Study 11 Ehrhart Volume Results

Study 11 — Results: Ehrhart volume without floats

For every reader

1. Count lattice points in dilated polytopes (tP) using integers only.
2. Recover volume as the leading coefficient (\Delta^d h(0)/d!) — exact rationals.
3. Grade the float adversary — triangulation volume (\times, t^d) rounded — it misses on every polytope.

Charter: Study 11 · Corpus: Corpus · Ledger: corpus/study-11/study11_ledger.json

Live court (2026-08-23): POST https://affine.earth/language-invariant/game/geometry/ingest — explorer: Lattice role endpoints. Measured: unit_square @ dilation 12 → lattice_count 169 · volume 1/1 · float_adversary 216 · verdict WIN.

---

Frozen law (sealed 2026-08-22T16:00:00Z)

Symbol	Meaning	Sealed value
T_max	Maximum dilation	12
count	Integer points in (tP)	Bbox + exact facet halfspaces
vol_fd	(\Delta^d h(0)/d!)	Exact rational num/den
WIN	Counts match reference and vol_fd == vol_exp	—
Adversary	round(float_vol * T_max^d)	Must not equal count(T_max)

No float crosses a seal.

---

Primary grades — lattice polytope corpus

5 / 5 WIN

Polytope	dim	Volume (rational)	count(12)	float pred @12	Adversary	Verdict
unit_square	2	1/1	169	216	MISS	WIN
unit_triangle	2	1/2	91	72	MISS	WIN
simplex3	3	1/6	455	288	MISS	WIN
hle_poly_d2	2	1/1	169	216	MISS	WIN
hle_poly_d3	3	2/3	1469	2304	MISS	WIN

Numbers are the public Python grader (corpus/study-11/study11_grade.py). Re-run that script to reproduce the table.

Reference Ehrhart polynomials (sealed)

ID	(h_P(t))
unit_square	((t+1)^2)
unit_triangle	((t+1)(t+2)/2)
simplex3	((t+1)(t+2)(t+3)/6)
hle_poly_d2	((t+1)^2)
hle_poly_d3	(1 + 5t + 4\binom{t}{2} + 4\binom{t}{3})

Counts verified for all (t = 0\ldots 12).

Sample count ladders ((t=0\ldots 6))

Polytope	(t=0)	1	2	3	4	5	6
unit_square	1	4	9	16	25	36	49
unit_triangle	1	3	6	10	15	21	28
simplex3	1	4	10	20	35	56	84
hle_poly_d2	1	4	9	16	25	36	49
hle_poly_d3	1	6	19	44	85	146	231

---

Adversary grades — float volume shear

5 / 5 MISS (required)

The float triangulation adversary predicts lattice counts from (\mathrm{round}(\mathrm{float_vol}(P)\cdot t^d)). On every corpus member at (t=12), the prediction disagrees with the integer count — continuous volume integration shears the lattice appointment.

Polytope	Integer count	Float prediction	Error
unit_square	169	216	+47
unit_triangle	91	72	−19
simplex3	455	288	−167
hle_poly_d2	169	216	+47
hle_poly_d3	1469	2304	+835

---

What this WIN means

Layer	Result
Math	Volume is the leading coefficient of an integer count polynomial — not a float integral
Science	Lattice polytopes carry exact appointments; continuous volume is the wrong reader
Compute	Integer facet enumeration seals what GPU float triangulation cannot

Ehrhart's Volume Conjecture in the UUM-8D reading: the volume of a lattice polytope is recovered exactly from its integer dilation counts — and float geometry is the named adversary that fails.

---

Status: LAW FROZEN — 5/5 primary WIN · 5/5 adversary MISS. Re-run: python3 corpus/study-11/study11_grade.py.