OK, so I continue my talk. So today, it's third part, so properties. So to prove some properties, more properties on character circle, so in the definition yesterday, I just used the local acyclicity of the familiar moistened curves. But now, to prove something more, I need a similar condition for more familiar moistened surfaces. So let me state this first. So the definition is quite similar. So I just replace curve by surface. So x will be as before. So yesterday, this was c, but today, it will be p. So this is smooth moistened. So yesterday, it was just flat, but now I assume smooth. So this is smooth moistened to smooth surface. So p stands for surface. So we say this is non-characteristic if. So the condition now looks the same. So the intersection is so yesterday, we just assume that the moistened is flat. So we didn't have the injection, but now we assume smooth so we can talk about the image. So image of df. So now, sorry, x cross p. So today, this is injection so we can talk about the image. So this is inside the 0 section. So this is one condition. And also, so we had the second condition yesterday and it looks quite the same. So the intersection is the 0 section. And we regard this as a closed subset of x. And this will be flat over p for every i. So this is a non-characteristic is left to do. But is it only a closed subset or is it a skin? Yeah, so I regard this as a reduced closed subset. And I define this as a fiber product. So this has some skin structure. Is it necessary for you to have flatness or is it enough that it's dominant? No, actually, I need this flatness. So this is a bit strong condition. No, but the problem is that when you make a talocalization, you make the components i into several branches. And then the condition of dominant, like what you have is preserved. But the flatness is not preserved under the composition to branches. And the problems you are studying are kind of talocal. So that is a natural to make a definition which is not sufficiently bears well under a talal. So SI is already fixed. So after talocalization, I don't change the numbering. Yes, but the condition looks not very stable by localization. Yeah, yeah, yeah, you're right, yes. Because originally, there are side reducible components. So again, this is not stable. Yeah, but I need to put some condition anyway. So this works for me. Maybe this is not the best definition, but I'll put it like this. So this is for one more. So as yesterday, I need to work with the families. So the second part is that I draw the same diagram. So now it looks the same. Yeah, so this is P. So this is the commutative diagram. So today, everything is really smooth, smooth, smooth, smooth, smooth schemes. And here, as yesterday, this will be smooth. As yesterday, so this will be et al. And so this is today, this will be smooth. So this is already smooth. So this is your cube dimension, too. So we consider such diagram. And we say f over v, if it is non-cast type, this is to s over v. So this looks the same as yesterday. So for every v closed point, the base change, this is non-cast type, this is to the pullback, this is. So I keep the same number in here. OK, so this is a non-cast type condition. Now I introduce SS2. So yesterday, we had SS1 because we had more smooth to curve. Now we have SS2. So SS2 is for commutative diagram. As above, so non-cast type implies over v, implies local type ct of the pullback of our complex. So this looks, I guess, similar to yesterday. So yesterday, we had SS1, and we had the same thing. And also universal locality. Oh, yeah, yeah. I think it follows from this theory of vanishing cycle that locality is usually equivalent. Yeah, yeah, yeah. So this doesn't change if I put this or not. But this is easier to think about. Thank you. OK, yeah. So this is the condition we really need. And the first statement I'm going to make is, so we study immersion. So x is our smooth variety. And y will be a smooth divisor. So this is the immersion of the smooth divisor. So I want to introduce another notion that this immersion is non-cast type. So this is maybe, so I need one more definition, 2.2. So OK. So we say this is non-cast type. I need to introduce this stronger condition, so strictly, so respectively, strictly non-cast type. If, so the condition is similar as a verb. So the intersection. So this time, so there we consider the image of the differential from the target. So this time, so we have immersion. So this is the regular immersion. So we look at the connoisseur. So this is the connoisseur. Yeah. No, no. Yeah. Yeah. So we consider this intersection inside the cotangent bound. So this is inside the zero section. So just the first condition looks similar. And then the second condition is that the second condition. So now this TI is a closed subset of X. So for this first, just non-cast type, so this is a divisor. And for this stronger condition, so we require that, so this intersection is irreducible and reduced divisor. So the first part is just non-cast type, but I need to define a little stronger condition. Cartier divisor, yes. Yeah, yeah. Yeah, before taking the intersection, it's divisor. But then if you localize the risk, if it's irreducible and reduced, it can become empty. And then, or maybe it's not empty because the zero section is, no, I'm confused. Yeah, so TI is the original intersection with the zero section. So this is not the risky local property, irreducible and reduced, because it can become empty. Yeah, yeah, yeah, yeah, yeah, yeah. OK, so more empty, yeah. Yeah, thank you. Yeah, now I can state the first term. Yeah, I think I need more space, yes. Now that's the first term, so I put two stuff because I need to assume both SS1 and SS2. OK, so I assume, so we have this single support satisfying SS1 and SS2. So by this definition, this thing is supposed to satisfy SS1, but I also require that it satisfies SS2. What do you mean? You mean that there is a way to associate single support in such a way it satisfies those properties? So do you mean for all schemes at once, or just? No, no, just for my x and k. Ah, OK. So SSk was already SS1, in some sense? Yeah, yeah, yes, yes. So it was very nice, and SSk that satisfied SS2, this is the condition. Yeah, so I fix x and k and SS, and I call this SSk, and I say this triple satisfies SS1 and SS2. This is more precise way to make this assumption. That OK? So yeah, so they are fixed. OK, and I take my regular imagination, so this is the imagination of a smooth