Question

Asymptotic notation O satisfies the transitive property i.e. if f(n)=O(g(n)) and g(n)=O(h(n)), then f(n)=O(h(n)). Now we know that 2ⁿ =O(2^n-1), 2^n-1 =O(2^n-2?),....... , 2ⁱ=O(2^i-1?),....... So using rule of transitivity, we can write 2ⁿ =O(2^i-1?).We can go extending this, so that finally 2ⁿ =O(2^k?), where k is constant.So we can write

2ⁿ =O(1?). Do you agree to what has been proved?If not,where is the fallacy? 6 marks (ALGORITHM ANALYSIS AND DESIGN based problem)

Answer 1

We say that f(n)=O(g(n)) when there exist constants c and k such that $\forall_{n > k} f(n) \leq c*g(n)$

In the above analysis, somewhere down the line, we state the following wrongly,

$an-b \text{ is } O(d) \text{ where a,b,d } > 0$ . This thing can never happen because, for any constant value of d we can always find a c such that, $an-b \text{ is } > c*d$ for some value of n.

This disproves the above claim that 2n is O(1).