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aarav0
aarav0
Asked: September 11, 20262026-09-11T09:52:04-05:00 2026-09-11T09:52:04-05:00In: Bit Manipulation & Low-Level Computing, Data Structures & Algorithms

How does Brian Kernighan’s bit algorithm work, and why does n & (n – 1) clear the lowest set bit?

In low-level systems programming and coding interviews, people always use the expression n & (n - 1) to count set bits (Hamming Weight) or check if a number is a power of 2.

I know it works, but what is the exact binary arithmetic proof behind why subtracting 1 from an integer flips all bits up to and including the lowest set bit?

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  1. Abhay Tiwari
    Abhay Tiwari Begginer
    2026-09-11T09:52:07-05:00Added an answer on September 11, 2026 at 9:52 am

    The trick n & (n - 1) is one of the most elegant one-liners in computer engineering. Let’s look at the exact bitwise mechanics so the mathematical proof becomes obvious.

    1. What happens when you subtract 1 in binary?

    Think about standard base-10 math: when you subtract 1 from 1000, what happens? The lowest non-zero digit (1) becomes 0, and all trailing zeroes become 9s: 0999.

    Binary works exactly the same way, but with 0s and 1s:

    Any positive binary integer can be written in this general form:

    n = (arbitrary prefix) 1 0 0 0 ... 0
    

    where the 1 shown is the lowest set bit (the rightmost 1), followed by zero or more 0s.

    When you compute n - 1:

    1. The arbitrary prefix before the lowest 1 remains completely untouched.
    2. That lowest 1 turns into a 0 (borrowing from the subtraction).
    3. All the trailing 0s flip into 1s!
        n     = (prefix) 1 0 0 0
      n - 1   = (prefix) 0 1 1 1
    

    2. The Bitwise AND Operation: n & (n – 1)

    Now perform a bitwise AND between n and n - 1:

        n     = (prefix) 1 0 0 0
    & n - 1   = (prefix) 0 1 1 1
    ----------------------------
      result  = (prefix) 0 0 0 0
    

    Look at what happened:

    • The prefix matched identically → remains preserved.
    • The rightmost 1 was paired with 0 → becomes 0!
    • The trailing 0s were paired with 1s → remain 0!

    Conclusion: The operation n & (n - 1) turns off the lowest set bit in n and leaves every other bit completely unchanged. Pure mathematical magic!


    3. Real-World Applications

    A. Counting Set Bits in O(k) time (where k is number of 1s)

    Instead of looping 32 or 64 times, Brian Kernighan’s algorithm loops only as many times as there are 1-bits:

    def count_set_bits(n: int) -> int:
        count = 0
        while n > 0:
            n &= (n - 1)  # Strips off one set bit per loop iteration
            count += 1
        return count
    

    If a 64-bit integer has only two set bits, this loop executes exactly twice and terminates!

    B. Instant Power of Two Check in O(1)

    A power of two in binary has exactly one set bit (e.g. 8 = 1000_2, 16 = 10000_2). If you strip that single bit and the result is 0, it was a power of 2:

    def is_power_of_two(n: int) -> bool:
        return n > 0 and (n & (n - 1)) == 0
    

    C. Hardware POPCNT Alternative

    On modern x86_64 CPUs, you have the dedicated hardware assembly instruction POPCNT (or __builtin_popcount in GCC/Clang), which computes set bits in a single CPU cycle. But when writing portable code or kernel routines without AVX/SSE guarantees, Brian Kernighan’s algorithm remains the golden standard.

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