pkuwm commented on a change in pull request #1037:
URL: https://github.com/apache/helix/pull/1037#discussion_r435058130



##########
File path: 
helix-core/src/main/java/org/apache/helix/controller/rebalancer/constraint/ExcessiveTopStateResolver.java
##########
@@ -0,0 +1,112 @@
+package org.apache.helix.controller.rebalancer.constraint;
+
+/*
+ * Licensed to the Apache Software Foundation (ASF) under one
+ * or more contributor license agreements.  See the NOTICE file
+ * distributed with this work for additional information
+ * regarding copyright ownership.  The ASF licenses this file
+ * to you under the Apache License, Version 2.0 (the
+ * "License"); you may not use this file except in compliance
+ * with the License.  You may obtain a copy of the License at
+ *
+ *   http://www.apache.org/licenses/LICENSE-2.0
+ *
+ * Unless required by applicable law or agreed to in writing,
+ * software distributed under the License is distributed on an
+ * "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
+ * KIND, either express or implied.  See the License for the
+ * specific language governing permissions and limitations
+ * under the License.
+ */
+
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+import org.apache.helix.api.rebalancer.constraint.AbnormalStateResolver;
+import org.apache.helix.controller.stages.CurrentStateOutput;
+import org.apache.helix.model.Partition;
+import org.apache.helix.model.StateModelDefinition;
+import org.slf4j.Logger;
+import org.slf4j.LoggerFactory;
+
+/**
+ * The abnormal state resolver that graceful fixes double-topstates issue for 
the single topstate
+ * state model.
+ * Note the regular Helix rebalance pipeline will also remove the excessive 
top state replica.
+ * However, the default rebalancer logic cannot guarantee a clean resolution. 
For example, if the
+ * double-topstates situation has already impact the data of the top state 
replicas, then the
+ * controller should reset both of them, then bring back one top state replica 
on the right
+ * allocation. For the application which has such a requirement, they should 
use this resolver or
+ * a more advanced resolver which check the application data to ensure the 
resolution is complete.
+ */
+public class ExcessiveTopStateResolver implements AbnormalStateResolver {
+  private static final Logger LOG = 
LoggerFactory.getLogger(ExcessiveTopStateResolver.class);
+
+  /**
+   * The current states are not valid if there are more than 2 top state 
replicas for a single top
+   * state state model.
+   */
+  @Override
+  public boolean isCurrentStatesValid(final CurrentStateOutput 
currentStateOutput,
+      final String resourceName, final Partition partition, 
StateModelDefinition stateModelDef) {
+    if (!stateModelDef.isSingleTopStateModel()) {
+      return true;
+    }
+    if (currentStateOutput.getCurrentStateMap(resourceName, 
partition).values().stream()
+        .filter(state -> state.equals(stateModelDef.getTopState())).count() > 
1) {
+      return false;

Review comment:
       @jiajunwang In CurrentStateOutput, could we add a top state counter map 
so we could cache the top state counter, like below? Then we could avoid that 
stream filter computation? Tradeoff is we need a bit more memory for the cache. 
But most of them are just references. 
   ```
     public void setCurrentState(String resourceName, Partition partition, 
String instanceName,
         String state) {
       (...... current code ......)
       // Counter number of top state replicas for a single top state model. 
       if (state.equals(stateModelDef.getTopState())) {
         Map<String, Integer> counterMap =
             _topStateCounter.computeIfAbsent(resourceName, k -> new 
HashMap<>())
                 .getOrDerfault(partition, new HashMap<>());
         counterMap.put(state, counterMap.getOrDefault(state, 0));
       }
     }
   ```
   
   Not sure if we need to optimize this. Maybe you could test it. It seems for 
this part, the time complexity is down from O(n) to O(1), but I am not sure 
what the actual time saving is, considering the whole pipeline. If the whole 
pipeline complexity is O(N^2), with this optimization, it is O(N), that may 
help. If the whole pipeline is O(2 * N), with this optimization, still O(N).




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