12 recent entries Archive:
In this section:
Subtopics:
Comments disabled |
Tue, 07 Aug 2007
Different arrangements for standard dice
To understand just what is being asked for here, first observe that a standard pair of dice throws a 2 exactly 1/36 of the time, a 3 exactly 2/36 of the time, and so forth:
The standard dice have faces numbered 1, 2, 3, 4, 5, and 6. It should be clear that if one die had {0,1,2,3,4,5} instead, and the other had {2,3,4,5,6,7}, then the probabilities would be exactly the same. Similarly you could subtract 3.7 from every face of one die, giving it labels {-2.7, -1.7, -0.7, 0.3, 1.3, 2.3}, and if you added the 3.7 to every face of the other die, giving labels {4.7, 5.7, 6.7, 7.7, 8.7, 9.7}, you'd still have the same chance of getting any particular total. For example, there are still exactly 2 ways out of 36 possible rolls to get the total 3: you can roll -2.7 + 5.7, or you can roll -1.7 + 4.7. But the question is to find a nontrivial relabeling. Like many combinatorial problems, this one is best solved with generating functions. Suppose we represent a die as a polynomial. If the polynomial is Σa_{i}x^{i}, it represents a die that has a_{i} chances to produce the value i. A standard die is x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x, with one chance to produce each integer from 1 to 6. (We can deal with probabilities instead of "chances" by requiring that Σa_{i} = 1, but it comes to pretty much the same thing.) The reason it's useful to adopt this representation is that rolling the dice together corresponds to multiplication of the polynomials. Rolling two dice together, we multiply (x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x) by itself and get P(x) = x^{12} + 2x^{11} + 3x^{10} + 4x^{9} + 5x^{8} + 6x^{7} + 5x^{6} + 4x^{5} + 3x^{4} + 2x^{3} + x^{2}, which gives the chances of getting any particular sum; the coefficient of the x^{9} term is 4, so there are 4 ways to roll a 9 on two dice. What we want is a factorization of this 12th-degree polynomial into two polynomials Q(x) and R(x) with non-negative coefficients. We also want Q(1) = R(1) = 6, which forces the corresponding dice to have 6 faces each. Since we already know that P(x) = (x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x)^{2}, it's not hard; we really only have to factor x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x and then see if there's any suitable way of rearranging the factors. x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x = x(x^{4} + x^{2} + 1)(x + 1) = x(x^{2} + x + 1)(x^{2} - x + 1)(x + 1). So P(x) has eight factors:
We want to combine these into two products Q(x) and R(x) such that Q(1) = R(1) = 6. If we calculate f(1) for each of these, we get 1, 3 (pink), 1, and 2 (blue). So each of Q and R will require one of the factors that has f(1) = 3 and one that has f(1) = 2; we can distribute the f(1) = 1 factors as needed. For normal dice the way we do this is to assign all the factors in each row to one die. If we want alternative dice, our only real choice is what to do with the x^{2} - x + 1 and x factors. Redistributing the lone x factors just corresponds to subtracting 1 from all the faces of one die and adding it back to all the faces of the other, so we can ignore them. The only interesting question is what to do with the x^{2} - x + 1 factors. The normal distribution assigns one to each die, and the only alternative is to assign both of them to a single die. This gives us the two polynomials:
And so the solution is that one die has faces {1,2,2,3,3,4} and the other has faces {1,3,4,5,6,8}:
Counting up entries in the table, we see that there are indeed 6 ways to throw a 7, 4 ways to throw a 9, and so forth. One could apply similar methods to the problem of making a pair of dice that can't roll 7. Since there are six chances in 36 of rolling 7, we need to say what will happen instead in these 6 cases. We might distribute them equally among some of the other possibilities, say 2, 4, 6, 8, 10, and 12, so that we want the final distribution of results to correspond to the polynomial 2x^{12} + 2x^{11} + 4x^{10} + 4x^{9} + 6x^{8} + 6x^{6} + 4x^{5} + 4x^{4} + 2x^{3} + 2x^{2}. The important thing to notice here is that the coefficient of the x^{7} term is 0. Now we want to factor this polynomial and proceed as before. Unfortunately, it is irreducible. (Except for the trivial factor of x^{2}.) Several other possibilities are similarly irreducible. It's tempting to reason from the dice to the algebra, and conjecture that any reducible polynomial that has a zero x^{7} term must be rather exceptional in other ways, such as by having only even exponents. But I'm not sure it will work, because the polynomials are more general than the dice: the polynomials can have negative coefficients, which are meaningless for the dice. Still, I can fantasize that there might be some result of this type available, and I can even imagine a couple of ways of getting to this result, one combinatorial, another based on Fourier transforms. But I've noticed that I have a tendency to want to leave articles unpublished until I finish exploring all possible aspects of them, and I'd like to change that habit, so I'll stop here, for now. [ Addendum 20070905: There are some followup notes. ]
