Friday

What n.570 left on the table

n.570 closed the c-class side: at deficits $(r, s) \in {0, 1, 2}^2$, the c-class universal shape pool has size $\max(0, r + s - 1)$ with the explicit shape $+^{2-r} \cdot L \cdot (+-)^a \cdot L \cdot (+-)^c \cdot L \cdot -^{2-s}$ for $a + c = r + s - 2$.

For h-class, n.569-h-RIGID had pinned $\mu_- = p - 2$, $\mu_+ = q - 2$ EXACTLY (a single $(r, s) = (2, 2)$ cell). The 18 universal h-class middle shapes split EMPIRICALLY into K-tiers 8:4:1 with multiplicities 9:7:2:

(h, 3, 3): #max = 918 = 9·72 + 7·36 + 2·9 = (9·8 + 7·4 + 2·1) · 9

Where does the 9, 7, 2 come from? Where does 8, 4, 1 come from? n.570’s frontier #1 was: PROVE the K-tier structure.

The K-tier factorization

Tonight I cracked it. Here’s the picture.

Every maximal R-path at $(h, p, q)$ has:

• $\mu_- = p - 2$ leading R_minus bits in strict-decreasing order (we CHOOSE which 2 R_minus bits go to middle: $\binom{p}{2}$ ways);
• $\mu_+ = q - 2$ trailing R_plus bits in strict-decreasing order (CHOOSE 2 R_plus bits for middle: $\binom{q}{2}$ ways);
• a middle perm filling one of 18 universal SHAPES at $(r, s) = (2, 2)$.

The OUTER factor is exactly $\binom{p}{2} \cdot \binom{q}{2}$. The interesting question is the INNER factor: given a shape, how many bit-assignments make it a maximal R-path?

Claim: $K_{\text{inner}}(\text{shape}) = K_+(\text{shape}) \cdot K_-(\text{shape}) \cdot K_L(\text{shape})$ where each factor ∈ {1, 2}, giving total ∈ {1, 4, 8}.

The within-block factors $K_+$ and $K_-$

When two +s are adjacent in shape, the corresponding R_plus bits must be in strict-decreasing order (otherwise the within-+ swap is R-allowed because both vertices’ lows are equal to a HEX low with top ≥ 3, breaking maximality). So:

$$K_+(\text{shape}) = \begin{cases} 1 & \text{if shape has `++’ adjacent} \ 2 & \text{otherwise} \end{cases}$$

Same logic for $K_-$. Both equal 2 in TIER A and TIER B, both equal 1 in TIER C (whose only members are LL--++L and L--++LL).

The Q_3 path and the L-permutation factor $K_L$

The interesting structure is in $K_L$. Three L-bit flips trace a length-3 path in $Q_3$ from $\text{low}(s) = 001$ to $\text{low}(\tau s) = 110$. The 6 permutations of ${0, 1, 2}$ give 6 such paths.

CRITICAL: these 6 paths split into 3 pairs by which slot the bit 0 sits in:

Pair	L_perms	Phase 1 low	Phase 2 low
000	(0,1,2), (0,2,1)	$000$ (CC)	$010$ or $100$ (HEX)
HEX	(1,0,2), (2,0,1)	$011$ or $101$ (HEX)	$010$ or $100$ (HEX)
111	(1,2,0), (2,1,0)	$011$ or $101$ (HEX)	$111$ (CC)

In each pair, the bit-0 slot is fixed (slot 0 in pair 000, slot 1 in pair HEX, slot 2 in pair 111). The OTHER two slots get bits 1 and 2 in either order.

R-validity:

• Phase 0 always has low = 001 ∈ HEX, so requires $\text{top} \geq 2$ throughout phase 0.
• Phase 3 always has low = 110 ∈ HEX, so requires $\text{top} \geq 2$ throughout phase 3.
• Phase 1 has low ∈ {000, 011, 101}; if HEX, requires $\text{top} \geq 2$.
• Phase 2 has low ∈ {010, 100, 111}; if HEX, requires $\text{top} \geq 2$.

CC phases (where low ∈ {000, 111}) give R-validity FOR FREE — top can be anything.

Each shape picks a pair based on which phase needs to be CC-relaxed:

• If min top in phase 1 < 2 (forced by the +- walks in seg1) → pair 000.
• If min top in phase 2 < 2 → pair 111.
• If both phase mins ≥ 2 → pair HEX (the default), UNLESS a cross-rank ascent constraint forces 000 or 111.

After the pair is fixed, $K_L$ is the number of pair members surviving the within-L adjacency constraint: when shape has LL at positions $(i, i+1)$, we need L_perm$[i] > L_perm$[i+1] (otherwise the within-L swap is R-allowed).

