 So far I have introduced the concept of Turing machines, Turing acceptable languages, Turing computable functions and for combination of Turing machines, for combining smaller Turing machines to construct a complex or larger Turing machines and we introduced the concept of machine schema and a machine schema represents a Turing machine, which is a composite Turing machine based on the some of the smaller Turing machines. So, this we have introduced and in the previous lecture I asked you to understand that when we are taking two components of same Turing machine and if you are combining in a particular way and if the states are common, some states are common what kind of problem that you encounter. So, this I asked you to look at and see what is what will happen in such a situation and why we are maintaining the state sets or disjoint. So, let me give some more some information on that looking at that exercise and then we will further discuss on some of the issues related to this complex Turing machines. In this example, when I said if you combine L with L and if you consider because you know this L is the Turing machine whose transition map as we have whatever the input letters you have whatever is the input letter say A 1, A 2, A n and in all those input letters you can simply take a left move and halt this is what we define. So, that is delta of q naught sigma is halt by moving to left for all sigma in sigma. Now, whenever we are combining like this if you consider here also q naught and here also q naught what is the situation and how this composite machine will be. So, in case I take this and now if I combine you see the new this composite Turing machine the states as per our this thing this is q naught will be there and one p naught you will be choosing. So, these are the states and sigma as original and delta we do it like this as per the definition because the new delta let me call this is delta L and the delta what we define here you have q naught. Let me assume for the time being a b blank or the input letters and see what will happen as per the original this thing when it is halting I have to put it in p naught and then I take a left move and here also I will put it in p naught and take a left move and I put it in p naught and take a left move. So, this is what when this is what is the first performance whenever it is halting the first component we have to do that and in p naught whatever that machine is doing that is what we have to do. So, that is what we have to do. So, in L now that is going to q naught. So, what are the performance of that q naught that we have to do. So, in p naught in p naught what we have to do whatever that this is doing, but as per the definition again here again here that is because it is going to H L there is only one you have considered. So, again this will be p naught L because it is going to halting configuration this is p naught L this is p naught L. So, what will happen with this what will happen with this kind of transitions it will take a left move in the beginning and it is connected to p naught and in p naught if you get any of these symbols again you are going to left and you are continuing in p naught that is what we are doing and there is no situation that you are getting halting. So, essentially this is when I am taking the same state set the situation is essentially this not this not as desired because here essentially you require only two left moves it is behaving like the Turing machine which always takes left moves on the input and of course eventually this will hang this is not the desired construction. So, for this for this this is not the way that if you consider same state set this is not the way that we have to do. So, in which case what you do for L L you are taking two copies of left machine let me denote this is number one and the second copy in which case say for example, if you take this is say q 1 and for this you may take q 2 and now you see that as when we are defining this q 1 is there in the definition what are the symbols let me assume a b blank for the time being. So, as it is halting you will connect you will put it in say in the first machine that is let me call p 1 is a state and take a left move and here in b again that I am putting in p 1 and left move p 1 and left move and for p 1 how do we define that is p 2 corresponding to this I am taking auxiliary state say the state set now is corresponding to this what are the states of this q 1 q 2 corresponding to the first one let me take p 1 for this p 2. So, this is p 2 L and this is p 2 L this is p 2 L because the second machine the definition is again it is connected to initial state and in the initial state of the second machine this L 2 the q 2 in the that is the initial state in q 2 when you put any symbol it is going to the left the definition is the same this is what the left machine. So, I will again have to put because it is a it is going to the halting state I have to put it in auxiliary state of the second machine that is q 2 here that is p 2 here. So, I will be putting in p 2 now you cross check for this machine there is no definition of eta and when you come to p 2 it will simply because as there is no definition whatever is a current symbol that will print the same symbol and go to the halting state. So, that is the construction halting state blank. So, in general here I have just assumed a b blank or the input symbols, but if you take any alphabet. So, we essentially copy the same and you look at now the difference between taking the same state set because when I am calling L because for the short notation I have just in the beginning I said L connected to L there are 2 copies of L, but you have to maintain the different states. And when you are taking a particular machine if you are using it for several times you take copies several copies of that machine that means essentially the state set you take different labels of the states. So, that you can maintain the disjoint state set in the components when you are combining then it pursues the property as desired. Otherwise there is a problem