Public static string reverse1(string s int n =

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Unformatted text preview: public static String reverse1(String s) { int N = s.length(); String rev = ""; for (int i = 0; i < N; i++) rev = s.charAt(i) + rev; return rev; } 24. In this problem you NOT allowed to use any of the theorems about Big-Oh stated in the lecture slides, the textbook, or the lab writeups. Your proof should rely only on the definition of Big-Oh. Prove that n √ n is not O ( n ). 25. For each of the following statements, if true prove using only the definition of big-Oh, if false give a counterexample. (As usual, the functions involved take strictly positive values.) (a) For any f ( n ), g ( n ) is O ( nf ( n )), where g ( n ) = ∑ n i =1 f ( i ). (b) For any f ( n ) ,g ( n ), if log f ( n ) is O (log g ( n )) then f ( n ) is O ( g ( n )). 26. Suppose 0 < f ( n ) < g ( n ) for all n ≥ 1000. For the following statement, if true , give a proof, and if false , give a counterexample. You can assume any of the theorems about Big-Oh stated in the lecture notes. n · f ( n ) + g ( n ) is O ( g ( n ) · log n ) 27. (20 pts) Assume that the concatenation of strings s1 + s2 takes O ( k ) time where k is the size of the longest of s 1 and s 2 . Assume also that the method call s.substring(i, j) takes O ( j- i ) time. Everything else without method calls, as usual, takes constant time. Find a Big-Oh bound for the worst-case running time of the method below by writing the recurrence relation and solving it for N = 2 m . Write the result in terms of N (not m ). public static String reverse2(String s) { int N = s.length(); if (N <= 1) return s; String left = s.substring(0, N/2); String right = s.substring(N/2, N); String revright = reverse2(right); String revleft = reverse2(left); return (revright+revleft); } 28. Consider the following: 9 static int foo(char a) { for (int i = a.length-1; i > 0; i=i/2) if (a[i] == ’@’) return 2; return 3; } static void bar(char b) { for (int j = 0; j < b.length; j=j+foo(b)) b[j]=0; } Analyze the worst-case running time of bar(b) as a function of n = b.length and give a Big-Theta (Θ( ... )) bound. For code fragments whose running time is a constant, do not give a specific number of steps, use symbolic constants c 1 ,c 2 , etc. instead. 29. In this problem you NOT allowed to use any of the theorems about Big-Oh stated in the lecture slides, the textbook, or the lab writeups. Your proof should rely only on the definition of Big-Oh. Prove that 2 2 n is not O (2 n ). 30. In this problem you NOT allowed to use any of the theorems about Big-Oh stated in the lecture slides, the textbook, or the lab writeups. Your proof should rely only on the definition of Big-Oh. Suppose 0 < f ( n ) < g ( n ) for all n > 1000. Suppose also that g ( n ) is O ( n 2 ). Let h ( n ) = f ( n ) p g ( n )....
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public static String reverse1(String s int N = s.length...

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