Relativesof the Big-Oh
Recall that f(n) is O(g(n)) if, for large n, f is not much bigger than g. That is g is some sort of upper bound on f. How about a definition for the case when g is (in the same sense) a lower bound for f?
Definition: Let f(n) and g(n) be real valued functions of an integer value. Then f(n) is Ω(g(n)) if g(n) is O(f(n)).
Definition: We write f(n) is Θ(g(n)) if both f(n) is O(g(n)) and f(n) is Ω(g(n)).
Remarks We pronounce f(n) is Θ(g(n)) as "f(n) is big-Theta of g(n)"
Examples to do on the board.
Recall that big-Oh captures the idea that for large n, f(n) is not much bigger than g(n). Now we want to capture the idea that, for large n, f(n) is tiny compared to g(n).
If you remember limits from calculus, what we want is that f(n)/g(n)→0 as n→∞. However, the definition we give does not use the word limit (it essentially has the definition of a limit built in).
Definition: Let f(n) and g(n) be real valued functions of an integer variable. We say f(n) is o(g(n)) if for any c>0, there is an n0 such that f(n)≤cg(n) for all n>n0. This is pronounced as "f(n) is little-oh of g(n)".
Definition: Let f(n) and g(n) be real valued functions of an integer variable. We say f(n) is ω(g(n) if g(n) is o(f(n)). This is pronounced as "f(n) is little-omega of g(n)".
Examples: log(n) is o(n) and x2 is ω(nlog(n)).
Homework: R-1.4. R-1.22
If the asymptotic time complexity is bad, say Ω(n8), or horrendous, say Ω(2n), then for large n, the algorithm will definitely be slow. Indeed for exponential algorithms even modest n's (say n=50) are hopeless.
Algorithms that are o(n) (i.e., faster than linear, a.k.a. sub-linear), e.g. logarithmic algorithms, are very fast and quite rare. Note that such algorithms do not even inspect most of the input data once. Binary search has this property. When you look up a name in the phone book you do not even glance at a majority of the names present.
Linear algorithms (i.e., Θ(n)) are also fast. Indeed, if the time complexity is O(nlog(n)), we are normally quite happy.
Low degree polynomial (e.g., Θ(n2), Θ(n3), Θ(n4)) are interesting. They are certainly not fast but speeding up a computer system by a factor of 1000 (feasible today with parallelism) means that a Θ(n3) algorithm can solve a problem 10 times larger. Many science/engineering problems are in this range.
It really is true that if algorithm A is o(algorithm B) then for large problems A will take much less time than B.
Definition: If (the number of operations in) algorithm A is o(algorithm B), we call A asymptotically faster than B.
Example:: The following sequence of functions are
ordered by growth rate, i.e., each function is
little-oh of the subsequent function.
log(log(n)), log(n), (log(n))2, n1/3, n1/2, n, nlog(n), n2/(log(n)), n2, n3, 2n.
Modest multiplicative constants (as well as immodest additive constants) don't cause too much trouble. But there are algorithms (e.g. the AKS logarithmic sorting algorithm) in which the multiplicative constants are astronomical and hence, despite its wonderful asymptotic complexity, the algorithm is not used in practice.
See table 1.10 on page 20.
This is hard to type in using html. The book is fine and I will write the formulas on the board.
Definition: The sigma notation: ∑f(i) with i going from a to b.
Theorem: Assume 0<a≠1. Then ∑ai i from 0 to n = (1-an+1)/(1-a).
Proof: Cute trick. Multiply by a and subtract.
Theorem: ∑i from 1 to n = n(n+1)/2.
Proof: Pair the 1 with the n, the 2 with the (n-1), etc. This gives a bunch of (n+1)s. For n even it is clearly n/2 of them. For odd it is the same (look at it).
Homework: R-1.14. (It would have been more logical to assign this last time right after I did R-1.13. In the future I will do so).
Recall that logba = c means that bc=a. b is called the base and c is called the exponent.
What is meant by log(n) when we don't specify the base?
I assume you know what ab is. (Actually this is not so obvious. Whatever 2 raised to the square root of 3 means it is not writing 2 down the square root of 3 times and multiplying.) So you also know that ax+y=axay.
Theorem: Let a, b, and c be positive real numbers. To ease writing, I will use base 2 often. This is not needed. Any base would do.
⌊x⌋ is the greatest integer not greater than x. ⌈x⌉ is the least integer not less than x.
⌊5⌋ = ⌈5⌉ = 5
⌊5.2⌋ = 5 and ⌈5.2⌉ = 6
⌊-5.2⌋ = -6 and ⌈-5.2⌉ = -5
To prove the claim that there is a positive n satisfying nn>n+n, we merely have to note that 33>3+3.
To refute the claim that all positive n satisfy nn>n+n, we merely have to note that 11<1+1.
"P implies Q" is the same as "not Q implies not P". So to show that in the world of positive integers "a2≥b2 implies that a≥b" we can show instead that "NOT(a≥b) implies NOT(a2≥b2)", i.e., that "a<b implies a2<b2", which is clear.
Assume what you want to prove is false and derive a contradiction.
Theorem: There are an infinite number of primes.
Proof: Assume not. Let the primes be p1 up to pk and consider the number A=p1p2…pk+1. A has remainder 1 when divided by any pi so cannot have any pi as a factor. Factor A into primes. None can be pi (A may or may not be prime). But we assumed that all the primes were pi. Contradiction. Hence our assumption that we could list all the primes was false.
The goal is to show the truth of some statement for all integers n≥1. It is enough to show two things.
Theorem: A complete binary tree of height h has 2h-1 nodes.
We write NN(h) to mean the number of nodes in a complete binary tree
of height h.
A complete binary tree of height 1 is just a root so NN(1)=1 and
21-1 = 1.
Now we assume NN(k)=2k-1 nodes for all k<h
and consider a complete
binary tree of height h.
It is just two complete binary trees of height
h-1 with new root to connect them.
So NN(h) = 2NN(h-1)+1 = 2(2h-1-1)+1 = 2h-1, as desired