Exercise 2.10

[gzs715]

In Exercise 2.10 --- More efficient prefix-free transformation.
Shall we suppose that F is a some (not necessarily prefix-free) one-to-one representation? Otherwise, if F is not one-to-one, we cannot prove that F' is a prefix-free representation of O.
For example, if O = {a,b} and F(a) = F(b), then G(|F(a)|) = G(|F(b)|) = G(1)
=> F'(a) = G(1)F(a) = G(1)F(b)=F'(b) => F' is not prefix-free