Erdős problem 352 — wave w041

Date: 2026-08-12 (UTC; initial run 2026-07-31)

Claim labels

• [a] elementary-rigorous: a complete non-computational argument is given

here.

• [b] rigorous modulo named theorem/source: the theorem and primary source

are identified.

• [c] plausible/structural-unverified: explicitly not asserted as a

theorem.

• [d] computational-only: a finite exact computation with a standalone

checker.

• [source-checked]: a bibliographic or live-site fact directly checked at

the linked source; this is not a mathematical-proof label.

2026-08-12 fifth update: exact values through f(26)

Three more finite values close exactly:

These are [d: computational-only] and still do not imply a uniform bound in \(N\).

The three lower witnesses pass every triple with repetition under the exact predicate \((A_2-2N)^2-4D_2>0\):

The upper audits exhaust every normalized diameter vector, no-high branch, and fixed-high branch, reconstructing and directly checking a proper coloring of each compatibility graph:

The strict runs took 4,094.44, 6,223.87, and 9,988.56 checker seconds, respectively. Their canonical certificate-stream SHA-256 values are:

N=24  4bdd9d9f9bad728c22531e753711fe126971ce3eb29d52937c818add3b15202b
N=25  da9e4abb06fc00a51d58ff40e255889c7d9c3cafe502276e97d52260be3f5a3a
N=26  a0a09b5ebee6bcdae77c14f296b0172c2586a95af83250ab0f5237fddb0d827a

The unmodified public upper implementation at commit 7a4a19468771766cb4f20e10dfa5cf770cc2487b independently reproduces all three global bounds and every one of the 243, 263, and 285 per-diameter rows. A separate hostile C++ implementation rebuilds the signed-vector/D4 census, lenses, branches, and full repeated-triple coexistence predicate; it also reproduces all three certificate streams byte-for-byte. The frozen evidence is in f24-exact, f25-exact, f26-exact, and f24-26-hostile. Run the assertion-based checkers with ordinary python3, never python3 -O. [d, source-checked]

2026-08-12 fourth update: a local-universality obstruction

The remaining fixed-high branches admit the following all-\(N\) structural lemma. It is a negative result about a natural coloring strategy, not the missing upper bound.

Let \(d\in\mathbb Z^2\setminus\{0\}\), write \(\delta=|d|\), and suppose \(100\delta\le N\). Then there is an oriented fixed-high branch with forced points \(O=0,B,H\) whose compatibility graph contains the arithmetic- progression clique

Here oriented means that \(B\) is an arbitrary integral diameter endpoint; this family has not been quotiented by the D4 symmetries used in the finite enumeration. In particular, the union of compatible differences over all oriented fixed-high branches contains every nonzero lattice vector of length at most \(N/100\). [a]

The construction is explicit. Put \(g=\gcd(d_x,d_y)\), let \(t\) be the least positive multiple of \(g\) with \(t\ge16\delta\), and choose \(v\in\mathbb Z^2\) such that

Such a normalized Bezout solution exists because all solutions differ by a multiple of \(d/g\). Moreover

With

take the point set \(O,B,H,H-d,\ldots,H-Ld\). Since \(v\cdot d\ge0\), the distance \(|rd+v|\) is nondecreasing for \(r\ge0\), so \(O,B\) are a diameter pair. Writing \(D=|B|\),

Every point other than \(B\) is collinear. A nonzero triangle therefore has area \(kt/2\), where \(k\) is a coefficient gap. Gaps inside the displayed cluster have \(k\le L\), hence

Gaps from \(O\) to the same cluster have \(k\ge M-L\), hence

Collinear and repeated triples have area zero and diameter at most \(D<N\). Thus every triple, with repetitions, satisfies the strict avoiding inequality. The gap from \(O\) to \(H\) also proves that \(H\) is high, and the remaining \(L\) points coexist simultaneously, proving the clique claim. [a]