that you know you may construct a wrong Turing machine with the machine schema. So, here the machine schema I hope it is clear to you that here you have 2 machines in this machine schema when you are looking at because this Turing machine as actually represented by the machine schema here the Turing machines are L 1, L 2 there are 2 copies of left machine. And the sigma the common alphabet for all that is what is here the eta the partial map will now indicate and this machine is the initial machine for this machine schema where eta of L 1 for all symbols for all sigma is connected to this second left machine this is for all sigma in sigma. And after L 2 there is no definition. So, this is a partial function for only the machine L 1 the L 1 I mean here the left the machine the copy first copy here. So, this is what is the definition of eta and with this machine schema we have constructed this machine and this machine precisely takes only 2 left moves on the input and the corresponding transition map is this because this machine schema represents the Turing machine whose state set is q and of course, alphabet is sigma and delta as indicated in the table and the initial state here is the initial state of the first machine that is q 1. So, this is the Turing machine this is the Turing machine that is represented by this machine schema. Now, let me give little complex example this is a very simple example that I have indicated, but this I hope clearly distinguishes that when you are combined when you are taking a particular Turing machine several copies if you do not distinguish the states states disjoint states that if you do not maintain what kind of problem will come that I hope it is through this example I hope you understand that. Now, let me recall the example that left shift machine that I have constructed in the previous lecture that is because you have to move all the things all the input one cell to it is left. So, remember that we have constructed like this blank it takes a right move and then sigma not equal to blank whatever is there you just take a left move and then print that sigma and then take a right move and then take a right move and now you take one more right move whenever you are getting blank you just take a left move and print blank there. So, this is how we have constructed and you see here there is a machine r here once and here is r and here is a left machine here is a left machine that you can clearly see that you mentioned l is here l is also here. Now, here is one typical situation that when you are reading sigma here under the sigma what are the sigma that you are reading at this place that this thing that you have to print after this much situation like you have to take you have to compose with this Turing machine and then you have to remember what sigma is taken over and you have to print that here. So, first one has to understand that under this sigma this component need to be constructed. So, this component let us construct it separately and call that machine one different machine schema and this composite machine has to be called depending on that sigma. So, thus here and of course here there are r here is one r there is another r and of course, l has you have only one and there is l and this I am anyway combining. So, thus you see what are the components here let me call this is one right machine and this is a second copy of that right machine. Thus in the machine schema the machines the basic machines if you call that is s this m the common alphabet whatever that you have sigma and yeta will define and the first machine here as you can see that is a lash. And where this m the set of machines having a lash as one component here and you have one machine r one let me write and similarly you require another copy this r 2 you call and now here you can see something like because this l and anyway I am going to combine this. So, one left machine you would require and printing blank this machine you would require this as indicated here union for all non blank symbol you require this machine that is union a machine l composed with machine l composed with p sigma this sigma and sigma minus blank of course. In fact, I can write these two together these two together and I can represent this at once anyway does not matter. So, these are the machines that we are seeing whatever the machines here I am writing first in this particular place these things you have to first declare that means l p sigma this is the machine schema for that machine that let me write this as m sigma. And you can also combine this for sigma in sigma and you can call that also here there is no problem. So, for sigma in sigma in which case I may just represent it like this l p blank. So, I do not require this here in that case here for all sigma and sigma. So, let us represent it this way and now see that how to write this machine schema is clear to you because there is one machine l and first fixed sigma you have to take p sigma and combine that. So, how many such machines that you have to declare for all for each symbol of sigma you have to have one such machine. So, you take this an exercise to write the machine schema for this write the machine schema for this. Now, those machines are now available in this particular list this is a finite set of two dimensions. Now, for this for this machine schema the eta now we will define like this the l hash for all symbols it should be connected to this one R 1. So, the first machine the machine schema has to be like this sigma is connected to the first for all sigma in sigma and now you look at here for R 1 for non blank symbols I am connecting to this for blank symbol I am connecting to this whatever it is for all sigma this will be connected to l p sigma. So, I can declare it like this. So, eta this R 1 whatever sigma that you are getting whatever sigma that you are getting you will connect to that l p sigma this is we named it as m sigma. So, this is what is m sigma that you have constructed separately. So, you connected to that machine because this for given sigma for sigma in sigma this