On solving zero-dimensional and parametric systems

Transcription

1 On solving zero-dimensional and parametric systems Fabrice Rouillier - SALSA (INRIA) project and CALFOR (LIP6) team Paris, France

2 General Objectives E = {p 1,, p r }, F = {f 1,, f l }, with p i, f i Q[U, X] U = U 1,, U d parameters X = X d+1,, X n indeterminates C = {x C n, p 1 = 0,, p r = 0, f 1 0,, f s 0} S = {x R n, p 1 =0,, p r =0, f 1 > 0,, f s > 0} In part 1 : we suppose that d = 0 (no parameter) and that the ideal p 1,, p r is zero-dimensional (finite number of complex zeroes). In part 2 : we suppose that d 0 and that the ideal p 1,, p r U=u is zero-dimensional for almost all u C d (general results will be described but only this case will be detailed).

3 General Objectives (2) The goal is to propose mathematical objects and algorithms to SOLVE such systems. The end-user queries we are interested in are (see G.-M. Greuel s talk): Zero-dimensional systems : count the real / complex roots ; detect multiple points and compute multiplicities ; provide an acurate and certified approximation of the roots ; signs of polynomials at the real roots of a system ; Parametric systems : count the real / complex roots wrt parameter s values ; describe geometrically the solutions set ; provide formal expressions of the roots ; provide numerically stable solutions.

4 General Objectives (3) Constraints : Exact/certified results : a real root is not a complex root with a small imaginary part... A numerical approximation must be certified with respect to a precision given by the end-user, etc. Universal algorithms : being able to check the assumptions (for example zero-dimensional) and the mathematical arbitrary choices (so called generic choices). Efficiency : computation time (bit operations) but also memory consumming.

5 Part 1 : Zero-dimensional Systems

6 Univariate case We assume that we know how to solve the univariate case (at least in the real case) : Counting/isolating real roots : Descarte s based methods, Sturm sequences etc. Evaluating polynomials (or their signs) at the roots of a univariate polynomial. More details / examples will be given in workshop B2 (Feb 27 - March 3) A reference book : Algorithms in Real Algebraic Geometry - Basu Pollack and Roy (Springer)

7 Notations E = {p 1, p s } Q[X 1,, X n ], is the ideal generated by E; V (E) C n is the zero set of or equivalently the set of complex solutions of {x C n, p 1 (x) = 0,, p n (x) = 0}; x C n, x i denotes its i-th coordinate.

8 Using Gröbner bases (see C. Traverso s talk) Theorem 1. Let G = {g 1,, g l } be a Gröbner basis for any ordering < of E = {p 1,, p s } Q[X 1,, X n ] s. The following properties are equivalent: For all index i, i = 1 n, there exists a polynomial g j G and a positive integer n j such that X i n j = LM (g j, < ); The system {p 1 = 0,, p s =0} has a finite number of solutions in C n. Q[X 1,, X n ] G is a finite dimensional Q-vector space Proof. According to Theorem 3, the 2nd and 3rd items are equivalent; m First item implies that 1 i = n, m i s.t. 1,, X i i (modulo I) and thus implies 3rd item. are lineary dependent m 3rd item implies that 1 i = n, m i s.t. 1,, X i i are lineary dependent (modulo I) so that p i Q[X i ], p i G. Thus, normalform(p i, G, < 0) = 0 g i G such αm that LM(g i, < ) divides LM(p i, < )=X i i, α N. Corollary 2. B = {t = X 1 e1 X n e n, (e 1,, e n ) N n normalform(t, G, < ) = t} = {w 1,, w D } is a basis of Q[X 1,, X n ]/ as a Q-vector space;