This rules out one broad but tempting approach. Fix \(C\), and suppose one homomorphism

is used unchanged to color every oriented fixed-high branch at that \(N\). Among the \((\lceil\sqrt{CN}\rceil+1)^2\) points of the corresponding square, two have the same image. Their nonzero difference lies in \(\ker\phi_N\) and has length \(O_C(\sqrt N)\), hence at most \(N/100\) for all sufficiently large \(N\). The lemma realizes that difference as a compatible edge, which the homomorphism colors monochromatically. Consequently no single small-image homomorphic coloring—in particular, no single lattice-coset coloring of index \(O(N)\)—works unchanged across the whole oriented family. [a]

The orientation qualifier is essential. This does not rule out one homomorphic coloring common only to chosen D4-normalized representatives, because normalizing a branch also transports the coloring kernel. It also does not rule out a coloring chosen adaptively from the particular \(B,H\), which remains the relevant possibility for a uniform upper bound.

The standalone exact checker tests 47,088 signed and pseudorandom instances, including every repeated triple in the five-point core; an independent hostile implementation additionally materializes 9,821,816 repeated triples in the full progression sets. Both pass. The amended producer and hostile audit are frozen in verification/352/local-universality. Run the assertion-based checker with ordinary python3, never python3 -O. [d]

2026-08-12 third update: exact values through f(23)

Five further finite values close exactly:

These are [d: computational-only] and do not imply a uniform bound in \(N\).

The existing lower witnesses pass every triple with repetition under the exact predicate

Their definition-level audits are:

No priority claim is made for these previously posted witnesses. [d, source-checked]

The upper audits exhaust every normalized diameter vector, its no-high branch, and every fixed-high branch. Each reconstructs and directly checks a proper coloring of the corresponding compatibility graph:

The runs took 392.15, 666.96, 1,096.64, 1,846.97, and 3,140.50 seconds respectively, with maximum resident sizes 21,444, 21,408, 21,332, 20,620, and 20,560 KiB. Their canonical certificate-stream SHA-256 values are:

N=19  2d8b6e3ef793869a9e6a05726e6e7b434efc7b3695d3307343e1b083a4e1822d
N=20  bafa932d0529410be48dd4fb8cbd31362ca00bb1188a41699107522f89588996
N=21  cd5c1d7f245d3412ee966bd4ffbf138f06f6d5213d57ab589b9563001ae9bca3
N=22  64858529dc55172b3d9abba0595768f3f7da04517be5129e51cb265f777f55d0
N=23  7ce7aaa06bfd68ad93f7ac06c79a6454337ba034efb81c9e1a0ffb1541d250f3

Thus every branch is bounded by the size of the corresponding witness, which proves (W1). [d]

The unmodified public upper-bound implementation at commit 7a4a19468771766cb4f20e10dfa5cf770cc2487b independently returns the same diameter counts, high-branch counts, global bounds, and tight directions in all five cases. The source witness JSON, strict checker output, and public reference output are frozen in f19-exact, f20-exact, f21-exact, f22-exact, and f23-exact. An independent hostile implementation reconstructs every branch through N=23, checks the full repeated-triple predicate inside each color class, and reproduces all five certificate digests byte-for-byte; its compact bundle is f19-23-hostile. Run the assertion-based checker with ordinary python3, never python3 -O. [d, source-checked]

2026-08-12 second update: the exact value f(18)=34

The finite computation has now been extended by one further exact value:

This is [d: computational-only] and still does not prove the uniform estimate f(N)=O(N) needed for the continuous problem.

The 34-point lower witness was posted previously and is frozen in the public witness repository cited below. The independent audit checks all \(\binom{34+2}{3}=7{,}140\) triples with repetition using

All pass; the minimum integer margin is 21. No priority claim is made for the witness. [d, source-checked]

For the upper bound, the exhaustive diameter-pair reduction has 140 normalized vectors, 140 no-high branches, and 12,594 fixed-high branches. The primary audit reconstructs 12,734 coloring certificates and directly checks 3,011,741 same-color pairs. Pair coloring bounds every branch by 34 except the no-high branch with B=(5,4), where its relaxation gives 35. [d]

That exceptional branch has 42 low candidates and 57 incompatibility edges. The following nine edges form a matching:

(-1,2)--(6,2)    (-1,3)--(2,-1)   (0,1)--(3,-2)
(0,4)--(3,-1)    (1,-1)--(4,5)    (1,5)--(4,-1)
(2,5)--(5,0)     (2,6)--(5,3)     (3,5)--(6,1)

If 33 candidates were pairwise compatible, their nine omitted vertices would be a vertex cover and would have to select exactly one endpoint from each matching edge. An exhaustive audit of all 2^9=512 endpoint choices finds exactly two vertex covers. The first surviving 33-set contains both forbidden triples

(1,-1), (1,4), (6,1)
(1,-1), (1,4), (6,2),

and the second contains both

(-1,2), (4,0), (4,5)
(-1,3), (4,0), (4,5).

