Fastest way to get the integer part of sqrt(n)? - c++

Question

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.

Answer 1

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 :)

Answer 2

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

Answer 3

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 const array 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).

Answer 4

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 :)

Answer 5

To do integer sqrt you can use this specialization of newtons method:
Def isqrt(N):
a = 1
b = N
while |a-b| > 1
b = N / a
a = (a + b) / 2
return a
Basically for any x the sqrt lies in the range (x ... N/x), so we just bisect that interval at every loop for the new guess. Sort of like binary search but it converges must faster.
This converges in O(loglog(N)) which is very fast. It also doesn't use floating point at all, and it will also work well for arbitrary precision integers.

Answer 6

This is so short that it 99% inlines:
static inline int sqrtn(int num) {
int i = 0;
__asm__ (
"pxor %%xmm0, %%xmm0\n\t" // clean xmm0 for cvtsi2ss
"cvtsi2ss %1, %%xmm0\n\t" // convert num to float, put it to xmm0
"sqrtss %%xmm0, %%xmm0\n\t" // square root xmm0
"cvttss2si %%xmm0, %0" // float to int
:"=r"(i):"r"(num):"%xmm0"); // i: result, num: input, xmm0: scratch register
return i;
}
Why clean xmm0? Documentation of cvtsi2ss
The destination operand is an XMM register. The result is stored in the low doubleword of the destination operand, and the upper three doublewords are left unchanged.
GCC Intrinsic version (runs only on GCC):
#include <xmmintrin.h>
int sqrtn2(int num) {
register __v4sf xmm0 = {0, 0, 0, 0};
xmm0 = __builtin_ia32_cvtsi2ss(xmm0, num);
xmm0 = __builtin_ia32_sqrtss(xmm0);
return __builtin_ia32_cvttss2si(xmm0);
}
Intel Intrinsic version (tested on GCC, Clang, ICC):
#include <xmmintrin.h>
int sqrtn2(int num) {
register __m128 xmm0 = _mm_setzero_ps();
xmm0 = _mm_cvt_si2ss(xmm0, num);
xmm0 = _mm_sqrt_ss(xmm0);
return _mm_cvtt_ss2si(xmm0);
}
^^^^ All of them require SSE 1 (not even SSE 2).
Note: This is exactly how GCC calculates (int) sqrt((float) num) with -Ofast. If you want higher accuracy for larger i, then we can calculate (int) sqrt((double) num) (as noted by Gumby The Green in the comments):
static inline int sqrtn(int num) {
int i = 0;
__asm__ (
"pxor %%xmm0, %%xmm0\n\t"
"cvtsi2sd %1, %%xmm0\n\t"
"sqrtsd %%xmm0, %%xmm0\n\t"
"cvttsd2si %%xmm0, %0"
:"=r"(i):"r"(num):"%xmm0");
return i;
}
or
#include <xmmintrin.h>
int sqrtn2(int num) {
register __v2df xmm0 = {0, 0};
xmm0 = __builtin_ia32_cvtsi2sd(xmm0, num);
xmm0 = __builtin_ia32_sqrtsd(xmm0);
return __builtin_ia32_cvttsd2si(xmm0);
}

Answer 7

The following solution computes the integer part, meaning floor(sqrt(x)) exactly, with no rounding errors.
Problems With Other Approaches
using float or double is neither portable nor precise enough
#orlp's isqrt gives insane results like isqrt(100) = 15
approaches based on huge lookup tables are not practical beyond 32 bits
using a fast inverse sqrt is very imprecise, you're better off using sqrtf
Newton's approach requires expensive integer division and a good initial guess
My Approach
Mine is based on the bit-guessing approach proposed on Wikipedia. Unfortunately the pseudo-code provided on Wikipedia has some errors so I had to make some adjustments:
// C++20 also provides std::bit_width in its <bit> header
unsigned char bit_width(unsigned long long x) {
return x == 0 ? 1 : 64 - __builtin_clzll(x);
}
template <typename Int, std::enable_if_t<std::is_unsigned<Int, int = 0>>
Int sqrt(const Int n) {
unsigned char shift = bit_width(n);
shift += shift & 1; // round up to next multiple of 2
Int result = 0;
do {
shift -= 2;
result <<= 1; // make space for the next guessed bit
result |= 1; // guess that the next bit is 1
result ^= result * result > (n >> shift); // revert if guess too high
} while (shift != 0);
return result;
}
bit_width can be evaluated in constant time and the loop will iterate at most ceil(bit_width / 2) times. So even for a 64-bit integer, this will be at worst 32 iterations of basic arithmetic and bitwise operations.
The compile output is only around 20 instructions.
Performance
I have benchmarked my methods against float-bases ones by generating inputs uniformly. Note that in the real world most inputs would be much closer to zero than to std::numeric_limits<...>::max().
for uint32_t this performs about 25x worse than using std::sqrt(float)
for uint64_t this performs about 30x worse than using std::sqrt(double)
Accuracy
This method is always perfectly accurate, unlike approaches using floating point math.
Using sqrtf can provide incorrect rounding in the [228, 232) range. For example, sqrtf(0xffffffff) = 65536, when the square root is actually 65535.99999.
Double precision doesn't work consistently for the [260, 264) range. For example, sqrt(0x3fff...) = 2147483648, when the square root is actually 2147483647.999999.
The only thing that covers all 64-bit integers is x86 extended precision long double, simply because it can fit an entire 64-bit integer.
Conclusion
As I said, this the only solution that handles all inputs correctly, avoids integer division and doesn't require lookup tables.
In summary, if you need a method that is independent of precision and doesn't require gigantic lookup tables, this is your only option.
It might be especially useful in a constexpr context where performance isn't critical and where it could be much more important to get a 100% accurate result.
Alternative Approach Using Newton's Method
Newton's method can be quite fast when starting with a good guess. For our guess, we will round down to the next power of 2 and compute the square root in constant time. For any number 2x, we can obtain the square root using 2x/2.
template <typename Int, std::enable_if_t<std::is_unsigned_v<Int>, int> = 0>
Int sqrt_guess(const Int n)
{
Int log2floor = bit_width(n) - 1;
// sqrt(x) is equivalent to pow(2, x / 2 = x >> 1)
// pow(2, x) is equivalent to 1 << x
return 1 << (log2floor >> 1);
}
Note that this is not exactly 2x/2 because we lost some precision during the rightshift. Instead it is 2floor(x/2).
Also note that sqrt_guess(0) = 1 which is actually necessary to avoid division by zero in the first iteration:
template <typename Int, std::enable_if_t<std::is_unsigned_v<Int>, int> = 0>
Int sqrt_newton(const Int n)
{
Int a = sqrt_guess(n);
Int b = n;
// compute unsigned difference
while (std::max(a, b) - std::min(a, b) > 1) {
b = n / a;
a = (a + b) / 2;
}
// a is now either floor(sqrt(n)) or ceil(sqrt(n))
// we decrement in the latter case
// this is overflow-safe as long as we start with a lower bound guess
return a - (a * a > n);
}
This alternative approach performs roughly equivalent to the first proposal, but is usually a few percentage points faster. However, it heavily relies on efficient hardware division and result can vary heavily.
The use of sqrt_guess makes a huge difference. It is roughly five times faster than using 1 as the initial guess.

