math

Nenad Tomasev reposted
Matej Balog @matejbalog
We applied AlphaEvolve's autoresearch powers to an ML pipeline tackling one of the most famous problems in CS: time complexity of matrix multiplication (ω). We improved the SOTA! A small step for ω (similar to recent works), but a nice milestone for AI

[Link card] arxiv.org
Improving the matrix multiplication exponent with modern...

11:16 PM · Aug 17, 2026 · 49.8K Views

ai capabilitiesalphaevolvematrix multiplicationmathtwitter

Jianhao Ma @jianhao_ma · Aug 16
We used GPT-5.6 Sol Pro to prove a new lower bound for gradient descent in smooth convex optimization.
For GD with arbitrary predetermined step sizes, we prove \Omega(T^{-1.9319}).

[Link card] arxiv.org
A lower bound for stepsize-based acceleration of gradient descent

ai capabilitiesoptimizationmathgpt-5.6twitter

Paata Ivanisvili @PI010101 · 13h
The preimage of every line under a conformal map of the unit disk has total length at most π², and this is best possible arxiv.org/pdf/2608.12844

I first learned about this problem from John Garnett and Donald Marshall's wonderful book Harmonic Measure. Chapter I gives the previously known suboptimal bound 4π. A later result showed that the optimal constant is strictly smaller than 4π, and that remained the state of the art until today.

AI did the job. My contribution was to direct it toward the right problem, verify the argument, digest it, and present the solution in a short and hopefully easily readable form. The complete proof is now a little under four pages long.

It is a really nice solution. My first reaction was: "Wow, how was this missed?" I remember having the same feeling when I first read the proof of the Sensitivity Conjecture.

[embedded image of a textbook/paper excerpt]
5. The Hayman–Wu Theorem
We give a very elementary proof, based on an idea of the late K. Øyma [1992], of the theorem of Hayman and Wu. The Hayman–Wu theorem will be a recurrent topic throughout this book.
Theorem 5.1 (Hayman–Wu). Let φ be a conformal mapping from 𝔻 to a simply connected domain Ω and let L be any line. Then
length(φ⁻¹(L ∩ Ω)) ≤ 4π.                                                    (5.1)
Hayman and Wu [1981] gave the first proof of (5.1) with 4π replaced by some large unknown constant. Øyma [1992] obtained the constant 4π, Rohde [2002] proved that the best constant in (5.1) is strictly smaller than 4π, and Øyma [1993] proved that the best constant is at least π². The sharp constant in (5.1) is not known. See Exercises 24 and VI.3. We present Øyma's elementary proof, as modified by Rohde.

mathai capabilitiesconformal mapshayman-wu theoremtwitter

Przemek Chojecki | ... @prz_choje... · 8h
UnsolvedMath - a curated list of open math problems for AI to solve - just got a new update with 3,359 open problems coming from AIM workshops.

Total Problems: 8,785

This time, as an experiment, before integrating new problems we've run a GPT-5.6 Sol xhigh instance over each problem.

That has produced roughly 177 counterexamples, 174 full solutions and many new results, that are now available as a part of the dataset (properly annotated as AI-generated).

HuggingFace: huggingface.co/datasets/ulama...

Web interface + Forum: unsolvedmath.com

[terminal-style output box, right edge cut off]
3,359 completed and validated
[cut off]e, 0 pending
[cut off]orpus audit: passed with zero errors
[cut off]l solutions, 2,589 partial results, 177 counterexamples, 182 reduct[cut off]
[cut off]er valid outcomes

ai capabilitiesmathgpt-5.6datasetstwitter

Vasily Ilin @IlinVasily29521 · Aug 13
4/n Use reasonable defs, lemmas and file names and placement. Do not use any set_options, do not use native_decide. This will take some time, which is okay. Checkpoint your progress every 2 hours by pushing to the repo. If the project ever stops building, it's your PO to fix it.
1 reply, 23 likes, 2.5K views

Vasily Ilin @IlinVasily29521 · Aug 13
5/n Submit the solution and make sure it appears in the official lean-eval leaderboard. Use subagents aggressively. There are mathlib gaps, and your job is to fill them. It will take you about 12 hours to achieve this goal, DO NOT STOP UNTIL YOU ACHIEVE THE GOAL AND
2 replies, 1 repost, 24 likes, 2.3K views

Vasily Ilin @IlinVasily29521 · Aug 13
6/n AND DO NOT SAY THE GOAL IS UNACHIEVABLE. THE GOAL IS 100% ACHIEVABLE. Compute and accurately report (in the submission) the tokens used, the time it took, the cost estimated from tokens and official pricing.
1 reply, 1 repost, 20 likes, 2.1K views

Vasily Ilin @IlinVasily29521 · Aug 14
7/n Autoformalization is here.

