Walkthrough — degree-8 T-curves

0. What was actually missing

Not a cleverer search — an independent verifier and the actual current record. The prompt's leftover was stale twice over: it said "2359 realized, 53 non-maximal schemes with no T-curve witness", but the April ICMS paper's 53 are maximal schemes, and the July v3 of the companion (arXiv:2602.06888) had already moved the counts to 2,367 realized, 38 of 89 M-schemes T-realized, and — the fact that reshapes the whole attack — Theorem 21: a maximal degree-8 T-curve has $(p,n)\in\{(19,3),(15,7),(11,11),(7,15)\}$ . Four of the six algebraically open M-schemes have $(p,n)=(3,19)$ . So combinatorial patchworking, the intended stack, is provably useless for four of the six targets, and the entire T-curve attack surface is: the two deep-nest schemes ⟨4⊔1⟨2⊔1⟨14⟩⟩⟩, ⟨14⊔1⟨2⊔1⟨4⟩⟩⟩, plus the 39 M-schemes and unknown non-maximal schemes missing from their census. Nobody knew this in the room until the papers were actually fetched.

1. Named false starts

The unit-coefficient conic. Sanity test asserted that all-plus signs in degree 2 give an empty curve, because $1+x+y+x^2+xy+y^2$ has negative discriminant. The engine said ⟨1⟩. The engine was right: Viro's polynomial is $1+t(x+y)+t^3(x^2+xy+y^2)$ with the lifting in the exponents, and on $y=0$ its discriminant $t^2(1-4t)$ is positive for small $t$ — the conic is nonempty.
Rooting the nesting tree at "the outside". The planar instinct — root the oval-containment forest at the unbounded region — has no meaning in $\mathbb{RP}^2$ ; there is no unbounded region and no a priori marked point. First plan (root at a corner of the rhombus) is simply wrong: that region can sit inside ovals.
Which Harnack sign rule? Itenberg–Viro print "− iff both coordinates even", other sources "− iff both odd". Ten minutes lost to the worry; the two differ by $(x,y)\mapsto(-x,-y)$ composed with a global sign flip, so they give the same curve up to isotopy. The engine reproduces ⟨18⊔1⟨3⟩⟩ from either.

2. The useful failure

The conic false start is the one that taught something: a T-curve is the curve of the deformed polynomial at small $t$ , and no intuition about unit-coefficient polynomials transfers. After that, every sanity expectation was taken from published statements (Prop 10 of arXiv:2602.06888 v3, Harnack component counts) instead of mental algebra.

3. The click

The double cover decides everything that has no planar analogue. Work on the sphere $S^2$ = two copies of the glued rhombus, with the antipodal boundary gluing crossing sheets. Then:

a curve component is an oval iff its preimage has two components (a pseudoline lifts connectedly);
a complement region is the outermost one iff its preimage is connected (= it contains a noncontractible loop; for even degree exactly one region is non-orientable);
the region-adjacency graph (regions as nodes, ovals as edges) of disjoint ovals in $\mathbb{RP}^2$ is a tree, and rooting it at the non-orientable region turns BFS into the nesting forest.

Everything is then two union-finds (triangles for the curve, corners for the regions) run once on the quotient and once on the double cover, plus an Euler-characteristic assertion $\chi=1$ that catches gluing bugs.

4. The argument, in the order it was found

Reflect the triangulation of $T_8$ into the rhombus $|x|+|y|\le 8$ , signs replicated by $\varepsilon(\pm i,\pm j)=\varepsilon(i,j)(\pm1)^i(\pm1)^j$ .
In each triangle with both signs, one midpoint segment; segments join across an edge iff the edge's endpoint signs differ.
Components: union-find over cut triangles. Regions: union-find over (triangle, vertex) corners — within a triangle the two same-sign corners merge; across every edge, corners at the same vertex merge; boundary edges merge antipodally.
Double cover: the same relations on two sheets, boundary gluing crossing sheets; lift counts classify ovals/pseudolines and find the root region.
Scheme: BFS the region tree, encode each oval as the sorted tuple of its children, print canonically (⟨18 u 1⟨3⟩⟩ style).
Convexity: a triangulation is certified convex by exhibiting heights whose triangle planes lie strictly below every other lifted lattice point — checked in exact Fraction arithmetic. Their published integer MIN_WEIGHTS pass this check verbatim, which is what makes the census replay a certificate replay, not a trust-me replay.
Search: random positive-definite quadratic + integer noise heights, float lower hull, exact re-certification, then simulated annealing on signs with a fast array-based evaluator (cross-checked against the exact engine on all 2,367 census certificates and 40 random cases; 1.4 ms/eval vs 6 ms).

