alex @alexthecamel@schelling.pt

@emilygorcenski
this was my experience moving from australia to the uk (probably works for any european country)

amusing watching freddie de boer try to put the crisis of trust in the state on the left back in its bottle

man all people wanted was a twitter clone that was more peaceful lol

almost up to mastodon following numbers lol

this is a fairly comprehensive migration

ok has to be a way to quickly find people (same handle)

ok i guess i can try this bluesky thing if ppl can get me an invite

what do you people think the base rate for empire longevity is anyway
we remember rome because it lasted, not because it fell

guys rome had a pretty good run having a couple centuries of peace and prosperity to work this out sounds like not the most pressing problem in the world

???
---
RT @elonmusk
@DurantQuotesBot Low birthrate is under-appreciated as causal in the fall of civilizations. Rome was having birth rate issues even during the reign of Caesar.
https://twitter.com/elonmusk/status/1647774353172447238

time to drag out the quip about whether you would trust a hotdog seller who thinks hotdogs are gross again

(other sources of comparative advantage in dating markets:
women should try dating poly men?
nerdy straight guys should try dating trans women?
these seem maybe harder to exploit though)

a shockingly high percentage of straight women would not date a bisexual guy?
i feel like there is definitely like ... alpha here

i think we underestimate the amount of bullshitting a layman with google can come up with

experiment i would like to see run: GPT-4 versus layman with Google
https://scottaaronson.blog/?p=7209

(possibly different tilings Tk for each k)

construct P_1, P_2, etc recursively so that P_k is minimal and there exists a tiling T with P_1(T)=P1, P2(T)=P2, ... Pk(T)=Pk
because we can construct an S so that T->S we know for any such T anything in Pk occurs infinitely often in T

(note that we're talking about extending the tiling to *cover a particular square*, so all the ways of extending are mutually exclusive + exhaustive)

now assume there is no T->T and there are countably many tilings
let's organise them into a list
T1, T2, ...

so we can always choose one so that our current patch (and any subpatches, and with these we'll eventually have all finite patches in S) occurs infinitely often in T, by the infinite pigeonhole principle
so lemma (2) done.