from sympy.core.add import Add
from sympy.core.assumptions import check_assumptions
from sympy.core.containers import Tuple
from sympy.core.exprtools import factor_terms
from sympy.core.function import _mexpand
from sympy.core.mul import Mul
from sympy.core.numbers import Rational
from sympy.core.numbers import igcdex, ilcm, igcd
from sympy.core.power import integer_nthroot, isqrt
from sympy.core.relational import Eq
from sympy.core.singleton import S
from sympy.core.sorting import default_sort_key, ordered
from sympy.core.symbol import Symbol, symbols
from sympy.core.sympify import _sympify
from sympy.functions.elementary.complexes import sign
from sympy.functions.elementary.integers import floor
from sympy.functions.elementary.miscellaneous import sqrt
from sympy.matrices.dense import MutableDenseMatrix as Matrix
from sympy.ntheory.factor_ import (
divisors, factorint, multiplicity, perfect_power)
from sympy.ntheory.generate import nextprime
from sympy.ntheory.primetest import is_square, isprime
from sympy.ntheory.residue_ntheory import sqrt_mod
from sympy.polys.polyerrors import GeneratorsNeeded
from sympy.polys.polytools import Poly, factor_list
from sympy.simplify.simplify import signsimp
from sympy.solvers.solveset import solveset_real
from sympy.utilities import numbered_symbols
from sympy.utilities.misc import as_int, filldedent
from sympy.utilities.iterables import (is_sequence, subsets, permute_signs,
signed_permutations, ordered_partitions)
# these are imported with 'from sympy.solvers.diophantine import *
__all__ = ['diophantine', 'classify_diop']
class DiophantineSolutionSet(set):
"""
Container for a set of solutions to a particular diophantine equation.
The base representation is a set of tuples representing each of the solutions.
Parameters
==========
symbols : list
List of free symbols in the original equation.
parameters: list
List of parameters to be used in the solution.
Examples
========
Adding solutions:
>>> from sympy.solvers.diophantine.diophantine import DiophantineSolutionSet
>>> from sympy.abc import x, y, t, u
>>> s1 = DiophantineSolutionSet([x, y], [t, u])
>>> s1
set()
>>> s1.add((2, 3))
>>> s1.add((-1, u))
>>> s1
{(-1, u), (2, 3)}
>>> s2 = DiophantineSolutionSet([x, y], [t, u])
>>> s2.add((3, 4))
>>> s1.update(*s2)
>>> s1
{(-1, u), (2, 3), (3, 4)}
Conversion of solutions into dicts:
>>> list(s1.dict_iterator())
[{x: -1, y: u}, {x: 2, y: 3}, {x: 3, y: 4}]
Substituting values:
>>> s3 = DiophantineSolutionSet([x, y], [t, u])
>>> s3.add((t**2, t + u))
>>> s3
{(t**2, t + u)}
>>> s3.subs({t: 2, u: 3})
{(4, 5)}
>>> s3.subs(t, -1)
{(1, u - 1)}
>>> s3.subs(t, 3)
{(9, u + 3)}
Evaluation at specific values. Positional arguments are given in the same order as the parameters:
>>> s3(-2, 3)
{(4, 1)}
>>> s3(5)
{(25, u + 5)}
>>> s3(None, 2)
{(t**2, t + 2)}
"""
def __init__(self, symbols_seq, parameters):
super().__init__()
if not is_sequence(symbols_seq):
raise ValueError("Symbols must be given as a sequence.")
if not is_sequence(parameters):
raise ValueError("Parameters must be given as a sequence.")
self.symbols = tuple(symbols_seq)
self.parameters = tuple(parameters)
def add(self, solution):
if len(solution) != len(self.symbols):
raise ValueError("Solution should have a length of %s, not %s" % (len(self.symbols), len(solution)))
super().add(Tuple(*solution))
def update(self, *solutions):
for solution in solutions:
self.add(solution)
def dict_iterator(self):
for solution in ordered(self):
yield dict(zip(self.symbols, solution))
def subs(self, *args, **kwargs):
result = DiophantineSolutionSet(self.symbols, self.parameters)
for solution in self:
result.add(solution.subs(*args, **kwargs))
return result
def __call__(self, *args):
if len(args) > len(self.parameters):
raise ValueError("Evaluation should have at most %s values, not %s" % (len(self.parameters), len(args)))
rep = {p: v for p, v in zip(self.parameters, args) if v is not None}
return self.subs(rep)
class DiophantineEquationType:
"""
Internal representation of a particular diophantine equation type.
Parameters
==========
equation :
The diophantine equation that is being solved.
free_symbols : list (optional)
The symbols being solved for.
Attributes
==========
total_degree :
The maximum of the degrees of all terms in the equation
homogeneous :
Does the equation contain a term of degree 0
homogeneous_order :
Does the equation contain any coefficient that is in the symbols being solved for
dimension :
The number of symbols being solved for
"""
name = None # type: str
def __init__(self, equation, free_symbols=None):
self.equation = _sympify(equation).expand(force=True)
if free_symbols is not None:
self.free_symbols = free_symbols
else:
self.free_symbols = list(self.equation.free_symbols)
self.free_symbols.sort(key=default_sort_key)
if not self.free_symbols:
raise ValueError('equation should have 1 or more free symbols')
self.coeff = self.equation.as_coefficients_dict()
if not all(_is_int(c) for c in self.coeff.values()):
raise TypeError("Coefficients should be Integers")
self.total_degree = Poly(self.equation).total_degree()
self.homogeneous = 1 not in self.coeff
self.homogeneous_order = not (set(self.coeff) & set(self.free_symbols))
self.dimension = len(self.free_symbols)
self._parameters = None
def matches(self):
"""
Determine whether the given equation can be matched to the particular equation type.
"""
return False
@property
def n_parameters(self):
return self.dimension
@property
def parameters(self):
if self._parameters is None:
self._parameters = symbols('t_:%i' % (self.n_parameters,), integer=True)
return self._parameters
def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet:
raise NotImplementedError('No solver has been written for %s.' % self.name)
def pre_solve(self, parameters=None):
if not self.matches():
raise ValueError("This equation does not match the %s equation type." % self.name)
if parameters is not None:
if len(parameters) != self.n_parameters:
raise ValueError("Expected %s parameter(s) but got %s" % (self.n_parameters, len(parameters)))
self._parameters = parameters
class Univariate(DiophantineEquationType):
"""
Representation of a univariate diophantine equation.
A univariate diophantine equation is an equation of the form
`a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are
integer constants and `x` is an integer variable.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import Univariate
>>> from sympy.abc import x
>>> Univariate((x - 2)*(x - 3)**2).solve() # solves equation (x - 2)*(x - 3)**2 == 0
{(2,), (3,)}
"""
name = 'univariate'
def matches(self):
return self.dimension == 1
def solve(self, parameters=None, limit=None):
self.pre_solve(parameters)
result = DiophantineSolutionSet(self.free_symbols, parameters=self.parameters)
for i in solveset_real(self.equation, self.free_symbols[0]).intersect(S.Integers):
result.add((i,))
return result
class Linear(DiophantineEquationType):
"""
Representation of a linear diophantine equation.
A linear diophantine equation is an equation of the form `a_{1}x_{1} +
a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are
integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import Linear
>>> from sympy.abc import x, y, z
>>> l1 = Linear(2*x - 3*y - 5)
>>> l1.matches() # is this equation linear
True
>>> l1.solve() # solves equation 2*x - 3*y - 5 == 0
{(3*t_0 - 5, 2*t_0 - 5)}
Here x = -3*t_0 - 5 and y = -2*t_0 - 5
>>> Linear(2*x - 3*y - 4*z -3).solve()
{(t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3)}
"""
name = 'linear'
def matches(self):
return self.total_degree == 1
def solve(self, parameters=None, limit=None):
self.pre_solve(parameters)
coeff = self.coeff
var = self.free_symbols
if 1 in coeff:
# negate coeff[] because input is of the form: ax + by + c == 0
# but is used as: ax + by == -c
c = -coeff[1]
else:
c = 0
result = DiophantineSolutionSet(var, parameters=self.parameters)
params = result.parameters
if len(var) == 1:
q, r = divmod(c, coeff[var[0]])
if not r:
result.add((q,))
return result
else:
return result
'''
base_solution_linear() can solve diophantine equations of the form:
a*x + b*y == c
We break down multivariate linear diophantine equations into a
series of bivariate linear diophantine equations which can then
be solved individually by base_solution_linear().
Consider the following:
a_0*x_0 + a_1*x_1 + a_2*x_2 == c
which can be re-written as:
a_0*x_0 + g_0*y_0 == c
where
g_0 == gcd(a_1, a_2)
and
y == (a_1*x_1)/g_0 + (a_2*x_2)/g_0
This leaves us with two binary linear diophantine equations.
For the first equation:
a == a_0
b == g_0
c == c
For the second:
a == a_1/g_0
b == a_2/g_0
c == the solution we find for y_0 in the first equation.
The arrays A and B are the arrays of integers used for
'a' and 'b' in each of the n-1 bivariate equations we solve.
'''
A = [coeff[v] for v in var]
B = []
if len(var) > 2:
B.append(igcd(A[-2], A[-1]))
A[-2] = A[-2] // B[0]
A[-1] = A[-1] // B[0]
for i in range(len(A) - 3, 0, -1):
gcd = igcd(B[0], A[i])
B[0] = B[0] // gcd
A[i] = A[i] // gcd
B.insert(0, gcd)
B.append(A[-1])
'''
Consider the trivariate linear equation:
4*x_0 + 6*x_1 + 3*x_2 == 2
This can be re-written as:
4*x_0 + 3*y_0 == 2
where
y_0 == 2*x_1 + x_2
(Note that gcd(3, 6) == 3)
The complete integral solution to this equation is:
x_0 == 2 + 3*t_0
y_0 == -2 - 4*t_0
where 't_0' is any integer.
Now that we have a solution for 'x_0', find 'x_1' and 'x_2':
2*x_1 + x_2 == -2 - 4*t_0
We can then solve for '-2' and '-4' independently,
and combine the results:
2*x_1a + x_2a == -2
x_1a == 0 + t_0
x_2a == -2 - 2*t_0
2*x_1b + x_2b == -4*t_0
x_1b == 0*t_0 + t_1
x_2b == -4*t_0 - 2*t_1
==>
x_1 == t_0 + t_1
x_2 == -2 - 6*t_0 - 2*t_1
where 't_0' and 't_1' are any integers.
Note that:
4*(2 + 3*t_0) + 6*(t_0 + t_1) + 3*(-2 - 6*t_0 - 2*t_1) == 2
for any integral values of 't_0', 't_1'; as required.
This method is generalised for many variables, below.
'''
solutions = []
for Ai, Bi in zip(A, B):
tot_x, tot_y = [], []
for j, arg in enumerate(Add.make_args(c)):
if arg.is_Integer:
# example: 5 -> k = 5
k, p = arg, S.One
pnew = params[0]
else: # arg is a Mul or Symbol
# example: 3*t_1 -> k = 3
# example: t_0 -> k = 1
k, p = arg.as_coeff_Mul()
pnew = params[params.index(p) + 1]
sol = sol_x, sol_y = base_solution_linear(k, Ai, Bi, pnew)
if p is S.One:
if None in sol:
return result
else:
# convert a + b*pnew -> a*p + b*pnew
if isinstance(sol_x, Add):
sol_x = sol_x.args[0]*p + sol_x.args[1]
if isinstance(sol_y, Add):
sol_y = sol_y.args[0]*p + sol_y.args[1]
tot_x.append(sol_x)
tot_y.append(sol_y)
solutions.append(Add(*tot_x))
c = Add(*tot_y)
solutions.append(c)
result.add(solutions)
return result
class BinaryQuadratic(DiophantineEquationType):
"""
Representation of a binary quadratic diophantine equation.
A binary quadratic diophantine equation is an equation of the
form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`, where `A, B, C, D, E,
F` are integer constants and `x` and `y` are integer variables.
