Solving quadratic equations in dimension 5 or more without factoring
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1 Solving quadratic equations in dimension 5 or more without factoring ANTS X UCSD July, Pierre Castel pierre.castel@unicaen.fr castel Laboratoire de Mathématiques Nicolas Oresme CNRS UMR 6139 Université de Caen (France)
2 Summary 1 Introduction 2 The algorithm 3 Complexity 4 Example
3 What s next: Introduction 1 Introduction
4 Quadratic equations... We consider homogenous quadratic equations with integral coefficients and search for a nontrivial and integral solution. Dimension 1: Equation: Solution: ax 2 = 0 x = 0 Dimension 2: Equation: ax 2 + bxy + cy 2 = 0 Solution: 1 Compute = b 2 4ac 2 If is a square, solutions are: x = b ± y 2a Pierre Castel 3 / 28
5 Quadratic equations... We consider homogenous quadratic equations with integral coefficients and search for a nontrivial and integral solution. Dimension 1: Equation: Solution: ax 2 = 0 x = 0 Dimension 2: Equation: ax 2 + bxy + cy 2 = 0 Solution: 1 Compute = b 2 4ac 2 If is a square, solutions are: x = b ± y 2a Pierre Castel 3 / 28
6 Quadratic equations... We consider homogenous quadratic equations with integral coefficients and search for a nontrivial and integral solution. Dimension 1: Equation: Solution: ax 2 = 0 x = 0 Dimension 2: Equation: ax 2 + bxy + cy 2 = 0 Solution: 1 Compute = b 2 4ac 2 If is a square, solutions are: x = b ± y 2a Pierre Castel 3 / 28
7 Minimisation and Reduction We use the matrix notation: Q is the n dimensional symmetric matrix containing the coefficients of the equation. The equation is now: t XQX = 0 with X Z n. Let Q be a quadratic form with determinant. Minimising Q: finding transformations for Q in order to get another quadratic form Q with same dimension as Q such that: Q and Q have the same solutions (up to a basis change), det(q ) divides. Reducing the form Q: it s finding a basis change B such that: det(b) = ±1, the coefficients of Q = t BQB are smaller than the ones of Q. Pierre Castel 4 / 28
8 Minimisation and Reduction We use the matrix notation: Q is the n dimensional symmetric matrix containing the coefficients of the equation. The equation is now: t XQX = 0 with X Z n. Let Q be a quadratic form with determinant. Minimising Q: finding transformations for Q in order to get another quadratic form Q with same dimension as Q such that: Q and Q have the same solutions (up to a basis change), det(q ) divides. Reducing the form Q: it s finding a basis change B such that: det(b) = ±1, the coefficients of Q = t BQB are smaller than the ones of Q. Pierre Castel 4 / 28
9 Minimisation and Reduction We use the matrix notation: Q is the n dimensional symmetric matrix containing the coefficients of the equation. The equation is now: t XQX = 0 with X Z n. Let Q be a quadratic form with determinant. Minimising Q: finding transformations for Q in order to get another quadratic form Q with same dimension as Q such that: Q and Q have the same solutions (up to a basis change), det(q ) divides. Reducing the form Q: it s finding a basis change B such that: det(b) = ±1, the coefficients of Q = t BQB are smaller than the ones of Q. Pierre Castel 4 / 28
10 Quadratic equations in dimensions 3, 4 and more: Simon s algorithm 1 Factor the determinant of Q, 2 Minimise Q relatively to each prime factor of det(q), 3 Reduce Q using the LLL algorithm, 4 Use number theory tools in order to end the minimisation of Q, 5 Considering intersections of some isotropic spaces of good dimension, deduce a solution for the form of the beginning. This algorithm: creates a link between factoring and solving quadratic equations can be generalised to forms of higher dimension Pierre Castel 5 / 28
