greater than n, there is no way the two of these elements will be the same. No two beta i1 and beta ij and beta ij prime will not be equal. So, what does this mean? This product is non-zero. I picked my beta such that the order of beta was greater than n. Once you have ordered greater than n, then I take this constitutive powers, there will be no repetitions. So, beta power 0 all the way to beta power n minus 1, there is no repetition. So, here I am taking simply difference and product. So, none of these differences will go to 0. So, all these differences are non-zero. So, product of them is non-zero. So, that is the way you show that if you take any b columns for b less than or equal to d minus 1, you can take a suitable sub matrix b by b sub matrix and show that it is determinant is non-zero because of this random on structure. So, how this random on structure came up is by construction. So, if you look at how I constructed my hedge, I picked consecutive powers of beta. I picked beta, beta squared, quantal beta by d minus 1. So, I did not pick it up as consecutive. If I change the sequence, if I pick like beta, then beta power 3, then beta power 10 like that, then I will not get the random on matrix. I have to pick it consecutively, only then I will get the random on matrix. Is that clear? That is why the structure came up. And I could use the random on matrix result and get the determinant easy. So, this is a specific construction. It is called the BCH construction. And what it gives you is powerful. It gives you this minimum distance property. This is the minimum number of columns that are linearly independent. So, I should be slightly careful. It says it gives you a lower bound on a minimum distance. So, once you have this, you can conclude that the minimum distance is greater than or equal to d. How do you show the minimum distance is equal to d? You have to actually find d columns which are linearly dependent. That you may or may not be able to do depending on what is said. But in this case, you can actually say that d columns will be linearly dependent. How do you say that? Number of rows itself is d minus 1. So, if you take d columns, clearly it will be linearly dependent. So, if you allow for coefficients from the field, then you can get d columns to be linearly dependent. But d minus 1 of few rows cannot be linearly dependent. So, that is the power of this construction. And it is very generic. I mean, you give me an n, you give me a d. I can always go to a large enough field and give you this. So, that is the power of how easy to use it is and how powerful it is in guaranteeing minimum distance. Is it okay? Beta power? What is the beta power? What is the? It is not beta power, it is the difference. Well, I mean, okay, so be slightly careful when you do these factors and all that. So, these are elements of a field, you know. So, anything will be a factor of anything else. You know, you are working with the field, right? You are not working in a ring or something where some things do not divide. Everything divides everything else. So, if you are working in a polynomial setting or something or if you are working with integers, then it is interesting to show factors. But here, all my elements are field elements, okay? So, everything divides everything else. So, there is a notion of factoring. Factors is not very nice in a field. There is no notion of factor. Everything divides everything else, okay? Only in rings this factorization becomes interesting, okay? So, here I am not thinking these betas as polynomial objects and all that. These are entries from a field and for me factors are irrelevant from a field, okay? Anything else? That is the determinant, okay? Actually, I can go back and check this if you like. Maybe you need one on the top is what you are saying. Okay? So, maybe the dynamite expression will have one on the top, okay? So, maybe there are some additional factors here. There could be some constant here or maybe some C not equal to 0. So, what if I want to write out an IV sum in this invisible plane? It doesn't matter, right? Power of beta will never be 0. See, beta is non-zero, right? It will only be 1. I mean, it doesn't matter. So, some constant is fine. Okay? So, maybe I should have just put that proportional, it should have equal, okay? See, non-zero constants up front doesn't make any difference. It will never make it 0. So, I am only interested in differences, okay? So, only the differences are important. Okay? It doesn't matter. So, yeah, in the dynamite expression, maybe there is like a beta or I1, I don't know. It doesn't matter. So, it's something up front which is non-zero. Okay? All right. So, let's send very nice statements about dynamite matrix. You can go and look it up. So, you can also view it as a, like, each of the term you can view it as a polynomial. If you think of beta as an unknown, you think of a polynomial, you can show that this essential polynomial you get in the determinant. The only factors it has are the form X bar I minus X bar. So, it's a very strong result. I mean, there's no way we can avoid it. The only way to make this determinant 0 is to have some two columns be identical. No other way in which you can make it 0. Okay? All right. So, okay. So, this takes a bit of digesting. So, let's see some examples. Okay? Let's see some examples. Okay? So, start with a very simple example. We're going to pick n equals 3 and we'll take beta as your g of 16. Okay? So, order of beta equals the same. There's no problem. 