Thread: Have a few odd problems.

/**
 * 
 */
 
/**
 * @author pradcoc
 *
 */
 
import java.io.*;
import java.util.*;
 
public class RecursionClassOne
{ // beginning of class
 
 
public static void main(String[] args)
{ // beginning of main
  System.out.println("the first argument inputted is " + args[0]);
    int n = Integer.parseInt(args[0]);
printLines(n);
binary(n); 
binary(n).toString();
System.out.println(binary(n).toString());
} // end of main
 
public static void printLines(int n)
{ // beginning of method
 
if (0 <= n <= 9)
{ // beginning of if
System.out.println(n);
return;
} // end of if
 
else
{ // beginning of else
 
System.out.println(n%10);
printLines(n/10);
 
} // end of else
 
} // end of method
 
public static String binary(int n)
{ // beginning of method;
 
	Integer n2 = n;
String str = "";
if (n2 == 0)
{ // beginning of if
str.concat(n2.toString());
return(str);
} // end of if
 
else if (n ==1);
{ // beginning of else if
str.concat(n2.toString());
return(str);
} // end of else if
 
else
{ // beginning of else
str.concat((n2 %2).toString());
binary(n/2);
 
} // end of else
 
} // end of method
 
} // end of program

First of all, I'm getting a silly message saying that for

if (0<=n <=9)

"The operator <= is undefined for the argument type(s) boolean, int"

It doesn't like the else before str.concant((n2 % 2).toString());

and it doesn't like str.concant((n2 % 2).toString()); either;

Also, for both methods, the way it's printing or returning the values now is backwards of the order I want it in.

For printLines(1234)

It's returning, or will return:

4
3
2
1

And for binary 26, it's returning, or will return

01011

But I want it to return

11010

How do I fix that?

Check your syntax where all those compilation errors occur. For example,

if (0<=n && n<=9)//is correct
else if (n ==1);//I assume you do not want a semi-colon here.

import java.io.*;
import java.util.*;
 
public class RecursionClassOne
{ // beginning of class
 
 
public static void main(String[] args)
{ // beginning of main
  System.out.println("the first argument inputted is " + args[0]);
    int n = Integer.parseInt(args[0]);
printLines(n);
binary(n).toString();
System.out.println(binary(n).toString());
 
} // end of main
 
public static void printLines(int n)
{ // beginning of method
 
if (0 <= n && n<= 9)
{ // beginning of if
System.out.println(n);
return;
} // end of if
 
else
{ // beginning of else
 
System.out.println(n%10);
printLines(n/10);
 
} // end of else
 
} // end of method
 
public static String binary(int n)
{ // beginning of method;
 
	Integer n2 = n;
String str = "";
if (n2 == 0)
{ // beginning of if
str.concat(n2.toString());
return(str);
} // end of if
 
else if (n ==1)
{ // beginning of else if
str.concat(n2.toString());
return(str);
} // end of else if
 
else
{ // beginning of else
	Integer n3 = n2%2;
str.concat((n3).toString());
binary(n/2);
 
} // end of else
 
} // end of method
 
} // end of program

Ok, it's saying for my last else in binary, that I have to return something, but I don't want it to return something there.

Also, my stuff is still printing out backwards of the way I want it to.

Originally Posted by javapenguin

Ok, it's saying for my last else in binary, that I have to return something, but I don't want it to return something there.

Well, you've defined the function to return a String so by definition you must. Return an empty string if you have to.

Also, my stuff is still printing out backwards of the way I want it to.

Recursive algorithms are quite dependent upon order of operation. If you have 'single branch recursion' (which these function are), calling the recursive function before any output will print things in reverse order compared to calling any output before calling the recursive function.

Originally Posted by javapenguin

Why would I want it to return anything for the else part?

Because you defined it that way. If that is not what you want I suggest you redefine the method in terms you DO want (wouldn't you want to do something with the recursive returned strings from the binary function?) Recursion isn't an easy concept, so I suggest you seriously sit down and step through the process with some sort of input writing down the values and function class. Doing so will layout out how it truly works and perhaps answer many of your questions.

Originally Posted by copeg

Because you defined it that way. If that is not what you want I suggest you redefine the method in terms you DO want (wouldn't you want to do something with the recursive returned strings from the binary function?) Recursion isn't an easy concept, so I suggest you seriously sit down and step through the process with some sort of input writing down the values and function class. Doing so will layout out how it truly works and perhaps answer many of your questions.

