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Can someone explain in a simple way why there are so few known exact Ramsey numbers? I guess it's because there are no efficient algorithms for this task, but are there so many combinations to test?

And an additional question: How are the bounds determined? Why do the bounds, that are known, have those values? Why not i.e. try to take a lower number for the upper bound for some 2-coloring?

Kaj Hansen
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Fred Funks
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    What constitutes a simple explanation depends on what you know. What is your mathematical background? – Ben Grossmann Jul 15 '14 at 00:35
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    This is not an answer, but a relevant story that people should know. Paul Erdos told this story: "Aliens invade the earth and threaten to obliterate it in a year's time unless human beings can find $R(5,5)$. We could marshal the world's best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded $R(6,6)$, however, we would have no choice but to launch a preemptive attack." – Gerry Myerson Jul 15 '14 at 02:59
  • @GerryMyerson Unfortunately the page is not found. Nice story though :) – Sirswagger21 Jul 07 '21 at 20:43
  • @FruDe the quote can also be found (at least, for the time being) at https://blogs.scientificamerican.com/roots-of-unity/moores-law-and-ramsey-numbers/ which cites a 1990 Scientific American article by Ronald Graham and Joel Spencer. – Gerry Myerson Jul 07 '21 at 23:05

1 Answers1

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Unfortunately, the original proof that Ramsey numbers exist (as in, are finite) was non-constructive, so mathematicians have been fighting an uphill battle from the beginning. Consider the diagonal Ramsey numbers $R(s, s)$. The smallest of these which remains unknown is $R(5, 5)$, so let's say we suspect $R(5, 5) = 43$—the current lower bound. We would need to confirm that the relevant property holds for every possible $2$-coloring on $K_{43}$. Since this graph has $\displaystyle \binom{43}{2} = 903$ edges, there are $2^{903}$ possible colorings. This is a $272$-digit number! For comparison, this number is many orders of magnitude larger than the total number of protons, neutrons, and electrons in the entire observable universe. Of course, it's possible we can whittle this number down a bit with some ingenuity (for instance, some colorings are equivalent up to symmetry), but it would still be unimaginably gargantuan. Brute force is out of the question, even for the most powerful supercomputers.

Bounds:

The work on lower bounds for diagonal Ramsey numbers improves upon the original lower bound Paul Erdős found using his probabilistic method$^\dagger$. With this method, it's possible to show (rather easily) that $\displaystyle R(s, s) \geq \lfloor 2^{\frac{s}{2}} \rfloor$. Of course, the general lower bound has since been improved, but it still has an exponential growth factor of $\sqrt{2}$.

A relatively decent upper bound for diagonal Ramsey numbers can be proven using the same approach as in the proof that $R(s, t) < \infty$. That is, we can show $\displaystyle R(s, t) \leq \binom{s+t-2}{s-1}$. When $s=t$, we get $\displaystyle R(s, s) \leq \binom{2s-2}{s-1}$, which grows exponentially with a growth factor of $4$. The current upper bound has been improved a bit, but still has the same growth factor.

$^\dagger$ I'd highly recommend checking out this proof; it's one of my favorites. A downright ingenious application of probability theory to this area of math. Find it here.

Kaj Hansen
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  • That's what I was looking for. Thank you a lot! – Fred Funks Jul 15 '14 at 07:12
  • @FredFunks, Glad I could help! – Kaj Hansen Jul 15 '14 at 07:53
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    "[W]e can whittle down this number significantly, but it would still be astronomical." Still bigger than astronomical, since it would still be much bigger than the number of atoms in the visible universe! – David Richerby Jul 15 '14 at 08:28
  • "Bigger than astronomical..." That's a phrase you'd be hard-pressed to hear outside of Ramsey theory. :) – Kaj Hansen Jul 15 '14 at 08:29
  • ...unless you have a look at some cryptographic problems, which are designed to have "bigger than astronomical" difficulty ;) – Massimo Jul 15 '14 at 13:36
  • Many problems in mathematics are "bigger than astronomical". Yet "bigger than astronomical" is not a phrase I've commonly heard..."[practically] impossible" is a common equivalent. We don't usually think of "astronomical" as having an upper limit, although in some literal sense it does. – Tim S. Jul 15 '14 at 14:58
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    If by "the original proof" you mean Ramsey's 1928 paper, to say that "it did nothing more than show that $R(s,t)\lt\infty$" seems unduly harsh. In fact Ramsey notes that his general argument gives the upper bound $R(n,n)\le2^{n(n-1)/2},$ which he then improves to $R(n,n)\le n!.$ – bof May 14 '18 at 22:40
  • Not just "unduly harsh". It is actually false. – Andrés E. Caicedo Jun 19 '18 at 15:16
  • Thanks for the critique bof & Andres. I went ahead and simply deleted the first bit as I think the rest stands on its own. – Kaj Hansen Jun 19 '18 at 20:34
  • While you are editing this answer, it's worth noting that the sentence on growth factor is no longer true; it may be down to 3.999 or even to 3.8. – Misha Lavrov Nov 15 '24 at 22:09
  • @MishaLavrov wow, thanks a bunch. I'll read into this soon & make an edit. I've been rather out of touch with the math world for the past 5 or so years. – Kaj Hansen Nov 21 '24 at 23:08