The propagation of error in classical geometry constructions

Posted on September 10, 2018 by Joel David Hamkins

I’d like to discuss the issue of error and error propagation in the constructions of classical geometry. How does error propagate in these constructions? How sensitive are the familiar classical constructions to small errors in the use of the straightedge or compass?

Let me illustrate what I have in mind by considering the classical construction of Apollonius of the perpendicular bisector of a line segment $AB$. One forms two circles, centered at $A$ and $B$ respectively, each with radius $AB$. These arcs intersect at points $P$ and $Q$, respectively, which form the perpendicular bisector, meeting the original segment at the midpoint $C$.

That is all fine and good, and one can easily prove that indeed $PQ$ is perpendicular to $AB$ and that $C$ is the midpoint of $AB$, as desired.

When carrying out such a construction in practice, however, there will inevitably be some small errors. We do not expect to implement it exactly, with infinite precision, but rather, we expect some small errors in the placement of the compass or straightedge, and perhaps these errors may accumulate and they propagate through the construction. What I would like to discuss is the sensitivity of this construction and the other classical constructions to these small errors.

For example, suppose that we are given points $A$ and $B$. When we seek to construct the circle centered at $A$ with radius $AB$, we place the point of the compass at $A$, and this placement may have some small error deviating from $A$, landing somewhere in the blue circle. Similarly, the writing (or etching) implement end of the compass is placed at $B$, with its own possibly different error, landing the orange circle at $B$. The arc actually resulting will be some arc arising with some such small errors, like this:

We may represent the space of all arcs that could arise in conformance with those error bounds as the blurred orange arc below. This image was created simply by drawing many dozens of such arcs in orange, with various choices for the center and radius within the error circles, and blending the results together.

We carry out the same construction with similar errors for the other arc, centered at $B$ and passing through $A$. These arcs overlap in the darker orange regions above and below, determining the points $P$ and $Q$.

The actual arcs we draw and the corresponding vertical will land somewhere inside these blurred regions, perhaps like this:

Note that in this particular case, the resulting line $PQ$ is noticably non-perpendicular to $AB$, and the resulting point $C$ is noticably not the midpoint. Consider the space of all the bisectors $PQ$ that might arise in conformance with our errors on $A$ and $B$, showing the result as the vertical red shaded region. The darker red region is the space of possible points $C$ that we might have constructed as the “midpoint” $C$, in conformance with the error estimates.

Given the size of the original error bounds on the points $A$ and $B$, it may be surprising that even such a standard simple construction as this — constructing the perpendicular bisector and midpoint of a segment — appears to have comparatively large error propagation, since the shaded red region $C$ is quite large and includes many points that one would not say are close to being the midpoint. In this sense, the Apollonius perpendicular bisector construction appears to be sensitive to the errors of compass placement.

Is there a better construction? For example, in terms of improving the accuracy at least of the perpendicularity of $PQ$ to $AB$, it would seem to help to use a much larger circle, which would lower the variation in the resulting “right” angle. But this is partly because we have so far assumed that compass error arises only with the placement of the points of the compass, and not during the course of actually drawing the arc. But of course, one can imagine that errors arise from a flexing of the compass during use, causing it to deviate from circularity, or from slippage, which might reasonably be expected to cause increasing error with the length or degree of the arc, and so on, and such a model of error might have greater errors with large circles.

One could in principle carry out similar analyses for any geometric construction, and use the corresponding results to compare the sensitivity of various methods for constructing the same object, as well as modeling different sources of error. The goal might be to mount a precise analysis of all the standard constructions and compare competing constructions for accuracy.

There is a literature of papers doing precisely this, and I will try to post some references later (or please do so in the comments, if you have some good ones).

Another approach to error estimation would be to think of the errors at points $A$ and $B$ as probability distributions, centered at $A$ and $B$ and with a certain variation; and one then gets corresponding distributions for the points $P$ and $Q$, which are not rigid shapes as in my diagrams, but qualitatively similar distributions spread out in that region, and a resulting probability distribution for the point $C$.

Finally, let me mention that one might hope to improve the accuracy of a construction, simply by repeating it and averaging the result, or by some other convergence algorithm. For example, as a first step, we might simply perform the Apollonius bisector construction twice, producing midpoint candidates $C_0$ and $C_1$, and we could proceed simply find the midpoint of $C_0C_1$ as a further presumably more accurate midpoint. Or we could iterate in some other manner and hope to converge to the actual midpoint. For example, we could produce seven midpoint candidates, and take the resulting median point.

Draw an infinite chessboard in perspective, using straightedge only

Posted on August 30, 2018 by Joel David Hamkins

I’d like to explain to you how to draw chessboards by hand in perfect perspective, using only a straightedge. In this post, I’ll explain how to construct chessboards of any size, starting with the size of the basic unit square.

This post follows up on the post I made yesterday about how to draw a chessboard in perspective view, using only a straightedge. That method was a subdivision method, where one starts with the boundary of the desired board, and then subdivides to make a chessboard. Now, we start with the basic square and build up. This method is actually quite efficient for quickly making very large boards in perspective view.

I want to emphasize that this is something that you can actually do, right now. It’s fun! All you need is a piece of paper, a pencil and a straightedge. I’ll wait right here while you gather your materials. Use a ruler or a chop stick (as I did) or the edge of a notebook or the lid of a box. Sit at your table and draw a huge chessboard in perspective. You can totally do this.

Start with a horizon, having two points at infinity (orange), at left and right, and a third point midway between them (brown), which we will call the diagonal infinity. Also, mark the front corner of your chess board (blue).

Extend the front corner to the points at infinity. And then mark off (red) a point that will be a measure of the grid spacing in the chessboard. This will the be size of the front square.

You can extend that point to infinity at the right. This delimits the first rank of the chessboard.

Next, extend the front corner of the board to the diagonal infinity.

The intersection of that diagonal with the previous line determines a point, which when extended to infinity at the left, produces the first square of the chessboard.

And that line determines a new point on the leading rank edge. Extend that point up to the diagonal infinity, which determines another point on the second rank line.

Extend that line to infinity at the left, which determines another point on the leading rank edge.

Continuing in this way, one can produce as many first rank squares as desired. Go ahead and do that. At each step, you extend up to the diagonal infinity, which determines a new point, which when extended to infinity at the left determines another point, and so on.

If you should now reflect on the current diagram, you may notice that we have actually determined many further points in the grid than we have mentioned — and thanks to my daughter Hypatia for noticing this simplification — for there is a whole triangle of further intersection points between the files and the diagonals.

We can use these points (and we do not need them all) to construct the rest of the board, by drawing out the lines to infinity at the right. Thus, we construct the whole chessboard:

One can construct a perspective chessboard of any size this way, and one can simply continue with the construction and make it larger, if desired.

It will look a little better if you add a point at infinity down below (and do so directly below the diagonal point at infinity, but a good distance down below the board), and extend the board downward one level. The corresponding diagram on yesterday’s post might be helpful.

You can now color the tile pattern, and you’ll have a chessboard in perfect perspective view.

If you keep going, you can make extremely large chessboards. In time, I hope that you will come to learn how to complete an infinite chess board in finite time.

Draw a chessboard in perspective view, using straightedge only

Posted on August 29, 2018 by Joel David Hamkins

Let me show you how to hand-draw an image of a chessboard in perspective, using only a straightedge. There is no need to measure any distances or to make calculations of any kind. All you need is a straightedge, paper and your pen or pencil.

Imagine the vast doodling potential! A person who happens to be stuck in a lecture on some other less interesting topic could easily make one or more elaborate chessboards in perfect perspective, using only a straightedge! Please carry out the construction and then share your resulting images. I would love to see them.

To begin, give yourself a horizon at the top of the page, with two points at infinity (in orange), and also two points (blue) that will become the front and back corner of your chessboard. You can play around with different arrangements of these points, which will lead ultimately to different perspective views.

Using your straightedge, draw the lines from those reference points to the points at infinity. The resulting enclosed region will ultimately become the main chess board. One can alternatively think of starting with the front edges of the desired board, and then setting a horizon and using that to determine the points at infinity, and then adding the back edges. If you like, you can add a third point at infinity down below, and a front bottom reference point (blue), to give the board some thickness. Construct the lines to the bottom point at infinity.

The result is now the main outline of your chessboard as a rectangular solid.

Next, construct the center point of the top face, by drawing the two diagonals and finding the intersection point. I call this the center point, because it is the point that represents in the perspective view the actual center point of the chessboard, even though this point is not in the “center” of the quadrilateral representing the board. It is remarkable that one can find that point without needing to measure or calculate anything, simply because the two straight lines of the diagonal of the chessboard intersect at that point, and this remains true in the perspective view. This is the key idea that enables the entire construction method to proceed with only a straightedge.

Construct the midpoint lines by drawing the lines from that centerpoint to the points at infinity.

Now you have the chessboard with the main midlines drawn.

Construct the center points of the two diagonal squares, by intersecting the diagonals of each of them.

Construct the lines that extend those points to the points at infinity.

Thus, you have constructed the main 4 x 4 grid.

You can extend the grid lines down to the bottom point at infinity like so:

One can stop with the 4 x 4 board, if desired. Simply add suitable shading, and you’ve completed the 4 x 4 chessboard in perspective.

Alternatively, one can continue with one more iteration to construct the 8 x 8 board. From the 4 x 4 grid (with no shading yet) simply construct the center points of the squares on the main diagonal, and extend those lines to infinity. This will enable you to draw the 8 x 8 grid lines, and after shading, you’ll have the complete chessboard.

