---------------------------- 3 ----------------------------
Growth of Functions
3-1a

‘’
漸近的
3.1 Asymptotic notation

“”
-notation: f(n) = (g(n))
3-2
f(n) = (g(n)) iff g(n)=(f(n)), Ex. n2=(3n2-6n)
O-notation: f(n) = O(g(n))
g(n) is an asymptotically upper bound for f(n).
a set
O(g(n))= {f(n)| there exist positive constants c
and n0 such that
0  f(n)cg(n)
for all n n0}
調大
g(n) is an asymptotically tight bound for f(n).
(g(n)) = {f(n)| there exist positive constants c1,
c2, and n0 such that
0 c1 g(n)f(n)c2 g(n)
for all n  n0}
調小
調大
when n is sufficiently large
f
g
g




g
Example: Prove that 3n2 - 6n = (n2).
Proof: To do so, we have to determine c1, c2, and
n0 such that
f
c1n2 3n2 - 6n c2n2,
(g(n))  O(g(n))
f(n) = (g(n)) implies f(n) = O(g(n))
6n = O(n), 6n = O(n2)
"The running time is O(n2)" means "the worstcase running time is O(n2)."
‘’
(for all nn0)
-notation: f(n) = (g(n))
g(n) is an asymptotically lower bound for f(n).
dividing which by n2 yields
c1 3 - 6/n c2.
3-1y
Clearly, by choosing c1=2, c2=3 and n0=6 we can
Q.E.D
verify that 3n2 - 6n = (n2).
There are many choices!
(g(n))= {f(n)| there exists positive constants c
and n0 such that
調小
0  cg(n) f(n)
for all n n0}

3-3
f(n) = (g(n)) iff (f(n)=O(g(n))) & (f(n)=(g(n)))
3-4
-notation: f(n) = (g(n)) (little-omega of g of n)
‘’
3-3x
(g(n)) = {f(n)| for(any) positive constant c, there
exists a constant n0 > 0 such that
0  cg(n) < f(n)
for all n n0}
tight bound
upper bound
o(g(n)) = Ο(g(n)) \ Θ(g(n)) ???
o-notation: f(n) = o(g(n)) (little-oh of g of n)
o(g(n))=




f(n) = (g(n)) iff lim
f (n )
= ∞.
g
(
n
)
n →∞
lower bound
3-3y
‘’
任意大
2n2 = (n), but 2n2  (n2).
{f(n)| for (any) positive constant c, there
exists a constant n0 > 0 such that
0  f(n) < cg(n)
for all n n0}
任意小
2n = o(n2), but 2n2  o(n2).
f(n) = o(g(n)) can also be defined as
f (n )
lim
= 0.
n → ∞g ( n )
Comparison of functions
 functions:
      O  
 real numbers:         

 Transitivity, Reflexivity, Symmetry, Transpose 3-4x
Symmetry

 Any two real numbers can be compared.
(trichotomy) But, not any two functions can be
compared.
Example:
f(n)=n and g(n)=n1+sin n
(e)~(h)
Homework: Problems 3-2, 3-3, 3-4.
lg* n = k
3-4y
3-5
k
lglg...lg n  1
3-6
Appendix A: Summation formulas
n
n
n
k 1
k 1
k 1
 (cak  bk )  c  ak   bk
1 + 2 + ... + n
n
n
1
k

 2 n(n  1)  (n 2 )
k 1
x0 + x1+ x2 + ... + xn
 x k  ( x n 1  1) /( x  1)
k 0
1
 log e n  O(1)
k
k 1
Hn  

 xk 
k 0

n
1
loge n = 1 x dx
n
(Harmonic series)
= O(lg n)
1
( x  1)
1 x
3-5x
n 1

 kx k 
k 0
x
(1  x )2
( x  1)
(differentiating and then
multiplying by x)
n 1 1
1
1
1




(
)
1


n
k 1 k ( k  1) k 1 k k  1
n
n
k 1
k 1
lg  ak   lg ak
*  n/a /b = n/ab
*  n/a /b = n/ab
* lga b = (lgc b)/(lg c a) * algc b = blgc a