def roots_quintic(f):
"""
Calulate exact roots of a solvable quintic
"""
result = []
coeff_5, coeff_4, p, q, r, s = f.all_coeffs()
# Eqn must be of the form x^5 + px^3 + qx^2 + rx + s
if coeff_4:
return result
if coeff_5 != 1:
l = [p / coeff_5, q / coeff_5, r / coeff_5, s / coeff_5]
if not all(coeff.is_Rational for coeff in l):
return result
f = Poly(f / coeff_5)
quintic = PolyQuintic(f)
# Eqn standardized. Algo for solving starts here
if not f.is_irreducible:
return result
f20 = quintic.f20
# Check if f20 has linear factors over domain Z
if f20.is_irreducible:
return result
# Now, we know that f is solvable
for _factor in f20.factor_list()[1]:
if _factor[0].is_linear:
theta = _factor[0].root(0)
break
d = discriminant(f)
delta = sqrt(d)
# zeta = a fifth root of unity
zeta1, zeta2, zeta3, zeta4 = quintic.zeta
T = quintic.T(theta, d)
tol = S(1e-10)
alpha = T[1] + T[2] * delta
alpha_bar = T[1] - T[2] * delta
beta = T[3] + T[4] * delta
beta_bar = T[3] - T[4] * delta
disc = alpha**2 - 4 * beta
disc_bar = alpha_bar**2 - 4 * beta_bar
l0 = quintic.l0(theta)
l1 = _quintic_simplify((-alpha + sqrt(disc)) / S(2))
l4 = _quintic_simplify((-alpha - sqrt(disc)) / S(2))
l2 = _quintic_simplify((-alpha_bar + sqrt(disc_bar)) / S(2))
l3 = _quintic_simplify((-alpha_bar - sqrt(disc_bar)) / S(2))
order = quintic.order(theta, d)
test = (order * delta.n()) - ((l1.n() - l4.n()) * (l2.n() - l3.n()))
# Comparing floats
# Problems importing on top
from sympy.utilities.randtest import comp
if not comp(test, 0, tol):
l2, l3 = l3, l2
# Now we have correct order of l's
R1 = l0 + l1 * zeta1 + l2 * zeta2 + l3 * zeta3 + l4 * zeta4
R2 = l0 + l3 * zeta1 + l1 * zeta2 + l4 * zeta3 + l2 * zeta4
R3 = l0 + l2 * zeta1 + l4 * zeta2 + l1 * zeta3 + l3 * zeta4
R4 = l0 + l4 * zeta1 + l3 * zeta2 + l2 * zeta3 + l1 * zeta4
Res = [None, [None] * 5, [None] * 5, [None] * 5, [None] * 5]
Res_n = [None, [None] * 5, [None] * 5, [None] * 5, [None] * 5]
sol = Symbol('sol')
# Simplifying improves performace a lot for exact expressions
R1 = _quintic_simplify(R1)
R2 = _quintic_simplify(R2)
R3 = _quintic_simplify(R3)
R4 = _quintic_simplify(R4)
# Solve imported here. Causing problems if imported as 'solve'
# and hence the changed name
from sympy.solvers.solvers import solve as _solve
a, b = symbols('a b', cls=Dummy)
_sol = _solve(sol**5 - a - I * b, sol)
for i in range(5):
_sol[i] = factor(_sol[i])
R1 = R1.as_real_imag()
R2 = R2.as_real_imag()
R3 = R3.as_real_imag()
R4 = R4.as_real_imag()
for i, root in enumerate(_sol):
Res[1][i] = _quintic_simplify(root.subs({a: R1[0], b: R1[1]}))
Res[2][i] = _quintic_simplify(root.subs({a: R2[0], b: R2[1]}))
Res[3][i] = _quintic_simplify(root.subs({a: R3[0], b: R3[1]}))
Res[4][i] = _quintic_simplify(root.subs({a: R4[0], b: R4[1]}))
for i in range(1, 5):
for j in range(5):
Res_n[i][j] = Res[i][j].n()
Res[i][j] = _quintic_simplify(Res[i][j])
r1 = Res[1][0]
r1_n = Res_n[1][0]
for i in range(5):
if comp(im(r1_n * Res_n[4][i]), 0, tol):
r4 = Res[4][i]
break
u, v = quintic.uv(theta, d)
sqrt5 = math.sqrt(5)
# Now we have various Res values. Each will be a list of five
# values. We have to pick one r value from those five for each Res
u, v = quintic.uv(theta, d)
testplus = (u + v * delta * sqrt(5)).n()
testminus = (u - v * delta * sqrt(5)).n()
