If there exists an independent set A, with neighbourhood at most 101|A|, remove it and the neighbourhood and induct down. The key claim is that given a vertex V, the vertices that are a distance d<=49 away from V form an independent set. Let p(x), denote the number of vertices at a distance x from V. So then we must have 101d(x) < d(x+1) + d(x-1) and so d(x+1) > 100d(x). Thus d(49) >= 100^(49) > 10^(100)/102.

R8kt wrote: If there exists an independent set A, with neighbourhood at most 101|A|, remove it and the neighbourhood and induct down. The key claim is that given a vertex V, the vertices that are a distance d<=49 away from V form an independent set. Let p(x), denote the number of vertices at a distance x from V. So then we must have 101d(x) < d(x+1) + d(x-1) and so d(x+1) > 100d(x). Thus d(49) >= 100^(49) > 10^(100)/102. your idea is interesting, but why 100^(49) > 10^(100)/102 ?