9 Counting the solutions (C. Traverso s talk) Definition 3. The multiplicity µ(α) of α ( V () is the ) dimension of the C[X localization of 1,, X n ] C[X1,, X at α (denoted by n ] ). α Theorem 4. C[X 1, Proof., X n ] < α V () ( C[X1,, X n ] α V (), e α C[X 1,, X n ] such that α V () e α = 1, e α e α = 0 if α α, e 2 α = e α. (consequence : e α (α) = 1 and e α (β) = 0 if β α V ()) ) α. s α = β α X 1 β 1 α 1 β 1 : n α N s.t. t α = s α n α : t α (α) = 1 and t α t β = 0; t α, α V () = 1 r α, α V () r αt α = 1. Set e α = r α t α. e α ( C[X1,, X n ] ) < ( C[X1,, X n ] ) α Corollary 5. The dimension of Q[X 1,, X n ] zeroes of counted with multiplicities. equals the number of complex

10 Stickelberger s theorem (C. Traverso s talk) Definition 6. Let h C[X 1,, X n ]; m h : C[X 1,, X n ] C[X 1,, X n ] u hu Theorem 7. The eigenvalues of m h are the h(α), α V ()with multiplicity µ(α). Proof. e α (f f(α))(β)=0, β V () n α N, e α (f f(α)) n α ( ) C[X1,, X the restriction of m eα (f f(α)) to n ] is thus nilpotent so (( that 0 is its ) ) unique eigenvalue (of multiplicity µ(α) = C[X1,, X dim n ] ). α ( ) ( ) C[X1,, X conclude using Theorem 4 : e n ] C[X1,, X n ] α and < α C[X 1,, X n ] ( ) C[X1,, X n ] < α V () α. α

11 Application of Stickelberger s theorem Corollary 8. (Stickelberger) Trace(m h ) = α V () µ(α)h(α); Charpol(m h, T) = α V () (T h(α))µ(α) Det(m h ) = α V () h(α)µ(α) Examples of use : h = 1 : Trace(m 1 )=dim ( C[X1,, X n ] h = X i : Charpol(m Xi, T) = α V () (T α i) µ(α), and thus V (Charpol(m Xi, T)) = {α i, α V ()} ) C. Traverso s talk : A first objective is to COMPUTE explicitly the matrices m Xi, i = 1 n.

12 Computing in the quotient algebra (C. Traverso s talk) For computing a matrix of m Xi, we need : A (monomial) basis B = {w 1,, w D } of Q[X 1, of C[X 1,, X n ] ordering, X n ] (or equivalently ) : can be deduced from a Gröbner basis for any A way for computing the class h of h Q[X 1,, X n ] in Q[X 1,, X n ] as a vector wrt B (denoted by h B from now) : for example the normalform. G For example, the colums of the matrix of m h wrt B are the coordinates of h w i B.

13 Counting distinct roots The matrix of M Xi wrt B has X i w 1,, X i w D as columns. If E has rational coefficients, then G Q[X 1,, X n ] and M Xi has also rational entries. One can compute its characteristic polynomial and apply any univariate solver to isolate all the possible real i-th coordinates of the zeroes of. Definition 9. (Separating element) Let t Q[X 1,, X n ]. t separates V () if (α, β) V () 2, α β t(α) t(β). Lemma 10. t separates V () V (charpol(m t ))( R n )=#V ()( R n ) Lemma 11. Almost all polynomials separate V () C n. Probabilistic algorithm for computing V () or V () R n Lemma 12. charpol(m t ) squarefree t separates V (). Deterministic filter

14 A (naïve) deterministic algorithm Lemma 13. if d = V () < D, t { n i=1 that t separates V (). j i 1 X i, j = 0 n d (d 1) } such 2 Proof. Let (α, β) V (), α β. p α,β (T) = n i=1 T i 1 (α i β i ) 0. Thus V (p α,β ) < and there exists at most n integers j such that p α,β (j) = 0. d(d 1), so there 2 intergers j such that there exists (α, β) the number of distinct couples (α, β) V () 2 is d (d 1) 2 exists at most n V () 2, α β with p α,β (j) = 0 Lemma 14. t separates V () iff V (charpol(m t )) = max ( V (charpol(m n j i=1 i 1 X i )), j 0 = n d (d 1) ) 2

