It’s been almost 24 hours since the mini-polymath4 project was opened in order to collaboratively solve Q3 from the 2012 IMO. In that time, the first part of the question was solved, but the second part remains open. In other words, it remains to show that for any sufficiently large and any , there is some such that the second player B in the liar’s guessing game cannot force a victory no matter how many questions he asks.
As the previous research thread is getting quite lengthy (and is mostly full of attacks on the first part of the problem, which is now solved), I am rolling over to a fresh thread (as is the standard practice with the polymath projects). Now would be a good time to summarise some of the observations from the previous thread which are still relevant here. For instance, here are some relevant statements made in previous comments:
- Without loss of generality we may take N=n+1; if B can’t win this case, then he certainly can’t win for any larger value of N (since A could simply restrict his number x to a number up to n+1), and if B can win in this case (i.e. he can eliminate one of the n+1 possibilities) he can also perform elimination for any larger value of N by partitioning the possible answers into n+1 disjoint sets and running the N=n+1 strategy, and then one can iterate one’s way down to N=n+1.
- In order to show that B cannot force a win, one needs to show that for any possible sequence of questions B asks in the N=n+1 case, it is possible to construct a set of responses by A in which none of the n+1 possibilities of x are eliminated, thus each x belongs to at least one of each block of k+1 consecutive sets that A’s answers would indicate.
- It may be useful to look at small examples, e.g. can one show that B cannot win when k=1, n=1? Or when k=2, n=3?
It seems that some of the readers of this blog have already obtained a solution to this problem from other sources, or from working separately on the problem, so I would ask that they refrain from giving spoilers for this question until at least one solution has been arrived at collaboratively.
Also, participants are encouraged to edit the wiki as appropriate with new developments and ideas, and participate in the discussion thread for any meta-discussion about the polymath project.
Chris Evans has proposed a new polymath project, namely to attack the “Hot Spots conjecture” for acute-angled triangles. The details and motivation of this project can be found at the above link, but this blog post can serve as a place to discuss the problem (and, if the discussion takes off, to start organising a more formal polymath project around it).
Richard Lipton has just proposed on his blog to discuss the following conjecture of Erdos as a polymath project: that there are no natural number solutions to the equation
with . Previous progress on this problem (including, in particular, a proof that any solution to this equation must have an extremely large value of , and specifically that ) can be found here.
For the time being this is an almost empty post, the main purpose of which is to provide a space for mathematical comments connected with the project of assessing whether it is possible to use the recent ideas of Sanders and of Bateman and Katz to break the barrier in Roth’s theorem. (In a few hours’ time I plan to write a brief explanation of what one of the main difficulties seems to be.)
Added later. Tom Sanders made the following remarks as a comment. It seems to me to make more sense to have them as a post, since they are a good starting point for a discussion. So I have taken the liberty of upgrading the comment. Thus, the remainder of this post is written by Tom.
This will hopefully be an informal post on one aspect of what we might need to do to translate the Bateman-Katz work into the setting.
One of the first steps in the Bateman-Katz argument is to note that if is a cap-set (meaning it is free of three-term progressions) of density then we can assume that there are no large Fourier coefficients, meaning
They use this to develop structural information about the large spectrum, , which consequently has size between and . This structural information is then carefully analysed in the `beef’ of the paper. (more…)
This post marks the official opening of the mini-polymath2 project to solve a problem from the 2010 IMO. I have selected the fifth question (which appears to be slightly more challenging than the sixth, for a change) as the problem to focus on:
In each of six boxes
there is initially one coin. There are two types of operation allowed:
- Type 1: Choose a nonempty box with . Remove one coin from and add two coins to .
- Type 2: Choose a nonempty box with . Remove one coin from and exchange the contents of (possibly empty) boxes and .
Determine whether there is a finite sequence of such operations that results in boxes
being empty and box
coins. (Note that
The comments to this post shall serve as the research thread for the project, in which participants are encouraged to post their thoughts and comments on the problem, even if (or especially if) they are only partially conclusive. Participants are also encouraged to visit the discussion thread
for this project, and also to visit and work on the wiki page
to organise the progress made so far.
- All are welcome. Everyone (regardless of mathematical level) is welcome to participate. Even very simple or “obvious” comments, or comments that help clarify a previous observation, can be valuable.
- No spoilers! It is inevitable that solutions to this problem will become available on the internet very shortly. If you are intending to participate in this project, I ask that you refrain from looking up these solutions, and that those of you have already seen a solution to the problem refrain from giving out spoilers, until at least one solution has already been obtained organically from the project.
- Not a race. This is not intended to be a race between individuals; the purpose of the polymath experiment is to solve problems collaboratively rather than individually, by proceeding via a multitude of small observations and steps shared between all participants. If you find yourself tempted to work out the entire problem by yourself in isolation, I would request that you refrain from revealing any solutions you obtain in this manner until after the main project has reached at least one solution on its own.
- Update the wiki. Once the number of comments here becomes too large to easily digest at once, participants are encouraged to work on the wiki page to summarise the progress made so far, to help others get up to speed on the status of the project.
- Metacomments go in the discussion thread. Any non-research discussions regarding the project (e.g. organisational suggestions, or commentary on the current progress) should be made at the discussion thread.
- Be polite and constructive, and make your comments as easy to understand as possible. Bear in mind that the mathematical level and background of participants may vary widely.
I am proposing the sixth question for the 2010 International Mathematical Olympiad (traditionally, the trickiest of the six problems) as a mini-polymath project for next month. Details and discussions are in this post on my other blog.
[Update, June 27: the project is scheduled to start on Thursday, July 8 16:00 UTC.]
Is taking place on Gowers’s blog!
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For a description of Erdos’ discrepancy problem and a large discussion see this blog on Gowers’s blog.
The decision for the next polymath project over Gowers’s blog will be between three projects: The polynomial DHJ problem, Littlewood problem, and the Erdos discrepency problem. To help making the decision four polls are in place!
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