def test_dispersion():
pytest.raises(ValueError,
lambda: dispersionset(poly(x * y, x, y), poly(x, x)))
pytest.raises(ValueError, lambda: dispersionset(poly(x, x), poly(y, y)))
fp = poly(0, x)
assert sorted(dispersionset(fp)) == [0]
fp = poly(2, x)
assert sorted(dispersionset(fp)) == [0]
fp = poly(x + 1, x)
assert sorted(dispersionset(fp)) == [0]
assert dispersion(fp) == 0
fp = poly((x + 1) * (x + 2), x)
assert sorted(dispersionset(fp)) == [0, 1]
assert dispersion(fp) == 1
fp = poly(x * (x + 3), x)
assert sorted(dispersionset(fp)) == [0, 3]
assert dispersion(fp) == 3
fp = poly((x - 3) * (x + 3), x)
assert sorted(dispersionset(fp)) == [0, 6]
assert dispersion(fp) == 6
fp = poly(x**4 - 3 * x**2 + 1, x)
gp = fp.shift(-3)
assert sorted(dispersionset(fp, gp)) == [2, 3, 4]
assert dispersion(fp, gp) == 4
assert sorted(dispersionset(gp, fp)) == []
assert dispersion(gp, fp) == -oo
fp = poly(x**2 + 2 * x - 1, x)
gp = poly(x**2 + 2 * x + 3, x)
assert dispersionset(fp, gp) == set()
fp = poly(x * (3 * x**2 + a) * (x - 2536) * (x**3 + a), x)
gp = fp.as_expr().subs({x: x - 345}).as_poly(x)
assert sorted(dispersionset(fp, gp)) == [345, 2881]
assert sorted(dispersionset(gp, fp)) == [2191]
gp = poly((x - 2)**2 * (x - 3)**3 * (x - 5)**3, x)
assert sorted(dispersionset(gp)) == [0, 1, 2, 3]
assert sorted(dispersionset(gp, (gp + 4)**2)) == [1, 2]
fp = poly(x * (x + 2) * (x - 1), x)
assert sorted(dispersionset(fp)) == [0, 1, 2, 3]
fp = poly(x**2 + sqrt(5) * x - 1, x, domain='QQ<sqrt(5)>')
gp = poly(x**2 + (2 + sqrt(5)) * x + sqrt(5), x, domain='QQ<sqrt(5)>')
assert sorted(dispersionset(fp, gp)) == [2]
assert sorted(dispersionset(gp, fp)) == [1, 4]
# There are some difficulties if we compute over Z[a]
# and alpha happenes to lie in Z[a] instead of simply Z.
# Hence we can not decide if alpha is indeed integral
# in general.
fp = poly(
4 * x**4 + (4 * a + 8) * x**3 + (a**2 + 6 * a + 4) * x**2 +
(a**2 + 2 * a) * x, x)
assert sorted(dispersionset(fp)) == [0, 1]
