Exercise 3: Why Reducing Only at the End Is Still Dramatically Slower — Possible Solution ==================================================================== THE KEY DISTINCTION: NUMBER OF MULTIPLICATIONS VS. COST PER MULTIPLICATION ------------------------------ This chapter's own O(log b) claim counts the NUMBER of multiplication operations performed - and that count genuinely stays the same whether or not the intermediate results are reduced along the way. But the count of operations isn't the whole cost picture: each individual multiplication also has its OWN cost, which depends on how large the numbers being multiplied actually are. WHAT THIS CHAPTER'S OWN DIGIT-LENGTH FINDING SHOWS ------------------------------ Without reducing after every squaring, the raw unreduced number's digit length DOUBLES at every single step - this chapter's own verified example showed a starting single-digit number growing to 978 digits after just 11 squarings. Multiplying two large numbers together is not a constant-time operation - the cost of a single multiplication grows with the size of the numbers involved (this connects directly to how real multiplication algorithms scale with input size, material covered more formally in Algorithms & Complexity). WHY THIS MAKES THE "SAME NUMBER OF STEPS" VERSION SLOWER ------------------------------ A version that reduces after every step performs the same O(log b) NUMBER of multiplications, but each individual multiplication stays small and cheap the whole way through - the numbers never exceed roughly the size of the modulus n. A version that only reduces at the very end performs the identical O(log b) number of multiplication OPERATIONS, but each one gets progressively more expensive as the intermediate numbers keep doubling in size - by the final squaring, it could be multiplying together numbers with hundreds or thousands of digits, even though the exponent itself might correspond to a modulus of only a few dozen digits. RESULT ------------------------------ "Same number of multiplications" is not the same claim as "same total cost." Reducing at every step keeps each individual multiplication cheap; deferring all reduction to the end keeps the COUNT of multiplications the same but lets each one grow dramatically more expensive, which is exactly why real implementations always reduce at every single step rather than just at the end. WHY THIS WORKS AS AN ANSWER ------------------------------ The explanation distinguishes explicitly between the NUMBER of multiplications (unchanged either way, per this chapter's own O(log b) claim) and the COST of each individual multiplication (which grows dramatically without per-step reduction, per this chapter's own verified digit-length doubling), rather than treating "O(log b) multiplications" as automatically meaning "fast" on its own.