Polyhedral Methods for Positive Dimensional Solution Sets

Transcription

1 Polyhedral Methods for Positive Dimensional Solution Sets Danko Adrovic and Jan Verschelde th International Conference on Applications of Computer Algebra The Hilton Southwest Hotel, Houston, USA Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 1 / 32

2 Introduction goal: extend polyhedral homotopies for isolated solutions to polyhedral methods to find space curves as solutions main objects: tropism Puiseux series as an illustration: cyclic n-roots polynomial systems, n = 4 through 7 emphasis on cyclic 8-roots polynomial system verify that the total degree is 144 we take symmetry into account assumptions on the space curves we can find: they are reduced, free of multiplicities in general position with respect to x 1 = 0 Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 2 / 32

3 Fundamental Theorem Theorem (Fundamental Theorem Of Tropical Algebraic Geometry) ω Trop(I ) Q n p V (I ) : val(p) = ω Q n. Rephrasing the Theorem: Every rational vector in the tropical variety corresponds to the leading powers of a Puiseux series, converging to a point in the algebraic variety. For a constructive proof of the Fundamental Theorem, we refer to Anders Nedergaard Jensen, Hannah Markwig, Thomas Markwig: An Algorithm for Lifting Points in a Tropical Variety. Collect. Math. vol. 59, no. 2, pages , Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 3 / 32

4 Mixed Volume For square (#equations = #variables) systems: Theorem (Bernshtein s Theorem A) The number of roots of a generic system equals the mixed volume of its Newton polytopes. For any system, the mixed volume bounds the number of isolated solutions in (C ) n, C = C \ {0}. Thm A bounds #solutions and Thm B tells when bound is exact. Theorem (Bernshtein s Theorem B) Consider F (x) = 0, F = (f 1, f 2,..., f n ), x = (x 1, x 2,..., x n ). If for all V 0: in V F (x) = 0 has no solutions in (C ) n, then F (x) = 0 has exactly as many isolated solutions in (C ) n as the mixed volume. We see Fundamental Theorem of Tropical Algebraic Geometry as a generalization of Bernshtein s Theorem B. We use Bernshtein s Theorem A &B as a way to solve polynomial systems with polyhedral methods. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 4 / 32

5 Polyhedral Homotopies The polyhedral method can be seen in the light of polyhedral homotopies. We can lift polynomials by introducing a parameter t. f (x) = a A c a x a f (x) = a A c a x a t ω (a) where ω is the lifting function, c a random constant. We look for solution curves of the form: x 1 = t x 2 = t v 2 (b 2 + O(t)) x 3 = t v 3 (b 3 + O(t)). x n = t vn (b n + O(t)) where t goes from 0 to 1. We want to generalize polyhedral homotopies for isolated solutions to polyhedral homotopies for space curves. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 5 / 32

6 Cyclic n-roots Polynomial Systems The cyclic n-roots polynomial systems are benchmark problems for polynomial system solvers. Denote x = (x 1, x 2,..., x n ). f 1 = x 0 + x x n = 0 f 2 = x 0 x 1 + x 1 x x n 1 x n + x n x 0 = 0 F (x) = n 1 i i = 3, 4,..., n 1 : x k mod n = 0 j=0 k=j f n = x 0 x 1 x 2... x n 1 = 0 cyclic n-roots polynomial systems: square systems: we expect isolated solutions but for cyclic 4- & 8-roots, we have space curves J. Backelin: Square multiples n give infinitely many cyclic n-roots. Reports, Matematiska Institutionen, Stockholms Universitet, Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 6 / 32

7 Cyclic n-roots Statistics Cyclic n-roots Polynomial Systems n Mixed Volume #V 0 #V #V 0 = number of isolated solutions #V 1 = degree of the space curve solution Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 7 / 32

