[math-fun] How many pieces? puzzle
Let the n-dimensional torus T^n be defined as the n-cube Q_n = [-1, 1]^n with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.) The vertices of Q_n are the 2^n points {-1, 1}^n. Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus. Puzzle: How many pieces do these planes cut the torus T^n into? E.g., for n = 2 there are two pieces. —Dan
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 . See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb WFL On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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A related question is, what is the max number of pieces you can cut a bagel into with 3 cuts? Answer: 13 For n cuts, see https://oeis.org/A003600 But what about a solid figure-8, or a 4-D bagel (or torus)? 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 Sun, Sep 16, 2018 at 8:07 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 .
See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb
WFL
On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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For n = 4 , I conjecture 72 pieces, comprising: 8 regular cross-topes (16-cell dual tesseracts); 64 di-pyramid 8-cells with tetrahedral bases. This claim would require checking for combinatorial consistency, and for sum of contents = 16 with edge-lengths = sqrt2 . More details provided on request. It should not be difficult to push such computations further using computer assistance: I have sketched an algorithm, though not implemented a program. The numbers of pieces grows exponentially with n ; however it is necessary to investigate only a much smaller (linear?) subset of distinct types. The geometry required seems much simpler in covering Euclidean n-space, where the configuration lifts to a lattice tesselation --- effectively an infinite (n+1)-dimensional polytope. Fred Lunnon Fred Lunnon<fred.lunnon@gmail.com> Mon, Sep 17, 2018 at 2:53 PM To: Andy Latto <andy.latto@pobox.com> Andy, Octahedra have edge-length sqrt2 < 2 , and vertices at mid-points of cube edges; they meet other octahedra only only along tiling edges and vertices. Tetrahedra fill the spaces remaining, for instance with vertices at [0,0,0], [0,1,1], [1,0,1], [1,1,0] . The corresponding Euclidean lattice got by "unwrapping" the 3-torus has 4 families of parallel planes; a cube is cut by 3 planes from each family. Consult the small exploded diagram near the start of the Wikipedia page --- though you need to imagine an edge-length 2 cube! Fred On 9/17/18, Neil Sloane <njasloane@gmail.com> wrote:
A related question is, what is the max number of pieces you can cut a bagel into with 3 cuts? Answer: 13 For n cuts, see https://oeis.org/A003600
But what about a solid figure-8, or a 4-D bagel (or torus)?
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 Sun, Sep 16, 2018 at 8:07 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 .
See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb
WFL
On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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Duh --- light dawns! Unwinding Dan's n-torus into its universal covering tesselation of Euclidean space yields simply what seems to be known only as " n-space analogue of the face-centred cubic crystal tiling" or such (maybe Neil can correct this?): the unit-sphere packing where centres have evenly many odd integer Cartesian components. From the combinatorics of the hypercube it is now straightforward that the number of regions in the torus equals 2^n 3^(n-1) ; also the number of vertices equals 2^(n-1) . I might flesh this out with more detail on request. Fred Lunnon On 9/19/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 4 , I conjecture 72 pieces, comprising: 8 regular cross-topes (16-cell dual tesseracts); 64 di-pyramid 8-cells with tetrahedral bases. This claim would require checking for combinatorial consistency, and for sum of contents = 16 with edge-lengths = sqrt2 . More details provided on request.
It should not be difficult to push such computations further using computer assistance: I have sketched an algorithm, though not implemented a program. The numbers of pieces grows exponentially with n ; however it is necessary to investigate only a much smaller (linear?) subset of distinct types. The geometry required seems much simpler in covering Euclidean n-space, where the configuration lifts to a lattice tesselation --- effectively an infinite (n+1)-dimensional polytope.
Fred Lunnon
Fred Lunnon<fred.lunnon@gmail.com> Mon, Sep 17, 2018 at 2:53 PM To: Andy Latto <andy.latto@pobox.com>
Andy,
Octahedra have edge-length sqrt2 < 2 , and vertices at mid-points of cube edges; they meet other octahedra only only along tiling edges and vertices. Tetrahedra fill the spaces remaining, for instance with vertices at [0,0,0], [0,1,1], [1,0,1], [1,1,0] .
The corresponding Euclidean lattice got by "unwrapping" the 3-torus has 4 families of parallel planes; a cube is cut by 3 planes from each family. Consult the small exploded diagram near the start of the Wikipedia page --- though you need to imagine an edge-length 2 cube!
Fred
On 9/17/18, Neil Sloane <njasloane@gmail.com> wrote:
A related question is, what is the max number of pieces you can cut a bagel into with 3 cuts? Answer: 13 For n cuts, see https://oeis.org/A003600
But what about a solid figure-8, or a 4-D bagel (or torus)?
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 Sun, Sep 16, 2018 at 8:07 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 .
See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb
WFL
On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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Corrected version, with customary apologies: Duh --- light dawns! Unwinding Dan's n-torus into its universal covering tesselation of Euclidean space yields simply what seems to be known only as " n-space analogue of the face-centred cubic crystal tiling" or some such (maybe Neil can correct this?): the unit-sphere packing where centres have evenly many odd integer Cartesian components. From the combinatorics of the hypercube it is now straightforward that the number of regions in the torus equals 2^(n-1) 3^(n-2) ; also the number of vertices equals 2^(n-1) . I might flesh this out with more detail on request. Fred Lunnon
On 9/19/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 4 , I conjecture 72 pieces, comprising: 8 regular cross-topes (16-cell dual tesseracts); 64 di-pyramid 8-cells with tetrahedral bases. This claim would require checking for combinatorial consistency, and for sum of contents = 16 with edge-lengths = sqrt2 . More details provided on request.
It should not be difficult to push such computations further using computer assistance: I have sketched an algorithm, though not implemented a program. The numbers of pieces grows exponentially with n ; however it is necessary to investigate only a much smaller (linear?) subset of distinct types. The geometry required seems much simpler in covering Euclidean n-space, where the configuration lifts to a lattice tesselation --- effectively an infinite (n+1)-dimensional polytope.
Fred Lunnon
Fred Lunnon<fred.lunnon@gmail.com> Mon, Sep 17, 2018 at 2:53 PM To: Andy Latto <andy.latto@pobox.com>
Andy,
Octahedra have edge-length sqrt2 < 2 , and vertices at mid-points of cube edges; they meet other octahedra only only along tiling edges and vertices. Tetrahedra fill the spaces remaining, for instance with vertices at [0,0,0], [0,1,1], [1,0,1], [1,1,0] .
The corresponding Euclidean lattice got by "unwrapping" the 3-torus has 4 families of parallel planes; a cube is cut by 3 planes from each family. Consult the small exploded diagram near the start of the Wikipedia page --- though you need to imagine an edge-length 2 cube!
Fred
On 9/17/18, Neil Sloane <njasloane@gmail.com> wrote:
A related question is, what is the max number of pieces you can cut a bagel into with 3 cuts? Answer: 13 For n cuts, see https://oeis.org/A003600
But what about a solid figure-8, or a 4-D bagel (or torus)?
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 Sun, Sep 16, 2018 at 8:07 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 .
See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb
WFL
On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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Neil Sloane