Fastest way to get the integer part of sqrt(n)?
As we know if n is not a perfect square, then sqrt(n) would not be an integer. Since I need only the integer part, I feel that calling sqrt(n) wouldn't be that fast, as it takes time to calculate the fractional part also.
So my question is,
Can we get only the integer part of sqrt(n) without calculating the actual value of sqrt(n)? The algorithm should be faster than sqrt(n) (defined in <math.h> or <cmath>)?
If possible, you can write the code in asm block also.
I would try the Fast Inverse Square Root trick.
It's a way to get a very good approximation of 1/sqrt(n) without any branch, based on some bit-twiddling so not portable (notably between 32-bits and 64-bits platforms).
Once you get it, you just need to inverse the result, and takes the integer part.
There might be faster tricks, of course, since this one is a bit of a round about.
EDIT: let's do it!
First a little helper:
// benchmark.h
#include <sys/time.h>
template <typename Func>
double benchmark(Func f, size_t iterations)
{
f();
timeval a, b;
gettimeofday(&a, 0);
for (; iterations --> 0;)
{
f();
}
gettimeofday(&b, 0);
return (b.tv_sec * (unsigned int)1e6 + b.tv_usec) -
(a.tv_sec * (unsigned int)1e6 + a.tv_usec);
}
Then the main body:
#include <iostream>
#include <cmath>
#include "benchmark.h"
class Sqrt
{
public:
Sqrt(int n): _number(n) {}
int operator()() const
{
double d = _number;
return static_cast<int>(std::sqrt(d) + 0.5);
}
private:
int _number;
};
// http://www.codecodex.com/wiki/Calculate_an_integer_square_root
class IntSqrt
{
public:
IntSqrt(int n): _number(n) {}
int operator()() const
{
int remainder = _number;
if (remainder < 0) { return 0; }
int place = 1 <<(sizeof(int)*8 -2);
while (place > remainder) { place /= 4; }
int root = 0;
while (place)
{
if (remainder >= root + place)
{
remainder -= root + place;
root += place*2;
}
root /= 2;
place /= 4;
}
return root;
}
private:
int _number;
};
// http://en.wikipedia.org/wiki/Fast_inverse_square_root
class FastSqrt
{
public:
FastSqrt(int n): _number(n) {}
int operator()() const
{
float number = _number;
float x2 = number * 0.5F;
float y = number;
long i = *(long*)&y;
//i = (long)0x5fe6ec85e7de30da - (i >> 1);
i = 0x5f3759df - (i >> 1);
y = *(float*)&i;
y = y * (1.5F - (x2*y*y));
y = y * (1.5F - (x2*y*y)); // let's be precise
return static_cast<int>(1/y + 0.5f);
}
private:
int _number;
};
int main(int argc, char* argv[])
{
if (argc != 3) {
std::cerr << "Usage: %prog integer iterations\n";
return 1;
}
int n = atoi(argv[1]);
int it = atoi(argv[2]);
assert(Sqrt(n)() == IntSqrt(n)() &&
Sqrt(n)() == FastSqrt(n)() && "Different Roots!");
std::cout << "sqrt(" << n << ") = " << Sqrt(n)() << "\n";
double time = benchmark(Sqrt(n), it);
double intTime = benchmark(IntSqrt(n), it);
double fastTime = benchmark(FastSqrt(n), it);
std::cout << "Number iterations: " << it << "\n"
"Sqrt computation : " << time << "\n"
"Int computation : " << intTime << "\n"
"Fast computation : " << fastTime << "\n";
return 0;
}
And the results:
sqrt(82) = 9
Number iterations: 4096
Sqrt computation : 56
Int computation : 217
Fast computation : 119
// Note had to tweak the program here as Int here returns -1 :/
sqrt(2147483647) = 46341 // real answer sqrt(2 147 483 647) = 46 340.95
Number iterations: 4096
Sqrt computation : 57
Int computation : 313
Fast computation : 119
Where as expected the Fast computation performs much better than the Int computation.
Oh, and by the way, sqrt is faster :)
Edit: this answer is foolish - use (int) sqrt(i)
After profiling with proper settings (-march=native -m64 -O3) the above was a lot faster.
Alright, a bit old question, but the "fastest" answer has not been given yet. The fastest (I think) is the Binary Square Root algorithm, explained fully in this Embedded.com article.
It basicly comes down to this:
unsigned short isqrt(unsigned long a) {
unsigned long rem = 0;
int root = 0;
int i;
for (i = 0; i < 16; i++) {
root <<= 1;
rem <<= 2;
rem += a >> 30;
a <<= 2;
if (root < rem) {
root++;
rem -= root;
root++;
}
}
return (unsigned short) (root >> 1);
}
On my machine (Q6600, Ubuntu 10.10) I profiled by taking the square root of the numbers 1-100000000. Using iqsrt(i) took 2750 ms. Using (unsigned short) sqrt((float) i) took 3600ms. This was done using g++ -O3. Using the -ffast-math compile option the times were 2100ms and 3100ms respectively. Note this is without using even a single line of assembler so it could probably still be much faster.
