# Source code for sympy.functions.combinatorial.numbers

"""
This module implements some special functions that commonly appear in
combinatorial contexts (e.g. in power series); in particular,
sequences of rational numbers such as Bernoulli and Fibonacci numbers.

Factorials, binomial coefficients and related functions are located in
the separate 'factorials' module.
"""

from sympy.core.function import Function, expand_mul
from sympy.core import S, Symbol, Rational, oo, Integer, C, Add, Dummy
from sympy.core.compatibility import as_int, SYMPY_INTS
from sympy.core.cache import cacheit
from sympy.functions.combinatorial.factorials import factorial

from sympy.mpmath import bernfrac
from sympy.mpmath.libmp import ifib as _ifib

def _product(a, b):
p = 1
for k in xrange(a, b + 1):
p *= k
return p

from sympy.utilities.memoization import recurrence_memo

# Dummy symbol used for computing polynomial sequences
_sym = Symbol('x')
_symbols = Function('x')

#----------------------------------------------------------------------------#
#                                                                            #
#                           Fibonacci numbers                                #
#                                                                            #
#----------------------------------------------------------------------------#

[docs]class fibonacci(Function): """ Fibonacci numbers / Fibonacci polynomials The Fibonacci numbers are the integer sequence defined by the initial terms F_0 = 0, F_1 = 1 and the two-term recurrence relation F_n = F_{n-1} + F_{n-2}. The Fibonacci polynomials are defined by F_1(x) = 1, F_2(x) = x, and F_n(x) = x*F_{n-1}(x) + F_{n-2}(x) for n > 2. For all positive integers n, F_n(1) = F_n. * fibonacci(n) gives the nth Fibonacci number, F_n * fibonacci(n, x) gives the nth Fibonacci polynomial in x, F_n(x) Examples ======== >>> from sympy import fibonacci, Symbol >>> [fibonacci(x) for x in range(11)] [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55] >>> fibonacci(5, Symbol('t')) t**4 + 3*t**2 + 1 References ========== .. [1] http://en.wikipedia.org/wiki/Fibonacci_number .. [2] http://mathworld.wolfram.com/FibonacciNumber.html See Also ======== bell, bernoulli, catalan, euler, harmonic, lucas """ @staticmethod def _fib(n): return _ifib(n) @staticmethod @recurrence_memo([None, S.One, _sym]) def _fibpoly(n, prev): return (prev[-2] + _sym*prev[-1]).expand() @classmethod def eval(cls, n, sym=None): if n.is_Integer: n = int(n) if n < 0: return S.NegativeOne**(n + 1) * fibonacci(-n) if sym is None: return Integer(cls._fib(n)) else: if n < 1: raise ValueError("Fibonacci polynomials are defined " "only for positive integer indices.") return cls._fibpoly(n).subs(_sym, sym)
[docs]class lucas(Function): """ Lucas numbers Lucas numbers satisfy a recurrence relation similar to that of the Fibonacci sequence, in which each term is the sum of the preceding two. They are generated by choosing the initial values L_0 = 2 and L_1 = 1. * lucas(n) gives the nth Lucas number Examples ======== >>> from sympy import lucas >>> [lucas(x) for x in range(11)] [2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123] References ========== .. [1] http://en.wikipedia.org/wiki/Lucas_number .. [2] http://mathworld.wolfram.com/LucasNumber.html See Also ======== bell, bernoulli, catalan, euler, fibonacci, harmonic """ @classmethod def eval(cls, n): if n.is_Integer: return fibonacci(n + 1) + fibonacci(n - 1) #----------------------------------------------------------------------------# # # # Bernoulli numbers # # # #----------------------------------------------------------------------------#
