 Okay, so this is the first question with the board reached for read millicodes. We are asked to make list of parameters of read millicodes of customers. M is 2 power M, which is 64. M would have been a 6. Some code we have summation of 0 to R. M2i, this case is 62i. Md is 2 power of M, and then you just make a table. I think most people have done this table. Okay, I mean almost everybody got 5 marks on this question. So, this is very few cases when I had to give partial credit. Almost everybody got most things correct. Okay, so second one was C, BPH code, binary BPH code, and it was 15. And the first part asked you to find, generate a polynomial for C and it's DL. Okay, so N equals 15 is G of 16. And I expected you to be quite intimately familiar with the minimal polynomials and everything in G of 16. Okay, so I mean I saw people writing some crazy expressions. I mean you should be very comfortable with the G of 16. Okay, so G of X will be, so it's 2 error correcting, right? So T equals 2. So it should be the LCM of minimal polynomial of alpha, alpha squared, alpha power 3, and alpha power 4. But alpha, alpha squared, and alpha power 4 all have a minimal polynomial which is X star 4 plus X plus 1. I'm assuming that's what you put to generate your C. And then for alpha power 3, you'll get X star 4 plus X star 3 plus X star 4 plus X star 4. So if you multiply this out and simplify, you will end up getting 1 plus X star 4 plus X star 6 plus X star 7 plus X star 8. Okay, so the simplification is useful for the remaining questions. For instance, to write down the generated polynomial, generated a matrix and all that. This is very useful. If you don't do the simplification, it will be complicated. Okay, and then for the deal, you have to find H of X. H of X will be X star 15 plus 1 by D of X. And that would basically all the other polynomials here. So what would be X plus 1? X star 15 plus 1 times X star 4 plus X star 3 plus 1. And this is not the generated polynomial of the deal. The generated polynomial of the deal will be X star degree of H of X which is 7 by 4 plus X star 1. Okay, so you can substitute that if you will get this D of X plus 1 again. X star plus X plus 1. And this here will become X star 4 plus X plus 1. Okay, so if you can multiply this out, it's not very hard to do that. You'll get 1 plus X plus X star 3 plus X star 7. And part B, you can see this multiplication form will help you. Okay, so you simply write D as GFX is the first row. Remember, length should be 15. Okay, so I saw some people had actually filtered out all the zeros. You don't have to necessarily do it. You give me the algorithm for generating the... Generative matrix, I'll accept it. It's not too bad. Okay, so then you keep on shifting at X point GFX till you go to the last bit. So similarly, H you should do. X times GX. Okay, so this will be basically... How many rows will there be? Degree is 8. So there will be 7 rows. 7 rows. There you lie. Okay. So part C asks you to find the dimension at the exact minimum distance. Dimension of this D and minimum degree of GFX. I don't know what it is. This D is 15 minus 4 H to the star. Okay. So the minimum distance, you know, D is given to the star from the DCH bound. Okay, but what is the weight of GFX itself? It's 5. Okay. So I should say here, it should be passing us. This is the code word corresponding to GFX as weight side. Okay, so that means D equals 5. Okay, so I've given a reasonable amount of partial credits. I have some logic set of partial credit also. Totally it's 10. I think some people have varying degrees of partial max in case you think you didn't get what you deserved. Come on talk to me. I have a platform which is a seller that I want to disturb it for you. I can disturb it in case you have any questions about partial credit. Okay, so third question. I think enough people don't do this well. I think they don't expect the rates element decoding kind of question. But there is a smart way to do it. So I think somebody will find the smart way and open the smart way. But anyway, that will not. Some people find the difficult way to enter the answer. But that's fine. I'm going to look at the answer in many ways. In fact, my partial credits is based on the standard PGZ kind of decoding. It's not based on any smart method. First is GFX for... I remember putting here is again D equals 2. RS over... N equals 7. RS over GF8. Okay, so first thing you should do is write down tables for GF8 and all that. Although it was very quick to do anyway. And I did that right there on the site. And see, it didn't take too long. I wrote down the table on that side and then it wrote RSX and found S1, S2, S3, S4. Then wrote down sigma effects and wrote SFX and sigma effects equal to degree 3, degree 4 to 0. Solve for sigma 1, sigma 2 and then found the 2 zeros. Inside the 2 zeros are just one and all side. So you just substitute. You'll see quickly you'll find the 2 zeros. Okay, after you find the 2 zeros, you'll find the... Then look at RS and then equate S1, S2 and get the RMA kills and you flip it again. It's quite simple, but the method is very straight forward. Okay? GFX is basically alpha. So you take alpha belonging to GF8 primitive. So in most cases, it's very obvious what alpha is when you write down but I would suggest that you mention these kind of things. Okay, so don't just generally write X because alpha and expect me to interpolate and know that alpha is a primitive element of GF8. No, that's true, that's still... it's good to write it. Or at least write it table so that I know what you're dealing with. Okay? So not correct to simply simplify it. Alpha is what I need it for. You need to go under the matrix. It's called standard. No need to simplify it, but you'll find it if you like. And then you're asked to decode out of X to alpha plus X plus X to alpha plus X to alpha plus X to alpha plus X to alpha plus X to alpha plus X to alpha plus X. Okay? So if you notice very carefully, see one thing which is always a code word most of the time is basically the all-ins vector. Okay, for many rich elements, BCH codes, all-ins vector will be a code word. Okay, why is that... why is that true? Well, you see X bar n plus 1, B n plus 1 times X plus alpha for all-ins, X plus alpha power n minus 1, right? And the GFX will always be something here, right? So there will be the remaining stuff