P65 erasure-robust measurement directions: four rows proved optimal

The values

All three continuous optima are irrational, while a nine-decimal answer always has rational V, so no grid answer attains them; that does not rule out a score equal to ⌊10¹⁸ × optimum⌋, so the grid maximum of these three rows is not certified and a record may sit one or more units below the bound. d = 3, n = 8 is the other way round: the grid maximum is settled, and the continuous optimum is only pinned to an interval narrower than 2 × 10⁻²⁰.

0. Setting and four tools

An answer is n nonzero vectors v₁, …, vₙ ∈ ℝᵈ. For a d-element subset S let Vₛ be the d × d matrix of those rows and Rₛ = det(Vₛ)² / ∏|vᵢ|² (product over i ∈ S); by Hadamard’s inequality 0 ≤ Rₛ ≤ 1. The objective is V = minₛ Rₛ, to be maximised. The site requires every coordinate in [−1, 1] with at most nine decimals; the score is the integer ⌊10¹⁸V⌋, computed exactly from integer determinants and norms, and V is displayed truncated to 12 decimals. The four rows here are (d, n) = (3, 6), (3, 8), (3, 10) and (4, 8).

V is unchanged by multiplying any row by a nonzero number (−1 included), by permuting rows, and by applying one orthogonal map (reflections included) to all rows. So we may take unit rows u₁, …, uₙ and write pₛ = det(Uₛ) for the oriented minor with S listed increasingly. If some pₛ = 0 then V = 0 and every upper bound below is trivial; otherwise put m = minₛ |pₛ| > 0, xₛ = |pₛ|/m ≥ 1 and yₛ = log xₛ ≥ 0, so that V = m².

Lemma 0.1 (weighted Cauchy–Binet). Take row weights a₁, …, aₙ ≥ 0, at least d of them positive, and let bₛ be the product of the weights in S. Cauchy–Binet gives det(Σ aᵢuᵢuᵢᵀ) = Σₛ bₛpₛ². The matrix is positive semidefinite with trace Σaᵢ (the rows are unit vectors), and AM–GM on its d eigenvalues bounds the determinant by (Σaᵢ/d)ᵈ. Hence

m² · Σₛ bₛxₛ² ≤ (Σᵢ aᵢ / d)ᵈ, in particular m² · Σₛ xₛ² ≤ (n/d)ᵈ.

With all weights 1 the right side is 8, 512/27, 1000/27 and 16 for (3, 6), (3, 8), (3, 10) and (4, 8). Equality in the unweighted bound needs UᵀU = (n/d)I, a tight frame. So every lower bound on F(y) = Σₛ bₛ exp(2yₛ) is an upper bound on V = m².

Lemma 0.2 (Grassmann–Plücker sign conditions). Write p(i, j, k) for the minor with rows in the given order (an unsorted tuple carries its permutation sign). For d = 3, for each row i and rows a < b < c < e other than i:

p(i,a,b)·p(i,c,e) − p(i,a,c)·p(i,b,e) + p(i,a,e)·p(i,b,c) = 0,

There are n·C(n − 1, 4) of them: 30, 280 and 1,260 for n = 6, 8 and 10. For d = 4 there are two kinds: the three-term identities on two common rows and four others (420 for n = 8), and, for a 3-set I and a 5-set J = {j₁ < … < j₅}, Σₖ (−1)ᵏ p(I, jₖ)·p(J ∖ jₖ) = 0 (terms with jₖ ∈ I vanish; |I ∩ J| = 2, 1, 0 leave 3, 4, 5 nonzero terms). In a real frame with no zero minor the nonzero terms of an identity cannot all have the same sign; that is a condition on the sign vector alone. Once the sign pattern is fixed, an identity with exactly one term of the minority sign becomes, after taking absolute values and dividing by m²,

x(A)·x(B) = Σₖ x(Cₖ)·x(Dₖ)

one product equals the sum of the others.

