How do I break out of this recursive call?

Question

I am a newbie to recursion and I am still learning it, so please tolerate my poor logic if it is bad. I have this function which has 5 parameters a,b,c,x,y. so what I essentially want to do is take an element out of either of these variables and add it to the other to finally get x , y. I want to try out this by myself and I have nearly done it, only i wanted to ask if there's any way i could get out of this recursive call, once I get the answer as "YES".

public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String in = sc.nextLine();
        int[] input = new int[10];
        for(int i=0; i < 10; i=i+2) {
            input[i] = Integer.parseInt(String.valueOf(in.charAt(i)));
        }
        int A = input[0];
        int B = input[2];
        int C = input[4];
        int X = input[6];
        int Y = input[8];
        persistent(A,B,C,X,Y);
    }

    private static void persistent(int a, int b, int c, int x, int y) {
        if(a == 0 || b == 0 || c == 0) {
            if((a == x && b == y) && (b == x && a == y)) {
                System.out.println("YES");
            }
            else if((b == x && c == y) || (c == x && b == y)) {
                System.out.println("YES");
            }
            else if((a == x && c == y) && (c == x && a == y)) {
                System.out.println("YES");
            }
            else {
            return;
            }
        }
        persistent(a-1,b+1,c,x,y);
        persistent(a-1,b,c+1,x,y);
        persistent(a+1,b-1,c,x,y);
        persistent(a,b-1,c+1,x,y);
        persistent(a+1,b,c-1,x,y);
        persistent(a,b+1,c-1,x,y);
    }

Simply `return;` after your `System.out.println("YES")` to halt execution of your function. In recursion, this is called a `terminating case`. Note that (at least in production-facing applications) it's a good idea to have a separate break condition as well, to prevent your method from going on forever. — nbokmans, Jul 20 '21 at 16:52
i did do that but it just returns to the next line....which is yet another call. I want it to skip those function calls and get out of the function altogether. — kshitij Sabale, Jul 20 '21 at 16:56
You could return a boolean in your recursive method, and once you find that one has returned true, don't execute any more statements. — khelwood, Jul 20 '21 at 16:57

score 1 · Accepted Answer · answered Jul 20 '21 at 21:40

Your code never does.

Recursive algorithms pretty much all boil down to this exact same style:

First, check if some edge case has been reached (generally, the very simplest case). In this case, return immediately - do not recurse. By tautology, if the answer is so easy you can just give it without needing to recurse, that defines 'edge case' / 'simple case'. There must be at least one such case, or a recursive algorithm cannot work.
Otherwise, provide your answer and feel free to employ as many recursive calls as you prefer, but every single last one of those calls must be simpler, defined by the idea that it is strictly closer to that simple case as mentioned in 1.
All state is conveyed via parameters. It is common to have a private helper method that does the actual work which has a bunch of extra parameters, and the public API that is a one-liner that calls the helper, providing initial values for those extra parameters. Only if you have no such state can you omit this one.

Your code isn't doing this. There is a simple case, which is if a or b or c is 0, but your 6 recursive calls are not clearly moving towards simplicity.

The fix is not obvious. Your algorithm cannot work as written and cannot be fixed without considerably rethinking it.

Hopefully the above will help you: Your recursive calls need to become simpler somehow, guaranteed. Right now it's not guaranteed: Yes, every call is moving one of the 3 numbers closer to the edge case (be 0), but there is no clear flow: If I call your code with, say, 3/4/5 as arguments, then it makes recurisve calls with 2/5/5, 2/4/6, etcetera. That first call (2/5/5) is not moving guaranteed closer to the edge case, because as part of its call stack, it'll be doing 3/4/5 again.

Noted ! Thanks for the answer. I'll look out for it. I am new to recursion , so have a long way to go. Thanks for the help! — kshitij Sabale, Jul 21 '21 at 03:42

How do I break out of this recursive call?

1 Answers1