[docs]class bernoulli(Function): r""" Bernoulli numbers / Bernoulli polynomials The Bernoulli numbers are a sequence of rational numbers defined by B_0 = 1 and the recursive relation (n > 0):: n ___ \ / n + 1 \ 0 = ) | | * B . /___ \ k / k k = 0 They are also commonly defined by their exponential generating function, which is x/(exp(x) - 1). For odd indices > 1, the Bernoulli numbers are zero. The Bernoulli polynomials satisfy the analogous formula:: n ___ \ / n \ n-k B (x) = ) | | * B * x . n /___ \ k / k k = 0 Bernoulli numbers and Bernoulli polynomials are related as B_n(0) = B_n. We compute Bernoulli numbers using Ramanujan's formula:: / n + 3 \ B = (A(n) - S(n)) / | | n \ n / where A(n) = (n+3)/3 when n = 0 or 2 (mod 6), A(n) = -(n+3)/6 when n = 4 (mod 6), and:: [n/6] ___ \ / n + 3 \ S(n) = ) | | * B /___ \ n - 6*k / n-6*k k = 1 This formula is similar to the sum given in the definition, but cuts 2/3 of the terms. For Bernoulli polynomials, we use the formula in the definition. * bernoulli(n) gives the nth Bernoulli number, B_n * bernoulli(n, x) gives the nth Bernoulli polynomial in x, B_n(x) Examples ======== >>> from sympy import bernoulli >>> [bernoulli(n) for n in range(11)] [1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66] >>> bernoulli(1000001) 0 References ========== .. [1] http://en.wikipedia.org/wiki/Bernoulli_number .. [2] http://en.wikipedia.org/wiki/Bernoulli_polynomial .. [3] http://mathworld.wolfram.com/BernoulliNumber.html .. [4] http://mathworld.wolfram.com/BernoulliPolynomial.html See Also ======== bell, catalan, euler, fibonacci, harmonic, lucas """ # Calculates B_n for positive even n @staticmethod def _calc_bernoulli(n): s = 0 a = int(C.binomial(n + 3, n - 6)) for j in xrange(1, n//6 + 1): s += a * bernoulli(n - 6*j) # Avoid computing each binomial coefficient from scratch a *= _product(n - 6 - 6*j + 1, n - 6*j) a //= _product(6*j + 4, 6*j + 9) if n % 6 == 4: s = -Rational(n + 3, 6) - s else: s = Rational(n + 3, 3) - s return s / C.binomial(n + 3, n) # We implement a specialized memoization scheme to handle each # case modulo 6 separately _cache = {0: S.One, 2: Rational(1, 6), 4: Rational(-1, 30)} _highest = {0: 0, 2: 2, 4: 4} @classmethod def eval(cls, n, sym=None): if n.is_Number: if n.is_Integer and n.is_nonnegative: if n is S.Zero: return S.One elif n is S.One: if sym is None: return -S.Half else: return sym - S.Half # Bernoulli numbers elif sym is None: if n.is_odd: return S.Zero n = int(n) # Use mpmath for enormous Bernoulli numbers if n > 500: p, q = bernfrac(n) return Rational(int(p), int(q)) case = n % 6 highest_cached = cls._highest[case] if n <= highest_cached: return cls._cache[n] # To avoid excessive recursion when, say, bernoulli(1000) is # requested, calculate and cache the entire sequence ... B_988, # B_994, B_1000 in increasing order for i in xrange(highest_cached + 6, n + 6, 6): b = cls._calc_bernoulli(i) cls._cache[i] = b cls._highest[case] = i return b # Bernoulli polynomials else: n, result = int(n), [] for k in xrange(n + 1): result.append(C.binomial(n, k)*cls(k)*sym**(n - k)) return Add(*result) else: raise ValueError("Bernoulli numbers are defined only" " for nonnegative integer indices.") #----------------------------------------------------------------------------# # # # Bell numbers # # # #----------------------------------------------------------------------------#
[docs]class bell(Function): r""" Bell numbers / Bell polynomials The Bell numbers satisfy B_0 = 1 and .. math:: B_n = \sum_{k=0}^{n-1} \binom{n-1}{k} B_k. They are also given by: .. math:: B_n = \frac{1}{e} \sum_{k=0}^{\infty} \frac{k^n}{k!}. The Bell polynomials are given by B_0(x) = 1 and .. math:: B_n(x) = x \sum_{k=1}^{n-1} \binom{n-1}{k-1} B_{k-1}(x). The second kind of Bell polynomials (are sometimes called "partial" Bell polynomials or incomplete Bell polynomials) are defined as .. math:: B_{n,k}(x_1, x_2,\dotsc x_{n-k+1}) = \sum_{j_1+j_2+j_2+\dotsb=k \atop j_1+2j_2+3j_2+\dotsb=n} \frac{n!}{j_1!j_2!\dotsb j_{n-k+1}!} \left(\frac{x_1}{1!} \right)^{j_1} \left(\frac{x_2}{2!} \right)^{j_2} \dotsb \left(\frac{x_{n-k+1}}{(n-k+1)!