o 8VaE@s ddlmZddlmZddlmZddlmZddlm Z ddl m Z ddl m Z mZmZddlmZdd lmZdd lmZmZdd lmZdd lmZdd lmZddlmZddlmZmZddl m!Z!ddZ"d ddZ#ddZ$ddZ%ddZ&ddddZ'dS)!)Add)ordered) expand_log)Pow)S)Dummy)LambertWexplog)root)roots)Polyfactor)_mexpand) separatevars)collect)powsimp)solve_invert)uniqcs^fdd|jD}t|D]}d|}||vr,||vr,|dtjur'|}||q|S)aprocess the generators of ``poly``, returning the set of generators that have ``symbol``. If there are two generators that are inverses of each other, prefer the one that has no denominator. Examples ======== >>> from sympy.solvers.bivariate import _filtered_gens >>> from sympy import Poly, exp >>> from sympy.abc import x >>> _filtered_gens(Poly(x + 1/x + exp(x)), x) {x, exp(x)} csh|] }|jvr|qS free_symbols).0gsymbolr9/usr/lib/python3/dist-packages/sympy/solvers/bivariate.py "sz!_filtered_gens..)genslistas_numer_denomrZOneremove)Zpolyrr rZagrrr_filtered_genss  r$NcsPfdd|D}t|dkr|dS|r&ttt|fdddSdS) aLReturns the term in lhs which contains the most of the func-type things e.g. log(log(x)) wins over log(x) if both terms appear. ``func`` can be a function (exp, log, etc...) or any other SymPy object, like Pow. If ``X`` is not ``None``, then the function returns the term composed with the most ``func`` having the specified variable. Examples ======== >>> from sympy.solvers.bivariate import _mostfunc >>> from sympy.functions.elementary.exponential import exp >>> from sympy.abc import x, y >>> _mostfunc(exp(x) + exp(exp(x) + 2), exp) exp(exp(x) + 2) >>> _mostfunc(exp(x) + exp(exp(y) + 2), exp) exp(exp(y) + 2) >>> _mostfunc(exp(x) + exp(exp(y) + 2), exp, x) exp(x) >>> _mostfunc(x, exp, x) is None True >>> _mostfunc(exp(x) + exp(x*y), exp, x) exp(x) cs4g|]}rjr|jvsjs|r|qSr)Z is_Symbolrhasrtmp)Xrr Gs  z_mostfunc..rrcs |SN)count)x)funcrrMs z_mostfunc..)keyN)atomslenmaxr!r)lhsr-r(Zftermsr)r(r-r _mostfunc,s  r4cCsddlm}||}||\}}|jr*|jr*t||\}}}|||||fS|js5d}||}}n |}t|j|dd\}}|rL| }| }|||fS)a(Return ``a, b, X`` assuming ``arg`` can be written as ``a*X + b`` where ``X`` is a symbol-dependent factor and ``a`` and ``b`` are independent of ``symbol``. Examples ======== >>> from sympy.functions.elementary.exponential import exp >>> from sympy.solvers.bivariate import _linab >>> from sympy.abc import x, y >>> from sympy import S >>> _linab(S(2), x) (2, 0, 1) >>> _linab(2*x, x) (2, 0, x) >>> _linab(y + y*x + 2*x, x) (y + 2, y, x) >>> _linab(3 + 2*exp(x), x) (2, 3, exp(x)) r) factor_termsFZas_Add) Zsympy.core.exprtoolsr5expandas_independentis_Mulis_Add_linabrcould_extract_minus_sign)argrr5ZindZdepabr,rrrr;Qs     r;c stt|}t|t|}|sgS||d}t| tr>||||jd}|jd}t|ts3gS| jd }||7}||jvrEgSt||\}}t |||}| |dusb|jvrdgS|jd}t||\}} | |krwgSt d} t | | |} ddg} g} || \}}|\}}t||}t d}fddt||||D}|D]+}| D]&}t||}|r|jsq| |}| D] }| || |qqq| S)z Given an expression assumed to be in the form ``F(X, a..f) = a*log(b*X + c) + d*X + f = 0`` where X = g(x) and x = g^-1(X), return the Lambert solution, ``x = g^-1(-c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(-f/a)))``. rNrhstcsg|] }|qSrr)rrBr>r?drrr)sz_lambert..)rrr4r subs isinstanceargsrr;rZas_coefficientrrr"Z as_coeff_Mulr r keysrZis_realappend)eqr,mainlogotherfZX2ZlogtermZlogargcZX1uZxusolnsZlambert_real_branchesZsolZnumZdenperBrGr=kwr@ZxurrCr_lambertxsV            (  rTcsfdd}|jdd\}}| }fddD}|s t|js&|jrtdij|fdd fd d }|jrp|rp|d }||} ||} | jso| ro| t j t j sot t | t | } || Sn#|jr|rt t |dd }t |}|r|jr||} || S|i}tt|dd }t} t|| \} }| | |i}g}|st|t }|r|jr|d krtt |t |}nG|jr||d }|r|jsfdd|tDr|st |t ||}n t ||t ||}tt |}nt||}|sxt|t}|rxt||}|jrE|d krEtt t |t |}n3|jrx||d }||}||}|ri|ri|d9}|d9}t |t |}tt |}|st|t}|rȈ|jjvrt||}|jr|d krtt t |t |}n!|jr||d }||}||}t |t |}tt |}|std|tt|S)aReturn solution to ``f`` if it is a Lambert-type expression else raise NotImplementedError. For ``f(X, a..f) = a*log(b*X + c) + d*X - f = 0`` the solution for ``X`` is ``X = -c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(f/a))``. There are a variety of forms for `f(X, a..f)` as enumerated below: 1a1) if B**B = R for R not in [0, 1] (since those cases would already be solved before getting here) then log of both sides gives log(B) + log(log(B)) = log(log(R)) and X = log(B), a = 1, b = 1, c = 0, d = 1, f = log(log(R)) 1a2) if B*(b*log(B) + c)**a = R then log of both sides gives log(B) + a*log(b*log(B) + c) = log(R) and X = log(B), d=1, f=log(R) 1b) if a*log(b*B + c) + d*B = R and X = B, f = R 2a) if (b*B + c)*exp(d*B + g) = R then log of both sides gives log(b*B + c) + d*B + g = log(R) and X = B, a = 1, f = log(R) - g 2b) if g*exp(d*B + h) - b*B = c then the log form is log(g) + d*B + h - log(b*B + c) = 0 and X = B, a = -1, f = -h - log(g) 3) if d*p**(a*B + g) - b*B = c then the log form is log(d) + (a*B + g)*log(p) - log(b*B + c) = 0 and X = B, a = -1, d = a*log(p), f = -log(d) - g*log(p) csLfdddD\}}t|}||kr |t|tt|S)aReturn the unique solutions of equations derived from ``expr`` by replacing ``t`` with ``+/- symbol``. Parameters ========== expr : Expr The expression which includes a dummy variable t to be replaced with +symbol and -symbol. symbol : Symbol The symbol for which a solution is being sought. Returns ======= List of unique solution of the two equations generated by replacing ``t`` with positive and negative ``symbol``. Notes ===== If ``expr = 2*log(t) + x/2` then solutions for ``2*log(x) + x/2 = 0`` and ``2*log(-x) + x/2 = 0`` are returned by this function. Though this may seem counter-intuitive, one must note that the ``expr`` being solved here has been derived from a different expression. For an expression like ``eq = x**2*g(x) = 1``, if we take the log of both sides we obtain ``log(x**2) + log(g(x)) = 0``. If x is positive then this simplifies to ``2*log(x) + log(g(x)) = 0``; the Lambert-solving routines will return solutions for this, but we must also consider the solutions for ``2*log(-x) + log(g(x))`` since those must also be a solution of ``eq`` which