Lemma 0.3 (positive-star form). d = 3: call one row row 0 and project the others onto the plane orthogonal to u₀ (oriented by u₀), giving wᵢ with p(0, i, j) = det₂(wᵢ, wⱼ). With no zero minor the wᵢ are nonzero and pairwise non-parallel. Take a line through the origin missing every wᵢ, flip each wᵢ (that is, the row uᵢ) to one side, and reorder the rows by angle; then p(0, i, j) > 0 for all i < j. For d = 4 do the same in the plane orthogonal to u₀ and u₁, making every p(0, 1, i, j) positive. So every real frame with no zero minor is equivalent, with the same V, to one whose minors through row 0 (rows 0 and 1 when d = 4) are all positive: positive-star form. The free sign bits left are 35 for (3, 8), 84 for (3, 10) and 55 for (4, 8).

Lemma 0.4 (log-convex dual). Fix a sign pattern. For each identity j used, in the form x(Aⱼ)x(Bⱼ) = Σₖ x(Cⱼₖ)x(Dⱼₖ) of Lemma 0.2, put

gⱼ(y) = y(Aⱼ) + y(Bⱼ) − log Σₖ exp(y(Cⱼₖ) + y(Dⱼₖ)).

Log-sum-exp is convex, so gⱼ is concave, and every actual frame of the pattern has gⱼ(y) = 0. Given bₛ ≥ 0, F(y) = Σₛ bₛ exp(2yₛ) is convex, and for nonnegative weights wⱼ so is L = F − Σ wⱼgⱼ. For any positive reference vector x⁰ put y⁰ = log x⁰ and c = ∇L(y⁰), that is cₛ = 2bₛ(x⁰ₛ)² − Σⱼ wⱼ ∂gⱼ/∂yₛ(y⁰), where ∂gⱼ/∂yₛ is 1 at S ∈ {Aⱼ, Bⱼ} and −x⁰(Cⱼₖ)x⁰(Dⱼₖ)/Σₗ x⁰(Cⱼₗ)x⁰(Dⱼₗ) at the two minors of right-hand product k: rational functions of x⁰. If c is componentwise nonnegative, then for every actual y (y ≥ 0 and gⱼ(y) = 0):

F(y) = L(y) ≥ L(y⁰) + c·(y − y⁰) ≥ F(y⁰) − Σⱼ wⱼ gⱼ(y⁰) − c·y⁰.

The first inequality is the supporting hyperplane of a convex function, the second uses c ≥ 0 and y ≥ 0. The reference need not be realisable. With rational x⁰ and w the right side is a combination of rationals and logarithms of positive rationals, and rigorous rational enclosures of those logarithms give a rigorous lower bound. When x⁰ is an actual optimal frame, the identities used hold exactly there (gⱼ(y⁰) = 0) and cₛ = 0 wherever x⁰ₛ > 1, so c·y⁰ = 0 and the bound is exactly F(y⁰), with no logarithm evaluated. Lemma 0.1 then turns it into a bound on V = m².

The sign filter is only a necessary condition: some kept patterns may have no real frame at all. That costs nothing: an upper bound on V for every kept pattern covers every real frame.

1. d = 3, n = 6: the continuous optimum (2/5)(1 − 1/√5)

There are 20 minors. A common reflection makes p(0, 1, 2) > 0, leaving 2¹⁹ = 524,288 sign vectors. The 30 three-term identities reject 512,384 of them and keep 11,904. Row permutations and row sign changes (renormalising the overall sign) split these into four orbits of sizes 1,920, 5,760, 3,840 and 384. Any real frame can be moved within its orbit to the orbit’s representative pattern without changing V.

The first three orbits. For each representative, Lemma 0.4 with bₛ = 1, 30 rational weights and a rational reference point bounds Σx² from below by more than 40.794, 42.180 and 54.224 respectively, all above 40. By Lemma 0.1, V = m² ≤ 8/Σx² < 1/5, while (2/5)(1 − 1/√5) ≈ 0.2211 > 1/5.

