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| '''Van der Waerden's theorem''' is a theorem in the branch of [[mathematics]] called [[Ramsey theory]]. Van der Waerden's theorem states that for any given positive [[integer]]s ''r'' and ''k'', there is some number ''N'' such that if the integers {1, 2, ..., ''N''} are [[Graph coloring|colored]], each with one of ''r'' different colors, then there are at least ''k'' integers in [[arithmetic progression]] all of the same color. The least such ''N'' is the [[Van der Waerden number]] ''W''(''r'', ''k''). It is named after the Dutch mathematician [[Bartel Leendert van der Waerden|B. L. van der Waerden]].<ref>{{cite journal |authorlink=Bartel Leendert van der Waerden |first=B. L. |last=van der Waerden |title={{lang|de|Beweis einer Baudetschen Vermutung}} |journal=Nieuw. Arch. Wisk. |volume=15 |year=1927 |issue= |pages=212–216 }}</ref>
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| For example, when ''r'' = 2, you have two [[:Category:Color|colors]], say [[red|<font color=red>red</font>]] and [[blue|<font color=blue>blue</font>]]. ''W''(2, 3) is bigger than 8, because you can color the integers from {1, ..., 8} like this:
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| '''1''' <u>2</u> <u>3</u> '''4''' '''5''' <u>6</u> <u>7</u> '''8'''
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| and no three integers of the same color form an [[arithmetic progression]]. But you can't add a ninth integer to the end without creating such a progression. If you add a [[red|<font color=red>red 9</font>]], then the [[red|<font color=red>red 3</font>]], [[red|<font color=red>6</font>]], and [[red|<font color=red>9</font>]] are in arithmetic progression. Alternatively, if you add a [[blue|<font color=blue>blue 9</font>]], then the [[blue|<font color=blue>blue 1</font>]], [[blue|<font color=blue>5</font>]], and [[blue|<font color=blue>9</font>]] are in arithmetic progression. In fact, there is no way of coloring 1 through 9 without creating such a progression. Therefore, ''W''(2, 3) is 9.
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| | :<math forcemathmode="mathml">E=mc^2</math> |
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| It is an open problem to determine the values of ''W''(''r'', ''k'') for most values of ''r'' and ''k''. The proof of the theorem provides only an upper bound. For the case of ''r'' = 2 and ''k'' = 3, for example, the argument given below shows that it is sufficient to color the integers {1, ..., 325} with two colors to guarantee there will be a single-colored arithmetic progression of length 3. But in fact, the bound of 325 is very loose; the minimum required number of integers is only 9. Any coloring of the integers {1, ..., 9} will have three evenly spaced integers of one color.
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| | :<math forcemathmode="png">E=mc^2</math> |
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| For ''r'' = 3 and ''k'' = 3, the bound given by the theorem is 7(2·3<sup>7</sup> + 1)(2·3<sup>7·(2·3<sup>7</sup> + 1)</sup> + 1), or approximately 4.22·10<sup>14616</sup>. But actually, you don't need that many integers to guarantee a single-colored progression of length 3; you only need 27. (And it is possible to color {1, ..., 26} with three colors so that there is no single-colored arithmetic progression of length 3; for example, RRYYRRYBYBBRBRRYRYYBRBBYBY.)
