Implementing x / 6 Using Only Bit Manipulations

This is an interesting question from one of the lab assignments in Introduction to Computer Systems, fall 2018 at Peking University.

Problem Description

Given a 32-bit integer \(x\)(in two‘s complement), implement a C function that returns \(\frac{x}{6}?\) using ONLY bit manipulations(operators like ~ ! | ^ & << >> +). Your function should behave exactly as the C expression x/6.

Hint: You can use the following formula(Formula 1)

\[ 2 = \frac{2+1}{2} \times \frac{2^2+1}{2^2} \times \frac{2^4+1}{2^4}\times\frac{2^8+1}{2^8}... \]

Inspiration

Since division is very slow using hardware, compilers often use optimizations to speed up division. For example, gcc will replace x/6 with x*171/1024 when x is relatively small, and implement x*171/1024 with shift left and shift right instructions. However, our function must cover all 32-bit two‘s complement integers, which means some other techniques are needed to make such replacement possible.

Resolution

We can change Formula 1 into the following form:

\[ \frac{1}{6} = \frac{1}{8} \times \frac{2^2+1}{2^2} \times \frac{2^4+1}{2^4}\times\frac{2^8+1}{2^8}... \]

Thus we can calculate this(Formula 2)

\[ p = \frac{x}{8} \times \frac{2^2+1}{2^2} \times \frac{2^4+1}{2^4}\times\frac{2^8+1}{2^8} \times \frac{2^{16}+1}{2^{16}} \]

Which can be implmented using a combination of shift-right and add operations(note that you must program carefully to avoid overflows). However, errors occur since expressions like x>>y return \(\lfloor x/2^y \rfloor\). We can counter the error by this(Formula 3)

\[ \frac{x}{6} = p + \frac{x}{6} - p = p + \frac{1}{6}(x-6p) \]

Since errors introduced by shift-rights will only cause \(p\) to be smaller than \(\frac{x}{6}\), we can deduce that \(x-6p > 0\). You can then approximate an upper bound of \(x-6p\), which depends on your implementation of Formula 2.

Suppose that \(x-6p < M\)(where M is small), then we can approximate \(\frac{1}{6}\) in Formula 3 using some \(X \approx \frac{1}{6}\) while keeping the equation true

\[ \lfloor \frac{1}{6} (x-6p)\rfloor = \lfloor X \cdot (x-6p) \rfloor \]

Choose a proper \(X = a/2^b\), and we are done!

/*
 * divSix - calculate x / 6 without using /
 *   Example: divSix(6) = 1,
 *            divSix(2147483647) = 357913941,
 *   Legal ops: ~ ! | ^ & << >> +
 *   Max ops: 40
 *   Rating: 4
 */
int divSix(int x) {
    int p;
    int q,y,t;
    x=x+(x>>31&5);
    p=x>>3;
    p=p+(p>>2);
    p=p+(p>>4);
    p=p+(p>>8);
    p=p+(p>>16);
    q=~p+1;
    t=x+(q<<1)+(q<<2);
    t=t+(t<<1)+(t<<3);
    return p+(t>>6);
}

Implementing x / 6 Using Only Bit Manipulations

原文：https://www.cnblogs.com/hehao98/p/10603028.html