User blog comment:KthulhuHimself/Plexation, TaN, and such/@comment-5529393-20151014102617

For one-dimensional strings, the base function (I'll call it B(n) to keep in general) gets applied n times to the second to last member, where n is the last member. So

...{1} m {1} n[  = ...{1} B^n (m) {1} 0[ = ...{1} B^n (m)

Remember if your base function is at the level w+1, then B^n(m) is at level w+2 in the FGH. So if you have a string of length n of m's, you get (say with m = 4):

]m {1} m {1} m {1} m[ = ]m {1} m {1} f_{w+2} (m)[ = ]m {1} f^2_{w+2} (m)[ = ]f^3_{w+2}(m)[

or for general n you get f^{n-1}_{w+2} (m)  (I assume that when you're down to a one-element string you just take that number as the final value.), which is one level higher in the FGH, or about f_{w+3}(n).

Now, if we go to two dimensions, then the same pattern happens: the function f_{w+3} gets applied n times to the second to last element, so the second to last element becomes about f_{w+4}(n). So an n-element string with {2} separators will apply f_{w+4} n-1 times, which will be about f_{w+5}.

Similarly each additional dimensions adds two layers of recursion, so an n-element string of dimension n is about f_{w+2n+1}(n), or roughly f_{w2}(n), not f_{w^3 2}(n) as stated in the article.

Your higher-order terminal array notation is incomplete because there is no base case, only the recursive case. Also, Z_n is defined as a function on one variable, so it is not clear which variable in ]a {b(c)} d[ is being used in Z_n and what to do with the other variables (other than c). If we assume though that {a {b(0)} c} = {a {b} c}, and just take a, b, and d equal to n, then we get

]n {n(0)} n[ ~ f_{w2} (n), as stated above.

]n {n(1)} n[ ~ f_{w2+1} (n), since we are just applying the previous function n times.

]n {n(2)} n[ ~ f_{w2+2} (n), ditto

In general, ]n {n(c)} n[ ~ f_{w2+c} (n).

For multiple-order levels, I have the same objections as before: no base case and it's not clear what variable is used in Z_n and what the other variables are doing. But if we make the same assumptions as before, then we get

]n {n(n,0)} n[ = ]n {n(n)} n[ ~ f_{w3} (n) from above

]n {n(n,1)} n[ ~ f_{w3+1}(n), since Z_n applies the previous function n times

]n {n(n,c)} n[ ~ f_{w3+c}(n), ditto

]n {n(n,n)} n[ ~ f_{w4} (n)

and continuing the same pattern, each additonal n in the order adds w to the FGH level, so

]n {n(n,n,n)} n[ ~ f_{w5} (n)

]n {n(n,n,n,n)} n[ ~ f_{w6} (n)

and so on; n variables in the order gets you f_{w(n+2)}(n), or about f_{w^2} (n) in the FGH.

Continuing on to the last notation, we have

]n {n(n,,1)} n[ ~ f_{w^2} (n)

then the hierarchy builds up the same way except with ,,1 at the end, so we have

]n {n(n,,2)} n[ = ]n {n((n,n,n,...,n),,1)} n[ ~ f_{w^2 2}(n)

]n {n(n,,c)} n[ ~ f_{w^2 c} (n)

]n {n(n,,n)} n[ ~ f_{w^3} (n)

To continue, I need to know exactly how the notation proceeds from here, e.g. does it go to ]n {n(n,,n,c) n[, or does it go next to ]n {n(n,,n,,c) n[, and whichever comes next, how does it resolve?

So that's my analysis - to sum up, your ordinals are too high.