How to make perfect power algorithm more efficient?

Question

I have the following code:

def isPP(n):
  pos = [int(i) for i in range(n+1)]
  pos = pos[2:] ##to ignore the trivial n** 1 == n case
  y = []
  for i in pos:
      for it in pos:
          if i** it == n:
              y.append((i,it))
              #return list((i,it))
      #break
  if len(y) <1:
      return None
  else:
      return list(y[0])

Which works perfectly up until ~2000, since I'm storing far too much in memory. What can I do to make it work efficiently for large numbers (say, 50000 or 100000). I tried to make it end after finding one case, but my algorithm is still far too inefficient if the number is large.

Any tips?

Answer 1

IIRC，迭代检查“它有平方根吗？有立方根吗？有第四个根吗？……”要容易得多，您很快就会得出推定的根必须在1之间的观点。和2 ，此时您可以停止。

Answer 2

A number n is a perfect power if there exists a b and e for which b ^ e = n . For instance 216 = 6^3 = 2^3 * 3^3 is a perfect power, but 72 = 2^3 * 3^2 is not.

The trick to determining if a number is a perfect power is to know that, if the number is a perfect power, then the exponent e must be less than log2 n , because if e is greater then 2^ e will be greater than n . Further, it is only necessary to test prime e s, because if a number is a perfect power to a composite exponent it will also be a perfect power to the prime factors of the composite component; for instance, 2^15 = 32768 = 32^3 = 8^5 is a perfect cube root and also a perfect fifth root.

The function isPerfectPower shown below tests each prime less than log2 n by first computing the integer root using Newton's method, then powering the result to check if it is equal to n . Auxiliary function primes compute a list of prime numbers by the Sieve of Eratosthenes, iroot computes the integer k th-root by Newton's method, and ilog computes the integer logarithm to base b by binary search.

def primes(n): # sieve of eratosthenes
    i, p, ps, m = 0, 3, [2], n // 2
    sieve = [True] * m
    while p <= n:
        if sieve[i]:
            ps.append(p)
            for j in range((p*p-3)/2, m, p):
                sieve[j] = False
        i, p = i+1, p+2
    return ps

def iroot(k, n): # assume n > 0
    u, s, k1 = n, n+1, k-1
    while u < s:
        s = u
        u = (k1 * u + n // u ** k1) // k
    return s

def ilog(b, n): # max e where b**e <= n
    lo, blo, hi, bhi = 0, 1, 1, b
    while bhi < n:
        lo, blo, hi, bhi = hi, bhi, hi+hi, bhi*bhi
    while 1 < (hi - lo):
        mid = (lo + hi) // 2
        bmid = blo * pow(b, (mid - lo))
        if n < bmid: hi, bhi = mid, bmid
        elif bmid < n: lo, blo = mid, bmid
        else: return mid
    if bhi == n: return hi
    return lo

def isPerfectPower(n): # x if n == x ** y, or False
    for p in primes(ilog(2,n)):
        x = iroot(p, n)
        if pow(x, p) == n: return x
    return False

There is further discussion of the perfect power predicate at my blog .

Answer 3

a relevant improvement would be:

import math

def isPP(n):

        # first have a look at the length of n in binary representation
        ln = int(math.log(n)/math.log(2)) + 1

        y = []
        for i in range(n+1):
                if (i <= 1):
                        continue
                # calculate max power

                li = int(math.log(i)/math.log(2))
                mxi = ln / li + 1
                for it in range(mxi):
                        if (it <= 1):
                                continue
                        if i ** it == n:
                                y.append((i,it))
                                # break if you only need 1

        if len(y) <1:
                return None
        else:
                return list(y[0])

Answer 4

I think a better way would be implementing this "hack":

import math

def isPP(n):
    range = math.log(n)/math.log(2)
    range = (int)(range)
    result = []
    for i in xrange(n):
        if(i<=1):
            continue
        exponent = (int)(math.log(n)/math.log(i))
        for j in [exponent-1, exponent, exponent+1]:
            if i ** j == n:
                result.append([i,j])
    return result

print isPP(10000)

Result:

[[10,4],[100,2]]

The hack uses the fact that:

if log(a)/log(b) = c,
    then power(b,c) = a

Since this calculation can be a bit off in floating points giving really approximate results, exponent is checked to the accuracy of +/- 1 .

You can make necessary adjustments for handling corner cases like n=1, etc.