| | '''source''' |
| | :<math forcemathmode="source">E=mc^2</math> --> |
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| Anyone who can reduce the general upper bound to any 'reasonable' function can win a large cash prize. [[Ronald Graham]] has offered a prize of [[US$]]1000 for showing ''W''(2,''k'')<2<sup>''k''<sup>2</sup></sup>.<ref>{{cite journal |authorlink=Ronald Graham |first=Ron |last=Graham |title=Some of My Favorite Problems in Ramsey Theory |journal=INTEGERS (The Electronic Journal of Combinatorial Number Theory |url=http://www.integers-ejcnt.org/vol7-2.html |volume=7 |issue=2 |year=2007 |pages=#A2 }}</ref> The best-known upper bound is due to [[Timothy Gowers]],<ref>{{cite journal |authorlink=Timothy Gowers |first=Timothy |last=Gowers |title=A new proof of Szemerédi's theorem |journal=Geom. Funct. Anal. |volume=11 |issue=3 |pages=465–588 |year=2001 |url=http://www.dpmms.cam.ac.uk/~wtg10/papers.html |doi=10.1007/s00039-001-0332-9 }}</ref> who establishes
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| : <math>W(r,k) \leq 2^{2^{r^{2^{2^{k + 9}}}}},</math>
| | ==Demos== |
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| by first establishing a similar result for [[Szemerédi's theorem]], which is a stronger version of Van der Waerden's theorem. The previously best-known bound was due to [[Saharon Shelah]] and proceeded via first proving a result for the [[Hales–Jewett theorem]], which is another strengthening of Van der Waerden's theorem.
| | Here are some [https://commons.wikimedia.org/w/index.php?title=Special:ListFiles/Frederic.wang demos]: |
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| The best-known lower bound for <math>W(2, k)</math> is that <math>W(2, k) > 2^k/k^\epsilon</math> for all positive <math>\epsilon</math>.<ref>{{cite book |title=Discrete Mathematics And Its Applications |editor=M. Sethumadhavan |last=Brown | first=Tom C. | pages=80 | chapter=A partition of the non-negative integers, with applications to Ramsey theory |authorlink= |coauthors= |year=2006 |publisher=Alpha Science Int'l Ltd. |location= |isbn=81-7319-731-8 }}</ref>
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| == Proof of Van der Waerden's theorem (in a special case) ==
| | * accessibility: |
| | ** Safari + VoiceOver: [https://commons.wikimedia.org/wiki/File:VoiceOver-Mac-Safari.ogv video only], [[File:Voiceover-mathml-example-1.wav|thumb|Voiceover-mathml-example-1]], [[File:Voiceover-mathml-example-2.wav|thumb|Voiceover-mathml-example-2]], [[File:Voiceover-mathml-example-3.wav|thumb|Voiceover-mathml-example-3]], [[File:Voiceover-mathml-example-4.wav|thumb|Voiceover-mathml-example-4]], [[File:Voiceover-mathml-example-5.wav|thumb|Voiceover-mathml-example-5]], [[File:Voiceover-mathml-example-6.wav|thumb|Voiceover-mathml-example-6]], [[File:Voiceover-mathml-example-7.wav|thumb|Voiceover-mathml-example-7]] |
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| | ** Orca: There is ongoing work, but no support at all at the moment [[File:Orca-mathml-example-1.wav|thumb|Orca-mathml-example-1]], [[File:Orca-mathml-example-2.wav|thumb|Orca-mathml-example-2]], [[File:Orca-mathml-example-3.wav|thumb|Orca-mathml-example-3]], [[File:Orca-mathml-example-4.wav|thumb|Orca-mathml-example-4]], [[File:Orca-mathml-example-5.wav|thumb|Orca-mathml-example-5]], [[File:Orca-mathml-example-6.wav|thumb|Orca-mathml-example-6]], [[File:Orca-mathml-example-7.wav|thumb|Orca-mathml-example-7]]. |
| | ** From our testing, ChromeVox and JAWS are not able to read the formulas generated by the MathML mode. |
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| The following proof is due to [[Ronald Graham|Ron Graham]] and B.L. Rothschild.<ref name="Graham1974">{{cite journal |authorlink=Ronald Graham |first=R. L. |last=Graham |first2=B. L. |last2=Rothschild |title=A short proof of van der Waerden's theorem on arithmetic progressions |journal=Proc. American Math. Soc. |volume=42 |issue=2 |year=1974 |pages=385–386 |doi=10.1090/S0002-9939-1974-0329917-8 }}</ref> [[A. Ya. Khinchin|Khinchin]]<ref>{{Cite document
| | ==Test pages == |
| | last1 = Khinchin | first1 = A. Ya.
