 Alright, so we're talking about decoding RM codes, okay? So in this context, it turns out we were looking at finite geometries, and what I'm going to do in this class is to begin with a simple example and talk about how lines and planes and all these things look in a very simple example so that you get a rough idea. And after that, we'll try and go into some more general things. Okay, so the example I'm going to take is EG4,2. So what is EG4,2? The points of this equation jump at 3. Basically, 0,1,4, right? And for some reason, this is not BW. The points are 0,1,4, so how many of them are there? There are 16 points, okay? So just to visualize, maybe we'll write these points down like this, okay? And maybe a little bit below here. So those are the 16 points in the geometry, okay? So you're definitely used to the real geometry where you have an infinite number of points and you always deal with equations. Here there are just 16 points, so this thing is pretty easy, okay? So for the subsets of the points, we're going to call them as lines and planes and all that. So one flat, two flat, three flats. The dimension is 4, right? So you'll have one flat, which would be the line, and then two flats, which you can think of as a plane, and three flat, which is called the hyper plane, one less than the dimension, and then whatever you think together is the last one, okay? So one point, which is all 0, is called the origin, just like the problem, okay? And it's easiest to think of these flats that go through the origin, okay? So because any flat that goes through the origin will be a subspace of 0, 1, m, okay? 0, 1, 4. If you give a proper subspace. So once it's a subspace, you know, subspaces are defined by bases or they're dual bases, right? So either the bases itself or the set of orthogonal things describe the subspace. So it's very nice, okay? So you start with lines and planes and all that that pass through the origin, okay? So how do I define a line if I want one of the points to be the origin? So usually if a line passes through the origin, what is its equation in the 2D plane around? It's some y equals mx, right? So you have some constant m and then you multiply with x, okay? So if I say the line passes through the origin, right? So what you need is actually just one more point on the line. Any other point, okay? So you have the origin and then if I give you any other point, say x1, y1, okay? So let me do it that way, okay? So if I give you the origin and then I give you some other point, x1, y1, what do you do to get the line? You basically take this vector and then multiply that vector by all possible scalars. So that will give you all the entire line, right? So that's what you do, okay? So there is a similar argument here, okay? So I give you the origin. If I give you any other point, say for instance, this point, what should I do to get the line passing through the origin and that point? I have to take that point and multiply by all possible scalars. But for me in the binary case, the scalars are only 0 and 1. So if I multiply by 0, I'll get 0. Multiply by 1, I'll get that point itself, okay? So as it turns out, the origin and any other point will form a line, okay? So how many such lines will we have? The pass through the origin, 15 lines will have passing through the origin, okay? So now how do you get lines? So is that clear? So for instance, lines through the origin are something like 0, 0, 0, 0, 0, 0 and then 0, 1, 1, 0. Then we have other lines, 0, 0, 0, 1, 1, 1, 1. So 15 such lines I got. Okay, these are the lines that pass through the origin. So how do you get a line that does not pass through the origin? Okay, so if you have a line that does not pass through the origin, you can always shift it by some constant, translate it so that it passes through the origin, okay? It's the same principle you can use here also. So you take a line through the origin and then add something to it. Add a constant number to it, that number to it. You'll get some other line that passes through some other two points, okay? If you already do, so if you want a line, so if you want any other line, so you take this guy and say add something, so 0, 0, 0, 1. So take this line and then add 0, 0, 0, 1. I'll get another line. It will be 0, 0, 0, 1 and then 0, 1, 1, 1. So that will be another line. So if you think of this as a line, this girl is another line, okay? So that's the idea, okay? So lines can be described that way, okay? So lines are described by two points that pass through one, right? So in the binary case, when this guy is two, the scale-ups are really only two. So we'll have only two points in the line. You can't have anything more than two. So there will be only two points in every line, okay? So another thing that we'll have to use is this notion of an incidence vector. So