Source code for sympy.ntheory.generate

"""
Generating and counting primes.

"""
import random
from bisect import bisect
# Using arrays for sieving instead of lists greatly reduces
# memory consumption
from array import array as _array

from sympy.core.compatibility import as_int
from .primetest import isprime


def _arange(a, b):
    ar = _array('l', [0]*(b - a))
    for i, e in enumerate(range(a, b)):
        ar[i] = e
    return ar


[docs]class Sieve: """An infinite list of prime numbers, implemented as a dynamically growing sieve of Eratosthenes. When a lookup is requested involving an odd number that has not been sieved, the sieve is automatically extended up to that number. >>> from sympy import sieve >>> from array import array # this line and next for doctest only >>> sieve._list = array('l', [2, 3, 5, 7, 11, 13]) >>> 25 in sieve False >>> sieve._list array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23]) """ # data shared (and updated) by all Sieve instances _list = _array('l', [2, 3, 5, 7, 11, 13]) def __repr__(self): return "<Sieve with %i primes sieved: 2, 3, 5, ... %i, %i>" % \ (len(self._list), self._list[-2], self._list[-1])
[docs] def extend(self, n): """Grow the sieve to cover all primes <= n (a real number). Examples ======== >>> from sympy import sieve >>> from array import array # this line and next for doctest only >>> sieve._list = array('l', [2, 3, 5, 7, 11, 13]) >>> sieve.extend(30) >>> sieve[10] == 29 True """ n = int(n) if n <= self._list[-1]: return # We need to sieve against all bases up to sqrt(n). # This is a recursive call that will do nothing if there are enough # known bases already. maxbase = int(n**0.5) + 1 self.extend(maxbase) # Create a new sieve starting from sqrt(n) begin = self._list[-1] + 1 newsieve = _arange(begin, n + 1) # Now eliminate all multiples of primes in [2, sqrt(n)] for p in self.primerange(2, maxbase): # Start counting at a multiple of p, offsetting # the index to account for the new sieve's base index startindex = (-begin) % p for i in range(startindex, len(newsieve), p): newsieve[i] = 0 # Merge the sieves self._list += _array('l', [x for x in newsieve if x])
[docs] def extend_to_no(self, i): """Extend to include the ith prime number. i must be an integer. The list is extended by 50% if it is too short, so it is likely that it will be longer than requested. Examples ======== >>> from sympy import sieve >>> from array import array # this line and next for doctest only >>> sieve._list = array('l', [2, 3, 5, 7, 11, 13]) >>> sieve.extend_to_no(9) >>> sieve._list array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23]) """ i = as_int(i) while len(self._list) < i: self.extend(int(self._list[-1] * 1.5))
[docs] def primerange(self, a, b): """Generate all prime numbers in the range [a, b). Examples ======== >>> from sympy import sieve >>> print([i for i in sieve.primerange(7, 18)]) [7, 11, 13, 17] """ from sympy.functions.elementary.integers import ceiling # wrapping ceiling in int will raise an error if there was a problem # determining whether the expression was exactly an integer or not a = max(2, int(ceiling(a))) b = int(ceiling(b)) if a >= b: return self.extend(b) i = self.search(a)[1] maxi = len(self._list) + 1 while i < maxi: p = self._list[i-1] if p < b: yield p i += 1 else: return
[docs] def search(self, n): """Return the indices i, j of the primes that bound n. If n is prime then i == j. Although n can be an expression, if ceiling cannot convert it to an integer then an n error will be raised. Examples ======== >>> from sympy import sieve >>> sieve.search(25) (9, 10) >>> sieve.search(23) (9, 9) """ from sympy.functions.elementary.integers import ceiling # wrapping ceiling in int will raise an error if there was a problem # determining whether the expression was exactly an integer or not test = int(ceiling(n)) n = int(n) if n < 2: raise ValueError("n should be >= 2 but got: %s" % n) if n > self._list[-1]: self.extend(n) b = bisect(self._list, n) if self._list[b - 1] == test: return b, b else: return b, b + 1
def __contains__(self, n): try: n = as_int(n) assert n >= 2 except (ValueError, AssertionError): return False if n % 2 == 0: return n == 2 a, b = self.search(n) return a == b def __getitem__(self, n): """Return the nth prime number""" n = as_int(n) self.extend_to_no(n) return self._list[n - 1] # Generate a global object for repeated use in trial division etc
sieve = Sieve()
