Dime Time

Everyone knows the shapes for one and two dimensions, far fewer know the shapes for three and four dimensions, let alone five, six and seven. And what shapes are those? The shapes that answer this question:

• How many equidistant points are possible in 1d, 2d, 3d, 4d…?

In one dimension it’s obvious that the answer is 2. In other words, you can get only two equidistant points, (a,b), on a straight line. Point a must be as far from point b as point b is from point a. You can’t add a third point, c, such that (a,b,c) are equidistant. Not on a straight line in 1d. But suppose you bend the line into a circle, so that you’re working in two dimensions. It’s easy to place three equidistant points, (a,b,c), on a circle.

equidistant points on a circle

Three equidistant points around a circle forming the vertices of an equilateral triangle


And it’s also easy to see that the three points will form the vertices of an equilateral triangle. Now try adding a fourth point, d. If you place it in the center of the triangle, it will be equidistant from (a,b,c), but it will be nearer to (a,b,c) than they are to each other. So you can have only 2 equidistant points in 1d and 3 equidistant points in 2d.

But what are the co-ordinates of the equidistant points in 1d and 2d? Suppose (a,b) in 1d are given the co-ordinates (0) and (1), so that a is 1 unit distant from b. When you move to 2d and add point c, the co-ordinates for (a,b) become (0,0) and (1,0). They’re still 1 unit distant from each other. But what are the co-ordinates for c? Start by placing c exactly midway between a and b, so that it has the co-ordinates (0.5,0) and is 0.5 units distant from both a and b. Now, if you move c in the first dimension, it will become nearer either to a or b: (0.49,0) or (0.51,0) or (0.48,0) or (0.52,0)…

But if you move c in the second dimension, it will always be equidistant from a and b, because (a,b) stay in the first dimension, as it were, and c moves equally away from both into the second dimension. So where in 2d will c be 1 unit distant from both a and b just as a and b are 1 unit distant from each other in 1d? You can see the answer here:

equilateral_triangle heightHeight of an equilateral triangle


The co-ordinates for c are (0.5,√3/2) or (0.5,0.8660254…), because the second co-ordinate satisfies the Pythagorean equation 1^2 = 0.5^2 + (√3/2)^2 = 0.25 + 0.75. That is, to find the second co-ordinate of c for 2d, you find the answer to √(1 – 0.5^2) = √(1-0.25) = √0.75 = √(3/4) = √3/2 = 0.8660254….

But you can’t add a fourth point, d, in 2d such that (a,b,c,d) are equidistant. So let’s move to 3d for the points (a,b,c,d). Begin with point d in the center of the triangle formed by (a,b,c), where it will have the co-ordinates (0.5,√3/6,0) = (0.5,0.28867513…,0) and will be equidistant from (a,b,c). But d will be nearer to (a,b,c) than they are to each other. However, if you move d in the third dimension, it will be moving equally away from (a,b,c). So where in 3d will d be 1 unit from (a,b,c)? By analogy with 2d, the third co-ordinate for d will satisfy the generalized Pythagorean equation √(1 – 0.5^2 – (√3/6)^2). And √6/3 = √(1 – 0.5^2 – (√3/6)^2) = 0.81649658… So point d will have the co-ordinates (0.5,√3/6,√6/3) = (0.5, 0.288675135…, 0.816496581…).

And the four points (a,b,c,d) will be the vertices of a three-dimensional shape called the tetrahedron:

Rotating tetrahedron

Rotating tetrahedron


But you can’t add a fifth point, e, in 3d such that (a,b,c,d,e) are equidistant. So let’s move to 4d, the fourth dimension, for the points (a,b,c,d,e). Begin with point e in the center of the tetrahedron formed by (a,b,c,d), where it will have the co-ordinates (0.5,√3/6,√6/12,0) = (0.5,0.28867513…, 0.2041241…, 0) and will be equidistant from (a,b,c,d). But e will be nearer to (a,b,c,d) than they are to each other. However, if you move e in the fourth dimension, it will be moving equally away from (a,b,c,e). So where in 4d will e be 1 unit from (a,b,c,d)? By analogy with 2d and 3d, the co-ordinate for 4d will satisfy the equation √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2). And √10/4 = √(1 – 0.5^2 – (√3/6)^2 – (√6/12)^2) = 0.79056941… So point e will have the co-ordinates (0.5,√3/6,√6/3,√10/4) = (0.5, 0.288675135…, 0.816496581…, 0.79056941…).

And the five points (a,b,c,d,e) will be the vertices of a four-dimensional shape called variously the hyperpyramid, the 5-cell, the pentachoron, the 4-simplex, the pentatope, the pentahedroid and the tetrahedral pyramid. It’s impossible for 3d creatures like human beings (at present) to visualize the hyperpyramid, but we can see its 3d shadow, as it were. And here is the 3d shadow of a rotating hyperpyramid:

Rotating hyperpyramid or 5-cell

Rotating hyperpyramid


N.B. Wikipedia reveals the mathematically beautiful fact that the “simplest set of coordinates [for a hyperpyramid] is: (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (φ,φ,φ,φ), with edge length 2√2, where φ is the golden ratio.”