what is an incidence vector? So I have 16 points, okay? So what I can do is I can think of, sorry, I have 16 points. I can think of putting these 16 points in a row, okay? So I'll put 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0. All of them in a row, okay? And then if I have a subset, okay? If I have a subset of these points, here the points appear up, okay? If I have a subset of these points, so the subset is the line, okay? So maybe it's the line, okay? 0, 0, 0, 0, 0, and then 0, 1, 1, 0. I can define an incidence vector for the subset. What is an incidence vector for the subset? If a particular point belongs to the subset, so that entry you put 1. If it doesn't belong, it would be, okay? So like this will be the 0, 1, 1, 0 from there, but the row will be 1. Everything else will be, okay? So what will be the length of the incidence vector? It will be 16, okay? So anytime you have a basic set, and then you have a bunch of subsets of all these sets. Yeah, this set and we have a lot of subsets. You can always think of incidence vectors, okay? So this is a very commonly used technique in common topics and all this. We can use incidence vectors to do a lot of counting. Yes, we can do that also here. So one nice thing with respect to each element codes is incidence vectors are vectors which look like they could be code ones, right? In a way. They are length, 2 param, length 16, they are the right length, and they have boundaries, 0 or 1. So incidence vectors of these subsets are, they belong in the proper space at least, okay? We will show eventually that incidence vectors of these one flats and two flats belong to each element codes. We will do that. But for now, these are just vectors of length 2 param, okay? So in general, let me hear in this case, vectors of length 16. So, okay? Eventually these incidence vectors will are the bridge between this finite geometry and read the rate miller codes, okay? So through these incidence vectors, we go from geometry to read miller codes. Okay? So that is lines. If I want planes, what should I do? Okay? So a plane is basically a two-dimensional subspace. Okay? So you have a two-dimensional subspace of these 16 points. You need two basis vectors. Okay? So maybe I take these two as some basis vectors. This guy, maybe this guy. Okay? It turns out once you have, I mean, any subspace will definitely have zero also in it, right? What will be the, what will be another one? See, I am defining a two-dimensional subspace with these two as the basis vectors. So what are the vectors in the subspace? Yeah. So it's the sum of those two, right? So it's going to be just 0, 0, 0, 1, which is this guy. Okay? So those four guys form a plane which passes through the origin. Okay? So it's a two-flag that goes through the origin. It's also a two-dimensional subspace. So it's nice. So if maybe you wanted to find some other plane that way. So all you have to come up with is two different points in this. Take that as the basis of your plane and then form the linear combinations. We'll get the plane. So how many points will that be in a plane? In a two-flag, you'll have four. Okay? So likewise in a three-flag, you'll have eight. Okay? So of course beyond three-flag, it's not really interesting in the three-dimensional thing, right? So four-flag is interacting and after that you won't get anything else. Okay? So okay, so for two-flag, there are four points. One easy thing is to take the basis and then form the four vectors which are there in the two-flag. It turns out you can also have planes that don't pass through the origin. How do you think of those planes? In R3 for instance, you can have a plane that does not pass through the origin. You can always translate it to make it pass through the origin. So in reverse, you can start with a plane that is in the, through the origin and then add something more. Same thing you can do here. So I'll take this plane and say add 10102. Okay? So I need another shape here. So I'll use the shape. Okay? I'll add 10102 each point of that plane. So what will I get? Okay? 1011. And then I'll get 0010. And then 0011, right? So that's it. So you can see where even in terms of there is a translation effect going on, I mean this is just the way to write down the points. Write down the points in any other way you like. But you can see it kind of makes sense to tell these things of planes and all that. It's not the point. So that would be another plane which does not pass through the origin. It's translated away from the origin. So it's not a subset. But still if you have four points. All right? So that's the, that's the idea. Okay? So you can have, for instance, incidence vector for both for a two-flag. Okay? So you have an incidence vector for the two-flag. How many ones will it have? For instance, if you take this