Computing Nearest Gcd with Certification G. Chze, A. Galligo, B. - PowerPoint PPT Presentation

Computing Nearest Gcd with Certification G. Chèze, A. Galligo, B. Mourrain, J.-C. Yakoubsohn Institut de Mathématiques de Toulouse, Université de Nice, Inria. RAIM 2011 G. Chèze (IMT) Computing Nearest Gcd with Certification 1 / 31

Plan The approximate gcd problem 1 A bisection approach 2 Complexity results 3 Some examples 4 G. Chèze (IMT) Computing Nearest Gcd with Certification 2 / 31

The problem Given f ( z ) and g ( z ) in C d [ z ] , find p ( z ) and q ( z ) in C d [ z ] , which are solutions of the minimization problem p , q ∈ C d [ z ] , Resultant ( p , q )= 0 || f − p || 2 + || g − q || 2 . min G. Chèze (IMT) Computing Nearest Gcd with Certification 3 / 31

Bibliography Algebraic approach via Euclid’s Algorithm, resultant 1 and subresultant : Schönhage, Noda-Sasaki, Corless-Gianni-Trager-Watt, Hribernig-Stetter, Emiris-Galligo-Lombardi, Rupprecht, Beckermann-Labahn. Padé approximation and structured matrices approach : 2 Bini-Pan, Bini-Boito, Zhi, Yang-Zhi. Rootfinding and cluster root approach : 3 Pan, Zeng, Graillat-Langlois. Optimization approach : 4 Karmarkar-Lakshman, Kaltofen-Yang-Zhi, Nie-Demmel-Gu, Terui. G. Chèze (IMT) Computing Nearest Gcd with Certification 4 / 31

Karmarkar and Lakshman’s formula The approximate gcd problem can be reduced to the minimization problem : f ( z ) f ( z ) + g ( z ) g ( z ) min . � d k = 0 z k z k z ∈ C If z 0 is a minimum we get p and q with this formula : d f ( z 0 ) k z k p ( z ) f ( z ) − z 0 � = × � d k z k k = 0 z 0 k = 0 0 d g ( z 0 ) k z k q ( z ) g ( z ) − z 0 � = × � d k z k k = 0 z 0 k = 0 0 G. Chèze (IMT) Computing Nearest Gcd with Certification 5 / 31

Our goal What do we want ? Give an algorithm to solve the Karmarkar and Lakshman’s 1 minimization problem in a certified way. Study the complexity of this algorithm. 2 G. Chèze (IMT) Computing Nearest Gcd with Certification 6 / 31

Some notations We set : X + iY z , = f ( z ) f ( z ) + g ( z ) g ( z ) F ( X , Y ) = . � d k = 0 z k z k G ( X , Y ) = G 1 ( X , Y ) , G 2 ( X , Y ) � � is the system of the numerators of the gradient ∇ F ( X , Y ) . G. Chèze (IMT) Computing Nearest Gcd with Certification 7 / 31

Some notations E is an exclusion test : E ( G i , S ) is true if G i has (in a certified way) no zero in the square S otherwise it returns false. I is an inclusion test : I ( G , S ) is true if G has (in a certified way) one zero in the square S otherwise it returns false. G. Chèze (IMT) Computing Nearest Gcd with Certification 8 / 31

An exclusion test Proposition Let S ( x , y , r ) be a square. We set : || D k G i ( x , y ) || r k . M ( G i , x , y , r ) = | G i ( x , y ) | − � k ! k ≥ 1 If M ( G i , x , y , r ) > 0 then S ( x , y , r ) does not contain any zero of G i ( X , Y ) . Definition E G i , S is true if M ( G i , x , y , r ) > 0 else E ( G i , S ) is false. � � G. Chèze (IMT) Computing Nearest Gcd with Certification 9 / 31

An exclusion test Proposition Let S ( x , y , r ) be a square. We set : || D k G i ( x , y ) || r k . M ( G i , x , y , r ) = | G i ( x , y ) | − � k ! k ≥ 1 If M ( G i , x , y , r ) > 0 then S ( x , y , r ) does not contain any zero of G i ( X , Y ) . Definition E G i , S is true if M ( G i , x , y , r ) > 0 else E ( G i , S ) is false. � � G. Chèze (IMT) Computing Nearest Gcd with Certification 9 / 31

An exclusion test Proposition Let S ( x , y , r ) be a square. We set : || D k P ( x , y ) || r k . M ( P , x , y , r ) = | P ( x , y ) | − � k ! k ≥ 1 If M ( P , x , y , r ) > 0 then S ( x , y , r ) does not contain any zero of P ( X , Y ) . Definition E + � P , S is true if M ( P , x , y , r ) > 0 and P ( x , y ) > 0 else � E + ( P , S ) is false. G. Chèze (IMT) Computing Nearest Gcd with Certification 10 / 31

An inclusion test Definition β := β ( G ; x , y ) = || DG ( x , y ) − 1 G ( x , y ) || � 1 � 1 / ( k − 1 ) γ := γ ( G ; x , y ) = sup k ≥ 2 k ! || DG ( x , y ) − 1 D k G ( x , y ) || α := α ( G ; x , y ) = βγ . G. Chèze (IMT) Computing Nearest Gcd with Certification 11 / 31

An inclusion test Theorem (Smale α -Theorem) If √ α ( G , x , y ) < ( 13 − 3 17 ) / 4 then G has one and only one zero ( x ∗ , y ∗ ) in the open ball B ( x , y , σ ( x , y )) with √ σ ( x , y ) = 1 + α − 1 − 6 α + α 2 4 γ and the Newton’s iteration from ( x , y ) converges quadratically to ( x ∗ , y ∗ ) . G. Chèze (IMT) Computing Nearest Gcd with Certification 12 / 31

