递归地将1 ^ 2到n ^ 2的整数相加

Question

i'm having some trouble recursively adding integers in java from 1^2 to n^2. I want to be able to recursively do this in the recurvMath method but all i'm getting is an infinite loop.

import java.util.Scanner;

public class Lab9Math {

int count = 0;
static double squareSum = 0;


public static void main(String[] args){

    int n = 0;
    Scanner scan = new Scanner(System.in);
    System.out.println("Please enter the value you want n to be: ");
    n = scan.nextInt();

    Lab9Math est = new Lab9Math();
    squareSum = est.recurvMath(n);
    System.out.println("Sum is: "+squareSum);
}

public int recurvMath(int n){

    System.out.println("N:" +n);
        if(n == 0){
            return 0; 
        }//end if
        if (n == 1){
            return 1;
        }//end if
        if (n > 1){
            return (recurvMath((int) ((int) n+Math.pow(n, 2))));
        }//end if
        return 0;
    }//end method       
}//end class

I'm not fully grasping the nature of defining this recursively, as i know that i can get to here:

return (int) (Math.pow(n, 2));

but i can't incorporate the calling of the recurvMath method correctly in order for it to work. Any help would be appreciated. Thanks!

Answer 1

In general, when trying to solve recursive problems, it helps to try to work them out in your head before programming them.

You want to sum all integers from 1 ² to n ² . The first thing we need to do is express this in a way that lends itself to recursion. Well, another way of stating this sum is:

• The sum of all integers from 1 ² to (n-1) ² , plus n ²

That first step is usually the hardest because it's the most "obvious". For example, we know that "a + b + c" is the same as "a + b", plus "c", but we have to take a leap of faith of sorts and state it that way to get it into a recursive form.

So, now we have to take care of the special base case, 0:

• When n is 0, the sum is 0.

So let's let recurvMath(n) be the sum of all integers from 1 ² to n ² . Then, the above directly translates to:

• recurvMath(n) = recurvMath(n-1) + n ²
• recurvMath(0) = 0

And this is pretty easy to implement:

public int recurvMath(int n){    
    System.out.println("N:" +n);
    if(n == 0){
        return 0; 
    } else {
        return recurvMath(n-1) + (n * n);
    }
}

Note I've chosen to go with n * n instead of Math.pow() . This is because Math.pow() operates on double , not on int .

By the way, you may also want to protect yourself against a user entering negative numbers as input, which could get you stuck. You could use if (n <= 0) instead of if (n == 0) , or check for a negative input and throw eg IllegalArgumentException , or even use Math.abs() appropriately and give it the ability to work with negative numbers.

---

Also, for completeness, let's take a look at the problem in your original code. Your problem line is:

recurvMath((int) ((int) n+Math.pow(n, 2)))

Let's trace through this in our head. One of your int casts is unnecessary but ignoring that, when n == 3 this is recurvMath(3 + Math.pow(3, 2)) which is recurvMath(12) . Your number gets larger each time. You never hit your base cases of 1 or 0, and so you never terminate. Eventually you either get an integer overflow with incorrect results, or a stack overflow.

Answer 2

Try this

import java.util.Scanner;

public class Lab9Math {

int count = 0;
static double squareSum = 0;


public static void main(String[] args){

    int n = 0;
    Scanner scan = new Scanner(System.in);
    System.out.println("Please enter the value you want n to be: ");
    n = scan.nextInt();

    Lab9Math est = new Lab9Math();
    squareSum = est.recurvMath(n);
    System.out.println("Sum is: "+squareSum);
}

public int recurvMath(int n){

    System.out.println("N:" +n);
        if(n == 1){
            return 1; 
        }//end if
        // More simplified solution
        return recurvMath(n-1) + (int) Math.pow(n, 2); // Here is made changes

    }//end method       
}//end class

Answer 3

instead of saying:

return (recurvMath((int) ((int) n+Math.pow(n, 2))));

i instead said:

return (int) ((Math.pow(n, 2)+recurvMath(n-1)));