} \right) ^{j_{n-k+1}}. * bell(n) gives the n^{th} Bell number, B_n. * bell(n, x) gives the n^{th} Bell polynomial, B_n(x). * bell(n, k, (x1, x2, ...)) gives Bell polynomials of the second kind, B_{n,k}(x_1, x_2, \dotsc, x_{n-k+1}). Notes ===== Not to be confused with Bernoulli numbers and Bernoulli polynomials, which use the same notation. Examples ======== >>> from sympy import bell, Symbol, symbols >>> [bell(n) for n in range(11)] [1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975] >>> bell(30) 846749014511809332450147 >>> bell(4, Symbol('t')) t**4 + 6*t**3 + 7*t**2 + t >>> bell(6, 2, symbols('x:6')[1:]) 6*x1*x5 + 15*x2*x4 + 10*x3**2 References ========== .. [1] http://en.wikipedia.org/wiki/Bell_number .. [2] http://mathworld.wolfram.com/BellNumber.html .. [3] http://mathworld.wolfram.com/BellPolynomial.html See Also ======== bernoulli, catalan, euler, fibonacci, harmonic, lucas """ @staticmethod @recurrence_memo([1, 1]) def _bell(n, prev): s = 1 a = 1 for k in xrange(1, n): a = a * (n - k) // k s += a * prev[k] return s @staticmethod @recurrence_memo([S.One, _sym]) def _bell_poly(n, prev): s = 1 a = 1 for k in xrange(2, n + 1): a = a * (n - k + 1) // (k - 1) s += a * prev[k - 1] return expand_mul(_sym * s) @staticmethod def _bell_incomplete_poly(n, k, symbols): r""" The second kind of Bell polynomials (incomplete Bell polynomials). Calculated by recurrence formula: .. math:: B_{n,k}(x_1, x_2, \dotsc, x_{n-k+1}) = \sum_{m=1}^{n-k+1} \x_m \binom{n-1}{m-1} B_{n-m,k-1}(x_1, x_2, \dotsc, x_{n-m-k}) where B_{0,0} = 1; B_{n,0} = 0; for n>=1 B_{0,k} = 0; for k>=1 """ if (n == 0) and (k == 0): return S.One elif (n == 0) or (k == 0): return S.Zero s = S.Zero a = S.One for m in xrange(1, n - k + 2): s += a * bell._bell_incomplete_poly( n - m, k - 1, symbols) * symbols[m - 1] a = a * (n - m) / m return expand_mul(s) @classmethod def eval(cls, n, k_sym=None, symbols=None): if n.is_Integer and n.is_nonnegative: if k_sym is None: return Integer(cls._bell(int(n))) elif symbols is None: return cls._bell_poly(int(n)).subs(_sym, k_sym) else: r = cls._bell_incomplete_poly(int(n), int(k_sym), symbols) return r #----------------------------------------------------------------------------# # # # Harmonic numbers # # # #----------------------------------------------------------------------------#
[docs]class harmonic(Function): r""" Harmonic numbers The nth harmonic number is given by \operatorname{H}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}. More generally: .. math:: \operatorname{H}_{n,m} = \sum_{k=1}^{n} \frac{1}{k^m} As n \rightarrow \infty, \operatorname{H}_{n,m} \rightarrow \zeta(m), the Riemann zeta function. * harmonic(n) gives the nth harmonic number, \operatorname{H}_n * harmonic(n, m) gives the nth generalized harmonic number of order m, \operatorname{H}_{n,m}, where harmonic(n) == harmonic(n, 1) Examples ======== >>> from sympy import harmonic, oo >>> [harmonic(n) for n in range(6)] [0, 1, 3/2, 11/6, 25/12, 137/60] >>> [harmonic(n, 2) for n in range(6)] [0, 1, 5/4, 49/36, 205/144, 5269/3600] >>> harmonic(oo, 2) pi**2/6 >>> from sympy import Symbol, Sum >>> n = Symbol("n") >>> harmonic(n).rewrite(Sum) Sum(1/_k, (_k, 1, n)) We can rewrite harmonic numbers in terms of polygamma functions: >>> from sympy import digamma, polygamma >>> m = Symbol("m") >>> harmonic(n).rewrite(digamma) polygamma(0, n + 1) + EulerGamma >>> harmonic(n).rewrite(polygamma) polygamma(0, n + 1) + EulerGamma >>> harmonic(n,3).rewrite(polygamma) polygamma(2, n + 1)/2 - polygamma(2, 1)/2 >>> harmonic(n,m).rewrite(polygamma) (-1)**m*(polygamma(m - 1, 1) - polygamma(m - 1, n + 1))/factorial(m - 1) Integer offsets in the argument can be pulled out: >>> from sympy import expand_func >>> expand_func(harmonic(n+4)) harmonic(n) + 1/(n + 4) + 1/(n + 3) + 1/(n + 2) + 1/(n + 1) >>> expand_func(harmonic(n-4)) harmonic(n) - 1/(n - 1) - 1/(n - 2) - 1/(n - 3) - 1/n Some limits can be computed as well: >>> from sympy import limit, oo >>> limit(harmonic(n), n, oo) oo >>> limit(harmonic(n, 2), n, oo) pi**2/6 >>> limit(harmonic(n, 3), n, oo) -polygamma(2, 1)/2 >>> limit(harmonic(m, n), m, oo) zeta(n) References ========== .. [1] http://en.wikipedia.org/wiki/Harmonic_number .. [2] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber/ .. [3] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber2/ See Also ======== bell, bernoulli, catalan, euler, fibonacci, lucas """ # Generate one memoized Harmonic number-generating function for each # order and store it in a dictionary _functions = {} nargs = (1, 2) @classmethod def eval(cls, n, m=None): if m is None: m = S.One if n == oo: return C.zeta(m) if n.is_Integer and n.is_nonnegative and m.is_Integer: if n == 0: return S.Zero if not m in cls._functions: @recurrence_memo([0]) def f(n, prev): return prev[-1] + S.One / n**m cls._functions[m] = f return cls._functions[m](int(n)) def _eval_rewrite_as_polygamma(self, n, m=1): from sympy.functions.special.gamma_functions import polygamma return S.NegativeOne**m/factorial(m - 1) * (polygamma(m - 1, 1) - polygamma(m - 1, n + 1)) def _eval_rewrite_as_digamma(self, n, m=1): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma) def _eval_rewrite_as_trigamma(self, n, m=1): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma) def _eval_rewrite_as_Sum(self, n, m=None): k = C.Dummy("k", integer=True) if m is None: m = S.One return C.Sum(k**(-m), (k, 1, n)) def _eval_expand_func(self, **hints): n = self.args[0] m = self.args[1] if len(self.args) == 2 else 1 if m == S.One: if n.is_Add: off = n.args[0] nnew = n - off if off.is_Integer and off.is_positive: result = [S.One/(nnew + i) for i in xrange(off, 0, -1)] + [harmonic(nnew)] return Add(*result) elif off.is_Integer and off.is_negative: result = [-S.One/(nnew + i) for i in xrange(0, off, -1)] + [harmonic(nnew)] return Add(*result) return self def _eval_rewrite_as_tractable(self, n, m=1): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma).rewrite("tractable", deep=True) #----------------------------------------------------------------------------# # # # Euler numbers # # # #----------------------------------------------------------------------------#
[docs]class euler(Function): r""" Euler numbers The euler numbers are given by:: 2*n+1 k ___ ___ j 2*n+1 \ \ / k \ (-1) * (k-2*j) E = I ) ) | | -------------------- 2n /___ /___ \ j / k k k = 1 j = 0 2 * I * k E = 0 2n+1 * euler(n) gives the n-th Euler number, E_n Examples ======== >>> from sympy import Symbol, euler >>> [euler(n) for n in range(10)] [1, 0, -1, 0, 5, 0, -61, 0, 1385, 0] >>> n = Symbol("n") >>> euler(n+2*n) euler(3*n) References ========== .. [1] http://en.wikipedia.org/wiki/Euler_numbers .. [2] http://mathworld.wolfram.com/EulerNumber.html .. [3] http://en.wikipedia.org/wiki/Alternating_permutation .. [4] http://mathworld.wolfram.com/AlternatingPermutation.html See Also ======== bell, bernoulli, catalan, fibonacci, harmonic, lucas """ nargs = 1 @classmethod def eval(cls, m, evaluate=True): if not evaluate: return if m.is_odd: return S.Zero if m.is_Integer and m.is_nonnegative: from sympy.mpmath import mp m = m._to_mpmath(mp.prec) res = mp.eulernum(m, exact=True) return Integer(res) def _eval_rewrite_as_Sum(self, arg): if arg.is_even: k = C.Dummy("k", integer=True) j = C.Dummy("j", integer=True) n = self.args[0] / 2 Em = (S.ImaginaryUnit * C.Sum( C.Sum( C.binomial(k, j) * ((-1)**j * (k - 2*j)**(2*n + 1)) / (2**k*S.ImaginaryUnit**k * k), (j, 0, k)), (k, 1, 2*n + 1))) return Em def _eval_evalf(self, prec): m = self.args[0] if m.is_Integer and m.is_nonnegative: from sympy.mpmath import mp from sympy import Expr m = m._to_mpmath(prec) oprec = mp.prec mp.prec = prec res = mp.eulernum(m) mp.prec = oprec return Expr._from_mpmath(res, prec) #----------------------------------------------------------------------------# # # # Catalan numbers # # # #----------------------------------------------------------------------------#
[docs]class catalan(Function): r""" Catalan numbers The n-th catalan number is given by:: 1 / 2*n \ C = ----- | | n n + 1 \ n / * catalan(n) gives the n-th Catalan number, C_n Examples ======== >>> from sympy import (Symbol, binomial, gamma, hyper, polygamma, ... catalan, diff, combsimp, Rational, I) >>> [ catalan(i) for i in range(1,10) ] [1, 2, 5, 14, 42, 132, 429, 1430, 4862] >>> n = Symbol("n", integer=True) >>> catalan(n) catalan(n) Catalan numbers can be transformed into several other, identical expressions involving other mathematical functions >>> catalan(n).rewrite(binomial) binomial(2*n, n)/(n + 1) >>> catalan(n).rewrite(gamma) 4**n*gamma(n + 1/2)/(sqrt(pi)*gamma(n + 2)) >>> catalan(n).rewrite(hyper) hyper((-n + 1, -n), (2,), 1) For some non-integer values of n we can get closed form expressions by rewriting in terms of gamma functions: >>> catalan(Rational(1,2)).rewrite(gamma) 8/(3*pi) We can differentiate the Catalan numbers C(n) interpreted as a continuous real funtion in n: >>> diff(catalan(n), n) (polygamma(0, n + 1/2) - polygamma(0, n + 2) + log(4))*catalan(n) As a more advanced example consider the following ratio between consecutive numbers: >>> combsimp((catalan(n + 1)/catalan(n)).rewrite(binomial)) 2*(2*n + 1)/(n + 2) The Catalan numbers can be generalized to complex numbers: >>> catalan(I).rewrite(gamma) 4**I*gamma(1/2 + I)/(sqrt(pi)*gamma(2 + I)) and evaluated with arbitrary precision: >>> catalan(I).evalf(20) 0.39764993382373624267 - 0.020884341620842555705*I References ========== .. [1] http://en.wikipedia.org/wiki/Catalan_number .. [2] http://mathworld.wolfram.com/CatalanNumber.html .. [3] http://functions.wolfram.com/GammaBetaErf/CatalanNumber/ .. [4] http://geometer.org/mathcircles/catalan.pdf See Also ======== bell, bernoulli, euler, fibonacci, harmonic, lucas sympy.functions.combinatorial.factorials.binomial """ @classmethod def eval(cls, n, evaluate=True): if n.is_Integer and n.is_nonnegative: return 4**n*C.gamma(n + S.Half)/(C.gamma(S.Half)*C.gamma(n + 2)) def fdiff(self, argindex=1): n = self.args[0] return catalan(n)*(C.polygamma(0, n + Rational(1, 2)) - C.polygamma(0, n + 2) + C.log(4)) def _eval_rewrite_as_binomial(self, n): return C.binomial(2*n, n)/(n + 1) def _eval_rewrite_as_gamma(self, n): # The gamma function allows to generalize Catalan numbers to complex n return 4**n*C.gamma(n + S.Half)/(C.gamma(S.Half)*C.gamma(n + 2)) def _eval_rewrite_as_hyper(self, n): return C.hyper([1 - n, -n], [2], 1) def _eval_evalf(self, prec): return self.rewrite(C.gamma).evalf(prec) ####################################################################### ### ### Functions for enumerating partitions, permutations and combinations ### #######################################################################
class _MultisetHistogram(tuple): pass _N = -1 _ITEMS = -2 _M = slice(None, _ITEMS) def _multiset_histogram(n): """Return tuple used in permutation and combination counting. Input is a dictionary giving items with counts as values or a sequence of items (which need not be sorted). The data is stored in a class deriving from tuple so it is easily recognized and so it can be converted easily to a list. """ if type(n) is dict: # item: count if not all(isinstance(v, int) and v >= 0 for v in n.values()): raise ValueError tot = sum(n.values()) items = sum(1 for k in n if n[k] > 0) return _MultisetHistogram([n[k] for k in n if n[k] > 0] + [items, tot]) else: n = list(n) s = set(n) if len(s) == len(n): n = [1]*len(n) n.extend([len(n), len(n)]) return _MultisetHistogram(n) m = dict(zip(s, range(len(s)))) d = dict(zip(range(len(s)), [0]*len(s))) for i in n: d[m[i]] += 1 return _multiset_histogram(d) def nP(n, k=None, replacement=False): """Return the number of permutations of n items taken k at a time. Possible values for n:: integer - set of length n sequence - converted to a multiset internally multiset - {element: multiplicity} If k is None then the total of all permutations of length 0 through the number of items represented by n will be returned. If replacement is True then a given item can appear more than once in the k items. (For example, for 'ab' permutations of 2 would include 'aa', 'ab', 'ba' and 'bb'.) The multiplicity of elements in n is ignored when replacement is True but the total number of elements is considered since no element can appear