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relation.py
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relation.py
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r"""
Symbolic Equations and Inequalities
Sage can solve symbolic equations and inequalities. For
example, we derive the quadratic formula as follows::
sage: a,b,c = var('a,b,c')
sage: qe = (a*x^2 + b*x + c == 0)
sage: qe
a*x^2 + b*x + c == 0
sage: print(solve(qe, x))
[
x == -1/2*(b + sqrt(b^2 - 4*a*c))/a,
x == -1/2*(b - sqrt(b^2 - 4*a*c))/a
]
The operator, left hand side, and right hand side
--------------------------------------------------
Operators::
sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.operator()
<built-in function ge>
sage: (x^3 + 2/3 < x - pi).operator()
<built-in function lt>
sage: (x^3 + 2/3 == x - pi).operator()
<built-in function eq>
Left hand side::
sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.lhs()
x^3 + 2/3
sage: eqn.left()
x^3 + 2/3
sage: eqn.left_hand_side()
x^3 + 2/3
Right hand side::
sage: (x + sqrt(2) >= sqrt(3) + 5/2).right()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).rhs()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).right_hand_side()
sqrt(3) + 5/2
Arithmetic
----------
Add two symbolic equations::
sage: var('a,b')
(a, b)
sage: m = 144 == -10 * a + b
sage: n = 136 == 10 * a + b
sage: m + n
280 == 2*b
sage: int(-144) + m
0 == -10*a + b - 144
Subtract two symbolic equations::
sage: var('a,b')
(a, b)
sage: m = 144 == 20 * a + b
sage: n = 136 == 10 * a + b
sage: m - n
8 == 10*a
sage: int(144) - m
0 == -20*a - b + 144
Multiply two symbolic equations::
sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi, hold=True)
sage: m * n
x*sin(x) == (5*x + 1)*sin(2*pi + x)
sage: m = 2*x == 3*x^2 - 5
sage: int(-1) * m
-2*x == -3*x^2 + 5
Divide two symbolic equations::
sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi, hold=True)
sage: m/n
x/sin(x) == (5*x + 1)/sin(2*pi + x)
sage: m = x != 5*x + 1
sage: n = sin(x) != sin(x+2*pi, hold=True)
sage: m/n
x/sin(x) != (5*x + 1)/sin(2*pi + x)
Substitution
------------
Substitution into relations::
sage: x, a = var('x, a')
sage: eq = (x^3 + a == sin(x/a)); eq
x^3 + a == sin(x/a)
sage: eq.substitute(x=5*x)
125*x^3 + a == sin(5*x/a)
sage: eq.substitute(a=1)
x^3 + 1 == sin(x)
sage: eq.substitute(a=x)
x^3 + x == sin(1)
sage: eq.substitute(a=x, x=1)
x + 1 == sin(1/x)
sage: eq.substitute({a:x, x:1})
x + 1 == sin(1/x)
Solving
-------
We can solve equations::
sage: x = var('x')
sage: S = solve(x^3 - 1 == 0, x)
sage: S
[x == 1/2*I*sqrt(3) - 1/2, x == -1/2*I*sqrt(3) - 1/2, x == 1]
sage: S[0]
x == 1/2*I*sqrt(3) - 1/2
sage: S[0].right()
1/2*I*sqrt(3) - 1/2
sage: S = solve(x^3 - 1 == 0, x, solution_dict=True)
sage: S
[{x: 1/2*I*sqrt(3) - 1/2}, {x: -1/2*I*sqrt(3) - 1/2}, {x: 1}]
sage: z = 5
sage: solve(z^2 == sqrt(3),z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.
We illustrate finding multiplicities of solutions::
sage: f = (x-1)^5*(x^2+1)
sage: solve(f == 0, x)
[x == -I, x == I, x == 1]
sage: solve(f == 0, x, multiplicities=True)
([x == -I, x == I, x == 1], [1, 1, 5])
We can also solve many inequalities::
sage: solve(1/(x-1)<=8,x)
[[x < 1], [x >= (9/8)]]
We can numerically find roots of equations::
sage: (x == sin(x)).find_root(-2,2)
0.0
sage: (x^5 + 3*x + 2 == 0).find_root(-2,2,x)
-0.6328345202421523
sage: (cos(x) == sin(x)).find_root(10,20)
19.634954084936208
We illustrate some valid error conditions::
sage: (cos(x) != sin(x)).find_root(10,20)
Traceback (most recent call last):
...
