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.