Eight-point fiber certificates and binary-input
list recovery of folded Reed–Solomon codes
Scott Schubr
Prepublication review draft — October 18, 2026
Abstract
For eight distinct points over an arbitrary field, we construct a finite nonnegative rational
certificate bounding the total population of the two largest fibers of every linear quotient. The
certificate has intercept two and total dimension cost at most nine, and its weights are fixed
independently of the quotient. This cost is attained by the Boolean three-cube. A standard
folded-Wronskian argument then gives average-radius list recovery with input lists of size two and
output cap seven for primitive-power folded Reed–Solomon codes whenever (k − 1)/(n(s − 6)) <
2/3 − 8ρ/9, with folding s ≥ 7. The asymptotic tradeoff is the corresponding case of the existing
Chen–Zhang conjecture. As supporting results, we prove that three arbitrary replacements of
vertices of a Boolean four-cube preserve certificate cost at most 56/3, and exhibit an all-field
family of aﬀine dimension six attaining that cost. The general sixteen-point bound remains open.
Priority relative to an unavailable manuscript of Chen, Cheraghchi, Si and Zhang is explicitly
left unresolved.
1 Introduction and statements
List recovery asks how many codewords can agree on many coordinates with a table of small symbol
lists. Chen and Zhang [ 1, Conjecture 3.6 and Theorem 3.7] identified a sharp candidate tradeoff
for folded Reed–Solomon codes when the output cap plus one is a power of the input list size, and
proved the binary-input, output-three case. Brakensiek, Chen, Dhar and Zhang [ 3, Theorem 1.6]
subsequently proved the general zero-error case and developed entropy and remainder Brascamp–
Lieb methods. Their version of October 11, 2026 retains the finer positive-error tradeoff as an open
problem beyond the known output-three case [ 3, Remark 1.7 and Section 1.3].
The principal result here is a finite aﬀine-configuration inequality supplying the eight-message,
positive-error case. The proof treats arbitrary dependence coeﬀicients over the original field. It
combines elementary incidence in small aﬀine dimension with a support analysis of the aﬀine de-
pendency space in large dimension. Subspace-design and local-profile methods [ 5, 4] provide the
established coding framework.
Attribution and unresolved manuscript overlap. We acknowledge the prior work of Chen,
Cheraghchi, Si, and Zhang on this conjecture. Their manuscript [ 2] is cited by Brakensiek–Chen–
Dhar–Zhang [3] as establishing the binary-input zero-error case. We have not obtained its full text,
so overlap with the present eight-point theorem or its supporting lemmas remains unresolved; we
make no claim of priority over any result established there. The conjectured tradeoff, Boolean-cube
obstruction, rank-budget framework and folded-Wronskian method are prior work. Our mathemat-
ical statements below do not depend on access to that manuscript.
1
Let S be a finite set of distinct points, and let V = dir( S) be the linear span of its pairwise
differences. Translate one point to zero when viewing S inside V . For K ≤ V , let b2(S/K) be the
sum of the two largest cardinalities of intersections of S with aﬀine K-cosets. If there is only one
nonempty coset, the second cardinality is zero.
Definition 1.1. A certificate for S is a finite collection of subspaces W ≤ V and weights yW ∈ Q≥0
such that
b2(S/K) ≤ 2 +
∑
W
yW dim(K ∩ W ) ( K ≤ V ). (1)
Its cost is ∑
W yW dim W . The weights may depend on S, but not on K. Write C(S) for the
infimum of certificate costs.
Theorem 1.2 (Eight-point certificate) . Every set S of eight distinct points over any field has a
certificate of cost at most nine. For aﬀine dimensions 1, . . . ,7, respectively, suﬀicient costs are
6, 8, 9, 9, 9, 8, 7.
For the Boolean three-cube, C(S) = 9 .
Theorem 1.3 (Folded Reed–Solomon consequence) . Let n ≥ 1, s ≥ 7 and 1 ≤ k ≤ sn be integers,
let q > sn be a prime power, and let γ generate F×
q . Encode deg f < k by the n folded blocks
(
f (γjs), . . . , f(γjs+s−1)
)
, 0 ≤ j < n.
