[math-fun] cutting up a cube
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
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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
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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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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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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This series starts 4, 96, … and does not seem to be in OEIS because of a bound I have on the next element (which is being improved as I write this). Consider an interior point of one of these regions: a d-tuple of real numbers between 0 and 1. If we re-order the d-tuple we get a different region because there are planes that distinguish the order of any consecutive pair of numbers in the d-tuple. We may therefore count the number of regions for a particular ordering and multiply by d!. Using Mathematica I generate random sorted d-tuples and assign them a binary code giving the side of each of the hyperplanes. Whenever a new code is found it is appended to the codebook of regions. Initially the codebook grows rapidly, and then, mercifully, the “curve flattens”. Reduced by d! the series starts 2, 16, … Right now the next number is at least 23450, with Mathematica discovering new “cases” at a rate of 21/10^5. Any guesses — insights? — where it will end? -Veit
On Apr 10, 2020, at 9:59 PM, Neil Sloane <njasloane@gmail.com> wrote:
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
How many hyperplanes do you get in the 4D case? I believe I see 68. 12 through 8 points (cutting the 4D cube in half), 32 through 6 points, and 24 through 4 points. I'm using a similar approach to yours, so I'm curious why I'm only seeing 18,432 regions for the 4D case. The number I get for the 5D case is 31,610,880. On Sat, Apr 11, 2020 at 9:08 AM Veit Elser <ve10@cornell.edu> wrote:
This series starts 4, 96, … and does not seem to be in OEIS because of a bound I have on the next element (which is being improved as I write this).
Consider an interior point of one of these regions: a d-tuple of real numbers between 0 and 1. If we re-order the d-tuple we get a different region because there are planes that distinguish the order of any consecutive pair of numbers in the d-tuple. We may therefore count the number of regions for a particular ordering and multiply by d!.
Using Mathematica I generate random sorted d-tuples and assign them a binary code giving the side of each of the hyperplanes. Whenever a new code is found it is appended to the codebook of regions. Initially the codebook grows rapidly, and then, mercifully, the “curve flattens”.
Reduced by d! the series starts 2, 16, … Right now the next number is at least 23450, with Mathematica discovering new “cases” at a rate of 21/10^5. Any guesses — insights? — where it will end?
-Veit
On Apr 10, 2020, at 9:59 PM, Neil Sloane <njasloane@gmail.com> wrote:
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
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Regarding the numbers and classes of hyperplanes: OEIS entry A007847 gives the hyperplane counts as: a(1) = 2 a(2) = 6 a(3) = 20 a(4) = 140 a(5) = 3254 ... This matches what I found (at least up to a(5)). The classes of hyperplanes are listed below for d = 2-5. Each class is shown below preceded by the number of instances of that class. I define two hyperplanes as being in the same class of the verticex set of one can be transformed to the vertex set of the other by some combination of (1) permuting the coordinates and (2) inverting some set of coordinates (1->0 and 0->1). For d=2 there are 2 classes hyperplanes (i.e., lines): 4: 00 01 2: 00 11 This gives a total of 6. The first class corresponds to the 4 perimeter slices. For d=3 there are 3 classes of hyperplanes (i.e., planes): 6: 000 001 010 011 6: 000 001 110 111 8: 000 011 101 This gives a total of 20. The first class corresponds to the 6 perimeter slices. For d=4 there are 6 classes of hyperplanes: 8: 0000 0001 0010 0011 0100 0101 