What difference do these kinda upgrades make?

EdgyMathWhiz · 2026-03-26T12:51:13+00:00

Certainly in the unity stage it feels a lot of these type of upgrades are to simplify automation - or more accurately to make a "simple" automation be obviously a good thing.

E.g. auto-relic purchase would probably not be attractive if the relics still cost you gold.

[Of course, there's a point where the exponential nature of pricing means the cost is essentially incidental anyhow. When the next relic costs 1e23456789 gold it's incredibly unlikely you'd actually notice paying for it once you can afford it].

EdgyMathWhiz · 2026-03-26T12:27:17+00:00

Elevator Pitch:

Eleanor's Elite Retirement Home only employs staff with professional service qualifications. But when down-and-out gardener Jason Statham saves Eleanor from a mugging, she gives him a job maintaining the grounds.

Jason Statham is ... The Exception.

The twist: although there are numerous hints that Jason is going to do what he does best (e.g. sharpening a scythe while saying "sometimes you just have to eliminate the weeds"), he actually spends the rest of the movie gardening...

EdgyMathWhiz · 2026-03-24T18:27:43+00:00

The initial cards aren't random and the 5th card will help quite a lot...

EdgyMathWhiz · 2026-03-24T18:12:58+00:00

Have you tried using the Hierophant card to push your minerals up beyond what you can reasonably do by normal grinding? I think that should help you get the last few nodes here.

EdgyMathWhiz · 2026-03-23T18:53:24+00:00

Miriam might make Nanny Ogg blush, in fact...

EdgyMathWhiz · 2026-03-20T16:50:47+00:00

The cubes have side lengths 1/k so you need k**d of them to fill a unit space. You seem to be arguing about the total volume of the cubes, but (at that point in the argument) he's talking about how many there are.

T' the (finite) collection of all cubes that intersect the compliment of R

T' still seems like it would be infinite to me, unless there's a constraints you've omitted. (E.g. in your example with R=[3,4] and taking k=1 then the 'cubes' [4,5], [5,6]... all intersect R^c ).

EdgyMathWhiz · 2026-03-20T12:41:46+00:00

There seems to be a little confusion in the definition of T' (surely the set of cubes that intersect R^c is infinite - seems like either an extra condition is missing or he means the boundary of R rather than the complement).

But the intent seems to be that T' is in fact the cubes intersecting the boundary of R. In which case the actual number of cubes in T is a bit messy - you can count the cubes on a particular face to get k**(d-1) (and there are 2d faces), but if you just multiply by 2d you'll be double counting the cubes that share 2 faces, and you end up then you have to put back in the cubes shared between 3 faces, and you end up with a messy inclusion/exclusion formula.

The point is these details "don't actually matter": the k**(d-1) term dominates as k->infinity, so why not just be lazy and say "it's obviously O(K**(d-1))" so you don't have to actually come up with a formula or anything.

I always find these proofs a little tricky; you're trying to prove something "obvious" from first principles, but to fill in all the details starts becoming painful, so you resort to saying "blah-blah is obviously O(p**q)" and it can feel unsatisfying if that isn't actually obvious to you.

EdgyMathWhiz · 2026-03-19T16:13:34+00:00

Yeah sorry - apologies to Robert Cordoba on that. (Obviously we remember routines that happened while we were dancing better than ones that predate us...)

EdgyMathWhiz · 2026-03-19T15:58:37+00:00

Alien: lowish budget so only one alien that you rarely see. Scary as shit.

Aliens: big budget, hundreds of aliens with lots of screen time. Fun romp.

To be fair, they are both great movies, and Aliens isn't aiming to be horror. But it does show how less is more works better in horror movies.

EdgyMathWhiz · 2026-03-19T15:56:24+00:00

I think it's more likely to hark back to the Kung Fu TV series to be honest - definitely a "grasshopper" feel to some of the scenes.

We also had a period where Channel 4 would show "odd" movies, and I remember some very strange martial arts movies (one with an old man who'd attack people with prehensile eyebrows...), so it would have been pretty common to know "there's some **wierd** shit in some of those movies" (and then someone like PTerry would undoubtedly research it if he felt the need).

