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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 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>
| | If you have the desire to procedure settings instantly, loading files instantly, nevertheless the body is logy plus torpid, what would you do? If you are a giant "switchboard" that is deficiency of efficient administration program plus effective housekeeper, what would you do? If you have send your exact commands to a mind, but the body could not perform correctly, what would you do? Yes! You require a full-featured repair registry!<br><br>If you registry gets cluttered up with a great deal of junk we don't use, your PC can run slower. Therefore it is actually prudent that you regularly get your registry cleaned.<br><br>One of the most overlooked factors a computer may slow down is because the registry has become corrupt. The registry is basically your computer's running system. Anytime you're running a computer, entries are being produced plus deleted from a registry. The effect this has is it leaves false entries inside a registry. So, the computer's resources should function around these false entries.<br><br>If you feel you don't have enough funds at the time to upgrade, then the number one choice is to free up some room by deleting a few of the unwelcome files and folders.<br><br>The final step is to create certain that you clean the registry of the computer. The "registry" is a large database which shops important files, settings & options, plus info. Windows reads the files it demands inside order for it to run programs through this database. If the registry gets damaged, afflicted, or clogged up, then Windows may not be capable to correctly access the files it needs for it to load up programs. As this arises, problems and mistakes like the d3d9.dll error happen. To fix this plus prevent future setbacks, you must download plus run a registry cleaning tool. The very suggested software is the "Frontline [http://bestregistrycleanerfix.com/fix-it-utilities fix it utilities]".<br><br>Why this problem occurs frequently? What are the causes of it? In truth, there are 3 major causes which can cause the PC freezing issue. To solve the issue, we should take 3 procedures in the following paragraphs.<br><br>Most likely in the event you are experiencing a slow computer it can be a couple years older. We moreover might not have been told which whilst you utilize your computer everyday; there are certain items which it requires to continue running in its ideal performance. We additionally will not even own any diagnostic tools that will get a PC running like hot again. So do not allow that stop we from getting a program cleaned. With access to the web we can find the tools that will assist you receive your program running like fresh again.<br><br>A program plus registry cleaner can be downloaded from the web. It's user friendly plus the task does not take long. All it does is scan and then whenever it finds errors, it will fix plus clean those errors. An error free registry might protect the computer from mistakes plus provide we a slow PC fix. |
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| For example, when ''r'' = 2, you have two colors, say <span style="color:red;">red</span> and <span style="color:blue;">blue</span>. ''W''(2, 3) is bigger than 8, because you can color the integers from {1, ..., 8} like this:
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| {|class="wikitable"
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| | 1
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| | 4
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| | 5
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| | 6
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| | 7
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| | '''<span style="color:blue;">B</span>'''
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| | '''<span style="color:red;">R</span>'''
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| | '''<span style="color:red;">R</span>'''
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| | '''<span style="color:blue;">B</span>'''
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| | '''<span style="color:blue;">B</span>'''
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| | '''<span style="color:red;">R</span>'''
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| | '''<span style="color:red;">R</span>'''
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| | '''<span style="color:blue;">B</span>'''
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| |}
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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 <span style="color:red;">red 9</span>, then the <span style="color:red;">red 3</span>, <span style="color:red;">6</span>, and <span style="color:red;">9</span> are in arithmetic progression. Alternatively, if you add a <span style="color:blue;">blue 9</span>, then the <span style="color:blue;">blue 1</span>, <span style="color:blue;">5</span>, and <span style="color:blue;">9</span> 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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| 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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| 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.)
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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 upper bound currently known 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>
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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.
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| The best lower bound currently known for <math>W(2, k)</math> is that for all positive <math>\varepsilon</math> we have <math>W(2, k) > 2^k/k^\varepsilon</math>, for all sufficiently large <math>k</math>.<ref>{{cite journal |authorlink=Zoltan Szábo|first=Zoltán |last=Szabó |title=An application of Lovász' local lemma -- a new lower bound for the van der Waerden number |journal=Random Struct. Algorithms |volume=1 | issue = 3 |pages=343-360 |year=1990 }}</ref>
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| == Proof of Van der Waerden's theorem (in a special case) ==
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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 | |
| | 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
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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 | |
| ::<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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If you have the desire to procedure settings instantly, loading files instantly, nevertheless the body is logy plus torpid, what would you do? If you are a giant "switchboard" that is deficiency of efficient administration program plus effective housekeeper, what would you do? If you have send your exact commands to a mind, but the body could not perform correctly, what would you do? Yes! You require a full-featured repair registry!
If you registry gets cluttered up with a great deal of junk we don't use, your PC can run slower. Therefore it is actually prudent that you regularly get your registry cleaned.
One of the most overlooked factors a computer may slow down is because the registry has become corrupt. The registry is basically your computer's running system. Anytime you're running a computer, entries are being produced plus deleted from a registry. The effect this has is it leaves false entries inside a registry. So, the computer's resources should function around these false entries.
If you feel you don't have enough funds at the time to upgrade, then the number one choice is to free up some room by deleting a few of the unwelcome files and folders.
The final step is to create certain that you clean the registry of the computer. The "registry" is a large database which shops important files, settings & options, plus info. Windows reads the files it demands inside order for it to run programs through this database. If the registry gets damaged, afflicted, or clogged up, then Windows may not be capable to correctly access the files it needs for it to load up programs. As this arises, problems and mistakes like the d3d9.dll error happen. To fix this plus prevent future setbacks, you must download plus run a registry cleaning tool. The very suggested software is the "Frontline fix it utilities".
Why this problem occurs frequently? What are the causes of it? In truth, there are 3 major causes which can cause the PC freezing issue. To solve the issue, we should take 3 procedures in the following paragraphs.
Most likely in the event you are experiencing a slow computer it can be a couple years older. We moreover might not have been told which whilst you utilize your computer everyday; there are certain items which it requires to continue running in its ideal performance. We additionally will not even own any diagnostic tools that will get a PC running like hot again. So do not allow that stop we from getting a program cleaned. With access to the web we can find the tools that will assist you receive your program running like fresh again.
A program plus registry cleaner can be downloaded from the web. It's user friendly plus the task does not take long. All it does is scan and then whenever it finds errors, it will fix plus clean those errors. An error free registry might protect the computer from mistakes plus provide we a slow PC fix.