def construct_spx(spx, max=Integer(1280)):
for r in range(3, max / 4 + 1):
spx[r] = {}
s = Integer(1)
m = Integer(4) * r
while m <= max:
c = sum([[(tuple(v), n) for n in Integers(r)]
for v in Integers(2)**Integer(s)], [])
c = [(v, n, Integer(1)) for v, n in c] + [(v, n, Integer(-1))
for v, n in c]
spx[r][s] = Graph([c, spx_adj])
s += 1
m *= 2
print "Finished r = %d, constructed %d graphs" % (r, len(spx[r]))
def __call__(self, n, modulus=0):
"""
Give the nth term of a binary recurrence sequence, possibly mod some modulus.
INPUT:
- ``n`` -- an integer (the index of the term in the binary recurrence sequence)
- ``modulus`` -- a natural number (optional -- default value is 0)
OUTPUT:
- An integer (the nth term of the binary recurrence sequence modulo ``modulus``)
EXAMPLES::
sage: R = BinaryRecurrenceSequence(3,3,2,1)
sage: R(2)
9
sage: R(101)
16158686318788579168659644539538474790082623100896663971001
sage: R(101,12)
9
sage: R(101)%12
9
"""
R = Integers(modulus)
F = matrix(
R, [[0, 1], [self.c, self.b]]
) # F*[u_{n}, u_{n+1}]^T = [u_{n+1}, u_{n+2}]^T (T indicates transpose).
v = matrix(R, [[self.u0], [self.u1]])
return list(F**n * v)[0][0]
def __init__(self, n):
"""
Args:
n (int): The number of elements in the finite ring.
"""
f = Integers(n)
operations = {'+': lambda x, y: x + y, '*': lambda x, y: x * y}
AlgebraicStructure.__init__(self, set(f), operations)
def construct_spx(r, s, multiedges=None, **kargs):
r"""
Construct a SPX(2, r, s) graph.
Return a tuple containing the data that can be used to construct
the requested SPX graph, and a boolean indicating whether the
constructed graph should be considered a multigraph.
INPUT:
- ``r`` - the ``r`` parameter indicating the range of the counter.
- ``s`` - the ``s`` parameter indicating the length of the string.
- ``multiedges`` - whether the constructed graph should be considered
a multigraph. If ``None`` (default), the second element of the output
will be set to ``True`` when ``r == 1``. If ``False`` and ``r == 1``,
a ``ValueError`` is raised. Otherwise, the second element of the
output will be ``multiedges``.
- any other named parameters are silently ignored.
"""
c = [tuple(x) for x
in cartesian_product([[tuple(y) for y
in Integers(2)**Integer(s)], Integers(r),
[Integer(1), Integer(-1)]])]
if r == 1:
if multiedges is False:
raise ValueError("A SPX graph with r = 1 has multiple edges")
data = sum([[((v, n, t), (v, n, -t)),
((v, n, t), (v[1:] + (Integer(0),), n, -t)),
((v, n, t), (v[1:] + (Integer(1),), n, -t))]
for v, n, t in c if t == 1], [])
multiedges = True
else:
data = [c, spx_adj]
return (data, multiedges)
def n_cylinders_edges(self, n):
r"""
EXAMPLES::
sage: from slabbe.markov_transformation import markov_transformations
sage: T = markov_transformations.Selmer()
sage: E = T.n_cylinders_edges(1)
sage: len(E)
39
"""
from sage.rings.finite_rings.integer_mod_ring import Integers
edges = set()
for w, cyl in self.n_cylinders_iterator(n):
cols = cyl.columns()
indices = Integers(len(cols))
edges.update(frozenset((cols[i], cols[i + 1])) for i in indices)
return edges
def __init__(self, n):
"""
Args:
n (int): The size of the underlying set.
"""
self.n = n
coeffs = product(range(n), repeat=n)
polys = frozenset(PolynomialRing(Integers(n), 'x')(r) for r in coeffs)
mats = set()
element_names = {}
for poly in polys:
m = matrix(n, sparse=True)
for i in range(n):
m[i, poly(i) % n] = 1
m.set_immutable()
mats.add(m)
element_names[m] = poly
operation = {'*': lambda x, y: x * y}
AlgebraicStructure.__init__(self, mats, operation, element_names)
def n_cylinders_edges(self, n):
r"""
Return the set of edges of the n-cylinders.
