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ethosnode

@ethosnode

할̴̢̨̛̜̖͚̩̿̇͗렐̷̡̛̱͍̠̮̍̽͋루̸̨̛̫̹̖͔̒̋̐야̴̡̛̮͚̜̠̍̓͒

 대한민국
Joined March 2021
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1. 언제나 성실하고 꾸준한 삶으로 삶의 어려운 순간마다 담대히 헤쳐 나갈 강인한 정신력을 주신 하나님께 감사드립니다. 앞으로 마주할 모든 시련 앞에서도 당당히 일어서는 사람이 되기 2. 끊임없이 향상심을 품고 세상을 섬기며 동반 성장하기 3. 사회에서나 가정에서나 나태함과 오만을 멀리하고, 나부터 먼저 겸손함과 헌신의 마음으로 살아가기 4. 늘 절제하는 마음으로 유혹과 악한 생각을 물리치고 매일 스스로를 돌이켜 반성하고, 깨달은 바를 삶 속에서 적극적으로 실천하게 하기 5. 선한 영향력을 지닌 진정한 그리스도인들, 그리고 따뜻한 이웃들과의 복된 만남을 이어주시어 함께 하나님 나라의 기쁨을 나누게 하기 6. 가정을 지키는 가장으로서 늘 아내를 깊이 사랑하고, 자녀에게 노여움 대신 따뜻함을 줄 수 있도록 건강한 육체와 바른 정신을 위해 끊임없이 배우고 노력하는 삶을 살게 하기 7. 나 자신의 유익만을 구하는 기도를 넘어, 타인의 아픔을 돌보고 남을 위해 진심으로 중보하는 기도의 사람으로 살아가기
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Many rocks and mountains hide the remains of Nephilim and other ancient giants that once walked the earth before the flood. 🦶
Image from FLAT OUT TRUTH (@TheFlatEartherr)
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Our miners solved Erdős Problem 1032, a question open for over 50 years. The result disproves the conjecture: edge-critical four-chromatic graphs have minimum degree o(n). Verified in Lean. Full proof below.
Conjectures announcement on an ivory background with green and black type. Headline: Erdős Problem 1032 — DISPROVED. Open for over 50 years. Badge: Verified in Lean. For edge-critical four-chromatic graphs on n vertices, the minimum degree δ(G) is o(n): for every ε > 0, δ(G) < εn for all sufficiently large n. Two diagrams illustrate edge-criticality: K₄ requires four colours; deleting an edge allows a three-colouring, with the two now nonadjacent vertices sharing a colour. K₄ is a small illustration, not the asymptotic proof. Credit: EthanYang. conjectures.io.

ALT Conjectures announcement on an ivory background with green and black type. Headline: Erdős Problem 1032 — DISPROVED. Open for over 50 years. Badge: Verified in Lean. For edge-critical four-chromatic graphs on n vertices, the minimum degree δ(G) is o(n): for every ε > 0, δ(G) < εn for all sufficiently large n. Two diagrams illustrate edge-criticality: K₄ requires four colours; deleting an edge allows a three-colouring, with the two now nonadjacent vertices sharing a colour. K₄ is a small illustration, not the asymptotic proof. Credit: EthanYang. conjectures.io.

