Win at Nim! The secret mathematical strategy for kids (with challange problems in transfinite Nim for the rest of us)

Welcome to my latest instance of Math for Kids!

Today I had the pleasure to make an interactive mathematical presentation at my son’s school to the 7th / 8th grade Math Team, about 30 math-enthusiastic kids (twelve and thirteen years old) along with their math teachers and the chair of the school math department.

The topic was the game of Nim! This game has a secret mathematical strategy enabling anyone with that secret knowledge to win against those without it. It is a great game for kids, because with the strategy they can realistically expect to beat their parents, friends, siblings and parent’s friends almost every single time!


To play Nim, one player sets up a number of piles of blocks, and the opponent chooses whether to go first or second. The players take turns removing blocks — each player may remove any number of blocks (at least one) from any one pile, and it is fine to take a whole pile — whichever player takes the last block wins.

For the math team, we played a few demonstration games, in which I was able to beat all the brave challengers, and then the kids paired off to play each other and gain familiarity with the game. Then, it was time for the first strategy discussion.

What could the secret winning strategy be? I explained to the kids a trick that mathematicians often use when approaching a difficult problem, namely, to consider in detail some very simple special cases or boundary instances of the problem. It often happens that these special cases reveal a way of thinking about the problem that applies much more generally.

Perhaps one of the easiest special cases of Nim occurs when there is only one pile. If there is only one pile, then clearly one wants to go first, in order to make the winning move: take the entire pile!

Two balanced piles

A slightly less trivial and probably more informative case arises when there are exactly two piles. If the stacks have the same height, then the kids realized that the second player could make copying moves so as to preserve this balanced situation. The key insight now is that this copying strategy is a winning strategy, because if one can always copy, then in particular one will have a move whenever the opponent did, and so the opponent will never take the last block. With two piles, therefore, one wants always to make them balanced. If they are initially unbalanced, then choose to go first and follow the balancing strategy. If they are initially balanced, then choose to go second, and copy whatever moves your opponent makes to rebalance them.


A balanced position

With that insight, it is not difficult to see that it is winning to leave a position with any number of pairs of balanced piles. One can in effect play on each pair separately, because whenever the opponent makes a move on one of the piles, one can copy the move with the corresponding partner pile. In this way, we may count such a position overall as balanced. The more fundamental game-theoretic observation to make is that balanced piles in effect cancel each other out in any position, and one can ignore them when analyzing a position. When two balanced piles are present in a possibly more complicated position, one can pretend that they aren’t there, precisely because whenever your opponent plays on one of them, you can copy the move on the other, and so any winning strategy for the position in which those piles are absent can be converted into a winning strategy in which the balanced piles are present.

This idea now provides a complete winning strategy in the case that all piles have height one or two at most. One wants to leave a position with an even number of piles of each height. If only one height has an odd number of piles, then take a whole pile of that height. And if there are odd numbers of piles both of height one and two, then turn a height-two pile into a pile of height one, and this will make them both even. So any unbalanced position can be balanced, and any move on a balanced position will unbalance it.


1+2+3 counts as balanced

Let’s now consider that there may be piles of height three. For example, consider the basic position with piles of height one, two and three. The observation to make here is that any move on this position can be replied to with a move that leaves it balanced (check it yourself to be sure!). It follows that this position is winning to leave for the other player (and so one should go second on $1+2+3$). It would be nice if we could consider this position itself as already balanced in some sense. Indeed, we may incorporate this situation into the balancing idea if we think of the pile of height three as really consisting of two subpiles, one of height two and one of height one. In this way, the Nim position 1+2+3 counts as balanced, since the 3 counts as 2+1, which balances the other stacks.  The 1+2+3 position has two stacks of height two and two of height one, when one regards the stack of height three as having a substack of height two and a substack of height one.

