# Transfinite epistemic logic puzzle challenge!

Can you solve my challenge puzzle?

Cheryl   Welcome, Albert and Bernard, to my birthday party, and I thank you for your gifts. To return the favor, as you entered my party, I privately made known to each of you a rational number of the form $$n-\frac{1}{2^k}-\frac{1}{2^{k+r}},$$ where $n$ and $k$ are positive integers and $r$ is a non-negative integer; please consider it my gift to each of you. Your numbers are different from each other, and you have received no other information about these numbers or anyone’s knowledge about them beyond what I am now telling you. Let me ask, who of you has the larger number?

Albert    I don’t know.

Bernard    Neither do I.

Albert    Indeed, I still do not know.

Bernard    And still neither do I.

Cheryl    Well, it is no use to continue that way! I can tell you that no matter how long you continue that back-and-forth, you shall not come to know who has the larger number.

Albert    What interesting new information! But alas, I still do not know whose number is larger.

Bernard    And still also I do not know.

Albert    I continue not to know.

Bernard    I regret that I also do not know.

Cheryl    Let me say once again that no matter how long you continue truthfully to tell each other in succession that you do not yet know, you will not know who has the larger number.

Albert    Well, thank you very much for saving us from that tiresome trouble! But unfortunately, I still do not know who has the larger number.

Bernard    And also I remain in ignorance. However shall we come to know?

Cheryl    Well, in fact, no matter how long we three continue from now in the pattern we have followed so far—namely, the pattern in which you two state back-and-forth that still you do not yet know whose number is larger and then I tell you yet again that no further amount of that back-and-forth will enable you to know—then still after as much repetition of that pattern as we can stand, you will not know whose number is larger! Furthermore, I could make that same statement a second time, even after now that I have said it to you once, and it would still be true. And a third and fourth as well! Indeed, I could make that same pronouncement a hundred times altogether in succession (counting my first time as amongst the one hundred), and it would be true every time. And furthermore, even after my having said it altogether one hundred times in succession, you would still not know who has the larger number!

Albert    Such powerful new information! But I am very sorry to say that still I do not know whose number is larger.

Bernard    And also I do not know.

Albert    But wait! It suddenly comes upon me after Bernard’s last remark, that finally I know who has the larger number!

Bernard    Really? In that case, then I also know, and what is more, I know both of our numbers!

Albert    Well, now I also know them!

Question. What numbers did Cheryl give to Albert and Bernard?

If you can determine the answer, make comments below or post a link to your solution. I shall post a solution in a few days time.

See my earlier transfinite epistemic logic puzzles, with solutions. These were inspired by Timothy Gowers’s example.

# Now I know!

Inspired by Timothy Gowers’s example, here is my transfinite epistemic logic problem.

First, let’s begin with a simple finite example.

Cheryl   Hello Albert and Bernard! I have given you each a different natural number ($0, 1, 2, \ldots$). Who of you has the larger number?

Albert   I don’t know.

Bernard   I don’t know either.

Albert    Even though you say that, I still don’t know.

Bernard    And still neither do I.

Albert    Alas, I continue not to know.

Bernard   And also I do not know.

Albert     Yet, I still do not know.

Bernard     Aha! Now I know which of us has the larger number.

Albert      In that case, I know both our numbers.

Bernard.  And now I also know both numbers.

Question: What numbers do Albert and Bernard have?

Click for the solution.

Now, let us consider a transfinite instance. Consider the following conversation.

Cheryl     I have given you each a different ordinal number, possibly transfinite, but possibly finite. Who of you has the larger ordinal?

Albert     I don’t know.

Bernard     I don’t know, either.

Albert     Even though you say that, I still don’t know.

Bernard     And still neither do I.

Albert     Alas, I still don’t know.

Bernard     And yet, neither do I.

Cheryl     Well, this is becoming boring. Let me tell you that no matter how much longer you continue that back-and-forth, you still will not know the answer.

