Consider the following recursive method: public int someFun (int n) { if (n < = 0) return 2; else return someFun (n-1) * someFun (n-1); } (a) When the program calls someFun (5), how many times will someFun (3) be called? (b) What does this method calculate when the input parameter n is a non-negative integer?

(a) someFunc (3) will be called 4 times.

(b) For non negative number n someFunc method calculates 2^2^n.

Explanation:

When you call someFunc (5) it will call someFunc (4) two time.

So now we have two someFunc (4) now each someFunc (4) will call someFunc (3) two times. Hence the call to someFun (3) is 4 times.

someFunc (n) calculates someFunc (n-1) two times and calculates it's product.

someFunc (n) = someFunc (n-1) ^2 ... (1)

someFunc (n-1) = someFunc (n-2) ^2 ... (2)

substituting the value form eq2 to eq 1.

someFunc (n) = someFunc (n-2) ^2^2

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= someFunc (n-n) ^2^n.

=2^2^n

2 raised to the power 2 raised to the power n.