Source code for sympy.solvers.diophantine
from __future__ import print_function, division
from sympy import (Add, ceiling, divisors, factor_list, factorint, floor, igcd,
ilcm, Integer, integer_nthroot, isprime, Matrix, Mul, nextprime,
perfect_power, Poly, S, sign, solve, sqrt, Subs, Symbol, symbols, sympify,
Wild)
from sympy.core.function import _mexpand
from sympy.simplify.radsimp import rad_rationalize
from sympy.utilities import default_sort_key, numbered_symbols
from sympy.core.numbers import igcdex
from sympy.ntheory.residue_ntheory import sqrt_mod
from sympy.core.compatibility import range
from sympy.core.relational import Eq
from sympy.solvers.solvers import check_assumptions
__all__ = ['base_solution_linear', 'classify_diop', 'cornacchia', 'descent',
'diop_bf_DN', 'diop_DN', 'diop_general_pythagorean',
'diop_general_sum_of_squares', 'diop_linear', 'diop_quadratic',
'diop_solve', 'diop_ternary_quadratic', 'diophantine', 'find_DN',
'partition', 'square_factor', 'sum_of_four_squares',
'sum_of_three_squares', 'transformation_to_DN']
[docs]def diophantine(eq, param=symbols("t", integer=True)):
"""
Simplify the solution procedure of diophantine equation ``eq`` by
converting it into a product of terms which should equal zero.
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()``.
Output of ``diophantine()`` is a set of tuples. Each tuple represents a
solution of the input equation. In a tuple, solution for each variable is
listed according to the alphabetic order of input variables. i.e. if we have
an equation with two variables `a` and `b`, first element of the tuple will
give the solution for `a` and the second element will give the solution for
`b`.
Usage
=====
``diophantine(eq, t)``: Solve the diophantine equation ``eq``.
``t`` is the parameter to be used by ``diop_solve()``.
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.solvers.diophantine import diophantine
>>> from sympy.abc import x, y, z
>>> diophantine(x**2 - y**2)
set([(-t_0, -t_0), (t_0, -t_0)])
#>>> diophantine(x*(2*x + 3*y - z))
#set([(0, n1, n2), (3*t - z, -2*t + z, z)])
#>>> diophantine(x**2 + 3*x*y + 4*x)
#set([(0, n1), (3*t - 4, -t)])
See Also
========
diop_solve()
"""
if isinstance(eq, Eq):
eq = eq.lhs - eq.rhs
eq = Poly(eq).as_expr()
if not eq.is_polynomial() or eq.is_number:
raise TypeError("Equation input format not supported")
var = list(eq.expand(force=True).free_symbols)
var.sort(key=default_sort_key)
terms = factor_list(eq)[1]
sols = set([])
for term in terms:
base = term[0]
var_t, jnk, eq_type = classify_diop(base)
solution = diop_solve(base, param)
if eq_type in ["linear", "homogeneous_ternary_quadratic", "general_pythagorean"]:
if merge_solution(var, var_t, solution) != ():
sols.add(merge_solution(var, var_t, solution))
elif eq_type in ["binary_quadratic", "general_sum_of_squares", "univariate"]:
for sol in solution:
if merge_solution(var, var_t, sol) != ():
sols.add(merge_solution(var, var_t, sol))
return sols
[docs]def merge_solution(var, var_t, solution):
"""
This is used to construct the full solution from the solutions of sub
equations.
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.
"""
l = []
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:
l.append(next(solution))
else:
l.append(next(params))
for val, symb in zip(l, var):
if check_assumptions(val, **symb.assumptions0) is False:
return tuple()
return tuple(l)
[docs]def diop_solve(eq, param=symbols("t", integer=True)):
"""
Solves the diophantine equation ``eq``.
Similar to ``diophantine()`` but doesn't try to factor ``eq`` as latter
does. Uses ``classify_diop()`` to determine the type of the eqaution and
calls the appropriate solver function.
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, -2*t_0 + 5)
>>> diop_solve(4*x + 3*y -4*z + 5)
(t_0, -4*t_1 + 5, t_0 - 3*t_1 + 5)
>>> diop_solve(x + 3*y - 4*z + w -6)
(t_0, t_0 + t_1, -2*t_0 - 3*t_1 - 4*t_2 - 6, -t_0 - 2*t_1 - 3*t_2 - 6)
>>> diop_solve(x**2 + y**2 - 5)
set([(-2, -1), (-2, 1), (2, -1), (2, 1)])
See Also
========
diophantine()
"""
var, coeff, eq_type = classify_diop(eq)
if eq_type == "linear":
return _diop_linear(var, coeff, param)
elif eq_type == "binary_quadratic":
return _diop_quadratic(var, coeff, param)
elif eq_type == "homogeneous_ternary_quadratic":
x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff)
return _parametrize_ternary_quadratic((x_0, y_0, z_0), var, coeff)
elif eq_type == "general_pythagorean":
return _diop_general_pythagorean(var, coeff, param)
elif eq_type == "univariate":
l = solve(eq)
s = set([])
for soln in l:
if isinstance(soln, Integer):
s.add((soln,))
return s
elif eq_type == "general_sum_of_squares":
return _diop_general_sum_of_squares(var, coeff)
[docs]def classify_diop(eq):
"""
Helper routine used by diop_solve() to find the type of the ``eq`` etc.
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 Integer(1). Type is an
element in the set {"linear", "binary_quadratic", "general_pythagorean",
"homogeneous_ternary_quadratic", "univariate", "general_sum_of_squares"}
Usage
=====
``classify_diop(eq)``: Return variables, coefficients and type of the
``eq``.
Details
=======
``eq`` should be an expression which is assumed to be zero.
