For S1---
|v|=8, |E|=13, There is theorem
CSPG with no K3 -------->e<=(2n-4), e is the no. of edges, n is no. of vertices and CSPG is connected simple planner graph
contrapositive of theorem is e>(2n-4)--------->CSPG with K3
so as in the question 13>(2*8-4) so this graph have K3, so here min 3 color required
So S1 is true
For S2---
r=(e-n+k+1)--------1 where e is the no. of edges, n is no. of vertices, k is no. of component in graph
as in the statement mentioned it is planner and have even degree 0 and 2 possible but not 4, 6, 8…. in that case it will be not planner because K5 will present
2+2+2+2…..n times = 2*e
e=n put in equation 1
r=n-n+k+1 = k+1, so all k component’s inner version can be colored with 1 color and one outer region required 1 color so it is 2 colorable
if we have few verices with 0 degree then it is isolated verex which can be colored with 1 color.
so this statement also true
Ans is c