The initial arrangement of points affects the nature of the perspective view. Having the points at infinity very far away will produce something closer to orthoprojection; having them close produces a more extreme perspective, which simulates a view from a vantage point extremely close to the board.

Please give the construction a try! All you need is paper, pencil and a straightedge! Provide links below in the comments to photos of your creations.

Meanwhile, let me point you towards my follow post, How to draw infinite chessboards by hand in perfect perspective, using only a straightedge. The difference between the methods is that the method of this post is about subdividing a given board, and the other method is about generating arbitrarily large chessboards from a given unit square.

I learned these construction method years ago from my CUNY geometer colleague Ilya Kofman.

Math for kids: fun with orthoprojections!

Posted on May 27, 2018 by Joel David Hamkins

I had the pleasure a few weeks ago to visit my daughter’s math class at her school and undertake some fun math activities with the sixth graders (11-12 years old, all girls). What a fun time we had!

The topic was orthoprojection and the problem of gaining insight about a three-dimensional object or arrangement by considering its various orthogonal projections.

I had prepared a collection of puzzles, which I hoped would challenge the students in spatial reasoning and abstract visualization. (The puzzle booklet is available at bottom.)

The puzzles involved orthogonal projections of arrangements of unit blocks.

Unit blocks are a certain kind of wooden block toy, which exhibit precise dimensions in the ratio $1\times 2\times 4$. Thus, two blocks oriented the shortest way combine to the same thickness as one block oriented another way, or half the long length. Thus, unit blocks can be flexibly combined in a huge variety of combinations, often leading to aesthetically pleasing and structurally sound creations. Furthermore, they are thought to help develop a child’s intuition for fractions in a completely natural and meaningful way, for often the child needs to build a support exactly $1/4$ unit or $1/2$ unit taller, or what have you, and in this way they are lead to see how the fractional units combine into wholes.

Unit blocks are commonly found in American elementary schools and kindergartens, and woodworkers will recognize that one can make a set of unit blocks from standard $2\times 4$ lumber, available at most hardware stores. You can also buy unit block sets online, and my opinion is that one hardly needs any other toy than a good unit block set for any child aged 2 to 102. Note: you do NOT need any of the fancy extra blocks, not following the unit block standard, that are sometimes available with these sets; the quality of play is best with just the unit blocks and half-units, thin units, double units and so on.

For the school visit, I brought sufficient blocks for all the girls, and explained that we were going to play with blocks the way that a mathematician or engineer might play with blocks. And I had prepared a puzzle booklets, one per child (available below).

For each puzzle, one sees the front view, top view and right side view of an arrangement of blocks, and the challenge is to assemble the blocks so as to realize those views. (In these puzzles, I had chosen not to show any hidden lines and edges, to make them slightly more challenging, although it would also make sense to me to show hidden lines; it would be customary to do so with a dashed line.)

Here are a few more easy examples with solutions.

It happens that these front/top/side projection view problems inspire some deep feelings in me, for they remind me of my father, who was a mechanical engineer. In my childhood, he would often bring home and show me blueprints of the machines or machine parts that he was designing or working with, and those blueprints were filled with front/top/side projection views of the various parts. In pre-computer design days, engineers would specify their machine parts by providing the various othogonal views. (Nowadays, of course, computers are used and one can compute orthogonal and perspective views from any angle.)

In my daughter’s classroom, the students took up the puzzles with a seriousness that shocked me. Once we had passed out the puzzle booklets and distributed the blocks, the girls just steamed through the puzzles, one after the other, totally absorbed. They didn’t want any hints or advice or help of any kind; they just went from each puzzle to the next, solving them one after another. There were a sufficient number of blocks for them all to work on the smaller puzzles on their own, but for the larger puzzles, they formed groups and combined blocks. It was a big success!

Without further ado, here are your puzzles. I’ve also included some photos below, out of order, and some puzzles do not match a photo and vice versa, but you can look at the photos if you need a hint.

After these puzzles, we moved on to the inverse problem. The girls made their own arrangement of blocks and then drew all six orthogonal projections: front, top, right, left, back, bottom. You can draw them on the net of a cube, so that you can imagine folding the cube so as to realize the projections.

And after this, we moved beyond unit blocks to more general shapes. Can you solve the following projection puzzles?

The most challenging puzzle was the following. Can you imagine a solid that appears as a square from the front, a circle from the top and a triangle from the right side?

The complete puzzle booklet is available here: Fun with orthoprojections!

(And here is an alternative version of the puzzles made by David Butler for use with Jenga blocks, which have the dimensional ratios $3\times 5\times 15$ rather than $1\times2\times 4$: Jenga Views; see also this Twitter thread.)

Although I made the puzzles with six-graders in mind, the puzzles are interesting for people of any age, and I have proof: a picture of some of my CUNY math-major college students solving the puzzles in my office.

And here are videos of some fascinating sculptures playing with orthoprojections:

Infinite Sudoku and the Sudoku game

Posted on April 16, 2018 by Joel David Hamkins

Consider what I call the Sudoku game, recently introduced in the MathOverflow question Who wins two-player Sudoku? posted by Christopher King (user PyRulez). Two players take turns placing numbers on a Sudoku board, obeying the rule that they must never explicitly violate the Sudoku condition: the numbers on any row, column or sub-board square must never repeat. The first player who cannot continue legal play loses. Who wins the game? What is the winning strategy?

The game is not about building a global Sudoku solution, since a move can be legal in this game even when it is not part of any global Sudoku solution, provided only that it doesn’t yet explicitly violate the Sudoku condition. Rather, the Sudoku game is about trying to trap your opponent in a maximal such position, a position which does not yet explicitly violate the Sudoku condition but which cannot be further extended.

In my answer to the question on MathOverflow, I followed an idea suggested to me by my daughter Hypatia, namely that on even-sized boards $n^2\times n^2$ where $n$ is even, then the second player can win with a mirroring strategy: simply copy the opponent’s moves in reflected mirror image through the center of the board. In this way, the second player ensures that the position on the board is always symmetric after her play, and so if the previous move was safe, then her move also will be safe by symmetry. This is therefore a winning strategy for the second player, since any violation of the Sudoku condition will arise on the opponent’s play.

This argument works on even-sized boards precisely because the reflection of every row, column and sub-board square is a totally different row, column and sub-board square, and so any new violation of the Sudoku conditions would reflect to a violation that was already there. The mirror strategy definitely does not work on the odd-sized boards, including the main $9\times 9$ case, since if the opponent plays on the central row, copying directly would immediately introduce a Sudoku violation.

After posting that answer, Orson Peters (user orlp) pointed out that one can modify it to form a winning strategy for the first player on odd-sized boards, including the main $9\times 9$ case. In this case, let the first player begin by playing $5$ in the center square, and then afterwards copy the opponent’s moves, but with the ten’s complement at the reflected location. So if the opponent plays $x$, then the first player plays $10-x$ at the reflected location. In this way, the first player can ensure that the board is ten’s complement symmetric after her moves. The point is that again this is sufficient to know that she will never introduce a violation, since if her $10-x$ appears twice in some row, column or sub-board square, then $x$ must have already appeared twice in the reflected row, column or sub-board square before that move.

This idea is fully general for odd-sized Sudoku boards $n^2\times n^2$, where $n$ is odd. If $n=2k-1$, then the first player starts with $k$ in the very center and afterward plays the $2k$-complement of her opponent’s move at the reflected location.

Conclusion.

On even-sized Sudoku boards, the second player wins the Sudoku game by the mirror copying strategy.
On odd-sized Sudoku boards, the first players wins the Sudoku game by the complement-mirror copying strategy.

Note that on the even boards, the second player could also play complement mirror copying just as successfully.

What I really want to tell you about, however, is the infinite Sudoku game (following a suggestion of Sam Hopkins). Suppose that we try to play the Sudoku game on a board whose subboard squares are $\mathbb{Z}\times\mathbb{Z}$, so that the full board is a $\mathbb{Z}\times\mathbb{Z}$ array of those squares, making $\mathbb{Z}^2\times\mathbb{Z}^2$ altogether. (Or perhaps you might prefer the board $\mathbb{N}^2\times\mathbb{N}^2$?)

One thing to notice is that on an infinite board, it is no longer possible to get trapped at a finite stage of play, since every finite position can be extended simply by playing a totally new label from the set of labels; such a move would never lead to a new violation of the explicit Sudoku condition.

For this reason, I should like to introduce the Sudoku Solver-Spoiler game variation as follows. There are two players: the Sudoku Solver and the Sudoku Spoiler. The Solver is trying to build a global Sudoku solution on the board, while the Spoiler is trying to prevent this. Both players must obey the Sudoku condition that labels are never to be explicitly repeated in any row, column or sub-board square. On an infinite board, the game proceeds transfinitely, until the board is filled or there are no legal moves. The Solver wins a play of the game, if she successfully builds a global Sudoku solution, which means not only that every location has a label and there are no repetitions in any row, column or sub-board square, but also that every label in fact appears in every row, column and sub-board square. That is, to count as a solution, the labels on any row, column and sub-board square must be a bijection with the set of labels. (On infinite boards, this is a stronger requirement than merely insisting on no repetitions.)

The Solver-Spoiler game makes sense in complete generality on any set $S$, whether finite or infinite. The sub-boards are $S^2=S\times S$, and one has an $S\times S$ array of them, so $S^2\times S^2$ for the whole board. Every row and column has the same size as the sub-board square $S^2$, and the set of labels should also have this size.