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: 0, y**2: 1, x*y: -1}, 'binary_quadratic')
"""
eq = eq.expand(force=True)
coeff = eq.as_coefficients_dict()
diop_type = None
var = []
if isinstance(eq, Symbol):
var.append(eq)
coeff[eq] = Integer(1)
elif isinstance(eq, Mul) and Poly(eq).total_degree() == 1:
var.append(eq.as_two_terms()[1])
coeff[eq.as_two_terms()[1]] = Integer(eq.as_two_terms()[0])
else:
var = list(eq.free_symbols)
var.sort(key=default_sort_key)
coeff = dict([reversed(t.as_independent(*var)) for t in eq.args])
for c in coeff:
if not isinstance(coeff[c], Integer):
raise TypeError("Coefficients should be Integers")
if Poly(eq).total_degree() == 1:
diop_type = "linear"
elif len(var) == 1:
diop_type = "univariate"
elif Poly(eq).total_degree() == 2 and len(var) == 2:
diop_type = "binary_quadratic"
x, y = var[:2]
if isinstance(eq, Mul):
coeff = {x**2: 0, x*y: eq.args[0], y**2: 0, x: 0, y: 0, Integer(1): 0}
else:
for term in [x**2, y**2, x*y, x, y, Integer(1)]:
if term not in coeff.keys():
coeff[term] = Integer(0)
elif Poly(eq).total_degree() == 2 and len(var) == 3 and Integer(1) not in coeff.keys():
for v in var:
if v in coeff.keys():
diop_type = "inhomogeneous_ternary_quadratic"
break
else:
diop_type = "homogeneous_ternary_quadratic"
x, y, z = var[:3]
for term in [x**2, y**2, z**2, x*y, y*z, x*z]:
if term not in coeff.keys():
coeff[term] = Integer(0)
elif Poly(eq).degree() == 2 and len(var) >= 3:
for v in var:
if v in coeff.keys():
diop_type = "inhomogeneous_general_quadratic"
break
else:
if Integer(1) in coeff.keys():
constant_term = True
else:
constant_term = False
non_square_degree_2_terms = False
for v in var:
for u in var:
if u != v and u*v in coeff.keys():
non_square_degree_2_terms = True
break
if non_square_degree_2_terms:
break
if constant_term and non_square_degree_2_terms:
diop_type = "inhomogeneous_general_quadratic"
elif constant_term and not non_square_degree_2_terms:
for v in var:
if coeff[v**2] != 1:
break
else:
diop_type = "general_sum_of_squares"
elif not constant_term and non_square_degree_2_terms:
diop_type = "homogeneous_general_quadratic"
else:
coeff_sign_sum = 0
for v in var:
if not isinstance(sqrt(abs(Integer(coeff[v**2]))), Integer):
break
coeff_sign_sum = coeff_sign_sum + sign(coeff[v**2])
else:
if abs(coeff_sign_sum) == len(var) - 2 and not constant_term:
diop_type = "general_pythagorean"
elif Poly(eq).total_degree() == 3 and len(var) == 2:
x, y = var[:2]
diop_type = "cubic_thue"
for term in [x**3, x**2*y, x*y**2, y**3, Integer(1)]:
if term not in coeff.keys():
coeff[term] == Integer(0)
if diop_type is not None:
return var, coeff, diop_type
else:
raise NotImplementedError("Still not implemented")
[docs]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 import diop_linear
>>> from sympy.abc import x, y, z, t
>>> from sympy import Integer
>>> 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, -6*t_0 - 4*t_1 + 3, 5*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)
if diop_type == "linear":
return _diop_linear(var, coeff, param)
def _diop_linear(var, coeff, param):
"""
Solves diophantine equations of the form:
a_0*x_0 + a_1*x_1 + ... + a_n*x_n == c
Note that no solution exists if gcd(a_0, ..., a_n) doesn't divide c.
"""
if len(var) == 0:
return None
if Integer(1) in coeff:
#coeff[] is negated because input is of the form: ax + by + c == 0
# but is used as: ax + by == -c
c = -coeff[Integer(1)]
else:
c = 0
# Some solutions will have multiple free variables in their solutions.
params = [str(param) + "_" + str(i) for i in range(len(var))]
params = [symbols(p, integer=True) for p in params]
if len(var) == 1:
if coeff[var[0]] == 0:
if c == 0:
return tuple([params[0]])
else:
return tuple([None])
elif divisible(c, coeff[var[0]]):
return tuple([c/coeff[var[0]]])
else:
return tuple([None])
"""
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 = []
no_solution = tuple([None] * len(var))
for i in range(len(B)):
tot_x, tot_y = 0, 0
if isinstance(c, Add):
# example: 5 + t_0 + 3*t_1
args = c.args
else: # c is a Mul, a Symbol, or an Integer
args = [c]
for j in range(len(args)):
if isinstance(args[j], Mul):
# example: 3*t_1 -> k = 3
k = args[j].as_two_terms()[0]
param_index = params.index(args[j].as_two_terms()[1]) + 1
elif isinstance(args[j], Symbol):
# example: t_0 -> k = 1
k = 1
param_index = params.index(args[j]) + 1
else: #args[j] is an Integer
# example: 5 -> k = 5
k = args[j]
param_index = 0
sol_x, sol_y = base_solution_linear(k, A[i], B[i], params[param_index])
if isinstance(args[j], Mul) or isinstance(args[j], Symbol):
if isinstance(sol_x, Add):
sol_x = sol_x.args[0]*params[param_index - 1] + sol_x.args[1]
elif isinstance(sol_x, Integer):
sol_x = sol_x*params[param_index - 1]
if isinstance(sol_y, Add):
sol_y = sol_y.args[0]*params[param_index - 1] + sol_y.args[1]
elif isinstance(sol_y, Integer):
sol_y = sol_y*params[param_index - 1]
else:
if sol_x is None or sol_y is None:
return no_solution
tot_x += sol_x
tot_y += sol_y
solutions.append(tot_x)
c = tot_y
solutions.append(tot_y)
return tuple(solutions)
[docs]def base_solution_linear(c, a, b, t=None):
"""
Return the base solution for a linear diophantine equation with two
variables.
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 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, -2*t + 5)
>>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0
(7*t, -5*t)
"""
d = igcd(a, igcd(b, c))
a = a // d
b = b // d
c = c // d
if c == 0:
if t != None:
return (b*t , -a*t)
else:
return (S.Zero, S.Zero)
else:
x0, y0, d = extended_euclid(int(abs(a)), int(abs(b)))
x0 = x0 * sign(a)
y0 = y0 * sign(b)
if divisible(c, d):
if t != None:
return (c*x0 + b*t, c*y0 - a*t)
else:
return (Integer(c*x0), Integer(c*y0))
else:
return (None, None)
[docs]def extended_euclid(a, b):
"""
For given ``a``, ``b`` returns a tuple containing integers `x`, `y` and `d`
such that `ax + by = d`. Here `d = gcd(a, b)`.
Usage
=====
``extended_euclid(a, b)``: returns `x`, `y` and `\gcd(a, b)`.
Details
=======
``a`` Any instance of Integer.
``b`` Any instance of Integer.