divisor. And I assume that this is a strictly non-diastatic, this strict one. OK, then the, so I will explain this notation in a minute. And I take the union with the zero section, so this is inside the cotangent bundle, all right? So this is the single support of the pullback, restitution of k, and we can compute the character cycle. So the character cycle of the restitution is given by the pullback of the character cycle. So here, what is this I seek? So to define I seek, so I have correspondence like this. So we have this restitution to y and the canonical projection to, yeah, so this is inclusion and this is projection. So first we take pullback by this inclusion. So here I get y, so this intersection. And this non-characteristic condition means that the restitution here is finite. So this implies that by using this correspondence, we can define that this image and also this cycle. So this non-characteristic is to assure this finiteness. And with this finiteness, we can define this and that. So this is first statement. And I prove this term at the same time as the next one. But so I cannot give a very precise statement of this because it will require the whole business with ramification theory. So I will be a little vague here. So we have an open machine. So u is a complement. This is a simple divisor with a simple number crossing. And I assume that my k is 0 extension of f. So f is a locally constant, constant active on u. And I need to assume one technical condition that so then we may have ramification of this shift f along this divisor d on the boundary. And my condition is that ramification of f along d is strongly non-designated. So I'm sorry that I don't have time to explain this technical terminology. So this is something close to what Kato introduced to cleanness condition, but in non-nogorizmic context. And this is a little bit stronger than the counterpart of his condition. But in any case, so this is some condition coming from ramification theory. Is this strongly related to the same strategy? No, no. I changed the slightly. So here it's strictly. Here it's strongly. So there's no relation. But the only problem that the generic can be fierce? It can be fierce, yes, yes. If this is just an irreducible divisor, does it mean that when you cut transversely, you get more or less the same kind of verification or not? But this is like clean. This is like clean. So this condition is always satisfied if you remove so outside of co-dimension. To this condition is always satisfied. OK, then the conclusion is that this. So now we have this characteristic cycle. So this is equal to what defined by the ramification theory. So I don't have no time to define. So if we assume this don't do it, the ramification theory tells you how to define a character cycle. And the claim is that this gives exactly what we are talking about. So this includes connormal bundles of higher co-dimension, at least two things? In fact, it doesn't contribute. So you consider only the zero section and the co-dimension. Yeah, if you have wide ramification, we get just something like line bundle over this whole component of divisor. In the case if you have wide ramification, so this condition implies that you get only something like a line bundle of divisor on the boundary. Oh, plus zero section, yes. It's the definition I've seen two days ago. OK, OK, OK, I remember it. But this is included in the theorem that SS1 is not satisfied. It's as if the singular support exists. Well, so in this case, SS1 and SS2 are satisfied. Yeah, so it's a part of this. Yeah, so I didn't put any star here. So this is in conditional. So in other words, this guy, this one satisfies the minimum formula. So if we put it this way, then there is no condition. So under this condition, we can define a character circle. And that one satisfies the minimum formula we were studying yesterday. So in this way, there is no condition, so we don't put any star there. So there is no isoclinical condition? Yes. Do you allow it many slots? Many slots, no problem, yeah. So because the minimum formula is at our local. The cleanness seems to be just for the luck. Ah, cleanness you can localize it also. No, no, but it is for just the worst slot. No, I think I'm allowed to make direct sum of the composition. If it's a local, you can get a direct sum of the composition. Each one has a step. And for strongly, nor the generated top local, you have a direct sum of the composition into isoclinical or not? Are you sorry? So strongly on the generator does have to compose the top locally to guys which have only one? Yeah, yeah, yeah. If already non-generated implies that. Non-generated condition, without strict, just non-generated implies that. And you didn't define non-generated? Non-generated, I'm sorry, I don't have it. Really, no? Yeah, it's similar to, it's similar. It's generalization of clean in this non-organism context. OK, so now I very briefly sketch how the proof work goes. And then I give some consequence of these two ceremonies. And then if time allows, I will go more detail of the proof. So outline of the proof. So this is very, here it's very sketchy. So first I put this 3.4 in dimension 2. And then I reduce 3.3 from this particular case of 3.4. And then the last part is easy, so 3.4 in general. And for this first step, I need a global argument using proper, normal scheme. And this step is another easy. So if I have time at the end, I will say something on this part. But before going there, I give you some consequence of these two theorems. The first is the corollary. So the first statement is the compatibility with the smooth, smooth, smooth. So if f is smooth, then the constructable character circle is compatible with the smooth pullback. So here, so this is just pullback. And this one is again by correspondence. So this time, we have a correspondence like this. And here it's just a smooth pullback. And here, because this is smooth, so this is injection. So in this case, there is no problem in defining the reference. Like I said. The assumption here that there are some axioms called for x or for both x and y? Just x. OK. Then it follows that the SS1 holds for y. I'm sorry? So if you know it for x, it follows that SS1 will hold for y. And then you can define the characteristic cycle and then you can state this. Yeah, yeah, yeah. SS, yeah, yeah. Yeah, so also this includes SS f star of k is f star of SS k. So once you know this, then you know that. OK, so this is the first consequence. And I