# How to make perfect power algorithm more efficient?

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?

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 es, 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 kth-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.