Examples
========
>>> from sympy.solvers.diophantine import extended_euclid
>>> extended_euclid(4, 6)
(-1, 1, 2)
>>> extended_euclid(3, 5)
(2, -1, 1)
"""
if b == 0:
return (1, 0, a)
x0, y0, d = extended_euclid(b, a%b)
x, y = y0, x0 - (a//b) * y0
return x, y, d
[docs]def divisible(a, b):
"""
Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise.
"""
return igcd(int(a), int(b)) == abs(int(b))
[docs]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 import diop_quadratic
>>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t)
set([(-1, -1)])
References
==========
.. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,[online],
Available: http://www.alpertron.com.ar/METHODS.HTM
.. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online],
Available: 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)
if diop_type == "binary_quadratic":
return _diop_quadratic(var, coeff, param)
def _diop_quadratic(var, coeff, t):
x, y = var[:2]
for term in [x**2, y**2, x*y, x, y, Integer(1)]:
if term not in coeff.keys():
coeff[term] = Integer(0)
A = coeff[x**2]
B = coeff[x*y]
C = coeff[y**2]
D = coeff[x]
E = coeff[y]
F = coeff[Integer(1)]
d = igcd(A, igcd(B, igcd(C, igcd(D, igcd(E, F)))))
A = A // d
B = B // d
C = C // d
D = D // d
E = E // d
F = F // d
# (1) Linear case: A = B = C = 0 ==> considered under linear diophantine equations
# (2) 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, http://www.alpertron.com.ar/METHODS.HTM#SHyperb
l = set([])
if A == 0 and C == 0 and B != 0:
if D*E - B*F == 0:
if divisible(int(E), int(B)):
l.add((-E/B, t))
if divisible(int(D), int(B)):
l.add((t, -D/B))
else:
div = divisors(D*E - B*F)
div = div + [-term for term in div]
for d in div:
if divisible(int(d - E), int(B)):
x0 = (d - E) // B
if divisible(int(D*E - B*F), int(d)):
if divisible(int((D*E - B*F)// d - D), int(B)):
y0 = ((D*E - B*F) // d - D) // B
l.add((x0, y0))
# (3) 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, http://www.alpertron.com.ar/METHODS.HTM#Parabol
elif B**2 - 4*A*C == 0:
if A == 0:
s = _diop_quadratic([y, x], coeff, t)
for soln in s:
l.add((soln[1], soln[0]))
else:
g = igcd(A, C)
g = abs(g) * sign(A)
a = A // g
b = B // g
c = C // g
e = sign(B/A)
if e*sqrt(c)*D - sqrt(a)*E == 0:
z = symbols("z", real=True)
roots = solve(sqrt(a)*g*z**2 + D*z + sqrt(a)*F)
for root in roots:
if isinstance(root, Integer):
l.add((diop_solve(sqrt(a)*x + e*sqrt(c)*y - root)[0], diop_solve(sqrt(a)*x + e*sqrt(c)*y - root)[1]))
elif isinstance(e*sqrt(c)*D - sqrt(a)*E, Integer):
solve_x = lambda u: e*sqrt(c)*g*(sqrt(a)*E - e*sqrt(c)*D)*t**2 - (E + 2*e*sqrt(c)*g*u)*t\
- (e*sqrt(c)*g*u**2 + E*u + e*sqrt(c)*F) // (e*sqrt(c)*D - sqrt(a)*E)
solve_y = lambda u: sqrt(a)*g*(e*sqrt(c)*D - sqrt(a)*E)*t**2 + (D + 2*sqrt(a)*g*u)*t \
+ (sqrt(a)*g*u**2 + D*u + sqrt(a)*F) // (e*sqrt(c)*D - sqrt(a)*E)
for z0 in range(0, abs(e*sqrt(c)*D - sqrt(a)*E)):
if divisible(sqrt(a)*g*z0**2 + D*z0 + sqrt(a)*F, e*sqrt(c)*D - sqrt(a)*E):
l.add((solve_x(z0), solve_y(z0)))
# (4) Method used when B**2 - 4*A*C is a square, is descibed in p. 6 of the below paper
# by John P. Robertson.
# http://www.jpr2718.org/ax2p.pdf
elif isinstance(sqrt(B**2 - 4*A*C), Integer):
if A != 0:
r = sqrt(B**2 - 4*A*C)
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)
sol = diop_solve(eq, t)
sol = list(sol)
for solution in sol:
s0 = solution[0]
t0 = solution[1]
x_0 = S(B*t0 + r*s0 + r*t0 - B*s0)/(4*A*r)
y_0 = S(s0 - t0)/(2*r)
if isinstance(s0, Symbol) or isinstance(t0, Symbol):
if check_param(x_0, y_0, 4*A*r, t) != (None, None):
l.add((check_param(x_0, y_0, 4*A*r, t)[0], check_param(x_0, y_0, 4*A*r, t)[1]))
elif divisible(B*t0 + r*s0 + r*t0 - B*s0, 4*A*r):
if divisible(s0 - t0, 2*r):
if is_solution_quad(var, coeff, x_0, y_0):
l.add((x_0, y_0))
else:
_var = var
_var[0], _var[1] = _var[1], _var[0] # Interchange x and y
s = _diop_quadratic(_var, coeff, t)
while len(s) > 0:
sol = s.pop()
l.add((sol[1], sol[0]))
# (5) 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 solution in solns_pell:
for X_i in [-solution[0], solution[0]]:
for Y_i in [-solution[1], solution[1]]:
x_i, y_i = (P*Matrix([X_i, Y_i]) + Q)[0], (P*Matrix([X_i, Y_i]) + Q)[1]
if isinstance(x_i, Integer) and isinstance(y_i, Integer):
l.add((x_i, y_i))
else:
# In this case equation can be transformed into a Pell equation
#n = symbols("n", integer=True)
fund_solns = solns_pell
solns_pell = set(fund_solns)
for X, Y in fund_solns:
solns_pell.add((-X, -Y))
a = diop_DN(D, 1)
T = a[0][0]
U = a[0][1]
if (isinstance(P[0], Integer) and isinstance(P[1], Integer) and isinstance(P[2], Integer)
and isinstance(P[3], Integer) and isinstance(Q[0], Integer) and isinstance(Q[1], Integer)):
for sol in solns_pell:
r = sol[0]
s = sol[1]
x_n = S((r + s*sqrt(D))*(T + U*sqrt(D))**t + (r - s*sqrt(D))*(T - U*sqrt(D))**t)/2
y_n = S((r + s*sqrt(D))*(T + U*sqrt(D))**t - (r - s*sqrt(D))*(T - U*sqrt(D))**t)/(2*sqrt(D))
x_n = _mexpand(x_n)
y_n = _mexpand(y_n)
x_n, y_n = (P*Matrix([x_n, y_n]) + Q)[0], (P*Matrix([x_n, y_n]) + Q)[1]
l.add((x_n, y_n))
else:
L = ilcm(S(P[0]).q, ilcm(S(P[1]).q, ilcm(S(P[2]).q,
ilcm(S(P[3]).q, ilcm(S(Q[0]).q, S(Q[1]).q)))))
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):
Z = P*Matrix([X, Y]) + Q
x, y = Z[0], Z[1]
if isinstance(x, Integer) and isinstance(y, Integer):
Xt = S((X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t +
(X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t)/ 2
Yt = S((X + sqrt(D)*Y)*(T_k + sqrt(D)*U_k)**t -
(X - sqrt(D)*Y)*(T_k - sqrt(D)*U_k)**t)/ (2*sqrt(D))
Zt = P*Matrix([Xt, Yt]) + Q
l.add((Zt[0], Zt[1]))
X, Y = X*T + D*U*Y, X*U + Y*T
return l
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.