squiggly ellipse incidence vector, this is going to be one. This is going to be one. And then you'll have a one at 10000. And then you'll have a one at 1001. So that would be the incidence vector for the two-flag. Okay? Already one nice thing you're observing is, if I have an n minus r-flag, what will be the rate of the incidence vector? Two-flag and minus r. Okay? So it's all the rates are powers of two. And that's something which is very nice where it will occur. Because we know powers of two is what really happens and occurs in the rate millipole definition. Okay? So all those things are nice connections which you can think of. Okay? All right? So this is the, this is kind of something that was, I was alluring to go to the end of last class how it works all day. Okay? So some interesting questions that we turn out that will be very important in final geometry at least is given a geometry like this, how many lines are there? So let me see. Take some time to think about it and answer this question. In this, 16.014, how many lines are there? Is that the correct answer? 16, 3, 3, 2? Is that okay? Will you not be 16, 3, 2, 2, correct? Everybody's happy with that? Okay? That would make sense? No? Maybe, yes. Okay? So, so, so there are various ways to do this counting. One easy method that I like is to do the, do the incidence vector kind of thing. So what you can do is, you can, you can make a, you can make a big, big matrix. Okay? You can make a big, big matrix. You put the points there. So how many talents will there be? 16 points. And then you put lines here. And you fill it up with incidence vectors for the lines. Okay? So maybe you have capital or lines. Okay? How many ones will there be in each line? Two. So there will be two ones. So where is this two? How many ones will there be in each column? 15? Is that okay? Everybody's happy? Is that correct? 15, the correct answer? Okay, so let's say 15. Okay? So if you say 15 ones. So two ones in each row. And then 15 ones in each column. So now this gives you a formula for L. Does it not? You count the total number of ones in the matrix, column wise and row wise. Okay? So you count it column wise. You have 16 times 15. That has to be equal to L times 2 at two and right. Okay? So L equals 16 choose two. Okay? So I wonder for lines it is really quite simple. I don't have to do it. So planes and all it can be a little bit more tricky. Okay? Not that difficult. But you can count the total number of planes. Okay? And then you might want to count things like, if I have a line, how many planes contain that line? Okay? So let's say this is the number of lines. What about the number of two flats? Number of two flats as well? 15 C2. 15 C2? Not 16 C2? 15 C2. Okay? So you can take any two but one of them cannot be 16 C3. Why did it go to C3 now? Won't you get repetitions? Okay? So let's do this once again. Okay? So I'm just doing this the fun. It's not really that useful. So let's say how many two flats are there in this? Okay? So how do you go about doing that? How many two flats are there? See lines were easy. Any two points between any two points over the line. But how do you think of, how do you think of two flats? Right? How do you, how do you do that? Okay? What's the question? Right? What's the question? So you have to be careful about ever counting. Okay? So because you can count planes in multiple ways. See for instance, if I take this quickly plane, I could have taken the basis to be these two, or these two, or these two. Okay? Any which way I take the basis, I will get the same plane. Okay? So I cannot just count 15 choose two. There I'm ever counting. Right? If I say 15 choose two, it gives you all possible choices. Right? So it would be slightly careful there. And then you have to divide by something. What should you divide by? So let's say a number of two flats through the origin. Give me the number of two flats through the origin. How will you add that? See if we divide by three. Yeah. So you have to divide by three. Right? Is it clear? So you have to take 15 choose two, and then you have to divide by three. Why do you have to divide by three? Because every plane has three different bases. Okay? And you're counting all the possible bases. That's why we divide by three. We're overcounted by three. Okay? So that will be the number of two flats that pass through the origin. Okay? So likewise, you can do. So if the number of two flats passing through the origin is 15 choose two by three. What is the number of two flats passing through any other point? So again, the same thing. Okay? So once you answer that question, you put this matrix. You put this matrix, you get their answer. Right? Each plane has four points. There are 16 points. And how many planes are there through each point? 15. 