Examples
========
>>> from sympy.abc import x, y
>>> from sympy.solvers.diophantine.diophantine import BinaryQuadratic
>>> b1 = BinaryQuadratic(x**3 + y**2 + 1)
>>> b1.matches()
False
>>> b2 = BinaryQuadratic(x**2 + y**2 + 2*x + 2*y + 2)
>>> b2.matches()
True
>>> b2.solve()
{(-1, -1)}
References
==========
.. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online],
Available: https://www.alpertron.com.ar/METHODS.HTM
.. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online],
Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf
"""
name = 'binary_quadratic'
def matches(self):
return self.total_degree == 2 and self.dimension == 2
def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet:
self.pre_solve(parameters)
var = self.free_symbols
coeff = self.coeff
x, y = var
A = coeff[x**2]
B = coeff[x*y]
C = coeff[y**2]
D = coeff[x]
E = coeff[y]
F = coeff[S.One]
A, B, C, D, E, F = [as_int(i) for i in _remove_gcd(A, B, C, D, E, F)]
# (1) Simple-Hyperbolic case: A = C = 0, B != 0
# In this case equation can be converted to (Bx + E)(By + D) = DE - BF
# We consider two cases; DE - BF = 0 and DE - BF != 0
# More details, https://www.alpertron.com.ar/METHODS.HTM#SHyperb
result = DiophantineSolutionSet(var, self.parameters)
t, u = result.parameters
discr = B**2 - 4*A*C
if A == 0 and C == 0 and B != 0:
if D*E - B*F == 0:
q, r = divmod(E, B)
if not r:
result.add((-q, t))
q, r = divmod(D, B)
if not r:
result.add((t, -q))
else:
div = divisors(D*E - B*F)
div = div + [-term for term in div]
for d in div:
x0, r = divmod(d - E, B)
if not r:
q, r = divmod(D*E - B*F, d)
if not r:
y0, r = divmod(q - D, B)
if not r:
result.add((x0, y0))
# (2) Parabolic case: B**2 - 4*A*C = 0
# There are two subcases to be considered in this case.
# sqrt(c)D - sqrt(a)E = 0 and sqrt(c)D - sqrt(a)E != 0
# More Details, https://www.alpertron.com.ar/METHODS.HTM#Parabol
elif discr == 0:
if A == 0:
s = BinaryQuadratic(self.equation, free_symbols=[y, x]).solve(parameters=[t, u])
for soln in s:
result.add((soln[1], soln[0]))
else:
g = sign(A)*igcd(A, C)
a = A // g
c = C // g
e = sign(B / A)
sqa = isqrt(a)
sqc = isqrt(c)
_c = e*sqc*D - sqa*E
if not _c:
z = Symbol("z", real=True)
eq = sqa*g*z**2 + D*z + sqa*F
roots = solveset_real(eq, z).intersect(S.Integers)
for root in roots:
ans = diop_solve(sqa*x + e*sqc*y - root)
result.add((ans[0], ans[1]))
elif _is_int(c):
solve_x = lambda u: -e*sqc*g*_c*t**2 - (E + 2*e*sqc*g*u)*t \
- (e*sqc*g*u**2 + E*u + e*sqc*F) // _c
solve_y = lambda u: sqa*g*_c*t**2 + (D + 2*sqa*g*u)*t \
+ (sqa*g*u**2 + D*u + sqa*F) // _c
for z0 in range(0, abs(_c)):
# Check if the coefficients of y and x obtained are integers or not
if (divisible(sqa*g*z0**2 + D*z0 + sqa*F, _c) and
divisible(e*sqc*g*z0**2 + E*z0 + e*sqc*F, _c)):
result.add((solve_x(z0), solve_y(z0)))
# (3) Method used when B**2 - 4*A*C is a square, is described in p. 6 of the below paper
# by John P. Robertson.
# https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf
elif is_square(discr):
if A != 0:
r = sqrt(discr)
u, v = symbols("u, v", integer=True)
eq = _mexpand(
4*A*r*u*v + 4*A*D*(B*v + r*u + r*v - B*u) +
2*A*4*A*E*(u - v) + 4*A*r*4*A*F)
solution = diop_solve(eq, t)
for s0, t0 in solution:
num = B*t0 + r*s0 + r*t0 - B*s0
x_0 = S(num) / (4*A*r)
y_0 = S(s0 - t0) / (2*r)
if isinstance(s0, Symbol) or isinstance(t0, Symbol):
if len(check_param(x_0, y_0, 4*A*r, parameters)) > 0:
ans = check_param(x_0, y_0, 4*A*r, parameters)
result.update(*ans)
elif x_0.is_Integer and y_0.is_Integer:
if is_solution_quad(var, coeff, x_0, y_0):
result.add((x_0, y_0))
else:
s = BinaryQuadratic(self.equation, free_symbols=var[::-1]).solve(parameters=[t, u]) # Interchange x and y
while s:
result.add(s.pop()[::-1]) # and solution <--------+
# (4) B**2 - 4*A*C > 0 and B**2 - 4*A*C not a square or B**2 - 4*A*C < 0
else:
P, Q = _transformation_to_DN(var, coeff)
D, N = _find_DN(var, coeff)
solns_pell = diop_DN(D, N)
if D < 0:
for x0, y0 in solns_pell:
for x in [-x0, x0]:
for y in [-y0, y0]:
s = P*Matrix([x, y]) + Q
try:
result.add([as_int(_) for _ in s])
except ValueError:
pass
else:
# In this case equation can be transformed into a Pell equation
solns_pell = set(solns_pell)
for X, Y in list(solns_pell):
solns_pell.add((-X, -Y))
a = diop_DN(D, 1)
T = a[0][0]
U = a[0][1]
if all(_is_int(_) for _ in P[:4] + Q[:2]):
for r, s in solns_pell:
_a = (r + s*sqrt(D))*(T + U*sqrt(D))**t
_b = (r - s*sqrt(D))*(T - U*sqrt(D))**t
x_n = _mexpand(S(_a + _b) / 2)
y_n = _mexpand(S(_a - _b) / (2*sqrt(D)))
s = P*Matrix([x_n, y_n]) + Q
result.add(s)
else:
L = ilcm(*[_.q for _ in P[:4] + Q[:2]])
k = 1
T_k = T
U_k = U
while (T_k - 1) % L != 0 or U_k % L != 0:
T_k, U_k = T_k*T + D*U_k*U, T_k*U + U_k*T
k += 1
for X, Y in solns_pell:
for i in range(k):
if all(_is_int(_) for _ in P*Matrix([X, Y]) + Q):
_a = (X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t
_b = (X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t
Xt = S(_a + _b) / 2
Yt = S(_a - _b) / (2*sqrt(D))
s = P*Matrix([Xt, Yt]) + Q
result.add(s)
X, Y = X*T + D*U*Y, X*U + Y*T
return result
class InhomogeneousTernaryQuadratic(DiophantineEquationType):
"""
Representation of an inhomogeneous ternary quadratic.
No solver is currently implemented for this equation type.
"""
name = 'inhomogeneous_ternary_quadratic'
def matches(self):
if not (self.total_degree == 2 and self.dimension == 3):
return False
if not self.homogeneous:
return False
return not self.homogeneous_order
class HomogeneousTernaryQuadraticNormal(DiophantineEquationType):
"""
Representation of a homogeneous ternary quadratic normal diophantine equation.
Examples
========
>>> from sympy.abc import x, y, z
>>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadraticNormal
>>> HomogeneousTernaryQuadraticNormal(4*x**2 - 5*y**2 + z**2).solve()
{(1, 2, 4)}
"""
name = 'homogeneous_ternary_quadratic_normal'
def matches(self):
if not (self.total_degree == 2 and self.dimension == 3):
return False
if not self.homogeneous:
return False
if not self.homogeneous_order:
return False
nonzero = [k for k in self.coeff if self.coeff[k]]
return len(nonzero) == 3 and all(i**2 in nonzero for i in self.free_symbols)
def solve(self, parameters=None, limit=None) -> DiophantineSolutionSet:
self.pre_solve(parameters)
var = self.free_symbols
coeff = self.coeff
x, y, z = var
a = coeff[x**2]
b = coeff[y**2]
c = coeff[z**2]
(sqf_of_a, sqf_of_b, sqf_of_c), (a_1, b_1, c_1), (a_2, b_2, c_2) = \
sqf_normal(a, b, c, steps=True)
A = -a_2*c_2
B = -b_2*c_2
result = DiophantineSolutionSet(var, parameters=self.parameters)
# If following two conditions are satisfied then there are no solutions
if A < 0 and B < 0:
return result
if (
sqrt_mod(-b_2*c_2, a_2) is None or
sqrt_mod(-c_2*a_2, b_2) is None or
sqrt_mod(-a_2*b_2, c_2) is None):
return result
z_0, x_0, y_0 = descent(A, B)
z_0, q = _rational_pq(z_0, abs(c_2))
x_0 *= q
y_0 *= q
x_0, y_0, z_0 = _remove_gcd(x_0, y_0, z_0)
# Holzer reduction
if sign(a) == sign(b):
x_0, y_0, z_0 = holzer(x_0, y_0, z_0, abs(a_2), abs(b_2), abs(c_2))
elif sign(a) == sign(c):
x_0, z_0, y_0 = holzer(x_0, z_0, y_0, abs(a_2), abs(c_2), abs(b_2))
else:
y_0, z_0, x_0 = holzer(y_0, z_0, x_0, abs(b_2), abs(c_2), abs(a_2))
x_0 = reconstruct(b_1, c_1, x_0)
y_0 = reconstruct(a_1, c_1, y_0)
z_0 = reconstruct(a_1, b_1, z_0)
sq_lcm = ilcm(sqf_of_a, sqf_of_b, sqf_of_c)
x_0 = abs(x_0*sq_lcm // sqf_of_a)
y_0 = abs(y_0*sq_lcm // sqf_of_b)
z_0 = abs(z_0*sq_lcm // sqf_of_c)
result.add(_remove_gcd(x_0, y_0, z_0))
return result
class HomogeneousTernaryQuadratic(DiophantineEquationType):
"""
Representation of a homogeneous ternary quadratic diophantine equation.
Examples
========
>>> from sympy.abc import x, y, z
>>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadratic
>>> HomogeneousTernaryQuadratic(x**2 + y**2 - 3*z**2 + x*y).solve()
{(-1, 2, 1)}
>>> HomogeneousTernaryQuadratic(3*x**2 + y**2 - 3*z**2 + 5*x*y + y*z).solve()
{(3, 12, 13)}
"""
name = 'homogeneous_ternary_quadratic'
def matches(self):
if not (self.total_degree == 2 and self.dimension == 3):
return False
if not self.homogeneous:
return False
if not self.homogeneous_order:
return False
nonzero = [k for k in self.coeff if self.coeff[k]]
return not (len(nonzero) == 3 and all(i**2 in nonzero for i in self.free_symbols))
def solve(self, parameters=None, limit=None):
self.pre_solve(parameters)
_var = self.free_symbols
coeff = self.coeff
x, y, z = _var
var = [x, y, z]
# Equations of the form B*x*y + C*z*x + E*y*z = 0 and At least two of the
# coefficients A, B, C are non-zero.
# There are infinitely many solutions for the equation.
# Ex: (0, 0, t), (0, t, 0), (t, 0, 0)
# Equation can be re-written as y*(B*x + E*z) = -C*x*z and we can find rather
# unobvious solutions. Set y = -C and B*x + E*z = x*z. The latter can be solved by
# using methods for binary quadratic diophantine equations. Let's select the
# solution which minimizes |x| + |z|
result = DiophantineSolutionSet(var, parameters=self.parameters)
def unpack_sol(sol):
if len(sol) > 0:
return list(sol)[0]
return None, None, None
if not any(coeff[i**2] for i in var):
if coeff[x*z]:
sols = diophantine(coeff[x*y]*x + coeff[y*z]*z - x*z)
s = sols.pop()
min_sum = abs(s[0]) + abs(s[1])
for r in sols:
m = abs(r[0]) + abs(r[1])
if m < min_sum:
s = r
min_sum = m
result.add(_remove_gcd(s[0], -coeff[x*z], s[1]))
return result
else:
var[0], var[1] = _var[1], _var[0]
y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff))
if x_0 is not None:
result.add((x_0, y_0, z_0))
return result
if coeff[x**2] == 0:
# If the coefficient of x is zero change the variables
if coeff[y**2] == 0:
var[0], var[2] = _var[2], _var[0]
z_0, y_0, x_0 = unpack_sol(_diop_ternary_quadratic(var, coeff))
else:
var[0], var[1] = _var[1], _var[0]
y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff))
else:
if coeff[x*y] or coeff[x*z]:
# Apply the transformation x --> X - (B*y + C*z)/(2*A)
A = coeff[x**2]
B = coeff[x*y]
C = coeff[x*z]
D = coeff[y**2]
E = coeff[y*z]
F = coeff[z**2]
_coeff = {}
_coeff[x**2] = 4*A**2
_coeff[y**2] = 4*A*D - B**2
_coeff[z**2] = 4*A*F - C**2
_coeff[y*z] = 4*A*E - 2*B*C
_coeff[x*y] = 0
_coeff[x*z] = 0
x_0, y_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, _coeff))
if x_0 is None:
return result
p, q = _rational_pq(B*y_0 + C*z_0, 2*A)
x_0, y_0, z_0 = x_0*q - p, y_0*q, z_0*q
elif coeff[z*y] != 0:
if coeff[y**2] == 0:
if coeff[z**2] == 0:
# Equations of the form A*x**2 + E*yz = 0.
A = coeff[x**2]
E = coeff[y*z]
b, a = _rational_pq(-E, A)
x_0, y_0, z_0 = b, a, b
else:
# Ax**2 + E*y*z + F*z**2 = 0
var[0], var[2] = _var[2], _var[0]
z_0, y_0, x_0 = unpack_sol(_diop_ternary_quadratic(var, coeff))
else:
# A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, C may be zero
var[0], var[1] = _var[1], _var[0]
y_0, x_0, z_0 = unpack_sol(_diop_ternary_quadratic(var, coeff))
else:
# Ax**2 + D*y**2 + F*z**2 = 0, C may be zero
x_0, y_0, z_0 = unpack_sol(_diop_ternary_quadratic_normal(var, coeff))
if x_0 is None:
return result
result.add(_remove_gcd(x_0, y_0, z_0))
return result
class InhomogeneousGeneralQuadratic(DiophantineEquationType):
"""
Representation of an inhomogeneous general quadratic.
