INDIAN INSTITUTE OF TECHNOLOGY ROORKEE

Data Structures CSN- 102 Dr. R. Balasubramanian Associate Professor Department of Computer Science and Engineering Indian Institute of Technology Roorkee Roorkee 247 667 [email protected] https://sites.google.com/site/balaiitr/

More Big-Oh Examples 7n-2 7n-2 is O(n) need c > 0 and n0  1 such that 7n-2  c•n for n  n0 this is true for c = 7 and n0 = 1



3n3 + 20n2 + 5

3n3 + 20n2 + 5 is O(n3) need c > 0 and n0  1 such that 3n3 + 20n2 + 5  c•n3 for n  n0 this is true for c = 4 and n0 = 21



3 log n + 5 3 log n + 5 is O(log n) need c > 0 and n0  1 such that 3 log n + 5  c•log n for n  n0 this is true for c = 8 and n0 = 2 2

Big-Oh and Growth Rate • The big-Oh notation gives an upper bound on the growth rate of a function • The statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n) • Useful to find the worst case of an algorithm

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• Prove that

2

f ( n )  5n  2 n  1

is

2

O (n )

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• Write an efficient program to find whether given number is prime or not.

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AKS algorithm • Agrawal, Kayal and Saxena from IIT Kanpur have come up with O(log k n) algorithm for a prime number program – PRIMES is in P, Annals of Mathematics, 160(2): 781-793, 2004 • Infosys Mathematics Prize, 2008. • Fulkerson Prize for the paper “PRIMES is in P”, 2006. • Gödel Prize for the paper “PRIMES is in P", 2006. • Several awards and prizes

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Programming Contest Sites • • • • •

http://icpc.baylor.edu(ACM ICPC) http://www.codechef.com (Online Contest) http://www.topcoder.com http://www.spoj.com http://www.interviewstreet.com

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Find the output of the following code

int n=32; steps=0; for (int i=1; i<=n;i*=2) steps++; cout<
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• A log10 (n) , B  log 2 (n) both are of O(log n), base doesn’t matter, • Suppose xm  log m n  f (n) and xk  log k n  g (n)

m

xm

m

xm

 n and

k

k

xk

n

xk

 x m  x k log m k  f ( n )  g ( n ) log m k  cg ( n ) where c  log m k  f ( n )  cg ( n )  f ( n ) and g ( n ) are of O( log n)

 -notation For function g(n), we define (g(n)), big-Omega of n, as the set: (g(n)) = {f(n) :  positive constants c and n0, such that n  n0, we have 0  cg(n)  f(n)} Intuitively: Set of all functions whose rate of growth is the same as or higher than that of g(n). g(n) is an asymptotic lower bound for f(n). 10

• Prove that

2

f ( n )  5n  2 n  1

is

2

(n )

11

Example (g(n)) = {f(n) :  positive constants c and n0, such that n  n0, we have 0  cg(n)  f(n)} • n = (log n). Choose c and n0. for c=1 and n0 =16,

c * log n  n

, n  16

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Omega • Omega gives us a LOWER BOUND on a function.  Big-Oh says, "Your algorithm is at least this good."  Omega says, "Your algorithm is at least this bad."

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-notation For function g(n), we define (g(n)), big-Theta of n, as the set: (g(n)) = {f(n) :  positive constants c1, c2, and n0, such that n  n0, we have 0  c1g(n)  f(n)  c2g(n)

} Intuitively: Set of all functions that have the same rate of growth as g(n).

g(n) is an asymptotically tight bound for f(n). 15

-notation For function g(n), we define (g(n)), big-Theta of n, as the set: (g(n)) = {f(n) :  positive constants c1, c2, and n0, such that n  n0, we have 0  c1g(n)  f(n)  c2g(n)

} Technically, f(n)  (g(n)). Older usage, f(n) = (g(n)). I’ll accept either…

f(n) and g(n) are nonnegative, for large n. 16

Example (g(n)) = {f(n) :  positive constants c1, c2, and n0, such that n  n0, 0  c1g(n)  f(n)  c2g(n)} • 10n2 - 3n = (n2) • What constants for n0, c1, and c2 will work? • Make c1 a little smaller than the leading coefficient, and c2 a little bigger. • To compare orders of growth, look at the leading term. • Exercise: Prove that n2/2-3n= (n2)

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Example (g(n)) = {f(n) :  positive constants c1, c2, and n0, such that n  n0, 0  c1g(n)  f(n)  c2g(n)}

• Is 3n3  (n4) ?? • How about 22n (2n)??

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Examples 3n2 + 17 • (1), (n), (n2)  lower bounds • O(n2), O(n3), ...  upper bounds • (n2)  exact bound

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Relations Between O, 

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o-notation For a given function g(n), the set little-o: o(g(n)) = {f(n):  c > 0,  n0 > 0 such that  n  n0, we have 0  f(n) < cg(n)}. f(n) becomes insignificant relative to g(n) as n approaches infinity:

lim [f(n) / g(n)] = 0 n

g(n) is an upper bound for f(n) that is not asymptotically tight. Observe the difference in this definition from previous ones. Why? 21

o-notation •

f(n)=3n+2 is o(n2)

•

f(n)=17n3 + n2 logn

is o(n4)

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 -notation For a given function g(n), the set little-omega:

(g(n)) = {f(n):  c > 0,  n0 > 0 such that

 n  n0, we have 0  cg(n) < f(n)}.

f(n) becomes arbitrarily large relative to g(n) as n approaches infinity:

lim [f(n) / g(n)] = .

n

g(n) is a lower bound for f(n) that is not asymptotically tight. 23

 -notation •

f(n)=3n+2 is

•

f(n)=17n3 +

n2

 (1) is

2

 (n )

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Comparison of Functions fg  ab f (n) = O(g(n)) f (n) = (g(n)) f (n) = (g(n)) f (n) = o(g(n)) f (n) =  (g(n))

    

a a a a a

  = < >

b b b b b

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