Bully算法
bully算法是一个分布式系统中动态选择master节点的算法,进程号最大的非失效的节点将被选为master。 算法用三种消息类型:
代码语言:javascript复制1)选举消息 (Election Message: Sent to announce election.)
2)应答消息(Answer (Alive) Message: Responds to the Election message.)
3)选举成功消息 (Coordinator (Victory) Message: Sent by winner of the election to announce victory.)
当一个进程P从失败中恢复,或者接收到主节点失效信息,进程P将做以下事情:
代码语言:javascript复制1)如果进程P有最大的进程ID,那么它则会向其他节点广播Coordinator (Victory) Message。否则进程P向进程号比它大的进程发送Election Message
2)如果进程P发送Election Message后,没有接收到应答,它就会向其他节点广播Coordinator (Victory) Message,并成为master。
3)如果进程P接收到比它进程号更高的进程的Answer (Alive) Message信息,那么它这次的选举就失败了,等待接收其他节点的Coordinator (Victory) Message。
4)如果进程P接收到比它进程号低的进程的Election message,那么它会向该节点返回一个Answer (Alive) Message,并启动选举进程,并向比它更高的进程发送Election Message。
5)如果进程P接收到Coordinator (Victory) Message,那么它就会把发送这条消息的节点看作为master进程。
ES Master选举过程
我看的源码是5.6版本的。因此以下的解释都是依据5.6的源码来说的。 当master节点失效之后,各个节点find master的过程如下: 1)每个节点会ping配置文件中discovery.zen.ping.unicast.hosts的IP,找到所有存活的节点并过滤 2)找到非本身的active master
代码语言:javascript复制 List<DiscoveryNode> activeMasters = new ArrayList<>();
for (ZenPing.PingResponse pingResponse : pingResponses) {
// We can't include the local node in pingMasters list, otherwise we may up electing ourselves without
// any check / verifications from other nodes in ZenDiscover#innerJoinCluster()
if (pingResponse.master() != null && !localNode.equals(pingResponse.master())) {
activeMasters.add(pingResponse.master());
}
}
3)找到所有的可成为master的节点集合masterCandidates ,包含自己
代码语言:javascript复制 // nodes discovered during pinging
List<ElectMasterService.MasterCandidate> masterCandidates = new ArrayList<>();
for (ZenPing.PingResponse pingResponse : pingResponses) {
if (pingResponse.node().isMasterNode()) {
masterCandidates.add(new ElectMasterService.MasterCandidate(pingResponse.node(), pingResponse.getClusterStateVersion()));
}
}
4)如果activeMasters 为空,也就是说不存在活着的master节点,同时当前活着的节点满足配置中discovery.zen.minimum_master_nodes的数量,那么就从masterCandidates 挑出ID最小的节点,让其成为master节点。如果activeMasters 不为空,则从中选择最小的ID成为Master节点即可。
代码语言:javascript复制if (activeMasters.isEmpty()) {
if (electMaster.hasEnoughCandidates(masterCandidates)) {
final ElectMasterService.MasterCandidate winner = electMaster.electMaster(masterCandidates);
logger.trace("candidate {} won election", winner);
return winner.getNode();
} else {
// if we don't have enough master nodes, we bail, because there are not enough master to elect from
logger.warn("not enough master nodes discovered during pinging (found [{}], but needed [{}]), pinging again",
masterCandidates, electMaster.minimumMasterNodes());
return null;
}
} else {
assert !activeMasters.contains(localNode) : "local node should never be elected as master when other nodes indicate an active master";
// lets tie break between discovered nodes
return electMaster.tieBreakActiveMasters(activeMasters);
}
electMaster.electMaster方法和electMaster.tieBreakActiveMasters方法则都是从集合中选取最小节点的ID:
代码语言:javascript复制 public MasterCandidate electMaster(Collection<MasterCandidate> candidates) {
assert hasEnoughCandidates(candidates);
List<MasterCandidate> sortedCandidates = new ArrayList<>(candidates);
sortedCandidates.sort(MasterCandidate::compare);
return sortedCandidates.get(0);
}
public DiscoveryNode tieBreakActiveMasters(Collection<DiscoveryNode> activeMasters) {
return activeMasters.stream().min(ElectMasterService::compareNodes).get();
}
如果当前不存在active master,那么activeMasters 一定为空,则从masterCandidates 从选出id最小的节点即可。 如果当前存在active master,且当前节点不是active maste,那么从activeMasters 中选出id最小的节点。 如果当前存在active master,且当前节点是active maste,那么activeMasters 为空,从masterCandidates 中选出id最小的节点即自己。
在我的感觉中,当前active master的个数要么为空,要么为1,这边不知道为什么要用一个链表。。。为了防止脑裂情况出现吗??