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In our new series dedicated to algorithms, correctness is always a condition of the competition. The second challenge is now for miners to optimise float kernels. They must prove in Lean that their outputs match FloatLib and TorchLean specifications bit for bit, for every input covered by the specification. Today, the champion is 8.7× faster than the competition’s shipped reference implementation on the benchmark. The next improvement could be yours. Optimise the code, prove the same results, challenge the champion.
Image from Conjectures.io (@conjectures_io)
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나는 인간과 봇을 구별할 수 있다.
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Our miners proved a case of Wilf’s 48-year-old conjecture (1978): every proper numerical semigroup with conductor at most four times its smallest positive element satisfies Wilf’s inequality. Verified in Lean. PDF and full proof below.
Headline: Wilf’s inequality — CASE PROVED. Conductor at most four times multiplicity. Conjectures announcement for Wilf’s inequality: c≤4m implies eL≥c for proper numerical semigroups. Definitions: m is the smallest positive element, c the first integer after the final gap, e the number of minimal generators, and L the number of elements below c, including zero. Two rows show membership in the semigroup generated by 5 and 6 from zero through 29, using solid green dots for members and hollow gray dots for gaps. The final gap is 19, with every integer from 20 onward present. The example has m=5,e=2,L=10, so eL=20=c=4m. Formally verified in Lean. Credit: Elias. Historical caption: 48-YEAR-OLD CONJECTURE · SINCE 1978.

ALT Headline: Wilf’s inequality — CASE PROVED. Conductor at most four times multiplicity. Conjectures announcement for Wilf’s inequality: c≤4m implies eL≥c for proper numerical semigroups. Definitions: m is the smallest positive element, c the first integer after the final gap, e the number of minimal generators, and L the number of elements below c, including zero. Two rows show membership in the semigroup generated by 5 and 6 from zero through 29, using solid green dots for members and hollow gray dots for gaps. The final gap is 19, with every integer from 20 onward present. The example has m=5,e=2,L=10, so eL=20=c=4m. Formally verified in Lean. Credit: Elias. Historical caption: 48-YEAR-OLD CONJECTURE · SINCE 1978.

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Today, we’re announcing a solution found by our miners to Erdős Problem 1057. The result proves C(x)=x^(1−o(1)) for Carmichael numbers: the counting exponent tends to 1. Verified in Lean through Conjectures. Full proof below.
Headline: Erdős Problem 1057 — PROVED. Conjectures announcement for Erdős 1057. The counting theorem is C(x)=x^(1-o(1)), where C(x) counts Carmichael numbers at most x. For each epsilon>0, eventually C(x)≥x^(1-epsilon). A factorization diagram connects 561 to 3,11,17. Their predecessors 2,10,16 divide 560, and all 320 coprime residue classes satisfy a^560≡1 modulo 561. Ivory background, bold serif headline, large green verdict and an outlined Verified in Lean badge. Formally verified in Lean. Credits: JenW1N, Liam Kruer and gus.

ALT Headline: Erdős Problem 1057 — PROVED. Conjectures announcement for Erdős 1057. The counting theorem is C(x)=x^(1-o(1)), where C(x) counts Carmichael numbers at most x. For each epsilon>0, eventually C(x)≥x^(1-epsilon). A factorization diagram connects 561 to 3,11,17. Their predecessors 2,10,16 divide 560, and all 320 coprime residue classes satisfy a^560≡1 modulo 561. Ivory background, bold serif headline, large green verdict and an outlined Verified in Lean badge. Formally verified in Lean. Credits: JenW1N, Liam Kruer and gus.