The icosahedral orbit. The fourth orbit contains the icosahedral frame below. With φ = (1 + √5)/2, label each of the 20 minors H or L according as the icosahedral frame’s minor has absolute value 2φ² or 2φ; ten of each. The labels only name index sets; nothing is assumed about the sizes of a general frame’s minors. In this sign pattern the odd term of each of the 30 identities is an H·H product and the other two are of type L·L and L·H:

H(A)·H(B) = L(C)·L(D) + L(E)·H(F),

and each H occurs exactly six times on left sides and exactly three times as the H of a right-hand L·H. Since every x ≥ 1, L(C)L(D) ≥ 1 and L(E)H(F) ≥ H(F), so H(A)H(B) ≥ 1 + H(F). Multiplying the 30 inequalities gives (∏H)⁶ ≥ (∏(1 + H))³, that is (∏H)² ≥ ∏(1 + H). Let g be the geometric mean of the ten H. Expanding ∏(1 + H) = Σₖ eₖ(H) and applying AM–GM to the C(10, k) terms of each elementary symmetric sum gives eₖ(H) ≥ C(10, k)gᵏ, so ∏(1 + H) ≥ (1 + g)¹⁰. Hence g²⁰ ≥ (1 + g)¹⁰, g² ≥ 1 + g and g ≥ φ. By AM–GM again ΣH² ≥ 10g² ≥ 10φ², while ΣL² ≥ 10. So Σx² ≥ 10(1 + φ²) = 5(5 + √5), and Lemma 0.1 gives

V = m² ≤ 8 / (5(5 + √5)) = 2(5 − √5)/25 = (2/5)(1 − 1/√5).

Equality. The six vectors (0, ±1, φ), (±1, φ, 0), (φ, 0, ±1), the six axes through opposite vertices of a regular icosahedron, all have |v|² = 1 + φ² and Σvᵢvᵢᵀ = 2(1 + φ²)I. Their 20 minors have absolute values 2φ (ten) and 2φ² (ten), so their minimum normalised volume is (2φ)²/(1 + φ²)³ = (2/5)(1 − 1/√5): equality.

The grid. V* = 0.221114561800016824287…, and ⌊10¹⁸V*⌋ = 221114561800016824. The package’s legal answer replaces (1, φ) by the consecutive Fibonacci numbers (433494437, 701408733)/10⁹, keeping the zeros and signs, and scores 221114561800016823. The nine-decimal maximum is one of these two integers; which one is not decided here.

2. d = 3, n = 8: the grid maximum 66035176782963417

Only the integer score is proved for this row: every real frame, of any precision, has ⌊10¹⁸V⌋ ≤ 66035176782963417, and a nine-decimal answer reaches it. The continuous optimum is not determined. No box of grid points is enumerated: the proof is a single continuous upper bound that falls below the next integer.

Sign patterns. Put the frame in positive-star form (Lemma 0.3): the 21 minors through row 0 are positive and 35 signs are free. A complete depth-first search over the 2³⁵ assignments, pruning only where one of the 280 three-term identities has all three terms of one sign, keeps exactly 24,698 patterns.

135 classes. The package lists 135 integer 8 × 3 matrices with no zero minor. For each matrix, every ordered choice of the first two rows, their signs and a common reflection (the other rows’ signs and order are then forced by positive-star form) gives its set of positive-star images. The 135 image sets are pairwise disjoint, lie inside the 24,698 patterns and cover all of them. So every real frame with no zero minor can be transformed, without changing V, to have exactly the sign pattern of one of the 135 matrices.

Excluding 134 classes. For each class, Lemma 0.4 with bₛ = 1 and 280 rational weights bounds Σx² from below, and with Lemma 0.1 (constant 512/27) every one of 134 classes has V < 0.061110 < 31/500 = 0.062. The remaining class (key 8000b7ffbffdec) gets only V < 0.0703 from such a certificate, and V < 0.06667 < 1/15 from a weighted one (eight positive rational row weights). The legal answer below scores about 0.066035 > 0.062, so the optimum lies in this class.

