 Okay, so the main topic for the day is we have seen most of the definitions already, it's not going to be a new definition for you, so let me make it once again, so we'll start with n being a block length, okay, and we'll pick once again element beta on some t of 2 per m such that number of beta is greater than or equal to n, okay, so we'll do that, okay, and then next we'll design a paradigm check matrix, okay, suppose the design error correcting capability say t, then you have 1 beta over beta squared all the way to beta squared minus 1, 1 beta squared, beta squared, all the way to beta squared minus 1, and down to beta part 2p, beta part 2p squared, all the way to beta part 2p squared minus 1, so I thought I've not departed much from the DCH codes, seems like it's the exact same thing that the Reed Solomon code, here RS code, the t error correcting RS code of length n is basically set of all c, here's the big change, the big change is that you won't expect your code words to come from gf2, you'll have your code words to come from gf2 per m, okay, such that h times c transpose, okay, so the big change is in that escape, okay, instead of gf2 here, I put gf2 per m, okay, so this simplifies a lot of the computations, it also adds to a lot of the computations, I'll show you how it works, okay, so that's the first, is there already any questions, okay, so have a paradigm check matrix from gf2 per m, and add the code also from gf2 per m, so everything is very easy to do now, it's not very hard, okay, so the block length is n, and to find the dimension, all I have to do is find the rank of h, and the rank of h will be equal to n minus k, okay, and it turns out the rank of h in this form is in fact equal to 2p, okay, so one can show rank of h is equal to 2p, okay, so in a way there's ways of showing it, in fact I'll determine the dimension directly for you, okay, so that's what I'm going to do, I'm not going to find the rank first and then find the dimension, I'll find the dimension directly and then from there we can go to the rank, okay, also we can, so how do you find the dimension directly, okay, so for that what you do is you look at this hc transpose equal to 0, okay, then look at the first row, what is the first row tell me, it says, right, so if you think of the code word cs, c0, c1, c2, I'll do it cn minus 1, remember each ci is from what? gf2 param, okay, so it's not g2, I can now write a code word for the novel, it's c0 plus c1x plus, so one can cn minus 1x and each coefficient is from gf2 param, okay, now the first row says c0 plus c1 beta plus c2 beta squared, so on till cn minus 1 beta per n minus 1 equals 0, which is the same as saying cf beta equals 0, okay, I have a polynomial with coefficients from gf2 param and it has a root beta n gf2 param, okay, so that implies 1, okay, cfx should have some factor, what should be the factor of cfx, cx plus beta provides cfx, okay, it should be actually minus beta, where is that put plus beta here, this characteristic is 2, so it doesn't matter whether it's minus or plus, actually it should put minus, minus is the same as this, now likewise I can do for the second row, what happens to the second row, c of beta squared is 0, it should be 0, x plus beta squared has to divide cfx, so likewise you go on the way to the 2p row, it says c of beta per 2p equals 0 equals, x plus beta per 2p divides cfx, okay, so you put all these things together, what can I say, the LCM of x plus beta, x plus beta squared, so on till x plus beta per 2p has to divide cfx, okay, what is the LCM of all these things, the product, because all of them are irreducible, degree 1 guys, so from there you see that implies x plus beta times x plus beta squared all the way to x plus beta per 2p has to divide cfx, in fact every step that we did is reversible, okay, so you can go back, you can go back here, you can go back here, you can go back here, you can go back here, okay, so hc transpose is 0 if and only if, this polynomial which I will call as gfx divides cfx, okay, so this is the, this is the gfx which is called generated polynomial, so there it is all of them, they are clear, so the problem in bch codes was we could just not say this, okay, so all these things are true for bch codes, but then this doesn't quite solve the problem because the polynomial on the side is actually classical efficient from gf2 par n and you don't want that, you want binary code words, okay, so that's, so we needed to go to minimal polynomials and all that, so each one encodes we have no such problems, if we are happy with code words from gf2 par n, okay, so this will be the difference, okay, so you have a polynomial, okay, so another description for the same league for element 4 that we had before, which basically cfx equals m of x times gfx, the gfx that we had before, okay, so that m of x has degree, what is the degree of gfx, first of all, degree equals 2b, right, I want multiples of this gfx and the cfx degree should be less than or equal to n minus 1, so what is the maximum degree of mfx that I can go to, n minus 2d minus 1, okay, so m k minus 1 x power k minus 1 and k equals n minus 2, correct, okay, so that gives you the