0110 0111 12: 0000 0001 0010 0011 1100 1101 1110 1111 32: 0000 0001 0110 0111 1010 1011 16: 0000 0011 0101 1001 8: 0000 0011 0101 1010 1100 1111 64: 0000 0011 0101 1110 This gives a total of 140. The first class corresponds to the 8 perimeter slices. For d=5 there are 15 classes of hyperplanes: 10: 00000 00001 00010 00011 00100 00101 00110 00111 01000 01001 01010 01011 01100 01101 01110 01111 20: 00000 00001 00010 00011 00100 00101 00110 00111 11000 11001 11010 11011 11100 11101 11110 11111 80: 00000 00001 00010 00011 01100 01101 01110 01111 10100 10101 10110 10111 80: 00000 00001 00110 00111 01010 01011 10010 10011 40: 00000 00001 00110 00111 01010 01011 10100 10101 11000 11001 11110 11111 320: 00000 00001 00110 00111 01010 01011 11100 11101 32: 00000 00011 00101 01001 10001 32: 00000 00011 00101 01001 10010 10100 10111 11000 11011 11101 80: 00000 00011 00101 01001 10110 11010 11100 11111 160: 00000 00011 00101 01001 11110 160: 00000 00011 00101 01010 01100 01111 10110 320: 00000 00011 00101 01110 10110 320: 00000 00011 00101 01110 11000 11011 11101 640: 00000 00011 00101 01110 11001 960: 00000 00011 01101 10101 11010 This gives a total of 3254. The first class corresponds to the 10 perimeter slices. Tom Tomas Rokicki writes:
How many hyperplanes do you get in the 4D case? I believe I see 68. 12 through 8 points (cutting the 4D cube in half), 32 through 6 points, and 24 through 4 points.
I'm using a similar approach to yours, so I'm curious why I'm only seeing 18,432 regions for the 4D case.
The number I get for the 5D case is 31,610,880.
On Sat, Apr 11, 2020 at 9:08 AM Veit Elser <ve10@cornell.edu> wrote:
This series starts 4, 96, … and does not seem to be in OEIS because of a bound I have on the next element (which is being improved as I write this).
Consider an interior point of one of these regions: a d-tuple of real numbers between 0 and 1. If we re-order the d-tuple we get a different region because there are planes that distinguish the order of any consecutive pair of numbers in the d-tuple. We may therefore count the number of regions for a particular ordering and multiply by d!.
Using Mathematica I generate random sorted d-tuples and assign them a binary code giving the side of each of the hyperplanes. Whenever a new code is found it is appended to the codebook of regions. Initially the codebook grows rapidly, and then, mercifully, the “curve flattens”.
Reduced by d! the series starts 2, 16, … Right now the next number is at least 23450, with Mathematica discovering new “cases” at a rate of 21/10^5. Any guesses — insights? — where it will end?
-Veit
On Apr 10, 2020, at 9:59 PM, Neil Sloane <njasloane@gmail.com> wrote:
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
I withdraw my results for 4D and higher. (Apparently I missed a hyperplane or two.) -tom On Tue, Apr 14, 2020 at 9:49 AM Tom Karzes <karzes@sonic.net> wrote:
Regarding the numbers and classes of hyperplanes:
OEIS entry A007847 gives the hyperplane counts as:
a(1) = 2 a(2) = 6 a(3) = 20 a(4) = 140 a(5) = 3254 ...
This matches what I found (at least up to a(5)).
The classes of hyperplanes are listed below for d = 2-5. Each class is shown below preceded by the number of instances of that class.
I define two hyperplanes as being in the same class of the verticex set of one can be transformed to the vertex set of the other by some combination of (1) permuting the coordinates and (2) inverting some set of coordinates (1->0 and 0->1).
For d=2 there are 2 classes hyperplanes (i.e., lines):
4: 00 01 2: 00 11
This gives a total of 6. The first class corresponds to the 4 perimeter slices.
For d=3 there are 3 classes of hyperplanes (i.e., planes):
6: 000 001 010 011 6: 000 001 110 111 8: 000 011 101
This gives a total of 20. The first class corresponds to the 6 perimeter slices.
For d=4 there are 6 classes of hyperplanes:
8: 0000 0001 0010 0011 0100 0101 0110 0111 12: 0000 0001 0010 0011 1100 1101 1110 1111 32: 0000 0001 0110 0111 1010 1011 16: 0000 0011 0101 1001 8: 0000 0011 0101 1010 1100 1111 64: 0000 0011 0101 1110
This gives a total of 140. The first class corresponds to the 8 perimeter slices.
