A bug in Sympy’s solve
Inspired by this ask.sagemath.orgquestion
<https://ask.sagemath.org/question/59063/weird-c-values-from-solving-system-of-equations/>
.
This question was about the names of the symbolic variables created by
Maxima’s solve to denote arbitrary constants. Exploring the example used in
this questions revealed a non-trivial problem with sympy’s solve.
Problem
Solve the system :
$$
\begin{align*}
- a*{1}^{3} a*{2} + a*{1} a*{2}^{2} &= 0 \
- 3 a*{1}^{2} a*{2} b*{1} + 2 a*{1} a*{2} b*{2} - a*{1} b*{2} + a*{2}^{2}
b*{2} &= 0 \
- a*{1}^{2} a*{2}^{2} + a_{2}^{3} &= 0 \
- 2 a*{1}^{2} a*{2} b*{2} - 2 a*{2}^{2} b*{1} + 3 a*{2}^{2} b_{2} &= 0
\end{align*}
$$
# Set up sympy (brutal version)import sympyfrom sympy import *
init_session()
init_printing(pretty_print=False)# System to solve
a1, a2, b1, b2 = symbols('a1 a2 b1 b2')
Unk = [a1, a2, b1, b2]
eq1 = a1 * a2**2 - a2 * a1**3
eq2 = 2*a1*a2*b2 + b2*a2**2 - 3*a2*a1**2*b1 - a1*b2
eq3 = a2**3 - a2**2*a1**2
eq4 = 3*a2**2*b2 - 2*a2*a1**2*b2 - 2*a2**2*b1
Sys = [eq1, eq2, eq3, eq4]
IPython console for SymPy 1.9 (Python 3.9.9-64-bit) (ground types: gmpy)
These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()
Documentation can be found at https://docs.sympy.org/1.9/
Attempt to use the “automatic” Sympy solver:
Sol = solve(Sys, Unk)
DSol = [dict(zip(Unk, u)) for u in Sol]
DSol
[{a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2: 0,
b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b2/2, b2: b2}, {a1: a1, a2: 0, b1: b1, b2:
0}, {a1: -sqrt(a2), a2: a2, b1: 0, b2: 0}, {a1: -2**(5/6)*sqrt(48*2**(1/3) +
624/(1499 + 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 +
3*sqrt(303)*I)**(1/3))/6, a2: 16/3 + 104*2**(2/3)/(3*(1499 +
3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3, b1: b2/2,
b2: b2}, {a1: -sqrt(6)*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))/6, a2: 2**(2/3)*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 -
sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)/((1
+ sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)), b1: b2/2, b2: b2}, {a1:
-sqrt(6)*sqrt(32 - 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))
- 2**(1/3)*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))/6, a2:
2**(2/3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)/((1 - sqrt(3)*I)*(1499
+ 3*sqrt(303)*I)**(1/3)), b1: b2/2, b2: b2}]
Something went sideways: the six first solutions are okay,but the last
three use expressions, some of them being polynomials in b2.
Attempt to check them formally :
Chk=[[u.subs(s) for u in Sys] for s in DSol]
Chk
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0,
0, 0], [-2**(5/6)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2*sqrt(48*2**(1/3) + 624/(1499 +
3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6 +
sqrt(2)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499
+ 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))**(3/2)/54, -2**(2/3)*b2*(16/3 +
104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/12 - 2**(5/6)*b2*(16/3 +
104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
3*sqrt(303)*I)**(1/3)/3)*sqrt(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/3 + b2*(16/3 + 104*2**(2/3)/(3*(1499
+ 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2 +
2**(5/6)*b2*sqrt(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6, -2**(2/3)*(16/3 +
104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
3*sqrt(303)*I)**(1/3)/3)**2*(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) +
3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/18 + (16/3 + 104*2**(2/3)/(3*(1499 +
3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**3,
-2**(2/3)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) +
2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 +
3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/9 +
2*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 +
3*sqrt(303)*I)**(1/3)/3)**2], [2**(1/6)*sqrt(3)*(-208/3 + (1 + sqrt(3)*I)*(32 +
(-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)*(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))**(3/2)/(18*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 2**(5/6)*sqrt(3)*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 -
sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)),
-2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*sqrt(32 - 2**(1/3)*(1
+ sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 +
sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 3*2**(2/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 -
sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))/(2*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) +
sqrt(6)*b2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3) -
416*2**(2/3)/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/6 +
2*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 +
sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)), -2*2**(1/3)*(-208/3 + (1 +
sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)) +
4*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**3/((1 +
sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)), -2*2**(2/3)*b2*(-208/3 + (1 +
sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) +
4*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 +
sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))], [-2**(5/6)*sqrt(3)*(-208/3 + (1
- sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)) +
2**(1/6)*sqrt(3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(32 - 416*2**(2/3)/((1
- sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3))**(3/2)/(18*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)), sqrt(6)*b2*sqrt(32 - 416*2**(2/3)/((1 -
sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3))/6 + 2*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2/((1 - sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))
- 3*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(16/3 -
208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 -
sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6)/(2*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 +
(-1 + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)*sqrt(32 - 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)),
4*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**3/((1 -
sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)) - 2*2**(1/3)*(-208/3 + (1 -
sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2*(16/3 - 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)),
-2*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 +
6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(16/3 -
208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 -
sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)*(1499 +
3*sqrt(303)*I)**(1/3)) + 4*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 +
sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 +
6*sqrt(303)*I)**(1/3)/12)**2/((1 - sqrt(3)*I)**2*(1499 +
3*sqrt(303)*I)**(2/3))]]
Some of these expressions are polynomial in b2 whose coefficients of
not-null dregree are nort *obviously* null.
