Industrious Dice |

I’m a professional mathematician. That means somebody pays me to do math. I’m also a recreational mathematician. That means you might have to pay me to get me to stop.

Wearing my recreational mathematician hat – quite literally, as you’ll see – I gave a talk earlier this year on some newfangled dice that do the same sorts of jobs as old-fashioned dice but with fewer pips (“pips” being the technical name for all those little black dots).

My G4G15 talk: “Industrious Dice, or Pip-Pip Hooray!”

Fewer pips? Why would anyone on earth care about that? Dice manufacturers certainly don’t; drilling the pips and coloring them in isn’t a major expense. But I’m not an applied mathematician. I’m a pure/recreational mathematician, and I’m allowed to care about anything I choose to. When I choose well, I find something new and interesting.

FEWER PIPS

The standard die (called a d6 in the gaming community) has one job: to generate a random number between 1 and 6. The d6 accomplishes this job in a straightforward fashion by having 1, 2, 3, 4, 5, and 6 pips on its six faces, using a total of 1 + 2 + 3 + 4 + 5 + 6 = 21 pips; because of the cube’s symmetries, each of the six faces has an equal chance of being the face that’s facing up when the die stops rolling.

But here’s a top view of a die that gets the same job done with far fewer pips:

Industrious Dice |
An industrious dodecahedron

This die is not a 6-sided solid but a 12-sided solid called a dodecahedron. To use this newfangled die, throw it, wait till it stops rolling, and count not just the pips on the top face but also the pips on the five surrounding faces; that is, count the pips on the six faces that are visible from directly above the die. For instance, in the picture above we see six pips, so the throw would have value 6. The pips are arranged in such a fashion that the value of a throw is equally likely to be 1, 2, 3, 4, 5, or 6.

When you roll such a die repeatedly, each face of the dodecahedron is an upper face one-half of the time. In contrast, when you roll a cubical die repeatedly, each face of the cube is the top face only one-sixth of the time. So the dodecahedral die is (from a pip-use perspective) three times as efficient as the cubical die (one-half is three times bigger than one-sixth), and it correspondingly does its job with only one-third as many pips: 7 rather than 21.

To design such a die, I had to solve a puzzle: Put non-negative integers into the twelve regions in the diagram below so that the twelve sums you get by adding each number to its five neighbors are 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6 in some order. (See Endnote #1 for one solution.) All but one of the regions are pentagons; the last one (the outer region) looks like a square with a pentagonal hole in it. To get you started, I’ve put the number 0 into the outer region.

A fill-in-the-numbers puzzle

FEWER DIMENSIONS

The link between my die-design problem and my fill-in-the-numbers puzzle goes by way of a form of polyhedral cartography pioneered by the 19th century German mathematician Victor Schlegel. He was a contemporary of the more widely-known mathematician and writer Charles Howard Hinton who coined the now-famous word “tesseract” to describe a four-dimensional analogue of the cube. Hinton and Schlegel worked in an era when the geometry of higher-dimensional space had seized the imaginations of mathematicians and non-mathematicians alike. Schlegel’s great contribution was finding a way to subdue n-dimensional hypersolids (or maybe just tame them a wee bit) by projecting them down into (n−1)-dimensional space. Such diagrams can help us navigate higher-dimensional objects. We’ll focus on the case n = 3 and see how Schlegel would have had us map the faces of the 3-dimensional dodecahedron into the 2-dimensional plane.

To picture Schlegel’s trick, imagine that it’s night-time in a modernistic children’s playground that features a dodecahedral climbing structure (something Hinton’s son Sebastian would have probably liked since Sebastian Hinton is now remembered as the inventor of the jungle gym). It’s not a solid dodecahedron but an airy assemblage of struts (thirty, if you care to count them). Imagine that you climb on top of the dodecahedron and shine a flashlight down on the structure beneath you, from a point just above the middle of the structure’s pentagonal top face; the shadows cast on the ground below form the Schlegel diagram.

A Schlegel diagram for a dodecahedron

(Here I’ve used thickness of the line segments to correspond to how close the respective struts are from the viewer’s perspective, or relatedly, how wide the shadows are.)

FINDING A STORY TO TELL

This dodecahedral die is an outgrowth of the Muñiz-Propp die I mentioned in my earlier essay “Let x equal x“. I decided a year ago that I wanted to give a talk on these two dice at an event held this past February called the Gathering for Gardner; it’s a biennial conclave of mathematicians, magicians, and other folks who admire the work of Martin Gardner, and since both mathematicians and magicians like dice, the topic seemed like a natural fit.

But then I thought, wouldn’t it be even better to do something similar involving dice with fifteen sides, since the 2024 meeting was the fifteenth of its kind?