For d=5 there are 15 classes of hyperplanes:
10: 00000 00001 00010 00011 00100 00101 00110 00111 01000 01001 01010 01011 01100 01101 01110 01111 20: 00000 00001 00010 00011 00100 00101 00110 00111 11000 11001 11010 11011 11100 11101 11110 11111 80: 00000 00001 00010 00011 01100 01101 01110 01111 10100 10101 10110 10111 80: 00000 00001 00110 00111 01010 01011 10010 10011 40: 00000 00001 00110 00111 01010 01011 10100 10101 11000 11001 11110 11111 320: 00000 00001 00110 00111 01010 01011 11100 11101 32: 00000 00011 00101 01001 10001 32: 00000 00011 00101 01001 10010 10100 10111 11000 11011 11101 80: 00000 00011 00101 01001 10110 11010 11100 11111 160: 00000 00011 00101 01001 11110 160: 00000 00011 00101 01010 01100 01111 10110 320: 00000 00011 00101 01110 10110 320: 00000 00011 00101 01110 11000 11011 11101 640: 00000 00011 00101 01110 11001 960: 00000 00011 01101 10101 11010
This gives a total of 3254. The first class corresponds to the 10 perimeter slices.
Tom
Tomas Rokicki writes:
How many hyperplanes do you get in the 4D case? I believe I see 68. 12 through 8 points (cutting the 4D cube in half), 32 through 6 points, and 24 through 4 points.
I'm using a similar approach to yours, so I'm curious why I'm only seeing 18,432 regions for the 4D case.
The number I get for the 5D case is 31,610,880.
On Sat, Apr 11, 2020 at 9:08 AM Veit Elser <ve10@cornell.edu> wrote:
This series starts 4, 96, … and does not seem to be in OEIS because of a bound I have on the next element (which is being improved as I write this).
Consider an interior point of one of these regions: a d-tuple of real numbers between 0 and 1. If we re-order the d-tuple we get a different region because there are planes that distinguish the order of any consecutive pair of numbers in the d-tuple. We may therefore count the number of regions for a particular ordering and multiply by d!.
Using Mathematica I generate random sorted d-tuples and assign them a binary code giving the side of each of the hyperplanes. Whenever a new code is found it is appended to the codebook of regions. Initially the codebook grows rapidly, and then, mercifully, the “curve flattens”.
Reduced by d! the series starts 2, 16, … Right now the next number is at least 23450, with Mathematica discovering new “cases” at a rate of 21/10^5. Any guesses — insights? — where it will end?
-Veit
On Apr 10, 2020, at 9:59 PM, Neil Sloane <njasloane@gmail.com> wrote:
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
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Thanks to everyone who responded. There are now two new sequences, the main one being A333539, and A333540 if you divide by d!. And A007847 has been expanded with Tom's list of hyperplanes. On Tue, Apr 14, 2020 at 1:53 PM Tomas Rokicki <rokicki@gmail.com> wrote:
I withdraw my results for 4D and higher. (Apparently I missed a hyperplane or two.)
-tom
On Tue, Apr 14, 2020 at 9:49 AM Tom Karzes <karzes@sonic.net> wrote:
Regarding the numbers and classes of hyperplanes:
OEIS entry A007847 gives the hyperplane counts as:
a(1) = 2 a(2) = 6 a(3) = 20 a(4) = 140 a(5) = 3254 ...
This matches what I found (at least up to a(5)).
The classes of hyperplanes are listed below for d = 2-5. Each class is shown below preceded by the number of instances of that class.
I define two hyperplanes as being in the same class of the verticex set of one can be transformed to the vertex set of the other by some combination of (1) permuting the coordinates and (2) inverting some set of coordinates (1->0 and 0->1).
