Transcript Review: Mathematical Induction

MCA 202: Discrete Structures
Instructor
Neelima Gupta
[email protected]
Table Of Contents
Growth Functions
Introduction
• For each problem to be solved, we may have multiple
algorithms as solution. However, the best must be
chosen.
• Comparison of the algorithms depends mainly on:
– Time Complexity
Time the algorithm takes to run.
– Memory Space
Memory the algorithm requires.
Thanks to Arti Seth, Roll No. - 4 (MCA
• Let us consider two algorithms for a same problem that run in
f1(n) and f2(n) respectively, ‘n’ being the input size (memory
words required for inputs).
• Consider the graph for f1(n) and f2(n). For smaller values
(n<n0), we don’t care, however we observe that for larger
values of n (n  n0), f1(n)  f2(n), i.e, f1(n) grows slower than
f2(n). Hence, we should prefer f1(n).
f2(n
)f (
1
n)
no
Thanks to Arti Seth, Roll No. - 4 (MCA
Asymptotic Notations
Big O Notation
• In general a function
– f(n) is O(g(n)) if there exist positive constants c and n0 such
that f(n)  c  g(n) for all n  n0
• Formally
– O(g(n)) = { f(n):  positive constants c and n0 such that f(n)
 c  g(n)  n  n0
• Intuitively, it means f(n) grows no faster than g(n).
Thanks to Arti Seth, Roll No. - 4 (MCA
Examples:
Q: f(n) = n2 g(n) = n2 – n Is f(n) = O(g(n))?
Sol 1: (by hit and trial method)
Let c = 1
Claim: n2  1(n2-n)
Proof:
To show: n2<=n(n-1)
T.S: n  n-1, which is not true for any value of n.
Let c = 2
Claim: n2  2(n2-n)
Proof:
T.S: n2  2n2-2n
T.S: 2n  n2, which is true for n  2.
Hence, f(n) = O(g(n))
Thanks to Arti Seth, Roll No. - 4 (MCA
Sol 2:
Let c = 2,
2 g(n) = 2(n2-n) = n2 + (n2-2n)
 n2 , for n2-2n  0
n2  2n
n2
= f(n)
Hence, f(n) = O(g(n)).
• If power and coefficient of leading terms of f(n) and g(n)
are same, then for simplicity ‘c’ can be taken as:
c = Number of negative terms + 1
Thanks to Arti Seth, Roll No. - 4 (MCA
Q: f(n) = n2 g(n) = n2 – n Is g(n) = O(f(n))?
Sol: g(n) = n2 – n
≤ n2 for all n ≥ 1
= f(n)
g(n) ≤ f(n) for all n ≥ 1
Hence, g(n) = O(f(n)).
Thanks to Arti Seth, Roll No. - 4 (MCA
Q: f(n) = n3 g(n) = n3 - n2 – n
Is f(n) = O(g(n))?
Sol: Let c=3
3 g(n) = 3(n3 - n2 - n)
= n3 + (n3 - 3n2 ) + (n3 - 3n)
≥ n3 , for (n3 - 3n2 ) ≥ 0 and (n3 - 3n) ≥ 0
n ≥ 3 and
n≥ 3
= f(n) , for n ≥ n0 , where n0 = max { 3, 3 }
=3
Hence, f(n) = O(g(n)).
Thanks to Arti Seth, Roll No. - 4 (MCA
Q: f(n) = n3 g(n) = n3 - n2 – n Is g(n) =
O(f(n))?
Sol: g(n) = n3 - n2 – n
≤ n3 for all n ≥ 1
= f(n)
g(n) ≤ f(n) for all n ≥ 1
Hence, g(n) = O(f(n)).
Thanks to Arti Seth, Roll No. - 4 (MCA
Omega Notation
• In general a function
– f(n) is (g(n)) if  positive constants c and n0 such that
0  cg(n)  f(n)  n  n0
• Intuitively, it means f(n) grows at least as fast as g(n).
f(n)
cg(n)
Thanks to Arti Seth, Roll No. - 4 (MCA
Theta Notation
• A function f(n) is O(g(n)) if  positive constants c1, c2
and n0, such that
c1g(n)  f(n)  c2g(n)  n  n0
n0 = max {n1 , n2}
c2g(
n)
f(
n)
c1g(
n)
n1
n2
Thanks to Arti Seth, Roll No. - 4 (MCA
f(n)
g(n)
c
n3
n3+ n2 -10n+5
f(n)=O(g(n))
>=1
n3
n3- 2n2 + 100n
F(n)=O(g(n))
>1
n5
n5+ 4n4 - 3n3+ 2n2 – n +1
F(n)=Ω (g(n))
<1
n5
n5- 4n4 + 3n3- 2n2 + n -1
F(n)=Ω (g(n))
<=1
Thanks Anika Garg Roll no.-1 MCA 2012
Assignment 0: Relations
Between Q, , O
For any two functions g(n) and f(n),
f(n) = Q(g(n)) iff
f(n) = O(g(n)) and f(n) = (g(n)).
