Source code for sympy.simplify.simplify

from __future__ import print_function, division

from collections import defaultdict

from sympy.core import (Basic, S, Add, Mul, Pow,
    Symbol, sympify, expand_mul, expand_func,
    Function, Dummy, Expr, factor_terms,
    symbols, expand_power_exp)
from sympy.core.compatibility import (iterable,
    ordered, range, as_int)
from sympy.core.numbers import Float, I, pi, Rational, Integer
from sympy.core.function import expand_log, count_ops, _mexpand, _coeff_isneg
from sympy.core.rules import Transform
from sympy.core.evaluate import global_evaluate
from sympy.functions import (
    gamma, exp, sqrt, log, exp_polar, piecewise_fold)
from sympy.core.sympify import _sympify
from sympy.functions.elementary.exponential import ExpBase
from sympy.functions.elementary.hyperbolic import HyperbolicFunction
from sympy.functions.elementary.integers import ceiling
from sympy.functions.elementary.complexes import unpolarify
from sympy.functions.elementary.trigonometric import TrigonometricFunction
from sympy.functions.combinatorial.factorials import CombinatorialFunction
from sympy.functions.special.bessel import besselj, besseli, besselk, jn, bessely

from sympy.utilities.iterables import has_variety

from sympy.simplify.radsimp import radsimp, fraction
from sympy.simplify.trigsimp import trigsimp, exptrigsimp
from sympy.simplify.powsimp import powsimp
from sympy.simplify.cse_opts import sub_pre, sub_post
from sympy.simplify.sqrtdenest import sqrtdenest
from sympy.simplify.combsimp import combsimp

from sympy.polys import (together, cancel, factor)


import mpmath



[docs]def separatevars(expr, symbols=[], dict=False, force=False): """ Separates variables in an expression, if possible. By default, it separates with respect to all symbols in an expression and collects constant coefficients that are independent of symbols. If dict=True then the separated terms will be returned in a dictionary keyed to their corresponding symbols. By default, all symbols in the expression will appear as keys; if symbols are provided, then all those symbols will be used as keys, and any terms in the expression containing other symbols or non-symbols will be returned keyed to the string 'coeff'. (Passing None for symbols will return the expression in a dictionary keyed to 'coeff'.) If force=True, then bases of powers will be separated regardless of assumptions on the symbols involved. Notes ===== The order of the factors is determined by Mul, so that the separated expressions may not necessarily be grouped together. Although factoring is necessary to separate variables in some expressions, it is not necessary in all cases, so one should not count on the returned factors being factored. Examples ======== >>> from sympy.abc import x, y, z, alpha >>> from sympy import separatevars, sin >>> separatevars((x*y)**y) (x*y)**y >>> separatevars((x*y)**y, force=True) x**y*y**y >>> e = 2*x**2*z*sin(y)+2*z*x**2 >>> separatevars(e) 2*x**2*z*(sin(y) + 1) >>> separatevars(e, symbols=(x, y), dict=True) {'coeff': 2*z, x: x**2, y: sin(y) + 1} >>> separatevars(e, [x, y, alpha], dict=True) {'coeff': 2*z, alpha: 1, x: x**2, y: sin(y) + 1} If the expression is not really separable, or is only partially separable, separatevars will do the best it can to separate it by using factoring. >>> separatevars(x + x*y - 3*x**2) -x*(3*x - y - 1) If the expression is not separable then expr is returned unchanged or (if dict=True) then None is returned. >>> eq = 2*x + y*sin(x) >>> separatevars(eq) == eq True >>> separatevars(2*x + y*sin(x), symbols=(x, y), dict=True) == None True """ expr = sympify(expr) if dict: return _separatevars_dict(_separatevars(expr, force), symbols) else: return _separatevars(expr, force)
def _separatevars(expr, force): if len(expr.free_symbols) == 1: return expr # don't destroy a Mul since much of the work may already be done if expr.is_Mul: args = list(expr.args) changed = False for i, a in enumerate(args): args[i] = separatevars(a, force) changed = changed or args[i] != a if changed: expr = expr.func(*args) return expr # get a Pow ready for expansion if expr.is_Pow: expr = Pow(separatevars(expr.base, force=force), expr.exp) # First try other expansion methods expr = expr.expand(mul=False, multinomial=False, force=force) _expr, reps = posify(expr) if force else (expr, {}) expr = factor(_expr).subs(reps) if not expr.is_Add: return expr # Find any common coefficients to pull out args = list(expr.args) commonc = args[0].args_cnc(cset=True, warn=False)[0] for i in args[1:]: commonc &= i.args_cnc(cset=True, warn=False)[0] commonc = Mul(*commonc) commonc = commonc.as_coeff_Mul()[1] # ignore constants commonc_set = commonc.args_cnc(cset=True, warn=False)[0] # remove