is how of course, this is for all sigma in sigma one another is same for all sigma in sigma we have to connect to m sigma this is the situation. Now, m sigma for non blank thing that will be connected to this, but not for blank. So, here for sigma in sigma minus blank this eta will be defined for m sigma for that m sigma only m sigma for all symbols. Now, let me use some other say zeta is. So, for non blank symbols this is connected to the second copy of R second copy of R this is for all this zeta in sigma because this is connected unconditionally this is connected to this unconditionally. So, but this is connected only when the sigma is non blank that is how is indicated here for non blank symbols we are connecting it this way. Now, this case let me indicate this here for all sigma in sigma as earlier. Now, this R the second copy of R will be connected to first copy of R unconditionally I mean for all symbols essentially. So, that is another definition we require the second copy of R is connected to the first copy R sorry for all symbols for all sigma in sigma and now there is no connection from this. So, this is a so this is a this is undefined here now the set of missions is declared and the initial mission is declared here and a sigma is the common alphabet as indicated earlier. Now, this eta is as defined here this is the partial map eta is given here. Now, this mission schema precisely represents here here you have to things to note one is taking R 2 copies this R 1 R 2 and because we are carrying some information on this particular particular edge this is a particular place what we have to do we have to first construct this composite mission and here you may call that as some m sigma what are the sigma non blank thing that you are carrying that m sigma will be connected to this. So, there are all the Turing missions here under consideration. So, this is essentially these are all m sigma. So, now how many Turing missions are there in this m there is a 1 2 3 and here mod sigma. So, essentially in the size of m is size of m is mod sigma plus 3. So, this many Turing missions are here and appropriately depending on the sigma that you are getting it here you see that you may see that here there 1 2 3 4 only 5 missions are there as a basic mission components here, but they are not actually 5 this depending on the sigma that what we whatever that that is connected to here this is represented in this variable. So, the Turing missions in this mission schema is number of symbols of sigma plus 3 these 3 missions are there. So, this is the mission schema of the left shift machine and in short notation we are representing it that way the way that I have constructed earlier. So, likewise what are the Turing missions we give we have to be very careful to look at the correspondence with the proper definition the basic definition the in terms of the states. Because the diagram intuitively everything will be clear and construction is very easy compact and very nice representation that we have, but the problem is when if you want to actually refer to the definition and what are the respective states if you want to write you have to be very careful with this mission schema. Because the components essentially you have to maintain the disjoint set of states and you have to be more and more careful how to represent this kind of situation this kind of situation when the composite mission which is called here. So, you can now treat this as one of the building block and accordingly you have to declare the mission schema. Otherwise you know if you if you separate it here in this present mission scheme it will be very difficult for you to represent it here it is in fact not possible. Because the sigma the information of sigma on this particular at this particular place has to be carried and therefore this has to be combined prior to declaring the mission schema of this SL. Now, let me give one more example there the sigma information may be quite. So, this is so called copying machine. So, this copying machine how it performs you give any string x as input this is the format that we are following we want this kind of situation output the same x by separated by one blank will be declared. So, that the copy is shown very clearly. So, this copying mission if I write the mission c how do we write because here the logic will be. So, the input may be given like this say a 1 a n this is where we are starting. So, finally, we should be left with this kind of take. So, the logic one may look at like this because here at this first place I have to have this a 1. So, that means I have to go all the way till this place carry this information and copy it here. And then you go to the second position from here all the way till a 2 and copy that and to a 2 position like this. So, this a 1 a 2 a n when they are there you from this position you go to this place you can understand that till go till blank and take a right move and what is the first symbol go till end of the tape and printed there and so on you keep doing this till you reach till this blank. So, this is a logic one can easily follow to construct such a Turing mission. So, in the beginning a lash you go till end of the tape left end take a right move. Now, what are the symbol that you are reading if it is not blank then let me print blank there and then I will take r hash 2 times the indicating r hash square I mean this the mission r hash I am applying 2 times. So, the abbreviation m a Turing mission m this abbreviation I am following m square is m m the Turing mission m m. So, you can take 2 right hash moves there you print whatever that you are carrying p sigma and then you go a lash square and at that place you print that same symbol back and then take a right move. Now, you see how complex is this when you are receiving blank that means after finishing the input given to you then you just take r hash there you can halt. Look here this is the copying mission as desire because x the input whenever it is given a 1 a 2 a n what it does in the beginning it comes of course, after finitely many steps you