For 0 ≤ ρ < 3/4, the strict condition
k − 1
n(s − 6) < 2
3 − 8ρ
9 (2)
implies average-radius (ρ, 2, 7) list recovery: against any table of at most two folded symbols per
block, no eight distinct codewords have average agreement at least 1 − ρ. In particular the code is
ordinarily (ρ, 2, 7) list recoverable.
The coding statement is proved in Section 5, including all finite-parameter and multiplicative-
order hypotheses. At fixed ρ, its asymptotic rate threshold is R < 2/3 − 8ρ/9. This is the ℓ = 2 ,
L = 7 case of the Chen–Zhang conjectured curve, not a newly proposed curve. For comparison, the
generic entropy analysis in [ 3] gives the suﬀicient rate (1−ρ) log(4(1−ρ))/ log 8 in this normalization.
At ρ = 1 /4, these thresholds are 4/9 and approximately 0.39624. No new decoding algorithm is
asserted; related algorithmic results appear in [ 6].
Theorem 1.4 (Supporting sixteen-point results) . Let C be an independent-edge aﬀine Boolean
four-cube over any field. If S consists of sixteen distinct points with |S ∩ C| ≥ 13, then S has a
certificate of cost at most 56/3.
Moreover, let U = dir( C), choose thirteen vertices T ⊂ C and t ∈ T , and choose a, b linearly
independent modulo U . Then
S = T ∪ {t + a, t + b, t + a + b}
has aﬀine dimension six and C(S) = 56 /3.
The second family lies within the first theorem’s three-replacement class. Its sharpness concerns
Definition 1.1; it does not construct a coding obstruction or disprove the formal Chen–Zhang
conjecture. We do not establish the universal sixteen-point inequality or an unconditional output-
fifteen coding guarantee.
2
2 Aﬀine tools
Lemma 2.1 (Selected-fiber rank bound) . Let S = {v1, . . . , vm} have aﬀine dimension d. On
E0 = {c ∈ Fm : ∑
i ci = 0} define T (c) = ∑
i civi, and put Λ = ker T . Then dim Λ = m − 1 − d. If
K < V , choose the two largest nonempty K-fibers A, B and set c = |A| + |B|. Let E ≤ E0 consist
of coeﬀicient vectors supported on A ∪ B with separate coeﬀicient sums zero on A and B. With
δ = dim(E ∩ Λ),
b2(S/K) − 2 ≤ dim K + δ. (3)
Proof. A proper K cannot contain all pairwise differences, so there are at least two nonempty fibers.
We have dim E = c − 2 and T (E) ≤ K. Rank–nullity gives c − 2 = dim T (E) + δ ≤ dim K + δ.
Lemma 2.2 (Adjoining aﬀine coloops) . Let T consist of m ≥ 3 distinct points with direction space
U . Adjoin j points so that S has aﬀine dimension dim U +j; equivalently, their homogenized vectors
are independent modulo the homogeneous span of T . For K ≤ dir(S),
b2(S/K) ≤ b2(T /(K ∩ U )) + dim K − dim(K ∩ U ) + 1U ⊆K. (4)
Consequently, if a certificate for the modified quantity b2(T /H) + 1H=U has ambient coeﬀicient at
least one and cost D, then S has an ordinary certificate of cost at most D + j.
Proof. Translate a core point to zero. The j outside points are then linearly independent in V /U.
The case K = V is immediate. Otherwise select two nonempty largest fibers, containing s outside
points and cT core points, and let b ∈ { 0, 1, 2} count the fibers meeting the core. The space of
coeﬀicients with zero sum on each selected fiber has image modulo U of dimension s − (2 − b).
Indeed a fiber not meeting the core imposes one independent zero-sum condition on its outside
coeﬀicients; a core coeﬀicient absorbs that condition in a fiber meeting the core. Independence in
V /U gives
s − (2 − b) ≤ dim K − dim(K ∩ U ).
If U ̸⊆ K, the core occupies at least two fibers, and cT + 2 − b ≤ b2(T /(K ∩ U )): when only one
selected fiber meets it, a second nonempty core fiber contributes at least one; when neither does,
the two core fibers contribute at least two. If U ⊆ K, the same inequality holds with an additional
one on the right. Adding proves ( 4). For the last assertion, add weight one on V and subtract one
from the core ambient coeﬀicient. The new coeﬀicients are nonnegative and the cost increases by
dim V − dim U = j.