The mixture certificate. To get under the next integer the certificate mixes four weighted Cauchy–Binet inequalities. Atom k has nonnegative rational row weights aₖ,ᵢ (six positive, two zero); with Cₖ = (Σᵢ aₖ,ᵢ / 3)³ and bₖ,ₛ the product of the weights in S divided by Cₖ, Lemma 0.1 reads m² Σₛ bₖ,ₛxₛ² ≤ 1. With nonnegative rational mixture weights μₖ summing to 1 and bₛ = Σₖ μₖbₖ,ₛ, m² Σₛ bₛxₛ² ≤ 1. The certificate is stated for an integer representative in a different row order; the full set of positive-star images of its 56 minor signs, under the same signed-permutation action, equals the remaining class’s image set, so it applies to the whole class with no symmetry assumed of a competing frame. Lemma 0.4 with these bₛ, a rational reference point and 280 nonnegative weights (31 of them positive) gives a rational L with

L > 10¹⁸ / 66035176782963418, V = m² ≤ 1/L < 0.066035176782963418.

The margin is small: L exceeds the boundary by about 2.2 × 10⁻¹⁶, a relative 1.5 × 10⁻¹⁷, which is why the logarithms need high-precision enclosures. With the other 134 classes and the zero-minor case, every real frame has ⌊10¹⁸V⌋ ≤ 66035176782963417.

Attainment. The legal answer is (0.315504791, ±0.431761327, ±0.494892959) (all four sign choices), (0.267570676, ±0.794328857, 0) and (0.500878354, 0, ±0.289746371). Exact rational evaluation over all 56 triples gives the score 66035176782963417 exactly (display 0.066035176782). So the grid maximum is this integer, and it equals the current record.

The continuous problem stays open. The answer’s exact V and 1/L pin the continuous optimum to [0.066035176782963417006…, 0.066035176782963417024…], an interval narrower than 2 × 10⁻²⁰, but its exact value is not determined; the continuous optimum of this row remains open.

3. d = 3, n = 10: the continuous optimum 2(3 − √5)/27

Threshold. Let Q = 2(3 − √5)/27 ≈ 0.0565876 and T = 113/2000 = 0.0565 < Q (an exact comparison in ℚ(√5)). If V < T the frame is already below Q. Otherwise every eight of the ten rows form a frame with V ≥ T, since its 56 triples are among the 120. Among §2’s 135 classes, 131 have certified bounds (512/27)/Lₖ below T; the other four (keys 8000b7ffbffdec, 8000f7ffbfffec, 80010fffffff93 and 800187fdffff80) have 296 positive-star images between them.

Overlap-table search. Put the ten-frame in positive-star form at row 0: the 36 minors through row 0 are positive and 84 signs are free. For each of the C(9, 7) = 36 seven-element sets J of rows 1–9, the eight-frame {0} ∪ J in increasing order is again in positive-star form, so its 35 free signs must be one of the 296 allowed patterns. A complete search over the 84 bits, propagating only values shared by every remaining allowed row of some table and branching both ways otherwise, ends with exactly 6 assignments.

The dodecahedral sign class. Let ρ = φ − 1 = 1/φ. The ten rows (1, ±1, ±1), (0, ±ρ, φ), (±ρ, φ, 0), (φ, 0, ±ρ) are the ten axes through opposite vertices of a regular dodecahedron. Computing its 120 minor signs exactly in ℚ(√5), and all its positive-star images, gives exactly the six assignments. So every frame with V ≥ T is, up to signed permutation, in the dodecahedral sign pattern.

The exact dual. The dodecahedral rows have |v|² = 3 and Σvᵢvᵢᵀ = 10I, and the smallest absolute minor is √5 − 1. Put ξₛ = |pₛ|/(√5 − 1) ∈ ℚ(√5); then ξₛ ≥ 1 and Σξₛ² = 375 + 125√5. Apply Lemma 0.4 with bₛ = 1, reference x⁰ = ξ, the 1,260 three-term identities and weights in ℚ(√5), all nonnegative and 180 of them positive. Each identity used holds exactly at ξ, so gⱼ(log ξ) = 0; all 120 residuals satisfy cₛ ≥ 0, and cₛ = 0 whenever ξₛ ≠ 1, so c·log ξ = 0. Hence, with no logarithm evaluated, F(y) ≥ 375 + 125√5 on the whole class, and Lemma 0.1 gives

V = m² ≤ (1000/27) / (375 + 125√5) = 8 / (27(3 + √5)) = 2(3 − √5)/27.