them for i, a in enumerate(args): c, nc = a.args_cnc(cset=True, warn=False) c = c - commonc_set args[i] = Mul(*c)*Mul(*nc) nonsepar = Add(*args) if len(nonsepar.free_symbols) > 1: _expr = nonsepar _expr, reps = posify(_expr) if force else (_expr, {}) _expr = (factor(_expr)).subs(reps) if not _expr.is_Add: nonsepar = _expr return commonc*nonsepar def _separatevars_dict(expr, symbols): if symbols: if not all((t.is_Atom for t in symbols)): raise ValueError("symbols must be Atoms.") symbols = list(symbols) elif symbols is None: return {'coeff': expr} else: symbols = list(expr.free_symbols) if not symbols: return None ret = dict(((i, []) for i in symbols + ['coeff'])) for i in Mul.make_args(expr): expsym = i.free_symbols intersection = set(symbols).intersection(expsym) if len(intersection) > 1: return None if len(intersection) == 0: # There are no symbols, so it is part of the coefficient ret['coeff'].append(i) else: ret[intersection.pop()].append(i) # rebuild for k, v in ret.items(): ret[k] = Mul(*v) return ret def _is_sum_surds(p): args = p.args if p.is_Add else [p] for y in args: if not ((y**2).is_Rational and y.is_real): return False return True
[docs]def posify(eq): """Return eq (with generic symbols made positive) and a dictionary containing the mapping between the old and new symbols. Any symbol that has positive=None will be replaced with a positive dummy symbol having the same name. This replacement will allow more symbolic processing of expressions, especially those involving powers and logarithms. A dictionary that can be sent to subs to restore eq to its original symbols is also returned. >>> from sympy import posify, Symbol, log, solve >>> from sympy.abc import x >>> posify(x + Symbol('p', positive=True) + Symbol('n', negative=True)) (_x + n + p, {_x: x}) >>> eq = 1/x >>> log(eq).expand() log(1/x) >>> log(posify(eq)[0]).expand() -log(_x) >>> p, rep = posify(eq) >>> log(p).expand().subs(rep) -log(x) It is possible to apply the same transformations to an iterable of expressions: >>> eq = x**2 - 4 >>> solve(eq, x) [-2, 2] >>> eq_x, reps = posify([eq, x]); eq_x [_x**2 - 4, _x] >>> solve(*eq_x) [2] """ eq = sympify(eq) if iterable(eq): f = type(eq) eq = list(eq) syms = set() for e in eq: syms = syms.union(e.atoms(Symbol)) reps = {} for s in syms: reps.update(dict((v, k) for k, v in posify(s)[1].items())) for i, e in enumerate(eq): eq[i] = e.subs(reps) return f(eq), {r: s for s, r in reps.items()} reps = dict([(s, Dummy(s.name, positive=True)) for s in eq.free_symbols if s.is_positive is None]) eq = eq.subs(reps) return eq, {r: s for s, r in reps.items()}
[docs]def hypersimp(f, k): """Given combinatorial term f(k) simplify its consecutive term ratio i.e. f(k+1)/f(k). The input term can be composed of functions and integer sequences which have equivalent representation in terms of gamma special function. The algorithm performs three basic steps: 1. Rewrite all functions in terms of gamma, if possible. 2. Rewrite all occurrences of gamma in terms of products of gamma and rising factorial with integer, absolute constant exponent. 3. Perform simplification of nested fractions, powers and if the resulting expression is a quotient of polynomials, reduce their total degree. If f(k) is hypergeometric then as result we arrive with a quotient of polynomials of minimal degree. Otherwise None is returned. For more information on the implemented algorithm refer to: 1. W. Koepf, Algorithms for m-fold Hypergeometric Summation, Journal of Symbolic Computation (1995) 20, 399-417 """ f = sympify(f) g = f.subs(k, k + 1) / f g = g.rewrite(gamma) g = expand_func(g) g = powsimp(g, deep=True, combine='exp') if g.is_rational_function(k): return simplify(g, ratio=S.Infinity) else: return None
[docs]def hypersimilar(f, g, k): """Returns True if 'f' and 'g' are hyper-similar. Similarity in hypergeometric sense means that a quotient of f(k) and g(k) is a rational function in k. This procedure is useful in solving recurrence relations. For more information see hypersimp(). """ f, g = list(map(sympify, (f, g))) h = (f/g).rewrite(gamma) h = h.expand(func=True, basic=False) return h.is_rational_function(k)