No solver is currently implemented for this equation type.
"""
name = 'inhomogeneous_general_quadratic'
def matches(self):
if not (self.total_degree == 2 and self.dimension >= 3):
return False
if not self.homogeneous_order:
return True
else:
# there may be Pow keys like x**2 or Mul keys like x*y
if any(k.is_Mul for k in self.coeff): # cross terms
return not self.homogeneous
return False
class HomogeneousGeneralQuadratic(DiophantineEquationType):
"""
Representation of a homogeneous general quadratic.
No solver is currently implemented for this equation type.
"""
name = 'homogeneous_general_quadratic'
def matches(self):
if not (self.total_degree == 2 and self.dimension >= 3):
return False
if not self.homogeneous_order:
return False
else:
# there may be Pow keys like x**2 or Mul keys like x*y
if any(k.is_Mul for k in self.coeff): # cross terms
return self.homogeneous
return False
class GeneralSumOfSquares(DiophantineEquationType):
r"""
Representation of the diophantine equation
`x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`.
Details
=======
When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be
no solutions. Refer [1]_ for more details.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import GeneralSumOfSquares
>>> from sympy.abc import a, b, c, d, e
>>> GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve()
{(15, 22, 22, 24, 24)}
By default only 1 solution is returned. Use the `limit` keyword for more:
>>> sorted(GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve(limit=3))
[(15, 22, 22, 24, 24), (16, 19, 24, 24, 24), (16, 20, 22, 23, 26)]
References
==========
.. [1] Representing an integer as a sum of three squares, [online],
Available:
https://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
"""
name = 'general_sum_of_squares'
def matches(self):
if not (self.total_degree == 2 and self.dimension >= 3):
return False
if not self.homogeneous_order:
return False
if any(k.is_Mul for k in self.coeff):
return False
return all(self.coeff[k] == 1 for k in self.coeff if k != 1)
def solve(self, parameters=None, limit=1):
self.pre_solve(parameters)
var = self.free_symbols
k = -int(self.coeff[1])
n = self.dimension
result = DiophantineSolutionSet(var, parameters=self.parameters)
if k < 0 or limit < 1:
return result
signs = [-1 if x.is_nonpositive else 1 for x in var]
negs = signs.count(-1) != 0
took = 0
for t in sum_of_squares(k, n, zeros=True):
if negs:
result.add([signs[i]*j for i, j in enumerate(t)])
else:
result.add(t)
took += 1
if took == limit:
break
return result
class GeneralPythagorean(DiophantineEquationType):
"""
Representation of the general pythagorean equation,
`a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import GeneralPythagorean
>>> from sympy.abc import a, b, c, d, e, x, y, z, t
>>> GeneralPythagorean(a**2 + b**2 + c**2 - d**2).solve()
{(t_0**2 + t_1**2 - t_2**2, 2*t_0*t_2, 2*t_1*t_2, t_0**2 + t_1**2 + t_2**2)}
>>> GeneralPythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2).solve(parameters=[x, y, z, t])
{(-10*t**2 + 10*x**2 + 10*y**2 + 10*z**2, 15*t**2 + 15*x**2 + 15*y**2 + 15*z**2, 15*t*x, 12*t*y, 60*t*z)}
"""
name = 'general_pythagorean'
def matches(self):
if not (self.total_degree == 2 and self.dimension >= 3):
return False
if not self.homogeneous_order:
return False
if any(k.is_Mul for k in self.coeff):
return False
if all(self.coeff[k] == 1 for k in self.coeff if k != 1):
return False
if not all(is_square(abs(self.coeff[k])) for k in self.coeff):
return False
# all but one has the same sign
# e.g. 4*x**2 + y**2 - 4*z**2
return abs(sum(sign(self.coeff[k]) for k in self.coeff)) == self.dimension - 2
@property
def n_parameters(self):
return self.dimension - 1
def solve(self, parameters=None, limit=1):
self.pre_solve(parameters)
coeff = self.coeff
var = self.free_symbols
n = self.dimension
if sign(coeff[var[0] ** 2]) + sign(coeff[var[1] ** 2]) + sign(coeff[var[2] ** 2]) < 0:
for key in coeff.keys():
coeff[key] = -coeff[key]
result = DiophantineSolutionSet(var, parameters=self.parameters)
index = 0
for i, v in enumerate(var):
if sign(coeff[v ** 2]) == -1:
index = i
m = result.parameters
ith = sum(m_i ** 2 for m_i in m)
L = [ith - 2 * m[n - 2] ** 2]
L.extend([2 * m[i] * m[n - 2] for i in range(n - 2)])
sol = L[:index] + [ith] + L[index:]
lcm = 1
for i, v in enumerate(var):
if i == index or (index > 0 and i == 0) or (index == 0 and i == 1):
lcm = ilcm(lcm, sqrt(abs(coeff[v ** 2])))
else:
s = sqrt(coeff[v ** 2])
lcm = ilcm(lcm, s if _odd(s) else s // 2)
for i, v in enumerate(var):
sol[i] = (lcm * sol[i]) / sqrt(abs(coeff[v ** 2]))
result.add(sol)
return result
class CubicThue(DiophantineEquationType):
"""
Representation of a cubic Thue diophantine equation.
A cubic Thue diophantine equation is a polynomial of the form
`f(x, y) = r` of degree 3, where `x` and `y` are integers
and `r` is a rational number.
No solver is currently implemented for this equation type.
Examples
========
>>> from sympy.abc import x, y
>>> from sympy.solvers.diophantine.diophantine import CubicThue
>>> c1 = CubicThue(x**3 + y**2 + 1)
>>> c1.matches()
True
"""
name = 'cubic_thue'
def matches(self):
return self.total_degree == 3 and self.dimension == 2
class GeneralSumOfEvenPowers(DiophantineEquationType):
"""
Representation of the diophantine equation
`x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`
where `e` is an even, integer power.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import GeneralSumOfEvenPowers
>>> from sympy.abc import a, b
>>> GeneralSumOfEvenPowers(a**4 + b**4 - (2**4 + 3**4)).solve()
{(2, 3)}
"""
name = 'general_sum_of_even_powers'
def matches(self):
if not self.total_degree > 3:
return False
if self.total_degree % 2 != 0:
return False
if not all(k.is_Pow and k.exp == self.total_degree for k in self.coeff if k != 1):
return False
return all(self.coeff[k] == 1 for k in self.coeff if k != 1)
def solve(self, parameters=None, limit=1):
self.pre_solve(parameters)
var = self.free_symbols
coeff = self.coeff
p = None
for q in coeff.keys():
if q.is_Pow and coeff[q]:
p = q.exp
k = len(var)
n = -coeff[1]
result = DiophantineSolutionSet(var, parameters=self.parameters)
if n < 0 or limit < 1:
return result
sign = [-1 if x.is_nonpositive else 1 for x in var]
negs = sign.count(-1) != 0
took = 0
for t in power_representation(n, p, k):
if negs:
result.add([sign[i]*j for i, j in enumerate(t)])
else:
result.add(t)
took += 1
if took == limit:
break
return result
# these types are known (but not necessarily handled)
# note that order is important here (in the current solver state)
all_diop_classes = [
Linear,
Univariate,
BinaryQuadratic,
InhomogeneousTernaryQuadratic,
HomogeneousTernaryQuadraticNormal,
HomogeneousTernaryQuadratic,
InhomogeneousGeneralQuadratic,
HomogeneousGeneralQuadratic,
GeneralSumOfSquares,
GeneralPythagorean,
CubicThue,
GeneralSumOfEvenPowers,
]
diop_known = {diop_class.name for diop_class in all_diop_classes}
def _is_int(i):
try:
as_int(i)
return True
except ValueError:
pass
def _sorted_tuple(*i):
return tuple(sorted(i))
def _remove_gcd(*x):
try:
g = igcd(*x)
except ValueError:
fx = list(filter(None, x))
if len(fx) < 2:
return x
g = igcd(*[i.as_content_primitive()[0] for i in fx])
except TypeError:
raise TypeError('_remove_gcd(a,b,c) or _remove_gcd(*container)')
if g == 1:
return x
return tuple([i//g for i in x])
def _rational_pq(a, b):
# return `(numer, denom)` for a/b; sign in numer and gcd removed
return _remove_gcd(sign(b)*a, abs(b))
def _nint_or_floor(p, q):
# return nearest int to p/q; in case of tie return floor(p/q)
w, r = divmod(p, q)
if abs(r) <= abs(q)//2:
return w
return w + 1
def _odd(i):
return i % 2 != 0
def _even(i):
return i % 2 == 0
def diophantine(eq, param=symbols("t", integer=True), syms=None,
permute=False):
"""
Simplify the solution procedure of diophantine equation ``eq`` by
converting it into a product of terms which should equal zero.
Explanation
===========
For example, when solving, `x^2 - y^2 = 0` this is treated as
`(x + y)(x - y) = 0` and `x + y = 0` and `x - y = 0` are solved
independently and combined. Each term is solved by calling
``diop_solve()``. (Although it is possible to call ``diop_solve()``
directly, one must be careful to pass an equation in the correct
form and to interpret the output correctly; ``diophantine()`` is
the public-facing function to use in general.)
Output of ``diophantine()`` is a set of tuples. The elements of the
tuple are the solutions for each variable in the equation and
are arranged according to the alphabetic ordering of the variables.
e.g. For an equation with two variables, `a` and `b`, the first
element of the tuple is the solution for `a` and the second for `b`.
Usage
=====
``diophantine(eq, t, syms)``: Solve the diophantine
equation ``eq``.
``t`` is the optional parameter to be used by ``diop_solve()``.
``syms`` is an optional list of symbols which determines the
order of the elements in the returned tuple.
By default, only the base solution is returned. If ``permute`` is set to
True then permutations of the base solution and/or permutations of the
signs of the values will be returned when applicable.
Details
=======
``eq`` should be an expression which is assumed to be zero.
``t`` is the parameter to be used in the solution.
Examples
========
>>> from sympy import diophantine
>>> from sympy.abc import a, b
>>> eq = a**4 + b**4 - (2**4 + 3**4)
>>> diophantine(eq)
{(2, 3)}
>>> diophantine(eq, permute=True)
{(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)}
>>> from sympy.abc import x, y, z
>>> diophantine(x**2 - y**2)
{(t_0, -t_0), (t_0, t_0)}
>>> diophantine(x*(2*x + 3*y - z))
{(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)}
>>> diophantine(x**2 + 3*x*y + 4*x)
{(0, n1), (3*t_0 - 4, -t_0)}
See Also
========
diop_solve
sympy.utilities.iterables.permute_signs
sympy.utilities.iterables.signed_permutations
"""
eq = _sympify(eq)
if isinstance(eq, Eq):
eq = eq.lhs - eq.rhs
try:
var = list(eq.expand(force=True).free_symbols)
var.sort(key=default_sort_key)
if syms:
if not is_sequence(syms):
raise TypeError(
'syms should be given as a sequence, e.g. a list')
syms = [i for i in syms if i in var]
if syms != var:
dict_sym_index = dict(zip(syms, range(len(syms))))
return {tuple([t[dict_sym_index[i]] for i in var])
for t in diophantine(eq, param, permute=permute)}
n, d = eq.as_numer_denom()
if n.is_number:
return set()
if not d.is_number:
dsol = diophantine(d)
good = diophantine(n) - dsol
return {s for s in good if _mexpand(d.subs(zip(var, s)))}
else:
eq = n
eq = factor_terms(eq)
assert not eq.is_number
eq = eq.as_independent(*var, as_Add=False)[1]
p = Poly(eq)
assert not any(g.is_number for g in p.gens)
eq = p.as_expr()
assert eq.is_polynomial()
except (GeneratorsNeeded, AssertionError):
raise TypeError(filldedent('''
Equation should be a polynomial with Rational coefficients.'''))