Lemma 2.3 (Modified planar certificate) . For m ≥ 3 points T spanning an aﬀine plane with
direction U , there are a ≥ 1, w ≥ 0 rational and a line L1 ≤ U with
b2(T /H) + 1H=U ≤ 2 + a dim H + w dim(H ∩ L1), 2a + w ≤ m.
Thus a planar core together with j aﬀine coloops has ordinary certificate cost at most m + j.
Proof. For a direction line L, set mL = b2(T /L) − 2. Positive values occur only in pair-difference
directions, so there are finitely many. The two largest values m1 ≥ m2 in distinct directions
are positive, because an aﬀine spanning triple supplies two directions with capture at least three.
Two unions of two parallel lines in different directions intersect in at most four points. Therefore
m1 + m2 ≤ m. Let L1 maximize mL and set
λ = max
{
1, m − 1
m1 + m2
}
, a = λm2, w = λ(m1 − m2).
3
Then a ≥ 1, w ≥ 0, and 2a + w = max {m − 1, m1 + m2} ≤ m. For H = L1 the increment
a + w ≥ m1; for every other line it is a ≥ m2. For H = U the required increment is m − 1, and for
H = 0 the bound is exact. Apply Lemma 2.2.
Lemma 2.4 (Two directions). Two distinct parallel L-lines and two distinct parallel M -lines, with
L ̸= M , have at most four cross-intersections. If there are at least three, all four lines lie in one
aﬀine plane. If there are at least two, their aﬀine span has dimension at most three.
Proof. Each line pair has at most one intersection. Three edges of the bipartite 2-by-2 intersection
graph connect its four vertices. An intersecting nonparallel pair determines a plane, and the parallel
mates reached by those edges also lie in it. With two intersections, either two matched pairs span
parallel planes of direction L + M , whose union has aﬀine dimension at most three, or three lines
lie in a plane and adjoining the fourth increases dimension by at most one.
3 Eight points in aﬀine dimensions at most four
Throughout, captures refer to b2(S/K). For dimension one, weight six on V works: the zero
kernel captures two and V captures eight. Dimension two follows from Lemma 2.3, discarding its
nonnegative indicator. Dimension seven will follow from aﬀine independence in Section 4.
Lemma 3.1 (Dimension three) . Eight distinct points of aﬀine dimension three admit cost at most
nine.
Proof. If every line direction captures at most five, take yV = 3. Lines get right-hand side five and
planes eight, so all kernels are covered. Otherwise choose a line direction L with capture cL ≥ 6. It
is at most seven: two parallel lines have aﬀine span at most two. For any plane P not containing L,
its two selected cosets capture at most four points on the selected two L-lines and at most 8 − cL
outsiders. Hence
b2(S/P ) ≤ 12 − cL. (5)
If cL = 7 and another direction M captured at least five, the selected subsets would meet in at
least four points. There are at most four cross-intersections, so Lemma 2.4 would make all eight
points coplanar, a contradiction. Thus other lines capture at most four. Weights yV = 2 , yL = 3
work: line L gets seven, other lines four, planes containing L get nine, and transverse planes get
six, enough by ( 5). Cost is nine.
If cL = 6, no other line can capture six: four common selected points would again make all eight
points coplanar. If another direction captures five, the selected subsets meet in three or four points.
Three intersections put their eight-point union in a plane, impossible. Four intersections give seven
coplanar points plus one aﬀine coloop, covered by Lemma 2.3 at cost at most eight. Otherwise
all other lines capture at most four. Take yV = 2 , yL = 2 , of cost eight. Planes containing L get
eight and transverse planes get six, again covering ( 5). Zero and full kernels satisfy every displayed
certificate directly.
Lemma 3.2 (Modified seven-point core) . Seven distinct points T of aﬀine dimension three, with
direction U , admit a certificate for b2(T /H) + 1H=U of cost at most eight and with coeﬀicient on
U at least one.
Proof. If every line captures at most four, weight 5/2 on U works. Otherwise a line L captures
five or six. If it captures six, another line capturing at least five would have four common selected
points, putting all seven points in a plane. Thus all other lines capture at most four. Use yU = 2 ,
4
yL = 2, of cost eight. A plane avoiding L captures at most five and gets six; a plane containing L
gets eight. The modified full-kernel value is eight and is covered.