dimension, so you see that this Reed-Selman code has dimension equals n minus 2t and from this you can go to the rank of h which will be 2p itself, okay, the next question is minimum distance, what about minimum distance, okay, so I am going to say minimum distance equals 2p plus 1, okay, so the proof is basically two things I have to show, I have to show that the minimum distance cannot be less than or equal to 2t, how do I show that, you use the same proof as you did before with Vandermonde matrices, suppose you have a weight w code word, where w is less than or equal to 2t, okay, you fix the locations of those w things, I1 through Iw, fix the corresponding columns, you know that that column has to have less than w column rank, but then you can take w rows of that column on the w columns and create a Vandermonde matrix out of it which will have full rank and that will be a contradiction, so w less than or equal to 2t is not possible, okay, we do not worry about binary code words, we only worry about rank there, right, so it is the same thing, okay, so it cannot be less than or equal to 2t, how do I know that it will be equal to 2t plus 1, how do I know that there will be a 2t plus 1 weight code word word, sorry, yeah, exactly, see now my code words are from gf2 pattern, so I can simply use linear algebra, symbol, whatever I want, I do not have to worry about the binary factor, right, so you take 2t plus 1, any 2t plus 1 columns of this tidal check matrix, the rank has to be only 2t, it means they are linearly dependent, so there will be a code word supported on any 2t plus 1 columns, okay, in fact that makes this code very, very special, okay, dimension is k equals n minus 2t, minimum distance 2t plus 1 is n minus k plus 1, okay, so what does that mean? rs codes are maximum distance separable in the sense that this satisfies the single tension, okay, so that implies rs codes have 30 meters and k equals n minus k plus 1 and that makes them maximum distance separable codes, this satisfies the single tension bar, okay, so they are very good codes in some sense, okay, over gf2 pattern, they are very good codes, okay, so couple of other questions and things I want to point out, dimension is k, so in the binary case how many code words would have been if dimension is k, would have been there, there would have been 2 power k, okay, but remember this is not binary, okay, the number of symbols each coordinate is 2 power n, okay, so the number of code words will be 2 power nk, it is not just 2 power k, is that okay, so it is a different thing here, suddenly everything is being raised to the power m, okay, so this parameter will actually be n, then k will be, remember k will be this, so let me write it down, when I say block length is n, it is n elements of elements that work, gf2 power m, okay, so that is a different thing, so when I say k, it is once again k elements of gf2 power m, okay, what is the motion of minimum distance again, what is the distance between 2 vectors in gf2 power m, number of places in which they decide, it is the same, so when I say minimum weight, it is number of non-zero positions, okay, those things remain the same as before, number of, so you cannot say number of places in which it is 1 nothing, minimum weight, for weight, right, say number of non-zero elements, anything non-zero is a weight component, okay, that is a weight component, alright, so this is how a general rate settlement code looks, and a lot of detail is given here, for instance you can encode very easily, how do you encode, take your message and then multiply by the generator matrix, okay, so you might wonder that this is not a systematic encoder, but I will get a systematic encoder, you do the x bar n minus k multiplication and do right by gfx, take the remainder and append it in the beginning, you will get a systematic encoder, so you can get a very simple systematic encoder for it, okay, but the only confusion here is, what do you do with elements of gf2 power m, okay, so how do you get to them, I mean ultimately you will only have bits in the relays, right, it is going to be bits, how do you go from bits to elements of gf2 power m, do the encoding, after you encode you have to transmit it using a similar digital communication kind of system, which will again expect bits, so how do you go from bits to symbols of gf2 power m and how do you go from symbols of gf2 power m back to bits, okay, those are the two things which is likely missing here, so I am going to draw a block diagram to illustrate how this will work, so if you have an nk 3th element, how the whole block diagram will look will be as follows, okay, so you will start with first m times k bits, this will be your original message, the binary message vector will be m times k bits, so what will I do, I will do some kind of a conversion to, okay, so I will convert to gf2 power m, okay, and I will get a message vector from 0 and 1, mk minus 1 and each m i belongs to gf2 power m, okay, in a way if you think about it, this is just a reorganizing of the k m vectors, instead of thinking of them as mk bits, I am going to put them in k rows, each row having m bits and each row instead of thinking