8 Basic Definitions Polynomial System f 1 = 0 f 2 = 0 F (x) =. f k = 0 x = (x 1, x 2,..., x n ), f i C[x] We define the support sets of F(x) = 0 to be: (A 1, A 2,..., A k ) = (Supp(f 1 ), Supp(f 2 ),..., Supp(f k )) with A i = ((a 11, a 12,..., a 1n ), (a 21, a 22,..., a 2n ),..., (a N1, a N2,..., a Nn )), where N is the number of monomials in f i, a j Z and (a j1, a j2,..., a jn ) are exponents of a monomial in f i. Support set A i of f i spans the Newton polytope as P i = ConvexHull(A i ) Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 8 / 32

9 The Cayley Embedding & Polytope We define the Cayley embedding C E of the set A as k 1 C E (A 1, A 2,... A k ) = (A 1 {0} k 1 ) (A i+1 e i ) where e i denotes the i th vector of the standard basis. We define the Cayley polytope of the Cayley embedding as i=1 C = conv(c E ) R n+k 1 NOTE The Cayley embedding results in a convex polytope of dimension m = n + k 1, where n denotes the number of variables and k denotes the number of polynomials in the system. We use cddlib of K. Fukuda for all facet normals of the Cayley polytope. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 9 / 32

10 Formal Power Series Definition (Puiseux Series) Puiseux series is a formal power series, in which the symbol for the indeterminate is allowed to carry negative and fractional exponents. + p=r c p t p q q N, r Z, c p C Our polyhedral method for space curve has its origin in the Newton-Puiseux algorithm for expansions of algebraic curves in the plane. polynomials and polygons slopes of polygon edges and exponents of the series For more information on algebraic curves in the plane, we refer to: R.J. Walker: Algebraic Curves, 1950, Princeton University Press. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 10 / 32

11 Tropisms and Space Curves Definition (Pretropism) A pretropism is a normal vector to at least an edge of each polytope. a pretropism might generate a Puiseux series expansion of a space curve Let V = (v 1, v 2,..., v n ) be a pretropism and v 1 > 0: x 1 = t v 1 (b 1 + c 1 t w ) Definition (Tropism) x 2 = t v 2 (b 2 + c 2 t w ) G(x, t) = x 3 = t v 3 (b 3 + c 3 t w ). x n = t vn (b n + c n t wn +... ) A tropism is a pretropism which is the leading exponent vector of a Puiseux series expansion for a curve, expanded about t = 0. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 11 / 32

12 Initial Ideal Definition (Initial Form) Let f be a polynomial with support A and let V be a pretropism. Then the initial form in V (f ) is the sum of all monomials in f, where the inner product a, V reaches its minimum at least twice over a A. Initial Form System For a system F (x) = 0, F = (f 1, f 2,..., f k ), and pretropism V, the initial form system is defined by in V (F ) = (in V (f 1 ), in V (f 2 ),..., in V (f k )). Solving initial form system leads to solutions at infinity that are isolated, or to the leading coefficients of the Puiseux expansion of a curve. Puiseux series expansion of a curve: solutions at infinity, denoted by b i Let V = (v 1, v 2,..., v n ) be a pretropism and let t denote a free variable: x i = t v i (b i + c i t w i + ), i = 1, 2,..., k. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 12 / 32

13 Power Transformation Power Transform advantages: acts on exponents works with exact data, i.e. integers leaves approximate coefficients of monomials unaffected We are interested in a particular kind of power transform: Definition (Unimodular Transform) Unimodular matrix M is a square, nonsingular, integer matrix with det(m) = ±1. Furthermore, unimodular matrices of order n form a group under matrix multiplication. The group is GL n (Z), the general linear group. Unimodular power transformation: transforms initial form systems to get to the canonical form for Puiseux series expansion Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 13 / 32

