this post was submitted on 16 Jul 2023
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Hi there - large numbers are fun and I was learning about the Busy Beaver function which (theoretically) produces unfathomably large numbers by finding the maximum number of 1s written on a blank Turing machine tape out of the set of all n-state Turing machines that halt.

I was wondering if a conceptually more obvious, but larger variation could count the maximum number of steps taken before halting out of all n-state Turing machines that halt?

Would these numbesr not grow faster than the traditional Busy Beaver, since the number of steps will always be greater (or equal?) to the number of 1s written?

Obviously, the halting problem shows that we can't know beforehand if the machines will actually halt, but that issue is common to both versions.

Just curious if there is a reason the problem is not considered this way?

Any googologists out there with insights?

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[โ€“] [email protected] 3 points 1 year ago (1 children)
[โ€“] [email protected] 1 points 1 year ago

Ah... Many thanks! This is really interesting stuff!