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Adrian Hernandez Islas
Asked: 2020-09-26 15:49:15 +0800 CST 2020-09-26 15:49:15 +0800 CST

Exercise with Recursion Search

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My problem is the following:

A certain type of bomb requires a detonator to activate. These bombs are made in such a way that only their detonator can activate them. This is done by making them have a different detonator and plug for each bomb.

You have in your possession 20 bombs with their respective 20 detonators. However, you don't know which detonator corresponds to which bomb and you need to know. You can't compare any detonator to any bomb, as for this you would need to plug them in and activate the bomb. You can only compare the detonators with each other, and the bombs with each other.

Take into account that in the locations from 1to 20are the sizes of the plugs of the bombs and from 21to to are the 40sizes of the detonators.

I have already solved it as follows (recursively):

public class DetonadorAndBomb {

    public static void bomb(int []a,int m,int n){
        if(a[m]==a[n+1]){
            System.out.println(a[n+1]);
            if(!(m==(n+1))){
                n=1;
                bomb(a,m+1,n);
            }

        } else{
            bomb(a,m,n+1);
        }
    }

    public static void main(String[] args) {
        int num[] = {1,3,3,1};
        bomb(num,0,1);
    }
}

And in the console it results:

1

3

3

It is not supposed to print the last value.

java-8

1 Answers

  1. Best Answer

    Let's see if I understood:

    to [ ] = {1, 2, 3, 3, 1, 2}

    Departure:

    Position 0 ( switch 1) corresponds to position 4 ( pump 1)

    Position 1 ( switch 2) corresponds to position 5 ( pump 2)

    Position 2 ( switch 3) corresponds to position 3 ( pump 3)

    Possible recursive solution:

    public class DetonadorAndBomb {

    public static void bomb(int []a,int m,int n)
    {
        int t = a.length,    // Tamaño del vector.
            h = t/2;         // Posición donde comienza la segunda parte del vector.
        // Control para que m se mantenga en la primera mitad del vector.
        if (m<h)
        {
            if(a[m]==a[n])    // Se encontró una coincidencia.
            {
                System.out.format("A la posicion %d (interruptor %d) le corresponde "+
                                  "la posicion %d (bomba %d)\n", m, a[m], n, a[n]);
                bomb(a,m+1,h);    /* La posición m ya tuvo coincidencia con
                                     la posición n, entonces se sigue buscando,
                                     para las siguientes posiciones, comenzando en m+1 */
            }
            else if (n < t-1)     /* No hubo coincidencia, entonces se controla que
                                     haya posiciones disponibles en la segunda parte
                                     del vector para seguir buscando posibles
                                     coincidencias para la posición m */
                bomb(a,m,n+1);
            else                  /* No hubo coincidencia y no quedan posiciones
                                     disponibles en la segunda mitad del vector,
                                     por lo que la búsqueda debe continuar desde m+1. */
                bomb(a,m+1,h);
        }
    }

    public static void main(String[] args) {
        int num[] = {1,2,3,3,1,2};

        bomb(num,0,num.length/2);
    }
}

    Code available with debugging messages to visually see how the recursive function works.

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