see that with a lash a 1 a 2 a n this is how the situation is and then it takes a right move it takes a right move that means you will be here and as this is not blank it prints blank here it prints blank here and then it takes 2 r hash missions r hash square. So, that means it comes to this end and then 1 more r hash. So, next blank it will look for that means at the first r hash it will come here and with another r hash let me probably write here the what are the missions that we are using instead of saying the there are finitely many number of steps. So, here l hash is applied and here r is applied and here r hash is applied and here also r hash is applied. So, that situation is as follows a 1 a 2 a n. So, next blank on the right. So, that is immediately there. So, at this place it will print that sigma what are the sigma is carried a 1 is carried and therefore, here what will happen. So, when you are applying that p sigma there. So, you have things like this a 1 a 2 a n blank at this place you are there and you print a 1. Now, you look at the mission it takes 2 l hash. So, first l hash comes here. So, apply a l hash and you will be at this place this is the situation apply once more l hash that mission. So, you will come to this blank this a 1 is printed. Now, at this place you print that sigma back. So, whatever that still it is carrying that a 1 that is the information is maintained. So, that is now a 1 a a 2 and so on a n a 1. Now, look at I connect I go to write and connect to the same loop from this p sigma. So, that means now I take 1 right move now it is a it is with now a 2. Now, a 2 will at this place you print blank look at print blank r r hash square p sigma and l hash square again p sigma. So, that means, what will happen after finitely many steps here you are of course, here printing blank and you are going till first blank and the next blank is here at the end that is here. So, here it will print a 2 it will come back here to this blank position and it will type this a 2 back and it will move to the position. And now a 3 will be copied to the end and so on till you receive this till you come to this particular blank. Then once you come to this blank it will simply take r hash and it will go to the right end. So, thus the output finitely will be so after print typing a n it will check now the here is blank. So, it will go to the right end that is r hash this is how the input is pursued to get x x what are the x input given. So, x x will be given as output with this machine. Now, look at the construction the composite machine that here we have for copying this particular blank what are the block I am under lining here this particular block. You see what are the sigma that I am carrying here in this composite machine in this composite machine I have to have I have to use the same information here and I have to use that information again here. So, that means here one may understand that this printing blank taking r hash square again you have to print sigma then l hash square and then print sigma. So, the sigma whenever I am carrying what are the non blank information that I am carrying in this particular machine I have to use it in this I have to use it this sigma has to be used in this particular at this particular place and again coming back and printing. So, this entire component first you have to construct separately for each sigma non blank you have to declare that machine and this machine need to be constructed first. So, once you construct this machine and if you call that as m sigma that m sigma can be used as a component of this construction and accordingly you can you can declare the machine schema of the copying machine. So, thus what I can say the copying machine in which case will be precisely for for all sigma in sigma minus blank if you construct that machine first then the copying machine when you are looking for the machine schema you have to construct it this way as r sigma non blank m sigma and then. So, take a right move and sigma if it is not this then you connect to m sigma and m sigma will be connected back to r and whenever you are receiving blank you are taking r hash. So, this is the simplest look how elegant is this machine copying machine. So, with this you know notation the copying machine can be seen this is a very elegant manner and now only thing is only thing you have to understand that m sigma you have to declare first and then you see there are 1, 2, 3, 4 components and in this machine schema you can you can declare these are the missions and this is l hash is 1 mission r r hash union this m sigma for sigma not blank. So, for sigma not blank sigma not equal to blank let me write like that in a short way sigma not equal to blank for all those. So, these are the Turing machines and what are the common alphabet you have and eta here essentially these definitions we have to have and the initial mission is l hash this is the machine schema and here the eta is defined l hash for all symbols sigma will be connected to r for all sigma in sigma and eta r now you see there are 2 situations that for non blank you are connecting to m sigma for blank you are connecting to. So, r for sigma this will be connected to m sigma for all sigma in sigma minus blank and the definition for blank is r hash this is how you are connecting and now this m sigma will be connected to for all sigma wherever it is wherever it is there m sigma for all symbols let me you say eta is connected to is connected to this r for all this zeta in sigma. So, that is how the construction is now you look at how actually we have to represent the machine schema and see how to maintain those things and to represent that now once you declare these things and for each component you have the states that you that are known to you and when I am combining these 4 missions you take 4 auxiliary states and as per the definition of the Turing mission composite Turing mission which is represented by the machine schema as I have given the definition you can write the state transition table and you see this is actually matching with the original