def signsimp(expr, evaluate=None): """Make all Add sub-expressions canonical wrt sign. If an Add subexpression, ``a``, can have a sign extracted, as determined by could_extract_minus_sign, it is replaced with Mul(-1, a, evaluate=False). This allows signs to be extracted from powers and products. Examples ======== >>> from sympy import signsimp, exp, symbols >>> from sympy.abc import x, y >>> i = symbols('i', odd=True) >>> n = -1 + 1/x >>> n/x/(-n)**2 - 1/n/x (-1 + 1/x)/(x*(1 - 1/x)**2) - 1/(x*(-1 + 1/x)) >>> signsimp(_) 0 >>> x*n + x*-n x*(-1 + 1/x) + x*(1 - 1/x) >>> signsimp(_) 0 Since powers automatically handle leading signs >>> (-2)**i -2**i signsimp can be used to put the base of a power with an integer exponent into canonical form: >>> n**i (-1 + 1/x)**i By default, signsimp doesn't leave behind any hollow simplification: if making an Add canonical wrt sign didn't change the expression, the original Add is restored. If this is not desired then the keyword ``evaluate`` can be set to False: >>> e = exp(y - x) >>> signsimp(e) == e True >>> signsimp(e, evaluate=False) exp(-(x - y)) """ if evaluate is None: evaluate = global_evaluate[0] expr = sympify(expr) if not isinstance(expr, Expr) or expr.is_Atom: return expr e = sub_post(sub_pre(expr)) if not isinstance(e, Expr) or e.is_Atom: return e if e.is_Add: return e.func(*[signsimp(a) for a in e.args]) if evaluate: e = e.xreplace({m: -(-m) for m in e.atoms(Mul) if -(-m) != m}) return e
[docs]def simplify(expr, ratio=1.7, measure=count_ops, fu=False): """ Simplifies the given expression. Simplification is not a well defined term and the exact strategies this function tries can change in the future versions of SymPy. If your algorithm relies on "simplification" (whatever it is), try to determine what you need exactly - is it powsimp()?, radsimp()?, together()?, logcombine()?, or something else? And use this particular function directly, because those are well defined and thus your algorithm will be robust. Nonetheless, especially for interactive use, or when you don't know anything about the structure of the expression, simplify() tries to apply intelligent heuristics to make the input expression "simpler". For example: >>> from sympy import simplify, cos, sin >>> from sympy.abc import x, y >>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2) >>> a (x**2 + x)/(x*sin(y)**2 + x*cos(y)**2) >>> simplify(a) x + 1 Note that we could have obtained the same result by using specific simplification functions: >>> from sympy import trigsimp, cancel >>> trigsimp(a) (x**2 + x)/x >>> cancel(_) x + 1 In some cases, applying :func:`simplify` may actually result in some more complicated expression. The default ``ratio=1.7`` prevents more extreme cases: if (result length)/(input length) > ratio, then input is returned unmodified. The ``measure`` parameter lets you specify the function used to determine how complex an expression is. The function should take a single argument as an expression and return a number such that if expression ``a`` is more complex than expression ``b``, then ``measure(a) > measure(b)``. The default measure function is :func:`count_ops`, which returns the total number of operations in the expression. For example, if ``ratio=1``, ``simplify`` output can't be longer than input. :: >>> from sympy import sqrt, simplify, count_ops, oo >>> root = 1/(sqrt(2)+3) Since ``simplify(root)`` would result in a slightly longer expression, root is returned unchanged instead:: >>> simplify(root, ratio=1) == root True If ``ratio=oo``, simplify will be applied anyway:: >>> count_ops(simplify(root, ratio=oo)) > count_ops(root) True Note that the shortest expression is not necessary the simplest, so setting ``ratio`` to 1 may not be a good idea. Heuristically, the default value ``ratio=1.7`` seems like a reasonable choice. You can easily define your own measure function based on what you feel should represent the "size" or "complexity" of the input expression. Note that some choices, such as ``lambda expr: len(str(expr))`` may appear to be good metrics, but have other problems (in this case, the measure function may slow down simplify too much for very large expressions). If you don't know what a good metric would be, the default, ``count_ops``, is a good one. For example: >>> from sympy import symbols, log >>> a, b = symbols('a b', positive=True) >>> g = log(a) + log(b) + log(a)*log(1/b) >>> h = simplify(g) >>> h log(a*b**(-log(a) + 1)) >>> count_ops(g) 8 >>> count_ops(h) 5 So you can see that ``h`` is simpler than ``g`` using the count_ops metric. However, we may not like how ``simplify`` (in this case, using ``logcombine``) has created the ``b**(log(1/a) + 1)`` term. A simple way to reduce this would be to give more weight to powers as operations in ``count_ops``. We can do this by using the ``visual=True`` option: >>> print(count_ops(g, visual=True)) 2*ADD + DIV + 4*LOG + MUL >>> print(count_ops(h, visual=True)) 2*LOG + MUL + POW + SUB >>> from sympy import Symbol, S >>> def my_measure(expr): ... POW = Symbol('POW') ... # Discourage powers by giving POW a weight of 10 ... count = count_ops(expr, visual=True).subs(POW, 10) ... # Every other operation gets a weight of 1 (the default) ... count = count.replace(Symbol, type(S.One)) ... return count >>> my_measure(g) 8 >>> my_measure(h) 14 >>> 15./8 > 1.7 # 1.7 is the default ratio True >>> simplify(g, measure=my_measure) -log(a)*log(b) + log(a) + log(b) Note that because ``simplify()`` internally tries many different simplification strategies and then compares them using the measure function, we get a completely different result that is still different from the input expression by doing this. """ expr = sympify(expr) try: return expr._eval_simplify(ratio=ratio, measure=measure) except AttributeError: pass original_expr = expr = signsimp(expr) from sympy.simplify.hyperexpand import hyperexpand from sympy.functions.special.bessel import BesselBase from sympy import Sum, Product if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack return expr if not isinstance(expr, (Add, Mul, Pow, ExpBase)): if isinstance(expr, Function) and hasattr(expr, "inverse"): if len(expr.args) == 1 and len(expr.args[0].args) == 1 and \ isinstance(expr.args[0], expr.inverse(argindex=1)): return simplify(expr.args[0].args[0], ratio=ratio, measure=measure, fu=fu) return expr.func(*[simplify(x, ratio=ratio, measure=measure, fu=fu) for x in expr.args]) # TODO: Apply different strategies, considering expression pattern: # is it a purely rational function? Is there any trigonometric function?... # See also https://github.com/sympy/sympy/pull/185. def shorter(*choices): '''Return the choice that has the fewest ops. In case of a tie, the expression listed first is selected.''' if not has_variety(choices): return choices[0] return min(choices, key=measure) expr = bottom_up(expr, lambda w: w.normal()) expr = Mul(*powsimp(expr).as_content_primitive()) _e = cancel(expr) expr1 = shorter(_e, _mexpand(_e).cancel()) # issue 6829 expr2 = shorter(together(expr, deep=True), together(expr1, deep=True)) if ratio is S.Infinity: expr = expr2 else: expr = shorter(expr2, expr1, expr) if not isinstance(expr, Basic): # XXX: temporary hack return expr expr = factor_terms(expr, sign=False) # hyperexpand automatically only works on hypergeometric terms expr = hyperexpand(expr) expr = piecewise_fold(expr) if expr.has(BesselBase): expr = besselsimp(expr) if expr.has(TrigonometricFunction) and not fu or expr.has( HyperbolicFunction): expr = trigsimp(expr, deep=True) if expr.has(log): expr = shorter(expand_log(expr, deep=True), logcombine(expr)) if expr.has(CombinatorialFunction, gamma): expr = combsimp(expr) if expr.has(Sum): expr = sum_simplify(expr) if expr.has(Product): expr = product_simplify(expr) short = shorter(powsimp(expr, combine='exp', deep=True), powsimp(expr), expr) short = shorter(short, factor_terms(short), expand_power_exp(expand_mul(short))) if short.has(TrigonometricFunction, HyperbolicFunction, ExpBase): short = exptrigsimp(short, simplify=False) # get rid of hollow 2-arg Mul factorization hollow_mul = Transform( lambda x: Mul(*x.args), lambda x: x.is_Mul and len(x.args) == 2 and x.args[0].is_Number and x.args[1].is_Add and x.is_commutative) expr = short.xreplace(hollow_mul) numer, denom = expr.as_numer_denom() if denom.is_Add: n, d = fraction(radsimp(1/denom, symbolic=False, max_terms=1)) if n is not S.One: expr = (numer*n).expand()/d if expr.could_extract_minus_sign(): n, d = fraction(expr) if d != 0: expr = signsimp(-n/(-d)) if measure(expr) > ratio*measure(original_expr): expr = original_expr return expr
def sum_simplify(s): """Main function for Sum simplification""" from sympy.concrete.summations import Sum from sympy.core.function import expand terms = Add.make_args(expand(s)) s_t = [] # Sum Terms o_t = [] # Other Terms for term in terms: if isinstance(term, Mul): other = 1 sum_terms = [] if not term.has(Sum): o_t.append(term) continue mul_terms = Mul.make_args(term) for mul_term in mul_terms: if isinstance(mul_term, Sum): r = mul_term._eval_simplify() sum_terms.extend(Add.make_args(r)) else: other = other * mul_term if len(sum_terms): #some simplification may have happened #use if so s_t.append(Mul(*sum_terms) * other) else: o_t.append(other) elif isinstance(term, Sum): #as above, we need to turn this into an add list r = term._eval_simplify() s_t.extend(Add.make_args(r)) else: o_t.append(term) result = Add(sum_combine(s_t), *o_t) return result def sum_combine(s_t): """Helper function for Sum simplification Attempts to simplify a list of sums, by combining limits / sum function's returns the simplified sum """ from sympy.concrete.summations import Sum used = [False] * len(s_t) for method in range(2): for i, s_term1 in enumerate(s_t): if not used[i]: for j, s_term2 in enumerate(s_t): if not used[j] and i != j: temp = sum_add(s_term1, s_term2, method) if isinstance(temp, Sum) or isinstance(temp, Mul): s_t[i] = temp s_term1 = s_t[i] used[j] = True result = S.Zero for i, s_term in enumerate(s_t): if not