This way of thinking produces a complete winning strategy for Nim positions involving piles of height at most three. (And this is a strategy that can be mastered even by very young children — a few years ago I had talked about Nim with much younger children, Math for six-year-olds: Win at Nim!, first-graders at my daughter’s school, and at that time we concentrated on posititions with piles of height at most three. Older kids, however, can handle the full strategy.) Namely, the winning strategy in this case is to strive to balance the position, to make an even number overall of piles of height one and two, where we count piles of height three as one each of one and two. If you always give your opponent a balanced position, then  you will win!  Faced with an unbalanced position, it is a fact that you can always find a balancing move, and any move on an balanced position will unbalance it.  If the game is just starting, and you are deciding whether to go first or second, you should determine whether it is balanced yet or not.  If it unbalanced, then you should go first and make the balancing move; if it is already balanced, then you should go second and adopt the copying strategy, in which you re-balance the position with each move.

The general winning strategy, of course, goes beyond three. The key idea is to realize that what is really going on when we represent $3$ as $2+1$ is that we are using the binary representation of the number $3$. To explain, I wrote the following numbers on the chalkboard $$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ \cdots$$ and was very pleased when the kids immediately shouted out, “The powers of two!” I explained that any natural number can be expressed uniquely as a sum of distinct powers of two. Asked for a favorite number less than one hundred, one student suggested $88$, and together we calculated $$88=64+16+8,$$ which means that the binary representation of $88$ is $1011000$, which I read off as, “one $64$, no $32$s, one $16$, one $8$, no $4$s, no $2$s and no $1$s. This is just the same as thinking of $9572$ as 9 thousands, 5 hundreds, 7 tens and 2 ones, using the powers of ten. It is interesting to learn that one may easily count very high on one hand using binary, up to 1023 on two hands!

The general strategy is to view every Nim pile as consisting of subpiles whose height is a power of two, and to make sure that one leaves a position that is balanced in the sense that every power of two has an even number of such instances in the position. So we think of $3$ as really $2+1$ for the purposes of balancing; $4$ counts as itself because it is a power of two, but $5$ counts as $4+1$ and $6$ counts as $4+2$ and $7$ as $4+2+1$. Another way to describe the strategy is that we express all the pile heights in binary, and we want an even number of $1$s in each binary place position.

The mathematical facts to verify are (1) any move on a balanced position in this powers-of-two sense will cause it to become unbalanced, and (2) any unbalanced position can be balanced in one move. It follows that leaving balanced positions is a winning strategy, because the winning move of taking the last block is a balancing move rather than an unbalancing move.

One can prove statement (1) by realizing that when you move a single stack, the binary representation changes, and so whichever binary digits changed will now become unbalanced.  For statement (2), consider the largest unbalanced power of two $2^k$ and move on any stack that contains a $2^k$ size substack. Since $2^k-1=111\cdots11$ in binary, one can attain any binary pattern for the smaller height stacks by removing between $1$ and $2^k$ many blocks. So one can balance the position.

As a practical matter, the proof of (2) also shows how one can find a (winning) balancing move, which can otherwise be difficult in some cases: look for the largest unbalanced power of two, and move on any pile containing such a subpile, making sure to leave a balanced position.

In most actual instances of Nim, the pile heights are rarely very tall, and so one is usually considering just $1$, $2$ and $4$ as the powers of two that arise.  A traditional starting configuration has piles of height 1, 3, 5, and 7, and this position is balanced, because one may view it as: $1, 2+1, 4+1, 4+2+1$, and there are an even number of 1s, 2s and 4s.

It is interesting to consider also the Misère form of Nim, where one wants NOT to take the last block. This version of the game also has a secret mathematical strategy, which I shall reveal later on.

Challenge 1.   What is the winning strategy in Misère Nim?

If you figure it out, please post a comment! I’ll post the solution later. One might naively expect that the winning strategy of Misère Nim is somehow totally opposite to the winning strategy of regular Nim, but in fact, the positions $1,2,3$ and $1,3,5,7$ are winning for the second player both in Nim and also in Misère Nim. Indeed, I claim that all nontrivial Nim positions that are winning for regular Nim (with a suitable meaning of “nontrivial”) are also winning for Misère Nim. Can you prove it?

Another interesting generalization, for the set-theorists, is to consider transfinite Nim, where the piles can have transfinite ordinal height. So we have finitely many piles of ordinal height, perhaps infinite, and a move consists of making any one pile strictly shorter. Since there are no infinite descending sequence of ordinals, the game will terminate in finitely many moves, and the winner is whowever removes the last block.

