[Someone]

There is no guarantee that the number of bits in 'all the bits' is finite. Because of that, there is no guarantee that such an algorithm finishes. The simplest counterexample is N=2, with a distance of 1/3 between the nodes, and shortest (and only) tour with length 2/3.

[mikeash]

You don't need to represent the actual length of the solution in a finite number of bits, you just need to represent the solution itself. The solution can be represented as the set of edges contained within it. Each edge is either present or not, so a solution can be represented as a bit set the size of the number of edges in the graph, or at most n^2 bits where n is the number of vertices.

For a practical idea of how you'd do this, I think you'd just modify your binary search based on the possible circuit values. Basically, figure out what number excludes half of the remaining solutions, rather than just half of the range, and use that in your binary search.

[Someone]

Yes, any algorithm that halves the set of potential solutions in each iteration will work, but I do not see how one can get such an iteration step from "a decider that can answer "is there an answer less than X?". If suspect (but cannot prove or even have reasonably solid arguments) that the "figure out what number excludes half of the remaining solutions" step will be at least as hard as the problem itself.

Now, if I had such an oracle, I would let it loose on all graphs equal to the target graph with one edge removed. I think (but cannot prove (yet?)) that that is a way to prove, for each edge, whether it is by necessity part of the shortest tour. I also supect that, for most graphs, the set of remaining "might be in the shortest tour" will be so small that an exhaustive search will be easy.