Warning: Cannot modify header information - headers already sent by (output started at /home/public/wp-content/themes/simplelin/functions.php:1) in /home/public/wp-includes/feed-rss2.php on line 8
ramsey – Ben Barber https://babarber.uk mathematical storytelling Fri, 31 Aug 2018 15:23:18 +0000 en-US hourly 1 https://wordpress.org/?v=5.9.2 The Namer-Claimer game, part 2 https://babarber.uk/457/the-namer-claimer-game-part-2/ https://babarber.uk/457/the-namer-claimer-game-part-2/#respond Fri, 31 Aug 2018 15:23:18 +0000 http://babarber.uk/?p=457 Discrete Mathematics, Volume 344, Issue 3, March 2021, 112256 https://doi.org/10.1016/j.disc.2020.112256 arXiv

I’ve previously written about the Namer-Claimer game.  I can now prove that the length of the game is \Theta(\log \log n) with optimal play from each side, matching the greedy lower bound.  The upper bound makes use of randomness, but in a very controlled way.  Analysing a truly random strategy still seems like it will be very difficult.

The proof brings up a surprising connection to the Ramsey theory of Hilbert cubes.

]]> https://babarber.uk/457/the-namer-claimer-game-part-2/feed/ 0 Partition regularity and other combinatorial problems https://babarber.uk/157/thesis/ https://babarber.uk/157/thesis/#respond Fri, 16 Dec 2016 14:17:16 +0000 http://babarber.uk/?p=157 This is the imaginative title of my PhD thesis.  It contains four unrelated pieces of work.  (I was warned off using this phrasing in the thesis itself, where the chapters are instead described as “self-contained”.)

The first and most substantial concerns partition regularity.  It is a coherent presentation of all of the material from Partition regularity in the rationals, Partition regularity with congruence conditions and Partition regularity of a system of De and Hindman.

The remaining three chapters are expanded versions of Maximum hitting for n sufficiently large, Random walks on quasirandom graphs and A note on balanced independent sets in the cube.1

Erratum: the form of Talagrand’s inequality quoted as Theorem 6 in Chapter 4 has an incorrect proof in the book cited.  The proof can be modified to prove a slightly weaker result, with 60 replaced by 90 and an assumption that the median of X is at least 100c^2 (a constant).  This is remarked in the published version of this chapter, Random walks on quasirandom graphs.

1 “A note … ” was fairly described by one of my examiners as a “potboiler”.  It was also my first submitted paper, and completed in my second of three years.  Perhaps this will reassure anybody in the first year of a PhD who is worrying that they have yet to publish.

]]> https://babarber.uk/157/thesis/feed/ 0 Distinguishing subgroups of the rationals by their Ramsey properties https://babarber.uk/108/distinguishing-subgroups-of-the-rationals-by-their-ramsey-properties/ https://babarber.uk/108/distinguishing-subgroups-of-the-rationals-by-their-ramsey-properties/#respond Wed, 31 Aug 2016 17:08:16 +0000 http://babarber.uk/?p=108 Ben Barber, , Neil Hindman, Imre Leader and Dona Strauss, Journal of Combinatorial Theory, Series A, Volume 129, January 2015, Pages 93–104 PDF

In Partition regularity in the rationals we (Barber, Hindman and Leader) showed that there are systems of equations that are partition regular over \mathbb Q but not over \mathbb Z. Here we show that this separation is very strong: there is an uncountable chain of subgroups from \mathbb Z to \mathbb Q such that each subgroup has a system that is partition regular over it but over no earlier subgroup. We use our new central sets approach, but this result could also have been proved using the original density method.

Most of the work in this paper is spent proving that the systems we construct are strongly partition regular, in the sense that the variables can be forced to take different values. If you only want to see the application then you can skip part of the argument without losing anything.

]]> https://babarber.uk/108/distinguishing-subgroups-of-the-rationals-by-their-ramsey-properties/feed/ 0 Partition regularity without the columns property https://babarber.uk/105/partition-regularity-without-the-columns-property/ https://babarber.uk/105/partition-regularity-without-the-columns-property/#respond Wed, 31 Aug 2016 17:01:35 +0000 http://babarber.uk/?p=105 Ben Barber, Neil Hindman, Imre Leader and Dona Strauss, Proc. Amer. Math. Soc. 143 (2015), 3387-3399 PDF

Rado’s theorem states that a finite matrix is partition regular if and only if it has the “columns property”. It is easy to write down infinite matrices with the columns property that are not partition regular, but all known examples of partition regular matrices do have the columns property. In this paper we describe a matrix that is partition regular but fails to have the columns property in the strongest possible sense.

The main contribution of this paper is a translation of the key lemma of “Partition regularity in the rationals” to work with central, rather than dense, sets. Central sets have very strong combinatorial properties; for example, they contain solutions to all finite partition regular systems. As a result, our theorems are harder to prove but easier to apply—for the application above we could have proved the partition regularity of the systems using density methods, but the argument would have been more involved.

