GSoC 2019 - Week 1111 Aug 2019
This was the eleventh week meeting with the GSoC mentors which was scheduled on
Sunday 11th August, 2019 between 11:30 - 12:30 PM (IST). Me, Yathartha and Amit
were the attendees of the meeting.
_solve_modular was discussed in this meeting.
Here is all the brief description about new solver
_solve_modular for solving
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.
In the starting of it there is a check if domain is a subset of 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)) 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
n = Dummy('n', integer=True) f_x, g_n = _invert_modular(modterm, rhs, n, symbol)
I am defining n in
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
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
if f_x is symbol: if domain is not S.Integers: return domain.intersect(g_n) return g_n
_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.
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
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
Hope I am able to clear out everything!!
Code improvement takes time!!Follow @jmig5776