ababa11e wrote: August 13th, 2024, 12:22 pm
im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)
edit: rutabaga do your worst
lmao cracker hierarchy is awesome, but you just gave me another new idea
the generalized version of cplus is capital C(x), but it can take any amount of arguments.
first, though, let's define cplus+++(x1,x2,x3,x4) to make sure everyone understands the pattern.
cplus+++(x1,x2,x3,x4)=cplus++(x1,x2,cplus++(x1,x2,cplus++(x1,x2...x4 times...cplus++(x1,x2,x3)...)
now we can tell that in general:
C(x1,x2,...x(n-1),x(n)) = C(x1,x2...x(n-2),C(x1,x2...x(n-2),C(x1,x2...x(n) times...C(x1,x2...x(n-1))...)
...but i just had ANOTHER idea, so i don't want to stop there. first, let's make C((x)) be C(x,x,...C(x) times...x).
now let's run through the whole entire hierarchy again. let's pretend that the old C((x)) is the new C(x), and we have to do the whole hierarchy thing again to get back to C((x)). doing this gives us C((1,x)). doing the C(x) to C((1,x)) procedure, C((1,x)) times, gets us to C((2,x)).
you can probably see where this is headed. doing the C(x) to C((n,x)) procedure, repeated C((n,x)) times, leads us to C((n+1,x)).
naturally, we'll eventually get more arguments in the function. first, C((C((...C((x,x)) times...C((x,x)),x...)),x)) is C((1,1,x)).
we'll grow the second argument in the same way as normal-- C(x) to C((1,1,x)) repeated C((1,1,x)) times is C((1,2,x)).
increasing the new first argument goes like this:
C((a,C((a,C((...C((a,C((x,x)),x)) times...a,C((x,x)),x...)),x)),x)) = C((a+1,1,x)).
when the "a" reaches C((C((C((...C((C((x,x,x)),x,x)) times...C((x,x,x)),x,x...)),x,x)),x,x)), we get a fourth argument.
we increase the 2nd-to-last argument the same way we did before. increasing the 3rd-to-last argument is mostly the same:
C((n1,a,C((n1,a,C((...C((n1,a,C((x,x,x)),x)) times... n1,a,C((x,x,x)),x...)),x)),x)) = C((n1,a+1,1,x)).
n1 is just placeholder for numbers that come before the a-argument.
when a = C((n1,C((n1,C((...C((n1,C((x,x,x)),x,x)) times...n1,C((x,x,x)),x,x...)),x,x)),x,x)), increase the 4th-to-last number by 1.
a pattern is faintly visible here, and while it'd be easier to see with more arguments, it'd take a lot of time to explain each step of adding each argument, so if this isn't well-defined enough, i can explain it more in another post because this one is already kinda long.
adding a 5th argument is just like adding a 4th was, and it behaves like the other arguments. you can put any amount of arguments in C((...)). to top this all off, C(((x))) is C((x,x,...C((x)) times...x)).
my number is C(((418))).
not sure if i did my worst or not
) as