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Homework – Chapter 1

Homework – Chapter 1. 作業解答. Problem 1. Given the Fibonacci number as 1 2 3 5 8 13 21 34… where the next Fibonacci number will be the sum of its previous two Fibonacci number. Example: The number of 8 is the sum of 3 and 5. The next number will be 55 since it is the sum of 21 and 34. .

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Homework – Chapter 1

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  1. Homework – Chapter 1 作業解答

  2. Problem 1 • Given the Fibonacci number as 1 2 3 5 8 13 21 34… where the next Fibonacci number will be the sum of its previous two Fibonacci number. • Example: • The number of 8 is the sum of 3 and 5. • The next number will be 55 since it is the sum of 21 and 34.

  3. Problem 1 • Write a recursive algorithm to generate the Fibonacci number N. int Fibonacci (int N) { if (N <= 2) return N; return Fibonacci(N-1) + Fibonacci(N-2); }

  4. Problem 1 • Write a non-recursive algorithm using loop structure to generate the Fibonacci number N. int Fibonacci2 (int N) { if (N <= 2) return N; Let N1 = 1, N2 = 2, Total = 0; for (i = 3 to N) { Total = N1 + N2; N1 = N2; N2 = Total; } return Total; }

  5. Problem 1 • Discuss the execution efficiency of these two programs. • The process of the recursive Fibonacci() can be illustrated by the following tree structure. • Space complexity of Fibonacci(): • At most N-1 stack frames (the depth of the tree) are pushed into function call stack when Fibonacci() is called. Therefore, the complexity is O(N). • Time complexity of Fibonacci(): • Therefore, the time complexity depends on how many nodes the tree has, which is O(2N).

  6. Space complexity of Fibonacci2(): • Only a fix number of automatic variables are required. Therefore, the complexity is O(1). • Time complexity of Fibonacci2(): • The for-loop runs in N-2 times. Its body perform addition in O(1) for each iteration. Therefore, the total computing time is O(N). • Conclusion: • Fibonacci2() is more efficient than Fibonacci(). Its space complexity and time complexity are relatively lower.

  7. Problem 2 • Ordering by asymptotic growth rates: • log(n!) • 4logn • (n-1)! • n．2n • (logn)logn. • Please write computation!

  8. Solution • log(n!) log(n!) < log(nn) = O(n logn) • 4logn Let S = 4logn logS = log4logn = log22logn = 2logn = logn2 ∴S = n2 = O(n2) • (n-1)! = O(n!) • n．2n <2n．2n = 22n = O(2n) ∴ (A) < (B) < (D) < (C)

  9. Solution • (logn)logn Let S = (logn)logn log S = (logn)(loglogn) = log(nloglogn) S = nloglogn < nlogn < 2n 2n is growing faster than nlogn. ∴ (A) < (B) < (E) < (D) < (C)

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