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Big-O Analysis
Definition: Suppose that f(n) and g(n) are nonnegative functions of n. Then we saythat f(n) is O(g(n)) provided that there are constants C > 0 and N > 0 such
that for all n > N, f(n) Cg(n).
Big-O expresses an upper bound on the growth rate of a function, for sufficiently large
values of n.
By the definition above, demonstrating that a function f is big-O of a function g requires
that we find specific constants C and N for which the inequality holds (and show that the
inequality does, in fact, hold).
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Big-O Example
Consider the following function: 25 5( ) 22 2
T n n n
We might guess that:
We could easily verify the guess by induction:
If n = 2, then T(2) = 17 which is less than 20, so the guess is valid if n = 2.
Assume that for some n 2, T(n) 5n2. Then:
2 2
2
2
2 2
5 5 5 5 5 5( 1) ( 1) ( 1) 2 5 2
2 2 2 2 2 2
5 52 5 5
2 2
5 5 5 by the inductive assumption
5 10 5 5( 1)
T n n n n n n
n n n
n n
n n n
Thus, by induction, T(n) 5n2 for all n 2. So, by definition, T(n) is O(n2).
2( ) 5 for all 2T n n n
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Making the Guess
The obvious question is "how do we come up with the guess in the first place"?
Here's one possible analysis (which falls a bit short of being a proof):
2
2 2
2
5 5( ) 2
2 2
5 52 2 (replace n with n^2, subtract 2)
2 2
5
T n n n
n n
n
The middle step seems sound since if n 2 then n n2, substituting n2 will thus add at
least 5 to the expression, so that subtracting 2 should still result in a larger value.
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Big-O Theorems
For all the following theorems, assume that f(n) is a non-negative function of n and that Kis an arbitrary constant.
Theorem 2: A polynomial is O(the term containing the highest power of n)
Theorem 3: K*f(n) is O(f(n)) [i.e., constant coefficients can be dropped]
Theorem 1: K is O(1)
)7(is1000537)( 424 nOnnnnf
)(is7)( 44 nOnng
Theorem 4: If f(n) is O(g(n)) and g(n) is O(h(n)) then f(n) is O(h(n)). [transitivity]
)(is1000537)( 424 nOnnnnf
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Big-O Theorems
Theorem 5: Each of the following functions is strictlybig-O of its successors:
K [constant]
logb(n) [always log base 2 if no base is shown]
n
n logb(n)n2
n to higher powers
2n
3n
larger constants to the n-th power
n! [n factorial]
nn
smaller
larger
)2(and)(and))log((is)log(3)( 2 nOnOnnOnnnf
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Big-O Theorems
Theorem 6: In general, f(n) is big-O of the dominant term of f(n), wheredominant may usually be determined from Theorem 5.
Theorem 7: For any base b, logb(n) is O(log(n)).
)(is10005)log(37)( 22 nOnnnnnf
)3(is100000037)( 4 nn Onng
)(is))log((7)(
2
nOnnnnh
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Big-Omega
In addition to big-O, we may seek a lower bound on the growth of a function:
Definition: Suppose that f(n) and g(n) are nonnegative functions of n. Then we say
that f(n) is (g(n)) provided that there are constants C > 0 and N > 0 such
that for all n > N, f(n) Cg(n).
Big- expresses a lower bound on the growth rate of a function, for sufficiently largevalues of n.
Analagous theorems can be proved for big-.
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Big-Theta
Finally, we may have two functions that grow at essentially the same rate:
Definition: Suppose that f(n) and g(n) are nonnegative functions of n. Then we say
that f(n) is (g(n)) provided that f(n) is O(g(n)) and also that f(n) is(g(n)).
If f is (g) then, from some point on, f is bounded below by one multiple of g andbounded above by another multiple of g (and vice versa).
So, in a very basic sense f and g grow at the same rate.
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Order and Limits
The task of determining the order of a function is simplified considerably by thefollowing result:
Theorem 8: f(n) is (g(n)) if
cc
ng
nf
n
0where
)(
)(lim
Recall Theorem 7 we may easily prove it (and a bit more) by applying Theorem 8:
)ln()2ln(
)ln()2ln(lim
)2ln(
1)ln(
1
lim)log()(loglim
bb
n
bnnn
nn
b
n
The last term is finite and positive, so logb(n) is (log(n)) by Theorem 8.
Corollary: if the limit above is 0 then f(n) is strictly O(g(n)), and
if the limit is then f(n) is strictly (g(n)).
