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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

481.0. "Cutting the infinite-dimensional cheese" by NMGV01::ASKSIMON (Don't upset the 'Deus ex Machina') Fri May 02 1986 12:34

				     n
	Given a space isomorphic to R  containing a "hypercheese",  and
	such a cheese in which it is possible to inscribe n mutually
	perpendicular line segments,  and moreover,  the cheese is of
  	such a shape that no straight line can pass through the cheese
  	more than once:  Suppose that we may cut this cheese with m
  	(n-1)-planes,  i.e.  surfaces of the general equation:

	    n
	    --
	    \
	    /	  a x	= 0
	    --     i i
	   i <> r
	for some r in {1,...,n}

	Let M:(Z+ x Z+) --> Z+ be the function mapping m and n to the
	maximum number pieces that can be produced using any m cuts
	(note that this also varies with n).  Prove that:

						       m
	    Given any m>0,	Lim       M(m,n)   =  2
			    n -> +infinity

	SDC.
  
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481.1ENGINE::ROTHSat May 03 1986 11:5916
    If the (convex) piece of cheese satisfies the conditions in .0, you
    may let it expand arbitrarily in size without changing the number of
    pieces it's sliced into.  So the problem really depends on how many
    pieces can one subdivide n dimensional Euclidean space into with m
    hyperplanes.

    Clearly, for any m <= n, the number of subdivisions of n-space will
    be 2^m, since you will need m > n to ever completely enclose any finite
    extents of n-space between hyperplanes (eg, with m = n+1 you enclose
    an n-dimensional simplex).

    An interesting combinatorial problem arises with n finite:

    How many pieces can you subdivide n-space into with m > n hyperplanes?

- Jim