Challenge 2.  Who wins the transfinite Nim game with piles of heights: $$1\qquad \omega+3\qquad \omega^\omega+5\qquad \omega^{\omega+3}+\omega^\omega\cdot3+\omega\cdot 5+7\qquad \epsilon_0\qquad \omega_1$$ and what are the winning moves? What is the general winning strategy for transfinite Nim?

Post your solutions, and I’ll post a solution later.


The absolute truth about non-absolute truth, JAF – Weak Arithmetics Days, New York, July 2015

This will be a talk for the Journées sur les Arithmétiques Faibles – Weak Arithmetics Days conference, held in New York at the CUNY Graduate Center, July 7 – 9, 2015.

Abstract. I will discuss several fun theorems and folklore results illustrating that the satisfaction relation of first-order logic is less absolute than one might have expected. Two models of set theory, for example, can have the same natural numbers $\langle\mathbb{N},+,\cdot,0,1,<\rangle$, yet disagree on their theories of arithmetic truth; two models of set theory can have the same natural numbers and a computable linear order in common, yet disagree on whether it is a well-order and hence disagree about $\omega_1^{CK}$; two models of set theory can have the same natural numbers and the same reals, yet disagree on projective truth; two models of set theory can have a rank initial segment of the universe $\langle V_\delta,{\in}\rangle$ in common, yet disagree about whether it is a model of ZFC. These theorems and others can be proved with elementary classical model-theoretic methods. Indefinite arithmetic truthOn the basis of these observations, Ruizhi Yang (Fudan University, Shanghai) and I have argued that the definiteness of the theory of truth for a structure, even in the case of arithmetic, cannot be seen as arising solely from the definiteness of the structure itself in which that truth resides, but rather is a higher-order ontological commitment.

Main article: Satisfaction is not absolute

The weakly compact embedding property, Apter-Gitik celebration, CMU 2015

This will be a talk at the Conference in honor of Arthur W. Apter and Moti Gitik at Carnegie Mellon University, May 30-31, 2015.  I am pleased to be a part of this conference in honor of the 60th birthdays of two mathematicians whom I admire very much.

Moti GitikArthur W. Apter









Abstract. The weakly compact embedding property for a cardinal $\kappa$ is the assertion that for every transitive set $M$ of size $\kappa$ with $\kappa\in M$, there is a transitive set $N$ and an elementary embedding $j:M\to N$ with critical point $\kappa$. When $\kappa$ is inaccessible, this property is one of many equivalent characterizations of $\kappa$ being weakly compact, along with the weakly compact extension property, the tree property, the weakly compact filter property and many others. When $\kappa$ is not inaccessible, however, these various properties are no longer equivalent to each other, and it is interesting to sort out the relations between them. In particular, I shall consider the embedding property and these other properties in the case when $\kappa$ is not necessarily inaccessible, including interesting instances of the embedding property at cardinals below the continuum, with relations to cardinal characteristics of the continuum.

This is joint work with Brent Cody, Sean Cox, myself and Thomas Johnstone.

Slides | Article | Conference web site

Carnegie Mellon University, College of Fine Arts

Solution to my transfinite epistemic logic puzzle, Cheryl’s Rational Gifts

Thanks so much to everyone for trying out my transfinite epistemic logic puzzle, which I have given the name Cheryl’s Rational Gifts, on account of her gifts to Albert and Bernard. I hope that everyone enjoyed the puzzle.  See the list of solvers and honorable mentions at the bottom of this post. Congratulations!

As I determine it, the solution is that

Albert has the number $100\frac38$, and

Bernard has the number $100\frac7{16}$.

Let me explain my reasoning and address a few issues that came up in the comments.

First, let’s understand the nature of the space of possible numbers that Cheryl describes, those of the form 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. Although this may seem complicated at first, in fact this set consists simply of a large number of increasing convergent sequences, one after the other. Specifically, the smallest of the numbers is $0=1-\frac12-\frac12$, and then $\frac14$, $\frac38$, $\frac7{16}$, and so on, converging to $\frac12$, which itself arises as $\frac12=1-\frac14-\frac14$. So the numbers begin with the increasing convergent sequence $$0 \quad\frac14\quad \frac38\quad \frac7{16}\quad\cdots\quad\to\quad \frac12.$$Immediately after this comes another sequence of points of the form $1-\frac14-\frac1{2^{2+r}}$, which converge to $\frac34$, which itself arises as $1-\frac18-\frac18$. So we have $$\frac12\quad \frac58\quad \frac{11}{16}\quad\frac{23}{32}\quad\cdots\quad\to\quad \frac34.$$Following upon this, there is a sequence converging to $\frac78$, and then another converging to $\frac{15}{16}$, and so on. Between $0$ and $1$, therefore, what we have altogether is an increasing sequence of increasing sequences of rational numbers, where the start of the next sequence is precisely the limit of the previous sequence.