15 choose two by three. So number of planes times four should be equal to 15 choose two by three times 16. You solve that and you'll get another place. Okay? So you can put the matrix like this. The critical question to answer is how many flats, how many two flats or three flats or whatever pass through a particular point? That you can do by this simple counting argument. Once you do that, you can answer the question of total number of planes, total number of two flats, three flats, etc. So it starts with counting. It will vary on overcounting itself. The only thing you have to be careful about is overcounting. Okay? We're okay. So these things are not too critical. Maybe there will be some assignment questions. But it's quite easy to do this calculation. Okay? They're okay. That's fine. So we can maybe check that. So number of two flats through the origin, number of two flats, two, two, zero, zero, zero, zero. It's going to be equal to, like you said, 15 choose two. That's correct. But then you'll be overcounting by three. So what is it? Seven times five, thirty-five. Okay? So this answer is thirty-five. So I don't know what this method is. Correct? Okay? So thirty-five planes that pass through every single point. Okay? So number of two flats, total number of two flats times four should be equal to 16 times thirty-five. So that gives you the number of two flats to be 140. Okay? That will be the total number of two flats that pass through the, that are in this geometry. Okay? So like I said, these are all finite geometry. Subjective will be finite. You can't have anything infinite. Okay? So number of planes in R2 is no, number of lines in R2 is no need to ask a question. Your answer is infinite. You know, immediately what it is. Here it's not like that. Okay? So you have a finite number and you have to play around with the combinatorics to correctly count. So it's not too hard that you have to do it carefully. How do I get this three again? Three comes from the, the answer to the question, how many different bases are there in a two-slide? In one two-dimensional subspace that passes through the origin, how many different bases are there? That answer is three. Okay? That number can be calculated. It's a little bit more tricky as you go larger and larger, but it can be calculated. Okay? Think about it. So the binary case is very easy. The non-binary cases, it becomes a little bit more tricky, but it can be done again. Okay? All right. So that's done. Another question I want to ask is, suppose I fix a line. Let's say a line through the origin. Okay? So I fix a line through the origin. How many planes will contain that line? That is my next question. Fourteen, right? Does it seem like a simple enough answer? Fourteen. Fourteen. Fourteen. Fourteen. What are we going to do? No. And then if these are fixed, then one more point is fixed. Because that's just above here. Some of these two is that itself. A fixing time. There's nothing more you got. So you take any other point. So suppose I fix this line 0000011, 00001. I have the liberty to pick one other point. So that gives me one more. So I can pick that plane in two different ways. I cannot pick every this point at that point. So it's going to be 14 by 2, so 7. 7 would be the correct answer. Okay? So you pick these two. You pick one point. So you pick 00001. That tells you that 0000 is also there. So you can't say all 14 will give you different planes. Two of the points will give you the same plane. So you divide by 2. You get 7. Okay? So these are simple calculations. But if you do it for the first time, it will be packed through the overcounting. Yes, watch out for the overcounting. Okay? So it's a simple question of dividing, but still you have to watch out for overcounting. Okay? So the number of planes that pass through a fixed plane is 7. Okay? So something like that. So those are the answers that you can do. So my point is, the main point I want to make is, in any finite geometry like this, all these questions have exact answers. Questions like how many planes, how many lines are contained in a plane, how many planes pass through a given line, how many planes pass through a point, how many free flats pass through a point, how many free flats are contained in how many free flats are contained in free flats. All these questions have fixed definite answers. Okay? So you can think about them, do the computation, and get the answer. Okay? So that's the point I wanted to make last class. I don't know if it was very clear to you that this is the way you count them. It's not very clear. Okay? So now, next comes the main result which I kind of informally stated for you. The main result which links these two things first. Incident vectors of M minus R flats in EGM comma 2 are probe worlds of... I don't know why I've been so precise last time because we