Walkthrough — Seymour's second-neighborhood conjecture
0. What was actually missing
The 2026 computational paper (Halkiewicz, arXiv:2601.21563v3) enumerates Pisa graphs — strongly connected oriented graphs with — through 7 vertices and stops. It conjectures that every Pisa graph is a directed cycle or an orientation of minus a matching. The next finite handle is the census. A new bound is a complete type list at that order, or even one certified Pisa graph whose missing graph is neither a matching nor (as their table actually found) regular.
The degree-threshold line (Kaneko–Locke , Sadhukhan–Sandeep–Sen , Brukhman ) is a different problem. Isolated random-graph statistics are an incomplete search.
1. Named false starts
Treat Halkiewicz Conjecture 5.1 as the published record. Their own Table 2 lists 720 Pisa graphs on 7 vertices whose missing graph is 2-regular. Guo–Kang–Zwaneveld's directed cycle-square is one of them. The matching conjecture is already false at the order they computed. What they actually established is a regular-missing census through .
Treat Eulerian as a bound. Brukhman's counting becomes equality precisely when the graph is Eulerian of degree on vertices (missing a 2-factor). CP-SAT says there is no counterexample for . But Kaneko–Locke already covers every oriented graph with a vertex of outdegree , so those UNSAT results are a check of the encoding, not a new threshold. The first Eulerian case not implied by Kaneko–Locke is , .
Cadical instant UNSAT. A first CNF encoding reported UNSAT in a millisecond with an empty DRAT file, through . Cardinality auxiliaries were colliding (CardEnc reused variable ids). After passing top_id, the same instance is SAT for “Eulerian only” and for (cycle-powers exist) and UNSAT for . The bogus proofs were discarded.
CEGAR that blocks whole orientations. Asking SAT for an irregular-missing Pisa graph and forbidding each non-strong model as a full assignment looped for 200 rounds. Switching to source-SCC cuts (force an incoming arc into the source component) found a witness immediately.
Labeled n=8 enumeration. is about 130 core-hours at the n=7 rate. Not tonight.
2. The useful failure
The Eulerian equality case is still the right picture, even if it is not a new bound at small . In a hypothetical Eulerian counterexample on ,
- every trap has capacity exactly 3,
- is forced to be the two missing neighbours of plus one in-neighbour ,
- is a fixed-point-free permutation with no 2-cycles,
- the two missing neighbours of together with form a directed triangle,
- so the missing 2-factor has girth at least 4.
That skeleton is what the SAT instances are killing. It is also why should first grow new missing types that are 2-regular or “2-regular plus a chord”.
Replaying n=6 and n=7 with an independent C enumerator (enum_pisa.c) matched Halkiewicz's labeled counts exactly, and added one fact he did not state: every Pisa graph on 7 vertices is Seymour-tight (all 4080), while only 270 of 1050 on 6 vertices are.
3. The click
Halkiewicz's n=7 rigidity (missing graph regular) is an accident. Already at n=6 the missing graph can be a matching, which is irregular as a degree sequence. The question at n=8 is whether non-matching irregular missing graphs occur.
They do. A CEGAR SAT model with source-component cuts produced a strong oriented graph on 8 vertices with
- missing edges ,
- missing-degree sequence ,
- margins ,
- .
The missing graph is a 4-cycle plus a , not a matching and not regular. seymour.py recomputes first and second neighbourhoods from the arc list and confirms Pisa. The same search produced a non-tight 2-regular-missing example (K_8 minus a 2-factor) and recovered the tight cycle and cycle-square.
So the n=8 census is not “the same two families plus matchings”. New types start immediately.
4. The argument, in the order it was found
- Fetch Espuny Díaz et al., Halkiewicz v3, Sadhukhan–Sandeep–Sen, Brukhman, Guo–Kang–Zwaneveld.
- Implement bitmask first/second neighbourhoods. Replay and lex products.
- Exhaust labeled oriented graphs for . Match Halkiewicz 1050 and 4080.
- Notice Conjecture 5.1 is already false at n=7; the live claim is regularity of the missing graph.
- Attempt Brukhman's next layer by SAT. Learn it is implied by Kaneko–Locke for .
- Search n=8 by missing-degree type. Tight cycle and cycle-square appear. Then a non-tight 2-factor. Then two irregular missing graphs.
- Run
geng -q 8through every undirected iso type (12346 graphs, 2.50 billion orientations of representatives). Exactly seven missing-degree types occur. SAT then supplied an explicit witness for the one type ( ) the CEGAR hunt had not been asked for.
The n=8 irregular witness is checked as follows. Vertices . Arcs:
0→5, 0→6,
1→0, 1→5, 1→6,
2→0, 2→1,
3→0, 3→1,
4→2, 4→3, 4→7,
5→4, 5→6, 5→7,
6→4, 6→7,
7→2, 7→3.
Missing: the C_4 and the diamond with chord . Every vertex has , the maximum margin is 0, and the digraph is strongly connected. Certificate: compute/certs/n8_irregular_pisa.json, ternary code 145923119419.
5. Computer search
- Labeled census:
certs/pisa_n{4,5,6,7}.json. n=7 took 217s, graphs. - Geng types n=6,7:
certs/pisa_geng_n{6,7}.json, types agree with the labeled census. - Geng n=8, finished:
certs/pisa_geng_n8.json,certs/pisa_n8_types.json. Seven types, 162 Pisa orientations of representatives. - SAT witnesses, independently rechecked:
certs/verified_witnesses.json. - Eulerian UNSAT (OR-Tools ; Cadical after the encoding fix). DIMACS in
certs/eulerian_n*_n2le*.cnf. This does not beat Kaneko–Locke. - OR-Tools unknown at 300s.
6. What is proved vs still open
Proved tonight (machine-checked, independently replayable).
- Halkiewicz's labeled Pisa counts for are correct. Every n=7 Pisa graph is Seymour-tight.
- There exist Pisa graphs on 8 vertices whose missing graph is not a matching and is not regular. Explicit certificates above.
- There exist non-tight Pisa graphs on 8 vertices whose missing graph is 2-regular.
- Census. Every Pisa orientation of an 8-vertex graph, up to undirected isomorphism of the underlying graph, has missing-degree sequence in . No tournaments, no 4-regular missing graphs, no incomplete matchings.
Still open.
- Seymour's conjecture.
- The full labeled n=8 count (162 is a representative count, not a labeled one).
- Eulerian for .
- Any hereditary class beyond those already in Fidler–Yuster / Ghazal / Dara / Daamouch.