a RG5d}'@sddlmZddlmZddlmZmZddlmZddl m Z ddl m Z ddl mZdd d d Zddd d dZddZddZddZdS)) annotations) factor_terms)IntegerRational)S)Dummy)_sympify)as_intlist)returncCst|}tdd|Dr|jr2t|ddS|jrHt|j|jdS|jrp|j t j urp|j jrptdd|j S|j rt|jdkr|jdjr|jdjr|jdj jr|jdj t j ur|j\}}td|j|j |jS|\}}|jr|jrt||S|jr(t|jdkr(|j\}}n t j}|}|jrt j}|j rdt|jdkrd|j\}}n|jrxtd}|}|jr|jr|j t j ur|j jr|j }t||||Std|dS)aReturn the continued fraction representation of a Rational or quadratic irrational. Examples ======== >>> from sympy.ntheory.continued_fraction import continued_fraction >>> from sympy import sqrt >>> continued_fraction((1 + 2*sqrt(3))/5) [0, 1, [8, 3, 34, 3]] See Also ======== continued_fraction_periodic, continued_fraction_reduce, continued_fraction_convergents css|] }|jVqdSN) is_Rational).0ir\/var/www/html/django/DPS/env/lib/python3.9/site-packages/sympy/ntheory/continued_fraction.py z%continued_fraction..rz4expecting a rational or quadratic irrational, not %sN)rallatoms is_Integercontinued_fraction_periodicr pqis_PowexprHalfbaseis_Mullenargsexpandas_numer_denomis_AddZeroNaNr ValueError)aebrdbccrrrcontinued_fraction s\         r/rcCsddlm}m}ttt||||g\}}}}|dkrBtd||dkrRtd|sZd}||}|jrttt ||||S|dkr| | | }}}||||}|dkr|| }| |} t d| } | d|d8<| S||d9}||9}||d|r:||d9}||9}||9}||9}g} i} ||f| vrt | | ||f<| |||| d||}||d|}qB| ||f} | d| | | dgS) aR Find the periodic continued fraction expansion of a quadratic irrational. Compute the continued fraction expansion of a rational or a quadratic irrational number, i.e. `\frac{p + s\sqrt{d}}{q}`, where `p`, `q \ne 0` and `d \ge 0` are integers. Returns the continued fraction representation (canonical form) as a list of integers, optionally ending (for quadratic irrationals) with list of integers representing the repeating digits. Parameters ========== p : int the rational part of the number's numerator q : int the denominator of the number d : int, optional the irrational part (discriminator) of the number's numerator s : int, optional the coefficient of the irrational part Examples ======== >>> from sympy.ntheory.continued_fraction import continued_fraction_periodic >>> continued_fraction_periodic(3, 2, 7) [2, [1, 4, 1, 1]] Golden ratio has the simplest continued fraction expansion: >>> continued_fraction_periodic(1, 2, 5) [[1]] If the discriminator is zero or a perfect square then the number will be a rational number: >>> continued_fraction_periodic(4, 3, 0) [1, 3] >>> continued_fraction_periodic(4, 3, 49) [3, 1, 2] See Also ======== continued_fraction_iterator, continued_fraction_reduce References ========== .. [1] https://en.wikipedia.org/wiki/Periodic_continued_fraction .. [2] K. Rosen. Elementary Number theory and its applications. Addison-Wesley, 3 Sub edition, pages 379-381, January 1992. r)sqrtfloorz(expected non-negative for `d` but got %szThe denominator cannot be 0.rrN) sympy.functionsr0r1r mapr r(rcontinued_fraction_iteratorrr/r!append)rrr,sr0r1sdnwfZone_ftermspqrrrrrKsH9       rcsddlm}gtdfdd}tj}t||D]}q8rtd}|t|g||}||d}|| }n|}|j rt |}|j r|j ddkr|j|j }|S)a2 Reduce a continued fraction to a rational or quadratic irrational. Compute the rational or quadratic irrational number from its terminating or periodic continued fraction expansion. The continued fraction expansion (cf) should be supplied as a terminating iterator