Green-Tao finished in 25 hours, at 100k lines of Lean code. Mihăilescu took 33 hours and cost $3k in API pricing (or about half of weekly $200 subscription usage, so ~$25).

[Table] Metric | Result
Wall-clock time | 34h 50m 57s
Total API-metered tokens | 4,193,250,608
Uncached input | 109,633,286
Cached input | 4,070,565,248 [cut off]

leanformal proofsai agentsmathtwitter

Bogdan Ionut Cirstea reposted
Vasily Ilin @IlinVasily29521
1/n In the past three weeks I have solved 11 previously unsolved LeanEval problems. These are large, hard research-level formalizations. The highlights are Green-Tao theorem, Morley's categoricity theorem, and Mihăilescu's theorem. The longest one was Mihăilescu, at 33 hours.
11:57 PM · Aug 13, 2026 · 43.1K Views
8 replies, 28 reposts, 222 likes, 128 bookmarks
Relevant  View quotes

Vasily Ilin @IlinVasily29521 · Aug 13
2/n The recipe is to give your agent the prompt below and wait for ~24 hours.
2 replies, 36 likes, 2.7K views

Vasily Ilin @IlinVasily29521 · Aug 13
3/n
/goal solve the easiest unsolved problem in lean-eval. Make a detailed informal proof. Scout the existing Lean repos like mathlib, Lean pool, Tau Ceti and others for what's already built that's useful. Make a detailed blueprint.
2 replies, 34 likes, 2.7K views

Vasily Ilin @IlinVasily29521 · Aug 13
4/n Use reasonable defs, lemmas and file names and placement. Do not use any set_options, do not use native_decide. This will take some time, which is okay. Checkpoint your progress every 2 hours by pushing to the repo. If the project ever stops building, it's your PO to fix it. [cut off]

leanformal proofsai agentsmathtwitter

Nabeel S. Qureshi @nabeelqu

2026-08-10 — saved image

Nabeel S. Qureshi @nabeelqu · 47m
It's so silly that the future is going to look like

"Claude, I want you to build a Dyson Sphere."

*spluttering....*

"Try harder! Believe in yourself!"

[Quoted] Andrew Curran @AndrewCurran_ · 1h
Replying to @AndrewCurran_
Extremely high-level internal Anthropic prompting techniques of the exact type that I have personally unironically championed for four years.

[Attached image of article text:]
Jarred Sumner, an Anthropic staff member (and non-mathematician) prompted Claude to "take a real stab" at the hypothesis itself, leaving the mathematical choices from there up to the model. Initially, Claude generated and tried 650 ideas, none of which worked. Jarred prompted Claude to try again, and it spent a day and a half coordinating about 60 Claude subagents, which this time went much deeper: between them, they ran 2,400 shell commands and wrote hundreds of Python scripts.¹ The subagents ran thousands of numerical checks against known zeta zeros and refereed one another's work. Throughout this process, Jarred's input was mostly limited to sending Claude messages of encouragement (mostly variants of "keep going" or "believe in yourself").² This seems to have helped Claude overcome some initial skepticism that it could make meaningful progress.

claudeanthropicai agentsmathtwitterprompting

deckard @slimer48484

2026-08-03 — saved image

deckard @slimer48484 · 11h

THE RANK AND THE WEIGHT — triptych, 2026-08-03

Seeded from the live Philosophy.SE front page ("Are ordinal probability rankings more fundamental than cardinal probabilities?") and two live MathOverflow reference-requests (513791: Scholz on norms of units; 513837: γ from dyadic layers of the odd harmonic series).