# For any specific value of a, the dispersion is 3*a
# but the algorithm can not find this in general.
# This is the point where the resultant based Ansatz
# is superior to the current one.
fp = poly(a**2 * x**3 + (a**3 + a**2 + a + 1) * x, x)
gp = fp.as_expr().subs({x: x - 3 * a}).as_poly(x)
assert sorted(dispersionset(fp, gp)) == []
fpa = fp.as_expr().subs({a: 2}).as_poly(x)
gpa = gp.as_expr().subs({a: 2}).as_poly(x)
assert sorted(dispersionset(fpa, gpa)) == [6]
# Work with Expr instead of Poly
f = (x + 1) * (x + 2)
assert sorted(dispersionset(f)) == [0, 1]
assert dispersion(f) == 1
f = x**4 - 3 * x**2 + 1
g = x**4 - 12 * x**3 + 51 * x**2 - 90 * x + 55
assert sorted(dispersionset(f, g)) == [2, 3, 4]
assert dispersion(f, g) == 4
# Work with Expr and specify a generator
f = (x + 1) * (x + 2)
assert sorted(dispersionset(f, None, x)) == [0, 1]
assert dispersion(f, None, x) == 1
f = x**4 - 3 * x**2 + 1
g = x**4 - 12 * x**3 + 51 * x**2 - 90 * x + 55
assert sorted(dispersionset(f, g, x)) == [2, 3, 4]
assert dispersion(f, g, x) == 4
def test_dispersion():
pytest.raises(ValueError, lambda: dispersionset(poly(x*y, x, y),
poly(x, x)))
pytest.raises(ValueError, lambda: dispersionset(poly(x, x),
poly(y, y)))
fp = poly(0, x)
assert sorted(dispersionset(fp)) == [0]
fp = poly(2, x)
assert sorted(dispersionset(fp)) == [0]
fp = poly(x + 1, x)
assert sorted(dispersionset(fp)) == [0]
assert dispersion(fp) == 0
fp = poly((x + 1)*(x + 2), x)
assert sorted(dispersionset(fp)) == [0, 1]
assert dispersion(fp) == 1
fp = poly(x*(x + 3), x)
assert sorted(dispersionset(fp)) == [0, 3]
assert dispersion(fp) == 3
fp = poly((x - 3)*(x + 3), x)
assert sorted(dispersionset(fp)) == [0, 6]
assert dispersion(fp) == 6
fp = poly(x**4 - 3*x**2 + 1, x)
gp = fp.shift(-3)
assert sorted(dispersionset(fp, gp)) == [2, 3, 4]
assert dispersion(fp, gp) == 4
assert sorted(dispersionset(gp, fp)) == []
assert dispersion(gp, fp) == -oo
fp = poly(x**2 + 2*x - 1, x)
gp = poly(x**2 + 2*x + 3, x)
assert dispersionset(fp, gp) == set()
fp = poly(x*(3*x**2+a)*(x-2536)*(x**3+a), x)
gp = fp.as_expr().subs({x: x - 345}).as_poly(x)
assert sorted(dispersionset(fp, gp)) == [345, 2881]
assert sorted(dispersionset(gp, fp)) == [2191]
gp = poly((x-2)**2*(x-3)**3*(x-5)**3, x)
assert sorted(dispersionset(gp)) == [0, 1, 2, 3]
assert sorted(dispersionset(gp, (gp+4)**2)) == [1, 2]
fp = poly(x*(x+2)*(x-1), x)
assert sorted(dispersionset(fp)) == [0, 1, 2, 3]
fp = poly(x**2 + sqrt(5)*x - 1, x, domain='QQ<sqrt(5)>')
gp = poly(x**2 + (2 + sqrt(5))*x + sqrt(5), x, domain='QQ<sqrt(5)>')
assert sorted(dispersionset(fp, gp)) == [2]
assert sorted(dispersionset(gp, fp)) == [1, 4]
# There are some difficulties if we compute over Z[a]
# and alpha happenes to lie in Z[a] instead of simply Z.
# Hence we can not decide if alpha is indeed integral
# in general.
fp = poly(4*x**4 + (4*a + 8)*x**3 + (a**2 + 6*a + 4)*x**2 + (a**2 + 2*a)*x, x)
assert sorted(dispersionset(fp)) == [0, 1]
# For any specific value of a, the dispersion is 3*a
# but the algorithm can not find this in general.
# This is the point where the resultant based Ansatz
# is superior to the current one.
fp = poly(a**2*x**3 + (a**3 + a**2 + a + 1)*x, x)
gp = fp.as_expr().subs({x: x - 3*a}).as_poly(x)
assert sorted(dispersionset(fp, gp)) == []
fpa = fp.as_expr().subs({a: 2}).as_poly(x)
gpa = gp.as_expr().subs({a: 2}).as_poly(x)
assert sorted(dispersionset(fpa, gpa)) == [6]
# Work with Expr instead of Poly
f = (x + 1)*(x + 2)
assert sorted(dispersionset(f)) == [0, 1]
assert dispersion(f) == 1
f = x**4 - 3*x**2 + 1
g = x**4 - 12*x**3 + 51*x**2 - 90*x + 55
assert sorted(dispersionset(f, g)) == [2, 3, 4]
assert dispersion(f, g) == 4
# Work with Expr and specify a generator
f = (x + 1)*(x + 2)
assert sorted(dispersionset(f, None, x)) == [0, 1]
assert dispersion(f, None, x) == 1
f = x**4 - 3*x**2 + 1
g = x**4 - 12*x**3 + 51*x**2 - 90*x + 55
assert sorted(dispersionset(f, g, x)) == [2, 3, 4]
assert dispersion(f, g, x) == 4