can also deduce formula for Euler number. So if x is a projective and smooth, then you can compute the Euler number as the intersection number. OK, so I will spend some time to explain this. That's the proof. So proofed by induction, proof of 3.6. So this is induction on dimension. So if dimension is 0, then there's nothing to prove, essentially. So I use the following lemma. So I take embedding as yesterday, satisfying this E, condition E as yesterday. Then there is a pencil, so satisfying. First, so this is the map defined by this pencil. So this has at most isolated characteristic point. So I take JL member of this pencil. So there is H in this pencil, such that the intersection, this embedding is strictly non-statistic. And I also need one more intersection, so z. So this is the intersection with the axis of the pencil. So this is smooth. So I'm taking JL member, so this is a smooth deviser of this y. And this is also sort of non-characteristic. And I need one more condition. So we may have a finite many character points. So these are character points, not in the image of this intersection with the axis. So I don't prove this same way, but this is another E. So you can prove this by using Bayton's theorem. OK. Then to prove this order of formula, it's just a computation. So we want to compute this order number. Then so I want to write the notation for the pullback. So we have this quality. And so we need to compute this guy. But we can apply the gotonic of the suffability formula here using this pencil. So to do the order number of L, so this is the kind of L. And the order number of the z-x fiber. And we have a total dimension. So u-lands are isolated character points. So this is gotonic of the suffability. Yeah, so everything is standard. So now I can apply induction hypothesis. So we have y here and z here. So we can apply induction hypothesis. And also theorem I just raised plus theorem 3.3. So sorry, I forgot to put minus sign there. Yeah, so I don't like this. But this will be minus sign here. I'm sorry for that. Yeah, so this is minus of my induction. Yeah, so here the co-dimension is 2, so it's even. So I don't have sign there. So I is the image here, and I prime it. I z is the image there. And I also have a minimal formula. So minus total dimension of this psi u is characterized by this intersection number at u. So we substitute this in this formula. And we compute the substitute. And we need a little more computation. So plus compute chances. Then we get the equality for all other numbers. In fact, you did not say that the SS2 condition is preserved by cutting with the device. And restriction is preserved. Yeah, yeah, it's preserved, yeah. Where is it used there? By the induction. So I cut x by y, and I need one more step here. I need to cut x two times. But it was implicitly there. So to state, yeah. Do you need here the vital or strictest non-characteristic? Yeah, strictly non-characteristic. So I play by this lemma, yeah. But then you have to cut it another way, OK? And one pencil works two times, yeah. OK, so remaining 15 minutes, I say more words on this part. I assume 3.4 in dimension two. And so in some sense, a reduction to dimension two. This part is, I don't say anything on this part. Yeah, so this part is basically not my dream, but this part is new, so I prefer to do this part. OK, so I take, so this is this formula. So I want to prove this formula. So to define this side, so basically the definition was, so let me remind you that the definition is made by looking at this marked pencil. So it's enough to prove this quality. So u is an isolated character point. So we let this u and prime under, we want to prove this type of formula. Then this will imply that quality. So we want to prove this formula. So for this left-hand side, so this is by definition, so this is computed, this is a millenium formula. So the total dimension of a vanishing cycle to u is this motion. So this is a millenium formula. Excuse me, so I want to prove that we have ss. Ah, I skipped that part, sorry. So you assume that ss1 is satisfied after cutting, right? Yeah, yeah. It means you assume that you have a millenium formula. Yeah, yeah, yeah, yeah, yes, yes. And so I introduced one notation. So I put the bar, so this is reduced to the character class. So this is the character class with the components of the zero section removed. Do you take, is it closed or just a locally closed? Is it? Ah, closed. So this is the linear combination. So this is just a matter of notation. So we can rewrite this as the character cycle of this vanishing cycle. Then there is a slight abuse of notation. So I just take the coefficient. Yeah, it's a number, so yeah. So you have to cut it with the zero section. Yeah, yeah, you can say that, yeah. Yeah, so you put the number up as a coefficient of the fiber. OK, so yeah. So now, so we have this, we are working with this map. So this was PL prime. But let me write this, go to this G. So I have, so I was a divisor of X, so I have a closed margin. So suppose we are given a map to surface. So I want to use SS2. So I take a map to P, so this is the surface. And the Cartesian, so surface and line. So this is also a regular margin. So suppose we are given such a diagram, Cartesian diagram. Then the construction of, so now we are using a formalism that you told us yesterday. So this, so I continue, so this same thing. So I can switch, so I estimate that the singularity here is quasi-finite. So that the formation of one single cycle committed with base change. So now I can put H first, like this. So this is the computation on the left-hand side. Now I go to the right-hand side. So right-hand side, so similarly. So in the same way of writing, so we can put back by this star of I3 of, yeah, star, star, cube, k bar. And here the switching is easy, but just by intersection theory. And so first here we need to prove a generalization of the mineral formula. So here, so this is the generalization of the mineral formula. But this part is easy, so I will skip this part. So we need to compare this and that. So the formula is reduced to find such a diagram satisfying that this is equal to that. So we need to prove, we need to find such a diagram satisfying this is equal to that. Then the idea is the same as yesterday. So I will argue same method as the proof of the mineral formula for PL star, so this was yesterday. So the point is that, so first we have the universal family and we prove that the flatness of the smart