Re: [math-fun] Prince Rupert's problem, cubes in cubes, Kay R.P.DV. Schulz
I stumbled upon https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before. It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side. f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier. She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries. Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4). She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution. But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion. Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n. I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected. Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.) Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P? (One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.) Perhaps Pechenik and Shultz already consider such questions? Jim Propp On Wednesday, April 20, 2016, Warren D Smith <warren.wds@gmail.com> wrote:
I stumbled upon
https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf
which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before.
It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side.
f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier.
She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix
f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter
She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries.
Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4).
She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution.
But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion.
Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n.
I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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The "stella octangula" (pair of reciprocal regular tetrahedra), and similar cube + octahedron, dodecahedron + icosahedron, are _almost_ counterexamples of such P,Q , though obstructed by just finitely many points. This suggests that any proof of impossibility is going to be a delicate matter ... WFL On 4/25/16, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P?
(One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.)
Perhaps Pechenik and Shultz already consider such questions?
Jim Propp
On Wednesday, April 20, 2016, Warren D Smith <warren.wds@gmail.com> wrote:
I stumbled upon
https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf
which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before.
It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side.
f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier.
She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix
f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter
She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries.
Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4).
She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution.
But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion.
Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n.
I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Hello math-fun, There are quite a lot of videos of a cube A passing through a copy B of itself -- but I was wondering about a third copy C (of A) through which B would pass simultaneously -- or a A-B-C-D-E...-A chain of cubes interacting in a kind of loop, or a grid made of A-cubes inter- acting with a "through-grid" of B's, which would slide in the 3rd dimension through a C grid, etc. Well, I dunno if this might be of interest, I'd say. My two CGI cents. É.
On Apr 25, 2016, at 11:36 AM, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
How about: Q=convex polyhedron approximation of oblate ellipsoid P=convex polyhedron approximation of prolate ellipsoid If their centers and symmetry axes coincide, and the short(long) diameter of Q is less(greater) than the long(short) diameter of P, then Q\P is a torus. -Veit
I think by "pierce each other", Jim means that P pierces Q *and* Q pierces P. Andy On Mon, Apr 25, 2016 at 2:05 PM, Veit Elser <ve10@cornell.edu> wrote:
On Apr 25, 2016, at 11:36 AM, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
How about:
Q=convex polyhedron approximation of oblate ellipsoid P=convex polyhedron approximation of prolate ellipsoid
If their centers and symmetry axes coincide, and the short(long) diameter of Q is less(greater) than the long(short) diameter of P, then Q\P is a torus.
-Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
This very interesting question of Jim's, below, seems closely related to John Conway's 1997 question about whether there exists a "holyhedron" (revised phrasing as of 2000): "The Holyhedron Problem. Is there a polyhedron in e-dimensional Euclidean space that has only finitely many plane faces, each of which is a closed connected subset of the appropriate plane whose relative interior in that plane is multiply connected?" Jade Vinson found the first example about two years after the problem was posed; it has over 78 million faces. Don Hatch found an example with only 492 faces in about 2003. This seems to be the current record for fewest faces. Conway offered a prize of $10,000 divided by the number of faces for an example of a holyhedron with a record low number of faces. It is unclear whether this offer still stands. ((( Varying the definition slightly to allow a) a noncompact polyhedron in R^3, there is an easy example consisting of a Z^3-translated collection of interpenetrating solid tetrahedra. One then takes the boundary of their union. and b) a holyhedron in S^3, based on 24 likewise interpenetrating spherical tetrahedra. ))) —Dan
On Apr 25, 2016, at 8:36 AM, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P?
(One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.)
Perhaps Pechenik and Shultz already consider such questions?
Maybe for P,Q extract any two tetrahedra from https://en.wikipedia.org/wiki/Compound_of_five_tetrahedra --- I'm having difficulty visualising this! WFL On 4/25/16, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P?
(One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.)
Perhaps Pechenik and Shultz already consider such questions?
Jim Propp
On Wednesday, April 20, 2016, Warren D Smith <warren.wds@gmail.com> wrote:
I stumbled upon
https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf
which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before.
It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side.
f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier.
She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix
f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter
She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries.
Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4).
She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution.
But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion.
Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n.
I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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The mist has cleared --- P - Q is disconnected --- withdrawn! I keep thinking WDS may have had an insight about this; but he lost me at the "recursion" ... WFL On 4/27/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Maybe for P,Q extract any two tetrahedra from https://en.wikipedia.org/wiki/Compound_of_five_tetrahedra --- I'm having difficulty visualising this!