"""
x, y = var[:2]
eq = x**2*coeff[x**2] + x*y*coeff[x*y] + y**2*coeff[y**2] + x*coeff[x] + y*coeff[y] + coeff[Integer(1)]
return _mexpand(Subs(eq, (x, y), (u, v)).doit()) == 0
[docs]def diop_DN(D, N, t=symbols("t", integer=True)):
"""
Solves the equation `x^2 - Dy^2 = N`.
Mainly concerned in the case `D > 0, D` is not a perfect square, which is
the same as generalized Pell equation. To solve the generalized Pell
equation this function Uses LMM algorithm. Refer [1]_ for more details on
the algorithm.
Returns one solution for each class of the solutions. Other solutions of
the class can be constructed according to the values of ``D`` and ``N``.
Returns a list containing the solution tuples `(x, y)`.
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 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:
http://www.jpr2718.org/pell.pdf
"""
if D < 0:
if N == 0:
return [(S.Zero, S.Zero)]
elif N < 0:
return []
elif N > 0:
d = divisors(square_factor(N))
sol = []
for divisor in d:
sols = cornacchia(1, -D, N // divisor**2)
if sols:
for x, y in sols:
sol.append((divisor*x, divisor*y))
return sol
elif D == 0:
if N < 0 or not isinstance(sqrt(N), Integer):
return []
if N == 0:
return [(S.Zero, t)]
if isinstance(sqrt(N), Integer):
return [(sqrt(N), t)]
else: # D > 0
if isinstance(sqrt(D), Integer):
r = sqrt(D)
if N == 0:
return [(r*t, t)]
else:
sol = []
for y in range(floor(sign(N)*(N - 1)/(2*r)) + 1):
if isinstance(sqrt(D*y**2 + N), Integer):
sol.append((sqrt(D*y**2 + N), y))
return sol
else:
if N == 0:
return [(S.Zero, S.Zero)]
elif abs(N) == 1:
pqa = PQa(0, 1, D)
a_0 = floor(sqrt(D))
l = 0
G = []
B = []
for i in pqa:
a = i[2]
G.append(i[5])
B.append(i[4])
if l != 0 and a == 2*a_0:
break
l = l + 1
if l % 2 == 1:
if N == -1:
x = G[l-1]
y = B[l-1]
else:
count = l
while count < 2*l - 1:
i = next(pqa)
G.append(i[5])
B.append(i[4])
count = count + 1
x = G[count]
y = B[count]
else:
if N == 1:
x = G[l-1]
y = B[l-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), True)
zs = [i for i in zs if i <= abs(m) // 2 ]
if abs(m) != 2:
zs = zs + [-i for i in zs]
if S.Zero in zs:
zs.remove(S.Zero) # Remove duplicate zero
for z in zs:
pqa = PQa(z, abs(m), D)
l = 0
G = []
B = []
for i in pqa:
a = i[2]
G.append(i[5])
B.append(i[4])
if l != 0 and abs(i[1]) == 1:
r = G[l-1]
s = B[l-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
l = l + 1
if l == length(z, abs(m), D):
break
return sol
[docs]def cornacchia(a, b, m):
"""
Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`.
Uses the algorithm due to Cornacchia. The method only finds primitive
solutions, i.e. ones with `\gcd(x, y) = 1`. So this method can't 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 = 1`, only the
solutions with `x \geq y` are found. For more details, see the References.
Examples
========
>>> from sympy.solvers.diophantine import cornacchia
>>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35
set([(2, 3), (4, 1)])
>>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25
set([(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
"""
sols = set([])
a1 = igcdex(a, m)[0]
v = sqrt_mod(-b*a1, m, True)
if v is None:
return None
if not isinstance(v, list):
v = [v]
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
if isinstance(sqrt(m1), Integer):
s = sqrt(m1)
sols.add((int(r), int(s)))
return sols
[docs]def PQa(P_0, Q_0, D):
"""
Returns useful information needed to solve the Pell equation.
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 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. 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(1):
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
[docs]def diop_bf_DN(D, N, t=symbols("t", integer=True)):
"""
Uses brute force to solve the equation, `x^2 - Dy^2 = N`.
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 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. http://www.jpr2718.org/pell.pdf
"""
sol = []
a = diop_DN(D, 1)
u = a[0][0]
v = a[0][1]
if abs(N) == 1:
return diop_DN(D, N)
elif N > 1:
L1 = 0
L2 = floor(sqrt(S(N*(u - 1))/(2*D))) + 1
elif N < -1:
L1 = ceiling(sqrt(S(-N)/D))
L2 = floor(sqrt(S(-N*(u + 1))/(2*D))) + 1
else:
if D < 0:
return [(S.Zero, S.Zero)]
elif D == 0:
return [(S.Zero, t)]
else:
if isinstance(sqrt(D), Integer):
return [(sqrt(D)*t, t), (-sqrt(D)*t, t)]
else:
return [(S.Zero, S.Zero)]
for y in range(L1, L2):
if isinstance(sqrt(N + D*y**2), Integer):
x = sqrt(N + D*y**2)
sol.append((x, y))
if not equivalent(x, y, -x, y, D, N):
sol.append((-x, y))
return sol
[docs]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.
