Source code for sympy.solvers.diophantine

from __future__ import print_function, division

from sympy import (degree_list, Poly, igcd, divisors, sign, symbols, S, Integer, Wild, Symbol, factorint,
    Add, Mul, solve, ceiling, floor, sqrt, sympify, Subs, ilcm, Matrix, factor_list, perfect_power,
    isprime, nextprime, integer_nthroot, Expr, Pow)

from sympy.simplify.simplify import rad_rationalize, _mexpand
from sympy.ntheory.modular import solve_congruence
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 xrange
from sympy.core.relational import Eq
from sympy.solvers.solvers import check_assumptions

__all__ = ['diophantine', 'diop_solve', 'classify_diop', 'diop_linear', 'base_solution_linear',
'diop_quadratic', 'diop_DN', 'cornacchia', 'diop_bf_DN', 'transformation_to_DN', 'find_DN',
'diop_ternary_quadratic',  'square_factor', 'descent', 'diop_general_pythagorean',
'diop_general_sum_of_squares', 'partition', 'sum_of_three_squares', 'sum_of_four_squares']


[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, -t), (t, -t)]) #>>> 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 - 5, -2*t + 5) >>> diop_solve(4*x + 3*y -4*z + 5) (3*t + 4*z - 5, -4*t - 4*z + 5, z) >>> diop_solve(x + 3*y - 4*z + w -6) (t, -t - 3*y + 4*z + 6, y, z) >>> 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) var = list(eq.free_symbols) var.sort(key=default_sort_key) coeff = {} diop_type = None coeff = dict([reversed(t.as_independent(*var)) for t in eq.args]) for v in coeff: if not isinstance(coeff[v], Integer): raise TypeError("Coefficients should be Integers") if len(var) == 1: diop_type = "univariate" elif Poly(eq).total_degree() == 1: diop_type = "linear" 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 - 5, -2*t - 5) Here x = -3*t - 5 and y = -2*t - 5 >>> diop_linear(2*x - 3*y - 4*z -3) (-3*t - 4*z - 3, -2*t - 4*z - 3, z) 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): x, y = var[:2] a = coeff[x] b = coeff[y] if len(var) == len(coeff): c = 0 else: c = -coeff[Integer(1)] if len(var) == 2: sol_x, sol_y = base_solution_linear(c, a, b, param) return (sol_x, sol_y) elif len(var) > 2: X = [] Y = [] for v in var[2:]: sol_x, sol_y = base_solution_linear(-coeff[v], a, b) X.append(sol_x*v) Y.append(sol_y*v) sol_x, sol_y = base_solution_linear(c, a, b, param) X.append(sol_x) Y.append(sol_y) l = [] if None not in X and None not in Y: l.append(Add(*X)) l.append(Add(*Y)) for v in var[2:]: l.append(v) else: for v in var: l.append(None) return tuple(l)
[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 xrange(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) 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 = 0 done = False T_k = T U_k = U while not done: k = k + 1 if (T_k - 1) % L == 0 and U_k % L == 0: done = True T_k, U_k = T_k*T + D*U_k*U, T_k*U + U_k*T for soln in solns_pell: X = soln[0] Y = soln[1] done = False for i in xrange(k): X_1 = X*T + D*U*Y Y_1 = X*U + Y*T x = (P*Matrix([X_1, Y_1]) + Q)[0] y = (P*Matrix([X_1, Y_1]) + Q)[1] if is_solution_quad(var, coeff, x, y): done = True x_n = S( (X_1 + sqrt(D)*Y_1)*(T + sqrt(D)*U)**(t*L) + (X_1 - sqrt(D)*Y_1)*(T - sqrt(D)*U)**(t*L) )/ 2 y_n = S( (X_1 + sqrt(D)*Y_1)*(T + sqrt(D)*U)**(t*L) - (X_1 - sqrt(D)*Y_1)*(T - sqrt(D)*U)**(t*L) )/ (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)) if done: break 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 xrange(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 xrange(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 xrange(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 xrange(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 xrange(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://homepages.ed.ac.uk/jkellehe/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 xrange(1, n): for t in partition(n, i): yield (t,) + (0,) * (k - i) else: yield tuple() else: a = [1 for i in xrange(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 xrange(1, k): for a in partition(n, m): yield tuple(a) + (0,) * (k - m) else: a = [0 for i in xrange(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 xrange(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 xrange(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 xrange(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