The Gateway to Computer Science Excellence
+1 vote
120 views
Consider the following statements:
$(1)$ Any two functions $f,g$ are always comparable under big Oh,that is $f=O(g)$ or $g=O(f)$
$(2)$ If $f=O(g)$ and $f=O(h)$ then $g(n)=\theta(h)$
$A)$ $(1)$ is true $(2)$ is false
$B)$ $(1)$ is false $(2)$ is true
$C)$ Both are false
$D)$ Both are true
in Algorithms by Veteran (59.5k points) | 120 views
0
I have a problem with statement $(1)?$
0
the second statement is also false right?
0
yeah answer should be C
0
but any two function we represent as F(n) = Og(n) or G(n) = O(f(n)) right ???
+1

no..consider f(n)= n and g(n)= nsinx+1 ..in this case the 2 functions are not comparable.

0
No @magma
0
see 3 scenarios

either f(n) >= g(n) or f(n) < = g(n) or f(n) = g(n)

when f(n) > = g(n) == > g(n) = O(f(n))

when g(n) > = f(n) === >f(n) = O(g(n))

when g(n) = f(n) ==== > then also it's right :3

# confused
0
F(n) = n and G(n) = n^2 , we can say f(n) = O(g(n)) bt cant represent G(n) = o(f(n))
0
when g(n) = f(n) then g(n) = theta(f(n)) and vice versa
0

F(n) = n and G(n) = n^2 , we can say f(n) = O(g(n)) bt cant represent G(n) = o(f(n))

 But in the question they mention "OR"

either of them can be satisfied :3

 

0
Ohk got it

thanks
0
Both are false
0
o sorry wrong example ........ example of somoshree is right
0

2 Answers

+2 votes

The first statement is false. 
Take two functions, $f(n) = 0.5$, $g(n) = sin(n)$.

You cannot compare these two functions using Big-Oh notation.

For more, read here: https://cs.stackexchange.com/questions/1780/are-the-functions-always-asymptotically-comparable

by Loyal (6.8k points)
0 votes

Statement 1 is false.

Consider f(n) = 0.5 and g(n) = sin(n)
or, consider f(n) = n and g(n) = 1 when n is even; $n^{2}$ when n is odd.

In both the above cases neither f(n) = Og(n) nor g(n) = Of(n)

 

Statement 2 is false

Consider g(n) = 2n and h(n) = n and f(n) = 10n

 

Other important statements

All the below cases are possible:-

$f(n) = Og(n)$ and $g(n) = Of(n)$ — Case 1

$f(n) = Og(n)$ and $g(n) \neq Of(n)$ — Case 2

$f(n) \neq Og(n)$ and $g(n) \neq Of(n)$ — Case 3

 

Case 1

When f(n) and g(n) are identical both functions can upper bound each other.

 

Case 2

f(n) = n and g(n) = $n^{2}$

 

Case 3

See "Statement 1 is false"

by Loyal (7k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,394 answers
198,594 comments
105,445 users