Stack-overflow exception on probabilistic calculation

Question

I was doing problem 14 on Project Euler (note: I'm not looking for a solution to the Project Euler problem) when I ran into an interesting stack overflow exception.

My non-probabilistic approach worked just fine but when I attempted the same problem with a probabilistic approach I ran into the stack overflow exception. The funny thing is that the exception only occurs about 17% of the times. A thousand run-throughs yielded 166 exceptions.

I know my probabilistic logic is flawed, but I'm more interested in the cause of the exceptions and ways to prevent them from occurring. Do I simply need to do some memory management, maybe set some variables to null after using them? If so where would the key-points be to do so?

The code is as follows:

public class Problem14_LongestCollatzSequence {

    private static final int STARTING_CHAIN_LENGTH = 1;
    private static final int PROBABLY_RIGHT = 100000;

    /**
     * Calculate and return the Collatz sequence of a given number.
     *
     * @param number The number for which the Collatz sequence is to be
     * calculated.
     * @param chainlength The length of the chain for the number. This should
     * start with an initial value of 1.
     * @return The Length of the Collatz sequence.
     */
    private static int getChainLength(long number, int chainlength) {
        // All chains should end with 1.
        if (number != 1) {
            // If the number is even, halve the number, otherwise multiply it by 3 and add 1.
            if (number % 2 == 0) {
                number = number / 2;
            } else {
                number = number * 3 + 1;
            }
            // Call this function again.
            return getChainLength(number, ++chainlength);
        }
        // Return the length of the chain.
        return chainlength;
    }

    /**
     * Determine and return the number below a maximum value that will result in
     * the longest Collatz chain.
     *
     * @param maxStartingNumber The maximum value (exclusive) of the numbers
     * that will be tested.
     * @return The number that will produce the longest Collatz sequence in the
     * given range.
     */
    private static int calculateLongestChain(int maxStartingNumber) {
        Random random = new Random();
        int probabilityCounter = 0;
        int currentChainNumber = 0;
        int longestChainNumber = 0;
        int currentChainLength = 0;
        int longestChainLength = 0;

        // Get the chain length of random numbers until a certain number of unsuccsessful attempts have been made.
        while (probabilityCounter <= PROBABLY_RIGHT) {
            currentChainNumber = random.nextInt(maxStartingNumber);
            currentChainLength = getChainLength(currentChainNumber, STARTING_CHAIN_LENGTH);
            // If the current chain-length is bigger than the previously calculated one, reset the counter and update the chain number, otherwise increase the counter.
            if (currentChainLength > longestChainLength) {
                probabilityCounter = 0;
                longestChainLength = currentChainLength;
                longestChainNumber = currentChainNumber;
            } else {
                ++probabilityCounter;
            }
        }
        return longestChainNumber;
    }

    private static int calculateLongestChainNP(int maxStartingNumber) {
        // Non-probabilistic way to calculate the longest Collatz sequence.
        int currentChainLength = 0;
        int longestChainLength = 0;
        int longestChainNumber = 0;
        // Simply loop through all the numbers in the range to calculate the one resulting in the longest sequence.
        for (int i = 1; i < maxStartingNumber; i++) {
            currentChainLength = getChainLength(i, STARTING_CHAIN_LENGTH);
            if (currentChainLength > longestChainLength) {
                longestChainLength = currentChainLength;
                longestChainNumber = i;
            }
        }
        return longestChainNumber;
    }

    public static void main(String[] args) {
        int exceptionCount = 0;
        for (int count = 0; count < 1000; count++) {
            try {
                int testNumber = 1000000;
                System.out.println("Probabilistic answer: " + calculateLongestChain(testNumber));
                System.out.println("Non-probabilistic answer: " + calculateLongestChainNP(testNumber) + "\n");
            } catch (java.lang.StackOverflowError soe) {
                exceptionCount++;
                System.err.println(soe + "\n");
            }
        }
        System.out.println("Exception count: " + exceptionCount);
    }
}

I wanted to provide the full output as well, but that puts me over the character limit.

Answer 1

You will see in your stackoverflow exception the cause of the exception. In this case it is too much recursion and you will see it by a repeating stackframes in the stacktrace.

Try to make your algorithm iterative instead of recursive and your problem is solved.

Answer 2

Your recursion is too deep. You can increase call stack on your JVM with -Xss 4096m , but this is brute force. Be more elegant and use a while loop instead of recursion in getChainLength() :

private static int getChainLength(long number, int chainlength) {
        // All chains should end with 1.
        while (number != 1) {
            // If the number is even, halve the number, otherwise multiply it by 3 and add 1.
            if (number % 2 == 0) {
                number = number / 2;
            } else {
                number = number * 3 + 1;
            }
            // Call this function again.
            ++chainlength;
        }
        // Return the length of the chain.
        return chainlength;
    }