按位和按位或运算符的巧合？

Question

So, while playing with the bitwise AND and bitwise OR operators, I noticed the following:

(a & b) + (a | b) = a + b

and there is a corresponding proof which mainly relies on the fact for any two bits x and y,

(a,b) = (0,0) --> (a&b, a|b) = (0, 0) = (a, b)
(a,b) = (0,1) --> (a&b, a|b) = (0, 1) = (a, b)
(a,b) = (1,0) --> (a&b, a|b) = (0, 1) = (b, a)
(a,b) = (1,1) --> (a&b, a|b) = (1, 1) = (b, a)

Now, I was wondering - is this a mere coincidence, or are these bitwise operations actually used in this way? I don't think that computers actually compute addition in this way, since it would be a recursive definition... but it seems too nice of a property to have been random!

Answer 1

AND and OR are idempotent and commutative:

a & a = a
a | a = a
a & b = b & a
a | b = b | a

They also absorb each other:

a & (a | b) = a
a | (a & b) = a

Thus:

a + b = (a | a) + (b & b)
a + b = (a | (a & b)) + (b & (a | b))

(a | a) + (b & b) = (a | (a & b)) + (b & (a | b))
a + b = (a & b) + (a | b)

Another way to see it is that, for two bits x and y , x + y = x | y x + y = x | y and x & y = 0 unless both bits are set, and then x & y adds the missing bit to x | y x | y .

Answer 2

when a and b value only 0 or 1 it is not surprising to have (a & b) + (a | b) == a + b

• having both a and b valuing 1 (a & b) and (a | b) value 1, so you do 1+1 on each side of ==

• else (a & b) is obviously 0 so you compare a|b and a+b while there is no possible carry on a+b and in that case a and b being a boolean do use + or | is the same

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Note when a and b value only 0 or 1 then a&b equals a*b in all cases

Answer 3

(a & b) + (a | b) = a + b is indeed true in general, not just for 0 and 1.

An other way to look at it, is that whenever the bits in a and b at a particular position are different (so one of them is zero and the other is one), then in (a & b) + (a | b) the 0 is put on the left hand side of the + and the 1 is put on the right hand side. If the bits are the same then it doesn't make any difference.

It's like a more granular form of min(a, b) + max(a, b) , at the bit level intead of the word level.

Reordering bits like that has no effect on the sum. Consider that both a and b are already sums, of the form a[0] + 2*a[1] + 4*a[2]... . a + b is a bigger sum, and (a & b) + (a | b) merely reordered the terms of that sum.