conductor. And I need one more point. So we construct this diagram using universal family and we show that both sides are some values of flat function by using the flatness of the smart conductor. Then we still need to do something more that is that. So once you prove flatness, we are reduced to show that we have equality on dense open. And so equality on dense open. So here we use the ramification theory plus the fact that we know theorem 3.4 in dimension 2. So we can use the theorem in dimension 2 because so here we are working on surface. So this is the tool to reduce problem in this situation. OK, but this part is another difficult. And this part is fairly similar to what we have seen yesterday. So I close by saying something on this point. What means that you need to know that you have a single or supporting dimension 2 and conditional? You need this result? That one I don't need. But I need to know that the dimension 1, the quality of this construction and the construction using ramification theory are the same. This part I need. So I need to construct the universal family. So yesterday I used the P check and D and G. So this one, so this P check is the dual-positive space. So we could write this as a grass man. And this D was the flag variety. And this G was. So this part was enough to play with SS1. But now I want to do something with SS2. So I will complete this in hexagon. So I put C here and F here, D here, D here. So B will be a base and C will be a curve and so on. And F will be a flag. I don't know what is E. So G3E, it's going the other way. Flag 2, 3E, flag 1, 1, 3E. And then in the middle I have flag 1, 2, 3E. So we have some grass man band. And they parametrize. So here, same, it's parametrized dimension 2 sub-space in dimension 3 space. And so they are dual to each other. And here we have P1 bundle. So the dimension here is 2, so it's just P1 bundle. And we have another P1 bundle here. So here we have line in plane. And this we have Cartesian diagram here. So by this commutative square, we have map here, this map. So over this base space, so this is P1 bundle. And this is P2 bundle. So this is P1 bundle, and this is P2 bundle. So this is P1 bundle. And this is P2 bundle. So this P1 bundle is a family of lines in this P2 bundle. So if you take for each point of B, we get line inside plane. And so here we have, so I need a little more space here. And we have from here to map G, and so, yeah, yeah, yeah, X. So this is what we have seen yesterday. So this is the universal family with z-section with axis. So this is codimension 2 in X. And here we have P2 bundle. So by taking Cartesian product, we get some blow-up of X cross G. So this is some universal version of universal family of some blow-up of X. And again, we take this. And so we regard this as universal family parameterized by B of hyperplane sections. So this is the construction of universal family for this picture. And yeah, I'm sorry, I'm running out of time. So using this construction, we show flatness and the applied conditions, sorry, to prove the quality here. So thank you very much. Are there any questions? There'll be one. So two of you, you talk of the surface and the locally constant lead sheet from the company's advisor. So no assumption on the ramification. And then as you put in this case, everything is OK. So you have singular support. In particular, you have the cacti cycle and the whole formula. So now you stated here that formation of the cacti cycle is just pulled back by smoothness. So you have your surface X. And suppose Y over X is maybe one-dimensional. Not really smooth, but maybe some isolated singularity, for example, in this curve. So what would be some transversality conditions that somehow the formation of the characteristic cycle would be compatible with the pullback? So can you weaken the smoothness by some kind of a nice transversality condition with respect to singular support or characteristic cycle? It's such a question. So yeah. Can you imagine a non-characteristic notion from morphisms? Yes. Yeah. But even yoyotubu dimension zero case, so you have open, inside you have etal covering. And but you may have ramification on the boundary. And it may kill ramification, but it may preserve ramification. So the situation must be rather complicated. So I don't know if I can form it. Yeah. So of course there should be some non-characteristic condition, but for the moment, I don't see how to put to such condition. Composition of vanishing cycles and powers. So you take pullback and then you take vanishing cycles. So this construction is very hard to understand. But with your formalism, it should come out. Is it something different from characteristic zero case that is for D-modules or for what you said holds or not? I have a souvenir of something like this in cashier. Because there is only a termification characteristic zero. Of course the D-modules is sometimes wild.  OK, so it must be that it is different from the characteristic zero. Oh yeah. In some sense you have the divisor case.  It's a smooth and it's a regular immersion. Since you already have the, you can see things as complex. So it's like you are able to define a non-characteristic. You said when a divisor is non-characteristic. Yeah. OK? You said when a smooth morphism is non-characteristic. But actually you have Y to X and you have Fencing on Y, you have Fencing on X and so the arithmetic cycles are both sides and then the relative conditions for that. Of course they can decompose like this, but usually it's not, this may not be non-characteristic. So you are writing for pair of morphisms, for pair of sheaths, morphisms. For pair of sheaths and morphisms you should have an ocean. So why do I say, instead of just taking a sheath on Y, a sheath on X and a map, more complex, and then you really have a relative notion of that. And the character is exactly that. You mean a map between the two? Yes, yes, of course. Between the complex and the complex. Yes, yes, yes, yes. You have a map above, yes. And suppose, it was skeptical. I'm too mad at me. Yeah, in any case there are many questions in this area. So are there more questions? So may I ask you just to clarify? So as far as I understand, for the case of a surface, you said that the line constructed the theory. But what did the line do exactly? Yeah, so he makes some assumptions. And the assumption he says that one should be able to define a character, a cycle, a satisfying meaning of formula. So this is what's stated