I think I can win Game 1 against almost anyone—in other words, I think I have a larger computable number than any sort of computable number I’ve seen anyone describe in these sorts of contests, where the top entries typically use the fast-growing hierarchy for large recursive ordinals, in contests where Busy Beaver and beyond aren’t allowed.
Okay I have to ask. Care to provide a brief description? You can assume familiarity with all the standard tricks if that helps.
I suspect that TREE(3) is the number he’s referring to, since I’ve seen it mentioned on LW a few times, and it seems to be larger than most other numbers that are considered notable for their largeness.
Of course, it’s easy to come up with a number larger than TREE(3).
As far as I know, the TREE function has no particular relationship to the busy beaver functions. The TREE function is computable, whereas the busy beaver functions are not.
I wonder how TREE(3) compares to Loader’s number. If I understand correctly, if Kruskal’s tree theorem can be proved in the calculus of constructions using a reasonable number of symbols (where 3^^^3 counts as a reasonable number, but TREE(3) does not), then Loader’s number is much larger.
Edit: Wikipedia states that Friedman’s special cases of Kruskal’s tree theorem can “easily” be proved in second-order arithmetic, which can be expressed in the calculus of constructions. I’m pretty sure this means that the TREE function can be written in the calculus of constructions using a reasonable number of symbols, meaning that Loader’s number is much larger than TREE(n) for any reasonable value of n.
And speaking of which, what kind of big number contest bans busy beaver? Srsly : <
A computable one. OP is not clear how his hypercomputer is solving the halting problem—does it have a ‘halts(s)’ function for programs that do not use ‘halts’ function, or what. The solution 2 in the OP is pretty much equivalent to the busy beaver, and it can not be done without use of halting oracle because some of those programs would loop forever, making the final print statement unreachable.
Okay I have to ask. Care to provide a brief description? You can assume familiarity with all the standard tricks if that helps.
I suspect that TREE(3) is the number he’s referring to, since I’ve seen it mentioned on LW a few times, and it seems to be larger than most other numbers that are considered notable for their largeness.
Of course, it’s easy to come up with a number larger than TREE(3).
I dunno, is that tree sequence related to the busy beaver sequence? And speaking of which, what kind of big number contest bans busy beaver? Srsly : <
As far as I know, the TREE function has no particular relationship to the busy beaver functions. The TREE function is computable, whereas the busy beaver functions are not.
I wonder how TREE(3) compares to Loader’s number. If I understand correctly, if Kruskal’s tree theorem can be proved in the calculus of constructions using a reasonable number of symbols (where 3^^^3 counts as a reasonable number, but TREE(3) does not), then Loader’s number is much larger.
Edit: Wikipedia states that Friedman’s special cases of Kruskal’s tree theorem can “easily” be proved in second-order arithmetic, which can be expressed in the calculus of constructions. I’m pretty sure this means that the TREE function can be written in the calculus of constructions using a reasonable number of symbols, meaning that Loader’s number is much larger than TREE(n) for any reasonable value of n.
A computable one. OP is not clear how his hypercomputer is solving the halting problem—does it have a ‘halts(s)’ function for programs that do not use ‘halts’ function, or what. The solution 2 in the OP is pretty much equivalent to the busy beaver, and it can not be done without use of halting oracle because some of those programs would loop forever, making the final print statement unreachable.
Yup, that makes sense.