5. Computer residue

compute/sanity.py: Harnack schemes degrees 1–8 exact (⟨J⟩, ⟨1⟩, ⟨J⊔1⟩, ⟨4⟩, ⟨J⊔6⟩, ⟨9⊔1⟨1⟩⟩, ⟨J⊔15⟩, ⟨18⊔1⟨3⟩⟩), component counts $g+1$ , all-plus degree-8 gives the depth-4 nest ⟨1⟨1⟨1⟨1⟩⟩⟩⟩, broken certificates rejected.
compute/replay_census.py: 2,367/2,367 PASS in 36 s; 22-oval subset = exactly the 38 bold Table-1 schemes; fast evaluator agreed on every certificate.
compute/record.py: the 89-row Orevkov/Geiselmann Table 1 as data; self-checks 89 = 38+12+39, Gudkov–Rokhlin on every row, per-(p,n) bold counts 5/11/14/8/0.
Search residue (8 workers × 25 min, finished 06:14 UTC; full counts in ATTACK.md and compute/found_schemes.json): 595 certified triangulations, 2,112,073 sign evaluations, 902 distinct schemes. All 902 survived exact re-verification — zero witnesses where the fast evaluator and the Fraction engine disagreed, which is the one thing this run does establish about the search code. All 902 are already in the census. Nothing new, no bugs.
The shape of the failure is the interesting part. The annealer saturates oval count at depth 2 (it found the two 22-oval M-schemes ⟨18⊔1⟨3⟩⟩ and ⟨17⊔1⟨1⟩⊔1⟨2⟩⟩, both depth 2 and both already known T-realizable) and saturates depth at 3, but never both at once: best depth-3 scheme was ⟨7⊔1⟨10⊔1⟨1⟩⟩⟩ at 20 ovals. The census has 22-oval depth-3 schemes, so this is the objective's fault, not the geometry's. Adding a nesting bonus (workers 6–8) widened the logged set (728/738/748 distinct vs 500–626) without reaching depth 4 — it buys diversity, not depth. A next attempt should anneal on a target-scheme edit distance, or construct the nest directly from the tropical side, rather than hoping a component-count objective wanders into ⟨4⊔1⟨2⊔1⟨14⟩⟩⟩.
Neither open deep-nest target appeared in any worker. Nor is either in the census. That is silence from ~2×10⁶ samples of a space of size $2^{45}$ per triangulation. It is residue, not a bound.

6. What is proved vs still open

Proved (replayed, certificates in-repo or re-derivable): the census counts of arXiv:2602.06888 v3 are real; every published degree-8 patchwork certificate is convex, primitive, and produces the claimed scheme under an independent implementation.

Still open, untouched by tonight: algebraic realizability of the six Orevkov M-schemes; T-curve realizability of the 39; the true count of degree-8 real schemes. The search re-found 902 of the 2,367 and added none; the sign space alone is $2^{45}$ per triangulation and the triangulation space is astronomically larger, so absence of a hit is not evidence of absence. Outcome: residue.

Walkthrough, second session — the maximal stratum

7. What was missing the first time

The first session read arXiv:2602.06888 v3 for its numbers — 2,367, 38, 12, 39, and Theorem 21 — and then went searching in $\{\pm1\}^{45}$ . Section 3 of the same paper says that search is the wrong space.

Their Theorem 13 is Haas' theorem: a T-curve is maximal if and only if some collection 𝔖 of compatible Harnack splits is valid for it, meaning the triangulation refines the zone decomposition and the signs are Harnack on each zone with neighbouring zones differing. And §3.5: $\mathcal{C}(\mathfrak{S})$ does not depend on which unimodular triangulation refines 𝔖.

So the M-curves are indexed by a finite combinatorial gadget. All the annealer was ever doing was blindly hunting for members of a stratum that has an exact description.

