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 On 8/13/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
And so to hunt missing relations when n = 4 . Via brute-force equation-solving I discovered that
(k0 + k5) - (k1 + k3) , (k0 + k5) - (k2 + k4) , [RWG]
(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) ,
all vanish (verified numerically).
But the Jacobian (again!) has rank 5, instead of 3. Flubber, flubber ...
Fred Lunnon
On 8/13/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Aaarghhh! Lost the mods --- should have read <<
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 + k0*k5 + k6*k5 = constant .
WFL
On 8/12/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Ahah! RWG's second set of Steiner 5-ring constraints failed my numerical tests because of a wrong sign for k0 --- I assume curvatures have sign built-in, allowing for circles to evert under transformation. I didn't initially suspect this problem, since by Murphy's law the dodgy sign passed when tested on the canonical concentric ring.
So the constraints for n = 5 should have read: 5 cyclic shifts (fixing k0,k6 ) of k1^2 + k3*k5 + k2*k4 + k2*k3 = k3^2 + k5*k2 + k4*k1 + k1*k2 ; and 5 shifts of 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5 = constant .
[My updated versions --- I can't be bothered to correct all the signs in his original! Could the first set also be pruned to some equivalent LHS = const form perhaps?]
Unfortunately, this makes no difference to the rank, which remains 5 rather than 4; I remain baffled by this discrepancy.
Fred Lunnon