User blog:B1mb0w/Strong D Function

Strong D Function

The strong D function is based on the weaker d function defined in User blog:B1mb0w/Deeply Nested Ackermann. The rules are similar with the significant change being that the D function:

\(D(x_1,x_2,x_3,x_4,...,x_n)\)

expands to this function:

\(D( x_1-1, D(x_1,x_2,x_3,x_4,...,x_n-1), ..., D(x_1,x_2,x_3,x_4,...,x_n-1))\)

The same expansion is used to replace each input parameter \(x_2\) to \(x_n\).

For 2 parameters, the D function is equivalent to the d function:

\(d(a,b)=d(a-1,d(a,b-1))=D(a,b)=D(a-1,D(a,b-1))\)

For 3 parameters, the D function quickly dominates the weaker d function:

\(d(a,b,c)=d(a-1,d(a,b-1),d(a,b,c-1))\)

\(D(a,b,c)=D(a-1,D(a,b,c-1),D(a,b,c-1))\)

Calculated Examples

\(D = 0\)  This is a null function that always returns zero.

\(D(3) = 4\)    This is the successor function

\(D(1,2) = 5\)   This is the same as d(1,2)

\(D(1,0,0)\)     expands as follows:

\(= D(0, D(0,1,1), D(0,1,1)) = D(4,4)\) This is the same as d(4,4) and is comparable to \(f_3(6)\)

In fact \(D(4,4) >> f_{\omega}(3)\)

\(D(1,0,1)\)     expands as follows:

\(= D(0, D(1,0,0), D(1,0,0)) = D(D(4,4),D(4,4)) >> D(f_{\omega}(3),f_{\omega}(3))\)

and is comparable to \(f_{\omega+1}(3)\)

My calculations show that \(D(1,0,n)\) is comparable to \(f_{\omega+n}(3)\)

More Examples with 3 parameters

\(D(1,1,0) = D(0,D(1,0,1),D(1,0,1)\) which is equal to \(D(1,0,2)\) and comparable to \(f_{\omega+2}(3)\)

Similarly

\(D(1,1,1) = D(0,D(1,0,2),D(1,0,2)\) which is equal to \(D(1,0,3)\) and comparable to \(f_{\omega+3}(3)\)

My calculations show that \(D(1,1,n)\) is comparable to \(f_{\omega+n+2}(3)\)

Next

\(D(1,2,0) = D(0,D(1,1,2),D(1,1,2)\) which is equal to \(D(1,1,3)\) and comparable to \(f_{\omega+5}(3)\)

Similarly

\(D(1,2,1) = D(0,D(1,1,3),D(1,1,3)\) which is equal to \(D(1,1,4)\) and comparable to \(f_{\omega+6}(3)\)

My calculations show that \(D(1,2,n)\) is comparable to \(f_{\omega+n+5}(3)\)

and \(D(1,3,n)\) is comparable to \(f_{\omega+n+8}(3)\)

and \(D(1,m,n)\) is comparable to \(f_{\omega+n+(m+1)!^+-1}(3)\) where \(!^+\) is the additive version of factorial.

D function examples with 3 parameters - continues

\(D(2,0,0)\) will grow significantly faster.

\(= D(1,D(1,2,2),D(1,2,2))\) is comparable to \(D(1,f_{\omega+7}(3),f_{\omega+7}(3)\)

d(1,3,z) >> d(6,z-1)

d(1,y,z) >> d(y+3, z-1)

d(2,0,z) = d(1, d(1,2), d(2,0,z-1))

>> d( d(1,2)+3, d(2,0,z-1) -1)

>> d(8,z)

Graham's Number

The d function can reach g1 where g4 is Graham's number relatively quickly. As shown above d(1,3,0) >> g1. But from my analysis we need the power of the d function with many more input parameters to reach G (Graham's Number).

d(6,0) >> g1

d( d(6,0), 0) >> g2

d(6,0,0) = d(5, d(5,6), !!!)  where !!! is a big number that we can ignore for now

= d(4, d(4, d(5,6)-1), !!!)

...

= d(0, d(1, d(2, d(3, d(4, d(5,6)-1)-1)-1)-1)-1), !!!) and the leading parameter = 0 can be dropped.

The value of the second parameter is big but does not equal d(6,0) therefore d(6,0,0) << g2 but if we expand

d(6,1,0) we will get d(6,0) nested in the long expansion and therefore d(6,1,0) >> g2 by a very great amount.

We can reach g3 by moving to 4 input parameters

d(6,1,1,0) >> g3

We finally reach G (Graham's Number) with 64 input parameters

d(6,1,1,1, ..., 1,0) >> G

Growth Rate

The growth rate of the d function is much faster than the Fast-growing hierarchy. Here are some comparisons:

d(x,y) = d(x-1, d(x,y-1)) = d^(y+2) (x-1,x)

fx (y) = f(x-1)^y (y)      even if y>>x the d function will exceed f because y+2 > y and the extra function recursion will always generate the largest number. If y >> >> x by a ridiculous amount then this may not be true.

and also

f6 (6) >> g1

d(6,6) >> f6 (6) >> d(6,0) >> g1

Next

My next blog post will introduce a new Alpha function that I have been thinking about. You can find it here: User blog:B1mb0w/Alpha Function