K‮ly‬e @khird@qoto.org

@trinsec fifty new users in the last two hours lol, you bet there'll be a spike

@s8n nah I'm happy to have them, just dealing with a crush of "someone on Poast used a racial slur!" reports right now so I figured a bit of education was in order

Looks like we are getting a lot of former Twitter users today - I hope you enjoy exploring QOTO and the rest of the Fediverse! Our rules are at https://qoto.org/about/more#rules and you will be expected to abide by them. Be aware that our rules are not the same as Twitter's, and other instances on the Fediverse will have different rules that their users are expected to follow.

If you're considering reporting a foreign user (i.e. from an instance other than QOTO), please take a minute to read that instance's rules. There's no point in filing reports over conduct that the foreign instance doesn't prohibit. You can block the user or his whole instance yourself, but QOTO won't intervene except where the offending user is evading your blocks e.g. by setting up alts to contact you.

If you think the user's conduct *does* violate his instance's rules, make sure you tick the little box to forward a copy of the report to the foreign moderators. They might hand down some kind of punishment to him if they agree with you. But in either case, no punishment will come to him from QOTO's team.

I'll do my best to answer questions if you tag me.

@bonifartius H2 gas has a reasonable energy density in J/kg, but since it has such a low density in kg/m³, it has a poor volumetric energy density in J/m³.

Another way to look at this is to say H2's enthalpy of combustion in kg/mol is very low (propane's is about eight times higher), so you need far more moles of hydrogen to store equal amounts of energy. But here you run into the ideal gas law P=ρRT. To get many moles into a reasonable amount of space, we need a high density ρ. R is a physical constant, and so you either need to drop T by refrigerating your storage, or raise P by physically reinforcing it. Both of those options are too heavy to be competitive for mobile applications - even if you don't have to carry around as many kg of fuel, you're losing more than the difference in carrying around the tank to hold it in. For static applications where you can tolerate the weight, there are other cheaper ways (notably, pumped-storage hydro) to store your energy.

@louisrcouture Is there something in Quebec's political system that disfavours broad coalition parties? The list you gave makes the Democrats seem very ideologically diverse, relatively speaking - four of the five parties' first listed equivalent is a Democrat. Similarly, the Conservatives have had everyone from Bernier to MacKay to Sloan to Poilievre challenging for the leadership in the past five years, so it happens elsewhere on both sides of the spectrum. Why not in Quebec?

@louisrcouture That'd be interesting - I don't really have a good sense for how the parties in QC match up with the political spectrum used in English-speaking North America, but it'd be neat to learn more

@valleyforge I dunno about that. By the time the US was expanding westward, sextants and marine chronometers were sufficiently accurate to get pretty accurate fixes, so the DC-London offset would've been well established. It's hard to see a practical benefit from subtracting that offset from all the longitudes to rebase them against the national capital.

On the other hand, the idea of time zones was still fairly new, influenced by the railways. Prior to that, each town set its own clock, and the idea of having an accurate clock time away from cities didn't exist - you could accurately measure that some event occurred a certain time after another, but it wasn't really meaningful to speak of N o'clock in rural areas. So the benefits of being on the same standard as the rest of the English-speaking world would also have been less obvious than they are today.

@valleyforge a lot of state boundaries are referenced to degrees longitude west of DC, from before everyone gave up their weird nationalistic need to make themselves the centre of the world and just accepted Greenwich would remain the standard

@Pat A bit cheeky, but guessing Rock for both wouldn't be wrong - he has two roles in the film, one a CIA agent and one his twin brother the streetwise punk.

More seriously though, I suspect you could identify them as Hopkins (64 years old) and Rock (37 years old) and have people draw similar inferences with no mention of race at all.

@Acer

A "loan shark" is a person who lends money illegally. Because loan sharks can't use the courts to force repayment of their illegal debt, they have a reputation for using threats, violence, and other illegal tactics to convince the borrower to repay them. Loan sharks typically lend to people who can't borrow from legal banks, like people whose credit is too poor or who want the existence of the debt to remain hidden to other banks or courts. The interest rates associated with illegal loans are stereotypically higher.

A "government sponsored loan shark" sounds like someone who lends money or collects debts under similar circumstances, except with the support of the government. The borrowers might have been required by law to borrow the money, or it might be at a very high interest rate. If the borrower fails to repay the debt, the bailiff might seize their property or the courts might order the debt repaid and jail the borrower for contempt of court until he complies.

Whoever wrote that intended you to have a negative opinion of the person he's describing.

@publiclewdness@linuxrocks.online

@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.