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

Question

I saw this question, and pop up this idea.

Answer 1

The fastest solution is either testing if n > 0 && 3**19 % n == 0 as given in another answer or perfect hashing (below). First I'm giving two multiplication-based solutions.

Multiplication

I wonder why everybody missed that multiplication is much faster than division:

for (int i=0, pow=1; i<=19, pow*=3; ++i) {
    if (pow >= n) {
        return pow == n;
    }
}
return false;

Just try all powers, stop when it grew too big. Avoid overflow as 3**19 = 0x4546B3DB is the biggest power fitting in signed 32-bit int.

Multiplication with binary search

Binary search could look like

int pow = 1;
int next = pow * 6561; // 3**8
if (n >= next) pow = next;
next = pow * 81; // 3**4
if (n >= next) pow = next;
next = pow * 81; // 3**4; REPEATED
if (n >= next) pow = next;
next = pow * 9; // 3**2
if (n >= next) pow = next;
next = pow * 3; // 3**1
if (n >= next) pow = next;
return pow == next;

One step is repeated, so that the maximum exponent 19 = 8+4+4+2+1 can exactly be reached.

Perfect hashing

There are 20 powers of three fitting into a signed 32-bit int, so we take a table of 32 elements. With some experimentation, I found the perfect hash function

def hash(x):
    return (x ^ (x>>1) ^ (x>>2)) & 31;

mapping each power to a distinct index between 0 and 31. The remaining stuff is trivial:

// Create a table and fill it with some power of three.
table = [1 for i in range(32)]
// Fill the buckets.
for n in range(20): table[hash(3**n)] = 3**n;

Now we have

table = [
     1162261467, 1, 3, 729, 14348907, 1, 1, 1,
     1, 1, 19683, 1, 2187, 81, 1594323, 9,
     27, 43046721, 129140163, 1, 1, 531441, 243, 59049,
     177147, 6561, 1, 4782969, 1, 1, 1, 387420489]

and can test very fast via

def isPowerOfThree(x):
    return table[hash(x)] == x