more times than the number of elements in n. Examples ======== >>> from sympy.functions.combinatorial.numbers import nP >>> from sympy.utilities.iterables import multiset_permutations, multiset >>> nP(3, 2) 6 >>> nP('abc', 2) == nP(multiset('abc'), 2) == 6 True >>> nP('aab', 2) 3 >>> nP([1, 2, 2], 2) 3 >>> [nP(3, i) for i in range(4)] [1, 3, 6, 6] >>> nP(3) == sum(_) True When replacement is True, each item can have multiplicity equal to the length represented by n: >>> nP('aabc', replacement=True) 121 >>> [len(list(multiset_permutations('aaaabbbbcccc', i))) for i in range(5)] [1, 3, 9, 27, 81] >>> sum(_) 121 References ========== .. [1] http://en.wikipedia.org/wiki/Permutation See Also ======== sympy.utilities.iterables.multiset_permutations """ try: n = as_int(n) except ValueError: return Integer(_nP(_multiset_histogram(n), k, replacement)) return Integer(_nP(n, k, replacement)) @cacheit def _nP(n, k=None, replacement=False): from sympy.functions.combinatorial.factorials import factorial from sympy.core.mul import prod if k == 0: return 1 if isinstance(n, SYMPY_INTS): # n different items # assert n >= 0 if k is None: return sum(_nP(n, i, replacement) for i in range(n + 1)) elif replacement: return n**k elif k > n: return 0 elif k == n: return factorial(k) elif k == 1: return n else: # assert k >= 0 return _product(n - k + 1, n) elif isinstance(n, _MultisetHistogram): if k is None: return sum(_nP(n, i, replacement) for i in range(n[_N] + 1)) elif replacement: return n[_ITEMS]**k elif k == n[_N]: return factorial(k)/prod([factorial(i) for i in n[_M] if i > 1]) elif k > n[_N]: return 0 elif k == 1: return n[_ITEMS] else: # assert k >= 0 tot = 0 n = list(n) for i in range(len(n[_M])): if not n[i]: continue n[_N] -= 1 if n[i] == 1: n[i] = 0 n[_ITEMS] -= 1 tot += _nP(_MultisetHistogram(n), k - 1) n[_ITEMS] += 1 n[i] = 1 else: n[i] -= 1 tot += _nP(_MultisetHistogram(n), k - 1) n[i] += 1 n[_N] += 1 return tot @cacheit def _AOP_product(n): """for n = (m1, m2, .., mk) return the coefficients of the polynomial, prod(sum(x**i for i in range(nj + 1)) for nj in n); i.e. the coefficients of the product of AOPs (all-one polynomials) or order given in n. The resulting coefficient corresponding to x**r is the number of r-length combinations of sum(n) elements with multiplicities given in n. The coefficients are given as a default dictionary (so if a query is made for a key that is not present, 0 will be returned). Examples ======== >>> from sympy.functions.combinatorial.numbers import _AOP_product >>> from sympy.abc import x >>> n = (2, 2, 3) # e.g. aabbccc >>> prod = ((x**2 + x + 1)*(x**2 + x + 1)*(x**3 + x**2 + x + 1)).expand() >>> c = _AOP_product(n); dict(c) {0: 1, 1: 3, 2: 6, 3: 8, 4: 8, 5: 6, 6: 3, 7: 1} >>> [c[i] for i in range(8)] == [prod.coeff(x, i) for i in range(8)] True The generating poly used here is the same as that listed in http://tinyurl.com/cep849r, but in a refactored form. """ from collections import defaultdict n = list(n) ord = sum(n) need = (ord + 2)//2 rv = [1]*(n.pop() + 1) rv.extend([0]*(need - len(rv))) rv = rv[:need] while n: ni = n.pop() N = ni + 1 was = rv[:] for i in range(1, min(N, len(rv))): rv[i] += rv[i - 1] for i in range(N, need): rv[i] += rv[i - 1] - was[i - N] rev = list(reversed(rv)) if ord % 2: rv = rv + rev else: rv[-1:] = rev d = defaultdict(int) for i in range(len(rv)): d[i] = rv[i] return d def nC(n, k=None, replacement=False): """Return the number of combinations of n items taken k at a time. Possible values for n:: integer - set of length n sequence - converted to a multiset internally multiset - {element: multiplicity} If k is None then the total of all combinations of length 0 through the number of items represented in n will be returned. If replacement is True then a given item can appear more than once in the k items. (For example, for 'ab' sets of 2 would include 'aa', 'ab', and 'bb'.) The multiplicity of elements in n is ignored when replacement is True but the total number of elements is considered