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 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. 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):
"""
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 addition of the legths 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 import length
>>> length(-2 , 4, 5) # (-2 + sqrt(5))/4
3
>>> length(-5, 4, 17) # (-5 + sqrt(17))/4
4
"""
x = P + sqrt(D)
y = Q
x = sympify(x)
v, res = [], []
q = x/y
if q < 0:
v.append(q)
res.append(floor(q))
q = q - floor(q)
num, den = rad_rationalize(1, q)
q = num / den
while 1:
v.append(q)
a = int(q)
res.append(a)
if q == a:
return len(res)
num, den = rad_rationalize(1,(q - a))
q = num / den
if q in v:
return len(res)
[docs]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.
This is used to solve the general quadratic equation by transforming it to
the latter form. Refer [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 import transformation_to_DN
>>> from sympy.solvers.diophantine import classify_diop
>>> 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, Subs
>>> 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(Subs(x**2 - 3*x*y - y**2 - 2*y + 1, (x, y), (u, v)).doit())
>>> 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.
http://www.jpr2718.org/ax2p.pdf
"""
var, coeff, diop_type = classify_diop(eq)
if diop_type == "binary_quadratic":
return _transformation_to_DN(var, coeff)
def _transformation_to_DN(var, coeff):
x, y = var[:2]
a = coeff[x**2]
b = coeff[x*y]
c = coeff[y**2]
d = coeff[x]
e = coeff[y]
f = coeff[Integer(1)]
g = igcd(a, igcd(b, igcd(c, igcd(d, igcd(e, f)))))
a = a // g
b = b // g
c = c // g
d = d // g
e = e // g
f = f // g
X, Y = symbols("X, Y", integer=True)
if b != Integer(0):
B = (S(2*a)/b).p
C = (S(2*a)/b).q
A = (S(a)/B**2).p
T = (S(a)/B**2).q
# 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, Integer(1): f*T*B}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, -S(C)/B, 0, 1])*B_0
else:
if d != Integer(0):
B = (S(2*a)/d).p
C = (S(2*a)/d).q
A = (S(a)/B**2).p
T = (S(a)/B**2).q
# 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, Integer(1): f*T - A*C**2}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [S(1)/B, 0, 0, 1])*A_0, Matrix(2, 2, [S(1)/B, 0, 0, 1])*B_0 + Matrix([-S(C)/B, 0])
else:
if e != Integer(0):
B = (S(2*c)/e).p
C = (S(2*c)/e).q
A = (S(c)/B**2).p
T = (S(c)/B**2).q
# 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, Integer(1): f*T - A*C**2}
A_0, B_0 = _transformation_to_DN([X, Y], coeff)
return Matrix(2, 2, [1, 0, 0, S(1)/B])*A_0, Matrix(2, 2, [1, 0, 0, S(1)/B])*B_0 + Matrix([0, -S(C)/B])
else:
# TODO: pre-simplification: Not necessary but may simplify
# the equation.
return Matrix(2, 2, [S(1)/a, 0, 0, 1]), Matrix([0, 0])
[docs]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 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.
http://www.jpr2718.org/ax2p.pdf
"""
var, coeff, diop_type = classify_diop(eq)
if diop_type == "binary_quadratic":
return _find_DN(var, coeff)
def _find_DN(var, coeff):
x, y = var[:2]
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[Integer(1)]
simplified = _mexpand(Subs(eq, (x, y), (u, v)).doit())
coeff = dict([reversed(t.as_independent(*[X, Y])) for t in simplified.args])
for term in [X**2, Y**2, Integer(1)]:
if term not in coeff.keys():
coeff[term] = Integer(0)
return -coeff[Y**2]/coeff[X**2], -coeff[Integer(1)]/coeff[X**2]
def check_param(x, y, a, t):
"""
Check if there is a number modulo ``a`` such that ``x`` and ``y`` are both
integers. If exist, then find a parametric representation for ``x`` and
``y``.
Here ``x`` and ``y`` are functions of ``t``.
"""
k, m, n = symbols("k, m, n", integer=True)
p = Wild("p", exclude=[k])
q = Wild("q", exclude=[k])
ok = False
for i in range(a):
z_x = _mexpand(Subs(x, t, a*k + i).doit()).match(p*k + q)
z_y = _mexpand(Subs(y, t, a*k + i).doit()).match(p*k + q)
if (isinstance(z_x[p], Integer) and isinstance(z_x[q], Integer) and
isinstance(z_y[p], Integer) and isinstance(z_y[q], Integer)):
ok = True
break
if ok == True:
x_param = x.match(p*t + q)
y_param = y.match(p*t + q)
if x_param[p] == 0 or y_param[p] == 0:
if x_param[p] == 0:
l1, junk = Poly(y).clear_denoms()
else:
l1 = 1
if y_param[p] == 0:
l2, junk = Poly(x).clear_denoms()
else:
l2 = 1
return x*ilcm(l1, l2), y*ilcm(l1, l2)
eq = S(m - x_param[q])/x_param[p] - S(n - y_param[q])/y_param[p]
lcm_denom, junk = Poly(eq).clear_denoms()
eq = eq * lcm_denom
return diop_solve(eq, t)[0], diop_solve(eq, t)[1]
else:
return (None, None)
[docs]def diop_ternary_quadratic(eq):
"""
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 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)
if diop_type == "homogeneous_ternary_quadratic":
return _diop_ternary_quadratic(var, coeff)
def _diop_ternary_quadratic(_var, coeff):
x, y, z = _var[:3]
var = [x]*3
var[0], var[1], var[2] = _var[0], _var[1], _var[2]
# 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
# unobviuos 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|
if coeff[x**2] == 0 and coeff[y**2] == 0 and coeff[z**2] == 0:
if coeff[x*z] != 0:
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:
if abs(r[0]) + abs(r[1]) < min_sum:
s = r
min_sum = abs(s[0]) + abs(s[1])
x_0, y_0, z_0 = s[0], -coeff[x*z], s[1]
else:
var[0], var[1] = _var[1], _var[0]
y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff)
return simplified(x_0, y_0, z_0)
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 = _diop_ternary_quadratic(var, coeff)
else:
var[0], var[1] = _var[1], _var[0]
y_0, x_0, z_0 = _diop_ternary_quadratic(var, coeff)
else:
if coeff[x*y] != 0 or coeff[x*z] != 0:
# 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 = dict()
_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 = _diop_ternary_quadratic(var, _coeff)
if X_0 == None:
return (None, None, None)
l = (S(B*y_0 + C*z_0)/(2*A)).q
x_0, y_0, z_0 = X_0*l - (S(B*y_0 + C*z_0)/(2*A)).p, y_0*l, z_0*l
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 = (S(-E)/A).p
a = (S(-E)/A).q
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 = _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 = _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 = _diop_ternary_quadratic_normal(var, coeff)
return simplified(x_0, y_0, z_0)
def transformation_to_normal(eq):
"""
Returns the transformation Matrix from general ternary quadratic equation
`eq` to normal form.
