Was ist eine Home Prime?

Als Beispiel nimm HP (4). Finden Sie zuerst die Primfaktoren. Die Primfaktoren von 4 ( in numerischer Reihenfolge vom kleinsten zum größten, immer ) sind 2, 2. Nehmen Sie diese Faktoren als wörtliche Zahl. 2, 2 wird 22. Dieser Faktorisierungsprozess wird fortgesetzt, bis Sie eine Primzahl erreichen.

number    prime factors
4         2, 2
22        2, 11
211       211 is prime

Sobald Sie eine Primzahl erreicht haben, endet die Sequenz. HP (4) = 211. Hier ist ein längeres Beispiel mit 14:

number    prime factors
14        2, 7
27        3, 3, 3
333       3, 3, 37
3337      47, 71
4771      13, 367
13367     13367 is prime

Ihre Herausforderung besteht darin, ein Programm zu erstellen, das HP (x) bei gegebenem x berechnet und dies so schnell wie möglich durchführt . Sie können beliebige Ressourcen außer einer Liste bekannter Home-Primzahlen verwenden.

Beachten Sie, dass diese Zahlen sehr schnell sehr groß werden. Bei x = 8 springt HP (x) bis auf 3331113965338635107. HP (49) muss noch gefunden werden.

Die Programmgeschwindigkeit wird auf einem Raspberry Pi 2 getestet, wobei die folgenden Eingaben gemittelt werden:

16
20
64
65
80

Wenn Sie einen Raspberry Pi 2 haben, messen Sie das Programm selbst und mitteln Sie die Zeiten.

Mathematica, HP (80) in ~ 0,88 s

NestWhile[
  FromDigits[
    Flatten[IntegerDigits /@ 
      ConstantArray @@@ FactorInteger[#]]] &, #, CompositeQ] &

Anonyme Funktion. Nimmt eine Zahl als Eingabe und gibt eine Zahl als Ausgabe zurück.

PyPy 5.4.1 64bit (Linux), HP (80) ~ 1,54s

Die 32bit Version läuft etwas langsamer.

Ich verwende vier verschiedene Faktorisierungsmethoden mit empirisch ermittelten Haltepunkten:

• n <2 ³⁰ → Faktorisierung durch Versuchsaufteilung
• n <2 ⁷⁰ → Pollards Rho (Brentsche Variante)
• sonst → Lenstra ECF (mit Montgomery-Kurven, Suyama-Parametern), ~ 100 Kurven testen
• Wenn ECF fehlgeschlagen ist → Multiple Polynomial Quadratic Sieve (unter Verwendung von Montgomery's Polynomen)

Ich habe eine Weile versucht, einen sauberen Bruch zwischen ECF und MPQS zu finden, aber es scheint keinen zu geben. wie auch immer, falls n jedoch einen kleinen Faktor enthält, wird ECF diesen normalerweise fast sofort finden. Daher habe ich mich dafür entschieden, nur einige Kurven zu testen, bevor ich zu MPQS übergehe.

Derzeit ist es nur ~ 2x langsamer als Mathmatica, was ich sicherlich als Erfolg betrachte.

home-prime.py

import math
import my_math
import mpqs

max_trial = 1e10
max_pollard = 1e22

def factor(n):
  if n < max_trial:
    return factor_trial(n)
  for p in my_math.small_primes:
    if n%p == 0:
      return [p] + factor(n/p)
  if my_math.is_prime(n):
    return [n]
  if n < max_pollard:
    p = pollard_rho(n)
  else:
    p = lenstra_ecf(n) or mpqs.mpqs(n)
  return factor(p) + factor(n/p)


def factor_trial(n):
  a = []
  for p in my_math.small_primes:
    while n%p == 0:
      a += [p]
      n /= p
  i = 211
  while i*i < n:
    for o in my_math.offsets:
      i += o
      while n%i == 0:
        a += [i]
        n /= i
  if n > 1:
    a += [n]
  return a


def pollard_rho(n):
  # Brent's variant
  y, r, q = 0, 1, 1
  c, m = 9, 40
  g = 1
  while g == 1:
    x = y
    for i in range(r):
      y = (y*y + c) % n
    k = 0
    while k < r and g == 1:
      ys = y
      for j in range(min(m, r-k)):
        y = (y*y + c) % n
        q = q*abs(x-y) % n
      g = my_math.gcd(q, n)
      k += m
    r *= 2
  if g == n:
    ys = (ys*ys + c) % n
    g = gcd(n, abs(x-ys))
    while g == 1:
      ys = (ys*ys + c) % n
      g = gcd(n, abs(x-ys))
  return g

