03/05/1422:10:55 120 CS 61B: Lecture 20 Monday, March 10, 2014Today’s reading: Goodrich & Tamassia, Chapter 4 (especially 4.2 and 4.3).ASYMPTOTIC ANALYSIS (bounds on running time or memory)===================Suppose an algorithm for processing a retail store’s inventory takes: - 10,000 milliseconds to read the initial inventory from disk, and then - 10 milliseconds to process each transaction (items acquired or sold).Processing n transactions takes (10,000 + 10 n) ms. Even though 10,000 >> 10,we sense that the "10 n" term will be more important if the number oftransactions is very large.We also know that these coefficients will change if we buy a faster computer ordisk drive, or use a different language or compiler. We want a way to expressthe speed of an algorithm independently of a specific implementation on aspecific machine--specifically, we want to ignore constant factors (which getsmaller and smaller as technology improves).Big-Oh Notation (upper bounds on a function’s growth)---------------Big-Oh notation compares how quickly two functions grow as n -> infinity.Let n be the size of a program’s _input_ (in bits or data words or whatever).Let T(n) be a function. For now, T(n) is the algorithm’s precise running time in milliseconds, given an input of size n (usually a complicated expression).Let f(n) be another function--preferably a simple function like f(n) = n.We say that T(n) is in O( f(n) ) IF AND ONLY IF T(n) <= c f(n) WHENEVER n IS BIG, FOR SOME LARGE CONSTANT c. * HOW BIG IS "BIG"? Big enough to make T(n) fit under c f(n). * HOW LARGE IS c? Large enough to make T(n) fit under c f(n).EXAMPLE: Inventory-------------------Let’s consider the function T(n) = 10,000 + 10 n, from our example above.Let’s try out f(n) = n, because it’s simple. We can choose c as large as wewant, and we’re trying to make T(n) fit underneath c f(n), so pick c = 20. c f(n) = 20 n ** ^ / ** | | / ** | | / ** | | / ** | | / ** T(n) = 10,000 + 10 n 30,000 + | / ** | | / ** | | / ** | |/** 20,000 + ** | **| | **/ | | ** / | 10,000 ** / | | / | | / | |/ | O-------+------------------------> n 1,000As these functions extend forever to the right, their asymptotes will nevercross again. For large n--any n bigger than 1,000, in fact--T(n) <= c f(n). *** THEREFORE, T(n) is in O(f(n)). ***Next, you must learn how to express this idea rigorously. Here is thecentral lesson of today’s lecture, which will bear on your entire career asa professional computer scientist, however abstruse it may seem now:|=============================================================================|| FORMALLY: O(f(n)) is the SET of ALL functions T(n) that satisfy: || || There exist positive constants c and N such that, for all n >= N, || T(n) <= c f(n) ||=============================================================================|Pay close attention to c and N. In the graph above, c = 20, and N = 1,000.Think of it this way: if you’re trying to prove that one function isasymptotically bounded by another [f(n) is in O(g(n))], you’re allowed tomultiply them by positive constants in an attempt to stuff one underneath theother. You’re also allowed to move the vertical line (N) as far to the rightas you like (to get all the crossings onto the left side). We’re onlyinterested in how the functions behave as n shoots off toward infinity.EXAMPLES: Some Important Corollaries-------------------------------------[1] 1,000,000 n is in O(n). Proof: set c = 1,000,000, N = 0. -> Therefore, Big-Oh notation doesn’t care about (most) constant factors. We generally leave constants out; it’s unnecessary to write O(2n), because O(2n) = O(n). (The 2 is not wrong; just unnecessary.)[2] n is in O(n^3). [That’s n cubed]. Proof: set c = 1, N = 1. -> Therefore, Big-Oh notation can be misleading. Just because an algorithm’s running time is in O(n^3) doesn’t mean it’s slow; it might also be in O(n). Big-Oh notation only gives us an UPPER BOUND on a function. c f(n) = n^3 ^ * / | * / | * / T(n) = n | * / | * / | * / | * / | */ 1 + * | /* | / * | / *| | / *| | / *| | / * | |/ ** | O***----+------------------------> n N = 1[3] n^3 + n^2 + n is in O(n^3). Proof: set c = 3, N = 1. -> Big-Oh notation is usually used only to indicate the dominating (largest and most displeasing) term in the function. The other terms become insignificant when n is really big. Here’s a table to help you figure out the dominating term.03/05/1422:10:55 220Table of Important Big-Oh Sets------------------------------Arranged from smallest to largest, happiest to saddest, in order of increasingdomination: function common name -------- ----------- O( 1 ) :: constant is a subset of O( log n ) :: logarithmic is a subset of O( log^2 n ) :: log-squared [that’s (log n)^2 ] is a subset of O( root(n) ) :: root-n [that’s the square root] is a subset of O( n ) :: linear is a subset of O( n log n ) :: n log n is a subset of O( n^2 ) :: quadratic is a subset of O( n^3 ) :: cubic is a subset of O( n^4 ) :: quartic is a subset of O( 2^n ) :: exponential is a subset of O( e^n ) :: exponential (but more so)Algorithms that run in O(n log n) time or faster are considered efficient.Algorithms that take n^7 time or more are
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