14 Cyclic 4-Roots System The only pretropism is (1, 1, 1, 1) Cyclic 4-Roots Initial Form In Direction (1, 1, 1, 1) x 1 + x 3 = 0 x 0 x 1 + x 0 x 3 + x 1 x 2 + x 2 x 3 = 0 in (1, 1,1, 1) (F )(x) = x 0 x 1 x 3 + x 1 x 2 x 3 = 0 x 0 x 1 x 2 x 3 1 = 0 Using U to transform in (1, 1,1, 1) (F ): U = x 0 = z 0 ; x 1 = z 1 z 0 ; x 2 = z 0 z 2 ; x 3 = z 3 z 0 in (1, 1,1, 1) (F )(z) = z 1 /z 0 + z 3 /z 0 = 0 z 1 z 2 + z 2 z 3 + z 1 + z 3 = 0 z 1 z 2 z 3 /z 0 + z 1 z 3 /z 0 = 0 z 1 z 2 z 3 1 = 0 Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 14 / 32

15 Cyclic 4-Roots System Cyclic 4-Root Polynomial System Transformed z 1 + z 3 = 0 z 1 z 2 + z 2 z 3 + z 1 + z 3 = 0 in (1, 1,1, 1) (F )(z) = z 1 z 2 z 3 + z 1 z 3 = 0 z 1 z 2 z 3 1 = 0 Solutions of the transformed initial form system are (z 1 = 1, z 2 = 1, z 3 = 1) and (z 1 = 1, z 2 = 1, z 3 = 1). Let z 0 = t: For cyclic 4-roots, the initial terms of the series are exact solutions x 0 = t 1 x 0 = t 1 x 1 = t 1 x 1 = t 1 x 2 = t 1 and x 2 = t 1 x 1 = t 1 x 1 = t 1 Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 15 / 32

16 Cyclic 5-Roots System Cyclic 5-Roots Polynomial System x 0 + x 1 + x 2 + x 3 + x 4 = 0 x 0 x 1 + x 0 x 4 + x 1 x 2 + x 2 x 3 + x 3 x 4 = 0 F (x) = x 0 x 1 x 2 + x 0 x 1 x 4 + x 0 x 3 x 4 + x 1 x 2 x 3 + x 2 x 3 x 4 = 0 x 0 x 1 x 2 x 3 + x 0 x 1 x 2 x 4 + x 0 x 1 x 3 x 4 + x 0 x 2 x 3 x 4 + x 1 x 2 x 3 x 4 = 0 x 0 x 1 x 2 x 3 x 4 1 = 0 has only isolated solutions all Newton polytopes are in generic position mixed volume is sharp and equals 70. We want to exploit the cyclic permutation symmetry. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 16 / 32

17 Cyclic 5-Roots System First 4 Equations of Cyclic 5-Roots Polynomial System x 0 + x 1 + x 2 + x 3 + x 4 = 0 x 0 x 1 + x 0 x 4 + x 1 x 2 + x 2 x 3 + x 3 x 4 = 0 F (x) = x 0 x 1 x 2 + x 0 x 1 x 4 + x 0 x 3 x 4 + x 1 x 2 x 3 + x 2 x 3 x 4 = 0 x 0 x 1 x 2 x 3 + x 0 x 1 x 2 x 4 + x 0 x 1 x 3 x 4 + x 0 x 2 x 3 x 4 + x 1 x 2 x 3 x 4 = 0 homogeneous system, i.e. embedded in projective space the solutions are lines tropism (1,1,1,1,1) or ( 1, 1, 1, 1, 1), via unimodular coordinate transformation: eliminate one variable from the system return the system to affine space Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 17 / 32

18 Cyclic 5-Roots System Unimodular matrix: U = Coordinate transformation: x 0 = z 0 x 1 = z 0 z 1 x 2 = z 0 z 2 x 3 = z 0 z 3 x 4 = z 0 z 4 z 1 + z 2 + z 3 + z = 0 z 1 z 2 + z 2 z 3 + z 3 z 4 + z 1 + z 4 = 0 in (1,1,1,1,1) (F )(z) = z 1 z 2 z 3 + z 2 z 3 z 4 + z 1 z 2 + z 1 z 4 + z 3 z 4 = 0 z 1 z 2 z 3 z 4 + z 1 z 2 z 3 + z 1 z 2 z 4 + z 1 z 3 z 4 + z 2 z 3 z 4 = 0 Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 18 / 32

