 So we've been looking at problems in many block codes, so we're going to look at some problems in finite fields. So I think there's a question about factors, we should come to that later, but for now, let's just go through these questions and see if there's anything that's interesting or something that I should comment on which you have not seen explicitly in class. So the first few questions are very standard questions are constructions of finite fields. So here again the method is quite standard, what do you need for constructing GFP power m, meter power of alpha, degree m, over musical polynomial from what? Zp alpha, right? These are coefficients from Zp. So you need that and then we'll simply construct this GFP power m as polynomial degree, less than or equal to m minus 1 and Zp alpha, okay? And then for addition we simply do addition, there's no problem, addition of polynomials, model of p as for the positions, and for multiplication it's not below power of alpha, okay? So one question that is usually asked in the construction is to make a table for this field. The table usually means you have to make a table between the power representation and the vector representation, okay? So that's a very standard table. So in this field, for instance, you make a table between alpha i and the vector representation, okay? So this will go from i equals 0, which is 1 alpha, alpha square, assuming alpha is primitive, of course. So to take a primitive element, then go all the way to, that's the last variable, okay? So p power m minus 2, okay? And then here you have a notation, okay? So how do you figure this out? So the way you do it is you start with 1 alpha, alpha square, till about alpha power m minus 1, it will be just itself. Then for alpha power m, what will you do? It will depend on power of alpha, right? And then for alpha power m plus 1, you have to keep doing that repeatedly, just multiply by alpha, do model of power of alpha all the time, okay? And then you can fill it out. So this is a table which represents the field in many ways, right? So if you have to represent this field in a computer program and do manipulations, this is the table that you would remember. So of course there's no alpha, alpha will simply be indexed by the power. So it goes 0, 1, 2, p power m minus 1. And here you have that test, that test will be coefficient from z, right? So each vector here, for instance, a generic alpha power i, you will have a 0, 0, 1, 2, p power m minus 1. Each case comes at p, right? So this vector will be from 0, 1, 2, p power m minus 1, okay? So once you start it in a computer program, you can do any financial operation you want. So this is a table which is very useful in construction. Yes, is that a question? So if you define the construction of field files, then the alpha will automatically denote the primitive element of the program. So see, the question was about when is alpha primitive, right? That is the question. So if you do this, if you do the time of alpha, when is alpha primitive? It's not all the time, okay? So it may be primitive, may not be primitive. So if you pick your degree and reducible polynomial carefully, you can make it primitive, okay? So it's not... So here I have just assumed that alpha itself is primitive. In some cases, it may not be primitive. In that case, some alpha plus 1 or alpha plus 1. Some other element will be primitive. And then you make the table with that element. Okay, the table is always made with any primitive element. There is no point in making the table with any other element because it wouldn't be complete. Okay, so you make the table only with the primitive element. Okay, so if you look at the questions, most of them are basically addition and multiplication tables. So instead of making addition and multiplication tables, it's enough if you make this table, then so any addition and multiplication can be done with this, okay? And there are various constructions, some with what's called a primitive reducible polynomial, another is non-primitive reducible polynomial. So all these things you can try. So let me just define primitive reducible polynomial, if you like. Okay, so if you have power effects being reducible, between n and r and px, then the power effect is primitive, power effect i for which, i for which, power effect r for x bar r minus 1 is equal to p bar n minus 1. Okay, so that's the definition for primitive reducible polynomial. First of all, how do you know that there will be some i for which power effects will divide x bar r minus 1? Okay, so the right way to answer that is you can set the finite field with this power effect as the reducible polynomial, right? And then the root of that reducible polynomial will definitely divide x bar p bar n minus 1 and then separately the effects will divide x bar i minus 1 for some i. Okay, so if i equals p bar n minus 1, it will definitely divide. So when I say it's primitive, I want the smallest i for which it divides x bar i minus 1 to be p bar n minus 1. So here's an example. To take power effects equal to x bar 4 plus x bar 1, it will be the small and primitive. You can check that. On the other hand, to take power effects to be x bar 4 plus x bar 3 plus