Talk:Graham's number

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One of the 500 most frequently viewed mathematics articles.

Improved bound[edit]

Look at this article. There are presented much stronger upper bound for that problem where Graham's number was used. 31.42.233.14 (talk) 22:23, 18 March 2013 (UTC)

It is a wiki article on Wikia, not a peer reviewed source. Probably not even considered reliable by Wikipedia guidelines. SpinningSpark 15:52, 17 August 2013 (UTC)

Up arrows[edit]

Just a suggestion. The article already uses → for Conway's notation. I wonder if it would be an idea to use the unicode up arrow too: ↑↑. Would make it easier for readers to copy / paste the formulae, I just wanted to post to facebook about Graham's number and remade one of the formulae as 2↑↑(2↑↑(2↑↑9)) which you can use anywhere that supports unicode.

I'm not going to actually do it because there are so many formulae in the article and for consistency of style you'd need to do the whole thing. But might be worth considering as something to do? Robert Walker (talk) 09:00, 16 August 2013 (UTC)

What does this mean?[edit]

I've read and re read this text. But I have no idea of what Grahams number explains. Could someone write something in plain english for the non professional what Graham's number is supposed to solve. — Preceding unsigned comment added by 92.32.104.205 (talk) 11:22, 18 October 2013 (UTC)

Maybe this will help ... Here's a formulation of the problem in terms of row-vectors (one dimensional arrays of n numbers), which I used in a program to study this problem with arbitrary colorings (OR):
A unit n-cube has 2n corners, corresponding to the 2n binary row-vectors of length n. (E.g. for n = 3, these are the 23 = 8 vectors [0 0 0], [0 0 1], [0 1 0], [0 1 1], [1 0 0], [1 0 1], [1 1 0], [1 1 1], and similarly for other values of n.) A coloring is just an assignment of either 'red' or 'blue' to each possible pair of these vectors. Let's call a given coloring eventful if there are four such vectors (say a,b,c,d) with the following two properties:
(1) the six possible pairs of the four vectors (i.e. the pairs {a, b} {a, c} {a, d} {b, c} {b, d} {c, d}) are all assigned the same color, and
(2) if a difference vector is formed from each of the six pairs (subtracting one vector of the pair from the other, in either order (e.g. {a, b} → a - b, {a, c} → c - a, etc.) then any two of these difference vectors is sufficient to generate the other four by linear combination. (This is an algebraic way of saying that the points corresponding to the four vectors are coplanar. There are well-known algorithms for checking this.)
NB: The number of pairs to be colored is 2n - 1(2n - 1), so the number of possible colorings is 22n - 1(2n - 1), which gets large very quickly.
Here's the behavior as n increases ...
n = 1:   0% of the possible colorings are eventful (because there are fewer than 4 corners),
n = 2:  ~3% of the possible colorings are eventful (2 of the 64 possible colorings),
n = 3:  ~30% of the possible colorings are eventful (estimated by random colorings),
n = 4:  ~95% of the possible colorings are eventful (estimated by random colorings),
5 ≤ n < N*:  the percentage of colorings that are eventful approaches (but does not reach) 100%,
n ≥ N*100% of the possible colorings are eventful, where N* is some unknown number greater than 12.
The existence of the number N* was proved by Graham & Rothschild, and "Graham's number" (G) was specially invented by Graham as an easily described upper bound on N*. (Maybe the number N* itself should be called the "Graham-Rothschild number" to distinguish it from G.)
r.e.s. (talk) 22:18, 23 October 2013 (UTC)
I think the IP was asking for something to go in to the article rather than an explanation here. Personally, I think the explanation already in the article is easier to understand than the above. SpinningSpark 15:30, 24 October 2013 (UTC)
You are right. This is even harder tounderstand. it is neither written in plain english or for the non professional. — Preceding unsigned comment added by 78.78.76.98 (talk) 07:33, 28 October 2013 (UTC)
Let me try in a thoroughly non-formal way that will probably annoy the hell out of every real mathematician reading this. For n=2 draw a square of dots (four dots). Draw lines joining every dot to every other dot. How many rectangles can be found in this figure? Only one of course, the original square. Now colour every line either black or red. Can you find a colouring scheme so that no rectangle is all the same colour? Trivially easy of course since there is only one rectangle. That one is a pass meaning Graham's number is more than 2. Now draw a cube of eight dots (n=3) and join every dot to every other dot. This is no longer trivial because besides the six squares of the cube faces there are also rectangles on the diagonals. It may be non-trivial but it is still pretty easy to find a colouring scheme so that no rectangle is all one colour. Pass that one too, Graham's number is greater than 3. For n=4 one must draw a four-dimensional cube, a tesseract or 4-cube. This is starting to get hard because rectangles can be found all over the place, but still possible to do. Carry on increasing n for higher and higher dimensions. Eventually you will come to a value of n where it is impossible to colour every rectangle in mixed colours. Graham manged to prove that this value of n cannot be greater than Graham's number (although it may well be a lot smaller). Graham's number is an upper bound on the maximum value of n. Regards, SpinningSpark 09:31, 28 October 2013 (UTC)
My reply above is not intended for the general reader, but is for anyone acquainted with (or willing to learn) the most-basic vector/array concepts. It is an explanation of the problem in terms of simple arrays of integers, completely avoiding the difficulty of visualizing hypercube geometry. Saying "just do with 4 & higher dimensional hypercubes what was done with a cube" does not convey a clear idea of *how* this is to be actually carried over to the higher dimensions, whereas the above explanation does describe the *how* (in terms of integer arrays), which makes it slightly technical.
r.e.s. (talk) 20:30, 30 October 2013 (UTC)
My question was if someone could write something in the article for the general reader. Not for the professional. — Preceding unsigned comment added by 92.32.106.60 (talk) 12:29, 6 November 2013 (UTC)

