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Discussion (15 comments)

Showing 25 comments

konmok

15 days ago

3 replies

I'm a big qntm fan. I highly recommend their "antimemetics" SCP stories and articles.

jdpage

15 days ago

2 replies

There's a new, professionally-published book version of "There Is No Antimemetics Division" out as well[1], if you want to support Sam's work that way. I have print copies of both the self-published V1 and the new V2. I'm very excited about the latter, though I haven't finished it yet.

[1]: https://qntm.org/antimemetics

Analemma_

14 days ago

1 reply

One small word of caution if you read the older version first: for what I assume are copyright reasons around using SCP in a professionally-published book, the new published version has had to strip out all the SCP references and change the names of all the characters, but it is otherwise very close to the old one. There are a handful of new scenes and some other small differences, but many pages and chapters are word-for-word identical apart from the aforementioned name changes.

This could just be a me thing, but I found this incredibly distracting after being so used to the old version, and just couldn't manage to enjoy it. Fortunately I bought the old one as well.

lencastre

14 days ago

I’ve read the older version and really liked it, strange ending and all, and I’ve gifted the new version for X-mas. My xmas wish list is for a 6 episode mini-series funded by the fruit company.

patleeman

15 days ago

I loved this book. The audiobook is available on spotify and was a great listen.

solid_fuel

15 days ago

4 replies

I really enjoyed one of their other stories - Ra https://qntm.org/ra

riffraff

15 days ago

2 replies

I'll add that Lena/MMAcevedo[0] is both a wonderful story and terrifying

[0] https://qntm.org/mmacevedo

lencastre

14 days ago

they’re so eerily prescient

moss_dog

15 days ago

One of my favorites!

gostsamo

14 days ago

Ra was a disappointment for me. If you end up rewriting your entire world at the end of the book, it is an intellectual failing to tackle the main issues straight on. Combine it with an mc who suddenly becomes just an idiot walked around and what you end up with is some SV eschatonism. Lots of preaching and ready conclusions, but little to return to later.

phildenhoff

14 days ago

I hope qntm has the chance to traditionally publish Ra and have it edited as well. I enjoyed the book a lot, but felt it needed a solid once over.

Really enjoyed the novel though! Planning to reread it in the spring.

LoganDark

15 days ago

I love Ra -- Fine Structure is also great!

willtemperley

14 days ago

Looks like they got a publishing deal: https://qntm.org/publ

analog8374

15 days ago

1 reply

Hey he does good scifi too

jimbobthrowawy

14 days ago

Multi talented. He also wrote the fastest standards compliant json library.

AnotherGoodName

15 days ago

1 reply

Fun fact with arrow notation, if you put it under a modulus it quickly converges to the same value no matter how high in exponents you go!

Eg. 2^2^2 = 2^4 mod 35 = 16

Let's go one higher

2^2^2^2 = 2^16 mod 35 = 16 too!

and once more for the record

2^2^2^2^2 = 2^65536 mod 35 = 16 as well. It'll keep giving this result no matter how high you go.

https://www.wolframalpha.com/input?i=2%5E2%5E2%5E2+mod+35 for a link of this to play with.

I could do this with any modulus and any exponent too.

2^3^3 = 2^3^3^3 = 7 mod 11 etc.

The reason is that the orders of powers are effected by the totient recursively and since totients always reduce, eventually the totient converges to 1. This is where the powers no longer matter under modulus. Eg. the totient of 35 is 12 (the effective modulo of the first order power), the totient of 12 is 2 (the effective modulo of the second order power), the totient of 2 is 1 (the effective modulo of the third order power) and so after 3 powers under mod 35 it converges.

ashivkum

15 days ago

1 reply

I'm pretty sure there was a project Euler problem premised on this property but I can't find it at the moment.

AnotherGoodName

15 days ago

A classic would be quickly computing such big numbers under a modulus. You just compute the carmichael totient recursively till it hits 1, disregard higher orders and then going backwards calculate the powers, reducing by the modulo of the current order (this way it never gets large enough to be a pain to calculate). The totients reduce in logn time and each step is logn so it’s merely logn^2 to calculate.

piskov

15 days ago

1 reply

Ah, for a second I hoped it is another novel.

If you haven’t read “There is no antimemetics division”, do it now. Easily one of the top science fiction out there.

However buy the Pinguin books 2025 edition, not the self-published free one — that version has a meh ending and suffers from not having an editor.

Yossarrian22

14 days ago

1 reply

Wait! The ending is improved in the new version!?

piskov

14 days ago

No more dead ghosts guiding from the astral plane.

Not that it is top-notch, mind you, but much more coherent.

The book is heavily rewritten.

Sharlin

15 days ago

1 reply

Is it buggy for at lesst 2^(n)^2? It gives 4 for any n, but surely for example 2^^2 = 2^(2^2) != 4?

OgsyedIE

15 days ago

1 reply

2^^2 != 2^(2^2). Instead, 2^^2 = 2^2.

This will make more sense if you look at how the inputs a,b,n in the toy (2,2,3) and (2,3,3) present differently.

Sharlin

14 days ago

Yeah, got it now, thanks!

112233

15 days ago

As someone from time to time peeking into googology.fandom.com , my favorite big number device probably still is loader.c, simply because of how concrete and unreachable it feels.

Too bad most Friedman's work has linkrotted by now...

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