It's a different covering structure, but the products of 2+-i, 2+-j, 2+-k generate a dense covering of all directions. (Divide the generators by sqrt5 to get units.) Except for the cancellations of conjugates, like (2+i)(2-i) = 5, every product is different. You can view the pre-space as a tree with node degree 5, except the root is degree 6. The post-space is points on the unit quaternion sphere. I assume the covering is near uniform, but haven't seen the confirming theorem. Rich PS: Does anyone else miss the XCT instruction? The 7094 & PDP6/10 had it, but it seems to have vanished. I guess it's an architectural nightmare for the hardware folks. --Rich --- Quoting Henry Baker <hbaker1@pipeline.com>:
Perhaps "space-filling curve" might be better than "random walk" ?
At 03:51 PM 3/15/2018, Henry Baker wrote:
We know that the circle group is enumerated uniformly by
cos(t)+i*sin(t), for i in [0,2*pi)
using a single (real) parameter t.
Is there a way of uniformly enumerating the unit quaternions * with 1 parameter (probably some sort of random walk) ? * with 2 parameters (probably some sort of random walk) ? * with 1 complex parameter ????? * with 3 parameters ???
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