EXAMPLES::
sage: from slabbe.matrix_cocycle import cocycles
sage: ARP = cocycles.ARP()
sage: ARP.n_cylinders_edges(1)
{frozenset({(1, 1, 0), (1, 1, 1)}),
frozenset({(0, 1, 0), (1, 1, 0)}),
frozenset({(1, 1, 1), (2, 1, 1)}),
frozenset({(0, 0, 1), (1, 0, 1)}),
frozenset({(0, 1, 0), (0, 1, 1)}),
frozenset({(0, 1, 1), (1, 0, 1)}),
frozenset({(1, 0, 0), (1, 1, 0)}),
frozenset({(1, 1, 0), (2, 1, 1)}),
frozenset({(1, 0, 1), (1, 1, 2)}),
frozenset({(1, 1, 0), (1, 2, 1)}),
frozenset({(1, 0, 1), (2, 1, 1)}),
frozenset({(0, 0, 1), (0, 1, 1)}),
frozenset({(1, 0, 1), (1, 1, 1)}),
frozenset({(0, 1, 1), (1, 2, 1)}),
frozenset({(0, 1, 1), (1, 1, 2)}),
frozenset({(1, 0, 0), (1, 0, 1)}),
frozenset({(1, 1, 1), (1, 2, 1)}),
frozenset({(1, 0, 1), (1, 1, 0)}),
frozenset({(0, 1, 1), (1, 1, 1)}),
frozenset({(0, 1, 1), (1, 1, 0)}),
frozenset({(1, 1, 1), (1, 1, 2)})}
"""
from sage.rings.finite_rings.integer_mod_ring import Integers
edges = set()
for w,cyl in self.n_cylinders_iterator(n):
cols = cyl.columns()
indices = Integers(len(cols))
edges.update(frozenset((cols[i], cols[i+1])) for i in indices)
return edges
def logarithmTable(a, p, base, trace=False):
"""
Compute the tables of logarithms with base a modulo a prime p
for the given factor base, as needed by the Index calculus algorithm.
Requires Sage to run.
"""
from sage.matrix.constructor import Matrix
from sage.rings.finite_rings.integer_mod_ring import Integers
v = []
s = set()
r, l = 0, len(base)
M = Matrix(nrows=0, ncols=l)
fields = [
Integers(q) for q in totalFactorization(p - 1, trace=descend(trace))
]
while r < l:
i = randint(1, p - 1)
if i in s:
continue
s.add(i)
x = pow(a, i, p)
f = factorizeByBase(int(x), base, p)
if not f:
continue
if trace:
print("found factorization %d^%d = %s (mod %d)" %
(a, i, ' * '.join("%d^%d" % (q, e)
for q, e in zip(base, f) if e != 0), p))
MM = Matrix(list(M) + [f])
if all(Matrix(F, MM).rank() > r for F in fields):
M = MM
r += 1
v.append(i)
elif trace:
print("rank not increased, discarding")
return [e for e, in M**-1 * Matrix(zip(v)) % (p - 1)]
def _is_p_power_mod(a, p, N):
"""
Determine if ``a`` is a ``p`` th power modulo ``N``.
By the CRT, this is equivalent to the condition that ``a`` be a ``p`` th power mod all
distinct prime powers dividing ``N``. For each of these, we use the strong statement of
Hensel's lemma to lift ``p`` th powers mod `q` or `q^2` or `q^3` to ``p`` th powers mod `q^e`.
INPUT:
- ``a`` -- an integer
- ``p`` -- a rational prime number
- ``N`` -- a positive integer
OUTPUT:
- True if ``a`` is a ``p`` th power modulo ``N``; False otherwise.
EXAMPLES::
sage: sage.combinat.binary_recurrence_sequences._is_p_power_mod(2**3,7,29)
False
sage: sage.combinat.binary_recurrence_sequences._is_p_power_mod(2**3,3,29)
True
"""
#By the chinese remainder theorem, we can answer this question by examining whether
#a is a pth power mod q^e, for all distinct prime powers q^e dividing N.
for q, e in N.factor():
#If a = q^v*x, with
v = a.valuation(q)
#then if v>=e, a is congruent to 0 mod q^e and is thus a pth power trivially.
if v >= e:
continue
#otherwise, it can only be a pth power if v is a multiple of p.
if v % p != 0:
return False
#in this cse it is a pth power if x is a pth power mod q^(e-v), so let x = aa,
#and (e-v) = ee:
aa = a / q**v
ee = e - v
#The above steps are equivalent to the statement that we may assume a and qq are
#relatively prime, if we replace a with aa and e with ee. Now we must determine when
#aa is a pth power mod q^ee for (aa,q)=1.
