When we ask the user to specify the parameter to be used, what should
be the input? should it be a symbol or a string which contain the symbol
to be used. What I am asking is whether we should use,
>>>var("t")
>>>diop_solve(3*x*y + 5*y -7, param=t)
or
>>>diop_solve(3*x*y + 5*y -7, param="t")
On Wed, Jun 19, 2013 at 10:35 AM, Thilina Rathnayake <thilina.r...@gmail.com
> wrote:
> Thank you Ondrej for the reply.
>
> I think that's the way we should go at it. I can implement the
> moduels to find the solution and return those naturally as a list.
> Later we consider about the high level API's.
>
>
> On Wed, Jun 19, 2013 at 1:28 AM, Ondřej Čertík <ondrej.cer...@gmail.com>wrote:
>
>> Hi Thilina,
>>
>> On Tue, Jun 18, 2013 at 12:39 PM, Thilina Rathnayake
>> <thilina.r...@gmail.com> wrote:
>> > Hi everyone,
>> >
>> > Before continuing further with the Diophantine module development (PR
>> #2168)
>> > I thought it would be better for me to get other people's views on the
>> > representation
>> > of solutions returned by diop_solve().
>>
>> Thanks for discussing it.
>>
>> >
>> > The main routine of the module is diop_solve(), which takes a
>> Diophantine
>> > equation
>> > as an argument and returns the solution of the equation. Currently the
>> > solution is
>> > returned as a dictionary. Ex:
>> >
>> >> >>>diop_solve(4*x + 6*y - 4)
>> >> {x: 6*t - 2, y: -4*t + 2}
>> >> >>>diop_solve(3*x - 5*y + 7*z -5)
>> >> {x: -25*t - 14*z + 10, y: -15*t - 7*z + 5, z: z}
>> >
>> >
>> > Everything works fine here because the solutions are parametric.
>>
>> Right. For these equations I think a dictionary is the best solution,
>> as it is simple and clear. You should allow the user to specify the
>> "t" symbol, e.g. something like:
>>
>> var("x y z t")
>> diop_solve(3*x - 5*y + 7*z -5, param=t)
>>
>> so that the user can specify other variables names besides "t" as well.
>>
>> >
>> > But when I was trying to solve quadratic Diophantine equation ( this
>> has the
>> > form
>> > Ax**2 + Bxy + Cy**2 + Dx + Ey + F), they involve solutions which are not
>> > parametric.
>> > For example, the equation 2*x*y + 5*x + 56*y + 7 = 0 (which is a special
>> > case of the
>> > quadratic equation) has 8 solution pairs (x, y). (-27, 64), (-29, -69),
>> > (-21, 7) and five more.
>> >
>> > To represent these in a dictionary which has the above form, we have to
>> > split the solution
>> > pair and put it in to two lists which are keyed under x and y in the
>> dict.
>> > if the user want
>> > to retrieve a solution pair he would have to find the x value and the y
>> > value of the solution
>> > separately. Returned value would look like,
>> >
>> >> {x: [-27, -29, -21, ...], y: [64, -69, 7, ...]}
>> >
>> >
>> > Is this a good way to cope with this situation? I personally feel that
>> it is
>> > not natural to
>> > split a solution pair and enable the access of it's elements separately.
>> >
>> > I would like to know what the others have to say on this.
>>
>>
>> So for this I agree with Aaron:
>>
>> > You may want to look at
>> > https://code.google.com/p/sympy/issues/detail?id=3560 and some of the
>> > ideas for a unified solve object.
>>
>> We definitely need a consistent interface to the solve() command.
>>
>> > Already you have the issue that you
>> > are returning a parameter, but there is no easy way to access that
>> > parameter (and what happens if t is one of the variables?).
>>
>> The user can specify his own symbols as "params", as I suggested above.
>>
>> I would also look how Mathematica does it:
>>
>> http://reference.wolfram.com/mathematica/guide/DiophantineEquations.html
>>
>> e.g.:
>>
>> http://reference.wolfram.com/mathematica/ref/Reduce.html
>>
>>
>>
>> In general, I would suggest you simply write the low level modules for
>> actually solving the equation.
>> Those can return pretty much any representation that you think is the
>> best for that particular type of the equation.
>>
>> Then we need a consistent high level API, and that will take some time
>> to get right. But no matter what API we settle on in the end, it
>> should be quite simple to call the low level solver and convert the
>> result if needed. What do you think Aaron?
>>
>> Ondrej
>>
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