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