Calculating Average of Two integer without Overflow
Let's calculate average of two integers and return the average as an INT.
The usual approach is integer divison and take floor of the of the floating point result.
avg = floor((a + b) / 2)
But if any of the two values of a or b is the INT_MAX for C\C++ or std::i32::MAX in Rust;
it will cause a overflow.
Rust compiler will show something as follows:
thread 'main' panicked at 'attempt to add with overflow',
So the following pice of code will give us the average without any overflow.
fn average_with_overflow(a: i32, b: i32) -> i32 {
return (a & b) + ((a ^ b) >> 1);
}How this trick works?
The sum of an integer can be written as follows:
sum = carries + sum without carries
To calculate the above we can do the following:
a + b == ((a & b) << 1) + (a ^ b) [Eq.1]
Now dividing by 2 is same as 1 bit right shift.
Therefore,
(a + b) / 2 = (a + b) >> 1 [Eq.2]
Combining [Eq.1] and [Eq.2]:
(a + b) / 2 = (a + b) >> 1
= ((( a & b ) << 1) + (a ^ b)) >> 1
= (a & b) + ((a ^ b) >> 1) [Eq.3]
Where >> is distributive.
Example:
Let's imagine we have a calculator that has only three registers to hande data. Therefore the max decimal digit it will be able to handle in 9 or 111 in binary.
Let's take two int a = 5 and b = 6
So their binaries are. Here b' denotes binary representation.
a = b'101
b = b'110
Their integer sum will be
a + b = 11 (decimal)
= b'1011 (binary. We can see it overflows.)
So we have no space to keep the most singnigican bit here. (Asume our number are big-endian).
Let's apply the [Eq.3]
a ^ b = b'101 ^ b'110
= b'011
a & b = b'101 & b'110
= b'100
(a ^ b) >> 1 = b'011 >> 1
= b'001 (so dividing 1 by 2 making the MSB 0)
Finally,
(a & b) (a ^ b) >> 1 = b'100 + b'001
= b'101
= 5
Hope this example helps to get the clear picture.
The Rust code can be found here. https://gist.github.com/eNipu/6e8c3917865a4eb1e8cb34184d954d28 It will also work for C/C++. But not python since Python ints are arbitrary precision. We can verify the correctnes of this trick using Python.