It isn’t the first time I’ve given a talk at a Gathering for Gardner that saluted the ordinal number of that particular Gathering; for instance, I prepared a talk for the fourteenth Gathering based on Stan Wagon’s terrific article “Fourteen Proofs of a Result About Tiling a Rectangle”. That particular Gathering had to be rescheduled because of the COVID-19 pandemic, and I ended up not presenting a talk at the rescheduled (and virtual) fourteenth Gathering for Gardner, but I still wrote a Mathematical Enchantments essay based on the talk I’d meant to give: “What Proof Is Best?”

Unfortunately, there’s a problem with fifteen-sided dice. If you go to the web in search of a “d15” die, you’ll find they all come with the warning “This die is not isohedral” (or words to that effect). Isohedral polyhedra are polyhedra with the pleasant property that for any two faces F and G of the polyhedron, there’s a symmetry of the polyhedron that carries F to G. Cubes are isohedral, and so are dodecahedra. Isohedral convex polyhedra make excellent dice because they’re just as likely to land on any face as on any other. But there are no n-sided isohedral convex polyhedra when n is an odd number (like the number 15).

On the other hand, there are 30-sided isohedral convex polyhedra, and indeed one of them – the triacontahedron – is the basis of the standard d30 die. Since 30 is twice 15, it’d be easy to use such a die to generate a random number between 1 and 15: have two faces showing each of the numbers from 1 to 15. But I don’t want to use numbers on the faces; the whole point is to use pips, and to use them as “efficiently” as possible. Of course we could use 1+1+2+2+…+15+15 = 240 pips and just count the pips on the top face, but who wants a die that has 15 pips on some of its faces? Either the pips would be tiny or the die would be huge. Instead, when we roll such a die, we should count the pips on the top face and the four neighboring faces. This increases the efficiency by a factor of five, so the total number of pips is just 240/5 or 48 — assuming, that is, that such an arrangement exists.

A triacontahedron

As in the case of my dodecahedral die for simulating a d6, the problem of arranging pips on a triacontahedral die to simulate a d15 can be recast as the puzzle of filling in numbers in a Schlegel digram so that all the possible sums we want (in this case, all the numbers between 1 and 15) are equally represented. But now there are 30 variables instead of just 12, and I found that my brute-force approach to solving the puzzle, which had succeeded for the dodecahedron, was too slow to yield even a single solution. So I reached out for help.

As it happens, I’m not the only person I know who walks the line between pure math and recreational math – and who, in the process of walking the line, tries to blur it a little. For several decades Rich Schroeppel has run an email forum called math-fun for folks like me who don’t draw a sharp distinction between the two kinds of math; such mathematical luminaries as Bill Thurston and John Conway were active in the group in its early days. Two current members are George Hart and Robin Houston, and they were able to find several solutions to my puzzle.

Among the solutions they found, one stands out as especially satisfying. If you want to see what their die looks like, print out the following image on some stiff paper, cut out the network of 30 rhombuses, fold along the edges joining adjacent rhombuses, and tape the whole thing up to form a triacontahedron.

An industrious triacontahedron

Especially pleasing to me are the following three properties that this die satisfies.

(1) No face has more than 3 pips. I don’t want too many pips to have to share a face – that’d force the pips to be small or the die to be big.

(2) If you count the pips visible from overhead and the pips visible from underneath (imagining that the die rests on a glass table), you always get a total of 16 pips. This property is reminiscent of the way the two opposite faces of a cubical die show numbers that add up to 7.

(3) There are equal numbers of faces with 1 pips, 2 pips, and 3 pips. This property lends more visual variety to the die.

But what I really like is that Houston and Hart showed that these three properties, taken together, single out the die shown above. As a mathematician, I’m attracted to systems of constraints that are delicately poised between being too lax and too restrictive. A problem that has only one solution is an especially satisfying kind of problem to come across or to invent.

TELLING THE STORY I FOUND

Once I knew there was a new and nifty way to generate a random number between 1 and 15, I had the seed for my Gathering for Gardner 15 (“G4G15“) talk. In planning the talk, I decided to dramatize my quirky design criterion by creating and inhabiting a pompous character who would care about the problem much more than I did, and who would view the efficiency of the die as not just an esthetic issue but a moral one.

You’ll see that this character wears a top hat. Although I borrowed the hat from one of the magicians attending the Gathering, it was intended to be more like a capitalist’s hat than a magician’s hat (think of the hat in the board game Monopoly). More specifically, he’s a ruthless industrialist who’s offended that the pips on a standard die don’t work hard enough.

