`zone_trade`'s inner prune, settled on every cell

2026-08-17. Hunt r_401bbf, run 2b9b08aa. Instrument: probe.py, data: results.json, pins: test_prune_discharge.py. Reads: hunts/frontier_math/k2_closure.py, hunts/r_a97060/RESULTS.md §4. Nothing here is evidence for or against RH (docs/08); nothing here claims k >= 3 or the unequal-depth quantifier, and no cap was widened and no zone re-tuned.

0. The answer

Outcome (a) of the brief: the delta is zero on every cell, and the prune is sound by argument as well as by measurement.

13,200 cell evaluations, every one of them audited, none sampled. Not one margin moves anywhere in its first twelve decimals. The largest deviation in the whole table is 1.1e-16, which is 3.4e-14 of the smallest margin, thirteen orders of magnitude below anything that could flip a cell.

That residual is float summation order, not a trade difference, and the table says so itself: 1467 cells in the signed pass and 965 in the unsigned one have a negative delta, which is arithmetically impossible for a genuine bite, since the exhaustive maximum is over a superset and can only ever come out higher. Its size agrees: a near total of order 0.16 has an ulp of 2.8e-17, so 1.1e-16 is four of them, which is what re-associating a sum of twenty-odd component trades costs. Cells whose delta is exactly 0.0 number 3428 and 3084.

So the published k=2 table does not depend on the prune. The assumption that hunts/r_a97060/RESULTS.md §4 recorded as load-bearing and unchecked is discharged, and the prune should stay in the code, documented, rather than be removed: it costs nothing and it is what makes the search affordable.

1. What "disabled" was taken to mean

The brief says: re-run with the inner prune disabled. Deleting only that line would have left the other cut standing:

if val + rem * caps[idx] <= best:      # outer branch-and-bound bound
    return
...
    if v > val - 1e-18 or m == 0:      # the inner prune, the brief's target
        dfs(idx + 1, ms + [m], v, rem - m)

and the outer bound's admissibility rests on the same hypothesis the inner prune needs. Auditing one while trusting the other would have left the question half open. The surviving components in this table are small (at most 8 zones carry a nonzero cap in any connected component of the whole table, at most 6 in the signed mode), so every admissible multiplicity vector can simply be enumerated: C(Z+10, Z) <= 43758 vectors, evaluated as one vectorised quadratic form. That is an exact maximum with no cuts at all, hence a strict upper bound on what the inner-prune-free search would find, so a zero delta against it settles the brief's question a fortiori and settles the outer bound for free.

Both trades run on the same zone data, component by component, inside one pass. Each cell's delta is therefore an exact difference of two solvers on one input, not a comparison of two independently scanned tables. Everything outside the trade (the far rows, C1, the budget floor) is identical by construction, so the margin delta is the near delta.

2. Why the prune was never going to bite, and the hypothesis that makes it so

The measurement is the deliverable, but it came with an argument that explains it, and the argument is worth more than the table because it covers cap values this table never produces.

Claim. If every pair charge is nonnegative and every cap is nonnegative, the inner prune removes nothing: the search still returns the exact maximum.

Proof. Write F(ms, idx, rem) for the best completion value from zone idx given the prefix ms and budget rem. Two monotonicities:

1. F is nondecreasing in rem: the feasible set of completions grows.
2. F is nonincreasing in each prefix multiplicity: a prefix atom enters a completion only through the subtraction -2*q_ji*ms_j*m_i, which is nonnegative when q >= 0.

Now take a node with running value val and a branch m whose value v does not improve it, v <= val. Its best completion is v + F(ms+[m], idx+1, rem-m), which by (2) is at most v + F(ms+[0], idx+1, rem-m), which by (1) is at most v + F(ms+[0], idx+1, rem) <= val + F(ms+[0], idx+1, rem). The right-hand side is exactly what the m = 0 branch can reach, and m = 0 is always descended into (the or m == 0 clause). So the pruned branch cannot beat a branch the search already takes. ∎

The same hypothesis makes the outer bound admissible: with q >= 0 no completion gains more than rem copies of the largest remaining cap, which is rem * caps[idx] once the caps are sorted descending.