Each has A_2=25, D_2=34, and exact margin

Thus the exceptional branch has at most 32 candidates, hence at most 34 points after restoring its diameter pair. This closes the upper half of (V1). [d]

The complete run took 216.13 seconds and 20,940 KiB maximum resident memory. Its canonical audited certificate-stream SHA-256 is b761add523bbe8a4f1d8f822268ddc28a893e9a1318ebc73807822d7ca09fea8. A separate focused checker rebuilds the witness, lens, incompatibility graph, matching, covers, and four failed triples directly in under a second. The versioned verification bundle is verification/352/f18-exact. Run the assertion-based checker with ordinary python3, never python3 -O. [d]

2026-08-12 first update: the exact value f(17)=32

The live problem, discussion, and proof-claim pages were fetched again through the Bright Data browser before this update. They still showed status OPEN, 17 comments, 0 claimed proofs, and Currently working on this problem: None. The dedicated proof page again said that no proof claims had been submitted. [source-checked]

The finite computation below has now been extended by one exact value:

This is [d: computational-only]. It does not prove the uniform estimate f(N)=O(N) and therefore does not solve the continuous problem.

The 32-point lower witness was posted previously by BorisAlexeev and appears in KentaKitamura's public witness repository. An independent definition-level check evaluates all \(\binom{32+2}{3}=5{,}984\) triples with repetition using the exact predicate

All pass, and the minimum integer margin is 5. No priority claim is made for this witness. [d, source-checked]

For the upper bound, the same diameter-pair reduction described below gives 125 normalized vectors B, 125 no-high branches, and 9,414 fixed-high branches. Deterministic proper colorings, independently audited by full triple enumeration within every color class, bound every branch by 32 except the no-high branch with B=(5,4). Across all 9,539 branches the audit checks 1,534,917 same-color pairs. [d]

The exceptional branch has 40 low candidates and an incompatibility graph with 48 edges. It contains the following matching of nine disjoint edges:

(-1,2)--(4,0)    (-1,3)--(3,0)    (0,3)--(4,-1)
(0,4)--(2,-1)    (1,-1)--(4,5)    (1,4)--(6,1)
(1,5)--(5,1)     (2,5)--(5,0)     (3,5)--(6,2)

A pairwise-compatible 31-subset would have to omit exactly one endpoint of each matching edge. The checker audits all 2^9=512 choices. Exactly two nine-point omissions cover all 48 incompatibility edges. The corresponding 31-subsets are respectively killed by the triples

(0,3), (3,-1), (4,5)
(1,-1), (2,5), (5,1).

For each triple, A_2=22 and D_2=37, so its exact margin is

Thus the exceptional branch contains at most 30 candidates, or 32 points after adjoining its diameter pair. This proves the upper half of (U1). [d]

Two standalone implementations and their logs are frozen in the versioned verification bundle. The primary checker reconstructs every branch and prints certificate-stream SHA-256 dd1b957a56174169a95cc31fd68d77b47d51dd38745f6252059954b74aad78e2. A hostile implementation uses a different shoelace/dot-product geometry path and set-based DSATUR coloring; it independently reproduces the global cap and all branch counts. A separate 80-decimal radical audit checks 5,984 witness triples and 21,119 lens triples. It finds 12 exact zero-margin lens triples, so the strict sign in (U2) is materially exercised rather than vacuous. [d]

The full primary run took 113.40 seconds and 20,292 KiB maximum resident memory; the independent full branch audit took 211.23 seconds and 19,428 KiB. The short matching audit takes under a second. Run the assertion-based checkers with ordinary python3, never python3 -O. [d]

Initial 2026-07-31 live-page check

I fetched the live problem page, its complete discussion thread, and its proof-claim page through the Bright Data browser on 2026-07-31. Direct extraction was allowed to finish before any mathematical work. [source-checked]

The live statement, verbatim apart from normalizing its visual line breaks, is:

Is there some 𝑐 >0 such that every measurable 𝐴 ⊆ℝ² of measure ≥𝑐 contains the vertices of a triangle of area 1?