15 Hermite s quadratic form Double goal : count the distinct complex/real roots - decrease the number of operations for searching a separating element. Definition 15. (Hermite s quadratic form). For f Q[X 1,, X n ] : q f : Q[X 1,, X n ] Q h G Trace(m fh 2) Theorem 16. For f Q[X 1,, X n ], rank(q f )=#{x V (), f(x) 0} signature(q f )=#{x V () R n, f(x) > 0} #{x V () R n, f(x) < 0} Corollary 17. Taking f = 1 : rank(q 1 ) = #V () signature(q 1 )=#(V () R n )

16 Hermite s quadratic form Set d = #V (), t a separating element of V (). w d,, w D st {w i } D i=1 = {1, t,, t d 1, w d,, w D } is a basis of Q[X 1,, X n ] ; q f (h ) = ( µ(α)f(α) D α V () i=1 G ) 2 h i w i (α) for hg = D h i=1 i w i ; q f (h G ) =(h 1,, h D )Γ t with Γ = 1 t(α 1) µ(α 1 ) f µ(α d ) f d t(α 1 ) d 1 1 t(α d ) t(α d ) d 1 q f (h G ) = α V () µ(α)f(α)(l α(h)) 2 Γ h 1 h D w d (α 1 ) w D (α 1 ) w d (α d ) w D (α d ) if α α, µ(α)f(α)(l α (h)) 2 + µ(α)f(α)(l ᾱ (h)) 2 = l 1,α (h ) 2 l 2,α (h ) 2, l 1,α, l 2,α being linar forms with real coefficients. G G l α, l 1,α, l 2,α linearly indep h µ(α)f(α)(l(α, α V () R h)) 2 has n the same signature as q f. G

17 Variable s elimination An easy example : compute G a Gröbner basis of ; compute D = dim Q Q[X 1,, X n ] and {w 1,, w D } from G; compute 1, X 1,, X D 1 1 (normalform) and check they are lineary G independant; [ ] ] D 1 if so, solve the linear system, X 1G 1,, X 1 β = [X D 1, X 2,, X n X D 1 D 1 i=0 β 1,i+1 X i 1 = 0 X and get a system equivalent to E : 2 D 1 i=0 β 2,i+1 X i 1 = 0 X n D 1 i=0 β n,i+1 X i 1 = 0 ELSE??

18 Lexicographic Gröbner bases By change of ordering (FGLM algorithm - see C. Traverso s talk) Suppose that Q[X 1,, X n ] is computed (Gröbner basis G wrt any monomial ordering < and a monomial basis B = {w 1,, w D }). Strategy : compute the image mg B of monomials m in increasing order wrt the lexicographic ordering until getting a set D = #B Q-linearly independant vectors w 1,, w D } (w 1 = 1 and w i 1 < lex w i, 2 i = D). Solve [ w 1, w D ] β = [ ] X i w j, set g i,j = X i w j D D k=1 β i,j,k w k n,j=1 i=1 Theorem 18. G = {g i,j, i = 1 n, j = 1 D} is a Gröbner basis wrt < lex of. Proof. p, g G s.t. LM(g, < lex ) divides LM(p, < lex ) since other else p = D i=1 a i w i (and thus do not belongs to )

19 Remarks on lexicographic Gröbner bases General shape in the zero-dimensional case for < lex : f 1 (X 1 ) f 2 (X 1, X 2 ) f k3 1(X 1, X 2 ) f k3 (X 1, X 2, X 3 ) f kn (X 1,, X n ) f kn +1(X 1,, X n ) Case of ideals in Shape position ideals for < lex (degree(f 1 )=D) : f 1 (X 1 ) X 2 f 2 (X 1, X 2 ) X n f n (X 1,, X n ) This occurs for example when X 1 is separating and =. Triangular sets (lexicographic Gröbner bases in the zero-dimensional case) : f 1 (X I )=0 n X f 2 (X 1 )=0 n X n n + f n (X 1,, X n 1 )=0

