Particular solution of Diophantine equation

[Faisal Riyaz]

How can I get a single solution of a diophantine equation when the solution is returned

in terms of independent parameters?

>>> from sympy.abc import x, y, z, t

>>> from sympy import diophantine, Symbols

>>> diophantine(x + y + z, t)

{(t_0, t_1, -t_0 - t_1)}

>>> x, y, z = _.pop()

Substitution does not work:

>>> t_0, t_1 = Symbols('t_0 t_1')

>>> x.subs(t_0, 1)
t_0
>>> y.subs(t_1, 1)
t_1
>>> z.subs({t_0: 1, t_1: 1})
-t_0 - t_1

Thanks

Faisal Riyaz

[Oscar Benjamin]

This is because symbols in sympy only match if they have the same assumptions:

In [23]: (x, y, z), = diophantine(x + y + z, t)

In [24]: x
Out[24]: t₀

In [25]: type(x)
Out[25]: sympy.core.symbol.Symbol

In [26]: x.name
Out[26]: 't_0'

In [28]: x.assumptions0
Out[28]:
{'integer': True,
'complex': True,
'commutative': True,
'rational': True,
'algebraic': True,
'extended_real': True,
'imaginary': False,
'real': True,
'irrational': False,
'hermitian': True,
'finite': True,
'infinite': False,
'transcendental': False,
'noninteger': False}

In [29]: x == Symbol('t_0')
Out[29]: False

In [30]: x == Symbol('t_0', integer=True)
Out[30]: True


If we use integer=True then the substitution will work:

In [31]: t_0, t_1 = symbols('t_0, t_1', integer=True)

In [32]: tuple(s.subs({t_0:0, t_1:0}) for s in (x, y, z))
Out[32]: (0, 0, 0)


Note that this kind of situation was discussed in a recent thread and
you can use free symbols:

In [1]: sol, = diophantine(x + y + z, t)

In [2]: sol
Out[2]: (t₀, t₁, -t₀ - t₁)

In [3]: syms = Tuple(*sol).free_symbols

In [4]: rep = {s: 0 for s in syms}

In [5]: tuple(s.subs(rep) for s in sol)
Out[5]: (0, 0, 0)


Personally I think it's a poor API that inserts new symbols without
also returning the list of those symbols and it would be better if it
was like:

(sol, params), = diophantine(x + y + z, t)

Here params would directly return (t_0, t_1) for further use.

Alternatively it should be possible to provide the symbols directly:

sol, = diophantine(x + y + z, [t_0, t_1])


Oscar

[Faisal Riyaz]

I think we should put this on the docs page of the diophantine.

Alternatively it should be possible to provide the symbols directly:

sol, = diophantine(x + y + z, [t_0, t_1])

If the user does not have any idea about the number of parameters in the solution,

this won't work.

Thanks

Faisal

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[Oscar Benjamin]

On Fri, 17 Jul 2020 at 15:02, Faisal Riyaz <faisalr...@gmail.com> wrote:
>
> I think we should put this on the docs page of the diophantine.

I agree

>> Alternatively it should be possible to provide the symbols directly:
>>
>> sol, = diophantine(x + y + z, [t_0, t_1])
>
> If the user does not have any idea about the number of parameters in the solution,
> this won't work.

If you have three unknowns is it possible to end up with more than
three parameters in the solution?

Another possibility is an iterator that gives symbols.

--
Oscar

[Faisal Riyaz]

 If you have three unknowns is it possible to end up with more than
three parameters in the solution?

Yes, the number of parameters cannot be more than 3.

 

Another possibility is an iterator that gives symbols.

Yes, returning an iterator seems like a good solution.

Faisal

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