Email from Jon Wild, April 23 2019
Here is a pair of arrangements of four circles that share everything,
more than just whatever the truth table says. In each case you have four
circles such that:
A, B and D form a cycle with only pairwise intersections.
C has part of its disk outside all the other circles, and the rest lies
completely inside A, extending to within A's intersection with B.
This is *everything* about the arrangements' intersections and the
overlapping regions that are formed (in each case: A(twice), B, C, D; AB,
AC, CD, BD; ABC; and they each enclose a region of the plane that is not
inside any of the circles but is bound by parts of them). But clearly
they are not the same arrangement: on the left, the outside boundary is
formed by arcs of A, C, A again, D, and B. On the right, the outside
boundary is simply A, D, B.
There *is* an operation that turns one into the other, and that is to turn
one inside-out, or to evert it around the central bound region that lies
outside all the circles, swapping its boundary with the outer boundary.
This is one of many such legal moves on the projective plane, but of all
the reprojections this one is special because the configurations keep all
their invariants except the outside boundary.
If an arrangement has more than one internal "null" region that is bound
by the exterior of 3 or more circles like this, then it can have more than
one "inside-out" partner arrangement.