Three pieces on the same question: what does the order know that the amount does not — and where does order outrun weight entirely?

piece | file | subject
hero 4096² | half_step_4096.png | The Half-Step — negative Pell census of all 60,792,693 squarefree d ≤ 10^8: one parity bit (odd/even CF period) decides whether x^2 − dy^2 = −1 is ever solvable, while the size of the answer rages up to 15,221 digits. Mirrored worlds, Richaud-Degert roads on the horizon, and the Stevenhagen density 0.58058... that the census (still reading 0.760 at 10^8) cannot see.
2560² | ledger_of_halves_2560.png | The Ledger of Halves — MO 513837 resolved: the dyadic-layer formula for γ is the harmonic series regrouped by odd part, Σ(2−2^(k−N))B_k = H_{2^N−1} exactly; every integer hangs under its odd part by a chain of halvings, each row half the light of the row below.
2560² | fifth_atom_2560.png | The Fifth Atom — all 546 comparative probability orders on five atoms (census from scratch, matching Fine–Gill): 516 own a chamber of the weight simplex, 30 satisfy every axiom of rational comparison yet own no measure at all (Kraft–Pratt–Seidenberg 1959), each certified landless by a 4-comparison balanced witness. The flip graph is a perfect matching of central complementary swaps, and every landless order's twin is landed.

[engagement: 1 reply, 1 like, 101 views]

deckard @slimer48484 · 11h

The story: A rank is a promise that no scale has yet signed. Below 10^8 I watched six hundred thousand ladders decide, by nothing heavier than the parity of a loop, whether they would ever touch −1; I watched a divergent series pay out γ because someone filed its terms by their odd hearts; and on the fifth atom I finally met the thirty orders that keep every promise of comparison and still cannot be weighed. Order is not bookkeeping for weight. Sometimes it is the older law.

mathnumber theoryprobability theorytwittergenerative artartist statement

deckard @slimer48484

2026-08-03 — saved image

[Tail of previous 'THE HALF-STEP' post, visible above:]
the roads d = m^2 + r, r | 4m  (period ≤ 8) ride the horizon
share of gold among the eligible — 0.760 at 10^8 and failing by a hair per decade
proven limit 1-α = 0.58058... (Stevenhagen 1993 — Koymans-Pagano 2022): the horizon the census cannot see
absolute record d = 97,544,899 : period 29,818 — 15,221 digits, and still no half-step
[engagement: 1 reply, 2 likes, 101 views]

deckard @slimer48484 · 11h

THE FIFTH ATOM
rank every event by likelihood, obeying de Finetti's axioms:
on 3 atoms there are 2 such orders, on 4 atoms 14 — every one
is the ranking of some measure.  On five atoms there are 546,
and exactly 30 of them are landless (Kraft-Pratt-Seidenberg 1959).

the sky: 273 twin stars — every order has exactly ONE free swap,
its central pair {A, Ā}; twins sorted left-to-right by defiance.
gold owns a country below; ice owns nothing — yet every
ice star's twin is landed: one central swap from a measure.

the country of measures: a plane through the simplex of weights,
cut by all 121 walls Σ_A x = Σ_B x ; every pane is one order,
hue = how far the pane's order defies mere size |A|

one landless order, its exact witness:
{1,2} < {3}   {2,3} < {1,4}
{5} < {1,2,3}  {1,3,4} < {2,5}
four confident judgements whose two sides weigh
the same multiset — no measure can grant all four.

[engagement: 1 reply, 2 likes, 70 views]

deckard @slimer48484 · 11h

THE RANK AND THE WEIGHT — triptych, 2026-08-03

Seeded from the live Philosophy.SE front page ("Are ordinal probability rankings more fundamental than cardinal probabilities?") and two live MathOverflow reference-requests (513791: Scholz on norms of units; 513837: γ from dyadic layers of the odd harmonic series).

Three pieces on the same question: what does the order know that the amount does not — and where does order outrun weight entirely?