Suppose the largest line capture is five. If another direction captures five, the selected subsets
meet in three or four points. Three would make all seven coplanar. Four give six coplanar points
and one aﬀine coloop. Lemma 2.3 supplies an ordinary certificate on U with ambient coeﬀicient
one and cost between six and seven: the modified planar core has cost at least m − 1 = 5 and
adjoining the coloop adds one. Thus its full-kernel right-hand side is at least eight, covering the
extra indicator as well. If L is the only direction capturing five, take yU = 2 , yL = 1 . Planes
avoiding L capture at most 4 + 2 = 6 ; those containing it get seven. The modified full-kernel value
eight is covered by nine. All other ranks are immediate.
Lemma 3.3 (Dimension four) . Eight distinct points of aﬀine dimension four admit cost at most
nine.
Proof. If a line direction captures at least six points, six points on two parallel lines span at most a
plane. The remaining two points force that plane to have dimension two and are independent aﬀine
coloops, since the total aﬀine dimension is four. Lemma 2.3 gives cost at most eight. Line captures
seven or eight are incompatible with dimension four. Similarly, if a plane direction captures at
least seven points, seven lie in an aﬀine space of dimension at most three; dimension four forces a
seven-point three-dimensional core and one coloop. Lemmas 3.2 and 2.2 give cost at most nine. A
plane cannot capture all eight.
We may assume line captures at most five and plane captures at most six. If no line captures
five, yV = 9 /4 works: ranks one, two and three get 17/4, 13/2, 35/4, respectively. If just one line
L captures five, yV = 2, yL = 1 works.
It remains to consider two different directions capturing five. Their selected subsets A, B meet
in t ∈ {2, 3, 4} points. If t = 2, their union is all eight and Lemma 2.4 bounds its aﬀine dimension
by three. If t = 3, seven points lie in a plane and the last point raises dimension to at most three.
Both contradict the hypothesis. If t = 4, their union is six planar points; dimension four forces two
independent aﬀine coloops, already covered. This exhausts the cases; zero and full kernels satisfy
the stated weights.
4 Dependency supports in larger aﬀine dimension
Use the notation of Lemma 2.1 with m = 8 . Distinctness excludes a nonzero aﬀine dependency
supported on fewer than three points. We will repeatedly use the following observation. Let C be
the exact support of a nonzero dependency, and suppose the dependency space restricted to C is
one-dimensional. Then its direction space UC has dimension |C| −2. If that dependency belongs to
E, all points of C lie in b ≤ 2 selected fibers, with coeﬀicient sum zero on each. The coeﬀicient space
on C with these zero-sum conditions has dimension |C| −b; its image has dimension |C| −b − 1 and
lies in K ∩ UC. This image is nonzero: for |C| = 3, a split into two nonempty groups would create
a singleton with nonzero coeﬀicient, so b = 1; for |C| ≥ 4, the dimension is at least |C| − 3 ≥ 1.
Lemma 4.1 (Dimension five) . Every eight-point set of aﬀine dimension five admits cost at most
nine.
Proof. Here dim Λ = 2 . Let C be the union of the supports of all dependencies, put t = |C| and
U = dir( C). All dependencies are supported on C, so u = dim U = t − 3. Distinctness implies
t ≥ 4. For proper K, put h = dim(K ∩ U ) and retain δ = dim(E ∩ Λ).
5
First suppose C is proper, so u ≤ 4. If δ > 0, then h ≥ 1: otherwise projection modulo K is
injective on the aﬀine span of C, so the two selected fibers contain at most two core points, too few
to support a dependency. If δ = 2, all dependencies belong to E, so all of C lies in the two selected
fibers. If b ≤ 2 of these meet C, the coeﬀicient space on C with zero sums on the groups has image
dimension t − b − 2. Thus
h ≥ t − b − 2 ≥ u − 1.
For u = 3, 4 these observations give δ ≤ h; take yV = yU = 1, of cost 5 + u ≤ 9. For u = 1, 2 they
give δ ≤ 2h; take yV = 1, yU = 2, of cost 5 + 2u ≤ 9. Lemma 2.1 proves the inequalities for proper
kernels, and the full kernel is covered because the cost is at least seven.
