6. Identify the basic operations and construct a recurrence relation C(n) that characterizes the time complexity of the algorithm. Determine the order of growth for C(n) either solving the recurrence relation or using the Master Theorem if appropriate. You may assume that n = 2<sup>k</sup> for some integer k. Foo6 (n) // Description: // Input: a positive integer n // Output: if n = 0 return 1 if n = 1 return 2 else if n % 2 = 1 return Foo6(n/2) * Foo6(n/2) * 2 else return Foo6(n/2) * Foo6(n/2)
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The Foo6 function has base cases for n=0 and n=1, and two recursive cases depending on whether n is odd or even. Show more…
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Given these methods: METHOD math1: public int math1(int n) { if (n <= 1) { return 1; } else { return (n * 2) + math1(n-1); } } METHOD math2: public int math2(int n) { if (n <= 1) { return 1; } else { return n + math1(n) * math2(n/2); } } Now set up a recurrence relation for the running time of the method math2 as a function of n. Solve your recurrence relation to specify the Big-Oh bound of math2. HINT: When doing this, the call to math1 can be replaced by the equation that you found when solving the recurrence relation for math1 in part a).
Sri K.
Recurrence Relations Solve using: Ex.6 (Expand, Guess, Verify method) S(1)= 1 S(n)= S(n-1)+(2n-1) for n>=2
Solve the recurrence relations together with the initial conditions given: a) a_n = a_{n-1} for n ≥ 1 and a_0 = 2. b) a_n = -4 a_{n-1} - 4 a_{n-2} for n ≥ 2 and a_0 = 0, a_1 = 1. c) a_n = 4 a_{n-2} for n ≥ 2 and a_0 = 0, a_1 = 4. a) The solution is given by: a_n = │ │ for all n. b) The solution is given by: a_n = │ │ for all n. c) The solution is given by: a_n = │ │ for all n.
Adi S.
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