def ec_add((x1, z1), (x2, z2), (x0, z0), n):
  t1, t2 = (x1-z1)*(x2+z2), (x1+z1)*(x2-z2)
  x, z = t1+t2, t1-t2
  return (z0*x*x % n, x0*z*z % n)

def ec_double((x, z), (a, b), n):
  t1 = x+z; t1 *= t1
  t2 = x-z; t2 *= t2
  t3 = t1 - t2
  t4 = 4*b*t2
  return (t1*t4 % n, t3*(t4 + a*t3) % n)

def ec_multiply(k, p, C, n):
  # Montgomery ladder algorithm
  p0 = p
  q, p = p, ec_double(p, C, n)
  b = k >> 1
  while b > (b & -b):
    b ^= b & -b
  while b:
    if k&b:
      q, p = ec_add(p, q, p0, n), ec_double(p, C, n)
    else:
      q, p = ec_double(q, C, n), ec_add(p, q, p0, n),
    b >>= 1
  return q

def lenstra_ecf(n, m = 5):
  # Montgomery curves w/ Suyama parameterization.
  # Based on pseudocode found in:
  # "Implementing the Elliptic Curve Method of Factoring in Reconfigurable Hardware"
  # Gaj, Kris et. al
  # http://www.hyperelliptic.org/tanja/SHARCS/talks06/Gaj.pdf
  # Phase 2 is not implemented.
  B1, B2 = 8, 13
  for i in range(m):
    pg = my_math.primes()
    p = pg.next()
    k = 1
    while p < B1:
      k *= p**int(math.log(B1, p))
      p = pg.next()
    for s in range(B1, B2):
      u, v = s*s-5, 4*s
      x = u*u*u
      z = v*v*v
      t = pow(v-u, 3, n)
      P = (x, z)
      C = (t*(3*u+v) % n, 4*x*v % n)
      Q = ec_multiply(k, P, C, n)
      g = my_math.gcd(Q[1], n)
      if 1 < g < n: return g
    B1, B2 = B2, B1 + B2


if __name__ == '__main__':
  import time
  import sys
  for n in sys.argv[1:]:
    t0 = time.time()
    i = int(n)
    f = []
    while len(f) != 1:
      f = sorted(factor(i))
      #print i, f
      i = int(''.join(map(str, f)))
    t1 = time.time()-t0
    print n, i
    print '%.3fs'%(t1)
    print

Beispiel-Timings

    $ pypy home-prime.py 8 16 20 64 65 80
8 3331113965338635107
0.005s

16 31636373
0.001s

20 3318308475676071413
0.004s

64 1272505013723
0.000s

65 1381321118321175157763339900357651
0.397s

80 313169138727147145210044974146858220729781791489
1.537s

Der Durchschnitt der 5 liegt bei ca. 0,39s.

Abhängigkeiten

mpqs.py wird direkt aus meiner Antwort auf Fastest semiprime factorization mit ein paar sehr geringfügigen Änderungen übernommen.

mpqs.py

import math
import my_math
import time

# Multiple Polynomial Quadratic Sieve
def mpqs(n, verbose=False):
  if verbose:
    time1 = time.time()

  root_n = my_math.isqrt(n)
  root_2n = my_math.isqrt(n+n)

  # formula chosen by experimentation
  # seems to be close to optimal for n < 10^50
  bound = int(5 * math.log(n, 10)**2)

  prime = []
  mod_root = []
  log_p = []
  num_prime = 0

  # find a number of small primes for which n is a quadratic residue
  p = 2
  while p < bound or num_prime < 3:

    # legendre (n|p) is only defined for odd p
    if p > 2:
      leg = my_math.legendre(n, p)
    else:
      leg = n & 1

    if leg == 1:
      prime += [p]
      mod_root += [int(my_math.mod_sqrt(n, p))]
      log_p += [math.log(p, 10)]
      num_prime += 1
    elif leg == 0:
      if verbose:
        print 'trial division found factors:'
        print p, 'x', n/p
      return p

    p = my_math.next_prime(p)