in his. Let's say just for a surface, is it the assumptions that he's trying? Are you satisfied, or? Yes. Yeah, so yes, yeah, yeah, yeah. For a surface, in your approach, so you said that your approach is so what is your approach for a surface? You have to prove these conditions, SS1 and SS2. And for a surface, this, as you said, is not difficult because you add the condom or weight. It's usually an aggregation theory. OK, so for surface, I can prove SS1 and SS2 using. For SS2, it's trivial because I put to some sense conditions. So the non-trivial part is SS1. So it's mapped to curve, and we need to prove local cycleity. And this is exactly the concept of semi-continuity of doing normal. And to apply this during normal, we need to study the restriction to curve. So this is a new input from our aggregation theory. So may I ask you also to, so in the simple case of Artein, Schreier, Ranquard, and Schieff, so I think that Carter studied it, OK, but let us, so what is the, so suppose you have a normal crossing device or define a tau local, your formally local, you are a subpart of the coordinates, and you have a function regular on the complement, so you write it as a Lorentz series, and that load the terms that are piece powers in response, so you can up to Artein-Schreier thing, OK. So what is the, so you bring it to some normal commons, and what is the, how do you see the condition of strictly non-degenerate and non-degenerate, OK. So in this case, so, so in Kato code it referenced one, but this is in logarithmic context, so we have some variant of this, so this is, so if your equation is t, p, t is f, so this is basically d of f, so f may have some pole, but you can regard this as some differential form, and this differential form define, gives you a cycle in the Kotanzen bundle, and if, I think actually it should write in the f over x, x to the n or something like that, and then you have, and this is in co-dimension one. Yeah, co-dimension one, yes, yeah. And this, and this, yeah, the best, yeah. And f is function of y, yeah, then you, you want to, OK, so in terms of this, yeah. So they will do the generic point of the divisor, but the condition of non-degenerate or strong, or strictly non-degenerate, so you can read it off from the function. Yeah, yeah, so you regard this as some differential form on the bound, on this divisor, and then you ask if there is zero or not, zero or pole, and if it's not on the basis, then you are in this non-degenerate situation. On each component? Yeah. And in this case, non-degenerate is the same as strictly non-degenerate or? This strictly non-degenerate is a little tricky condition, so you may have some component that may tame, tame the non-degenerate, and some component that may tame the non-degenerate, and I don't want that they meet. I need to remove the intersection of the tame the non-degenerate component and the non-degenerate component. So rank one, shape? Yeah, for any, any, any, any, any, any length. But if it is just out in stride, there is no tame the ramified. Yeah, but, yeah. And for p equal to, maybe you get some non-degenerate. Ah, yeah, yeah, p equal to two is exceptional, yeah. Ah, okay, so...


Hello and welcome to the series of today lectures on the subject digital techniques for secondary IT students. I am Dr. Rishal Gajbar and in this today lecture we are going to implement forest 1 multiplexer circuit by using veriloc sdl language. At the end of this session you will be able to implement forest 1 multiplexer circuit by using veriloc sdl. You will also be able to write the test bench in veriloc to verify the module correctness. So, the software that we are going to use for the simulation purpose is model sim student edition. It is a free of cost software and you can download this by using the link which is provided here. This is the workflow for our session. We will first discuss the veriloc module for a circuit which is forest 1 multiplexer in our case. Then we will discuss the test bench for this circuit for verifying its correctness. Then we will verify the simulation results in model sim finally. So, the diagram here shows the forest 1 multiplexer. The forest 1 multiplexer is a combinational circuit which has 4 inputs and only one output and to select any of these 4 inputs we require 2 select inputs which are represented here using s 1 and s 0. S 1 represent the MSB whereas, s 0 represent the LSB of the select inputs and depending on the select combination which you provide here the particular input will be. So, the output will be connected to particular input and that input will be available at the output. So, this is the function table which represent the operation. So, here you can see for the input combination 0 0 at the select input the y is equal to i 0. So, what is the meaning that whenever you provide s 1 equal to 0 and s 0 equal to 0 this y will be connected to the i 0 and whatever is available at the i 0 that will be available for the y which is our output and similarly you can also see the other input combinations also. So, for 0 1 you get y 1. So, for 1 0 y is equal to i 2 and so on. The same function table you can represent by using this logical expression. Now, let us discuss the variable of module for 4 s to 1 multiplexer. So, the module definition will start with the module keyword, MUX 4 1 is the name of the module and inside the bracket we provide the list of input and output. So, in this case I am providing input underscore lines comma select underscore lines comma y. So, input lines comma select lines comma y. So, these 2 are our inputs input lines and select lines whereas, we have only one output which is nothing but our y. The select line in this case. So, since we have 2 select line we are going to represent it as a vector. So, this vector select line is represented as a vector here. So, this vector representation is shown here. So, here you can see 1 colon 0 inside the square bracket and input is the keyword which represent that the select lines are the inputs and this one represent that the first bit is the MSB of the select line that is what the meaning of 1 colon 0 in this case. So, similarly the input lines there are 4 input lines. So, that is also represented using the vector in this case. So, input space in the square bracket 3 colon 0. The meaning here is that the third bit will be the MSB and the 0th bit will be the LSB of this select lines sorry for this input lines. So, here we have defined inputs and output is defined as output