So here’s the puzzle:

Take the Myer’s polyhex tile:

Can you:

1. Find rules to avoid easy draws (for example wandering off in one direction to infinity).
2. Find a winning strategy?

I personally have no idea, so this is a challenge problem!

If you have access to a Laser Cutter or other fancy computer device the cut files are on Thingiverse.

So, I’m teaching a new course, Math 2033, Mathematical Thought, and it’s going great! I’d like to take a moment to write about it!

(This is one reason the MF has been kinda slow lately; another is that I’m chair) When it’s fully up and running, we’ll have about 150 students in one large section each semester (we’re starting with about 100). In a nutshell, it’s the Math Factor, as a course.

So, the list of topics is pretty familiar; from the podcast you are pretty well acquainted with the kinds of things I like to share: game theory, encryption, a little number theory, theory of computation & godel’s theorem, cardinality/infinity; plus more visual hands on things like topology, graph theory, symmetry, four-dimensional geometry, and so forth, and some baby programming in a playground IDE (scratch.mit.edu)

The real thing though is that my co-teacher and I have taken a kind of radical approach to the structure of the course—and our crazy ideas are working out great! In a nutshell, the students are guaranteed a C just for showing up and doing what they’re told (more on that in a sec), but to get an A, they have to become active collaborators in the building of the course, adding to the long-term infrastructure. I didn’t expect to give out very many A’s at all, but a surprising number of students seem up to the challenge. Part of the point is that this reward structure aligns the interests of the course directly with the interest of the students. Another nice thing is that it is much more like the reward structure of Real Life, far more so than most academic experiences: you can coast and do ok, but to really succeed, initiative and imagination are required. Interestingly, 20% of the class can’t even rise to the minimal standard of showing up, and will fail. 

Another nice thing is that students can bring to bear any of their own interests and abilities; we need such a wide range of things done—photography, writing, editing, leaders on our discussion board, organizers, all kinds of stuff. It is in fact possible to get an A by dragooning other students into harvesting, trimming and delivering a huge load of bamboo for some math sculptures the class will be making soon. The fact is, I have large ambitions for this, and no way to do more than a fraction of the work; students that help bring this off will be the ones that get an A.

SO, how does it actually work? The basic daily rhythm is that we give a lecture, usually with some sort of hands-on fun and games component. We then post a prompt or two on the class discussion board (hidden to the outside world). The students have 24 hours to post, and then another 24 hours to comment on each others ideas. This is the real heart of the course and the activity has steadily grown, reaching 1700 posts a couple of weeks ago. Wow!  (I shouldn’t exaggerate this though: some are really into it, many are trying to get by with as little as possible. I am aiming for a culture where slacking is gently disgraced, and we’re on track to get there)

As you can guess, this has completely lifted out of my ability to monitor; we have a number of ways this is digested and managed. For example, about half a dozen of the more thoughtful students are responsible for reading all of posts and trying to raise the level of discourse, and for creating useful summaries of the best ideas.

BUT that’s just the “internal” part of the course. Externally, open to the world, is a wiki, math2033.uark.edu which is going pretty well. I view this as a multi-year project, so this is a pretty good start. Most of what you see there is the product of about twenty students, and a few really do A LOT of work, including having developed the basic organizational framework. (So they get A’s for sure)

I’ve been spending a lot of time developing solid materials for use in the course, such as this sample handout, on the Halting Problem.

The students seem pretty pumped. It’s working!!! 

(We passed an important milestone last week; several students told me they ended up fooling around way too much with one of the optional assignments, messing them up in other classes! Perfect!)

If every guest finds their book then the whole group win a trip to Paris.

What is their best strategy?

This puzzle comes from a Peter Winkler book, Mathematical Mind-Benders.   It was called Names in Boxes and was about prisoners trying to escape the death penalty.

This is the guests best strategy:

Associate each box with one of the guests.  Each guest first opens their own box.  If they find their own book they can stop.  Otherwise they open the box associated with the owner of the book.  They keep doing this until they find their own book or have opened 10 boxes.