Now suppose C = S. Any two independent dependencies have supports whose union is all eight
points, because they span Λ. Call a one-dimensional dependency subspace small if its support has
size at most four. There are three possibilities: no small line; one small line of support size three
or four; or two small lines with disjoint four-point supports. Indeed two different small lines must
have union of supports all eight, hence both supports have size four and are disjoint. A combination
using both has support eight, so there is no third small line.
For a proper K of dimension k ≥ 2, we have δ ≤ k/2. This is immediate for δ ≤ 1. If δ = 2, all
eight points occur in the selected fibers, so dim E = 6 and dim T (E) = 4 , forcing k ≥ 4. Therefore
b2(S/K) ≤ 2 + 3
2 dim K (k ≥ 2). (6)
For a line kernel, δ = 2 is impossible. If δ = 1 , Lemma 2.1 gives capture at most four, so the
dependency in E spans a small line.
With no small line, weight 3/2 on V works, of cost 15/2. With one small support C0, take
yV = 3 /2, yUC0
= 1 /2, of cost at most 17/2. The observation preceding the lemma supplies a
nonzero intersection for any exceptional line kernel, and ( 6) handles the larger proper kernels.
With two small lines, use their disjoint four-point supports C1, C2 and Ui = dir( Ci), each of
dimension two. Take yV = yU1 = yU2 = 1 , of cost nine. If δ = 2 , both dependencies belong to E,
giving nonzero intersections with both Ui. If δ = 1 and its line is one of these two, one nonzero
intersection suﬀices. Otherwise its generator uses both supports and hence all eight points. Then
dim E = 6 and dim T (E) = 5 , forcing K = V , contrary to propriety. Thus in every proper case
δ ≤ dim(K ∩ U1) + dim(K ∩ U2),
as required by Lemma 2.1. In all three full-core cases, zero and full kernels satisfy the displayed
certificates directly.
Lemma 4.2 (Dimensions six and seven) . In aﬀine dimension six the suﬀicient cost is at most
eight; in dimension seven it is seven.
Proof. In dimension six the dependency space has dimension one. Let its nonzero support C have
size t ≥ 3, so U = dir(C) has dimension t − 2. If δ = 1 for a proper kernel, the observation at the
start of the section gives
dim(K ∩ U ) ≥ max{1, t − 3} =: a.
Thus yV = 1, yU = 1/a works and has cost 6 + (t − 2)/a ≤ 8. For δ = 0 ambient weight one suﬀices.
The full kernel is covered since the cost is at least six.
In dimension seven the points are aﬀinely independent, so Λ = 0 . Lemma 2.1 gives b2(S/K) ≤
2 + dim K for proper K. Weight one on V also covers the full kernel and has cost seven.
6
Proof of Theorem 1.2. The upper bounds follow from the planar and low-dimensional arguments
and Lemmas 4.1–4.2. All displayed weights are rational, nonnegative, finitely supported and fixed
for S.
For sharpness let S = {0, 1}3 in a three-dimensional space. Let Hi be its three coordinate
hyperplanes. Then b2(S/Hi) = 8 . For every W , the coordinate functionals restricted to W span
W ∗, so at least dim W restrictions are nonzero. Consequently
3∑
i=1
dim(W ∩ Hi) ≤ 2 dim W.
Summing any certificate’s inequalities at these three kernels gives 18 ≤ ∑
W yW
∑
i dim(W ∩ Hi) ≤
2 ∑
W yW dim W . Every certificate therefore costs at least nine, proving equality.
5 From certificates to folded Reed–Solomon codes
We give the standard polynomial rank-budget argument explicitly to fix its characteristic, order and
finite-folding conventions. This is established subspace-design machinery [ 5, 4], not an additional
claim of originality.
Lemma 5.1 (Folded-Wronskian budget). Under the hypotheses of Theorem 1.3, let Pk = {f ∈
Fq[X] : deg f < k } and let Kj ≤ P k be the kernel of the jth folded evaluation block. For W ≤ P k
with 1 ≤ u = dim W ≤ s,
n−1∑
j=0
dim(W ∩ Kj) ≤ u(k − 1)
s − u + 1 .