Upon reflection, one realizes that what matters about $S$ is just its cardinality, and we really have for every cardinal $\kappa$ the $\kappa$-Sudoku Solver-Spoiler game, whose board is $\kappa^2\times\kappa^2$, a $\kappa\times\kappa$ array of $\kappa\times\kappa$ sub-boards. In particular, the game $\mathbb{Z}^2\times\mathbb{Z}^2$ is actually isomorphic to the game $\mathbb{N}^2\times\mathbb{N}^2$, despite what might feel initially like a very different board geometry.

What I claim is that the Solver has a winning strategy in the all the infinite Sudoku Solver-Spoiler games, in a very general and robust manner.

Theorem. For every infinite cardinal $\kappa$, the Solver has a winning strategy to win the $\kappa$-Sudoku Solver-Spoiler game.

The strategy will win in $\kappa$ many moves, producing a full Sudoku solution.
The Solver can win whether she goes first or second, starting from any legal position of size less than $\kappa$.
The Solver can win even when the Spoiler is allowed to play finitely many labels at once on each turn, or fewer than $\kappa$ many moves (if $\kappa$ is regular), even if the Solver is only allowed one move each turn.
In the countably infinite Sudoku game, the Solver can win even if the Spoiler is allowed to make infinitely many moves at once, provided only that the resulting position can in principle be extended to a full solution.

Proof. Consider first the countably infinite Sudoku game, and assume the initial position is finite and that the Spoiler will make finitely many moves on each turn. Consider what it means for the Solver to win at the limit. It means, first of all, that there are no explicit repetitions in any row, column or sub-board. This requirement will be ensured since it is part of the rules for legal play not to violate it. Next, the Solver wants to ensure that every square has a label on it and that every label appears at least once in every row, every column and every sub-board. If we think of these as individual specific requirements, we have countably many requirements in all, and I claim that we can arrange that the Solver will simply satisfy the $n^{th}$ requirement on her $n^{th}$ play. Given any finite position, she can always find something to place in any given square, using a totally new label if need be. Given any finite position, any row and any particular label $k$, since can always find a place on that row to place the label, which has no conflict with any column or sub-board, since there are infinitely many to choose from and only finitely many conflicts. Similarly with columns and sub-boards. So each of the requirements can always be fulfilled one-at-a-time, and so in $\omega$ many moves she can produce a full solution.

The argument works equally well no matter who goes first or if the Spoiler makes arbitrary finite play, or indeed even infinite play, provided that the play is part of some global solution (perhaps a different one each time), since on each move the Solve can simply meet the requirement by using that solution at that stage.

An essentially similar argument works when $\kappa$ is uncountable, although now the play will proceed for $\kappa$ many steps. Assuming $\kappa^2=\kappa$, a consequence of the axiom of choice, there are $\kappa$ many requirements to meet, and the Solve can meet requirement $\alpha$ on the $\alpha^{th}$ move. If $\kappa$ is regular, we can again allow the Spoiler to make arbitrary size-less-than-$\kappa$ size moves, so that at any stage of play before $\kappa$ the position will still be size less than $\kappa$. (If $\kappa$ is singular, one can allow Spoiler to make finitely many moves at once or indeed even some uniform bounded size $\delta<\kappa$ many moves at once. $\Box$

I find it interesting to draw out the following aspect of the argument:

Observation. Every finite labeling of an infinite Sudoku board that does not yet explicitly violate the Sudoku condition can be extended to a global solution.

Similarly, any size less than $\kappa$ labeling that does not yet explicitly violate the Sudoku condition can be extended to a global solution of the $\kappa$-Sudoku board for any infinite cardinal $\kappa$.

What about asymmetric boards? It has come to my attention that people sometimes look at asymmetric Sudoku boards, whose sub-boards are not square, such as in the six-by-six Sudoku case. In general, one could take Sudoku boards to consist of a $\lambda\times\kappa$ array of sub-boards of size $\kappa\times\lambda$, where $\kappa$ and $\lambda$ are cardinals, not necessarily the same size and not necessarily both infinite or both finite. How does this affect the arguments I’ve given?

In the finite $(n\times m)\times (m\times n)$ case, if one of the numbers is even, then it seems to me that the reflection through the origin strategy works for the second player just as before. And if both are odd, then the first player can again play in the center square and use the mirror-complement strategy to trap the opponent. So that analysis will work fine.

In the case $(\kappa\times\lambda)\times(\lambda\times\kappa)$ where $\lambda\leq\kappa$ and $\kappa=\lambda\kappa$ is infinite, then the proof of the theorem seems to break, since if $\lambda<\kappa$, then with only $\lambda$ many moves, say putting a common symbol in each of the $\lambda$ many rectangles across a row, we can rule out that symbol in a fixed row. So this is a configuration of size less than $\kappa$ that cannot be extended to a full solution. For this reason, it seems likely to me that the Spoiler can win the Sudoko Solver-Spoiler game in the infinite asymmetric case.

Finally, let’s consider the Sudoku Solver-Spoiler game in the purely finite case, which actually is a very natural game, perhaps more natural than what I called the Sudoku game above. It seems to me that the Spoiler should be able to win the Solver-Spoiler game on any nontrivial finite board. But I don’t yet have an argument proving this. I asked a question on MathOverflow: The Sudoku game: Solver-Spoiler variation.

Escape!

Posted on December 24, 2017 by Joel David Hamkins

Here is an interesting game I heard a few days ago from one of my undergraduate students; I’m not sure of the provenance.

The game is played with stones on a grid, which extends indefinitely upward and to the right, like the lattice $\mathbb{N}\times\mathbb{N}$. The game begins with three stones in the squares nearest the origin at the lower left. The goal of the game is to vacate all stones from those three squares. At any stage of the game, you may remove a stone and replace it with two stones, one in the square above and one in the square to the right, provided that both of those squares are currently unoccupied.

For example, here is a sample play.

Question. Can you play so as completely to vacate the yellow corner region?

One needs only to move the other stones out of the way so that the corner stones have room to move out. Can you do it? It isn’t so easy, but I encourage you to try.

Here is an online version of the game that I coded up quickly in Scratch: Escape!

My student mentioned the problem to me and some other students in my office on the day of the final exam, and we puzzled over it, but then it was time for the final exam. So I had a chance to think about it while giving the exam and came upon a solution. I’ll post my answer later on, but I’d like to give everyone a chance to think about it first.

Solution. Here is the solution I hit upon, and it seems that many others also found this solution. The main idea is to assign an invariant to the game positions. Let us assign weights to the squares in the lattice according to the following pattern. We give the corner square weight $1/2$, the next diagonal of squares $1/4$ each, and then $1/8$, and so on throughout the whole playing board. Every square should get a corresponding weight according to the indicated pattern.

The weights are specifically arranged so that making a move in the game preserves the total weight of the occupied squares. That is, the total weight of the occupied squares is invariant as play proceeds, because moving a stone with weight $1/2^k$ will create two stones of weight $1/2^{k+1}$, which adds up to the same. Since the original three stones have total weight $\frac 12+\frac14+\frac14=1$, it follows that the total weight remains $1$ after every move in the game.

Meanwhile, let us consider the total weight of all the squares on the board. If you consider the bottom row only, the weights add to $\frac12+\frac14+\frac18+\cdots$, which is the geometric series with sum $1$. The next row has total weight $\frac14+\frac18+\frac1{16}+\cdots$, which adds to $1/2$. And the next adds to $1/4$ and so on. So the total weight of all the squares on the board is $1+\frac12+\frac14+\cdots$, which is $2$. Since we have $k$ stones with weight $1/2^k$, another way to think about it is that we are essentially establishing the sum $\sum_k\frac k{2^k}=2$.

The subtle conclusion is that after any finite number of moves, only finitely many of those other squares are occupied, and so some of them remain empty. So after only finitely many moves, the total weight of the occupied squares off of the original L-shape is strictly less than $1$. Since the total weight of all the occupied squares is exactly $1$, this means that the L-shape has not been vacated.

So it is impossible to vacate the original L-shape in finitely many moves. $\Box$

Suppose that we relax the one-stone-per-square requirement, and allow you to stack several stones on a single square, provided that you eventually unstack them. In other words, can you play the stacked version of the game, so as to vacate the original three squares, provided that all the piled-up stones eventually are unstacked?

No, it is impossible! And the proof is the same invariant-weight argument as above. The invariance argument does not rely on the one-stone-per-square rule during play, since it is still an invariant if one multiplies the weight of a square by the number of stones resting upon it. So we cannot transform the original stones, with total weight $1$, to any finite number of stones on the rest of the board (with one stone per square in the final position), since those other squares do not have sufficient weight to add up to $1$, even if we allow them to be stacked during intermediate stages of play.

Meanwhile, let us consider playing the game on a finite $n\times n$ board, with the rule modified so that stones that would be created in row or column $n+1$ in the infinite game simply do not materialize in the $n\times n$ game. This breaks the proof, since the weight is no longer an invariant for moves on the outer edges. Furthermore, one can win this version of the game. It is easy to see that one can systematically vacate all stones on the upper and outer edges, simply by moving any of them that is available, pushing the remaining stones closer to the outer corner and into oblivion. Similarly, one can vacate the penultimate outer edges, by doing the same thing, which will push stones into the outer edges, which can then be vacated. By reverse induction from the outer edges in, one can vacate every single row and column. Thus, for play on this finite board with the modified rule on the outer edges, one can vacate the entire $n\times n$ board!