19 Cyclic 5-Roots System z 1 + z 2 + z 3 + z = 0 z 1 z 2 + z 2 z 3 + z 3 z 4 + z 1 + z 4 = 0 in (1,1,1,1,1) (F )(z) = z 1 z 2 z 3 + z 2 z 3 z 4 + z 1 z 2 + z 1 z 4 + z 3 z 4 = 0 z 1 z 2 z 3 z 4 + z 1 z 2 z 3 + z 1 z 2 z 4 + z 1 z 3 z 4 + z 2 z 3 z 4 = 0 has 14 isolated solutions, e.g.: z 1 = c 1, z 2 = c 2, z 3 = c 3, z 4 = c 4. For z 0 = t, in the original coordinates we have x 0 = t, x 1 = tc 1, x 2 = tc 2, x 3 = tc 3, x 4 = tc 4, as representations for the 14 solution lines. Substituting into the omitted equation x 0 x 1 x 2 x 3 x 4 1 = 0, yields a univariate polynomial in t of the form kt 5 1 = 0, where k is a constant. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 19 / 32

20 Cyclic 5-Roots System Out of the 14 solutions: 10 are of the form t 5 1 accounting for 10 5 = 50 solutions 2 are of the form ( )t 5 1 accounting for 2 5 = 10 solutions 2 are of the form ( )t 5 1 accounting for 2 5 = 10 solutions Accounting for all 70 solutions of the cyclic 5-roots system. NOTE Additional symmetry: 1 ( ) Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 20 / 32

21 Cyclic 8-Roots System Cyclic 8-roots system: 831 facet normals (computed with cddlib) 29 pretropism generators 5 lead to initial forms with solutions (1, 1, 0, 1, 0, 0, 1, 0) (1, 1, 1, 1, 1, 1, 1, 1) (1, 0, 1, 0, 0, 1, 0, 1) (1, 0, 1, 1, 0, 1, 0, 0) (1, 0, 0, 1, 0, 1, 1, 0) For the initial form solutions we used the blackbox solver of PHCpack. Symbolic manipulations for the computation of the second term of the Puiseux series were done with Sage. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 21 / 32

22 Cyclic 8-Roots System For the pretropism V = (1, 1, 0, 1, 0, 0, 1, 0), the initial form system is x 1 + x 6 = 0 x 1 x 2 + x 5 x 6 + x 6 x 7 = 0 x 4 x 5 x 6 + x 5 x 6 x 7 = 0 x 0 x 1 x 6 x 7 + x 4 x 5 x 6 x 7 = 0 in V (F )(x) = x 0 x 1 x 2 x 6 x 7 + x 0 x 1 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 5 x 6 x 7 + x 0 x 1 x 4 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 x 5 x 6 = 0 x 0 x 1 x 2 x 4 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 1 = 0 V defines the unimodular coordinate transformation: x 0 = z 0, x 1 = z 1 /z 0, x 2 = z 2, x 3 = z 0 z 3, x 4 = z 4, x 5 = z 5, x 6 = z 6 /z 0, x 7 = z 7. Using the new coordinates, we transform the initial form system in V (F )(x). Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 22 / 32

23 Cyclic 8-Roots System z 1 + z 6 = 0 z 1 z 2 + z 5 z 6 + z 6 z 7 = 0 z 4 z 5 z 6 + z 5 z 6 z 7 = 0 z 4 z 5 z 6 z 7 + z 1 z 6 z 7 = 0 in V (F )(z) = z 1 z 2 z 6 z 7 + z 1 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 + z 1 z 2 z 5 z 6 z 7 + z 1 z 4 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 z 7 + z 1 z 2 z 4 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 z 7 1 = 0 Solving in V (F )(z), we obtain 8 solutions (all in the same orbit). We select z 0 = t, z 1 = I, z 2 = 1 2 I 2, z 3 = 1, z 4 = 1 + I, z 5 = I 2, z 6 = I, z 7 = 1 I, I = 1. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 23 / 32

