 So, in the previous lecture, I had discussed about what is basically is meant by the switching system and I explicitly stated that it is the information switching about which we are going to discuss in the course and then I actually gave the history that how the manual telephony actually started. Manual telephony basically means the manual switches whereby the human operators were used for doing the switching function at the exchanges and we required those exchanges because we wanted to reduce the cost of laying the cables because we never wanted to have full mesh connectivity where every user is going to be connected to every other user. So, and then of course, there was all kind of a problem which were there with the manual system and I had discussed about the protocol or the complete sequence of procedure which was used to set up a call or tear down a call and how the call details were recorded in a register which was a paper register in the earlier days and then which was then later on used for to generate the bills by the companies. And then of course, I mentioned that a gentleman an inventor named A. B. Strauser actually came up with the idea of having an automatic exchange and was purely electromechanical it was using mechanical components basically rotary arms which will change the connection point they can connect to different outgoing ports from an incoming port and these were all essentially being controlled through electromagnets and using electromagnetic electrical pulses. So, these magnets the arm electromagnets will cause the arm to actually move the connection arm and this therefore, is changing the connection between input and output port. And I actually discussed about pictorially how a unisector and how a two motion selector will usually used to look like, but these are all old stuff now because I think none of the exchange all across the world are working on this thing except probably in some museum perhaps and, but you can still find pictures on the net. And then I had come up with I discussed the concept of cross point a cross point was invented because of the minterance issues which was involved in case of the electromechanical switches or Strauser exchanges and which is basically gears being worn out with time the connections are basically wherever the two contacts two points metallic points are coming into contact those getting weird out and there is dust oxidation all kind of issues because these are all exposed to air. So, ultimately people came up with an idea of having two metallic contacts which actually get either in touch with each other or they are separated in a glass bulb which is sealed glass container and having some inert gas that is basically the cross point then I also mentioned there is going to be a two electromagnets which are required. So, that you both of them are energized you will have the cross point snapped in cross point will have a connection you can actually work even with one, but with one you cannot build up a switch. So, let me start with having only one electromagnet doing this job. So, you will actually have a glass bulb and then there will be one cross point there will be another cross point this actually is one cross point, but they are two metallic contacts. So, I can actually push an electromagnet like this and whenever you are going to energize this at this point at this control input this will armature will actually move down and in turn the connection will be made. So, the input port will be connected to the output port. Now, if I build up this kind of thing and I want to create a cross bar basically a switch this is going to be slightly tricky it is not that easy. So, let me take up a four point and I want to connect it to say four outgoing ports and how I am going to create a switch. So, I require a point here. So, this is essentially a cross point I am showing it to a bulb and this is getting connected here. So, whenever this snapped in the connection will be made here. Now, I can create another bulb here another bulb here. Now, how these will be actually control that is usually the issue. So, similarly I can make four of these kind of the second branch and then connect them to only thing I have to ensure is that for any output line either this bulb or this bulb and they will be actually similarly there will be more bulbs here. So, only one of them have to be operated. So, there will be four lines only signal should come from only one of the incoming lines not any not more than one because then information cannot be separated out it will get mixed up actually. So, this technically is nothing but is a cross point this I mentioned in my earlier lecture also. So, this is one cross point this is equivalent to this is the next one which is equivalent this one third one which is this one and fourth and similarly this one corresponds to this this one corresponds to this and so on. So, all these cross points and which is obvious way I will require 16 cross points of this kind for connecting four inputs to four outputs and of course, as I mentioned I need not I need not only I need to only maintain about this upper half of the contacts because one will usually will never like to connect to one because this one and this one are connected to the same user actually and this two is connected to second user. So, you only technically require n into n minus 1 by 2. So, in this case this will turn out to be 6 4 into 3 by 2 which is 6. So, I have 6 cross point 1 2 3 4 5 and 6. Now, the problem is I need to have how many control lines to actually set up the connection because I will have some control circuitry because that is required additionally which will control each one of those cross points then I have to also have some some logic which I for example, maintaining that only one signal should come from to any output port. So, this output port for example, this one can get a signal from this input this input and this input. So, only one of these three should be activated. So, one in each column only one of the cross point should be activated not more