#If q != p, then by Hensel's lemma, we may lift a pth power mod q, to a pth power
#mod q^2, etc.
if q != p:
#aa is necessarily a pth power mod q if p does not divide the order of the multiplicative
#group mod q, ie if q is not 1 mod p.
if q % p == 1:
#otherwise aa if a pth power mod q iff aa^(q-1)/p == 1
if GF(q)(aa)**((q - 1) / p) != 1:
return False
#If q = p and ee = 1, then everything is a pth power p by Fermat's little theorem.
elif ee > 1:
#We use the strong statement of Hensel's lemma, which implies that if p is odd
#and aa is a pth power mod p^2, then aa is a pth power mod any higher power of p
if p % 2 == 1:
#ZZ/(p^2)ZZ^\times is abstractly isomorphic to ZZ/(p)ZZ cross ZZ/(p-1)ZZ. then
#aa is a pth power mod p^2 if (aa)^(p*(p-1)/p) == 1, ie if aa^(p-1) == 1.
if Integers(p**2)(aa)**(p - 1) != 1:
return False
#Otherwise, p=2. By the strong statement of Hensel's lemma, if aa is a pth power
#mod p^3, then it is a pth power mod higher powers of p. So we need only check if it
#is a pth power mod p^2 and p^3.
elif ee == 2:
#all odd squares a 1 mod 4
if aa % 4 != 1:
return False
#all odd squares are 1 mod 8
elif aa % 8 != 1:
return False
return True
def period(self, m):
"""
Return the period of the binary recurrence sequence modulo
an integer ``m``.
If `n_1` is congruent to `n_2` modulu ``period(m)``, then `u_{n_1}` is
is congruent to `u_{n_2}` modulo ``m``.
INPUT:
- ``m`` -- an integer (modulo which the period of the recurrence relation is calculated).
OUTPUT:
- The integer (the period of the sequence modulo m)
EXAMPLES:
If `p = \\pm 1 \\mod 5`, then the period of the Fibonacci sequence
mod `p` is `p-1` (c.f. Lemma 3.3 of [BMS06]).
::
sage: R = BinaryRecurrenceSequence(1,1)
sage: R.period(31)
30
sage: [R(i) % 4 for i in range(12)]
[0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1]
sage: R.period(4)
6
This function works for degenerate sequences as well.
::
sage: S = BinaryRecurrenceSequence(2,0,1,2)
sage: S.is_degenerate()
True
sage: S.is_geometric()
True
sage: [S(i) % 17 for i in range(16)]
[1, 2, 4, 8, 16, 15, 13, 9, 1, 2, 4, 8, 16, 15, 13, 9]
sage: S.period(17)
8
Note: the answer is cached.
"""
#If we have already computed the period mod m, then we return the stored value.
if m in self._period_dict:
return self._period_dict[m]
else:
R = Integers(m)
A = matrix(R, [[0, 1], [self.c, self.b]])
w = matrix(R, [[self.u0], [self.u1]])
Fac = list(m.factor())
Periods = {}
#To compute the period mod m, we compute the least integer n such that A^n*w == w. This necessarily
#divides the order of A as a matrix in GL_2(Z/mZ).
#We compute the period modulo all distinct prime powers dividing m, and combine via the lcm.
#To compute the period mod p^e, we first compute the order mod p. Then the period mod p^e
#must divide p^{4e-4}*period(p), as the subgroup of matrices mod p^e, which reduce to
#the identity mod p is of order (p^{e-1})^4. So we compute the period mod p^e by successively
#multiplying the period mod p by powers of p.
for i in Fac:
p = i[0]
e = i[1]
#first compute the period mod p
if p in self._period_dict:
perp = self._period_dict[p]
else:
F = A.change_ring(GF(p))
v = w.change_ring(GF(p))
FF = F**(p - 1)
p1fac = list((p - 1).factor())
#The order of any matrix in GL_2(F_p) either divides p(p-1) or (p-1)(p+1).
#The order divides p-1 if it is diagonalizable. In any case, det(F^(p-1))=1,
#so if tr(F^(p-1)) = 2, then it must be triangular of the form [[1,a],[0,1]].