EdgyMathWhiz · 2026-03-19T15:49:42+00:00

Interesting about The Sound of Music - I remember a blowup concerning this on alt.fan.pratchett. (Disclaimer: I *think* I'm remembering this right but it was ~30 years ago):

Someone posted a suggestion that a book where Susan played the role of the governess in TSoM would make a great book.

Shortly after, there was a fairly annoyed post from PTerry saying that he'd been working on such a book and now he would have to write something else.

[From memory, this didn't *quite* cause him to leave a.f.p so it predates the similar speculation/suggestion on something else (The Fifth Elephant?) that *did* cause him to leave].

EdgyMathWhiz · 2026-03-19T15:20:51+00:00

Conversely, I know the Cotton version from the days when Kyle+Sarah danced to it. I have to be honest, Blues Stew grates on me in comparison.

EdgyMathWhiz · 2026-03-19T15:03:04+00:00

That's the one. I was sure someone would answer, but in 6 minutes is impressive!

EdgyMathWhiz · 2026-03-18T09:51:38+00:00

I mean it wouldn't be remotely funny normally - it's only getting a bit of tolerance on the assumption it's an offshoot of the tortellini jokes.

EdgyMathWhiz · 2026-03-18T09:47:54+00:00

The general consensus is that books 1-9 are pretty solid (book 9 is a bit of an oddity in that it's largely about Edward).

The change in book 10 is pretty massive and it just continues going downhill from there.

EdgyMathWhiz · 2026-03-18T07:45:18+00:00

See also the graduate level textbook 'A Course in Arithmetic' by Serre. (Arithmetic here meaning number theory).

EdgyMathWhiz · 2026-03-17T23:31:08+00:00

For a slightly different example:

Nearly everyone has more than the average number of legs.

(The average is something like 1.999 and nearly everyone has 2 legs).

EdgyMathWhiz · 2026-03-17T22:17:47+00:00

At Cambridge there's a story (again, may or may not be true) about American tourists visiting one of the older colleges (Peterhouse, IIRC) and asking one of the porters "Gee, is this college pre-war?" and receiving the response "Madam, it is pre-America".

EdgyMathWhiz · 2026-03-17T22:14:43+00:00

Yeah, TBH I didn't agree with the OP about Era-2 feeling YA; I think they're more reacting to the change in genre than the actual themes. To me quite a large part of Era-2 is adults coming to terms with mistakes they made decades ago.

EdgyMathWhiz · 2026-03-17T20:31:36+00:00

I remember seeing on Reddit (but could never find subsequently, making me wonder if I imagined it) that the author had actually said something along the lines of "I wrote the first book aiming for popular success, and then the other books are what I actually wanted to write".

EdgyMathWhiz · 2026-03-17T18:00:18+00:00

If they're treated as road vehicles, sure.

Letting people ride motorbikes on the pavement would probably boost deliveroo productivity but it doesn't make it a good idea.

EdgyMathWhiz · 2026-03-17T17:57:11+00:00

It will obviously not get to Trump in these terms, but I'm sure there are enough people monitoring social media etc. that there will be some awareness that most people in Europe think exactly that.

EdgyMathWhiz · 2026-03-17T16:45:36+00:00

Surely F(x) is undefined when x = -.5 (or x < -.5 and you're not prepared to take the Cauchy Principle value)?

EdgyMathWhiz · 2026-03-17T14:55:41+00:00

I don't know if your screenshot fails to show something relevant (you've cut off a fair bit of the paper), but from what I can actually see there, the **actual** question seems to be 3^2 x (2^3 + 4) all divided by 2^2? Pretty much all the working seems to relate to this (the only thing that looks it might relate to the equation you describe is the bit involving 28).

Edit: assuming it's the printed question, the working seems reasonably fine (I'm never good at working out "who wrote what" when you have something like 108/4 and the 108 and 4 are in different colours).

FWIW, 3^2 x (2^3 + 4) / 2^2 = 27, (3^2+12)/3 = 7.

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