There were some conflicting answers to my question. What is the answer as far as we know it? The number of pieces when a unit n-dimensional cube is cut by all the hyperplanes defined by any n of the vertices? a(1) = 1 a(2) = 4 a(3) = 96 (?) a(4) = ? What are the values, and to whom should they be credited? Then I'll create an entry in the OEIS Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Mon, Apr 13, 2020 at 12:02 PM Veit Elser <ve10@cornell.edu> wrote:
I now know the actual counts (not just bounds) up to d=4.
Starting with the simplex, e.g. for d=3,
000 001 011 111
I iteratively generate sets of convex pieces by slicing through all the pieces with another hyperplane to get the pieces in the next iteration.
I checked my hyperplane generator against A007847 (you can drop the 2d hyperplanes that bound the hypercube). Here are the counts for the number of pieces having d+1, d+2, … vertices:
d=2 {{3, 2}}
total pieces: 2
d=3 {{4, 12}, {5, 4}}
total pieces: 16
d=4
{{5,6784},{6,4024},{7,4936},{8,2704},{9,1912},{10,936},{11,824},{12,496},{13,360},{14,352},{15,256},{16,72},{17,48},{18,16},{19,16},{20,16}}
total pieces: 23752
To get the number of pieces in the hypercube just multiply these by d!.
-Veit
On Apr 12, 2020, at 11:58 AM, Tomas Rokicki <rokicki@gmail.com> wrote:
I misspoke; it's the cuts through the center of the cube that cut the cube into 24 regions (no center region) and then the corner cuts (that go through three points) cut each of these regions into 4.
There is indeed a center region created by just the corner cuts (it's an octahedron, as in a birectified cube).
I'm sorry for the confusion.
On Sun, Apr 12, 2020 at 12:52 AM Tom Karzes <karzes@sonic.net> wrote:
Are you sure about these numbers? The corner cuts should divide the cube into an odd number of regions, since we know it leaves a single region in the center, and there should be an even number of remaining regions by symmetry.
Even if the 8 corner cuts did divide it into 24 regions, I don't think the 6 cuts that go through the center divide each of these into 4 regions. All 6 of those cuts go through the center, so clearly they divide the center region into more than 4 regions.
Tom
Tomas Rokicki writes:
For the cube the answer is 96 regions. There are 14 cuts through the cube; six cut the cube in half along a face diagonal, and eight cut off a corner with a triangle through the three adjacent corners. The corner cuts alone divide the cube in 24 regions, and then the cuts through the center of the cube further divide each of these into four regions.
In four dimensions I believe the answer is 18432 regions but I'm not completely sure I've got the details exactly right.
On Fri, Apr 10, 2020 at 10:23 PM Tom Karzes <karzes@sonic.net> wrote:
That's an interesting question, and difficult to visualize even in 3 dimensions.
A closely related question is this: If you extend the slices to infinity, into how many regions do they divide the entire space? I.e., include the regions exterior to the square/cube/hypercube. This makes the boundary cuts more interesting. I don't know if it makes the problem harder or easier, but it does create more regions to count.
In 2 dimensions, the number of exterior regions is 12. Adding the 4 interior regions gives a total of 16 regions for the entire plane.
In 3 dimensions, well, not sure.
By the way, even in 3 dimensions, I believe there are regions interior to the cube that do not intersect any vertices or edges of the cube: For each vertex, consider the plane defined by the three vertices that differ from that vertex in exactly one coordinate. For instance, for vertex (0, 1, 0) we get vertices (1, 1, 0), (0, 0, 0), and (0, 1, 1). These three vertices define a plane, and there are 8 such sets of them. Together, these 8 planes create an octahedron-shaped region in the center of the cube. You can think of it as a tiny internal dual of the cube. The vertices of the octahedron are the centers of the faces of the cube. If you then proceed to add the remaining planar slices, this interior octahedron gets sliced up into even more pieces.
Do there exist interior regions that do not intersect any of the cube faces? How about in 4 dimensions?
Tom
Neil Sloane writes:
Take a unit square and cut along the lines joining any two vertices. This cuts the square into 4 pieces.
Now take a unit cube and make plane cuts though any three vertices: how many pieces are produced? (I don't know)
Same question for a unit d-dimensional cube, where the cuts are along hyperplanes through any d vertices.
Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
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