# Evaluated numbers suffixed with _n
# We will use evaluated numbers for calculation. Much faster.
r4_n = r4.n()
r2 = r3 = None
for i in range(5):
r2temp_n = Res_n[2][i]
for j in range(5):
# Again storing away the exact number and using
# evaluated numbers in computations
r3temp_n = Res_n[3][j]
if (comp(r1_n * r2temp_n**2 + r4_n * r3temp_n**2 - testplus, 0,
tol) and comp(
r3temp_n * r1_n**2 + r2temp_n * r4_n**2 - testminus,
0, tol)):
r2 = Res[2][i]
r3 = Res[3][j]
break
if r2:
break
# Now, we have r's so we can get roots
x1 = (r1 + r2 + r3 + r4) / 5
x2 = (r1 * zeta4 + r2 * zeta3 + r3 * zeta2 + r4 * zeta1) / 5
x3 = (r1 * zeta3 + r2 * zeta1 + r3 * zeta4 + r4 * zeta2) / 5
x4 = (r1 * zeta2 + r2 * zeta4 + r3 * zeta1 + r4 * zeta3) / 5
x5 = (r1 * zeta1 + r2 * zeta2 + r3 * zeta3 + r4 * zeta4) / 5
result = [x1, x2, x3, x4, x5]
# Now check if solutions are distinct
result_n = []
for root in result:
result_n.append(root.n(5))
result_n = sorted(result_n, key=default_sort_key)
prev_entry = None
for r in result_n:
if r == prev_entry:
# Roots are identical. Abort. Return []
# and fall back to usual solve
return []
prev_entry = r
return result
def roots_quintic(f):
"""
Calulate exact roots of a solvable quintic
"""
result = []
coeff_5, coeff_4, p, q, r, s = f.all_coeffs()
# Eqn must be of the form x^5 + px^3 + qx^2 + rx + s
if coeff_4:
return result
if coeff_5 != 1:
l = [p/coeff_5, q/coeff_5, r/coeff_5, s/coeff_5]
if not all(coeff.is_Rational for coeff in l):
return result
f = Poly(f/coeff_5)
quintic = PolyQuintic(f)
# Eqn standardised. Algo for solving starts here
if not f.is_irreducible:
return result
f20 = quintic.f20
# Check if f20 has linear factors over domain Z
if f20.is_irreducible:
return result
# Now, we know that f is solvable
for _factor in f20.factor_list()[1]:
if _factor[0].is_linear:
theta = _factor[0].root(0)
break
d = discriminant(f)
delta = sqrt(d)
# zeta = a fifth root of unity
zeta1, zeta2, zeta3, zeta4 = quintic.zeta
T = quintic.T(theta, d)
tol = S(1e-10)
alpha = T[1] + T[2]*delta
alpha_bar = T[1] - T[2]*delta
beta = T[3] + T[4]*delta
beta_bar = T[3] - T[4]*delta
disc = alpha**2 - 4*beta
disc_bar = alpha_bar**2 - 4*beta_bar
l0 = quintic.l0(theta)
l1 = _quintic_simplify((-alpha + sqrt(disc)) / S(2))
l4 = _quintic_simplify((-alpha - sqrt(disc)) / S(2))
l2 = _quintic_simplify((-alpha_bar + sqrt(disc_bar)) / S(2))
l3 = _quintic_simplify((-alpha_bar - sqrt(disc_bar)) / S(2))
order = quintic.order(theta, d)
test = (order*delta.n()) - ( (l1.n() - l4.n())*(l2.n() - l3.n()) )
# Comparing floats
# Problems importing on top
from sympy.utilities.randtest import comp
if not comp(test, 0, tol):
l2, l3 = l3, l2
# Now we have correct order of l's
R1 = l0 + l1*zeta1 + l2*zeta2 + l3*zeta3 + l4*zeta4
R2 = l0 + l3*zeta1 + l1*zeta2 + l4*zeta3 + l2*zeta4
