K‮ly‬e @khird@qoto.org

@freemo @jahnke

I think handwriting is likely less similar hand-to-hand than you think. I'm right-handed and learning to handwrite Persian (a right-to-left language) gave me a lot of appreciation for the difficulty lefties experience in writing English - the tip of your pen tends to be at an angle where it digs into the paper rather than smoothly brushing against it, you drag your hand through the ink you've just written, etc. It seems probable that the muscle movements to guide a pen along a certain path with one hand are going to be quite different from those to guide it along the same path with the other hand, so much so that it's essentially a different skill.

@realcaseyrollins

This sort of thing has been likened to evaporative cooling[1]. From a collection of particles of varying temperature, you can allow the hottest ones to escape, and the average temperature of the remaining particles will be lower even if no individual particle has cooled.

Similarly, if people aren't restricted from being hostile and insulting on a certain instance, eventually the moderates who don't like being insulted seek a less hostile environment elsewhere - but since it's not the extremists that leave, the average of the remaining members becomes more extremist. You can start with free speech and wind up with Nazis depressingly easily.

A balance has to be struck to maintain a stable community. On one hand, enough dissent must be tolerated that you're not constantly retreating from the advancing frontier of forbidden expression. On the other, you have to be willing to expel the troublemakers before they drive off your moderate members. I think @freemo and the gang are doing a pretty good job here of striking that balance.

1: https://www.lesswrong.com/posts/ZQG9cwKbct2LtmL3p/evaporative-cooling-of-group-beliefs

@design_RG

You have to think through how it would scale. An image you post could appear in arbitrarily many people's timelines simultaneously due to federation. If several thousand Mastodon clients all of a sudden try to load the same image from the webserver, that's a big spike it would have to handle.

One alternative would be for the client's instance to make a local copy on the first request, and rewrite the url so its users don't have to keep going back to the original. You'd get only one hit per instance instead of per user, but then you're back to storing a local copy anyway, so what's the gain?

@dragfyre

Thanks! I found Discord nearly unusable in a mobile browser, but the website itself is very nice. A suggestion: it might be worth linking from the Study Hall page to the official bookstores, which I find have content not present in the free repositories.

@Absinthe @alex_ @freemo I have been learning Persian, where Q is used to transcribe a certain letter (ق) that corresponds to a sound absent from English, so that's how I read it. Kind of like the French R but with the throat closed.

@Absinthe I posted a couple, too, but they didn't get any nibbles so I didn't really keep at it. @donovank What particular topics are interesting to you as ground for challenges?

@pschwede

Sounds like you aren't implementing this line from the problem description : "When he returns, he notices that all Xmasium-atoms in the pot now are of the same type."

@pschwede

Let a, b, c be the count of each species in no particular order. Three operations exist:
1. a++; b--; c--
2. a--; b++; c--
3. a--; b--; c++

Note that a' = (b+c) either decreases by two or is unchanged. This implies two properties:
a' cannot change from odd to even.
a' cannot increase.

The surviving species at the end (a' = 0) has to be γ, since the others have odd initial a'. And no more than 29 can be of this type, limited by the initial a' of the α species.

The problem recently posted by @freemo reminded me of this one from my freshman analysis class in university.

I claim that, for all natural numbers N that are not perfect squares, √N is irrational. Determine, and prove, whether my claim is true or false.

Maybe you've seen it before, if not; I encourage you to have a go at solving it.

@freemo

The case where a is an integer but not a natural number can be excluded because primes are a subset of the natural numbers. If a < 1, then p = a³ - 1 < 0 which contradicts p's primality.

@freemo

Assume the opposite: some cube a³ = p + 1 for some natural number a ≠ 2 and prime p.

The difference of cubes formula shows that:
p = a³ - 1 = (a² + a + 1)(a - 1)

Both terms (a² + a + 1) and (a - 1) are integers. Since p is prime, it follows that exactly one of the two must be equal to unity.

The first possibility can be ruled out because a² + a + 1 = 1 has no solution in the natural numbers.

The second is impossible because a - 1 = 1 contradicts the assumption a ≠ 2.

@freemo

My thoughts on the matter, having read and considered the existing discussion but not replying to any particular part of it:

Enlisting a new mod for every language spoken here doesn't make sense. A monolingual server with five mods and a thousand users is very different from a server with five languages, each with two hundred users and one mod - each language may have a different rate of abuse, mods can't pick up one another's slack when they go on vacation, etc. Moreover, evaluating a foreign-language candidate's suitability is unlikely to be better than guesswork, and rogue mods can cause a lot of damage very quickly. I do not think you will find this an effective solution.

You might reason something like the following:
1. In order that other servers need not block us to avoid being spammed, and to improve the experience of our own users, all content posted here is moderated.
2. As discussed above, content in languages other than English cannot be practicably moderated here.
3. Therefore, the set of content posted here (which, by statement 1, *is* guaranteed moderated) must be disjoint from the set of content in languages other than English (which, by statement 2, *cannot be* guaranteed moderated).

Ergo, content posted here must be in English.

If you think that user-submitted translations on the honour system are suffucient to make content locally moderatable, you can amend statements 2 and 3 accordingly - but I tend to think that you shouldn't consider such posts actively moderated, because you rely on user reports to find out that someone's cheating.

The appropriate solution is *not* for QOTO to try to bodge together a Swiss-army-knife mod team that can moderate any content, but for users who post content that QOTO cannot moderate to use another server that can (or recruit their own moderators and start their own). Then we federate with them and everyone is happy.

@design_RG @Sphinx

I've seen this said a couple of times and don't fully understand it. What's the downside to letting them follow you, even assuming it's a malicious bot? It doesn't leak any information beyond what an unregistered third party can gain by looking at your profile in a web browser as I understand things.

On my part I allow a couple such followers, and it helped me get listed in the QOTO directory to get real followers. So there's an upside.

@Zyxer
Given \(x^2 + y^2 = 4z + 3\) with \((w, x, y, z)\in\Bbb{N}\), \(x\) is odd if and only if \(y\) is even. This results from the observation that \(4z + 3\) must be odd and the fact that, if \(k^n\) contains a factor of two for \((k, n)\in\Bbb{N}\), a factor of two must also necessarily be present in \(k\). It is not possible, then, for \(y^x = x^w\) to be true, because one side of the equation is an odd integer while the other is an even integer.

@Absinthe

function sorted = snail(array)
if numel(array) <= 1
sorted = array;
else
sorted = [array(1, 1:(end-1)),...
array(1:(end-1), end)',...
array(end, end:-1:2),...
array(end:-1:2, 1)',...
snail(array(2:(end-1), 2:(end-1))) ]; end; end;