has the same value when the ``x`` in ``x**2`` is negated. If `g(x)` does not have even powers of symbol then we don't want to replace the ``x`` there with ``-x``. So the role of the ``t`` in the expression received by this function is to mark where ``+/-x`` should be inserted before obtaining the Lambert solutions. csg|] }|iqSr)xreplace)rZsgnexprrrBrrr) szC_solve_lambert.._solve_even_degree_expr..)rAr)_solve_lambertextendr!r)rWrBrZnlhsZplhsZsols)r rVr_solve_even_degree_exprs*  z/_solve_lambert.._solve_even_degree_exprTr6cs0g|]}|jttfvs|jr|jjvr|qSr)r-r r is_Powrr&rrrr)s   z"_solve_lambert..rBcs|jo |jko |jjSr*)r[baser Zis_evenirrrr.(sz _solve_lambert..cs |jSr*)r r]rBrrr.*s r)Zforce)Zdeepcsg|] }|jvr|qSrrr&rrrr)\s rAz:%s does not appear to have a solution in terms of LambertWNr_)r8NotImplementedErrorr:r9rZ assumptions0replacer%rErZComplexInfinityZNaNrr rUrrrr4rTr0rr rr<rr!r)rMrr rZZnrhsr3r@ZlamcheckZt_indepZt_termZ_rhsrJrr^ZsolnrKrLZdiffZmainexpZmaintermZmainpowr)r rrBrrXs "6                  rXTfirstcs8tddd}|rFt|}|}t}t}tt|||i||||dd}|rD||i} |d| |d| |dfSd S|}|}t|} g} | D]} t| |} | j } | vsn| vrpn| | qWt| |fSfd d }g} | }| |krt | ||} t | ||}||||| } | d ur| || |fSg} | }| |krtdD]C}t | |||} t | ||}||||| } | d ur| || |fSqd Sd S) aGiven an expression, f, 3 tests will be done to see what type of composite bivariate it might be, options for u(x, y) are:: x*y x+y x*y+x x*y+y If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and equating the solutions to ``u(x, y)`` and then solving for ``x`` or ``y`` is equivalent to solving the original expression for ``x`` or ``y``. If ``x`` and ``y`` represent two functions in the same variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p`` can be solved for ``t`` then these represent the solutions to ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``. Only positive values of ``u`` are considered. Examples ======== >>> from sympy.solvers.solvers import solve >>> from sympy.solvers.bivariate import bivariate_type >>> from sympy.abc import x, y >>> eq = (x**2 - 3).subs(x, x + y) >>> bivariate_type(eq, x, y) (x + y, _u**2 - 3, _u) >>> uxy, pu, u = _ >>> usol = solve(pu, u); usol [sqrt(3)] >>> [solve(uxy - s) for s in solve(pu, u)] [[{x: -y + sqrt(3)}]] >>> all(eq.subs(s).equals(0) for sol in _ for s in sol) True rOT)ZpositiveFrcrrNcs.t|||}|j}|vs|vrdS|Sr*)rrEr)rMvrNnewfreer,yrrokszbivariate_type..ok)rr Zas_exprbivariate_typerErUrZ make_argsrrrIZdegreer Zcoeff_monomialrange)rMr,rjrdrOrPZ_xZ_yZrvZrepsrGrgr>rhrkrDr?ZitryrrirrlsV ' & "      rlr*)(Zsympy.core.addrZsympy.core.compatibilityrZsympy.core.functionrZsympy.core.powerrZsympy.core.singletonrZsympy.core.symbolrZ&sympy.functions.elementary.exponentialrr r Z(sympy.functions.elementary.miscellaneousr Zsympy.polys.polyrootsr Zsympy.polys.polytoolsr rrZsympy.simplify.simplifyrZsympy.simplify.radsimprrZsympy.solvers.solversrrZsympy.utilities.iterablesrr$r4r;rTrXrlrrrrs.             %'Ia