11 Quadratic equations in dimensions 3, 4 and more: Simon s algorithm 1 Factor the determinant of Q, 2 Minimise Q relatively to each prime factor of det(q), 3 Reduce Q using the LLL algorithm, 4 Use number theory tools in order to end the minimisation of Q, 5 Considering intersections of some isotropic spaces of good dimension, deduce a solution for the form of the beginning. This algorithm: creates a link between factoring and solving quadratic equations can be generalised to forms of higher dimension Pierre Castel 5 / 28
12 The problem: Pro: As soon as the factorisation of the determinant is known, Simon s algorithm is very efficient. Cons: But as soon as the size of the determinant reaches 50 digits, the factorisation becomes prohibitively slow. So, we are given the following problem: Problem: Let Q be a dimension 5 quadratic form. We assume that det(q) cannot be factored (in a reasonable amount of time). Find a non zero vector X Z 5 such that: t XQX = 0 Pierre Castel 6 / 28
13 The problem: Pro: As soon as the factorisation of the determinant is known, Simon s algorithm is very efficient. Cons: But as soon as the size of the determinant reaches 50 digits, the factorisation becomes prohibitively slow. So, we are given the following problem: Problem: Let Q be a dimension 5 quadratic form. We assume that det(q) cannot be factored (in a reasonable amount of time). Find a non zero vector X Z 5 such that: t XQX = 0 Pierre Castel 6 / 28
14 The problem: Pro: As soon as the factorisation of the determinant is known, Simon s algorithm is very efficient. Cons: But as soon as the size of the determinant reaches 50 digits, the factorisation becomes prohibitively slow. So, we are given the following problem: Problem: Let Q be a dimension 5 quadratic form. We assume that det(q) cannot be factored (in a reasonable amount of time). Find a non zero vector X Z 5 such that: t XQX = 0 Pierre Castel 6 / 28
15 What s next: The algorithm 2 The algorithm Principle Completion Computing a solution Minimisations
16 Principle Simon s algorithm is very efficient as soon as the factorization of det(q) is known. Idea: 1 Build another quadratic form Q 6 starting from Q for which computing a solution is easy, 2 Use Simon s algorithm to find a solution for Q 6, 3 Deduce a solution for Q. Pierre Castel 7 / 28
17 Principle Simon s algorithm is very efficient as soon as the factorization of det(q) is known. Idea: 1 Build another quadratic form Q 6 starting from Q for which computing a solution is easy, 2 Use Simon s algorithm to find a solution for Q 6, 3 Deduce a solution for Q. Pierre Castel 7 / 28
18 Principle Simon s algorithm is very efficient as soon as the factorization of det(q) is known. Idea: 1 Build another quadratic form Q 6 starting from Q for which computing a solution is easy, 2 Use Simon s algorithm to find a solution for Q 6, 3 Deduce a solution for Q. Pierre Castel 7 / 28
19 Principle Simon s algorithm is very efficient as soon as the factorization of det(q) is known. Idea: 1 Build another quadratic form Q 6 starting from Q for which computing a solution is easy, 2 Use Simon s algorithm to find a solution for Q 6, 3 Deduce a solution for Q. Pierre Castel 7 / 28
20 How to build Q 6? If Q designs the matrix of the quadratic form Q, we build Q 6 in the following way: Q 6 = Q t X X z Where X Z 5 is randomly chosen and z Z. So we have: det(q 6 ) = det(q)z t X Co (Q)X And we choose z such that: det(q 6 ) = t X Co (Q)X (mod det(q)). Pierre Castel 8 / 28
21 How to build Q 6? If Q designs the matrix of the quadratic form Q, we build Q 6 in the following way: Q 6 = Q t X X z Where X Z 5 is randomly chosen and z Z. So we have: det(q 6 ) = det(q)z t X Co (Q)X And we choose z such that: det(q 6 ) = t X Co (Q)X (mod det(q)). Pierre Castel 8 / 28
22 The way to the solution... As the value of det(q 6 ) is known in advance, we try some vector X until we have det(q 6 ) prime. Principle: det(q 6 ) being prime, it is possible to use Simon s algorithm in order to find a vector T Z 6 such that: t TQ 6 T = 0 Pierre Castel 9 / 28