space y then output y is defined as a reg data type. So, reg space y the select lines are represented as wire data type in these 2 lines. Why? Because here we here in these 3 lines we are just representing what is input and what is output. So, input and output we have defined explicitly here and the data types are represented in these 3 lines. So, output is defined as a reg data type whereas the select line and input lines are represented as the wire data type. Then we will write down the always block. So, always will be written using the always keyword. Always block is like a while one it is going to execute continuously. So, always at the rate inside the bracket we are going to provide the OR operation between input underscore lines or select underscore lines. The meaning here is that for any valued input these input lines or select lines take this block is going to get executed. Then inside the bracket we are going to write down the case right. So, case statement. So, the case statement will be written as follows. So, case is the keyword that will be used. So, all the bold letters in this case are nothing but the keywords bold and lower case all these are the keywords of the V-relock. So, case inside bracket you are providing the select lines which is the select input in our case and select lines. How many select lines are there? Two select lines are there right. So, we have to define the 2 bit binary number in this case. So, the first case will be represented as the 2 dash B 0 0 here 2 represent that it is a 2 bit binary number dash B represent it is a binary and 0 0 is the value. So, the first combination is 0 0 and we know that for the select line having the value of 0 0 that is S 1 equal to 0 and S 0 equal to 0 we have to assign output Y as the I 0. So, that is why it is represented as it is. So, it is a vector notation. So, by using the indexing we are accessing the I 0 in this case. So, we have written Y equal to input underscore lines in inside the square bracket the index is 0 for I 0 right. So, and so on. So, for in for the input combination 1 1 what should be the output Y? The output Y should be 3 right I 3. So, that is why it is written as Y is equal to input underscore lines inside the square bracket 3. So, this case will end with the end case keyword whereas, the module will be completed with the end module keyword. So, this is the verilog module definition for 4 S to 1 multiplexer. Now, let us discuss the test bench for 4 S to 1 multiplexer for verifying its correctness. So, it will also be a module starting with the module keyword this is the name of the test bench that we have given test marks 4 1. Now, here the roles of the inputs and outputs will be reversed than its verilog definition. So, here the inputs will be defined as a reg data type. So, reg inside bracket 1 colon 0 select lines the meaning is that select lines is nothing but a vector having the two bits and it is defined as a reg data type. Similarly, output Y similarly, input lines are defined as a vector of 4 bits using the reg data type and output is defined as the wire data type using wire keyword. So, here we are instantiating our 4 S to 1 multiplexer. So, MUX 4 1 space MO MO is the name of the object and inside the bracket we are providing the list of input and outputs. Then we are going to write the initial begin statement. So, this initial will start the initial block and there are multiple inputs that we have to provide. So, since there are multiple statement we are going to enclose them inside the begin and end block. So, and here you can see the first input combination is. So, only few combinations are shown because of the space limitation you can take in your actual program you can take all possible input combination and verify the correctness of your module. So, the input lines in the first line are defined as 4 bit binary number. So, 4 dash B 0 0 0 0 and select lines are defined as since it is a 2 bit binary number 2 dash B 0 0 and I have given the delay of 100 units. So, I have given 4 combinations you can take all possible combinations. So, this begin will end using the end keyword whereas, the module definition of our test bench will end with the end module keyword. So, before going to the actual simulation pause the video for 1 minute and write down the answer of the given question. I hope you have written the answer. So, the third bit is the MSB since the input vector is declared as input space. So, input is the keyword in this case inside the square bracket 3 colon 0 input lines the meaning here is that the third bit will be the MSB bit whereas, the 0th bit will be the LSB bit. Now, let us see the simulation I have already written the program. So, this is the module definition that we have discussed in the slides this is the test bench that we have discussed the first job is to compile the program. So, go to compile select test max 4 1 dot v compile. So, just check whether your compile is successful or not and one can see that the compilation is successful. Then our next job is to go to work library go to test max 4 1 go to simulate. So, this window will open here then go to the we have to add the waveform. So, go to add 2 wave all items in region here you can see that the select lines as well as the input lines are vectors select lines have 2 bits whereas, the input lines have 4 bits and there is one output and the next job is to run and since I have only provided the 4 combinations in the input. So, only 4 combinations will be here. So, let us discuss one of the input combination. So, for the first input combination the select lines have the value of 0 0 whereas, the input lines have the value of all 0s and in this case since test select lines are 0 0. So, whatever is available at the i 0 that should go to y and since i 0 is equal to 0 in this case i 0 is equal to 0 in this case you can see y is also equal to 0. In the next combination the select lines have the value of 0 1. So, in this case whatever is available at the i 1 that should come at the y and in this case you can see y 1 input line that is i 1 is taking the value of 1. So, that is why y is also equal to 1 in this case and in the next in the last combination you can see that the select input is 1 1. So, whatever is available at the i 3 that should be available at y the i 3 in this case is represented as 1. So, that is why your output y should also be equal to 1 the same you can observe here. These are the references. Thank you very much.