Albert      Well, thank you, Cheryl, for that new information. However, I still do not know who has the larger ordinal.

Bernard     And yet still neither do I.

Albert     Alas, even now I do not know!

Bernard      And neither do I!

Cheryl     Excuse me; you two can go back and forth like this again, but let me tell you that no matter how much longer you continue in that pattern, you will not know.

Albert      Well, ’tis a pity, since even with this further information, I still do not know.

Bernard     Aha! Now at last I know who of us has the larger ordinal.

Albert     In that case, I know both our ordinals.

Bernard. And now I also know both ordinals.

Question:   What ordinals do they have?

Click for a solution.

See my next transfinite epistemic logic puzzle challenge!

Solutions.

1. For the first problem, with natural numbers, let us call the numbers $a$ and $b$, respectively, for Albert and Bernard. Since Albert doesn’t know at the first step, it means that $a\neq 0$, and so $a$ is at least $1$. And since Bernard can make this conclusion, when he announces that he doesn’t know, it must mean that $b$ is not $0$ or $1$, for otherwise he would know, and so $b\geq 2$. On the next round, since Albert still doesn’t know, it follows that $a$ must be at least $3$, for otherwise he would know; and then, because Bernard still doesn’t know, it follows that $b$ is at least $4$. The next round similarly yields that $a$ is at least $5$ and then that $b$ is at least $6$. Because Albert can undertake all this reasoning, it follows that $a$ is at least $7$ on account of Albert’s penultimate remark. Since Bernard announces at this point that he knows who has the larger number, it must be that Bernard has $6$ or $7$ and that Albert has the larger number. And since Albert now announces that he knows the numbers, it must be because Albert has $7$ and Bernard has $6$.
2. For the transfinite problem, let us call the ordinal numbers $\alpha$ and $\beta$, respectively, for Albert and Bernard. Since Albert doesn’t know at the first step, it means that $\alpha\neq 0$ and so $\alpha\geq 1$. Similarly, $\beta\geq 2$ after Bernard’s remark, and then $\alpha\geq 3$ and $\beta\geq 4$ and this would continue for some time. Because Cheryl says that no matter how long they continue, they will not know, it follows that both $\alpha$ and $\beta$ are infinite ordinals, at least $\omega$. But since Albert does not know at this stage, it means $\alpha\geq\omega+1$, and then $\beta\geq \omega+2$. Since Cheryl says again that no matter how long they continue that, they will not know, it means that $\alpha$ and $\beta$ must both exceed $\omega+k$ for every finite $k$, and so $\alpha$ and $\beta$ are both at least $\omega\cdot 2$. Since Albert still doesn’t know after that remark, it means $\alpha\geq\omega\cdot 2+1$. But now, since Bernard knows at this point, it must be that $\beta=\omega\cdot 2$ or $\omega\cdot 2+1$, since otherwise he couldn’t know. So Albert’s ordinal is larger. Since at this point Albert knows both the ordinals, it must be because Albert has $\omega\cdot 2+1$ and Bernard has $\omega\cdot 2$.

It is clear that one may continue in this way through larger transfinite ordinals. When the ordinals become appreciable in size, then it will get harder to turn it into a totally finite conversation, by means of Cheryl’s remarks, but one may succeed at this for quite some way with suitably obscure pronouncements by Cheryl describing various limiting processes of the ordinals. To handle any given (possibly uncountable) ordinal, it seems best that we should consider conversations of transfinite length.

# Math for eight-year-olds: graph theory for kids!

This morning I had the pleasure to be a mathematical guest in my daughter’s third-grade class, full of inquisitive eight- and nine-year-old girls, and we had a wonderful interaction. Following up on my visit last year (math for seven-year-olds), I wanted to explore with them some elementary ideas in graph theory, which I view as mathematically rich, yet accessible to children.

My specific aim was for them to discover on their own the delightful surprise of the Euler characteristic for connected planar graphs.

We began with a simple example, counting together the number of vertices, edges and regions. For counting the regions, I emphasized that we count the “outside” region as one of the regions.