Indeed, in the finite $n\times n$ version of the game, there is no way to lose! If one simply continues making legal moves as long as this is possible, then the board will eventually be completely vacated. To see this, notice first that if there are stones on the board, then there is at least one legal move. Suppose that we can make an infinite sequence of legal moves on the $n\times n$ board. Since there are only finitely many squares, some of the squares must have been moved-upon infinitely often. If you consider such a square closest to the origin (or of minimal weight in the scheme of weights above), then since the lower squares are activated only finitely often, it is clear that eventually the given square will replenished for the last time. So it cannot have been activated infinitely often. (Alternatively, argue by induction on squares from the lower left that they are moved-upon at most finitely often.) Indeed, I claim that the number of steps to win, vacating the $n\times n$ board, does not depend on the order of play. One can see this by thinking about the path of a given stone and its clones through the board, ignoring the requirement that a given square carries only one stone. That is, let us make all the moves in parallel time. Since there is no interaction between the stones that would otherwise interfere, it is clear that the number of stones appearing on a given square in total is independent of the order of play. A tenacious person could calculate the sum exactly: each square is becomes occupied by a number of stones that is equal to the number of grid paths to it from one of the original three stones, and one could use this sum to calculate the total length of play on the $n\times n$ board.

Still don’t know, an epistemic logic puzzle

Posted on December 20, 2017 by Joel David Hamkins

Here is a epistemic logic puzzle that I wrote for my students in the undergraduate logic course I have been teaching this semester at the College of Staten Island at CUNY. We had covered some similar puzzles in lecture, including Cheryl’s Birthday and the blue-eyed islanders.

Bonus Question. Suppose that Alice and Bob are each given a different fraction, of the form $\frac{1}{n}$, where $n$ is a positive integer, and it is commonly known to them that they each know only their own number and that it is different from the other one. The following conversation ensues.

JDH: I privately gave you each a different rational number of the form $\frac{1}{n}$. Who has the larger number?

Alice: I don’t know.

Bob: I don’t know either.

Alice: I still don’t know.

Bob: Suddenly, now I know who has the larger number.

Alice: In that case, I know both numbers.

What numbers were they given?

Give the problem a try! See the solution posted below.

Meanwhile, for a transfinite epistemic logic challenge — considerably more difficult — see my puzzle Cheryl’s rational gifts.

Solution.
When Alice says she doesn’t know, in her first remark, the meaning is exactly that she doesn’t have $\frac 11$, since that is only way she could have known who had the larger number. When Bob replies after this that he doesn’t know, then it must be that he also doesn’t have $\frac 11$, but also that he doesn’t have $\frac 12$, since in either of these cases he would know that he had the largest number, but with any other number, he couldn’t be sure. Alice replies to this that she still doesn’t know, and the content of this remark is that Alice has neither $\frac 12$ nor $\frac 13$, since in either of these cases, and only in these cases, she would have known who has the larger number. Bob replies that suddenly, he now knows who has the larger number. The only way this could happen is if he had either $\frac 13$ or $\frac 14$, since in either of these cases he would have the larger number, but otherwise he wouldn’t know whether Alice had $\frac 14$ or not. But we can’t be sure yet whether Bob has $\frac 13$ or $\frac 14$. When Alice says that now she knows both numbers, however, then it must be because the information that she has allows her to deduce between the two possibilities for Bob. If she had $\frac 15$ or smaller, she wouldn’t be able to distinguish the two possibilities for Bob. Since we already ruled out $\frac 13$ for her, she must have $\frac 14$. So Alice has $\frac 14$ and Bob has $\frac 13$.

Many of the commentators came to the same conclusion. Congratulations to all who solved the problem! See also the answers posted on my math.stackexchange question and on Twitter:

Epistemic logic puzzle: Still Don’t Know.

Another question from the logic final exam. Epistemic logic. https://t.co/UBpQPTyVPc

— Joel David Hamkins (@JDHamkins) December 20, 2017

Buckets of fish!

Posted on March 4, 2017 by Joel David Hamkins

Let me tell you about the game Buckets of fish.

This is a two-player game played with finitely many buckets in a line on the beach, each containing a finite number of fish. There is also a large supply of additional fish available nearby, fresh off the boats.

Taking turns, each player selects a bucket and removes exactly one fish from it and then, if desired, adds any finite number of fish from the nearby supply to the buckets to the left.

For example, if we label the buckets from the left as 1, 2, 3 and so on, then a legal move would be to take one fish from bucket 4 and then add ten fish to bucket 1, no fish to bucket 2, and ninety-four fish to bucket 3. The winner is whoever takes the very last fish from the buckets, leaving them empty.

Since huge numbers of fish can often be added to the buckets during play, thereby prolonging the length of play, a skeptical reader may wonder whether the game will necessarily come to an end. Perhaps the players can prolong the game indefinitely? Or must it always come to an end?

Question. Does every play of the game Buckets of fish necessarily come to an end?

The answer is yes, every game must eventually come to a completion. I shall give several arguments.

Theorem. Every play of the game Buckets of fish ends in finitely many moves. All the fish in the buckets, including all the new fish that may have been added during play, will eventually run out by some finite stage during play.

That is, no matter how the players add fish to the buckets during play, even with an endless supply of fish from the boats, they will eventually run out of fish in the buckets and one of the players will take the last fish.

First proof. We prove the claim by (nested) induction on the number of buckets. If there is only one bucket, then there are no buckets to the left of it, and so there is no possibility in this case to add fish to the game. If the one bucket contains $k$ fish, then the game clearly ends in $k$ moves. Assume by induction that all plays using $n$ buckets end in finitely many moves, and suppose that we have a game situation with $n+1$ buckets, with $k$ fish in bucket $n+1$. We now prove by induction on $k$ that all such games terminate. This argument is therefore an instance of nested induction, since we are currently inside our proof by induction on $n$, in the induction step of that proof, and in order to complete it, we are undertaking a separate full induction on $k$. If $k=0$, then there are no fish in bucket $n+1$, and so the game amounts really to a game with only $n$ buckets, which terminates in finitely many steps by our induction hypothesis on $n$. So, let us assume that all plays with $k$ fish in bucket $n+1$ terminate in finitely many moves. Consider a situation where there are $k+1$ many fish in that bucket. I claim that eventually, one of those fish must be taken, since otherwise all the moves will be only on the first $n$ buckets, and all plays on only $n$ buckets terminate in finitely many moves. So at some point, one of the players will take a fish from bucket $n+1$, possibly adding additional fish to the earlier buckets. But this produces a situation with only $k$ fish in bucket $n+1$, which by our induction assumption on $k$ we know will terminate in finitely many steps. So we have proved that no matter how many fish are in bucket $n+1$, the game will end in finitely many moves, and so the original claim is true for $n+1$ buckets. Thus, the theorem is true for any finite number of buckets. QED

A second proof. Let me now give another proof, following an idea arising in a conversation with Miha Habič. We want to prove that there is no infinitely long play of the game Buckets of fish. Suppose toward contradiction that there is a way for the players to conspire to produce an infinite play, starting from some configuration of some finite number $n$ of buckets, each with finitely many fish in them. Fix the particular infinitely long play. Let $m$ be the right-most bucket from which a fish was taken infinitely often during that infinite course of play. It follows, for example, that $m<n$, since the top bucket can be used only finitely often, as it never gets replenished. Since bucket $m$ starts with only finitely many fish in it, and each time it is replenished, it is replenished with only finitely many fish, it follows that in order to have been used infinitely many times, it must also have been replenished infinitely often. But each time it was replenished, it was because there was some bucket further to the right that had been used. Since there are only finitely many buckets to the right of bucket $m$, it follows that one of them must have been used infinitely often. This contradicts the choice of $m$ as the right-most bucket that was used infinitely often. QED

A third proof. Let me now give a third proof, using ordinals. We shall associate with each Buckets-of-fish position a certain ordinal. With the position $$7\quad 2\quad 5\quad 24,$$ for example, we associate the ordinal $$\omega^3\cdot 24+\omega^2\cdot 5+\omega\cdot 2+7.$$ More generally, the number of fish in each bucket of a position becomes the coefficient of the corresponding power of $\omega$, using higher powers for the buckets further to the right. The key observation to make is that these associated ordinals strictly descend for every move of the game, since one is reducing a higher-power coefficient and increasing only lower-power coefficients. Since there is no infinite descending sequence of ordinals, it follows that there is no infinite play in the game Buckets of fish. This idea also shows that the ordinal game values of positions in this game are bounded above by $\omega^\omega$, and every ordinal less than $\omega^\omega$ is realized by some position. QED

OK, fine, so now we know that the game always ends. But how shall we play? What is the winning strategy? Say you are faced with buckets having fish in the amounts: $$4\quad 5\quad 2\quad 0\quad 7\quad 4$$ What is your winning move? Please give it some thought before reading further.

The winning strategy turns out to be simpler than you might have expected.

Theorem. The winning strategy in the game Buckets of fish is to play so as to ensure that every bucket has an even number of fish.

Proof. Notice first, as a warm-up, that in the case that there is only one bucket containing an even number of fish, then the second player will win, since the first player will necessarily make it odd, and then the second player will make it even again, and so on. So it will be the second player who will make it zero, winning the game. So with one bucket, the player who can make the bucket even will be the winner.

Next, notice that if you play so as to give your opponent an even number of fish in every bucket, then whatever move your opponent makes will result in an odd number of fish in the bucket from which he or she takes a fish (and possibly also an odd number of fish in some of the earlier buckets as well, if they happen to add an odd number of fish to some of them). So if you give your opponent an all-even position, then they cannot give you back an all-even position.

Finally, notice that if you are faced with a position that is not all-even, then you can simply take a fish from the right-most odd bucket, thereby making it even, and add fish if necessary to the earlier buckets so as to make them all even. In this way, you can turn any position that is not all-even into an all-even position in one move.