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| | title = Three Pearls of Number Theory
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| | publisher = Dover
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| | location = Mineola, NY
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| | date = 1998
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| | isbn = 978-0-486-40026-6
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| | postscript = .}}
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| </ref> gives a fairly simple proof of the theorem without estimating ''W''(''r'', ''k'').
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| We will prove the special case mentioned above, that ''W''(2, 3) ≤ 325. Let ''c''(''n'') be a coloring of the integers {1, ..., 325}. We will find three elements of {1, ..., 325} in arithmetic progression that are the same color.
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| Divide {1, ..., 325} into the 65 blocks {1, ..., 5}, {6, ..., 10}, ... {321, ..., 325}, thus each block is of the form {''b'' ·5 + 1, ..., ''b'' ·5 + 5} for some ''b'' in {0, ..., 64}. Since each integer is colored either red or blue, each block is colored in one of 32 different ways. By the [[pigeonhole principle]], there are two blocks among the first 33 blocks that are colored identically. That is, there are two integers ''b''<sub>1</sub> and ''b''<sub>2</sub>, both in {0,...,32}, such that
| | *[[Inputtypes|Inputtypes (private Wikis only)]] |
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| : ''c''(''b''<sub>1</sub>·5 + ''k'') = ''c''(''b''<sub>2</sub>·5 + ''k'')
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| for all ''k'' in {1, ..., 5}. Among the three integers ''b''<sub>1</sub>·5 + 1, ''b''<sub>1</sub>·5 + 2, ''b''<sub>1</sub>·5 + 3, there must be at least two that are the same color. (The [[pigeonhole principle]] again.) Call these ''b''<sub>1</sub>·5 + ''a''<sub>1</sub> and ''b''<sub>1</sub>·5 + ''a''<sub>2</sub>, where the ''a''<sub>''i''</sub> are in {1,2,3} and ''a''<sub>1</sub> < ''a''<sub>2</sub>. Suppose (without loss of generality) that these two integers are both red. (If they are both blue, just exchange 'red' and 'blue' in what follows.)
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| Let ''a''<sub>3</sub> = 2·''a''<sub>2</sub> − ''a''<sub>1</sub>. If ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is red, then we have found our arithmetic progression: ''b''<sub>1</sub>·5 + ''a''<sub>''i''</sub> are all red.
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| Otherwise, ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is blue. Since ''a''<sub>3</sub> ≤ 5, ''b''<sub>1</sub>·5 + ''a''<sub>3</sub> is in the ''b''<sub>1</sub> block, and since the ''b''<sub>2</sub> block is colored identically, ''b''<sub>2</sub>·5 + ''a''<sub>3</sub> is also blue.
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| Now let ''b''<sub>3</sub> = 2·''b''<sub>2</sub> − ''b''<sub>1</sub>. Then ''b''<sub>3</sub> ≤ 64. Consider the integer ''b''<sub>3</sub>·5 + ''a''<sub>3</sub>, which must be ≤ 325. What color is it?
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| If it is red, then ''b''<sub>1</sub>·5 + ''a''<sub>1</sub>, ''b''<sub>2</sub>·5 + ''a''<sub>2</sub>, and ''b''<sub>3</sub>·5 + ''a''<sub>3</sub> form a red arithmetic progression. But if it is blue, then ''b''<sub>1</sub>·5 + ''a''<sub>3</sub>, ''b''<sub>2</sub>·5 + ''a''<sub>3</sub>, and ''b''<sub>3</sub>·5 + ''a''<sub>3</sub> form a blue arithmetic progression. Either way, we are done.