I don’t think the triacontahedral d15 has much of a future in the gaming world, despite the existence of a small but enthusiastic community of gamers who collect novelty dice. Nor does the dodecahedral d6. This gloomy conclusion is a consequence of the geometry of gaming in the physical world. Think about a game like Monopoly that uses dice: you have two or more players sitting around a board and all players need to be able to see what value each player has rolled. The more efficient (or, as my character would say, “industrious”) a die is, the narrower the range of viewing-angles is for the players who want to confirm what value was rolled, since each player must see all the relevant faces at once. Two people sitting on opposite sides of a board wouldn’t be able to see the top five faces of the dodecahedral d6 at the same time; they’d have to take turns viewing it from above. The same goes for people using a triacontahedral d15. Imagine two players trying to get a view of the die at the same time. Hilarity or romance might ensue, but the possibility of skull trauma cannot be ignored. So: fabricate and use industrious dice at your own risk!

Still, the process of coming up with these dice offered me some of the same satisfactions as doing a piece of “serious” research, so I don’t view it as time wasted. And if my talk on the topic amused a few hundred people, all the better.

Here are a couple of concluding puzzles for you:

First, suppose we chamfer the twelve edges of a blank die and put pips on the chamfered edges; then when we roll the die, four edges are always on top (the four edges that bound the top face). Can we arrange pips on the edges so that when we roll the die and count the pips on the top four edges, we get a random number between 1 and 6? Find such an arrangement or prove that none exists. See Endnote #2.

Second, suppose we truncate the eight corners instead and put pips on the truncated corners; then when we roll the die, four corners are always on top (the four corners of the top face). Can we arrange pips at the corners so that when we roll the die and count the pips on the top four corners, we get a random number between 1 and 6? Find such an arrangement or prove that none exists. See Endnote #3.

Thanks to George Hart and Robin Houston without whose help I couldn’t have given the talk and wouldn’t have written this essay. Great thanks to the organizers of the 15th Gathering for Gardner. Thanks also to Tom Roby for suggesting that I wear a top hat and to The Time-Traveling Wizard (aka Roger Manderscheid) for providing one.

ENDNOTES

#1. Here’s one solution:

On the left I show a solution to the puzzle; on the right I show the twelve sums that verify that the solution satisfies the requirement of yielding each sum between 1 and 6 exactly twice.

#2. No such die exists. Informally, we can argue that since each pip on such a die “counts” (that is, contributes to the value of the throw) one-third of the time (unlike the pips of a standard die which count only one-sixth of the time), the proposed die must be exactly twice as efficient as a standard die, meaning the number of pips must be not 21 but 21/2. But 21/2 is not an integer. More rigorously, we could assign twelve variables a,b,…,l to represent the number of pips on the twelve chamfered edges. For each of the six possible ways the die could land, the number of pips shown by the surrounding edges is a sum of four of the variables. If we add up all six of these sums, we get 2a+2b+…+2l, where each variable occurs with the coefficient 2 because each edge belongs to exactly two faces. But the six sums are supposed to add up to 1+2+3+4+5+6, or 21; and 2a+2b+…+2l is even whereas 21 is odd, so no such assignment of pips exists.

#3. Many such dice exist. My favorite is one in which no corner-face has more than 2 pips (a reasonable thing to require since the corner-faces are not very big). Here is what the assignment of pips looks like on a Schlegel diagram of the cube:

If one makes the corner-faces as big as possible, the result is a solid called a cuboctahedron with 8 triangular faces and 6 square faces. One could make a cuboctahedral d6 that has 7 pips distributed among the triangular faces, but a player might have to roll such a die repeatedly until it lands on a square face.

An industrious cuboctahedron

Hmm, is a thrown cuboctahedron likelier to end up resting on a triangular face or on a square face? A square face, I think; but how much likelier? If any of you have a sturdy cuboctahedron lying around and do some experiments, let me know what you find!

LEAVE A REPLY

Please enter your comment!
Please enter your name here

More like this

Reintroducing the IMplementation Reflection Tool

By Claire Neely, Senior Implementation Specialist Illustrative Mathematics’ redesigned IMplementation Reflection Tool (IRT) is a powerful, non-evaluative resource...

Unizor – Creative Mind through Art of Mathematics: Geometry+...

https://www.youtube.com/watch?v=hAXtrAWNVuI Notes to a video lecture on http://www.unizor.com Geometry+ 10Problem AGiven two parallel lines and a segment AB...

A Little AIME Problem | mathgarage

A Little AIME Problem Posted: May 16, 2022 in Mathematics Tags: AIME, math, Math Contest Solution In general, let By Pythagoras Identity, Solve this...