The same page displayed all of the following:

• status: OPEN;
• last edited: 23 January 2026;
• 17 comments on this problem;
• 0 claimed proofs for this problem; the dedicated page says “No proof

claims have been submitted yet”;

• Currently working on this problem: None;
• **Interested in collaborating: Vjeko_Kovac, JineonBaek,

J_Koizumi_144**.

Thus neither mandatory stop condition applied. In particular, “interested in collaborating” is recorded but is not the page's “currently working” marker. [source-checked]

Known results displayed on the live page

The page lists the following facts.

1. Erdős proved (unpublished) that an infinite-measure planar set, or an

unbounded positive-measure set, contains a triangle of every prescribed positive area. The cited Erdős sources say this follows easily from the Lebesgue density theorem. [b, source-checked]

1. Erdős suggested the sharp constant

\[ c_0=\frac{4\pi}{\sqrt{27}}\approx2.418. \] An open disk of radius \(<2\cdot3^{-3/4}\) is the obstruction: its largest inscribed triangle is equilateral and has area \(<1\). [a for the disk calculation; source-checked for Erdős's conjecture]

1. Freiling and Mauldin proved that outer measure \(>c_0\) forces a triangle

of area \(>1\). The page notes that this gives the original threshold for a compact convex set. [b: Mauldin 2002, source-checked]

1. It suffices to settle the question for sets that are unions of the

interiors of finitely many compact convex sets; Freiling and Mauldin proved the sharp \(c_0\) result for unions of at most three such sets. [b: Mauldin 2013, source-checked]

What the 17 live comments add

The site warns that comments are user-supplied and not verified. I read all 17 and used them only after the checks described below. [source-checked]

• The comments supply the older Erdős references and the Mauldin 2002/2013

references, now incorporated into the main page.

• J_Koizumi_144 gives a precise lattice reduction. A finite

\(S\subset\mathbb Z^2\) is called \(N\)-avoiding when \[ \bigl|\operatorname{area}(\triangle pqr)-N\bigr| >\operatorname{diam}(\triangle pqr) \tag{1} \] for every triple \(p,q,r\in S\), with repetitions allowed. If \(f(N)\) is the maximum size of such a set, the continuous problem is equivalent to \(f(N)=O(N)\). A linked note contains the proof.

• BorisAlexeev posted candidate lower-bound witnesses through \(N=27\).

KentaKitamura's public repository rechecks those and supplies additional witnesses through \(N=40\).

• KentaKitamura posted a diameter-pair/compatibility-graph upper-bound scheme

and exact values only for \(N=1,\ldots,7\): \[ 1,4,4,7,9,10,12. \] The comment says the readable exact-chromatic implementation became too slow beyond \(7\).

• A separate comment claims and argues that the all-low subproblem has sharp

asymptotic \((4\pi/\sqrt{27})N+O(\sqrt N)\). Other comments discuss dyadic discretization, horizontal sections, toy grid computations, disk-like extremizers, and widely separated “local extremizers.” These are useful structural context, not claimed proofs of the full problem.

• The newest substantive continuous result cited in the thread is the May

2026 preprint of Bulj and Kovač discussed below.

