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I knew what NAF was in a vague 'read it once in a crypto paper' kind of way but had never actually implemented it under pressure.
Start by defining NAF and explaining the greedy algorithm that processes bits from least significant to most, using a carry to ensure no two consecutive non-zero digits. Then prove minimality by showing any signed binary representation with no adjacent non-zeros has minimal weight, and analyze the time complexity as O(log n) since the algorithm processes each bit once.
Pro tip: Mention that NAF is unique and has the minimal Hamming weight among all signed binary representations, and that it's used in elliptic curve cryptography for efficient scalar multiplication. Also, note that the algorithm can be implemented with simple bitwise operations, making it practical.
Explain that NAF is a signed binary representation with digits in {-1, 0, 1} where no two consecutive digits are non-zero. State that it is unique and has minimal Hamming weight.
Outline the greedy algorithm: while n > 0, if n is odd, set digit to 2 - (n mod 4) (which gives 1 or -1), then n = n - digit; else set digit to 0; then n = n / 2. This ensures no adjacent non-zero digits.
Argue that any signed binary representation with no adjacent non-zeros has weight at most that of any other representation, using an exchange argument: replacing any block of consecutive non-zero digits with a single non-zero digit reduces weight, and NAF achieves this optimally.
Show that the algorithm processes each bit of n exactly once, performing constant-time operations, so time complexity is O(log n) where n is the integer. Space complexity is also O(log n) for the output.
Mention that NAF is used in cryptography (e.g., ECC) to reduce the number of point additions. Note that while NAF minimizes weight, it may increase the length by at most one digit compared to binary, which is a minor trade-off.
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