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Thread: Recursion: Fibonacci Number Question

  1. #1
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    Default Recursion: Fibonacci Number Question

    How do I use rFibNum and start the loop of trials with n= 40 and determine what value of n the program fails to complete. And how would I do this with memoizedFibonocc?

    Thanks for the help
    //Recursion: Fibonacci Number

    import java.io.*;
     
    public class FibonacciNumberTest3
    {
       static BufferedReader keyboard = new
                 BufferedReader(new InputStreamReader(System.in));
       static int callCountint = 0;
       static long callCountlong = 0;
       public static void main(String[] args) throws IOException
        {
        int n;
        long t;
            long nthFib;
     
            String callCountStr ;
     
        for(n = 2; n < 50; n++ ){
     
     
            //stop if the size of the fib number exceeds size of int 
            // this point was determined by trial and error
            if(n>44){
                System.out.print("type enter to continue.");
                keyboard.readLine();
            }
     
     
            String nStr = String.format("%1$3s", String.valueOf(n));
     
            //initialize the array of memoized results
            for(int i = 0; i< 100; i++)
                        solvedFibs[i] = -1;
             solvedFibs[0] = 0;
             solvedFibs[1] = 1;
     
     
            //remember the start time and calculate the term
            t = System.currentTimeMillis();
     
            // pick one of the following recursion method calls
            nthFib =rFibNum( n);
            //nthFib =memoizedRecursion( n);
     
            //how long did it take
            t = System.currentTimeMillis() - t;
     
            //output the results
            System.out.print("The Fibonacci number "
                       + nStr + " is: " + String.format("%1$12s", nthFib) );
            if(callCountlong > 1000000000){
                long oneBillion = 1000000000;
                callCountlong = callCountlong / oneBillion;
                callCountStr = Long.toString(callCountlong)+" billion ";
            }else{
                if(callCountlong >= 1000000){
                    callCountlong = callCountlong / 1000000;
                    callCountStr = Long.toString(callCountlong)+" million ";
                }else callCountStr = Long.toString(callCountlong);
     
            }
            //format the data into nice even columns 
            //expand the strings by adding spaces to fill the column width
            // %1$24s will add spaces on the left making a 24 character column.
     
            String fibCalls = String.format("%1$12s",callCountStr);
            String callCountintStr = String.format("%1$12s",callCountint);
            System.out.println(" making ("+callCountintStr + ") "+fibCalls
                                                +" function calls taking "+t+" ms.");
            callCountint = 0;
            callCountlong = 0;
        }// end for loop
        }// end main
     
        public static long rFibNum(int n)
        {
            FibonacciNumberTest3.callCountint++;
            FibonacciNumberTest3.callCountlong++;
            if(n < 1)
                    return 0;
            else if(n == 1)
                    return 1;
            else
            return rFibNum( n - 1) + rFibNum( n - 2);
        }// end rFibNum
     
     
        /**
         * An array which stores calculated fibonacci numbers.
         */
        static long[] solvedFibs = new long[100];
     
        /**
         * Computes the nth fibonacci number using memoization 
         */
        static long memoizedRecursion(int n) 
        {
                //some counters for the printout of results
                FibonacciNumberTest3.callCountint++;
                FibonacciNumberTest3.callCountlong++;
     
                if (n < 1) return solvedFibs[1]; // f(0) = 0
                if (n == 1) return solvedFibs[2]; // f(1) = 1
                // If the nth fibonacci number has not been calculated we calculate it. 
                if (solvedFibs[n-2] < 0)
                        solvedFibs[n-2] = memoizedRecursion(n-2);
                if (solvedFibs[n-1] < 0)
                        solvedFibs[n-1] = memoizedRecursion(n-1);
                solvedFibs[n] = solvedFibs[n-1] + solvedFibs[n-2];
                return solvedFibs[n];
        }//end memoizedRecursion
     
     
    }


  2. #2

  3. The Following 2 Users Say Thank You to Darryl.Burke For This Useful Post:

    copeg (October 5th, 2013), newbie (October 5th, 2013)

  4. #3
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    Default Re: Recursion: Fibonacci Number Question

    F(n) = F(n-1) + F(n-2);

    F(0) = 0;
    F(1) = 1;


    At which value of n?

    add a throws clause (throws InputMismatchException) to your two methods (rfib and memorized).

    Then, when you call them in your for loop, put a try/catch block around each iteration like this


    try
    {


    }

    catch(InputMismatchException ime)
    {
    System.out.println(n);
    break;
    }


    It will stop your for loop though when it throws the exception. If you don't want it to stop when you reach that point and instead just tell them to hit enter, remove the "break;"
    Then it will just print out the n value where it stopped working.

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    Default Re: Recursion: Fibonacci Number Question

    not what I'm asking for exactly. How do I use rFibNum and start the loop of trials with n= 40

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    Default Re: Recursion: Fibonacci Number Question

    for(n = 40; n < 50; n++ )
    {
    ...........

    }

  7. #6
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    Default Re: Recursion: Fibonacci Number Question

    public class F {


    int fibo (int f) { if (f=0)return 0;
    else if (f=1) return 1;
    else return fibo(f-1)+fibo(f-2);
    }
    System.out.print(fibo(3));

    }

    What is incorrect in this code as the compiler show 5 errors in System.out.print..-some identifiers expected, illegal start of type

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