Modular algorithm to compute Hermite normal forms of integer matrices¶
AUTHORS:
Clement Pernet and William Stein (2008-02-07): initial version
- sage.matrix.matrix_integer_dense_hnf.add_column(B, H_B, a, proof)[source]¶
The add column procedure.
INPUT:
B
– a square matrix (may be singular)H_B
– the Hermite normal form of Ba
– an n x 1 matrix, where B has n rowsproof
– boolean; whether to prove result correct, in case we use fallback method
OUTPUT:
x – a vector such that H’ = H_B.augment(x) is the HNF of A = B.augment(a)
EXAMPLES:
sage: B = matrix(ZZ, 3, 3, [1,2,5, 0,-5,3, 1,1,2]) sage: H_B = B.echelon_form() sage: a = matrix(ZZ, 3, 1, [1,8,-2]) sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: x = hnf.add_column(B, H_B, a, True); x [18] [ 3] [23] sage: H_B.augment(x) [ 1 0 17 18] [ 0 1 3 3] [ 0 0 18 23] sage: B.augment(a).echelon_form() [ 1 0 17 18] [ 0 1 3 3] [ 0 0 18 23]
>>> from sage.all import * >>> B = matrix(ZZ, Integer(3), Integer(3), [Integer(1),Integer(2),Integer(5), Integer(0),-Integer(5),Integer(3), Integer(1),Integer(1),Integer(2)]) >>> H_B = B.echelon_form() >>> a = matrix(ZZ, Integer(3), Integer(1), [Integer(1),Integer(8),-Integer(2)]) >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> x = hnf.add_column(B, H_B, a, True); x [18] [ 3] [23] >>> H_B.augment(x) [ 1 0 17 18] [ 0 1 3 3] [ 0 0 18 23] >>> B.augment(a).echelon_form() [ 1 0 17 18] [ 0 1 3 3] [ 0 0 18 23]
- sage.matrix.matrix_integer_dense_hnf.add_column_fallback(B, a, proof)[source]¶
Simplistic version of add_column, in case the powerful clever one fails (e.g., B is singular).
INPUT:
B
– a square matrix (may be singular)a
– an n x 1 matrix, where B has n rowsproof
– boolean; whether to prove result correct
OUTPUT: x; a vector such that
H' = H_B.augment(x)
is the HNF ofA = B.augment(a)
EXAMPLES:
sage: B = matrix(ZZ,3, [-1, -1, 1, -3, 8, -2, -1, -1, -1]) sage: a = matrix(ZZ,3,1, [1,2,3]) sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.add_column_fallback(B, a, True) [-3] [-7] [-2] sage: matrix_integer_dense_hnf.add_column_fallback(B, a, False) [-3] [-7] [-2] sage: B.augment(a).hermite_form() [ 1 1 1 -3] [ 0 11 1 -7] [ 0 0 2 -2]
>>> from sage.all import * >>> B = matrix(ZZ,Integer(3), [-Integer(1), -Integer(1), Integer(1), -Integer(3), Integer(8), -Integer(2), -Integer(1), -Integer(1), -Integer(1)]) >>> a = matrix(ZZ,Integer(3),Integer(1), [Integer(1),Integer(2),Integer(3)]) >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.add_column_fallback(B, a, True) [-3] [-7] [-2] >>> matrix_integer_dense_hnf.add_column_fallback(B, a, False) [-3] [-7] [-2] >>> B.augment(a).hermite_form() [ 1 1 1 -3] [ 0 11 1 -7] [ 0 0 2 -2]
- sage.matrix.matrix_integer_dense_hnf.add_row(A, b, pivots, include_zero_rows)[source]¶
The add row procedure.