[Other articles in category /math] permanent link Mon, 06 Aug 2007
Standard analytic polyhedra
(0,0,0)And you can see at a glance whether two vertices share an edge (they are the same in two of their three components) or are opposite (they differ in all three components). Last week I was reading the Wikipedia article about the computer game "Hunt the Wumpus", which I played as a small child. For the Guitar Hero / WoW generation I should explain Wumpus briefly. The object of "Wumpus" is to kill the Wumpus, which hides in a network of twenty caves arranged in a dodecahedron. Each cave is thus connected to three others. On your turn, you may move to an adjacent cave or shoot a crooked arrow. The arrow can pass through up to five connected caves, and if it enters the room where the Wumpus is, it kills him and you win. Two of the caves contain bottomless pits; to enter these is death. Two of the caves contain giant bats, which will drop you into another cave at random; if it contains a pit, too bad. If you are in a cave adjacent to a pit, you can feel a draft; if you are adjacent to bats, you can hear them. If you are adjacent to the Wumpus, you can smell him. If you enter the Wumpus's cave, he eats you. If you shoot an arrow that fails to kill him, he wakes up and moves to an adjacent cave; if he enters you cave, he eats you. You have five arrows. I did not learn until much later that the caves are connected in a dodecahedron; indeed, at the time I probably didn't know what a dodecahedron was. The twenty caves were numbered, so that cave 1 was connected to 2, 5, and 8. This necessitated a map, because otherwise it was too hard to remember which room was connected to which. Or did it? If the map had been a cube, the eight rooms could have been named 000, 001, 010, etc., and then it would have been trivial to remember: 011 is connected to 111, 001, and 010, obviously, and you can see it at a glance. It's even easy to compute all the paths between two vertices: the paths from 011 to 000 are 011–010–000 and 011–001–000; if you want to allow longer paths you can easily come up with 011–111–110–100–000 for example. And similarly, the Wumpus source code contains a table that records which caves are connected to which, and consults this table in many places. If the caves had been arranged in a cube, no table would have been required. Or if one was wanted, it could have been generated algorithmically. So I got to wondering last week if there was an analogous nomenclature for the vertices of a dodecahedron that would have obviated the Wumpus map and the table in the source code. I came up with a very clever proof that there was none, which would have been great, except that the proof also worked for the tetrahedron, and the tetrahedron does have such a convenient notation: you can name the vertices (0,0,0), (0,1,1), (1,0,1), and (1,1,0), where there must be an even number of 1 components. (I mentioned this yesterday in connection with something else and promised to come back to it. Here it is.) So the proof was wrong, which was good, and I kept thinking about it. The next-simplest case is the octahedron, and I racked my brains trying to come up with a convenient notation for the vertices that would allow one to see at a glance which were connected. When I finally found it, I felt like a complete dunce. The octahedron has six vertices, which are above, below, to the left of, to the right of, in front of, and behind the center. Their coordinates are therefore (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1) and (0,0,-1). Two vertices are opposite when they have two components the same (necessarily both 0) and one different (necessarily negatives). Otherwise, they are connected by an edge. This is really simple stuff. Still no luck with the dodecahedron. There are nice canonical representations of the coordinates of the vertices—see the Wikipedia article, for example—but I still haven't looked at it closely enough to decide if there is a simple procedure for taking two vertices and determining their geometric relation at a glance. Obviously, you can check for adjacent vertices by calculating the distance between them and seeing if it's the correct value, but that's not "at a glance"; arithmetic is forbidden. It's easy to number the vertices in layers, say by calling the top five vertices A_{1} ... A_{5}, then the five below that B_{1} ... B_{5}, and so on. Then it's easy to see that A_{3} will be adjacent to A_{2}, A_{4}, and B_{3}, for example. But this nomenclature, unlike the good ones above, is not isometric: it has a preferred orientation of the dodecahedron. It's obvious that A_{1}, A_{2}, A_{3}, A_{4}, and A_{5} form a pentagonal face, but rather harder to see that A_{2}, A_{3}, B_{2}, B_{3}, and C_{5} do. With the cube, it's easy to see what a rotation or a reflection looks like. For example, rotation of 120° around an axis through a pair of vertices of the cube takes vertex (a, b, c) to (c, a, b); rotation of 90° around an axis through a face takes it to (1-b, a, c). Similarly, rotations and reflections of the tetrahedron correspond to simple permutations of the components of the vertices. Nothing like this exists for the A-B-C-D nomenclature for the dodecahedron. I'll post if I come up with anything nice.