2 and minus. Okay? Suppose I want to say d equals 3. Okay? d equals 3 is a bit interesting. But we know that this is not very hard. But still, let's just start with d equals 3. Okay? So, what will you have? In the first row, you have 1, beta, beta square, I'll go into beta, 14. Second row, we'll have 1, beta square, beta star, I'll go into beta star, 28. Okay? So, here it's guaranteed that no two, no column is clearly 0. Okay? Minimum distance is not 0. Okay? Next, you have the question of can two columns be linearly dependent on this? Okay? So, it cannot be. You can prove it in various ways. One way to see it is beta star i, beta star 2i and beta star 3, beta star 3. So, things like that. I mean, these two cannot be linearly dependent. Right? So, what you're doing, you're doing, fundamentally, a nonlinear operation to go from here to there. You're spoiling each element to go from here to there. There's no way there will be a linear dependence. Okay? So, it cannot happen. Okay? So, that way, you can kind of intuitively also argue that it has to be a chord. That's to a chord. But how do you show that three columns will be linearly dependent? Okay? So, that is easy. So, there are only two rows. So, if you take any three columns, the rank cannot be three. It has to be, anyway. It cannot, it has to be two. Right? So, only two things can be linearly dependent. Third one has to be a linear combination of these two. So, you can find that chord. Okay? So, next important thing to worry about is, so, H is a target check matrix. You want to increase some gf-16. Okay? So, I'm defining a chord over gf-16 now. Okay? So, what if I want a binary chord? Right? That might be the next question you might want to ask. I don't want the, I don't want a chord over gf-16. I have a, actually, a bit skew transfer. What do we do with gf-16? I want a binary chord. So, how do I get a binary chord? This is the next question, which is also equally important. Okay? So, we'll address that. That's why the idea of binary dc-h chord versus non-binary dc-h chords will come. Okay? So, right now we're thinking of just the five-digit matrix and properties and we're not worried about this, whether the chord you have is binary or non-binary, etc. Let's say if I'm seeing this chord over gf-16, it doesn't matter. There's some alpha, beta and all showing up in this chord. We'll think about how to deal with it later. Okay? So, let's go to d. So, I could do d equals 4. The d equals 4 is not very interesting because I want one more error correction capability, right? The next step is to go to d equals 5. So, here you're going to have 4 rows, right? If I go to d equals 5, I'll have 4 rows. What are the 4 rows? The first two rows would be the same as d equals 3. So, the third row is going to be... So, if you want, you can simplify these things. It's not so crucial. Is that okay? So, that would be the characteristic matrix that you get and it has minimum distance equals 5. Okay? So, that's how we go about constructing it. So, if you want, you can keep doing this. If you want d equals 7, then you have 6 rows in your biotech matrix and you have d equals 7. Okay? So, like this, you can keep proceeding for any other case. Is that okay? Any questions? So, this is the essentially DCS construction. And like I said, it's quite a powerful idea and we'll see how powerful it is as we go along. But before we proceed, there is this confusing notion of how do you go to a binary code from this code. So, it seems like the idea of a finite field was quite important here. So, you have to pick an element deta, order of deta greater than or equal to n. So, you need at least n to be ordered. And you're going to a larger and larger field. So, how do you go to a binary field from it? So, that's an important question. And it turns out there are two ways of doing it. How to come back to a binary code from this code over gf2 power n. So, there are two ways of doing it. And those two ideas give you two different kinds of constructions. So, the first idea we'll see is what's called a binary BCH code. The next idea will be the non-binary