than one that is one rule it is possible to activate more than one in a row. So, it is possible that I can activate this scenario this usually is known as multicasting a multicasting technically means from user one I can send it to a group of user the same information. So, one can send it to 2 3 4 5 it is not sending to 6 7 8 for example. So, this multicasting if it is sending it to everybody this is broadcasting. So, that is a different between difference between multicast and broadcast. So, only for multicast whenever multicast need to be implemented by a switch more than one will be activated cross points will be activated within a row. So, this actually means I require 6 actually control signals. So, there will be 6 wires going out from here and as you say this actually complexity is O n square. So, as my switch size increases I require larger and larger number of control wires being going out from the control from the controller unit. Now, how to actually handle this scenario can I make it better. So, if you remember in the earlier lecture actually I have drawn modified the picture there was a modified version this was not the one I had also something called an electromagnet sitting in here there were 2 control lines. So, one is a horizontal I call it other one we call it vertical control line when this is snapped in this actually there is a half way somewhere. So, you will end up in getting this is will be the new position when this electromagnet will get activated this will further move and you will get this position. So, then the snap will happen. So, unless you activate both horizontal and vertical the connection will not be made between the 2 points. So, cross point will only operate if both the electromagnets are operated with one it will not be. Now, how to solve this whole problem actually is an issue. Now, let me try to set up a certain connection if I do it this way. So, I am drawing a 4 by 4 cross bar. So, that technically means again I can actually have 4 lines, but that actually means for all the 16 cross points I require 16 to 32 actually control wires, but this is actually being doubled, but may be this is possible let me try it out. I can build up a control one control for the complete horizontal line and this is controlling all the electromagnets here. Then for the second row again I may have one single control line controlling all the electromagnets. So, this is the way actually this can be done and I can use similarly the vertical lines for controlling the vertical component of the electromagnet. So, number of control wires are required are here is 4 into 2 which is 8 which is o n the complexity is linear here it is not n square. Now, can we actually work with this kind of situation. So, if for example, I want to set up a connection between say 1 2 2 prime 2 2 3 prime. So, may be probably let me check if I can make all connections or not. So, and remember these are not symmetric usually whenever you make connection from 1 2 2 the other side of connection will be from 2 2 1 because the voice is always bidirectional and I am assuming this connection to be unidirectional, but I am actually taking it as generalized which I am not putting the symmetry conditions which are usually true for most of the voice circuits. So, voice circuit is always set up in both the directions. So, that when any one of the two users are talking other person can listen to it. So, it is actually a bidirectional communication between two end points. So, if I want to set up 1 2 2 what I will do is I will excite. So, I am going to actually put a take here I will excite 2 and I will excite 1. So, this will snap in and this connection will be done. I want to set up 2 and 3. So, I will excite 2 here and a 3 here and this will snap in, but there is going to be a problem here. So, the problem here it will be when I am going to excite 2. So, this will also excite this thing because column is already excited and whenever I am going to excite 3 this is also going to get excited. Now, this is a complication. So, maybe we need to put up some other rule. So, what we will do is for any cross point. So, I am going to have two control wires, but I can put up a logic circuit. This will then create I am now going to create a logic circuit here in between and one of them will be creating a controlling horizontal line other one will be controlling a vertical line of the cross point. Now, this logic circuit will ensure that if the horizontal line is done activated first and vertical line activated second then only h and v both will be activated on the outgoing line. If only h if v is done first and h is done later on then this will not be activated. So, this will got the activation this will not cause the activation. So, once you do this what will happen to the same switch. So, let me see what is going to happen. So, if 1, 2 connection I am going to set up. So, 1 is this I am now doing horizontal first and 2 I will do second. So, horizontal vertical the connection should snap in. So, connection will snap in this will work 2 and 3 if I do 2 here 2 being activated. So, this line gets activated and then what I will find is the here the vertical this 2 was activated first. So, this cannot be activated actually. So, this particular connection will not snap in only this will snap in. But, when I am going to do go for 3 when I will activate 3 now for this particular connection I will find that now the condition h v because 1 has been activated first and 3 has been later. So, this connection as well as this connection will snap in. So, from 4 I am able to reduce it to 3. So, I have to do further innovation if I want this thing to actually work otherwise this system will not work. So, I need to modify the logic circuit and may be what I can do is one of the options is I actually activate a horizontal line first then vertical line first then you put down the horizontal line connection will remain connection will set up and horizontal and vertical both are up and once you put horizontal line down connection will be maintained and when