A bit of hit-and-miss trials leads to this attempt at *numerical* check :
def chz(x):
r = x.factor().is_zero
if r is None: return x.coeff(b2).n()
return r
[[chz(u) for u in v] for v in Chk]
[[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, -223.427427555427 + 0.e-25*I, True, True],
[True, -0.388012232966408 - 5.25038392589403e-29*I, True, True],
[True, -0.e-137 + 1.12445821450677e-139*I, True, True]]
Solution #8 may be exactr, but #6 and #7 cannot.
Manual solution.
eq1 and eq3 give us solutions for a1 and a2 :
S13=solve([eq3, eq1], [a2, a1], dict=True) ; S13
[{a1: -sqrt(a2)}, {a1: sqrt(a2)}, {a2: 0}]
which we prefer to rewrite as :
S13 = [{a2:a1**2}, {a2:0}]; S13
[{a2: a1**2}, {a2: 0}]
Substituting these values in eq4 gives us :
E4 = [eq4.subs(s) for s in S13] ; E4
[-2*a1**4*b1 + a1**4*b2, 0]
The second solution tells us that S13[1] is also,a solution to [eq1, eq3,
eq4].
S134=S13[1:] ; S134
[{a2: 0}]
Substituting S13[0] ineq4and solving for the variables of the resulting
polynomial augments the setS134of solutions of[eq1, eq3`, eq4]’ :
V4=E4[0].free_symbols
S4=[solve(E4[0], v, dict=True) for v in V4] ; S4for S in S4:
for s in S:
S0={u:S13[0][u].subs(s) for u in S13[0].keys()}
S134 += [S0.copy()|s]
S134
[{a2: 0}, {a2: a1**2, b1: b2/2}, {a1: 0, a2: 0}, {a2: a1**2, b2: 2*b1}]
Again, we prefer to rewrite it in a simpler (and shorter) fashion :
S134=[{a2:0}, {a2:a1**2, b2:2*b1}] ; S134
[{a2: 0}, {a2: a1**2, b2: 2*b1}]
Substituting in eq3 gives :
[eq2.subs(s) for s in S134]
[-a1*b2, -a1**4*b1 + 4*a1**3*b1 - 2*a1*b1]
Solving these equations for their free symbols a,d merging with the
previous partial solutions gives us the solutions of the full system :
S1234=[]for S in S134:
# print("S=",S)
E=eq2.subs(S)
# print("E=",E)
S1=[solve(E, v, dict=True) for v in E.free_symbols]
# print("S1=",S1)
for s in flatten(S1):
# print("s=",s)
S0={u:S[u].subs(s) if "subs" in dir(S[u]) else S[u] for u in S.keys()}
S1234+=[S0.copy()|s]
S1234
[{a1: 0, a2: 0}, {a2: 0, b2: 0}, {a2: a1**2, b1: 0, b2: 0}, {a1: 0, a2: 0, b2:
2*b1}, {a1: 4/3 + (-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) +
16/(9*(-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)), a2: (4/3 + (-1/2 -
sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) + 16/(9*(-1/2 - sqrt(3)*I/2)*(37/27
+ sqrt(303)*I/9)**(1/3)))**2, b2: 2*b1}, {a1: 4/3 + 16/(9*(-1/2 +
sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(37/27 +
sqrt(303)*I/9)**(1/3), a2: (4/3 + 16/(9*(-1/2 + sqrt(3)*I/2)*(37/27 +
sqrt(303)*I/9)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(37/27 +
sqrt(303)*I/9)**(1/3))**2, b2: 2*b1}, {a1: 4/3 + 16/(9*(37/27 +
sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3), a2: (4/3 +
16/(9*(37/27 + sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3))**2, b2:
2*b1}]
Again, some solutions, substituted in eq2, give first-degree monomials in b1
whose oefficient cannot be shownt to be null byis_zero :
[[e.subs(s).is_zero for e in Sys] for s in S1234]
[[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, True, True, True],
[True, None, True, True],
[True, None, True, True],
[True, None, True, True]]
But, this time, the numerical check points to a probably null result :
[Sys[1].subs(S1234[u]).coeff(b1).n() for u in range(3,6)]
[0, 0.e-125 + 0.e-127*I, 0.e-125 - 0.e-127*I]
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