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This article seems to leave out a key caveat: They only proved this conjecture for sufficiently large n, not for all n. [Edit: Oops, apparently I was wrong, it

This article seems to leave out a key caveat: They only proved this conjecture for sufficiently large n, not for all n. [Edit: Oops, apparently I was wrong, it does list that caveat -- but doesn't go into the further detail I discuss below.] Now, if they have a bound on n, or if it's possible to derive a bound from their proof, then this could eventually become a proof of the full conjecture, by computer checking all smaller n. But if it's nonconstructive, then it can't be. To be clear, this is still an interesting and impressive result either way! But let's make sure we've correctly stated just what it is. From the description in the article, it sounds like it ought to be possible to derive a bound on n from their proof. But from what I've seen, it doesn't sound like they've done so. So, now some proof-miner will have to go and do that for them, I guess, and possibly prove the full conjecture as a result...

sohkamyung 6y ago
It does say this at the end of the article. > Because of the probabilistic elements, their proof only works for large enough hypergraphs — those with more than a specific number of vertices. This approach is common in combinatorics, and mathematicians consider it a nearly complete proof since it only omits a finite number of hypergraphs.
- Sniffnoy 6y ago
  Ah, I missed that, thanks. That said, I think there's still a big difference between "only omits a finite number of hypergraphs" and "only omits a finite number of hypergraphs of bounded size". I'd regard the latter as nearly a complete proof, not so much the former. (E.g., I don't think anyone would say that we have a nearly complete proof of Artin's primitive root conjecture.) Now, it sounds like this is the latter in the sense that it can be made so, but it also sounds like it's not explicitly the latter and someone needs to actually still do that work.
  - pmiller2 6y ago
    For any n, there are only a finite number of hypergraphs on n vertices, much less linear hypergraphs. It is therefore correct, and equivalent to say that the proof omits "only a finite number of hypergraphs," when the result is shown for sufficiently large n. I haven't read the paper through, but, I suspect it's even possible to extract some explicit value of n for which the conjecture is true. In any case, it's enough to know that there is such an n. I looked up Artin's primitive root conjecture on Wiki because I wasn't familiar with the result. I think you may be referencing the result that there are at most 2 primes for which the conjecture fails, no? If so, I would agree that's not close to a complete proof. The difference here is that, in principle, since hypergraphs are just finite combinatorial objects, one could, in theory just run a generic coloring algorithm over the finite number of things left out of the proof to obtain the complete result with a bow tied around it. In the case of Artin's conjecture, you have two difficulties: identifying 1 or 2 candidate primes for which it may fail, and verifying the conjecture for those primes. Even if you can somehow overcome the first difficulty and identify the candidate primes for which it may fail, in the absence of machinery that lets us make such a conclusion for those two primes, we are reduced to calculating an infinite number of discrete logarithms, which, obviously doesn't fly. So, TL;DR: the difference here is literally the difference between finite and infinite. There are a finite number of hypergraphs left out, which can, in theory be tested using a finite number of operations to verify the conjecture. Not so with Artin's conjecture. https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primit...
    - obastani 6y ago
      I don't think that's quite right. There are two possible precise statements for such claims: 1) There exists some n such that all integers >= n satisfy the desired property. 2) For n = [a specific constant], all integers >= n satisfy the desired property. I'm not familiar with Artin's conjecture, but from your description, it satisfies (1) but not (2). The reason is that if there are at most 2 such primes, we can take n to be the larger of the two primes plus one. Since all primes are finite, this choice of n is also finite. I think the key question is whether the paper described in this article proves (1) or (2). From the discussion, it sounds like it proves (2), which is the stronger result.
  - exactlysolved 6y ago
    There is not a big difference between those two statements. In fact they are identical in meaning.
    - obastani 6y ago
      There is a subtle but important difference. To be more precise, consider the following two statements: 1) There exists some n such that all integers >= n satisfy the desired property. 2) For n = [a specific constant], all integers >= n satisfy the desired property. These two statements are not the same, but both imply that there are a finite number of counterexamples. The second one is stronger, since we could prove the statement by enumerating all k < n and checking the statement for each such k; if all these checks pass, then the statement is correct. This strategy does not work for the first strategy since we do not know n, only that such an n exists. In particular, there could be a non-constructive proof that establishes existence of such an n without providing any way to compute such an n. From the discussion, it does sound like this paper is proving (2), not (1).
      - Sniffnoy 6y ago
        Yup. You came in and said it before I could get back to this...
js8 6y ago
> Now, if they have a bound on n, or if it's possible to derive a bound from their proof, then this could eventually become a proof of the full conjecture, by computer checking all smaller n. But if it's nonconstructive, then it can't be. Actually, if they could somehow embed the small hypergraphs into the large ones so that the coloring on the small ones is the same, they could prove it completely. I wonder why is this not the case, but I suspect they are gonna do it, see: “There is still the assumption in the paper that the number of nodes must be very large, but that’s probably just some additional work,” said Lovász. “Essentially, the conjecture is now proved.”

