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)
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:
Thus we can calculate this(Formula 2)
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
floor). We can counter the error by this(Formula 3)
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
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);
}