WFL
On 4/25/16, James Propp <jamespropp@gmail.com> wrote:
Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected.
Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.)
Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P?
(One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.)
Perhaps Pechenik and Shultz already consider such questions?
Jim Propp
On Wednesday, April 20, 2016, Warren D Smith <warren.wds@gmail.com> wrote:
I stumbled upon
https://www.kitp.ucsb.edu/sites/default/files/preprints/2013/13-142.pdf
which is a 123-page manuscript / note-collection by Kay R. Pechenick DeVicci Shultz, partly typed and partly handwritten, mostly dating to 1996 or before.
It seems she knew most everything I knew about the cube problem and more. Here is a summary. She defines f(m,n) = the edgelength of the largest m-dimensional cube (or m-cube) which fits into an n-cube of unit side.
f(m,n) = MAX_Q 1 / MAX(j=1..n) SUM(i=1..m) |Q_{i j}| where Q is an nXn orthogonal matrix, for the Rupert problem m=n-1, and the outermost max is over all such orthogonal matrices Q. I'd been calling this "Huber's magic formula" thinking it was due to Greg Huber, but actually it seems K.R.P.DV.S. knew it earlier.
She then gives f(m,n)<f(m,n+1) because: add column of 0s f(m,n)<f(m-1,n) because: delete row f(k,m)*f(m,n)<=f(k,n) because: put k-cube in m-cube and m-cube in n-cube f(m,m+1)>1 (which implies the first two inequalities can use < not merely <=) because: she considers infinitesimal rotations and solves a linear program thus proving existence of an infinitesimal rotation that does better than the identity matrix
f(k*m,k*n)>=f(m,n) because: kronecker matrix product f(m1+m2+...+mk, n1+n2+...+nk) >= min( f(m1,n1), f(m2,n2), ... f(mk,nk) ) f(m,n) <= sqrt(n/m) because of: diameter
She claims as a major result: f(m,n) = sqrt(n/m) if and only if m divides n. Any orthogonal matrix Q accomplishing this contains only +1, -1, and 0 as entries, each row containing exactly n/m nonzero entries, and her example arises from a Kronecker product of mXm identity matrices with an all-1's vector with (n/m) entries.
Finally, she claims in the handwritten part of the manuscript via an immense case analysis to have found the exact optimum to establish the value of f(3,4).
She notes on p.54 that many authors assume without proof that the problem of finding f(n-1,n) and Prince Rupert's problem are equivalent, but she is herself carefully does not claim that, and I agree with her caution.
But... The only reason the two might fail to be equivalent is that perhaps the optimum Rupert way of passing an n-cube A thru another B, happens NEVER during the motion, to have an (n-1)-cube cross-section of A, which lies entirely inside B. But if so, then there must exist some "tilted" cross-section of A which does lie entirely inside B, and this tilted thing is topologically the same as an (n-1)-cube, and indeed a linear transformation of it which as a map is necessarily expansive or length-preserving for every point-pair. Therefore, it seems to me, this tilted cross-section must include as a SUBSET, a set geometrically congruent to the untilted cross-section. If so, that proves the two problems really ARE equivalent -- any Rupert-motion must involve a moment when an (n-1)-cube cross-section of A lies inside B; and conversely the largest such inscribing yields a Rupert-motion.
Meanwhile... I have been working on the Prince Rupert problem but not for cubes, rather for other regular polytopes, especially simplexes and octahedra. My current claim about that (which unfortunately differs from what I used to think) is the n-octahedral Rupert problem actually has the SAME expansion factor as the n-cube Rupert problem (!). For the n-simplex Rupert problem I have a formula involving orthogonal matrix, and my computer by applying that formula to random orthogonal matrices has rediscovered the known results for n=2 and n=3. (Also for the cube problem, for n=2, 3, and 4.) But it so far has failed to find a simplex solution for n=4, which failure I find surprising since I used to think it was clear a solution existed for all n.
I think Kay R.P.DV.Schultz's idea about linear program and infinitesimal rotations is an excellent insight which could be applied to any other polytope besides n-cube to determine whether any slight rotation exists that would work. I.e. it will rapidly decide Rupertability for any convex polytope, provided the Rupert is achievable via an infinitesimally tiny rotation angle.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (7)
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Andy Latto -
Dan Asimov -
Eric Angelini -
Fred Lunnon -
James Propp -
Veit Elser -
Warren D Smith