[math-fun] Ruperting the hypercube
Take an N-dimensional cube. Rotate it randomly. The chance the original hypercube will be able to pass thru the result ("drilled hole" in "Prince Rupert's problem") by motion along the x-axis is (says Monte Carlo): N chance 2 99.999% 3 1.295% =approx= 1/77 =approx= exp(-4.35) 4 1 / (4*10^8) =approx= exp(-20)
=5 seems below 10^(-10)
so empirically it looks as though the chance depends exponentially on a quadratic or cubic of N, for example chance =approx= N^(-((N-1)*N-2)) or something. That in turn would suggest that the maximum expansion factor indeed behaves like 1 + C^(-N) for some C>1, meaning the lower bound 1+const/9^N already has the right behavior (aside from changing the value of 9, and possible subexponential adjustments). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
participants (1)
-
Warren D Smith