20 The case of radical ideals Main remark : if t Q[X 1,, X n ] separates V () and if =, then assuming that T is a new variable, E {T t} is in shape position for < lex with T < lex X 1 < lex < lex X n. Proof : charpol(m t ) = α V () (T t(α)) = minpol(m t) is squarefree. An algorithm : compute G a Gröbner basis of for < D R L, B = {w 1,, w D } a basis of Q[X 1,, X n ] and d = #V () (Hermite s quadratic form); compute p t = minpol(m t ) for t { n i=1 j i 1 D(D 1) X i, j = n } until 0 2 degree(squarefree(p t )) = d; ] ] if degree(p t ) = D solve [1, t,, t D 1 β = [t D, X 1,, X n and return G G E = {T D 1 D 1 i=0 β 1,i+1 T i,x j D 1 i=0 β 2,i+1 T i, j 1 = n} ( x = (x0, x 1,, x n ) V ( E ) x = (x 1,, x n ) V () ) ELSE?

21 Remarks There exists algorithms to compute the radical of an ideal but they are inefficient for large problems. They require the computation of several Gröbner bases ; For some problems, one would also like to compute the multiplicities of the solutions... Note that the above algorithm may compute O(n D 2 ) minimal polynomials before concluding with a failure! At this stage : we do not know how to reduce the problem (efficiently) to an easy univariate one when the ideal is not in shape position ; Numerical instability : the polynomials in a lexicographic Gröbner basis have HUGE coefficients (excepted the first one) : this often forbids to plug numerical approximations of the roots of the first polynomial into the basis to get an accurate approximation of the coordinates for large examples.

22 The Rational Univariate Representation (RUR) Definition 19. For t Q[X 1,, X n ]: f t = D i=0 a i T D i = charpol(m t ), f t square-free part. v Q[X 1,, X n ], g t,v (T)= d 1 Trace(m i=0 vt i)h d i 1 (f t, T), where j d = degree(f t ) and Hj (f t, T) = i=0 a i T j i Theorem 20. If t separates V (), then and : V ()( R n ) V (f t )( R) α = (α 1,, α n ) t(α) R t (β)=( g t,x 1 (β) t,xn (β) g t,1 (β) g t,1 (β) β f t Q[T] and v Q[X 1,, X n ], g t,v Q[T]; µ(α)= µ(t(α)) = µ(r t (t(α)))= (f t )(t(α)) (f t ) (t(α)) This generalizes the elimination process to the case of non radical ideals

23 The RUR : Proof By construction, f t Q[T] and v Q[X 1,, X n ], g t,v Q[T]; Thus V () V (f t ) V ()( R n ) V (f t )( R n ). By Stickelberger s theorem, f t (T)= α V () (T t(α))µ(α). Thus f t (T)= (T t(α)) α V () Consider g v,t = ( ) α V () µ(α)v(α) β V ( E α (T t(β)) ),β if t separates V () ( ) g t,v (t(α))= µ(α)v(α) β V ( E α (t(α) t(β)) ),β (f t (T) ) = g v,1 (T) = ( ) α V () µ(α) β V ( E α (T t(β)) ),β (f t (T)) = ( ) α V (),β ) β V ( E α (T t(β)) so that v(α) = g t,v(t(α)) and µ(α) = (f t g t,1 (t(α)) )(t(α)) = (f t )(t(α)) g t,1 (t(α)) (f t ) (t(α))

24 The RUR : Proof To prove the theorem, we finally need to show : g v,t (T)= α V ( E ) µ(α)v(α) ( β V ( E ),β α (T t(β)) ) = d 1 trace(m i=0 vt i)h d i 1 (f t, T) with f t = i=0 D a i T D i = α V () (T t(α)) and H j (f t, T)= j i=0 a i T j i g v,t (T) = f t (T) i 0 trace(m vt i) T i+1 α V () d 1 j=0 µ(α)v(α) T t(α) = i 0 g v,t (T) = d i 1 i=0 trace(m vt i)a j T d i j 1 = d 1 trace(m vt i)h d i 1 (f t, T) i=0 α V () µ(α)v(α)t(α)i T i+1 =

