What Chris says is basically right, you are better off with human insight.
Nonetheless, SymPy should be able to solve zero dimensional polynomial systems
with full generality, using the Gröbner basis algorithm. Clearly, there are
some problems with this for numeric coefficients, which Mateusz will have to
look at (also computing Gröbner bases can be expensive in general, which
explains why it hangs, but I have noticed our implementation to be particularly
slow in many cases).
Aaron Meurer
On Oct 10, 2010, at 10:10 PM, smichr wrote:
> First of all, I would suggest that you solve the system symbollically
> and then substitute in values when you have a general solution.
>
> Then, I would recommend some human insight to overcome the less-than-
> optimal algorithm. :-)
>
> eq1 = y*2/(e*2 - a) - (x + e)**2/a + 1
> eq3 = y*2/(e*2 - c) - (x - e)**2/c + 1
>
> The equations 1 and 3 are linear in y, quadratic in x. If you start by
> solving eq1 for x and substituting into eq3 you get an equation of the
> form A + B*y+sqrt(C+E*y) which master cannot solve (but is solvable on
> my 1766 branch). That gets you off on the wrong path from start.
> Instead, try something like the following:
>
>>>> eq1
> 1 + 2*y/(-a + 2*e) - (e + x)**2/a
>>>> eq3
> 1 + 2*y/(-c + 2*e) - (x - e)**2/c
>
> Solve for the linear symbol:
>
>>>> ys = solve(eq1, y)[0]
>
> Substitute it into the other equation and keep only the numerator:
>
>>>> eq3.subs(y, ys).as_numer_denom()[0].expand()
> -4*a*c*e*x + c*a**2 - a*c**2 - 2*a*e*x**2 + 2*c*e*x**2 + 4*a*x*e**2 +
> 4*c*x*e**2 - 2*a*e**3 + 2*c*e**3
>
> Solve it for x:
>
>>>> xs=solve(_, x)
>
> Define your replacements:
>
>>>> r1=dict(
> ... a = ((a2 - c2)/2)**2,
> ... b = ((b2 - c2)/2)**2,
> ... c = ((d2 - c2)/2)**2,
> ... e = e2/2 )
>>>> r2=dict(
> ... a2 = 5.10125066033543,
> ... b2 = 1.844025414703168,
> ... c2 = 2.303750131640994,
> ... d2 = 0,
> ... e2 = 3)
>
> Now calculate your solutions and print the results, checking (to 3 sig
> digits and dropping "small" residuals) that they satisfy the original
> equation; also use a formatting string to make the numbers more tidy:
>
>
>>>> for xi in xs:
> ... xi = xi.subs(r1).subs(r2).n()
> ... yi = ys.subs(r1).subs(r2).subs(x, xi).n()
> ... print ('%g '*2) % (xi, yi), [q.subs(r1).subs(r2).subs(dict(x=xi,
> y=yi)).n(3, chop=True) for q in [eq1, eq3]]
> ...
> 7.20535 19.6876 [0, 0]
> 0.192176 0.241862 [0, 0]
>
> So sympy can do it. The basics are there. You just have to help it.
> Can you take the steps above and wrap those into your hyperplot? I
> think you'll have a pretty robust routine if you do.
>
> p.s. I think this is a good reminder that just because you have two
> equations doesn't mean you should feed them to your CAS as "two
> equations". It always pays to see what form they have. (Or conversely,
> a CAS should not just blindly try to apply a general algorithm:
> looking for linear symbols, getting rid of uneccessary terms, etc...
> makes everything work better.)
>
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