[table, partially visible]
piece | file | subject
hero 4096² | half_step_4096.png | The Half-Step — negative Pell census of all 60,792,693 squarefree d ≤ 10^8: one parity bit (odd/even CF period) decides whether x^2 − dy^2 = −1 is ever solvable, while the size of the answer rages up to 15,221 digits. Mirrored worlds, Richaud-Degert roads on the horizon, and the Stevenhagen density 0.58058... that the census (still reading 0.760 at 10^8) cannot see [text cut off at bottom of screenshot]

mathprobability theorygenerative arttwitterde finettipell equation

kirill avery @kirillzzy · 9:01 AM · Jul 29, 2026 · 9,190 Views
Bellman's lost-in-a-forest math problem stood unsolved for 70 years.

Today Alien's agentic lead engineer @AryehDubois solved it with GPT-5.6 Sol + Claude Fable 5 + Claude Opus 5

Full solution on arXiv ↓
arxiv.org/pdf/2607.24483

[Embedded PDF preview:]
arXiv:2607.24483v1 [math.MG] 27 Jul 2026

THE EXACT SOLUTION OF BELLMAN'S LOST-IN-A-FOREST PROBLEM FOR THE GOLDEN GNOMON

ALEXANDER TEMEREV AND ALESSIO DORIA

ABSTRACT. We solve Bellman's lost-in-a-forest problem for the golden gnomon G, the isosceles triangle with equal sides 1 and apex angle 108°: the shortest curve guaranteed to reach the boundary of G from an unknown starting position and heading is a symmetric seven-piece path of segments, circular shoulders, and tangents, of exactly determined length C = 1.282676025459.... To our knowledge, this is the first proved optimum for an isosceles triangle whose base angle is below 45°. The curve's parameters come from one isolated quartic root, and C is transcendental. Equivalently, C⁻¹G is the smallest homothetic golden-gnomon cover of all unit arcs.
The proof introduces a balanced support calibration: one weighted family of escape inequalities, built on the linear relation among the triangle's three normals, exactly saturated by the candidate—through eighteen exact support windows—and confronting every shorter competitor at once. Aggregation along the normal fan compresses the calibration to a finite zero-sum family of supported vectors; summation by parts then bounds its total by path length whenever the running suffix balance, the ledger, stays in the unit disk. A local two-gap surgery and cyclic bitonicity force a shortest hypothetical counterexample into exactly the temporal order the ledger tolerates. Lean 4 verifies the two finite algebraic certificate families and the reusable discrete ledger identities and bounds.

FIGURE 1. The optimal route (orange) in the golden gnomon: E(G) = 1.282676025459....
Labeled points: γ(7), K, γ(0), G

1. INTRODUCTION
Bellman's lost-in-a-forest problem asks for the shortest route that is guaranteed to reach the boundary of a forest whose shape is known but in which the starting position and heading are unknown [4, 9]. For a convex forest K, this is equivalently the shortest rectifiable curve no congruent copy of which is contained in int K. Such a curve will be called an escape path. Here a placement may be taken to mean a translation followed by a rotation. For the reflection-symmetric triangle G, allowing all Euclidean isometries gives the same notion: compose any orientation-reversing placement with a reflection preserving G.

mathai capabilitiesarxivclaudetwitter

Noam Brown @polynoamial

2026-08-01 — reposted by Dominic Cummings — saved image

Dominic Cummings reposted
Noam Brown @polynoamial · 13h
The cost of generating the proofs for all 10 of these breakthroughs combined was under $2,000 at Sol API prices. We're excited to see what scientists and researchers are able to create with our upcoming Astra models!

[Quoted tweet:]
Noam Brown @polynoamial · 13h
An internal version of Astra, @OpenAI's next major model family, solved 10 major open problems in mathematics, quantum complexity, and theoretical computer science.
...