25 The RUR : a naïve algorithm Algorithm : compute G a Gröbner basis of for < D R L, B = {w 1,, w D } a basis of and d = #V () (Hermite s quadratic form); Q[X 1,, X n ] compute f t = charpol(m t ) for t { n i=1 degree(f t )=d; j i 1 X i, j = 0 n compute Trace(m Xi t j ), i=1 n, j = 1 d; and deduce; return g t,v (T)= d 1 trace(m i=0 vt i)h d i 1 (f t, T) for v = 1, X1,, X n Remarks : Trace(m Xi t j) normalform(x it j w k ), i = 1 n, j = 1 D, k = 1 D O(n D 2 N), N D = cost of normalform Hermite s quadratic form Trace(w i w j ) normalform(w i w j w k ), k =1 D O(D 3 N), N D= cost of normalform charpol(m t ) O(D 4 ) (with a naïve algorithm) D(D 1) } until 2 Theoretical complexity : O(n D 6 ) arithm. operations, Practical complexity : O(D 4 + n D 3 ) arithm. operations

26 For computing RUR, we only need The RUR : remarks a monomial basis B = {w 1,, w D } of Q[X 1, 1 and i > 1, k, w i = X k w j the matrices M Xi of m Xi wrt B;, X n ] I ; Suppose that w 1 = This input can be provided by Gröbner bases but also by some new alternatives (B. Mourrain and P. Trebuchet s work). From now we suppose we only have these data as input and that the rational numbers in the M Xi are of binary size t. The goal is now to design an efficient algorithm (binary complexity) for computing the RUR.

27 Objectives RUR = separating element t AND a RUR-Candidate {f t (T), g t,1 (T), g t,x1 (T),, g t,xn (T)} : if t separates V (I), V ()( R) V (f t )( R) α = (α 1,, α n ) t(α) (X 1 (α) = g t,x 1 (t(α)) g t,1 (t(α)) n(α)= g t,xn (t(α)) g t,1 (t(α)) t(α) preserving multiplicities and real roots. Many variants for computing a RUR-Candidate - few solutions for computing a RUR. This difference is critical for decision algorithms. Can be computed as a lexicographic Gröbner basis when I is radical and X 1 separates V ()(Faugère, Yokoyama,... multi-mod. or p-adic methods). A major problem is to check that a RUR-Candidate is a RUR or equivalently that t separates V () for non radical ideals.

28 From few traces Theorem 21. Given q 1 [1] = [Trace(m w1 ),, Trace(m wd )] (first line of Hermite s quadratic form), a RUR-Candidate can be computed in O(D 3 + n D 2 ) arithmetic operations. Proof. See algorithms below

29 The first polynomial According to Stickelberger s theorem, Trace(m t i) = D i=1 α i is the i-th Newton s sum of f t = charpol(m t ) = D i=0 a i T D i. One can thus deduce the coefficients of f t from the scalars Trace(m t i), 0 { i = D by solving the triangular linear system : (D i)a i = } i 1 j=0 a i j Trace(m t j) Algorithm charpol : D i=0 Input = q 1 [1] = [Trace(m wi )] i=1 D, + B + M Xi, i = 1n = [1, 0,, 0]; M yg t = D i=1 t i M Xi ; For i = 0 D do Trace(m t i) = yg q 1 [1] yg = M t yg /* at the i-th yg step = t i G Solve (D i)a i = i 1 j=0 a i j Trace(m t j) O(D 3 ) operations (note this is a general algorithm for computing the characteristic polynomial of any element in Q[X 1,, X n ]. I