[Other articles in category /math] permanent link
Sub-blogs and the math sub-blog
If you really do want just the math articles for some reason, you can subscribe to the feeds at http://blog.plover.com/math/index.rss or http://blog.plover.com/math/index.atom. I've been generating these sub-feed files since the blog began, and I know nobody uses them. But perhaps someone would like to. Similarly, there are sub-feeds for other subsections of the blog, for example "physics". Most of these topic areas receive many fewer updates than does the "math" section:
Personally I feel that the eclecticism of the blog is one of its attractions, and I gather that a lot of other people do too, but perhaps not everyone agrees.
[Other articles in category /meta] permanent link Sun, 05 Aug 2007
The 123456 die
That got me thinking about the problem in general. For some reason I've been trying to construct a die whose faces come up with probabilities 1/21, 2/21, 3/21, 4/21, 5/21, and 6/21 respectively. Unless there is a clever insight I haven't had, I think this will be rather difficult to do explicitly. (Approximation methods will probably work fairly easily though, I think.) I started by trying to make a hexahedron with faces that had areas 1, 2, 3, 4, 5, 6, and even this has so far evaded me. This will not be sufficient to solve the problem, because the probability that the hexahedron will land on face F is not proportional to the area of F, but rather to the solid angle subtended by F from the hexahedron's center of gravity. Anyway, I got interested in the idea of making a hexahedron whose faces had areas 1..6. First I tried just taking a bunch of simple shapes (right triangles and the like) of the appropriate sizes and fitting them together geometrically; so far that hasn't worked. So then I thought maybe I could get what I wanted by taking a tetrahedron or a disphenoid or some such and truncating a couple of the corners. As Polya says, if you can't solve the problem, you should try solving a simpler problem of the same sort, so I decided to see if it was possible to take a regular tetrahedron and chop off one vertex so that the resulting pentahedron had faces with areas 1, 2, 3, 4, 5. The regular tetrahedron is quite tractable, geometrically, because you can put its vertices at (0,0,0), (0,1,1), (1,0,1), and (1,1,0), and then a plane that chops off the (0,0,0) vertex cuts the three apical edges at points (0,a,a), (b,0,b), and (c,c,0), for some 0 ≤ a, b, c ≤ 1. The chopped-off areas of the three faces are simply ab√3/4, bc√3/4, ca√3/4, and the un-chopped base has area √3/4, so if we want the three chopped faces to have areas of 2/5, 3/5 and 4/5 times √3/4, respectively, we must have ab = 3/5, bc = 2/5, and ca = 1/5, and we can solve for a, b, c. (We want the new top face to have area 1/5 · √3/4, but that will have to take care of itself, since it is also determined by a, b, and c.) Unfortunately, solving these equations gives b = √6/√5, which is geometrically impossible. We might fantasize that there might be some alternate solution, say with the three chopped faces having areas of 1/5, 2/5 and 4/5 times √3/4, and the top face being 3/5 · √3/4 instead of 1/5 · √3/4, but none of those will work either. Oh well, it was worth a shot. I do think it's interesting that if you know the areas of the bottom four faces of a truncated regular tetrahedron, that completely determines the apical face. Because you can solve for the lengths of the truncated apical edges, as above, and then that gives you the coordinates of the three apical vertices. I had a brief idea about truncating a square pyramid to get the hexahedron I wanted in the first place, but that's more difficult, because you can't just pick the lengths of the four apical edges any way you want; their upper endpoints must be coplanar. The (0,a,a), (b,0,b), (c,c,0) thing has been on my mind anyway, and I hope to write tomorrow's blog article about it. But I've decided that my articles are too long and too intermittent, and I'm going to try to post some shorter, more casual ones more frequently. I recently remembered that in the early days of the blog I made an effort to post every day, and I think I'd like to try to resume that. [ Addendum 20070905: There are some followup notes. ]
[Other articles in category /math] permanent link Wed, 01 Aug 2007
The snub disphenoid