By following this strategy, a player will ensure that he or she will take the last fish, since the winning move is to make the all-zero position, which is an all-even position, and the opponent cannot produce an all-even position. QED

In the particular position of the game mentioned before the theorem, therefore, the winning move is to take a fish from the bucket with 7 fish and add an odd number of fish to the bucket with 5 fish, thereby producing an all-even position.

Finally, let’s consider a few variations of the game. It is clear that the all-even strategy works in the versions of the game where one is limited to add at most one fish to each of the earlier buckets, and this version of the game is actually playable, since the number of fish does not grow too much. A similar variation arises where one can either or add or remove any number of fish (or just at most one) from any of the earlier buckets, or where one can, say, add either 5 or 6 fish only to each of the earlier buckets. What is important in the argument is simply that one should be able to ensure the all-even nature of the buckets.

For a more interesting variation, consider what I call the Take 3 version of the game, where one can take either one, two or three fish from any bucket and then add any number of fish to the earlier buckets. The game must still eventually end, but what is the winning strategy?

Question. What is your strategy in the Take 3 variation of Buckets of fish?

Please post your answers in the comments, and I’ll post an answer later. One can generalize this to the Take $n$ variation, where on each turn, the player is allowed to take between 1 and $n$ fish from any bucket, and add as many fish as desired to the earlier buckets.

Another puzzling variation is where each player can take any number of fish from a bucket, and then add any number of fish to earlier buckets. Can you find a strategy for this version of the game? Please post in the comments.

All triangles are isosceles

Posted on February 11, 2017 by Joel David Hamkins

Let me share a mathematical gem with you, the following paradoxical “theorem.”

Theorem. Every triangle is isosceles.

Proof. Consider an arbitrary triangle $\triangle ABC$. Let $Q$ be the intersection of the angle bisector (blue) at $\angle A$ and the perpendicular bisector (green) of $BC$ at midpoint $P$.

Drop perpendiculars from $Q$ to $AB$ at $R$ and to $AC$ at $S$. Because $P$ is the midpoint of $BC$ and $PQ$ is perpendicular, we deduce $BQ\cong CQ$ by the Pythagorean theorem. Since $AQ$ is the angle bisector of $\angle A$, the triangles $AQR$ and $AQS$ are similar, and since they share a hypotenuse, they are congruent. It follows that $AR\cong AS$ and also $QR\cong QS$. Therefore $\triangle BQR$ is congruent to $\triangle CQS$ by the hypotenuse-leg congruence theorem. So $RB\cong SC$. And therefore,
$$AB\cong AR+RB\cong AS+SC\cong AC,$$
and so the triangle is isosceles, as desired. QED

Corollary. Every triangle is equilateral.

Proof. The previous argument proceeded from an arbitrary vertex of the triangle, and so any pair of adjacent sides in the triangle are congruent. So all three are congruent, and therefore it is equilateral. QED

Perhaps you object to my figure, because depending on the triangle, perhaps the angle bisector of $A$ passes on the other side of the midpoint $P$ of $BC$, which would make the point $Q$ lie outside the triangle, as in the following figure.

Nevertheless, essentially the same argument works also in this case. Namely, we again let $Q$ be the intersection of the angle bisector at $\angle A$ with the perpendicular bisector of $BC$ at midpoint $P$, and again drop the perpendiculars from $Q$ to $R$ and $S$. Again, we get $BQ\cong CQ$ by the Pythagorean theorem, using the green triangles. And again, we get $\triangle ARQ\cong\triangle ASQ$ since these are similar triangles with the same hypotenuse. So again, we conclude $\triangle BQR\cong\triangle CQS$ by hypotenuse-leg. So we deduce $AB\cong AR-BR\cong AS-CS\cong AC$, by subtracting rather than adding as before, and so the triangle is isosceles.

Question. What is wrong with these arguments?

Please post your answers in the comments below.

The argument is evidently due to W. W. Rouse Ball, Mathematical Recreations and Essays (1892). I first heard it years ago, when I was in graduate school. Shortly afterward, my advisor W. Hugh Woodin happened to be a little late to seminar, and so I leaped to the chalkboard and gave this proof, leaving the distinguished audience, including R. Solovay, scratching their heads for a while. Woodin arrived, but Solovay wouldn’t let him start the seminar, since he wanted to resolve the triangle argument. What fun!

There are no regular polygons in the hexagonal lattice, except triangles and hexagons

Posted on December 8, 2016 by Joel David Hamkins

I should like to follow up my post last week about the impossibility of finding regular polygons in the integer lattice. This time, however, we shall consider the hexagonal and triangular lattices. It is easy to find lattice points that form the vertices of an equilateral triangle or a regular hexagon.

And similarly, such figures can be found also in the triangular lattice.

Indeed, the triangular and hexagonal lattices are each refinements of the other, so they will allow exactly the same shapes arising from lattice points.

But can you find a square? How about an octagon?

Question. Which regular polygons can be formed by vertices of the hexagonal or triangular lattices?

The answer is that in fact there are no other types of regular polygons to be found! I’d like to prove this by means of some elementary classical reasoning.

Theorem. The only non-degenerate regular polygons that arise from vertices in the hexagonal or triangular lattices are the equilateral triangles and regular hexagons.

This theorem extends the analysis I gave in my post last week for the integer lattice, showing that there are no regular polygons in the integer lattice, except squares.

Proof. The argument for the hexagonal and triangular lattices uses a similar idea as with the integer lattice, but there is a little issue with the square and pentagon case. We can handle both the hexagonal and triangular lattices at the same time. The crucial fact is that both of these lattices are invariant by $120^\circ$ rotation about any lattice vertex.

To get started, suppose that we can find a square in the hexagonal lattice. We may rotate this square by $120^\circ$ about the vertex $a$, and the square vertices will all land on lattice-vertex points. Next, we may rotate the resulting square about the point $b$, and again the vertices will land on lattice points. So we have described how to transform any square vertex point $a$ to another lattice point $c$ which is strictly inside the square. By applying that operation to each of the four vertices of the square, we thereby arrive by symmetry at a strictly smaller square. Thus, for any square in the hexagonal or triangular lattice, there is a strictly smaller square. But if there were any square in the lattice at all, there would have to be a square with smallest side-length, since there are only finitely many lattice distances less than any given length. So there can be no square in the hexagonal or triangular lattice.

The same construction works with pentagons.

If there is a pentagon in the lattice, then we may rotate it about point $a$, and then again about point $b$, resulting in point $c$ strictly inside the pentagon, which leads to an infinite sequence of strictly smaller pentagon, whose sizes (by similarity) go to zero. So there can be no pentagon in the hexagonal or triangular lattices.

If we attempt to apply this argument with triangles or hexagons, then the process simply leads back again to points in the original figure. But of course, since triangles and hexagons do arise in these grids, we didn’t expect the construction to work with them.

But also, this particular two-step rotation construction does not succeed with the heptagon (7-sides) or larger polygons, since the resulting point $c$ ends up outside the original heptagon, which means that the new heptagon we construct ends up being larger, rather than smaller than the original.

Fortunately, for the 7-gon and higher, we may fall back on the essential idea used in the square lattice case. Namely, because the interior angles of these polygons are now larger than $120^\circ$, we may simply rotate each side of the polygon by $120^\circ$ and thereby land at a lattice point. In this way, we construct a proportionally smaller instance of the same regular $n$-gon, and so there can be no smallest instance of such a polygon.

In summary, in every case of a regular polygon except the equilateral triangle and the regular hexagon, we found that by means of $120^\circ$ rotations we were able to find a strictly smaller instance of the polygon. Therefore, there can be no instances of such polygons arising from lattice points in the hexagonal or triangular tilings. QED

See my earlier post: there are no regular polygons in the integer lattice, except squares.

There are no nondegenerate regular polygons in the integer lattice, except for squares

Posted on December 4, 2016 by Joel David Hamkins

Consider a regular square lattice, formed by a grid of intersection points of uniformly spaced parallel horizontal and vertical lines in the plane. One may easily find points that form the vertices of a square in this lattice.

But can one find points that form the vertices of regular hexagon? Or a regular pentagon? How about an equilateral triangle? And so on.

The answer is that one cannot find these shapes using vertices in the square lattice.

Theorem. There is no regular pentagon in the square lattice, and no regular hexagon, no regular heptagon, and so on. Indeed, the only nondegenerate regular polygons to be found using vertices in a square lattice are squares themselves.

I think this must be a classical result. I was inspired by Vaughn Climenhaga’s beautiful image in his Proof without words answer on MathOverflow, which handled the case of a hexagon. Reflecting upon it, I realized that the same idea works with other regular polygons, and I endeavored to produce the corresponding images, below.

Proof. Suppose that we could find five vertices in the square lattice that formed a regular pentagon. Construct on each side a perpendicular of the same length, as pictured in brown below. Since the lattice is invariant under $90^\circ$ rotations centered at a lattice point, each of these new points is still a lattice point. And by symmetry, they form the vertices of a (proportionally smaller) regular pentagon. Therefore, there can be no regular pentagon at all in the square lattice, since if there is one, it is clear that there would have to be a smallest instance.

An alternative way to argue is: by similarity, the size of the smaller pentagon shrinks by the same factor with each step, and so in the limit the size approaches zero; but clearly, we cannot have a lattice-point regular pentagon whose size is smaller than the square lattice spacing itself. So there is no regular pentagon in the square lattice.

The same argument works with larger regular polygons. The main point to realize is that for all regular $n$-gons, where $n>4$, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular $n$-gon is strictly smaller than the original. This completes the proof for all $n$-gons for $n>4$.