  # size of the sieve
  x_max = bound*8

  # maximum value on the sieved range
  m_val = (x_max * root_2n) >> 1

  # fudging the threshold down a bit makes it easier to find powers of primes as factors
  # as well as partial-partial relationships, but it also makes the smoothness check slower.
  # there's a happy medium somewhere, depending on how efficient the smoothness check is
  thresh = math.log(m_val, 10) * 0.735

  # skip small primes. they contribute very little to the log sum
  # and add a lot of unnecessary entries to the table
  # instead, fudge the threshold down a bit, assuming ~1/4 of them pass
  min_prime = int(thresh*3)
  fudge = sum(log_p[i] for i,p in enumerate(prime) if p < min_prime)/4
  thresh -= fudge

  sieve_primes = [p for p in prime if p >= min_prime]
  sp_idx = prime.index(sieve_primes[0])

  if verbose:
    print 'smoothness bound:', bound
    print 'sieve size:', x_max
    print 'log threshold:', thresh
    print 'skipping primes less than:', min_prime

  smooth = []
  used_prime = set()
  partial = {}
  num_smooth = 0
  prev_num_smooth = 0
  num_used_prime = 0
  num_partial = 0
  num_poly = 0
  root_A = my_math.isqrt(root_2n / x_max)

  if verbose:
    print 'sieving for smooths...'
  while True:
    # find an integer value A such that:
    # A is =~ sqrt(2*n) / x_max
    # A is a perfect square
    # sqrt(A) is prime, and n is a quadratic residue mod sqrt(A)
    while True:
      root_A = my_math.next_prime(root_A)
      leg = my_math.legendre(n, root_A)
      if leg == 1:
        break
      elif leg == 0:
        if verbose:
          print 'dumb luck found factors:'
          print root_A, 'x', n/root_A
        return root_A

    A = root_A * root_A

    # solve for an adequate B
    # B*B is a quadratic residue mod n, such that B*B-A*C = n
    # this is unsolvable if n is not a quadratic residue mod sqrt(A)
    b = my_math.mod_sqrt(n, root_A)
    B = (b + (n - b*b) * my_math.mod_inv(b + b, root_A))%A

    # B*B-A*C = n <=> C = (B*B-n)/A
    C = (B*B - n) / A

    num_poly += 1

    # sieve for prime factors
    sums = [0.0]*(2*x_max)
    i = sp_idx
    for p in sieve_primes:
      logp = log_p[i]

      inv_A = my_math.mod_inv(A, p)
      # modular root of the quadratic
      a = int(((mod_root[i] - B) * inv_A)%p)
      b = int(((p - mod_root[i] - B) * inv_A)%p)

      amx = a+x_max
      bmx = b+x_max

      ax = amx-p
      bx = bmx-p

      k = p
      while k < x_max:
        sums[k+ax] += logp
        sums[k+bx] += logp
        sums[amx-k] += logp
        sums[bmx-k] += logp
        k += p

      if k+ax < x_max:  
        sums[k+ax] += logp
      if k+bx < x_max:
        sums[k+bx] += logp
      if amx-k > 0:
        sums[amx-k] += logp
      if bmx-k > 0:
        sums[bmx-k] += logp
      i += 1

    # check for smooths
    x = -x_max
    for v in sums:
      if v > thresh:
        vec = set()
        sqr = []
        # because B*B-n = A*C
        # (A*x+B)^2 - n = A*A*x*x+2*A*B*x + B*B - n
        #               = A*(A*x*x+2*B*x+C)
        # gives the congruency
        # (A*x+B)^2 = A*(A*x*x+2*B*x+C) (mod n)
        # because A is chosen to be square, it doesn't need to be sieved
        sieve_val = (A*x + B+B)*x + C

        if sieve_val < 0:
          vec = {-1}
          sieve_val = -sieve_val

        for p in prime:
          while sieve_val%p == 0:
            if p in vec:
              # keep track of perfect square factors
              # to avoid taking the sqrt of a gigantic number at the end
              sqr += [p]
            vec ^= {p}
            sieve_val = int(sieve_val / p)

        if sieve_val == 1:
          # smooth
          smooth += [(vec, (sqr, (A*x+B), root_A))]
          used_prime |= vec
        elif sieve_val in partial:
          # combine two partials to make a (xor) smooth
          # that is, every prime factor with an odd power is in our factor base
          pair_vec, pair_vals = partial[sieve_val]
          sqr += list(vec & pair_vec) + [sieve_val]
          vec ^= pair_vec
          smooth += [(vec, (sqr + pair_vals[0], (A*x+B)*pair_vals[1], root_A*pair_vals[2]))]
          used_prime |= vec
          num_partial += 1
        else:
          # save partial for later pairing
          partial[sieve_val] = (vec, (sqr, A*x+B, root_A))
      x += 1

