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我有一堆日志文件,其中包含事件日志及其中记录的矢量时钟。现在在比较任意两个事件的矢量时钟时,对矢量时钟的每个分量的平方和求根并用结果与另一个比较,然后得出结论是否正确较小的值先于另一个?

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不,如果有办法将它减少到一个值,我们将使用它而不是向量!

要比较矢量时钟,您需要逐段比较整个矢量。

class VectorClock {
    private long[] clocks;
    ...
    /**
     * This is before other iff both conditions are met:
     * - each process's clock is less-than-or-equal-to its own clock in other; and
     * - there is at least one process's clock which is strictly less-than its
     *   own clock in other
     */
    public boolean isBefore(VectorClock other) {
        boolean isBefore = false;
        for (int i = 0; i < clocks.length; i++) {
            int cmp = Long.compare(clocks[i], other.clocks[i]);
            if (cmp > 0)
              return false; // note, could return false even if isBefore is true
            else if (cmp < 0)
              isBefore = true;
        }
        return isBefore;
    }
}

您可以只使用最小值和最大值进行不太准确的传递:

class VectorClockSummary {
    private long min, max;
    ...
    public tribool isBefore(VectorClockSummary other) {
        if (max < other.min)
            return true;
        else if (min > other.max)
            return false;
        else
            return maybe;
    }
}
于 2013-03-16T04:46:33.707 回答