Incredibly this gives them more than a 33% chance of success.  Even more incredibly this strategy will always have a better than 30% chance of success regardless of the number of guests.

To see how this works lets imagine that Stan repeats the procedure forever.  At some point he will find his book.  Next he will open his own box again.  He will repeat the same loop endlessly.  

If this loop contains up to ten boxes then all of the guests whose boxes lie in the loop will find their book.

The procedure splits the boxes into separate loops.  If no loop is bigger than ten then all of the guests will succeed, otherwise most of them will fail.

It is still true each individual guest has a fifty-fifty chance, the procedure means that they tend to succeed or fail together.

So what is the chance of success?

We have come across these sorts of loops before in Follow Up: Loops and the Harmonic Series. In the comments we noted that on average there will be 1/r loops of length r. As we are interested in loops containing more than half the boxes there can only be one such loop, so we can say the chance of a loop of length r is 1/r.

First we calculate the number of loops of length r if we have n boxes, in the puzzle n =10.  There are n!/r!(n-r)! ways of choosing r boxes and there are r! ways of ordering them.   We could have chosen any box in the loop to be the first one so to avoid double counting we must divide by r.  This gives (n!/r!(n-r)!)r!/r = n!/r(n-r)!

The other n-r boxes could be arranged in (n-r)! ways.  In total there are n! arrangements of all n boxes.  So a random arrangment will have on average (n-r)!/n! occurrences of a particular loop.  

A random arrangement will have on average (n!/r(n-r)!)((n-r)!/n!) = 1/r loops of length r.

Since there can only be one loop larger than 10 this means the chance of success is 1 -1/11 -1/12 -… -1/20 which is about 0.331229…

As the number of guests increases this tends to 1 – integral[ N<x<2N ]( 1/x ) = 1 – ln 2N + ln N = 1 – ln 2 = 0.3068528…

However many guests there are they will always have a better than 30% chance of success!

You have eight batteries and know that four are good and four are dead, but don’t know which are which.  Your only method of testing them is to insert two into a device that will work if you’ve put in two good batteries and not otherwise.  How many such “tests” are required in order to be sure that you’ve located two good batteries?

As of this posting, the answer to this question is not yet on the POTW website, but if you come to this later, the spoiler may be there, so be careful to avoid spoilers if you want to work this through.

Janine is one of twenty guests at a Christmas party.  Each guest is given a book as a present.  Janines’s book is called ‘Living with Crazy Buttocks’.  She isn’t sure what to make of that.

The guests are invited to play a game.  Each book is put into an identical cardboard box.  The boxes can be opened and closed without leaving a mark.  The twenty boxes are piled up around the Christmas Tree.

The guests are told that they will each have the opportunity to open half of the boxes.  Their objective is to find their own book.  If they all succeed the group wins and they will win a trip to Paris.  If any one of them fails then the group fails but they will each get a Twinkie to keep for life.

The guests are taken to another room and then taken to the tree one at a time.  They cannot see what any other guest does at the tree.  They are not able to communicate once  the game starts.  The boxes are put back after each guest, as though they had never been there.

You would think that the chance of the group succeeding was 1/2^20 but they can do much better than that.

The group must come up with a strategy before the game starts.  What is the best strategy to get the group to Paris, and let Janine keep her ’Crazy Buttocks’?

These books are all real.  They will be helpful if you have ever had any of the following thoughts:

We all know the Nazis killed millions of innocent people but what were they like on ecological issues?

I would like to speak Italian but can’t be bothered to learn any Italian words, can you help?

Aubergines are very flushed, just how angry are they?

I think I’m dead, how can I tell for certain?

I am rich but dead.  How should I pimp my coffin?

I am worried about running into large, slow moving objects; can you suggest any strategies to avoid this?

Just how boring was 1587?

I live thousands of miles from Versailles.  Will I get a good view?

I am English, am I human?

My buttocks are insane.