What? W takes an array and returns an integer.
W([1, 1]) = W([W([1, 0]), 0]) + 1.
Yes, but in your explanation it looked like it has the same array as you started with.testitemqlstudop wrote: ↑November 14th, 2019, 8:18 pmWhat? W takes an array and returns an integer.
W([1, 1]) = W([W([1, 0]), 0]) + 1.
But wait! there's more naïve extensions to be done!Moosey wrote: ↑November 26th, 2019, 7:40 pmwhat if...
1 ((()))
2 (()())(()())
3 (())(())(())(()())
4 ()()()()(())(())(()())
5 ()()()(())(())(()())
6 ()()(())(())(()())
7 ()(())(())(()())
8 (())(())(()())
9 ()()()()()()()()()(())(()())
10 ()()()()()()()()(())(()())
11 ()()()()()()()(())(()())
12 ()()()()()()(())(()())
13 ()()()()()(())(()())
14 ()()()()(())(()())
15 ()()()(())(()())
16 ()()(())(()())
17 ()(())(()())
18 (())(()())
19 ()()()()()()()()()()()()()()()()()()()(()())
20 ()()()()()()()()()()()()()()()()()()(()())
21 ()()()()()()()()()()()()()()()()()(()())
...
38 (()())
39 (()) x39
40 ()x40 (()) x38
...
80 (()) x38
...
~80*2^38 ()
~80*2^38
kirby-paris but you append n copies of the grandparent too
you can do it where you append n copies of all x-parents too
1 (((())))
2 ((()())(()()))((()())(()()))
...
probably comparable to 2^^n for some smallish n
Define ×(n) -- is it n×n = n^2?PkmnQ wrote: ↑January 13th, 2020, 1:08 pmHmm, what if I were to plug a function into itself?
¤(f,a,n,k)
f is the function
a is the amount of arguments for the function
x is just a number
¤(f,1,n,k) = f(n) × k
otherwise, ¤(f(...,kn),a-1,f(n,n,n...)×k,kn)
2×2 = 4
¤(×,2,2,2) = ¤(×4,1,8,4) = 128
¤(¤×,2,2,2) = ¤(¤×k4,1,128,4) = ¤(×,2,128,4) × 4 = ¤(×512,1,16384,512) × 4 = 16384 × 512 × 512 × 4 = 17179869184 (please check for me)
0th layer is 2^2, 1st layer is 2^7, and 2nd layer is 2^34.
How much does this grow proportional to x? I don't know, I'm not a googologist.
dcf1 has tetrational or maybe barely rate. Probably > f_3(n), but definitely < f_4(n)nolovoto wrote: ↑January 13th, 2020, 11:21 amdon't actually know too much about programming functions
Here's my dumb functions I invented
dumbcantorfunction1(1)= 1 (binary) = 1
dumbcantorfunction1(2)= 101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(1) times) = 5
dumbcantorfunction1(3)= 101000101000000000101000101000000000000000000000000000101000101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(2) (5) times) = 1534774961612150361293125 1.53477*10^24
dumbcantorfunction1(4)= (1 digit replaced with 101 and 0 digit replaced with 000 iterated dumbcantorfunction1(3) (1534774961612150361293125) times) = idunno a really big number
dumbcantorfunction2(n) = (1 digit replaced with binary of dumbcantorfunction2(n-1) or 101 if n<=2 and 0 replaced with 0's of the same digit number as dumbcantorfunction2(n-1) or 000 if n<=2 iterated dumbcantorfunction2(n-1) times)
so basically the same number s until it gets bigger at n=4
dumbcantorfunction3(n) = dumbcantorfunction1(n) but instead of always 101 and 000 one zero is added symmetrically for each side of the center zero of the replacement for each (1 zero for 101 = 10001, 1 zero for 000 = 00000). The number of zeros to add for each replacement is (dumbcantorfunction1(n-1)) with (dumbcantorfunction2(n-1)) number of factorials
at this point, i dunno if somethings broken here
if any of this stuff actually works name the number dumbcantorfunction3(121412181214121) after me
dunno if this is computable. dunno if graham's notation laughs at this pathetically small number. Just wanted to use this to inspire myself to do gud in college
Code: Select all
x = 5, y = 9, rule = B3-jqr/S01c2-in3
3bo$4bo$o2bo$2o2$2o$o2bo$4bo$3bo!
Code: Select all
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
Well, first, it's not much stronger than the strongestHdjensofjfnen wrote: ↑January 25th, 2020, 3:10 amKind of cheating, but you could define a function P(n) which is the product of all functions in this thread.
Are you talking about the ×4 inMoosey wrote: ↑January 13th, 2020, 4:45 pmDefine ×(n) -- is it n×n = n^2?PkmnQ wrote: ↑January 13th, 2020, 1:08 pmHmm, what if I were to plug a function into itself?
¤(f,a,n,k)
f is the function
a is the amount of arguments for the function
x is just a number
¤(f,1,n,k) = f(n) × k
otherwise, ¤(f(...,kn),a-1,f(n,n,n...)×k,kn)
2×2 = 4
¤(×,2,2,2) = ¤(×4,1,8,4) = 128
¤(¤×,2,2,2) = ¤(¤×k4,1,128,4) = ¤(×,2,128,4) × 4 = ¤(×512,1,16384,512) × 4 = 16384 × 512 × 512 × 4 = 17179869184 (please check for me)
0th layer is 2^2, 1st layer is 2^7, and 2nd layer is 2^34.
How much does this grow proportional to x? I don't know, I'm not a googologist.
No, I'm talking about ×(n)
This is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann functionPkmnQ wrote: ↑February 6th, 2020, 7:55 amA*(0,n)=n+1, n>=0
A*(m,0)=A*(m-1,1)+1, m>0
A*(m,n)=A*(m-1,A*(m,n-1)+1)+1, m>0 & n>0 (my addition)
I’ll let the 1’s slide, because they’re 0+1.
Also, the first line was originally 2 lines, but they simplify to that.
Python doesn’t like that this recurses more than 1000 times, so it calls quits and says “DO IT YOURSELF!”
No, Python, I AM NOT CALCULATING THIS BY HAND!
It's actually a modification of a modification that you did before, when somebody posted about the Ackermann function.Moosey wrote: ↑February 8th, 2020, 10:15 amThis is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann functionPkmnQ wrote: ↑February 6th, 2020, 7:55 amA*(0,n)=n+1, n>=0
A*(m,0)=A*(m-1,1)+1, m>0
A*(m,n)=A*(m-1,A*(m,n-1)+1)+1, m>0 & n>0 (my addition)
I’ll let the 1’s slide, because they’re 0+1.
Also, the first line was originally 2 lines, but they simplify to that.
Python doesn’t like that this recurses more than 1000 times, so it calls quits and says “DO IT YOURSELF!”
No, Python, I AM NOT CALCULATING THIS BY HAND!