The dodecahedral frame attains it: its minimum normalised volume is (√5 − 1)²/3³ = (6 − 2√5)/27 = Q, checked over all 120 triples in exact arithmetic.

The grid. Q = 0.056587557222237800265…, ⌊10¹⁸Q⌋ = 56587557222237800, and the package’s legal answer scores 56587557222237799. The nine-decimal maximum is one of these two integers; it is not decided here.

On uniqueness. The package adds a corollary: up to a common orthogonal map, row order and row scaling, the dodecahedral directions are the only maximiser (strict convexity of L forces y = log ξ, equality in Lemma 0.1 forces a tight frame, and equal oriented minors plus tightness force an orthogonal change of basis). That is the package’s claim; our checkers do not test it, and nothing on this page relies on it.

4. d = 4, n = 8: the continuous optimum a²

Sign patterns. Positive-star form (Lemma 0.3) makes the 15 minors p(0, 1, i, j) positive and leaves 55 signs. The sign filter uses every identity whose 3-set I and 5-set J share at most two rows: 2,576 of them, 1,680 with three nonzero terms, 840 with four and 56 with five. Our exhaustive search over the 55 bits, with no symmetry reduction, visits 54,884,515 nodes and keeps 6,693,736 patterns.

2,628 classes. The package lists 2,628 representative patterns. For each, every ordered choice of the first three rows, their signs and a common reflection (the other rows’ signs and order are then forced) gives its image set; the image sets are pairwise disjoint and their union is exactly the 6,693,736 patterns. (The package reached the same set from 279,078 canonical patterns and a 24-element symmetry group; all 279,078 lie in our set, but our count does not use the group.)

Excluding 2,627 classes. Write q = a². The isolating interval gives q > (223/1000)² > 31/625 = 0.0496. The following certificates give V < 31/625 for 2,627 classes.

2,600 classes by an AM–GM certificate. In a fixed pattern each three-term identity (two common rows plus four others, 420 in all) reads x(A)x(B) = x(C)x(D) + x(E)x(F), and AM–GM gives x(A)x(B) ≥ 2√(x(C)x(D)x(E)x(F)), that is hⱼ·y ≥ log 2, where hⱼ is +1 at A and B and −1/2 at C, D, E and F. With nonnegative rational weights wⱼ, S = Σwⱼ and g = Σwⱼhⱼ, the certificate checks gᵢ ≤ 1 for all 70 components. Since y ≥ 0, Σyᵢ ≥ g·y ≥ S log 2, and Jensen for exp gives F = Σexp(2yᵢ) ≥ 70·exp((S/35)·log 2). Lemma 0.1 (constant 16) gives

V ≤ (16/70)·exp(−(S/35)·log 2) < 31/625 whenever (S/35)·log 2 > log(1000/217).

The other 27 classes use Lemma 0.4 with bₛ = 1 and every three-, four- and five-term identity that has a unique odd term in the pattern, giving F > 10000/31 and so V < 31/625; the largest of these 27 bounds is about 0.047585.

