 us planar graphs Euler's criterion to decide whether a graph is planar or not. Now so far we have talked about graphs and we have been drawing diagrams of graphs on a plane however we have not checked the issue that whether I can draw a graph on a plane without its edges overlapping on each other. For example I can think of a graph like this where these are the vertices and I have got two edges crossing each other on the other hand I could have drawn it on the plane in almost same way just by drawing an edge like this now we see that this graph G and the other one G prime both are essentially same although if you look at this graph no edge is overlapping with another I mean cutting another now our question is that when a graph can be drawn on a plane in this way and when it cannot be drawn in the in this way. So we come to the formal definition of planarity of a graph we write a graph G is said to be planar if there exists some geometric representation of G which can be drawn on a plane such that no two of its edges intersect a graph that cannot be drawn on a plane without such a crossover is said to be a non planar graph now the question is that can we create some examples of non planar graphs in this context there are two famous graphs called Kuratov-Kuratovsky is two graphs which are shown to be non planar the first graph is the complete graph with five vertices or written as K sub five and the second graph is written as K three three which is a bipartite graph a complete bipartite graph with parameters three three so the complete graph with five vertices looks like this we have got five vertices V1 V2 V3 V4 V5 we have got a cycle here and then we have to connect V1 and V3 we connect that we have to connect V3 and V5 we connect that now we have to connect V5 and V2 we connect that and we have to connect V3 V2 and V4 we connect that and now we have to connect V4 and V1 and now we see that neither can we go in this direction nor in this direction or nor in this direction without cutting one of the edges so I can write probably like this and like this and we know that there is an intersection over here so so we see that we are unable to draw a complete graph with five vertices without intersecting the edges so I will draw it again so let us see the graph again we have got V1 then V2 we draw like this this is V2 then we draw like this this is V3 and then we come back to here this is V4 then we have got V5 and we join V5 and V1 and now V1 and V3 are joined then V2 and V5 are joined then V2 and V4 has to be joined we can join like this and then we join V3 and V5 and then we see that suppose I have to join as since we have to join V4 and V1 we have no other way other than cutting or intersecting one of the edges so we may write like this or whichever we go we will intersect the edges so this is Kudatowski's first graph and let us look at the second graph which is as I said that a complete bipartite graph 33 and K33 it has got the set of vertices is partition into two subsets each of three vertices and then vertex from each vertex from one subset is connected to vertices of the other subset so I will have like this then here like this and then like this in this context we define bipartite graph in general graph a graph G VE a graph G V, E is said to be a bipartite graph if V is partitioned into two subsets V1 and V2 where V1 intersection V2 is empty and all the edges are connecting vertices of V1 to vertices of V2 and there is no edge connecting vertices of V1 so we write that where V1 V2 is the empty set and no vertex of VI is adjacent to another vertex of VI for I is 1 and 2 so the vertices of V1 are not adjacent to each other vertices of V2 are not adjacent to each other so now when we look at this graph what we denote as K33 we see that the three vertices over here are not adjacent to each other and the three vertices over here are not adjacent to each other whereas there are edges from one subset to the other now this graph K33 is something more here that cardinality of the set of vertices is 6 and cardinality of V1 is equal to cardinality of V2 is equal to 3 and if we note that each vertex of V1 is connected or is adjacent to each of the vertices of V2 and converse so we have K33 is a complete bipartite graph we draw it again now what we would like to prove here is that K33 is not planar now it is not difficult to check that K33 is isomorphic to a graph like this so this I leave for exercise that these two graphs are isomorphic so if I call this K33 and this simply H K33 and H are isomorphic now if I start drawing H all over again then here then we will start drawing the cycle over here and then I can of course draw this one and then see that we can choose to draw this edge from the top but then if I want to draw an edge from here to here then I am forced to intersect over here this is the place where the intersection will occur so this graph also looks like a graph which cannot be drawn on a plane but the proofs that I have given right now are are intuitive proofs and we would move on to prove to a proof which is more analytic for that we would like to have more definitions so first of all we would like to mention that whenever we have a simple planar graph we can embed it on a plane by using only straight lines we do not have to have any crooked lines to embed a planar graph if it can be embedded it can be embedded by using a straight line we do not give a proof of that but it is very intuitive and after that we define a very important concept called regions in the context of planar planar graphs so a plane representation of a graph a plane representation of a graph divides the plane into regions a region is characterized by the set of edges forming its boundary. Now let us look at some examples let us look at a planar graph like this which is reasonably straight forward kind of graph this is a planar graph and we see that its edges are forming regions and there is another region which is outside this graph so I can have regions which are both finite and infinite so if the region is enclosed by the edges then it is finite but of course I will have one one region which is infinite which is in intuitively outside the graph so here we have got three regions let us call them R1 R2 and R3 now we can remove this distinction between finite and infinite region by embedding a graph on a sphere on the surface of a sphere by using stereographic projection now I will quickly give an idea of the stereographic projection of a plane onto a surface of a sphere so suppose I have got a sphere like this where this is the south pole this is the north pole and I put the sphere on a plane let us call this p and what I do is that I take a point x on a on the plane and then connect x to the