Question

How to solve this Python problem?

Calling all units, B-and-E in progress def is..kerfectbeker(n): A positive integer n is said to be a perfect power if it can be expressed as the power b**e for some two integers band e that are both greater than one. (Any positive integer n can always be expressed as the trivial power n**1, so we don't care about that.) For example, the integers 32, 125 and 441 are perfect powers since they equal 2**5,5**3 and 21**2 respectively. This function should determine whether the positive integer n given as argument is some perfect power. To do this, your code needs to somehow iterate through a sufficient number of possible combinations of b and e that could work, returning True as soon as it finds some b and e that satisfy b**e == n. Since n can get pretty large, your function should not examine too many combinations of b and e above and beyond those that are necessary and sufficient to determine the answer. Achieving this efficiency is the central educational point of this particular problem. Expected result 42 False 441 True 469097433 True 12**34 True False 128*34 - 1 (The automated tester for this problem is based on the mathematical theorem about perfect powers that says that after the special case of two consecutive perfect powers 8 and 9, whenever the positive integer n is a perfect power, n-1 cannot be a perfect power. This theorem makes it really easy to generate random test cases with known correct answers, both positive and negative. For example, we would not need to try out all possible ways to express the number as an integer power to know right away that the humongous integer 1234**5678 - 1 is not a perfect power.)

Answer 1

import math

def is_perfect_power(n):

  if (n <= 1):

return True

for x in range(2, (int)(math.sqrt(n)) + 1):

p = x

while (p <= n) :

p = p * x

if (p == n) :

return True

return False

print(is_perfect_power(42))

print(is_perfect_power(441))

print(is_perfect_power(469097433))

print(is_perfect_power(12**34))

print(is_perfect_power(12**34-1))

import math def is_perfect_power(n): if (n <= 1): return True for x in range(2, (int) (math.sqrt(n)) + 1): P = x while (p <= n) : P = P * X if (p = n) : return True return false 16 17 18 print(is_perfect_power (42) print(is_perfect_power(441)) print(is_perfect_power ( 469097433) print(is_perfect_power(12**34)) print(is_perfect_power(12**34-1))