Averaging problem

[franktaw at netscape.net]

Correction: if you take a number in the middle, the weight 2^-1 is not
possible.  Thus for n>2, the value n-2 is always obtainable.

I still don't see any way to get a value strictly between 1 and n-2.

Franklin T. Adams-Watters

-----Original Message-----
From: franktaw at netscape.net

I don't think that all possible values can be obtained.  For n > 1, any
given number can be given any power of 2 weight between 2^-1
and 2^-(n-1).  This means that operands all the same except for one
different value will produce n-1 distinct values.  While 1 value is of
course always obtainable (just take all operands equal), I don't see
any way to get strictly between 1 and n-1 distinct values.  I don't
know if there are other unobtainable values between n-1 and C(n-1),
but I expect there are.

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