9

An evil wizard plays the following game with two dwarfs $A$ and $B$: he thinks of a function $f:\mathbb{R}\to\mathbb{R}$ (which is not required to have any regularity properties, such as measurability, ..) and asks $A$ and $B$ to guess it.

$A$ and $B$ play in two separate moments.
$A$ begins and he is allowed to ask the values of $f$ on some subset $S_1\subset\mathbb{R}$.
Then he can ask the values of $f$ on some $S_2$, and so on.
He must guarantee that he will only ask about a finite number of subsets, say $N$, and that $$ \bigcup_{i=1}^N S_i\subsetneq\mathbb{R}. $$ Note that $N$ is not fixed; $A$ must only guarantee that eventually he will stop posing questions and that, at that moment, there are some values of $f$ which the wizard has not yet revealed him.
When he stops, he has to guess the remaining values of the function.

Then it's $B$'s turn, and everything proceeds exactly in the same manner.

$A$ and $B$ can not communicate with each other (except before the game begins, to decide the strategy), so one can also think of $A$ and $B$ playing at the same time but in separate rooms.

$A$ and $B$ are both freed by the wizard if at least one of them guesses the correct function, otherwise he kills both of them.

Is there a winning strategy for the two dwarfs?

Mizar
  • 5,911
  • 21
  • 47
  • 1
    There are no restrictions on the $S_i$, right? So why would the dwarf want to take $N$ different subsets and not just the union? – Jolien Apr 11 '15 at 18:36
  • 1
    @Math1000: That’s not permitted. $S_1$ must be a proper subset of $\Bbb R$. – Brian M. Scott Apr 11 '15 at 18:48
  • Sorry, missed that part. – Math1000 Apr 11 '15 at 19:01
  • What's the point of having two dwarves if they can't communicate? – Kitegi Apr 11 '15 at 19:08
  • @Jolien: the choice of $S_2$ occurs after the wizard reveals the value of $f$ on $S_1$. So the strategy of the dwarf might ask him to choose $S_2$ depending on the preceding answer of the wizard. – Mizar Apr 11 '15 at 19:52
  • @Farnight: that could give the dwarfs better chances to rescue themselves (of course if the strategy of $A$ is different from that of $B$..) – Mizar Apr 11 '15 at 19:53
  • @Mizar So, each one must be able to narrow it to 2 possibilities at most? – Kitegi Apr 11 '15 at 20:20
  • 1
    Funny that you happen to be setting a bounty for this -- I just came across the question today and started to think about it :-) – joriki Jul 22 '15 at 21:05
  • I'm still not sure I understand this. Can the first dwarf ask for the values of $f$ on, say, $\mathbb{R}-{0}$? If the subsets must be connected, he could just set $S_1 = {x \mid x < 0}$ and $S_2 = {x \mid x > 0}$. – Brian Tung Jul 22 '15 at 21:08
  • Does the wizard say anything at all to the second dwarf about the exchange with the first dwarf? – Brian Tung Jul 22 '15 at 21:10
  • The wizard says nothing to the second dwarf about the exchange with $A$ (this also answers your previous question, right?). – Mizar Jul 22 '15 at 21:15
  • Maybe I'm obtuse, but I don't see how it answers my first question about whether the first dwarf can ask for the values of $f$ on $\mathbb{R}-{0}$. Also, does the second dwarf find out (via the fact that the dwarfs have not been freed) that the first dwarf did not guess correctly? Also, do you have any reason to believe that the answer to the question is in the affirmative? – Brian Tung Jul 22 '15 at 21:17
  • Yes, $A$ can ask for the values of $f$ on $\mathbb{R}-{0}$, but no, $B$ doesn't know if $A$ was correct (but this is really irrelevant: $B$ can always assume that $A$ was wrong). I was told this problem at second hand and the answer should be yes (!). – Mizar Jul 22 '15 at 21:32
  • @BrianTung: I think the point is that their later queries depend on the results of their earlier queries. Without that, it would seem even more strange that this should work than it already does. – joriki Jul 22 '15 at 21:34
  • The problem is somewhat reminiscent of http://math.stackexchange.com/questions/1032928 -- I'd suspect that it likewise requires the axiom of choice. – joriki Jul 22 '15 at 21:43
  • Yes, I was also thinking that we need AC. – Brian Tung Jul 22 '15 at 21:49
  • What is the point of adaptive queries when the function is arbitrary? The value of $f$ on some $S$ says nothing about the value of $f$ on $\mathbb{R}\setminus S$. – Reinstate Monica Jul 22 '15 at 21:52
  • @Solomonoff'sSecret: Look at the question I linked to. Infinitely many prisoners are saved despite never gaining any information about themselves. Weird things are possible. – joriki Jul 22 '15 at 21:55
  • @Solomonoff'sSecret: try to tackle the puzzle linked by joriki in his last comment or look at its solution. This should serve as an example of what can happen when there are more than one player. – Mizar Jul 22 '15 at 21:56
  • Yes, although there are at least an infinite number of players in that one. There are only two here. – Brian Tung Jul 22 '15 at 21:58
  • @BrianTung: The deeds of the few outweigh the deeds of the many ;-) – joriki Jul 22 '15 at 22:02
  • @joriki This situation is different. In the other problem all participants have all but finitely much information. Here there is uncountably much information missing from the participant. I'm not saying it's impossible, but it fails my intuition. – Reinstate Monica Jul 22 '15 at 22:02
  • @Solomonoff'sSecret: I'm not sure what you mean by "uncountably much information". They only have to guess a single number (since they can ask for all but one number in their final query), so that's a countably infinite number of bits. – joriki Jul 22 '15 at 22:06
  • @joriki I apologize, you are right. It is countably infinitely much information. But how does that help? – Reinstate Monica Jul 22 '15 at 22:21
  • 3
    Assuming the Axiom of Choice, they can reduce the problem to this question http://math.stackexchange.com/questions/371184/predicting-real-numbers/372495#372495 by both picking $S_1 = \Bbb R \setminus \Bbb N$ and discarding the info the wizard tells them. – mercio Jul 23 '15 at 14:22

1 Answers1

2

Assuming the Axiom of Choice, they can reduce the problem to this question by both picking $S_1= \Bbb R \setminus \Bbb N$ and discarding the info the wizard tells them.

Community
  • 1
mercio
  • 50,180