24 Cyclic 8-Roots System Taking solution at infinity, we build a series of the form: z0 = t z1 = I + c 1 t z2 = 1 2 I 2 + c 2t z3 = 1 + c 3 t z4 = 1 + I + c 4 t z5 = I 2 + c 5t z6 = I + c 6 t z7 = ( 1 I ) + c 7 t Plugging series form into transformed system, collecting all coefficients of t 1, solving yields c 1 = 1 I c 2 = 1 2 c 3 = 0 c 4 = 1 c 5 = 1 2 c 6 = 1 + I c 7 = 1 The second term in the series, still in the transformed coordinates: z0 = t z1 = I + ( 1 I )t z2 = 1 2 I t z3 = 1 z4 = 1 + I t z5 = I t z6 = I + (1 + I )t z7 = ( 1 I ) + t Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 24 / 32

25 Cyclic 8-Roots System In certain instances one term in the Puiseux series satisfies the entire system. The initial form in direction V = (1, 1, 1, 1, 1, 1, 1, 1) is in V (F )(x) = x 1 + x 3 + x 5 + x 7 = 0 x 0 x 1 + x 0 x 7 + x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 5 + x 5 x 6 + x 6 x 7 = 0 x 0 x 1 x 7 + x 1 x 2 x 3 + x 3 x 4 x 5 + x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 + x 0 x 1 x 2 x 7 + x 0 x 1 x 6 x 7 + x 0 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 + x 2 x 3 x 4 x 5 + x 3 x 4 x 5 x 6 + x 4 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 x 7 + x 0 x 1 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 x 5 + x 3 x 4 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 x 4 x 5 + x 0 x 1 x 2 x 3 x 4 x 7 + x 0 x 1 x 2 x 3 x 6 x 7 + x 0 x 1 x 2 x 5 x 6 x 7 + x 0 x 1 x 4 x 5 x 6 x 7 + x 0 x 3 x 4 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 x 5 x 6 + x 2 x 3 x 4 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 x 4 x 5 x 7 + x 0 x 1 x 2 x 3 x 5 x 6 x 7 + x 0 x 1 x 3 x 4 x 5 x 6 x 7 + x 1 x 2 x 3 x 4 x 5 x 6 x 7 = 0 x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 1 = 0 Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 25 / 32

26 Cyclic 8-Roots System The unimodular matrix U = and its corresponding coordinate transformation: x 0 = z 0, x 1 = z 1 /z 0, x 2 = z 0 z 2, x 3 = z 3 /z 0, x 4 = z 0 z 4, x 5 = z 5 /z 0, x 6 = z 0 z 6, x 7 = z 7 /z 0. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 26 / 32

27 Cyclic 8-Roots System Initial form for V = (1, 1, 1, 1, 1, 1, 1, 1) after transformation, in V (F )(z) z 1 + z 3 + z 5 + z 7 = 0 z 1 z 2 + z 2 z 3 + z 3 z 4 + z 4 z 5 + z 5 z 6 + z 6 z 7 + z 1 + z 7 = 0 z 1 z 2 z 3 + z 3 z 4 z 5 + z 5 z 6 z 7 + z 1 z 7 = 0 z 1 z 2 z 3 z 4 + z 2 z 3 z 4 z 5 + z 3 z 4 z 5 z 6 + z 4 z 5 z 6 z 7 + z 1 z 2 z 3 + z 1 z 2 z 7 + z 1 z 6 z 7 + z 5 z 6 z 7 = 0 = z 1 z 2 z 3 z 4 z 5 + z 3 z 4 z 5 z 6 z 7 + z 1 z 2 z 3 z 7 + z 1 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 + z 2 z 3 z 4 z 5 z 6 z 7 + z 1 z 2 z 3 z 4 z 5 + z 1 z 2 z 3 z 4 z 7 + z 1 z 2 z 3 z 6 z 7 + z 1 z 2 z 5 z 6 z 7 + z 1 z 4 z 5 z 6 z 7 + z 3 z 4 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 z 7 + z 1 z 2 z 3 z 4 z 5 z 7 + z 1 z 2 z 3 z 5 z 6 z 7 + z 1 z 3 z 4 z 5 z 6 z 7 = 0 z 1 z 2 z 3 z 4 z 5 z 6 z 7 1 = 0 We then solve in V (F )(z). Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 27 / 32