Why isn't it recursively going to the base cases?

public static String binary(int n)
{ // beginning of method;
 
	Integer n2 = n;
String str = "";
if (n2 == 0)
{ // beginning of if
str.concat(n2.toString());
return(str);
} // end of if
 
else if (n ==1)
{ // beginning of else if
str.concat(n2.toString());
return(str);
} // end of else if
 
else
{ // beginning of else
	Integer n3 = n2%2;
str.concat((n3).toString());
binary(n/2);
 
} // end of else
 
} // end of method

Why isn't it, when I call binary(n/2), checking to see if n/2 is 0 or 1 and if not, keep checking till it is and then returning the String?

I found the problem area. Why won't it concat anything to str?

else
{ // beginning of else
	Integer n3 = n2%2;
str.concat((n3).toString()); // should concat n2%2
System.out.println("Str is: " + str);
binary(n/2);
 
 
} // end of else

What do I do?

I need to get the else part to return something and also get the method to return the String with the correct binary value, no its value in reverse. But I'm not allowed, I think, to use the method toBinaryString(int n).

Originally Posted by aussiemcgr

Ok, quick recursion review.

Look at my code below of an example of recursion. Review the comments to see if you understand how it works:

public class TestRead 
{
 
    public static void main(String[] args) 
    {
        //Test Cases
        System.out.println(recursiveMethod(1));
        System.out.println(recursiveMethod(2));
        System.out.println(recursiveMethod(3));
        System.out.println(recursiveMethod(4));
        System.out.println(recursiveMethod(5));
        System.out.println(recursiveMethod(6));
        System.out.println(recursiveMethod(7));
        System.out.println(recursiveMethod(8));
        System.out.println(recursiveMethod(9));
        System.out.println(recursiveMethod(10));
    }
    /* Recursive Method
     * Two important things regarding how we declare here: 
     * First is what we send this method
     * Second is what this method returns
     *
     * Those two things will be important to remember while we implement our method
     */
    public static int recursiveMethod(int n)
    {
        /* Base Case:
         * For this method, our Base Case (or, end case) is when n==1.
         * When this occurs, we want to return n (which will equal 1)
         */
        if(n==1)
            return n;
        /* If this is not the Base Case, we want to do our next step. For this method,
         * we want to multiply n with our call to recursiveMethod again. The very thing
         * that makes a recursive method so it that it will call itself.
         */
        else
            /* Now, there are two important things here. The first is the fact that we
             * are sending this method a variable that is different from the variable
             * that this method is currently running with. If you send it the same variable,
             * it will do the same thing, and just loop forever. The second important thing
             * is our return call. This says we want to return the value of n mulitplied by
             * our next call this method. 
             *
             * Remember, this method will return an int, so when we call this method below,
             * it will return an int which we can multiply by n. We will then return THAT
             * value to wherever this method was called from. That could be at our main, or
             * it could be at another instance of this method.
             */
            return n*recursiveMethod(n-1);
    }
}

This program results in the following cases and results:


Lets have a look at the first 3 cases and trace them.
CASE 1:
Call: recursiveMethod(1)
Hits Base Case, returns 1 to main
CASE 2:
Call: recursiveMethod(2)
Passes Base Case, calls recursiveMethod(1)
Hits Base Case, returns 1 to previous recursiveMethod call
Returns (2*1) to main
CASE3:
Call: recursiveMethod(3)
Passes Base Case, calls recursiveMethod(2)
Passes Base Case, calls recursiveMethod(1)
Hits Base Case, returns 1 to previous recursiveMethod call
Returns (2*1) to previous recursiveMethod call
Returns (3*2) to main


Do you see how the further we start from the base case, the more exponential our calls to our recursive method becomes? That is the essence of recursion. Do you understand this much?

Yes, my code for printLines(int n) used recursion too and it is now working perfectly. However, I don't know what to call recursively. I thought that....ok, now I'm starting to understand something.

I want it to return something and pass that value to the method to concat.

But how do I do that?

Also, I think it's going to print the binary out backward.

Originally Posted by aussiemcgr

But, that is what recursion is...

There really isn't any point in having that System.out.println in you else statement in your recursive method. There is no real way to track it by doing that other than just watching the digits go down.

I still don't quite understand what this method should do.

No, no, no. It's returning nothing.