the vertical line will be now put down then the connection will be closed. So, this is the period when the connection will snap in. So, I can use this particular logic and this can be programmed in this logic circuit actually. So, once you do this let me see what is going to happen. So, when 1 is going to go up which is horizontal 2 is going to go up and then of course is like 1 is going up 2 is going up 2 prime and then 1 is going down. So, once this happens this connection will snap in now when you are going to activate 2 after this. So, there has to be all these lines are they are active and now only depending on which line I am going to now go will go up it will depend on that. Now, I am going to actually put 3 prime will go up, but 1 is already down. So, 1 up and 3 up condition will not happen. So, 1 up and 3 up condition does not happen. So, this will not snap in this will not snap in only this particular part will snap in when 2 3 will happen okay. The other condition is for this 2 prime is up, but 2 prime is not going up after 2 this 2 is going up later than 2 prime. So, this cannot this connection also cannot snap in only these 2 connection will snap in and they will remain till the time 1 and till the time 2 prime or 3 prime which are up. So, these will actually control the connections okay. So, from any other input you will not be able to set up the connection to these outgoing ports. So, only 1 of these columns can remain active at any point of time only thing which you have to ensure is you should never do 1 1 up you should never do than 2 up and then if you are going to do 2 prime up then there is a problem actually. So, if you do 1 up and 2 up and then 2 prime up even if you then actually let down 1 you go down to in this case these both these particular points and these point both of them will remain snapped in because that is what they will be happening at this point. So, every cross point will be snapped 1 at a time that is a condition if you maintain this algorithm can work very well to set up the connection in case of a cross bar matrix and you require only 4 horizontal control line and 4 vertical control lines and complexity is going to be O n as far as the control lines are concerned. Only thing is that with every cross point you require some additional logic circuit which is nothing but technically a commutative logic circuit is a sorry is a sequential logic circuit and this can be built using some flip flop and other things. So, I think all fundamentals of basic electronics this electronics can be used here to build up the system. So, that is how the cross bar used to actually work, but so far what I have done actually very smartly what I have done is I have always said horizontal goes up vertical goes up and horizontal goes down and the connection remains on when vertical goes down the connection is closed. You should ask a question why it cannot happen this way that vertical goes up horizontal goes up vertical goes down connection remains and when horizontal goes down connection gets closed. Let us see what will happen if I am going to use this particular strategy I am going to take same 4 by 4 same connection sequence and let us see if I can do that I can handle this connection. So, I am going to now do the vertical thing first. So, my connection set up pattern is 1 2 2 prime and 2 2 3 prime. So, that is a pair which I am trying to set up. So, 1 2 2 prime if I want to set up as per rule that I have to activate 2 first. So, if I do activation of 2 first 2 prime goes up 1 goes up 2 prime goes down connection should remain on till 1 goes down. Now, at this point I am going to now put when this connection is up that time let me try out 3 prime goes up 2 goes up 3 prime goes down the next connection set up. Let us see what happens in this scenario. So, when 2 is going 2 prime is going to be activated 2 prime gets activated 1 prime will remain activated at this point of time in this row. So, this is control lines. So, you will do 2 prime first as you can observe from here and then 1. So, this connection will snap in and this 1 line is now active this is active at this point of time 2 prime then goes down when 3 prime goes up and. So, this and this all thing get activated, but 3 prime has not been activated first. So, this line will have nothing to do this will not get activated and once 3 prime is not activated 3 prime is activated 1 prime has 1 has not been activated first it has to be activated later on if this point has cross point has to be activated when now you will excite 2. So, this point will snap in nothing else can actually snap in this line is not active only 1 is active here now 2 is got activated 3 prime you will put down and this line will remain on hold. So, both these connections will work. So, this actually does work there is no issues in this case, but only problem in this case is you cannot implement a multicast thing suppose I want to set up 1 to 2 prime and 1 to 3 prime kind of scenario I want to set up both these connections now let us see whether this is possible or not possible. So, in this case I need to activate remember I need to activate vertical first. So, I will activate 2 prime I have to activate 3 prime I have to activate 1 prime and then of course 2 and 3 will be activated first when 1 prime is activated these both will snap in let me show it with a different color and then 2 prime can go down 3 prime can go down and so far 1 is there being on hold the connection will both of these will remain snapped in, but these both have to be done simultaneously. Now, if I am going to use horizontal vertical horizontal then what will happen the same scenario I have to now activate 1 first 2 prime 3 prime and do a control. So, both of them will snap in and then 2 and the 1 has to go down and then these 2 points will remain snapped in till the 2 vertical lines which are going to be there you need to actually hold on to 2 vertical lines which are there. So, I think these both of these systems actually do work without any problem except in certain scenarios, but