General form of the ternary quadratic equation is `ax^2 + by^2 cz^2 + dxy +
eyz + fxz`. This function returns a 3X3 transformation Matrix which
transforms the former equation to the form `ax^2 + by^2 + cz^2 = 0`. This
is not used in solving ternary quadratics. Only implemented for the sake
of completeness.
"""
var, coeff, diop_type = classify_diop(eq)
if diop_type == "homogeneous_ternary_quadratic":
return _transformation_to_normal(var, coeff)
def _transformation_to_normal(var, coeff):
_var = [var[0]]*3
_var[1], _var[2] = var[1], var[2]
x, y, z = var[:3]
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
else:
# 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 = dict()
_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 = _transformtion_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(3, 3, [1, 0, 0, 0, 1, 0, 0, 0, 1])
[docs]def simplified(x, y, z):
"""
Simplify the solution `(x, y, z)`.
"""
if x == None or y == None or z == None:
return (x, y, z)
g = igcd(x, igcd(y, z))
return x // g, y // g, z // g
[docs]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.abc import x, y, z
>>> from sympy.solvers.diophantine import parametrize_ternary_quadratic
>>> parametrize_ternary_quadratic(x**2 + y**2 - z**2)
(2*p*q, p**2 - q**2, p**2 + q**2)
Here `p` and `q` are two co-prime integers.
>>> parametrize_ternary_quadratic(3*x**2 + 2*y**2 - z**2 - 2*x*y + 5*y*z - 7*y*z)
(2*p**2 - 2*p*q - q**2, 2*p**2 + 2*p*q - q**2, 2*p**2 - 2*p*q + 3*q**2)
>>> parametrize_ternary_quadratic(124*x**2 - 30*y**2 - 7729*z**2)
(-1410*p**2 - 363263*q**2, 2700*p**2 + 30916*p*q - 695610*q**2, -60*p**2 + 5400*p*q + 15458*q**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)
if diop_type == "homogeneous_ternary_quadratic":
x_0, y_0, z_0 = _diop_ternary_quadratic(var, coeff)
return _parametrize_ternary_quadratic((x_0, y_0, z_0), var, coeff)
def _parametrize_ternary_quadratic(solution, _var, coeff):
x, y, z = _var[:3]
x_0, y_0, z_0 = solution[:3]
v = [x]*3
v[0], v[1], v[2] = _var[0], _var[1], _var[2]
if x_0 == None:
return (None, None, None)
if x_0 == 0:
if y_0 == 0:
v[0], v[2] = v[2], v[0]
z_p, y_p, x_p = _parametrize_ternary_quadratic((z_0, y_0, x_0), v, coeff)
return x_p, y_p, z_p
else:
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[:3]
r, p, q = symbols("r, p, q", integer=True)
eq = x**2*coeff[x**2] + y**2*coeff[y**2] + z**2*coeff[z**2] + x*y*coeff[x*y] + y*z*coeff[y*z] + z*x*coeff[z*x]
eq_1 = Subs(eq, (x, y, z), (r*x_0, r*y_0 + p, r*z_0 + q)).doit()
eq_1 = _mexpand(eq_1)
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 x, y, z
[docs]def diop_ternary_quadratic_normal(eq):
"""
Solves the quadratic ternary diophantine equation,
`ax^2 + by^2 + cz^2 = 0`.
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 satisifes 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 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)
if diop_type == "homogeneous_ternary_quadratic":
return _diop_ternary_quadratic_normal(var, coeff)
def _diop_ternary_quadratic_normal(var, coeff):
x, y, z = var[:3]
a = coeff[x**2]
b = coeff[y**2]
c = coeff[z**2]
if a*b*c == 0:
raise ValueError("Try factoring out you equation or using diophantine()")
g = igcd(a, igcd(b, c))
a = a // g
b = b // g
c = c // g
a_0 = square_factor(a)
b_0 = square_factor(b)
c_0 = square_factor(c)
a_1 = a // a_0**2
b_1 = b // b_0**2
c_1 = c // c_0**2
a_2, b_2, c_2 = pairwise_prime(a_1, b_1, c_1)
A = -a_2*c_2
B = -b_2*c_2
# If following two conditions are satisified then there are no solutions
if A < 0 and B < 0:
return (None, None, None)
if (sqrt_mod(-b_2*c_2, a_2) == None or sqrt_mod(-c_2*a_2, b_2) == None or
sqrt_mod(-a_2*b_2, c_2) == None):
return (None, None, None)
z_0, x_0, y_0 = descent(A, B)
if divisible(z_0, c_2) == True:
z_0 = z_0 // abs(c_2)
else:
x_0 = x_0*(S(z_0)/c_2).q
y_0 = y_0*(S(z_0)/c_2).q
z_0 = (S(z_0)/c_2).p
x_0, y_0, z_0 = simplified(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)
l = ilcm(a_0, ilcm(b_0, c_0))
x_0 = abs(x_0*l//a_0)
y_0 = abs(y_0*l//b_0)
z_0 = abs(z_0*l//c_0)
return simplified(x_0, y_0, z_0)
[docs]def square_factor(a):
"""
Returns an integer `c` s.t. `a = c^2k, \ c,k \in Z`. Here `k` is square
free.
Examples
========
>>> from sympy.solvers.diophantine import square_factor
>>> square_factor(24)
2
>>> square_factor(36)
6
>>> square_factor(1)
1
"""
f = factorint(abs(a))
c = 1
for p, e in f.items():
c = c * p**(e//2)
return c
[docs]def pairwise_prime(a, b, c):
"""
Transform `ax^2 + by^2 + cz^2 = 0` into an equivalent equation
`a'x^2 + b'y^2 + c'z^2 = 0` where `a', b', c'` are pairwise relatively
prime.