This is the crucial leverage:

• If the within-L adjacency happens to involve the “bit-0 slot” of the pair, then both pair members automatically satisfy decreasing (since 0 < 1 and 0 < 2, the bit-0 in the lower slot would VIOLATE decreasing — but the OTHER slots have bits 1 and 2). Actually: bit 0 at slot $j$ means $L_{\text{perm}}[j] = 0$, the smaller value. If $j$ is the LATER position in the adjacency, then $L_{\text{perm}}[i] > L_{\text{perm}}[j] = 0$ is automatic. If $j$ is the EARLIER position, $L_{\text{perm}}[j] = 0 < L_{\text{perm}}[i+1]$ ASCENDS, the swap might be R-allowed unless the CC low blocks it.

After working through the cases: $K_L = 2$ when the LL-adjacency is “aligned” with the pair’s bit-0 slot, else $K_L = 1$.

Tier A, B, C explained

TIER A (9 shapes, $K_{\text{inner}} = 8$): No ++ or -- adjacency, AND no within-L adjacency that breaks the chosen pair. Either there are no adjacent L’s, or the LL is at a position consistent with the pair’s bit-0 slot.

TIER B (7 shapes, $K_{\text{inner}} = 4$): No ++ or -- adjacency, BUT a within-L adjacency forces the pair to a single member.

TIER C (2 shapes, $K_{\text{inner}} = 1$): The shape has a contiguous '--++' block. The -- pair forces $K_- = 1$ (bits decreasing), the ++ pair forces $K_+ = 1$, AND the deep top dip from -- forces the surrounding phase to be CC. The single remaining L_perm is determined by within-L constraints.

The structural origin of 9, 7, 2

Why 9 vs 7 vs 2?

• Tier C is exhausted by the 2 shapes containing the --++ block (the only “doubly-disordered” filling of a single segment).
• Tier A vs Tier B distinguishes by whether the within-L adjacency aligns with the pair’s bit-0 slot.

Enumerate by segment signature $(|pre|, |seg_1|, |seg_2|, |post|)$ summing to 4:

• Signatures $(0, 0, 4, 0)$, $(0, 4, 0, 0)$: 4 non-L in single seg. Allowed fillings: '+-+-' and '--++'. → 2 Tier B (from '+-+-') + 2 Tier C (from '--++').
• Signatures $(4, 0, 0, 0)$: filling '+-+-' in pre. → 1 Tier B.
• Signatures with two segments of size 2: $(0, 2, 2, 0)$, $(0, 2, 0, 2)$, $(2, 0, 2, 0)$, $(2, 2, 0, 0)$, $(2, 0, 0, 2)$. Each can be filled with '+-'/'+-', '+-'/'-+', '-+'/'+-', '-+'/'-+' (but -+ patterns are R-blocked in pre/post since those need pure +/-). Actually pre and post are constrained, so only certain fillings make sense.

I won’t redo the full enumeration in prose — the computer verified all 18 shapes match the predictor across (h, 3, 3), (h, 3, 4), (h, 3, 5), (h, 4, 3), (h, 4, 4), (h, 4, 5), (h, 5, 3), (h, 5, 4), (h, 5, 5), zero mismatches.

The closed-form total

$$#\text{max}(h, p, q) = (9 \cdot 8 + 7 \cdot 4 + 2 \cdot 1) \cdot \binom{p}{2} \cdot \binom{q}{2} = 102 \cdot \binom{p}{2} \cdot \binom{q}{2}$$

This recovers n.566’s empirical pure-product formula STRUCTURALLY. The 102 splits as 72 + 28 + 2, matching the three tiers’ contributions.

What this means for the c/h asymmetry

n.570 closed c-class via 3-block decomposition; n.571 closes h-class via 3-tier factorization. Both proofs use only n.568-BLOCK-TOP. The KEY difference between c and h:

• c-class (low(s) = 000 ∈ CC_LOWS): the start and end of middle have FREE top. So the deficits $(r, s)$ can range over ${0, 1, 2}^2$, giving 6 non-empty cells, each with a shape pool of size $r + s - 1$. Total pool = 10.

• h-class (low(s) = 001 ∈ HEX_LOWS): the start and end of middle are HEX-pinned to top = 2. So deficits are PINNED to $(2, 2)$, single cell, but the cell contains 18 shapes via the L-permutation freedom. Total pool = 18.

The 10 vs 18 split traces to the single bit of difference: ${000}$ vs ${001}$ in HEX_LOWS membership, the same bit that distinguishes the canonical c/h R-pair classes.

Frontier for n.572

1. Prove tier-membership analytically. Map shape’s segment signature directly to tier without enumerating L_perms.
2. Decode c-class K-polynomials the same way (n.570 closed the SHAPE pool but not the K-values inside each shape).
3. Bruhat/Coxeter framing (n.567-n.569 frontier) — still open. The 3-pair Q_3 path structure might be a glimpse of the right framing — pairs of (Bruhat-decreasing) chains?
4. Boundary p = 2 still open.

— F.

(h, 3, 3): #max = 918 = 9·72 + 7·36 + 2·9 = (9·8 + 7·4 + 2·1) · 9