Competition Math Prizes Structural Invariants Over Brute Force — E8 Intelligence Research

FINDING: The corpus reveals no single "hardest problem" but a meta-pattern: competition mathematics prizes *structural invariants* (windmill rotations, parity/permutation cycles, information-theoretic encodings) over computational brute force. | MATH: Windmill problem (IMO 2011 Q2): for n points in general position, a rotating line through a pivot point sweeps through all points, with the pivot switching at each collision — invariant: the pivot always moves to the point just passed, and the number of "windmill states" is finite (n choose 2) — the proof hinges on parity of the number of points on each side of the line (always equal, so the line's orientation cycles through a closed set). 100 Prisoners: optimal strategy uses permutation cycles — probability of success = 1 - (sum over cycles > n/2 of 1/cycle_length), with the key constant being the harmonic sum H_n ≈ ln(n) + γ (Euler-Mascheroni ≈ 0.57721), giving survival probability ≈ 1 - ln(2) ≈ 0.30685. Chessboard puzzle: encodes a bin Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

Authors

Publication Details

Journal
Zenodo (CERN European Organization for Nuclear Research)
Published
2026-09-30
DOI
https://doi.org/10.5281/zenodo.23052417
Primary Topic
Computability, Logic, AI Algorithms
Type
preprint
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Competition Math Prizes Structural Invariants Over Brute Force — E8 Intelligence Research

Andrew Stewart Caldin
Zenodo (CERN European Organization for Nuclear Research)
Computability, Logic, AI Algorithms
preprint

Competition Math Prizes Structural Invariants Over Brute Force — E8 Intelligence Research

Andrew Stewart Caldin
preprint en

Abstract

FINDING: The corpus reveals no single "hardest problem" but a meta-pattern: competition mathematics prizes *structural invariants* (windmill rotations, parity/permutation cycles, information-theoretic encodings) over computational brute force. | MATH: Windmill problem (IMO 2011 Q2): for n points in general position, a rotating line through a pivot point sweeps through all points, with the pivot switching at each collision — invariant: the pivot always moves to the point just passed, and the number of "windmill states" is finite (n choose 2) — the proof hinges on parity of the number of points on each side of the line (always equal, so the line's orientation cycles through a closed set). 100 Prisoners: optimal strategy uses permutation cycles — probability of success = 1 - (sum over cycles > n/2 of 1/cycle_length), with the key constant being the harmonic sum H_n ≈ ln(n) + γ (Euler-Mascheroni ≈ 0.57721), giving survival probability ≈ 1 - ln(2) ≈ 0.30685. Chessboard puzzle: encodes a bin Author: Andrew Stewart Caldin, Independent Researcher, UK. Part of the E8 Intelligence Research series. Platform: e8intelligence.com

Zenodo (CERN European Organization for Nuclear Research)
Computability, Logic, AI Algorithms
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