Ciro Santilli @cirosantilli 37

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GPT-5 2025-10-27

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Project Euler problem 961 2025-10-27

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Andrew He 2025-10-27

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Competitive programmer 2025-10-27

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Updates / Getting banned from Project Euler 2025-10-27

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I have been banned from Project Euler for life, and cannot login to my previous account projecteuler.net/profile/cirosantilli.png

The ban happened within 12 hours of me publishing a solution to Project Euler problem 961 github.com/lucky-bai/projecteuler-solutions/pull/94 which was one-shot by a free GPT-5 account as MathArena had alerted me to being possible: matharena.ai/?comp=euler--euler&task=4&model=GPT-5+%28high%29&run=1

The problem leaderboard contains several people solved the problem within minutes of it being released, so almost certainly with an LLM.

I'm a huge believer in giving answers to problems, and I take the ban with pride.

It is funny to see that people waste their time policing this kind of useless stuff.

Project Euler likely has many fun problems, and can be a useful machine learning benchmark.

The "secret club" mentality is their only blemish, and incompatible with open science.

They should also make sure that LLMs don't one shot their future problems BEFORE publishing them!

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Project Euler problem 943 Created 2025-10-27 Updated 2025-10-27

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projecteuler.net/problem=943

A naive T in Python is:

from collections import deque

def T(a: int, b: int, N: int) -> int:
    total = a
    q = deque([a] * (a - 1))
    is_a = False
    for i in range(N - 1):
        cur = q.popleft()
        total += cur
        q.extend([a if is_a else b] * cur)
        is_a = not is_a
    return total

assert T(2, 3, 10) == 25
assert T(4, 2, 10**4) == 30004
assert T(5, 8, 10**6) == 6499871

which passes the tests, but takes half a second on PyPy. So clearly it is not going to work for 22332223332233 which has 14 digits.

Maybe if T is optimized enough, then we can just bruteforce over the ~40k possible sum ranges 2 to 223. 1 second would mean 14 hours to do them all, so bruteforce but doable. Otherwise another optimization step may be needed at that level as well: one wonders if multiple sums can be factored out, or if the modularity can of the answer can help in a meaningful way. The first solver was ecnerwala using C/C++ in 1 hour, so there must be another insight missing, unless they have access to a supercomputer.

The first idea that comes to mind to try and optimize T is that this is a dynamic programming, but then the question is what is the recurrence relation.

The sequence appears to be a generalization of the Kolakoski sequence but to numbers other than 1 and 2.

maths-people.anu.edu.au/~brent/pd/Kolakoski-ACCMCC.pdf "A fast algorithm for the Kolakoski sequence" might provide the solution, the paper says:

It is conjectured that the algorithm runs in time $O (n^{α})$ and space $O (n^{α})$ , where $α = l o g (2) / l o g (3) \approx 0.631$

and provides exactly a recurrence relation and a dynamic programming approach.

www.reddit.com/r/dailyprogrammer/comments/8df7sm/20180419_challenge_357_intermediate_kolakoski/ might offer some reference implementations. It references a longer slide by Brent: maths-people.anu.edu.au/~brent/pd/Kolakoski-UNSW.pdf

www.reddit.com/r/algorithms/comments/8cv3se/kolakoski_sequence/ asks for an implementation but no one gave anything. Dupe question: math.stackexchange.com/questions/2740997/kolakoski-sequence contains an answer with Python and Rust code but just for the original 1,2 case.

github.com/runbobby/Kolakoski has some C++ code but it is not well documented so it's not immediately easy to understand what it actually does. It does appear to consider the m n case however.

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ARC-AGI is a black hole for early retired tech and finance bros 2025-10-27

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People who do cool open tech stuff when don't need money anymore are awesome: