How to check if an integer is a power of 3?

Question

I saw this question, and pop up this idea.

Answer 1

Recursively divide by 3, check that the remainder is zero and re-apply to the quotient.

Note that 1 is a valid answer as 3 to the zero power is 1 is an edge case to beware.

Answer 2

Simple and constant-time solution:

return n == power(3, round(log(n) / log(3)))

Answer 3

You can do better than repeated division, which takes O(lg(X) * |division|) time. Essentially you do a binary search on powers of 3. Really we will be doing a binary search on N, where 3^N = input value). Setting the Pth binary digit of N corresponds to multiplying by 3^(2^P), and values of the form 3^(2^P) can be computed by repeated squaring.

Algorithm

• Let the input value be X.
• Generate a list L of repeated squared values which ends once you pass X.
• Let your candidate value be T, initialized to 1.
• For each E in reversed L, if T*E <= X then let T *= E.
• Return T == X.

Complexity:

O(lg(lg(X)) * |multiplication|) - Generating and iterating over L takes lg(lg(X)) iterations, and multiplication is the most expensive operation in an iteration.

Answer 4

Your question is fairly easy to answer by defining a simple function to run the check for you. The example implementation shown below is written in Python but should not be difficult to rewrite in other languages if needed. Unlike the last version of this answer, the code shown below is far more reliable.

Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 08:06:12) [MSC v.1900 64 bit (AMD64)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> import math
>>> def power_of(number, base):
    return number == base ** round(math.log(number, base))

>>> base = 3
>>> for power in range(21):
    number = base ** power
    print(f'{number} is '
          f'{"" if power_of(number, base) else "not "}'
          f'a power of {base}.')
    number += 1
    print(f'{number} is '
          f'{"" if power_of(number, base) else "not "}'
          f'a power of {base}.')
    print()


1 is a power of 3.
2 is not a power of 3.

3 is a power of 3.
4 is not a power of 3.

9 is a power of 3.
10 is not a power of 3.

27 is a power of 3.
28 is not a power of 3.

81 is a power of 3.
82 is not a power of 3.

243 is a power of 3.
244 is not a power of 3.

729 is a power of 3.
730 is not a power of 3.

2187 is a power of 3.
2188 is not a power of 3.

6561 is a power of 3.
6562 is not a power of 3.

19683 is a power of 3.
19684 is not a power of 3.

59049 is a power of 3.
59050 is not a power of 3.

177147 is a power of 3.
177148 is not a power of 3.

531441 is a power of 3.
531442 is not a power of 3.

1594323 is a power of 3.
1594324 is not a power of 3.

4782969 is a power of 3.
4782970 is not a power of 3.

14348907 is a power of 3.
14348908 is not a power of 3.

43046721 is a power of 3.
43046722 is not a power of 3.

129140163 is a power of 3.
129140164 is not a power of 3.

387420489 is a power of 3.
387420490 is not a power of 3.

1162261467 is a power of 3.
1162261468 is not a power of 3.

3486784401 is a power of 3.
3486784402 is not a power of 3.

>>> 

NOTE: The last revision has caused this answer to become nearly the same as TMS' answer.

Answer 5

This is a constant time method! Yes. O(1). For numbers of fixed length, say 32-bits.

Given that we need to check if an integer n is a power of 3, let us start thinking about this problem in terms of what information is already at hand.

1162261467 is the largest power of 3 that can fit into an Java int.
1162261467 = 3^19 + 0

The given n can be expressed as [(a power of 3) + (some x)]. I think it is fairly elementary to be able to prove that if x is 0(which happens iff n is a power of 3), 1162261467 % n = 0.

The general idea is that if X is some power of 3, X can be expressed as Y/3a, where a is some integer and X < Y. It follows the exact same principle for Y < X. The Y = X case is elementary.

So, to check if a given integer n is a power of three, check if n > 0 && 1162261467 % n == 0.