which you take together and multiply with GFX. Okay, except for this, you should not take this. You take the remaining guys here and multiply them with GFX. You will definitely get the all-ins code word. So all-ins vector is a very high likelihood to be a code word in your code for BCH reads all-ins. So why is that useful here? So if you look at the all-ins, that is a very good guess as the transmitted code word because it's clear enough to this. You see a lot of runs and then only two things are changing. Now you see T is equal to 2, which means minimum distance is at least 5. Exactly equal to 5. So clearly this can not be closer to any other code word other than all-ins because it's a distance 2 other than the all-ins code word. So that's the idea here. So if you pick up this as R, how many distance between R and 1, 1, 1, 1, 1, 1 equals 2. Okay? In the first place it's alpha, first place it's alpha. Okay? And the 2 is then it's equal to D, right? And it's less than or equal to D by 2. Okay? So it's strictly less than D by 2. Okay? That implies there can be no other code word which is closer than this all-ins. So you can definitely say the maximum likelihood you would be going is all-ins. Okay? So this is a quick way of getting to the answer but even if you do it, it's not a lot of work. You can quickly do it. It's, I think if you spend a lot of time in the readable network systems and you spend even more time in the DCH4, you won't have time for this. But if you did those things very fast, I think those things are quite automatic. If you did those things very fast, you'll have enough time to complete the syndrome and make sure your errors are not there. You make one or two errors, definitely. You go back and fix it, all those you can do. But even if you did that time, if you knew that the valence problem was going to be there, you can quickly guess the answer. Okay? That's one way. So this is a quick guess. Okay? If you do it the BGZ way, you'll get that the first syndrome is out of that time. The second syndrome of valence problem, I think a lot of people may mistake for that thing. If this was the valence, these kind of syndromes is just write it down, have the table there and then write each term and then cancel. Don't cancel in your head because there are too many terms which will cancel. Okay? Don't cancel in your head. Write every term and then cancel at the end. That's much better. Okay? So S3 is alpha of 4, we'll go for alpha of 4. And then, so you start in BGZ with two errors, right? So two errors when the sigma effects is complete, 1 plus sigma 1x plus sigma 2x squared. And then when you write down equations, so I mean I don't try to remember the matrix equations. Some people remember that. We don't necessarily do that. So remember what happens? S of x times sigma of x should have no coefficients, no term, x part p plus 1, x part mu plus 1, number of errors we're guessing. So x part 3, x part 4, must not be there in S of x times sigma of x. Okay? So if you write S of x times sigma of x, you get alpha part power x plus alpha part power x plus alpha part power 3 plus alpha part power x plus alpha part power x plus alpha part power x plus sigma 2. Okay? What is the coefficient of alpha part 3? It's very easy to see that. So we have alpha plus sigma 1 plus alpha part power sigma 2 plus mu equal to 0. What's the coefficient of x part 4? Alpha part 4 plus alpha sigma 1 plus sigma 2 equals 0. These are the two equations. Okay? So you have to see if it's solvable. Okay? If these two are not solvable, then your guess was wrong. You have to go down to one method. Okay? But here it's clearly solvable. I mean you can quickly see that the determinant here is non-zero. So you solve for it. So you solve for it. It's very easy to solve for it. We'll get sigma 1 equals alpha square. Sigma 2 equals alpha part 6. Okay? So you don't have to remember the matrix. In small cases, it's very easy. I mean it's not very hard to come up with this. So the sigma of x is actually 1 plus alpha part power x plus alpha part 6 x squared. So you start guessing 0. Okay? So you start something. You cannot avoid it. But luckily, interestingly, there's 0 for 1 and alpha. Okay? You can quickly guess. The first thing you're going to guess is 1. You'll see 1 plus alpha square plus alpha part 6 is 0. The next thing you're going to guess is alpha. And you'll see 1 plus alpha square plus alpha part 6 is 0. So 1 and alpha is 0. So here there is one more trick. You should be careful here. This is not the error locator. What is the error locator? It's 1 for the first one. But what's the second error locator? Alpha part minus 1, which will be alpha part 6. Okay? This is the first and last error locator. Okay? That can immediately give you a mean test to the error locator. If you're aware of the possibility, you don't have to really evaluate the error magnitude. You know, immediately the first and last error locator is a problem. You can quickly get to the answer if you like. But if you want to evaluate it, you have to write down the two equations. So if you do, look at S1. S1 is alpha part 5. Alpha part 5 is... So E0 plus E6. X2 part 6, right? Alpha part 6. Okay, remember if you have this, E of X is of the sum. Some E0 plus E6 X power 6. Right? So you just put X equals alpha here. You'll get the first syndrome. Okay? I'll do the second syndrome. I'm going to put X equals alpha squared. So that will be E6 times alpha part 5. Okay? So alpha B2, you'll get E0 equals alpha part 3. And E6 equals alpha part 3 again. Okay? And then you put it back in, you add up, you'll get the answer. Okay? Alpha plus alpha part 3 will be 1. Okay? That's our minimal polynomial. So you see the error magnitude. Okay, so it's good to do. So don't think of Reed Solomon as one of your laboratories. I think you look at the pressure and think, ah, it's painful. Then you can quickly get to it. It's just nothing in it. I mean, it's just some... Well, it's a big deal in the equation. It's not a very major set of equations. Just two substitutions. GFA is such a small field, very easy to do. You don't need any major calculation. Okay? That's it. I've given partial marks once again. I have a switch here. I'm not going to describe that to you, but in case you feel you didn't get enough, comment out for me. I can explain what I did. So that's it. As far as the quiz is concerned.