Contents
Where this sits
The channels take the step, the clock and the existing Zero-Point Tri-Engine, and feed pairing and memory and the simulation page. The other two components on this page have charts of their own below.
From engines to channels
Every mechanism of the tri-engine pages keeps its timing and gets a solver meaning.
| Tri-engine | Solver channel | Status |
|---|---|---|
| Engine I, II, III on branches 0, 1, 2 (0°, 120°, 240°) | Channels A, B, C: three particle populations | DEFINITION |
| One instruction per joint cycle, onload at k = 2, offload twelve frames later at k = 6 | One hyperbolic step per joint cycle, carried by that packet | DESIGN |
| Exact round trip of each packet (lift, ghost map, fold) | H1: every corner and state checked on the sheet exactly | DESIGN |
| Latch a result, pair at the zero point (n = 0) | Each channel’s incumbent is certified and compared at every zero point | DESIGN |
| Timing clutch: hold a finished engine | Rest a stalled channel and stop spending its evaluations | DESIGN |
| Orthogonal membrane: remove the differential z, keep the common s | Recentre the channels on their common incumbent, logging the differential removed | DESIGN |
| Memory on the outer and inner manifolds, with parity and a Merkle chain | Best answers written as words; the parity recovers one erased channel record | DESIGN |
| The 720 → 72 → 10 pump on each engine’s gear | The pump on each channel’s exact trajectory of Lorentz displacements (H7) | DESIGN |
One step in one packet
A packet lasts twelve frames, from its branch’s onload (k = 2) to its offload (k = 6), one and a half turns of the eight-step clock. The solver’s work is laid on those frames so each stage of the clock does what its name says.
The step itself is done in the first half (d = 0 to 3): lift, corners, costs, centroid. The second half does the bookkeeping that makes the step trustworthy: the schedules update at the index stage, the new state is checked on the sheet at the zero point, the next rotation is seeded, Helix B certifies the incumbent during the second pass, and the offload writes the state and its displacement into the register.
The 24-frame schedule
The joint frame is n = F mod 24 (Proposition 3.7). Channel A onloads at n = 18, B at 10 and C at 2, each offloading twelve frames later, so every channel takes exactly one step per joint cycle: 30 per register, 90 across the three.
At most two packets are ever in flight, and onloads and offloads never share a frame, so each zero point sees three freshly certified incumbents without any two channels writing at once. Whether this beats a plain round-robin is the paper’s Open Problem 3, and it is tested here as conjecture C3.
Pairing and Helix B
The paper’s machine splits work between Helix A, which proposes, and Helix B, which certifies in exact arithmetic. The channels are Helix A. Nothing they propose is believed until Helix B has checked it.
Helix B’s checks are cheap and exact: decoding a point to a colour or a truth assignment is a sign or angle test, counting monochromatic edges or unsatisfied clauses is integer arithmetic, and the sheet check is one exact Lorentz form. A memory word holds the three channel records and the best, with the parity that lets one erased record be rebuilt (Prop 3.10) and the Merkle chain that locates an altered one, as on the convergence calculator.
Clutch and membrane
The convergence calculator’s three clutch modes carry over.
| Mode | On the calculator | For the solver |
|---|---|---|
| hold | A finished engine stops turning until the others land | A channel whose incumbent has not improved for m joint cycles rests: its packets still run on schedule but spend no evaluations, and its share of the budget goes to the others |
| free | Every engine turns every frame; the held results shear | Every channel steps every joint cycle regardless; the channels drift apart |
| membrane | Free-run, then at the zero point the differential z is removed and the common rotation s kept | At the zero point each channel’s population is moved by a boost toward the common incumbent (the Klein centroid of the three, weighted by certified cost); the differential removed is logged |
The membrane trades diversity for agreement. Too strong, and the three channels collapse onto one search; too weak, and they never share what they find. Its strength is a preregistered setting of benchmark B1, with an ablation that turns it off.
The pump over a register
A register is 720 frames: frames \(720(r-1)+1\) to \(720r\), as on the Zero-Point page.
On each frame a channel’s state either stays (the increment is the identity) or, at its offload, moves by its step’s displacement \(\Delta q_n = L_{x_n}L_{x_{n-1}}^{-1}\), a rational Lorentz matrix after the snap. At the register’s end the pump groups the 720 increments into 72 ten-frame blocks and then the inner blocks of sizes 8, 7, 7, 7, 7, 8, 7, 7, 7, 7, and returns
$$P = \Delta q_{720}\cdots\Delta q_1,\qquad Q^{(10)} = Q^{(8)} = P^{-1},\qquad H = Q^{(8)}Q^{(10)}P = P^{-1}$$exactly, for every channel, in non-commutative exact arithmetic (H7; the Zero-Point page’s self-test Z4 already checks the same identity on random rational frames). A channel that rested all register closes with P = 1.
That the exact record of every step multiplies out: no step was lost, duplicated or corrupted. It is an audit of the bookkeeping, not evidence of progress toward a solution.
Portfolio and partition
Portfolio (first)
All three channels attack the same instance from different starts and rotations. The zero point picks the best certified answer. Reuses the calculator’s pairing exactly, so it is built first.
Partition (second)
The channels split the variables, like polystep’s blockwise mode: each moves its own block while the others are frozen at their last offload. The zero point assembles one full answer from the three blocks for Helix B.
The evaluation budget
Comparisons are fair only at equal budgets, so the budget is counted the way polystep counts it: one evaluation is one candidate scored.
$$\text{evaluations per register} = \underbrace{30}_{\text{steps per channel}} \times \sum_{\text{channels}} (\text{particles}) \times 2d \times (\text{probes}).$$For example, three channels of 4 particles in \(\mathbb{H}^{8}\) with one probe spend \(30 \times 12 \times 16 = 5{,}760\) evaluations per register. A rested channel spends none, which is how the clutch moves budget to the channels still improving.