x bar plus x bar 5, it will be the small but not primitive. Do you see that? So what is the smallest i for which this polynomial will divide x bar i plus 1? It's 5, right? So this power effects divides x bar i plus 1. Now we are in binary, so the benefit and that's kind of the same. So it divides x bar 5 plus 1. So here's a non-binary example with 9 for instance. So you can take... So if you want a non-binary example, so another example, to take power effects to be x bar plus 1 instead of primitive. Yeah, I think so. There's one example like that. So x bar plus 1 in, it's at 3x. This is reducible. Am I right? How do I check that it's reducible? You can substitute x equals 0, x equals 1 and x equals 2. We never get 0. So it's reducible but not primitive. What is the smallest i for which this will divide? 4, right? So x bar 4 minus 1 is x bar plus 1 times. x bar minus 1, and then so x bar plus 1 divides. x bar 4 plus 1 and 4 minus 1 is 8. In particular, in these 3x and 4 we divide. So if it were primitive, what is the smallest i for which it should divide? 8. 8 minus 1. So there are other examples. So there's a 3x for which it will be... for instance, x bar plus 2 and the 3x for which it will be primitive or not. So these are primitive or not. So there are other examples that are personal. So around this construction, there can be questions like this. Of course, to understand it, another question is the isomorphism between two fields. So you have to identify corresponding polynomials and find roots for the polynomials and once you map it to the primitive elements that way, it's the isomorphism. So that's the idea of isomorphism. Identify which elements? The primitive element. So the question is how do you identify a primitive element? So usually, it's best to pick a primitive polynomial and construct your field. Okay? I see one. Okay. Can we map... We explicitly find the... We find explicitly the isomorphism. Okay. So we basically find the primitive polynomial, we find the primitive polynomial, find the roots of the primitive polynomial and build the fields. There will be roots. And map those to that to each other. You know that element is primitive and all the powers will map the elements and that gives you the isomorphism. Okay, that's the way of doing it. I think your solutions must talk about it. Do you have enough more? It's not straight to the isomorphism only. So you have to look for roots for the element. But the ways of doing it beyond the isomorphism is usually what the isomorphism is. Just try to define the elements till you get a primitive polynomial. Okay? So the isomorphism in all is not too critical from a coding point of view. Okay? The order is a little bit important. So let's take maybe for instance the third question. Okay? So the question here is it says, you take gf9, okay? And what gf9 is going to be? 0,1,0,5,4,5,6,7. The alpha is some primitive element. So the order of alpha is, okay? You have to find the order of the other elements. So you have to find the order of the other elements. So that is the standard formula I gave you. What is the order of alpha, right? It's what? 8,5,6,7. Yeah. So the order of alpha is going to be dcp of i comma alpha. So that's the formula to keep in mind. So you just apply that very, very easily. So it's also a little intuitive and nice formula to think about. Alpha itself, power of 8 goes to 1. Any other power depending on how it factors into alpha. So for instance, the order of alpha square would do up. It would be clear, right? 3 always factors, 3 already is a factor of 8. So you never have to raise it 2 again. So already the 2 is there. When you raise it, the 4 will be there. 1. Okay? So what about 3? 3 you can't do anything because 3 doesn't have any common factors with 8. So you have to raise it to 8. Okay? What about 4? It's going to be 2. Okay? So 4 is already there. 5. 8 again. 6. 6 is 4, right? And then the 2 is already there. If you raise it to the power 4, that's enough. Okay? And then what about 7? 8. 8. So you have this kind of a construction. All right? So there are some irreducible polynomials. I think there are 3 irreducible polynomials. And then there are 3 over 2 over 3. 3 or 4. 4, right? No, no, no. Yeah. Okay. So, no, no, no. So degree 2 would be 3. Degree 1 is 3 of them, right? So that's x minus 1 and x minus 2. So you have 3 of them. So in fact, one of these alpha powers will be equal to 2. Right? So in any construction you make in GF9, remember GF9 also contains GF3. You have 0 and 1 should also have 2. Some power of alpha will be in fact equal to 2. Okay, so if you do an action computation, you'll see it will be equal to 2. It's not very hard to do that. So we have 3 degree 1 irreducible polynomials and then 3 degree 2 irreducible polynomials and you can do that. Okay? So it's a little unusual to do these things in coding because usually we are obsessed with characteristic 2. Okay? Characteristic 2 is what is most interesting from a coding theory point. Okay? So the same exercise we will repeat for GF16 which is very easy. Then we also ask for GF32. So what will happen in GF32? Yeah, so if you have alpha primitive, the order of alpha is 31. Okay? So 31 cannot have a common factor