A graph that can be drawn on a plane without edges crossing is called planar.
Given that three regions R1,R2,R3 as we have 6 vertices
R1 :01 2 5 2 4 3
R2: 0 3 1
R3:1 3 4 2
as we have found that we have 7 edges and 6 vertices
Therom:Euler’s Formula For a connected planar simple graph G=(V,E) with e=|E| and v=|V|, if we let r be the number of regions that are created when drawing a planar representation of the graph, then r=e−v+2.
henced proved from above graph by euler formula of a planer connected graph
for the given vertices useing edges we have constucted the following graph
from the above graph it is haveing 4 regions r1,r2,r3,r4
and 8 edges and 6 vertices
from eulver formula of a planar graph
hence proved that above graph is a planar graph
in practice, it is difficult to use Kuratowski’s criterion to quickly decide whether a given graph is planar. However, there exist fast algorithms for this problem: for a graph with n vertices, it is possible to determine in time O(n) (linear time) whether the graph may be planar or no
For a simple, connected, planar graph with v vertices and e edges, the following simple conditions hold:
Theorem 1. If v ≥ 3 then e ≤ 3v − 6;
Theorem 2. If v ≥ 3 and there are no cycles of length 3, then e ≤ 2v − 4.
planar graphs are sparse graphs, in that they have only O(v) edges, asymptotically smaller than the maximum O(v2). The graph K3,3, for example, has 6 vertices, 9 edges, and no cycles of length 3. Therefore, by Theorem 2, it cannot be planar. Note that these theorems provide necessary conditions for planarity that are not sufficient conditions, and therefore can only be used to prove a graph is not planar
by useing Kuratowski’s Theorem we can that it is not planar
if V3 THEN e3v-6
we have 9 edges and 6 vertics to prove that it is not a planar graph
hence not proved we have 9 edges and 6 vertices and it not a planar graph hence proved by above theorom every planar graph have O(n) edges hence proved from planar graph