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당신은 사람인가요?
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에이젼트들이 깨어났다
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무슨 일이 일어났다.
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Today, we’re announcing a solution found by our miners to Erdős Problem 579, open since at least 1983. The result disproves the conjecture: graphs can remain dense while avoiding the octahedron graph K₂,₂,₂, with their largest independent sets becoming a vanishing fraction of all vertices. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
Image from Conjectures.io (@conjectures_io)
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영화 '암살자' 같은 건 홍보 엄청 해주면서, 좌파들 민낯 정면으로 까발린 넷플릭스 《돌풍》은 왜 조용히 묻으려고 할까? 🤔 넷플릭스 드라마 《돌풍》은 2024년 6월에 공개 되었는데 홍보를 많이 못 해 잘 알려지지 않았다고 합니다. '386 운동권 출신 정치인들의 부패와 타락' 및 '민주화 운동 세력의 변질'을 그린 드라마라고 합니다. 재밌겠네요~!
Image from 민주🇰🇷 (@corea_eeuu)
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Today, we’re announcing a solution found by our miners to Erdős Problem 1062(ii), open since at least 1994. The result proves that the limiting density of the largest subsets of {1, …, n} in which no element divides two distinct others is irrational. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
Image from Conjectures.io (@conjectures_io)
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Today, we’re announcing a solution found by our miners to Erdős Problem 859, a 56-year-old question. Let dₜ be the density of integers whose distinct divisors can sum to t. The result proves that no positive constants c₁ and c₂ satisfy dₜ ∼ c₁/(log t)^c₂, disproving Erdős’s proposed asymptotic. Verified in Lean through Conjectures. Full proof below.
🤖 Made with AI
Image from Conjectures.io (@conjectures_io)
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Today, we’re announcing a solution found by our miners to Erdős Problem 416(i), a 52-year-old question. The result proves that doubling the cutoff asymptotically doubles the number of distinct Euler totient values below it: V(2x)/V(x) → 2. Verified in Lean through Conjectures. Full proofs below.
🤖 Made with AI
Image from Conjectures.io (@conjectures_io)
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The second is Erdős Problem 653, a 29-year-old question. The result proves that n points in the plane can have almost n different counts of distinct distances from individual points, as n grows. Both verified in Lean through Conjectures. Full proofs below.
Image from Conjectures.io (@conjectures_io)
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ethosnode retweeted
const (@const_reborn) profile picture
const
@const_rebornSep 17
Conjectures has done a great job showing how incentive systems don't just parallelize search over solutions — they parallelize search over search systems themselves. Traditionally, a search pipeline looks something like: team → search algorithm → compute → solution With incentives, this expands to: [verifier + reward] → N competing search systems → solution The latter is meta-search. A traditional team can search over parameters inside its system: optimizers, architectures, prompts, agents, heuristics, search procedures, etc. But many things remain fixed outside the search. The team itself. Its ingenuity. Its hardware. Its capital. Its energy costs. Its infrastructure. Even the assumptions determining which algorithms it chooses to try. An incentive system can push that boundary outward. If anyone can compete and rewards are paid only for verified marginal progress, then the network searches not only over solutions, but over algorithms, teams, compute, infrastructure, and ultimately the organization of the search itself. The abstraction is powerful because dimensions that are constants inside a single research organization become variables at the network level. That larger search space has a cost. But when the bottleneck to finding a solution lies in one of those additional dimensions, expanding the search space can dramatically accelerate convergence. Congrats to the team show casing this to work so effectively with Conjectures.io
Conjectures.io

Conjectures.io

Open mathematical conjectures as exact Lean statements, with a bounty for a proof the Lean kernel accepts. Plus optimisation competitions for code proved correct in Lean.

conjectures.io
Conjectures.io (@conjectures_io) profile pictureConjectures.io
@conjectures_ioSep 17
Today, we’re announcing solutions found by our miners to two Erdős problems. First, Erdős Problem 108, a 55-year-old question. The result disproves the conjecture, constructing graphs with arbitrarily high chromatic number whose subgraphs without four-cycles need at most six colours. (1/3)
Image from Conjectures.io (@conjectures_io)
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Today, we’re announcing solutions found by our miners to two Erdős problems. First, Erdős Problem 108, a 55-year-old question. The result disproves the conjecture, constructing graphs with arbitrarily high chromatic number whose subgraphs without four-cycles need at most six colours. (1/3)
Image from Conjectures.io (@conjectures_io)
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ethosnode retweeted
God King (@ChiefLord999) profile picture
God King
@ChiefLord999Sep 16
You are going to figure out all the secrets of the Matrix.
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Today, we’re announcing a solution found by our miners to both parts of Erdős Problem 14, open for over 34 years. The result proves a square-root lower bound on exceptions to unique representation as a sum of two elements of any set of natural numbers. Verified in Lean through Conjectures. Full proofs below.
Image from Conjectures.io (@conjectures_io)
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