INPUT:
A
– a matrix in Hermite normal form with n columnb
– an n x 1 row matrixpivots
– sorted list of integers; the pivot positions of A
OUTPUT:
H
– the Hermite normal form of A.stack(b)new_pivots
– the pivot columns of H
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: A = matrix(ZZ, 2, 3, [-21, -7, 5, 1,20,-7]) sage: b = matrix(ZZ, 1,3, [-1,1,-1]) sage: hnf.add_row(A, b, A.pivots(), True) ( [ 1 6 29] [ 0 7 28] [ 0 0 46], [0, 1, 2] ) sage: A.stack(b).echelon_form() [ 1 6 29] [ 0 7 28] [ 0 0 46]
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> A = matrix(ZZ, Integer(2), Integer(3), [-Integer(21), -Integer(7), Integer(5), Integer(1),Integer(20),-Integer(7)]) >>> b = matrix(ZZ, Integer(1),Integer(3), [-Integer(1),Integer(1),-Integer(1)]) >>> hnf.add_row(A, b, A.pivots(), True) ( [ 1 6 29] [ 0 7 28] [ 0 0 46], [0, 1, 2] ) >>> A.stack(b).echelon_form() [ 1 6 29] [ 0 7 28] [ 0 0 46]
- sage.matrix.matrix_integer_dense_hnf.benchmark_hnf(nrange, bits=4)[source]¶
Run benchmark program.
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: hnf.benchmark_hnf([10,25],32) ('sage', 10, 32, ...), ('sage', 25, 32, ...),
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> hnf.benchmark_hnf([Integer(10),Integer(25)],Integer(32)) ('sage', 10, 32, ...), ('sage', 25, 32, ...),
- sage.matrix.matrix_integer_dense_hnf.benchmark_magma_hnf(nrange, bits=4)[source]¶
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: hnf.benchmark_magma_hnf([50,100],32) # optional - magma ('magma', 50, 32, ...), ('magma', 100, 32, ...),
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> hnf.benchmark_magma_hnf([Integer(50),Integer(100)],Integer(32)) # optional - magma ('magma', 50, 32, ...), ('magma', 100, 32, ...),
- sage.matrix.matrix_integer_dense_hnf.det_from_modp_and_divisor(A, d, p, z_mod, moduli, z_so_far=1, N_so_far=1)[source]¶
This is used for internal purposes for computing determinants quickly (with the hybrid \(p\)-adic / multimodular algorithm).
INPUT:
A
– a square matrixd
– a divisor of the determinant of Ap
– a primez_mod
– values of det/d (mod …)moduli
– the moduli so farz_so_far
– for a modulus p in the list moduli, (z_so_far mod p) is the determinant of A modulo pN_so_far
– N_so_far is the product over the primes in the list moduli
OUTPUT:
A triple (det bound, new z_so_far, new N_so_far).
EXAMPLES:
sage: a = matrix(ZZ, 3, [6, 1, 2, -56, -2, -1, -11, 2, -3]) sage: factor(a.det()) -1 * 13 * 29 sage: d = 13 sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.det_from_modp_and_divisor(a, d, 97, [], []) (-377, -29, 97) sage: a.det() -377
>>> from sage.all import * >>> a = matrix(ZZ, Integer(3), [Integer(6), Integer(1), Integer(2), -Integer(56), -Integer(2), -Integer(1), -Integer(11), Integer(2), -Integer(3)]) >>> factor(a.det()) -1 * 13 * 29 >>> d = Integer(13) >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.det_from_modp_and_divisor(a, d, Integer(97), [], []) (-377, -29, 97) >>> a.det() -377
- sage.matrix.matrix_integer_dense_hnf.det_given_divisor(A, d, proof=True, stabilize=2)[source]¶
Given a divisor d of the determinant of A, compute the determinant of A.
INPUT:
A
– square integer matrixd
– nonzero integer that is assumed to divide the determinant of Aproof
– boolean (default:True
); compute det modulo enough primes so that the determinant is computed provably correctly (via the Hadamard bound). It would be VERY hard fordet()
to fail even whenproof
isFalse
.stabilize
– integer (default: 2); if proof = False, then compute the determinant modulo \(p\) untilstabilize
successive modulo determinant computations stabilize.