[docs]def prime(nth): """ Return the nth prime, with the primes indexed as prime(1) = 2, prime(2) = 3, etc.... The nth prime is approximately n*log(n) and can never be larger than 2**n. References ========== - http://primes.utm.edu/glossary/xpage/BertrandsPostulate.html Examples ======== >>> from sympy import prime >>> prime(10) 29 >>> prime(1) 2 See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range primepi : Return the number of primes less than or equal to n """ n = as_int(nth) if n < 1: raise ValueError("nth must be a positive integer; prime(1) == 2"); return sieve[n]
[docs]def primepi(n): """ Return the value of the prime counting function pi(n) = the number of prime numbers less than or equal to n. Examples ======== >>> from sympy import primepi >>> primepi(25) 9 See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range prime : Return the nth prime """ n = int(n) if n < 2: return 0 else: n = int(n) return sieve.search(n)[0]
[docs]def nextprime(n, ith=1): """ Return the ith prime greater than n. i must be an integer. Notes ===== Potential primes are located at 6*j +/- 1. This property is used during searching. >>> from sympy import nextprime >>> [(i, nextprime(i)) for i in range(10, 15)] [(10, 11), (11, 13), (12, 13), (13, 17), (14, 17)] >>> nextprime(2, ith=2) # the 2nd prime after 2 5 See Also ======== prevprime : Return the largest prime smaller than n primerange : Generate all primes in a given range """ n = int(n) i = as_int(ith) if i > 1: pr = n j = 1 while 1: pr = nextprime(pr) j += 1 if j > i: break return pr if n < 2: return 2 if n < 7: return {2: 3, 3: 5, 4: 5, 5: 7, 6: 7}[n] nn = 6*(n//6) if nn == n: n += 1 if isprime(n): return n n += 4 elif n - nn == 5: n += 2 if isprime(n): return n n += 4 else: n = nn + 5 while 1: if isprime(n): return n n += 2 if isprime(n): return n n += 4
[docs]def prevprime(n): """ Return the largest prime smaller than n. Notes ===== Potential primes are located at 6*j +/- 1. This property is used during searching. >>> from sympy import prevprime >>> [(i, prevprime(i)) for i in range(10, 15)] [(10, 7), (11, 7), (12, 11), (13, 11), (14, 13)] See Also ======== nextprime : Return the ith prime greater than n primerange : Generates all primes in a given range """ from sympy.functions.elementary.integers import ceiling # wrapping ceiling in int will raise an error if there was a problem # determining whether the expression was exactly an integer or not n = int(ceiling(n)) if n < 3: raise ValueError("no preceding primes") if n < 8: return {3: 2, 4: 3, 5: 3, 6: 5, 7: 5}[n] nn = 6*(n//6) if n - nn <= 1: n = nn - 1 if isprime(n): return n n -= 4 else: n = nn + 1 while 1: if isprime(n): return n n -= 2 if isprime(n): return n n -= 4
[docs]def primerange(a, b): """ Generate a list of all prime numbers in the range [a, b). If the range exists in the default sieve, the values will be returned from there; otherwise values will be returned but will not modifiy the sieve. Notes ===== Some famous conjectures about the occurence of primes in a given range are [1]: - Twin primes: though often not, the following will give 2 primes an infinite number of times: primerange(6*n - 1, 6*n + 2) - Legendre's: the following always yields at least one prime primerange(n**2, (n+1)**2+1) - Bertrand's (proven): there is always a prime in the range primerange(n, 2*n) - Brocard's: there are at least four primes in the range primerange(prime(n)**2, prime(n+1)**2) The average gap between primes is log(n) [2]; the gap between primes can be arbitrarily large since sequences of composite numbers are arbitrarily large, e.g. the numbers in the sequence n! + 2, n! + 3 ... n! + n are all composite. References ========== 1. http://en.wikipedia.org/wiki/Prime_number 2. http://primes.utm.edu/notes/gaps.html Examples ======== >>> from sympy import primerange, sieve >>> print([i for i in primerange(1, 30)]) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] The Sieve method, primerange, is generally faster but it will occupy more memory as the sieve stores values. The default instance of Sieve, named sieve, can be used: >>> list(sieve.primerange(1, 30)) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] See Also ======== nextprime : Return the ith prime greater than n prevprime : Return the largest prime smaller than n randprime : Returns a random prime in a given range primorial : Returns the product of primes based on condition Sieve.primerange : return range from already computed primes or extend the sieve to contain the requested range. """ from sympy.functions.elementary.integers import ceiling # if we already have the range, return it if b <= sieve._list[-1]: for i in sieve.primerange(a, b): yield i return # otherwise compute, without storing, the desired range if a >= b: return # wrapping ceiling in int will raise an error if there was a problem # determining whether the expression was exactly an integer or not a = int(ceiling(a)) - 1 b = int(ceiling(b)) while 1: a = nextprime(a) if a < b: yield a else: return