So, I'm going to talk to everybody about piranhas root bore and pecans, something we've been seeing down here in Southwest Alabama, and I think the rest of the state should be aware of it if they're not already. I'd like to shout out to Brian Wilkins for really helping me out with this study and understand these insects better he's the boots on the ground he's a research associate. Down here in Southwest Alabama working at the fair hope research station and he is my pecan guru he's my, my go to person for all my pecan questions. And a real wealth of knowledge that he has he's been in the industry for a long time so kudos to Brian. And thank him for all the these data that he has shared with me and pictures and taking me out to see these orchards with traps in them and all that he's been really educational for my learning. But we'll go ahead and move on. If I can get my slide to move. There we go. So this is what I'm talking about the promise root bore beetle. These beetles can put an orchard in decline, causing limb by limb, a die back until the country falls over or dies due to lack of water or nutrient uptake. These beetles feed on the root systems of a number of tree species. And because of all those alternate host it can make them very difficult to remedy. These bills were first discovered in 1969 on pecan trees in Albany, Georgia, and those folks in Albany, Georgia have been gracing gracious enough to share them with the rest of us across the southeast and parts of the Midwest. Adults are very robust broad but somewhat flatten blackish to reddish brown beetles with antenna roughly half the length of their bodies on females and the entire length of the body on males. The larvae can be really large fleshy elongate grubs creamy white to yellowish in color with three sets of small legs and a swelling behind their head capsule with really strong black mandibles really creepy looking creatures crawling around in these soils. And as you can see there's several species of the insects can complete their life cycle on pecans hickory oaks. I mean how many oaks do you see out there every day there's a billion of them just down the street. So, their host range is wide they like poplars peaches chestnuts apples blueberries. Basically if it's a deciduous fruiting crop. They're likely going to be able to live off of its roots. There's a population spike in a con orchard when land is clear cut nearby so when they're taking out a forest nearby you might see them start coming on to your property. And for those of us in Southwest Alabama and South Alabama in general when a hurricane rolls through and pushes a bunch of trees over in the woods, or on the con orchard. That will cause these insects to move to a more suitable location like your orchard, they walk up and they see above a tasty pecan roots. So, these insects have quite a long life cycle lasting several years eggs are laid in the late spring after the adults emerge. So typically these eggs are laid near the host plant so once the larvae hatches it begins at search for tree roots within that wide host range and the adult does a really good job of making sure that that offspring is close to food. Once they encounter the tree root, they begin to feed and may feed for up to three to five years before crawling within a foot of the soil surface to begin pupation. And this happens around May is when they're crawling, you know, up closer to the surface. Adult emergence seems to be staggered, and there can be one or two peaks of adults adult populations in a given growing season time of emergence is quite tricky to nail down because it fluctuates year by year and location by location. The sex ratio of these insects is typically six females per male and once in the adult form they get down to business pretty quickly because they only live for about a week. Adults likely do not feed, but they for sure are going to breed and lay up 300 to 500 eggs in the soil. A lot of times I'll lay these singly or in clutches. Good luck finding the eggs in the soil. So the damage that's going to occur is they're going to the larvae is the damaging portion of these insects. They're going to be eating the barks off the roots, girdling them and then severing them. And I've got some good pictures to illustrate what happens when they are in the ground and munching on your pecan roots. But a sustained infestation may result in limb by limb death and you're going to see a picture of that here soon. So my first picture this is after a hurricane pecan tree blew over. There you can see what looks like roots sticking up out of the base of that plant. Most of them are severed off and this is due to prioness root bores. This tree fell because it had such a bad infestation of root bores that it had no structure to stand on. And this caused the wind to blow that tree right over. Maybe it would have survived if the root bores had not been in the area. It may have had a full root system and been able to withstand the winds. However, that was not the case. You can see these roots are kind of clubby and they just eat right on through them. Here's another picture. I'll let you guess which side of the tree the root bores were on. If you guessed the top side of the tree, you're correct. There's hardly any roots over there and that's because they were munching away having a field day on this pecan tree and we didn't realize it until the tree fell. The left side has got a nice set of roots. Not the best looking. It probably had some damage too, but there's still roots there as opposed to the top side of that tree. Here's what I mean by limb by limb death. As you can see there's some green canopy off to the left there, but that main central leader has died back. We've got suckers and low limbs coming out all from the base of the tree. That's a good indication that the root system has been tampered with in some sort of way. This is due to prying root bores in this case. You'll see limb by limb death just like this picture here. We can see there's a thin canopy. The tree is around it, especially this tree where I got part of it in the very top left hand corner. That's a really healthy tree. You look at all the leaf material and then you look at the center of the picture where this tree has limb die back. There's very few leaves. The canopy is real sparse. You can see a lot of wood in there. That's an issue. This tree will end up dying if the root bores are not taken care of and we don't remove all of that dead plant material or dying plant material. Sometimes