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.