R3 = l0 + l2*zeta1 + l4*zeta2 + l1*zeta3 + l3*zeta4
R4 = l0 + l4*zeta1 + l3*zeta2 + l2*zeta3 + l1*zeta4
Res = [None, [None]*5, [None]*5, [None]*5, [None]*5]
Res_n = [None, [None]*5, [None]*5, [None]*5, [None]*5]
sol = Symbol('sol')
# Simplifying improves performace a lot for exact expressions
R1 = _quintic_simplify(R1)
R2 = _quintic_simplify(R2)
R3 = _quintic_simplify(R3)
R4 = _quintic_simplify(R4)
# Solve imported here. Causing problems if imported as 'solve'
# and hence the changed name
from sympy.solvers.solvers import solve as _solve
a, b = symbols('a b', cls=Dummy)
_sol = _solve( sol**5 - a - I*b, sol)
for i in range(5):
_sol[i] = factor(_sol[i])
R1 = R1.as_real_imag()
R2 = R2.as_real_imag()
R3 = R3.as_real_imag()
R4 = R4.as_real_imag()
for i, root in enumerate(_sol):
Res[1][i] = _quintic_simplify(root.subs({ a: R1[0], b: R1[1] }))
Res[2][i] = _quintic_simplify(root.subs({ a: R2[0], b: R2[1] }))
Res[3][i] = _quintic_simplify(root.subs({ a: R3[0], b: R3[1] }))
Res[4][i] = _quintic_simplify(root.subs({ a: R4[0], b: R4[1] }))
for i in range(1, 5):
for j in range(5):
Res_n[i][j] = Res[i][j].n()
Res[i][j] = _quintic_simplify(Res[i][j])
r1 = Res[1][0]
r1_n = Res_n[1][0]
for i in range(5):
if comp(im(r1_n*Res_n[4][i]), 0, tol):
r4 = Res[4][i]
break
u, v = quintic.uv(theta, d)
sqrt5 = math.sqrt(5)
# Now we have various Res values. Each will be a list of five
# values. We have to pick one r value from those five for each Res
u, v = quintic.uv(theta, d)
testplus = (u + v*delta*sqrt(5)).n()
testminus = (u - v*delta*sqrt(5)).n()
# Evaluated numbers suffixed with _n
# We will use evaluated numbers for calculation. Much faster.
r4_n = r4.n()
r2 = r3 = None
for i in range(5):
r2temp_n = Res_n[2][i]
for j in range(5):
# Again storing away the exact number and using
# evaluated numbers in computations
r3temp_n = Res_n[3][j]
if( comp( r1_n*r2temp_n**2 + r4_n*r3temp_n**2 - testplus, 0, tol) and
comp( r3temp_n*r1_n**2 + r2temp_n*r4_n**2 - testminus, 0, tol ) ):
r2 = Res[2][i]
r3 = Res[3][j]
break
if r2:
break
# Now, we have r's so we can get roots
x1 = (r1 + r2 + r3 + r4)/5
x2 = (r1*zeta4 + r2*zeta3 + r3*zeta2 + r4*zeta1)/5
x3 = (r1*zeta3 + r2*zeta1 + r3*zeta4 + r4*zeta2)/5
x4 = (r1*zeta2 + r2*zeta4 + r3*zeta1 + r4*zeta3)/5
x5 = (r1*zeta1 + r2*zeta2 + r3*zeta3 + r4*zeta4)/5
result = [x1, x2, x3, x4, x5]
# Now check if solutions are distinct
result_n = []
for root in result:
result_n.append(root.n(5))
result_n = sorted(result_n)
prev_entry = None
for r in result_n:
if r == prev_entry:
# Roots are identical. Abort. Return []
# and fall back to usual solve
return []
prev_entry = r
return result
def N_equals(a, b):
"""Check whether two complex numbers are numerically close"""
return comp(a.n(), b.n(), 1.e-6)