used[i]: result = Add(result, s_term) return result def factor_sum(self, limits=None, radical=False, clear=False, fraction=False, sign=True): """Helper function for Sum simplification if limits is specified, "self" is the inner part of a sum Returns the sum with constant factors brought outside """ from sympy.core.exprtools import factor_terms from sympy.concrete.summations import Sum result = self.function if limits is None else self limits = self.limits if limits is None else limits #avoid any confusion w/ as_independent if result == 0: return S.Zero #get the summation variables sum_vars = set([limit.args[0] for limit in limits]) #finally we try to factor out any common terms #and remove the from the sum if independent retv = factor_terms(result, radical=radical, clear=clear, fraction=fraction, sign=sign) #avoid doing anything bad if not result.is_commutative: return Sum(result, *limits) i, d = retv.as_independent(*sum_vars) if isinstance(retv, Add): return i * Sum(1, *limits) + Sum(d, *limits) else: return i * Sum(d, *limits) def sum_add(self, other, method=0): """Helper function for Sum simplification""" from sympy.concrete.summations import Sum from sympy import Mul #we know this is something in terms of a constant * a sum #so we temporarily put the constants inside for simplification #then simplify the result def __refactor(val): args = Mul.make_args(val) sumv = next(x for x in args if isinstance(x, Sum)) constant = Mul(*[x for x in args if x != sumv]) return Sum(constant * sumv.function, *sumv.limits) if isinstance(self, Mul): rself = __refactor(self) else: rself = self if isinstance(other, Mul): rother = __refactor(other) else: rother = other if type(rself) == type(rother): if method == 0: if rself.limits == rother.limits: return factor_sum(Sum(rself.function + rother.function, *rself.limits)) elif method == 1: if simplify(rself.function - rother.function) == 0: if len(rself.limits) == len(rother.limits) == 1: i = rself.limits[0][0] x1 = rself.limits[0][1] y1 = rself.limits[0][2] j = rother.limits[0][0] x2 = rother.limits[0][1] y2 = rother.limits[0][2] if i == j: if x2 == y1 + 1: return factor_sum(Sum(rself.function, (i, x1, y2))) elif x1 == y2 + 1: return factor_sum(Sum(rself.function, (i, x2, y1))) return Add(self, other) def product_simplify(s): """Main function for Product simplification""" from sympy.concrete.products import Product terms = Mul.make_args(s) p_t = [] # Product Terms o_t = [] # Other Terms for term in terms: if isinstance(term, Product): p_t.append(term) else: o_t.append(term) used = [False] * len(p_t) for method in range(2): for i, p_term1 in enumerate(p_t): if not used[i]: for j, p_term2 in enumerate(p_t): if not used[j] and i != j: if isinstance(product_mul(p_term1, p_term2, method), Product): p_t[i] = product_mul(p_term1, p_term2, method) used[j] = True result = Mul(*o_t) for i, p_term in enumerate(p_t): if not used[i]: result = Mul(result, p_term) return result def product_mul(self, other, method=0): """Helper function for Product simplification""" from sympy.concrete.products import Product if type(self) == type(other): if method == 0: if self.limits == other.limits: return Product(self.function * other.function, *self.limits) elif method == 1: if simplify(self.function - other.function) == 0: if len(self.limits) == len(other.limits) == 1: i = self.limits[0][0] x1 = self.limits[0][1] y1 = self.limits[0][2] j = other.limits[0][0] x2 = other.limits[0][1] y2 = other.limits[0][2] if i == j: if x2 == y1 + 1: return Product(self.function, (i, x1, y2)) elif x1 == y2 + 1: return Product(self.function, (i, x2, y1)) return Mul(self, other) def _nthroot_solve(p, n, prec): """ helper function for ``nthroot`` It denests ``p**Rational(1, n)`` using its minimal polynomial """ from sympy.polys.numberfields import _minimal_polynomial_sq from sympy.solvers import solve while n % 2 == 0: p = sqrtdenest(sqrt(p)) n = n // 2 if n == 1: return p pn = p**Rational(1, n) x = Symbol('x') f = _minimal_polynomial_sq(p, n, x) if f is None: return None sols = solve(f, x) for sol in sols: if abs(sol - pn).n() < 1./10**prec: sol = sqrtdenest(sol) if _mexpand(sol**n) == p: return sol