Proof. Choose a basis g1, . . . , gu of distinct degrees di < k by degree elimination. The polynomial
D(X) = det
(
gb(γaX)
)
0≤a<u, 1≤b≤u
is nonzero: its coeﬀicient in degree ∑
b db is a product of the nonzero leading coeﬀicients and the
Vandermonde determinant in the distinct elements γdb. They are distinct because γ has order
q − 1 ≥ sn and 0 ≤ db < k ≤ sn. Also deg D ≤ u(k − 1).
Put hj = dim(W ∩ Kj). Change basis so the first hj polynomials vanish on block j. For every
0 ≤ t ≤ s − u, all entries in those columns vanish at X = γjs+t. Each such column is divisible by
X − γjs+t, so the determinant has root multiplicity at least hj. A basis change multiplies D by
a nonzero scalar. All these roots, across j, t, are distinct by the same order condition. Summing
multiplicities gives (s − u + 1) ∑
j hj ≤ deg D ≤ u(k − 1).
Proof of Theorem 1.3. All sn evaluation points are distinct and k ≤ sn, so the encoding is injective.
Suppose eight distinct messages formed a forbidden average-radius list. Their aﬀine-direction space
V ≤ P k has dimension at most seven. At block j, the evaluation fibers are aﬀine cosets of Kj ∩V , so
the number of messages whose folded symbol is in the two-symbol table is at most b2(S/(Kj ∩ V )).
Apply Theorem 1.2. For each certificate support W ≤ V , its dimension is at most seven, and
Lemma 5.1 gives
1
n
∑
j
dim(W ∩ Kj) ≤ k − 1
n(s − 6) dim W =: r dim W.
The mean number of captured messages is therefore at most 2 + 9 r. A verage agreement at least
1 − ρ would require it to be at least 8(1 − ρ), giving r ≥ 2/3 − 8ρ/9, contrary to ( 2). If each of
eight messages individually has agreement at least 1 − ρ, its average does too, proving ordinary list
recovery.
7
Corollary 5.2 (Asymptotic form) . Fix 0 ≤ ρ < 3/4, set B = 2 /3 − 8ρ/9, and take 0 < η < B .
Under the remaining hypotheses of Theorem 1.3, rate R = k/(sn) ≤ B −η and folding s ≥ 7+ ⌈12/η⌉
suﬀice.
Proof. Here R ≤ B ≤ 2/3, and
r < Rs
s − 6 = R + 6R
s − 6 < R + η ≤ B.
The integer and field constraints still apply. The theorem does not include equality r = B.
Remark 5.3. Sharp cost nine does not establish exact finite-code optimality. The asymptotic target
and its coding obstruction originate in [ 1]. Our proof establishes a suﬀicient inequality through a
particular class of nonnegative certificates. We make no claim about optimal extraction complexity,
and do not infer new code constructions from established transfer theorems.
6 Three replacements and a noncube sharp family
The following proof treats the structured sixteen-point statements independently of any universal
sixteen-point assertion.
Lemma 6.1 (Cube intersection bound) . An aﬀine h-flat meets an independent-edge Boolean a-cube
in at most 2h points.
Proof. Inside the cube’s aﬀine span, choose coordinate functionals whose restrictions give an injec-
tive projection of the intersection flat into Fr, where r ≤ h is that flat’s dimension. Each chosen
coordinate of a cube vertex has at most two values, so there are at most 2r ≤ 2h such vertices. The
argument uses neither an ordering nor a characteristic assumption.
Proof of the first assertion of Theorem 1.4. Translate the cube’s aﬀine span to its direction space
U , of dimension four. At least thirteen retained vertices span U , since a proper aﬀine subspace
contains at most eight cube vertices. For K ≤ V = dir( S) put H = K ∩ U , h = dim H, and
δ = dim K − h. Core fibers are its H-fibers. Let j be the number of points of S outside U . We
have 0 ≤ j ≤ 3.
