# Implementing x / 6 Using Only Bit Manipulations

Published:

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);
}


Tags: