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我正在查看和弦算法,但我不明白它如何声称具有容错性。
据我了解,给定的键值对基于算法精确存储在一个节点上。所以,我的问题是,如果那个节点失败了,算法失败的容忍度如何?如果我查询那个键,我不会得到这个值,因为它对应的节点已经关闭