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| from sympy.core import Symbol, S, oo | |
| from sympy.functions.elementary.miscellaneous import sqrt | |
| from sympy.polys import poly | |
| from sympy.polys.dispersion import dispersion, dispersionset | |
| def test_dispersion(): | |
| x = Symbol("x") | |
| a = Symbol("a") | |
| fp = poly(S.Zero, x) | |
| assert sorted(dispersionset(fp)) == [0] | |
| fp = poly(S(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) is -oo | |
| 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 | |