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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 2012)
• 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).
no
f2(n)f1(n)
Thanks to Arti Seth, Roll No. - 4 (MCA 2012)
Asymptotic NotationsBig 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 2012)
Examples:Q: f(n) = n2 g(n) = n2 – n Is f(n) = O(g(n))?
Sol 1: (by hit and trial method)
Let c = 1Claim: 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 = 2Claim: n2 2(n2-n)Proof:T.S: n2 2n2-2nT.S: 2n n2, which is true for n 2.Hence, f(n) = O(g(n))
Thanks to Arti Seth, Roll No. - 4 (MCA 2012)
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 2012)
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 2012)
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
= 3Hence, f(n) = O(g(n)).
Thanks to Arti Seth, Roll No. - 4 (MCA 2012)
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 2012)
Omega Notation
• In general a function– f(n) is (g(n)) if positive constants c and n0 such
that 0 cg(n) f(n) n n0
• Intuitively, it means f(n) grows at least as fast as g(n).
Thanks to Arti Seth, Roll No. - 4 (MCA 2012)
f(n)
cg(n)
Theta Notation• A function f(n) is O(g(n)) if positive constants c1, c2 and n0,
such that c1g(n) f(n) c2g(n) n n0
n0 = max n1 , n2
n1 n2
c2g(n)f(n)c1g(n)
Thanks to Arti Seth, Roll No. - 4 (MCA 2012)
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 , , O
For any two functions g(n) and f(n),
f(n) = (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,
– o() is like < – O() is like
() is like > () is like
() 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
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 n & n2
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 n2 & n3
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_ci.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 & n1/2
• For the time being we can assume the result log ( n ) = o(n) log ( n ) < n we will prove it later.
Relation between n & 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
Relation between n1/2 & log n
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
• 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 n2 & 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 nn 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 & nlog 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
Relation between n & n/log 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