All names, dates, links, tables, and the previous \(N\le7\) exact range in this inventory were checked in the rendered thread. [source-checked]

Primary-source literature audit

1. Paul Erdős, *Set-theoretic, measure-theoretic, combinatorial, and

number-theoretic problems concerning point sets in Euclidean space*, Real Analysis Exchange 4 (1978/79), 113–138, author archive PDF. Pages 121–123 state the infinite-measure observation, the finite constant question, the disk obstruction, and the proposed value \(4\pi/\sqrt{27}\). [b, source-checked]

1. R. Daniel Mauldin, Some Problems in Set Theory, Analysis and Geometry,

in Paul Erdős and His Mathematics I, Bolyai Society Mathematical Studies 11 (2002), 493–506, author-uploaded PDF. Section 1 states the Freiling–Mauldin outer-measure theorem and the convex/covering reductions used on the live page. [b, source-checked]

1. R. Daniel Mauldin, *Some Problems and Ideas of Erdős in Analysis and

Geometry, in Erdős Centennial* (2013), 365–376, author-hosted PDF, DOI. Problems 5.1 and 5.2 give the continuous problem and finite-union reduction; the following paragraphs prove the sharp constant for one, two, and three compact convex bodies. [b, source-checked]

1. J. Koizumi, Triangles in a Planar Measurable Set,

linked note. Theorem 2 states and proves the equivalence with \(f(N)=O(N)\). This is an informal linked note rather than a journal publication, so I independently reconstruct the reduction below. [source-checked; mathematical use is [a]]

1. Aleksandar Bulj and Vjekoslav Kovač, *On hyperbolic corners and unit-area

triangles in planar sets of large measure*, arXiv:2605.30033v1, submitted 28 May

1. The identifier, title, authors, date, and full 25-page paper all

exist. Its Theorem 2 says that if \(M_\triangle(R)\) is the supremum of measures of unit-area-triangle-avoiding measurable subsets of \([0,R]^2\), then for \(R\ge10\) \[ M_\triangle(R)\lesssim R^2\left(\frac{\log\log R}{\log R}\right)^{1/2}. \tag{2} \] The paper explicitly says the desired \(O(1)\) bound remains open. [b, source-checked]

Targeted searches for the exact statement, “triangle of area one,” “unit-area triangles” with measurable sets, Freiling–Mauldin, and “Erdős Problem #352” found no later primary source claiming the full problem or the exact lattice values below. This is a search miss and not a proof of novelty. [c, source-checked]

Initial finite result through N=16

Let \(f(N)\) have the precise meaning in (1).

Theorem

The exact values for \(8\le N\le16\) are

[d: exhaustive integer certificates]

The lower-bound point sets were already present in the live thread/witness repository; this run independently verifies them. The new contribution of this run is the matching exhaustive upper bound for every \(N=8,\ldots,16\). Thus it extends the posted exact range \(N\le7\), but it does not give a new continuous counterexample or solve the uniform problem. [d, source-checked]

The July run's standalone checker was runs/erdos352_wavew041_reverify.py. It used only the Python standard library, embedded every lower witness, constructed every upper certificate from scratch, and then audited the certificates independently. Its SHA-256 at the time of that initial run was

3f79b4cfca1db2ada15e38ce9dac317336ae92ffb5b5a816464bd6ce41f11791

[d]

Proof of the finite upper bounds

1. An exact integer predicate

For lattice points \(p,q,r\), let

Thus the triangle area is \(A_2/2\) and its diameter is \(\sqrt{D_2}\). Both sides of (1) are nonnegative, so multiplying by two and squaring shows that (1) is equivalent to

Every comparison in the checker is therefore an exact integer comparison. [a]

2. Exhausting diameter pairs

Repetitions in (1) matter. Applying it to \((p,p,q)\) gives

Every finite avoiding set has a diameter pair. Translate one endpoint to \(A=(0,0)\), and apply a sign change, coordinate swap, and reflection to put the other endpoint at

These are all lattice-preserving isometries and preserve absolute triangle area. The checker enumerates every vector in (6). An independent audit starts with every signed vector in the square \([-(N-1),N-1]^2\), normalizes its full dihedral orbit, and checks that the two lists coincide. [a for completeness; d for the list comparison]

Write \(D=|A-B|\). Since \(A,B\) are a diameter pair, every further point \(q\) lies in the finite lens

In particular every coordinate of \(q\) lies in \([-(N-1),N-1]\). The checker exhausts that entire integer box, imposes (7), and tests \((A,B,q)\) using (4). There is no truncation heuristic. [a,d]

3. The exhaustive low/high branch split

For a surviving \(q\), the diameter of \(\triangle ABq\) is \(D\), so (1) gives exactly one of

Call these candidates low and high.