For d=2 there are 2 classes hyperplanes (i.e., lines):
4: 00 01 2: 00 11
This gives a total of 6. The first class corresponds to the 4 perimeter slices.
For d=3 there are 3 classes of hyperplanes (i.e., planes):
6: 000 001 010 011 6: 000 001 110 111 8: 000 011 101
This gives a total of 20. The first class corresponds to the 6 perimeter slices.
For d=4 there are 6 classes of hyperplanes:
8: 0000 0001 0010 0011 0100 0101 0110 0111 12: 0000 0001 0010 0011 1100 1101 1110 1111 32: 0000 0001 0110 0111 1010 1011 16: 0000 0011 0101 1001 8: 0000 0011 0101 1010 1100 1111 64: 0000 0011 0101 1110
This gives a total of 140. The first class corresponds to the 8 perimeter slices.
For d=5 there are 15 classes of hyperplanes:
10: 00000 00001 00010 00011 00100 00101 00110 00111 01000 01001 01010 01011 01100 01101 01110 01111 20: 00000 00001 00010 00011 00100 00101 00110 00111 11000 11001 11010 11011 11100 11101 11110 11111 80: 00000 00001 00010 00011 01100 01101 01110 01111 10100 10101 10110 10111 80: 00000 00001 00110 00111 01010 01011 10010 10011 40: 00000 00001 00110 00111 01010 01011 10100 10101 11000 11001 11110 11111 320: 00000 00001 00110 00111 01010 01011 11100 11101 32: 00000 00011 00101 01001 10001 32: 00000 00011 00101 01001 10010 10100 10111 11000 11011 11101 80: 00000 00011 00101 01001 10110 11010 11100 11111 160: 00000 00011 00101 01001 11110 160: 00000 00011 00101 01010 01100 01111 10110 320: 00000 00011 00101 01110 10110 320: 00000 00011 00101 01110 11000 11011 11101 640: 00000 00011 00101 01110 11001 960: 00000 00011 01101 10101 11010
This gives a total of 3254. The first class corresponds to the 10 perimeter slices.
Tom
Tomas Rokicki writes:
How many hyperplanes do you get in the 4D case? I believe I see 68. 12 through 8 points (cutting the 4D cube in half), 32 through 6 points, and 24 through 4 points.
I'm using a similar approach to yours, so I'm curious why I'm only seeing 18,432 regions for the 4D case.
The number I get for the 5D case is 31,610,880.
On Sat, Apr 11, 2020 at 9:08 AM Veit Elser <ve10@cornell.edu> wrote:
This series starts 4, 96, … and does not seem to be in OEIS because of a bound I have on the next element (which is being improved as I write this).
Consider an interior point of one of these regions: a d-tuple of real numbers between 0 and 1. If we re-order the d-tuple we get a different region because there are planes that distinguish the order of any consecutive pair of numbers in the d-tuple. We may therefore count the number of regions for a particular ordering and multiply by d!.
Using Mathematica I generate random sorted d-tuples and assign them a binary code giving the side of each of the hyperplanes. Whenever a new code is found it is appended to the codebook of regions. Initially the codebook grows rapidly, and then, mercifully, the “curve flattens”.
Reduced by d! the series starts 2, 16, … Right now the next number is at least 23450, with Mathematica discovering new “cases” at a rate of 21/10^5. Any guesses — insights? — where it will end?
-Veit
On Apr 10, 2020, at 9:59 PM, Neil Sloane <njasloane@gmail.com> wrote:
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
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Sorry for being slow to reply. I was having “office hours”. -Veit
On Apr 14, 2020, at 2:44 PM, Neil Sloane <njasloane@gmail.com> wrote:
Thanks to everyone who responded. There are now two new sequences, the main one being A333539, and A333540 if you divide by d!. And A007847 has been expanded with Tom's list of hyperplanes.
participants (4)
-
Neil Sloane -
Tom Karzes -
Tomas Rokicki -
Veit Elser