The case of an equilateral triangle requires special care. If one attempts the same construction idea as above, building the perpendicular on the edges of a triangle, then the resulting triangle becomes larger, rather than smaller, and so the proof method breaks down.

Nevertheless, one can reduce the equilateral triangle case to a hexagon: if you could find an equilateral triangle in the square lattice, then since the lattice is invariant under translation via a lattice-point line segment, it follows that we could build a regular hexagon. But we have already showed that we cannot find a regular hexagon in the square lattice, and so we cannot find an equilateral triangle.

triangle-hexagon

So we’ve completed the proof for all nondegenerate regular polygons, except the square. QED

For those who might be interested, here is the tex code I used to generate the nested polygons. The code accepts input $n$ to produce a regular $n$-gon with the perpendiculars constructed.

\documentclass{minimal} \usepackage[rgb]{xcolor} \usepackage{tikz} \usetikzlibrary{positioning,calc} \usepackage{ifthen}

\begin{document}
\newcommand\nestedpolygon[1]{ \begin{tikzpicture}[scale=4] \pgfmathsetmacro\n{#1} \pgfmathsetmacro\m{\n-1} \pgfmathsetmacro\shrink{sqrt((sin(180/\n))^2+(cos(180/\n)-2*sin(180/\n))^2)} \pgfmathsetmacro\OffSet{acos((sin(180/\n)^2+cos(180/\n)*(cos(180/\n)-2*sin(180/\n)))/\shrink)} \pgfmathsetmacro\gc{360+100*exp((5-\n)/3.5)} \pgfmathsetmacro\gcc{640+120*exp((5-\n)/3)} \definecolor{GonColor}{wave}{\gc} \definecolor{GonColorC}{wave}{\gcc} \pgfmathsetmacro\d{8} \foreach \k in {0,...,\d} { \pgfmathsetmacro\R{\shrink^\k}; \pgfmathsetmacro\c{100*exp(-\k/6)}; \pgfmathsetmacro\a{2*\R*sin(180/\n)} \pgfmathsetmacro\f{\k*\OffSet} \foreach \i in {0,...,\m} { \coordinate (x) at (360*\i/\n+\f:\R); \coordinate (y) at (360*\i/\n+360/\n+\f:\R); % the perp indicator \ifthenelse{\k=0\AND\i=\m\AND\n<20}{\draw[very thin,black!\c!white] ($(x)!.85!(y)$) node (p) {} -- ($(p)!1!90:(y)$) node (q) {} -- ($(q)!1!90:(p)$);}{} % connecting to lower level \draw [line width=.5*exp(-\k/2),GonColorC!\c!white] (y) -- ($(y)!1!-90:(x)$); % edge and vertices of current gon \draw [line width=1.5*exp(-\k/2),GonColor!\c!white] (x) node [circle,scale=.3*\R,black!\c!white,fill=black!\c!white] {} -- (y); }

\draw (\f:\R) node [circle,scale=.3*\R,black!\c!white,fill=black!\c!white] {}; } \end{tikzpicture}}

\foreach\n in {5,6,7,8,9,10,12,16,20} { \eject $$\nestedpolygon{\n}$$ }
\end{document}

See my follow-up post on the regular polygons arising in the hexagonal and triangular lattices.

A lifetime of hot air

Posted on November 14, 2016 by Joel David Hamkins

We’ve been making some Fermi estimations in the Math for Liberal Arts class I am teaching, and today we discuss the following question:

Question. If you collect all the hot-air that you have breathed in your life, what would the volume be? If you made a hot-air balloon, would it be able to lift you and all your possessions?

To answer, let’s start with the first part. How much do I breathe? If I imagine inhaling and then exhaling a deep, big breath, I figure I could inflate a small paper bag, perhaps well over one liter, but probably not as much as two liters. But my passive resting breathing is probably much less than a big deep breath. So let’s figure a half liter per ordinary passive breath. How often do I breathe? Well, in the swimming pool, I can hold my breath under water for a minute or even two minutes (in my younger swim-team days); but if I hold my breath right now, I can say that it does start to feel a little unnatural, like I should take my next breath, even after just about five or ten seconds, even though this impulse could be resisted longer. It seems to me that my body wants to take another breath in about that time. If we breathe every five seconds, that would mean 12 breaths per minute, so let’s say ten breaths per minute, which would mean a volume of 5 liters per minute. That makes $5\times 60=300$ liters per hour, or $300*24=7200$ liters per day. In a year, this would be $7200\times 365$, which is less than 7000 times 400, which is 2,800,000 liters per year. Let’s say 2.5 million liters per year of hot air. Times 50 years would make $125$ million liters of hot air in all.

Now, each liter of air fills a cube 10 cm on a side, and one thousand of these fit into a cubic meter. So we’ve got 125,000 cubic meters of hot air. This is the same as a cube 50 meters by 50 meters by 50 meters. That is my hot-air balloon! Filled with a lifetime of hot air. (This is considerably larger, more than one hundred times as large, as a typical recreational hot-air balloon, which I understand are usually under 1000 cubic meters. From this point of view, it would seem likely that it could lift me and all my possessions, although body temperature may be much less than is achieved in those balloons.)

If the air was at my body temperature ($98.6^\circ$ F), then would it be able to lift me and all my possessions? Well, let’s see how much it would lift. Hot air expands in proportion to temperature (from absolute zero). If it is a day like today, about $50^\circ$ degrees F, which is about a 50 degree F difference, and absolute zero is minus 460 F. So this is about a 10 per cent increase in temperature. (In metric: we have body temperature of about 37 degrees, and it is about 10 degrees Celsius today, so a difference of 27 degrees, and absolute zero is minus 270, so about ten per cent increase, as I had said.)

So the heat of the hot air caused it to expand in volume by ten percent. The buoyant force of the hot-air balloon is exactly the weight of this displaced air, by the Archimedean principle. Thus, the lifting force of my hot-air balloon will be equal to the mass of air filling ten percent of the volume we computed. How much does air weigh? I happen to remember from my high school science class that one mole of air at one atmosphere of pressure is about 22 liters (my teacher had a cube of exactly that size sitting on his desk, to help us to visualize it). And I also know that air is mainly nitrogen, which forms the molecule $N_2$, and since nitrogen in the periodic table has an atomic number of 14, the molecule $N_2$ has a mass of 28 grams per mole. So air weighs about 28 grams per 22 liters, which is about 1.3 grams per liter. Each cubic meter is one thousand liters, and so 1.3 kilograms per cubic meter (this is much larger than I had expected—air weighs more than one kilogram per cubic meter!). My hot air in total was 125,000 cubic meters, and we said that because of the temperature difference, the volume expanded by ten percent, or 12,500 cubic meters. This expansion would displace an equal volume of air, which weighs 1.3 kg per cubic meter. Thus, the displaced air weighs $12,500\times 1.3\approx 16,000$ kilograms, or about 16 metric tons. So all my hot air, at body temperature in a giant hot air balloon on a chilly day, would have a lifting force able to lift 16 metric tons.

Would this lift me and all my possessions? Do I own 16 tons of stuff? Well, thankfully, I don’t own a car, which would be a ton or more by itself. But I do own a lot of books, a piano, an oven, a dishwasher, some heavy furniture, paintings, and various other items, as well as a collection of large potted plants on my terrace. It seems likely to me that I could fit most if not all my possessions within 16 tons. To gain a little confidence in this, let’s estimate the mass of my books. My wife and I have about twelve large shelves filled with books, each about 2 meters, and then I have about 3 more such shelves filled with books in my offices at the university. If we count half of the home books as mine, plus my office books, that makes 9 shelves times two meters, for about 20 meters of books. If the books are about 25 cm tall on average, and 15 cm deep, that makes $20\times .25\times .15=.75$ cubic meters of books. Let’s round up to one cubic meter of books. How much does a cubic meter of paper weigh? Well, one ream of copy paper weighs about 2 kilograms, and that is a volume of 8.5 by 11 by 2 inches. One meter is about 33 inches, and so we could fill one cubic meter with a pile of reams of copy paper 3 by 3 by 17, which would be 163 reams, or about 300 kilograms. So not even a half ton of books! So I can definitely lift all my most important possessions within the 16 tons.

Final answer: Yes, if we filled a giant balloon with all the hot-air I have breathed in my life, at body temperature, then it would lift me and all my possessions.

On number sense — Would one day of NYC coffee fill the Statue of Liberty? And other fun questions…

Posted on November 8, 2016 by Joel David Hamkins

Today I gave a lecture on what I call number sense, using a process of estimation and approximation in order to calculate various unknown quantities, including a few fantastical ones. How much coffee is made per day in New York City? Would it fill up the Statue of Liberty? Approximately how many babies are born in New York City each day? If you made a stack of quarters to reach the distance to the moon, what would the dollar-value be? If you piled those quarters in a heap, would it fit in Central Park? How much does the Empire State Building weigh?

These kinds of back-of-the-envelope calculations, in my view, have at their heart the idea that one can solve a difficult or seemingly impossible problem by breaking it into more manageable pieces. We don’t just pull a final answer out of the air, but rather make simplifying assumptions and informed estimates for related quantities that we are more familiar with or have knowledge about, and then use that information to derive a better estimate for the main quantity. For most of these questions, at the outset we may have little idea what would be a reasonable answer, but by the end, we gain some insight and find ourselves a little closer to the truth.

