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Asymptotic Complexity. CS2111 CS2110 – Fall 2014. Readings, Homework. I ssues How to look at a program and calculate, formally or informally, its execution time. Determine whether some function f(n) is O(g(n)). Worst case for selection / insertion sorts. Selection sort b[0..n-1] - PowerPoint PPT Presentation

Asymptotic ComplexityCS2111CS2110 Fall 2014*

Readings, HomeworkIssuesHow to look at a program and calculate, formally or informally, its execution time.Determine whether some function f(n) is O(g(n))*

- Worst case for selection / insertion sorts*Selection sort b[0..n-1]//inv b[0..i-1] sorted, b[0..i-1]
- Worst case for selection / insertion sorts*Selection sort b[0..n-1]//inv b[0..i-1] sorted, b[0..i-1]
Find first occurrence of r in s (indexOf)*/** = position of first occurrence of r in s (-1 if not in) */public static int find(String r, String s) { int nr= r.length(); int ns= s.length(); // inv: r is not in s[0..i-1+nr-1] for (int i= 0; i < ns nr; i= i+1) { if (s.substring(i, i+nr).equals(r)) return i; } return -1;}O(nr) Executed how many times --worst case?ns nr + 1 Therefore worst-case time is O(nr *(ns nr + 1))nr = 1: O(ns). nr = ns: O(ns). nr = ns/2: O(ns*ns)

Dealing with nested loopsint c = 0;for (int i= 0; i < n; i++) { for (int j= 0; j < n; j++) { if ((j % 2) == 0) { for (int k= i; k < n; k++) c= c+1; }

else { for (int h= 0; h < j; h++) c= c+1; } }}*n iterationsn iterationsTrue n*n/2 timesLoop is executed n/2 times, with i = 0, 1, 2, , n-1It has n-i iterations. Thats 1 + 2 + n = n*(n+1)/2 its. Thats O(n*n*n)

Dealing with nested loopsint i= 0; int c= 0;while (i < n) { int k= i; while (k < n && b[k] == 0) { c= c+1; k= k + 1; } i= k+1;}*What is the execution time?It is O(n). It looks atEach element of b[0..n-1]ONCE.

Using Big-O to Hide ConstantsWe say f(n) is order of g(n) if f(n) is bounded by a constant times g(n)Notation: f(n) is O(g(n))Roughly, f(n) is O(g(n)) means that f(n) grows like g(n) or slower, to within a constant factor"Constant" means fixed and independent of n*Formal definition: f(n) is O(g(n)) if there exist constants c and N such that for all n N, f(n) cg(n)

A Graphical View*To prove that f(n) is O(g(n)):Find N and c such that f(n) c g(n) for all n > NPair (c, N) is a witness pair for proving that f(n) is O(g(n))*

- Big-O Examples*Let f(n) = 3n2 + 6n 7Prove that f(n) is O(n2)f(n) is O(g(n)) if there exist constants c and N such that for all n N, f(n) cg(n)3n2 + 6n 7 = 1, n
- Big-O Examples*Let f(n) = 3n2 + 6n + 7Prove that f(n) is O(n2)f(n) is O(g(n)) if there exist constants c and N such that for all n N, f(n) cg(n)3n2 + 6n + 7 = 1, n
Big-O Examples*Let f(n) = 3n2 + 6n - 7Prove that f(n) is O(n3)f(n) is O(g(n)) if there exist constants c and N such that for all n N, f(n) cg(n) 3n2 + 6n 7< 3n2 + 6n= 1 = 9n2= 1

Choose N = 1 and c = 9

So, f(n) isO(n2) O(n3), O(n4),

Big-O Examples*T(0) = 1T(n) = 2 * T(n-1)Give a closed formula (no recursion) for T(n)

T(0) = 1T(1) = 2T(2) = 4T(3) = 8One idea:Look at all small cases and find a patternT(n) = 2^n

Big-O Examples*For quicksort in best case, i.e. two partitions are same size.T(0) = 1T(1) = 1T(n) = K*n + 2 * T(n/2) // The Kn is to partition arrayT(0) = K //Simplify computation: assume K > 1T(1) = K // And use K instead of 1T(2^1) = T(2) = 2K + 2K = 4KT(2^2) = T(4) = 4K + 2(4K) = 12K = 3*(2^2)K T(2^3) = T(8) = 8K + 2(12K) = 32K = 4*(2^3)KT(2^4) = T(16) = 16K + 2(32K) = 80K = 5*(2^4)KT(2^n) = (n+1)*(2^n)*KT(m) = log(2m)*m*K

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