definition given to you. But using this notation you can construct a very elegant complex Turing mission in an elegant manner. Now, in order to look at more and more complex works that we can pursue with Turing missions let me first introduce one more notion that essentially computing with multiple inputs. So, in which case let me first give an example where you have one string that you are taking from say sigma 1 star and I am taking another string y from say sigma 2 star and the function what I want to pursue is I take 2 strings. So, then the function will be like this and now let me call sigma 1 union sigma 2 star here how it is defined if you take 2 strings I wanted to simply I wanted to concatenate them. So, given 2 strings concatenating this is a very basic operation that we can easily pursue using your usual computer. Now, you see in order to capture that notion first the question is essentially whether it is Turing computable function. Now, in which case the function will be it is not just taking 1 input it will take 2 inputs. So, I am representing this a function from 2 sets that is sigma 1 cross sigma 1 star cross sigma 2 star union 2 sigma 1 union sigma 2 star because what are the symbols of sigma 1 sigma 2 may occur here as you see. So, this is the function we see this is Turing computable function in which case my notation here is you will be giving input in this format this is the initial this thing after finitely many steps I should be left with this kind of situation that is the expectation. So, using this format this input format and this is the output format. So, when 2 strings are given this is how I will take when 3 strings are given then I will separate because this is the blank symbol is a special symbol that special symbol I will put in between and you see that this whatever we are taking the input we are not allowing blanks in the beginning as you have noted earlier. Now, this blank can be used as special symbol if you are already using somehow or other in the input the blank also in which situation you can introduce some other special symbol because essentially you should distinguish when I am saying I am giving 2 strings as input you should be able to distinguish like what is the first string what is the second string on the input tape. So, that way I am I do not require because I have reserved this blank symbol as a special symbol I will just use it. So, this blank in the beginning and the second blank because still between these 2 blanks this block whatever is that is the first input I can say and between the second blocks of blanks I will I will say this is the second input that is how if you want 3 strings as input then I will separate with one more blank and I will give say for example, x, y, z I can give. So, this is the this is how we give the input and we expect the output as here I have mentioned one string if again if you want to have 2 strings as output again you can maintain such a format and so on by separating blanks. So, let me fix the convention that the inputs if n number of inputs are given then they will be separated by the blanks and if m number of outputs are expected they will be separated by blanks. So, in general if a function is considered that say for example, sigma 1 star cross sigma 2 star cross and so on cross sigma n star consider a function take n inputs and for example, gamma 1 star cross gamma 2 star and so on cross gamma m star some alphabet sigma s and gamma s are alphabets m number of n number of inputs are taken say x 1, x 2, x n is the input taken which is ascending to say for example, y 1, y 2, y m this is the this is the thing then the input will be of this x 1 blank x 2 blank and so on blank x n. So, at this place you are reading a writing head initially this is the input and after finitely many steps the machine should halt with the output y 1, y 2 and so on y m this is the expectation. Now, as per this format we say a function f is Turing computable if there is a Turing machine compute this function. Now, I will say a function so this function f is computable if there is a Turing machine computing this function. Now, the function what I have mentioned in the just earlier with two inputs if you want to concatenate them. Now, you see the input x y this is how it is given the expectation is after finitely many step the machine should halt by concatenating these two in this way. Can you look at that how can I how can I pursue this essentially from the original position of the input you just go to the this first blank to its left and keep copying the symbols to its left. So, that y can be shifted one cell to its left and then at the end whatever is the cell that you can make it blank. So, when I say a 1 a 2 a n is x and say b 1 b 2 b m is y from here you go here move b 1 then move b 2 move b 3 and so on this b m once it is moved then you can make it blank here and you stop here. So, this process I need not now pursue it separately what are the machine s l the left shift machine what it does you give any string with one blank in the beginning. So, it simply shifts this x one cell to its left this is the operation of s l. Now, this function f is computed by s l precisely because s l if you just connect then automatically from here it will shift y and then the resultant string is x y because we are halting at the end of y and now you see thus f is a Turing computable function computed by s l. S l the left shift machine looks very simple, but you see such a thing that it will now if you want to come concatenate say for example, 4 strings are given then you can use this s l for that many times and see that it is now let me look at that example. The example x 1 x 2 x 3 x 4 say 4 components are there now g is a function this is appropriately from for that sigma i star you can write now concatenating this if you want to pursue what is the Turing machine. Now, the input is like this x 1 x 2 x 3 x 4 this is the input now I apply left shift machine once then what will happen that x 1 x 2 this x 3 x 4 will be combined at this place once I apply a left shift machine this x 