Source code for diofant.simplify.simplify

from collections import defaultdict

import mpmath

from ..core import (Add, Basic, Dummy, E, Expr, Float, I, Integer, Mul, Pow,
                    Rational, S, Symbol, count_ops, expand_func, expand_log,
                    expand_mul, expand_power_exp, factor_terms, oo, pi,
                    sympify)
from ..core.compatibility import as_int, iterable, ordered
from ..core.evaluate import global_evaluate
from ..core.function import _coeff_isneg, _mexpand
from ..core.rules import Transform
from ..functions import (besseli, besselj, besselk, bessely, ceiling, exp,
                         exp_polar, gamma, jn, log, piecewise_fold, root, sqrt,
                         unpolarify)
from ..functions.combinatorial.factorials import CombinatorialFunction
from ..functions.elementary.hyperbolic import HyperbolicFunction
from ..functions.elementary.trigonometric import TrigonometricFunction
from ..polys import cancel, factor, together
from ..utilities import has_variety
from .combsimp import combsimp
from .cse_opts import sub_post, sub_pre
from .powsimp import powsimp
from .radsimp import fraction, radsimp
from .sqrtdenest import sqrtdenest
from .trigsimp import exptrigsimp, trigsimp


[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 diofant.abc import alpha >>> 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 ret = {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 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_extended_real): return False return True def _nthroot_solve(p, n, prec): """ helper function for ``nthroot`` It denests ``root(p, n)`` using its minimal polynomial """ from ..polys.numberfields import _minimal_polynomial_sq from ..solvers import solve while n % 2 == 0: p = sqrtdenest(sqrt(p)) n = n // 2 if n == 1: return p pn = root(p, n) x = Symbol('x') f = _minimal_polynomial_sq(p, n, x) if f is None: return sols = solve(f, x) for sol in sols: sol = sol[x] if abs(sol - pn).n() < 1./10**prec: sol = sqrtdenest(sol) if _mexpand(sol**n) == p: return sol
[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`` Notes ===== 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 ======== >>> nthroot(90 + 34*sqrt(7), 3) sqrt(7) + 3 """ expr = sympify(expr) n = sympify(n) p = root(expr, 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 = root(-expr, 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 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. >>> posify(x + Symbol('p', positive=True) + Symbol('n', negative=True)) (n + p + _x, {_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) [{x: -2}, {x: 2}] >>> eq_x, reps = posify([eq, x]); eq_x [_x**2 - 4, _x] >>> solve(*eq_x) [{_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({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 = {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. Notes ===== The algorithm [1]_ 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. References ========== .. [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') g = combsimp(g) if g.is_rational_function(k): return simplify(g, ratio=oo) else: return
[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. See Also ======== 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 ======== >>> 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 >>> signsimp(_) -(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) E**(-(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 Diofant. 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: >>> 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: >>> 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:`~diofant.core.function.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. :: >>> 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: >>> 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 >>> 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 .hyperexpand import hyperexpand from ..functions.special.bessel import BesselBase from ..concrete import Sum, Product if not isinstance(expr, Basic) or not expr.args: # XXX: temporary hack return expr if not isinstance(expr, (Add, Mul, Pow, exp_polar)): 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 sympy/sympy#6829 expr2 = shorter(together(expr, deep=True), together(expr1, deep=True)) if ratio is oo: 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, exp_polar) or any(a.base is E for a in short.atoms(Pow))): 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 _real_to_rational(expr, tolerance=None): """ Replace all reals in expr with rationals. >>> nsimplify(.76 + .1*x**.5, rational=True) sqrt(x)/10 + 19/25 """ p = expr reps = {} reduce_num = None if tolerance is not None and tolerance < 1: reduce_num = ceiling(1/tolerance) for float in p.atoms(Float): key = float if reduce_num is not None: r = Rational(float).limit_denominator(reduce_num) elif (tolerance is not None and tolerance >= 1 and float.is_Integer is False): r = Rational(tolerance*round(float/tolerance)).limit_denominator(int(tolerance)) else: r = nsimplify(float, rational=False) # e.g. log(3).n() -> log(3) instead of a Rational if float and not r: r = Rational(float) elif not r.is_Rational: if float < 0: float = -float d = Pow(10, int((mpmath.log(float)/mpmath.log(10)))) r = -Rational(str(float/d))*d elif float > 0: d = Pow(10, int((mpmath.log(float)/mpmath.log(10)))) r = Rational(str(float/d))*d else: r = Integer(0) reps[key] = r return p.subs(reps, simultaneous=True)
[docs]def nsimplify(expr, constants=[], tolerance=None, full=False, rational=None): """ 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). Examples ======== >>> 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]) E**(-pi/2) >>> nsimplify(pi, tolerance=0.01) 22/7 See Also ======== diofant.core.function.nfloat """ try: return sympify(as_int(expr)) except (TypeError, ValueError): pass expr = sympify(expr) expr = sympify(expr).xreplace({Float('inf'): oo, Float('-inf'): -oo}) if expr in (oo, -oo): return expr if rational or expr.free_symbols: return _real_to_rational(expr, tolerance) # Diofant'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, strict=False) 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.findpoly(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 xv not 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) return expr rv = re + im*I # 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)
[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 diofant.abc import a >>> 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_extended_real or force and a.is_extended_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 isinstance(a, 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 isinstance(ai, 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 assert isinstance(l, log) log1[(c,)].append(([], l)) # logs that have the same coefficient can multiply for k in list(log1): 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)): if k not 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 trigonometric 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 diofant.abc import nu >>> besselsimp(besselj(nu, z*polar_lift(-1))) E**(I*pi*nu)*besselj(nu, z) >>> besselsimp(besseli(nu, z*polar_lift(-I))) E**(-I*pi*nu/2)*besselj(nu, z) >>> besselsimp(besseli(Rational(-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) == Rational(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
def sum_simplify(s): """Main function for Sum simplification""" from ..concrete import Sum terms = Add.make_args(s) s_t = [] # Sum Terms o_t = [] # Other Terms for term in terms: if isinstance(term, Mul): constant = 1 other = 1 s = 0 n_sum_terms = 0 for j in range(len(term.args)): if isinstance(term.args[j], Sum): s = term.args[j] n_sum_terms = n_sum_terms + 1 elif term.args[j].is_number: constant = constant * term.args[j] else: other = other * term.args[j] if other == 1 and n_sum_terms == 1: # Insert the constant inside the Sum s_t.append(Sum(constant * s.function, *s.limits)) elif other != 1 and n_sum_terms == 1: o_t.append(other * Sum(constant * s.function, *s.limits)) else: o_t.append(term) elif isinstance(term, Sum): s_t.append(term) else: o_t.append(term) 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): s_t[i] = temp s_term1 = s_t[i] used[j] = True result = Add(*o_t) for i, s_term in enumerate(s_t): if not used[i]: result = Add(result, s_term) return result def sum_add(self, other, method=0): """Helper function for Sum simplification""" from ..concrete import Sum if type(self) == type(other): if method == 0: if self.limits == other.limits: return Sum(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 Sum(self.function, (i, x1, y2)) elif x1 == y2 + 1: return Sum(self.function, (i, x2, y1)) return Add(self, other) def product_simplify(s): """Main function for Product simplification""" from ..concrete 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 ..concrete 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 clear_coefficients(expr, rhs=Integer(0)): """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 ======== >>> 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 Integer(0), 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