Question

I´ve made a program that does the (usually called) hailstone sequence, The program basically does this:

creates an int (value) and assign it a value.

If the int is even, divide it by two.

If the int is odd, multiply it by three and add one. Continue this process until n is equal to one.

it seems to be working fine with most numbers, but this number 99888769, the app hangs on a negative integer. Why is this?, they say that no-one has yet been able to proove that it stops, I´m not expecting I have solved that. But would be interesting to know why my app stops. –

    private void hailStoneSequence(){
    int value = 99888769;
    int i = 0;
    boolean trueOrFalse = isOddOrEven (value);
    while (value != 1){
        while (trueOrFalse == true && value != 1){
            i++;
            int previousValue = value;
            value = value / 2;
            println( previousValue +" is even, so I take half: "+value);
            trueOrFalse = isOddOrEven (value); // returning true or false, and inserting the newly divided number. So that it breaks loop when nescesary.
        }
            while (trueOrFalse == false && value != 1){
                i++;
                int previousValue = value;
                value = (value * 3) + 1;
                println (previousValue +" is odd, so I make 3n+1: "+value); 
                trueOrFalse = isOddOrEven (value);  
            }
                }       
    println ("\n\nThe process took "+i+" to reach "+value);
}

private boolean isOddOrEven(int value){
    /*
     * Takes an value and returns true, if that number is even.
     * Else it returns false.
     */
    if (value % 2 != 0){
    return false;
    }else{
        return true;
    }
}

}

Answer 1

As you keep increasing ints, they will eventually (in what might seem like a startling behavior) become negative because you are surpassing the maximum value of the int type (2^31-1), i.e. you end up changing the bit (of the int’s binary representation) that is used to store the sign of the number. Use long instead.