of them as m bits, I am going to think of them as elements of gf2 power m, okay, what is the difference between just a vector of m bits and thinking of it as elements of gf2 power m, what is the additional when you think of it as gf2 power m, you can multiply two m bit vectors, okay, so if you just think of them as m bits, the only thing you will do is maybe add, you will XR them together, you wouldn't know how to multiply them, when I think of two m bit vectors as elements of gf2 power m, I can also multiply them and get what, get another m bit vector as answer, okay, so that's what the added notion of m means, okay, so once you get this message vector, you can also think of it as a polynomial of mfx, okay, what do you do, we will do encoding, so you encode both gfx, if you like you can do systematic, otherwise you do not systematic and you get a code, okay, cfx and then this will be c0, c1, I am going to cn minus 1, okay, once again remember this will be how many bits, okay, I have to convert this into bits, so I will convert to bits again and I will get how many bits, n times m bits, okay, so maybe I should write n times n, okay, this is the final code word vector in primary, okay, alright, is this clear, this is how the encoder will operate, okay, so if you pick for instance k to b, so for instance for example, if you pick, so very common charge for m is 8, okay, so what you are doing operations over your 256, what do you think we like 8, because 8 bits is a byte and people like to think of it as bytes and a computer for instance, a lot of things are stored as bytes, it's easy to count and bunch, okay, so here for instance you might pick to be 239, okay, so this is a very common situation and you might pick n to be 255, okay, this is a very very common code, okay, so how many bits do you need here, so you can encode 8 times 239, how much is that, let me see, should be something in the 2, 655, so do the multiplication, so 39 into 8, what is that, 19, 912, so 1912, okay, so something like, it will be in the 2,000 range, it cannot go to the 6,000 range at all, okay, don't get carried away, okay, so here you would have n times k to be equal to 1912, okay, so what will be n times n, 2040, okay, so that's the many bits you have, so internally you will be thinking of this n times k bits of actually 239 elements of gf2 power n and you have to think gf2 is 256, you have to think that way because the encoding involves a multiplication, so you have to multiply two 8 bit vectors to get there, you get that throughout, you will have some circuitry for it and all that and it involves multiplication like that, so you have to think of it internally as 239 bytes and these bytes come from gf2 256, they also have multiplication capability, okay, but to the outside world you actually have a 1912 bit input and 2040 bit output, okay, so that is the other point here which you should keep in mind, alright, okay, so this is the overall code, alright, so now there are some subtle points here about minimum distance, binary, what does it mean for the overall binary thing and so forth, we will come back to it later on but for now we will, for the next few lectures we are going to talk about this binary to gf2 power n and gf2 power n back to binary, it can always be done, we know that it always can be done, so we will simply treat everything as elements of gf2 power n and we will do encoding and decoding at that level and finally we will come back and comment on what happens in the binary, okay, so let's just accept this kind of operation and move on to more interesting properties, okay, so there are a lot of properties for each element code, a lot of things are known about these nbs codes in general and each element code in particular, so we won't go into all that detail here and then we quickly jump to the decoding part, the only thing I emphasize is the nbs part and then also the fact that gfx is the end of the polynomial, so let me just point out some connection to the cyclic property, okay, so if you pick n to be equal to the order of theta, this will make the rs code rs code will be cyclic, okay, the same proof as before goes through, there is nothing that you have to change, n becomes equal to order of theta, the rs code becomes cyclic, so you simply rotate the code word, you get another code word, there is no problem, okay, so as far as cyclic codes are concerned, we saw only binary cyclic codes, we never saw non-binary cyclic codes, so we can also define non-binary cyclic codes pretty much in the same fashion as I did before, okay, I had f2x and then I had r2 which was f2x modulo x bar n plus 1, you do the same thing with, so the f2 you put f2 prm, you will get the exact same formulation and you can show that these rs codes are in fact ideals in r2 prm should have r2, okay, so all those things will go through and the generator polynomial that we had is actually the generator for the ideal inside r2 prm, also condition, these all those things are true but you are not going to do that, okay, you are not going to do that here, we will just skip to that, okay, so what is our most interest is decoding, as you can see encoding seems like you can do