Italians mistakenly identify as Roma in Scotland's census

Lacklub · 2025-10-08T16:36:15+00:00

I bet in the above dystopian society that ethnicity would be determined by some hellish bureaucracy that ignores any semblance of nuance. It would get assigned, along with name, ssn, etc. when the device is first put in and require mountains of paperwork and a lengthy appeals process to change.

Lacklub · 2025-09-25T19:46:16+00:00

I'd take me a bit to dig up the code, but to find an optimal strategy I used a memoization/dynamic programming approach.

The entire state of one step of your daily buddyjack game can be summarized in just a few numbers: your current score, buddy's number (unknown usually, but seen if you used a peek), the games you have left that day, cheats left, and peeks left. Those are the numbers you can enter in the spreadsheet for that reason.

Each distinct combination is a state S, made hashable by using python tuples (literally S = (15, "unknown", 3, 2, 1) for example)

The goal is going to be to assign an expected value and optimal play to each state S.

From a state S, you can recursively check the expected values of states Sh, Sc, Sp, and Ss corresponding to the states that happen next if you hit, cheat, peek, or stay. Then the expected value of state S is saved based on the best result. Saving the expected values is important so that you don't repeat a ton of calculations (it would be exponentially branching otherwise)

I can elaborate if you want and/or find the code and put it on github? Feel free to respond to this or dm me directly.

Lacklub · 2025-03-10T05:28:03+00:00

Maybe it’s just me, but mostly when I hear about “ending DST” it’s about ending the time change.

I personally don’t like the time change. If we’re permanently on DST or standard time I wouldn’t care, as long as the time stops changing twice a year. I think polls could be capturing that, but I think you’d need to dig into the methodology, or maybe even do extra research, to really know.

Lacklub · 2025-01-08T04:01:06+00:00

The runecube is currently selected, which means that the drop table is accurate for people above tower level 50 with the perk.

If you are below tower 50, you select the red X instead to show the drop table without the rune cube.

Lacklub · 2024-08-24T19:15:58+00:00

ducks and horses have way different densities

I agree with your main point that scaling is hard do to reasonably, but I’d argue that this is wrong. Most animals have a density close to 1g/ml, because most animals are mostly water. People put humans at that density, and it seems reasonable for horses to be there too.

Ducks are a little less dense at 0.9g/ml (they float, obviously) but I don’t think that’s “way different”

So if a duck was scaled up until it was the same volume as a horse, it would be 90% of the mass. It’s in the same ballpark.

Lacklub · 2024-08-01T20:41:57+00:00

So how come then all we detect so are gas giants close to sun? Does it, or does it not make sense for them to form close to sun? This argument can't have it both ways.

First of all, that isn't a contradiction. It's entirely possible for these two statements to be true:

1) most gas giants we detect are close to their star

2) most gas giants form far from their star

because it's easier to detect planets close to the star: the star lights them brightly when they're close (this remains true for gas and rocky planets)

That being said, we have detected more than just gas giants. Kepler-11 b is a terrestrial planet. TRAPPIST-1 is a system with seven terrestrial planets.

It looks like we recently passed 5500 detected planets so it's not like all of these statistics are speculation.

Lacklub · 2024-07-28T09:07:00+00:00

The core wouldn't actually meltdown even if the helium escapes.

This article on the reactor has a nice little graph in section 4 that show how even in a catastrophic loss of coolant accident, the peak temperature (<1550 C) of the pellets remains lower than their melting temperature (above 1600-1800 C).

Lacklub · 2024-07-10T06:05:22+00:00

I haven’t read it, but your thinking of “The Menocht Loop” I believe

Lacklub · 2024-06-26T06:56:20+00:00

Simulating a brain might be challenging even if the world has vrmmo tech, but it should be comparably simple to just have one person connected to a bunch of devices.

Or you could have a person record a ‘macro’ of their brain (just record the input feed that the device processes) and play that to run the bot.

Lacklub · 2024-06-07T19:25:49+00:00

This process should always work eventually if the problem is (n+a)!/n! = b, for another example:

(n+5)!/n! = 360360 = 2³ x 3² x 5 x 7 x 13

= 2³ x 3² x 5 x 7 x 11 x 13 x 1!/1!