]]> https://babarber.uk/105/partition-regularity-without-the-columns-property/feed/ 0 Partition regularity of a system of De and Hindman https://babarber.uk/102/partition-regularity-of-a-system-of-de-and-hindman/ https://babarber.uk/102/partition-regularity-of-a-system-of-de-and-hindman/#respond Wed, 31 Aug 2016 16:57:53 +0000 http://babarber.uk/?p=102 INTEGERS 14 (2014) #A31 PDF

De and Hindman proposed that a particular system should be partition regular but not partition regular near zero. With Neil Hindman and Imre Leader I found a different example; in this paper I show that De and Hindman’s original system also works.

]]> https://babarber.uk/102/partition-regularity-of-a-system-of-de-and-hindman/feed/ 0 Partition regularity with congruence conditions https://babarber.uk/99/partition-regularity-with-congruence-conditions/ https://babarber.uk/99/partition-regularity-with-congruence-conditions/#respond Wed, 31 Aug 2016 16:55:20 +0000 http://babarber.uk/?p=99 Ben Barber and Imre Leader, Journal of Combinatorics, Volume 4 (2013), Number 3 PDF

Does a partition regular system remain partition regular if we ask that each variable x_i is divisible by d_i? Not necessarily. This answers several open questions from Hindman, Leader and Strauss’s 2003 survey.

The proof of Proposition 5 in the journal version is not entirely clear; I recommend reading the updated pdf linked above. My thanks to Boaz Tsaban for pointing this out.

]]> https://babarber.uk/99/partition-regularity-with-congruence-conditions/feed/ 0 Partition regularity in the rationals https://babarber.uk/95/partition-regularity-in-the-rationals/ https://babarber.uk/95/partition-regularity-in-the-rationals/#respond Wed, 31 Aug 2016 16:51:15 +0000 http://babarber.uk/?p=95 Ben Barber, Neil Hindman, Imre Leader, Journal of Combinatorial Theory, Series A, Volume 120, Issue 7, September 2013, Pages 1590–1599 PDF

A system of linear equations is partition regular if, whenever the natural numbers are finitely coloured, the system of equations has a monochromatic solution. Partition regularity can also be defined over the rationals, and if the system of equations is finite then these notions coincide. We construct an example of an infinite system which is partition regular over the rationals but not the naturals. The proof is based on examining what happens when you take iterated sumsets and difference sets of subsets of the integers with positive upper density.

]]> https://babarber.uk/95/partition-regularity-in-the-rationals/feed/ 0 Ultrafilter quantifiers and Hindman’s theorem https://babarber.uk/38/ultrafilter-quantifiers-and-hindmans-theorem/ https://babarber.uk/38/ultrafilter-quantifiers-and-hindmans-theorem/#respond Sat, 07 May 2016 23:47:37 +0000 http://babarber.uk/?p=38 A filter on \mathbb N is a consistent notion of largeness for subsets of \mathbb N. “Largeness” has the following properties.

  • if A is large and A \subseteq B then B is large
  • if A and B are large then A \cap B is large
  • the empty set is not large

At most one of A and A^\text{c} is large; an ultrafilter is a filter which always has an opinion about which it is.

  • either A or A^\text{c} is large

The usual notation for “A is large” is A \in \U, where \U is the ultrafilter. This casts ultrafilters as sets, rather than notions of largeness. To bring the notion of largeness to the foreground we can use ultrafilter quantifiers; we write \U(x)\ A(x), read “for \U-most x, A(x) holds” (where we have also identified A with the predicate “is a member of A“).

  • if \U(x)\ \phi(x) and \phi \implies \psi then \U(x)\ \psi(x)
  • if \U(x)\ \phi(x) and \U(x)\ \psi(x) then \U(x)\ \phi(x) \wedge \psi(x)
  • \forall x\ \phi(x) \implies \U(x)\ \phi(x) \implies \exists x\ \phi(x)
  • \neg [\U(x)\ \phi(x)] \iff \U(x)\ \neg \phi(x)

From this point of view \forall x\ \phi(x) says that the set of elements with property \phi is everything, and \exists x\ \phi(x) that the set of elements with property \phi is non-empty. \U behaves like a mixture of \forall and \exists, with the considerable advantage that logical operations pass through \U unchanged without having to worry about De Morgan’s laws.

Adding ultrafilters

It turns out that the set of ultrafilters on \mathbb N is a model for the Stone–Čech compactification \beta \mathbb N of (\mathbb N, +). \mathbb N is embedded as the set of principal ultrafilters (“A is large if and only if n \in A“), and the addition on \mathbb N extends to \beta \mathbb N:

    \[\U + \V = \{A : \{x : A-x \in \V\} \in \U\}.\]

(Here A-x should be interpreted as (A-x) \cap \mathbb N.) But to use this addition with quantifiers all we need to know is that (\U+\V)\ \phi(x) \iff \U(x)\ \V(y)\ \phi(x+y). If you can see that from the first definition then your brain is wired differently from mine.

