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Growth of functions. Function Growth. The running time of an algorithm as input size approaches infinity is called the asymptotic running time We study different notations for asymptotic efficiency. In particular, we study tight bounds, upper bounds and lower bounds.
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Function Growth • The running time of an algorithm as input size approaches infinity is called the asymptotic running time • We study different notations for asymptotic efficiency. • In particular, we study tight bounds, upper bounds and lower bounds.
The “sets” and their use – big Oh • Big “oh” - asymptotic upper bound on the growth of an algorithm • When do we use Big Oh? • Theory of NP-completeness • To provide information on the maximum number of operations that an algorithm performs • Insertion sort is O(n2) in the worst case • This means that in the worst case it performs at most cn2 operations • Insertion sort is also O(n6) in the worst case since it also performs at most dn6 operations
The “sets” and their use – Omega Omega - asymptotic lower bound on the growth of an algorithm or a problem* When do we use Omega? 1. To provide information on the minimum number of operations that an algorithm performs • Insertion sort is (n) in the best case • This means that in the best case its instruction count is at least cn, • It is (n2) in the worst case • This means that in the worst case its instruction count is at least cn2
The “sets” and their use – Omega cont. 2. To provide information on a class of algorithms that solve a problem • Sort algorithms based on comparison of keys are (nlgn) in the worst case • This means that all sort algorithms based only on comparison of keys have to do at least cnlgn operations • Any algorithm based only on comparison of keys to find the maximum of n elements is (n) in every case • This means that all algorithms based only on comparison of keys to find maximum have to do at least cn operations
The “sets” and their use - Theta • Theta - asymptotic tight bound on the growth rate of an algorithm • Insertion sort is (n2) in the worst and average cases • The means that in the worst case and average cases insertion sort performs cn2 operations • Binary search is (lg n) in the worst and average cases • The means that in the worst case and average cases binary search performs clgn operations • Note: We want to classify an algorithm using Theta. • In Data Structures used Oh • Little “oh” - used to denote an upper bound that is not asymptotically tight. n is in o(n3). n is not in o(n)
The functions • Let f(n) and g(n) be asymptotically nonnegative functions whose domains are the set of natural numbers N={0,1,2,…}. • A function g(n) is asymptotically nonnegative, if g(n)³0 for all n³n0 where n0ÎN
Asymptotic Upper Bound: big O c g (n) Graph shows that for all n N, f(n) c*g(n) f (n) Why only for n N ? What is the purpose of multiplying by c > 0? N f (n) = O ( g ( n ))
Asymptotic Upper Bound: O Definition: Let f (n) and g(n) be asymptotically non-negative functions. We sayf (n) is in O( g ( n )) if there is a real positive constant c and a positive Integer N such that for every n ³ N 0 £f (n)£ cg (n ). Or using more mathematical notation O( g (n) ) = { f (n )| there exist positive constant c and a positive integer N such that 0 £f( n) £ cg (n ) for all n³ N}
n2 + 10 nO(n2) Why? take c = 2N= 10 2n2 > n2 + 10 n for all n>=10 1400 1200 1000 n2 + 10n 800 600 2 n2 400 200 0 0 10 20 30
Does 5n+2 ÎO(n)? Proof: From the definition of Big Oh, there must exist c>0 and integer N>0 such that 0 £ 5n+2£cn for all n³N. Dividing both sides of the inequality by n>0 we get: 0 £ 5+2/n£c. 2/n £ 2, 2/n>0 becomes smaller when n increases There are many choices here for c and N. If we choose N=1 then c³ 5+2/1= 7. If we choose c=6, then 0 £ 5+2/n£6. So N ³ 2. In either case (we only need one!) we have a c>o and N>0 such that 0 £ 5n+2£cn for all n ³ N. So the definition is satisfied and 5n+2 ÎO(n)
Does n2Î O(n)? No. We will prove by contradiction that the definition cannot be satisfied. Assume that n2Î O(n). From the definition of Big Oh, there must exist c>0 and integer N>0 such that 0 £n2£cn for all n³N. Dividing the inequality by n>0 we get 0 £n£ c for all n³N. n £c cannot be true for any n >max{c,N }, contradicting our assumption So there is no constant c>0 such that n£c is satisfied for all n³N, and n2 O(n)
O( g (n) ) = { f (n )| there exist positive constant c and positive integer N such that 0 £f( n) £ cg (n ) for all n³ N} • 1,000,000 n2O(n2) why/why not? • (n - 1)n / 2 O(n2) why /why not? • n / 2 O(n2) why /why not? • lg (n2) O( lgn ) why /why not? • n2O(n) why /why not?