probably didn't mention these incident vectors and all that. This is the first. So you take an M minus R flat in EGM comma 2. Okay? R is the same R, okay? It's not some other arbitrary thing. R flat in EGM comma 2. And then take its incidence vector. So its incidence vector will have a weight of 2 power M minus R. It's a M minus R flat. It's going to have a M minus R basis vector. So you pick 2 power M minus R for a different point step. So that will have a weight 2 power M minus R. So in fact, the non-linear code words, they are minimum weight code words. They are minimum weight code words. Since you do see it, they have a weight 2 power M minus R. And 2 power M minus R is the minimum distance of RMR comma M. Okay? So the incidence vectors of M minus R flats are the minimum weight code words of the RMR code R comma M. Okay? So that is the connection. And this gives you an easy way of coming up with the minimum weight code words of the RMR code. Okay? And like I said, the rules of RMR codes are also RMR codes. So if you want to come up with parity checks for decoding, you only have to come up with incidence vectors of some flats. Right? Because ultimately, if you're interested in orthogonal checks, you're going to be interested in lower weight guys. Right? You don't want very high weight vectors. Right? Because when orthogonality is going to be difficult, you want lower weight guys. Right? And minimum weight vectors of RMR code were simply incidence vectors of flats. And due to this, RMR code is again some other RMR code. So you only have to come up with incidence vectors of R flats to think about methodically coding it. That's the idea. Okay? Okay. So let me throw this to you. This is not very hard. It's probably very easy to code. Okay? So if you have an M minus R flat, okay? So far I've been talking about the basis vectors. M minus R basis vectors completely fit further. An equivalent picture is the dual space. Okay? So an M minus R flat, once you go to the dual space, it's defined by R equations. R equations of the term summation aij vj sbi the i is 0 so we'll go from 1 to m. aij i is from 1 to R aij is 0 at 1 not at 0. I mean, of course, I mean not at 0 is not the really big thing. It has to be linearly independent and all that. So this is this defines an I said this defines an M minus R flat so it cannot be dependent and all that. So it'd be perfect. And di also is 0. Okay? So any M minus R flat is defined by R equations. And remember, the degree of each equation is 1. Okay? So now I want to show this this belongs to R comma m. Okay? So what does R comma m have? It contains evaluations of of what? Evaluations of polynomials of degree less than or equal to R. Okay? So this is like a polynomial of degree 1. Okay? But my points are those which evaluate to which evaluate to 0 when I plug in both. Okay? So I have to move from 0 to 1. Right? These points that satisfy the problem where M minus R flat have to evaluate to 1 when I plug it in. Okay? And then I'll get the proper code load. Okay? Do you understand what I'm saying? Okay? So maybe I should write it down. Okay? So suppose I want to make an incidence vector for our M minus R flat. What would be the incidence vector? So I have my points on 0 and I'll go to R1. Okay? What do I do? I take each of this V1 and Vm. I plug it into the R equations. If all of them evaluate to 0 then I know I should put a 1 here. Okay? But I have to move from this world into a polynomial evaluation world. How does a polynomial evaluation world work? I have a polynomial. I first V1 to Vm. It's a Boolean polynomial. I simply plug in the V1 to Vm here and I put that value there. Okay? If I want the incidence vector as this way to belong to the real number 4 I should come up with a F of degree less than or equal to R which will evaluate to 1 whenever a point is in the M minus R flat which will evaluate to 0 otherwise. So the whole trick is to come up with that construction. How do you come up with an F of V1 to Vm which has degree R or less than or equal to R and then it should evaluate to 1 whenever I plug in a point of 2 flat M minus R flat it should evaluate to 0 otherwise. Is it okay? Okay? That trick is simple. It's not a very difficult trick. All you have to do is this. Okay? You take this and then you make a product of I equals 1 to R and these guys summation a i j v j plus b i but then you add a 1 to it. Okay? Don't just do it that way. You add a 1 to it. Okay? And then this j only goes from 1 to M. Okay? So that's the trick. Okay? Now I have a polynomial of degree what? Equal to R. Okay? So evaluation of this polynomial at all the points will definitely belong to the Read-Miller code, R M O F, R M I M. What is so nice about this construction is for every point in the M minus R flat what will happen to this expression for every i? It