At cathedrals around the world there is a tradition known as the "ringing of the changes", which consists of ringing the N bells in every one of the N! permutations. This has led bell-ringers to consider the various possible orderings of N numbers. Often the patterns used in bell-ringing consist of simple swaps and shifts of bell positions between adjacent peels, so that they are simple to play. It's interesting to visualize geometrically the shape of the patterns as they are being played. If we take each of the N! permutations of the numbers (1, 2,..., N) as the coordinates of a point in N-space, then obviously all the N! points are equi-distant from the origin. For N=4, the "shape" made by joining each point to its nearest neighbours is what's called a "truncated octahedron", one of the 13 Archimedian solids. It has six square faces and eight hexagonal faces, making a total of 14 faces. The number of edges is 36, as required by Euler's formula F + V - E = 2. If your web browser supports Java applets, you can see a rotating image of this solid here. To determine the 3D coordinates of the "4! solid" given in 4D space by the permutations of [1,2,3,4] it might be easiest to subtract 5/2 from each coordinate so as to center the solid at the origin. Then for each of the 4D vertices [W,X,Y,Z] we have W+X+Y+Z=0, which of course is the condition that ensures us we are really dealing with a 3D solid that happens to be rotated into 4D space. We know the 4D coordinates of the vertices are the permutations of V1 = [-3/2, -1/2, +1/2, +3/2] so each vertex is a distance of d = sqrt(5) from the origin. We want to find a rotation that brings all these points into the form [x,y,z,0]. We're free to place the point V1 on the x axis at [d, 0, 0, 0], and we can place the perpindicular point V2 = [+1/2, -3/2, +3/2, -1/2] on the y axis at [0, d, 0, 0]. (Note that V1 and V2 are orthogonal, as can be seen by computing V1 dot V2 = 0.) Then, for lack of any better ideas, let's consider the vertex given by permuting the last two coordinates of V1, i.e., V3 = [-3/2, -1/2, +3/2, +1/2] From the known distances of this vertex to V1, V2, and the origin, we know the 3D coordinates of V3 satisfy the conditions x^2 + y^2 + z^2 = 5 (x-d)^2 + y^2 + z^2 = 2 x^2 + (y-d)^2 + z^2 = 6 which implies that x = 4/d y = 2/d z = +-1 so we can assign it the 3D coordinates [4/d, 2/d, 1, 0]. This is our third independent vector in our 3D space, so we'll need something in 4D to complete the conditions. Let's say the 4D vector V4 = [0, 0, 0, 1] goes to [x,y,z,w] when it is subjected to our rotation. To ensure the determinant of our rotation matrix is 1 we need w=1/2, and then preservation of the distances to V4 leads to the conditions x^2 + y^2 + z^2 = 1 - (1/2)^2 (x-d)^2 + y^2 + z^2 = 3 - (1/2)^2 x^2 + (y-d)^2 + z^2 = 7 - (1/2)^2 which implies x = 3/(2d), y = -1/(2d), and z = +-1/2. Combining all these conditions we can solve for the rotation matrix R using the equation _ _ _ _ | d 0 4/d 3/(2d) || -3/2 1/2 -3/2 0 |-1 R = | 0 d 2/d -1/(2d) || -1/2 -3/2 -1/2 0 | | 0 0 1 1/2 || 1/2 3/2 3/2 0 | |_ 0 0 0 1/2 _||_ 3/2 -1/2 1/2 1 _| _ _ | -3/d -1/d 1/d 3/d | = (1/2)| 1/d -3/d 3/d -1/d | | 1 3 3 1 | |_ 1 1 1 1 _| Applying this transformation to the original 4D "permutation points" (and noting that we need consider only 12 points because they come in opposite pairs), we have the 3D coordinates 2W 2X 2Y 2Z x y z --- --- --- --- ----- ----- ----- -3 -1 +1 +3 d 0 0 -3 -1 +3 +1 4/d 2/d 1 -3 +3 -1 +1 2/d -4/d 1 -3 +3 +1 -1 1/d -2/d 2 -3 +1 -1 +3 4/d -3/d 0 -3 +1 +3 -1 2/d 1/d 2 -1 -3 +1 +3 4/d 2/d -1 -1 -3 +3 +1 3/d 4/d 0 -1 +1 -3 +3 2/d -4/d -1 -1 +1 +3 -3 -1/d 2/d 2 -1 +3 +1 -3 -2/d -1/d 2 -1 +3 -3 +1 0 -d 0 The other 12 points are just the negatives of these (x,y,z) vectors. As mentioned above, these 24 points constitute the vertices of a truncated octahedron. One way of characterizing the (n-1) dimensional polytopes is in terms of the distribution of distances between vertices. For example, with n=4 we have the 24 vertices are at following squared distances from any given vertex d^2: 0 2 4 6 8 10 12 14 16 18 20 #: 1 3 1 4 2 2 2 4 1 3 1 With n=5, the 120 distances relative to any given vertex are d^2: 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 #: 1 4 3 6 7 6 4 10 6 10 6 10 6 10 4 6 7 6 3 4 1 In general the n! squared distances from any given vertex in the n-permutation polytope are even integers and symmetrically distributed about the squared radius R^2 = n(n^2 - 1)/12. Thus we really only need to list the squared distances up to R^2/2, as shown below for the case n=6: d^2: 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 #: 1 5 6 9 16 12 14 24 20 21 23 28 24 34 20 32 42 29 As n increases the "valency" of the squared distances approaches a continuous distribution, with a maximum valence that seems to increase at roughly the same rate as C(m,m/2) whith m=3(n-4). A plot of the distribution of distances for the n-permutation polytope with n=7, 8, 9, and 10 is shown below, with each normalized to the same width and height. The actual max squared distances and corresponding values of C(m,m/2) are listed below: max n valence C(m,m/2) --- ------- --------- 7 184 126 0.6848 8 1066 924 0.8668 9 6688 6435 0.9622 10 49062 48620 0.9910 As n increases the density evidently approaches a normal distribution / ((x-u)/s)^2 \ V(x) = A exp( ------------- ) \ 2 / where u = n(n^2 - 1)/12 and empirically the parameters s and A seem to be roughly given by s ~ 2u sqrt(n/3) A ~ C(m,m/2) with m = 3(n-4) Comparisons of the normal distribution and the distribution of squared distances for n=8, 9, 10, and 11 are shown below

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