Returns a tuple containing `a', b', c'`. `\gcd(a, b, c)` should equal `1`
for this to work. 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 import pairwise_prime
>>> pairwise_prime(6, 15, 10)
(5, 2, 3)
See Also
========
make_prime(), reocnstruct()
"""
a, b, c = make_prime(a, b, c)
b, c, a = make_prime(b, c, a)
c, a, b = make_prime(c, a, b)
return a, b, c
[docs]def make_prime(a, b, c):
"""
Transform the equation `ax^2 + by^2 + cz^2 = 0` to an equivalent equation
`a'x^2 + b'y^2 + c'z^2 = 0` with `\gcd(a', b') = 1`.
Returns a tuple `(a', b', c')` which satisfies above conditions. Note that
in the returned tuple `\gcd(a', c')` and `\gcd(b', c')` can take any value.
Examples
========
>>> from sympy.solvers.diophantine import make_prime
>>> make_prime(4, 2, 7)
(2, 1, 14)
See Also
========
pairwaise_prime(), reconstruct()
"""
g = igcd(a, b)
if g != 1:
f = factorint(g)
for p, e in f.items():
a = a // p**e
b = b // p**e
if e % 2 == 1:
c = p*c
return a, b, c
[docs]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 a transformed version of the above
equation.
"""
g = igcd(a, b)
if g != 1:
f = factorint(g)
for p, e in f.items():
if e %2 == 0:
z = z*p**(e//2)
else:
z = z*p**((e//2)+1)
return z
[docs]def ldescent(A, B):
"""
Uses Lagrange's method to find a non trivial solution to
`w^2 = Ax^2 + By^2`.
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 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,
Mathematics of Computation, Volume 00, Number 0.
"""
if abs(A) > abs(B):
w, y, x = ldescent(B, A)
return w, x, y
if A == 1:
return (S.One, S.One, 0)
if B == 1:
return (S.One, 0, S.One)
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:
if isinstance(sqrt(abs(Q) // i), Integer):
B_0, d = sign(Q)*i, sqrt(abs(Q) // i)
break
if B_0 != None:
W, X, Y = ldescent(A, B_0)
return simplified((-A*X + r*W), (r*X - W), Y*(B_0*d))
# In this module Descent will always be called with inputs which have solutions.
[docs]def descent(A, B):
"""
Lagrange's `descent()` with lattice-reduction to find solutions to
`x^2 = Ay^2 + Bz^2`.
Here `A` and `B` should be square free and pairwise prime. Always should be
called with suitable ``A`` and ``B`` so that the above equation has
solutions.
This is more faster than the normal Lagrange's descent algorithm because
the gaussian reduction is used.
Examples
========
>>> from sympy.solvers.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 == -1:
return (None, None, None)
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 simplified(x_0*x_1 + A*z_0*z_1, z_0*x_1 + x_0*z_1, t_1*t_2*y_1)
[docs]def gaussian_reduce(w, a, b):
"""
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:
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):
"""
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[:2]
v_1, v_2 = v[:2]
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):
"""
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[:2]
return sqrt(dot((u_1, u_2), (u_1, u_2), w, a, b))
[docs]def holzer(x_0, y_0, z_0, a, b, c):
"""
Simplify the solution `(x_{0}, y_{0}, z_{0})` of the equation
`ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z_{0}^2 \geq \mid ab \mid` to
a new reduced solution `(x, y, z)` such that `z^2 \leq \mid ab \mid`.
"""
while z_0 > sqrt(a*b):
if c % 2 == 0:
k = c // 2
u_0, v_0 = base_solution_linear(k, y_0, -x_0)
else:
k = 2*c
u_0, v_0 = base_solution_linear(c, y_0, -x_0)
w = -(a*u_0*x_0 + b*v_0*y_0) // (c*z_0)
if c % 2 == 1:
if w % 2 != (a*u_0 + b*v_0) % 2:
w = w + 1
x = (x_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*u_0*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k
y = (y_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*v_0*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k
z = (z_0*(a*u_0**2 + b*v_0**2 + c*w**2) - 2*w*(a*u_0*x_0 + b*v_0*y_0 + c*w*z_0)) // k
x_0, y_0, z_0 = x, y, z
return x_0, y_0, z_0
[docs]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 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)
if diop_type == "general_pythagorean":
return _diop_general_pythagorean(var, coeff, param)
def _diop_general_pythagorean(var, coeff, t):
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] * -1
n = len(var)
index = 0
for i, v in enumerate(var):
if sign(coeff[v**2]) == -1:
index = i
m = symbols(str(t) + "1:" + str(n), integer=True)
l = []
ith = 0
for m_i in m:
ith = ith + m_i**2
l.append(ith - 2*m[n - 2]**2)
for i in range(n - 2):
l.append(2*m[i]*m[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:
lcm = ilcm(lcm, sqrt(coeff[v**2]) if sqrt(coeff[v**2]) % 2 else sqrt(coeff[v**2]) // 2)
for i, v in enumerate(var):
sol[i] = (lcm*sol[i]) / sqrt(abs(coeff[v**2]))
return tuple(sol)
[docs]def diop_general_sum_of_squares(eq, limit=1):
"""
Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`.
Returns at most ``limit`` number of solutions. Currently there is no way to
set ``limit`` using higher level API's like ``diophantine()`` or
``diop_solve()`` but that will be fixed soon.
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`. At most ``limit`` number of
solutions are returned.
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 import diop_general_sum_of_squares
>>> from sympy.abc import a, b, c, d, e, f
>>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345)
set([(0, 48, 5, 4, 0)])
Reference
=========
.. [1] Representing an Integer as a sum of three squares, [online],
Available:
http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares
"""
var, coeff, diop_type = classify_diop(eq)
if diop_type == "general_sum_of_squares":
return _diop_general_sum_of_squares(var, coeff, limit)
def _diop_general_sum_of_squares(var, coeff, limit=1):
n = len(var)
k = -int(coeff[Integer(1)])
s = set([])
if k < 0:
return set([])
if n == 3:
s.add(sum_of_three_squares(k))
elif n == 4:
s.add(sum_of_four_squares(k))
else:
m = n // 4
f = partition(k, m, True)
for j in range(limit):
soln = []
try:
l = next(f)
except StopIteration:
break
for n_i in l:
a, b, c, d = sum_of_four_squares(n_i)
soln = soln + [a, b, c, d]
soln = soln + [0] * (n % 4)
s.add(tuple(soln))
return s
## Functions below this comment can be more suitably grouped under an Additive number theory module
## rather than the Diophantine equation module.
[docs]def partition(n, k=None, zeros=False):
"""
Returns a generator that can be used to generate partitions of an integer
`n`.
