The standard way when using ordinal arithmetic is:
Take the ordinal 1, which is {{}}.
Replace each element with a ordered pair of the form {{a},{a,b}} with second element being 0 (that is {}).
Repeat with second element 1.
Take a union.
Take find the ordinal with this order.
Overall:
otp({
{{{}},{{},{}}},
{{{}},{{},{{}}}}
})
Or simplified
I saw somewhere that someone had decoded how an AI had learnt to do basic arithmetic. And it appeared to be using a massive expression containing lots of sin & cosines to do basic addition
12 Comments
theKalash@feddit.ch · 9 pts · 2y
abs(e^iπ^) + abs(i^2^)
Kuba@programming.dev · 4 pts · 2y
Thanks
rasensprenger@feddit.de · 8 pts · 2y
rasensprenger@feddit.de · 10 pts · 2y
TheOrs@lemmy.world · 8 pts · 2y
The standard way when using ordinal arithmetic is: Take the ordinal 1, which is {{}}. Replace each element with a ordered pair of the form {{a},{a,b}} with second element being 0 (that is {}). Repeat with second element 1. Take a union. Take find the ordinal with this order. Overall: otp({ {{{}},{{},{}}}, {{{}},{{},{{}}}} }) Or simplified
otp({ {{{}}}, {{{}},{{},{{}}}} })
slazer2au@lemmy.world · 7 pts · 2y
How complex you looking? https://en.m.wikipedia.org/wiki/Principia_Mathematica
MonkderZweite@feddit.ch · 3 pts · 2y
Most of them.
dbaner@lemmy.world · 2 pts · 2y
I saw somewhere that someone had decoded how an AI had learnt to do basic arithmetic. And it appeared to be using a massive expression containing lots of sin & cosines to do basic addition
Spzi@lemm.ee · 2 pts · 2y
(10^googol^)^0^ + (TREE(3))^0^
Although that's fairly easy to write. It's hard to calculate, if you calculate the brackets first.
pancake@lemmygrad.ml · 1 pts · 2y
(fix add a b := match a with O => b | S x => add x (S b) end) (S O) (S O)
Artisian@lemmy.world · 1 pts · 2y
Perhaps: (lim_{n->\infty} \sum_{m=1}^n 1/2^m ) + dim(Im(matrix([1,3,4],[2,6,8],[3,9,12])))
electrogamerman@lemmy.world · -1 pts · 2y
Pi/pi + pi/pi