Six outputs suffice: unique-Bayes counterexamples to Fisher consistency of the structured-SVM surrogate on modular metrics

k* <= 6: a modular metric on six outputs (the shortest-path metric of K_(3,3)), a full-support q = (1,1,2,2,3,3)/12 whose Bayes set is the single output a₃, and an exactly optimal score vector v = (-1,-1,-1,0,0,0) – optimality certified by an explicit self-coupling through weak duality – whose maximisers b₁,b₂,b₃ are all non-Bayes; this improves the source's own bound k* <= 8. The metric is modular but not median

A second six-output witness on the same K_(3,3) metric, q = (1,1,1,2,1,1)/7 and v = (1,1,1,0,0,0), whose single Bayes output b₁ lies in the HEAVY part and whose argmax is the whole light part – so the failure is not a knife-edge choice of q and the Bayes output may sit on either side

The source's four-output unit star (its Corollary 3.4) re-certified: v = (-1,0,0,0) is exactly optimal by a three-cycle self-coupling, Bayes = o, argmax = a,b,c, disjoint – and its q has a zero, so it does not bear on the full-support question

The source's five-output K_(2,3) witness (its Theorem 5.1 at m = 2, n = 3) re-certified, together with the proof that its Bayes set a₁,a₂ has two elements and therefore equals no singleton – the exact defect that leaves k* open

The source's three-dimensional Hamming-cube witness (its Theorem 5.4) re-certified, giving k* <= 8, and the cube metric proved median over all 512 triples – the only known median witness

Negative controls: no witness on 0 or 1 outputs; the three-point equilateral metric is a metric that fails the common-geodesic condition; at the witness distribution the embedded report of the Bayes output is ALSO an exact minimiser and IS decoded correctly (so the witness is about some minimiser, not all of them); the embedded report of a non-Bayes output is not optimal; and at the uniform q on K_(3,3) every output is Bayes

The certificate layer for the self-coupling dual, for an arbitrary finite metric: weak duality (the easy half of the source's Proposition 2.1), the optimality certificate, complementary slackness (its equation (4)), the embedded-report identity S_M(phi(y),x) = 2 L(y,x) (its Lemma 2.2), and the diagonal criterion – a positive diagonal entry in any optimal self-coupling forces argmax v = y with y Bayes

Exhaustive decision over ALL rational q and ALL score vectors v: no five-point modular metric among the five arising from the 21 connected graphs on five vertices, and none among the seventeen with integer distances at most four (from 228,502 labelled metrics), admits a full-support unique-Bayes disjoint-argmax witness – so k* = 6 conditional on the five-point classification

The counts of connected graphs on n vertices whose shortest-path metric is modular / median, n = 1..8: 1,1,1,3,5,16,41,156 and 1,1,1,3,4,11,23,69; the median row is OEIS A292623 and the modular row is absent from OEIS. Plus the engineering fact that decide cannot evaluate rational arithmetic in the Lean kernel (Rat addition normalises through Nat.gcd), while the identical table over the integers decides fine – the reason this family carries integer certificates

A seven-output witness: K_(3,4) with the tie broken at a₁, q = (2,1,1,1,1,1,1)/8 and v = (0,0,0,1,1,1,1), single Bayes output a₁ and argmax the whole four-element part – so witnesses are now kernel-certified at cardinalities 6, 7 and 8

The general tie-broken complete-bipartite family: for every m >= 3 and n >= m, K_(m,n) with q(a₁) = 2/(m+n+1), all other weights 1/(m+n+1), and v = 0 on the light part and 1 on the heavy part, is a witness – hence there is a witness on k outputs for EVERY k >= 6, and m = 2 (the source's own case) is exactly where the construction fails