23 The vector T is isotropic for Q 6. So, in a basis whose first vector is T, Q 6 has the form: Q 6 = 0 Pierre Castel 10 / 28
24 Decomposition Q 6 = H Q 4 The vector T is a solution for Q 6 so there exists an hyperbolic plane which contains it. With linear algebra (GCD), we get a correct basis. In such a basis, Q 6 has the shape: Q 6 = α Where α {0, 1} and Q 4 is a dimension 4 quadratic form, with determinant det(q 6 ). So it s prime again... Q 4 Pierre Castel 11 / 28
25 Decomposition Q 6 = H H Q 2... so we do it again : Simon s algorithm and linear algebra with Q 4. In the new basis, Q 6 has the following shape: α Q 6 = β Q where α, β {0, 1} and Q 2 is a dimension 2 quadratic form. Pierre Castel 12 / 28
26 If we denote by e 1 and e 3 the following basis vectors: Q 6 = α β Q Then e 1 and e 3 are both isotropics and orthogonals. The solution: consider a linear combinaison whose last coordinate is zero. Example: S = e 3 (6) e 1 e 1 (6) e 3 Pierre Castel 13 / 28
27 If we denote by e 1 and e 3 the following basis vectors: e 1 e α Q 6 = 0 β Q Then e 1 and e 3 are both isotropics and orthogonals. The solution: consider a linear combinaison whose last coordinate is zero. Example: S = e 3 (6) e 1 e 1 (6) e 3 Pierre Castel 13 / 28
28 If we denote by e 1 and e 3 the following basis vectors: e 1 e α Q 6 = 0 β Q Then e 1 and e 3 are both isotropics and orthogonals. The solution: consider a linear combinaison whose last coordinate is zero. Example: S = e 3 (6) e 1 e 1 (6) e 3 Pierre Castel 13 / 28
29 So S has the shape: S = S 0 with S Z5 Assuming that all of the basis changes have been applied, we have: t SQ S 6 = î t S 0 ó Q t X X z S 0 = t SQS = 0 We have then: S is a solution to our problem. Pierre Castel 14 / 28
30 The algorithm: 1 Complete Q in Q 6 in such a way that det(q 6 ) is prime, 2 Use Simon s algorithm for Q 6, 3 Using linear algebra, decompose Q 6 in Q 6 = H Q 4 (H hyperbolic plane), 4 Do step 2 for Q 4, 5 Using linear algebra, decompose Q 6 in Q 6 = H H Q 2 (H, H hyperbolic planes), 6 Deduce a solution for Q. Pierre Castel 15 / 28
31 Smith Normal Form: SNF Decomposition Let A be a k k matrix with integer entries and non zero determinant. There exists a unique matrix in Smith Normal Form D such that UAV = D with U and V unimodular and integer entries. If we denote by d i = d i,i, the d i are the elementary divisors of the matrix A, and we have : d d.. UAV = d k with d i+1 d i for 1 i < k Pierre Castel 16 / 28
32 The problem: In the algorithm, we are looking for X Z 5 such that det(q 6 ) is prime. However : Lemma Let Q be a dimension 5 quadratic form with determinant. Then for all X Z 5 and z Z, d 2 (Q) divides det(q 6 ). Problem If d 2 (Q) 1, det(q 6 ) will never be a prime! Pierre Castel 17 / 28
33 The problem: In the algorithm, we are looking for X Z 5 such that det(q 6 ) is prime. However : Lemma Let Q be a dimension 5 quadratic form with determinant. Then for all X Z 5 and z Z, d 2 (Q) divides det(q 6 ). Problem If d 2 (Q) 1, det(q 6 ) will never be a prime! Pierre Castel 17 / 28
34 The solution: Solution Do minimisations on Q to be in the case where d 2 (Q) = 1. We have the different cases: 1 Case d 5 (Q) 1, 2 Case d 4 (Q) 1 and d 5 (Q) = 1, 3 Case d 3 (Q) 1 and d 4 (Q) = 1, 4 Case d 2 (Q) 1 and d 3 (Q) = 1. Pierre Castel 18 / 28
35 The solution: Solution Do minimisations on Q to be in the case where d 2 (Q) = 1. We have the different cases: 1 Case d 5 (Q) 1, 2 Case d 4 (Q) 1 and d 5 (Q) = 1, 3 Case d 3 (Q) 1 and d 4 (Q) = 1, 4 Case d 2 (Q) 1 and d 3 (Q) = 1. Pierre Castel 18 / 28