will be 0. So that will be 1. Overall the polynomial is lower than 1 and you'll get a 1. For every point which is not in the M minus R flat what will happen? So at least one equation this will evaluate to 1 and that will make the entire polynomial 0 and you'll get the 0. Okay? So the evaluation of F is the same as the incidence vector. And it has degree R which means the incidence vector belongs to the Read-Miller code, R M O F. So it's a very simple course but not really that fancy. But it's a nice little construction. This trick was very interesting. How do you go from some equations that are equal to 0 to a polynomial whose evaluation is something that can be controlled? It is not a very difficult trick but it's an important trick to get this answer. Is that okay? All right. So if you have... So there are few other things that you should keep in mind now. Okay? So you have a lot of inclusions there which should be mind to laugh. Read-Miller code of R, M is contained in what? Read-Miller code of R plus 1 comma M. Right? It's fully contained in Read-Miller code of R plus 1 comma M. So if you have M minus R flats belonging to R M of R comma M will also belong to... The incidence vector of M minus R flats will also belong to R M of R plus 1 comma M. Okay? So once you fix M minus R R plus 1 all the way to M the incidence vectors will still belong. That's because of this inclusion. Okay? All right. So that's the first point. Another way to think about it is what about M minus R minus 1 flats? If I have an incidence vector of M minus R minus 1 flat that will be R M of R comma M. Is that okay? Is that correct? I didn't make a mistake there. Yeah, that's okay. All right? M minus R minus 1 into brackets this is M minus R plus 1. Okay? Because larger weight. Okay? So if you have M minus R plus 1 flat that will also belong to this scale. Okay? Is that okay? So let me write that down very carefully. Maybe with an example. Okay? So let's take an example. Let's take the 4 comma 2. So let's take R M of 2 comma 4. So both will contain incidence vectors of the robot. So first of all... Okay? So incidence vectors of M minus R right? So 2 flats in EG 4 comma 2. Is that okay? What else will it have? It will also have incidence vectors of 3 flats. It will also have incidence vectors of 4 flats but 4 flats are very trivial. This is the entire all ones. All ones is there. Is that okay? Is that clear? If you consider... If you contain the incidence vectors of 2 flats and 3 flats. So if you look at any other readmiller code, let's say readmiller code 3 comma 4. This will contain incidence vectors. I don't have to keep repeating the incidence vectors. When I say it contains 2 flats, I mean it's clear that it has to be the incidence vectors. So it will be 1 flat, 2 flats and okay, so that's how you have to think about it. So it will have all these things. Whatever the readmiller code is 1 comma 4 3 flats. Okay? So what is the dual? The dual of this case okay, right? Am I right? So if I want to do majority logic decoding of r m of 1 comma 4 okay, suppose that's what I want to do and I have to look for orthogonal parity checks in in the dual, right? I have to find the orthogonal parity checks for r m of 1 comma 4 which will be 3rd words of r m of 2 comma 4 okay? So what you do there the strategy is as follows okay, so like I said there was a main result what is the main result that I stated? r m of r comma m can be decoded by r plus 1 step majority logic decoding right? That was the you know lessons up to how many r of 2 bar m minus r minus 1 minus okay? Okay? What does that mean? So when I say r plus 1 step majority logic decoding remember these steps right? So the hamming code I illustrated that right? So you take the first you find orthogonal checks on a subset of the not just e0 but say e0 plus e1 e0 plus e2 and then you recursively repeat it repeat it repeat it until you get to just e0 alone and then you repeat the same process okay? So if I say half of 2 bar m minus r minus 1 in each step so what does that mean? j j has to be equal to 2 bar m minus r minus 1 okay? It has to be that right? Only then you will have you will have this many okay? So so in terms of what you have to do in r and r 2 and r 4 is to first if I am not wrong you have to start with two flats okay? right? Okay? So r m is 1.4 is two-step so you have to start with all the two flats okay? And then ask the question can I find so let me rephrase it okay? So you should start with trying to find orthogonal parity checks on a one-file okay? So that's the idea so let me write that down so here's an example m number is decodable m number is majority logic so let me write that down so two-step majority logic decodable okay good? j equals what? 7 right? No discussion number 7 is coming okay? So we have to answer the question 7 So