The above code works for both C and C++ and with minor syntax changes also for Java.
What works even better for a limited range is a binary search. On my machine this blows the version above out of the water by a factor 4. Sadly it's very limited in range:
#include <stdint.h>
const uint16_t squares[] = {
0, 1, 4, 9,
16, 25, 36, 49,
64, 81, 100, 121,
144, 169, 196, 225,
256, 289, 324, 361,
400, 441, 484, 529,
576, 625, 676, 729,
784, 841, 900, 961,
1024, 1089, 1156, 1225,
1296, 1369, 1444, 1521,
1600, 1681, 1764, 1849,
1936, 2025, 2116, 2209,
2304, 2401, 2500, 2601,
2704, 2809, 2916, 3025,
3136, 3249, 3364, 3481,
3600, 3721, 3844, 3969,
4096, 4225, 4356, 4489,
4624, 4761, 4900, 5041,
5184, 5329, 5476, 5625,
5776, 5929, 6084, 6241,
6400, 6561, 6724, 6889,
7056, 7225, 7396, 7569,
7744, 7921, 8100, 8281,
8464, 8649, 8836, 9025,
9216, 9409, 9604, 9801,
10000, 10201, 10404, 10609,
10816, 11025, 11236, 11449,
11664, 11881, 12100, 12321,
12544, 12769, 12996, 13225,
13456, 13689, 13924, 14161,
14400, 14641, 14884, 15129,
15376, 15625, 15876, 16129,
16384, 16641, 16900, 17161,
17424, 17689, 17956, 18225,
18496, 18769, 19044, 19321,
19600, 19881, 20164, 20449,
20736, 21025, 21316, 21609,
21904, 22201, 22500, 22801,
23104, 23409, 23716, 24025,
24336, 24649, 24964, 25281,
25600, 25921, 26244, 26569,
26896, 27225, 27556, 27889,
28224, 28561, 28900, 29241,
29584, 29929, 30276, 30625,
30976, 31329, 31684, 32041,
32400, 32761, 33124, 33489,
33856, 34225, 34596, 34969,
35344, 35721, 36100, 36481,
36864, 37249, 37636, 38025,
38416, 38809, 39204, 39601,
40000, 40401, 40804, 41209,
41616, 42025, 42436, 42849,
43264, 43681, 44100, 44521,
44944, 45369, 45796, 46225,
46656, 47089, 47524, 47961,
48400, 48841, 49284, 49729,
50176, 50625, 51076, 51529,
51984, 52441, 52900, 53361,
53824, 54289, 54756, 55225,
55696, 56169, 56644, 57121,
57600, 58081, 58564, 59049,
59536, 60025, 60516, 61009,
61504, 62001, 62500, 63001,
63504, 64009, 64516, 65025
};
inline int isqrt(uint16_t x) {
const uint16_t *p = squares;
if (p[128] <= x) p += 128;
if (p[ 64] <= x) p += 64;
if (p[ 32] <= x) p += 32;
if (p[ 16] <= x) p += 16;
if (p[ 8] <= x) p += 8;
if (p[ 4] <= x) p += 4;
if (p[ 2] <= x) p += 2;
if (p[ 1] <= x) p += 1;
return p - squares;
}
A 32 bit version can be downloaded here: https://gist.github.com/3481770
If you don't mind an approximation, how about this integer sqrt function I cobbled together.
int sqrti(int x)
{
union { float f; int x; } v;
// convert to float
v.f = (float)x;
// fast aprox sqrt
// assumes float is in IEEE 754 single precision format
// assumes int is 32 bits
// b = exponent bias
// m = number of mantissa bits
v.x -= 1 << 23; // subtract 2^m
v.x >>= 1; // divide by 2
v.x += 1 << 29; // add ((b + 1) / 2) * 2^m
// convert to int
return (int)v.f;
}
It uses the algorithm described in this Wikipedia article. On my machine it's almost twice as fast as sqrt :)
While I suspect you can find a plenty of options by searching for "fast integer square root", here are some potentially-new ideas that might work well (each independent, or maybe you can combine them):
- Make a
static constarray of all the perfect squares in the domain you want to support, and perform a fast branchless binary search on it. The resulting index in the array is the square root. - Convert the number to floating point and break it into mantissa and exponent. Halve the exponent and multiply the mantissa by some magic factor (your job to find it). This should be able to give you a very close approximation. Include a final step to adjust it if it's not exact (or use it as a starting point for the binary search above).