And the hypothesis holds by construction, not by geometry. The charge is q_ab = Kpair(min(dmax, 6))/200 with Kpair(u) = Re(ghat(0,u))**2, a square divided by 200. It is nonnegative for every input, so both cuts are sound for any cap vector whatsoever, including the enclosure pass's caps in hunts/r_a97060/, which this run did not re-enumerate. That is a strictly stronger statement than "measured 0.0 on 6600 cells", and it is why the recommended disposition is to document the prune rather than delete it.

This is a different situation from the other assumption r_a97060 §4 named. The pair-charge clamp really is sound only by accident of geometry (Kpair has a root near 6.65 and the widest component in the table is 1.94, corroborated here at 1.9404 over a coarse tau sweep). The prune is sound by the form of the objective. Two unstated assumptions, two different kinds of luck, and only one of them was luck.

One caveat kept in view: the 1e-18 tolerance in v > val - 1e-18 makes the test more permissive than the exact rule, so it descends into slightly-losing branches too. That is the safe direction and it cannot cut anything the exact rule keeps.

3. Controls

The planted bite is the control that gives the zero delta its meaning. Break exactly the hypothesis of §2, one negative off-diagonal charge, an attractive pair instead of a repulsive one, and the cuts fail at once:

The published search understates the true trade by 0.042 (18.6%) on that instance, so the probe is a genuine detector and the table's 1e-16 is a real negative rather than a blind one. The middle row is also the answer to why this run enumerated instead of just deleting the line the brief named: with the hypothesis broken, disabling the inner prune alone gives 0.146, further from the truth than leaving both cuts in. A half-audit would have been worse than none.

4. Cost

713 s wall for the whole thing on the container's CPU: 255 s for the signed pass, 450 s for the unsigned one, both at the published table's own scan resolution (x-step 0.01, three tau-samples per cell, zones 0.25). The exhaustive trade is not the expensive part: it is a single (N, Z) matrix product per component, so auditing every cell cost roughly what running the table once costs, and the "sample the binding cells" shortcut bought nothing worth having. The census was affordable at 6600 cells; that is a fact about this table's component sizes, not a general licence.

5. Honest scope

• This settles the trade solver, on the measured pass's zone data, plus (by §2) the solver for any nonnegative caps. It says nothing about the zone caps themselves, the component geometry, the far rows, C1, or the O8 floor. The enclosure grade of those is hunts/r_a97060/'s business, not this run's.
• The delta is zero to double precision, not symbolically: both solvers are float, and the residual 1e-16 is their differing summation order. A reader who wants the trade in exact arithmetic still has that to do, and r_a97060 already documented a 1e-9 float slack in the trade that dominates this by seven orders of magnitude.
• The sum m <= 10 atom budget is inherited from k2_closure.py and was not re-derived here. Both solvers use it, so it cannot show up in the delta; if it is wrong, it is wrong in both columns.
• No number in the published table changes. Nothing here moves the composite k=2 claim's grade, which still takes it from the v-convexity transfer (G3), and nothing here is evidence about RH.

Loose threads

1. The atom budget sum m <= 10 is an inherited constant. Both solvers respect it, so this audit is blind to it by construction. If the true adversary can place an eleventh atom in a component, every column of the table shifts together and no delta of this kind would reveal it. First step: check whether P's optimal multiplicity m* = 1/2 + cap/(2q) can reach 10 for the largest zone cap the table builds; if it cannot, the budget is provably slack and the constant can be retired to a lemma.
2. The clamp's monotonicity is still geometric. r_a97060 replaced it with a running minimum in its own pass, but k2_closure.py still ships Kpair(min(dmax, 6)) and is still sound only because the widest component is 1.94 against a first root near 6.65. That margin is not enforced anywhere. First step: assert the widest component width in hunts/frontier_math/test_k2_closure.py so a geometry change that crosses the root fails a test instead of silently over-crediting.
3. The exhaustive trade is cheap enough to be the default. At most 8 zones per component and ~0.4 ms per component, the enumeration could replace the branch-and-bound in k2_closure.zone_trade outright, deleting the question this run had to answer rather than documenting it. First step: swap it in behind a keyword argument and time the full table; if the cost is within a factor of two, the heuristic has no reason to survive.