usually what is preferred is this horizontal vertical horizontal and the output line is what is going to be used for controlling in because once you have snapped the connection you are not going to set up any connection to the vertical line. Now remember in this scenario of multicast which I had actually used earlier where vertical has been used first I have been setting both multicast simultaneously suppose the first point has been already set up wherever I have done 2 I have done 1 and 2 prime has been gone down connection is up and now I want to set up from 1 to 3 also another multicast connection now that will not be feasible in this scenario, but that will be feasible if I am going to use this particular case. So, usually most of the implementations were actually have been built with this kind of thing of course the way to handle this particular situation is if you are using lower kind of control system is that if you want to set up if you have already set up from 1 to 2 prime already a connection and now you want to also set up 1 to 3 prime when this connection is active you dismantle both the connections and now do a setup simultaneously. So, you have to break up an earlier connection that was the only way this could have been done here while in the second case or with the first this particular control strategy this can be done without any issues without breaking the earlier connection. So, this was a most popular variant which was actually used. So, this is how the cross point or cross bars usually used to work and this forms the basic unit of the switch. So, usually the switches are defined in this way there is a input port there is output port and important property of this particular switches it is strictly non-blocking. Now, I have to actually now define these terms because in the whole this lecture series you will be actually listening to these terms very frequently. So, before going on further so we define something called strictly non-blocking that is the first thing. So, meaning of this is if you are having an input port which is free if you are having an output port which is free irrespective of whatever other connections which have been set up between other input output pairs I will be able to always set up a connection or a path between this free input port and free output port without disturbing any other connection. So, a formal definition will come once we will define a clause theorem which is forestry blocking thing is basically says that if there is a set of free inputs here and set of free outputs here. So, it is talking terms of set. So, there is a I prime if I is the total number total set which is there total number of inputs O is the output. So, O prime is free outputs I prime is free inputs. So, I prime is a subset of I O prime is going to be subset of O. So, I should be able to create a legitimate multicast tree from any input which belongs to I prime to any subset of O prime any subset of O prime. So, there is a legitimate multicast tree which can be created. So, that is a most general definition of strictly non-blocking and this has to be done without disturbing any existing connections that is what the strictly non-blocking means. Now, second variety of a kind of switch of property which is known as rearrangeably non-blocking. So, this says actually the same thing can be done, but the condition that I will not disturb the existing connections cannot be maintained. So, existing connections might have to be disturbed, but we will make set up the same connection they will be broken for some time then I will again set up the connections and once a new connection is set up the input to output mapping will still be maintained same only there will be disturbances which will happen. So, I might I might be rearranging the structure in between and then this become rearrangeably non-blocking. So, in that sense if you see when I was looking at this particular case if I you are going to use a strategy of V H V V up H up V down to set up a connection this is actually a rearrangeably non-blocking switch because I am going to disconnect and then connect for such kind for creating certain multicast trees actually. If from one input you are connected to one output connection is already on functional and then you want to create the same input to another output free output then you have to disturb the earlier connection and then you have to connect again. So, in that sense yes it is a rearrangeably non-blocking. So, usually whenever I am going to use a block I am going to use call it cross bar I am actually assuming that I am going to use a strategy of horizontal control line first sorry horizontal control line first vertical then and then horizontal down then connection is set up till the vertical is also pulls down actually. So, that means we will be making them strictly non-blocking. So, rearrangeably non-blocking also I think now is clear we have to do the rearrangements, but any legitimate multicast tree between input and output pairs can always be created, but it requires disturbance of existing connections, but existing connection mapping from input to output will be still can be made, but connection will be broken for some time, but not necessary all the time, but some of time it will be done it cannot be guaranteed initially non-blocking it will be guaranteed. Now, third thing which is going to be blocking probability blocking switches. So, blocking switches means there is input is free output is free, but as something is not there in between the switch it is not possible to set up the path in cross bar of course you would not see this, but cross bars we do not use I will come to the reason why we do not want to use this, because cross point complexity here is O n square and of course next question comes is can I make up a switch which is strictly non-blocking like the cross bar and still have a cross point complexity less than O n square. So, I will give an hint later on we will actually prove through closed theorem yes it is actually thus possible. So, blocking switch are those