bitwise operators for finding less than in c

Question

This is a homework assignment which requires me to come up with a function to determine if x < y, if it is I must return 1, using only bitwise o

Answer 1

I think it can be achived doing following:

1. Xor both numbers, that will give you bits that differs
2. Get the most significant bit that differs, as this one will make a difference
3. Check if that most significant bit belongs to bigger value

Code:

int less(int x, int y) {
    //Get only diffed bits
    int msb = x ^ y;
    //Get first msb bit
    msb |= (msb >>  1);
    msb |= (msb >>  2);
    msb |= (msb >>  4);
    msb |= (msb >>  8);
    msb |= (msb >> 16);
    msb = msb - (msb >> 1);
    //check if msb belongs to y;
    return !((y & msb) ^ msb);
}

Answer 2

The idea of implementing subtraction is good.

int sub = x + (~y+1);

From here, you just need to check whether sub is negative, i.e. extract its sign bit. This is where the technical difficulty appears.

Let's assume x and y are unsigned 32-bit integers. Then the result of subtraction can be represented by a signed 33-bit integer (check its minimum and maximum values to see that). That is, using the expression x + (~y+1) directly doesn't help, because it will overflow.

What if x and y were 31-bit unsigned numbers? Then, x + (~y+1) can be represented by a 32-bit signed number, and its sign bit tells us whether x is smaller than y:

answer(x, y) := ((x + (~y+1)) >> 31) & 1

To convert x and y from 32 bits to 31 bits, separate their MSB (most significant bit) and all the other 31 bits into different variables:

msb_x = (x >> 31) & 1;
msb_y = (y >> 31) & 1;
x_without_msb = x << 1 >> 1;
y_without_msb = y << 1 >> 1;

Then consider the 4 possible combinations of their MSB:

if (msb_x == 0 && msb_y == 0)
    return answer(x_without_msb, y_without_msb);
else if (msb_x == 1 && msb_y == 0)
    return 0; // x is greater than y
else if (msb_x == 0 && msb_y == 1)
    return 1; // x is smaller than y
else
    return answer(x_without_msb, y_without_msb);

Converting all these if-statements to a single big ugly logical expression is "an exercise for the reader".

Answer 3

In order to know if x < y, you can simply ask if x - y < 0.

In other words, what is the sign of the result of x - y.

Since you stated you are to assume 32 bit integers, following will provide the correct result:

((x - y) >> 31) & 0x1

Answer 4

Here was my attempt (compiling results, x > y then 0, x < y then 1, x == y then 1):

((((x + ~y) >> 31) + 1))^1