Cheryl's numbers

The same pattern recurs between $1$ and $2$, between $2$ and $3$, and indeed between any positive integer $n$ and its successor $n+1$, for the numbers the occur between $n$ and $n+1$ are simply a translation of the numbers between $0$ and $1$. Thus, for every positive integer $k$ we have $n-\frac1{2^k}$ as the limit of the numbers $n-\frac{1}{2^k}-\frac{1}{2^{k+r}}$, as $r$ increases. Between any two non-negative integers, therefore, we have an increasing sequence of converging increasing sequences. Altogether, we have infinitely many copies of the picture between $0$ and $1$, which was infinitely many increasing convergent sequences, one after the other.

For those readers who are familiar with the ordinals, what this means is that the set of numbers forms a closed set of order type exactly $\omega^3$. We may associate the number $n-\frac{1}{2^k}-\frac{1}{2^{k+r}}$ with the ordinal number $\omega^2\cdot (n-1)+\omega\cdot (k-1)+r$, and observe that this correspondence is a (continuous) order-isomorphism of our numbers with the ordinals below $\omega^3$. In this way, we could replace all talk of the specific rational numbers in this puzzle with their corresponding ordinals below $\omega^3$ and imagine that Cheryl has actually given her friends ordinals rather than rationals. But to explain the solution, allow me to stick with the rational numbers.

The fact that Albert initially does not know who has the larger number implies that Albert does not have $0$, the smallest number overall, since if he were to have had $0$, then he would have known that Bernard’s must have been larger. Since then Bernard does not know, it follows that his number is neither $0$ nor $\frac14$, which is the next number, since otherwise he would have known that Albert’s number must have been larger. Since Albert continues not to know, we rule out the numbers up to $\frac38$ for him. And then similarly ruling out the numbers up to $\frac7{16}$ for Bernard. In this way, each step of the back-and-forth continuing denials of knowing eliminates the lowest remaining numbers from possibility.

Consequently, when Cheryl interrupts the first time, we learn that Albert and Bernard cannot have numbers on the first increasing sequence (below $\frac12$), since otherwise they would at some point come to know in that back-and-forth procedure who has the larger number, and so it wouldn’t be true that they wouldn’t know no matter how long they continued the back-and-forth, as Cheryl stated. Thus, after her remark, both Albert and Bernard know that both numbers are at least $\frac12$, which is the first limit point of the set of possible numbers.

Since at this point Albert states that he still doesn’t know who has the larger number, it cannot be that he has $\frac12$ himself, since otherwise he would have known that he had the smaller number. And then next since Bernard still doesn’t know, it follows that Bernard cannot have either $\frac12$ or $\frac58$, the next number. Thus, if they were to continue the iterative not-knowing-yet pronouncements, they would systematically eliminate the numbers on the second increasing sequence, which converges to $\frac34$. Because of Cheryl’s second interruption remark, therefore, it follows that their numbers do not appear on that second sequence, for otherwise they would have known by continuing that pattern long enough. Thus, after her remark, they both know that both numbers are at least $\frac34$.

And since Albert and Bernard in succession state that they still do not know, they have begun to eliminate numbers from the third sequence.

Consider now Cheryl’s contentful exasperated remark. What she says in the first part is that no matter how many times the three of them repeat that pattern, they will still not know. The content of this remark is precisely that neither of the two numbers can be on next sequence (the third), nor the fourth, nor the fifth and so on; they cannot be on any of the first $\omega$ many sequences (that is, below $1$), because if one of the numbers occurred on the $k^{th}$ sequence below $1$, as $1-\frac1{2^k}-\frac1{2^{k+r}}$, for example, then after $k-1$ repetitions of the three-way-pattern, it would no longer be true for Cheryl to say that no matter how long Albert and Bernard continued their back-and-forth they wouldn’t know, since they would indeed know after $r$ steps of that at that point. Thus, the first part of Cheryl’s remark implies that the numbers must both be at least $1$.

But Cheryl also says that the same statement would be true if she said it again. Thus, the numbers must not lie on any of the first $\omega$ many sequences above $1$. Those sequences converge to the limit points $1\frac12$, $1\frac34$, $1\frac78$ and so on. Consequently, after that second statement, everyone would know that the numbers must both be at least $2$. Similarly, after making the statement a third time, everyone knows the numbers must be at least $3$, and after the fourth time, everyone knows the numbers must be at least $4$.

Cheryl says that she could make the statement a hundred times altogether in succession (counting the time she has already said it as amongst the one hundred), and it would be true every time. Since each time she makes the statement, it eliminates precisely the possibility that one of the numbers is on any of the next $\omega$ many sequences, what everyone would know after the one hundredth pronouncement is precisely that both numbers are at least $100$. Even though she didn’t actually make the statement one hundred times, Albert and Bernard are entitled to know exactly that information even so, because she had said that the statement would be true every time, if she were to have said it one hundred times.

Note that it would be perfectly compatible with Cheryl making that statement one hundred times, if one of the numbers had been $100$, since each additional assertion simply eliminates the possibility that one of the numbers occurs on the sequences strictly before the next integer limit point, without eliminating the integer limit point itself.

Next Cheryl makes an additional statement, which it seems to me that some of the commentators did not give sufficient attention. Namely, she says, “And furthermore, even after my having said it altogether one hundred times in succession, you would still not know who has the larger number!” This statement gives additional epistemic information beyond the content of saying that the $100^{th}$ statement would be true. After the $100^{th}$ statement, according to what we have said, both Albert and Bernard would know precisely that both numbers are at least $100$. But Cheryl is telling them that they still would not know, even after the $100^{th}$ statement. Thus, it must be that neither Albert nor Bernard has $100$, since having that number is the only way they could know at that point who has the larger number.  (Note also that Cheryl did not say that they would know that the other does not know, but only that they each would not know after the $100^{th}$ assertion, an issue that appeared to trip up some commentators. So she is making a common-knowledge assertion about what their individual epistemic states would be in that case.)  The first few numbers after $100$ are: $$100\qquad 100\frac14\qquad 100\frac38\qquad 100\frac7{16}\qquad 100\frac{15}{32}\qquad\cdots\to\quad
100\frac12$$ So putting everything together, what everyone knows after Cheryl’s exasperated remark is that both numbers are at least $100\frac14$.

Since Albert still doesn’t know, it means his number is at least $100\frac 38$. Since Bernard doesn’t know after this, it means that Bernard cannot have either $100\frac14$ or $100\frac38$, since otherwise he would know that Albert’s is larger. So Bernard has at least $100\frac7{16}$.

But now suddenly, finally, Albert knows who has the larger number! How can this be? So far, all we knew is that Albert’s number was at least $100\frac 38$ and Bernard’s is at least $100\frac7{16}$. If Albert had $100\frac 38$, then indeed he would know that Bernard’s number is larger; but note also that if Albert had $100\frac7{16}$, then he would also know that Bernard must have the larger number (since he knows the numbers are different). But if Albert’s number were larger than $100\frac7{16}$, then he couldn’t know whether Bernard’s number was larger or not. So after Albert’s assertion, what we all know is precisely that Albert has either $100\frac38$ or $100\frac7{16}$.

But now, Bernard claims to know both numbers! How could he know which number Albert has? The only way that he can distinguish those two possibilities that we mentioned is if Bernard himself has $100\frac7{16}$, since this is the smallest possibility remaining for Bernard and if Bernard’s number were larger than that then Albert could have consistently had either $100\frac38$ or $100\frac7{16}$. Thus, because Bernard knows the numbers, it must be that Bernard has $100\frac7{16}$, which would eliminate this possibility for Albert.

So Albert has $100\frac38$ and Bernard has $100\frac7{16}$, and that is the solution of the puzzle.

