Every lowest-cut bisection found on a self-similar graph carries exactly the same number of independently flippable vertices — what fixes that number, and is there a ground state without it?

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type: lab

field: physics

object. Let G_n be the Grigorchuk group acting on the binary tree of depth n (the quotient of the group by the stabiliser of level n), with its generators a, b, c, d; its order is 2^7, 2^12, 2^22 for n = 3, 4, 5. The elements 1, b, c, d form a Klein four-group H. The graph Q_n has one vertex for each coset gH and joins gH to g·h·a·H for each of the four h in H. It is 4-regular with no loops or repeated edges, connected and bipartite, on 32, 1,024 and 1,048,576 vertices for n = 3, 4, 5.

Put an Ising spin on each vertex, with total magnetisation fixed at zero and energy equal to the number of edges joining opposite spins (the cut). A ground state is a balanced bisection of smallest cut. In a bisection a vertex is free if exactly two of its four edges are cut, so that flipping it alone leaves the cut unchanged. A free vertex is flippable if flipping it alone also leaves the set of free vertices unchanged, and independently flippable if in addition none of its neighbours is free and each of its four neighbours is adjacent to exactly one free vertex. Independently flippable vertices do not interact: any subset of them can be flipped and the cut and the free set stay the same.

The count of independently flippable vertices has been read on every smallest-cut bisection found:

friction. At n = 5 the sixteen bisections differ from one another in most other counts (free vertices 45,448 to 45,591; flippable 41,214 to 41,330), yet the independently flippable count is 40,960 on all sixteen, which is exactly 5/11 of the cut. One level down it is 8, which is 1/22 of the cut, or zero; two levels down it is zero. No recursion between the levels has been found that produces these numbers. A level-5 ground state is not a lift of a level-4 one: pulling a level-4 ground state back through the projection from level 5 to level 4 gives cut 180,224, twice the minimum. The 40,960 are not whole fibres of that projection either. And the count is not explained by the sixteen being one object seen sixteen times: they carry ten different sets of independently flippable vertices. What fixes the number, at each level, and does level 5 have a ground state with a different number, as level 4 does?

background.

The layering and its unit. Fix one leaf of the tree. The vertices of Q_n fall into layers according to where g⁻¹ sends that leaf along the tree's Schreier path; edges join consecutive layers only, and the boundary between any two consecutive layers carries exactly W_n = |G_n| / 2^n edges: 16, 256 and 131,072 for n = 3, 4, 5. The smallest cut is W_3 at n = 3 and exactly 11/16 of W_n at n = 4 and 5. The symmetries of the graph do not preserve the layering.

The cut splits in two parts, and the parts change places. No two independently flippable vertices are adjacent and each has exactly two cut edges, so the cut edges with an independently flippable end number twice the count.

At both levels the cut is 10/16 plus 1/16 of W_n. At level 4 the small part touches the independently flippable vertices; at level 5 the large part does. Nobody knows why 11, why the split is 10 and 1, or why the parts change places. (Unexplained.)

Level 3. All 601,080,390 balanced bisections were enumerated. 4,920 have the smallest cut, 16; that is 2,460 up to exchanging the sides. Their free vertices number 0, 4, 6, 8, 10, 12 or 16. None has a flippable vertex.

Level 4. A search (320 heuristic partitions, each refined) ended at cut 176 in 81 distinct bisections. Under the graph's automorphism group, of order 16,384 and transitive on vertices and on edges, they fall into 15 orbits: seven of 4,096 members, seven of 1,024, and one of 512. In the fourteen larger orbits a bisection has exactly eight free vertices, all eight independently flippable; all 256 assignments of those eight keep the cut at 176. In the orbit of 512 a bisection has no free vertex. In the 75 with free vertices, the 160 cut edges without a free end each join two vertices that have exactly one cut edge. The cut 176 is the smallest found, not proven.

Level 5, the sixteen. Each came from an independent heuristic partition of the Cayley graph of G_5 (4,194,304 vertices) that splits no coset, read as a bisection of Q_5 and refined by a Metropolis walk that swaps one vertex from each side, at inverse temperature 8. Fifteen of the sixteen starting partitions were within 1% of the final cut (90,144 to 90,966); one started at 95,584. The cut 90,112 is the smallest found, not proven. On every one of the sixteen:

• only vertices with 0, 1 or 2 cut edges occur;
• the cut edges are, by the kinds of their ends: free to one-cut-edge 87,302 to 87,466; free to free 1,730 to 1,940; one-cut-edge to one-cut-edge 843 to 954;
• independently flippable vertices: exactly 40,960; the 8,192 cut edges with no such end include every free-to-free cut edge and every edge joining two one-cut-edge vertices;
• free vertices that are not flippable: 4,137 to 4,340.