It is a fact that there exist idempotent ultrafilters, with \U + \U = \U. Given such a \U, we can play the following game. Suppose that \U(x)\ \phi(x). Then (\U+\U)(x)\ \phi(x), so \U(x)\ \U(y)\ \phi (x+y) and therefore \U(x)\ \U(y)\ \phi(x) \wedge \phi(y) \wedge \phi (x+y) (by ANDing with the original assertion, and the original assertion with the dummy variable x replaced by y). In particular, \exists x\ \U(y)\ \phi(x) \wedge \phi(y) \wedge \phi (x+y). But now we can fix a good choice of x and repeat the whole process with \psi(y) = \phi(x) \wedge \phi(y) \wedge \phi (x+y) in place of \phi to get that

    \[\exists x\ \exists y\ \U(z)\ \phi(x) \wedge \phi(y) \wedge \phi (x+y) \wedge \phi(z) \wedge \phi(x+z) \wedge \phi(y+z) \wedge \phi (x+y+z).\]

Iterating, we eventually obtain an infinite sequence x_1, x_2, \ldots such that \phi holds for every sum of finitely many of the x_i. Together with the observation that whenever \mathbb N is finitely coloured exactly one of the colour classes is \U-large this proves Hindman’s theorem, that we can always find an infinite sequence x_1, x_2, \ldots such that every sum of finitely many terms has the same colour.

A version of this argument with more of the details filled in is on the Tricki.

]]> https://babarber.uk/38/ultrafilter-quantifiers-and-hindmans-theorem/feed/ 0 Piecewise syndetic and van der Waerden https://babarber.uk/14/piecewise-syndetic-and-van-der-waerden/ https://babarber.uk/14/piecewise-syndetic-and-van-der-waerden/#respond Fri, 06 May 2016 17:35:35 +0000 http://babarber.uk/?p=14 Joel Moreira has just proved that whenever the natural numbers are finitely coloured we can find x and y such that x, x+y and xy are all the same colour.  He actually proves a much more general result via links to topological dynamics, but he includes a direct proof of this special case assuming only a consequence of van der Waerden’s theorem related to piecewise syndetic sets.  I want to record the details here so that I remember them in future.

Syndetic, thick and piecewise syndetic

A set of natural numbers is thick if it contains (arbitrarily) long intervals, and syndetic if it has bounded gaps. It is piecewise syndetic if it has bounded gaps on long intervals (the same bound for each interval).

Piecewise syndetic sets are partition regular:

Claim. If S = A \cup B is piecewise syndetic then either A or B is. (This extends to partitions into more parts by induction.)

Proof. Choose intervals I_n of length n on which the gaps in S are bounded by c. Suppose that A is not piecewise syndetic. Then for every d there is an n_d such that A \cap I_{n_d} has a gap more than d points wide. So B \cap I_{n_d} contains an interval of length d on which its gaps are bounded by c. \ \Box

It follows that, whenever the natural numbers are finitely coloured, one of the colour classes is piecewise syndetic. So to prove van der Waerden’s theorem it would suffice to show that every piecewise syndetic set has long arithmetic progressions. We’ll prove this apparent strengthening, assuming van der Waerden’s theorem itself.

Claim. If S is piecewise syndetic, then it contains long APs.

Proof. If S has long intervals with gaps bounded by c, then the union of the c translates S+1, \ldots, S+c of S contains long intervals. So S+1, \ldots, S+c can be viewed as a c-colouring of a set of long intervals, hence one of them contains long APs by van der Waerden’s theorem. But they are all translates of S, so S contains long APs. \ \Box

In fact we can say much more: for every k there is a d such that the set of a such that the AP of length k and common difference d starting at a is contained in S is piecewise syndetic. To see this, let B be a block of consecutive integers sufficiently long that whenever it is c-coloured it contains an AP of length k. Pack translates of B greedily into the intervals in S+1, \ldots, S+c. For each translate B_i of B there is a triple (a_i, d, j) such that B_i \cap S_j contains the AP of length k and common difference d starting at a_i. Now the set of a_i is piecewise syndetic (it has long intervals with gaps bounded by 2|B|). And there are only finitely many choices for d (at most |B|/(k-1)) and j (at most c). So by the first claim there is some choice of (d,j) such that the corresponding a_i are piecewise syndetic. That is, the set S \cap (S - d) \cap \cdots \cap (S-(k-1)d) is piecewise syndetic. This is Theorem 5.1 in Moreira’s paper, with F = [k] and n = d. The case of general F follows by taking k = \max F.

]]> https://babarber.uk/14/piecewise-syndetic-and-van-der-waerden/feed/ 0