8. The click: twists commute, so the stratum is a linear subspace

Their surgical twist (Lemma 14) is stated as $\sigma \mapsto \epsilon + \sigma\circ s_{ij}$ on $Z^+$ , which reads like an involution you have to apply in some order. But the extension rule (4) gives $\sigma(s_{ab}(x,y)) = \sigma(x,y) + ax + by$ , so the twist of a split $S$ is nothing but

$\sigma \;\longmapsto\; \sigma + \mathbb{1}_{Z^+_S}\cdot(\epsilon_S + i_Sx + j_Sy) \pmod 2 .$

Adding a fixed vector $\delta_S \in \mathbb{F}_2^{45}$ . Twists therefore commute, order is irrelevant, and a whole collection acts by $\sum_{S\in\mathfrak{S}}\delta_S$ . Combined with Haas' theorem: on a fixed triangulation $\mathcal{T}$ , every maximal sign distribution lies in

$\eta \;+\; \operatorname{span}_{\mathbb{F}_2}\{\delta_S : S\text{ a Harnack split all of whose edges are edges of }\mathcal{T}\}.$

Ranks for the 184 census triangulations are 6–26. A rank-18 subspace is 262,144 sign vectors — a few minutes. Sweeping it is an exhaustive classification of the M-curves that triangulation supports, which no amount of annealing can deliver.

The measured payoff is blunt. Annealing: $2.1\times10^6$ sign vectors, 22 ovals reached twice. Subspace sweep, first pilot: 20,480 evaluations, 20,480 of them 22-oval M-curves.

9. Three checks before believing any of it

$\eta(x,y)=(x+1)(y+1)$ on a random unimodular refinement gives ⟨18 ⊔ 1⟨3⟩⟩ — Proposition 10, and it does.
Each of the 1,254 Harnack splits of $T_8$ , applied alone, must give an M-curve. All 1,254 give 22 ovals; the six schemes produced are ⟨18⊔1⟨3⟩⟩, ⟨17⊔1⟨1⟩⊔1⟨2⟩⟩, ⟨13⊔1⟨1⟩⊔1⟨6⟩⟩, ⟨14⊔1⟨7⟩⟩, ⟨17⊔1⟨2⊔1⟨1⟩⟩⟩, ⟨10⊔1⟨11⟩⟩ — all bold in their Table 1.
The real test: take the 38 published 22-oval .pcom certificates and solve $\eta+\sum\delta_S=\sigma$ over $\mathbb{F}_2$ (up to the eight equivalent sign functions). All 38 decompose, with collections of size 3–15. Their data and this implementation of Haas' theorem agree everywhere they can.

10. Named false starts, second session

Refine first, ask about regularity later. The first refinement routine took a maximal crossing-free set of primitive segments containing the split edges — a perfectly good unimodular triangulation, and Haas' theorem needs nothing more. Then twelve attempts to find a lifting for one, via Agmon–Motzkin–Schoenberg relaxation on the local-convexity system, returned nothing. The relaxation was fine: run on the standard triangulation it recovers $h=i^2+ij+j^2$ with slack exactly 1. Random unimodular triangulations of $T_8$ are simply almost never regular, and only regular ones are algebraic curves. Fix: stop refining and start from triangulations that are already certified — the census's own 184, and gen_triang's certified random ones — and read the splits off them. Every witness in runs3/ is then regular by construction.
Enumerating collections directly. 1,254 splits, 381,957 compatible pairs, maximal compatible collections of size 15–24: the clique space is hopeless and sampling it (mode haas) was the worst performer in the session, ~6 distinct schemes per hour. The subspace sweep on a fixed certified triangulation dominates it, because the triangulation does the compatibility bookkeeping for free — two edges of a triangulation never cross.

11. Second-session residue

A dent. ⟨4 ⊔ 1⟨5⟩ ⊔ 1⟨10⟩⟩, 21 ovals, one sign flip at $(6,1)$ from their own (5v1(5)v1(10)).pcom — certified convex on their integer MIN_WEIGHTS, scheme recomputed from scratch, and absent both from our replayed census and from their archive's file names. Their §5.3 lower bound 2,367 becomes ≥ 2,368. (compute/certs/new_schemes.json, compute/verify_new.py.)
No M-scheme dent. Nothing outside their 38 has appeared, and neither open $(19,3)$ deep nest. Not an obstruction: the sweeps cover a prefix of the 184 census triangulations, and the census triangulations are exactly the ones their own search already favoured, so re-finding their 38 is the expected outcome. What is new is that the answer per triangulation is now exhaustive rather than sampled, and the tool to ask the question of any triangulation now exists.

