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lecture4 - Measuring Functions Recurrences and Recursion...

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Unformatted text preview: Measuring Functions Recurrences and Recursion IE170: Algorithms in Systems Engineering: Lecture 4 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University January 22, 2007 Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See Everyone Gets an A! Go Bears! Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See Taking Stock Last Time Θ , O and Ω Recursion. See recursion. Analyzing Recurrences This Time Analyzing a simple algorithm The impact of data structures Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See A Canonical Problem Example: The Sorting Problem Input: A sequence of numbers a 1 , a 2 , . . . , a n Output: A reordering a 1 , a 2 , . . . , a n such that a 1 ≤ a 2 ≤ ··· ≤ a n . Sorting “in place” : No new memory is allocated (or at least a constant amount of memory is allocated). (The input is usually overwritten by the output as the algorithm executes.) Sorting “out of place” : New memory must be allocated Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See Sample. Reverse-Out-Of-Place In this case, we allocate a new array B public static void reverseOP(int A) { int n = A.length; int B = new int[n]; for (int j = 0; j < n; j++) { B[n-1-j] = A[j]; } System.arraycopy(B,0,A,0,n); } Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See Sample. Reverse-In-Place Here everything is done directly on A public static void reverseIP(int A) { int n = A.length; for(int j = 0; j < n/2; j++){ // Swap A[j] and A[n-j-1] int t = A[j]; A[j] = A[n-j-1]; A[n-j-1] = t; } } Jeff Linderoth IE170:Lecture 4 Measuring Functions Recurrences and Recursion Θ ,O, Ω Some Functions You’ll See Sorting: Some Java Code public static void iSortMe(int A) { for(int j = 1; j < A.length; j++) { int key = A[j]; int i = j-1; while(i >= 0 && A[i] > key) { A[i+1] = A[i]; i = i-1;...
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