# permute only sign
do_permute_signs = False
# permute sign and values
do_permute_signs_var = False
# permute few signs
permute_few_signs = False
try:
# if we know that factoring should not be attempted, skip
# the factoring step
v, c, t = classify_diop(eq)
# check for permute sign
if permute:
len_var = len(v)
permute_signs_for = [
GeneralSumOfSquares.name,
GeneralSumOfEvenPowers.name]
permute_signs_check = [
HomogeneousTernaryQuadratic.name,
HomogeneousTernaryQuadraticNormal.name,
BinaryQuadratic.name]
if t in permute_signs_for:
do_permute_signs_var = True
elif t in permute_signs_check:
# if all the variables in eq have even powers
# then do_permute_sign = True
if len_var == 3:
var_mul = list(subsets(v, 2))
# here var_mul is like [(x, y), (x, z), (y, z)]
xy_coeff = True
x_coeff = True
var1_mul_var2 = (a[0]*a[1] for a in var_mul)
# if coeff(y*z), coeff(y*x), coeff(x*z) is not 0 then
# `xy_coeff` => True and do_permute_sign => False.
# Means no permuted solution.
for v1_mul_v2 in var1_mul_var2:
try:
coeff = c[v1_mul_v2]
except KeyError:
coeff = 0
xy_coeff = bool(xy_coeff) and bool(coeff)
var_mul = list(subsets(v, 1))
# here var_mul is like [(x,), (y, )]
for v1 in var_mul:
try:
coeff = c[v1[0]]
except KeyError:
coeff = 0
x_coeff = bool(x_coeff) and bool(coeff)
if not any((xy_coeff, x_coeff)):
# means only x**2, y**2, z**2, const is present
do_permute_signs = True
elif not x_coeff:
permute_few_signs = True
elif len_var == 2:
var_mul = list(subsets(v, 2))
# here var_mul is like [(x, y)]
xy_coeff = True
x_coeff = True
var1_mul_var2 = (x[0]*x[1] for x in var_mul)
for v1_mul_v2 in var1_mul_var2:
try:
coeff = c[v1_mul_v2]
except KeyError:
coeff = 0
xy_coeff = bool(xy_coeff) and bool(coeff)
var_mul = list(subsets(v, 1))
# here var_mul is like [(x,), (y, )]
for v1 in var_mul:
try:
coeff = c[v1[0]]
except KeyError:
coeff = 0
x_coeff = bool(x_coeff) and bool(coeff)
if not any((xy_coeff, x_coeff)):
# means only x**2, y**2 and const is present
# so we can get more soln by permuting this soln.
do_permute_signs = True
elif not x_coeff:
# when coeff(x), coeff(y) is not present then signs of
# x, y can be permuted such that their sign are same
# as sign of x*y.
# e.g 1. (x_val,y_val)=> (x_val,y_val), (-x_val,-y_val)
# 2. (-x_vall, y_val)=> (-x_val,y_val), (x_val,-y_val)
permute_few_signs = True
if t == 'general_sum_of_squares':
# trying to factor such expressions will sometimes hang
terms = [(eq, 1)]
else:
raise TypeError
except (TypeError, NotImplementedError):
fl = factor_list(eq)
if fl[0].is_Rational and fl[0] != 1:
return diophantine(eq/fl[0], param=param, syms=syms, permute=permute)
terms = fl[1]
sols = set()
for term in terms:
base, _ = term
var_t, _, eq_type = classify_diop(base, _dict=False)
_, base = signsimp(base, evaluate=False).as_coeff_Mul()
solution = diop_solve(base, param)
if eq_type in [
Linear.name,
HomogeneousTernaryQuadratic.name,
HomogeneousTernaryQuadraticNormal.name,
GeneralPythagorean.name]:
sols.add(merge_solution(var, var_t, solution))
elif eq_type in [
BinaryQuadratic.name,
GeneralSumOfSquares.name,
GeneralSumOfEvenPowers.name,
Univariate.name]:
for sol in solution:
sols.add(merge_solution(var, var_t, sol))
else:
raise NotImplementedError('unhandled type: %s' % eq_type)
# remove null merge results
if () in sols:
sols.remove(())
null = tuple([0]*len(var))
# if there is no solution, return trivial solution
if not sols and eq.subs(zip(var, null)).is_zero:
sols.add(null)
final_soln = set()
for sol in sols:
if all(_is_int(s) for s in sol):
if do_permute_signs:
permuted_sign = set(permute_signs(sol))
final_soln.update(permuted_sign)
elif permute_few_signs:
lst = list(permute_signs(sol))
lst = list(filter(lambda x: x[0]*x[1] == sol[1]*sol[0], lst))
permuted_sign = set(lst)
final_soln.update(permuted_sign)
elif do_permute_signs_var:
permuted_sign_var = set(signed_permutations(sol))
final_soln.update(permuted_sign_var)
else:
final_soln.add(sol)
else:
final_soln.add(sol)
return final_soln
def merge_solution(var, var_t, solution):
"""
This is used to construct the full solution from the solutions of sub
equations.
Explanation
===========
For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`,
solutions for each of the equations `x - y = 0` and `x^2 + y^2 - z^2` are
found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But
we should introduce a value for z when we output the solution for the
original equation. This function converts `(t, t)` into `(t, t, n_{1})`
where `n_{1}` is an integer parameter.
"""
sol = []
if None in solution:
return ()
solution = iter(solution)
params = numbered_symbols("n", integer=True, start=1)
for v in var:
if v in var_t:
sol.append(next(solution))
else:
sol.append(next(params))
for val, symb in zip(sol, var):
if check_assumptions(val, **symb.assumptions0) is False:
return ()
return tuple(sol)
def _diop_solve(eq, params=None):
for diop_type in all_diop_classes:
if diop_type(eq).matches():
return diop_type(eq).solve(parameters=params)
def diop_solve(eq, param=symbols("t", integer=True)):
"""
Solves the diophantine equation ``eq``.
Explanation
===========
Unlike ``diophantine()``, factoring of ``eq`` is not attempted. Uses
``classify_diop()`` to determine the type of the equation and calls
the appropriate solver function.
Use of ``diophantine()`` is recommended over other helper functions.
``diop_solve()`` can return either a set or a tuple depending on the
nature of the equation.
Usage
=====
``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t``
as a parameter if needed.
Details
=======
``eq`` should be an expression which is assumed to be zero.
``t`` is a parameter to be used in the solution.
Examples
========
>>> from sympy.solvers.diophantine import diop_solve
>>> from sympy.abc import x, y, z, w
>>> diop_solve(2*x + 3*y - 5)
(3*t_0 - 5, 5 - 2*t_0)
>>> diop_solve(4*x + 3*y - 4*z + 5)
(t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5)
>>> diop_solve(x + 3*y - 4*z + w - 6)
(t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6)
>>> diop_solve(x**2 + y**2 - 5)
{(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)}
See Also
========
diophantine()
"""
var, coeff, eq_type = classify_diop(eq, _dict=False)
if eq_type == Linear.name:
return diop_linear(eq, param)
elif eq_type == BinaryQuadratic.name:
return diop_quadratic(eq, param)
elif eq_type == HomogeneousTernaryQuadratic.name:
return diop_ternary_quadratic(eq, parameterize=True)
elif eq_type == HomogeneousTernaryQuadraticNormal.name:
return diop_ternary_quadratic_normal(eq, parameterize=True)
elif eq_type == GeneralPythagorean.name:
return diop_general_pythagorean(eq, param)
elif eq_type == Univariate.name:
return diop_univariate(eq)
elif eq_type == GeneralSumOfSquares.name:
return diop_general_sum_of_squares(eq, limit=S.Infinity)
elif eq_type == GeneralSumOfEvenPowers.name:
return diop_general_sum_of_even_powers(eq, limit=S.Infinity)
if eq_type is not None and eq_type not in diop_known:
raise ValueError(filldedent('''
Although this type of equation was identified, it is not yet
handled. It should, however, be listed in `diop_known` at the
top of this file. Developers should see comments at the end of
`classify_diop`.
''')) # pragma: no cover
else:
raise NotImplementedError(
'No solver has been written for %s.' % eq_type)
def classify_diop(eq, _dict=True):
# docstring supplied externally
matched = False
diop_type = None
for diop_class in all_diop_classes:
diop_type = diop_class(eq)
if diop_type.matches():
matched = True
break
if matched:
return diop_type.free_symbols, dict(diop_type.coeff) if _dict else diop_type.coeff, diop_type.name
# new diop type instructions
# --------------------------
# if this error raises and the equation *can* be classified,
# * it should be identified in the if-block above
# * the type should be added to the diop_known
# if a solver can be written for it,
# * a dedicated handler should be written (e.g. diop_linear)
# * it should be passed to that handler in diop_solve
raise NotImplementedError(filldedent('''
This equation is not yet recognized or else has not been
simplified sufficiently to put it in a form recognized by
diop_classify().'''))
classify_diop.func_doc = ( # type: ignore
'''
Helper routine used by diop_solve() to find information about ``eq``.
Explanation
===========
Returns a tuple containing the type of the diophantine equation
along with the variables (free symbols) and their coefficients.
Variables are returned as a list and coefficients are returned
as a dict with the key being the respective term and the constant
term is keyed to 1. The type is one of the following:
* %s
Usage
=====
``classify_diop(eq)``: Return variables, coefficients and type of the
``eq``.
Details
=======
``eq`` should be an expression which is assumed to be zero.
``_dict`` is for internal use: when True (default) a dict is returned,
otherwise a defaultdict which supplies 0 for missing keys is returned.
Examples
========
>>> from sympy.solvers.diophantine import classify_diop
>>> from sympy.abc import x, y, z, w, t
>>> classify_diop(4*x + 6*y - 4)
([x, y], {1: -4, x: 4, y: 6}, 'linear')
>>> classify_diop(x + 3*y -4*z + 5)
([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear')
>>> classify_diop(x**2 + y**2 - x*y + x + 5)
([x, y], {1: 5, x: 1, x**2: 1, y**2: 1, x*y: -1}, 'binary_quadratic')
''' % ('\n * '.join(sorted(diop_known))))
def diop_linear(eq, param=symbols("t", integer=True)):
"""
Solves linear diophantine equations.
A linear diophantine equation is an equation of the form `a_{1}x_{1} +
a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are
integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables.
Usage
=====
``diop_linear(eq)``: Returns a tuple containing solutions to the
diophantine equation ``eq``. Values in the tuple is arranged in the same
order as the sorted variables.
Details
=======
``eq`` is a linear diophantine equation which is assumed to be zero.
``param`` is the parameter to be used in the solution.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_linear
>>> from sympy.abc import x, y, z
>>> diop_linear(2*x - 3*y - 5) # solves equation 2*x - 3*y - 5 == 0
(3*t_0 - 5, 2*t_0 - 5)
Here x = -3*t_0 - 5 and y = -2*t_0 - 5
>>> diop_linear(2*x - 3*y - 4*z -3)
(t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3)
See Also
========
diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(),
diop_general_sum_of_squares()
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == Linear.name:
parameters = None
if param is not None:
parameters = symbols('%s_0:%i' % (param, len(var)), integer=True)
result = Linear(eq).solve(parameters=parameters)
if param is None:
result = result(*[0]*len(result.parameters))
if len(result) > 0:
return list(result)[0]
else:
return tuple([None]*len(result.parameters))
def base_solution_linear(c, a, b, t=None):
"""
Return the base solution for the linear equation, `ax + by = c`.
Explanation
===========
Used by ``diop_linear()`` to find the base solution of a linear
Diophantine equation. If ``t`` is given then the parametrized solution is
returned.
Usage
=====
``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients
in `ax + by = c` and ``t`` is the parameter to be used in the solution.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import base_solution_linear
>>> from sympy.abc import t
>>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5
(-5, 5)
>>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0
(0, 0)
>>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5
(3*t - 5, 5 - 2*t)
>>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0
(7*t, -5*t)
"""
a, b, c = _remove_gcd(a, b, c)
if c == 0:
if t is not None:
if b < 0:
t = -t
return (b*t, -a*t)
else:
return (0, 0)
else:
x0, y0, d = igcdex(abs(a), abs(b))
x0 *= sign(a)
y0 *= sign(b)
if divisible(c, d):
if t is not None:
if b < 0:
t = -t
return (c*x0 + b*t, c*y0 - a*t)
else:
return (c*x0, c*y0)
else:
return (None, None)
def diop_univariate(eq):
"""
Solves a univariate diophantine equations.
Explanation
===========
A univariate diophantine equation is an equation of the form
`a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are
integer constants and `x` is an integer variable.
Usage
=====
``diop_univariate(eq)``: Returns a set containing solutions to the
diophantine equation ``eq``.
Details
=======
``eq`` is a univariate diophantine equation which is assumed to be zero.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_univariate
>>> from sympy.abc import x
>>> diop_univariate((x - 2)*(x - 3)**2) # solves equation (x - 2)*(x - 3)**2 == 0
{(2,), (3,)}
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == Univariate.name:
return {(int(i),) for i in solveset_real(
eq, var[0]).intersect(S.Integers)}
def divisible(a, b):
"""
Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise.
"""
return not a % b
def diop_quadratic(eq, param=symbols("t", integer=True)):
"""
Solves quadratic diophantine equations.
i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a
set containing the tuples `(x, y)` which contains the solutions. If there
are no solutions then `(None, None)` is returned.
Usage
=====
``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine
equation. ``param`` is used to indicate the parameter to be used in the
solution.
Details
=======
``eq`` should be an expression which is assumed to be zero.
``param`` is a parameter to be used in the solution.
