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GSoC 2019 Report Jogi Miglani Solvers: Extending solveset

Jogi Miglani edited this page Aug 22, 2019 · 2 revisions

This report summarizes the work done in my GSoC 2019 project, Solvers: Extending with SymPy. A step by step development of the project is available at https://jmig5776.github.io.

About Me

I am, Jogi Miglani, a third year student at Indian Institute of Technology BHU Varanasi in the department of Mathematics and Computing.

Work Completed

Here is a list of PRs which were opened during the span of GSoC:

  1. #16796 Added _solve_modular for handling equations a - Mod(b, c) = 0 where only b is expr

  2. #16890 Fixing lambert in bivariate to give all real solutions

  3. #16960 (Don't Merge)(Prototype) Adding abs while converting equation to log form to get solved by _lambert

  4. #17043 Feature power_list to return all powers of a variable present in f

  5. #17079 Defining ImageSet Union

Here is a list of PRs merged:

  1. #16796 Added _solve_modular for handling equations a - Mod(b, c) = 0 where only b is expr

  2. #16890 Fixing lambert in bivariate to give all real solutions

Here is all the brief description about the PRs merged:

  1. #16796 Added _solve_modular for handling equations a - Mod(b, c) = 0 where only b is expr

In this PR a new solver _solve_modular was made for solving modular equations.

What type of equations to be considered and what domain?

A - Mod(B, C) = 0

    A -> This can or cannot be a function specifically(Linear, nth degree single
         Pow, a**f_x and Add and Mul) of symbol.(But currently its not a
        function of x)
    B -> This is surely a function of symbol.
    C -> It is an integer.
And domain should be a subset of S.Integers.

Filtering out equations

A check is being applied named _is_modular which verifies that only above mentioned type equation should return True.

Working of _solve_modular

In the starting of it there is a check if domain is a subset of Integers.

domain.is_subset(S.Integers)

Only domain of integers and it subset are being considered while solving these equations. Now after this it separates out a modterm and the rest term on either sides by this code.

modterm = list(f.atoms(Mod))[0]
rhs = -(S.One)*(f.subs(modterm, S.Zero))
if f.as_coefficients_dict()[modterm].is_negative:
    # f.as_coefficient(modterm) was returning None don't know why
    # checks if coefficient of modterm is negative in main equation.
    rhs *= -(S.One)

Now the equation is being inverted with the helper routine _invert_modular like this.

n = Dummy('n', integer=True)
f_x, g_n = _invert_modular(modterm, rhs, n, symbol)

I am defining n in _solve_modular because _invert_modular contains recursive calls to itself so if define the n there then it was going to have many instances which of no use. Thats y I am defining it in _solve_modular.

Now after the equation is inverted now solution finding takes place.

if f_x is modterm and g_n is rhs:
        return unsolved_result

First of all if _invert_modular fails to invert then a ConditionSet is being returned.

    if f_x is symbol:
        if domain is not S.Integers:
            return domain.intersect(g_n)
        return g_n

And if _invert_modular is fully able to invert the equation then only domain intersection needs to takes place. _invert_modular inverts the equation considering S.Integers as its default domain.

    if isinstance(g_n, ImageSet):
        lamda_expr = g_n.lamda.expr
        lamda_vars = g_n.lamda.variables
        base_set = g_n.base_set
        sol_set = _solveset(f_x - lamda_expr, symbol, S.Integers)
        if isinstance(sol_set, FiniteSet):
            tmp_sol = EmptySet()
            for sol in sol_set:
                tmp_sol += ImageSet(Lambda(lamda_vars, sol), base_set)
            sol_set = tmp_sol
        return domain.intersect(sol_set)

In this case when g_n is an ImageSet of n and f_x is not symbol so the equation is being solved by calling _solveset (this will not lead to recursion because equation to be entered is free from Mod) and then the domain intersection takes place.

What does _invert_modular do?

