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The five complexity classes can be imprecisely described by these phrases: f(n) = o(g(n)) means that f(n) grows more slowly than g(n). f(n) = O(g(n)) means that f(n) grows more slowly or at the same rate as g(n). f(n) = (g(n)) means that f(n) grows at the same rate g(n). f(n) = (g(n)) means that f(n) grows faster or at the same rate as g(n). f(n) = (g(n)) means that f(n) grows faster than g(n). EXAMPLE A.3 Asymptotic Growth Classes
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250 lgn = o(n), because 250 lgn grows more slowly than n. 0.086 n lgn = (n), because 0.086 n lgn grows faster than n.
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Keep in mind that these functions f(n), g(n), and so on, are usually used to describe how long it takes to run an algorithm. So if f(n) grows more slowly than g(n), then the algorithm with complexity f(n) is generally faster than the algorithm with complexity g(n). Less time is better. THE FIRST PRINCIPLE OF MATHEMATICAL INDUCTION The First Principle of Mathematical Induction, also called weak induction, is often used to prove formulas about positive integers. Theorem A.3 The First Principle of Mathematical Induction If {P1 , P2, P3 , . . . } is a sequence of statements such that: P1 is true. Each statement Pn can be deduced from its predecessor Pn 1 . Then all of the statements P1, P2 , P3, . . . are true. EXAMPLE A.4 Weak Induction
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Suppose we want to prove that the inequality 2n (n + 1)! is true for every n 1. Then the sequence of statements is 21 2! P1 : 22 3! P2 : 23 4! P3 : etc. The first few statements can be verified explicitly: 21 = 2 2 = 2! 22 = 4 6 = 3! 23 = 8 24 = 4! In particular, P1 is true, satisfying the first of the two requirements for weak induction. This is called the base of the induction. To verify the second requirement, we have to show that each statement Pn can be deduced from its predecessor Pn 1. So we examine the two general statements Pn 1 and Pn , and look for a connection: Pn 1: 2n 1 n! 2n (n + 1)! Pn : To derive Pn from Pn 1, we note that 2n = (2)(2n 1) and (n + 1)! = (n + 1)(n!). Thus, if we assume that Pn 1 is true, then we have 2n = (2)(2n 1) (2)(n!) (n + 1)(n!) = (n + 1)!, because n + 1 > 2. Verifying the second requirement of mathematical induction is called the inductive step.
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THE SECOND PRINCIPLE OF MATHEMATICAL INDUCTION The Second Principle of Mathematical Induction, also called strong induction, is nearly the same as the first principle. The only difference is in the inductive step. Theorem A.4 The Second Principle of Mathematical Induction If {P1 , P2, P3 , . . . } is a sequence of statements such that: P1 is true. Each statement Pn can be deduced from its predecessors {P1 , P2, P3 , . . . , Pn 1 }. Then all of the statements P1, P2 , P3, . . . are true. So to verify the inductive step with strong induction, we may assume that all n 1 statements P1 , P2 , P3, . . . , Pn 1 are true. EXAMPLE A.5 Strong Induction
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Prove that the Fibonacci numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, . . . are asymptotically exponential. More precisely, we prove that Fn = O(2 n ), where the Fibonacci numbers Fn are defined as F0 = 0, F1 = 1, and Fn = Fn 1 + Fn 2. So our sequence of statements is P1 : P2 : F1 F2 21 22
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P3 : F3 23 etc. These first few are true because of the following relationships: P1 : F1 = 1 2 F2 = 2 4 P2 : F3 8 P3 : For the inductive step, we assume that n 1 statements P1 , P2, P3 , . . . , Pn 1 are true and compare them with the nth statement Pn: P1 : P2 : P3 : : Pn 2: Pn 1: F1 F2 F3 : Fn 2 Fn 1 2n 2 2n 1 21 22 23
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Pn : Fn 2n Comparing the nth statement with the two that precede it, we see that Fn = Fn 1 + Fn 2 2n = (2)(2n 1) = 2n 1 + 2n 1 > 2n 1 + 2n 2 So we can derive Pn from Pn 1 and Pn 2 like this: Fn = Fn 1 + Fn 2 2n 1 2n 2 < 2n 2n for all n). This proves that all the statements are true (i.e., Fn
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