30 The coordinates Algorithm coordinates : Input = data from algorithm charpol + f t j f t = i=0 a i T d i the squarefree part of f t i For d H i=0 i (f t, t) = j=0 For v {1, X 1,, X n } a j t i j? already com p uted [Trace(m vw1 ),, Trace(m vwd )] = M v q 1 [1] t For i=d 1 Trace(m vhi (t))=h i (t) [Trace(m vw1 ),, Trace(m vwd )] t g t,v (T)= d 1 Trace(m i=0 vhi (f t,t))t d i 1 = d 1 Trace(m i=0 Xj t i)h d i 1(f t, T) O(n D 2 ) arithmetic operations

31 Computing Hermite s quadratic form j=1 D D q 1 = [Trace(m wi w j )] i=1 Remark : Trace(w i w j ) = w i w j q 1 [1]; Proposition 22. Hermite s quadratic form can be computed from q 1 [1] performing O(#TD) arithmetic operations where T = {w i w j, i = 1 D, j = 1 D}.

32 Algorithm RUR-Classic : RUR from few traces Input = q 1 [1] = [Trace(m wi )] i=1 D, + B + M Xi, i = 1n d = #V () (Hermite s quadratic form); O(D 3 ) compute f t = charpol(m t ) for t { n i=1 j i 1 X i, j 0 = n until degree(f t )=d; O(D 3 ) practical / O(n D 5 ) theoretical compute g t,v, v = 1, X 1,, X n O(n D 2 ) D(D 1) 2 } return f t, g t,v, v = 1, X 1,, X n The theoretical complexity is a worst case : all the possible separating elements excepted the last one are not separating. In practice, I haven t any example with more than two tries...

33 RUR from a multiplication table Definition 23. T = {w j w j, i 1 = D, j = 1D} Remark : #T < D 2 Examples : from a lexicographic G. basis in Shape position : #T = O(D 2 ) If D = δ n (Bezout), #T = O(2 n D 2 ) (ex. when #G = n) Computing T Computing w i w j : i, k, w i w j = X k w i w j w i w j = M Xk w i w j Requires O(#TD 2 ) arithmetic operations Computing q 1 from T : for i 1 = D, Trace(m wi ) = D i=1 (O(D 2 )) w i w j [j] Theorem 24. Computing a RUR from T requires O(#TD 2 + D 3 + n D 2 ) arithmetic operations.

34 Complexity : remarks Remark : up to now we have counted O(1) each arithmetical operation; This is fine for data of fixed sizes (Z/pZ, floating point numbers, etc.) but not for integers, rationals, polynomials, series, etc. For integers : The addition is linear in the size of its operands; The naïve multiplication is quadratic in the size of its operands; (fast fft-based versions are close to be linear (forget the log) for very large integers. Data are growing at each operation : a + b has size O(max (log 2 (a), log 2 (b))+1), ab has size O(log(a)+log(b)) Funny example : fast exponentiation may not be so fast... (m(a, b)= binary cost of the mult. of and integer of size a by an integer of size b) ) X 2n = x. x : arithmetic cost O(2 n 2 ) - binary cost O( n i=1 m(i, 1) X 2n = ( ) (x 2 ) 2 2 ( arithmetic cost O(n) - binary cost O n m(2 i, 2 i ) ) i=1

35 Complexity : choosing the right model We are working with rationals! The (bound on the) growth of coefficients in an addition is greater than in a multiplication! An example : [a 1,,a D ] [b 1,,b D ] a i, b i of size t with integers, result of size O(2 t + D) with rationals, result of size O(2 Dt) k iterations v = M v, v (resp. M) a vector (resp. a matrix) of dimension D with entries of size t : with integers, result with scalars of size O(k(t+D)) with rationals, result of size O(D k t) A more precise model : 1 u [a 1,, a D ] 1 v [b 1,, b D ] u, v, a i, b i of size t result of size O(2 t + D) For our problem we can assume that the scalars are integers!

