[math-fun] Upending the sphere — the discrete version: Puzzle
Label each unit grid square Q_(K,L) = {(x,y) in R^2 | |x-K| <= 1/2 and |y-L| <= 1/2} in the plane by the pair of integers (K, L) that are the coordinates of its center. Think of the plane R^2 as R^2 x {0} in R^3, an infinite tabletop. At time t = 0, place a unit cube C on Q(0,0). Assume the cube C rests on Q(K,L) at time t = N. A legal move, taken once per unit time step, is to rotate C by an angle of 90º about one of its 4 edges that are touching the tabletop at time t = N, so that it ends up at time t = N+1 resting on one of Q(K+1,L), Q(K-1,L), Q(K,L+1), or Q(K,L-1), with the side down at t = N+1 being one of the 4 sides of C that were vertical at t = N. Puzzle: ------- a) What is the smallest time t at which C can end up resting on Q(0,0) with its initially down side now facing up? Now what if we had used instead b) an octahedron or c) an icosahedron, on the triangular grid, with the corresponding question? (I.e., on each move the solid is rotated by π - (dihedral), so that it ends up on an adjacent triangle, resting on one of its faces adjacent to the one it had just been resting on.) d) The same question for a dodecahedron! (It doesn't matter that there is no tiling of the plane by regular pentagons; the question makes sense anyway.) —Dan
I could have sworn I had done this puzzle before. Maybe we've even discussed it here. But I don't remember the details of the solution. I can see that one can certainly do it in [answer redacted, but it is less than 105] steps, but I'm not sure how to prove that that is minimal. Using extremely similar tricks I have upper bounds for all the polyhedra listed. On Wed, Nov 7, 2018 at 6:40 PM Dan Asimov <dasimov@earthlink.net> wrote:
Label each unit grid square
Q_(K,L) = {(x,y) in R^2 | |x-K| <= 1/2 and |y-L| <= 1/2}
in the plane by the pair of integers (K, L) that are the coordinates of its center.
Think of the plane R^2 as R^2 x {0} in R^3, an infinite tabletop.
At time t = 0, place a unit cube C on Q(0,0).
Assume the cube C rests on Q(K,L) at time t = N.
A legal move, taken once per unit time step, is to rotate C by an angle of 90º about one of its 4 edges that are touching the tabletop at time t = N, so that it ends up at time t = N+1 resting on one of Q(K+1,L), Q(K-1,L), Q(K,L+1), or Q(K,L-1), with the side down at t = N+1 being one of the 4 sides of C that were vertical at t = N.
Puzzle: -------
a) What is the smallest time t at which C can end up resting on Q(0,0) with its initially down side now facing up?
Now what if we had used instead
b) an octahedron or
c) an icosahedron,
on the triangular grid, with the corresponding question? (I.e., on each move the solid is rotated by π - (dihedral), so that it ends up on an adjacent triangle, resting on one of its faces adjacent to the one it had just been resting on.)
d) The same question for a dodecahedron! (It doesn't matter that there is no tiling of the plane by regular pentagons; the question makes sense anyway.)
—Dan
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Wait, I don't have an answer for the octohedron, and suspect it can't be done. On Wed, Nov 7, 2018 at 7:05 PM Allan Wechsler <acwacw@gmail.com> wrote:
I could have sworn I had done this puzzle before. Maybe we've even discussed it here. But I don't remember the details of the solution. I can see that one can certainly do it in [answer redacted, but it is less than 105] steps, but I'm not sure how to prove that that is minimal. Using extremely similar tricks I have upper bounds for all the polyhedra listed.
On Wed, Nov 7, 2018 at 6:40 PM Dan Asimov <dasimov@earthlink.net> wrote:
Label each unit grid square
Q_(K,L) = {(x,y) in R^2 | |x-K| <= 1/2 and |y-L| <= 1/2}
in the plane by the pair of integers (K, L) that are the coordinates of its center.
Think of the plane R^2 as R^2 x {0} in R^3, an infinite tabletop.
At time t = 0, place a unit cube C on Q(0,0).
Assume the cube C rests on Q(K,L) at time t = N.
A legal move, taken once per unit time step, is to rotate C by an angle of 90º about one of its 4 edges that are touching the tabletop at time t = N, so that it ends up at time t = N+1 resting on one of Q(K+1,L), Q(K-1,L), Q(K,L+1), or Q(K,L-1), with the side down at t = N+1 being one of the 4 sides of C that were vertical at t = N.
Puzzle: -------
a) What is the smallest time t at which C can end up resting on Q(0,0) with its initially down side now facing up?
Now what if we had used instead
b) an octahedron or
c) an icosahedron,
on the triangular grid, with the corresponding question? (I.e., on each move the solid is rotated by π - (dihedral), so that it ends up on an adjacent triangle, resting on one of its faces adjacent to the one it had just been resting on.)
d) The same question for a dodecahedron! (It doesn't matter that there is no tiling of the plane by regular pentagons; the question makes sense anyway.)
—Dan
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On Wed, Nov 7, 2018 at 5:10 PM Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I don't have an answer for the octohedron, and suspect it can't be done.
I think you're right, since you can color both the octahedron and the triangular tiling with two colors such that any move also changes the color of the tangent faces. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
The rhombic dodecahedron also can't be done, but for entirely different reasons. I think. On Wed, Nov 7, 2018 at 7:17 PM Mike Stay <metaweta@gmail.com> wrote:
On Wed, Nov 7, 2018 at 5:10 PM Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I don't have an answer for the octohedron, and suspect it can't be done.
I think you're right, since you can color both the octahedron and the triangular tiling with two colors such that any move also changes the color of the tangent faces.
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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I am pretty sure the cuboctohedron can be done, though. On Wed, Nov 7, 2018 at 7:24 PM Allan Wechsler <acwacw@gmail.com> wrote:
The rhombic dodecahedron also can't be done, but for entirely different reasons. I think.
On Wed, Nov 7, 2018 at 7:17 PM Mike Stay <metaweta@gmail.com> wrote:
On Wed, Nov 7, 2018 at 5:10 PM Allan Wechsler <acwacw@gmail.com> wrote:
Wait, I don't have an answer for the octohedron, and suspect it can't be done.
I think you're right, since you can color both the octahedron and the triangular tiling with two colors such that any move also changes the color of the tangent faces.
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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participants (3)
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Allan Wechsler -
Dan Asimov -
Mike Stay