These kind of calculations are also known as Fermi estimations, in light of Fermi’s remarkable ability to make surprisingly accurate estimations on the basis of little or no hard data. The wikipedia page (thanks to Artie Prendergrast-Smith for mentioning this link in the comments) emphasizes that even in a case where the estimate is significantly off the true value, nevertheless we may still find value in the Fermi calculation, because it focusses our attention to the reasons for the divergence. In discovering which of the assumptions underlying our calculations was wrong, we come to a deeper understanding of the true situation.

In the lecture, I began with some very easy cases. For example, how many seats were in the auditorium? The students estimated that there were approximately 12 seats per row and about 10 rows, so 120 seats in all. How old was one of the students, in seconds? Well, he was 18 years old, and so we could simply multiply each year by 365 days, times 24 hours per day, times 60 minutes per hour, times 60 seconds per minute, to get
$$18\times 365\times 24\times 60\times 60\approx 600,000,000 \text{ seconds}.$$
One student objected about leap days, since 365 should be 365.25 or so. But I pointed out that this difference was not as important as it might seem, since already we had made far larger rounding assumptions. For example, the student was not exactly 18 years old, but 18 years old and some several months; by using 18 years only, we made a bigger difference in the answer than caused by the leap-day issue, which would be a difference of only five days or so over 18 years. For the same reason, we should feel free to round the numbers to make the calculation easier. We are aiming at a ballpark estimate rather than an exact answer.

Let’s now do some more interesting cases.

Coffee in New York. How much coffee is made each day in New York? Would it fill the Statue of Liberty? First, let me say that I really don’t have any definite information about how much coffee is made each day in New York, and I fear that my own coffee-obsessed perspective will lead me to over-estimate the amount, but let’s give it a try. New York City has a population of approximately 10 million people. Some of those people, like myself, drink a large amount of coffee each day, but many of the others do not drink coffee at all. I would think that a sizable percentage of the NYC population does drink coffee, perhaps as much as a third or even half consumes coffee daily. Many of those coffee-drinkers have more than one cup per day, and also surely more coffee is made than consumed. So it seems reasonable to me to estimate that we have approximately one medium cup of coffee per person on average per day in New York. Basically, we’re saying that the heavy coffee drinkers and the made-but-not-sold coffee approximately makes up for those who abstain, making the average about one cup per person. So we are talking about 10 million cups of coffee per day. A medium cup of brewed coffee at Starbucks is I think about 12-16 ounces, a little less than a pint, and so let’s say about 3 cups per liter. This amounts to roughly 3 million liters of coffee.

Would it fill the Statue of Liberty? The statue itself is, I estimate, about twenty stories tall, counting the base, and if each story is 15 ft, or 5 meters, that would mean 100 meters tall, counting the base. But I think that the base is about half the height, so let’s say 50 meters for the actual statue itself. I’ve never been inside the statue, but my students say that it is about 10 meters across inside, a little more at the bottom than near the top. If we approximated it as a rectangular solid, that would give a volume of $10\times 10\times 50$ cubic meters, or 5000 cubic meters. But since the statue tapers as you go up, particularly in the arm holding the torch, it really is more like a cone than a rectangular solid, and so we should divide by three. But let’s divide just by two, because she isn’t quite as tapered as a cone. So the Statue of Liberty has a volume of approximately 2500 cubic meters. One cubic meter can be thought of as a 10 by 10 by 10 array of little 10cm cubes, and each of those is exactly one liter. So a cubic meter is 1000 liters, and therefore the Statue of Liberty has a volume of $2500\times 1000=2.5$ million liters. But since we had 3 million liters of coffee, the answer our calculation arrives at is: Yes, one day’s worth of New York coffee would fill up the Statue of Liberty!

Well, we do not have perfect confidence in our estimates and assumptions — for example, perhaps there are many fewer coffee drinkers in New York than we estimated or perhaps we underestimated the volume of the Statue of Liberty. Since the estimated volumes were of basically similar magnitudes, we aren’t really entitled to say that definitely the coffee would fill up the Statue of Liberty. Rather, what we have come to know is that those two volumes are comparably similar in size; they are in the same ballpark.

Elevator trips. While riding downtown last weekend with my son on the subway, a crowded 4 train, we overheard the group standing next to us talking about elevators. One lady said, “My elevator company serves as many elevator trips in New York City in five days as the population of the entire world,” and the rest of her group, impressed, nodded affirmatively in reply. But my thoughts, upon hearing that, were to make a quick calculation. Suppose all 10 million NYC residents rode an elevator 10 times every day, which is way too high (probably one trip per person per day is more reasonable, since many people live and work in buildings without elevators). Even in this extreme case of ten trips per person per day, it would mean only 100 million trips total per day, or 500 million trips over 5 days. This is much less than the world population, and so no way is that person’s claim true, especially since there are also many elevator companies. I thought of mentioning my calculation to those people on the subway, but decided against it. Walking out of the subway in the East Village, however, I asked my son (14 years old) whether he heard those people talking about elevators, and he replied, “Oh, yes, and when they said that, I calculated it in my head: no way is that true.” He then proceeded to explain his calculation, similar to mine. Yay, Horatio!

The Chicago marathon. In the run-up to the Chicago marathon this year, on a route that would wind through the windy city streets, Newsweek magazine reported, “Chicago Marathon organizers expect 1.7 million fans to line the route this year.” (Thanks to the critical math commentary of Mark Iris for bringing this example to my attention.) The organizers had emphasized the economic impact of these spectators, many of whom would presumably be eating in Chicago restaurants and staying at Chicago hotels. But is this a reasonable number?

Let’s calculate. A marathon is approximately 26 miles, and the route has two sides for spectators, so let’s round it to 50 miles of spectator viewing spots. Each mile is about 1800 yards, so we have $50\times 1800=90000$ yards of viewing spots. Each spectator, standing shoulder-to-shoulder, with all their stuff, takes up about a yard of space. So if the marathon was lined on both sides for the entire route with spectators standing shoulder-to-shoulder, this would amount to about 90,000 spectators. In order to have 1.7 million spectators, therefore, they would have to be lined up behind each other. But even if the spectators were 10 people deep on each side for the entire route, which is a vast crowd, it would still amount only to 900,000 people. We would have basically to double this to get to 1.7 million. So, if the live event really had 1.7 million spectators lining the route, then it would mean that the race was lined 20 people deep on each side for the entire route. No way is this number correct! I have never had the chance to attend the Chicago marathon, but at the New York marathon, which I assume is comparable, I know that while there are thick crowds at the finish line in Central Park and at some of the other prominent or especially interesting race locations, most of the rest of the route is much thinner, and at the typical nothing-special location, one could expect easily to have a front-row spot.

How many bricks on the college campus? Our campus, covered with some lovely woods and green meadows, has a number of brick Georgian buildings. Most of these are the same standard size as the mathematics department, but there are also some larger buildings such as the library, the performing arts center, the campus center and the gymnasium. At my lecture, the students and I agreed that altogether it amounted to about 30 buildings of the size of the math building. This building is approximately 30 meters by 70 meters, which would make a perimeter of 200 meters, but because the building has wings and is not purely rectangular, let’s add another 100 meters for the folds in the walls, so about 300 meters around the base of the building. The building is two stories tall, about 10 meters tall, making the area of the walls about 3000 square meters. Of course, there are windows and doors that cut out of these walls, but let’s say that they are approximately accounted for by the extra bricks used in the trimming flourishes at the corners and top of the building. So we have 3000 square meters of brick. Each brick is approximately 10 cm by 30 cm, and so one square meter of brick would have about ten rows of a little more than three bricks across, or about 20 bricks. So each building has about $3000\times 20=60,000$ bricks. And with 30 buildings, this amounts to $30\times 60,000=1,800,000$ bricks in the buildings on campus. There are also some bricks in the central fountain, a few small brick walls and some bricks lining some of the walkways. So let’s add another ten percent for those extra bricks, arriving at about 2,000,000 bricks on campus in all.

Positive test result for a rare disease. Suppose that as part of your check-up, your doctor randomly administers a clinical test for a certain rare medical condition. The test is 99 percent accurate, in terms of false positives and false negatives, in the sense that 99 percent of people taking the test have an accurate result with the test. Suppose also that the disease itself is rare, held by only one in a million people. If your test comes back positive, what is the likelihood that you actually have the disease?

This is a subtle question. It might seem to make sense initially that there is a 99 percent chance that you have the disease, since that is the accuracy of the test. But this isn’t actually right, because it doesn’t account for the fact that the disease itself is extremely rare, and so the total number of false positives will actually far outweigh the true positive results. For example, suppose that one million people take the test. About one of these people actually has the disease, and that person is 99 percent likely to test positive. So we expect about one true positive result. And for the others, who don’t have the disease, 99 percent of them will test negative. In other words, about 990,000 people will test negative. The remaining 10,000 people, however, one percent of the total, will have false positive results! So you are in this group of 10,000 people who tested positive, with only one of them actually having the disease. So the odds that you actually have the disease are only about one in ten thousand.

Quarters to the moon. How many quarters would you have to stack to reach the moon? How much would it be worth? how much would it weigh? More than the earth? More than the Empire state building? If you put all those quarters into a pile, would it fit in Central Park?

Well, OK, the question is a little absurd, and there are all kinds of problematic issues: we couldn’t ever actually make such a tower of quarters, as it would topple over; it doesn’t make sense to build a tower to the moon, since the moon is in orbit around the earth and it is moving; the quarters would begin to orbit the earth themselves, or fall back down to the earth or to the moon. But let’s just try to ignore all those problematic issues, and just try to answer how many quarters it would take to make a pile that high.