switches where even if input and output is free sometimes you will not be able to set up the connection and of course the fourth category is wide sense non-blocking. So, we will see an example of this kind of switch also and you will formally prove that yes this is done through actually building up what we call state diagrams and state transition diagram and saying figuring out that if I can avoid certain state this switch will always remain non-blocking switch. So, what happens is basically the meaning is if while setting up the connection I follow certain algorithm and that algorithm is well defined and which ensures that I do not get into blocking states I always the avoid the blocking state I will set up all possible connection patterns between free input and output ports this usually happens because free input and output port they can be connected together in more than one possible ways you have to choose judiciously only those ways by which switch will always goes into a state which will always a non-blocking state it does not create a blocking structure. So, those kind of switching structures are wide sense non-blocking switches we will also see as an example of this kind of thing. Now, I have built up this switch across bar which is O n square complexity number of cross points which are required are n c 2 if it is only a upper half triangle of the full square then it is going to be O n square and it is strictly non-blocking switch. So, theoretically now the question is can I actually have something better than this or not yes it is possible to build up something better a class network provide O n 3 by 2 O n to power 1.5 is the complexity which will come. So, anyway I am not going into that as of now the only way that O n square now can be further reduced how that class network comes I am now coming to that thing that if I build up only one switch which is strictly non-blocking may be it may not work. So, idea is can I actually create n of course, if n becomes large say 10,000 or 20,000 say 1 lakh or 1 million or 1 crore how you will are going to set up a switch. So, in fact here also one should understand the concept of network. So, what you do is you never use a very large size switch as such you then will start always going to use multiple smaller switches and then create interconnection between them. So, you can actually put multiple switches like this and create subscriber to that. So, these are subscribers attached and I can create an interconnection between these switches. So, this what forms the network. So, all these switching matrices can be there in one building or they can be distributed all across the country all across the globe and that what forms that actually network. But as far as user is concerned whether you are in a Kanpur here and you are connecting to somebody in Bombay you see that whole network is nothing but single gigantic switch. So, internal network is usually not visible to the subscriber. So, in fact network is technically is nothing but single gigantic distributed switch which is built in redundancy mechanism and built in maintenance mechanism. So, part of it can fail, but the remaining part will always work even if some part fails your connections can be routed from other routes. So, the larger distributed structures are going to be random usually it will be a mesh, but what we can do is we can within a small dimension for within a building if I want to create or within an organization I want a larger dimension switch. Maybe it is a good idea that I can take a smaller switching elements switching crossbars and then join them in certain structured fashion to create a bigger gigantic switch. And that is a regular connection pattern we call it a switching network. So, next step is then that we have to see that how this switching structure is going to be built. So, now these using small switching elements or switching matrices I need to create a larger switch that is basically is the problem first and I need to retain certain properties. So, I think natural thing is that I would like to create a very similar property like we have with the crossbar strictly non-blocking property and it will create a larger switch using smaller crossbars. So, let us see if that can actually be done or not done. So, maybe common sense will tell I can going to get a crossbar some values n by n and I can use multiple of them and I am going to try with regular structure how this will be done. So, simple idea is that there will be n input and there are 1 to k. So, now total size which is going to be available is n into k. Now, but if I actually make this some input here cannot be connected to this output there is no path between them. So, I need to do something and secondly I also want is the path length from here to here input to output has to be exactly same almost. So, 1 possible ways I can now because I want n k inputs and n k outputs or let me also have similarly the other crossbars. So, which also have n outputs and they are k such things which are there. So, there is 1 stage 1 which is input stage 2 which is output and I want to create a connection between them. So, that everybody can talk to everybody else, but I want to keep them strictly non-blocking that is basically is the idea. So, this is a kind of a switching network which we are going to create, but it is a structured entity. Now, each of this input in this particular switch should be able to connect to any one of the outgoing port which usually is the condition. So, what I can do is I have I can create it as n by n. So, there are n outputs and there n such switches. So, I can just distribute them. So, 1 wire going to each one of them. So, I can do same thing here. So, perfect I have done this and then all free ports either are used as input port and output port. Now, this switch the question is whether it is a blocking switch or strictly non-blocking switch whether it is a rearrangeable non-blocking switch. So, we can actually observe if somebody has already set up a connection from here to here this path is busy. This