OUTPUT: integer; determinant
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: a = matrix(ZZ,3,[-1, -1, -1, -20, 4, 1, -1, 1, 2]) sage: matrix_integer_dense_hnf.det_given_divisor(a, 3) -30 sage: matrix_integer_dense_hnf.det_given_divisor(a, 3, proof=False) -30 sage: matrix_integer_dense_hnf.det_given_divisor(a, 3, proof=False, stabilize=1) -30 sage: a.det() -30
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> a = matrix(ZZ,Integer(3),[-Integer(1), -Integer(1), -Integer(1), -Integer(20), Integer(4), Integer(1), -Integer(1), Integer(1), Integer(2)]) >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(3)) -30 >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(3), proof=False) -30 >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(3), proof=False, stabilize=Integer(1)) -30 >>> a.det() -30
Here we illustrate proof=False giving a wrong answer:
sage: p = matrix_integer_dense_hnf.max_det_prime(2) sage: q = previous_prime(p) sage: a = matrix(ZZ, 2, [p, 0, 0, q]) sage: p * q 70368442188091 sage: matrix_integer_dense_hnf.det_given_divisor(a, 1, proof=False, stabilize=2) 0
>>> from sage.all import * >>> p = matrix_integer_dense_hnf.max_det_prime(Integer(2)) >>> q = previous_prime(p) >>> a = matrix(ZZ, Integer(2), [p, Integer(0), Integer(0), q]) >>> p * q 70368442188091 >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(1), proof=False, stabilize=Integer(2)) 0
This still works, because we do not work modulo primes that divide the determinant bound, which is found using a \(p\)-adic algorithm:
sage: a.det(proof=False, stabilize=2) 70368442188091
>>> from sage.all import * >>> a.det(proof=False, stabilize=Integer(2)) 70368442188091
3 primes is enough:
sage: matrix_integer_dense_hnf.det_given_divisor(a, 1, proof=False, stabilize=3) 70368442188091 sage: matrix_integer_dense_hnf.det_given_divisor(a, 1, proof=False, stabilize=5) 70368442188091 sage: matrix_integer_dense_hnf.det_given_divisor(a, 1, proof=True) 70368442188091
>>> from sage.all import * >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(1), proof=False, stabilize=Integer(3)) 70368442188091 >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(1), proof=False, stabilize=Integer(5)) 70368442188091 >>> matrix_integer_dense_hnf.det_given_divisor(a, Integer(1), proof=True) 70368442188091
- sage.matrix.matrix_integer_dense_hnf.det_padic(A, proof=True, stabilize=2)[source]¶
Return the determinant of A, computed using a \(p\)-adic/multimodular algorithm.
INPUT:
A
– a square matrixproof
– booleanstabilize
– (default: 2) if proof False, number of successive primes so that CRT det must stabilize
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as h sage: a = matrix(ZZ, 3, [1..9]) sage: h.det_padic(a) 0 sage: a = matrix(ZZ, 3, [1,2,5,-7,8,10,192,5,18]) sage: h.det_padic(a) -3669 sage: a.determinant(algorithm='ntl') -3669
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as h >>> a = matrix(ZZ, Integer(3), (ellipsis_range(Integer(1),Ellipsis,Integer(9)))) >>> h.det_padic(a) 0 >>> a = matrix(ZZ, Integer(3), [Integer(1),Integer(2),Integer(5),-Integer(7),Integer(8),Integer(10),Integer(192),Integer(5),Integer(18)]) >>> h.det_padic(a) -3669 >>> a.determinant(algorithm='ntl') -3669
- sage.matrix.matrix_integer_dense_hnf.double_det(A, b, c, proof)[source]¶
Compute the determinants of the stacked integer matrices A.stack(b) and A.stack(c).
INPUT:
A
– an (n-1) x n matrixb
– a 1 x n matrixc
– a 1 x n matrixproof
– whether or not to compute the det modulo enough times to provably compute the determinant
OUTPUT: a pair of two integers
EXAMPLES:
sage: from sage.matrix.matrix_integer_dense_hnf import double_det sage: A = matrix(ZZ, 2, 3, [1,2,3, 4,-2,5]) sage: b = matrix(ZZ, 1, 3, [1,-2,5]) sage: c = matrix(ZZ, 1, 3, [8,2,10]) sage: A.stack(b).det() -48 sage: A.stack(c).det() 42 sage: double_det(A, b, c, False) (-48, 42)
>>> from sage.all import * >>> from sage.matrix.matrix_integer_dense_hnf import double_det >>> A = matrix(ZZ, Integer(2), Integer(3), [Integer(1),Integer(2),Integer(3), Integer(4),-Integer(2),Integer(5)]) >>> b = matrix(ZZ, Integer(1), Integer(3), [Integer(1),-Integer(2),Integer(5)]) >>> c = matrix(ZZ, Integer(1), Integer(3), [Integer(8),Integer(2),Integer(10)]) >>> A.stack(b).det() -48 >>> A.stack(c).det() 42 >>> double_det(A, b, c, False) (-48, 42)
- sage.matrix.matrix_integer_dense_hnf.extract_ones_data(H, pivots)[source]¶
Compute ones data and corresponding submatrices of H.
