Talk:Cantor

Is this turing complete? Is it even close to being turing complete? Gggfr (talk) 13:32, 18 July 2024 (UTC)

I think this works as a proof using Underload (except for ! and a, which is fine because only (x), :, and ^ are required), treating each stack as 1 element of a bigger stack:

~ -> >-2#2>1#-1>1#-1

: -> >-1=1>2

* -> >-1#-1>-1+>1

(x) -> [x]>1

^ -> >-1!

S -> >-1^

Excluding *, there's always at most 1 element on each stack, so 7 elements is more than enough. I'm wondering if it might be Turing-complete with a limited number of stacks, since + allows for arbitrarily long strings, so I might try that as well. -PkmnQ (talk) 19:09, 18 July 2024 (UTC)