28 Cyclic 8-Roots System Initial form system in V (F )(z) has 72 solutions. In particular z 0 = t z 1 = 1 z 2 = I z 3 = I z 4 = 1 z 5 = 1 z 6 = I z 7 = I I = 1 x 0 = t x 1 = 1/t x 2 = It x 3 = I /t x 4 = t x 5 = 1/t x 6 = It x 7 = I /t satisfies the entire cyclic 8-roots polynomial system. Applying symmetry, we can find the remaining 7 as well. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 28 / 32

29 Cyclic 8-Roots System Definition (Branch Degree) Let V = (v 1, v 2,..., v m ) be a tropism and let R be the set of initial roots of the initial form system in V (F )(z). Then the degree of the branch is m #R max v i min m v i i=1 i=1 Tropisms, their cyclic permutations, and degrees: (1, 1, 1, 1, 1, 1, 1, 1) 8 2 = 16 (1, 1, 0, 1, 0, 0, 1, 0) (1, 0, 0, 1, 0, 1, 1, 0) = 32 (1, 0, 1, 0, 0, 1, 0, 1) (1, 0, 1, 1, 0, 1, 0, 0) = 32 (1, 0, 1, 1, 0, 1, 0, 0) (1, 0, 1, 0, 0, 1, 0, 1) = 32 (1, 0, 0, 1, 0, 1, 1, 0) (1, 1, 0, 1, 0, 0, 1, 0) = 32 TOTAL = is the degree of the solution curve of the cyclic 8-root system. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 29 / 32

30 Cyclic 8-Roots System-Zero Dimensional Part To find the isolated cyclic 8-roots exploiting symmetry we proceed in a similar way as for the cyclic 5-roots system The first 7 equations of the cyclic 8-roots system are a homogeneous system, i.e.: as in projective space the isolated solutions correspond to lines tropism (1,1,1,1,1,1,1,1) or ( 1, 1, 1, 1, 1, 1, 1, 1), via unimodular coordinate transformation: eliminate one variable from the system return the system to affine space Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 30 / 32

31 Cyclic 8-Roots System-Zero Dimensional Part After unimodular coordinate transformation, we find 144 isolated solutions of the first 7 equations. For a solution (c 1, c 2,..., c 7 ), the line in the original coordinates is x 0 = t, x 1 = c 1 t, x 2 = c 2 t, x 3 = c 3 t, x 4 = c 4 t, x 5 = c 5 t, x 6 = c 6 t, x 7 = c 7 t. Substituting into the omitted equation x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 1 = 0, yields a univariate polynomial in t of the form kt 8 1 = 0, where k is a constant. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 31 / 32

32 Cyclic 8-Roots System Among the 144 solutions, univariate polynomial k i t 8 1 = 0, for a constant k i, may occur numerous times, as in the cyclic 5-roots case. If k i t 8 1 = 0 occurs R times: it contributes 8 R solutions to = 1152 is the number of isolated cyclic 8-roots We repeat this calculation for each unique occurrence of k i t 8 1 = 0 and so obtain all the 1152 isolated solutions of the cyclic 8-roots system. Danko Adrovic, Jan Verschelde (UIC) Polyhedral Methods for Algebraic Sets ACA 2011, June 32 / 32