north pole and it is bound to cut the surface of the sphere on another point let us say x now it is not difficult to see that whatever line or whatever set of points are there on the plane in this way I can map it on the sphere and this map is one to one and on to except that I will have the all the infinite points getting mapped to the north pole rest of the all the points are mapped to a single point only they are where whichever direction I go it will map the north pole now it is again intuitively very clear that if we have a graph on a plane I can use the stereographic projection to map it on on this on a sphere spherical surface and vice versa and this gives us the theorem like this which I will state without a concrete proof but which is intuitively clear from what I have already told this is theorem a graph a graph can be embedded in the surface of a sphere if and only if it can be embedded in a plane well and the next theorem is again intuitive that will prove we will not prove but state that is a planar graph may be embedded in a plane such that any specified region can be made the infinite region made the infinite can be made the infinite region now we are in a position to start looking at a surprisingly elegant theorem by Euler which connects the regions number of vertices and edges in a planar graph so this is called Euler's formula it states that a connected planar graph with n vertices a connected planar graph with n vertices and e edges e – n 2 regions now indeed this is a very surprising result and would like to give a proof of this result so first of all we will observe that it is enough to prove this result for simple graphs the reason is that suppose F is the number of regions then if you if you see that if I start increasing the number of edges over a simple graph by introducing more parallel edges or self loops we will see that each edge will generate an extra face and therefore if I have something like that this that f is equal to e – n 2 for a simple graph suppose I have got a simple graph and for this this is true if I introduce one self loop then I am introducing one edge so e will increase by one and one region will also increase so this will increase by one again I introduce one parallel edge then again e will increase by one and f also will be increased by one so if this formula is true for simple graph then it is true for any graph so therefore we have only deal with simple graphs now let the polygonal net representing the given graph has f regions which we have already told let Kp be the number of p sided regions so what we are saying here that we can write a planar graph if at all it can be drawn on a plane by the vertices and the joining edges as straight lines therefore I will always be able to represent a polygonal represent a graph by a polygonal net so the faces will be polygons and let us say that Kp be the number of p sided regions that is p sided polygons as regions now if this happens then we see that 3 x K3 that is there are K3 many 3 sided regions and so the number of edges associated will be 3 x K3 then similarly 4 x K4 and similarly if we go on R x Kr and we know that each edge is going to be present in two regions therefore I will have two times e and if I sum up all these face sides so they will give me ultimately number of faces because K3 is the number of faces or regions with 3 edges and 4 and so on up to some are so therefore if I add all of them I am going to get the number of regions or which is sometimes called faces so I have got two expressions over here and I also know that the sum of the angles subtended at each vertex is 2 pi n sum of the angles subtended each vertex is equal to 2 pi n next sum of the interior angles of a p sided polygon is equal to pi p minus 2 and sum of the external exterior angles is equal to pi p plus 2 so in that network that polygonal network of the graph when drawn on a plane we will have f minus 1 bounded regions or finite regions and 1 infinite region and therefore if I sum up all the interior and exterior angles and I am going to get a sum like this pi into 3 minus 2 K3 plus pi into 4 minus 2 K4 plus and so on up to pi into r minus 2 Kr plus 4 pi and this is going to be equal to 2 pi e minus f plus 2 this is by using these two expressions and this is equal to 2 pi n and from this we get e minus f plus 2 is equal to n which implies that f is equal to e minus n plus 2 which is Euler's formula once we have done this we will check one corollary to this formula which says that in any simple graph with f regions n vertices and e edges where e is greater than 2 e is greater than or equal to 3 by 2 f and e is less than 3 n minus 6 now let us try to give a proof of this result now what we observe here is that suppose I have got f many regions each region will have at least three edges and so the total number of edges is 3f and we know that each edge will always be in two regions therefore twice of e has to be greater than or equal to this which proves this result now if on the other hand we put the Euler's formula over here we get f is equal to e minus n plus 2 then I get this is equal to thrice e minus thrice n plus 6 and which gives me after reduction e is less than or equal to 3n minus 6 now if I now look at k5 that is complete graph with five vertices here I will have e is equal to the n is equal to 5 n is equal to 5 e is equal to 10 and therefore I will have e one side 10 and the other side 3 into 5 minus 6 which gives me 9 so I have got 10 is less than or equal to 9 which is a contradiction and therefore it cannot be a planar graph if I now look at the second graph of Puratovsky we will see that this is k33 and so I will have e is equal to in this case e is equal to 9 and n is equal to 6 so if I now put in this value when we will get 9 is less than or equal to this is 3 into 6 minus 6 and this gives me 12 so there is no contradiction over here but I think a little more I look at the graph again so I have this graph and we observe that no region in this graph can be bounded by three sides the reason is that it is a it is a bipartite graph so if I start from any vertex if I come to another vertex this is on the other it is on the other set and then I go back it is again the same set and now I can never have a connection like this so I will never have a face which is bounded by three edges therefore I will have twice e is greater than or equal to 4 times f and which implies that twice e is greater than or equal to 4 times e minus n plus 2 now if I put e equal to 9 here I get 18 and if I put e equal to 9 and n equal to 6 plus 2 then I will get here 4 into 5 that is equal to 20 now this is a contradiction and therefore k33 also cannot be a planar graph with this I end today's lecture thank you