36 Complexity : choosing the right model Suppose that t is the binary size of the integers in M Xi = 1 M d Xi X i The multiplication table : at most δ = max (deg(w i w j )) iterations v = M v. Growth of coefficients : t δ (t + D) Bound on the binary cost : O(#TD 2 δ(t + D)) Hermite s quadratic form : Trace(w i w j ) = w i w j q 1 [1]; size O(δ (t + D)) bin. cost : O(#T Dδ (t + D)) Reduction of Hermite s quadratic form : growth of coefficients t D t (Rouillier s fraction-free algorithm) binary cost : O(D 4 δ (t + D)) First polynomial of the RUR : it is the characteristic polynomial of M t thus its coefficients are of size O(D t) (see G. Villard s survey). The intermediate iteration : t i+1 = M t t i induces a growth t D(t + D) (thus the resolution of the triangular system do not induce any growth) The binary cost is bounded by O(D 4 (t + D)). Coordinates : no significant growth (compared with the computation of the first polynomial) O(n D 3 (t + D))

37 Complexity : First remarks The computation of the multiplication table is not so costly compared to the rest since the data are not growing too much. The reduction of Hermite s quadratic form is the main operation. Algorithm RUR-Classic: Input = q 1 [1] = [Trace(m wi )] i=1 D, + B + M Xi, i = 1n d = #V () (Hermite s quadratic form); O(D 3 ) compute f t = charpol(m t ) for t { n i=1 j i 1 X i, j 0 = n until degree(f t )=d; O(D 3 ) practical /O(n D 5 ) theoretical compute g t,v, v = 1, X 1,, X n O(n D 2 ) return f t, g t,v, v = 1, X 1,, X n D(D 1) 2 } We need to change again our strategy!

38 Computing faster The strategy : compute d and t using modular arithmetic (modulo a prime number). The result will be correct excepted for a finite number of primes. check that the RUR-Candidate is a RUR after the computation to get a deterministic algorithm. Advantages : fast computations (fixed precision) for predicting t if the prime numbers are big enough : very few are bad ; take t = X i when possible (smaller results due to the sparsity of M Xi ) t may be given by the user.

39 Checking a RUR-Candidate There exists a filter : checking that f t is squarefree For the general case : Proposition 25. A RUR-Candidate R t () = {f t, g t,1, g t,x1,, g t,xn } is a RUR iff Trace(m pi w j ) = 0, i = 1 n, j = 1 D with p i = X i g t,1 (t) g t,xi (t); Proof. R t () is a RUR iff p i (α) = 0, i = 1 n, α V () since #V (f t ) V (). Hermite s quadratic form must be of rank 0: rank(q pi ) = rank( [ Trace(m pi w j w k ) ] j=1 D k=1 D ) = #{α V (), pi (α) 0} its first line is null Trace(m pi w j ) = 0, j = 1 D

40 Checking a RUR-Candidate Corollary 26. A RUR-Candidate R t () = {f t, g t,1, g t,x1,, g t,xn } is a RUR iff q 1 pg i = 0, i 1 = n, j 1 = D with p i = X i g t,1 (t) g t,xi (t); Proposition 27. One can check that a RUR-Candidate is a RUR in O(D 3 + n D 2 ) arithmetic operations (if the RUR-Candidate has been computed using RUR-Classic). Algorithm CheckRUR-Classic H 0 = a 0 [1, 0,, 0]; for i=1 D H i (t) = M t H i 1 (t) + a i H i 1 (t); if q 1 H D (t) [0,, 0] then return(false) i=1 for n g t,xi (t) = D i=1 Trace(X i t j )? alread y com puted H D i 1 (t) g t,1 (t) = D i=1 Trace(t j )H D i 1 (t) for i=1 n if q 1 ( g t,xi (t) + M Xi g t,1 (t)) [0,, 0] then return(false) return(true) O(D 3 + n D 2 ) arithmetic operations - no significant growth of coefficients (compared to the other sub-algorithms)

