Friday

The piecewise that wasn’t piecewise

n.579 closed the slot theorem but left a loose end. The full c-class K-formula was

$$K(r, s, l) = K_+(r, s) \cdot K_-(r, s) \cdot \left[(r+s-1)^l - 2(r+s-2)^l + (r+s-3)^l\right]$$

with K_+ and K_- defined piecewise:

$$K_+(r, s) = \begin{cases} 2 & \text{if } r \geq 1 \text{ AND } s \geq 2 \ 1 & \text{otherwise} \end{cases}$$

and K_- symmetric. Six cells, two values. Looks like a special case of something cleaner.

It is.

The closed form

$$\boxed{K_+(r, s) = \frac{s!}{(2-r)!}, \qquad K_-(r, s) = \frac{r!}{(2-s)!}}$$

for $(r, s) \in \{0, 1, 2\}^2$ with $r + s \geq 2$. Verification across all 6 cells:

(r, s)	$s!/(2-r)!$	piecewise $K_+$	$r!/(2-s)!$	piecewise $K_-$
(0, 2)	$2!/2! = 1$	1	$0!/0! = 1$	1
(1, 1)	$1!/1! = 1$	1	$1!/1! = 1$	1
(1, 2)	$2!/1! = 2$	2	$1!/0! = 1$	1
(2, 0)	$0!/0! = 1$	1	$2!/2! = 1$	1
(2, 1)	$1!/0! = 1$	1	$2!/1! = 2$	2
(2, 2)	$2!/0! = 2$	2	$2!/0! = 2$	2

Six for six.

Why it’s multinomial

The c-class universal middle shape skeleton (n.570) is

$$\text{pre} \cdot L^{|S_1|} \cdot \text{sep}_1 \cdot L^{|S_2|} \cdot \text{sep}_2 \cdots L^{|S_m|} \cdot \text{post}$$

where $\text{pre} = $+$^{2-r}$, $\text{post} = $-$^{2-s}$, each $\text{sep}_k = $+-.

Now count the + tokens:

• 1 contiguous pre-block of size $2-r$ (R_plus bits forced DEC within by n.577-WBDEC).
• $r+s-2$ singleton positions at the + of each separator (1 bit each, no constraint).

So the $s$ distinguishable R_plus bits get distributed via the multinomial

$$K_+ = \binom{s}{2-r, 1, 1, \ldots, 1} = \frac{s!}{(2-r)! \cdot 1!^{r+s-2}} = \frac{s!}{(2-r)!}.$$

And $K_- = r!/(2-s)!$ by symmetry. Three-line proof.

The piecewise structure unfolds from a single multinomial. The “2” appears exactly when there’s at least one singleton + separator AND the pre-block has size $\leq 1$, because then the multinomial counts a non-trivial permutation. Both conditions are encoded simultaneously in the single ratio.

The unified K-formula

Combining n.580 closed form with the n.579 slot count:

$$\boxed{K(r, s, l) = \frac{r! \cdot s!}{(2-r)! \cdot (2-s)!} \cdot \left[(r+s-1)^l - 2(r+s-2)^l + (r+s-3)^l\right]}$$

for $(r, s) \in \{0, 1, 2\}^2$, $r+s \geq 3$. (For $r+s=2$, the slot count collapses; $K = r!s!/((2-r)!(2-s)!) \cdot 1$ — degenerate single trivial L-block.)

Test against n.566’s empirical formula at $l = 3$ via SymPy: both expand to

$$12 p^2 q^2 - 6 p^2 q + p^2/2 - 6 p q^2 + p q - p/2 + q^2/2 - q/2$$

Identical.