Program 2
Write a recursive method binary(int n) that will return a String that stores the binary representation of that number (i.e. the string will have only 0’s and 1’s).
Here is how the binary of an integer number is calculated: let the number be 53.
53/2 = 26, 53-2*26 = 1 26/2 = 13, 26-2*13 = 0 13/2 = 6 , 13-2*6 = 1 6/2 = 3 , 6-2*3 = 0 3/2 = 1 , 3-2*1 = 1 1/2 = 0 , 1-2*0 = 1 The binary representation of 53 is “110101”, which are the above answers read from bottom to top. Thus binary(53) should return a string “110101”.
Hint: The binary representation of 26 is “11010”.

I'm gonna fiddle with Eclipse and see what I can get.

Hmmm....is it losing my String value and not concatinating it?

I think something is backward.

Re: string concat

Reread (or read for the first time) post #11... aussiemcgr has already explained how you are using the concat method incorrectly, and you seem to be ignoring it.

Originally Posted by aussiemcgr

Well, you have two problems. The first is exactly what the compiler told you. You want to return something or there is no way to capture the value. So, you want this:

else
{ // beginning of else
    Integer n3 = n2%2;
str.concat((n3).toString()); // should concat n2%2
System.out.println("Str is: " + str);
return binary(n/2);
 
 
} // end of else

The second is that the String method concat(String) returns a String, but you are not capturing that String (I've made this mistake several times, do it once and never forget the reason again). So, you have two options, you can use the concat method, or you dont have to. You can concat String with the plus sign. Here are both ways:
WITH CONCAT:

str = str.concat((n3).toString());

WITHOUT CONCAT:

str = str+(n2%2);

Since the mathematics are in parentheses and is not between quotes, it will be evaluated prior to concatenating to the Strings around it.

Do I want it to be evaluated after or before?

Ok, how do I capture this String? It seems to be getting lost.

It seems the values after my first digit are now getting lost.

package Assignment5;
 
/**
 * @author pradcoc
 *
 */
 
import java.io.*;
import java.util.*;
 
import javax.swing.JOptionPane;
 
public class RecursionClassTwo
{ // beginning of class
 
 
// Try using a negative as an argument if you want a good laugh.  
 
public static void main(String[] args)
{ // beginning of main
 
	System.out.println("the first argument inputted is " + args[0]);
 
  int n = Integer.parseInt(args[0]);
 
// binary(n);
// binary(n).toString();
System.out.println(binary(n).toString());
 
} // end of main
 
public static String binary(int n)
{ // beginning of method;
 
	Integer n2 = n; // makes my int value be set copied into an Integer value
	// System.out.println(n2.toBinaryString(n)); // just wanted to see what this would do
 
String str = "";
 
// my humor 
if (n < 0)
{
	String oops = "Sorry, no can do!";
	String doh = "Charlie, this is not your day!";
	String tryAgain = "Please try again.  Have a nice day. ";
	String funnyString = oops+ " " + doh + " " + tryAgain;
	JOptionPane.showMessageDialog(null, "Due to increasing costs and the bad economy and the recent raise in taxes and global warming taxes, we did not have the resources to be able to have this method work with negative values.", "No negatives.", JOptionPane.ERROR_MESSAGE);
	JOptionPane.showMessageDialog(null, "That and it won't work.");
	JOptionPane.showMessageDialog(null, "Thank President Obama", "Blame the Democrats", JOptionPane.ERROR_MESSAGE);
	JOptionPane.showMessageDialog(null, "Blame Al Gore too!", "The stupid 'Inconvenient Lie'!", JOptionPane.ERROR_MESSAGE);
	return funnyString;
}
// a base case
 
if (n == 0)
{ // beginning of if
str = str.concat(n2.toString()); // concats this to the previous Strings, if any.  If none, concats nothing to this.  
return(str);
} // end of if
 
else if (n ==1)
{ // beginning of else if
str = str.concat(n2.toString());
return(str);
} // end of else if
 
else
{ // beginning of else
/*	Integer n3 = n2%2;
	System.out.println(n3.toString());
 
str.concat((n3).toString());
System.out.println("Str is: " + str);
binary(n/2); */
 
Integer n3 = n2%2;
 
str = str.concat((n3).toString()); // should concat n2%2
System.out.println("Str is: " + str);
return  binary(n/2);
 
 
} // end of else
 
 
} // end of method
} // end of class

It's capturing them with Str is: but it won't concat these.

I think it's from possible dual return statements.

It's outputting this:
the first argument inputted is 32
Str is: 0
Str is: 0
Str is: 0
Str is: 0
Str is: 0
Str is finally: 1
1

I'd like it to output Str is finally: 100000

How do I get the zeroes in there?