Recent submissions

Comment · 6 months ago
This is never going to be resolved through speculation. There are parallels with the Zodiac killer's identity where there's a group of very likely candidates but no conclusive proof to definitively settle the question one way or the other. Outright admissions mean nothing (Wright). Even cryptographic proof could leave some margin of doubt (speculations of an ECDLP break or key theft).
Comment · 7 months ago
Resist hypes and just use whatever you feel like. Torvalds uses a 40 year old EMACS implementation and that seems to be working for him.
Comment · 7 months ago
I left Spotify years ago. Youtube is so much nicer in terms of content alone. But Youtube, with its insane backlog of video's not available elsewhere, is straight up a monopoly, so they too will start squeezing customers at some point. In anticipation of that I've been collecting flacs again. It's actually kind of a nice hobby.
Comment · 8 months ago
I'm just so glad I'm past the phase where I'd worry about these things. It's very exhausting.
Comment · 8 months ago
I doubt talking heads on the TV can move markets.
Comment · 8 months ago
How can the models be impressive if they switch to Chinese mid-sentence? I've observed those bizarre bugs too. Even GPT-3 didn't have those. Maybe GPT-2 did. It's actually impressive that they managed to botch it so badly.

Google is great at some things, but this isn't it.

Comment · 8 months ago
The Weezer video was quirky, funny, creative, catchy and appealing to multiple generations. A stroke of genius by Microsoft.
Comment · 9 months ago
Probably a prank or an ambush set up by the anti-billionaires.
Comment · 9 months ago
FWIW, I just gave Deepseek the same prompt and it solved it too (much faster than the 41m of ChatGPT). I then gave both proofs to Opus and it confirmed their equivalence.

The answer is yes. Assume, for the sake of contradiction, that there exists an \(\epsilon > 0\) such that for every \(k\), there exists a choice of congruence classes \(a_1^{(k)}, \dots, a_k^{(k)}\) for which the set of integers not covered by the first \(k\) congruences has density at least \(\epsilon\).

For each \(k\), let \(F_k\) be the set of all infinite sequences of residues \((a_i)_{i=1}^\infty\) such that the uncovered set from the first \(k\) congruences has density at least \(\epsilon\). Each \(F_k\) is nonempty (by assumption) and closed in the product topology (since it depends only on the first \(k\) coordinates). Moreover, \(F_{k+1} \subseteq F_k\) because adding a congruence can only reduce the uncovered set. By the compactness of the product of finite sets, \(\bigcap_{k \ge 1} F_k\) is nonempty.

Choose an infinite sequence \((a_i) \in \bigcap_{k \ge 1} F_k\). For this sequence, let \(U_k\) be the set of integers not covered by the first \(k\) congruences, and let \(d_k\) be the density of \(U_k\). Then \(d_k \ge \epsilon\) for all \(k\). Since \(U_{k+1} \subseteq U_k\), the sets \(U_k\) are decreasing and periodic, and their intersection \(U = \bigcap_{k \ge 1} U_k\) has density \(d = \lim_{k \to \infty} d_k \ge \epsilon\). However, by hypothesis, for any choice of residues, the uncovered set has density \(0\), a contradiction.

Therefore, for every \(\epsilon > 0\), there exists a \(k\) such that for every choice of congruence classes \(a_i\), the density of integers not covered by the first \(k\) congruences is less than \(\epsilon\).

\boxed{\text{Yes}}

Comment · 9 months ago
Rather the opposite. A vibe-coded startup cannot survive if it can be trivially duplicated. The proof will be in observing the inverse phenomenon; (pure) software companies disappearing.
Comment · 9 months ago
It's certainly morally and legally dubious to facilitate attacks on things that others choose to use in within their own private domains, just because you disagree with that choice. But that's how these people roll.
Comment · 9 months ago
I honestly don't understand why a terminal emulator needs to be performant. Seems like peak bikeshedding to me.
Comment · 9 months ago
Very, very few people care about privacy to a meaningful degree, even if they (loudly) profess that they do. For communication apps specifically, there's also the network effect problem and having to deal with the increasing proliferation of government speech laws (Online Safety Act and what have you).

If you just want to make a buck, build a ChatGPT wrapper where people will pay you for the privilege of uploading their deepest secrets and intellectual property to your servers.

If you're ideologically motivated, forget about the profit motive and go FOSS.

Comment · 1 year ago
Many people (not the EFF or ACLU, lol) saw it coming years ago and put in a monumental effort to try to turn to tide, only to be relentlessly repressed and dehumanized by bolsheviks and idiots.

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