    prev_num_smooth = num_smooth
    num_smooth = len(smooth)
    num_used_prime = len(used_prime)

    if verbose:
      print 100 * num_smooth / num_prime, 'percent complete\r',

    if num_smooth > num_used_prime and num_smooth > prev_num_smooth:
      if verbose:
        print '%d polynomials sieved (%d values)'%(num_poly, num_poly*x_max*2)
        print 'found %d smooths (%d from partials) in %f seconds'%(num_smooth, num_partial, time.time()-time1)
        print 'solving for non-trivial congruencies...'

      used_prime_list = sorted(list(used_prime))

      # set up bit fields for gaussian elimination
      masks = []
      mask = 1
      bit_fields = [0]*num_used_prime
      for vec, vals in smooth:
        masks += [mask]
        i = 0
        for p in used_prime_list:
          if p in vec: bit_fields[i] |= mask
          i += 1
        mask <<= 1

      # row echelon form
      col_offset = 0
      null_cols = []
      for col in xrange(num_smooth):
        pivot = col-col_offset == num_used_prime or bit_fields[col-col_offset] & masks[col] == 0
        for row in xrange(col+1-col_offset, num_used_prime):
          if bit_fields[row] & masks[col]:
            if pivot:
              bit_fields[col-col_offset], bit_fields[row] = bit_fields[row], bit_fields[col-col_offset]
              pivot = False
            else:
              bit_fields[row] ^= bit_fields[col-col_offset]
        if pivot:
          null_cols += [col]
          col_offset += 1

      # reduced row echelon form
      for row in xrange(num_used_prime):
        # lowest set bit
        mask = bit_fields[row] & -bit_fields[row]
        for up_row in xrange(row):
          if bit_fields[up_row] & mask:
            bit_fields[up_row] ^= bit_fields[row]

      # check for non-trivial congruencies
      for col in null_cols:
        all_vec, (lh, rh, rA) = smooth[col]
        lhs = lh   # sieved values (left hand side)
        rhs = [rh] # sieved values - n (right hand side)
        rAs = [rA] # root_As (cofactor of lhs)
        i = 0
        for field in bit_fields:
          if field & masks[col]:
            vec, (lh, rh, rA) = smooth[i]
            lhs += list(all_vec & vec) + lh
            all_vec ^= vec
            rhs += [rh]
            rAs += [rA]
          i += 1

        factor = my_math.gcd(my_math.list_prod(rAs)*my_math.list_prod(lhs) - my_math.list_prod(rhs), n)
        if 1 < factor < n:
          break
      else:
        if verbose:
          print 'none found.'
        continue
      break

  if verbose:
    print 'factors found:'
    print factor, 'x', n/factor
    print 'time elapsed: %f seconds'%(time.time()-time1)
  return factor

if __name__ == "__main__":
  import argparse
  parser = argparse.ArgumentParser(description='Uses a MPQS to factor a composite number')
  parser.add_argument('composite', metavar='number_to_factor', type=long, help='the composite number to factor')
  parser.add_argument('--verbose', dest='verbose', action='store_true', help="enable verbose output")
  args = parser.parse_args()

  if args.verbose:
    mpqs(args.composite, args.verbose)
  else:
    time1 = time.time()
    print mpqs(args.composite)
    print 'time elapsed: %f seconds'%(time.time()-time1)

my_math.pystammt aus dem selben Beitrag wie mpqs.py, habe aber auch den unbeschränkten Primzahlengenerator hinzugefügt, den ich in meiner Antwort verwendet habe die größte Lücke zwischen guten Primzahlen zu finden .

my_math.py

# primes less than 212
small_primes = [
    2,  3,  5,  7, 11, 13, 17, 19, 23, 29, 31, 37,
   41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,
   97,101,103,107,109,113,127,131,137,139,149,151,
  157,163,167,173,179,181,191,193,197,199,211]

# pre-calced sieve of eratosthenes for n = 2, 3, 5, 7
indices = [
    1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
   53, 59, 61, 67, 71, 73, 79, 83, 89, 97,101,103,
  107,109,113,121,127,131,137,139,143,149,151,157,
  163,167,169,173,179,181,187,191,193,197,199,209]