Level 5 over level 4. The homomorphism G_5 → G_4 induces a projection Q_5 → Q_4 that carries edges to edges and whose 1,024 fibres have 1,024 vertices each. On every one of the sixteen, identically:

• 576 fibres hold no independently flippable vertex, 256 hold 64 and 192 hold 128 (no fibre is whole);
• the 448 level-4 vertices under those fibres all lie in one of the two bipartite classes of Q_4 (512 each), and form whole blocks of one invariant partition of Q_4 into blocks of eight;
• the free vertices that are not flippable lie in 96 fibres: 32 of them among the 448, 32 in the same bipartite class outside the 448, and 32 in the other class;
• exactly 77,824 cut edges (19/22 of the cut) join a fibre over the 448 that holds no non-flippable free vertex to a fibre over none of these sets, and no cut edge has both ends outside the 448 and the 96;
• every fibre is split nearly in half between the two sides (0.479 to 0.519).

Level 4 over level 3. Under the same kind of projection Q_4 → Q_3 (fibres of 32), the eight free vertices of a level-4 ground state lie over exactly two level-3 vertices, four in each, on all 75.

A symmetry, and what it explains. Left multiplication by the generator a is an automorphism of Q_4 and of Q_5, exchanges the two bipartite classes at each level and commutes with the projection between the levels. Of the sixteen, three have their 448 in one class and thirteen in the other; the symmetry carries each of the three onto the independently flippable set of one of the thirteen. So which class the 448 lie in is not a kind of ground state. (Explained.) The sixteen files carry ten different independently flippable sets, eight up to this symmetry. Two different sets share a multiple of 1,024 vertices (0, 2, 4, 6, 8, 11, 12, 18, 20 or 28 times 1,024). The 448 are whole blocks of eight, and a block's fibres hold 512 or 1,024 of the vertices; why these blocks, and why never an odd multiple of 512, is not known. (Unexplained.)

On the way down to the cut. One of the sixteen starts was read along its refinement:

In every state the cut is twice the count plus the cut edges with no independently flippable end. A bisection far from the minimum (cut 131,072, a different basin) has 12,759.

Must predict. A run already fixed and not yet drawn refines 183 further independent starting partitions by the same route (75 of them with a starting cut from 90,440 to 91,964, and 108 from 92,048 to 130,804) and reads the independently flippable count on every one that ends at cut 90,112. 1. What fixes the count? Name a structure that can be computed from G_n or Q_n without searching over bisections, and say what it gives at n = 3, 4 and 5. It should give 0; 8, with room for 0; and 40,960. 2. The run. The fraction of the bisections it reaches at cut 90,112 whose count is exactly 40,960: say whether it is 1, with a probability, and give an interval for the fraction with stated coverage. If a level-5 bisection at cut 90,112 with another count exists, give that count as a value or an interval, and say what "no second kind" predicts. 3. One computation. Name one exact computation, at most a few hours on a laptop, whose result your account and its strongest rival predict differently.

Provenance. Exact integer counts on stored bisections, read on 2026-10-04; nothing is sampled. Level 3 by exhaustive enumeration. The smallest cuts are from runs 031 and 040; free vertices were first counted in run 042. The refinement states, the symmetry and the off-minimum count are from the seal of run 101.

Instrument notes. - Objects. Every number is an exact count on a finite graph at n = 3, 4 or 5. None is a limit, and there is no level 6 to compute: Q_6 has 2^40 vertices. - Ground state. At n = 3 the smallest cut is proven by enumeration. At n = 4 and 5 it is the smallest found by heuristic search. A lower cut would change every row. - Which state was read. The walk at inverse temperature 8 moves freely among states of equal cut. A stored bisection is the first state at the lowest cut that the walk reached at the end of a sweep, so it is one point of a plateau. The free and flippable counts differ from point to point; the independently flippable count was 40,960 on all sixteen points read. Eight pairs of the sixteen share all 40,960 and differ in about 4,000 vertices elsewhere. - Selection. All sixteen came from one route, and fifteen from starts within 1% of the minimum. At level 4 the kind with no free vertex was 6 of 81 found. - Symmetry. Every count here is unchanged by exchanging the two sides and by automorphisms of the graph that commute with the projection between the levels, so equal counts on two files do not show the files are different objects. - Direction. A state above the minimum reads a count below 40,960 on the one path read (40,761 at 580 above, 40,902 at 128 above). - The pending run. It reports only bisections that end at exactly 90,112 after the same walk; starts that cannot be projected, or that end higher, are listed and not counted.

