The beauty of bitwise AND

Re:The beauty of bitwise AND by thedufer on Friday August 05, 2016 @01:09AM (#12230174)

Excellent. I'm glad we agree. You'll notice that the stated identity f(x,x) = x fails to hold over x < 0. I hope this is sufficient to make it clear that the equation does not consider the negatives?

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Re:The beauty of bitwise AND by Strilanc on Sunday August 07, 2016 @04:07PM (#12243417)

Seems to work fine for -1:

    f(-1, -1)
    = sum_{n=0}^b 2^n (floor(-1/2^n) mod 2) (floor(-1/2^n) mod 2)
    = sum_{n=0}^b 2^n * 1 * 1
    = 1 + 10 + 100 + 1000 + ... [in binary]
    = ...11111 [in the 2-adics]
    = -1

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