Every avoiding set for a fixed \(B\) belongs to one of these exhaustive branches:

• it contains no high point, in which case \(A,B\) are fixed and only low

candidates remain;

• it contains a high point \(H\), in which case the checker tries every high

candidate as \(H\) and fixes \(A,B,H\).

In either branch, candidates incompatible with the fixed points or with the chosen diameter are first removed. [a,d]

4. Checkable colorings, not chromatic optimization

For a fixed set \(F\), where \(F=(A,B)\) or \(F=(A,B,H)\), form a graph on the remaining candidates. Two candidates are adjacent precisely when they can coexist with \(F\): their mutual distance does not exceed \(D\), and every triple in \(F\cup\{p,q\}\) passes (4).

Every actual avoiding extension of \(F\) is a clique in this compatibility graph. Consequently, any proper \(k\)-coloring—not necessarily an optimal one—proves that at most \(k\) candidates can be added:

The program uses a deterministic degree-ordered greedy coloring. The slow exact-chromatic search in the posted reference implementation is therefore unnecessary for these upper bounds. [a,d]

The crucial certification direction is checked independently. For each color class and every same-color pair \(p,q\), the audit verifies directly that either \(|p-q|>D\), or a full combinations-with-repetition scan of \(F\cup\{p,q\}\) finds a failed inequality (4). Hence no compatible pair has the same color, which is exactly the property needed in (9). [a,d]

5. Taking all maxima

For every normalized \(B\), the checker takes the larger of the no-high bound and all fixed-\(H\) bounds. It then takes the maximum over all \(B\). Sections 2–4 show that every \(N\)-avoiding lattice set occurs in one of the audited cases, up to an allowed lattice isometry. The resulting upper bound matches the embedded witness size for each row of (3), proving equality. [a,d]

Independent lower-bound and certificate audit

For a witness of size \(m\), the checker evaluates all \(\binom{m+2}{3}\) triples with repetition. The following table records the smallest exact value of the left side of (4), so every positive entry is a strict certificate. It also gives the exhaustive upper-search counts. [d]

The script has deliberately separate producer and audit paths:

1. the producer computes area with a based-vector determinant; the audit uses

the shoelace formula;

1. the producer computes distances from coordinate differences; the audit

uses a dot-product expansion;

1. the producer lists normalized \(B\)'s directly; the audit normalizes every

signed raw vector;

1. candidate pools are regenerated with the audit predicate and compared

exactly;

1. the audit does not trust the compatibility graph: it checks every

same-color pair by full triple enumeration and separately checks the fixed diameter.

Thus a search error can at worst produce a certificate that fails; no optimizer's assertion is accepted as a proof. [d]

As a cross-implementation check, I also ran the public upper-bound repository at commit 7a4a19468771766cb4f20e10dfa5cf770cc2487b with exact chromatic minimization disabled (--max-color-vertices 0). Its separate implementation returned the same upper totals and diameter-case counts for \(N=8,\ldots,16\). The embedded lower points were compared with the public witness repository at commit 09e5c5393943c881ab60118aa32fe5660539c8d5. The standalone checker imports neither repository. [d, source-checked]

For reproducibility, the SHA-256 digests of the canonical audited certificate streams are:

N= 8  82a246077f367d140d7e977167a203c15fa2fa0c3e158de41727bac94317ad41
N= 9  8c4251f00f2c14475832a7568dde89080796a126fae4a80b96a0ad549388e225
N=10  0a0c50ab0e12bca0fa487ba6f77beaaed3a64d5e78d8541f595c2d66133cbf3a
N=11  8687754c9fcf1433a3bfdc1435c3bffb1a097ae89374a44b842b209980c883c3
N=12  395fb6971b90dd2b0463323e02382890b75b4b36f91ebb621b8774e067cf2ae0
N=13  903a23b95da5c4ac678570123f410ed88138ab5fb9777bfd5244524672a52e3a
N=14  d9aa00efb67719feb1b8254925b3a321e338d52e56db068a0bfe1ae193168d9b
N=15  31650a5d06b64a1c161e117318bdab9d3103d277ab293790c27f82db7d60453f
N=16  839008d566ff41e4801c2f63fe61af5beb3081ad475a8b94864db84b7c5085d3