[docs]def logcombine(expr, force=False): """ Takes logarithms and combines them using the following rules: - log(x) + log(y) == log(x*y) if both are not negative - a*log(x) == log(x**a) if x is positive and a is real If ``force`` is True then the assumptions above will be assumed to hold if there is no assumption already in place on a quantity. For example, if ``a`` is imaginary or the argument negative, force will not perform a combination but if ``a`` is a symbol with no assumptions the change will take place. Examples ======== >>> from sympy import Symbol, symbols, log, logcombine, I >>> from sympy.abc import a, x, y, z >>> logcombine(a*log(x) + log(y) - log(z)) a*log(x) + log(y) - log(z) >>> logcombine(a*log(x) + log(y) - log(z), force=True) log(x**a*y/z) >>> x,y,z = symbols('x,y,z', positive=True) >>> a = Symbol('a', real=True) >>> logcombine(a*log(x) + log(y) - log(z)) log(x**a*y/z) The transformation is limited to factors and/or terms that contain logs, so the result depends on the initial state of expansion: >>> eq = (2 + 3*I)*log(x) >>> logcombine(eq, force=True) == eq True >>> logcombine(eq.expand(), force=True) log(x**2) + I*log(x**3) See Also ======== posify: replace all symbols with symbols having positive assumptions """ def f(rv): if not (rv.is_Add or rv.is_Mul): return rv def gooda(a): # bool to tell whether the leading ``a`` in ``a*log(x)`` # could appear as log(x**a) return (a is not S.NegativeOne and # -1 *could* go, but we disallow (a.is_real or force and a.is_real is not False)) def goodlog(l): # bool to tell whether log ``l``'s argument can combine with others a = l.args[0] return a.is_positive or force and a.is_nonpositive is not False other = [] logs = [] log1 = defaultdict(list) for a in Add.make_args(rv): if a.func is log and goodlog(a): log1[()].append(([], a)) elif not a.is_Mul: other.append(a) else: ot = [] co = [] lo = [] for ai in a.args: if ai.is_Rational and ai < 0: ot.append(S.NegativeOne) co.append(-ai) elif ai.func is log and goodlog(ai): lo.append(ai) elif gooda(ai): co.append(ai) else: ot.append(ai) if len(lo) > 1: logs.append((ot, co, lo)) elif lo: log1[tuple(ot)].append((co, lo[0])) else: other.append(a) # if there is only one log at each coefficient and none have # an exponent to place inside the log then there is nothing to do if not logs and all(len(log1[k]) == 1 and log1[k][0] == [] for k in log1): return rv # collapse multi-logs as far as possible in a canonical way # TODO: see if x*log(a)+x*log(a)*log(b) -> x*log(a)*(1+log(b))? # -- in this case, it's unambiguous, but if it were were a log(c) in # each term then it's arbitrary whether they are grouped by log(a) or # by log(c). So for now, just leave this alone; it's probably better to # let the user decide for o, e, l in logs: l = list(ordered(l)) e = log(l.pop(0).args[0]**Mul(*e)) while l: li = l.pop(0) e = log(li.args[0]**e) c, l = Mul(*o), e if l.func is log: # it should be, but check to be sure log1[(c,)].append(([], l)) else: other.append(c*l) # logs that have the same coefficient can multiply for k in list(log1.keys()): log1[Mul(*k)] = log(logcombine(Mul(*[ l.args[0]**Mul(*c) for c, l in log1.pop(k)]), force=force)) # logs that have oppositely signed coefficients can divide for k in ordered(list(log1.keys())): if not k in log1: # already popped as -k continue if -k in log1: # figure out which has the minus sign; the one with # more op counts should be the one num, den = k, -k if num.count_ops() > den.count_ops(): num, den = den, num other.append(num*log(log1.pop(num).args[0]/log1.pop(den).args[0])) else: other.append(k*log1.pop(k)) return Add(*other) return bottom_up(expr, f)
def bottom_up(rv, F, atoms=False, nonbasic=False): """Apply ``F`` to all expressions in an expression tree from the bottom up. If ``atoms`` is True, apply ``F`` even if there are no args; if ``nonbasic`` is True, try to apply ``F`` to non-Basic objects. """ try: if rv.args: args = tuple([bottom_up(a, F, atoms, nonbasic) for a in rv.args]) if args != rv.args: rv = rv.func(*args) rv = F(rv) elif atoms: rv = F(rv) except AttributeError: if nonbasic: try: rv = F(rv) except TypeError: pass return rv
[docs]def besselsimp(expr): """ Simplify bessel-type functions. This routine tries to simplify bessel-type functions. Currently it only works on the Bessel J and I functions, however. It works by looking at all such functions in turn, and eliminating factors of "I" and "-1" (actually their polar equivalents) in front of the argument. Then, functions of half-integer order are rewritten using strigonometric functions and functions of integer order (> 1) are rewritten using functions of low order. Finally, if the expression was changed, compute factorization of the result with factor(). >>> from sympy import besselj, besseli, besselsimp, polar_lift, I, S >>> from sympy.abc import z, nu >>> besselsimp(besselj(nu, z*polar_lift(-1))) exp(I*pi*nu)*besselj(nu, z) >>> besselsimp(besseli(nu, z*polar_lift(-I))) exp(-I*pi*nu/2)*besselj(nu, z) >>> besselsimp(besseli(S(-1)/2, z)) sqrt(2)*cosh(z)/(sqrt(pi)*sqrt(z)) >>> besselsimp(z*besseli(0, z) + z*(besseli(2, z))/2 + besseli(1, z)) 3*z*besseli(0, z)/2 """ # TODO # - better algorithm? # - simplify (cos(pi*b)*besselj(b,z) - besselj(-b,z))/sin(pi*b) ... # - use contiguity relations? def replacer(fro, to, factors): factors = set(factors) def repl(nu, z): if factors.intersection(Mul.make_args(z)): return to(nu, z) return fro(nu, z) return repl def torewrite(fro, to): def tofunc(nu, z): return fro(nu, z).rewrite(to) return tofunc def tominus(fro): def tofunc(nu, z): return exp(I*pi*nu)*fro(nu, exp_polar(-I*pi)*z) return tofunc orig_expr = expr ifactors = [I, exp_polar(I*pi/2), exp_polar(-I*pi/2)] expr = expr.replace( besselj, replacer(besselj, torewrite(besselj, besseli), ifactors)) expr = expr.replace( besseli, replacer(besseli, torewrite(besseli, besselj), ifactors)) minusfactors = [-1, exp_polar(I*pi)] expr = expr.replace( besselj, replacer(besselj, tominus(besselj), minusfactors)) expr = expr.replace( besseli, replacer(besseli, tominus(besseli), minusfactors)) z0 = Dummy('z') def expander(fro): def repl(nu, z): if (nu % 1) == S(1)/2: return exptrigsimp(trigsimp(unpolarify( fro(nu, z0).rewrite(besselj).rewrite(jn).expand( func=True)).subs(z0, z))) elif nu.is_Integer and nu > 1: return fro(nu, z).expand(func=True) return fro(nu, z) return repl expr = expr.replace(besselj, expander(besselj)) expr = expr.replace(bessely, expander(bessely)) expr = expr.replace(besseli, expander(besseli)) expr = expr.replace(besselk, expander(besselk)) if expr != orig_expr: expr = expr.factor() return expr
[docs]def nthroot(expr, n, max_len=4, prec=15): """ compute a real nth-root of a sum of surds Parameters ========== expr : sum of surds n : integer max_len : maximum number of surds passed as constants to ``nsimplify`` Algorithm ========= First ``nsimplify`` is used to get a candidate root; if it is not a root the minimal polynomial is computed; the answer is one of its roots. Examples ======== >>> from sympy.simplify.simplify import nthroot >>> from sympy import Rational, sqrt >>> nthroot(90 + 34*sqrt(7), 3) sqrt(7) + 3 """ expr = sympify(expr) n = sympify(n) p = expr**Rational(1, n) if not n.is_integer: return p if not _is_sum_surds(expr): return p surds = [] coeff_muls = [x.as_coeff_Mul() for x in expr.args] for x, y in coeff_muls: if not x.is_rational: return p if y is S.One: continue if not (y.is_Pow and y.exp == S.Half and y.base.is_integer): return p surds.append(y) surds.sort() surds = surds[:max_len] if expr < 0 and n % 2 == 1: p = (-expr)**Rational(1, n) a = nsimplify(p, constants=surds) res = a if _mexpand(a**n) == _mexpand(-expr) else p return -res a = nsimplify(p, constants=surds) if _mexpand(a) is not _mexpand(p) and _mexpand(a**n) == _mexpand(expr): return _mexpand(a) expr = _nthroot_solve(expr, n, prec) if expr is None: return p return expr
[docs]def nsimplify(expr, constants=(), tolerance=None, full=False, rational=None, rational_conversion='base10'): """ Find a simple representation for a number or, if there are free symbols or if rational=True, then replace Floats with their Rational equivalents. If no change is made and rational is not False then Floats will at least be converted to Rationals. For numerical expressions, a simple formula that numerically matches the given numerical expression is sought (and the input should be possible to evalf to a precision of at least 30 digits). Optionally, a list of (rationally independent) constants to include in the formula may be given. A lower tolerance may be set to find less exact matches. If no tolerance is given then the least precise value will set the tolerance (e.g. Floats default to 15 digits of precision, so would be tolerance=10**-15). With full=True, a more extensive search is performed (this is useful to find simpler numbers when the tolerance is set low). When converting to rational, if rational_conversion='base10' (the default), then convert floats to rationals using their base-10 (string) representation. When rational_conversion='exact' it uses the exact, base-2 representation. Examples ======== >>> from sympy import nsimplify, sqrt, GoldenRatio, exp, I, exp, pi >>> nsimplify(4/(1+sqrt(5)), [GoldenRatio]) -2 + 2*GoldenRatio >>> nsimplify((1/(exp(3*pi*I/5)+1))) 1/2 - I*sqrt(sqrt(5)/10 + 1/4) >>> nsimplify(I**I, [pi]) exp(-pi/2) >>> nsimplify(pi, tolerance=0.01) 22/7 >>> nsimplify(0.333333333333333, rational=True, rational_conversion='exact') 6004799503160655/18014398509481984 >>> nsimplify(0.333333333333333, rational=True) 1/3 See Also ======== sympy.core.function.nfloat """ try: return sympify(as_int(expr)) except (TypeError, ValueError): pass expr = sympify(expr).xreplace({ Float('inf'): S.Infinity, Float('-inf'): S.NegativeInfinity, }) if expr is S.Infinity or expr is S.NegativeInfinity: return expr if rational or expr.free_symbols: return _real_to_rational(expr, tolerance, rational_conversion) # SymPy's default tolerance for Rationals is 15; other numbers may have # lower tolerances set, so use them to pick