ValueError: Symbolic equation must be an equality.
sage: (SR(3)==SR(2)).find_root(-1,1)
Traceback (most recent call last):
...
RuntimeError: no zero in the interval, since constant expression is not 0.
There must be at most one variable::
sage: x, y = var('x,y')
sage: (x == y).find_root(-2,2)
Traceback (most recent call last):
...
NotImplementedError: root finding currently only implemented in 1 dimension.
Assumptions
-----------
Forgetting assumptions::
sage: var('x,y')
(x, y)
sage: forget() #Clear assumptions
sage: assume(x>0, y < 2)
sage: assumptions()
[x > 0, y < 2]
sage: (y < 2).forget()
sage: assumptions()
[x > 0]
sage: forget()
sage: assumptions()
[]
Miscellaneous
-------------
Conversion to Maxima::
sage: x = var('x')
sage: eq = (x^(3/5) >= pi^2 + e^i)
sage: eq._maxima_init_()
'(_SAGE_VAR_x)^(3/5) >= ((%pi)^(2))+(exp(0+%i*1))'
sage: e1 = x^3 + x == sin(2*x)
sage: z = e1._maxima_()
sage: z.parent() is sage.calculus.calculus.maxima
True
sage: z = e1._maxima_(maxima)
sage: z.parent() is maxima
True
sage: z = maxima(e1)
sage: z.parent() is maxima
True
Conversion to Maple::
sage: x = var('x')
sage: eq = (x == 2)
sage: eq._maple_init_()
'x = 2'
Comparison::
sage: x = var('x')
sage: (x>0) == (x>0)
True
sage: (x>0) == (x>1)
False
sage: (x>0) != (x>1)
True
Variables appearing in the relation::
sage: var('x,y,z,w')
(x, y, z, w)
sage: f = (x+y+w) == (x^2 - y^2 - z^3); f
w + x + y == -z^3 + x^2 - y^2
sage: f.variables()
(w, x, y, z)
LaTeX output::
sage: latex(x^(3/5) >= pi)
x^{\frac{3}{5}} \geq \pi
When working with the symbolic complex number `I`, notice that comparisons do not
automatically simplify even in trivial situations::
sage: I^2 == -1
-1 == -1
sage: I^2 < 0
-1 < 0
sage: (I+1)^4 > 0
-4 > 0
Nevertheless, if you force the comparison, you get the right answer (:trac:`7160`)::
sage: bool(I^2 == -1)
True
sage: bool(I^2 < 0)
True
sage: bool((I+1)^4 > 0)
False
More Examples
-------------
::
sage: x,y,a = var('x,y,a')
sage: f = x^2 + y^2 == 1
sage: f.solve(x)
[x == -sqrt(-y^2 + 1), x == sqrt(-y^2 + 1)]
::
sage: f = x^5 + a
sage: solve(f==0,x)
[x == 1/4*(-a)^(1/5)*(sqrt(5) + I*sqrt(2*sqrt(5) + 10) - 1), x == -1/4*(-a)^(1/5)*(sqrt(5) - I*sqrt(-2*sqrt(5) + 10) + 1), x == -1/4*(-a)^(1/5)*(sqrt(5) + I*sqrt(-2*sqrt(5) + 10) + 1), x == 1/4*(-a)^(1/5)*(sqrt(5) - I*sqrt(2*sqrt(5) + 10) - 1), x == (-a)^(1/5)]
You can also do arithmetic with inequalities, as illustrated
below::
sage: var('x y')
(x, y)
sage: f = x + 3 == y - 2
sage: f
x + 3 == y - 2
sage: g = f - 3; g
x == y - 5
sage: h = x^3 + sqrt(2) == x*y*sin(x)
sage: h
x^3 + sqrt(2) == x*y*sin(x)
sage: h - sqrt(2)
x^3 == x*y*sin(x) - sqrt(2)
sage: h + f
x^3 + x + sqrt(2) + 3 == x*y*sin(x) + y - 2
sage: f = x + 3 < y - 2
sage: g = 2 < x+10
sage: f - g
x + 1 < -x + y - 12
sage: f + g
x + 5 < x + y + 8
sage: f*(-1)
-x - 3 < -y + 2
TESTS:
We test serializing symbolic equations::
sage: eqn = x^3 + 2/3 >= x
sage: loads(dumps(eqn))
x^3 + 2/3 >= x
sage: loads(dumps(eqn)) == eqn
True
AUTHORS:
- Bobby Moretti: initial version (based on a trick that Robert
Bradshaw suggested).