A partition of `n` is a set of positive integers which add upto `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 there are no partions of `n` with size `k` then an empty tuple
is 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 import partition
>>> f = partition(5)
>>> next(f)
(1, 1, 1, 1, 1)
>>> next(f)
(1, 1, 1, 2)
>>> g = partition(5, 3)
>>> next(g)
(3, 1, 1)
>>> next(g)
(2, 2, 1)
Reference
=========
.. [1] Generating Integer Partitions, [online],
Available: http://jeromekelleher.net/partitions.php
"""
if n < 1:
yield tuple()
if k is not None:
if k < 1:
yield tuple()
elif k > n:
if zeros:
for i in range(1, n):
for t in partition(n, i):
yield (t,) + (0,) * (k - i)
else:
yield tuple()
else:
a = [1 for i in range(k)]
a[0] = n - k + 1
yield tuple(a)
i = 1
while a[0] >= n // k + 1:
j = 0
while j < i and j + 1 < k:
a[j] = a[j] - 1
a[j + 1] = a[j + 1] + 1
yield tuple(a)
j = j + 1
i = i + 1
if zeros:
for m in range(1, k):
for a in partition(n, m):
yield tuple(a) + (0,) * (k - m)
else:
a = [0 for i in range(n + 1)]
l = 1
y = n - 1
while l != 0:
x = a[l - 1] + 1
l -= 1
while 2*x <= y:
a[l] = x
y -= x
l += 1
m = l + 1
while x <= y:
a[l] = x
a[m] = y
yield tuple(a[:l + 2])
x += 1
y -= 1
a[l] = x + y
y = x + y - 1
yield tuple(a[:l + 1])
[docs]def prime_as_sum_of_two_squares(p):
"""
Represent a prime `p` which is congruent to 1 mod 4, as a sum of two
squares.
Examples
========
>>> from sympy.solvers.diophantine import prime_as_sum_of_two_squares
>>> prime_as_sum_of_two_squares(5)
(2, 1)
Reference
=========
.. [1] Representing a number as a sum of four squares, [online],
Available: http://www.schorn.ch/howto.html
"""
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 (b, a % b)
[docs]def sum_of_three_squares(n):
"""
Returns a 3-tuple `(a, b, c)` such that `a^2 + b^2 + c^2 = n` and
`a, b, c \geq 0`.
Returns (None, None, None) if `n = 4^a(8m + 7)` for some `a, m \in Z`. See
[1]_ for more details.
Usage
=====
``sum_of_three_squares(n)``: Here ``n`` is a non-negative integer.
Examples
========
>>> from sympy.solvers.diophantine import sum_of_three_squares
>>> sum_of_three_squares(44542)
(207, 37, 18)
References
==========
.. [1] Representing a number as a sum of three squares, [online],
Available: http://www.schorn.ch/howto.html
"""
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)
while n % 4 == 0:
v = v + 1
n = n // 4
if n % 8 == 7:
return (None, None, None)
if n in special.keys():
x, y, z = special[n]
return (2**v*x, 2**v*y, 2**v*z)
l = int(sqrt(n))
if n == l**2:
return (2**v*l, 0, 0)
x = None
if n % 8 == 3:
l = l if l % 2 else l - 1
for i in range(l, -1, -2):
if isprime((n - i**2) // 2):
x = i
break
y, z = prime_as_sum_of_two_squares((n - x**2) // 2)
return (2**v*x, 2**v*(y + z), 2**v*abs(y - z))
if n % 8 == 2 or n % 8 == 6:
l = l if l % 2 else l - 1
else:
l = l - 1 if l % 2 else l
for i in range(l, -1, -2):
if isprime(n - i**2):
x = i
break
y, z = prime_as_sum_of_two_squares(n - x**2)
return (2**v*x, 2**v*y, 2**v*z)
[docs]def sum_of_four_squares(n):
"""
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 import sum_of_four_squares
>>> sum_of_four_squares(3456)
(8, 48, 32, 8)
>>> sum_of_four_squares(1294585930293)
(0, 1137796, 2161, 1234)
References
==========
.. [1] Representing a number as a sum of four squares, [online],
Available: http://www.schorn.ch/howto.html
"""
if n == 0:
return (0, 0, 0, 0)
v = 0
while n % 4 == 0:
v = v + 1
n = n // 4
if n % 8 == 7:
d = 2
n = n - 4
elif n % 8 == 6 or n % 8 == 2:
d = 1
n = n - 1
else:
d = 0
x, y, z = sum_of_three_squares(n)
return (2**v*d, 2**v*x, 2**v*y, 2**v*z)
def power_representation(n, p, k, zeros=False):
"""
Returns a generator for finding k-tuples `(n_{1}, n_{2}, . . . n_{k})` such
that `n = n_{1}^p + n_{2}^p + . . . n_{k}^p`.
Here `n` is a non-negative integer. StopIteration exception is raised after
all the solutions are generated, so should always be used within a try-
catch block.
Usage
=====
``power_representation(n, p, k, zeros)``: Represent number ``n`` as a sum
of ``k``, ``p``th powers. If ``zeros`` is true, then the solutions will
contain zeros.
Examples
========
>>> from sympy.solvers.diophantine import power_representation
>>> f = power_representation(1729, 3, 2) # Represent 1729 as a sum of two cubes
>>> next(f)
(12, 1)
>>> next(f)
(10, 9)
"""
if p < 1 or k < 1 or n < 1:
raise ValueError("Expected: n > 0 and k >= 1 and p >= 1")
if k == 1:
if perfect_power(n):
yield (perfect_power(n)[0],)
else:
yield tuple()
elif p == 1:
for t in partition(n, k, zeros):
yield t
else:
l = []
a = integer_nthroot(n, p)[0]
for t in pow_rep_recursive(a, k, n, [], p):
yield t
if zeros:
for i in range(2, k):
for t in pow_rep_recursive(a, i, n, [], p):
yield t + (0,) * (k - i)
def pow_rep_recursive(n_i, k, n_remaining, terms, p):
if k == 0 and n_remaining == 0:
yield tuple(terms)
else:
if n_i >= 1 and k > 0 and n_remaining >= 0:
if n_i**p <= n_remaining:
for t in pow_rep_recursive(n_i, k - 1, n_remaining - n_i**p, terms + [n_i], p):
yield t
for t in pow_rep_recursive(n_i - 1, k, n_remaining, terms, p):
yield t
