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Quiz Entry - updated: 2026.07.14

For pen-and-paper arithmetic, what is the computational complexity of addition vs multiplication of two n-digit numbers, and why is that relevant to crypto?

Addition is O(n) — work scales linearly with digit count. Multiplication (long-form) is O(n²) — work scales quadratically. Both are polynomial. Factoring, however, is conjectured super-polynomial — exponentially harder than multiplication, which is exactly what makes RSA possible.

The pen-and-paper observation:

Operation 3-digit example Work
345 + 889 4 column additions (incl. carries) ~n + 1 ops
24 × 53 4 single-digit multiplications + 4 additions ~ ops

So multiplication is n²/n = n times harder than addition for n-digit numbers. Still polynomial — completely tractable for modern computers even with thousand-digit numbers.

Factoring is on a different complexity tier entirely:

Algorithm Complexity for n-bit integer
Trial division O(2^(n/2)) — exponential
Pollard's rho O(2^(n/4)) — exponential, but faster than trial
Quadratic sieve sub-exponential
General Number Field Sieve sub-exponential (best classical, but still way more than polynomial)

For a 3072-bit RSA key, GNFS needs roughly 2¹²⁸ operations — beyond anything any classical computer can do.

The cryptographic asymmetry:

  • Defender (key generation, encryption, decryption) → polynomial-time operations only. Fast.
  • Attacker (factor the public key) → super-polynomial. Infeasible.

Doubling the key size only doubles defender cost but raises attacker cost exponentially. That's why "use a longer key" is always the right response to "computers got faster."

Tip: This gap is what every PK system depends on. Quantum computers (Shor's algorithm) would also solve factoring in polynomial time, collapsing the gap — that's why post-quantum crypto is being deployed before quantum computers actually arrive.

From Quiz: ISF / Asymmetric Cryptography | Updated: Jul 14, 2026