#The order of the subgroup of matrices of this form is p, so the order must divide
#p(p-1) -- in fact it must be a multiple of p. If this is not the case, then the
#order divides (p-1)(p+1). As the period divides the order of the matrix in GL_2(F_p),
#these conditions hold for the period as well.
#check if the order divides (p-1)
if FF * v == v:
M = p - 1
Mfac = p1fac
#check if the trace is 2, then the order is a multiple of p dividing p*(p-1)
elif (FF).trace() == 2:
M = p - 1
Mfac = p1fac
F = F**p #replace F by F^p as now we only need to determine the factor dividing (p-1)
#otherwise it will divide (p+1)(p-1)
else:
M = (p + 1) * (p - 1)
p2fac = list(
(p + 1).factor()
) #factor the (p+1) and (p-1) terms separately and then combine for speed
Mfac_dic = {}
for i in list(p1fac + p2fac):
if i[0] not in Mfac_dic:
Mfac_dic[i[0]] = i[1]
else:
Mfac_dic[i[0]] = Mfac_dic[i[0]] + i[1]
Mfac = [(i, Mfac_dic[i]) for i in Mfac_dic]
#Now use a fast order algorithm to compute the period. We know that the period divides
#M = i_1*i_2*...*i_l where the i_j denote not necessarily distinct prime factors. As
#F^M*v == v, for each i_j, if F^(M/i_j)*v == v, then the period divides (M/i_j). After
#all factors have been iterated over, the result is the period mod p.
Mfac = list(Mfac)
C = []
#expand the list of prime factors so every factor is with multiplicity 1
for i in range(len(Mfac)):
for j in range(Mfac[i][1]):
C.append(Mfac[i][0])
Mfac = C
n = M
for i in Mfac:
b = Integer(n / i)
if F**b * v == v:
n = b
perp = n
#Now compute the period mod p^e by stepping up by multiples of p
F = A.change_ring(Integers(p**e))
v = w.change_ring(Integers(p**e))
FF = F**perp
if FF * v == v:
perpe = perp
else:
tries = 0
while True:
tries += 1
FF = FF**p
if FF * v == v:
perpe = perp * p**tries
break
Periods[p] = perpe
#take the lcm of the periods mod all distinct primes dividing m
period = 1
for p in Periods:
period = lcm(Periods[p], period)
self._period_dict[m] = period #cache the period mod m
return period
def residue(self, absprec=1, field=None, check_prec=True):
r"""
Reduces this element modulo `p^{\mathrm{absprec}}`.
INPUT:
- ``absprec`` -- a non-negative integer (default: ``1``)
- ``field`` -- boolean (default ``None``). Whether to return an element of GF(p) or Zmod(p).
- ``check_prec`` -- boolean (default ``True``). Whether to raise an error if this
element has insufficient precision to determine the reduction.
OUTPUT:
This element reduced modulo `p^\mathrm{absprec}` as an element of
`\ZZ/p^\mathrm{absprec}\ZZ`
EXAMPLES::
sage: R = ZpLC(7,4)
sage: a = R(8)
sage: a.residue(1)
1
TESTS::
sage: R = ZpLC(7,4)
sage: a = R(8)
sage: a.residue(0)
0
sage: a.residue(-1)
Traceback (most recent call last):
...
ValueError: cannot reduce modulo a negative power of p.
sage: a.residue(5)
Traceback (most recent call last):
...
PrecisionError: not enough precision known in order to compute residue.
sage: a.residue(5, check_prec=False)
8
sage: a.residue(field=True).parent()
Finite Field of size 7
"""
if not isinstance(absprec, Integer):
absprec = Integer(absprec)
if check_prec and absprec > self.precision_absolute():
raise PrecisionError(
"not enough precision known in order to compute residue.")
elif absprec < 0:
raise ValueError("cannot reduce modulo a negative power of p.")
if self.valuation() < 0:
raise ValueError(
"element must have non-negative valuation in order to compute residue."
)
if field is None:
field = (absprec == 1)
elif field and absprec != 1:
raise ValueError("field keyword may only be set at precision 1")
p = self._parent.prime()
if field:
from sage.rings.finite_rings.finite_field_constructor import GF
ring = GF(p)
else:
from sage.rings.finite_rings.integer_mod_ring import Integers
ring = Integers(p**absprec)
return ring(self.value())
def suspenders(self):
# from sage.rings import Integers
return Integers()