I've read the famous Erdos quote about R(5,5) vs R(6,6) and he seemed to think that it was at least theoretically possible if the whole human race had to find it quickly or face destruction.
"Suppose aliens invade the earth and threaten to obliterate it in a year's time unless human beings can find the Ramsey number for red five and blue five. We could marshal the world's best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded the Ramsey number for red six and blue six, however, we would have no choice but to launch a preemptive attack."
https://blogs.scientificamerican.com/roots-of-unity/moores-l...
I once spent some time calculating the effort (but don't have the results handy), a rough number for a naive approach to exhaustively searching the problem space would involve something like 10^200 graphs.
The number of graphs of size n is 2^(n choose 2). (There are n choose 2 pairs of points, each of which could be or not be an edge.) If the answer is 43, that's 2^903 which is roughly 6.762 * 10^271. If the answer is 48, that's 2^1128 which is roughly 3.646 x 10^339.
Naive plus better computers is not enough to tackle this problem.
This is overcounting by many orders of magnitude. For example, there's only one graph with one edge, not (n choose 2). For n = 43 you're overcounting these graphs by a factor of 903.
Similarly there are only two graphs with two edges (either the two edges are connected or not), not ((n choose 2) choose 2). For n = 43 you're overcounting these graphs by a factor of ~200k.
For graphs with three edges you can have (1) a triangle, (2) three edges connected end to end forming a single path, (3) three connected edges forming a star, (4) two connected edges and one single, or (5) three unconnected edges. Compared to your count of ((n choose 2) choose 3), you're overcounting by a factor of about 24 million for n=43.
The total (over)count is going to be dominated by graphs with approximately (n choose 2)/2 edges, which intuitively is where I also expect the overcounting factor to peak.
But suppose that we are able to do so. How much does this do for us? Well, it can help us by a factor of at most n!, because you generate isomorphic graphs by permuting the vertices. Which for 43 is around 6.04152630633738e+52. For 48 is around 1.24139155925361e+61. Those are big savings to be sure, but are still dwarfed by the size of the search space.
So the next thing to do is to not only try to look at each graph up to isomorphism once, but to somehow generate them in an order that makes it likely that you find cliques or independent sets early. Thereby letting you prune out big chunks of the search space. With the ability to start different computers out in different ranges so we can parallelize the search. But by now we're well down on the path to something that is very much not a naive search.
As others have said, brute force is just plain impossible. But we can limit the search space significantly by proving sub-results about the structures of the possible graphs. Somewhat trivial example to give the flavor: let's state the problem as "find the smallest number of vertices needed such that any graph must have either a 5-connected component (a "pentagram") or 5 independent vertices". We know that the number is at least 43. So if we are looking for a counterexample, a graph on 43 vertices that does not have either of these two subgraphs, what can we say about the possible number of edges that each vertex must have? (the degree, for graph theorists). We can immediately say that if we have 5 or more vertices with no edges (degree 0) then we have our independent set already, so any counterexample can have at most 4 vertices of degree 0.
By the way, my favorite Explain-like-I'm-5 version of Ramsey's theorem is "Total disorder is impossible": If a structure is large enough, there must be some substructure that is ordered.