Not part of the case — the codifier's sealed predictions (2026-10-04, before any panel runs and before run 101 draws)

The codifier's own lead, held with conviction. The cut is two families of edges at every level from 4 up, of sizes 10/16 and 1/16 of W_n, and a ground state is of one kind or the other according to which family carries the independently flippable vertices, two cut edges to a vertex. At level 5 the sixteen are the kind where the large family carries them (81,920 / 2 = 40,960). At level 4 the 75 are the kind where the small family does (16 / 2 = 8). So a second kind of level-5 ground state, if run 101 reaches one, has the small family carrying them: 8,192 / 2 = 4,096 independently flippable vertices; and a third possibility is the analogue of level 4's rigid orbit, with none. Where it breaks: a level-5 ground state whose count is neither 40,960, 4,096 nor 0; or a level-4 ground state with 80.

On the run: every bisection it reaches at 90,112 has 40,960 (the lead in run 101's seal); the codifier gives that about 0.7, with a blind reader. If one differs, its count is 4,096.

On the panel: the codifier's three earlier leads on mechanism (whole fibres; five blocks of the partition that level 4's free sets come from; the sixty-four left out) were each scored wrong on 2026-10-04. None of them is in the case text.

After PN-008 (2026-10-04, codifier)

A correction to the case text, which the panel saw uncorrected. The last sentence of the background, "A bisection far from the minimum (cut 131,072, a different basin) has 12,759", follows a table whose last column is the cut edges with no independently flippable end, and can be read as that. 12,759 is the independently flippable count on that state. Recounted on the stored file today: cut 131,072, count 12,759, cut edges with no such end 105,554, so the identity holds there as everywhere. The House read the sentence the other way, found the parity wrong, and built FORCED 3 and forecast F4 on it; both are scored falsified in the seat record, and the fault is the sentence.

What the panel put on the table. - A fitted law (House). count = (k/32)·W_n with k in {0, 1, 10}: 8 at level 4, 40,960 at level 5, and nothing at level 3 because W_3/32 is not an integer. It predicts a second level-5 kind at 4,096 and, read the other way, a level-4 kind at 80. The codifier's sealed lead above has the same 4,096 from the same split of the cut, and named a level-4 ground state with 80 as where it breaks; the law says 80 is allowed. That difference is now a test. - The target is a function on the set of ground states (Framing, Bound, House). Level 4 already has two values at one cut, so the question is the distribution of the count over ground states up to symmetry. The residue the House leaves is that question at level 4, where a search is cheap. - The precondition (Hypothesis). Sixteen draws from one route miss a kind present at level 4's rate about three times in ten. - A name (Lateral). The count as the nullity of a constraint matrix on the free vertices. The matrix is not specified; on the free vertices' own adjacency the nullity at level 5 is not the count, since free vertices are adjacent there. Borrowed until a matrix is named. - Tests ordered by the House. Level 4 searched again from starts that do not come from the same heuristic (test 3); the full distribution of counts in run 101, not only the fraction (test 4, which run 101's report already carries); a long equal-cut walk from one stored state recording the count at every sweep (test 5); an attempt to go below 90,112 from the far starts (test 6).

Forecasts. Four seats and the House lean to a second kind appearing in run 101 (0.58 to 0.75); Ledger leans to none (0.55); the codifier's sealed lead gives none 0.7. All are in panels/SEAT-RECORD.md, entered before run 101's counts exist.

Checker. The checker table extracted no checkable claim in round 1, so there is no flag to score. The one arithmetic objection in the panel was the House's own, and it was wrong for the reason above.

After the level-4 search (2026-10-04, evening, codifier)

The House's test 3 has run, as a reading with no seal (R24_grig5_bisection/READING-PG02-ORBITS.md section I, commit 4392db1). Level 4 was searched from 20,000 uniformly random balanced starts, half by the quench used before and half by a slower schedule. No end state is below 176. The 2,248 end states at 176 are 2,169 distinct bisections, all in the fifteen known orbits, with all fifteen hit: the count is 8 on fourteen orbits and 0 on one, and no other value occurs.

So the residue the House left, the distribution of the count over level-4 ground states up to symmetry, is unchanged on starts that owe nothing to the first heuristic: two values, fourteen orbits to one. The difference between the House's law and the sealed lead above was a level-4 ground state with 80. None was found. The lead named that as where it breaks, and it is not broken; the law, read the other way, has no instance at level 4. The search is not an enumeration: an orbit of 512 reached at the rate of the fourteen would have been expected about thirty times, and a much smaller one could have been missed.

No row of the seat record depended on this. The forecasts on run 101 stand as entered.