Answer 8

In many cases, even exact integer sqrt value is not needed, enough having good approximation of it. (For example, it often happens in DSP optimization, when 32-bit signal should be compressed to 16-bit, or 16-bit to 8-bit, without loosing much precision around zero).
I've found this useful equation:
k = ceil(MSB(n)/2); - MSB(n) is the most significant bit of "n"
sqrt(n) ~= 2^(k-2)+(2^(k-1))*n/(2^(2*k))); - all multiplications and divisions here are very DSP-friendly, as they are only 2^k.
This equation generates smooth curve (n, sqrt(n)), its values are not very much different from real sqrt(n) and thus can be useful when approximate accuracy is enough.

Answer 9

Why nobody suggests the quickest method?
If:
the range of numbers is limited
memory consumption is not crucial
application launch time is not critical
then create int[MAX_X] filled (on launch) with sqrt(x) (you don't need to use the function sqrt() for it).
All these conditions fit my program quite well.
Particularly, an int[10000000] array is going to consume 40MB.
What's your thoughts on this?

Answer 10

On my computer with gcc, with -ffast-math, converting a 32-bit integer to float and using sqrtf takes 1.2 s per 10^9 ops (without -ffast-math it takes 3.54 s).
The following algorithm uses 0.87 s per 10^9 at the expense of some accuracy: errors can be as much as -7 or +1 although the RMS error is only 0.79:
uint16_t SQRTTAB[65536];
inline uint16_t approxsqrt(uint32_t x) {
const uint32_t m1 = 0xff000000;
const uint32_t m2 = 0x00ff0000;
if (x&m1) {
return SQRTTAB[x>>16];
} else if (x&m2) {
return SQRTTAB[x>>8]>>4;
} else {
return SQRTTAB[x]>>8;
}
}
The table is constructed using:
void maketable() {
for (int x=0; x<65536; x++) {
double v = x/65535.0;
v = sqrt(v);
int y = int(v*65535.0+0.999);
SQRTTAB[x] = y;
}
}
I found that refining the bisection using further if statements does improve accuracy, but it also slows things down to the point that sqrtf is faster, at least with -ffast-math.

Answer 11

Or just do a binary search, cant write a simpler version imo:
uint16_t sqrti(uint32_t num)
{
uint16_t ret = 0;
for(int32_t i = 15; i >= 0; i--)
{
uint16_t temp = ret | (1 << i);
if(temp * temp <= num)
{
ret = temp;
}
}
return ret;
}

Answer 12

If you need performance on computing square root, I guess you will compute a lot of them.
Then why not caching the answer? I don't know the range for N in your case, nor if you will compute many times the square root of the same integer, but if yes, then you can cache the result each time your method is called (in an array would be the most efficient if not too large).

Answer 13

This is an addition for those in need of a precide square root for very large integers. The trick is to leverage the fast floating point square root of modern processors and to fix round-off errors.
template <typename T>
T preciseIntegerSqrt(T n)
{
if (sizeof(T) <= 4)
{
return std::sqrt((double)n);
}
else if (sizeof(T) <= 8)
{
T r = std::sqrt((double)n);
return r - (r*r-1 >= n);
}
else
{
if (n == 0) return 0;
T r = 0;
for (T b = (T(1)) << ((std::bit_width(n)-1) / 2); b != 0; b >>= 1)
{
T const k = (b + 2*r) * b;
r |= (n >= k) * b;
n -= (n >= k) * k;
}
return r;
}
}
Explanation: Integers of up to 32 bits do not need a correction, since they can be represented precisely as double-precision floating point numbers. 64-bit integers get along with a very cheap correction. For the general case, refer to Jan Schultke's excellent answer. The code provided here is very slightly faster that that one (10% on my machine, may vary with integer type and hardware).