it very easily, you go through the channel there is no problem, you will receive on the other side, how do you decode, that's the problem, okay, so once again if you want to do ideal ml decoding it is going to be impossible, okay, so if you have to write the syndrome decoder with all possible symptoms, you won't get anywhere, so what is the simplification that we will do, bounded distance decoding, okay, so we will say we will give up on ideal full-fledged decoding, ml decoding, we will simply take a bounded distance decoder, so what will the bounded distance decoder do, if there are less than or equal to TRS in the channel, I will try to correct it, the greater than TRS I will give up, okay, I won't even try to correct the idea on the bounded distance decoder, okay, so let's talk about these bounded distance decoders, okay, also algebraic and they are also very similar to the bch code except that you have this gf2 parameter, you will see what I mean by that when I describe the decoder, okay, so let's let's look at the picture again I have a message problem on my lamar parts, it's getting encoded by TRS, connected TRS, what do I mean, okay, okay, TRS TRS correct it, bch code and let's throw this problem mihr from TRS2 parameter, okay, and this is TRS correcting each element code over gf2 parameter, okay, and this is going to produce a code word cfx, okay, once again crr from gf2 parameter, this is going to work for the channel and the channel I am going to model as, similar to before I will have another polynomial cfx and where will the eis come from gf2 parameter, okay, is this a reasonable model, can I say that the error vector or the error polynomial has coordinates from gf2 parameter exactly, so it's same as the binary situation, right, so I would have actually converted the cfx into n times n bits and it would have gone through a binary symmetric channel, right, so binary error vector would have been added, I can just group m of those bits together and I can think of the error vector as actually being from gf2 parameter it's only adding, it's not multiplying so really there is no real big change here it's only adding and it's consistent with the addition in gf2 parameter, okay, so I can think of vfx also as coming from gf2 parameter, okay but there is a complication here, think about the error model a little bit more closely yeah, it will be different, so the error model is very different, okay, so I am going to talk about that right now, okay, so this ei being in gf2 parameter, okay but remember, underlying all this is a binary symmetric kind of channel which has n times m bits going through and n times m bit error vector being added, I just group it together and think of each error vector coming from gf2 parameter, okay, so now what do I mean by saying t errors, okay what do I mean by saying w errors w errors means what means what it means beta this guy v0, v1 vn minus 1 is equal to w, okay so if I say w errors actually are correct it means weight is equal to w, okay what is weight of a see remember each of these eis also gf2 parameter so what do I mean by saying weight is w number of non-zero coordinates when I view them as elements of gf2 parameter n equals w, okay does it mean that actually the number of errors binary errors of the introduced is it equal to w what should be the lowest number of binary errors that I introduced by the channel to give me this w it should be w, right you need at least w binary errors when you expand it and think of it as n times m bits only if at least w binary errors occur there can be w but there can be a maximum n times w n times w n times w binary errors resulting only in w errors in gf2 parameter, okay so that's a bit of a complication in the error model that you have to watch out for so in the binary case when I had n bits going through the channel if I have a weight w error vector what's the probability of the error vector it's p power w 1 minus p power n minus w in the gf2 power n case if I say I have a weight w error vector what's the probability of that particular error vector I can't say anything because I'm not giving you enough information it can be as high as p power w times 1 minus p power n m minus w it can be as high as that or it can be as low as p power m w times 1 minus p power n minus m m minus m w it can range over a wide range of possibilities you only can average it out and give you a number but it's not fixed so if I say a weight w error vector probability in the bsc model is not that fixed but anyway it's relevant to us because anyway we're doing bounded distance decoding finally we'll compute some probabilities and we'll see it matters a little bit but we're doing bounded distance decoding we don't have to really worry too much about these things we'll just assume some w errors at that and those w errors less than or equal to p so remember this there is an underlying binary symmetric channel which introduces big problems then there is an abstracted gf2 power n channel which introduces errors in every element so suppose we say let me do one more computation so what is the probability that let's say e is 0 with node equal to 0 1 minus 1 minus p power m but here