So that’s the hyperpyramid, with 5 points in 4d. But you can’t add a sixth point, f, in 4d such that (a,b,c,d,e,f) are equidistant. You have to move to 5d. And it should be clear by now that in any dimension nd, the maximum possible number of equidistant points, p, in that dimension will be p = n+1. And here are the co-ordinates for p in dimensions 1 to 10 (the co-ordinates are given in full for 1d to 4d, then for 5d to 10d only the co-ordinates of the additional point are given):

d1: (0), (1)
d2: (0,0), (1,0), (0.5,0.866025404)
d3: (0,0,0), (1,0,0), (0.5,0.866025404,0), (0.5,0.288675135,0.816496581)
d4: 0.5, 0.288675135, 0.204124145, 0.790569415
d5: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.774596669
d6: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.763762616
d7: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.755928946
d8: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.75
d9: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.745355992
d10: 0.5, 0.288675135, 0.204124145, 0.158113883, 0.129099445, 0.109108945, 0.0944911183, 0.0833333333, 0.0745355992, 0.741619849

In each dimension d, the final co-ordinate, cd+1, of the additional point satisfies the generalized Pythagorean equation cd+1 = √(1 – c1^2 – c2^2 – … cd^2).


Readers’ advisory: I am not a mathematician and the discussion above cannot be trusted to be free of errors, whether major or minor.

Live and Let Dice

How many ways are there to die? The answer is actually five, if by “die” you mean “roll a die” and by “rolled die” you mean “Platonic polyhedron”. The Platonic polyhedra are the solid shapes in which each polygonal face and each vertex (meeting-point of the edges) are the same. There are surprisingly few. Search as long and as far as you like: you’ll find only five of them in this or any other universe. The standard cubic die is the most familiar: each of its six faces is square and each of its eight vertices is the meeting-point of three edges. The other four Platonic polyhedra are the tetrahedron, with four triangular faces and four vertices; the octahedron, with eight triangular faces and six vertices; the dodecahedron, with twelve pentagonal faces and twenty vertices; and the icosahedron, with twenty triangular faces and twelve vertices. Note the symmetries of face- and vertex-number: the dodecahedron can be created inside the icosahedron, and vice versa. Similarly, the cube, or hexahedron, can be created inside the octahedron, and vice versa. The tetrahedron is self-spawning and pairs itself. Plato wrote about these shapes in his Timaeus (c. 360 B.C.) and based a mathemystical cosmology on them, which is why they are called the Platonic polyhedra.

An animated gif of a tetrahedron

Tetrahedron


An animated gif of a hexahedron

Hexahedron

An animated gif of an octahedron

Octahedron


An animated gif of a dodecahedron

Dodecahedron

An animated gif of an icosahedron

Icosahedron

They make good dice because they have no preferred way to fall: each face has the same relationship with the other faces and the centre of gravity, so no face is likelier to land uppermost. Or downmost, in the case of the tetrahedron, which is why it is the basis of the caltrop. This is a spiked weapon, used for many centuries, that always lands with a sharp point pointing upwards, ready to wound the feet of men and horses or damage tyres and tracks. The other four Platonic polyhedra don’t have a particular role in warfare, as far as I know, but all five might have a role in jurisprudence and might raise an interesting question about probability. Suppose, in some strange Tycholatric, or fortune-worshipping, nation, that one face of each Platonic die represents death. A criminal convicted of a serious offence has to choose one of the five dice. The die is then rolled f times, or as many times as it has faces. If the death-face is rolled, the criminal is executed; if not, he is imprisoned for life.

The question is: Which die should he choose to minimize, or maximize, his chance of getting the death-face? Or doesn’t it matter? After all, for each die, the odds of rolling the death-face are 1/f and the die is rolled f times. Each face of the tetrahedron has a 1/4 chance of being chosen, but the tetrahedron is rolled only four times. For the icosahedron, it’s a much smaller 1/20 chance, but the die is rolled twenty times. Well, it does matter which die is chosen. To see which offers the best odds, you have to raise the odds of not getting the death-face to the power of f, like this:

3/4 x 3/4 x 3/4 x 3/4 = 3/4 ^4 = 27/256 = 0·316…

5/6 ^6 = 15,625 / 46,656 = 0·335…

7/8 ^8 = 5,764,801 / 16,777,216 = 0·344…

11/12 ^12 = 3,138,428,376,721 / 8,916,100,448,256 = 0·352…

19/20 ^20 = 37,589,973,457,545,958,193,355,601 / 104,857,600,000,000,000,000,000,000 = 0·358…

Those represent the odds of avoiding the death-face. Criminals who want to avoid execution should choose the icosahedron. For the odds of rolling the death-face, simply subtract the avoidance-odds from 1, like this:

1 – 3/4 ^4 = 0·684…

1 – 5/6 ^6 = 0·665…

1 – 7/8 ^8 = 0·656…

1 – 11/12 ^12 = 0·648…

1 – 19/20 ^20 = 0·642…

So criminals who prefer execution to life-imprisonment should choose the tetrahedron. If the Tycholatric nation offers freedom to every criminal who rolls the same face of the die f times, then the tetrahedron is also clearly best. The odds of rolling a single specified face f times are 1/f ^f:

1/4 x 1/4 x 1/4 x 1/4 = 1/4^4 = 1 / 256

1/6^6 = 1 / 46,656

1/8^8 = 1 / 16,777,216

1/12^12 = 1 / 8,916,100,448,256

1/20^20 = 1 / 104,857,600,000,000,000,000,000,000

But there are f faces on each polyhedron, so the odds of rolling any face f times are 1/f ^(f-1). On average, of every sixty-four (256/4) criminals who choose to roll the tetrahedron, one will roll the same face four times and be reprieved. If a hundred criminals face the death-penalty each year and all choose to roll the tetrahedron, one criminal will be reprieved roughly every eight months. But if all criminals choose to roll the icosahedron and they have been rolling since the Big Bang, just under fourteen billion years ago, it is very, very, very unlikely that any have yet been reprieved.