Asymptotic Lower Bound, Omega: W f (n) c * g (n) N f(n) =( g ( n ))
Asymptotic Lower Bound: W Definition: Let f (n) and g(n) be asymptotically non-negative functions. We sayf (n) is W( g ( n )) if there is a positive real constant c and a positive integer N such that for every n ³ N0 £ c* g (n ) £ f ( n). Or using more mathematical notation W( g ( n )) = { f (n)| there exist positive constant c and a positive integer N such that 0 £c* g (n ) £f ( n) for all n³ N}
Is 5n-20ÎW (n)? Proof: From the definition of Omega, there must exist c>0 and integer N>0 such that 0 £ cn£ 5n-20 for all n³N Dividing the inequality by n>0 we get: 0 £ c £ 5-20/n for all n³N. 20/n £ 20, and 20/n becomes smaller as n grows. There are many choices here for c and N. Since c > 0, 5 – 20/n >0 and N >4 For example, if we choose c=4, then 5 – 20/n ³ 4 and N ³ 20 In this case we have a c>o and N>0 such that 0 £ cn£ 5n-20 for all n ³ N. So the definition is satisfied and 5n-20 Î W (n)
W( g ( n )) = { f (n)| there exist positive constant c and a positive integer N such that 0 £c* g (n ) £ f ( n) for all n³ N} • 1,000,000 n2W (n2) why /why not? • (n - 1)n / 2W (n2) why /why not? • n / 2W (n2) why /why not? • lg (n2)W ( lg n ) why /why not? • n2W (n) why /why not?
Asymptotic Tight Bound: Q d g (n) f (n) c g (n) f (n) =Q( g ( n )) N
Asymptotic Bound Theta: Q Definition: Let f (n) and g(n) be asymptotically non-negative functions. We say f (n) is Q( g ( n )) if there are positive constants c, d and a positive integer N such that for every n ³ N0 £cg (n ) £ f ( n) £ dg ( n ). Or using more mathematical notation Q( g ( n )) = { f (n)| there exist positive constants c, d and a positive integer N such that 0 £ cg (n ) £ f ( n) £ dg ( n ). for alln³ N}
More on Q • We will use this definition:Q (g (n)) =O(g (n) ) Ç W (g (n) )
More Q • 1,000,000 n2Q(n2) why /why not? • (n - 1)n / 2Q(n2) why /why not? • n / 2Q(n2) why /why not? • lg (n2)Q( lg n ) why /why not? • n2Q(n) why /why not?
Qualified Notation Ω Omega Notation • best case Ω - describes a lower bound for all input (it can't get any better than this). Example: the array is already correctly sorted. • worst case Ω - describes a lower bound for worst case input, possibly greater than best case. Example: the array is sorted in reverse order. • just Ω - same as best case Ω
Qualified Notation O Big-O Notation • best case O - describes an upper bound for best case input, possibly lower than worst case. Example: the array is already correctly sorted. • worst case O - describes an upper bound for all input (it can't get any worse than this). Example: the array is sorted in reverse order. • just O - same as worst case O
Qualified Notation Θ Theta Notation • best case Θ - not used • worst case Θ - describes asymptotic bounds for worst case input • just Θ - same as worst case Θ
Comparing functions • Similarly for • Reflexivity • Symmetry • Transpose symmetry