Assignment No 1
• Self study
a0 + a1 + … + an = (an+1 - 1)/(a - 1) for all a  1
– What is the sum for a = 2/3 as n  infinity? Is it O(1)?
Is it big or small?
– For a = 2, is the sum = O(2^n)? Is it big or small?
Q1 Show that a polynomial of degree k = theta(n^k).
Other Asymptotic Notations
• A function f(n) is o(g(n)) if for every positive
constant c, there exists a constant n0 > 0 such that
f(n) < c g(n)  n  n0
• A function f(n) is (g(n)) if for every positive
constant c, there exists a constant n0 > 0 such that
c g(n) < f(n)  n  n0
• Intuitively,
– () is like >
– Q() is like =
– o() is like <
– () is like 
– O() is like 
Arrange some functions
• f(n) = O(g(n)) => f(n) = o(g(n)) ?
• Is the converse true?
• Let us arrange the following functions in
ascending order (assume log n = o(n) is
known)
– n, n^2, n^3, sqrt(n), n^epsilon, log n, log^2 n, n
log n, n/log n, 2^n, 3^n
Relation between n &
2
n
Intuitively, n appears to be smaller than n2 .
Lets prove it now.
T.P. For any constant c > 0
n < c n2
i.e. 1 < c n
i.e. n > 1 / c
Hence, n < c n2 for n > 1 / c
i.e. n = o( n2)
we can also write it as n < n2.
Relation between n2 & n3
Intuitively, n2 appears to be smaller than n3 .
Lets prove it now.
T.P. For any constant c > 0
n2 < c n3
i.e. 1 < c n
i.e. n > 1 / c
Hence, n2 < c n3 for n > 1 / c
i.e. n2 = o( n3)
we can also write it as n2 < n3.
Relation between n & n1/2
Since n = o (n2 ), we have,
For every constant c > 0, there exists n_c s.t.
n < c n2 for all n >= n_c
Thus sqrt(n) < c n for all sqrt(n) >= n_c
i.e. sqrt(n) < c n for all n >= n_c^2.
Thus,
n1/2 = o( n)
we can also write it as n1/2 < n. Combining the
previous result
n1/2 < n < n2 < n3
Relation between n & log n
• For the time being we can assume the result
log ( n ) = o(n)
log ( n ) < n
we will prove it later.
Relation between n1/2 & log n
Assume log n = o(n)
let c > 0 be any constant
for c/2 > 0 there exists m > 0 such that
log n < (c/2) n for n > m
changing variables from n to n1/2 we get
log(n1/2 ) < (c/2) n1/2 for n1/2 > m
½ log( n ) < (c/2) n1/2 for n > m2
Contd..
let m2 = k
log( n ) < c n1/2 for n > k
Since c > 0 was chosen arbitrarily hence
log n = o( n1/2 )
or
log n < n1/2
Combining the results we get
log n < n1/2 < n < n2 < n3
Relation between
2
n
& nlog n
• Since log n = o(n)
for c > 0,  n0 > 0 such that  n  n0, we
have
log n < c n
Multiplying by n on both sides we get
n log( n ) < c n2  n  n0
 nlog n = o( n2 )
 nlog n < n2
Relation between n & nlog n
Solution:
let c> 0 be any constant such that
n < c n log (n)
 1 < c log( n )
 log( n) > 1 / c
 n > e1/c
i.e. n < c n log n n > e1/c
Since c was chosen arbitrarily
\ n =o(n log n )or n < n log n
Combining the results we can get
log n < n1/2 < n < n logn < n2 < n3
Relation between n & n/log n
• We know that n = o(nlogn)
for c > 0,  n0 > 0 such that  n  n0, we
have
n < c n log n
dividing both sides by log n we get
n/ log( n) < c n  n  n0
Þ n / logn = o(n)
i.e. n / logn < n
Assignment No 2
• Show that log^M n = o(n^epsilon) for all
constants M>0 and epsilon > 0. Assume that
log n = o(n). Also prove the following
Corollary: log n = o(n/log n)
• Show that n^epsilon = o(n/logn ) for every
0 < epsilon < 1 .
Hence we have,
log n < n/log n < n1/2 < n < n logn < n2 < n3
Assignment No 3
• Show that
– lim f(n)/g(n) = 0 => f(n) = o(g(n)).
n→∞
– lim f(n)/g(n) = c => f(n) = θ(g(n)).
n → ∞, where c is a positive constant.
• Show that log n = o(n).
• Show that n^k = o(2^n) for every
positive constant k.
• Show by definition of ‘small o’ that
a^n = o(b^n) whenever a < b , a and b are
positive constants.
Hence we have,
log n < n/log n < n1/2 < n < n logn < n2 < n3
<2n < 3n