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| A similar argument can be advanced to show that ''W''(3, 3) ≤ 7(2·3<sup>7</sup>+1)(2·3<sup>7·(2·3<sup>7</sup>+1)</sup>+1). One begins by dividing the integers into 2·3<sup>7·(2·3<sup>7</sup> + 1)</sup> + 1 groups of 7(2·3<sup>7</sup> + 1) integers each; of the first 3<sup>7·(2·3<sup>7</sup> + 1)</sup> + 1 groups, two must be colored identically.
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| Divide each of these two groups into 2·3<sup>7</sup>+1 subgroups of 7 integers each; of the first 3<sup>7</sup> + 1 subgroups in each group, two of the subgroups must be colored identically. Within each of these identical subgroups, two of the first four integers must be the same color, say red; this implies either a red progression or an element of a different color, say blue, in the same subgroup.
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| Since we have two identically-colored subgroups, there is a third subgroup, still in the same group that contains an element which, if either red or blue, would complete a red or blue progression, by a construction analogous to the one for ''W''(2, 3). Suppose that this element is yellow. Since there is a group that is colored identically, it must contain copies of the red, blue, and yellow elements we have identified; we can now find a pair of red elements, a pair of blue elements, and a pair of yellow elements that 'focus' on the same integer, so that whatever color it is, it must complete a progression.
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| The proof for ''W''(2, 3) depends essentially on proving that ''W''(32, 2) ≤ 33. We divide the integers {1,...,325} into 65 'blocks', each of which can be colored in 32 different ways, and then show that two blocks of the first 33 must be the same color, and there is a block coloured the opposite way. Similarly, the proof for ''W''(3, 3) depends on proving that
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| : <math>W(3^{7(2 \cdot 3^7+1)},2) \leq 3^{7(2 \cdot 3^7+1)}+1.</math>
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| By a double [[mathematical induction|induction]] on the number of colors and the length of the progression, the theorem is proved in general.
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| == Proof == | |
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| A [[Generalized arithmetic progression|''D-dimensional arithmetic progression'']] consists of
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| numbers of the form:
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| ::<math> a + i_1 s_1 + i_2 s_2 ... + i_D s_D </math>
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| where a is the basepoint, the s's are the different step-sizes, and the i's range from 0 to L-1. A d-dimensional AP is ''homogenous'' for some coloring when it is all the same color.
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| A ''D-dimensional arithmetic progression with benefits'' is all numbers of the form above, but where you add on some of the "boundary" of the arithmetic progression, i.e. some of the indices i's can be equal to L. The sides you tack on are ones where the first k i's are equal to L, and the remaining i's are less than L.
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| The boundaries of a D-dimensional AP with benefits are these additional arithmetic progressions of dimension d-1,d-2,d-3,d-4, down to 0. The 0 dimensional arithmetic progression is the single point at index value (L,L,L,L...,L). A D-dimensional AP with benefits is ''homogenous'' when each of the boundaries are individually homogenous, but different boundaries do not have to necessarily have the same color.
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| Next define the quantity MinN(L, D, N) to be the least integer so
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| that any assignment of N colors to an interval of length MinN or more
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| necessarily contains a homogenous D-dimensional arithmetical progression with benefits.
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| The goal is to bound the size of MinN. Note that MinN(L,1,N) is an upper bound for Van-Der-Waerden's
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| number. There are two inductions steps, as follows:
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| 1. Assume MinN is known for a given lengths L for all dimensions of arithmetic progressions with benefits up to D. This formula gives a bound on MinN when you increase the dimension to D+1:
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| let <math> M = {\mathrm MinN}(L,D,n)</math>
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| ::<math> {\mathrm MinN}(L, D+1 , n) \le M*{\mathrm MinN}(L,1,n^M)</math>
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| Proof:
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| First, if you have an n-coloring of the interval 1...I, you can define a ''block coloring'' of k-size
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| blocks. Just consider each sequence of k colors in each k block to define a unique color. Call this ''k-blocking'' an n-coloring. k-blocking an n coloring of length l produces an n^k coloring of length l/k.