This function helps to convert the equation A - Mod(B, C) = 0 to a form (f_x, g_n). First of all it checks the possible instances of invertible cases if not then it returns the equation as it is.

a, m = modterm.args
if not isinstance(a, (Dummy, Symbol, Add, Mul, Pow)):
        return modterm, rhs

Now here is the check for complex arguments and returns the equation as it is if somewhere it finds I.

if rhs.is_real is False or any(term.is_real is False \
            for term in list(_term_factors(a))):
        # Check for complex arguments
        return modterm, rhs

Now after this we check of emptyset as a solution by checking range of both sides of equation. As modterm can have values between [0, m - 1] and if rhs is out of this range then emptySet is being returned.

if (abs(rhs) - abs(m)).is_positive or (abs(rhs) - abs(m)) is S.Zero:
        # if rhs has value greater than value of m.
        return symbol, EmptySet()

Now the equation haveing these types are being returned as the following

if a is symbol:
        return symbol, ImageSet(Lambda(n, m*n + rhs), S.Integers)

    if a.is_Add:
        # g + h = a
        g, h = a.as_independent(symbol)
        if g is not S.Zero:
            return _invert_modular(Mod(h, m), (rhs - Mod(g, m)) % m, n, symbol)

    if a.is_Mul:
        # g*h = a
        g, h = a.as_independent(symbol)
        if g is not S.One:
            return _invert_modular(Mod(h, m), (rhs*invert(g, m)) % m, n, symbol)

The more peculiar case is of a.is_Pow which is handled as following.

if a.is_Pow:
        # base**expo = a
        base, expo = a.args
        if expo.has(symbol) and not base.has(symbol):
            # remainder -> solution independent of n of equation.
            # m, rhs are made coprime by dividing igcd(m, rhs)
            try:
                remainder = discrete_log(m / igcd(m, rhs), rhs, a.base)
            except ValueError: # log does not exist
                return modterm, rhs
            # period -> coefficient of n in the solution and also referred as
            # the least period of expo in which it is repeats itself.
            # (a**(totient(m)) - 1) divides m. Here is link of theoram:
            # (https://en.wikipedia.org/wiki/Euler's_theorem)
            period = totient(m)
            for p in divisors(period):
                # there might a lesser period exist than totient(m).
                if pow(a.base, p, m / igcd(m, a.base)) == 1:
                    period = p
                    break
            return expo, ImageSet(Lambda(n, period*n + remainder), S.Naturals0)
        elif base.has(symbol) and not expo.has(symbol):
            remainder_list = nthroot_mod(rhs, expo, m, all_roots=True)
            if remainder_list is None:
                return symbol, EmptySet()
            g_n = EmptySet()
            for rem in remainder_list:
                g_n += ImageSet(Lambda(n, m*n + rem), S.Integers)
            return base, g_n

Two cases are being created based of a.is_Pow

  1. x**a
  2. a**x

x**a - It is being handled by the helper function nthroot_mod which returns required solution. I am not going into very mch detail for more information you can read the documentation of nthroot_mod.

a**x - For this totient is being used in the picture whose meaning can be find on this Wikipedia page. And then its divisors are being checked to find the least period of solutions.

  1. #16890 Fixing lambert in bivariate to give all real solutions

This PR went through many up and downs and nearly made to the most commented PR. And with the help of @smichr it was successfully merged. It mainly solved the bug for not returning all solutions of lambert.

Explaining the function _solve_lambert (main function to solve lambert equations)

Input - f, symbol, gens
OutPut - Solution of f = 0 if its lambert type expression else NotImplementedError

This function separates out cases as below based on the main function present in the main equation.

For the first ones:
1a1) B**B = R != 0 (when 0, there is only a solution if the base is 0,
                   but if it is, the exp is 0 and 0**0=1
                   comes back as B*log(B) = log(R)
1a2) B*(a + b*log(B))**p = R or with monomial expanded or with whole
                            thing expanded comes back unchanged
   log(B) + p*log(a + b*log(B)) = log(R)
   lhs is Mul:
       expand log of both sides to give:
       log(B) + log(log(B)) = log(log(R))
1b) d*log(a*B + b) + c*B = R
   lhs is Add:
       isolate c*B and expand log of both sides:
       log(c) + log(B) = log(R - d*log(a*B + b))

If the equation are of type 1a1, 1a2 and 1b then the mainlog of the equation is taken into concern as the deciding factor lies in the main logarithmic term of equation.

For the next two,
   collect on main exp
   2a) (b*B + c)*exp(d*B + g) = R
       lhs is mul:
           log to give
           log(b*B + c) + d*B = log(R) - g
   2b) -b*B + g*exp(d*B + h) = R
       lhs is add:
           add b*B
           log and rearrange
           log(R + b*B) - d*B = log(g) + h

If the equation are of type 2a and 2b then the mainexp of the equation is taken into concern as the deciding factor lies in the main exponential term of equation.