This is used to optimized the
add_row()
function.INPUT:
H
– a matrix in HNFpivots
– list of all pivot column positions of H
OUTPUT:
C, D, E, onecol, onerow, non_onecol, non_onerow where onecol, onerow, non_onecol, non_onerow are as for the ones function, and C, D, E are matrices:
C
– submatrix of all non-onecol columns and onecol rowsD
– all non-onecol columns and other rowsE
– inverse of D
If D is not invertible or there are 0 or more than 2 non onecols, then C, D, and E are set to None.
EXAMPLES:
sage: H = matrix(ZZ, 3, 4, [1, 0, 0, 7, 0, 1, 5, 2, 0, 0, 6, 6]) sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.extract_ones_data(H, [0,1,2]) ( [0] [5], [6], [1/6], [0, 1], [0, 1], [2], [2] )
>>> from sage.all import * >>> H = matrix(ZZ, Integer(3), Integer(4), [Integer(1), Integer(0), Integer(0), Integer(7), Integer(0), Integer(1), Integer(5), Integer(2), Integer(0), Integer(0), Integer(6), Integer(6)]) >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.extract_ones_data(H, [Integer(0),Integer(1),Integer(2)]) ( [0] [5], [6], [1/6], [0, 1], [0, 1], [2], [2] )
- Here we get None’s since the (2,2) position submatrix is not invertible.
sage: H = matrix(ZZ, 3, 5, [1, 0, 0, 45, -36, 0, 1, 0, 131, -107, 0, 0, 0, 178, -145]); H [ 1 0 0 45 -36] [ 0 1 0 131 -107] [ 0 0 0 178 -145] sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.extract_ones_data(H, [0,1,3]) (None, None, None, [0, 1], [0, 1], [2], [2])
- sage.matrix.matrix_integer_dense_hnf.hnf(A, include_zero_rows=True, proof=True)[source]¶
Return the Hermite Normal Form of a general integer matrix A, along with the pivot columns.
INPUT:
A
– an n x m matrix A over the integersinclude_zero_rows
– boolean (default:True
); whether or not to include zero rows in the output matrixproof
– whether or not to prove the result correct
OUTPUT: tuple of:
matrix
– the Hermite normal form of Apivots
– the pivot column positions of A
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: a = matrix(ZZ,3,5,[-2, -6, -3, -17, -1, 2, -1, -1, -2, -1, -2, -2, -6, 9, 2]) sage: matrix_integer_dense_hnf.hnf(a) ( [ 2 0 26 -75 -10] [ 0 1 27 -73 -9] [ 0 0 37 -106 -13], [0, 1, 2] ) sage: matrix_integer_dense_hnf.hnf(a.transpose()) ( [1 0 0] [0 1 0] [0 0 1] [0 0 0] [0 0 0], [0, 1, 2] ) sage: matrix_integer_dense_hnf.hnf(a.transpose(), include_zero_rows=False) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] )
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> a = matrix(ZZ,Integer(3),Integer(5),[-Integer(2), -Integer(6), -Integer(3), -Integer(17), -Integer(1), Integer(2), -Integer(1), -Integer(1), -Integer(2), -Integer(1), -Integer(2), -Integer(2), -Integer(6), Integer(9), Integer(2)]) >>> matrix_integer_dense_hnf.hnf(a) ( [ 2 0 26 -75 -10] [ 0 1 27 -73 -9] [ 0 0 37 -106 -13], [0, 1, 2] ) >>> matrix_integer_dense_hnf.hnf(a.transpose()) ( [1 0 0] [0 1 0] [0 0 1] [0 0 0] [0 0 0], [0, 1, 2] ) >>> matrix_integer_dense_hnf.hnf(a.transpose(), include_zero_rows=False) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] )
- sage.matrix.matrix_integer_dense_hnf.hnf_square(A, proof)[source]¶
INPUT:
A
– a nonsingular n x n matrix over the integers
OUTPUT: the Hermite normal form of A
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: A = matrix(ZZ, 3, [-21, -7, 5, 1,20,-7, -1,1,-1]) sage: hnf.hnf_square(A, False) [ 1 6 29] [ 0 7 28] [ 0 0 46] sage: A.echelon_form() [ 1 6 29] [ 0 7 28] [ 0 0 46]
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> A = matrix(ZZ, Integer(3), [-Integer(21), -Integer(7), Integer(5), Integer(1),Integer(20),-Integer(7), -Integer(1),Integer(1),-Integer(1)]) >>> hnf.hnf_square(A, False) [ 1 6 29] [ 0 7 28] [ 0 0 46] >>> A.echelon_form() [ 1 6 29] [ 0 7 28] [ 0 0 46]
- sage.matrix.matrix_integer_dense_hnf.hnf_with_transformation(A, proof=True)[source]¶
Compute the HNF H of A along with a transformation matrix U such that U*A = H.