- William Stein: second version
- William Stein (2007-07-16): added arithmetic with symbolic equations
"""
from __future__ import print_function
from six.moves import range
import operator
def test_relation_maxima(relation):
"""
Return True if this (in)equality is definitely true. Return False
if it is false or the algorithm for testing (in)equality is
inconclusive.
EXAMPLES::
sage: from sage.symbolic.relation import test_relation_maxima
sage: k = var('k')
sage: pol = 1/(k-1) - 1/k -1/k/(k-1);
sage: test_relation_maxima(pol == 0)
True
sage: f = sin(x)^2 + cos(x)^2 - 1
sage: test_relation_maxima(f == 0)
True
sage: test_relation_maxima( x == x )
True
sage: test_relation_maxima( x != x )
False
sage: test_relation_maxima( x > x )
False
sage: test_relation_maxima( x^2 > x )
False
sage: test_relation_maxima( x + 2 > x )
True
sage: test_relation_maxima( x - 2 > x )
False
Here are some examples involving assumptions::
sage: x, y, z = var('x, y, z')
sage: assume(x>=y,y>=z,z>=x)
sage: test_relation_maxima(x==z)
True
sage: test_relation_maxima(z<x)
False
sage: test_relation_maxima(z>y)
False
sage: test_relation_maxima(y==z)
True
sage: forget()
sage: assume(x>=1,x<=1)
sage: test_relation_maxima(x==1)
True
sage: test_relation_maxima(x>1)
False
sage: test_relation_maxima(x>=1)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()
sage: assume(x>0)
sage: test_relation_maxima(x==0)
False
sage: test_relation_maxima(x>-1)
True
sage: test_relation_maxima(x!=0)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()
TESTS:
Ensure that ``canonicalize_radical()`` and ``simplify_log`` are not
used inappropriately, :trac:`17389`. Either one would simplify ``f``
to zero below::
sage: x,y = SR.var('x,y')
sage: assume(y, 'complex')
sage: f = log(x*y) - (log(x) + log(y))
sage: f(x=-1, y=i)
-2*I*pi
sage: test_relation_maxima(f == 0)
False
sage: forget()
Ensure that the ``sqrt(x^2)`` -> ``abs(x)`` simplification is not
performed when testing equality::
sage: assume(x, 'complex')
sage: f = sqrt(x^2) - abs(x)
sage: test_relation_maxima(f == 0)
False
sage: forget()
If assumptions are made, ``simplify_rectform()`` is used::
sage: assume(x, 'real')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()
But not if ``x`` itself is complex::
sage: assume(x, 'complex')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
False
sage: forget()
If assumptions are made, then ``simplify_factorial()`` is used::
sage: n,k = SR.var('n,k')
sage: assume(n, 'integer')
sage: assume(k, 'integer')
sage: f1 = factorial(n+1)/factorial(n)
sage: f2 = n + 1
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()
In case an equation is to be solved for non-integers, ''assume()''
is used::
sage: k = var('k')
sage: assume(k,'noninteger')
sage: solve([k^3==1],k)
[k == 1/2*I*sqrt(3) - 1/2, k == -1/2*I*sqrt(3) - 1/2]
sage: assumptions()
[k is noninteger]
"""
m = relation._maxima_()
#Handle some basic cases first
if repr(m) in ['0=0']:
return True
elif repr(m) in ['0#0', '1#1']:
return False
if relation.operator() == operator.eq: # operator is equality
try:
s = m.parent()._eval_line('is (equal(%s,%s))'%(repr(m.lhs()),repr(m.rhs())))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
elif relation.operator() == operator.ne: # operator is not equal
try:
s = m.parent()._eval_line('is (notequal(%s,%s))'%(repr(m.lhs()),repr(m.rhs())))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
else: # operator is < or > or <= or >=, which Maxima handles fine
try:
s = m.parent()._eval_line('is (%s)'%repr(m))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
if s == 'true':
return True
elif s == 'false':
return False # if neither of these, s=='unknown' and we try a few other tricks
if relation.operator() != operator.eq:
return False
difference = relation.lhs() - relation.rhs()
if repr(difference) == '0':
