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Finally, observe that the −m2 term is negative, and thus only serves to lower the value within. Since the Big Oh gives an upper bound, we can drop any negative term without invalidating the upper bound. That n ≥ m implies that mn ≥ m2 , so the negative term is not big enough to cancel any other term which is left. Thus we can simply express the worst-case running time of this algorithm as O(nm). After you get enough experience, you will be able to do such an algorithm analysis in your head without even writing the algorithm down.

N]) var int i, j for i from n to 1 for j from 1 to i − 1 if (A[j] > A[j + 1]) swap the values of A[j] and A[j + 1] Induction 1-10. [3] Prove that 1-11. [3] Prove that 1-12. [3] Prove that n i=n(n + 1)/2 for n ≥ 0, by induction. i=1 n i2 =n(n + 1)(2n + 1)/6 for n ≥ 0, by induction. i=1 n i3 =n2 (n + 1)2 /4 for n ≥ 0, by induction. i=1 1-13. 7 EXERCISES 1-14. [5] Prove by induction on n ≥ 1 that for every a = 1, n ai = i=0 an+1 − 1 a−1 1-15. [3] Prove by induction that for n ≥ 1, n i=1 1 n = i(i + 1) n+1 1-16.

Amateur algorists tend to draw a big messy instance and then stare at it helplessly. The pros look carefully at several small examples, because they are easier to verify and reason about. • Think exhaustively – There are only a small number of possibilities for the smallest nontrivial value of n. For example, there are only three interesting ways two intervals on the line can occur: (1) as disjoint intervals, (2) as overlapping intervals, and (3) as properly nesting intervals, one within the other.

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