INPUT:
A
– an n x m matrix A over the integersproof
– whether or not to prove the result correct
OUTPUT: tuple of:
matrix
– the Hermite normal form H of AU
– a unimodular matrix such that U * A = H
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: A = matrix(ZZ, 2, [1, -5, -10, 1, 3, 197]); A [ 1 -5 -10] [ 1 3 197] sage: H, U = matrix_integer_dense_hnf.hnf_with_transformation(A) sage: H [ 1 3 197] [ 0 8 207] sage: U [ 0 1] [-1 1] sage: U*A [ 1 3 197] [ 0 8 207]
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> A = matrix(ZZ, Integer(2), [Integer(1), -Integer(5), -Integer(10), Integer(1), Integer(3), Integer(197)]); A [ 1 -5 -10] [ 1 3 197] >>> H, U = matrix_integer_dense_hnf.hnf_with_transformation(A) >>> H [ 1 3 197] [ 0 8 207] >>> U [ 0 1] [-1 1] >>> U*A [ 1 3 197] [ 0 8 207]
- sage.matrix.matrix_integer_dense_hnf.hnf_with_transformation_tests(n=10, m=5, trials=10)[source]¶
Use this to randomly test that hnf with transformation matrix is working.
EXAMPLES:
sage: from sage.matrix.matrix_integer_dense_hnf import hnf_with_transformation_tests sage: hnf_with_transformation_tests(n=15, m=10, trials=10) 0 1 2 3 4 5 6 7 8 9
>>> from sage.all import * >>> from sage.matrix.matrix_integer_dense_hnf import hnf_with_transformation_tests >>> hnf_with_transformation_tests(n=Integer(15), m=Integer(10), trials=Integer(10)) 0 1 2 3 4 5 6 7 8 9
- sage.matrix.matrix_integer_dense_hnf.interleave_matrices(A, B, cols1, cols2)[source]¶
INPUT:
A, B – matrices with the same number of rows
cols1, cols2 – disjoint lists of integers
OUTPUT:
construct a new matrix C by sticking the columns of A at the positions specified by cols1 and the columns of B at the positions specified by cols2.