return True
# Try to apply some simplifications to see if left - right == 0.
#
# TODO: If simplify_log() is ever removed from simplify_full(), we
# can replace all of these individual simplifications with a
# single call to simplify_full(). That would work in cases where
# two simplifications are needed consecutively; the current
# approach does not.
#
simp_list = [difference.simplify_factorial(),
difference.simplify_rational(),
difference.simplify_rectform(),
difference.simplify_trig()]
for f in simp_list:
try:
if repr( f() ).strip() == "0":
return True
break
except Exception:
pass
return False
def string_to_list_of_solutions(s):
r"""
Used internally by the symbolic solve command to convert the output
of Maxima's solve command to a list of solutions in Sage's symbolic
package.
EXAMPLES:
We derive the (monic) quadratic formula::
sage: var('x,a,b')
(x, a, b)
sage: solve(x^2 + a*x + b == 0, x)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]
Behind the scenes when the above is evaluated the function
:func:`string_to_list_of_solutions` is called with input the
string `s` below::
sage: s = '[x=-(sqrt(a^2-4*b)+a)/2,x=(sqrt(a^2-4*b)-a)/2]'
sage: sage.symbolic.relation.string_to_list_of_solutions(s)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]
"""
from sage.categories.all import Objects
from sage.structure.sequence import Sequence
from sage.calculus.calculus import symbolic_expression_from_maxima_string
v = symbolic_expression_from_maxima_string(s, equals_sub=True)
return Sequence(v, universe=Objects(), cr_str=True)
###########
# Solving #
###########
def solve(f, *args, **kwds):
r"""
Algebraically solve an equation or system of equations (over the
complex numbers) for given variables. Inequalities and systems
of inequalities are also supported.
INPUT:
- ``f`` - equation or system of equations (given by a
list or tuple)
- ``*args`` - variables to solve for.
- ``solution_dict`` - bool (default: False); if True or non-zero,
return a list of dictionaries containing the solutions. If there
are no solutions, return an empty list (rather than a list containing
an empty dictionary). Likewise, if there's only a single solution,
return a list containing one dictionary with that solution.
There are a few optional keywords if you are trying to solve a single
equation. They may only be used in that context.
- ``multiplicities`` - bool (default: False); if True,
return corresponding multiplicities. This keyword is
incompatible with ``to_poly_solve=True`` and does not make
any sense when solving inequalities.
- ``explicit_solutions`` - bool (default: False); require that
all roots be explicit rather than implicit. Not used
when solving inequalities.
- ``to_poly_solve`` - bool (default: False) or string; use
Maxima's ``to_poly_solver`` package to search for more possible
solutions, but possibly encounter approximate solutions.
This keyword is incompatible with ``multiplicities=True``
and is not used when solving inequalities. Setting ``to_poly_solve``
to 'force' (string) omits Maxima's solve command (useful when
some solutions of trigonometric equations are lost).
EXAMPLES::
sage: x, y = var('x, y')
sage: solve([x+y==6, x-y==4], x, y)
[[x == 5, y == 1]]
sage: solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y)
[[x == -1/2*I*sqrt(3) - 1/2, y == -sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == -1/2*I*sqrt(3) - 1/2, y == sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == -sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 0, y == -1],
[x == 0, y == 1]]
sage: solve([sqrt(x) + sqrt(y) == 5, x + y == 10], x, y)
[[x == -5/2*I*sqrt(5) + 5, y == 5/2*I*sqrt(5) + 5], [x == 5/2*I*sqrt(5) + 5, y == -5/2*I*sqrt(5) + 5]]
sage: solutions=solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y, solution_dict=True)
sage: for solution in solutions: print("{} , {}".format(solution[x].n(digits=3), solution[y].n(digits=3)))
-0.500 - 0.866*I , -1.27 + 0.341*I
-0.500 - 0.866*I , 1.27 - 0.341*I
-0.500 + 0.866*I , -1.27 - 0.341*I
-0.500 + 0.866*I , 1.27 + 0.341*I