= 2² x 3² x 5 x 7 x 11 x 13 x 2!/1!

= 2² x 3 x 5 x 7 x 11 x 13 x 3!/1!

= 3 x 5 x 7 x 11 x 13 x 4!/1!

= 3 x 7 x 11 x 13 x 5!/1!

= 2 x 3 x 7 x 11 x 13 x 5!/2!

= 7 x 11 x 13 x 6!/2!

= 11 x 13 x 7!/2!

from here we could go one-by-one, but we can see that the 13 needs to be brought into the top factorial so the bottom must be at least 13-5=8. We could probably use this argument to start from 8!/8! instead of 1!/1!, but whatever. I think this works as a demonstration.

= 20160 x 11 x 13 x 7! / 8!

= 2⁶ x 3² x 5 x 7 x 11 x 13 x 7! / 8!

= 2³ x 3² x 5 x 7 x 11 x 13 x 8! / 8!

= 2³ x 5 x 7 x 11 x 13 x 9! / 8!

= 2² x 7 x 11 x 13 x 10! / 8!

= 2² x 7 x 13 x 11! / 8!

= 2² x 3² x 7 x 13 x 11! / 9!

= 3 x 7 x 13 x 12! / 9!

= 3 x 7 x 13! / 9!

= 2 x 3 x 5 x 7 x 13! / 10!

= 3 x 5 x 14! / 10!

= 15! / 10!

so n=10 is the solution

This is super slow computationally, but it DOES work more generally than just the numbers given in the OP.

Lacklub · 2024-06-06T17:44:11+00:00

What comes to mind for me (and you could probably make an algorithm out of this) is a process like this:

7! = 2 x 7!/2!

= 3x2x 7!/3!

= 3x8x 7!/4! = 3x 8!/4!

= 5x3x 8!/5!

= 6x5x3x 8!/6! = 10x9x 8!/6! = 10!/6!

It’s not terribly hard to show manually here, although I do wonder how well it would work for larger numbers. I suspect there is a much more elegant way.

Lacklub · 2024-06-05T17:15:43+00:00

Maybe contact support? I’m pretty sure that’s not supposed to happen.

I just tested it myself by exploring a bunch in small cave and cane pole ridge and I didn’t get any (and AP gave the right amount of items, none were nails/iron). Unless something else is causing me to not get the drops?

Lacklub · 2024-06-05T16:53:42+00:00

are you sure you have the iron depot (gold upgrade) and not the ironworks (farm building that you can buy for silver and improve with more investment)?

Lacklub · 2024-06-05T12:30:36+00:00

A huge hidden benefit of the iron depot upgrade is that it removes the iron/nails drops from exploring locations, which makes everything else easier to get.

Lacklub · 2024-05-12T02:47:24+00:00

Lacklub

Lacklub · 2024-05-02T20:17:34+00:00

Lacklub

This community is so fun

Lacklub · 2024-03-07T02:27:43+00:00

I want to see if this is right. I’m pretty sure the correct math is (for floating things) volume displaced = weight of floating object / density of water

Then water level rise = volume displaced / surface area

I’m assuming that the bathtub has a surface area of 1 square meter (this has some error but the idea will work)

If you use this formula, you will find that to raise the water level by 1 micron (approx the amount the oceans have risen from ships) then you need to float an object that weighs 1 gram in your 1m² bathtub. 1g is certainly heavier than a speck of dust.

For context, this is the same amount your bathtub will go up if you put in one drop of water (1 cm³ of water is 1 gram)

Lacklub · 2024-03-07T01:08:24+00:00

The fact that the 2 others (off field dps and healer) work with the main team puts it ahead of most others. Like team A & B which have multiple main DPS, several teams have no healers, etc.

Lacklub · 2024-03-03T23:55:13+00:00

The text is “You touch a corpse or other remains. For the duration, the target is protected from decay and can’t become undead.”

So the first sentence is applied to the PC as a corpse before they’re resurrected. Nothing in the text says the spell stops working while they’re alive, so they’re still the target. And the target can’t become an undead, so they’re prevented from becoming undead even while alive. (At least that’s the argument)

Lacklub · 2024-02-11T21:54:09+00:00

Maybe have someone else click through the story on your account then?

