问题
I'm looking for fast code for 64-bit (unsigned) cube roots. (I'm using C and compiling with gcc, but I imagine most of the work required will be language- and compiler-agnostic.) I will denote by ulong a 64-bit unisgned integer.
Given an input n I require the (integral) return value r to be such that
r * r * r <= n && n < (r + 1) * (r + 1) * (r + 1)
That is, I want the cube root of n, rounded down. Basic code like
return (ulong)pow(n, 1.0/3);
is incorrect because of rounding toward the end of the range. Unsophisticated code like
ulong
cuberoot(ulong n)
{
ulong ret = pow(n + 0.5, 1.0/3);
if (n < 100000000000001ULL)
return ret;
if (n >= 18446724184312856125ULL)
return 2642245ULL;
if (ret * ret * ret > n) {
ret--;
while (ret * ret * ret > n)
ret--;
return ret;
}
while ((ret + 1) * (ret + 1) * (ret + 1) <= n)
ret++;
return ret;
}
gives the correct result, but is slower than it needs to be.
This code is for a math library and it will be called many times from various functions. Speed is important, but you can't count on a warm cache (so suggestions like a 2,642,245-entry binary search are right out).
For comparison, here is code that correctly calculates the integer square root.
ulong squareroot(ulong a) {
ulong x = (ulong)sqrt((double)a);
if (x > 0xFFFFFFFF || x*x > a)
x--;
return x;
}
回答1:
The book "Hacker's Delight" has algorithms for this and many other problems. The code is online here. EDIT: That code doesn't work properly with 64-bit ints, and the instructions in the book on how to fix it for 64-bit are somewhat confusing. A proper 64-bit implementation (including test case) is online here.
I doubt that your squareroot function works "correctly" - it should be ulong a for the argument, not n :) (but the same approach would work using cbrt instead of sqrt, although not all C math libraries have cube root functions).
回答2:
You could try a Newton's step to fix your rounding errors:
ulong r = (ulong)pow(n, 1.0/3);
if(r==0) return r; /* avoid divide by 0 later on */
ulong r3 = r*r*r;
ulong slope = 3*r*r;
ulong r1 = r+1;
ulong r13 = r1*r1*r1;
/* making sure to handle unsigned arithmetic correctly */
if(n >= r13) r+= (n - r3)/slope;
if(n < r3) r-= (r3 - n)/slope;
A single Newton step ought to be enough, but you may have off-by-one (or possibly more?) errors. You can check/fix those using a final check&increment step, as in your OQ:
while(r*r*r > n) --r;
while((r+1)*(r+1)*(r+1) <= n) ++r;
or some such.
(I admit I'm lazy; the right way to do it is to carefully check to determine which (if any) of the check&increment things is actually necessary...)
回答3:
If pow is too expensive, you can use a count-leading-zeros instruction to get an approximation to the result, then use a lookup table, then some Newton steps to finish it.
int k = __builtin_clz(n); // counts # of leading zeros (often a single assembly insn)
int b = 64 - k; // # of bits in n
int top8 = n >> (b - 8); // top 8 bits of n (top bit is always 1)
int approx = table[b][top8 & 0x7f];
Given b and top8, you can use a lookup table (in my code, 8K entries) to find a good approximation to cuberoot(n). Use some Newton steps (see comingstorm's answer) to finish it.
回答4:
I've adapted the algorithm presented in 1.5.2 (the kth root) in Modern Computer Arithmetic (Brent and Zimmerman). For the case of (k == 3), and given a 'relatively' accurate over-estimate of the initial guess - this algorithm seems to out-perform the 'Hacker's Delight' code above.
Not only that, but MCA as a text provides theoretical background as well as a proof of correctness and terminating criteria.
Provided that we can provide a 'relatively' good initial over-estimate, I haven't been able to find a case that exceeds (7) iterations. (Is this effectively related to 64-bit values having 2^6 bits?) Either way, it's an improvement on the (21) iterations in the HacDel code!
The initial estimate I've used is based on a 'rounding up' of the number of significant bits in the value (x). Given (b) significant bits in (x), we can say: 2^(b - 1) <= x < 2^b. I state without proof (though it should be relatively easy to demonstrate) that: 2^ceil(b / 3) > x^(1/3)
Here's my code as it currently is...
static inline uint32_t u64_cbrt (uint64_t x)
{
#if (0) /* an exact IEEE-754 evaluation: */
if (x <= (UINT64_C(1) << (53)))
return (uint32_t) cbrt((double) x);
#endif
int bits_x = (64) - __builtin_clzll(x);
if (bits_x == 0)
return (0); /* cbrt(0) */
int exp_r = (bits_x + 2) / 3;
/* initial estimate: 2 ^ ceil(b / 3) */
uint64_t est_r = UINT64_C(1) << exp_r, r;
do /* quadratic convergence (?) */
{
r = est_r;
est_r = (2 * r + x / (r * r)) / 3;
}
while (est_r < r);
return ((uint32_t) r); /* floor(cbrt(x)) */
}
The crbt call probably isn't all that useful - unlike the sqrt call which can be implemented with extreme efficiency on modern hardware. That said, I've seen gains for sets of values under 2^53 (exactly represented in IEEE-754 doubles), which surprised me.