36 Cases 1, 2 and 3 We apply the basis change given by the matrix V of the SNF of Q: if d 5 (Q) 1: we just have to divide the matrix by d 5, we have divided det(q) by (d 5 ) 5. if d 4 (Q) 1 and d 5 (Q) = 1: we multiply the last row and column by d 4, we divide the matrix by d 4, we have multiplied det(q) by (d 4 ) 2 and divided by (d 4 ) 5. if d 3 (Q) 1 and d 4 (Q) = 1: we multiply the two last rows and columns by d 3, we divide the matrix by d 3, we have multiplied det(q) by (d 3 ) 4 and divided by (d 3 ) 5. Pierre Castel 19 / 28
37 Case d 2 (Q) 1 and d 3 (Q) = 1 We first apply the basis change given by the matrix V of the SNF of Q. In such a base, Q has the form : d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 We would like to multiply the 3 lasts rows and columns by d 2 and divide the matrix by d 2. But if we do this, we multiply the determinant by d2 6 and we divide it by d Solution: Solve a quadratic equation modulo d 2 such that: Q 3,3 0 (mod d 2 ) and do the desired operation on the two lasts rows and columns. Pierre Castel 20 / 28
38 Case d 2 (Q) 1 and d 3 (Q) = 1 We first apply the basis change given by the matrix V of the SNF of Q. In such a base, Q has the form : d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 We would like to multiply the 3 lasts rows and columns by d 2 and divide the matrix by d 2. But if we do this, we multiply the determinant by d2 6 and we divide it by d Solution: Solve a quadratic equation modulo d 2 such that: Q 3,3 0 (mod d 2 ) and do the desired operation on the two lasts rows and columns. Pierre Castel 20 / 28
39 Case d 2 (Q) 1 and d 3 (Q) = 1 We first apply the basis change given by the matrix V of the SNF of Q. In such a base, Q has the form : d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 We would like to multiply the 3 lasts rows and columns by d 2 and divide the matrix by d 2. But if we do this, we multiply the determinant by d2 6 and we divide it by d Solution: Solve a quadratic equation modulo d 2 such that: Q 3,3 0 (mod d 2 ) and do the desired operation on the two lasts rows and columns. Pierre Castel 20 / 28
40 Case d 2 (Q) 1 and d 3 (Q) = 1 We first apply the basis change given by the matrix V of the SNF of Q. In such a base, Q has the form : d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 We would like to multiply the 3 lasts rows and columns by d 2 and divide the matrix by d 2. But if we do this, we multiply the determinant by d2 6 and we divide it by d Solution: Solve a quadratic equation modulo d 2 such that: Q 3,3 0 (mod d 2 ) and do the desired operation on the two lasts rows and columns. Pierre Castel 20 / 28
41 How to get Q 3,3 0 (mod d 2 )? We begin by a Gram Schmidt orthogonalisation on the 3 3 block modulo d 2. In that basis, the block Q 3 has the form: a 0 b (mod d 2 ) 0 c It remains to solve the equation: How? 1 Simon s algorithm? 2 CRT? ax 2 + by 2 + cz 2 0 (mod d 2 ) 3 Pollard Schnorr s algorithm. Pierre Castel 21 / 28
42 How to get Q 3,3 0 (mod d 2 )? We begin by a Gram Schmidt orthogonalisation on the 3 3 block modulo d 2. In that basis, the block Q 3 has the form: a 0 b (mod d 2 ) 0 c It remains to solve the equation: How? 1 Simon s algorithm? 2 CRT? ax 2 + by 2 + cz 2 0 (mod d 2 ) 3 Pollard Schnorr s algorithm. Pierre Castel 21 / 28
43 How to get Q 3,3 0 (mod d 2 )? We begin by a Gram Schmidt orthogonalisation on the 3 3 block modulo d 2. In that basis, the block Q 3 has the form: a 0 b (mod d 2 ) 0 c It remains to solve the equation: How? 1 Simon s algorithm? 2 CRT? ax 2 + by 2 + cz 2 0 (mod d 2 ) 3 Pollard Schnorr s algorithm. Pierre Castel 21 / 28
44 How to get Q 3,3 0 (mod d 2 )? We begin by a Gram Schmidt orthogonalisation on the 3 3 block modulo d 2. In that basis, the block Q 3 has the form: a 0 b (mod d 2 ) 0 c It remains to solve the equation: How? 1 Simon s algorithm? 2 CRT? ax 2 + by 2 + cz 2 0 (mod d 2 ) 3 Pollard Schnorr s algorithm. Pierre Castel 21 / 28
45 How to get Q 3,3 0 (mod d 2 )? We begin by a Gram Schmidt orthogonalisation on the 3 3 block modulo d 2. In that basis, the block Q 3 has the form: a 0 b (mod d 2 ) 0 c It remains to solve the equation: How? 1 Simon s algorithm? 2 CRT? ax 2 + by 2 + cz 2 0 (mod d 2 ) 3 Pollard Schnorr s algorithm. Pierre Castel 21 / 28
