'Optimizing Fixed-Point Sqrt

I made what I think is a good fixed-point square root algorithm:

template<int64_t M, int64_t P>
typename enable_if<M + P == 32, FixedPoint<M, P>>::type sqrt(FixedPoint<M, P> f)
{
    if (f.num == 0)
        return 0;
    //Reduce it to the 1/2 to 2 range (based around FixedPoint<2, 30> to avoid left/right shift branching)
    int64_t num{ f.num }, faux_half{ 1 << 29 };
    ptrdiff_t mag{ 0 };
    while (num < (faux_half)) {
        num <<= 2;
        ++mag;
    }

    int64_t res = (M % 2 == 0 ? SQRT_32_EVEN_LOOKUP : SQRT_32_ODD_LOOKUP)[(num >> (30 - 4)) - (1LL << 3)];
    res >>= M / 2 + mag - 1; //Finish making an excellent guess
    for (int i = 0; i < 2; ++i)
        //                            \    |   /
        //                             \   |  /
        //                              _| V L
        res = (res + (int64_t(f.num) << P) / res) >> 1; //Use Newton's method to improve greatly on guess
        //                               7 A r
        //                              /  |  \
        //                             /   |   \
        //                       The Infamous Time Eater
    return FixedPoint<M, P>(res, true);
}

However, after profiling (in release mode) I found out that the division here takes up 83% of the time this algorithm spends. I can speed it up 6x by replacing the division with multiplication, but that's just wrong. I found out that integer division is much slower than multiplication, unfortunately. Is there any way to optimize this?

In case this table is necessary.

const array<int32_t, 24> SQRT_32_EVEN_LOOKUP = {
     0x2d413ccd, //magic numbers calculated by taking 0.5 + 0.5 * i / 8 from i = 0 to 23, multiplying by 2^30, and converting to hex
     0x30000000,
     0x3298b076,
     0x3510e528,
     0x376cf5d1,
     0x39b05689,
     0x3bddd423,
     0x3df7bd63,
     0x40000000,
     0x41f83d9b,
     0x43e1db33,
     0x45be0cd2,
     0x478dde6e,
     0x49523ae4,
     0x4b0bf165,
     0x4cbbb9d6,
     0x4e623850,
     0x50000000,
     0x5195957c,
     0x532370b9,
     0x54a9fea7,
     0x5629a293,
     0x57a2b749,
     0x59159016
};

SQRT_32_ODD_LOOKUP is just SQRT_32_EVEN_LOOKUP divided by sqrt(2).

c++algorithmperformancefixed-pointsqrt


Solution 1:[1]

Reinventing the wheel, really, and not in a good way. The correct solution is to calculate 1/sqrt(x) using NR, and then multiply once to get x/sqrt(x) - just check for x==0 up front.

The reason why this is so much better is that the NR step for y=1/sqrt(x) is just y = (3-x*y*y)*y/2. That's all straightforward multiplication.

Sources

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Source: Stack Overflow

Solution Source
Solution 1 BlackBeans

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