o jgã@s@dZddlmZmZddlmZddlmZdd„Zdd„Z d S) z Recurrences é)ÚSÚsympify)Úiterable)Úas_intcCsÒ|stjSt|ƒs tdƒ‚t|ƒstdƒ‚t|ƒ}|dkr!tdƒ‚dd„|Dƒ}dd„|Dƒ}t|ƒ}t|ƒ|kr=tdƒ‚|tjg|t|ƒ7}||krQ||Sd d„tt||ƒ|ƒDƒ}t |d d …|d ƒS) aŽ Evaluation of univariate linear recurrences of homogeneous type having coefficients independent of the recurrence variable. Parameters ========== coeffs : iterable Coefficients of the recurrence init : iterable Initial values of the recurrence n : Integer Point of evaluation for the recurrence Notes ===== Let `y(n)` be the recurrence of given type, ``c`` be the sequence of coefficients, ``b`` be the sequence of initial/base values of the recurrence and ``k`` (equal to ``len(c)``) be the order of recurrence. Then, .. math :: y(n) = \begin{cases} b_n & 0 \le n < k \\ c_0 y(n-1) + c_1 y(n-2) + \cdots + c_{k-1} y(n-k) & n \ge k \end{cases} Let `x_0, x_1, \ldots, x_n` be a sequence and consider the transformation that maps each polynomial `f(x)` to `T(f(x))` where each power `x^i` is replaced by the corresponding value `x_i`. The sequence is then a solution of the recurrence if and only if `T(x^i p(x)) = 0` for each `i \ge 0` where `p(x) = x^k - c_0 x^(k-1) - \cdots - c_{k-1}` is the characteristic polynomial. Then `T(f(x)p(x)) = 0` for each polynomial `f(x)` (as it is a linear combination of powers `x^i`). Now, if `x^n` is congruent to `g(x) = a_0 x^0 + a_1 x^1 + \cdots + a_{k-1} x^{k-1}` modulo `p(x)`, then `T(x^n) = x_n` is equal to `T(g(x)) = a_0 x_0 + a_1 x_1 + \cdots + a_{k-1} x_{k-1}`. Computation of `x^n`, given `x^k = c_0 x^{k-1} + c_1 x^{k-2} + \cdots + c_{k-1}` is performed using exponentiation by squaring (refer to [1_]) with an additional reduction step performed to retain only first `k` powers of `x` in the representation of `x^n`. Examples ======== >>> from sympy.discrete.recurrences import linrec >>> from sympy.abc import x, y, z >>> linrec(coeffs=[1, 1], init=[0, 1], n=10) 55 >>> linrec(coeffs=[1, 1], init=[x, y], n=10) 34*x + 55*y >>> linrec(coeffs=[x, y], init=[0, 1], n=5) x**2*y + x*(x**3 + 2*x*y) + y**2 >>> linrec(coeffs=[1, 2, 3, 0, 0, 4], init=[x, y, z], n=16) 13576*x + 5676*y + 2356*z References ========== .. [1] https://en.wikipedia.org/wiki/Exponentiation_by_squaring .. [2] https://en.wikipedia.org/w/index.php?title=Modular_exponentiation§ion=6#Matrices See Also ======== sympy.polys.agca.extensions.ExtensionElement.__pow__ z6Expected a sequence of coefficients for the recurrencezFExpected a sequence of values for the initialization of the recurrencerz@Point of evaluation of recurrence must be a non-negative integercSóg|]}t|ƒ‘qS©©r©Ú.0ÚargrrúR/var/www/html/zoom/venv/lib/python3.10/site-packages/sympy/discrete/recurrences.pyÚ gózlinrec..cSrrrr rrr r hrzECount of initial values should not exceed the order of the recurrencecSsg|]\}}||‘qSrr)r ÚuÚvrrr r ssNéÿÿÿÿ) rÚZerorÚ TypeErrorrÚ ValueErrorÚlenÚzipÚ linrec_coeffsÚsum)ÚcoeffsÚinitÚnÚcÚbÚkÚtermsrrr Úlinrec s&M r cs.tˆƒ‰‡‡fdd„‰‡‡‡fdd„‰ˆ|ƒS)a– Compute the coefficients of n'th term in linear recursion sequence defined by c. `x^k = c_0 x^{k-1} + c_1 x^{k-2} + \cdots + c_{k-1}`. It computes the coefficients by using binary exponentiation. This function is used by `linrec` and `_eval_pow_by_cayley`. Parameters ========== c = coefficients of the divisor polynomial n = exponent of x, so dividend is x^n cs¸tjgdt|ƒd|}t|ƒD]\}}t|ƒD]\}}||||||7<qqtt|ƒdˆddƒD]}tˆƒD]}|||d||ˆ|7<q@q:|dˆ…S)Néér)rrrÚ enumerateÚrange)rÚoffsetÚwÚiÚpÚjÚq)rrrr Ú_square_and_reduce‹sÿ &ÿ z)linrec_coeffs.._square_and_reducecsF|ˆkrtjg|tjgtjgˆ|dSˆˆ|dƒ|dƒS)Nr"r!)rrÚOne)r)Ú _final_coeffsr+rrr r-œs(z$linrec_coeffs.._final_coeffs)r)rrr)r-r+rrr rws rN) Ú__doc__Ú sympy.corerrÚsympy.utilities.iterablesrÚsympy.utilities.miscrr rrrrr Ús    m