我有一堆日志文件,其中包含事件日志以及其中记录的矢量时钟。现在,在比较任何两个事件的向量时钟时,取向量时钟每个分量的平方和的根并使用结果与另一个分量的和进行比较,然后得出结论,值较小的一个在另一个之前,是否正确?
不,如果有办法将其减少到一个值,我们将使用它而不是向量!
要比较矢量时钟,您需要分段比较整个矢量。
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;
}
}
您可以仅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;
}
}