a. Suppose that Event(i) is a boolean function that returns true for about n^0.2 values of i < n. In other words, the probability of the if being true is roughly n^0.2/n=1/n^0.8.

   for(int i=0;i<n;i++)
    {
            if(Event(i))
            {
                    for(int j=0;j<i*i;j++)
                    {
                            g(n);
                    }
            }
    }

            What is the order of execution   (i) if g(n) is order n?        (ii) if g(n) is order log(n)?      (iii) if g(n) is order nlog(n)?

b.  Solve the recurrence T(n)=7T(n/2)+n²   (As usual, we want only an order for the solution.)
c.  Find the recurrence relationship for the following function. (You can be fairly "approximate" about it.)

    int f(int k, int m)
    {
        if(k<m)
        {
                int sum=1;
                int mid1=k+(m-k+1)/3;
                int mid2=k+2*(m-k+1)/3;
                sum=sum+f(k,mid1)+f(mid2,m);
                return(sum);
        }
       else return(1);

    }

d.  Solve the recurrence for f  letting  T_f(n) be the time required for f to run when n=(m-k+1), i.e., the range of values in f.

e.  Find f(0,1), f(0,2), f(2,8), and f(0,12).

f.   Here is a simple procedure that is designed to find the second largest integer in integer array A. Analyze the algorithm.

int FindSecond(int [] A)
{
        int largest, secondlargest;
        if(A[0]>A[1])
        {
                largest=A[0];
                secondlargest=A[1];
        }
        else
        {
                largest=A[1];
                secondlargest=A[0];
        }
        for(int i=2;i<A.length;i++)
        {
                if(A[i]>largest)
                {
                        secondlargest=largest;
                        largest=A[i];
                }
                else if(A[i]>secondlargest)
                {
                        secondlargest=A[i];
                }
        }
        return(secondlargest);
}

g.  Do a thorough analysis to find T(n) for the following function.

        double power(double x, int n)
        {
                if(n==0) return(1.0);
                else if(n==1) return(x);
                else if((n%2)==0) return(square(power(x,n/2)));
                else return(square(power(x,n/2))*x);
        }

h.  Analyze the following code segment:

    for(int i=0;i<n;i++) {
        for(int j=0;j<i;j++) {
            for(int k=0;k*k<n;k++) {
                if(k>5) then f(n);
                else g(n);
             }
          }
       }
    }

when: a. f(n) takes time n and g(n) takes time n*n*n    b. f(n) takes constant time and g(n) takes time n*n