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