BCH code. We'll go to the non-binary a little bit later. It's traditional to see the binary BCH code first. So, we're going to see a binary BCH code. So, the main idea and the binary BCH code is to say it's to do the selling. So, you have binary codes from non-binary parity check matrices. That's the main notion that we're worried about here. There are multiple ways of doing it. Like I said, the first way we'll look at is a very simple, it's a class key, the first idea that came about. And for BCH codes, it gives you a very nice handle of those. Suppose I have H being a n minus k plus n matrix over gf2 power n. So, we'll restrict down to usually p equals 2. So, for p not equal to 2 also, you can do similar things. But usually p equals 2 is the case that's of interest in practice. So, we'll pitch it there. So, you can first define a code over gf2 power n. How do you do that? Set up our code words in gf2 power n and what is gf2 power mn? What is this quantity? What does this get? Interface from gf2 power n. Interface from gf2 power n or the n-dimensional vector space over gf2 power n. So, for x times c trans pro equals 0. I could do that. If this gives me a code over gf2 power n. So, one way to do a code over gf2 is what's known as a subfield subcode idea. It's a subfield code. The words suggest themselves. So, this idea is used in the BCS code construction. I'm going to say I'm going to note that all c's are only in gf2m. I won't go to this gf2 power mn. I want the code words only in that. But then, the condition they have to satisfy is the same. I won't change the condition that they have to satisfy. h times c trans pro equals 0. That's the same. There's no change in that. But then, I'm going to only look at code words in gf2m. The reason why we have subfield this. gf2 is a subfield of gf2 power m. Why do we have subcode? Because this will be a subcode of that. It cannot be the entire code. It satisfies the same constraints. Clearly, it belongs to that code also. It's a subfield. So, it belongs to that code also. But then, it will not be equal to the code. It will be only a subcode. So, some questions that are important is, So, this code, this will be an mk code. Lock length will be n. Dimension will be k. What about for the subfield subcode? n is all right. But what about k? It's not clear what k is. What is the dimension of this subfield subcode? That's not clear. So, when you define a binary code with a Taldichick matrix over a larger extension field, when you think of the bigger code over a large extension field, everything is very clear. It's a nice linear black code with the Taldichick matrix also coming from the same thing. It's fine. But when you want to look at its subfield subcode to define a binary code, lock length will be easy, but you have to worry about dimension. It's not clear what the dimension will be. Yes. What's the question? Hello. There are Taldichick matrix over gf code already, but what is the definition of a linear black code? It's a subspace of 0, 1, n, right? Subspace of n dimension of the binary code. Is this a subspace of the n dimension of the binary code? Yes. So, that's your answer. Really, I love to have. No, nobody is going to stop you. There's nothing to stop you from that. As long as I have a subspace of 0, 1, n, I have a linear binary black code. There's no problem. However, I construct it. It's up to me. I'm constructing here through an extension field. What does it give me? What is one of the properties of the subcode? Subcode. Minimum distance is still greater than or equal to the original minimum distance, right? Right? If this is nkd, minimum distance is still greater than or equal to. Is it okay? Even for the subcode. Those are the subcodes, right? The entire code has minimum distance d. Really, the subcode will also have minimum distance d. Okay? So, what this subfield subcode idea gives me? I mean, you might ask me, why do you go to all this trouble, go to that big field, and then come back? It gives me a guaranteed minimum distance. Okay? So, in primary world, I couldn't do this. Right? I couldn't directly construct a d equals 5. So, here I have a back tile digit matrix. It is over g of 16. Yes, I mean, it's a bit of an inconvenience because I can't find the dimension easily. I'll show you how to do it because all these problems have been overcome easily, at least for the BCS codes. It's easy to overcome this problem. It's not very hard. Okay? But then, I had a guaranteed minimum distance, and that's much more important too. Right? I can correct two errors now. Previously, I could never come up with a code that would correct two errors. I can correct two errors. Okay? So, we'll start here for now. And the next class, which is tomorrow, we'll look at how to find dimensions of the pink code in these problems.