7 ms·

_the second form surely exists_ Is this true for the BB function though? What if there is a beaver that never halts or loops, and has behavior sufficiently co

by Aeium 2y ago

_the second form surely exists_ Is this true for the BB function though? What if there is a beaver that never halts or loops, and has behavior sufficiently complex, such that it's impossible to prove it will never halt. Then for rules of that length, the second form doesn't exist.

School-Cotton 2y ago
Whether it’s possible to prove it halts or not is irrelevant. It either does halt, or not. Whether a human can prove that a function has a particular value doesn’t change whether that function is computable in the technical sense being used here.
mananaysiempre 2y ago
If you’re using ZFC, there is (TFA mentions the state of the art is BB(745); Yedida and Aaronson’s original work on BB(8000)[1] is quite fun to read from a programmer’s point of view). But the second form still exists (if you accept excluded middle)—you just can’t prove which one it is! Specifically, ZFC is consistent iff ZFC+“Y&A’s machine does halt” is consistent iff ZFC+“Y&A’s machine never halts” is consistent (a theorem in a fairly weak ambient metalogic). So you can take a stronger set theory that does prove the answer, it’s just that thus far we have no reason to prefer theories that answer yes to theories that answer no. (You don’t have to accept excluded middle, and it can on occasion be useful not to[2], but pragmatically you’re going to have a lot of difficulties even with first-year calculus unless you do.) [1] https://scottaaronson.blog/?p=2725 https://scottaaronson.blog/?p=2725 [2] https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/S0273-0979-2016-01556-4.pdf https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016...
feoren 2y ago
> What if there is a beaver that never halts or loops A Turing machine with finite states must eventually either halt or loop. Those are the only options, because there are only finitely many configurations it can be in, and each configuration completely determines the next. A "beaver" is defined to not loop. All "beavers" must halt, because otherwise they're just not considered for BB(n). All the challenge is in proving whether a given Turing machine does (or does not) halt, and therefore must not (or must) loop. Proving "halt" or "loop" proves the other one. Yes, the function `busy_beaver_6() = 576125642131574254..." must exist.
- School-Cotton 2y ago
  I disagree unless you state what you mean by "loop". If it's just "repeat a state" then any 6-state TM "loops" or halts after at most 6 turns... and many that "loop" will eventually halt. There are infinitely many configurations if you consider the tape. It is still true, of course, that every Turing machine either halts on a given input, or doesn't.
  - feoren 2y ago
    You're right: my argument is flawed. I had thought TFA relied on that argument, but the machine that writes 1 and moves right forever is a counterexample. Yet something about that still seems extremely "loopy". Is there something that must stop increasing after a while? Kolmogorov complexity? Or is that begging the question, since that's basically measuring the smallest TM that can produce that output?
    - School-Cotton 2y ago
      You’re right that it “feels loopy”, intuitively. I also don’t know how to formalize this notion.
      - feoren 2y ago
        How about this: given any arbitrarily large window size W, we can find an infinite number of timestamps (it may even be enough to say we can find two) in which the tape within the range of Head-W and Head+W is identical. So if you say 10, I can give you an infinite list of step-counts at which the 21 symbols on the tape centered at the head is identical to all the other times in that list. Assuming we can do that for any arbitrary window size, then the TM is in a "looping" state. Of course, a halted TM also has this property. So perhaps this is true of all Turing Machines?
  - samatman 2y ago
    I don't understand what you're disagreeing with. "loop" has a well-understood meaning here: return to an identical state. Not a similar one, identical. Because if it does that once, being a deterministic automaton, it will do so an infinite number of times without halting.

by Aeium 2y ago

School-Cotton 2y ago

Whether it’s possible to prove it halts or not is irrelevant. It either does halt, or not. Whether a human can prove that a function has a particular value doesn’t change whether that function is computable in the technical sense being used here.

mananaysiempre 2y ago

If you’re using ZFC, there is (TFA mentions the state of the art is BB(745); Yedida and Aaronson’s original work on BB(8000)[1] is quite fun to read from a programmer’s point of view). But the second form still exists (if you accept excluded middle)—you just can’t prove which one it is! Specifically, ZFC is consistent iff ZFC+“Y&A’s machine does halt” is consistent iff ZFC+“Y&A’s machine never halts” is consistent (a theorem in a fairly weak ambient metalogic). So you can take a stronger set theory that does prove the answer, it’s just that thus far we have no reason to prefer theories that answer yes to theories that answer no. (You don’t have to accept excluded middle, and it can on occasion be useful not to[2], but pragmatically you’re going to have a lot of difficulties even with first-year calculus unless you do.) [1] https://scottaaronson.blog/?p=2725 https://scottaaronson.blog/?p=2725 [2] https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/S0273-0979-2016-01556-4.pdf https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016...

feoren 2y ago

> What if there is a beaver that never halts or loops A Turing machine with finite states must eventually either halt or loop. Those are the only options, because there are only finitely many configurations it can be in, and each configuration completely determines the next. A "beaver" is defined to not loop. All "beavers" must halt, because otherwise they're just not considered for BB(n). All the challenge is in proving whether a given Turing machine does (or does not) halt, and therefore must not (or must) loop. Proving "halt" or "loop" proves the other one. Yes, the function `busy_beaver_6() = 576125642131574254..." must exist.

School-Cotton 2y ago
I disagree unless you state what you mean by "loop". If it's just "repeat a state" then any 6-state TM "loops" or halts after at most 6 turns... and many that "loop" will eventually halt. There are infinitely many configurations if you consider the tape. It is still true, of course, that every Turing machine either halts on a given input, or doesn't.
- feoren 2y ago
  You're right: my argument is flawed. I had thought TFA relied on that argument, but the machine that writes 1 and moves right forever is a counterexample. Yet something about that still seems extremely "loopy". Is there something that must stop increasing after a while? Kolmogorov complexity? Or is that begging the question, since that's basically measuring the smallest TM that can produce that output?
  - School-Cotton 2y ago
    You’re right that it “feels loopy”, intuitively. I also don’t know how to formalize this notion.
    - feoren 2y ago
      How about this: given any arbitrarily large window size W, we can find an infinite number of timestamps (it may even be enough to say we can find two) in which the tape within the range of Head-W and Head+W is identical. So if you say 10, I can give you an infinite list of step-counts at which the 21 symbols on the tape centered at the head is identical to all the other times in that list. Assuming we can do that for any arbitrary window size, then the TM is in a "looping" state. Of course, a halted TM also has this property. So perhaps this is true of all Turing Machines?
- samatman 2y ago
  I don't understand what you're disagreeing with. "loop" has a well-understood meaning here: return to an identical state. Not a similar one, identical. Because if it does that once, being a deterministic automaton, it will do so an infinite number of times without halting.