these jokers can make mistakes. They think they're chewing on a root when they're chewing on your drip irrigation for your pecan trees. They're going to cause leaks and all kinds of problems. They can clog these irrigation systems if the worms get big enough. All kinds of issues, not just with the plant roots. You'll see a gradual decline of the tree over time. The foliage may show symptoms of nutrient or drought stress. Usually these trees have these insects feeding on their roots. If they have the insects feeding on the roots, they're going to be thinner at the canopy like I stated earlier. With that thinner canopy, you're going to see a reduction in crop load and crop quality. This is not only going to impact the canopy of that tree, but it's going to impact the quality of those nuts you're trying to sell at market or wherever you're trying to sell them at. There is some hope because we can't nail down a specific date for emergence or time of year. I mean, we know it's during the growing season, but it can be different from year to year. Our best course of action is to monitor for these little monsters. So we can monitor for them starting in May and we're going to do that through August. When we start to catch adults in our trap, it's time to consider insecticide applications depending on your tolerance level.  You may see adults move around just before dark during June, July and August. During the day, they're going to be hiding much like the rest of us in South Alabama. We're hiding from the heat and humidity. They're hiding from predators. You can find them hiding in the leaf litter under logs, other debris, any kind of pile of anything where they can fit their little bodies into to get out of the sunlight, get out of sight. So birds and other things can't come up and get them. Light traps are very effective traps because these insects are mostly not eternal. I like the pheromone baited traps because they're bringing in the insect we want and not just everything that's attracted to a light. Males are attracted to pronic acid baited pheromone traps and females are attracted to alpha pine baited pitfall traps. The males are going to be flying around. The females are mostly going to be crawling. You can find these black panel traps like the ones in the picture there there on the right for about 30 bucks online and you're going to be able to find those lures as well. And that's really our best means of defense against these things. We monitor for them. When we, when we see a population, we start to spray and try to get rid of them. It's going to take several years to knock out these populations of brown or fruit borers because they take so long because they live so long in the soil they're ever present. So the best means of prevention is to grow a healthy tree, keep it stress free. If we experience dead trees in the orchards, we want to remove them as much as that plant material as we can. As much of the root ball as we can, we want to get it all out of there so they don't have a nice comfy home to feed in. If you see dead limbs, you'll want to remove those to help restore the balance between roots and shoots. We got to keep that balance and check the less roots are the less shoots we need. You also want to get dead plant material out of the orchard because it's a really good point for diseases and other insects to capitalize on your tree's weakness. They can sense when these plants are stressed. A lot of times woody plants when they're stressed they exude ethylene and all kinds of insects can tap into that and they'll find that stress tree before we ever would.  Try to keep a weed-free strip around the base of those trees. That's going to decrease the hiding locations for these insects and increase insecticide effectiveness. When the adults arise in the late spring, we need to be monitoring for them with these pheromone traps. As soon as we start to see them, we need to start making applications of chloropyrifos at four pints per acre. That's the insecticide we're going to need to use. As always, you're going to want to read the label on that insecticide bottle because it's the law. I can tell you things all day long, but if you're going against that label, you're breaking the law. So be sure to read that label in its entirety. This is currently our only means of control. There's ongoing research. I know Brian's working on some stuff. Currently he's got a three-year study going on four years with these insects just trying to nail down what we can do to keep them at bay and grow good quality healthy pecan trees down in Alabama. Hopefully he will have a eureka moment here in the next year or so, and I'll be able to give you an update and tell you all the great and wonderful things I've learned. When we remove these insects from the orchards, some of these trees will recover. Roots will regenerate, a callus will form, you know, just like when we're grafting a callus will form. And that's an indication that there's some healing going on. And so large tap roots and lateral roots that have been damaged will form that callus and regenerate new roots and shoots. So not all is lost.  Over time you will see a gradual rebound of your orchard. You know, it's not all well and good.  It's going to be added work, but you won't necessarily lose your entire orchard if you act. So with that, I'm done with my piece. If you have any questions, don't hesitate to call or email me at any time. I'm usually always available and I'll try to answer your questions as best I can. Thanks for having me. So Jacob, I have a quick question. If somebody needs the identification, do they need to send samples or how do you send samples? How do you collect samples? For these insects, you can always, if you have these insects or you think you have these insects, you can always collect some in a plastic bag and freeze them and keep them cold and bring them to your local extension office. And from there, a local agent can try to figure out what's going on or you can always send them off for identification to Auburn. If you've got alcohol lying around, that's always the best way to store these insects for identification. I've got a prioness beetle right here on my desk that's been in alcohol for over a year that I was using just the other day when I was making this presentation. So if you think you have them, come find one of us, one of your commercial horticulture agents, and we'll be sure to ID that thing for you so you can start your management practices.