of anything. So the order of every element becomes equal to 31. So the order of every alpha power i is i equals 1 to 30 equals 31. Of course, the order of 1 is 1 itself. Of course, 0 is not the function. Okay? So that's about all that. Okay? So let's move on to the web question. Okay? So then there are some simultaneous equations. So I want to look at that real quick just to get you a feel for what's happening here. So let's look at the 7th question first. First question. So you have alpha belonging to GF16 being primitive. This is the 7th question. Let's go over and decide the order of alpha first one. That's given to you. And then you give them two equations. Of course, y equals alpha power 14. And x power 3 equals alpha. Okay? And you will find x and y. Okay? So the first thing you need to do such problems is you need the table of GF16. Okay? Only then you can do any computation. So the first thing you should do when you see a problem involving GF16 in your exam or anything else, is to make that table. There are 16 elements in the table and you need to make that table. There's no other way. If you don't make the table, you can never do the computation. You quickly make the table. It's not very hard. If you practice it enough, it will come out very quickly. In fact, you can even mark it up if you like. Okay? So that you have the table ready with you. Okay? All right? So you quickly make the table on the side and then you can do the computation. Okay? So solving these kind of equations, what would you do if it's not finite field? If it was some real number or something, x plus y I give you, x plus y plus y plus y plus y, I give you. How do you do it? Yeah. So the trick is to factor x plus y plus y plus y, right? So how do you factor x plus y plus y plus y plus y? X plus y times? X squared minus x squared plus y plus y. You know that factor. So why is that factor useful? Well, x plus y is only given to you. Okay? And then that gives you x squared minus x squared plus y squared into a real number. And after that, what do you do? You write x plus y, x squared plus y squared as? X plus y times squared minus 2xy. Okay? That x plus y is again given to you. So ultimately, you get x squared. So once you find x plus y and xy, you can go to a quadratic equation and you'll get x and y individually. Okay? So such methods are, I mean, they're specifically elimination. If you're trying to eliminate one equation from the other, one variable from the other, how you do it is this thing. So you can also do it in the most straightforward way. Simply take x equals alpha plus 13 minus y and substitute it with the next one. You get a cubic equation and y. That might be quadratic directly. So that also might be easy enough, right? So that's something you can straight forward try out. If you want to do more fanciful position, you can do that also. Okay? Try one of the two. You will get one equation involving y or x. And then what do you do? Yeah. So in GF 16, that is a bit easier. It's a bit mechanical there. Okay? What do you do? One element after the other and find, which is easy. Okay? So in real numbers and all, you might need complicated formula for finding the roots. In GF 16, there's no, nearly no in the raw formula, but it's as good as simply substituting one element after the other and then figuring out what the root is. Is that clear? Okay? So let's try like the first attempt. Okay? So from the first equation, you see that y equals alpha plus 14 plus x. And then we substitute that into this. You put x for 3 plus alpha for 14 plus x whole power 3 equals alpha. So what happens if I expand this? Then alpha for 42. What is alpha for 42? Alpha for 12 plus 3 times, 3 is the same as 1, right? 3 times alpha for 14 squared, that will be alpha for 28, which is alpha for 13 times x plus. Another 3 times alpha for 14 times x squared. So that's just alpha for 14 x squared. That's what? x for 3. Okay? And that equals alpha. And then x for 3 cancels. And you get like a primary equation. Okay? So alpha for 14 x squared plus alpha for 14 x plus alpha plus alpha plus 12. Okay? So whatever that was. So that's what the table will help you to give you a simple answer there. Okay? So you might take the quadratic, quadratic, quadratic, quadratic, polynomial roots formula, right? So that might apply. It actually applies. But the problem is, characteristic 2 doesn't really apply. Okay? Because we will derive by 2. Right? So that's the formula. It's minus b plus or minus 5 root of b square minus 2. I see it for a c by 2 a. Okay? And we can't derive by 2. Okay? So that formula will not apply for a primary equation. It would be nice if that formula would apply, then you can actually compute it. So in the formula that will apply, you can't compute it directly. Okay? So you have to substitute one root of b square minus 2. Okay? So it's enough if you find one root. The other root is easy to find. Okay? So why? Okay? So you might know some other simple formula for product of root, sum of roots. That's the formula to apply. Okay? It's not that they go away. So if you write it as alpha 1, alpha 2, you get the same roots applying. Okay? So you simplify it a little bit. And find one root that can