[algorithm] Bitwise and in place of modulus operator

We know that for example modulo of power of two can be expressed like this:

  x % 2 inpower n == x & (2 inpower n - 1).

Examples:

x % 2 == x & 1
x % 4 == x & 3
x % 8 == x & 7 

What about general nonpower of two numbers?

Let's say:

x % 7==?

Not using the bitwise-and (&) operator in binary, there is not. Sketch of proof:

Suppose there were a value k such that x & k == x % (k + 1), but k != 2^n - 1. Then if x == k, the expression x & k seems to "operate correctly" and the result is k. Now, consider x == k-i: if there were any "0" bits in k, there is some i greater than 0 which k-i may only be expressed with 1-bits in those positions. (E.g., 1011 (11) must become 0111 (7) when 100 (4) has been subtracted from it, in this case the 000 bit becomes 100 when i=4.) If a bit from the expression of k must change from zero to one to represent k-i, then it cannot correctly calculate x % (k+1), which in this case should be k-i, but there is no way for bitwise boolean and to produce that value given the mask.

This is specifically a special case because computers represent numbers in base 2. This is generalizable:

(number)_base % base^x

is equivilent to the last x digits of (number)_base.

In this specific case (mod 7), we still can replace %7 with bitwise operators:

// Return X%7 for X >= 0.
int mod7(int x)
{
  while (x > 7) x = (x&7) + (x>>3);
  return (x == 7)?0:x;
}

It works because 8%7 = 1. Obviously, this code is probably less efficient than a simple x%7, and certainly less readable.

There is only a simple way to find modulo of 2^i numbers using bitwise.

There is an ingenious way to solve Mersenne cases as per the link such as n % 3, n % 7... There are special cases for n % 5, n % 255, and composite cases such as n % 6.

For cases 2^i, ( 2, 4, 8, 16 ...)

n % 2^i = n & (2^i - 1)

More complicated ones are hard to explain. Read up only if you are very curious.

Using bitwise_and, bitwise_or, and bitwise_not you can modify any bit configurations to another bit configurations (i.e. these set of operators are "functionally complete"). However, for operations like modulus, the general formula would be necessarily be quite complicated, I wouldn't even bother trying to recreate it.

Modulo "7" without "%" operator

int a = x % 7;

int a = (x + x / 7) & 7;

This only works for powers of two (and frequently only positive ones) because they have the unique property of having only one bit set to '1' in their binary representation. Because no other class of numbers shares this property, you can't create bitwise-and expressions for most modulus expressions.

There are moduli other than powers of 2 for which efficient algorithms exist.

For example, if x is 32 bits unsigned int then x % 3 = popcnt (x & 0x55555555) - popcnt (x & 0xaaaaaaaa)