# distances between sieve values
offsets = [
  10, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6,
   6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4,
   2, 4, 8, 6, 4, 6, 2, 4, 6, 2, 6, 6,
   4, 2, 4, 6, 2, 6, 4, 2, 4, 2,10, 2]

# tabulated, mod 105
dindices =[
  0,10, 2, 0, 4, 0, 0, 0, 8, 0, 0, 2, 0, 4, 0,
  0, 6, 2, 0, 4, 0, 0, 4, 6, 0, 0, 6, 0, 0, 2,
  0, 6, 2, 0, 4, 0, 0, 4, 6, 0, 0, 2, 0, 4, 2,
  0, 6, 6, 0, 0, 0, 0, 6, 6, 0, 0, 0, 0, 4, 2,
  0, 6, 2, 0, 4, 0, 0, 4, 6, 0, 0, 2, 0, 6, 2,
  0, 6, 0, 0, 4, 0, 0, 4, 6, 0, 0, 2, 0, 4, 8,
  0, 0, 2, 0,10, 0, 0, 4, 0, 0, 0, 2, 0, 4, 2]

max_int = 2147483647


# returns the index of x in a sorted list a
# or the index of the next larger item if x is not present
# i.e. the proper insertion point for x in a
def binary_search(a, x):
  s = 0
  e = len(a)
  m = e >> 1
  while m != e:
    if a[m] < x:
      s = m
      m = (s + e + 1) >> 1
    else:
      e = m
      m = (s + e) >> 1
  return m


# divide and conquer list product
def list_prod(a):
  size = len(a)
  if size == 1:
    return a[0]
  return list_prod(a[:size>>1]) * list_prod(a[size>>1:])


# greatest common divisor of a and b
def gcd(a, b):
  while b:
    a, b = b, a%b
  return a


# extended gcd
def ext_gcd(a, m):
  a = int(a%m)
  x, u = 0, 1
  while a:
    x, u = u, x - (m/a)*u
    m, a = a, m%a
  return (m, x, u)


# legendre symbol (a|m)
# note: returns m-1 if a is a non-residue, instead of -1
def legendre(a, m):
  return pow(a, (m-1) >> 1, m)


# modular inverse of a mod m
def mod_inv(a, m):
  return ext_gcd(a, m)[1]


# modular sqrt(n) mod p
# p must be prime
def mod_sqrt(n, p):
  a = n%p
  if p%4 == 3:
    return pow(a, (p+1) >> 2, p)
  elif p%8 == 5:
    v = pow(a << 1, (p-5) >> 3, p)
    i = ((a*v*v << 1) % p) - 1
    return (a*v*i)%p
  elif p%8 == 1:
    # Shank's method
    q = p-1
    e = 0
    while q&1 == 0:
      e += 1
      q >>= 1

    n = 2
    while legendre(n, p) != p-1:
      n += 1

    w = pow(a, q, p)
    x = pow(a, (q+1) >> 1, p)
    y = pow(n, q, p)
    r = e
    while True:
      if w == 1:
        return x

      v = w
      k = 0
      while v != 1 and k+1 < r:
        v = (v*v)%p
        k += 1

      if k == 0:
        return x

      d = pow(y, 1 << (r-k-1), p)
      x = (x*d)%p
      y = (d*d)%p
      w = (w*y)%p
      r = k
  else: # p == 2
    return a


#integer sqrt of n
def isqrt(n):
  c = n*4/3
  d = c.bit_length()

  a = d>>1
  if d&1:
    x = 1 << a
    y = (x + (n >> a)) >> 1
  else:
    x = (3 << a) >> 2
    y = (x + (c >> a)) >> 1

  if x != y:
    x = y
    y = (x + n/x) >> 1
    while y < x:
      x = y
      y = (x + n/x) >> 1
  return x


# integer cbrt of n
def icbrt(n):
  d = n.bit_length()

  if d%3 == 2:
    x = 3 << d/3-1
  else:
    x = 1 << d/3

  y = (2*x + n/(x*x))/3
  if x != y:
    x = y
    y = (2*x + n/(x*x))/3
    while y < x:
      x = y
      y = (2*x + n/(x*x))/3
  return x