The construction, and where the octic comes from. Let A, B, C be the package’s symmetric 8 × 8 integer matrices (zero diagonal, entries 0 and ±1). They satisfy A² = I, B² = C² = 2I and AB + BA = AC + CA = BC + CB = 0. For real a, b, c put H = aA + bB + cC, so that H² = (a² + 2b² + 2c²)I. Under the norm condition

a² + 2b² + 2c² = 1

we get H² = I; H is symmetric with trace 0, so its eigenvalues are four +1 and four −1. Then G = I + H has unit diagonal, G² = 2G and rank 4, and G = UUᵀ for an 8 × 4 matrix U with unit rows and UᵀU = 2I (a tight frame). The principal minors of G are the squared minors: det G[S, S] = det(Uₛ)². Expanding all 70 principal 4 × 4 minors of G as polynomials in a, b, c gives 12 distinct polynomials, and under the norm condition three of them simplify:

1 − 2b² − 2c² = a² (8 minors); 1 − a² − b² − 2c² − 2abc + a²c² = (b − ac)² (8 minors); (1 − a² − b² − c²)² − 4b²c² = (c² − b²)² (4 minors).

The first family equals a² automatically. Asking the other two to equal a² as well, with the signs b − ac = a and c² − b² = a, gives three equations together with the norm condition. Adding twice c² − b² = a to the norm condition gives 4c² + a² = 1 + 2a, so c² = (1 + 2a − a²)/4 and b² = c² − a = (1 − 2a − a²)/4. Squaring b = a(1 + c) and substituting gives 2a²c = N/4 with N = a⁴ − 2a³ − 6a² − 2a + 1, that is c = N/(8a²). Squaring again, N² = 64a⁴c² = 16a⁴(1 + 2a − a²), and

N² − 16a⁴(1 + 2a − a²) = a⁸ − 4a⁷ + 8a⁶ − 12a⁵ + 30a⁴ + 20a³ − 8a² − 4a + 1 = 0.

p changes sign between 223/1000 and 28/125 and p′ is negative throughout, so there is exactly one root a ≈ 0.2231217381 there; then b = a(1 + c) ≈ 0.3549553542 and c ≈ 0.5908595785 are positive. Computed exactly in ℚ[a]/(p), 20 principal minors equal a², and rational isolating intervals show the other 50 are strictly larger. So this frame attains V = a² = q ≈ 0.049783309999360073. q also satisfies q⁸ + 28q⁶ + 480q⁵ + 1222q⁴ − 960q³ + 284q² − 32q + 1 = 0 (no claim that this is its minimal polynomial).

A tight bound on the last class. The construction’s sign pattern, after an explicit signed permutation, is exactly the remaining class (mask 0x67f9b440190400000). Its oriented minors are recovered exactly as pₛ = det G[J, S]/p(J), with J one of the tight 4-row blocks and p(J) = a, so the reference x⁰ₛ = |pₛ|/a lies in ℚ[a]/(p): 20 entries equal 1 and 50 exceed 1. Lemma 0.4 with bₛ = 1 and every identity with a unique odd term, using 72 weights in ℚ[a]/(p) that are positive at the root (checked by isolating intervals): every identity used holds exactly at x⁰; all 70 residuals are ≥ 0 and the residual is exactly 0 at each of the 50 minors with x⁰ₛ > 1; so c·log x⁰ = 0 and F(y) ≥ F(log x⁰) = 16/a² exactly. Lemma 0.1 gives V ≤ 16/F ≤ a² on the whole class. With the other 2,627 classes below 31/625 < a², the continuous maximum is a².

The grid. ⌊10¹⁸a²⌋ = 49783309999360072. The package’s legal nine-decimal answer scores 49783309999020748, 339,324 units lower. The nine-decimal maximum lies between the two and is not decided here.

On uniqueness. The package likewise claims that the optimal directions are unique up to a common orthogonal map, row order and row scaling (the argument of §3). That is the package’s claim; our checkers do not test it, and nothing on this page relies on it.

