Integer cube root

20

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;
}
c
optimization
math
gcc
numerical-analysis
asked on Stack Overflow Dec 2, 2010 by Charles • edited Dec 2, 2010 by Charles

6 Answers

12

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).

answered on Stack Overflow Dec 2, 2010 by Fabian Giesen • edited Dec 11, 2010 by Fabian Giesen
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 produce 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 over the (21) iterations in the HacDel code - with linear O(b) convergence, despite having a loop body that is evidently much faster.

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)


static inline uint32_t u64_cbrt (uint64_t x)
{
    uint64_t r0 = 1, r1;

    /* IEEE-754 cbrt *may* not be exact. */

    if (x == 0) /* cbrt(0) : */
        return (0);

    int b = (64) - __builtin_clzll(x);
    r0 <<= (b + 2) / 3; /* ceil(b / 3) */

    do /* quadratic convergence: */
    {
        r1 = r0;
        r0 = (2 * r1 + x / (r1 * r1)) / 3;
    }
    while (r0 < r1);

    return ((uint32_t) r1); /* floor(cbrt(x)); */
}

A 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.

It's interesting that Intel's 'Icelake' microarchitecture will significantly improve integer division - an operation that seems to have been neglected for a long time. I simply won't trust the 'Hacker's Delight' answer until I can find a sound theoretical basis for it. And then I have to work out which variant is the 'correct' answer.

answered on Stack Overflow Jun 24, 2019 by Brett Hale • edited Aug 30, 2020 by Brett Hale
3

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...)

answered on Stack Overflow Dec 2, 2010 by comingstorm • edited Dec 2, 2010 by comingstorm
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.

answered on Stack Overflow Dec 2, 2010 by Keith Randall
1
// 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;
} 
answered on Stack Overflow Sep 24, 2013 by P_P
0

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.

answered on Stack Overflow Dec 2, 2010 by Karl Knechtel

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