so if you think of in terms of is it okay it's fine right so if you think in terms of gf2 power m channel what's the probability that a particular coordinate of the error vector is non-zero it is 1 minus 1 minus p power m right and this is what is the probability for e0 what will be the probability for e1 again the same thing is e0 even independent it will be independent so now if I ask you the question what's the probability of w errors in this how do you answer the question you can do m choose w hence this gate power w then 1 minus this gate power that is like the averaging out of the w errors so you can do that and get an answer so this you can call as something like ps which is the probability of symbol error so you can call it the probability of symbol error okay and I want to distinguish between ps and p p is the probability of each bit being in error ps is the probability that the symbol is in error and there is a complicated relationship between the two but if p becomes really really really small what will be the dominant term so it is going to be the dominant term so it can be useful in some approximations if you want to get a quick handle on what is happening m times p is a good approximation for ps okay and remember all your this case for instance you want to find the probability that the data p0,3 and minus 1 equals w what will be the probability of that you have to put n choose w then what ps power w 1 minus ps power n minus w is correct okay is this correct why am I worried about this 1 minus ps I think alright so you will get something like this in your answer and you have to watch out for this little bit in your computations it is not as simple as the binary case little bit more complicated not too much okay so after all this should be about the other vector finally you will get the received polynomial where each r is from gf to param this is the model we will use okay so now once I assume w errors in efx what will be the sum of efx w errors so efx can be taken in this terms you can take x power i1 plus x power i2 plus so on to x power iw but is that enough now I have to do coefficients so that is the big computation so I have to take practically 1 here particle e2 here particle ew here and each of these ea's will come from gf to param in the bch case when we were doing bounded distance decoding we were only interested in finding i1, i2, i3, iw once I find that I am done I just go to those corresponding locations and flip the bits when I say a bit is in error there is only one thing that can happen but when I say a symbol is in error it can happen in 2 power m minus 1 the same way the same things can add to it and so many different things can happen so not only do I have to find i1 through iw I should also find e1 through ew so we are decoding pass to find i1 through iw and e1 through ew of course this there is the same problem of not knowing w that we had before but like I said we will back a little later there is a iteratively solving it because the w problem gets taken care of automatically so we have to find both i1 through iw and e1 through ew so other than this complication the method is very similar to the method we had for the bch code so what is the first step that you can do you compute the syndrome so you do an rfx you compute the syndrome and the syndrome is basically evaluating rfx at beta, beta squared all the way to beta per 2p so let's do that I will compute s1 equals rf beta but what will be rf beta cf beta plus e of beta, what is cf beta 0 so it's the same as e of beta, what is e of beta e1 beta power i1 plus e2 beta power i2 all the way to dw beta power iw what is s2 explicitly or can I simply say s2 is s1 squared no, be very careful so if the coefficients are binary if it's even e2, ew are binary then of course s2 will be the same as s1 squared but now we have two coefficients which are from gf2 power n so when I square it what will happen I will get even squared even squared is not the same as even so you get a different polynomial a different expression and s2 is not dependent on s1 so it's definitely a different syndrome and it's a very very valid row and you have to evaluate it so this will be e1 power beta power 2i1 plus e2 times e1 times beta power 2i1 e2 times beta power 2i2 all the way to dw times beta power 2iw so you do this all the way down to the last syndrome which we have s2p are evaluated as beta power 2p e1 so beta power 2p i1 4 on 2 dw and all these 2p syndromes have to be evaluated having said that I should also point out simplification here so as soon as rfx comes in you don't have to necessarily try to evaluate rfx at beta, beta square, etc you can have a division circuit so dividing with x plus beta, x plus beta square, x plus beta power 3, x plus beta power 4 so it will be the same as before but except that division is a well understood thing from our BLSA point of view but remember what we divided gf2 power n so everything will be 8 bits in general so it's a bit more complicated but you can divide, once you divide the remainder you get is equal to rf beta I mean right if you divide rfx by x plus beta you get is equal to rf beta think about it so that's the idea so this