12. Named false start, third session: laminar depth is not nesting depth

The deep-nest search changed the search space, and that part was right. Every earlier pass swept sign vectors on a fixed triangulation; Haas' theorem (their §3.5) says the scheme of a maximal T-curve depends only on the compatible collection $\mathfrak{S}$ of Harnack splits, not on which unimodular refinement you pick. So the triangulation stops being an input and becomes a by-product, and the search runs over collections — which is the level at which nesting is supposed to be visible.

The structural hook: pairwise non-crossing chords have nested-or-disjoint zones, so the $Z^+$ zones of a compatible collection form a laminar family. Its depth bounds the nesting depth of $C(\mathfrak{S})$ . The plan followed immediately — to build a triple nest, drive the laminar forest deeper — and both beam and anneal in dn_sweep.py scored on exactly that.

Then the bound was actually measured, on the 38 published 22-oval certificates whose collections are already recorded:

zone depth	1	2	3	4	5	6	7	8	9
nest depth 2	6	5	3	7	3	1	—	1	—
nest depth 3	—	1	3	2	3	—	1	1	1

Zone depth runs 1 to 9. Nesting depth is 2 or 3 and never more. The two rows overlap almost everywhere: ⟨6 ⊔ 1⟨10 ⊔ 1⟨4⟩⟩⟩ gets a triple nest out of four splits at zone depth 2, while ⟨12 ⊔ 2⟨1⟩ ⊔ 1⟨5⟩⟩ spends fifteen splits reaching zone depth 8 and stays at depth 2. The bound is true and it is useless — it is not monotone in any direction the search can push, so a score built on it is close to a random walk. That is the whole explanation for the searcher's lack of traction; more compute would not have fixed it.

What the same afternoon's arithmetic gave instead was a much sharper picture of the target. Write a depth-3 M-scheme as ⟨ $a$ ⊔ 1⟨ $b$ ⊔ 1⟨ $c$ ⟩⟩⟩. Then $a+b+c=20$ , and counting ovals by depth gives $p=a+c+1$ , $n=b+1$ , so Rokhlin's $p-n\equiv k^2\equiv 0 \pmod 8$ at $k=4$ collapses to $20-2b\equiv 0$ , that is $b\equiv 2 \pmod 4$ . The published 38 realize $b=2,6,10,14$ — every admissible value except $b=18$ , and $b=18$ is $(p,n)=(3,19)$ , the class their Theorem 21 proves is not T-realizable at all. So the depth-3 family is saturated: the census already has every $b$ a T-curve can have.

Which means the two open nests, both at $b=2$ , are not asking for a new congruence class or a deeper nest. They are asking to move $a$ from 10 or 17 (published) to 4 or 14 (open) inside a row that is already occupied. Nothing here decides that — but it is a far better-posed question than "find a deeper nest", and it is the one the next attempt should score against.

13. Named false start, q1: even twists cannot leave the five

The 23 August note said to score a walk by $\lvert a-4\rvert$ or $\lvert a-14\rvert$ with $b$ pinned at 2, using only even Harnack splits (paper Theorem 17: even splits keep $(p,n)$ ). That walk was run. It is the wrong handle.

Start from any of the five published (19,3) collections, including the Prop 31 bow-tie that produces ⟨17⊔1⟨2⊔1⟨1⟩⟩⟩. Add, drop, or swap even splits only, and never drop an odd split. Two hundred thousand collections later the schemes are exactly those five:

⟨18⊔1⟨3⟩⟩, ⟨17⊔1⟨1⟩⊔1⟨2⟩⟩, ⟨17⊔1⟨2⊔1⟨1⟩⟩⟩, ⟨16⊔3⟨1⟩⟩, ⟨10⊔1⟨2⊔1⟨8⟩⟩⟩.

No $a=11,\dots,16$ . One even drop from the $a=10$ nest jumps straight to the $a=17$ nested box. The bow-tie path itself is even — every split's vertex parities include $(0,0)$ — so it was already inside the even-walk graph. The open nests are not one even move from any published (19,3) collection.

The first BFS that did drop odd splits left the (19,3) row immediately. That is Theorem 17 in the other direction, and it is why the pin matters.

14. The useful failure: the $a=10$ nest already has seven odds

Odd collections of size at most 3 were enumerated exhaustively (189 odd splits, 368,936 compatible collections). They produce twelve known M-schemes and neither open nest. That looked like a wall until the published $a=10$ collection was opened: it is twelve splits, of which seven are odd. Size 3 cannot see the one published depth-3 (19,3) T-curve, so silence there is not evidence about $a=4$ or $a=14$ .

