K‮ly‬e @khird@qoto.org

@louisrcouture The font is fine, but:
- beginning sentences with capital letters would greatly help, especially as French is my second language.
- the webpage breaks completely if you don't have JavaScript enabled, instead of falling back gracefully to a less-ornate version.

@dragfyre Sure, there are alternatives to proof-of-work schemes, most notably proof-of-stake. It's a different take on the same concept - a witness's reliability is linked to some scarce resource, either computing power in the case of proof-of-work, or previously earned currency in the case of proof-of-stake. A malicious witness must accumulate a significant quantity of that resource to make a false update to the ledger, which makes it cost-prohibitive to perform this type of attack. As long as honest witnesses cumulatively control more of the resource, the ledger is reliable.

You can imagine other such proof-of-X schemes where you base it on some other resource that's not environmentally destructive, but the underlying concept applies.

@dragfyre Something like, "maintain a reliable transaction ledger given only unreliable witnesses to the transactions"

@bonifartius you mean UIs like the Common Desktop Environment? I used that at a workplace, and then at home for a bit when it was open sourced a bit later. It was fun for a bit, but I do think modern UIs have improved on it since then.

@Gargron Will it expose the moderation tools? That's the biggest lack I find in the existing apps.

@Shamar Privacy advice from a Chick tract? Don't see that every day

@leah

@dragfyre Okay so here are my notes so far:

prob that 1 specific item has been chosen in a single round
9/k

prob that 3 specific items have all been chosen in a single round
9/k * 8/(k-1) * 7/(k-2) = 504/(k³-3k²+2k)

9 choose 3
84

prob that an event, with prob P of happening in any given way, happened in any of x possible ways
1 - (1-P)^x

prob that any 3 of 9 items have all been chosen in a single round
1 - [1 - 504/(k³-3k²+2k)]^84

This lets us compute k(2) = 41. I was surprised at how large k had to be for such a small t, but I couldn't find any errors in my arithmetic or reasoning.

I'm stuck on extending this to multiple previous rounds though. The straightforward approach would be to do 1 - (1-P)^(t²-t), substituting the above expression for P, but I think the probabilities aren't independent - if you draw three times, you have to do three comparisons: 1v2, 1v3, 2v3 - and once you know that 1v2 and 1v3 passed, the probability of 2v3 passing is no longer its a priori value.

@dragfyre Just to make sure I'm thinking about the problem correctly, a couple more questions:
1. You have k items in a bin
2. Draw exactly nine of these items and record them
3. Count how many of these items are being recorded for the second time or later
4. Replace all nine in the bin
5. Repeat steps 2, 3, and 4 until a total of t drawings have been completed & counted

When you say the items are chosen "with replacement", do you mean internally in step 2 (a single choice can then contain duplicate items), or just between drawings in step 4 (items can only be duplicated across choices, not within them)?

When you say "on average", do you mean that the average *choice* in a run contains no more than three previously-seen elements, or that in an average *run* no choice contains more than three previously-seen elements? A concrete example: if t=3, say the expected counts are 0, 2, and 4 duplicates in each successive round - the average choice contains two dupes, but the average run contains a choice with four dupes.

@dragfyre do you mean the minimum k? As you increase the choices available, the average number of collisions will decrease. So once you find a k with sufficiently low collisions that no more than three repeats occur on average, any number k' > k will also satisfy the same criterion.

@trinsec it sounds like you're experiencing chromatic aberration. Near the edges of your lenses the front and back surfaces aren't parallel, so the lenses act like prisms around the rim, separating colours of light. Shorter (bluer) wavelengths get moved further to the centre of the lens (for nearsighted wearers at least).

It took me a long time to get used to that as my prescription increased.

@kreyren http://esr.ibiblio.org/?p=8106 is an Open Source Definition author's statement on the controversy.

It appears I slightly misremembered the details - the ban actually targetted companies who did business with border services in a sort of second-order boycott, not the government itself.

@kreyren There was similar controversy during the Trump presidency, regarding the use of free software by the U.S. border services. I'm not sure how deeply split the "free software" and "open source" camps are, but the open source camp concluded that the 5th criterion of the OSD would not be satisfied by any licence banning that agency's use of the software. I imagine the same reasoning applies to bans on Russia's use equally.

@Acer

"upper limit" as a noun
"set an upper limit" as a verb

So the first link says the 45min/wk rule *sets an upper limit* of 1h30 for the retrospective of a two-week sprint, and the second advocates an *upper limit* of three bugs per developer.

@lurker anything in particular I should be looking forward to with the 5.16 series?

@freemo thank you!

@freemo could you please check in on QOTO's infrastructure when you get a chance? I'm seeing symptoms of federation failing in both Mastodon (missing posts from last night) and Matrix (can't join foreign channels).

@kreyren oops, never mind. I looked more closely and I think it's actually Sweden. Bosnia's only a candidate, so it'd be surprising to see her flag there. Would be easier on all concerned if they hung the flags vertically haha

@kreyren I think that's Bosnia