[d]

Historical N<=16 reproduction record

From the repository root:

python3 runs/erdos352_wavew041_reverify.py

The completed run printed:

N= 8 f(N)=14 witness_triples= 560 min_gap= 24 B_cases= 30 high_branches=   98 certs=  128 same_color_pairs=    28 tight_B=3,3 sha256=82a246077f367d140d7e977167a203c15fa2fa0c3e158de41727bac94317ad41 seconds=0.053
N= 9 f(N)=16 witness_triples= 816 min_gap= 29 B_cases= 38 high_branches=  268 certs=  306 same_color_pairs=    50 tight_B=4,1;3,3 sha256=8c4251f00f2c14475832a7568dde89080796a126fae4a80b96a0ad549388e225 seconds=0.163
N=10 f(N)=18 witness_triples=1140 min_gap= 13 B_cases= 46 high_branches=  518 certs=  564 same_color_pairs=   383 tight_B=4,3 sha256=0a0c50ab0e12bca0fa487ba6f77beaaed3a64d5e78d8541f595c2d66133cbf3a seconds=0.409
N=11 f(N)=21 witness_triples=1771 min_gap= 28 B_cases= 55 high_branches=  926 certs=  981 same_color_pairs=  2806 tight_B=4,2 sha256=8687754c9fcf1433a3bfdc1435c3bffb1a097ae89374a44b842b209980c883c3 seconds=1.045
N=12 f(N)=23 witness_triples=2300 min_gap=  9 B_cases= 64 high_branches= 1466 certs= 1530 same_color_pairs= 10809 tight_B=4,4 sha256=395fb6971b90dd2b0463323e02382890b75b4b36f91ebb621b8774e067cf2ae0 seconds=2.363
N=13 f(N)=24 witness_triples=2600 min_gap=  5 B_cases= 75 high_branches= 2348 certs= 2423 same_color_pairs= 50654 tight_B=5,1;5,2;6,1 sha256=903a23b95da5c4ac678570123f410ed88138ab5fb9777bfd5244524672a52e3a seconds=6.082
N=14 f(N)=26 witness_triples=3276 min_gap= 16 B_cases= 87 high_branches= 3572 certs= 3659 same_color_pairs=148701 tight_B=5,2;5,3 sha256=d9aa00efb67719feb1b8254925b3a321e338d52e56db068a0bfe1ae193168d9b seconds=13.828
N=15 f(N)=28 witness_triples=4060 min_gap= 17 B_cases= 99 high_branches= 5094 certs= 5193 same_color_pairs=373340 tight_B=4,4;5,3;5,4 sha256=31650a5d06b64a1c161e117318bdab9d3103d277ab293790c27f82db7d60453f seconds=28.996
N=16 f(N)=32 witness_triples=5984 min_gap= 16 B_cases=112 high_branches= 7094 certs= 7206 same_color_pairs=814708 tight_B=5,3 sha256=839008d566ff41e4801c2f63fe61af5beb3081ad475a8b94864db84b7c5085d3 seconds=69.480
ALL CHECKS PASSED for N=8..16; exact table=[14, 16, 18, 21, 23, 24, 26, 28, 32]; total_seconds=122.418

/usr/bin/time independently reported 122.48 seconds elapsed, 91% CPU, and 21,796 KiB maximum resident memory. python3 -m py_compile also passed. [d]

Why this is relevant to the continuous problem

For completeness, here is a reconstruction of Koizumi's equivalence.

Continuous statement implies \(f(N)=O(N)\)

Assume some constant \(c\) forces a unit-area triangle in every measurable set of measure \(>c\). Given an \(N\)-avoiding \(S\), put a disk of radius \(1/(3\sqrt N)\) around each \(q/\sqrt N\), \(q\in S\). The disks are disjoint by (5).