Then, I injected a little mystery by mentioning that Euler had discovered something peculiar about calculating $V-E+R$.  Could they find out what it was that he had noticed?

Each student had her own booklet and calculated the Euler characteristic for various small graphs, as I moved about the room helping out.

Eventually, the girls noticed the peculiar thing — they kept getting the number two as the outcome!  I heard them exclaim, why do we keep getting two?  They had found Euler’s delightful surprise!

The teachers also were very curious about it, and one of them said to me in amazement, “I really want to know why we always get two!”

With the students, I suggested that we try a few somewhat more unusual cases, to see how robust the always-two situation really was. But in these cases, we still got two as the result.

The girls made their own graphs and tested the hypothesis.

Eventually, of course, I managed to suggest some examples that do in fact test the always-two phenomenon, first by looking at disconnected graphs, and then by considering graphs drawn with crossing edges.

In this way, we were led to refine the $V-E+R=2$ hypothesis to the case of connected planar graphs.

Now, it was time for proof.  I was initially unsure whether I should actually give a proof, since these were just third-graders, after all, and I thought it might be too difficult for the students. But when the teachers had expressed their desire to know why, they had specifically encouraged me to give the argument, saying that even if some students didn’t follow it, there was still value merely in seeing that one can mount an argument like that.  Excellent teachers!

The idea of the proof is that $V-E+R=2$ is true at the start, in the case of a graph consisting of one vertex and no edges. Furthermore, it remains true when one adds one new vertex connected by one new edge, since the new vertex and new edge cancel out.  Also, it remains true when one carves out a new region from part of an old region with the addition of a single new edge, since in this case there is one new edge and one new region, and these also balance each other. Since any connected planar graph can be built up in this way by gradually adding new vertices and edges, this argument shows $V-E+R=2$ for any connected planar graph. This is a proof by mathematical induction on the size of the graph.

Next, we moved on to consider some three-dimensional solids and their surfaces.  With various polyhedra, and the girls were able to verify further instances of $V-E+R=2$.

The girls then drew their own solids and calculated the Euler characteristic.  I taught them how to draw a cube and several other solids pictured here; when the shape is more than just a simple cube, this can be a challenge for a child, but some of the children made some lovely solids:

In the end, each child had a nice little booklet to take home. The images above are taken from one of the students in the class.

What a great day!

You can find a kit of the booklet here: Math for eight-year-olds: graph theory for kids!

See also the account of my previous visit:  Math for seven-year-olds: graph coloring and Eulerian paths.

My new book:

# A mathematician’s year in Japan

• J. D. Hamkins, A Mathematician’s Year in Japan, author-published, via Amazon Kindle Direct Publishing, 2015. (ASIN:B00U618LM2, 156 pages, http://www.amazon.com/dp/B00U618LM2)
@BOOK{Hamkins2015:AMathematiciansYearInJapan,
author = {Joel David Hamkins},
title = {A {Mathematician's} {Year} in {Japan}},
publisher = {author-published, via Amazon Kindle Direct Publishing},
year = {2015},
month = {March},
url = {http://www.amazon.com/dp/B00U618LM2},
note = {ASIN:B00U618LM2, 156 pages, http://www.amazon.com/dp/B00U618LM2},
}

Years ago, when I was still a junior professor, I had the pleasure to live for a year in Japan, working as a research fellow at Kobe University. During that formative year, I recorded brief moments of my Japanese experience, and every two weeks or so—this was well before the current blogging era—I sent my descriptive missives by email to friends back home. I have now collected together those vignettes of my life in Japan, each a morsel of my experience. The book is now out!

A Mathematician’s Year in Japan
Joel David Hamkins

Glimpse into the life of a professor of logic as he fumbles his way through Japan.

A Mathematician’s Year in Japan is a lighthearted, though at times emotional account of how one mathematician finds himself in a place where everything seems unfamiliar, except his beloved research on the nature of infinity, yet even with that he experiences a crisis.

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