0.000 , -1.00
0.000 , 1.00
Whenever possible, answers will be symbolic, but with systems of
equations, at times approximations will be given, due to the
underlying algorithm in Maxima::
sage: sols = solve([x^3==y,y^2==x],[x,y]); sols[-1], sols[0]
([x == 0, y == 0], [x == (0.3090169943749475 + 0.9510565162951535*I), y == (-0.8090169943749475 - 0.5877852522924731*I)])
sage: sols[0][0].rhs().pyobject().parent()
Complex Double Field
If ``f`` is only one equation or expression, we use the solve method
for symbolic expressions, which defaults to exact answers only::
sage: solve([y^6==y],y)
[y == 1/4*sqrt(5) + 1/4*I*sqrt(2*sqrt(5) + 10) - 1/4, y == -1/4*sqrt(5) + 1/4*I*sqrt(-2*sqrt(5) + 10) - 1/4, y == -1/4*sqrt(5) - 1/4*I*sqrt(-2*sqrt(5) + 10) - 1/4, y == 1/4*sqrt(5) - 1/4*I*sqrt(2*sqrt(5) + 10) - 1/4, y == 1, y == 0]
sage: solve( [y^6 == y], y)==solve( y^6 == y, y)
True
Here we demonstrate very basic use of the optional keywords for
a single expression to be solved::
sage: ((x^2-1)^2).solve(x)
[x == -1, x == 1]
sage: ((x^2-1)^2).solve(x,multiplicities=True)
([x == -1, x == 1], [2, 2])
sage: solve(sin(x)==x,x)
[x == sin(x)]
sage: solve(sin(x)==x,x,explicit_solutions=True)
[]
sage: solve(abs(1-abs(1-x)) == 10, x)
[abs(abs(x - 1) - 1) == 10]
sage: solve(abs(1-abs(1-x)) == 10, x, to_poly_solve=True)
[x == -10, x == 12]
.. note::
For more details about solving a single equation, see
the documentation for the single-expression
:meth:`~sage.symbolic.expression.Expression.solve`.
::
sage: from sage.symbolic.expression import Expression
sage: Expression.solve(x^2==1,x)
[x == -1, x == 1]
We must solve with respect to actual variables::
sage: z = 5
sage: solve([8*z + y == 3, -z +7*y == 0],y,z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.
If we ask for dictionaries containing the solutions, we get them::
sage: solve([x^2-1],x,solution_dict=True)
[{x: -1}, {x: 1}]
sage: solve([x^2-4*x+4],x,solution_dict=True)
[{x: 2}]
sage: res = solve([x^2 == y, y == 4],x,y,solution_dict=True)
sage: for soln in res: print("x: %s, y: %s" % (soln[x], soln[y]))
x: 2, y: 4
x: -2, y: 4
If there is a parameter in the answer, that will show up as
a new variable. In the following example, ``r1`` is an arbitrary
constant (because of the ``r``)::
sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
[[x == -r1 + 3, y == r1]]
Especially with trigonometric functions, the dummy variable may
be implicitly an integer (hence the ``z``)::
sage: solve([cos(x)*sin(x) == 1/2, x+y == 0],x,y)
[[x == 1/4*pi + pi*z79, y == -1/4*pi - pi*z79]]
Expressions which are not equations are assumed to be set equal
to zero, as with `x` in the following example::
sage: solve([x, y == 2],x,y)
[[x == 0, y == 2]]
If ``True`` appears in the list of equations it is
ignored, and if ``False`` appears in the list then no
solutions are returned. E.g., note that the first
``3==3`` evaluates to ``True``, not to a
symbolic equation.
::
sage: solve([3==3, 1.00000000000000*x^3 == 0], x)
[x == 0]
sage: solve([1.00000000000000*x^3 == 0], x)
[x == 0]
Here, the first equation evaluates to ``False``, so
there are no solutions::
sage: solve([1==3, 1.00000000000000*x^3 == 0], x)
[]
Completely symbolic solutions are supported::
sage: var('s,j,b,m,g')
(s, j, b, m, g)
sage: sys = [ m*(1-s) - b*s*j, b*s*j-g*j ];
sage: solve(sys,s,j)
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,(s,j))
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,[s,j])
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
Inequalities can be also solved::
sage: solve(x^2>8,x)
[[x < -2*sqrt(2)], [x > 2*sqrt(2)]]
We use ``use_grobner`` in Maxima if no solution is obtained from
Maxima's ``to_poly_solve``::
sage: x,y=var('x y'); c1(x,y)=(x-5)^2+y^2-16; c2(x,y)=(y-3)^2+x^2-9
sage: solve([c1(x,y),c2(x,y)],[x,y])
[[x == -9/68*sqrt(55) + 135/68, y == -15/68*sqrt(11)*sqrt(5) + 123/68], [x == 9/68*sqrt(55) + 135/68, y == 15/68*sqrt(11)*sqrt(5) + 123/68]]
TESTS::
sage: solve([sin(x)==x,y^2==x],x,y)
[sin(x) == x, y^2 == x]
sage: solve(0==1,x)
Traceback (most recent call last):
...