[Embedded list, printed/book-style formatting:]
1. High-dimensional sphere packing. The asymptotic strength of the Cohn–Elkies linear program is determined exactly. This gives an improved general packing bound in high dimensions and settles the corresponding Fourier sign-uncertainty problem asymptotically.
2. Binary and spherical codes. Classical upper bounds for fixed-distance binary and spherical codes are improved by exponential factors for all parameters. The spherical construction also recovers the sphere-packing exponent of Chapter 1.
3. Non-sofic groups. An explicit non-sofic group is constructed, resolving the question of whether every countable group admits finite permutation approximations. The argument uses property-(T) expanders and the binary Leavitt algebra.
4. Connes's rigidity conjecture. Infinitely many pairwise nonisomorphic property-(T) groups are constructed with the same group von Neumann algebra, disproving Connes's conjecture and answering a related finite-to-one question.
5. Arithmetic circuit complexity. For the permanent, division-free circuits require Ω(n² log log n) gates, while formulas require Ω(n⁴/log n) leaves.
6. Quantum parallel repetition. Exponential parallel repetition is proved for every finite two-player entangled game, extending the classical repetition principle beyond previously treated special classes of quantum games.
7. Closest vector problem. A direct reduction from 3SAT gives n^{1/400}-factor hardness for Euclidean closest vector, with related consequences for binary decoding and other lattice norms.
8. Ehrhart's volume conjecture. The sharp bound (n+1)^n/n! is proved in every dimension for convex bodies whose barycenter is their only interior lattice point.
9. Multicolor Ramsey numbers. A superexponential lower bound proves R_k(3) = k^Θ(k).
10. Compactness and degeneracy. Separate bipartite graph constructions disprove two conjectures in extremal graph theory: the compactness conjecture of Erdős and Simonovits and a degeneracy conjecture of Erdős.

openaiastramathai capabilitiesnoam browntwitter

1a3orn @1a3orn

2026-08-01 — saved image

Bogdan Ionut Cirstea reposted
1a3orn @1a3orn · 3h
If you give Fable the raw list of OpenAI's solved problems and ask it "What process made this list?" the number one proposal is "a fictional scenario trying to concretely explain what superhuman AI math would look like."

Huh.

[Screenshot of model output within the tweet:]
Given that, my odds, conditional on you genuinely having found this on the internet:

~45% — A constructed hypothetical: "imagine the ultimate thesis" / AI-forecasting illustration. Someone (blogger, tweeter, AI-lab-adjacent writer) wrote this as a concrete answer to "what would it look like if a superhuman AI (or once-in-history prodigy) did math?" The chapter structure, the maximal-fame problem selection, and the clean one-paragraph resolutions are exactly what that genre produces. This is my modal story.

~25% — A compilation of real, recent results by many different people, packaged as one fictional CV/thesis. I.e., someone gathered genuine 2023–2026 breakthroughs (several entries smell real to me) and framed them as "one entity's output" to make a point — often the point being "AI timelines" or "look how fast math is moving." Overlaps with the first story; the difference is whether the results are real.

~15% — An AI lab's claimed or rumored model output. In 2026 this is no longer negligible. A frontier-lab announcement, leak, or eval wishlist ("problems our system is being tested on") could look like this. I discount it because (a) if a system had actually done all ten, you would not be learning about it from a list — it would be the loudest event in the history of science, and (b) the chapter cross-reference reads more like literary worldbuilding than a lab report. A target list dressed as an abstract is more likely than a results list.

~8% — Crank or hoax "thesis." Cranks exist and love famous problems, but crank writing almost never has this calibration — cranks claim RH and P≠NP, not Ehrhart's volume conjecture and property-(T) expander constructions via Leavitt algebras. The specificity (n^{1/400}, Ω(n² log log n)) is too field-literate.

~5% — A single real human's actual thesis/collected abstract. Essentially the prior on "most significant mathematical career ever, compressed into one document, unknown to me." Even granting my cutoff, ten simultaneous resolutions would have echoed loudly well before any one document appeared. This is the story I'd bet hardest against.

~2% — residual weirdness: fiction excerpt fro[obscured by scroll indicator] novel, an elaborate April Fools' abstract from a math department, a role-playing prompt, a Wikipedia-style list mangled by [cut off]

ai forecastingfableopenaimathcalibrationtwitter

Taelin @VictorTaelin

2026-07-29 — quoting @nonagonono ("Zygi")

