Modular algorithm to compute Hermite normal forms of integer matrices.¶
AUTHORS:
- Clement Pernet and William Stein (2008-02-07): initial version
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sage.matrix.matrix_integer_dense_hnf.
add_column
(B, H_B, a, proof)¶ The add column procedure.
INPUT:
- B – a square matrix (may be singular)
- H_B – the Hermite normal form of B
- a – an n x 1 matrix, where B has n rows
- proof – bool; 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]
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sage.matrix.matrix_integer_dense_hnf.
add_column_fallback
(B, a, proof)¶ 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 rows proof – bool; whether to prove result correctOUTPUT:
x – a vector such that H’ = H_B.augment(x) is the HNF of A = 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]
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sage.matrix.matrix_integer_dense_hnf.
add_row
(A, b, pivots, include_zero_rows)¶ The add row procedure.
INPUT:
- A – a matrix in Hermite normal form with n column
- b – an n x 1 row matrix
- pivots – 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]
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sage.matrix.matrix_integer_dense_hnf.
benchmark_hnf
(nrange, bits=4)¶ Run benchmark program.
EXAMPLES:
sage: import sage.matrix.matrix_integer_dense_hnf as hnf sage: hnf.benchmark_hnf([50,100],32) ('sage', 50, 32, ...), ('sage', 100, 32, ...),
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sage.matrix.matrix_integer_dense_hnf.
benchmark_magma_hnf
(nrange, bits=4)¶ 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, ...),
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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)¶ This is used for internal purposes for computing determinants quickly (with the hybrid p-adic / multimodular algorithm).
INPUT:
- A – a square matrix
- d – a divisor of the determinant of A
- p – a prime
- z_mod – values of det/d (mod …)
- moduli – the moduli so far
- z_so_far – for a modulus p in the list moduli, (z_so_far mod p) is the determinant of A modulo p.
- N_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
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sage.matrix.matrix_integer_dense_hnf.
det_given_divisor
(A, d, proof=True, stabilize=2)¶ Given a divisor d of the determinant of A, compute the determinant of A.
INPUT:
A
– a square integer matrixd
– a nonzero integer that is assumed to divide the determinant of Aproof
– bool (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 with proof=False.stabilize
– int (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
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
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
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
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sage.matrix.matrix_integer_dense_hnf.
det_padic
(A, proof=True, stabilize=2)¶ 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
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sage.matrix.matrix_integer_dense_hnf.
double_det
(A, b, c, proof)¶ Compute the determinants of the stacked integer matrices A.stack(b) and A.stack(c).
INPUT:
- A – an (n-1) x n matrix
- b – an 1 x n matrix
- c – an 1 x n matrix
- proof – 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)
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sage.matrix.matrix_integer_dense_hnf.
extract_ones_data
(H, pivots)¶ Compute ones data and corresponding submatrices of H.
This is used to optimized the
add_row()
function.INPUT:
- H – a matrix in HNF
- pivots – 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 rows
- D – all non-onecol columns and other rows
- E – 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] )
- 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])
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sage.matrix.matrix_integer_dense_hnf.
hnf
(A, include_zero_rows=True, proof=True)¶ 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 integers.
- include_zero_rows – bool (default: True) whether or not to include zero rows in the output matrix
- proof – whether or not to prove the result correct.
OUTPUT:
- matrix – the Hermite normal form of A
- pivots – 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] )
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sage.matrix.matrix_integer_dense_hnf.
hnf_square
(A, proof)¶ INPUT:
- a nonsingular n x n matrix A 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]
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sage.matrix.matrix_integer_dense_hnf.
hnf_with_transformation
(A, proof=True)¶ 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 integers.
- proof – whether or not to prove the result correct.
OUTPUT:
- matrix – the Hermite normal form H of A
- U – 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]
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sage.matrix.matrix_integer_dense_hnf.
hnf_with_transformation_tests
(n=10, m=5, trials=10)¶ 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
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sage.matrix.matrix_integer_dense_hnf.
interleave_matrices
(A, B, cols1, cols2)¶ 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]
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sage.matrix.matrix_integer_dense_hnf.
is_in_hnf_form
(H, pivots)¶ 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
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sage.matrix.matrix_integer_dense_hnf.
max_det_prime
(n)¶ Return the largest prime so that it is reasonably efficient to compute modulo that prime with n x n matrices in LinBox.
INPUT:
n
– a 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
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sage.matrix.matrix_integer_dense_hnf.
ones
(H, pivots)¶ 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 form
- pivots – 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])
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sage.matrix.matrix_integer_dense_hnf.
pad_zeros
(A, nrows)¶ Add zeros to the bottom of A so that the resulting matrix has nrows.
INPUT:
- A – a matrix
- nrows – an 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]
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sage.matrix.matrix_integer_dense_hnf.
pivots_of_hnf_matrix
(H)¶ Return the pivot columns of a matrix H assumed to be in HNF.
INPUT:
- H – a matrix that must be HNF
OUTPUT:
- list – 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]
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sage.matrix.matrix_integer_dense_hnf.
probable_hnf
(A, include_zero_rows, proof)¶ 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 matrix
- include_zero_rows – bool
- proof – bool
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] )
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sage.matrix.matrix_integer_dense_hnf.
probable_pivot_columns
(A)¶ 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)
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sage.matrix.matrix_integer_dense_hnf.
probable_pivot_rows
(A)¶ 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)
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sage.matrix.matrix_integer_dense_hnf.
sanity_checks
(times=50, n=8, m=5, proof=True, stabilize=2, check_using_magma=True)¶ 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 shape
- n – number of rows
- m – number of columns
- proof – test with proof true
- stabilize – parameter to pass to hnf algorithm when proof is False
- check_using_magma – if True 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)
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sage.matrix.matrix_integer_dense_hnf.
solve_system_with_difficult_last_row
(B, a)¶ Solve B*x = a when the last row of \(B\) contains huge entries using a clever trick that reduces the problem to solve C*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 row.
- a – 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]