what you do is you start with some one flat okay? So let me say the number 2 comma 4 and it contains contains what? two flats okay? So you start with any one flat okay? some one flat so that passes through the origin okay? And then ask the question then ask then what you do is you collect all all two flats passing through this okay? So you start with the one flat so that passes through the origin 0 0 0 0 0 0 0 0 0 0 1 something like that and then you collect all the two flats that not passing through I am sorry they say that contain this one flat right? getting confused but contain this specific one flat the answer to that is what? how many two flats will there be? we just do that calculation it's going to be 7 okay? you take the incidence vectors of all these two flats you know that they are parity checks for my code how do I know that how do I know that the parity checks for Rm1,4 what's the deal? Rm2,4 all the two flats belong to Rm2,4 the next thing is they will all be after the NL on this one flat okay? think about that it's a relatively easy thing to think about so you can go back and think about this geometry if you like okay? so if I take this line and then look at all the planes that contain that line they will be after the NL and that line okay? all of them will check that line well, all of them contain that line and no two of those planes can have anything in common it cannot happen okay? then it won't do after the plane because if they have something in common the other thing will also do the same it cannot be different right? because this line is fixed you have one other point you have two other points in this plane if there is some other plane which overlaps in one point what will happen? what is the other point? is it this overlapping in anything else? okay? think of this is true in okay? this is true in R3 also okay? if you have two planes that have a line in common they cannot have anything else in common because they are overlapping in that line they cannot overlap any models otherwise they will be on top of each other okay? think of R3 if you have two planes that are intersecting in a line okay? they cannot intersect anywhere else they will intersect anywhere else then the two planes okay? the same proof will be used here and show that these will form orthogonal part of each side the one-flat okay? that's fine reasonably clear okay? so so you take so these form an orthogonal part let's set up okay? on the one-flat is it okay? right? hope it's clear it's not okay too fast but that's all the two-flats that contain this one-flat there are seven of them so I can find seven that are orthogonal on the one-flat okay? so that's the first step so now what do I need? so that's the first step in my method logic decoder okay? now I can do two-step in the next step what do you do? you just repeat this process for several one-flats that pass through the origin okay? so this is one-flat whereas the same of them we just take any seven okay? no point in going beyond seven if you want you can go beyond seven also if you want to correct more than where a correcting capability you can go beyond seven so you take all the flats that approve origin for each of those flats you repeat the same exercise okay? so you take any other one-flat so that's the first step the first step will contain one-flat how many such one-flats do you need? you take seven such one-flats okay? repeat for seven such seven or more I'll say seven or more such one-flats okay? one-flats through the origin right? so all of these are one-flats to the origin what does that mean? these one-flats to the origin can overlap only with zero they cannot overlap anything else so all these one-flats to the origin form an orthogonal set for E0 for the origin itself okay? then in the second step we use both sets to be quite easy okay? okay? in the first step we have found the the the subset corresponding to the one-flat we have found okay? now we repeat that for every one-flat that you have say you have seven of them they all of them are passing through the origin remember that okay? all of them are passing through the origin so these one-flats now form an orthogonal no I can't say a parity check but they form an orthogonal subset for E0 why can't I say it's a parity check one-flats are not an RM2 come up they're not regular parity checks and one-step checks out okay? now in the second step use one-flats to find but recently clear okay? so this is for the origin what I did for the origin I can do for any other point okay? it doesn't matter origin is just some one point you just translate it you can just keep repeating that's the nice thing about geometry the origin is not successful simply translate repeat so we'll do the suitable adjustments in your translation okay? so maybe we'll do this explicitly for the RM1 