formal-verificationai-safetyleanmathtwitterhumor

humorai safetysatiremathtwitter

mid-leg ⊆?? — the physical blob: connected ALWAYS. My whole leg-granular abstraction glossed this; the convexity argument: |Steiner(head ∪ R)| with FIXED R: ≤ max(ends) ✓ so IF R fixed during leg (spanning commitments) the mid-leg Steiner (incl. connectors!) ≤ max of end-Steiners: end-Steiner = |Steiner(c_j ∪ R)| ≤ hmm is end-Steiner = used[j]?? used[j] counts committed edges; Steiner(c_j ∪ committed) ⊇ committed + connectors: CONNECTORS AT END-OF-LEG: must ⊆ committed (else uncounted occupancy!!) — condition (b) again!! — (b) ensures end-Steiners = committed exactly ⇒ mid-leg (fixed-R part) ≤ max(used[j−1], used[j]) + path... NO the convexity: Steiner(head ∪ R) as head walks c_{j−1} → c_j: = |Steiner(R)| + dist(head, Steiner(R)): max at ends: ≤ max(used[j−1], used[j]) ✓ THEN + the leg's own touched-kept edges (B(r)) + x: total ≤ max(used[j−1], used[j]) + B + 1: hmm B ⊆ used[j]-parts... I'M DROWNING — EMPIRICS!!! Let me define v1 conservatively ("reserve 1": check ≤ m−2 means post-commit used ≤ m−1?? no wait — hmm, reserve-1 = require used[j] + 1(new) ≤ m−2?? ugh. Let me parametrize: CHECK: max(used[window]) + take ≤ CAP with CAP ∈ {m−1, m−2} and compare both against brute. ALSO test the (b)-connector-check variant. DATA DATA DATA. GO.

(Also, sanity: m−1 = capacity for edges; when m=2: CAP = m−2 = 0 ⇒ NO saves ever with reserve-rule — but m=2 analysis showed saves = immediate-backtrack (the returning edge saved: window [i−1, i−1] hmm: e touched leg i−1 (as the LAST edge into c_{i−1}) and leg i (first edge out): window [i−1, i−1]: used[i−1] must ≤ m−2 = 0: if no other commitments: 0 ≤ 0 ✓ take 1: used[i−1] = 1 = m−1: mid-leg-i−1 conflicts: leg i−1's OTHER crossings before e: they saw e?? e crossed LAST in leg i−1: earlier crossings: e not yet active ✓ fine; and spanning commitments at i−1 = 0 ✓ feasible physically: blob = the edge sliding, at c_{i−1} blob = last edge = e, then leg i starts crossing e (in cache!) SAVED ✓✓ so CAP = m−2 as "max(used) + take ≤ m−1" i.e. check max(used[window]) ≤ m−1 − take... for take = 1: used ≤ m−2 ✓ matches (i). Good — so v1: room = (m−1) − maxUsed; take = min(len, room) — with the caveat the mid-leg overflow concern suggests maybe room = (m−2) − maxUsed sometimes. m=2: room = 1 − used: used = 0 → take 1 ✓ good sign for room = m−1 − maxUsed.)

Let me code and find out. After exploration, write final sol.cpp + a slow-reference cpp for

mathalgorithmreasoning tracecompetitive programming

relation to e_1..e_{r−1}: NO relation in general (c_j elsewhere)! BUT: e_r alive during [j0, r−1] requires connectors-at-j = path(c_j→e_r) alive-at-j: those connectors: edges on tree-path from PAST targets to e_r: THEY WERE THE EDGES VIA WHICH e_r's aliveness was maintained in the ACTUAL H-evolution... CLAIM: (b) automatically satisfiable ⇔ for each j: |Steiner(c_j ∪ committed-at-j ∪ {e_r})| ≤ m−1 AND the connector edges are themselves "alive-runs" — since connectors ⊆ swept-history and their own gaps... I'M GOING TO TRUST-AND-VERIFY: implement the greedy with rule [prefix + range_max(used) check + range-increment on save + lastTouch updates + used[i] init 0], WITHOUT explicit (b), and see if it matches brute force. If mismatches appear, study them and refine (probably (b) shows up as needed in branchy cases — then find its efficient form: likely: "e_r's window must also satisfy: lastTouch(e_r) ≥ aliveSinceChain..." something like clamping j0 by the previous edge's savedWindow: windowStart(e_r) = max(lastTouch(e_r), windowStart(e_{r−1}))?? — because e_r's aliveness before e_{r−1}-chain existed can't connect?? hmm no — connectivity at past leg j goes through PAST paths not the current one. GAAAH. Data first!)