be found easily by computation. All right? So that's the way where we're solving these things. Okay? So there's also a similar question, question number 8, which is the same. Two cases. The exact same setup. Okay? As before. Well, they cannot give an equation like this. Okay? Instead of x power 3, it is x square. Okay? Okay? Here's something curious will happen. Okay? So I'm characterized in two phase. Squaring will not really give you an independent equation. Okay? So you square and x plus y will be right. Okay? So you square and x plus y will be right. X square plus y square. Well, it has to be equal to alpha plus 6. So this equation is really useless. Okay? So this just repeats what the first equation is. So the only equation you have to solve is x plus y equals alpha power 3. How many different solutions are there? You can take x to be an arbitrary element of g of 16. So there are 16 different solutions to x plus y equals alpha power 3. So that's what we do there. Okay? So only this. So this is repeated. Okay? So it's not right there. So it's same as this one. So you only have to solve this. So there are 16 parts. Okay? So if you add them you have the 16 parts of the solution. Okay? Part b is a very important question. So if you only have an equation x plus y equals alpha power 3 but then x plus y equals alpha. What will happen if you try and solve this? Because that's why it's alpha power 3. What should happen to x plus y squared? It should be alpha power 6. So here there is an inconsistency. Okay? So if there is no solution then it's g of 6. Okay? Of course there are real numbers and all are complex. So you can definitely find solutions. Okay? So it's not a problem but in g of 16 it's not going to work. Okay? And then there are other methods. You can use repeated spreading till you get to something like that. So I'm going to skip that. Let's move on to Okay? So questions 12 to 16 don't really apply. Okay? 12 to 16 don't apply. It's not for the quiz at least. There are more general problems. It's not going to work. Okay? So let me talk about the questions that are interesting. So 15 for instance I'll see the factor of the form x power n plus 1 over g of 2. Okay? So when I say g of 2 it means I want for each factor to be have only binary coefficients. Okay? So how do you factor x power n plus 1 over g of 2 based on what I use? I'm sorry? Yeah, x plus 1 will be a factor always. So those things you can find, but I want to factor completely. Okay? So the first step is to factor x power n plus 1 into linear factors over some extension field. Okay? So that's the first step. Okay? So you factor into linear factors and some g of 2 power n. Okay? So how do you go about finding this n such that you factor in g of 2 power n? Yeah. Yeah, that's the idea. So you have to find some root for x power n plus 1 in this g of 2 power n. Okay? So remember I have a root for see, if I take a primitive element of this alpha and primitive element here what is the order of alpha? Alpha of 1 2 power n minus 1 equals 1. Okay? So which means this primitive element is a root of okay? So primitive root I'll tell you what is primitive root root of x power 2 power n minus 1 plus 1. Right? So what do you mean the primitive root? So not only is alpha a root all the powers of alpha are exactly the distant roots of x power 2 power n minus 1 plus 1. So that's okay? Power of alpha are distant roots. Okay? So you know all this right? These are the results of finite fields. Okay? So you take the primitive element take all its powers they give you all the distant roots of x power 2 power n minus 1. Okay? So similarly if you find an element of order n in 2 power n gf2 power n okay? Suppose you find so in this gf2 power n suppose beta of 1 gf2 power n has x plus beta times x plus beta squared so 1, 2 x plus beta power n minus 1 then you have x plus beta power n the value of x plus beta power n it will actually x plus 1 okay? So that will matter So even k m that this has to work okay? Right? This has to be 2n gf2 power n Why is that? You know beta is the root of x power n plus 1 beta squared also a root beta power 3 is also a root of x there cannot be any repetition there right? beta is order n if there is a repetition in the order we did it so that cannot happen okay? So the trick is to factor x power n plus 1 into linear factors you have to find an element of order n in gf2 power n okay? So how do we do that? What are the elements? What are the elements of gf2 power n? Yeah What is that better than? So you have to sign the m such that n derives 2 power n minus 1 So that is the first step okay? So you find n sub n n sub n derives 2 power n minus 1 and then I can gain so alpha is a primitive element of in gf2 power n okay? Alpha is a primitive element of gf2 power n I can gain but alpha power what? 2 power n minus 1 divided by n that is order equals order equal to n why will the 2 power m minus 1 by n be a proper position? We have chosen m such that n has to divide 2 power m minus 1 okay? So let me ask you this question for every n are you guaranteed that n will divide 2 power m minus 1 1 minus condition you need on n n has to be 2 power m minus 1 this is follows from this so much little bit