def saturation(A, proof=True, p=0, max_dets=5):
"""
Compute a saturation matrix of A.
INPUT:
A -- a matrix over ZZ
proof -- bool (default: True)
p -- int (default: 0); if not 0
only guarantees that output is p-saturated
max_dets -- int (default: 4) max number of dets of
submatrices to compute.
OUTPUT:
matrix -- saturation of the matrix A.
EXAMPLES:
sage: from sage.matrix.matrix_integer_dense_saturation import saturation
sage: A = matrix(ZZ, 2, 2, [3,2,3,4]); B = matrix(ZZ, 2,3,[1,2,3,4,5,6]); C = A*B
sage: C
[11 16 21]
[19 26 33]
sage: C.index_in_saturation()
18
sage: S = saturation(C); S
[11 16 21]
[-2 -3 -4]
sage: S.index_in_saturation()
1
sage: saturation(C, proof=False)
[11 16 21]
[-2 -3 -4]
sage: saturation(C, p=2)
[11 16 21]
[-2 -3 -4]
sage: saturation(C, p=2, max_dets=1)
[11 16 21]
[-2 -3 -4]
"""
# Find a submatrix of full rank and instead saturate that matrix.
r = A.rank()
if A.is_square() and r == A.nrows():
return identity_matrix(ZZ, r)
if A.nrows() > r:
P = []
while len(P) < r:
P = matrix_integer_dense_hnf.probable_pivot_rows(A)
A = A.matrix_from_rows(P)
# Factor out all common factors from all rows, just in case.
A = copy(A)
A._factor_out_common_factors_from_each_row()
if A.nrows() <= 1:
return A
A, zero_cols = A._delete_zero_columns()
if max_dets > 0:
# Take the GCD of at most num_dets randomly chosen determinants.
nr = A.nrows()
nc = A.ncols()
d = 0
trials = min(binomial(nc, nr), max_dets)
already_tried = []
while len(already_tried) < trials:
v = random_sublist_of_size(nc, nr)
tm = verbose('saturation -- checking det condition on submatrix')
d = gcd(d, A.matrix_from_columns(v).determinant(proof=proof))
verbose('saturation -- got det down to %s' % d, tm)
if gcd(d, p) == 1:
return A._insert_zero_columns(zero_cols)
already_tried.append(v)
if gcd(d, p) == 1:
# already p-saturated
return A._insert_zero_columns(zero_cols)
# Factor and p-saturate at each p.
# This is not a good algorithm, because all the HNF's in it are really slow!
#
#tm = verbose('factoring gcd %s of determinants'%d)
#limit = 2**31-1
#F = d.factor(limit = limit)
#D = [p for p, e in F if p <= limit]
#B = [n for n, e in F if n > limit] # all big factors -- there will only be at most one
#assert len(B) <= 1
#C = B[0]
#for p in D:
# A = p_saturation(A, p=p, proof=proof)
# This is a really simple but powerful algorithm.
# FACT: If A is a matrix of full rank, then hnf(transpose(A))^(-1)*A is a saturation of A.
# To make this practical we use solve_system_with_difficult_last_row, since the
# last column of HNF's are typically the only really big ones.
B = A.transpose().hermite_form(include_zero_rows=False, proof=proof)
B = B.transpose()
# Now compute B^(-1) * A
C = solve_system_with_difficult_last_row(B, A)
return C.change_ring(ZZ)._insert_zero_columns(zero_cols)
def saturation(A, proof=True, p=0, max_dets=5):
"""
Compute a saturation matrix of A.
INPUT:
- A -- a matrix over ZZ
- proof -- bool (default: True)
- p -- int (default: 0); if not 0 only guarantees that output is
p-saturated
- max_dets -- int (default: 4) max number of dets of submatrices to
compute.
OUTPUT:
matrix -- saturation of the matrix A.
EXAMPLES::
sage: from sage.matrix.matrix_integer_dense_saturation import saturation
sage: A = matrix(ZZ, 2, 2, [3,2,3,4]); B = matrix(ZZ, 2,3,[1,2,3,4,5,6]); C = A*B
sage: C
[11 16 21]
[19 26 33]
sage: C.index_in_saturation()
18
sage: S = saturation(C); S
[11 16 21]
[-2 -3 -4]
sage: S.index_in_saturation()
1
sage: saturation(C, proof=False)
[11 16 21]
[-2 -3 -4]
sage: saturation(C, p=2)
[11 16 21]
[-2 -3 -4]
sage: saturation(C, p=2, max_dets=1)
[11 16 21]
[-2 -3 -4]
"""
# Find a submatrix of full rank and instead saturate that matrix.
r = A.rank()
if A.is_square() and r == A.nrows():
return identity_matrix(ZZ, r)
if A.nrows() > r:
P = []
while len(P) < r:
P = matrix_integer_dense_hnf.probable_pivot_rows(A)
A = A.matrix_from_rows(P)
# Factor out all common factors from all rows, just in case.
A = copy(A)
A._factor_out_common_factors_from_each_row()
if A.nrows() <= 1:
return A
A, zero_cols = A._delete_zero_columns()
if max_dets > 0:
# Take the GCD of at most num_dets randomly chosen determinants.
nr = A.nrows()
nc = A.ncols()
d = 0
trials = min(binomial(nc, nr), max_dets)
already_tried = []
while len(already_tried) < trials:
v = random_sublist_of_size(nc, nr)
tm = verbose("saturation -- checking det condition on submatrix")
d = gcd(d, A.matrix_from_columns(v).determinant(proof=proof))
verbose("saturation -- got det down to %s" % d, tm)
if gcd(d, p) == 1:
return A._insert_zero_columns(zero_cols)
already_tried.append(v)
if gcd(d, p) == 1:
# already p-saturated
return A._insert_zero_columns(zero_cols)
# Factor and p-saturate at each p.
# This is not a good algorithm, because all the HNF's in it are really slow!
#
# tm = verbose('factoring gcd %s of determinants'%d)
# limit = 2**31-1
# F = d.factor(limit = limit)
# D = [p for p, e in F if p <= limit]
# B = [n for n, e in F if n > limit] # all big factors -- there will only be at most one
# assert len(B) <= 1
# C = B[0]
# for p in D:
# A = p_saturation(A, p=p, proof=proof)
# This is a really simple but powerful algorithm.
# FACT: If A is a matrix of full rank, then hnf(transpose(A))^(-1)*A is a saturation of A.
# To make this practical we use solve_system_with_difficult_last_row, since the
# last column of HNF's are typically the only really big ones.
B = A.transpose().hermite_form(include_zero_rows=False, proof=proof)
B = B.transpose()
# Now compute B^(-1) * A
C = solve_system_with_difficult_last_row(B, A)
return C.change_ring(ZZ)._insert_zero_columns(zero_cols)