4 will be shifted one cell to its left and this is the resultant string now this is one single string x 3 x 4 is a single string say some y. Now, once again you apply left shift machine then what will happen that is x 1 this x 2 x 3 x 4 this is the situation now because these 2 strings will be concatenated and now once again if I apply left shift machine the resultant string is x 1 x 2 x 3 x 4 you see how the left shift machine if you use it for 3 times. So, that means to pursue this g this can be computed by S L cube that is all this is the machine. So, S L you compose with S L and compose it with S L remember these 3 S L's you have to maintain 3 different copies if you compose S L with same copy again it will now pursue in a different manner that is it will keep trying to shift infinitely many times and you never know what will happen at the situation because infinitely many times it has to pursue then it will try keep shifting that it will it will hang. So, when I am writing S L cube here precisely 3 copies of S L's you take and concatenate them I mean compose them one to another all the 3 copies. So, this S L is helping you to concatenate this strings now if because you know one of the previous exercises I gave you one function let me write for example here that is say x the input string I ask you to give x x for this case what you do. So, that means the input will be given this format and the output is expected this way this is what we need now you can quickly understand that what are the so far machines we have constructed what are the machines so far we have constructed first I will start with this input like this and first apply copying machine just we have declared this copying machine will prepare a copy of x and halt at this position. If you concatenate if you compose this with S L now the resultant string is this will move one cell to its left. So, it will become x x so that means to pursue this to pursue this what you have to do you just conquer compose c with S L and of course, the starting machine here is c. So, this Turing machine computes this Turing machine computes h the function h as h of x is x x that you can pursue using this Turing machine. Now likewise as we have asked earlier I think this function is also asked to x x power r now you can try that what are the x given to you you have to first prepare the reversal and once you apply left shift machine on that those two strings will be concatenated. So, now you have to construct a machine first that it has to give x power r next to the given input without arising that and then you can concatenate them likewise other problems that we can look into. Now let me introduce the notion of computing with natural numbers. So, natural numbers let me write this is the set n I include 0 1 2 and so on. Now how do how do because we have discussed computing with strings now when I am talking about computing with numbers now I have to represent this numbers in a as a string because Turing machine we have understood that we are computing over strings. Now the representation here is any natural number you give me that natural number we represent in the muonary format with one symbol as i power n. So, that means this natural numbers we will now consider the alphabet singleton i and we take this correspondence with natural numbers. So, 0 with epsilon if it is 1 we are using 1 i and if it is 2 we will print 2 i's. So, this is the unary representation 3 3 i's and so on. So, that is how we represent and when I want to pursue certain basic computation with natural numbers that we can demonstrate and see that how to compose the basic Turing machines in order to compute the functions of natural numbers. Let me start with very basic function then afterwards you can say for example, addition this is a function from n plus n to n you know that given 2 natural numbers add 2 numbers. So, the input I will take say i power m blank i power n. So, what should be the result addition function this addition what I would require is i power m plus n as output. So, now you can quickly answer that what is the Turing machine that performs addition because only thing is input is different here natural numbers I say you can quickly see that if you just shift the second block then it will quickly give you the addition. So, that means, SL is. So, sometime back we have observed that to concatenate 2 strings I am using SL now you see this SL the same Turing machine if the input is natural numbers in this format it performs addition. So, this will be this computes the addition function of natural numbers SL computes addition on natural numbers. Now, one can try multiplication of course, this is again 2 natural numbers you give as input and this multiplication is m n is m multiplied by n that means the input if you are taking like this you are you have to give output this way. So, you can think of this and not only this multiplication may be you know the quotient you know the integral part of 2 numbers or may be subtracting. But of course, when I am saying that function from natural numbers to natural numbers we can perform monos that means you know when from a bigger number when you are subtracting you will subtract if you are if you want to subtract from a smaller number then what you have to do you will result 0. So, this kind of monos function or many other arithmetic operations are natural numbers many computable functions that you can represent through Turing machines there the input will be taken this way. So, if it is having one number as input and asking you to pursue something then you will you will you will take just one block of eyes if 2 natural numbers are inputs then as for addition and multiplication we are taking it will be like this and you are taking 3 numbers as input then again as per the format as 3 strings you have to take the input and so on. So, now you can try for quotient and other simple basic arithmetic operations that you can you normally do with your computer you can try constructing a Turing machine and some more examples in this direction we will discuss in the next lecture.