is the first step so you know the syndrome and this is very similar to what we had before so the steps are also exactly the same the first step is to define something where I look at xj this beta power ij 4 and these eis are actually called other values so they are called other values so once you do this the equations you get will look like ours, I'm somewhat quick to write it e1x1 plus e2x2 ewxw s2 is e1x2 star e2x2 star where I'll go I'll scoot through this e1x1 will be equal so that is your set of equations and you have a bunch of parasyne equations left hand side is known to you I mean that this path is known what is not known is x1 and e you don't know the x you don't know the e the method that we use is very very similar so what is the method that we use we did a transformation what was the transformation we did in the BCHD code we defined something what did we define sigma of x it was called the rare locator polynomial so how did I define that what was the definition for sigma of x 1 plus x1x times 1 plus x2x 1 plus xw x is a locate right so this is the syndrome the rare locator polynomial and I argued for how if you think of the sigma as simply 1 plus sigma 1x plus so on to sigma wxpaw you are actually doing basically setting sigma 1 and sigma w be the symmetric polynomials and x1 and xw the coefficients of the polynomial so that's what we are writing here that what is the next step after this we do an ssx and then we multiply ssx by sigma of x and observe that from xpaw w plus 1 to xpaw 2t there will be 0s and that same observation can be made here also with these evens and e2s and ew it doesn't affect anything remember the column was what mattered right we had s1x, s2x hpaw 2t and the column even factors over the first column you multiply 1 plus x1x there everything will go away we will never have w plus 1 all the way to xpaw 2t so the same argument as before can be used in the multiplication and you can show, if you define xpx to be s1x for the school except for hpaw 2t hpaw 2t and if you do the product ssx by sigma of x coefficients of xpaw w plus 1 over 2 xpaw 2t r is 0 okay so the same method as before I did it a bit laboriously last time I am going to just skip it this time you write it and then multiply column guys okay look at the first column multiply 1 plus x1x everything will cancel you will not have anything but then you have w minus 1 terms following so this can come in the only thing you can have 2 is w w plus 1 to 2t you will not have anything now for the next column what do you do you multiply 1 plus x2x first and then you do the same argument okay so I am going to kind of stop the decoder here because after this the method is exactly same as before so if you look at the equations from xpaw w plus 1 to xpaw 2w you will get a proper invertible matrix if that matrix is full rank you can find sigma 1 through sigma 2 so use these equations to find so use these equations to find sigma 1 sigma 2 sigma 2 okay so of course you will know w so what is the trick there you have to start with t and then go to t minus 1 t minus 2 etc till you get a full rank matrix so whenever you get a full rank matrix you can invert it and you can find sigma 1 through sigma w okay so of course we are not done here what should we do next find roots of sigma of x right okay I will come to the eis I will come to the slide soon enough find roots of sigma of x so do the proof show step there is a condition here when you find roots of sigma of x in gf2 power m what is the condition something has to happen for you to declare success what has to happen when you find roots of sigma of x in gf2 power m what should you get you should get w distinct roots if you don't get w distinct roots something went wrong okay so if there are w distinct roots okay so you can put like an algorithmic question here and say w distinct roots okay no means what declared failure s means xj is basically inverse of the root okay so you take the inverse of the root set it to be equal to xj okay so that's the way to find xj once you find xj I will find the ej see once the xj is unknown what do these equations become linear equations in the eis okay so you take the first w rows and that's to argue that you give me a linearly independent set of equations and that will be linearly independent again you can use random one if you look at the equation x1 to xw you will see it's a random one matrix so you do w by w it will be invertible so use the first w equations do a linear equation solution you can find finding the error locations is quite easy finding the error values is easy once you find the error locations find the error values is it okay so I am going to leave it at that and then after that the last step is use syndrome equations do w syndrome equations okay so they are now linear okay so for the read assignment we call that in principle the methods are very similar to the bchd code of this nothing really that is different only thing is the error value and that can be found ultimately by some simple linear okay so I am going to draw a block diagram here and let's see what happens here so you have rfx first step is to find error of p power power i so you find the