come up so with one guy maybe you can just repeat the same process okay? so I don't have to do it for every single point I have to do it maybe for the origin and then do it for one other point then do cyclic shifts after that in a suitable notation okay? so maybe we'll do this explicitly with one guy maybe you'll like that a little bit okay? so let me just check this guy okay? so this thing is not going away okay? so this was our so let's take the one flat zero zero zero zero zero zero zero one okay? like I said there are seven two flats that contain this one flat okay? so this so what are the two flats? so I'm going to write the equations directly okay? I'm looking for E0 plus E1 so I'll go with B in the parity set what else will that be? so it's just a question of taking one note so suppose I take zero zero one zero what will be the fourth point? zero zero one one so one parity check is E3 equals zero then I have E0 plus E1 what else can I take? I cannot take zero zero one one again because it's already there so maybe I'll take zero one zero zero so what will I get? zero one zero one what else? zero plus E1 plus I cannot take E1 again I'll take E6 E6 plus E7 equals zero and then not zero I'm sorry what am I saying? sorry so we will all do the syndromes right? S1 S2 S3 so on okay so how do I find this S1? R0 plus R1 plus R2 so E8 plus E9 equals S4 and then you have E0 plus E1 plus E10 plus E11 equals S5 then E0 plus E1 plus E12 plus E14 there was six and then the last one E0 plus E1 plus E13 plus E13 equals so we are computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer computer E0 plus E1, E0 plus E2, E0 plus E3, E0 plus E4, so on. Okay, so you repeat for E1 plus EI, E1 equals to E3 plus E2, let's say 7, just 7, you can do more if you like, but 7 is enough, okay, once you start at 7 then what do you do in the second step? Okay, in the second step I have E0 plus E1, E0 plus E2, so until, E0 plus E7, just simply take majority again, break, there will be no tie as such, majority will be, okay, so all these are caps, okay, so that's how you do the decode, okay, so in general if I go to RM as r comma m, okay, the general logic will follow as such, okay, so if you go to, so this is here, right, in this example, so if you have any code, any ridmila code reasonable size, you should be able to repeat the same process, start with a suitable flat and then you keep on doing one more, one more, one more. So in general what will this picture look like? If you want to decode RM of r comma m, the dual is what? Dual is RM of m minus r minus 1 comma m, right, that's the dual, okay, so what you do is, in the first step to start with an r flat, okay, you can find all the r plus 1 flats that pass through this r flat, that number will be exactly 2 power m minus r minus 1, so the same logic, it's not very hard to do that computation, okay, you will see the number of r plus 1 flats that pass through a fixed r flat or the container fixed r flat will be 2 power m minus r minus 1 and they will all be orthogonal on the particular r flat, okay, so after the first step you find r flats, so how many such r flats do you do? You take an r minus 1 flat and find how many ever all the r flats that are contained that contain that, okay, so you take r minus 1, r minus 2, so on till you get 2, finally 1 and then 1, okay, so you take r plus 1 flat that contain 1, okay, and proceed, okay, so you do one particular r flat, you will have to do a lot of r flats, right, so how do you take all those r flats? You take some other specific r minus 1 flat which is contained in this r flat, okay, and you pick several r flats which are all contained in the r minus 1 flat, okay, so after the second step you will have found the r minus 1 flat, now you will repeat the second step for several r minus 2 flats inside the container in the r minus 1 flat, so on and so on and so forth till you come back to the final step where you have many and then you have the points, okay, so that's the step, when in every step you will have at least 2 power m minus r minus 1, the limiting side will be the last, first one, okay, you will have 2 power m minus r minus 1 checks which are after the next, you can do the computation, okay, so that is the majority logic decoding for an arbitrary read mode of code, and there are improvements to those, there are alkanes of refinements, usually people don't decode the read mode of code directly, they always puncture it and decode the punctured version, why is that nice? Because that's the clip, once you find something for the first bit, the same thing shifted around will work for all the bits, and as many steps as you take, and at least that will work, the same thing works, okay, so that is the nice thing to do, okay, so we will stop here for today, hopefully this point was reasonably across.