The only downside is the division by: (r * r) - this can be slow, as the latency of integer division continues to fall behind other advances in ALUs. The division by a constant: (3) is handled by reciprocal methods on any modern optimising compiler.
回答5:
// On my pc: Math.Sqrt 35 ns, cbrt64 <70ns, cbrt32 <25 ns, (cbrt12 < 10ns)
// cbrt64(ulong x) is a C# version of:
// http://www.hackersdelight.org/hdcodetxt/acbrt.c.txt (acbrt1)
// cbrt32(uint x) is a C# version of:
// http://www.hackersdelight.org/hdcodetxt/icbrt.c.txt (icbrt1)
// Union in C#:
// http://www.hanselman.com/blog/UnionsOrAnEquivalentInCSairamasTipOfTheDay.aspx
using System.Runtime.InteropServices;
[StructLayout(LayoutKind.Explicit)]
public struct fu_32 // float <==> uint
{
[FieldOffset(0)]
public float f;
[FieldOffset(0)]
public uint u;
}
private static uint cbrt64(ulong x)
{
if (x >= 18446724184312856125) return 2642245;
float fx = (float)x;
fu_32 fu32 = new fu_32();
fu32.f = fx;
uint uy = fu32.u / 4;
uy += uy / 4;
uy += uy / 16;
uy += uy / 256;
uy += 0x2a5137a0;
fu32.u = uy;
float fy = fu32.f;
fy = 0.33333333f * (fx / (fy * fy) + 2.0f * fy);
int y0 = (int)
(0.33333333f * (fx / (fy * fy) + 2.0f * fy));
uint y1 = (uint)y0;
ulong y2, y3;
if (y1 >= 2642245)
{
y1 = 2642245;
y2 = 6981458640025;
y3 = 18446724184312856125;
}
else
{
y2 = (ulong)y1 * y1;
y3 = y2 * y1;
}
if (y3 > x)
{
y1 -= 1;
y2 -= 2 * y1 + 1;
y3 -= 3 * y2 + 3 * y1 + 1;
while (y3 > x)
{
y1 -= 1;
y2 -= 2 * y1 + 1;
y3 -= 3 * y2 + 3 * y1 + 1;
}
return y1;
}
do
{
y3 += 3 * y2 + 3 * y1 + 1;
y2 += 2 * y1 + 1;
y1 += 1;
}
while (y3 <= x);
return y1 - 1;
}
private static uint cbrt32(uint x)
{
uint y = 0, z = 0, b = 0;
int s = x < 1u << 24 ? x < 1u << 12 ? x < 1u << 06 ? x < 1u << 03 ? 00 : 03 :
x < 1u << 09 ? 06 : 09 :
x < 1u << 18 ? x < 1u << 15 ? 12 : 15 :
x < 1u << 21 ? 18 : 21 :
x >= 1u << 30 ? 30 : x < 1u << 27 ? 24 : 27;
do
{
y *= 2;
z *= 4;
b = 3 * y + 3 * z + 1 << s;
if (x >= b)
{
x -= b;
z += 2 * y + 1;
y += 1;
}
s -= 3;
}
while (s >= 0);
return y;
}
private static uint cbrt12(uint x) // x < ~255
{
uint y = 0, a = 0, b = 1, c = 0;
while (a < x)
{
y++;
b += c;
a += b;
c += 6;
}
if (a != x) y--;
return y;
}
回答6:
I would research how to do it by hand, and then translate that into a computer algorithm, working in base 2 rather than base 10.
We end up with an algorithm something like (pseudocode):
Find the largest n such that (1 << 3n) < input.
result = 1 << n.
For i in (n-1)..0:
if ((result | 1 << i)**3) < input:
result |= 1 << i.
We can optimize the calculation of (result | 1 << i)**3 by observing that the bitwise-or is equivalent to addition, refactoring to result**3 + 3 * i * result ** 2 + 3 * i ** 2 * result + i ** 3, caching the values of result**3 and result**2 between iterations, and using shifts instead of multiplication.
来源：https://stackoverflow.com/questions/4331820/integer-cube-root