EXAMPLES:
sage: A = matrix(ZZ, 2, [1,2,3,4]); B = matrix(ZZ, 2, [-1,5,2,3]) sage: A [1 2] [3 4] sage: B [-1 5] [ 2 3] sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: hnf.interleave_matrices(A, B, [1,3], [0,2]) [-1 1 5 2] [ 2 3 3 4]
>>> from sage.all import * >>> A = matrix(ZZ, Integer(2), [Integer(1),Integer(2),Integer(3),Integer(4)]); B = matrix(ZZ, Integer(2), [-Integer(1),Integer(5),Integer(2),Integer(3)]) >>> A [1 2] [3 4] >>> B [-1 5] [ 2 3] >>> import sage.matrix.matrix_integer_dense_hnf as hnf >>> hnf.interleave_matrices(A, B, [Integer(1),Integer(3)], [Integer(0),Integer(2)]) [-1 1 5 2] [ 2 3 3 4]
- sage.matrix.matrix_integer_dense_hnf.is_in_hnf_form(H, pivots)[source]¶
Return whether the matrix
H
is in Hermite normal form with given pivot columns.INPUT:
H
– matrixpivots
– sorted list of integers
OUTPUT: boolean
EXAMPLES:
sage: a = matrix(ZZ,3,5,[-2, -6, -3, -17, -1, 2, -1, -1, -2, -1, -2, -2, -6, 9, 2]) sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.is_in_hnf_form(a,range(3)) False sage: e = a.hermite_form(); p = a.pivots() sage: matrix_integer_dense_hnf.is_in_hnf_form(e, p) True
>>> from sage.all import * >>> a = matrix(ZZ,Integer(3),Integer(5),[-Integer(2), -Integer(6), -Integer(3), -Integer(17), -Integer(1), Integer(2), -Integer(1), -Integer(1), -Integer(2), -Integer(1), -Integer(2), -Integer(2), -Integer(6), Integer(9), Integer(2)]) >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.is_in_hnf_form(a,range(Integer(3))) False >>> e = a.hermite_form(); p = a.pivots() >>> matrix_integer_dense_hnf.is_in_hnf_form(e, p) True
- sage.matrix.matrix_integer_dense_hnf.max_det_prime(n)[source]¶
Return the largest prime so that it is reasonably efficient to compute modulo that prime with n x n matrices in LinBox.
INPUT:
n
– positive integer
OUTPUT: a prime number
EXAMPLES:
sage: from sage.matrix.matrix_integer_dense_hnf import max_det_prime sage: max_det_prime(10000) 8388593 sage: max_det_prime(1000) 8388593 sage: max_det_prime(10) 8388593
>>> from sage.all import * >>> from sage.matrix.matrix_integer_dense_hnf import max_det_prime >>> max_det_prime(Integer(10000)) 8388593 >>> max_det_prime(Integer(1000)) 8388593 >>> max_det_prime(Integer(10)) 8388593
- sage.matrix.matrix_integer_dense_hnf.ones(H, pivots)[source]¶
Find all 1 pivot columns of the matrix H in Hermite form, along with the corresponding rows, and also the non 1 pivot columns and non-pivot rows. Here a 1 pivot column is a pivot column so that the leading bottom entry is 1.
INPUT:
H
– matrix in Hermite formpivots
– list of integers (all pivot positions of H)
OUTPUT:
4-tuple of integer lists: onecol, onerow, non_oneol, non_onerow
EXAMPLES:
sage: H = matrix(ZZ, 3, 5, [1, 0, 0, 45, -36, 0, 1, 0, 131, -107, 0, 0, 0, 178, -145]); H [ 1 0 0 45 -36] [ 0 1 0 131 -107] [ 0 0 0 178 -145] sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.ones(H, [0,1,3]) ([0, 1], [0, 1], [2], [2])
>>> from sage.all import * >>> H = matrix(ZZ, Integer(3), Integer(5), [Integer(1), Integer(0), Integer(0), Integer(45), -Integer(36), Integer(0), Integer(1), Integer(0), Integer(131), -Integer(107), Integer(0), Integer(0), Integer(0), Integer(178), -Integer(145)]); H [ 1 0 0 45 -36] [ 0 1 0 131 -107] [ 0 0 0 178 -145] >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.ones(H, [Integer(0),Integer(1),Integer(3)]) ([0, 1], [0, 1], [2], [2])
- sage.matrix.matrix_integer_dense_hnf.pad_zeros(A, nrows)[source]¶
Add zeros to the bottom of A so that the resulting matrix has nrows.
INPUT:
A
– a matrixnrows
– integer that is at least as big as the number of rows of A
OUTPUT: a matrix with nrows rows
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: a = matrix(ZZ, 2, 4, [1, 0, 0, 7, 0, 1, 5, 2]) sage: matrix_integer_dense_hnf.pad_zeros(a, 4) [1 0 0 7] [0 1 5 2] [0 0 0 0] [0 0 0 0] sage: matrix_integer_dense_hnf.pad_zeros(a, 2) [1 0 0 7] [0 1 5 2]
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> a = matrix(ZZ, Integer(2), Integer(4), [Integer(1), Integer(0), Integer(0), Integer(7), Integer(0), Integer(1), Integer(5), Integer(2)]) >>> matrix_integer_dense_hnf.pad_zeros(a, Integer(4)) [1 0 0 7] [0 1 5 2] [0 0 0 0] [0 0 0 0] >>> matrix_integer_dense_hnf.pad_zeros(a, Integer(2)) [1 0 0 7] [0 1 5 2]
- sage.matrix.matrix_integer_dense_hnf.pivots_of_hnf_matrix(H)[source]¶
Return the pivot columns of a matrix H assumed to be in HNF.