since no element can appear more times than the number of elements in n. Examples ======== >>> from sympy.functions.combinatorial.numbers import nC >>> from sympy.utilities.iterables import multiset_combinations >>> nC(3, 2) 3 >>> nC('abc', 2) 3 >>> nC('aab', 2) 2 When replacement is True, each item can have multiplicity equal to the length represented by n: >>> nC('aabc', replacement=True) 35 >>> [len(list(multiset_combinations('aaaabbbbcccc', i))) for i in range(5)] [1, 3, 6, 10, 15] >>> sum(_) 35 If there are k items with multiplicities m_1, m_2, ..., m_k then the total of all combinations of length 0 hrough k is the product, (m_1 + 1)*(m_2 + 1)*...*(m_k + 1). When the multiplicity of each item is 1 (i.e., k unique items) then there are 2**k combinations. For example, if there are 4 unique items, the total number of combinations is 16: >>> sum(nC(4, i) for i in range(5)) 16 References ========== .. [1] http://en.wikipedia.org/wiki/Combination .. [2] http://tinyurl.com/cep849r See Also ======== sympy.utilities.iterables.multiset_combinations """ from sympy.functions.combinatorial.factorials import binomial from sympy.core.mul import prod if isinstance(n, SYMPY_INTS): if k is None: if not replacement: return 2**n return sum(nC(n, i, replacement) for i in range(n + 1)) assert k >= 0 if replacement: return binomial(n + k - 1, k) return binomial(n, k) if isinstance(n, _MultisetHistogram): N = n[_N] if k is None: if not replacement: return prod(m + 1 for m in n[_M]) return sum(nC(n, i, replacement) for i in range(N + 1)) elif replacement: return nC(n[_ITEMS], k, replacement) # assert k >= 0 elif k in (1, N - 1): return n[_ITEMS] elif k in (0, N): return 1 return _AOP_product(tuple(n[_M]))[k] else: return nC(_multiset_histogram(n), k, replacement) @cacheit def _stirling1(n, k): if n == k == 0: return S.One if 0 in (n, k): return S.Zero n1 = n - 1 # some special values if n == k: return S.One elif k == 1: return factorial(n1) elif k == n1: return C.binomial(n, 2) elif k == n - 2: return (3*n - 1)*C.binomial(n, 3)/4 elif k == n - 3: return C.binomial(n, 2)*C.binomial(n, 4) # general recurrence return n1*_stirling1(n1, k) + _stirling1(n1, k - 1) @cacheit def _stirling2(n, k): if n == k == 0: return S.One if 0 in (n, k): return S.Zero n1 = n - 1 # some special values if k == n1: return C.binomial(n, 2) elif k == 2: return 2**n1 - 1 # general recurrence return k*_stirling2(n1, k) + _stirling2(n1, k - 1)
[docs]def stirling(n, k, d=None, kind=2, signed=False): """Return Stirling number S(n, k) of the first or second (default) kind. The sum of all Stirling numbers of the second kind for k = 1 through n is bell(n). The recurrence relationship for these numbers is:: {0} {n} {0} {n + 1} {n} { n } { } = 1; { } = { } = 0; { } = j*{ } + { } {0} {0} {k} { k } {k} {k - 1} where j is:: n for Stirling numbers of the first kind -n for signed Stirling numbers of the first kind k for Stirling numbers of the second kind The first kind of Stirling number counts the number of permutations of n distinct items that have k cycles; the second kind counts the ways in which n distinct items can be partitioned into k parts. If d is given, the "reduced Stirling number of the second kind" is returned: S^{d}(n, k) = S(n - d + 1, k - d + 1) with n >= k >= d. (This counts the ways to partition n consecutive integers into k groups with no pairwise difference less than d. See example below.) To obtain the signed Stirling numbers of the first kind, use keyword signed=True. Using this keyword automatically sets kind to 1. Examples ======== >>> from sympy.functions.combinatorial.numbers import stirling, bell >>> from sympy.combinatorics import Permutation >>> from sympy.utilities.iterables import multiset_partitions, permutations First kind (unsigned by default): >>> [stirling(6, i, kind=1) for i in range(7)] [0, 120, 274, 225, 85, 15, 1] >>> perms = list(permutations(range(4))) >>> [sum(Permutation(p).cycles == i for p in perms) for i in range(5)] [0, 6, 11, 6, 1] >>> [stirling(4, i, kind=1) for