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CS 61B: Lecture 20
Wednesday, October 13, 2010
Today’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 of
transactions is very large.
We also know that these coefficients will change if we buy a faster computer or
disk drive, or use a different language or compiler. We want a way to express
the speed of an algorithm independently of a specific implementation on a
specific machine--specifically, we want to ignore constant factors (which get
smaller and smaller as technology improves).
Big-Oh Notation (upper bounds on a function’s growth)
---------------
We use Big-Oh notation to say how slowly code might run as its input grows.
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 precisely equal to the algorithm’s
running time, 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 we
want, 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
1000
As these functions extend forever to the right, their asymptotes will never
cross again. For large n--any n bigger than 1000, 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 the
central lesson of today’s lecture, which will bear on your entire career as
a 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 = 1000.
Think of it this way: if you’re trying to prove that one function is
asymptotically bounded by another [f(n) is in O(g(n))], you’re allowed to
multiply them by positive constants in an attempt to stuff one underneath the
other. You’re also allowed to move the vertical line (N) as far to the right
as you like (to get all the crossings onto the left side). We’re only
interested in how the functions behave as they shoot 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).
(It’s 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.