Remark 20 says nested odd zones cancel, so a nested odd pair can stay on (19,3) while changing the odd skeleton. Adding every such pair to each of the five published (19,3) collections is a 122-curve check. The $a=10$ nest admits zero compatible nested odd pairs. The other four seeds produce only known M-schemes. So the open nests are not a canceling pair away from a published (19,3) collection either.

A stronger negative, still not an obstruction: the one-split add/drop/swap neighbourhood of all 38 published M-collections is exactly those 38 schemes (28,861 evaluations). Collection-distance from the published M-set to either open nest is at least 2.

15. What the leftover thicken actually said through rank 20

The 23 August leftover was four rank-16 triangulations. Cheap first, then the rest of the intended leftover: every census triangulation of twist-rank at most 20, radius 1 around the whole Haas maximal stratum. That is 164 of 164, 1,438,512,000 sign distributions, exactly $46\cdot 2^r$ on each. Every scheme outside the published 2,367 is already among the seventeen. Rank 20 is fifteen triangulations, including the two published (19,3) liftings (17v1(1)v1(2)).pcom and (17v1(2v1(1))).pcom. Both of those finished; neither open nest is on that neighbourhood of either lifting. One other rank-20 certificate, (6v1(6v1(8))).pcom, rediscovered seven of the seventeen and nothing new.

That is a neighbourhood statement, not an obstruction. The open nests may sit at Hamming distance greater than 1 from the maximal Haas stratum, or on a triangulation that is not among the 184 census ones. Ranks 21–26 (twenty triangulations) were never in the leftover.

16. Rank 21 of that leftover, then the next collection moves

The five leftover triangulations of twist-rank 21 are now finished the same way: radius 1 around the whole Haas maximal stratum, $46\cdot 2^{21}$ sign distributions on each, 482,344,960 in all. Novel versus the published census is empty. None of the seventeen reappears either — those five certificates simply do not see the schemes that sit off the census. The two published (19,3) liftings were already rank 20. The only (19,3) certificate in ranks 21–26 is the Harnack triangulation at rank 23.

So the leftover thicken is no longer "ranks 21–26 were never tried." It is "ranks 21–26 are finished neighbourhoods and added nothing." That is still not an obstruction. Hamming distance greater than 1, or a triangulation outside the 184, remains possible.

The Harnack triangulation itself, radius 1 around the whole Haas maximal stratum, produced 26 schemes already in the published census. The other five rank-23 leftover certificates, all three rank-24 leftover certificates, and the one rank-26 leftover certificate did the same.

Collection space was pushed one step past q1 and stayed quiet. Four hundred thousand even twists, then 1.2 million, still produce exactly the five published (19,3) M-schemes; the queue is still open. Dropping at most one split and adding three, from every published depth-3 M-collection, now finishes: twenty-nine known M-schemes and neither open nest. That includes the five published (19,3) collections that q2 left under a time cap. Odd collections of size 4 finished and stayed on the same twelve M-schemes as size ≤ 3. The published $a=10$ nest still has seven odd splits; size 4 can see a larger odd part than size 3, but it did not see a new M-scheme, and silence there is still not evidence about $a=4$ or $a=14$ .

17. Named false start, leftover even components: the hole is skipped

Size 5 odd collections are now finished: 37,632,123 of them, the same twelve published M-schemes as size 4, neither open nest. The published $a=10$ nest still has seven odds, so size 5 was never going to see it. What size 5 does show is that no new M-scheme has an odd skeleton of size exactly 5.

The even components that can be enumerated say something sharper. Fix the seven odd splits of the published $a=10$ nest and take every compatible even subset: 126,336 collections, and the schemes are exactly $\langle 10\sqcup 1\langle 2\sqcup 1\langle 8\rangle\rangle\rangle$ and $\langle 17\sqcup 1\langle 2\sqcup 1\langle 1\rangle\rangle\rangle$ . Fix the nested-box odds instead: 25,292,736 collections, and the scheme is only the nested box. There is no intermediate $a$ .