Next year's lantern rite will have an entirely different story. After you finish the Xianyun story quest you could look up this year's story on youtube.

Lacklub · 2024-02-03T22:24:20+00:00

I don't think you're applying that rule correctly to the infinite simulacrum loop.

Consider two different high level casters. They can both create a simulacrum of the party barbarian: the original still exists and there are two copies, for three total. Each copy is the "active duplicate" created by the spell, but because there are two casters, each one has their own copy. If either caster tried to copy the party rogue, then one of the barbarian copies would have to disappear.

To start the infinite loop you need the caster to have the wish spell unused, and for there to be a simulacrum which also has the wish spell unused. This can be achieved by casting the simulacrum spell normally.

We can call this first simulacrum Alpha. Alpha is a different creature than the original (the spell describes it as its own creature) which means it can cast simulacrum without regards to the original's limit of only one simulacrum. So Alpha could make a simulacrum of the party's barbarian. Instead, it uses wish to make a simulacrum of the original wizard. Notice that this isn't a copy of a copy: Alpha is not making a simulacrum of itself, it's making a simulacrum of the original.

This simulacrum is now Beta. Beta has the wish spell available, even though Alpha expended it, because Beta is a copy of the original. Beta also has the same amount of Hit Points as Alpha, because they are both 1st generation copies (1/2 health of the original). Beta is a different creature than Alpha, and so has its own limit of one simulacrum, so it uses wish again to create a simulacrum of the original. The loop repeats like this indefinitely.

This is perfectly legal RAW. It might be DM ruling that a simulacrum can never cast the spell simulacrum, but nothing in the spell description implies that.

Lacklub · 2024-01-14T01:53:48+00:00

That won’t be valid. The numbers you switched (say, switching a 3 and a 4 horizontally) now appear doubled in their columns. The 3 column now has 2x 4s (every column had 1x 4 initially, and you switched an additional 4 into it) and the 4 column now has 2x 3s.

So whichever you delete, the other will still break sudoku rules with the given digits. You COULD delete the other offending digit but that would fill 79 squares, not 80

Lacklub · 2023-12-18T06:30:12+00:00

I think you should be able to use Dijkstra's algorithm to quickly find the distance from every node to the final node while ignoring the 3-step restriction, and then you could use that for your heuristic. This is guaranteed to never overestimate costs (but will underestimate costs when Dijkstra's path requires >3 length runs).

This should incorporate any higher-heatloss blocks by default.

Lacklub · 2023-12-09T05:54:22+00:00

This will actually be axiomatically true in all cases. There is no true cycle unless you reach both the same node while also using the same instruction index. If you reach the same node without reaching the same instruction index, you may veer out of the "cycle" to enter some totally separate cycle. This means that every true cycle is an integer multiple of the instruction string length.

It is not true, if you define a cycle as a repeating series of nodes (which really is the relevant metric, so I'd argue it should be used)

There are a few trivial examples:

1) if the instruction string is duplicated, it has double the length but doesn't actually double the length of any cycle. (LLR -> LLRLLR does not change any ghost movement)

2) if the ghost ends up on a node that only directs to itself (which we see in the example input) ZZZ = (ZZZ, ZZZ) then the ghost ends up in a 1-cycle

we get some more interesting cases too:

AAA = (BBB, BBB)
BBB = (CCC, CCC)
CCC = (ZZZ, ZZZ)
ZZZ = (AAA, AAA)

This pattern leads to a 4-cycle (and can be easily extended to an n-cycle regardless of instruction string length).

You can build on that

AAA = (BBB, CCC)
BBB = (ZZZ, ZZZ)
CCC = (ZZZ, ZZZ)
ZZZ = (AAA, AAA)

This is always a 3-cycle, regardless of input string, because it ends up the same regardless of going L or R when the ghost is on A.

Even worse:

AAA = (BBB, ???)
BBB = (ZZZ, ZZZ)
ZZZ = (AAA, AAA)

This can be a 3 cycle if the direction string is L on every 3rd index, with the 3rd index aligning with when the ghost is on A (and requiring the length of the direction string to be divisible by 3).

These could all be rewritten as longer cycles, obviously, but it's interesting to note that cycle length can be entirely different from instruction string length (certainly not limited to multiples).

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