INPUT:
H
– a matrix that must be HNF
OUTPUT: list of pivots
EXAMPLES:
sage: H = matrix(ZZ, 3, 5, [1, 0, 0, 45, -36, 0, 1, 0, 131, -107, 0, 0, 0, 178, -145]); H [ 1 0 0 45 -36] [ 0 1 0 131 -107] [ 0 0 0 178 -145] sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.pivots_of_hnf_matrix(H) [0, 1, 3]
>>> from sage.all import * >>> H = matrix(ZZ, Integer(3), Integer(5), [Integer(1), Integer(0), Integer(0), Integer(45), -Integer(36), Integer(0), Integer(1), Integer(0), Integer(131), -Integer(107), Integer(0), Integer(0), Integer(0), Integer(178), -Integer(145)]); H [ 1 0 0 45 -36] [ 0 1 0 131 -107] [ 0 0 0 178 -145] >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.pivots_of_hnf_matrix(H) [0, 1, 3]
- sage.matrix.matrix_integer_dense_hnf.probable_hnf(A, include_zero_rows, proof)[source]¶
Return the HNF of A or raise an exception if something involving the randomized nature of the algorithm goes wrong along the way.
Calling this function again a few times should result it in it working, at least if proof=True.
INPUT:
A
– a matrixinclude_zero_rows
– booleanproof
– boolean
OUTPUT:
the Hermite normal form of A. cols – pivot columns
EXAMPLES:
sage: a = matrix(ZZ,4,3,[-1, -1, -1, -20, 4, 1, -1, 1, 2,1,2,3]) sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.probable_hnf(a, True, True) ( [1 0 0] [0 1 0] [0 0 1] [0 0 0], [0, 1, 2] ) sage: matrix_integer_dense_hnf.probable_hnf(a, False, True) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] ) sage: matrix_integer_dense_hnf.probable_hnf(a, False, False) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] )
>>> from sage.all import * >>> a = matrix(ZZ,Integer(4),Integer(3),[-Integer(1), -Integer(1), -Integer(1), -Integer(20), Integer(4), Integer(1), -Integer(1), Integer(1), Integer(2),Integer(1),Integer(2),Integer(3)]) >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.probable_hnf(a, True, True) ( [1 0 0] [0 1 0] [0 0 1] [0 0 0], [0, 1, 2] ) >>> matrix_integer_dense_hnf.probable_hnf(a, False, True) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] ) >>> matrix_integer_dense_hnf.probable_hnf(a, False, False) ( [1 0 0] [0 1 0] [0 0 1], [0, 1, 2] )
- sage.matrix.matrix_integer_dense_hnf.probable_pivot_columns(A)[source]¶
INPUT:
A
– a matrix
OUTPUT: a tuple of integers
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: a = matrix(ZZ,3,[0, -1, -1, 0, -20, 1, 0, 1, 2]) sage: a [ 0 -1 -1] [ 0 -20 1] [ 0 1 2] sage: matrix_integer_dense_hnf.probable_pivot_columns(a) (1, 2)
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> a = matrix(ZZ,Integer(3),[Integer(0), -Integer(1), -Integer(1), Integer(0), -Integer(20), Integer(1), Integer(0), Integer(1), Integer(2)]) >>> a [ 0 -1 -1] [ 0 -20 1] [ 0 1 2] >>> matrix_integer_dense_hnf.probable_pivot_columns(a) (1, 2)
- sage.matrix.matrix_integer_dense_hnf.probable_pivot_rows(A)[source]¶
Return rows of A that are very likely to be pivots.
This really finds the pivots of A modulo a random prime.
