Count number of multiples against every power of 4 - c++

Question

Given a number, n, I need to efficiently find how many times this number is a multiple of all powers of 4 less than the given number.
For examples:
16 is a multiple of 4, and 16, so the result would be 2.
64 is a multiple of 4, 16, and 64, so the result would be 3.
256 is a multiple of 4, 16, 64, and 256, so the result would be 4.
14 is not a multiple of any power of 4, so the result would be 0.
35 is not a multiple of any power of 4, so the result would be 0.
Bitwise operations are preferred, and this is in a very tight loop so it is inside of a bottleneck that needs to be efficient. My code at the moment is the obvious answer, but I have to believe there is something more mathematical that can figure out the result in less steps:
power = 4;
while (power < n) {
result += !(n & (power - 1));
power *= 4;
}

Answer 1

You could use logarithms. A quick Google search for "fast log2 c++" brought up a pretty long list of ideas. Then your answer is log2(x)/2, and you'd have to find some way to make sure that your result is a whole number if you only want an answer for exact powers of 4.
If you are programming for an x86 processor, you can use BitScanForward & BitScanReverse to find the set bit, and use it to compute log2. The following code works in Visual Studio, for GCC or others, there are other ways to do inline assembly.
uint32_t exact_power_of_4_scan(uint32_t num)
{
unsigned long reverse;
unsigned long forward;
if (!_BitScanReverse(&reverse, num)) return 0;
_BitScanForward(&forward, num);
if (reverse != forward) return 0; // makes sure only a single bit is set
if (reverse & 0x1) return 0; // only want every other power of 2
return reverse / 2;
}
If you need a portable solution, table lookup might be the way to go, but is more complicated.
uint8_t not_single_bit[256] = {
1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
};
uint8_t log2_table[256] = {
0, 0, 1, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0,
4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
};
uint32_t exact_power_of_2(uint32_t num)
{
auto a = not_single_bit[num & 0xff];
auto b = not_single_bit[(num >> 8) & 0xff];
auto c = not_single_bit[(num >> 16) & 0xff];
auto d = not_single_bit[(num >> 24) & 0xff];
if (a + b + c + d != 3) {
return 0;
}
if (!a) {
return log2_table[num & 0xff];
}
if (!b) {
return log2_table[(num >> 8) & 0xff] + 8;
}
if (!c) {
return log2_table[(num >> 16) & 0xff] + 16;
}
return log2_table[(num >> 24) & 0xff] + 24;
}
uint32_t exact_power_of_4(uint32_t num)
{
auto ret = exact_power_of_2(num);
if (ret & 0x1) return 0;
return ret / 2;
}
Both are linear algorithms. The first will probably beat out looping for almost any value of num, but I haven't tested it. The second is probably only good for largish nums.

Answer 2

The mathematics would be to keep dividing by 4 until the result is no longer divisible by 4.
If you really want to do it with bitwise operations, techniques here can be used to count the number of trailing zero bits (i.e. the number of times a value is divisible by 2). Those can be adjusted to count pairs of trailing bits (i.e. divisibility by a power of 4 rather than 2).
Note that you will need to work with unsigned values to avoid certain cases of undefined or unspecified behaviours.
I would dispute your assertion that bitwise operations will make for a more efficient solution. It is not a given without testing, particularly with modern compilers.