sohkamyung 6y ago

It does say this at the end of the article. > Because of the probabilistic elements, their proof only works for large enough hypergraphs — those with more than a specific number of vertices. This approach is common in combinatorics, and mathematicians consider it a nearly complete proof since it only omits a finite number of hypergraphs.

Sniffnoy 6y ago
Ah, I missed that, thanks. That said, I think there's still a big difference between "only omits a finite number of hypergraphs" and "only omits a finite number of hypergraphs of bounded size". I'd regard the latter as nearly a complete proof, not so much the former. (E.g., I don't think anyone would say that we have a nearly complete proof of Artin's primitive root conjecture.) Now, it sounds like this is the latter in the sense that it can be made so, but it also sounds like it's not explicitly the latter and someone needs to actually still do that work.
- pmiller2 6y ago
  For any n, there are only a finite number of hypergraphs on n vertices, much less linear hypergraphs. It is therefore correct, and equivalent to say that the proof omits "only a finite number of hypergraphs," when the result is shown for sufficiently large n. I haven't read the paper through, but, I suspect it's even possible to extract some explicit value of n for which the conjecture is true. In any case, it's enough to know that there is such an n. I looked up Artin's primitive root conjecture on Wiki because I wasn't familiar with the result. I think you may be referencing the result that there are at most 2 primes for which the conjecture fails, no? If so, I would agree that's not close to a complete proof. The difference here is that, in principle, since hypergraphs are just finite combinatorial objects, one could, in theory just run a generic coloring algorithm over the finite number of things left out of the proof to obtain the complete result with a bow tied around it. In the case of Artin's conjecture, you have two difficulties: identifying 1 or 2 candidate primes for which it may fail, and verifying the conjecture for those primes. Even if you can somehow overcome the first difficulty and identify the candidate primes for which it may fail, in the absence of machinery that lets us make such a conclusion for those two primes, we are reduced to calculating an infinite number of discrete logarithms, which, obviously doesn't fly. So, TL;DR: the difference here is literally the difference between finite and infinite. There are a finite number of hypergraphs left out, which can, in theory be tested using a finite number of operations to verify the conjecture. Not so with Artin's conjecture. https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primit...
  - obastani 6y ago
    I don't think that's quite right. There are two possible precise statements for such claims: 1) There exists some n such that all integers >= n satisfy the desired property. 2) For n = [a specific constant], all integers >= n satisfy the desired property. I'm not familiar with Artin's conjecture, but from your description, it satisfies (1) but not (2). The reason is that if there are at most 2 such primes, we can take n to be the larger of the two primes plus one. Since all primes are finite, this choice of n is also finite. I think the key question is whether the paper described in this article proves (1) or (2). From the discussion, it sounds like it proves (2), which is the stronger result.
- exactlysolved 6y ago
  There is not a big difference between those two statements. In fact they are identical in meaning.
  - obastani 6y ago
    There is a subtle but important difference. To be more precise, consider the following two statements: 1) There exists some n such that all integers >= n satisfy the desired property. 2) For n = [a specific constant], all integers >= n satisfy the desired property. These two statements are not the same, but both imply that there are a finite number of counterexamples. The second one is stronger, since we could prove the statement by enumerating all k < n and checking the statement for each such k; if all these checks pass, then the statement is correct. This strategy does not work for the first strategy since we do not know n, only that such an n exists. In particular, there could be a non-constructive proof that establishes existence of such an n without providing any way to compute such an n. From the discussion, it does sound like this paper is proving (2), not (1).
    - Sniffnoy 6y ago
      Yup. You came in and said it before I could get back to this...

js8 6y ago

> Now, if they have a bound on n, or if it's possible to derive a bound from their proof, then this could eventually become a proof of the full conjecture, by computer checking all smaller n. But if it's nonconstructive, then it can't be. Actually, if they could somehow embed the small hypergraphs into the large ones so that the coloring on the small ones is the same, they could prove it completely. I wonder why is this not the case, but I suspect they are gonna do it, see: “There is still the assumption in the paper that the number of nodes must be very large, but that’s probably just some additional work,” said Lovász. “Essentially, the conjecture is now proved.”