If their union contained a unit-area triangle \(p_1p_2p_3\), set \(q_i'=\sqrt N\,p_i\). Then \(\operatorname{area}(\triangle q_1'q_2'q_3')=N\) and \(|q_i-q_i'|\le1/3\). The elementary determinant perturbation bound

with \(\varepsilon=1/3\) contradicts (1); the lattice diameter is at least one unless all \(q_i\) coincide, and the latter is impossible because one such small disk cannot contain a unit-area triangle. Hence the disk union avoids unit area and

[a]

\(f(N)=O(N)\) implies the continuous statement

Conversely, assume \(f(N)\le CN\). By inner regularity, it is enough to treat a compact \(A\) of measure \(>4\pi C\). For each integer \(N\), let \(S_N\) contain those \(q\in\mathbb Z^2\) for which the disk of radius \(2/\sqrt N\) about \(q/\sqrt N\) meets \(A\). These disks cover \(A\), and the sum of their areas gives

Thus \(S_N\) has a triple failing (1). After scaling by \(1/\sqrt N\), its triangle area differs from \(1\) by at most its diameter divided by \(\sqrt N\). Moving the vertex opposite a diameter side within its radius \(2/\sqrt N\) disk changes the area continuously through an interval large enough to hit \(1\).

This gives a unit-area triangle in the disk union for every \(N\). Each union lies within distance \(4/\sqrt N\) of compact \(A\); compactness and area continuity give a convergent subsequence whose limiting three vertices lie in \(A\) and span area \(1\). Therefore the continuous problem is equivalent to \(f(N)=O(N)\). The live page uses a threshold \(\ge c\), while the note uses \(>c\); existence of some threshold is unchanged by increasing the constant. [a]

Exact remaining obstruction

The finite table (3) supplies no uniformity in \(N\). It therefore cannot, even in principle, close the continuous problem. [a]

There is, however, a useful all-\(N\) reduction for the fixed-high cases. Write \(D=|B|<N\) for the normalized diameter and rotate Euclidean coordinates just for this argument so that \(AB\) is horizontal. Every candidate lies in the intersection of the radius-\(D\) disks about \(A\) and \(B\), whose greatest altitude above \(AB\) is \(\sqrt3D/2\). A high point of altitude \(h\) has

Consequently a high point can exist only if

For each fixed \(K\), all cases with \(D\le K\sqrt N\) already have

Thus, after the known all-low bound, any sequence obstructing a uniform linear estimate must come from fixed-high branches with \(D/\sqrt N\to\infty\) while still \(D<N\). This is an analytic reduction, not an extrapolation from the finite table. It does not control that remaining long-diameter regime. [a]

One tempting strengthening fails immediately beyond the current exact table. In every fixed-high computation through \(N=22\), each affine fiber \(2x-y=c\) happened to need at most three compatibility colors. At \(N=24\), however, the valid fixed set

has four surviving points on the fiber \(2x-y=19\), namely

and all six pairs are compatible with \(A,B,H\). Thus this fiber induces \(K_4\), so the observed three-color pattern is false rather than a hidden uniform lemma. The counterexample is checked by the full repeated-triple predicate, independently of greedy coloring. [d]

Within this diameter-pair method, a sufficient missing lemma is:

There is an absolute \(C\) such that, for every \(N\), every normalized diameter vector \(B\), and every no-high or fixed-high branch \(F\), the associated compatibility graph has a proper coloring with at most \(CN-|F|\) colors.

Such a uniform coloring lemma would imply \(f(N)=O(N)\) by (9) and hence solve the continuous problem by (10)–(11). The live discussion's all-low argument does not control all fixed-high branches; those branches are the precise remaining obstacle in this framework. [a for the implication; c for the prospect that such a coloring exists]

The computational growth already shows why extending a table is not a substitute. At \(N=18\) the full audit processed 12,594 high branches and 3,011,741 same-color pairs in 216.13 seconds. At \(N=23\) those figures rose to 40,566 high branches and 34,861,434 same-color pairs in 3,140.50 seconds. No finite extension, regardless of cost, would prove the required bound for all \(N\). [d]

PARTIAL: Exact exhaustive certificates prove f(N)=14,16,18,21,23,24,26,28,32,32,34,38,39,41,44,45,48,52,52 for N=8,...,26 in Koizumi's equivalent lattice model; an all-N construction also rules out any one O(N)-image homomorphic coloring used unchanged across every oriented fixed-high branch, but adaptive and normalized-family colorings remain and no uniform f(N)=O(N) bound is proved.