No external points: j = 0 . Each two-line capture is at most 4 + 3 = 7 and each two-plane
capture at most 8 + 3 = 11 . There is at most one line direction of capture seven. To see this, two
such selected covers each contain four cube vertices and all three noncube points. Their common
three points force all four support lines into a plane by Lemma 2.4. There are at most four cross-
intersections, so their cube subsets share at most one point. The plane would contain at least seven
cube vertices, contradicting Lemma 6.1.
If there is no direction of capture seven, use yV = 14/3. Captures at ranks one and two are at
most six and eleven, respectively, and rank three has right-hand side sixteen. If there is a unique
heavy direction L, use yV = 9/2, yL = 1/2, of cost 37/2. Lines and planes are covered. A rank-three
kernel containing L gets sixteen. If it does not contain L, two of its aﬀine cosets meet the selected
two L-lines in at most four points and cannot cover all seven points there. Its capture is at most
fifteen, below its right-hand side 31/2. Zero and full kernels are immediate.
8
One external point: j = 1 . The fifteen-point core contains at least thirteen cube vertices, so
its captures at h = 1, 2, 3 are bounded by 6, 10, 15. At h = 4 its modified capture is sixteen. Hence
b2(T /H) + 1H=U ≤ 2 + 13
3 h.
The external point is an aﬀine coloop. Lemma 2.2 gives yV = 1, yU = 10/3, of cost 55/3.
T wo external points: j = 2 . The fourteen-point core contains at least thirteen cube vertices.
Its single-fiber and two-fiber bounds are min(14, 2h +1) and min(14, 2h+1 +1); at h = 0 the two-fiber
capture is exactly two. If the external quotient vectors have rank two, they are independent aﬀine
coloops. The modified core coeﬀicient four suﬀices, and Lemma 2.2 gives yV = 1 , yU = 3 , of cost
eighteen.
If their rank is one, use yV = 2, yU = 2, also of cost eighteen. Its right-hand side is 2 + 4h + 2δ.
For δ = 1 the core two-fiber bound plus two gives, at h = 0, . . . ,4, the bounds 4, 7, 11, 16, 16, each at
most 4 + 4h. For δ = 0, no outsider shares a fiber with a core point. At h = 1, . . . ,4, the maximum
of two core fibers, one core fiber plus two outsiders, or two outsiders is bounded by 6, 10, 14, 16,
respectively, all at most 2 + 4h. At h = 0 distinctness gives two.
Three external points: j = 3. The core T consists of thirteen cube vertices. A single core fiber
has size at most min(13, 2h) and two have total at most min(13, 2h+1). Let r ∈ {1, 2, 3} be the rank
of the three nonzero quotient vectors in V /U.
If r = 3, the outsiders are independent aﬀine coloops. The modified core coeﬀicient 11/3 suﬀices,
and Lemma 2.2 gives yV = 1, yU = 8/3, of cost 53/3.
For r = 1 use yV = 3 , yU = 2 /3; for r = 2 use yV = 2 , yU = 5 /3. In either case the coeﬀicient
of h is 11/3. The following bounds cover all possible h and δ:
r δ h = 0 h = 1 h = 2 h = 3 h = 4
1 0 2 5 8 13 16
1 1 5 7 11 16 16
2 0 2 4 8 13 15
2 1 4 6 10 15 16
2 2 5 7 11 16 16
Each entry is at most 2 + (11/3)h + 3δ in the rank-one case, or 2 + (11/3)h + 2δ in the rank-two
case. For completeness, we justify the counts rather than relying on the table alone. When δ = 0,
no outsider joins a core fiber. Bound the capture by the maximum of two core fibers, one core fiber
plus outsiders, and outsiders alone. At rank two at most two outsiders can occupy one fiber, since
three in one fiber would have equal quotient vectors, of rank one. These observations give the δ = 0
rows, with h = 0 handled by distinctness. For rank one and δ = 1 , or rank two and δ = 2 , add
three to the core two-fiber bound.
In the remaining case r = 2 , δ = 1 , if both selected fibers meet the core, the selected outsider
quotient vectors lie in the line π(K) ≤ V /U. At most two of the three lie there, since their total
rank is two. Thus use the two-core-fiber bound plus two. If only one selected fiber meets the core,
use the single-core-fiber bound plus three. If neither does, use three. Their maxima give the middle
rank-two row. This includes oblique kernels as well as coordinate kernels.
