Question

The number of operations executed by algorithms A and B is 40n2 and 2n3, respectively. Determine n0 such that A is better than B for n ≥ n0.

Answer 1

Answer:

Given that:

A= 40n^2

B = 2n^3

By given scenario:

40n^2=2n^3

dividing both sides by 2

20n^2=n^3

dividing both sides by n^2 we get

20 = n

Now putting n=20 in algorithms A and B:

A=40n^2

= 40 (20)^2

= 40 * (400)

A= 16000

B= 2n^3

= 2 (20)^3

= 2(8000)

B= 16000

Now as A and B got same on n = 20, then as given:

n0 <20 for n =20

Let us take n0 = 19, it will prove A is better than B.

We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.