# Seemingly complex logic/set-theoretic puzzle

I got this puzzle some time ago and it has been bugging me since, I cant solve it - but it is supposedly solvable, I am interested in a solution or any tips on how to proceed.

In front of you is an entity named Adam. Adam is a solid block with a single speaker, through which he hears and communicates. For all propositions (statements that are either true or false) $p$, if $p$ is true and logically knowable to Adam, then Adam knows that $p$ is true. Adam is confined to his physical form, cannot move, and only has the sense of hearing. The only sounds Adam can make are to play one of two pre-recorded audio messages. One message consists of a very high note played for one second, and the other one a very low note played for one second.

Adam has mentally chosen a specific subset of the Universe of ordinary mathematics. The Universe of ordinary mathematics is defined as follows:

Let $S_0$ be the set of natural numbers:

$$S_0 = \{1,2,3,\ldots\}$$

$S_0$ has cardinality $\aleph_0$, the smallest and only countable infinity.

The power set of a set $X$, denoted $2^X$, is the set of all subsets of $X$. The power set of a set always has a cardinality larger than the set itself, $$|2^X| = 2^{|X|}$$

Let $S_1 = S_0 \cup 2^{S_0}$. $S_1$ has cardinality $2^{\aleph_0} = \beth_1$.

Let $S_2 = S_1 \cup 2^{S_1}$. $S_2$ has cardinality $2^{\beth_1} = \beth_2$.

In general, let $S_{n+1} = S_n \cup 2^{S_n}$. $S_{n+1}$ has cardinality $2^{\beth_n} = \beth_{n+1}$.

The Universe of ordinary mathematics is defined as $$\bigcup_{i=0}^\infty S_i$$

This Universe contains all sets of natural numbers, all sets of real numbers, all sets of complex numbers, all ordered $n$-tuples for all $n$, all functions, all relations, all Euclidean spaces, and virtually anything that arises in standard analysis.

The Universe of ordinary mathematics has cardinality $\beth_\omega$.

Adam has given you an infinite amount of time to accomplish your task. More specifically, the set of both questions asked by you and notes played by Adam can be of any cardinality. If in your strategy this set is uncountably large, for any number of possibilities of Adam's chosen subset, you must describe the order that the elements of this set take place in as completely as possible.

During your questioning, you are keeping track of the following numbers:

$B_1 =$ The number of questions in which Adam had the option of truthfully responding in the affirmative. (This number and the following numbers can of course be cardinal numbers.)

$B_2 =$ The number of questions in which Adam had the option of truthfully responding in the negative.

$B_3 =$ The number of questions in which Adam had the option of falsely responding in the affirmative.

$B_4 =$ The number of questions in which Adam had the option of falsely responding in the negative.

$B_5 =$ The number of questions in which Adam responded with the high note.

$B_6 =$ The number of questions in which Adam responded with the low note.

$B_7 =$ The number of questions.

Let $C = B_1+B_2+B_3+B_4+B_5+B_6+B_7$

A strategy exists which will eventually allow you to determine Adam's chosen subset. Describe such a strategy in which $C$ is as small as possible, for all possibilities of Adam's chosen subset.

• It's fine that you post here as well, but at least acknowledge that you've posted it on MSE beforehand. math.stackexchange.com/q/17688/622 – Asaf Karagila Jan 16 '11 at 18:42
• I don't get it. It seems to me that the rules permit "Adam" to, for instance, play the high note in response to every question, no matter what subset he has chosen. Did you mean to impose some consistency requirement on his answers? (In particular, I'm not sure what is meant by "contradicting himself", since it seems he can answer each individual question however he wants.) – Pete L. Clark Jan 16 '11 at 20:10
• Do you have the original source for this puzzle? – Not Mike Jan 16 '11 at 21:11
• The problem is part of Unigeg World's Smartest Person Contest 2010, available at psiq.org/human_intelligence_test.pdf The instructions say, "discussing contents of test with others is prohibited...discussing answers with others is strictly prohibited...publishing test in full or part thereof is prohibited," but there's no indication of any prize on offer or any enforcement mechanism. – Gerry Myerson Jan 16 '11 at 21:57
• In my opinion this is not a research-level math question, because (i) it has been taken from a list of puzzles almost verbatim (but without attribution), so (ii) it is not possible to get the OP to clarify the meaning of the question. Having somewhat cryptic instructions may be okay for a puzzle, if you like that sort of thing. It is not okay for a math question. – Pete L. Clark Jan 17 '11 at 3:27

First, observe that you can get around the difficulty that you don't know if high means yes or low in the following way. If you really want to ask the question $\varphi$, you should instead ask the question "high means yes for this round if and only if $\varphi$". If high means yes, then this is the same as asking $\varphi$. But if high means no, then it is like asking $\neg\varphi$, and so we may interpret a high anwer to this question as yes to $\varphi$. This transformation therefore ensures that we can in effect know that high means yes. (I mentioned a similar trick in this MO answer about guessing a number, when there can be wrong answers.)

For the lying issue, let me assume that by lying, you mean that Adam first decides whether to lie or tell the truth, and then calculates what a truthful answer would be, and then when telling the truth plays the appropriate tone, but if lying plays the opposite tone. With this interpretation, a similar trick allows us to extract the desired information. Namely, if you want to ask $\psi$, instead ask "you have decided to be truthful for this question iff $\psi$". If Adam decides to be truthful, then this question is answered the same as $\psi$. If he decides to lie, then he calculates what a truthful answer would be, given that he has already decided to lie, which is the opposite of $\psi$, and so he says the opposite of this. In this way, the double negation of the transformation allows us to get the desired information.

