For the problem of "Ruperting" the n-dimensional cube, octahedron, and regular simplex (i.e. -- what is the max expansion factor E so an E-scaled cube can pass thru a hole drilled in a congruent-copy cube?) Here is a quick summary of my computer's findings so far (expansion factor is a lower bound found by computer; prob is based on Monte Carlo; it tries random rotation matrices then applies a crude optimizer to the more promising candidates): CUBE (& octahedron): n expansion factor theoretical opt prob random orthog matrix works 2 1.41421 sqrt(2) 0.99999 3 1.06066017 sqrt(9/8) 0.01295 4 1.00732 1.00743475688 1 / [ (2.5+-0.4)*10^8 ] 5 1.00074 1.000839446859 below 10^(-11) 6 1.0000627 ? below 10^(-11) SIMPLEX: n expansion factor theoretical opt prob random orthog matrix works 2 1.15470 2/sqrt(3) 0.99999 3 1.01461187 sqrt(6)/(1+sqrt(2))=(2-sqrt(2))*sqrt(3) 0.001282(3) 4 1.000715 ? below 10^(-11) 5 1.00002711 ? below 10^(-11) 6 1.0000006807 ? below 10^(-11) I have proven the cube & octahedron Emax are equal; this is a special case of the more general Rupert-duality theorem. I also have proofs E>1 in all dimensions n>1 for all 3 problems. My proofs yield E>1+const/9^n for cube & oct (same proof found by Greg Huber earlier), and E>1+const^(2^(n-1)) for the simplex. Both proofs involve a rotation matrix which is a small perturbation of the identity matrix. For the simplex, however, the best rotation probably is NOT a small perturbation of the identity, in which case my proof is being intentionally stupid. The best upper bound on E that I have is sqrt(n/(n-1)) so there is a large gap between lower & upper bounds, and the asymptotic behavior of E for large n remains very unknown. The empirical results for the cube do indeed suggest the 1+const/9^n lower bound is approximately tight (up to altering the value of 9 and perhaps some subexponential factor). For the simplex, empirical results suggest the falloff of E-1 is faster than any simple exponential, but not as fast as double exponential. Perhaps for example E-1 =approx= 10^(1-n) / (n-2)!. Kay R. Shultz (and perhaps others?) had introduced the function f(m,n) defined as the largest edgelength for an m-cube which will fit inside a unit n-cube, 0<m<=n. This function can be generalized to apply to any convex n-polytope P_n, not just the n-cube, as follows: f(m,n) is the supremum of the scale factors s for a rotated P_m, such that the cartesian set-product s*P_m x R^(n-m) (which is an "infinite prism with base P_m") can "pass through" a rotated P_n having scale=1. That is, such that there exist mXm and nXn orthogonal matrices Q1 and Q2, such that for each point of s*Q1 P_m, n real numbers x1,x2,...,xn exist (the first m of which are independent of the point) so that this point plus (x1,x2,...,xn), lies strictly inside Q2 P_n. Note, if Q1 and Q2 were somehow known then the maximum possible s and the (x1,x2,...,xn) could be deduced by solving a linear program. If Q1 and Q2 were both demanded to be infinitesimal perturbations of the identity matrix, Q=Ident+epsilon*AntiSymMatrix, then the question of whether s>1 would be answerable by solving a linear program in n dimensions. It can be argued using Hadamard's determinant bounds (or the cruder factorial determinant bound arising by expansion by minors) and Cramer's formula for matrix inverse, that for ANY class of convex polytopes describable using polynomial(n)-bounded integer coefficients (e.g. cubes & octahedra, regular simplex by going to n+1 dimensional representation, permutohedron), this latter linear program shows that if s>1, then s-1 can be made larger than 1/FactoriallyGrowingFunction(n). This immediately improves upon my crude lower bound on E for the n-simplex, which involved double-exponential falloff of E-1 for large n, to show factorial style falloff, specifically k^n/(c*n)! for some positive constants c, k. (I could work out explicit bounds on all these constants if I were less lazy.) And the empirical computer findings for the simplex suggest this sort of behavior actually is the truth. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)