INPUT:
A
– a matrix
OUTPUT: a tuple of integers
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: a = matrix(ZZ,3,[0, -1, -1, 0, -20, 1, 0, 1, 2]) sage: a [ 0 -1 -1] [ 0 -20 1] [ 0 1 2] sage: matrix_integer_dense_hnf.probable_pivot_rows(a) (0, 1)
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> a = matrix(ZZ,Integer(3),[Integer(0), -Integer(1), -Integer(1), Integer(0), -Integer(20), Integer(1), Integer(0), Integer(1), Integer(2)]) >>> a [ 0 -1 -1] [ 0 -20 1] [ 0 1 2] >>> matrix_integer_dense_hnf.probable_pivot_rows(a) (0, 1)
- sage.matrix.matrix_integer_dense_hnf.sanity_checks(times=50, n=8, m=5, proof=True, stabilize=2, check_using_magma=True)[source]¶
Run random sanity checks on the modular \(p\)-adic HNF with tall and wide matrices both dense and sparse.
INPUT:
times
– number of times to randomly try matrices with each shapen
– number of rowsm
– number of columnsproof
– test with proof truestabilize
– parameter to pass to hnf algorithm when proof is Falsecheck_using_magma
– ifTrue
use Magma instead of PARI to check correctness of computed HNF’s. Since PARI’s HNF is buggy and slow (as of 2008-02-16 non-pivot entries sometimes are not normalized to be nonnegative) the default is Magma.
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf sage: matrix_integer_dense_hnf.sanity_checks(times=5, check_using_magma=False) small 8 x 5 0 1 2 3 4 (done) big 8 x 5 0 1 2 3 4 (done) small 5 x 8 0 1 2 3 4 (done) big 5 x 8 0 1 2 3 4 (done) sparse 8 x 5 0 1 2 3 4 (done) sparse 5 x 8 0 1 2 3 4 (done) ill conditioned -- 1000*A -- 8 x 5 0 1 2 3 4 (done) ill conditioned -- 1000*A but one row -- 8 x 5 0 1 2 3 4 (done)
>>> from sage.all import * >>> import sage.matrix.matrix_integer_dense_hnf as matrix_integer_dense_hnf >>> matrix_integer_dense_hnf.sanity_checks(times=Integer(5), check_using_magma=False) small 8 x 5 0 1 2 3 4 (done) big 8 x 5 0 1 2 3 4 (done) small 5 x 8 0 1 2 3 4 (done) big 5 x 8 0 1 2 3 4 (done) sparse 8 x 5 0 1 2 3 4 (done) sparse 5 x 8 0 1 2 3 4 (done) ill conditioned -- 1000*A -- 8 x 5 0 1 2 3 4 (done) ill conditioned -- 1000*A but one row -- 8 x 5 0 1 2 3 4 (done)
- sage.matrix.matrix_integer_dense_hnf.solve_system_with_difficult_last_row(B, a)[source]¶
Solve
B*x = a
when the last row of \(B\) contains huge entries using a clever trick that reduces the problem to solveC*x = a
where \(C\) is \(B\) but with the last row replaced by something small, along with one easy null space computation. The latter are both solved \(p\)-adically.INPUT:
B
– a square n x n nonsingular matrix with painful big bottom rowa
– an n x 1 column matrix
OUTPUT: the unique solution to
B*x = a
EXAMPLES:
sage: from sage.matrix.matrix_integer_dense_hnf import solve_system_with_difficult_last_row sage: B = matrix(ZZ, 3, [1,2,4, 3,-4,7, 939082,2930982,132902384098234]) sage: a = matrix(ZZ,3,1, [1,2,5]) sage: z = solve_system_with_difficult_last_row(B, a) sage: z [ 106321906985474/132902379815497] [132902385037291/1329023798154970] [ -5221794/664511899077485] sage: B*z [1] [2] [5]
>>> from sage.all import * >>> from sage.matrix.matrix_integer_dense_hnf import solve_system_with_difficult_last_row >>> B = matrix(ZZ, Integer(3), [Integer(1),Integer(2),Integer(4), Integer(3),-Integer(4),Integer(7), Integer(939082),Integer(2930982),Integer(132902384098234)]) >>> a = matrix(ZZ,Integer(3),Integer(1), [Integer(1),Integer(2),Integer(5)]) >>> z = solve_system_with_difficult_last_row(B, a) >>> z [ 106321906985474/132902379815497] [132902385037291/1329023798154970] [ -5221794/664511899077485] >>> B*z [1] [2] [5]