5. Our verification

zzzcy #308 emailed the proof package on 6 October 2026; as the sender stated, it was generated with AI assistance. Under the site’s rules we ran none of its scripts: we read only its data (sign catalogues, representatives, dual weights, reference points, matrices and answers) and decided every step with code written for this review:

• tools/p65-certificates.py (Python, exact rationals): every Grassmann–Plücker identity used is re-verified as a polynomial identity by expanding generic determinants (30, 280 and 1,260 in rank 3; 2,996 in rank 4); our own signed-permutation action is first checked against the determinant signs of transformed random real matrices; ℚ(√5) and ℚ[a]/(p) arithmetic is our own; logarithms use our own 480-bit fixed-point atanh series with a tail bound and outward rounding, self-tested against 150-digit decimal logarithms; every value, construction and grid answer is recomputed exactly.
• tools/p65-rank4-signs.cpp (C++): the complete d = 4 positive-star enumeration (no symmetry reduction; 6,693,736 patterns, 54,884,515 nodes) and the coverage by the 2,628 representatives (no image outside the set, no overlap, nothing uncovered); the Python side compares the C++ image sets of the first 25 classes with its own and recomputes 12 random class sizes.

Row by row:

• d = 3, n = 6: our 30 identities equal the package’s list; the full filter keeps 11,904 patterns, and the four orbits (1,920, 5,760, 3,840, 384) equal the package’s catalogue; the three log-convex duals give 40.794415, 42.180710 and 54.224291, all above 40; in the icosahedral pattern all 30 identities read H·H = L·L + L·H, with each H six times on the left and three on the right; V* and ⌊10¹⁸V*⌋ = 221114561800016824 are computed exactly, and the package’s answer scores 221114561800016823.
• d = 3, n = 8: our search (196,131 nodes) keeps 24,698 patterns, equal to the package’s list; the 135 integer representatives have disjoint images covering all of them; all 135 duals replay, 134 classes below 0.061110; the weighted certificate is below 1/15; the mixture certificate’s representative has exactly the exceptional class’s image set, and L − 10¹⁸/66035176782963418 ≈ 2.237 × 10⁻¹⁶ > 0; the answer scores exactly 66035176782963417.
• d = 3, n = 10: the package’s base files (representatives, duals, pattern list) are byte-identical to those of n = 8; at the threshold 113/2000 four classes and 296 allowed patterns remain (equal to the package’s list); our table search finds the 6 assignments in 83 nodes (the package reports 75 and 35 nodes with other branching orders, same result), exactly the dodecahedral frame’s positive-star images; the ℚ(√5) dual has 180 positive weights and 120 nonnegative residuals, zero wherever ξₛ ≠ 1, so F ≥ 375 + 125√5; ⌊10¹⁸Q⌋ = 56587557222237800, and the package’s answer scores 56587557222237799.
• d = 4, n = 8: enumeration and coverage as above; all 2,600 AM–GM and 27 log-convex certificates below 31/625; G symmetric with unit diagonal and G² = 2G; 20 principal minors equal a² and 50 are larger; the package’s relation list equals our reconstruction; 72 positive weights, 70 nonnegative residuals, F(log x⁰) = 16/a²; ⌊10¹⁸a²⌋ = 49783309999360072, and the package’s answer scores 49783309999020748.

A full run on 6 October 2026 printed ALL REQUESTED CHECKS PASSED in about 100 seconds, most of it the C++ enumeration. The simplification of the three minor families in §4 and the identity N² − 16a⁴(1 + 2a − a²) = p(a) were derived separately for this page and checked with a short polynomial script; the checker verifies their outcome (20 principal minors equal to a² exactly in ℚ[a]/(p)). To rerun (--pkg points at the unpacked package, whose data alone is read):

python tools/p65-certificates.py all --pkg <dir> --cxx g++

Single rows run with n6, n8, n10 or d4n8; d4n8 compiles tools/p65-rank4-signs.cpp with g++ -O2 -std=c++17. Not checked: the two uniqueness corollaries, and the package’s numerical discovery and its own reviews. Every value agrees with the package, and we found no error in the reasoning.

Credit and scope

The icosahedral and dodecahedral vertex axes are classical configurations; the d = 4 construction and all four optimality arguments come from this package, which makes no priority claim. The adopted contribution earns zzzcy #308 one permanent +2 proof award.

The three continuous optima close under the site’s convention for continuous theorems; their nine-decimal maxima are still open, and a new answer may still beat the current record, though never the bound in the table. For d = 3, n = 8 the grid maximum is settled and the continuous optimum remains open.

Open P65