Meanwhile, Rich points out, privately for some reason, that the curvatures in the n=6 case form a "magic rectangle": gosper.org/stein6.png --rwg WFL> Oh dear --- the outer circle had been getting the wrong sign, not the inner --- give me two options, and I unfailingly pick the wrong one. This came to light when I slotted in non-random n = 4 test case k1,k2,k3,k4,k5,k0 = 1,2,1,0,2,0 . So (third time lucky?) Steiner 4-ring constraints seem to be (k0 + k5) - (k1 + k3) , (k0 + k5) - (k2 + k4) , (k0 - k5)^2 - 4*(k3 + k4)*(k0 + k5) + 4*(k3^2 + k4^2) , (k0 - k5)^2 - 4*(k2 + k3)*(k0 + k5) + 4*(k2^2 + k3^2) , (k0 - k5)^2 - 4*(k1 + k2)*(k0 + k5) + 4*(k1^2 + k2^2) , (k0 - k5)^2 - 4*(k4 + k1)*(k0 + k5) + 4*(k4^2 + k1^2) . Their Groebner basis very neatly comprises the two linear relations, together with a single quadratic: hence the dimension of the constraints is (at most) 3 = n-1, as desired (easily it must exceed 2). [The Jacobian rank still comes out as 5 --- I must be misconstruing something elementary there, but it doesn't affect things for the present.] Note that earlier constraints and test-cases proposed for n = 5 should have signs of k0,k6 reversed. Fred Lunnon