User blog:Ecl1psed276/Growth rate of oracle Turing machines

It is known that the Busy Beaver function BB(n) outgrows all recursive functions, so we could say that it has a growth rate of \(f_{\omega_1^{CK}}\) in the FGH, for a suitable system of fundamental sequences below \(\omega_1^{CK}\) (Kleene's O may be used to do this).

However, consider if we make a special type of Turing machine, called the Oracle Turing Machine (OTM), that has another special state in addition to the "halt" state. This new state is called the "oracle" state. If the turing machine is put into this state, then we let N be the total number of 1's currently on the tape. We then determine whether Turing Machine #N terminates, given a well-ordered enumeration of all (non-oracle) Turing machines. (It doesn't matter exactly how this enumeration is defined, the result will be the same, except the enumeration should conform to the condition shown below). The current tape cell is set to 1 if machine #N halts, and 0 otherwise. This oracle operation is impossible to perform with a regular TM, because a regular TM cannot solve the halting problem.

Given access to this oracle, it is certainly possible to create an OTM with, say, less than 1,000,000 states that will output a number of 1's greater than \(BB^{100}(100)\). We can then define a "level-2" busy beaver function: BB2(n). It outputs the maximum possible number of 1's on the tape after simulating an OTM with exactly n states. My question is this: What is the growth rate of BB2(n)? Would it be \(\omega_1^{CK}*2\)? Or would it be \(\omega_2^{CK}\)? If it's the former, is it possible to modify the definition of the oracle to make the growth rate be \(\omega_2^{CK}\)?

Condition for the well-ordering of Turing machines: The condition is that if \(T_1\) and \(T_2\) are TM's, and \(T_1\) has more states than \(T_2\), then \(T_1\) must be assigned a larger number than \(T_2\).