robertk

Joined 18 years ago

Karma
2,680
Submissions
630
Member since
Feb 24
https://robertk.at.hn

[ my public key: https://keybase.io/robertzk; my proof: https://keybase.io/robertzk/sigs/MMuXENMT-5k_cTFWUS71BLbqSaYKTQhCxNURSaOMgkQ ]

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Recent submissions

Comment · 8 months ago
Comment · 9 months ago
Why not just open it inside of and print to a static image output within a fully sandboxed Docker container?
Comment · 9 months ago
Why not leak a dataset of N full text paraphrasings of the material, together with a zero-knowledge proof of how to take one of the paraphrasings and specifically "adjust" it to the real document (revealed in private to trusted asking parties)? Then the leaker can prove they released "at least the one true leak" without incriminating themselves. There is a cryptographic solution to this issue.
Comment · 9 months ago
It’s slightly biased. ( P(even) = 0.5702; Bias = +0.0702 (about 7 percentage points toward heads) ). You can use this Claude Code prompt to determine how much:

Use your web search tool call. Fetch a list of English words and find their incident frequency in common text (as a proxy for likelihood of someone knowing or thinking of the word on the fly). Take all words 10 characters or longer. Consider their parity (even number of letters or odd). What is the likelihood a coin comes up heads if and only if a word is even when sampled by incidence rate? You can compute this by grouping even and odd words, and summing up their respective incident rates in numerator and denominator. Report back how biased away this is from 0.5. Then do the same for words at least 9 characters to avoid “even start bias” given slight Zipf distribution statistics by word length. Average the two for a “fair sample” of the bias. Then run a bootstrap estimator with random choice of “at least N chars” (8 <= N <= 15) and random subsets of the dictionary (say 50% of words or whatever makes statistical sense). Report back the estimate of the bias with confidence interval (multiple bootstrap methods). How biased is this method from exactly random bits (0.5 prob heads/tails) at various confidence intervals?

Comment · 9 months ago
“Slightly fringe”
Comment · 10 months ago
I am sorry for your loss, Aella. I sobbed with you.

“Each passing minute is a greater percentage of the final minutes we have,” and yet “these [final] seconds are so soft”.

Death needs to die, some future dying day, not yet.

from everyone who’s had a mom, we join you: “Momma, I love you”.

Comment · 1 year ago
Comment · 1 year ago
Comment · 1 year ago
If I read a comment that has any probability of changing my mind about a fact or opinion, I always go to the user page to check their registration date. No hard cut-off date but I usually discount or ignore any account >= 2020.
Comment · 2 years ago
Comment · 2 years ago
No, it doesn’t. This concerns a corporation subject to legitimate national security concerns, not “a person, or a group of people.”
Comment · 2 years ago
Very cool result but the title is overselling the "AI" contribution. It seems like they trained a few standard binary classifiers (Naive Bayes, decision trees, kNN). The novelty is the independent variable coming from an attribute precomputed for many known elliptic curves in the LMFDB database, namely the Dirichlet coefficients of the associated L-function; and the dependent variable being whether or not the elliptic curve has complex multiplication (CM), an important theoretical property for which lots of flashy theorems begin with assuming whether or not the curve has CM. They go on to train another binary classifier (and a separate size k classifier) to determine a curve's Sato-Tate identity component using the Euler coefficients and group-theoretic information about the Sato-Tate group (constructed by randomly sampling elements and representing the two non-trivial coefficients of their characteristic polynomials as independent variables in the classifier). They also run a PCA: https://arxiv.org/pdf/2010.01213.pdf

The cool part is that they then stepped back and scratched their heads wondering why the classifier was so good at achieving separation for these dependent variables in the first place, and plotting the points showed them to be (non-linearly) separable due to a visually clear pattern! The punchline and the reason it's so important to understand these data points, the Euler coefficients for elliptic curves, is because they contain all the relevant number-theoretic information about the curve. With some major handwaving, understanding them perfectly would lead to things like the Langlands program (and some analogues of the Riemann hypothesis) getting resolved. These wide reaching conjectures are ultimately structural assertions about L-functions, and L-functions are uniquely specified by their Euler coefficients (the a_p term in their Euler factors). Will murmurations help with that? Who knows, but the more patterns the better for forming precise conjectures.

Relevant intersectional credentials: I have lead ML engineering teams in industry and also did my doctorate work in this area of math, including using the LMFDB database referenced in the article for my research (which was much smaller back then and has grown a lot, so very neat to see it's still a force for empirical findings!).

Comment · 3 years ago
Costco only takes cash and debit, not credit.

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