Revision Session1 - What is the complexity s = 0...

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1 Revision Session 3 Oct 2007 3pm What is the complexity? ……… 2n i s = 0; for (i=1;i<=2n;i++) s = s+i; O(n) O(n 2 ) ……… ……… ……… ……… ……… ……… ……… ……… ……… 2n 2n i j inner loop of unique indices (i,j) s = 0; for (i=1; i <= 2n; i++) for (j=1; j <= i; j++) s = s + j; 2 Number of shaded squares = (n 2 )(n 2 )/2 = O(n 4 ) ……… ……… ……… ……… ……… ……… ……… ……… ……… n 2 n 2 i j s = 0; for (i=1;i<=n*n;i++) for (j=1;j<=i;j++) s = s+j; Number of shaded squares = ½ (n * n) = O(n 2 ) ……… ……… ……… ……… ……… ……… ……… ……… ……… n n i j s = 0; for (i=1;i<=n;i++){ if ((i%2)== 0) for (j=1;j<=n;j++) s = s+j; else s++; } s = 0; for (i=1;i<=2n; i = i * 2) s = s+i; s = 0; for (i=2n ;i > 0; i = i / 2) s = s+i; 3 s = 0; for (i=2n ;i > 0; i = i / 13) s = s+i; • Single recursion, easy: – n recursive calls, since n decreases by 1 each time, so O(n). int getMin(int[] A, n) { if (n==1) return A[0]; int min_int = getMin(A,n-1); int x = A[n-1]; if (x < min_int) min_int = x; return min_int; } int f(int n) { if (n<=1) return 1; else return f(n-1)+f(n-1); } • Multiple recursion, not so easy: – Each f(n) spawns two f(n-1) calls. – Total time = 1+2+4+…+2 n-1 = (2 n -1) = O(2 n ). f(n) f(n-1) f(n-1) f(n-2) f(n-2) f(n-2) f(n-2) 1=2 0 2=2 1 4=2 2 #recursive calls per level 8=2 3 …… f(1) f(1) f(1) f(1) f(1) f(1) 2 n-1 4 int f(int n) { if (n<=1) return 1; else return f(n/2)+f(n/2); } int f(int n) { if (n<=1) return 1; else { int s = 0; for (int i=0; i<n; i++) s++; return f(n/2)+f(n/2); } } int f(int n) { if (n<=1) return 1; else { int s = 0; for (j = 0; j < n; j++) for (int i=0; i<n; i++) s++; return f(n/2)+f(n/2); } } Which of the following statements about Binary Search is TRUE? A. The values need not be sorted for Binary Search to work. B. In the worst case, the number of comparisons in Binary Search is about the same as the number of comparisons in Linear Search. C. Binary Search can only be implemented using recursion.
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Revision Session1 - What is the complexity s = 0...

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