# strong probable prime
def is_sprp(n, b=2):
  if n < 2: return False
  d = n-1
  s = 0
  while d&1 == 0:
    s += 1
    d >>= 1

  x = pow(b, d, n)
  if x == 1 or x == n-1:
    return True

  for r in xrange(1, s):
    x = (x * x)%n
    if x == 1:
      return False
    elif x == n-1:
      return True

  return False


# lucas probable prime
# assumes D = 1 (mod 4), (D|n) = -1
def is_lucas_prp(n, D):
  P = 1
  Q = (1-D) >> 2

  # n+1 = 2**r*s where s is odd
  s = n+1
  r = 0
  while s&1 == 0:
    r += 1
    s >>= 1

  # calculate the bit reversal of (odd) s
  # e.g. 19 (10011) <=> 25 (11001)
  t = 0
  while s:
    if s&1:
      t += 1
      s -= 1
    else:
      t <<= 1
      s >>= 1

  # use the same bit reversal process to calculate the sth Lucas number
  # keep track of q = Q**n as we go
  U = 0
  V = 2
  q = 1
  # mod_inv(2, n)
  inv_2 = (n+1) >> 1
  while t:
    if t&1:
      # U, V of n+1
      U, V = ((U + V) * inv_2)%n, ((D*U + V) * inv_2)%n
      q = (q * Q)%n
      t -= 1
    else:
      # U, V of n*2
      U, V = (U * V)%n, (V * V - 2 * q)%n
      q = (q * q)%n
      t >>= 1

  # double s until we have the 2**r*sth Lucas number
  while r:
    U, V = (U * V)%n, (V * V - 2 * q)%n
    q = (q * q)%n
    r -= 1

  # primality check
  # if n is prime, n divides the n+1st Lucas number, given the assumptions
  return U == 0


## Baillie-PSW ##
# this is technically a probabalistic test, but there are no known pseudoprimes
def is_bpsw(n):
  if not is_sprp(n, 2): return False

  # idea shamelessly stolen from Mathmatica's PrimeQ
  # if n is a 2-sprp and a 3-sprp, n is necessarily square-free
  if not is_sprp(n, 3): return False

  a = 5
  s = 2
  # if n is a perfect square, this will never terminate
  while legendre(a, n) != n-1:
    s = -s
    a = s-a
  return is_lucas_prp(n, a)


# an 'almost certain' primality check
def is_prime(n):
  if n < 212:
    m = binary_search(small_primes, n)
    return n == small_primes[m]

  for p in small_primes:
    if n%p == 0:
      return False

  # if n is a 32-bit integer, perform full trial division
  if n <= max_int:
    p = 211
    while p*p < n:
      for o in offsets:
        p += o
        if n%p == 0:
          return False
    return True

  return is_bpsw(n)


# next prime strictly larger than n
def next_prime(n):
  if n < 2:
    return 2

  # first odd larger than n
  n = (n + 1) | 1
  if n < 212:
    m = binary_search(small_primes, n)
    return small_primes[m]

  # find our position in the sieve rotation via binary search
  x = int(n%210)
  m = binary_search(indices, x)
  i = int(n + (indices[m] - x))

  # adjust offsets
  offs = offsets[m:] + offsets[:m]
  while True:
    for o in offs:
      if is_prime(i):
        return i
      i += o


# an infinite prime number generator
def primes(start = 0):
  for n in small_primes[start:]: yield n
  pg = primes(6)
  p = pg.next()
  q = p*p
  sieve = {221: 13, 253: 11}
  n = 211
  while True:
    for o in offsets:
      n += o
      stp = sieve.pop(n, 0)
      if stp:
        nxt = n/stp
        nxt += dindices[nxt%105]
        while nxt*stp in sieve:
          nxt += dindices[nxt%105]
        sieve[nxt*stp] = stp
      elif n < q:
        yield n
      else:
        sieve[q + dindices[p%105]*p] = p
        p = pg.next()
        q = p*p


# true if n is a prime power > 0
def is_prime_power(n):
  if n > 1:
    for p in small_primes:
      if n%p == 0:
        n /= p
        while n%p == 0: n /= p
        return n == 1

    r = isqrt(n)
    if r*r == n:
      return is_prime_power(r)

    s = icbrt(n)
    if s*s*s == n:
      return is_prime_power(s)

    p = 211
    while p*p < r:
      for o in offsets:
        p += o
        if n%p == 0:
          n /= p
          while n%p == 0: n /= p
          return n == 1

    if n <= max_int:
      while p*p < n:
        for o in offsets:
          p += o
          if n%p == 0:
            return False
      return True

    return is_bpsw(n)
  return False

Python 2 + Primefac 1.1

Ich habe keinen Raspberry Pi zum Testen.