Examples
========
>>> from sympy.abc import x, y, t
>>> from sympy.solvers.diophantine.diophantine import diop_quadratic
>>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t)
{(-1, -1)}
References
==========
.. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online],
Available: https://www.alpertron.com.ar/METHODS.HTM
.. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online],
Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf
See Also
========
diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(),
diop_general_pythagorean()
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == BinaryQuadratic.name:
if param is not None:
parameters = [param, Symbol("u", integer=True)]
else:
parameters = None
return set(BinaryQuadratic(eq).solve(parameters=parameters))
def is_solution_quad(var, coeff, u, v):
"""
Check whether `(u, v)` is solution to the quadratic binary diophantine
equation with the variable list ``var`` and coefficient dictionary
``coeff``.
Not intended for use by normal users.
"""
reps = dict(zip(var, (u, v)))
eq = Add(*[j*i.xreplace(reps) for i, j in coeff.items()])
return _mexpand(eq) == 0
def diop_DN(D, N, t=symbols("t", integer=True)):
"""
Solves the equation `x^2 - Dy^2 = N`.
Explanation
===========
Mainly concerned with the case `D > 0, D` is not a perfect square,
which is the same as the generalized Pell equation. The LMM
algorithm [1]_ is used to solve this equation.
Returns one solution tuple, (`x, y)` for each class of the solutions.
Other solutions of the class can be constructed according to the
values of ``D`` and ``N``.
Usage
=====
``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and
``t`` is the parameter to be used in the solutions.
Details
=======
``D`` and ``N`` correspond to D and N in the equation.
``t`` is the parameter to be used in the solutions.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_DN
>>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4
[(3, 1), (393, 109), (36, 10)]
The output can be interpreted as follows: There are three fundamental
solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109)
and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means
that `x = 3` and `y = 1`.
>>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1
[(49299, 1570)]
See Also
========
find_DN(), diop_bf_DN()
References
==========
.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.
Robertson, July 31, 2004, Pages 16 - 17. [online], Available:
https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf
"""
if D < 0:
if N == 0:
return [(0, 0)]
elif N < 0:
return []
elif N > 0:
sol = []
for d in divisors(square_factor(N)):
sols = cornacchia(1, -D, N // d**2)
if sols:
for x, y in sols:
sol.append((d*x, d*y))
if D == -1:
sol.append((d*y, d*x))
return sol
elif D == 0:
if N < 0:
return []
if N == 0:
return [(0, t)]
sN, _exact = integer_nthroot(N, 2)
if _exact:
return [(sN, t)]
else:
return []
else: # D > 0
sD, _exact = integer_nthroot(D, 2)
if _exact:
if N == 0:
return [(sD*t, t)]
else:
sol = []
for y in range(floor(sign(N)*(N - 1)/(2*sD)) + 1):
try:
sq, _exact = integer_nthroot(D*y**2 + N, 2)
except ValueError:
_exact = False
if _exact:
sol.append((sq, y))
return sol
elif 1 < N**2 < D:
# It is much faster to call `_special_diop_DN`.
return _special_diop_DN(D, N)
else:
if N == 0:
return [(0, 0)]
elif abs(N) == 1:
pqa = PQa(0, 1, D)
j = 0
G = []
B = []
for i in pqa:
a = i[2]
G.append(i[5])
B.append(i[4])
if j != 0 and a == 2*sD:
break
j = j + 1
if _odd(j):
if N == -1:
x = G[j - 1]
y = B[j - 1]
else:
count = j
while count < 2*j - 1:
i = next(pqa)
G.append(i[5])
B.append(i[4])
count += 1
x = G[count]
y = B[count]
else:
if N == 1:
x = G[j - 1]
y = B[j - 1]
else:
return []
return [(x, y)]
else:
fs = []
sol = []
div = divisors(N)
for d in div:
if divisible(N, d**2):
fs.append(d)
for f in fs:
m = N // f**2
zs = sqrt_mod(D, abs(m), all_roots=True)
zs = [i for i in zs if i <= abs(m) // 2 ]
if abs(m) != 2:
zs = zs + [-i for i in zs if i] # omit dupl 0
for z in zs:
pqa = PQa(z, abs(m), D)
j = 0
G = []
B = []
for i in pqa:
G.append(i[5])
B.append(i[4])
if j != 0 and abs(i[1]) == 1:
r = G[j-1]
s = B[j-1]
if r**2 - D*s**2 == m:
sol.append((f*r, f*s))
elif diop_DN(D, -1) != []:
a = diop_DN(D, -1)
sol.append((f*(r*a[0][0] + a[0][1]*s*D), f*(r*a[0][1] + s*a[0][0])))
break
j = j + 1
if j == length(z, abs(m), D):
break
return sol
def _special_diop_DN(D, N):
"""
Solves the equation `x^2 - Dy^2 = N` for the special case where
`1 < N**2 < D` and `D` is not a perfect square.
It is better to call `diop_DN` rather than this function, as
the former checks the condition `1 < N**2 < D`, and calls the latter only
if appropriate.
Usage
=====
WARNING: Internal method. Do not call directly!
``_special_diop_DN(D, N)``: D and N are integers as in `x^2 - Dy^2 = N`.
Details
=======
``D`` and ``N`` correspond to D and N in the equation.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import _special_diop_DN
>>> _special_diop_DN(13, -3) # Solves equation x**2 - 13*y**2 = -3
[(7, 2), (137, 38)]
The output can be interpreted as follows: There are two fundamental
solutions to the equation `x^2 - 13y^2 = -3` given by (7, 2) and
(137, 38). Each tuple is in the form (x, y), i.e. solution (7, 2) means
that `x = 7` and `y = 2`.
>>> _special_diop_DN(2445, -20) # Solves equation x**2 - 2445*y**2 = -20
[(445, 9), (17625560, 356454), (698095554475, 14118073569)]
See Also
========
diop_DN()
References
==========
.. [1] Section 4.4.4 of the following book:
Quadratic Diophantine Equations, T. Andreescu and D. Andrica,
Springer, 2015.
"""
# The following assertion was removed for efficiency, with the understanding
# that this method is not called directly. The parent method, `diop_DN`
# is responsible for performing the appropriate checks.
#
# assert (1 < N**2 < D) and (not integer_nthroot(D, 2)[1])
sqrt_D = sqrt(D)
F = [(N, 1)]
f = 2
while True:
f2 = f**2
if f2 > abs(N):
break
n, r = divmod(N, f2)
if r == 0:
F.append((n, f))
f += 1
P = 0
Q = 1
G0, G1 = 0, 1
B0, B1 = 1, 0
solutions = []
i = 0
while True:
a = floor((P + sqrt_D) / Q)
P = a*Q - P
Q = (D - P**2) // Q
G2 = a*G1 + G0
B2 = a*B1 + B0
for n, f in F:
if G2**2 - D*B2**2 == n:
solutions.append((f*G2, f*B2))
i += 1
if Q == 1 and i % 2 == 0:
break
G0, G1 = G1, G2
B0, B1 = B1, B2
return solutions
def cornacchia(a, b, m):
r"""
Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`.
Explanation
===========
Uses the algorithm due to Cornacchia. The method only finds primitive
solutions, i.e. ones with `\gcd(x, y) = 1`. So this method cannot be used to
find the solutions of `x^2 + y^2 = 20` since the only solution to former is
`(x, y) = (4, 2)` and it is not primitive. When `a = b`, only the
solutions with `x \leq y` are found. For more details, see the References.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import cornacchia
>>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35
{(2, 3), (4, 1)}
>>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25
{(4, 3)}
References
===========
.. [1] A. Nitaj, "L'algorithme de Cornacchia"
.. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's
method, [online], Available:
http://www.numbertheory.org/php/cornacchia.html
See Also
========
sympy.utilities.iterables.signed_permutations
"""
sols = set()
a1 = igcdex(a, m)[0]
v = sqrt_mod(-b*a1, m, all_roots=True)
if not v:
return None
for t in v:
if t < m // 2:
continue
u, r = t, m
while True:
u, r = r, u % r
if a*r**2 < m:
break
m1 = m - a*r**2
if m1 % b == 0:
m1 = m1 // b
s, _exact = integer_nthroot(m1, 2)
if _exact:
if a == b and r < s:
r, s = s, r
sols.add((int(r), int(s)))
return sols
def PQa(P_0, Q_0, D):
r"""
Returns useful information needed to solve the Pell equation.
Explanation
===========
There are six sequences of integers defined related to the continued
fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`},
{`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns
these values as a 6-tuple in the same order as mentioned above. Refer [1]_
for more detailed information.
Usage
=====
``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding
to `P_{0}`, `Q_{0}` and `D` in the continued fraction
`\\frac{P_{0} + \sqrt{D}}{Q_{0}}`.
Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import PQa
>>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4
>>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0)
(13, 4, 3, 3, 1, -1)
>>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1)
(-1, 1, 1, 4, 1, 3)
References
==========
.. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P.
Robertson, July 31, 2004, Pages 4 - 8. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf
"""
A_i_2 = B_i_1 = 0
A_i_1 = B_i_2 = 1
G_i_2 = -P_0
G_i_1 = Q_0
P_i = P_0
Q_i = Q_0
while True:
a_i = floor((P_i + sqrt(D))/Q_i)
A_i = a_i*A_i_1 + A_i_2
B_i = a_i*B_i_1 + B_i_2
G_i = a_i*G_i_1 + G_i_2
yield P_i, Q_i, a_i, A_i, B_i, G_i
A_i_1, A_i_2 = A_i, A_i_1
B_i_1, B_i_2 = B_i, B_i_1
G_i_1, G_i_2 = G_i, G_i_1
P_i = a_i*Q_i - P_i
Q_i = (D - P_i**2)/Q_i
def diop_bf_DN(D, N, t=symbols("t", integer=True)):
r"""
Uses brute force to solve the equation, `x^2 - Dy^2 = N`.
Explanation
===========
Mainly concerned with the generalized Pell equation which is the case when
`D > 0, D` is not a perfect square. For more information on the case refer
[1]_. Let `(t, u)` be the minimal positive solution of the equation
`x^2 - Dy^2 = 1`. Then this method requires
`\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small.
Usage
=====
``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in
`x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions.
Details
=======
``D`` and ``N`` correspond to D and N in the equation.
``t`` is the parameter to be used in the solutions.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_bf_DN
>>> diop_bf_DN(13, -4)
[(3, 1), (-3, 1), (36, 10)]
>>> diop_bf_DN(986, 1)
[(49299, 1570)]
See Also
========
diop_DN()
References
==========
.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.
Robertson, July 31, 2004, Page 15. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf
"""
D = as_int(D)
N = as_int(N)
sol = []
a = diop_DN(D, 1)
u = a[0][0]
if abs(N) == 1:
return diop_DN(D, N)
elif N > 1:
L1 = 0
L2 = integer_nthroot(int(N*(u - 1)/(2*D)), 2)[0] + 1
elif N < -1:
L1, _exact = integer_nthroot(-int(N/D), 2)
if not _exact:
L1 += 1
L2 = integer_nthroot(-int(N*(u + 1)/(2*D)), 2)[0] + 1
else: # N = 0
if D < 0:
return [(0, 0)]
elif D == 0:
return [(0, t)]
else:
sD, _exact = integer_nthroot(D, 2)
if _exact:
return [(sD*t, t), (-sD*t, t)]
else:
return [(0, 0)]
for y in range(L1, L2):
try:
x, _exact = integer_nthroot(N + D*y**2, 2)
except ValueError:
_exact = False
if _exact:
sol.append((x, y))
if not equivalent(x, y, -x, y, D, N):
sol.append((-x, y))
return sol
def equivalent(u, v, r, s, D, N):
"""
Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N`
belongs to the same equivalence class and False otherwise.
Explanation
===========
Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same
equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by
`N`. See reference [1]_. No test is performed to test whether `(u, v)` and
`(r, s)` are actually solutions to the equation. User should take care of
this.
Usage
=====
``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions
of the equation `x^2 - Dy^2 = N` and all parameters involved are integers.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import equivalent
>>> equivalent(18, 5, -18, -5, 13, -1)
True
>>> equivalent(3, 1, -18, 393, 109, -4)
False
References
==========
.. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P.
Robertson, July 31, 2004, Page 12. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf
"""
return divisible(u*r - D*v*s, N) and divisible(u*s - v*r, N)
def length(P, Q, D):
r"""
Returns the (length of aperiodic part + length of periodic part) of
continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`.
It is important to remember that this does NOT return the length of the
periodic part but the sum of the lengths of the two parts as mentioned
above.
Usage
=====
``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to
the continued fraction `\\frac{P + \sqrt{D}}{Q}`.
Details
=======
``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction,
`\\frac{P + \sqrt{D}}{Q}`.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import length
>>> length(-2, 4, 5) # (-2 + sqrt(5))/4
3
>>> length(-5, 4, 17) # (-5 + sqrt(17))/4
4
See Also
========
sympy.ntheory.continued_fraction.continued_fraction_periodic
"""
from sympy.ntheory.continued_fraction import continued_fraction_periodic
v = continued_fraction_periodic(P, Q, D)
if isinstance(v[-1], list):
rpt = len(v[-1])
nonrpt = len(v) - 1
else:
rpt = 0
nonrpt = len(v)
return rpt + nonrpt
def transformation_to_DN(eq):
"""
This function transforms general quadratic,
`ax^2 + bxy + cy^2 + dx + ey + f = 0`
to more easy to deal with `X^2 - DY^2 = N` form.