The same skip happens on the (7,15) hole that the hole-map had marked as both-neighbours-realised. The even component of $\langle 5\sqcup 1\langle 5\rangle\sqcup 1\langle 10\rangle\rangle$ contains $b=2,4,5,7$ and not $b=6$ . Even twists jump over the missing rung rather than filling it. That is why a BFS that only adds and drops even splits, however long, cannot be the construction that realises $\langle 5\sqcup 1\langle 6\rangle\sqcup 1\langle 9\rangle\rangle$ or the two open (19,3) nests from the published collections that share their odd skeleton.

18. One diagonal changes the available scheme

The missing degree of freedom was the triangulation. The earlier completed sweeps varied signs or compatible split collections in fixed neighborhoods. This time we kept a much smaller sign ball and changed one diagonal first. The interrupted attempt had written that route down but had left no saved search outputs.

The first candidate appeared after 82 balls. Start with the paper's ⟨5 ⊔ 1⟨3⟩ ⊔ 1⟨12⟩⟩ certificate, replace the diagonal joining (2,3) to (6,0) with the one joining (3,2) to (5,1), and change the signs at (0,0), (0,1) and (7,0). The recomputed scheme is ⟨3 ⊔ 1⟨3⟩ ⊔ 1⟨12⟩⟩. It has two fewer exterior ovals and is absent from the paper's archive and our seventeen additions.

There was still a gap between that picture and an algebraic curve. The old lifting violates two inequalities after the flip. The projection method in haas.regularize then failed to find a new lifting in 200,000 steps. That failure was useful because it isolated the remaining question: could this particular mesh carry a strict convex lifting? It did not answer that question negatively.

A linear feasibility solver found one. Clearing its denominators produced small integer heights, and exact arithmetic checked all 2,688 inequalities with slack at least 4. At that point Viro's patchworking hypothesis was satisfied. A second implementation in Rust recovered the same nesting tree by connecting monochromatic vertices and identifying the outside region, independently of the Python segment-tracing code.

The certificate proves that the number of nonempty degree-eight T-curve schemes is at least 2,385. The search log records 1,248,532 evaluations, but the proof needs only this one witness. We stopped there, leaving 1,108 planned balls unsearched. The construction has twenty ovals; neither open maximal deep nest is decided.

19. The leftover balls were the same move, not a new idea

What was missing was not another neighbourhood of signs on a fixed mesh. The q8 search had written down 1,190 one-flip balls and then stopped at the first new scheme. The leftover 1,108 balls were already a finite handle: the same plan hash, the same seeds, just the tasks that had not been run.

The useful failure was the extra domain. Flipping the new q8 certificate itself, then changing at most three signs, produced 513 schemes and nothing outside the 2,385. Iterating on the new mesh was the wrong next step. The productive direction was the rest of the original list, starting with the unused flips of nearby (7,15) M-certificates.

The click is that the a=3 hole is a pattern, not a one-off. ⟨a ⊔ 1⟨3⟩ ⊔ 1⟨12⟩⟩ was missing a=3 and present at a=0,1,2,4,5. ⟨a ⊔ 1⟨5⟩ ⊔ 1⟨10⟩⟩ was missing a=3 in exactly the same way. Task 107, from the paper's ⟨5 ⊔ 1⟨5⟩ ⊔ 1⟨10⟩⟩ certificate, fills it. Two other meshes, from ⟨8 ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨7⟩⟩ and ⟨8 ⊔ 2⟨1⟩ ⊔ 1⟨9⟩⟩, produce four further schemes that are likewise absent from the census and the prior additions, including a three-step family ⟨5 ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩, ⟨4 ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩, ⟨3 ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩ on one triangulation.

Computer search: six workers finished all 1,108 leftover balls (16,870,408 evaluations) and a seventh finished the 21 extra flips of the q8 certificate. Eleven tasks emitted a scheme outside the 2,385; five of those schemes are distinct. One-flip balls of those five certificates (115 balls) then filled ⟨2 ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩ and a parallel family ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨9⟩⟩ at a=2,3,4. Each got an integer lifting from a linear feasibility solve, then the same two independent topology checks as q8. The (19,3) block of the leftover plan (123 balls) was among the finished tasks and hit neither open nest.

What is proved: nine further nonempty degree-eight T-curve schemes, so the lower bound is at least 2,394, and the 1,190-ball q8 domain is now completely searched. What remains open: Hilbert 16(a) in degree eight, and both undecided (19,3) deep nests.