from primefac import primefac

def HP(n):
    factors = list(primefac(n))

    #print n, factors

    if len(factors) == 1 and n in factors:
        return n

    n = ""
    for f in sorted(factors):
        n += str(f)
    return HP(int(n))

Probieren Sie es online aus

Die primefacFunktion liefert eine Liste aller Primfaktoren von n. In seiner Definition nennt esisprime(n) , die eine Kombination aus Versuchsteilung, Eulers Methode und dem Miller-Rabin-Primalitätstest verwendet. Ich würde empfehlen, das Paket herunterzuladen und die Quelle anzuzeigen.

Ich habe versucht, n = n * 10 ** int(floor(log10(f))+1) + fanstelle der Verkettung Zeichenfolgen zu verwenden, aber es ist viel langsamer.

C #

using System;
using System.Linq;

public class Program
{
    public static void Main(string[] args) {

        Console.Write("Enter Number: ");

        Int64 n = Convert.ToInt64(Console.ReadLine());

        Console.WriteLine("Start Time: " + DateTime.Now.ToString("HH:mm:ss.ffffff"));
        Console.WriteLine("Number, Factors");

        HomePrime(n);

        Console.WriteLine("End Time: " + DateTime.Now.ToString("HH:mm:ss.ffffff"));
        Console.ReadLine();
    }

    public static void HomePrime(Int64 num) {
        string s = FindFactors(num);
        if (CheckPrime(num,s) == true) {
            Console.WriteLine("{0} is prime", num);
        } else {
            Console.WriteLine("{0}, {1}", num, s);
            HomePrime(Convert.ToInt64(RemSp(s)));
        }
    }

    public static string FindFactors(Int64 num) {
        Int64 n, r, t = num;
        string f = "";
        for (n = num; n >= 2; n--) {
            r = CalcP(n, t);
            if (r != 0) {
                f = f + " " + r.ToString();
                t = n / r;
                n = n / r + 1;
            }
        }
        return f;
    }

    public static Int64 CalcP(Int64 num, Int64 tot) {
        for (Int64 i = 2; i <= tot; i++) {
            if (num % i == 0) {
                return i;
            } 
        }
        return 0;
    }

    public static string RemSp(string str) {
        return new string(str.ToCharArray().Where(c => !Char.IsWhiteSpace(c)).ToArray());
    }

    public static bool CheckPrime(Int64 num, string s) {
        if (s == "") {
            return false;
        } else if (num == Convert.ToInt64(RemSp(s))) {
            return true;
        }
        return false;
    }

}

Dies ist eine optimierte Version des vorherigen Codes, bei der einige unnötige redundante Teile entfernt wurden.

Ausgabe (auf meinem i7 Laptop):

Enter Number: 16
Start Time: 18:09:51.636445
Number, Factors
16,  2 2 2 2
2222,  2 11 101
211101,  3 11 6397
3116397,  3 163 6373
31636373 is prime
End Time: 18:09:51.637954

Online testen

Perl + ntheory, HP (80) in 0,35 s auf dem PC

Kein Himbeer-Pi zur Hand.

use ntheory ":all";
use feature "say";
sub hp {
  my $n = shift;
  while (!is_prime($n)) {
    $n = join "",factor($n);
  }
  $n;
}
say hp($_) for (16,20,64,65,80);

Der Primalitätstest ist ES BPSW plus ein einzelnes MR auf Zufallsbasis für größere Zahlen. In dieser Größe könnten wir verwendenis_provable_prime (n-1 und / oder ECPP) ohne merklichen Geschwindigkeitsunterschied verwenden, aber das würde sich für Werte mit mehr als 300 Stellen ohne wirklichen Nutzen ändern. Das Factoring umfasst je nach Größe Trial, Power, Rho-Brent, P-1, SQUFOF, ECM und QS.

Für diese Eingaben wird ungefähr dieselbe Geschwindigkeit wie für Charles 'Pari / GP-Code auf der OEIS-Site ausgeführt. ntheory hat ein schnelleres Factoring für kleine Zahlen, und mein P-1 und ECM sind ziemlich gut, aber das QS ist nicht großartig, daher würde ich erwarten, dass Pari irgendwann schneller sein wird.