syndrome the next step is find sigma of x okay so we will do that that is a tricky step we have to do some solution and this is the most complicated step so once you find the sigma of x you find roots of sigma of x once you find the roots of sigma of x find okay so once you find easier what should we do we should go back here and then this gives you remember this gives you e-card effects come here and add it to the scale okay and assuming it is systematic encoding simply take the first k elements of c of x that will be your message is that okay so finding the syndrome like I said is not really a complicated step you can do it in so many ways finding sigma of x is the crucial complicated step and I think some of the project students are going to be doing some methods for this one is the pellicamp massie method which extends to the reach element code also you can do it for bch codes something called uk name domain decoder for it frequency domain decoder all kinds of methods are there for finding the sigma after that finding roots of sigma of x how do you do that what is the method for finding roots of sigma of x trial and error so you have to just try one after the other there is a more fancy method which is called gm search people will talk about basically it is trial and error so you have to try one after the other and find the roots and then once you find the roots finding ej need not involve the linear equation solving and all so in fact finding sigma effects also nobody will do the method that they describe and see critical methods which will nicely give you the solution and like this once you find these guys finding ej there is something that probably is algorithm which is what I have actually used okay so nobody uses some linear equations solving so it is simple substitution method it is not very hard think about it so it is a method that is used for finding ej so at the end of the day you can find the error vector and you can find the c cap of x of course there is also a probability that you will fail okay whenever the number of errors is greater than 30 the error correcting capability will also fail it is less than or equal to t you will never fail exactly right okay so that is the weak settlement code I mean it is inside I should tell you that it is a very very celebrated code it is one of the most widely deployed codes okay and at this stage in our course we are able to finish it as in one lecture but of course it required a lot of preparation before we could come to this point and all the ideas we have seen before two idea that was in this lecture we have seen before it is just a recasting in the gf2 param framework that is the only thing we have to do but like I said it is a very very celebrated code it is there in every single hard drive where the settlement code is used in every single hard drive I mean any hard drive you have in any device it will have that installment code it is used in optical communications for instance there are these cables that run under the oceans under the Pacific ocean, under the Atlantic ocean and all the carrying, carrying all your internet data from one continent to the other okay so they have huge re-solving code ships sitting there on both sides okay so you have the what else is re-solving code used in those are the two big applications but there are also several other problems that are still routinely used in re-solving codes okay so and in fact in computer science and some CS theory algorithms and few instance how many places re-solving code is a very very very famous code and it is for long it was considered as one of the great achievements of coding here it is a very simple code there are so many other descriptions this one that senses is very nice so maybe we will start here for this lecture I will let you digest this for a while and like I said the project people doing the projects are doing some implementation of this decoder maybe when they present you will see some more ideas on how this is done okay so in the next class I am going to go back to the binary world and interpret the re-solving code in the binary world what is it actually doing so it can correct T error vectors any weight T error vector that the error vector is viewed as belonging to GF2 parent so actually viewed as a binary error vector what is the weight that we can correct guaranteed it is a green T that is a bit of a disappointment cannot go to empty some empty but if you want to guarantee it can do only T so if I have T plus 1 binary errors I can put them into T plus 1 symbol and create T plus 1 symbol errors which cannot be corrected so it can only correct T bit errors but if you think in terms of bits what is the block length it is not N and K it is what K times N and N times N so you are losing a lot in that using a lot of block length to correct just the T error so there is a certain T error but compared to BCH codes each element codes will be poor binary codes they will not be that good binary codes compared to BCH codes but to each element codes have some huge advantages which is why they are used I will point out these things in the next class in the next class I think is the last class before Shastra