On deterministic weighted graph isomorphism from randomized

Question

Is there a $O(n^2)$ algorithm to resolve isomorphism between two weighted $n$-vertex graphs? This is a much easier problem than graph isomorphism.

Basically take an real edge weight set $\{w_1,\dots,w_s\}$ where $s=\Omega(n^\alpha)$ where $\alpha\in[0,2]$ and the edge weights are picked uniformly. We can think of the second graph as being generated as follows: Input the weighted first graph to a black box and the black box returns the second graph based on some mechanism.

For every $n$ fix a set of weights.

For every input graphs you assign weights uniformly from a distribution. Then you feed in the input graph, weights and the second graph to the black box and the black box returns you the weighted second graph. Then you have to decide isomorphism on these two weighted graphs. So the probability is over all assignments of weights (the weight set is fixed for every $n$ in some worst case way). If this has an algorithm that is in $\mathsf{BPP}$ then in theory it should be conjecturally derandomizable.

Answer 1

Consider first the case of two colors. We can identify with each vertex $x$ its color distribution $a$, which is the number of edges of color 1. The distribution of $a$ is $\mathrm{Bin}(n,1/2)$, and so if we divide the vertices into bins according to their color distribution, the heaviest bin will contain $O(\sqrt{n})$ vertices with high probability. If we repeat the process, we the heaviest will contain $O(\log n)$ vertices with high probability. Once more, and we expect all bins to have size 1.

What this sketch suggests is that even if you have a constant number of colors (in this case, 8) then with high probability, every vertex will have a unique color distribution (histogram of colors of the edges adjacent to it). This makes graph isomorphism very easy for such graphs.

Of course, this sketch isn't rigorous. To know for sure, you will have to make it rigorous, and find whether it indeed works; perhaps a constant number of rounds isn't enough, but $\alpha\log n$ rounds (as in your case) are.