Combining the two transformations allows us to get answers to any desired question.

Now, we simply proceed as follows. Since it seems permissible in the world of your question, let us enumerate all the elements of what you call the universe of ordinary mathematics, and ask of each such element whether it is in Adam's set, using the transformations above. In this way, we find out exactly the set of which he is thinking.

The end result is $\beth_\omega$ many questions. This is the optimal in the sense that any smaller bound on the number of questions would be less than $\beth_n$ for some $n$, with only $\beth_{n+1}$ many possible patterns of answers, but there are $\beth_{\omega+1}$ many sets that Adam might be considering.

Incidently, what you call the universe of ordinary mathematics is closely related to what is known in set theory as $V_{\omega+\omega}$, which is a model of the Zermelo axioms of set theory, one of the first axiomatizations of set theory. The $V$ hierarchy begins with $V_0$ being the empty set, and $V_{\alpha+1}=P(V_\alpha)$ and $V_\lambda=\bigcup_{\alpha\lt\lambda} V_\alpha$ for limit ordinals $\lambda$. Your universe is contained within $V_{\omega+\omega}$, but is actually missing huge parts of $V_{\omega+1}$, because you started only with the natural numbers, rather than the hereditary finite sets. For example, the set $\{\ a_k\mid k\in\mathbb{N}\ \}$, where $a_k=\{\{\{\cdots\}\}\}$ has depth $k$, is missing from your universe, but exists in $V_{\omega+1}$. It follows that your world of mathematics does not have the set HF consisting of all hereditary finite sets, or any similar set with unbounded finite depths. From this, it follows that your world does not satisfy some of the very elementary axioms of set theory, which would allow you to construct HF from the natural numbers. For example, the set mentioned above is the result of a very simple induction on finite-depth finte sets.

The fact that $V_{\omega+\omega}$ itself has no sets of size $\beth_\omega$ is precisely what led to the realization that the Zermelo axioms are too weak to prove even that $\beth_\omega$ exists. This realization led directly to the addition of the Replacement axiom to the axioms of set theory, resulting in the theory now known as ZFC.

I wonder why Adam cannot reason as follows:

If the answer to the question is yes, then I will answer truthfully and use the high note to mean "yes". Thus I will play the high note.

If the answer to the question is no, then I will answer truthfully and use the high note to mean "no".
Thus I will play the high note.

Suppose I get asked a question of the form "Does the high note mean 'yes' iff $\varphi$?" Well, I don't know whether the high note means yes until I know what the answer to the question is. But if $\varphi$ is true and I assume that the high note means 'yes', then the answer is 'yes', which I will signify by playing the high note. So this is at least a consistent answer. If $\varphi$ is true and I assume that the high note means "no", then the answer to the question is "no", which I will signify by playing the high note. Thus, so long as $\varphi$ is true, it doesn't matter whether I think the high note means 'yes' or 'no', since both assumptions are consistent and lead to the same answer: play the high note.

Now suppose $\varphi$ is false. Assume first that the high note means 'yes'. Then the answer to the question is 'no', so since I will answer truthfully and play the high note, the answer is no. That's a contradiction. So now let me assume that the high note means 'no'. Then the answer to the question is 'yes', so then the high note signifies 'yes', also a contradiction! What do I do in this situation? Ah, the rules say that if there is no way to answer without contradicting myself, then I can answer however I want. I will play the high note, since I am after all supposed to be giving away as little information as possible and if I play the high note every time then I am obviously giving away no information whatsoever.

• Pete, I assume your answer is a response to Joel's answer? In his solution, you don't simply ask "Does high mean 'yes' iff $\varphi$?", you have to combine it with the strategy to deal with lying, so you'd ask: "Is it the case that this is a truth-telling round iff (this is a high-means-'yes' round iff $\varphi$)?" – Amit Kumar Gupta Jan 17 '11 at 21:15
• Pete, my solution interprets the procedure as: first the question is asked, then Adam decides on meaning of high/low and whether to tell truth/lie, and then he answer accordingly. In particular, the truth conditions for the statement should be determined after the choices are made, and clearly my solution depends on this. If we allow Adam to determine the truth conditions before making the decisions, then I agree with you that we can learn nothing from him. In this case, the true nature of Adam would have to be clarified. A similar issue arises in the answer I linked to in my answer. – Joel David Hamkins Jan 17 '11 at 22:24
• @Amit: my response is motivated by Joel's answer. I argue that there is an interpretation of the question under which Adam can play the high tone every time, which clearly gives away no information. Because Joel gave a winning strategy under a different interpretation, I decided to clarify why in my interpretation the questions in Joel's strategy don't work. To simplify things, I made Adam's strategy such that he always tells the truth, so questions which ask about truth do not need to be considered. – Pete L. Clark Jan 17 '11 at 22:45
• @Pete It appears that Joel's answer was not the authors intended answer. I have found a newer version of the puzzle which seems to disallow such questions. seti.weebly.com/uploads/1/8/2/4/1824936/… – fastforward Jan 18 '11 at 21:22
• This seems to throw doubt on the credentials of those trying to evaluate the "World's Smartest Person"... – Cam McLeman Jan 28 '11 at 17:38

Based on the answers proposed to this question, I think this problem is ill-defined. The set of possible questions and assignment of truth valuations is not well defined. A fair definition of a "question" would be to specify a sset $X \subseteq \beth_\omega$, where yes means that the chosen set is inside $X$ and no means it is not. In this case, Pete's answer obviously proves that Adam always has a winning strategy.