switch per say the element itself is strictly non-blocking, but this path gets occupied next free input port and next free outgoing port they can they connection between them cannot be made because this path is not available this path is already occupied. So, this makes it a blocking switch. So, may be an idea can be instead of using n by n here can I use n by 2 n. So, let there be 2 n lines. So, I can put 2 lines all the way coming to everybody I can do it. Now, this is a different kind of cross bar cross bar earlier time I have told you was only having 4 incoming and 4 outgoing. So, I can actually have now 8 outgoing and 4 incoming. So, I can put that cross bar now I can set up 2 connections, but when I try setting up third connection. So, it is not possible actually. So, ultimately what you will do is if I want all n connection want to go to this n connection I require how many lines here actually that is very important. So, in worst case if all these ports are connecting to all these outgoing ports on switch 1 from switch k I require here n connections. So, you will be building up n n square here because you require n for this n for the next switch n for this n square outputs. So, you will end up in n into n square switch n you require 2 k such switches. So, number of cross points will be required is n cube 2 k now the whole thing can be equivalently represented by a switch of n k by n k. So, which will have a complexity of o n square k square in this case my switch complexity is 2 n cube k and of course, if it turns out to be that n is equal to k both of them are going to be technically same and both will be strictly non blocking switches. So, even with 2 stage actually it is not possible because there is no other way you can make the interconnection and this is only way you can build up a strictly non blocking switch. So, complexity is going to remain same it does not change. So, because if you put k is equal to n this will turn out to be o of n k square here also it is o n k whole square. So, maybe you have to go to a still better mechanism. So, let us see how this actually can be done. So, maybe with 2 stage this is not possible, but third stage is certainly is possible and that is where the cross network actually does come in. So, idea is this that I will have a switch I will have multiple of them and then I can create many such switches. Now, this switch can route the calls in this way there is exactly one path to each one of the switches and I can connect similarly this thing to each one of them and so on. So, idea is that you will actually have m 1 by there will be I will define actually m 1 by n 1 switch here k 1 of them k 2 switches and k 3 switches. So, because there are n 1 outputs. So, k 2 should be equal to n 1 that is how you can give no input port or no output port here should be left free only input ports are free here and output ports are free only here. So, that is a condition n of course total number of input ports will be m 1 into k 1 these are the number of incoming ports outgoing ports are n 3 into k 3. So, this switches are m 2 by n 2 these switches are m 3 by n 3 and since we are talking about symmetric switches these both terms m 1 into k 1 should be equal to n 3 into k 3. So, these switches should be same and you can see that k 2 has to be equal to n 1. So, that each one of these output port is connected to one switch in next stage similarly you can see that m 2 should be equal to the number of incoming ports has to be equal because one line is coming from each one of them. So, k 1 is equal to m 2 similarly should be equal to n 2 should be equal to k 3 and of course, k 2 is equal to n 1 is equal to m 3. So, this condition should be satisfied. So, this kind of structure known as class network and of course, intuitively or we will actually formally prove it this switch will be non-blocking if I am taking a case where m 1 is equal to n 3 it is all symmetric thing and when m 1 is equal to n 3 and k 1 is equal to k 3. So, that case when you will have k 2 greater than or equal to 2 into m 1 minus 1. If this condition is satisfied this switch will be strictly non-blocking switch. The logic is pretty simple that there is a free input port here which I want to connect I will show it with a different color. There is a free input port here which want to get connected to this one. So, remember there are only these many links which are there which will be nothing but. So, if number all output ports are occupied except this one which I want to connect you already have because I have not taken m 1 is equal to n 3. So, m 1 minus 1 these links are occupied similarly on this side also m 1 minus 1 links are occupied in worst case these might be choosing a set of nodes here these might be choosing another set of nodes here and there is no overlap between them. So, I need only one more extra which can be used to set up this particular free path to this and this will always guarantee that in worst case I am able to set up the path it becomes strictly non-blocking that is basically is the philosophy which will actually means m 1 minus 1 plus m 1 minus 1 plus 1 k 2 has to be greater than or equal to that that will give me the strictly non-blocking condition. This is also what is the class theorem technically is in fact there is a generalized form for that it says that k 2 has to be greater than or equal to m 1 plus n 3 minus 1 actually. So, this is a more refined form. So, I have taken it as taken a very simplified version of the same thing now with this if you can actually compute the cross point complexity. So, here I am actually now finishing and I think I urge all of you to try to make an estimate of what is going to be the cross point complexity for this configuration. This is known as cross network configuration and then you will actually figure out and appreciate yet this yes this is going to be have a smaller complexity than a cross bar and I am still able to create a strictly non-blocking switch. So, we will continue with the same thing this particular thing onward in the next lecture.