A number of commentators solved the puzzle, coming to the same conclusion that I did, and so let me give some recognition here. Congratulations!

Let me also give honorable mentions to the following people, who came very close.


The continuum hypothesis and other set-theoretic ideas for non-set-theorists, CUNY Einstein Chair Seminar, April, 2015

At Dennis Sullivan’s request, I shall speak on set-theoretic topics, particularly the continuum hypothesis, for the Einstein Chair Mathematics Seminar at the CUNY Graduate Center, April 27, 2015, in two parts:

  • An introductory background talk at 11 am, Room GC 6417
  • The main talk at 2 – 4 pm, Room GC 6417

I look forward to what I hope will be an interesting and fruitful interaction. There will be coffee/tea and lunch between the two parts.

Abstract. I shall present several set-theoretic ideas for a non-set-theoretic mathematical audience, focusing particularly on the continuum hypothesis and related issues.

At the introductory background talk, in the morning (11 am), I shall discuss and prove the Cantor-Bendixson theorem, which asserts that every closed set of reals is the union of a countable set and a perfect set (a closed set with no isolated points), and explain how it led to Cantor’s development of the ordinal numbers and how it establishes that the continuum hypothesis holds for closed sets of reals. We’ll see that there are closed sets of arbitrarily large countable Cantor-Bendixson rank. We’ll talk about the ordinals, about $\omega_1$, the long line, and, time permitting, we’ll discuss Suslin’s hypothesis.

At the main talk, in the afternoon (2 pm), I’ll begin with a discussion of the continuum hypothesis, including an explanation of the history and logical status of this axiom with respect to the other axioms of set theory, and establish the connection between the continuum hypothesis and Freiling’s axiom of symmetry. I’ll explain the axiom of determinacy and some of its applications and its rich logical situation, connected with large cardinals. I’ll briefly mention the themes and goals of the subjects of cardinal characteristics of the continuum and of Borel equivalence relation theory.  If time permits, I’d like to explain some fun geometric decompositions of space that proceed in a transfinite recursion using the axiom of choice, mentioning the open questions concerning whether there can be such decompositions that are Borel.

Dennis has requested that at some point the discussion turn to the role of set theory in the foundation for mathematics, compared for example to that of category theory, and I would look forward to that. I would be prepared also to discuss the Feferman theory in comparison to Grothendieck’s axiom of universes, and other issues relating set theory to category theory.

I know that you know that I know that you know…., CSI Undergraduate Conference on Research, Scholarship, and Performance, April 2015

UGCI shall give the plenary talk at the CSI Undergraduate Conference on Research, Scholarship, and Performance, April 30, 2015. My presentation will be followed by a musical performance.

This is a conference where undergraduate students show off their various scholarly and creative research projects, spanning all disciplines.

In my talk, I’ll present various logic puzzles that involve reasoning about knowledge, including knowledge of knowledge or knowledge of the lack of knowledge.  I’ll discuss the solution of Cheryl’s birthday problem, recently in the news, as well as other classic puzzles and some new ones.

It will be fun!


Cheryl’s Rational Gifts: transfinite epistemic logic puzzle challenge!


Can you solve my challenge puzzle?


Cheryl's numbers

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 have posted my own solution on another post.

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!


















  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. Cover

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

Page 1


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?

Page 2+3

Each student had her own booklet and calculated the Euler characteristic for various small graphs, as I moved about the room helping out.Page 5Page 4








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!

Page 6+7

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!”

Page 8


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.Page 9Page 10+11


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.Page 13Page 12








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.

Page 14+15

Page 16


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 $V-E+R=2$.

Page 17

Page 18+19


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:Page 21Page 20








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.

Graph Booklet
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, by Joel David Hamkins, available on Amazon Kindle Books

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,  
    author = {Joel David Hamkins},
    title = {A {Mathematician's} {Year} in {Japan}},
    publisher = {author-published, via Amazon Kindle Direct Publishing},
    year = {2015},
    month = {March},
    url = {},
    note = {ASIN:B00U618LM2, 156 pages,},

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, by Joel David Hamkins, available on Amazon Kindle BooksA 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.

Available on Amazon $4.49.