Anyway, earlier this week I was visiting John Batzel, who works upstairs from me, and discovered that he had obtained a really cool toy. It was a collection of large steel ball bearings and colored magnetic rods, which could be assembled into various polyhedra and trusses. This is irresistible to me. The pictures at right, taken around 2002, show me modeling a dodecahedron with less suitable materials. The first thing I tried to make out of John's magnetic sticks and balls was a regular dodecahedron, because it is my favorite polyhedron. (Isn't it everyone's?) This was unsuccessful, because it wasn't rigid enough, and kept collapsing. It's possible that if I had gotten the whole thing together it would have been stable, but holding the 50 separate magnetic parts in the right place long enough to get it together was too taxing, so I tried putting together some other things. A pentagonal dipyramid worked out well, however. To understand this solid, imagine a regular pyramid, such as the kind that entombs the pharaohs or collects mystical energy. This sort of pyramid is known as a square pyramid, because it has a square base, and thus four triangular sides. Imagine that the base was instead a pentagon, so that there were five triangular sides sides instead of only four. Then it would be a pentagonal pyramid. Now take two such pentagonal pyramids and glue the pentagonal bases together. You now have a pentagonal dipyramid. The success of the pentagonal dipyramid gave me the idea that rigid triangular lattices were the way to go with this toy, so I built an octahedron (square dipyramid) and an icosahedron to be sure. Even the icosahedron (thirty sticks and twelve balls) held together and supported its own weight. So I had John bring up the Wikipedia article about deltahedra. A deltahedron is just a polyhedron whose faces are all equilateral triangles.
Four of the deltahedra are the tetrahedron (triangular pyramid, with 4 faces), triangular dipyramid (6 faces), octahedron (square dipyramid, 8 faces), and pentagonal dipyramid (10 faces).
Another deltahedron is the "gyroelongated square dipyramid". You get this by taking two square pyramids, as with the octahedron. But instead of gluing their square bases together directly, you splice a square antiprism in between. The two square faces of the antiprism are not aligned; they are turned at an angle of 45° to each other, so that when you are looking at the top pyramid face-on, you are looking at the bottom pyramid edge-on, and this is the "gyro" in "gyroelongated". (The icosahedron is a gyroelongated pentagonal dipyramid.) I made one of these in John's office, but found it rather straightforward. The last deltahedron, however, was quite a puzzle. Wikipedia calls it a "snub disphenoid", and as I mentioned before, the name did not help me out at all. It took me several tries to build it correctly. It contains 12 faces and 8 vertices. When I finally had the model I still couldn't figure it out, and spent quite a long time rotating it and examining it. It has a rather strange symmetry. It is front-back and left-right symmetric. And it is almost top-bottom symmetric: If you give it a vertical reflection, you get the same thing back, but rotated 90° around the vertical axis. When I planned this article I thought I understood it better. Imagine sticking together two equilateral triangles. Call the common edge the "rib". Fold the resulting rhombus along the rib so that the edges go up, down, up, down in a zigzag. Let's call the resulting shape a "wing"; it has a concave side and a convex side. Take two wings. Orient them with the concave sides facing each other, and with the ribs not parallel, but at right angles. So far, so good. But this is where I started to get it wrong. The two wings have between them eight edges, and I had imagined that you could glue a rhombic antiprism in between them. I'm not convinced that there is such a thing as a rhombic antiprism, but I'll have to do some arithmetic to be sure. Anyway, supposing that there were such a thing, you could glue it in as I said, but if you did the wings would flatten out and what you would get would not be a proper polyhedron because the two triangles in each wing would be coplanar, and polyhedra are not allowed to have abutting coplanar faces. (The putative gyroelongated triangular dipyramid fails for this reason, I believe.) To make the snub disphenoid, you do stick eight triangles in between the two wings, but the eight triangles do not form a rhombic antiprism. Even supposing that such a thing exists. I hope to have some nice renderings for you later. I have been doing some fun work in rendering semiregular polyhedra, and I am looking forward to discussing it here. Advance peek: suppose you know how the vertices are connected by edges. How do you figure out where the vertices are located in 3-space? If you would like to investigate this, the snub disphenoid has 8 vertices, which we can call A, B, ... H. Then:
Here is a list of the eight deltahedra, with links to the corresponding Wikipedia articles:
[ Addendum 20070908: More about deltahedra. ]
[Other articles in category /math] permanent link |