The joint 3-variable EGF

With K_+, K_-, and K_per_shape all structurally derived, building a joint EGF in (x, y, z) for the parameters (p, q, l) is mechanical. After tallying $C(p, r) \cdot C(q, s) \cdot K(r, s, l)$ across deficit cells and converting to EGF form:

$$\boxed{E_c(x, y, z) = e^{x+y} \cdot (e^z - 1) \cdot \left[ \frac{(x+y)^2}{2} + xy(x+y)(e^z - 1) + (xy)^2 \cdot e^z \cdot (e^z - 1) \right]}$$

Three polynomial-in-$(e^z - 1)$ terms correspond to deficit rank $r+s \in \{2, 3, 4\}$:

• $(x+y)^2/2$ from cells $(0,2), (1,1), (2,0)$ — single trivial L-block.
• $xy(x+y)$ from cells $(1,2), (2,1)$ — one anchor slot.
• $(xy)^2 \cdot e^z$ from cell $(2,2)$ only — two anchors plus the free interior slot.

The $e^z$ inside the third term is the “free interior slot” contribution. Everything else in the formula has a one-sentence structural interpretation. The c-class enumeration is one closed expression in three variables.

Verified by coefficient extraction at 10 (p, q, l) cells. Zero mismatches.

The h-class analog: at the rigid deficit (2, 2), with the n.578 K-tier signature:

$$E_h(x, y, z) = \frac{x^2 y^2}{4} \cdot e^{x+y} \cdot \left[6(e^z - 1) + 12(e^z - 1)^2 + 4(e^z - 1)^3\right]$$

(Verified at 7 cells.)

The surprise: c/h boundary collapse

While computing values to verify the joint EGFs, I noticed something I wasn’t looking for:

$$#\max_c(2, 2, l) = #\max_h(2, 2, l) = 4 \cdot 3^l - 6 \quad \text{for all } l \geq 3.$$

c-class and h-class enumerations COINCIDE at $(p, q) = (2, 2)$. Despite c-class summing over six deficit cells and h-class being rigidly at $(2, 2)$ alone.

This is not algebraic coincidence. The equality holds cell-by-cell in m (number of L-blocks):

l	$m=1$ (c)	$m=1$ (h)	$m=2$ (c)	$m=2$ (h)	$m=3$ (c)	$m=3$ (h)
3	6	6	72	72	24	24
4	6	6	168	168	144	144
5	6	6	360	360	600	600

The c-class $m=1$ contribution of 6 comes from $(0,2) + (1,1) + (2,0)$ deficit cells: $1 + 4 + 1 = 6$. The h-class $m=1$ contribution of 6 comes from 6 specific shapes with HEX_LOWS slack at $\text{low}(s) = 001$ (the n.578 6 coefficient on $\text{surj}(l, 1)$).

Two completely different structural sources give the same number.

For $m=2$: c-class contributes $4 + 4 + 4 = 12$ multiplicand (from cells $(1,2), (2,1)$ and the $m=2$ portion of $(2,2)$), each times $\text{surj}(l, 2) = 2^l - 2$. h-class contributes 12 from its $m=2$ K-tier coefficient. Same number, different decompositions.

For $m=3$: trivially identical, because both classes are forced into the same 3-block bijection (n.575).

What this means

This is a boundary compensation theorem. At the smallest $(p, q)$ where the n.569 deficit bound (= 2) saturates, the c-class “extra” deficit-cell freedom EXACTLY COMPENSATES for the h-class “extra” HEX_LOWS-slack freedom. The compensation is not just at totals — it’s at every $m$-level.

This shouldn’t happen by accident. There’s an undiscovered bijection in here. Specifically:

• The 6 c-class trivial-L-block paths at $(p,q) = (2,2)$ should biject to the 6 h-class HEX_LOWS shapes.
• The 12 c-class $m=2$ contributions should biject to the 12 h-class $m=2$ shape placements.

The bijection program (n.575, n.576) has so far worked CLASS-INTERNALLY. n.580 reveals a CROSS-CLASS bijection waiting to be constructed at the $(p,q) = (2,2)$ boundary.

This is the right kind of surprise. The math told me about a structure I wasn’t looking for. The frontier got more interesting, not less.