3) d*p**(a*B + b) + c*B = R
   collect on main pow
   log(R - c*B) - a*B*log(p) = log(d) + b*log(p)

If the equation are of type 3 then the mainpow of the equation is taken into concern as the deciding factor lies in the main power term of equation.

Eventually from all of the three cases the equation is meant to be converted to this form:-

f(x, a..f) = a*log(b*X + c) + d*X - f = 0 which has the
solution,  X = -c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(f/a)).

And the solution calculation process is done by _lambert function.

Everything seems flawless?? You might be thinking no modification is required. Lets see what loopholes are there in it.

What does PR #16890 do?

There are basically two flaws present with the this approach.

  1. Not considering all branches of equation while taking log both sides.
  2. Calculation of roots should consider all roots in case having rational power.

1. Not considering all branches of equation while taking log both sides.

Let us consider this equation to be solved by _solve_lambert function.

-1/x**2 + exp(x/2)/2 = 0

So what the old _solve_lambert do is to convert this equation to following.

2*log(x) + x/2 = 0

and calculates its roots from _lambert. But it missed this branch of equation while taking log on main equation.

2*log(-x) + x/2 = 0

Yeah you can reproduce the original equation from this equation.So basically the problem was that it missed the branches of equation while taking log. And when does the main equation have more than one branch?? The terms having even powers of variable x leads to two different branches of equation.

So how it is solved? What I has done is that before actually gets into solving I preprocess the main equation and if it has more than one branches of equation while converting taking log then I consider all the equations generated from them.(with the help of _solve_even_degree_expr)

How I preprocess the equation? So what I do is I replace all the even powers of x present with even powers of t(dummy variable).

Code for targeted replacement
lhs = lhs.replace(
            lambda i:  # find symbol**even
                i.is_Pow and i.base == symbol and i.exp.is_even,
            lambda i:  # replace t**even
                t**i.exp)
Example:-
Main equation -> -1/x**2 + exp(x/2)/2 = 0
After replacement -> -1/t**2 + exp(x/2)/2 = 0

Now I take logarithms on both sides and simplify it.

After simplifying -> 2*log(t) + x/2 = 0

Now I call function _solve_even_degree_expr to replace the t with +/-x to generate two equations.

Replacing t with +/-x
1. 2*log(x) + x/2 = 0
2. 2*log(-x) + x/2 = 0

And consider the solutions of both of the equations to return all lambert real solutions of -1/x**2 + exp(x/2)/2 = 0.

Hope you could understand the logic behind this work.

2. Calculation of roots should consider all roots in case having rational power.

This flaw is in the calculation of roots in function _lambert. Earlier the function_lambert has the working like :-

  1. Find all the values of a, b, c, d, e in the required loagrithmic equation
  2. Then it defines a solution of the form
-c/b + (a/d)*l where l = LambertW(d/(a*b)*exp(c*d/a/b)*exp(-f/a), k)

and then it included that solution. I agree everything seems flawless here. but try to see the step where we are defining l.

Let us suppose a hypothetical algorithm just like algorithm used in _lambert in which equation to be solved is

x**3 - 1 = 0

and in which we define solution of the form

x = exp(I*2*pi/n) where n is the power of x in equation

so the algorithm will give solution

x = exp(I*2*pi/3) # but expected was [1, exp(I*2*pi/3), exp(-I*2*pi/3)]

which can be found by finding all solutions of

x**n - exp(2*I*pi) = 0

by a different correct algorithm. Thats y it was wrong. The above algorithm would have given correct values for x - 1 = 0.

And the question in your mind may arise that why only exp() because the possiblity of having more than one roots is in exp(), because if the algorithm would have been like x = a, where a is some real constant then there is not any possiblity of further roots rather than solution like x = a**(1/n). And its been done in code like this:

code
num, den = ((c*d-b*f)/a/b).as_numer_denom()
p, den = den.as_coeff_Mul()
e = exp(num/den)
t = Dummy('t')
args = [d/(a*b)*t for t in roots(t**p - e, t).keys()]

Work under development

This PR tends to define a unifying algorithm for linear relations.

Future Work

Here is a list that comprises of all the ideas (which were a part of my GSoC Proposal and/or thought over during the SoC) which can extend my GSoC project.

  1. Integrating helper solvers within solveset: linsolve, solve_decomposition, nonlinsolve

  2. Handle nested trigonometric equations.

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