2026-09-20 — leftover seeds, not leftover balls

What was missing was not another 1,190-ball plan. q9 had already finished that domain and then one-flip of its first five certificates. The four schemes that followup produced were never used as seeds. That is a smaller copy of the q8 stop: a finite list of unimodular flips, written down, half-run.

The useful observation is the a-hole again. ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨9⟩⟩ occupied a=2,3,4 after q9 and not a=1. Task 15, from ⟨2 ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩, fills a=1. The first ball, on the same seed, also opens a new family ⟨a ⊔ 2⟨1⟩ ⊔ 1⟨10⟩⟩ at a=1,2,3.

The click is that feeding the new certificates back as seeds is the same move q9 made once and then dropped. One-flip of the six leftover certificates fills a=0 of ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨2⟩ ⊔ 1⟨9⟩⟩, a=4 of ⟨a ⊔ 2⟨1⟩ ⊔ 1⟨10⟩⟩, a=1 of ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨4⟩ ⊔ 1⟨9⟩⟩, and ⟨2 ⊔ 1⟨1⟩ ⊔ 1⟨5⟩ ⊔ 1⟨9⟩⟩. The seven followup certificates were not fed back; that leftover is recorded, not searched.

Computer search: two workers finished all 106 leftover balls (1,613,956 evaluations). A second pair finished 169 one-flip balls of the six leftover certificates (2,573,194 evaluations). Thirteen distinct schemes outside the 2,394, each with an integer lifting and the same two independent topology checks as q8/q9.

What is proved: thirteen further nonempty degree-eight T-curve schemes, so the lower bound is at least 2,407. What remains open: Hilbert 16(a) in degree eight, and both undecided (19,3) deep nests.

2026-09-22 — the seven unused followups

The missing degree of freedom was again the unused followup certificates, not a larger radius. q10 finished one-flip of six leftover meshes and left seven schemes unseeded. That list is 205 unimodular flips, written down before the search.

A false start is treating search failure on either open nest as information: every ball here is 16–19 ovals. The useful failure is that the a-holes keep filling from the same one-flip move. ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨9⟩⟩ occupied a=1,2,3,4 after q10; task 17 fills a=0. The same mesh opens ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨10⟩⟩ at a=1,2.

The click is feeding those five leftover certificates back as seeds, the move q10 made once and then dropped. One-flip of the five fills ⟨1 ⊔ 1⟨1⟩ ⊔ 1⟨4⟩ ⊔ 1⟨10⟩⟩, a=0 of ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨4⟩ ⊔ 1⟨9⟩⟩, and a=0 of ⟨a ⊔ 1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨10⟩⟩. The three followup certificates were not fed back; that leftover is recorded, not searched.

Computer search: two workers finished all 205 leftover balls (3,121,330 evaluations). A second pair finished 150 one-flip balls of the five leftover certificates (2,283,900 evaluations). Eight distinct schemes outside the 2,407, each with an integer lifting and the same two independent topology checks as q8–q10.

What is proved: eight further nonempty degree-eight T-curve schemes, so the lower bound is at least 2,415. What remains open: Hilbert 16(a) in degree eight, and both undecided (19,3) deep nests.

2026-09-26 — the three unused followups finish empty

The missing degree of freedom was again the unused followup certificates. q11 finished one-flip of five leftover meshes and left three schemes unseeded. That list is 93 unimodular flips, written down before the search.

A false start is expecting the a-hole pattern to continue automatically. q9–q11 kept filling a=0 of a neighbouring family by feeding leftover certificates back as seeds. Here the same move on ⟨1 ⊔ 1⟨1⟩ ⊔ 1⟨4⟩ ⊔ 1⟨10⟩⟩, ⟨1⟨1⟩ ⊔ 1⟨4⟩ ⊔ 1⟨9⟩⟩, and ⟨1⟨1⟩ ⊔ 1⟨3⟩ ⊔ 1⟨10⟩⟩ produced 552 schemes, all already among the 2,415. No new a-hole filled.

The useful failure is that this one-flip followup chain now has no unused leftover certificate. Two-flip, radius four, or a different seed family would be a new handle, not a leftover of this one.

Computer search: one worker finished all 93 leftover balls (1,416,018 evaluations). Zero schemes outside the 2,415. No integer lifting was required.

What is proved: the 93-ball neighbourhood is empty of new nonempty schemes. The lower bound stays at least 2,415. What remains open: Hilbert 16(a) in degree eight, and both undecided (19,3) deep nests.