46 Pollard Schnorr s algorithm (1987) Solves equations of type: x 2 + ky 2 = m (mod n) Principle: Without factoring n Based on the property of multiplicativity of the norm in quadratic extensions: (x ky 2 1 )(x2 2 + ky 2 2 ) = X 2 + ky 2 Variables changes to decrease the size of the coefficients To be in the case where: (k, m) {(1, 1), ( 1, 1), ( 1, 1)} Pierre Castel 22 / 28
47 Pollard Schnorr s algorithm (1987) Solves equations of type: x 2 + ky 2 = m (mod n) Principle: Without factoring n Based on the property of multiplicativity of the norm in quadratic extensions: (x ky 2 1 )(x2 2 + ky 2 2 ) = X 2 + ky 2 Variables changes to decrease the size of the coefficients To be in the case where: (k, m) {(1, 1), ( 1, 1), ( 1, 1)} Pierre Castel 22 / 28
48 Using Pollard Schnorr We d like to solve: ax 2 + by 2 + cz 2 = 0 (mod d 2 ) We are going to use Pollard Schnorr to solve: x 2 + b a y 2 = c a (mod d 2 ) Taking z = 1 gives us a vector as we wish. ie in the basis containing the founded vector, Q has exactly the form: d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 Pierre Castel 23 / 28
49 Using Pollard Schnorr We d like to solve: ax 2 + by 2 + cz 2 = 0 (mod d 2 ) We are going to use Pollard Schnorr to solve: x 2 + b a y 2 = c a (mod d 2 ) Taking z = 1 gives us a vector as we wish. ie in the basis containing the founded vector, Q has exactly the form: d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 Pierre Castel 23 / 28
50 Finishing the minimisation Now that Q has the right form, we are able to minimise: d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 1 We multiply the two lasts rows and columns by d 2 2 We divide the matrix by d 2 Result: We have multiplied det(q) by d 4 2 and divided it by d 5 2, we have gained a factor d 2. Pierre Castel 24 / 28
51 Finishing the minimisation Now that Q has the right form, we are able to minimise: d 2 d 2 d 2 d2 2 d 2 2 d 2 d 2 d 2 d2 2 d 2 2 d 2 d 2 d 2 d 2 d 2 d2 2 d 2 2 d 2 d2 2 d 2 2 d2 2 d 2 2 d 2 d2 2 d We multiply the two lasts rows and columns by d 2 2 We divide the matrix by d 2 Result: We have multiplied det(q) by d 4 2 and divided it by d 5 2, we have gained a factor d 2. Pierre Castel 24 / 28
52 Finishing the minimisation Now that Q has the right form, we are able to minimise: d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 1 We multiply the two lasts rows and columns by d 2 2 We divide the matrix by d 2 Result: We have multiplied det(q) by d 4 2 and divided it by d 5 2, we have gained a factor d 2. Pierre Castel 24 / 28
53 Finishing the minimisation Now that Q has the right form, we are able to minimise: d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 d 2 1 We multiply the two lasts rows and columns by d 2 2 We divide the matrix by d 2 Result: We have multiplied det(q) by d 4 2 and divided it by d 5 2, we have gained a factor d 2. Pierre Castel 24 / 28
54 What s next: Complexity 3 Complexity
55 Complexity We write g = O (f ) if there exists α R, α 0 such that g = O (f log(f ) α ). Complexity Minimisation steps: O Ä log ( 5 ) 7ä Completion step: O Ä log ( 5 ) 5ä End of the algorithm: O (P (log ( 5 ))) P: non explicit polynomial given by the complexity of Simon s algorithm in dimensions 6 and 4. Global complexity: Probabilistic under GHR in O Ä log ( 5 ) 7 + P (log ( 5 )) ä Pierre Castel 25 / 28
56 Comparison 6 C. Simon Time in secs Size of the determinant in digits (Average over 1000 random matrices) Pierre Castel 26 / 28
57 What s next: Example 4 Example
58 A small example: Q = det(q) = ( 300 digits) Pierre Castel 27 / 28
59 A small example: Q = det(q) = ( 300 digits) Pierre Castel 27 / 28
60 Thanks for your attention. Pierre Castel castel
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