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3 votes
3 votes

Which of the following set is empty?

  1. $o (g(n)) \cap \omega (g(n))$
  2. $O (g(n)) \cap \Omega (g(n))$
  3. $o (g(n)) \cap O( g(n))$
  4. $\omega (g(n)) \cap \Omega (g(n))$
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1 Answer

5 votes
5 votes
Best answer

f(n) = O(g(n)) if f(n) <= c.g(n)

f(n) = Ώ(g(n)) if f(n) >= c.g(n)

therefore there is possibility of intersection there for option b,c,d will not be empty 

but 

f(n) = o(g(n)) if f(n) < c.g(n)

f(n) = ω(g(n)) if f(n) > c.g(n)

there is no possiblity of intersection therefore it will be empty

Answer is A

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4 Comments

let's me take your example itself 

f=x and g=x+1 and  c=2
So x< 2 (x+1)  here also it is true f(n) < c.g(n)   (here f should be strictly smaller ,equals condition not possible)

and   f=x and g=x+1 and  c=1/2

x > 1/2(x+1) here also it is true f(n)c.g(n) (here f should be strictly greater ,equals condition not possible)

So i'm not getting where your are finding the intersection between these 2 

for the second question

O( g(n) ) ∩ Ω( g(n) ) here it can't be empty always 

for example f=x and g= 2x

so for big-oh   x < = 1/2(2x) (where c=1/2)  ----->    x<= x   

  for big-omega  x>= 1/2(2x)  ----> x>=x so there is a possibility of intersection 

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So, for my example,

$\frac{1}{2}(x+1)$ < area under consideration < 2(x+1)

 

Is the "area under consideration" empty ???
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edited by
Your second example is correct.
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