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Suppose you with had two algorithms, A and B, with growth functions fa(n)=200n2 and fb(n)=2n4. If you were to do an exact analysis on these growth functions, i.e., not simplify them with Big-Oh or tilde approximation, what algorithm would you recommend for which values of n? (Hint: you should be partitioning the domain of n > 0). Explain how you arrived at your answer and why it is correct.

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kendrich

Answer:

For values when n<32 use fb(n) else use fa(n).

See explaination for details

Explanation:

Earlier when n=1 fa(n)=2000 while fb(n)=2. for n=2 fa(n)=2000*22 =2000*4=8000 while fb(n)=2*24=2*16=32. It is observed that fa(n) requires more time than fb(n) for small values.

Now, we will see when fb(n) crosses fa(n). This can happen only when fb(n) values equals or greater than fa(n)

therefore,

2000n2<=2n4

Solving equation we get n2>=1000 which can happen when n>=32.

So for values when n<32 use fb(n) else use fa(n)

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