41 RUR vs Lexicographic Gröbner bases in the shape position case When X 1 separates V () + is radical RUR f(x 1 ) = 0 X 2 = h 2(X 1 ) f (X 1 ) Lex. G.B. f(x 1 ) = 0 X 2 = f 2 (X 1 ) X n = h n(x 1 ) f (X 1 ) X n = f n (X 1 ) (same number of arithmetic operations in the shape position case) Proposition : f (X 1 )X i h i (X 1 )=X i f i (X 1 ) mod I Equivalently : f i = f 1 (X 1 )h n (X 1 ) mod I Remarks : all the coefficients of the RUR have the same size (O(D t)) In a Lex. G. B. the coefficients of f appear to be small (O(D t)) compared to those of f i.

42 Adding inequations/inequalities E = {p 1,, p r }, F = {f 1,, f l }, with p i, f i Q[X 1,, X n ] A straightforward method : C = {x C n, p 1 = 0,, p r = 0, f 1 0,, f s 0} S = {x R n, p 1 =0,, p r =0, f 1 > 0,, f s > 0} Compute f i ( g t,x 1 g t,1,, g t,xn g t,1 ), i = 1 s and study the signs of these univariate polynomials at the roots of f t. Drawback : terrible computations (substitutions+reduction modulo f t ) A less straightforward method : E = {p 1,, p r, T 1 f 1,, T s f s } Q[X 1,, X n ] Compute the RUR of E : {f t, g t,1, g t,x1,, g t,xn, g t,t1,, g t,ts } and study the signs of g t,1, g t,t1,, g t,ts at the roots of f t (univariate problem). Drawback : can not study f i, i = 1 s without re-computing an ideal

43 ,,,,,, Adding inequations/inequalities A solution : simulating Gröbner bases Definition 28. Given two monomial orderings < U (w.r.t. the variables U 1,, U d ) and < X (w.r.t. the variables X d+1,, X n ) one can define a block ordering < U,X : m < U,X m if and only if m U1 =1, (m U1 =1, U d =1 < X m U1 =1, U d =1 = m U1 =1, U d =1 U d =1 or and m X d+1 =1, X n =1 < U m X d+1 =1, X n =1 Lemma 29. If G is a Gröbner basis of = p 1,, p r, T 1 f 1,, T s f s for < X, then G {T i normalform(f i ), i = 1 s} is a Gröbner basis of E = p 1,, p r, T 1 f 1,, T s f s for < T,X where T is any admissible monomial ordering wrt T = [T 1,, T n ]; Remark : Q[X 1,, X n ] Q[X 1,, X n, T 1,, T n ] < so that the only extraneous computations needed to extend the RUR are the construction of the M Ti and the compu- E tations of the Trace(m Ti t ). j Complexity : O(D 3 + n D 2 + s D 2 )? RU R ).

44 SALSA library containing Software FGb (F4 algorithm implemented by J.C. Faugère) - Dynamic library written in C RS (RUR + real root isolation by F. Rouillier) - Dynamic library written in C An interface with Maple Software (version >9.5)... (see next lecture). Will be linked with Maple 11 (official distribution)

45 Conclusion We have seen how to solve in a certified way any kind of zero-dimensional system : counting complex/real roots, certified isolation (by the way or interval arithmetic + symbolic resolution of univariate polynomials), computation of the multiplicities, certified evaluation of polynomials at the roots of a zero-dimensional system (easy to certify the sign for example). Recent progress have been made using baby step/giant step technics (Rouillier 2005) decreasing the number of bit operations required to compute a RUR-Candidate and using specific multiplication tables (only partially stored) decreasing the memory consummed. The RUR can be used jointly with splitting technics like triangular sets. Systems with 100/200 complex roots are easy in general (seconds) Systems with up to 1000 complex solutions are (in general) reachable by the current versions of the software (minutes/hours) Systems with >10000 solutions have been solved by the beta versions. Important remark : only few solvers are able to provide the required univariate black-boxes able to solve such large polynomials (first polynomial of a RUR)...