The costs for r = 1, 2 are 3 · 5 + (2/3) · 4 = 53 /3 and 2 · 6 + (5/3) · 4 = 56 /3. All cases therefore
satisfy the required bound.
9
Proof of the sharp-family assertion of Theorem 1.4. Translate the cube to U . The thirteen vertices
span U by Lemma 6.1. Independence of a, b modulo U makes the added points distinct and gives
aﬀine dimension six. The preceding theorem supplies yV = 2, yU = 5/3, of cost 56/3.
For a matching lower bound, let Hi ≤ U be the four coordinate hyperplanes, and A1 = span(a),
A2 = span( b). For every X ≤ V put H = X ∩ U , h = dim H. At least h of the four restricted
coordinate functionals are nonzero, since they span H ∗. Hence
∑
i
dim(X ∩ Hi) ≤ 3h.
Also U, A1, A2 form a direct sum, so
dim(X ∩ A1) + dim(X ∩ A2) + 1
3
∑
i
dim(X ∩ Hi) ≤ dim X. (7)
For each Hi, the core occupies two fibers of sizes between five and eight. The three new points have
distinct nonzero quotient vectors modulo U and are singleton Hi-fibers. Thus b2(S/Hi) = 13 . In
direction A1, the square t + {0, a, b, a + b} supplies two double fibers and all other core points are
singleton fibers. Thus b2(S/A1) = 4 , and similarly for A2.
Multiply any valid certificate inequalities at A1, A2 by one and at the Hi by 1/3. Their total
left-hand excess over the intercept is
(4 − 2) + (4 − 2) + 4(13 − 2)
3 = 56
3 .
Applying (7) to each certificate support shows that this is at most its cost. This finite weak-duality
argument is valid over infinite fields as well. The matching upper certificate proves the exact
optimum.
Scope and further questions. The sharp family is assembled from a thirteen-vertex cube
core and an attached square. It rules out a cube-only equality classification for this certificate
problem, but no published uniqueness assertion is being contradicted. Proving cost at most 56/3 for
arbitrary sixteen-point configurations, and understanding all equality configurations, are separate
open tasks. A certificate lower witness by itself does not realize the simultaneous degree, evaluation
and agreement constraints needed for a coding lower bound.
Proof and computational status. All theorems in this paper have analytic proofs. Exact finite
computations served as diagnostics during development and are not premises of the arguments.
Separate AI-assisted adversarial reviews reconstructed the proofs; these are not external human
refereeing or proof-assistant verification.
AI assistance. OpenAI assistants were used extensively in developing arguments, literature dis-
covery, verification scripts, independent review attempts and drafting. Human authorship and
responsibility remain with the named author. The source-bound review records distinguish fresh
computations from authenticated reuse of earlier checks.
References
[1] Y. Chen and Z. Zhang, Explicit Folded Reed–Solomon and Multiplicity Codes Achieve Relaxed
Generalized Singleton Bounds , STOC 2025. arXiv:2408.15925v3, April 14, 2025; first version
August 28, 2024.
10
[2] Y. Chen, M. Cheraghchi, K.-S. Si and Z. Zhang, Evidence Toward a Conjecture in List Recov-
erability of Folded Reed–Solomon Codes , manuscript, 2025. Bibliographic entry and reported
zero-error scope follow [ 3]; full text unavailable to the present author at preparation.
[3] J. Brakensiek, Y. Chen, M. Dhar and Z. Zhang, Combinatorial Bounds for List Recov-
ery via Discrete Brascamp–Lieb Inequalities , full version of STOC 2026 extended abstract.
arXiv:2510.13775v2, October 11, 2026; first version October 15, 2025.
[4] J. Brakensiek, Y. Chen, M. Dhar and Z. Zhang, From Random to Explicit via Subspace Designs
With Applications to Local Properties and Matroids , STOC 2026. arXiv:2510.13777v1, October
15, 2025.
[5] V. Guruswami and S. Kopparty, Explicit subspace designs , Combinatorica 36 (2016), 161–185.
doi:10.1007/s00493-014-3169-1.
[6] R. Goyal and V. Guruswami, Structure Theorems (and Fast Algorithms) for List Recovery of
Subspace-Design Codes, arXiv:2512.08017v1, December 8, 2025.
11