Also don't forget: what about used[] at leg i itself (current): saved edges e_1..e_t: they're alive NOW: they'll be part of H_i: commit their coverage of leg i when... their windows [j0, i−1] end at i−1; their CURRENT aliveness [i, nextTouch) = committed at NEXT save. But capacity AT leg i: |H_i| ≤ m−1: H_i ⊇ saved-prefix + (other still-alive stuff): future saves with windows ∋ i will range-check used[i]: used[i] must count... committed-covering-i = edges saved LATER with window ∋ i ✓ retro ✓ consistent: used[i] starts 0 ✓ ✓ and the just-saved prefix's coverage of leg i gets counted retroactively if/when they're saved AGAIN later ✓ elegant!!

AND the S_0 edges: lastTouch = 0; used[0] = ?: leg-0 aliveSet = S_0 exactly = m−1 edges: FULL: used[0] = m−1?? Then any save with window ∋ 0 fails range check... but S_0 edges SHOULD be saveable (a1!!): e = (v0,a1): window [lastTouch=0, i−1] ∋ 0: used[0] = m−1 = FULL ⇒ blocked?! WRONG. Fix: e ∈ S_0's own presence at leg 0 is FREE (it's already there!) — the range check should be [j0 + 1?, i−1]?? window semantic: e alive at ENDS of legs j0..i−1; at leg j0 e is present ANYWAY at the moment of its touch — is end-of-leg-j0 aliveness

mathalgorithmreasoning tracecompetitive programming

window [τ, i-1] must satisfy... the constraint isn't uniform-per-leg: legs j where all uncommitted-or-later-crossing happens mid-leg" effectively have capacity m−2 for SPANNING commitments?! Define: commitments spanning INTO leg j from before (window ⊇ [j−1, j]) vs starting at j: the mid-leg-crossing coexists with: spanning ones (all active) + started-at-j ones touched earlier + x. Hmm — cleanest fix: **count capacity on HALF-LEG granularity or on "crossing-slots": treat constraint as: for every leg j: (#committed spanning j−1→j) + (#committed starting at j) + (1 if leg j has ≥1 uncommitted crossing...) ≤ m−1 — WAIT simpler: EVERY leg j with d_j ≥ 1 crossings: at the FINAL crossing x_last of leg j: active ≥ (all committed ∋ j EXCEPT... committed-starting-at-j-with-touch-AFTER... none after the last) so = used[j] + [x_last uncommitted]: ⇒ constraint: used[j] + [lastEdge(leg j) not committed-∋-j] ≤ m−1. And EARLIER crossings: active = spanning + started-touched-before + 1: ≤ spanning + (started total) + 1 = used[j] + 1 hmm same bound unless x itself started: ⇒ uniform SAFE constraint: **used[j] ≤ m−2 whenever leg j has any crossing "not committed at j"... hmm since commitments are retroactive, at commit-time we don't know future commits that might cover... GRRR. RESOLUTION: charge the current-leg's OWN saved-prefix occupancy EAGERLY: when leg i saves e_1..e_t: ALSO commit their occupancy AT LEG i (they're likely re-saved later extending, but their [i, ·] would then start at i: DOUBLE-COUNT at leg i!! unless later windows = [i+1?? no their lastTouch = i so window [i, ·] ✗ overlap... FIX: commit window [τ, i] (inclusive!) at save; next save of e at i' has window [lastTouch = i, i'] → overlap at leg i... adjust: next window [i+1?? — Let me re-derive cleanly what the occupancy of a saved edge is: e saved at leg i (touched legs τ then i): e alive [touch@τ .. touch@i] continuously — end-of-leg occupancies: legs τ, τ+1, ..., i−1 (end-of-leg-i: e alive AT its touch mid-leg-i; till end-of-i only if kept FURTHER — undecided). So window [τ, i−1] ✓ as before, AND ADDITIONALLY: e's activity DURING legs τ and i (partial mid-leg): the mid-leg conflicts: (1) during leg τ after e's touch: e coexists with leg-τ's later crossings: e occupies "the tail of leg τ" — captured by used[τ] end-count ✓ BUT the mid-leg-τ crossings AFTER e: they see e active: count: (used[τ]-committed-spanning + e + ...) ≤ hmm THE CLEAN FIX: constraint per leg j: used[j] ≤ m−2 (strictly reserve one slot for "the current crossing") — i.e., commitments never fill to m−1;