The Biggest number (with a name)[edit]

It seems that Graham's number is the biggest number with a name, because numbers bigger than it don't have a name, or do they?--RicHard-59 (talk) 12:05, 16 January 2014 (UTC)

I removed it because it was cited to Wiki Answers. It might be true, but answers from anonymous people on the internet is the epitome of an unreliable source. Besides, these things change very rapidly, it is quite easy to find sources that say the largest number ever needed to solve a genuine problem is the much smaller Skewes' number. Another question is what counts as naming a number? I am pretty sure that Ronald Graham did not name the number in his original paper. If just one person referred to one of these larger numbers by the name of the original author do we then consider it named? It is not a definite thing like christening; a name gradually develops by consensus. Thus, we need reliable sources to provide the synthesis for us. SpinningSpark 13:27, 16 January 2014 (UTC)

"all sorts of poorly chosen words"[edit]

I have reverted most of this edit which changed "Graham's number is unimaginably larger than other well-known large numbers such as a googol, googolplex, and even larger than Skewes' number and Moser's number" to "Graham's number is much larger than other large numbers such as a googol, googolplex, Skewes' number and Moser's number." While this might seem like just dealing with a case of WP:WTW, in fact it ends up with the sentence saying nothing at all, or even being wrong. "Graham's number is much larger than other large numbers"? Really? Larger than any of them? Even the ones larger than Graham's number? Obviously, that isn't true. It is only larger than large numbers that are commonly discussed. "Unimaginably larger" in most circumstances would be making unwarranted assumptions about our readers imagination, but in this case the epithet is truly appropriate. SpinningSpark 17:01, 3 August 2014 (UTC)

The sentence says exactly what it said before, just without the subjective terms "well-known" and "unimaginably". It did not previously say that Graham's number is larger than numbers that are larger than Graham's number, and it doesn't say that now. And no, "unimaginably" is not appropriate. You could try fine tuning the wording - "vastly" or "enormously" perhaps; simply reverting to restore unverifiable claims is not at all helpful. 190.161.182.12 (talk) 21:16, 3 August 2014 (UTC)
And what the hell would "beyond useless" mean, anyway? That is not the kind of tone an encyclopaedia should be written in. 190.161.182.12 (talk) 21:17, 3 August 2014 (UTC)
I don't agree. What you have written gives no sense at all of how much larger this number is. You might also want to read WP:BRD on the right way to behave in disputes. SpinningSpark 21:55, 3 August 2014 (UTC)
So first of all you said that the sentence implied that Graham's number was larger than itself. Now you say it doesn't give a sense of how much larger it is than other numbers. Which is it? In neither case does my removal of the words "well known" and "unimaginably" make any difference. What do you want the article to say - "mind-bogglingly"? "humungously"? Probably either of these would actually be better than "unimaginably" which makes unwarranted judgements about the limits of the human imagination. I am sure we can converge on a more suitable adverb than either of those though.
WP:BRD is an essay that leads many people to feel entitled to simply undo any edit they don't like without feeling the need to explain themselves. It is not policy and I happen to think it is an immensely damaging idea. Better is to continue with the dialogue we are having right now. 190.161.182.12 (talk) 22:58, 3 August 2014 (UTC)
Well first of all I did explain myself. Then you undid my edit, so where does that leave us? SpinningSpark 23:27, 3 August 2014 (UTC)
I didn't mean to imply that you didn't explain yourself - I appreciate the time you took to do so. I just meant that WP:BRD is widely used by people as an excuse to revert without explanation, and I don't view simply reverting on sight edits that you don't like as "the right way to behave".
So anyway, let's recap. What you said about my edit was that "in fact it ends up with the sentence saying nothing at all, or even being wrong". But the information contained in the sentence was identical to the previous version. Then you said the your problem was actually just that it didn't give the sense of scale that you considered necessary. So I asked what adverb you'd prefer to use. You haven't answered. So where this has left us is that I don't know what your complaint actually is right now. 190.161.182.12 (talk) 00:03, 4 August 2014 (UTC)
I thought I made that clear. I think the original wording of "unimaginably larger" nicely summed it up. And my point about your revert is that you are just as guilty of an "I don't like it edit". Even more so, as you have now twice inserted your favourite version. If I had replied in kind would you have reverted again? SpinningSpark 00:38, 4 August 2014 (UTC)
You didn't make it clear, no - your first revert undid everything I did except the removal of the word "easily". If it's just "unimaginably" you were fighting for, why remove all the other changes?
The word is a poor word. "Vastly" would be fine, "enormously" would be fine, some sort of attempt at actual quantification would be even better. Like saying for example how much bigger than the observable universe a representation of the number would be with each digit taking up a planck volume, and comparing that to the equivalent "size" of other large numbers, if such things have been discussed in reliable sources. "Unimaginably" just sounds like hyperbole. 190.161.182.12 (talk) 01:36, 4 August 2014 (UTC)