概率计算的堆栈跨流异常

Question

我在项目欧拉（Euler）上做问题14（注意：我不是在寻找欧拉（Euler）问题的解决方案），当我遇到一个有趣的堆栈溢出异常时。

我的非稳态方法效果很好，但是当我尝试使用概率方法遇到同样的问题时，我会遇到堆栈溢出异常。有趣的是，例外仅发生大约17％的时间。一千个跑步产生了166个例外。

我知道我的概率逻辑存在缺陷，但是我对防止它们发生的例外和方法更感兴趣。我是否只需要进行一些内存管理，也许在使用后将一些变量设置为无效？如果是这样，关键点在哪里？

代码如下：

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);
    }
}

我也想提供完整的输出，但这使我超过了角色限制。

Answer 1

您的递归太深。您可以使用-Xss 4096m增加JVM上的通话堆栈，但这是蛮力。更优雅，使用while循环代替getChainLength()中的递归：

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;
    }

Answer 2

您将在stackoverflow异常中看到异常的原因。在这种情况下，它太多了递归，您会通过堆栈中的重复堆叠式框架看到它。

尝试使您的算法迭代而不是递归，并解决了问题。