An inclusion test Definition I ( G , S ) is true if : √ α ( G ; x , y ) < 13 − 3 17 , 1 4 S ( x , y , r ) ⊂ B x , y , σ ( x , y ) , i.e. r < σ ( x , y ) . � � 2 G. Chèze (IMT) Computing Nearest Gcd with Certification 13 / 31

A bisection approach S 0 := S ( x 0 , y 0 , r 0 ) a square containing all the global minima of F := N / D . L := { S 0 } , µ = F ( x 0 , y 0 ) . While L is not empty do For each square of L do E + ( N − µ D , S ) , E ( G 1 , S ) , E ( G 2 , S ) . If at least one of these tests is true then remove S from L , Else do I ( G , S ) : If I ( G , S ) is false then divide S in 4 equal squares S 1 , . . . , S 4 and substitute S by { S 1 , . . . , S 4 } in L . If I ( G , S ) is true then we have an approximate zero of G . G. Chèze (IMT) Computing Nearest Gcd with Certification 14 / 31

A bisection approach If I ( G , S ) is true then compute the local minimum ( x ∗ , y ∗ ) of F in S , µ ∗ := F ( x ∗ , y ∗ ) . If µ ∗ > µ then remove S . Else µ ∗ ≤ µ then remove the bad squares, and µ := µ ∗ . G. Chèze (IMT) Computing Nearest Gcd with Certification 15 / 31

“Proof of the algorithm” Computation of an initial square Lemma (Karmarkar-Lakshman) Let F be the rational bivariate function corresponding to the approximate gcd problem. Let ( x , y ) be a global minimum of F then � ( x , y ) � ≤ 5 max ( � f � 2 , � g � 2 ) . S 0 = S ( 0 , 0 ) , 5 max ( � f � 2 , � g � 2 ) � � ⇒ G. Chèze (IMT) Computing Nearest Gcd with Certification 16 / 31

“Proof of the algorithm” We can satisfy the inclusion test. Proposition Let ( x ∗ , y ∗ ) be a zero of G and suppose that DG ( x ∗ , y ∗ ) is inversible. If 0 . 08 � ( x , y ) − ( x ∗ , y ∗ ) � ≤ γ ( G , x ∗ , y ∗ ) then √ α ( G , x , y ) ≤ 13 − 3 17 . 4 G. Chèze (IMT) Computing Nearest Gcd with Certification 17 / 31

Complexity results Theorem Assume that all the global minima of G ( X , Y ) = 0 are regular and are contained in a square S 0 := S ( x 0 , y 0 , r 0 ) . √ 2 Let C d := 1 + log 2 ( 4 d − 2 ) and � 3 r 0 C d γ ( G , A F ) � J := , where 0 . 07 ≤ δ 0 ≤ 0 . 08 , and log 2 K δ 0 K is a constant related to the geometry of the problem, A F is the set of global minima of F. Denote by R j the set of retained squares at step j. Then, for j ≥ J, 1- The Newton’s iteration applied to G from a point in R j converges quadratically to a minimum of F. G. Chèze (IMT) Computing Nearest Gcd with Certification 18 / 31

Complexity results Theorem 2- When the approximate gcd problem has a unique solution the number of exclusion steps belongs to A d 3 log 3 ( d ) � � + B d 3 � O , min A F d F DG ( x , y ) − 1 , Σ � where A , B are constants related to the geometry of the problem. G 1 G 2 G 2 G 1 Small A , B . Big A , B . G. Chèze (IMT) Computing Nearest Gcd with Certification 19 / 31

An example Input : f ( z ) = 9 x 3 − 18 x 2 + 3 x + 1, g ( z ) = − 37 x 3 + 63 x 2 − 30 x + 3. Output : z = 0 . 479, p ( z ) = 9 . 059 x 3 − 17 . 875 x 2 + 3 . 26 x + 1 . 544, q ( z ) = − 36 . 916 x 3 + 63 . 174 x 2 − 29 . 636 x + 3 . 758. G. Chèze (IMT) Computing Nearest Gcd with Certification 20 / 31

An example Graph of F ( X , Y ) 20 15 N/D 10 5 0 0.1 1 0.8 0.05 0.6 0.4 0.2 0 0 y x G. Chèze (IMT) Computing Nearest Gcd with Certification 21 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 22 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 23 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 24 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 25 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 26 / 31

An example z = 0 . 479 G. Chèze (IMT) Computing Nearest Gcd with Certification 27 / 31

Random examples 12 10 8 cpu 6 4 2 0 0 5 10 15 20 25 30 d F IG .: f = ( z − 1 ) . f 1 + 10 − 3 ǫ 1 , g = ( z − 1 ) . g 1 + 10 − 3 ǫ 2 Random means that we use the Gaussian distribution with mean 0 and variance 1. Our algorithm gives an approximate gcd near z − 1. G. Chèze (IMT) Computing Nearest Gcd with Certification 28 / 31

Solution at infinity f ( x ) = z 2 − 4 z + 3 , and g ( z ) = z 2 + 4 z + 3 . The global minimum is reached at infinity ! � ( x , y ) �→∞ F ( x , y ) = F ( ∞ ) = 2 , lim N ( X , Y ) − 2 D ( X , Y ) = 16 + 42 X 2 + 18 Y 2 > 0 . Remark : We compute F ( ∞ ) in the following way : F ( ∞ ) = F r ( 0 ) , where F r ( z ) = z deg ( F ) F ( 1 / z ) . G. Chèze (IMT) Computing Nearest Gcd with Certification 29 / 31