Magic Money coins

[Eric Hughes]

In thinking about my own averaging technique for finding
near-identities, I realize it needs some modification.

Remember the example that 3 was a near-identity near n/2.  Well so is
5, and 7, and -3, -5, -7, etc.  Even though 3 (or -1) seems to be the
best of the near-inverses, any one whose action is sufficiently
bounded will do.

The new observation is that the candidates for near-inverses will be
clustered and not distributed flatly over the ring.  There will also
be more than one cluster.  So you've got two choices.  First make a
histogram of the candidate choices and only average by clusters.
Secondly, one might also be able to transpose the clusters onto each
other and average them all.  The inverse image of this transposition
may also yield more near-inverses.

I think that averaging can be made to work, but it's not obvious to me
exactly what the technique will be.

Eric