it's not too difficult to prove you can also prove it if you like okay? for every order n there will be n such that n divides 2 power m minus 1 n we run a even of course there's no question of n dividing 2 power m minus 1 for any n but then if n is even what can we do? okay? so you have repeated so now you have the factor is only x star k plus 1 so if 2 is also even what do you do? you keep repeating this till you get to n odd power when you get to n odd power you see this result and do the factorization okay? so now we've not done we have only factor over gf2 power m however we go from gf2 power m factors to gf2 factors what is the platomic process? so you find also the conjugates of beta okay? so you know beta's conjugates all will be in this set only right? you are just raising beta with different powers it will all be within this set so you do all of the common conjugates multiply them together you are guaranteed that the polynomial will be minimum so it will be inside the coefficient from gf so that's h for n plus 1 you can put linear factor in gf2 power m okay? and then we combine linear factors conjugates okay? factors over gf2 okay? so this is very similar to the previous idea that we had okay? except that now instead of m being itself p power m minus 1 it is some other number and you have to go to a suitable okay? so again condense yourself but once you find the beta like this the conjugates of beta will always be inside this set right? so you always find conjugates I square beta and then I square it again so it only powers of beta I can never leave any of these factors so you can prove these factors together to go to a smaller so let me just show you really quickly interesting so let's start over I'm not doing it there so let's for 5 plus 1 okay? so what is the m? right? n equals 4 right? 5 2 power number 1 which is 15 okay? so if you take alpha doing a primitive element of gf2 power gf15 what is the alpha power 3 x plus alpha power 6 x plus alpha power 1 x plus alpha power 9 and then x plus 1 okay? I know now 369 and 12 other are circled on the cross side element so all those 4 will multiply together in gg x power 4 plus x power 3 plus x power 3 plus x power 6 x power 5 plus 1 equals x plus 1 times x power 4 plus x power 3 plus x power 4 okay? so that's the idea in that transition let me show you x power 9 plus 1 okay? so what do we do for m? so the times of 6 is the correct answer mg of 64 from above other of 9 7 is 9 so this will factor us you know alpha star 7 so on okay? so what do we do the last some factor alpha x plus alpha power 63 right? so there I'll level that so maybe I should write it back so let's just do this let's just do this x plus alpha power 14 x plus alpha power 21 x plus alpha power 28 married x plus alpha power 35 x plus alpha power 42 x plus alpha power 48 49 and so you can also do the third alpha and do it in an organized way so we'll show this x plus alpha power 56 times x plus 1 okay? so that's the factor in our gf64 so now we should start combining conjugates so let's look at conjugates first 7 okay? so if you look at 7 alpha power 7 alpha power 14 alpha power 28 alpha power 56 what's the next one? 1, 1, 2, B sorry 39 and then that's it one more unique 39 is 98 that should be 35 right? so that's it those are the conjugates the next one will come back to alpha power 7 right? 2, where this left? alpha power 21 alpha power 42 32 go back to 21 again okay? so that's the combination of conjugates so you see this will give you x plus x plus 1 that's easy to find out this will give you some 6 degree minimal polynomial okay? so it depends on the table okay? so if you have a table you have to multiply it and simplify it it will give you some 6 degree minimal polynomial again it will factor as x plus 1 times x plus x plus 1 times minimal polynomial alpha power 7 okay? so you can write it in terms of that okay? that's yeah that's it so there's another way to quickly find that out if you like so in fact this minimal polynomial can be easily found so to do this factorization do actually 1 plus x power 3 you know that again so what will be left? it will actually be 1 plus x power 3 plus x power 6 okay? so you can quickly find it in reverse like that if you like but anyway so that's the that's the okay? so anything else will factor the same way okay? so the smallest field over which x power n plus 1 factors into linear factors is called the splitting field okay? for any polynomial that is true any polynomial if there is a field over which it splits there is a splitting field the complex num field is a splitting field that's all for any polynomial it won't be a splitting there is no one unifying splitting field right? so you have to find the difference splitting fields okay so I think that kind of brings us to the end of most problems that I wanted to discuss there are some other problems and we are welcome to look at it and the question the exam is tomorrow right? so the exam is for 25 marks done these two assignments okay? so we meet where are we meeting for the exam right up there do you know the room? so anyway find out the room and then from that I will see you guys okay? thanks I am sorry it's a 50 minute exam standard course