TypeError: The first argument must be a symbolic expression or a list of symbolic expressions.
Test if the empty list is returned, too, when (a list of)
dictionaries (is) are requested (:trac:`8553`)::
sage: solve([SR(0)==1],x)
[]
sage: solve([SR(0)==1],x,solution_dict=True)
[]
sage: solve([x==1,x==-1],x)
[]
sage: solve([x==1,x==-1],x,solution_dict=True)
[]
sage: solve((x==1,x==-1),x,solution_dict=0)
[]
Relaxed form, suggested by Mike Hansen (:trac:`8553`)::
sage: solve([x^2-1],x,solution_dict=-1)
[{x: -1}, {x: 1}]
sage: solve([x^2-1],x,solution_dict=1)
[{x: -1}, {x: 1}]
sage: solve((x==1,x==-1),x,solution_dict=-1)
[]
sage: solve((x==1,x==-1),x,solution_dict=1)
[]
This inequality holds for any real ``x`` (:trac:`8078`)::
sage: solve(x^4+2>0,x)
[x < +Infinity]
Test for user friendly input handling :trac:`13645`::
sage: poly.<a,b> = PolynomialRing(RR)
sage: solve([a+b+a*b == 1], a)
Traceback (most recent call last):
...
TypeError: The first argument to solve() should be a symbolic expression or a list of symbolic expressions, cannot handle <type 'bool'>
sage: solve([a, b], (1, a))
Traceback (most recent call last):
...
TypeError: 1 is not a valid variable.
sage: solve([x == 1], (1, a))
Traceback (most recent call last):
...
TypeError: (1, a) are not valid variables.
Test that the original version of a system in the French Sage book
now works (:trac:`14306`)::
sage: var('y,z')
(y, z)
sage: solve([x^2 * y * z == 18, x * y^3 * z == 24, x * y * z^4 == 6], x, y, z)
[[x == 3, y == 2, z == 1], [x == (1.337215067... - 2.685489874...*I), y == (-1.700434271... + 1.052864325...*I), z == (0.9324722294... - 0.3612416661...*I)], ...]
"""
from sage.symbolic.expression import is_Expression
if is_Expression(f): # f is a single expression
ans = f.solve(*args,**kwds)
return ans
if not isinstance(f, (list, tuple)):
raise TypeError("The first argument must be a symbolic expression or a list of symbolic expressions.")
if len(f)==1:
# f is a list with a single element
if is_Expression(f[0]):
# if its a symbolic expression call solve method of this expression
return f[0].solve(*args,**kwds)
# otherwise complain
raise TypeError("The first argument to solve() should be a symbolic "
"expression or a list of symbolic expressions, "
"cannot handle %s"%repr(type(f[0])))
# f is a list of such expressions or equations
from sage.symbolic.ring import is_SymbolicVariable
if len(args)==0:
raise TypeError("Please input variables to solve for.")
if is_SymbolicVariable(args[0]):
variables = args
else:
variables = tuple(args[0])
for v in variables:
if not is_SymbolicVariable(v):
raise TypeError("%s is not a valid variable."%repr(v))
try:
f = [s for s in f if s is not True]
except TypeError:
raise ValueError("Unable to solve %s for %s"%(f, args))
if any(s is False for s in f):
return []
from sage.calculus.calculus import maxima
m = maxima(f)
try:
s = m.solve(variables)
except Exception: # if Maxima gave an error, try its to_poly_solve
try:
s = m.to_poly_solve(variables)
except TypeError as mess: # if that gives an error, raise an error.