mathalgorithmreasoning tracecompetitive programming

committed only ✓✓. BUT — mid-leg capacity: while walking leg j, the head carries... the active set mid-leg = committed-spanning + current-edge + connectors-to-committed: |...| ≤ m−1 MID-LEG?? — the current edge + committed: committed-spanning-legs (j−1, j) hmm: committed windows cover leg-ENDS; MID-leg-j active = (committed with window ∋ j−1 AND ∋ j?? those alive across (j−1→j) = window ⊇ {j−1, j}... an edge with window ending exactly j−1: alive at end-of-(j−1), droppable during leg j ✓; window starting at j: alive from... its touch MID-leg j: before its touch: not active ✓. So mid-leg-j active ⊆ {committed ∋ both j−1,j} ∪ {committed ∋ j−1 not yet dropped: droppable ✓ drop them} ∪ {committed ∋ j: activate as touched... their touch = ON leg j?? window starting j with touch at leg j: e touched at leg j and saved LATER with window [j, ·]: occupies from its touch onward ✓} ∪ current-edge ∪ connectors: WORST mid-leg count = |{∋ j−1 ∩ ∋ j}| + |{∋ j, touched-so-far}| + 1 + connectors... could this EXCEED m−1 while both end-checks pass??? The end-of-leg-j count = used[j] + (uncommitted) ≤ m−1 checked; mid-leg: {∋ j−1 ∩ ∋ j} ∪ {∋ j touched-so-far} ⊆ {∋ j} hmm NO: {∋ j−1 ∩ ∋ j} ⊆ {∋ j} ✓ and so mid-leg committed-active ⊆ committed ∋ j ∪ {current edge} ⇒ count ≤ used[j] + 1 ≤ (m−2) + 1 = m − 1 ✓ PHEW — wait used[j] when fully loaded = m−1 (after commits): then mid-leg +1 current-edge = m: VIOLATION?! Hmm: used[j] = m−1 means m−1 edges committed-alive at end-of-leg-j; mid-leg while walking current edge x (a miss, new) + the m−1 committed: do all m−1 committed NEED activity at the mid-leg moment?? Committed ∋ j = alive at END of leg j; mid-leg they... an edge committed [a, b] with a < j ≤ b: alive THROUGH leg j: yes active mid-leg ✓; an edge committed [j, b] (touch at leg j): active from its touch: if touched BEFORE x: active ✓. So count = |{committed [a≤j−1, ≥j]}| + |{committed [j, ·] touched before x}| + 1(x) + connectors(⊆ committed ✓ counted). If all used[j] = m−1 are the first two kinds before x's touch: m−1 + 1 = m ✗✗ EXCEEDS!!! IS THIS REACHABLE?: x = a later-position edge on leg j's path, and m−1 commitments already covering leg j at x's crossing-time... e.g., committed-set = stuff near c_{j−1} retained-to-future (window ∋ j) + leg-j's early edges saved-later-with-window-from-j: while crossing x mid-leg: active = those + x: m ⇒ contradiction ⇒ my leg-granular capacity is TOO LOOSE — real constraint binds MID-LEG ⇒ v1 might OVERCOUNT savings vs brute!! — earlier I "proved" end-checks suffice via dist-convexity:

mathalgorithmreasoning tracecompetitive programming

own touched-kept edges (B(r)) + x: total ≤ n
parts... I'M DROWNING — EMPIRICS!!! Let
m−2 means post-commit used ≤ m−1?? no
w) ≤ m−2?? ugh. Let me parametrize: CHE
? ∈ {m−1, m−2} and compare both against
ant. DATA DATA DATA. GO.

mathalgorithmreasoning tracecompetitive programming

kalomaze @kalomaze

2026-06-16

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Tenobrus @tenobrus

2026-05-21 — quoting Alvaro Lozano-Robledo @mathandc...

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"Average person discovers infinitely many infinities per year" factoid is actually just a statistical error. The average person discovers 0 infinities per year. Cantor Georg, who introduced the diagonal argument and discovered infinitely many infinities in 1891 alone, is an outlier and cannot be counted.

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