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| So given a n-coloring of an interval I of size M*MinN(L,1,n^M)) you can M-block it into an n^M coloring
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| of length MinN(L,1,n^M). But that means, by the definition of MinN, that you can find a 1-dimensional arithmetic sequence (with benefits) of length L in the block coloring, which is a sequence of blocks equally spaced, which are all the same block-color, i.e. you have a bunch of blocks of length M in the original sequence, which are equally spaced, which have exactly the same sequence of colors inside.
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| Now, by the definition of M, you can find a d-dimensional arithmetic sequence with benefits in any one of these blocks, and since all of the blocks have the same sequence of colors, the same d-dimensional AP with benefits appears in all of the blocks, just by translating it from block to block. This is the definition of a d+1 dimensional arithmetic progression, so you have a homogenous d+1 dimensional AP. The new stride parameter s_{D+1} is defined to be the distance between the blocks.
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| But you need benefits. The boundaries you get now are all old boundaries, plus their translations into identically colored blocks, because i_{D+1} is always less than L. The only boundary which is not like this is the 0 dimensional point when <math>i_1=i_2=...=i_{D+1}=L</math>. This is a single point, and is automatically homogenous.
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| 2. Assume MinN is known for one value of L and all possible dimensions D. Then you can bound MinN for length L+1.
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| ::<math>{\mathrm MinN}(L+1,D,n) \le 2{\mathrm MinN}(L,n,n)</math>
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| proof:
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| Given an n-coloring of an interval of size MinN(L,n,n), by definition, you can find an arithmetic sequence with benefits of dimension n of length L. But now, the number of "benefit" boundaries is equal to the number of colors, so one of the homogenous boundaries, say of dimension k, has to have the same color as another one of the homogenous benefit boundaries, say the one of dimension p<k. This allows a length L+1 arithmetic sequence (of dimension 1) to be constructed, by going along a line inside the k-dimensional boundary which ends right on the p-dimensional boundary, and including the terminal point in the p-dimensional boundary. In formulas:
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| if
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| ::<math> a+ L s_1 +L s_2... + L s_{D-k}</math> has the same color as
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| ::<math> a + L s_1 +L s_2 ... +L s_{D-p}</math>
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| then
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| ::<math> a + L*(s_1 ... +s_{D-k}) + u *(s_{D-k+1} ... +s_p) </math> have the same color
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| ::<math> u = 0,1,2,...,L-1,L </math> i.e. u makes a sequence of length L+1.
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| This constructs a sequence of dimension 1, and the "benefits" are automatic, just add on another point of whatever color. To include this boundary point, one has to make the interval longer by the maximum possible value of the stride, which is certainly less than the interval size. So doubling the interval size will definitely work, and this is the reason for the factor of two. This completes the induction on L.
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| Base case: MinN(1,d,n)=1, i.e. if you want a length 1 homogenous d-dimensional arithmetic sequence, with or without benefits, you have nothing to do. So this forms the base of the induction. The VanDerWaerden theorem itself is the assertion that MinN(L,1,N) is finite, and it follows from the base case and the induction steps.<ref name="Graham1974" />
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| ==See also==
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| * [[Van der Waerden number]]s for all known values for ''W''(''n'',''r'') and the best-known bounds for unknown values
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| ==References==
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| {{reflist}}
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| ==External links==
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| * [http://www.math.uga.edu/~lyall/REU/ramsey.pdf Proof of Van der Waerden's theorem]
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| [[Category:Ramsey theory]]
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| [[Category:Theorems in discrete mathematics]]
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| [[Category:Articles containing proofs]]
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| [[de:Satz von Van der Waerden]]
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| [[hu:Van der Waerden-tétel]]
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| [[ja:ファン・デル・ヴェルデンの定理]]
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| [[uk:Теорема ван дер Вардена]]
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| [[zh:范德瓦尔登定理]]
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