The method below attempts to calculate xn economically:

public double pow(double x, int n)

{

double y;

if (n == 1)

y = x;

else

y = pow(x, n/2) * pow(x, n - n/2); // Line 7

return y;

}

(a) How many multiplications will be executed when pow(1.234, 5) is called?

(b) How many multiplications will be executed if we replace Line 7 above with the following statements?

{ y = pow(x, n/2); y *= y; if (n % 2 != 0) y *= x; }

(c) How many multiplications will Version (a) above take to calculate pow(1.234, 9)? Version (b)?