I need help with a Script for the Fast Growing Hierarchy.

[LuxiusGOL]

I need to complete this script

Code: Select all

#GoogologyFGH.py
#Experimental
Number1 = input("What is X in F(X,Y)?")
Number2 = input("What is Y in F(X,Y)?")
Omega = Number2

if Number1 = 0:
    print(Number2+1)
    elif Number1 >= 1:
        for Number1 in Number2:
            

but how do you repeat Y Times F(X-1,N) where N is F(X-1,N) . . . Where N is F(X-1,Y)

[LuxiusGOL]

After this, I would want to make a script for an array-like notation I made.
Q(x) = x^x

Q(x,y) = Q(Q(Q(Q...y...(Q(x))...y...))) = x↑↑y

Q(x,y,z) = Q(x,Q(x,Q(x,...Q(x,y)) repeated z times

Q(y)(x)=Q(x,y) Qz(x,y)=Q(x,y,z)

Q(<a,b>,<c,d>) = Q(Q(a,b))(c,d) 2d array

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) = Q(Q(<a,b>,<c,d>))(<e,f>,<g,h>) 3d array
Repeat with [[x]] 4d, [[[x]]] 5d, and so on.

[Moosey]

I need to complete this script

Code: Select all

#GoogologyFGH.py
#Experimental
Number1 = input("What is X in F(X,Y)?")
Number2 = input("What is Y in F(X,Y)?")
Omega = Number2

if Number1 = 0:
    print(Number2+1)
    elif Number1 >= 1:
        for Number1 in Number2:
            

but how do you repeat Y Times F(X-1,N) where N is F(X-1,N) . . . Where N is F(X-1,Y)

It would help if I knew which of those was the input and which was the subscript, but:
You'd probably want to do something like, say, creating an iteration function thus:
It(x,y,n+1) = It(f_y(x),y,n); It(x,y,0) = x
And then construct the successor fgh as
f_(y+1)(x) = It(x,y,x)

After this, I would want to make a script for an array-like notation I made.
Q(x) = x^x

Q(x,y) = Q(Q(Q(Q...y...(Q(x))...y...))) = x↑↑y

Q(x,y,z) = Q(x,Q(x,Q(x,...Q(x,y)) repeated z times

Q(y)(x)=Q(x,y) Qz(x,y)=Q(x,y,z)

Q(<a,b>,<c,d>) = Q(Q(a,b))(c,d) 2d array

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) = Q(Q(<a,b>,<c,d>))(<e,f>,<g,h>) 3d array
Repeat with [[x]] 4d, [[[x]]] 5d, and so on.

Q(x,y,z) is only pentational level. Q(<a,b>,<c,d>) = Q(Q(a,b))(c,d) by your 2D rule = Q(c,d,Q(a,b)) = still only around pentation level-- specifically, Q(<n,n>,<n,n>) is around n^^^n^^n which is stronger than pentation but significantly weaker than n^^^^n.

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) = Q(Q(<a,b>,<c,d>))(<e,f>,<g,h>) By your 4D rule = Q(<e,f>,<g,h>,Q(<a,b>,<c,d>)), which is illdefined (Different dimensional entries in the array); therefore Arrays above 2D are illdefined.

[LuxiusGOL]

therefore Arrays above 2D are illdefined.

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) = Q(Q(<a,b>,<c,d>))(<e,f>,<g,h>) = Q(Q(Q(a,b))(c,d))(Q(Q(e,f))(g,h). Did I miss any brackets? I kinda fixed it.

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) is a 3D array, not a 4D one.









am I familiar to you?

[Moosey]

am I familiar to you?

yes

Q([<a,b>,<c,d>],[<e,f>,<g,h>]) = Q(Q(<a,b>,<c,d>))(<e,f>,<g,h>) = Q(Q(Q(a,b))(c,d))(Q(Q(e,f))(g,h

eh? not what the rules say.

[BlinkerSpawn]

It would help if I knew which of those was the input and which was the subscript, but:
You'd probably want to do something like, say, creating an iteration function thus:
It(x,y,n+1) = It(f_y(x),y,n); It(x,y,0) = x
And then construct the successor fgh as
f_(y+1)(x) = It(x,y,x)

Piggybacking:

Code: Select all

def FGH(ordinal,n):
	#let collapse(a,n) be a function returning a[n]
	#(let a+1[n] = a for all n)
	if ordinal: #presumably your zero ordinal is an empty list/string
		for i in range(n): n=FGH(collapse(ordinal,n),n)
	#limits act a bit stronger than in true FGH
	#if you want exact equivalence just add a temporary variable
	#but I like short code
	return n+1