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Order and Limits
The converse of Theorem 8 is false. However, it is possible to prove:
Theorem 9: If f(n) is (g(n)) then
provided that the limit exists.
cc
ng
nf
n
0where
)(
)(lim
A similar extension of the preceding corollary also follows.
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More Theorems
Many of the big-O theorems may be strengthened to statements about big-:
Theorem 11: A polynomial is (the highest power of n).
Theorem 10: If K > 0 is a constant, then K is (1).
kkk
kknk
k
k
naa
n
a
n
a
n
a
n
nanaa
1
1
1010 limlim
proof: Suppose a polynomial of degree k. Then we have:
Now ak> 0 since we assume the function is nonnegative. So by Theorem 8, the
polynomial is (nk).
QED
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More Theorems
Theorems 3, 6 and 7 can be similarly extended.
Theorem 12: K*f(n) is (f(n)) [i.e., constant coefficients can be dropped]
Theorem 13: In general, f(n) is big- of the dominant term of f(n), wheredominant may usually be determined from Theorem 5.
Theorem 14: For any base b, logb(n) is (log(n)).
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Strict Comparisons
For convenience, we will say that:- f is strictly O(g) if and only if f is O(g) but f is not (g)
- f is strictly (g) if and only if f is (g) but f is not (g)
For example, n log n is strictly O(n2) by Theorem 8 and its corollary, because:
0
1
/1lim
loglim
loglim
2
n
n
n
n
nn
nnn
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Big- Is an Equivalence Relation
Theorem 17: If f(n) is (g(n)) and g(n) is (h(n)) then f(n) is (h(n)). [transitivity]
Theorem 16: If f(n) is (g(n)) then g(n) is (f(n)). [symmetry]
Theorem 15: If f(n) is (f(n)). [reflexivity]
By Theorems 1517, is an equivalence relation on the set of positive-valued functions.
The equivalence classes represent fundamentally different growth rates.
Algorithms whose complexity functions belong to the same class are essentially
equivalent in performance (up to constant multiples, which are not unimportant inpractice).
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Applications to Algorithm Analysis
Ex 1: An algorithm with complexity function
is (n2) by Theorem 10.32
5
2
3)( 2 nnnT
Ex 2: An algorithm with complexity function
is O(n log(n)) by Theorem 5.
Furthermore, the algorithm is also (n log(n)) by Theorem 8 since:
2log4log3)( nnnnT
3log
243lim
log
)(lim
nnnnn
nT
nn
For most common complexity functions, it's this easy to determine the big-O and/or
big- complexity using the given theorems.
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Complexity of Linear Storage
For a contiguous list of N elements, assuming each is equally likely to be the target of asearch:
- average search cost is (N) if list is randomly ordered
- average search cost is (log N) is list is sorted
- average random insertion cost is (N)
- insertion at tail is (1)
For a linked list of N elements, assuming each is equally likely to be the target of asearch:
- average search cost is (N), regardless of list ordering
- average random insertion cost is (1), excluding search time
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Most Common Complexity Classes
Theorem 5 lists a collection of representatives of distinct big- equivalence classes:
K [constant]
logb(n) [always log base 2 if no base is shown]
n
n logb(n)
n2
n to higher powers
2n
3n
larger constants to the n-th power
n! [n factorial]nn
Most common algorithms fall into one of these classes.
Knowing this list provides some knowledge of how to compare and choose the right algorithm.
The following charts provide some visual indication of how significant the differences are
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Graphical Comparison
Common Growth Curves
0
200
400
600
800
1000
1200
1 2 3 4 5 6 7 8 9 10 11
n (input size)
log n
n
n log n
n^2
n^3
2^n
10^n
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Lower-order Classes
Low-order Curves
0
20
40
60
80
100
120
1 3 5 7 9
1
1
1
3
1
5
1
7
1
9
2
1
2
3
n (input size )
log n
n
n log n
n2
For significantly largevalues of n, only these
classes are truly
practical, and whether
n2 is practical is
debated.
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Proof of Theorem 8
Theorem 8: f(n) is
(g(n)) if
cc
ng
nf
n0where
)(
)(lim
Suppose that f and g are non-negative functions of n, and that
ccngnf
n0where
)()(lim
Then, from the definition of the limit, for every > 0 there exists an N > 0 such that
whenever n > N:
( )from which we have ( ) ( ) ( ) ( ) ( )
( )
f nc c g n f n c g n
g n
Let = c/2, then we have that:3
( ) ( ) whence is ( )2
cf n g n f g
( ) ( ) whence is ( )2
cg n f n f g
Therefore, by definition, f is (g).
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