if "Error executing code in Maxima" in str(mess):
raise ValueError("Sage is unable to determine whether the system %s can be solved for %s"%(f,args))
else:
raise
if len(s)==0: # if Maxima's solve gave no solutions, try its to_poly_solve
try:
s = m.to_poly_solve(variables)
except Exception: # if that gives an error, stick with no solutions
s = []
if len(s)==0: # if to_poly_solve gave no solutions, try use_grobner
try:
s = m.to_poly_solve(variables,'use_grobner=true')
except Exception: # if that gives an error, stick with no solutions
s = []
sol_list = string_to_list_of_solutions(repr(s))
# Relaxed form suggested by Mike Hansen (#8553):
if kwds.get('solution_dict', False):
if len(sol_list)==0: # fixes IndexError on empty solution list (#8553)
return []
if isinstance(sol_list[0], list):
sol_dict=[dict([[eq.left(),eq.right()] for eq in solution])
for solution in sol_list]
else:
sol_dict=[{eq.left():eq.right()} for eq in sol_list]
return sol_dict
else:
return sol_list
def solve_mod(eqns, modulus, solution_dict = False):
r"""
Return all solutions to an equation or list of equations modulo the
given integer modulus. Each equation must involve only polynomials
in 1 or many variables.
By default the solutions are returned as `n`-tuples, where `n`
is the number of variables appearing anywhere in the given
equations. The variables are in alphabetical order.
INPUT:
- ``eqns`` - equation or list of equations
- ``modulus`` - an integer
- ``solution_dict`` - bool (default: False); if True or non-zero,
return a list of dictionaries containing the solutions. If there
are no solutions, return an empty list (rather than a list containing
an empty dictionary). Likewise, if there's only a single solution,
return a list containing one dictionary with that solution.
EXAMPLES::
sage: var('x,y')
(x, y)
sage: solve_mod([x^2 + 2 == x, x^2 + y == y^2], 14)
[(4, 2), (4, 6), (4, 9), (4, 13)]
sage: solve_mod([x^2 == 1, 4*x == 11], 15)
[(14,)]
Fermat's equation modulo 3 with exponent 5::
sage: var('x,y,z')
(x, y, z)
sage: solve_mod([x^5 + y^5 == z^5], 3)
[(0, 0, 0), (0, 1, 1), (0, 2, 2), (1, 0, 1), (1, 1, 2), (1, 2, 0), (2, 0, 2), (2, 1, 0), (2, 2, 1)]
We can solve with respect to a bigger modulus if it consists only of small prime factors::
sage: [d] = solve_mod([5*x + y == 3, 2*x - 3*y == 9], 3*5*7*11*19*23*29, solution_dict = True)
sage: d[x]
12915279
sage: d[y]
8610183
For cases where there are relatively few solutions and the prime
factors are small, this can be efficient even if the modulus itself
is large::
sage: sorted(solve_mod([x^2 == 41], 10^20))
[(4538602480526452429,), (11445932736758703821,), (38554067263241296179,),
(45461397519473547571,), (54538602480526452429,), (61445932736758703821,),
(88554067263241296179,), (95461397519473547571,)]
We solve a simple equation modulo 2::
sage: x,y = var('x,y')
sage: solve_mod([x == y], 2)
[(0, 0), (1, 1)]
.. warning::
The current implementation splits the modulus into prime
powers, then naively enumerates all possible solutions
(starting modulo primes and then working up through prime
powers), and finally combines the solution using the Chinese
Remainder Theorem. The interface is good, but the algorithm is
very inefficient if the modulus has some larger prime factors! Sage
*does* have the ability to do something much faster in certain
cases at least by using Groebner basis, linear algebra
techniques, etc. But for a lot of toy problems this function as
is might be useful. At least it establishes an interface.
TESTS:
Make sure that we short-circuit in at least some cases::
sage: solve_mod([2*x==1], 2*next_prime(10^50))
[]
Try multi-equation cases::
sage: x, y, z = var("x y z")
sage: solve_mod([2*x^2 + x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 12)
[(0, 0), (4, 4), (0, 3), (4, 7)]
sage: eqs = [-y^2+z^2, -x^2+y^2-3*z^2-z-1, -y*z-z^2-x-y+2, -x^2-12*z^2-y+z]
sage: solve_mod(eqs, 11)
[(8, 5, 6)]
Confirm that modulus 1 now behaves as it should::
sage: x, y = var("x y")
sage: solve_mod([x==1], 1)
[(0,)]
sage: solve_mod([2*x^2+x*y, -x*y+2*y^2+x-2*y, -2*x^2+2*x*y-y^2-x-y], 1)
[(0, 0)]
"""
from sage.rings.all import Integer, Integers, crt_basis
from sage.symbolic.expression import is_Expression
from sage.misc.all import cartesian_product_iterator
from sage.modules.all import vector
from sage.matrix.all import matrix
if not isinstance(eqns, (list, tuple)):