How far is the moon? I don’t really know, and we could look it up to see what the astronomers say about it, but that would go against the back-of-the-envelope spirit of these problems. So I am just going to use two facts that I picked up years ago: first, I know that the speed of light is about 300,000 km per second; and second, I happen to know that it takes radio signals traveling at the speed of light about one second to get to the moon (eight minutes to the sun), and so the moon is about one light-second away. I remember this fact from having learned it in my childhood, because it was important in a science fiction story I had read, in which radio communication from earth to the moon base had a two-second delay: you send a message, it takes a second to get there, and then a second for the answer to come back. So the moon is one light second away, and light travels 300,000 km in one second. So the moon is about 300,000 kilometers distant (actual value: 387400 km).

Let’s now stack up the quarters. By eyeballing a quarter I had in my pocket, I’d say each quarter is about 2 mm thick, which means 500 quarters per meter, or 500,000 quarters per km. So altogether, we have $300,000\times500,000=150,000,000,000$ many quarters in the stack. A hundred and fifty billion quarters to the moon! It takes four quarters to make a dollar, so that is worth about $40 billion dollars, which might be less than you would have expected initially.

Let’s put those quarters in a big pile, packed as tightly as possible. Each quarter is a little over 2 cm in diameter, and so the volume of the rectangular solid bounding the quarter is a little more than 2 cm by 2 cm by 2 mm, or about 800 cubic mm, which we can round up to 1 cubic centimeter. That makes sense, because we can imagine folding a quarter up into a little 1 cm cube. So 1000 quarters takes up about one liter. Thus, our quarters-to-the-moon have a volume of about 150 million liters. There are 1000 liters in a cubic meter, and so this is 150,000 cubic meters. Since a cube 10 meters on a side, has a volume of 1000 cubic meters, we can fit all those quarters into 150 such cubes. I can easily imagine a giant art installation, with 150 such 10 meter cubes placed in Sheep Meadow in Central Park. They would definitely fit.

How much would the quarters weigh? Some of the students in my lecture had worked as cashiers, and so they were familiar with quarter rolls from the bank, which fit in your palm and have ten dollars worth of quarters, or forty quarters. I can imagine a quarter roll in one hand balancing the weight of a half-pound of gouda cheese at the deli in my other hand. So 40 quarters is .5 pound, which makes 100 quarters about 2.2 pounds, which is about 1 kg. We had 150 billion quarters, which is 1.5 billion times 100 quarters, so 1.5 billion kilograms.

Since we already imagined the quarters in Sheep Meadow in those 150 large boxes, clearly they weigh far less than the earth and perhaps comparable to the Empire State Building.

How much does the Empire State Building weigh? After posing this question, I found out that it is also evidently a famous Google interview question, and you can easily find other blog posts about it. But let me tell you how I thought about it. The Empire State Building is about 100 stories tall, and if we assume 12-15 feet per floor, or about 4 meters, this makes the total height (without tower) about 400 meters. The CUNY Graduate Center where I work is across the corner on Fifth Avenue and 34th street, and I have walked past the Empire State Building many times. I estimate that the base is about 75 meters on Fifth Avenue by 150 meters on 34th Street. But because of the step-backs, the middle part of the tower is considerably less than that, probably 30 by 80 near the top. Let’s say the average cross section of the building is 50 by 100 meters. Now, how much does that weigh? Of course, there is a lot of steel and masonary in the floors and walls, and concrete floors, and also all the contents of the building, with desks and paper files and books and interior walls and doors. Those things are heavy. I really have no idea how much those things weigh, but let me imagine that we build a virtual copy of a complete floor from the Empire State Building, without any of those floor beams or concrete or desks or walls or anything, and instead we flood that virtual room with water, until it has the same weight. Of course, the metal and concrete and masonary is much heavier than water, but the actual space is mostly empty air. How much water would it take to have the same weight? Well, let me just guess that we’d have to flood the virtual room about one-quarter deep with water to have the same weight as the actual materials. At 50 by 100 meters by 4 meters tall for each floor, this would mean a pool of water 50 by 100 meters, one meter deep, for a volume of 5000 cubic meters of water. We already said that one cubic meter of water is 1000 liters, each of which weighs one kilogram, so we are talking about $5000\times 1000$ kilograms or 5 million kilograms per floor. With 100 floors, this is 500 million kilograms, or 500,000 metric tons. So according to this estimate, the Empire State Building with all its contents weighs about 500,000 metric tons.

So it weighs less than the quarters to the moon (1.5 billion kilograms) —the Empire state building weighs about one-third as much as the quarters to the moon — but these two quantities are in the same ball park.

How many babies are born each day in New York City? The population of New York is approximately ten million people. And those people live about 70 years. Let’s imagine spreading those births out uniformly over the 70 year period. That means one million people born in 7 years. This is about 150,000 people born in one year, which is about 3000 births per week, or about 400 births per day. This estimate would be accurate if the city population were in a steady state, with births balancing deaths, but of course the population is increasing. On the other hand, most of that increase is from people moving here, not just being born here. But then again, the people moving to New York I expect are mainly young adults looking for a career, and so they will contribute to a heightened birth rate as they have children. Shall we add 25 percent more to account for the fact that the city is growing and not in a steady state? In that case, our estimate is that there are about 500 births each day in New York City.

There is an endless supply of such questions that can be calculated. I have been talking about them in the lectures for my course Math for Liberal Arts, an undergraduate course aimed at non-math students at my college. In this course, we are fairly free to cover some imaginative topics, and I’ve covered some game theory, some elementary graph theory, a little bit logic, and now this number sense topic, which I use to try to develop the students ability to attack a technical problem by breaking it down into more managable problems. But meanwhile, I also look upon it just as plain fun.

So here are a few more questions that you might enjoy thinking about. And please make your own.

What is the total volume of air that you have breathed in your lifetime? If you filled a balloon with all your hot air, how big would it be?
If you used that balloon as a hot-air balloon (with the hot air at your body temperature), would it have enough bouyency to lift you and all your possessions? See my post A lifetime of hot air for a detailed answer.
How many NYC metro cards exist? (Each lasts about a year or two; they get thrown out, but still exist in the landfill; the system has used metro cards for about twenty years; there is a large stock of new not-yet-used cards.)
How many doorknobs are there in your building?
How many subway tiles are there in the NYC subway system?
How many riders were on my crowded 4 train downtown this morning?
How many lightswitches are there on your university campus?
How much water does NYC use each day?
What are the odds that you drank a molecule of water that was once also drank by Julius Caesar?
How many people have there been? What fraction are currently alive?

Please try to figure them out, and post your solutions in the comments below!

Math for nine-year-olds: fold, punch and cut for symmetry!

Posted on February 22, 2016 by Joel David Hamkins

IMG_20160222_071711

Today I had the pleasure to visit my daughter’s fourth-grade classroom for some fun mathematical activities. The topic was Symmetry! I planned some paper-folding activities, involving hole-punching and cutting, aiming to display the dynamism that is present in the concept of symmetry. Symmetry occurs when a figure can leap up, transforming itself through space, and land again exactly upon itself in different ways. I sought to have the students experience this dynamic action not only in their minds, as they imagined various symmetries for the figures we considered, but also physically, as they folded a paper along a line of symmetry, checking that this fold brought the figure exactly to itself.

The exercises were good plain fun, and some of the kids wanted to do them again and again, the very same ones. Here is the pdf file for the entire project.

To get started, we began with a very simple case, a square with two dots on it, which the girls folded and then punched through with a hole punch, so that one punch would punch through both holes at once.

Fold punch and cut-page-001

Fold punch and cut-page-002 Next, we handled a few patterns that required two folds. I told the kids to look for the lines of symmetry and fold on them. Fold the paper in such a way that with ONE punch, you can make exactly the whole pattern.

Don’t worry if the holes you punch do not line up exactly perfectly with the dots — if the holes are all touching their intended target and there are the right number of holes, then I told the kids it is great!

Fold punch and cut-page-003 The three-fold patterns are a bit more challenging, but almost all of the kids were able to do it. They did need some help punching through, however, since it sometimes requires some strength to punch through many layers.

Fold punch and cut-page-004 With these further patterns, some of the folds don’t completely cover the paper. So double-check and make sure that you won’t end up with unwanted extra holes!

The second half of the project involved several cutting challenges. To begin, let’s suppose that you wanted to cut a square out of the middle of a piece of paper. How would you do it? Perhaps you might want to poke your scissors through and then cut around the square in four cuts. But can you do it in just ONE cut? Yes! If you fold the paper on the diagonals of the square, then you can make one quick snip to cut out exactly the square, leaving the frame completely intact.

Fold punch and cut-page-005

Fold punch and cut-page-006 Similarly, one can cut out many other shapes in just one cut. The rectangle and triangle are a little trickier than you might think at first, since at a middle stage of folding, you’ll find that you end up folding a shorter line onto a longer one, but the point is that this is completely fine, since one cut will go through both. Give it a try!

Fold punch and cut-page-007

Here are a few harder shapes, requiring more folds.

Fold punch and cut-page-008 It is an amazing mathematical fact, the fold-and-cut theorem, that ANY shape consisting of finitely many straight line segments can be made with just one cut after folding. Here are a few challenges, which many of the fourth-graders were able to do during my visit.

What a lot of fun the visit was! I shall look forward to returning to the school next time.

In case anyone is interested, I have made available a pdf file of this project. I would like to thank Mike Lawler, whose tweet put me onto the topic. And see also the awesome Numberphile video on the fold-and-cut theorem, featuring mathematician Katie Steckles.

See more of my Math for Kids!

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

Posted on May 22, 2015 by Joel David Hamkins

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! You can also see my solution and further discussion.

Joel David Hamkins

mathematics and philosophy of the infinite

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