Explanation
===========
This is used to solve the general quadratic equation by transforming it to
the latter form. Refer to [1]_ for more detailed information on the
transformation. This function returns a tuple (A, B) where A is a 2 X 2
matrix and B is a 2 X 1 matrix such that,
Transpose([x y]) = A * Transpose([X Y]) + B
Usage
=====
``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be
transformed.
Examples
========
>>> from sympy.abc import x, y
>>> from sympy.solvers.diophantine.diophantine import transformation_to_DN
>>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1)
>>> A
Matrix([
[1/26, 3/26],
[ 0, 1/13]])
>>> B
Matrix([
[-6/13],
[-4/13]])
A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B.
Substituting these values for `x` and `y` and a bit of simplifying work
will give an equation of the form `x^2 - Dy^2 = N`.
>>> from sympy.abc import X, Y
>>> from sympy import Matrix, simplify
>>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x
>>> u
X/26 + 3*Y/26 - 6/13
>>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y
>>> v
Y/13 - 4/13
Next we will substitute these formulas for `x` and `y` and do
``simplify()``.
>>> eq = simplify((x**2 - 3*x*y - y**2 - 2*y + 1).subs(zip((x, y), (u, v))))
>>> eq
X**2/676 - Y**2/52 + 17/13
By multiplying the denominator appropriately, we can get a Pell equation
in the standard form.
>>> eq * 676
X**2 - 13*Y**2 + 884
If only the final equation is needed, ``find_DN()`` can be used.
See Also
========
find_DN()
References
==========
.. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0,
John P.Robertson, May 8, 2003, Page 7 - 11.
https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == BinaryQuadratic.name:
return _transformation_to_DN(var, coeff)
def _transformation_to_DN(var, coeff):
x, y = var
a = coeff[x**2]
b = coeff[x*y]
c = coeff[y**2]
d = coeff[x]
e = coeff[y]
f = coeff[1]
a, b, c, d, e, f = [as_int(i) for i in _remove_gcd(a, b, c, d, e, f)]
X, Y = symbols("X, Y", integer=True)
if b:
B, C = _rational_pq(2*a, b)
A, T = _rational_pq(a, B**2)
# eq_1 = A*B*X**2 + B*(c*T - A*C**2)*Y**2 + d*T*X + (B*e*T - d*T*C)*Y + f*T*B
coeff = {X**2: A*B, X*Y: 0, Y**2: B*(c*T - A*C**2), X: d*T, Y: B*e*T - d*T*C, 1: f*T*B}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [S.One/B, -S(C)/B, 0, 1])*A_0, Matrix(2, 2, [S.One/B, -S(C)/B, 0, 1])*B_0
else:
if d:
B, C = _rational_pq(2*a, d)
A, T = _rational_pq(a, B**2)
# eq_2 = A*X**2 + c*T*Y**2 + e*T*Y + f*T - A*C**2
coeff = {X**2: A, X*Y: 0, Y**2: c*T, X: 0, Y: e*T, 1: f*T - A*C**2}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [S.One/B, 0, 0, 1])*A_0, Matrix(2, 2, [S.One/B, 0, 0, 1])*B_0 + Matrix([-S(C)/B, 0])
else:
if e:
B, C = _rational_pq(2*c, e)
A, T = _rational_pq(c, B**2)
# eq_3 = a*T*X**2 + A*Y**2 + f*T - A*C**2
coeff = {X**2: a*T, X*Y: 0, Y**2: A, X: 0, Y: 0, 1: f*T - A*C**2}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [1, 0, 0, S.One/B])*A_0, Matrix(2, 2, [1, 0, 0, S.One/B])*B_0 + Matrix([0, -S(C)/B])
else:
# TODO: pre-simplification: Not necessary but may simplify
# the equation.
return Matrix(2, 2, [S.One/a, 0, 0, 1]), Matrix([0, 0])
def find_DN(eq):
"""
This function returns a tuple, `(D, N)` of the simplified form,
`x^2 - Dy^2 = N`, corresponding to the general quadratic,
`ax^2 + bxy + cy^2 + dx + ey + f = 0`.
Solving the general quadratic is then equivalent to solving the equation
`X^2 - DY^2 = N` and transforming the solutions by using the transformation
matrices returned by ``transformation_to_DN()``.
Usage
=====
``find_DN(eq)``: where ``eq`` is the quadratic to be transformed.
Examples
========
>>> from sympy.abc import x, y
>>> from sympy.solvers.diophantine.diophantine import find_DN
>>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1)
(13, -884)
Interpretation of the output is that we get `X^2 -13Y^2 = -884` after
transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned
by ``transformation_to_DN()``.
See Also
========
transformation_to_DN()
References
==========
.. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0,
John P.Robertson, May 8, 2003, Page 7 - 11.
https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == BinaryQuadratic.name:
return _find_DN(var, coeff)
def _find_DN(var, coeff):
x, y = var
X, Y = symbols("X, Y", integer=True)
A, B = _transformation_to_DN(var, coeff)
u = (A*Matrix([X, Y]) + B)[0]
v = (A*Matrix([X, Y]) + B)[1]
eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[1]
simplified = _mexpand(eq.subs(zip((x, y), (u, v))))
coeff = simplified.as_coefficients_dict()
return -coeff[Y**2]/coeff[X**2], -coeff[1]/coeff[X**2]
def check_param(x, y, a, params):
"""
If there is a number modulo ``a`` such that ``x`` and ``y`` are both
integers, then return a parametric representation for ``x`` and ``y``
else return (None, None).
Here ``x`` and ``y`` are functions of ``t``.
"""
from sympy.simplify.simplify import clear_coefficients
if x.is_number and not x.is_Integer:
return DiophantineSolutionSet([x, y], parameters=params)
if y.is_number and not y.is_Integer:
return DiophantineSolutionSet([x, y], parameters=params)
m, n = symbols("m, n", integer=True)
c, p = (m*x + n*y).as_content_primitive()
if a % c.q:
return DiophantineSolutionSet([x, y], parameters=params)
# clear_coefficients(mx + b, R)[1] -> (R - b)/m
eq = clear_coefficients(x, m)[1] - clear_coefficients(y, n)[1]
junk, eq = eq.as_content_primitive()
return _diop_solve(eq, params=params)
def diop_ternary_quadratic(eq, parameterize=False):
"""
Solves the general quadratic ternary form,
`ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`.
Returns a tuple `(x, y, z)` which is a base solution for the above
equation. If there are no solutions, `(None, None, None)` is returned.
Usage
=====
``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution
to ``eq``.
Details
=======
``eq`` should be an homogeneous expression of degree two in three variables
and it is assumed to be zero.
Examples
========
>>> from sympy.abc import x, y, z
>>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic
>>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2)
(1, 0, 1)
>>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2)
(1, 0, 2)
>>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2)
(28, 45, 105)
>>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y)
(9, 1, 5)
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type in (
HomogeneousTernaryQuadratic.name,
HomogeneousTernaryQuadraticNormal.name):
sol = _diop_ternary_quadratic(var, coeff)
if len(sol) > 0:
x_0, y_0, z_0 = list(sol)[0]
else:
x_0, y_0, z_0 = None, None, None
if parameterize:
return _parametrize_ternary_quadratic(
(x_0, y_0, z_0), var, coeff)
return x_0, y_0, z_0
def _diop_ternary_quadratic(_var, coeff):
eq = sum([i*coeff[i] for i in coeff])
if HomogeneousTernaryQuadratic(eq).matches():
return HomogeneousTernaryQuadratic(eq, free_symbols=_var).solve()
elif HomogeneousTernaryQuadraticNormal(eq).matches():
return HomogeneousTernaryQuadraticNormal(eq, free_symbols=_var).solve()
def transformation_to_normal(eq):
"""
Returns the transformation Matrix that converts a general ternary
quadratic equation ``eq`` (`ax^2 + by^2 + cz^2 + dxy + eyz + fxz`)
to a form without cross terms: `ax^2 + by^2 + cz^2 = 0`. This is
not used in solving ternary quadratics; it is only implemented for
the sake of completeness.
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type in (
"homogeneous_ternary_quadratic",
"homogeneous_ternary_quadratic_normal"):
return _transformation_to_normal(var, coeff)
def _transformation_to_normal(var, coeff):
_var = list(var) # copy
x, y, z = var
if not any(coeff[i**2] for i in var):
# https://math.stackexchange.com/questions/448051/transform-quadratic-ternary-form-to-normal-form/448065#448065
a = coeff[x*y]
b = coeff[y*z]
c = coeff[x*z]
swap = False
if not a: # b can't be 0 or else there aren't 3 vars
swap = True
a, b = b, a
T = Matrix(((1, 1, -b/a), (1, -1, -c/a), (0, 0, 1)))
if swap:
T.row_swap(0, 1)
T.col_swap(0, 1)
return T
if coeff[x**2] == 0:
# If the coefficient of x is zero change the variables
if coeff[y**2] == 0:
_var[0], _var[2] = var[2], var[0]
T = _transformation_to_normal(_var, coeff)
T.row_swap(0, 2)
T.col_swap(0, 2)
return T
else:
_var[0], _var[1] = var[1], var[0]
T = _transformation_to_normal(_var, coeff)
T.row_swap(0, 1)
T.col_swap(0, 1)
return T
# Apply the transformation x --> X - (B*Y + C*Z)/(2*A)
if coeff[x*y] != 0 or coeff[x*z] != 0:
A = coeff[x**2]
B = coeff[x*y]
C = coeff[x*z]
D = coeff[y**2]
E = coeff[y*z]
F = coeff[z**2]
_coeff = {}
_coeff[x**2] = 4*A**2
_coeff[y**2] = 4*A*D - B**2
_coeff[z**2] = 4*A*F - C**2
_coeff[y*z] = 4*A*E - 2*B*C
_coeff[x*y] = 0
_coeff[x*z] = 0
T_0 = _transformation_to_normal(_var, _coeff)
return Matrix(3, 3, [1, S(-B)/(2*A), S(-C)/(2*A), 0, 1, 0, 0, 0, 1])*T_0
elif coeff[y*z] != 0:
if coeff[y**2] == 0:
if coeff[z**2] == 0:
# Equations of the form A*x**2 + E*yz = 0.
# Apply transformation y -> Y + Z ans z -> Y - Z
return Matrix(3, 3, [1, 0, 0, 0, 1, 1, 0, 1, -1])
else:
# Ax**2 + E*y*z + F*z**2 = 0
_var[0], _var[2] = var[2], var[0]
T = _transformation_to_normal(_var, coeff)
T.row_swap(0, 2)
T.col_swap(0, 2)
return T
else:
# A*x**2 + D*y**2 + E*y*z + F*z**2 = 0, F may be zero
_var[0], _var[1] = var[1], var[0]
T = _transformation_to_normal(_var, coeff)
T.row_swap(0, 1)
T.col_swap(0, 1)
return T
else:
return Matrix.eye(3)
def parametrize_ternary_quadratic(eq):
"""
Returns the parametrized general solution for the ternary quadratic
equation ``eq`` which has the form
`ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`.
Examples
========
>>> from sympy import Tuple, ordered
>>> from sympy.abc import x, y, z
>>> from sympy.solvers.diophantine.diophantine import parametrize_ternary_quadratic
The parametrized solution may be returned with three parameters:
>>> parametrize_ternary_quadratic(2*x**2 + y**2 - 2*z**2)
(p**2 - 2*q**2, -2*p**2 + 4*p*q - 4*p*r - 4*q**2, p**2 - 4*p*q + 2*q**2 - 4*q*r)
There might also be only two parameters:
>>> parametrize_ternary_quadratic(4*x**2 + 2*y**2 - 3*z**2)
(2*p**2 - 3*q**2, -4*p**2 + 12*p*q - 6*q**2, 4*p**2 - 8*p*q + 6*q**2)
Notes
=====
Consider ``p`` and ``q`` in the previous 2-parameter
solution and observe that more than one solution can be represented
by a given pair of parameters. If `p` and ``q`` are not coprime, this is
trivially true since the common factor will also be a common factor of the
solution values. But it may also be true even when ``p`` and
``q`` are coprime:
>>> sol = Tuple(*_)
>>> p, q = ordered(sol.free_symbols)
>>> sol.subs([(p, 3), (q, 2)])
(6, 12, 12)
>>> sol.subs([(q, 1), (p, 1)])
(-1, 2, 2)
>>> sol.subs([(q, 0), (p, 1)])
(2, -4, 4)
>>> sol.subs([(q, 1), (p, 0)])
(-3, -6, 6)
Except for sign and a common factor, these are equivalent to
the solution of (1, 2, 2).
References
==========
.. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart,
London Mathematical Society Student Texts 41, Cambridge University
Press, Cambridge, 1998.