i in range(5)] [0, 6, 11, 6, 1] First kind (signed): >>> [stirling(4, i, signed=True) for i in range(5)] [0, -6, 11, -6, 1] Second kind: >>> [stirling(10, i) for i in range(12)] [0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 0] >>> sum(_) == bell(10) True >>> len(list(multiset_partitions(range(4), 2))) == stirling(4, 2) True Reduced second kind: >>> from sympy import subsets, oo >>> def delta(p): ... if len(p) == 1: ... return oo ... return min(abs(i[0] - i[1]) for i in subsets(p, 2)) >>> parts = multiset_partitions(range(5), 3) >>> d = 2 >>> sum(1 for p in parts if all(delta(i) >= d for i in p)) 7 >>> stirling(5, 3, 2) 7 References ========== .. [1] http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind .. [2] http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind See Also ======== sympy.utilities.iterables.multiset_partitions """ # TODO: make this a class like bell() n = as_int(n) k = as_int(k) if n < 0: raise ValueError('n must be nonnegative') if k > n: return S.Zero if d: # assert k >= d # kind is ignored -- only kind=2 is supported return _stirling2(n - d + 1, k - d + 1) elif signed: # kind is ignored -- only kind=1 is supported return (-1)**(n - k)*_stirling1(n, k) if kind == 1: return _stirling1(n, k) elif kind == 2: return _stirling2(n, k) else: raise ValueError('kind must be 1 or 2, not %s' % k)
@cacheit def _nT(n, k): """Return the partitions of n items into k parts. This is used by nT for the case when n is an integer.""" if k == 0: return 1 if k == n else 0 return sum(_nT(n - k, j) for j in range(min(k, n - k) + 1)) def nT(n, k=None): """Return the number of k-sized partitions of n items. Possible values for n:: integer - n identical items sequence - converted to a multiset internally multiset - {element: multiplicity} Note: the convention for nT is different than that of nC andnP in that here an integer indicates n *identical* items instead of a set of length n; this is in keepng with the partitions function which treats its integer-n input like a list of n 1s. One can use range(n) for n to indicate n distinct items. If k is None then the total number of ways to partition the elements represented in n will be returned. Examples ======== >>> from sympy.functions.combinatorial.numbers import nT Partitions of the given multiset: >>> [nT('aabbc', i) for i in range(1, 7)] [1, 8, 11, 5, 1, 0] >>> nT('aabbc') == sum(_) True (TODO The following can be activated with >>> when taocp_multiset_permutation is in place.) >> [nT("mississippi", i) for i in range(1, 12)] [1, 74, 609, 1521, 1768, 1224, 579, 197, 50, 9, 1] Partitions when all items are identical: >>> [nT(5, i) for i in range(1, 6)] [1, 2, 2, 1, 1] >>> nT('1'*5) == sum(_) True When all items are different: >>> [nT(range(5), i) for i in range(1, 6)] [1, 15, 25, 10, 1] >>> nT(range(5)) == sum(_) True References ========== .. [1] http://undergraduate.csse.uwa.edu.au/units/CITS7209/partition.pdf See Also ======== sympy.utilities.iterables.partitions sympy.utilities.iterables.multiset_partitions """ from sympy.utilities.iterables import multiset_partitions if isinstance(n, SYMPY_INTS): # assert n >= 0 # all the same if k is None: return sum(_nT(n, k) for k in range(1, n + 1)) return _nT(n, k) if not isinstance(n, _MultisetHistogram): try: # if n contains hashable items there is some # quick handling that can be done u = len(set(n)) if u == 1: return nT(len(n), k) elif u == len(n): n = range(u) raise TypeError except TypeError: n = _multiset_histogram(n) N = n[_N] if k is None and N == 1: return 1 if k in (1, N): return 1 if k == 2 or N == 2 and k is None: m, r = divmod(N, 2) rv = sum(nC(n, i) for i in range(1, m + 1)) if not r: rv -= nC(n, m)//2 if k is None: rv += 1 # for k == 1 return rv if N == n[_ITEMS]: # all distinct if k is None: return bell(N) return stirling(N, k) if k is None: return sum(nT(n, k) for k in range(1, N + 1)) tot = 0 for p in multiset_partitions( [i for i, j in enumerate(n[_M]) for ii in range(j)]): tot += len(p) == k return tot