Exercise Lover

July 9, 2007

Exercises 15.1-5

Filed under: 15.1 — yuhanlyu @ 3:12 am

Exercises 15.1-4

Filed under: 15.1 — yuhanlyu @ 3:11 am

Solution wanted!

Exercises 15.1-3

Filed under: 15.1 — yuhanlyu @ 3:09 am

\sum_{i=1}^2 \sum_{j=1}^n r_i(j) = 2 \sum_{j=1}^n r_i(j) = 2^{n+1} - 2.

Exercises 15.1-2

Filed under: 15.1 — yuhanlyu @ 3:08 am

We can consider r_1(j) = r_2(j) for all j. Let T(j) = r_1(n-j), then T(n) = 2T(n-1), T(0) = 1. We can solve this recurrence relation with T(n-j) = 2^{n-j}. Hence r_1(j) = 2^{n-j}.

Exercises 15.1-1

Filed under: 15.1 — yuhanlyu @ 3:00 am

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