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type in (
"homogeneous_ternary_quadratic",
"homogeneous_ternary_quadratic_normal"):
x_0, y_0, z_0 = list(_diop_ternary_quadratic(var, coeff))[0]
return _parametrize_ternary_quadratic(
(x_0, y_0, z_0), var, coeff)
def _parametrize_ternary_quadratic(solution, _var, coeff):
# called for a*x**2 + b*y**2 + c*z**2 + d*x*y + e*y*z + f*x*z = 0
assert 1 not in coeff
x_0, y_0, z_0 = solution
v = list(_var) # copy
if x_0 is None:
return (None, None, None)
if solution.count(0) >= 2:
# if there are 2 zeros the equation reduces
# to k*X**2 == 0 where X is x, y, or z so X must
# be zero, too. So there is only the trivial
# solution.
return (None, None, None)
if x_0 == 0:
v[0], v[1] = v[1], v[0]
y_p, x_p, z_p = _parametrize_ternary_quadratic(
(y_0, x_0, z_0), v, coeff)
return x_p, y_p, z_p
x, y, z = v
r, p, q = symbols("r, p, q", integer=True)
eq = sum(k*v for k, v in coeff.items())
eq_1 = _mexpand(eq.subs(zip(
(x, y, z), (r*x_0, r*y_0 + p, r*z_0 + q))))
A, B = eq_1.as_independent(r, as_Add=True)
x = A*x_0
y = (A*y_0 - _mexpand(B/r*p))
z = (A*z_0 - _mexpand(B/r*q))
return _remove_gcd(x, y, z)
def diop_ternary_quadratic_normal(eq, parameterize=False):
"""
Solves the quadratic ternary diophantine equation,
`ax^2 + by^2 + cz^2 = 0`.
Explanation
===========
Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the
equation will be a quadratic binary or univariate equation. If solvable,
returns a tuple `(x, y, z)` that satisfies the given equation. If the
equation does not have integer solutions, `(None, None, None)` is returned.
Usage
=====
``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form
`ax^2 + by^2 + cz^2 = 0`.
Examples
========
>>> from sympy.abc import x, y, z
>>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic_normal
>>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2)
(1, 0, 1)
>>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2)
(1, 0, 2)
>>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2)
(4, 9, 1)
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == HomogeneousTernaryQuadraticNormal.name:
sol = _diop_ternary_quadratic_normal(var, coeff)
if len(sol) > 0:
x_0, y_0, z_0 = list(sol)[0]
else:
x_0, y_0, z_0 = None, None, None
if parameterize:
return _parametrize_ternary_quadratic(
(x_0, y_0, z_0), var, coeff)
return x_0, y_0, z_0
def _diop_ternary_quadratic_normal(var, coeff):
eq = sum([i * coeff[i] for i in coeff])
return HomogeneousTernaryQuadraticNormal(eq, free_symbols=var).solve()
def sqf_normal(a, b, c, steps=False):
"""
Return `a', b', c'`, the coefficients of the square-free normal
form of `ax^2 + by^2 + cz^2 = 0`, where `a', b', c'` are pairwise
prime. If `steps` is True then also return three tuples:
`sq`, `sqf`, and `(a', b', c')` where `sq` contains the square
factors of `a`, `b` and `c` after removing the `gcd(a, b, c)`;
`sqf` contains the values of `a`, `b` and `c` after removing
both the `gcd(a, b, c)` and the square factors.
The solutions for `ax^2 + by^2 + cz^2 = 0` can be
recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import sqf_normal
>>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11)
(11, 1, 5)
>>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True)
((3, 1, 7), (5, 55, 11), (11, 1, 5))
References
==========
.. [1] Legendre's Theorem, Legrange's Descent,
https://public.csusm.edu/aitken_html/notes/legendre.pdf
See Also
========
reconstruct()
"""
ABC = _remove_gcd(a, b, c)
sq = tuple(square_factor(i) for i in ABC)
sqf = A, B, C = tuple([i//j**2 for i,j in zip(ABC, sq)])
pc = igcd(A, B)
A /= pc
B /= pc
pa = igcd(B, C)
B /= pa
C /= pa
pb = igcd(A, C)
A /= pb
B /= pb
A *= pa
B *= pb
C *= pc
if steps:
return (sq, sqf, (A, B, C))
else:
return A, B, C
def square_factor(a):
r"""
Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square
free. `a` can be given as an integer or a dictionary of factors.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import square_factor
>>> square_factor(24)
2
>>> square_factor(-36*3)
6
>>> square_factor(1)
1
>>> square_factor({3: 2, 2: 1, -1: 1}) # -18
3
See Also
========
sympy.ntheory.factor_.core
"""
f = a if isinstance(a, dict) else factorint(a)
return Mul(*[p**(e//2) for p, e in f.items()])
def reconstruct(A, B, z):
"""
Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2`
from the `z` value of a solution of the square-free normal form of the
equation, `a'*x^2 + b'*y^2 + c'*z^2`, where `a'`, `b'` and `c'` are square
free and `gcd(a', b', c') == 1`.
"""
f = factorint(igcd(A, B))
for p, e in f.items():
if e != 1:
raise ValueError('a and b should be square-free')
z *= p
return z
def ldescent(A, B):
"""
Return a non-trivial solution to `w^2 = Ax^2 + By^2` using
Lagrange's method; return None if there is no such solution.
.
Here, `A \\neq 0` and `B \\neq 0` and `A` and `B` are square free. Output a
tuple `(w_0, x_0, y_0)` which is a solution to the above equation.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import ldescent
>>> ldescent(1, 1) # w^2 = x^2 + y^2
(1, 1, 0)
>>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2
(2, -1, 0)
This means that `x = -1, y = 0` and `w = 2` is a solution to the equation
`w^2 = 4x^2 - 7y^2`
>>> ldescent(5, -1) # w^2 = 5x^2 - y^2
(2, 1, -1)
References
==========
.. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart,
London Mathematical Society Student Texts 41, Cambridge University
Press, Cambridge, 1998.
.. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,
[online], Available:
https://nottingham-repository.worktribe.com/output/1023265/efficient-solution-of-rational-conics
"""
if abs(A) > abs(B):
w, y, x = ldescent(B, A)
return w, x, y
if A == 1:
return (1, 1, 0)
if B == 1:
return (1, 0, 1)
if B == -1: # and A == -1
return
r = sqrt_mod(A, B)
Q = (r**2 - A) // B
if Q == 0:
B_0 = 1
d = 0
else:
div = divisors(Q)
B_0 = None
for i in div:
sQ, _exact = integer_nthroot(abs(Q) // i, 2)
if _exact:
B_0, d = sign(Q)*i, sQ
break
if B_0 is not None:
W, X, Y = ldescent(A, B_0)
return _remove_gcd((-A*X + r*W), (r*X - W), Y*(B_0*d))
def descent(A, B):
"""
Returns a non-trivial solution, (x, y, z), to `x^2 = Ay^2 + Bz^2`
using Lagrange's descent method with lattice-reduction. `A` and `B`
are assumed to be valid for such a solution to exist.
This is faster than the normal Lagrange's descent algorithm because
the Gaussian reduction is used.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import descent
>>> descent(3, 1) # x**2 = 3*y**2 + z**2
(1, 0, 1)
`(x, y, z) = (1, 0, 1)` is a solution to the above equation.
>>> descent(41, -113)
(-16, -3, 1)
References
==========
.. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,
Mathematics of Computation, Volume 00, Number 0.
"""
if abs(A) > abs(B):
x, y, z = descent(B, A)
return x, z, y
if B == 1:
return (1, 0, 1)
if A == 1:
return (1, 1, 0)
if B == -A:
return (0, 1, 1)
if B == A:
x, z, y = descent(-1, A)
return (A*y, z, x)
w = sqrt_mod(A, B)
x_0, z_0 = gaussian_reduce(w, A, B)
t = (x_0**2 - A*z_0**2) // B
t_2 = square_factor(t)
t_1 = t // t_2**2
x_1, z_1, y_1 = descent(A, t_1)
return _remove_gcd(x_0*x_1 + A*z_0*z_1, z_0*x_1 + x_0*z_1, t_1*t_2*y_1)
def gaussian_reduce(w, a, b):
r"""
Returns a reduced solution `(x, z)` to the congruence
`X^2 - aZ^2 \equiv 0 \ (mod \ b)` so that `x^2 + |a|z^2` is minimal.
Details
=======
Here ``w`` is a solution of the congruence `x^2 \equiv a \ (mod \ b)`
References
==========
.. [1] Gaussian lattice Reduction [online]. Available:
https://web.archive.org/web/20201021115213/http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404
.. [2] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,
Mathematics of Computation, Volume 00, Number 0.
"""
u = (0, 1)
v = (1, 0)
if dot(u, v, w, a, b) < 0:
v = (-v[0], -v[1])
if norm(u, w, a, b) < norm(v, w, a, b):
u, v = v, u
while norm(u, w, a, b) > norm(v, w, a, b):
k = dot(u, v, w, a, b) // dot(v, v, w, a, b)
u, v = v, (u[0]- k*v[0], u[1]- k*v[1])
u, v = v, u
if dot(u, v, w, a, b) < dot(v, v, w, a, b)/2 or norm((u[0]-v[0], u[1]-v[1]), w, a, b) > norm(v, w, a, b):
c = v
else:
c = (u[0] - v[0], u[1] - v[1])
return c[0]*w + b*c[1], c[0]
def dot(u, v, w, a, b):
r"""
Returns a special dot product of the vectors `u = (u_{1}, u_{2})` and
`v = (v_{1}, v_{2})` which is defined in order to reduce solution of
the congruence equation `X^2 - aZ^2 \equiv 0 \ (mod \ b)`.
"""
u_1, u_2 = u
v_1, v_2 = v
return (w*u_1 + b*u_2)*(w*v_1 + b*v_2) + abs(a)*u_1*v_1
def norm(u, w, a, b):
r"""
Returns the norm of the vector `u = (u_{1}, u_{2})` under the dot product
defined by `u \cdot v = (wu_{1} + bu_{2})(w*v_{1} + bv_{2}) + |a|*u_{1}*v_{1}`
where `u = (u_{1}, u_{2})` and `v = (v_{1}, v_{2})`.
"""
u_1, u_2 = u
return sqrt(dot((u_1, u_2), (u_1, u_2), w, a, b))
def holzer(x, y, z, a, b, c):
r"""
Simplify the solution `(x, y, z)` of the equation
`ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z^2 \geq \mid ab \mid` to
a new reduced solution `(x', y', z')` such that `z'^2 \leq \mid ab \mid`.
The algorithm is an interpretation of Mordell's reduction as described
on page 8 of Cremona and Rusin's paper [1]_ and the work of Mordell in
reference [2]_.
References
==========
.. [1] Efficient Solution of Rational Conices, J. E. Cremona and D. Rusin,
Mathematics of Computation, Volume 00, Number 0.
.. [2] Diophantine Equations, L. J. Mordell, page 48.
"""
if _odd(c):
k = 2*c
else:
k = c//2
small = a*b*c
step = 0
while True:
t1, t2, t3 = a*x**2, b*y**2, c*z**2
# check that it's a solution
if t1 + t2 != t3:
if step == 0:
raise ValueError('bad starting solution')
break
x_0, y_0, z_0 = x, y, z
if max(t1, t2, t3) <= small:
# Holzer condition
break
uv = u, v = base_solution_linear(k, y_0, -x_0)
if None in uv:
break
p, q = -(a*u*x_0 + b*v*y_0), c*z_0
r = Rational(p, q)
if _even(c):
w = _nint_or_floor(p, q)
assert abs(w - r) <= S.Half
else:
w = p//q # floor
if _odd(a*u + b*v + c*w):
w += 1
assert abs(w - r) <= S.One
A = (a*u**2 + b*v**2 + c*w**2)
B = (a*u*x_0 + b*v*y_0 + c*w*z_0)
x = Rational(x_0*A - 2*u*B, k)
y = Rational(y_0*A - 2*v*B, k)
z = Rational(z_0*A - 2*w*B, k)
assert all(i.is_Integer for i in (x, y, z))
step += 1
return tuple([int(i) for i in (x_0, y_0, z_0)])
def diop_general_pythagorean(eq, param=symbols("m", integer=True)):
"""
Solves the general pythagorean equation,
`a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`.
Returns a tuple which contains a parametrized solution to the equation,
sorted in the same order as the input variables.
Usage
=====
``diop_general_pythagorean(eq, param)``: where ``eq`` is a general
pythagorean equation which is assumed to be zero and ``param`` is the base
parameter used to construct other parameters by subscripting.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_general_pythagorean
>>> from sympy.abc import a, b, c, d, e
>>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2)
(m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2)
>>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2)
(10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4)
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == GeneralPythagorean.name:
if param is None:
params = None
else:
params = symbols('%s1:%i' % (param, len(var)), integer=True)
return list(GeneralPythagorean(eq).solve(parameters=params))[0]
def diop_general_sum_of_squares(eq, limit=1):
r"""
Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`.
