Timeline for answer to Sandbox for Proposed Challenges by l4m2
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| Feb 5 at 15:06 | comment | added | aeh5040 | @l4m2 I am not convinced. I don't know what you mean by "a value promised large enough". Every individual integer is trivially computable. It is sequences that may not be. The sequence mapping n to BB(n) is not because it would require determining which of the n-state TMs halt for each n, which requires solving the halting problem. The (n-1)-state machines are in effect a subset of the n-state ones, but just knowing the max 1s produced by a halting n-state does not help knowing which n-1 state ones halt. | |
| Feb 4 at 15:16 | comment | added | l4m2 | @aeh5040 BB(n-1) is uncomputable because we don't have a value promised large enough, but here BB(n) is provided | |
| Feb 3 at 12:04 | comment | added | aeh5040 | Isn't this impossible? The BB(n) sequence is not computable. But if there was a program that solved your problem then wouldn't there be a simple modification of it that computed BB(n)? | |
| Jan 30 at 1:37 | history | edited | l4m2 | CC BY-SA 4.0 |
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| Jan 9 at 13:38 | comment | added | Wheat Wizard Mod | You need to explain your definition of Turing machine a little more. For one I assume you probably mean deterministic Turing machine, but this is worth specifying. Some definitions require that a Turning machine move the tape head on every transition, some definitions allow the tape head to remain fixed. Some definitions require a specific halt state, some halt when there is no transition available. etc. These differences are usually trivial but BB(k) is highly sensitive to them. | |
| Dec 30, 2025 at 5:16 | history | answered | l4m2 | CC BY-SA 4.0 |