3 ms·

Yeah, that’s the shift function / number of steps. I’m talking about the length of the contents on the tape when the machine enters a halt state. The notation “

• aphantastic 2y ago
  The length of the contents of the tape of any halting TM will be less than or equal to the number of steps it takes to run. Quite trivially: you can’t consume tape without taking a step. For BB(5), 4098 1’s are written to the tape (this is referred to as Σ(5)). As for the length of the contents, I’m not sure, but again you can’t so sume tape without a step, and anecdotally the BBs are generally “back and forth”, at least the Collatz-like ones (which BB(5) is).
  • Xcelerate 2y ago
    Sure, I’m just referring to the fact that there doesn’t seem to be a widespread standard on what “Busy Beaver” refers to in the literature. Scott Aaronson tends to reference the shift function, but older papers consider Σ as the “Busy Beaver function” or the “ones function” as you point out. My number came from running the machine itself and considering the largest contiguous section of output (since you typically start with a tape consisting of infinite zeros). If we think of the nth prime as the output of a Turing machine that computes the prime numbers where n is on the input tape, then if people are saying the output aloud of such a machine, it would make sense to do so similarly here as well. Ultimately, it doesn’t really matter. All these functions basically grow at the same rate asymptotically.