Returns at most ``limit`` number of solutions.
Usage
=====
``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which
is assumed to be zero. Also, ``eq`` should be in the form,
`x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`.
Details
=======
When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be
no solutions. Refer to [1]_ for more details.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_squares
>>> from sympy.abc import a, b, c, d, e
>>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345)
{(15, 22, 22, 24, 24)}
Reference
=========
.. [1] Representing an integer as a sum of three squares, [online],
Available:
https://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == GeneralSumOfSquares.name:
return set(GeneralSumOfSquares(eq).solve(limit=limit))
def diop_general_sum_of_even_powers(eq, limit=1):
"""
Solves the equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`
where `e` is an even, integer power.
Returns at most ``limit`` number of solutions.
Usage
=====
``general_sum_of_even_powers(eq, limit)`` : Here ``eq`` is an expression which
is assumed to be zero. Also, ``eq`` should be in the form,
`x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_even_powers
>>> from sympy.abc import a, b
>>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4))
{(2, 3)}
See Also
========
power_representation
"""
var, coeff, diop_type = classify_diop(eq, _dict=False)
if diop_type == GeneralSumOfEvenPowers.name:
return set(GeneralSumOfEvenPowers(eq).solve(limit=limit))
## Functions below this comment can be more suitably grouped under
## an Additive number theory module rather than the Diophantine
## equation module.
def partition(n, k=None, zeros=False):
"""
Returns a generator that can be used to generate partitions of an integer
`n`.
Explanation
===========
A partition of `n` is a set of positive integers which add up to `n`. For
example, partitions of 3 are 3, 1 + 2, 1 + 1 + 1. A partition is returned
as a tuple. If ``k`` equals None, then all possible partitions are returned
irrespective of their size, otherwise only the partitions of size ``k`` are
returned. If the ``zero`` parameter is set to True then a suitable
number of zeros are added at the end of every partition of size less than
``k``.
``zero`` parameter is considered only if ``k`` is not None. When the
partitions are over, the last `next()` call throws the ``StopIteration``
exception, so this function should always be used inside a try - except
block.
Details
=======
``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size
of the partition which is also positive integer.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import partition
>>> f = partition(5)
>>> next(f)
(1, 1, 1, 1, 1)
>>> next(f)
(1, 1, 1, 2)
>>> g = partition(5, 3)
>>> next(g)
(1, 1, 3)
>>> next(g)
(1, 2, 2)
>>> g = partition(5, 3, zeros=True)
>>> next(g)
(0, 0, 5)
"""
if not zeros or k is None:
for i in ordered_partitions(n, k):
yield tuple(i)
else:
for m in range(1, k + 1):
for i in ordered_partitions(n, m):
i = tuple(i)
yield (0,)*(k - len(i)) + i
def prime_as_sum_of_two_squares(p):
"""
Represent a prime `p` as a unique sum of two squares; this can
only be done if the prime is congruent to 1 mod 4.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import prime_as_sum_of_two_squares
>>> prime_as_sum_of_two_squares(7) # can't be done
>>> prime_as_sum_of_two_squares(5)
(1, 2)
Reference
=========
.. [1] Representing a number as a sum of four squares, [online],
Available: https://schorn.ch/lagrange.html
See Also
========
sum_of_squares()
"""
if not p % 4 == 1:
return
if p % 8 == 5:
b = 2
else:
b = 3
while pow(b, (p - 1) // 2, p) == 1:
b = nextprime(b)
b = pow(b, (p - 1) // 4, p)
a = p
while b**2 > p:
a, b = b, a % b
return (int(a % b), int(b)) # convert from long
def sum_of_three_squares(n):
r"""
Returns a 3-tuple $(a, b, c)$ such that $a^2 + b^2 + c^2 = n$ and
$a, b, c \geq 0$.
Returns None if $n = 4^a(8m + 7)$ for some `a, m \in \mathbb{Z}`. See
[1]_ for more details.
Usage
=====
``sum_of_three_squares(n)``: Here ``n`` is a non-negative integer.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import sum_of_three_squares
>>> sum_of_three_squares(44542)
(18, 37, 207)
References
==========
.. [1] Representing a number as a sum of three squares, [online],
Available: https://schorn.ch/lagrange.html
See Also
========
sum_of_squares()
"""
special = {1:(1, 0, 0), 2:(1, 1, 0), 3:(1, 1, 1), 10: (1, 3, 0), 34: (3, 3, 4), 58:(3, 7, 0),
85:(6, 7, 0), 130:(3, 11, 0), 214:(3, 6, 13), 226:(8, 9, 9), 370:(8, 9, 15),
526:(6, 7, 21), 706:(15, 15, 16), 730:(1, 27, 0), 1414:(6, 17, 33), 1906:(13, 21, 36),
2986: (21, 32, 39), 9634: (56, 57, 57)}
v = 0
if n == 0:
return (0, 0, 0)
v = multiplicity(4, n)
n //= 4**v
if n % 8 == 7:
return
if n in special.keys():
x, y, z = special[n]
return _sorted_tuple(2**v*x, 2**v*y, 2**v*z)
s, _exact = integer_nthroot(n, 2)
if _exact:
return (2**v*s, 0, 0)
x = None
if n % 8 == 3:
s = s if _odd(s) else s - 1
for x in range(s, -1, -2):
N = (n - x**2) // 2
if isprime(N):
y, z = prime_as_sum_of_two_squares(N)
return _sorted_tuple(2**v*x, 2**v*(y + z), 2**v*abs(y - z))
return
if n % 8 in (2, 6):
s = s if _odd(s) else s - 1
else:
s = s - 1 if _odd(s) else s
for x in range(s, -1, -2):
N = n - x**2
if isprime(N):
y, z = prime_as_sum_of_two_squares(N)
return _sorted_tuple(2**v*x, 2**v*y, 2**v*z)
def sum_of_four_squares(n):
r"""
Returns a 4-tuple `(a, b, c, d)` such that `a^2 + b^2 + c^2 + d^2 = n`.
Here `a, b, c, d \geq 0`.
Usage
=====
``sum_of_four_squares(n)``: Here ``n`` is a non-negative integer.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import sum_of_four_squares
>>> sum_of_four_squares(3456)
(8, 8, 32, 48)
>>> sum_of_four_squares(1294585930293)
(0, 1234, 2161, 1137796)
References
==========
.. [1] Representing a number as a sum of four squares, [online],
Available: https://schorn.ch/lagrange.html
See Also
========
sum_of_squares()
"""
if n == 0:
return (0, 0, 0, 0)
v = multiplicity(4, n)
n //= 4**v
if n % 8 == 7:
d = 2
n = n - 4
elif n % 8 in (2, 6):
d = 1
n = n - 1
else:
d = 0
x, y, z = sum_of_three_squares(n)
return _sorted_tuple(2**v*d, 2**v*x, 2**v*y, 2**v*z)
def power_representation(n, p, k, zeros=False):
r"""
Returns a generator for finding k-tuples of integers,
`(n_{1}, n_{2}, . . . n_{k})`, such that
`n = n_{1}^p + n_{2}^p + . . . n_{k}^p`.
Usage
=====
``power_representation(n, p, k, zeros)``: Represent non-negative number
``n`` as a sum of ``k`` ``p``\ th powers. If ``zeros`` is true, then the
solutions is allowed to contain zeros.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import power_representation
Represent 1729 as a sum of two cubes:
>>> f = power_representation(1729, 3, 2)
>>> next(f)
(9, 10)
>>> next(f)
(1, 12)
If the flag `zeros` is True, the solution may contain tuples with
zeros; any such solutions will be generated after the solutions
without zeros:
>>> list(power_representation(125, 2, 3, zeros=True))
[(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)]
For even `p` the `permute_sign` function can be used to get all
signed values:
>>> from sympy.utilities.iterables import permute_signs
>>> list(permute_signs((1, 12)))
[(1, 12), (-1, 12), (1, -12), (-1, -12)]
All possible signed permutations can also be obtained:
>>> from sympy.utilities.iterables import signed_permutations
>>> list(signed_permutations((1, 12)))
[(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1), (12, -1), (-12, -1)]
"""
n, p, k = [as_int(i) for i in (n, p, k)]
if n < 0:
if p % 2:
for t in power_representation(-n, p, k, zeros):
yield tuple(-i for i in t)
return
if p < 1 or k < 1:
raise ValueError(filldedent('''
Expecting positive integers for `(p, k)`, but got `(%s, %s)`'''
% (p, k)))
if n == 0:
if zeros:
yield (0,)*k
return
if k == 1:
if p == 1:
yield (n,)
else:
be = perfect_power(n)
if be:
b, e = be
d, r = divmod(e, p)
if not r:
yield (b**d,)
return
if p == 1:
for t in partition(n, k, zeros=zeros):
yield t
return
if p == 2:
feasible = _can_do_sum_of_squares(n, k)
if not feasible:
return
if not zeros and n > 33 and k >= 5 and k <= n and n - k in (
13, 10, 7, 5, 4, 2, 1):
'''Todd G. Will, "When Is n^2 a Sum of k Squares?", [online].
Available: https://www.maa.org/sites/default/files/Will-MMz-201037918.pdf'''
return
if feasible is not True: # it's prime and k == 2
yield prime_as_sum_of_two_squares(n)
return
if k == 2 and p > 2:
be = perfect_power(n)
if be and be[1] % p == 0:
return # Fermat: a**n + b**n = c**n has no solution for n > 2
if n >= k:
a = integer_nthroot(n - (k - 1), p)[0]
for t in pow_rep_recursive(a, k, n, [], p):
yield tuple(reversed(t))
if zeros:
a = integer_nthroot(n, p)[0]
for i in range(1, k):
for t in pow_rep_recursive(a, i, n, [], p):
yield tuple(reversed(t + (0,)*(k - i)))
sum_of_powers = power_representation
def pow_rep_recursive(n_i, k, n_remaining, terms, p):
# Invalid arguments
if n_i <= 0 or k <= 0:
return
# No solutions may exist
if n_remaining < k:
return
if k * pow(n_i, p) < n_remaining:
return
if k == 0 and n_remaining == 0:
yield tuple(terms)
elif k == 1:
# next_term^p must equal to n_remaining
next_term, exact = integer_nthroot(n_remaining, p)
if exact and next_term <= n_i:
yield tuple(terms + [next_term])
return
else:
# TODO: Fall back to diop_DN when k = 2
if n_i >= 1 and k > 0:
for next_term in range(1, n_i + 1):
residual = n_remaining - pow(next_term, p)
if residual < 0:
break
yield from pow_rep_recursive(next_term, k - 1, residual, terms + [next_term], p)
def sum_of_squares(n, k, zeros=False):
"""Return a generator that yields the k-tuples of nonnegative
values, the squares of which sum to n. If zeros is False (default)
then the solution will not contain zeros. The nonnegative
elements of a tuple are sorted.
* If k == 1 and n is square, (n,) is returned.
* If k == 2 then n can only be written as a sum of squares if
every prime in the factorization of n that has the form
4*k + 3 has an even multiplicity. If n is prime then
it can only be written as a sum of two squares if it is
in the form 4*k + 1.
* if k == 3 then n can be written as a sum of squares if it does
not have the form 4**m*(8*k + 7).
* all integers can be written as the sum of 4 squares.
* if k > 4 then n can be partitioned and each partition can
be written as a sum of 4 squares; if n is not evenly divisible
by 4 then n can be written as a sum of squares only if the
an additional partition can be written as sum of squares.
For example, if k = 6 then n is partitioned into two parts,
the first being written as a sum of 4 squares and the second
being written as a sum of 2 squares -- which can only be
done if the condition above for k = 2 can be met, so this will
automatically reject certain partitions of n.
Examples
========
>>> from sympy.solvers.diophantine.diophantine import sum_of_squares
>>> list(sum_of_squares(25, 2))
[(3, 4)]
>>> list(sum_of_squares(25, 2, True))
[(3, 4), (0, 5)]
>>> list(sum_of_squares(25, 4))
[(1, 2, 2, 4)]
See Also
========
sympy.utilities.iterables.signed_permutations
"""
yield from power_representation(n, 2, k, zeros)
def _can_do_sum_of_squares(n, k):
"""Return True if n can be written as the sum of k squares,
False if it cannot, or 1 if ``k == 2`` and ``n`` is prime (in which
case it *can* be written as a sum of two squares). A False
is returned only if it cannot be written as ``k``-squares, even
if 0s are allowed.
"""
if k < 1:
return False
if n < 0:
return False
if n == 0:
return True
if k == 1:
return is_square(n)
if k == 2:
if n in (1, 2):
return True
if isprime(n):
if n % 4 == 1:
return 1 # signal that it was prime
return False
else:
f = factorint(n)
for p, m in f.items():
# we can proceed iff no prime factor in the form 4*k + 3
# has an odd multiplicity
if (p % 4 == 3) and m % 2:
return False
return True
if k == 3:
if (n//4**multiplicity(4, n)) % 8 == 7:
return False
# every number can be written as a sum of 4 squares; for k > 4 partitions
# can be 0
return True