supplying the terms of the expansion. For terminating continued fractions, this is equivalent to ``list(continued_fraction_convergents(cf))[-1]``, only a little more efficient. If the expansion has a repeating part, a list of the repeating terms should be returned as the last element from the iterator. This is the format returned by continued_fraction_periodic. For quadratic irrationals, returns the largest solution found, which is generally the one sought, if the fraction is in canonical form (all terms positive except possibly the first). Examples ======== >>> from sympy.ntheory.continued_fraction import continued_fraction_reduce >>> continued_fraction_reduce([1, 2, 3, 4, 5]) 225/157 >>> continued_fraction_reduce([-2, 1, 9, 7, 1, 2]) -256/233 >>> continued_fraction_reduce([2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]).n(10) 2.718281835 >>> continued_fraction_reduce([1, 4, 2, [3, 1]]) (sqrt(21) + 287)/238 >>> continued_fraction_reduce([[1]]) (1 + sqrt(5))/2 >>> from sympy.ntheory.continued_fraction import continued_fraction_periodic >>> continued_fraction_reduce(continued_fraction_periodic(8, 5, 13)) (sqrt(13) + 8)/5 See Also ======== continued_fraction_periodic r)solvexc3s2|D](}t|tr&|Vq.|VqdSr ) isinstancer extend)cfnxtperiodr?rr untillists   z,continued_fraction_reduce..untillistyr2)Z sympy.solversr>rrr&continued_fraction_convergentscontinued_fraction_reducesortsubsradsimpr%rr r"func)rBr>rFr)rGZsolnsZpurervrrDrrIs&+  rIccs6ddlm}||}|V||8}|s(q2d|}q dS)a Return continued fraction expansion of x as iterator. Examples ======== >>> from sympy import Rational, pi >>> from sympy.ntheory.continued_fraction import continued_fraction_iterator >>> list(continued_fraction_iterator(Rational(3, 8))) [0, 2, 1, 2] >>> list(continued_fraction_iterator(Rational(-3, 8))) [-1, 1, 1, 1, 2] >>> for i, v in enumerate(continued_fraction_iterator(pi)): ... if i > 7: ... break ... print(v) 3 7 15 1 292 1 1 1 References ========== .. [1] https://en.wikipedia.org/wiki/Continued_fraction r)r1rN)r3r1)r?r1rrrrr5s" r5ccsbtjtj}}tjtj}}|D]<}||||||}}||}}||}}||Vq dS)av Return an iterator over the convergents of a continued fraction (cf). The parameter should be an iterable returning successive partial quotients of the continued fraction, such as might be returned by continued_fraction_iterator. In computing the convergents, the continued fraction need not be strictly in canonical form (all integers, all but the first positive). Rational and negative elements may be present in the expansion. Examples ======== >>> from sympy.core import pi >>> from sympy import S >>> from sympy.ntheory.continued_fraction import continued_fraction_convergents, continued_fraction_iterator >>> list(continued_fraction_convergents([0, 2, 1, 2])) [0, 1/2, 1/3, 3/8] >>> list(continued_fraction_convergents([1, S('1/2'), -7, S('1/4')])) [1, 3, 19/5, 7] >>> it = continued_fraction_convergents(continued_fraction_iterator(pi)) >>> for n in range(7): ... print(next(it)) 3 22/7 333/106 355/113 103993/33102 104348/33215 208341/66317 See Also ======== continued_fraction_iterator N)rr&One)rBZp_2Zq_2Zp_1Zq_1r)rrrrrrH/s*  rHN)rr) __future__rZsympy.core.exprtoolsrsympy.core.numbersrrsympy.core.singletonrsympy.core.symbolrsympy.core.sympifyrsympy.utilities.miscr r/rrIr5rHrrrrs      AmK,