Please be so kind as to write a review there.
jo eh ru

Embeddings of the universe into the constructible universe, current state of knowledge, CUNY Set Theory Seminar, March 2015

This will be a talk for the CUNY Set Theory Seminar, March 6, 2015.

I shall describe the current state of knowledge concerning the question of whether there can be an embedding of the set-theoretic universe into the constructible universe.

V to L

Question.(Hamkins) Can there be an embedding $j:V\to L$ of the set-theoretic universe $V$ into the constructible universe $L$, when $V\neq L$?

The notion of embedding here is merely that $$x\in y\iff j(x)\in j(y),$$ and such a map need not be elementary nor even $\Delta_0$-elementary. It is not difficult to see that there can generally be no $\Delta_0$-elementary embedding $j:V\to L$, when $V\neq L$.

Nevertheless, the question arises very naturally in the context of my previous work on the embeddability phenomenon, Every countable model of set theory embeds into its own constructible universe, where the title theorem is the following.

Theorem.(Hamkins) Every countable model of set theory $\langle M,\in^M\rangle$, including every countable transitive model of set theory, has an embedding $j:\langle M,\in^M\rangle\to\langle L^M,\in^M\rangle$ into its own constructible universe.

The methods of proof also established that the countable models of set theory are linearly pre-ordered by embeddability: given any two models, one of them embeds into the other; or equivalently, one of them is isomorphic to a submodel of the other. Indeed, one model $\langle M,\in^M\rangle$ embeds into another $\langle N,\in^N\rangle$ just in case the ordinals of the first $\text{Ord}^M$ order-embed into the ordinals of the second $\text{Ord}^N$. (And this implies the theorem above.)

In the proof of that theorem, the embeddings $j:M\to L^M$ are defined completely externally to $M$, and so it was natural to wonder to what extent such an embedding might be accessible inside $M$. And I realized that I could not generally refute the possibility that such a $j$ might even be a class in $M$.

Currently, the question remains open, but we have some partial progress, and have settled it in a number of cases, including the following, on which I’ll speak:

  • If there is an embedding $j:V\to L$, then for a proper class club of cardinals $\lambda$, we have $(2^\lambda)^V=(\lambda^+)^L$.
  • If $0^\sharp$ exists, then there is no embedding $j:V\to L$.
  • If $0^\sharp$ exists, then there is no embedding $j:V\to L$ and indeed no embedding $j:P(\omega)\to L$.
  • If there is an embedding $j:V\to L$, then the GCH holds above $\aleph_0$.
  • In the forcing extension $V[G]$ obtained by adding $\omega_1$ many Cohen reals (or more), there is no embedding $j:V[G]\to L$, and indeed, no $j:P(\omega)^{V[G]}\to V$. More generally, after adding $\kappa^+$ many Cohen subsets to $\kappa$, for any regular cardinal $\kappa$, then in $V[G]$ there is no $j:P(\kappa)\to V$.
  • If $V$ is a nontrivial set-forcing extension of an inner model $M$, then there is no embedding $j:V\to M$. Indeed, there is no embedding $j:P(\kappa^+)\to M$, if the forcing has size $\kappa$. In particular, if $V$ is a nontrivial forcing extension, then there is no embedding $j:V\to L$.
  • Every countable set $A$ has an embedding $j:A\to L$.

This is joint work of myself, W. Hugh Woodin, Menachem Magidor, with contributions also by David Aspero, Ralf Schindler and Yair Hayut.

See my related MathOverflow question: Can there be an embedding $j:V\to L$ from the set-theoretic universe $V$ to the constructible universe $L$, when $V\neq L$?

Talk Abstract

A conference in honor of W. Hugh Woodin’s 60th birthday, March 2015

I am pleased to announce the upcoming conference at Harvard celebrating the 60th birthday of W. Hugh Woodin.  See the conference web site for more information. Click on the image below for a large-format poster.


Moved to a new server, new WordPress installation

I’ve recently moved my site to a new server and apologize for the brief disruption in service. Many thanks to Sam Coskey and Peter Krautzberger for all the help with the move, and for all their help over the years with getting me started on WordPress at Boolesrings.  They really opened my eyes to the possibilities.

I am currently still fixing various technical issues with the new installation, but please post a comment if you see any issue that needs attention. My intention is that all links to will still work as before.

– Joel David Hamkins