the largest tolerance if None # was given if tolerance is None: tolerance = 10**-min([15] + [mpmath.libmp.libmpf.prec_to_dps(n._prec) for n in expr.atoms(Float)]) # XXX should prec be set independent of tolerance or should it be computed # from tolerance? prec = 30 bprec = int(prec*3.33) constants_dict = {} for constant in constants: constant = sympify(constant) v = constant.evalf(prec) if not v.is_Float: raise ValueError("constants must be real-valued") constants_dict[str(constant)] = v._to_mpmath(bprec) exprval = expr.evalf(prec, chop=True) re, im = exprval.as_real_imag() # safety check to make sure that this evaluated to a number if not (re.is_Number and im.is_Number): return expr def nsimplify_real(x): orig = mpmath.mp.dps xv = x._to_mpmath(bprec) try: # We'll be happy with low precision if a simple fraction if not (tolerance or full): mpmath.mp.dps = 15 rat = mpmath.pslq([xv, 1]) if rat is not None: return Rational(-int(rat[1]), int(rat[0])) mpmath.mp.dps = prec newexpr = mpmath.identify(xv, constants=constants_dict, tol=tolerance, full=full) if not newexpr: raise ValueError if full: newexpr = newexpr[0] expr = sympify(newexpr) if x and not expr: # don't let x become 0 raise ValueError if expr.is_finite is False and not xv in [mpmath.inf, mpmath.ninf]: raise ValueError return expr finally: # even though there are returns above, this is executed # before leaving mpmath.mp.dps = orig try: if re: re = nsimplify_real(re) if im: im = nsimplify_real(im) except ValueError: if rational is None: return _real_to_rational(expr, rational_conversion=rational_conversion) return expr rv = re + im*S.ImaginaryUnit # if there was a change or rational is explicitly not wanted # return the value, else return the Rational representation if rv != expr or rational is False: return rv return _real_to_rational(expr, rational_conversion=rational_conversion)
def _real_to_rational(expr, tolerance=None, rational_conversion='base10'): """ Replace all reals in expr with rationals. >>> from sympy import Rational >>> from sympy.simplify.simplify import _real_to_rational >>> from sympy.abc import x >>> _real_to_rational(.76 + .1*x**.5) sqrt(x)/10 + 19/25 If rational_conversion='base10', this uses the base-10 string. If rational_conversion='exact', the exact, base-2 representation is used. >>> _real_to_rational(0.333333333333333, rational_conversion='exact') 6004799503160655/18014398509481984 >>> _real_to_rational(0.333333333333333) 1/3 """ expr = _sympify(expr) inf = Float('inf') p = expr reps = {} reduce_num = None if tolerance is not None and tolerance < 1: reduce_num = ceiling(1/tolerance) for fl in p.atoms(Float): key = fl if reduce_num is not None: r = Rational(fl).limit_denominator(reduce_num) elif (tolerance is not None and tolerance >= 1 and fl.is_Integer is False): r = Rational(tolerance*round(fl/tolerance) ).limit_denominator(int(tolerance)) else: if rational_conversion == 'exact': r = Rational(fl) reps[key] = r continue elif rational_conversion != 'base10': raise ValueError("rational_conversion must be 'base10' or 'exact'") r = nsimplify(fl, rational=False) # e.g. log(3).n() -> log(3) instead of a Rational if fl and not r: r = Rational(fl) elif not r.is_Rational: if fl == inf or fl == -inf: r = S.ComplexInfinity elif fl < 0: fl = -fl d = Pow(10, int((mpmath.log(fl)/mpmath.log(10)))) r = -Rational(str(fl/d))*d elif fl > 0: d = Pow(10, int((mpmath.log(fl)/mpmath.log(10)))) r = Rational(str(fl/d))*d else: r = Integer(0) reps[key] = r return p.subs(reps, simultaneous=True) def clear_coefficients(expr, rhs=S.Zero): """Return `p, r` where `p` is the expression obtained when Rational additive and multiplicative coefficients of `expr` have been stripped away in a naive fashion (i.e. without simplification). The operations needed to remove the coefficients will be applied to `rhs` and returned as `r`. Examples ======== >>> from sympy.simplify.simplify import clear_coefficients >>> from sympy.abc import x, y >>> from sympy import Dummy >>> expr = 4*y*(6*x + 3) >>> clear_coefficients(expr - 2) (y*(2*x + 1), 1/6) When solving 2 or more expressions like `expr = a`, `expr = b`, etc..., it is advantageous to provide a Dummy symbol for `rhs` and simply replace it with `a`, `b`, etc... in `r`. >>> rhs = Dummy('rhs') >>> clear_coefficients(expr, rhs) (y*(2*x + 1), _rhs/12) >>> _[1].subs(rhs, 2) 1/6 """ was = None free = expr.free_symbols if expr.is_Rational: return (S.Zero, rhs - expr) while expr and was != expr: was = expr m, expr = ( expr.as_content_primitive() if free else factor_terms(expr).as_coeff_Mul(rational=True)) rhs /= m c, expr = expr.as_coeff_Add(rational=True) rhs -= c expr = signsimp(expr, evaluate = False) if _coeff_isneg(expr): expr = -expr rhs = -rhs return expr, rhs