Otj>dZddlmZmZddlmZddlmZdZdZ dS)z Recurrences )Ssympify)iterable)as_intc^|s tjSt|stdt|stdt |}|dkrt dd|D}d|D}t |}t ||krtd|tjg|t |z zz }||kr||Sdtt|||D}t|d d |d S) a 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 integerc,g|]}t|Sr.0args X/var/www/html/CA-Chatbot/venv/lib/python3.11/site-packages/sympy/discrete/recurrences.py zlinrec..gs(((#(((c,g|]}t|Sr r r s rrzlinrec..hs&&&#&&&rzECount of initial values should not exceed the order of the recurrencecg|] \}}||z Sr r )r uvs rrzlinrec..ss 9 9 9TQQqS 9 9 9rN) rZeror TypeErrorr ValueErrorlenzip linrec_coeffssum)coeffsinitncbktermss rlinrecr$ s\Z v F  +*++ + D>>.-.. . q A1uu/00 0 )((((A&&&&&A AA 1vvzz233 3 afXq3q66z ""1uut 9 9Sq!!4!4a88 9 9 9E uSbSz59 % %%rcXtfdfd|S)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 ctjgdt|zdz |zz}t|D]3\}}t|D]\}}|||z|zxx||zz cc<4t t|dz dz dD]9}t D]'}|||z dz xx|||zz cc<(:|dS)Nr)rrr enumeraterange) roffsetwipjqr r"s r_square_and_reducez)linrec_coeffs.._square_and_reduces VHaAhlV+ ,aLL ) )DAq!!  ) )1&1*q.!!!QqS(!!!! )s1vvz1q5"-- * *A1XX * *!a%!) !QqT )  *!u rc|kr5tjg|ztjgztjg|z dz zzS|dz|dzS)Nr(r')rrOne)r _final_coeffsr1r"s rr4z$linrec_coeffs.._final_coeffss` q55F8A:'16(AEAI*>> >%%mmAF&;&;QUCC Cr)r)r rr4r1r"s` @@@rrrwsm$ AA"DDDDDDD =  rN) __doc__ sympy.corerrsympy.utilities.iterablesrsympy.utilities.miscrr$rr rrr9sy"!!!!!!!......''''''j&j&j&Z/////r