[docs]def randprime(a, b): """ Return a random prime number in the range [a, b). Bertrand's postulate assures that randprime(a, 2*a) will always succeed for a > 1. References ========== - http://en.wikipedia.org/wiki/Bertrand's_postulate Examples ======== >>> from sympy import randprime, isprime >>> randprime(1, 30) #doctest: +SKIP 13 >>> isprime(randprime(1, 30)) True See Also ======== primerange : Generate all primes in a given range """ if a >= b: return a, b = list(map(int, (a, b))) n = random.randint(a - 1, b) p = nextprime(n) if p >= b: p = prevprime(b) if p < a: raise ValueError("no primes exist in the specified range") return p
[docs]def primorial(n, nth=True): """ Returns the product of the first n primes (default) or the primes less than or equal to n (when ``nth=False``). >>> from sympy.ntheory.generate import primorial, randprime, primerange >>> from sympy import factorint, Mul, primefactors, sqrt >>> primorial(4) # the first 4 primes are 2, 3, 5, 7 210 >>> primorial(4, nth=False) # primes <= 4 are 2 and 3 6 >>> primorial(1) 2 >>> primorial(1, nth=False) 1 >>> primorial(sqrt(101), nth=False) 210 One can argue that the primes are infinite since if you take a set of primes and multiply them together (e.g. the primorial) and then add or subtract 1, the result cannot be divided by any of the original factors, hence either 1 or more new primes must divide this product of primes. In this case, the number itself is a new prime: >>> factorint(primorial(4) + 1) {211: 1} In this case two new primes are the factors: >>> factorint(primorial(4) - 1) {11: 1, 19: 1} Here, some primes smaller and larger than the primes multiplied together are obtained: >>> p = list(primerange(10, 20)) >>> sorted(set(primefactors(Mul(*p) + 1)).difference(set(p))) [2, 5, 31, 149] See Also ======== primerange : Generate all primes in a given range """ if nth: n = as_int(n) else: n = int(n) if n < 1: raise ValueError("primorial argument must be >= 1") p = 1 if nth: for i in range(1, n + 1): p *= prime(i) else: for i in primerange(2, n + 1): p *= i return p
[docs]def cycle_length(f, x0, nmax=None, values=False): """For a given iterated sequence, return a generator that gives the length of the iterated cycle (lambda) and the length of terms before the cycle begins (mu); if ``values`` is True then the terms of the sequence will be returned instead. The sequence is started with value ``x0``. Note: more than the first lambda + mu terms may be returned and this is the cost of cycle detection with Brent's method; there are, however, generally less terms calculated than would have been calculated if the proper ending point were determined, e.g. by using Floyd's method. >>> from sympy.ntheory.generate import cycle_length This will yield successive values of i <-- func(i): >>> def iter(func, i): ... while 1: ... ii = func(i) ... yield ii ... i = ii ... A function is defined: >>> func = lambda i: (i**2 + 1) % 51 and given a seed of 4 and the mu and lambda terms calculated: >>> next(cycle_length(func, 4)) (6, 2) We can see what is meant by looking at the output: >>> n = cycle_length(func, 4, values=True) >>> list(ni for ni in n) [17, 35, 2, 5, 26, 14, 44, 50, 2, 5, 26, 14] There are 6 repeating values after the first 2. If a sequence is suspected of being longer than you might wish, ``nmax`` can be used to exit early (and mu will be returned as None): >>> next(cycle_length(func, 4, nmax = 4)) (4, None) >>> [ni for ni in cycle_length(func, 4, nmax = 4, values=True)] [17, 35, 2, 5] Code modified from: http://en.wikipedia.org/wiki/Cycle_detection. """ nmax = int(nmax or 0) # main phase: search successive powers of two power = lam = 1 tortoise, hare = x0, f(x0) # f(x0) is the element/node next to x0. i = 0 while tortoise != hare and (not nmax or i < nmax): i += 1 if power == lam: # time to start a new power of two? tortoise = hare power *= 2 lam = 0 if values: yield hare hare = f(hare) lam += 1 if nmax and i == nmax: if values: return else: yield nmax, None return if not values: # Find the position of the first repetition of length lambda mu = 0 tortoise = hare = x0 for i in range(lam): hare = f(hare) while tortoise != hare: tortoise = f(tortoise) hare = f(hare) mu += 1 if mu: mu -= 1 yield lam, mu