 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-224/4/c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

x

symbols each have probability at least

P (x)

, so

x P (x) \leq 1

. Hence

L^{*} (x) = ⌊ lo g_{2} x ⌋ \leq lo g_{2} x \leq - lo g_{2} P (x) .

 New to topics? Read the docs here!