Make variable expansion work for environment variables in k8s

core/commons/che-core-commons-lang/src/main/java/org/eclipse/che/commons/lang/TopologicalSort.java

@@ -0,0 +1,252 @@
+/*
+ * Copyright (c) 2012-2018 Red Hat, Inc.
+ * This program and the accompanying materials are made
+ * available under the terms of the Eclipse Public License 2.0
+ * which is available at https://www.eclipse.org/legal/epl-2.0/
+ *
+ * SPDX-License-Identifier: EPL-2.0
+ *
+ * Contributors:
+ *   Red Hat, Inc. - initial API and implementation
+ */
+package org.eclipse.che.commons.lang;
+
+import static com.google.common.collect.Maps.newLinkedHashMapWithExpectedSize;
+import static java.util.Collections.emptyList;
+
+import java.util.ArrayList;
+import java.util.Collection;
+import java.util.Comparator;
+import java.util.HashSet;
+import java.util.Iterator;
+import java.util.LinkedHashMap;
+import java.util.List;
+import java.util.Map;
+import java.util.Map.Entry;
+import java.util.Set;
+import java.util.TreeSet;
+import java.util.function.Function;
+import java.util.stream.Collectors;
+
+/**
+ * This is an implementation of a stable topological sort on a directed graph.
+ *
+ * <p>The sorting does not pose any requirements on the types being sorted. Instead, the
+ * implementation merely requires a function to provide it with a set of predecessors of a certain
+ * "node". The implementation of the function is completely in the hands of the caller.
+ *
+ * <p>Additionally, a function to extract an "ID" from a node is required. The reasoning behind this
+ * is that it usually is easier to work with identifiers when establishing the predecessors than
+ * with the full node instances. That said, nothing prevents the caller from using the actual node
+ * instance as its ID if the caller so wishes. The consequence of this is that, as a side-effect of
+ * the sorting, the duplicates, as determined by the equality of {@code ID} instances, are removed
+ * from the resulting sorted list.
+ *
+ * @param <N> the type of nodes
+ * @param <ID> the type of a node ID
+ */
+public final class TopologicalSort<N, ID> {
+
+  private final Function<N, ID> identityExtractor;
+  private final Function<N, Set<ID>> directPredecessorsExtractor;
+
+  /**
+   * @param identityExtractor a function to extract some kind of value uniquely identifying a node
+   *     amongst the others.
+   * @param directPredecessorsExtractor a function returning a list of ids of direct predecessors of
+   *     a node
+   */
+  public TopologicalSort(
+      Function<N, ID> identityExtractor, Function<N, Set<ID>> directPredecessorsExtractor) {
+    this.identityExtractor = identityExtractor;
+    this.directPredecessorsExtractor = directPredecessorsExtractor;
+  }
+
+  /**
+   * Given the function for determining the predecessors of the nodes, return the list of the nodes
+   * in topological order. I.e. all predecessors will be placed sooner in the list than their
+   * successors. Note that the input collection is assumed to contain no duplicate entries as
+   * determined by the equality of the {@code ID} type. If such duplicates are present in the input
+   * collection, the output list will only contain the first instance of the duplicates from the
+   * input collection.
+   *
+   * <p>The implemented sort algorithm is stable. If there is no relationship between 2 nodes, they
+   * retain the relative position to each other as they had in the provided collection (e.g. if "a"
+   * preceded "b" in the original collection and there is no relationship between them (as
+   * determined by the predecessor function), the "a" will still precede "b" in the resulting list.
+   * Other nodes may be inserted in between them though in the result).
+   *
+   * <p>The cycles in the graph determined by the predecessor function are ignored and nodes in the
+   * cycle are placed into the output list in the source order.
+   *
+   * @param nodes the collection of nodes
+   * @return the list of nodes sorted in topological order
+   */
+  public List<N> sort(Collection<N> nodes) {
+    // the linked hashmap is important to retain the original order of elements unless required
+    // by the dependencies between nodes
+    LinkedHashMap<ID, NodeInfo<ID, N>> nodeInfos = newLinkedHashMapWithExpectedSize(nodes.size());
+    List<NodeInfo<ID, N>> results = new ArrayList<>(nodes.size());
+
+    int pos = 0;
+    boolean needsSorting = false;
+    for (N node : nodes) {
+      ID nodeID = identityExtractor.apply(node);
+      // we need the set to be modifiable, so let's make our own
+      Set<ID> preds = new HashSet<>(directPredecessorsExtractor.apply(node));
+      needsSorting = needsSorting || !preds.isEmpty();
+
+      NodeInfo<ID, N> nodeInfo = nodeInfos.computeIfAbsent(nodeID, __ -> new NodeInfo<>());
+      nodeInfo.id = nodeID;
+      nodeInfo.predecessors = preds;
+      nodeInfo.sourcePosition = pos++;
+      nodeInfo.node = node;
+
+      for (ID pred : preds) {
+        // note that this means that we're inserting the nodeinfos into the map in an incorrect
+        // order and will have to sort them in the source order before we do the actual topo sort.
+        // We take that cost because we gamble on there being no dependencies in the nodes as a
+        // common case.
+        NodeInfo<ID, N> predNode = nodeInfos.computeIfAbsent(pred, __ -> new NodeInfo<>());
+        if (predNode.successors == null) {
+          predNode.successors = new HashSet<>();
+        }
+        predNode.successors.add(nodeID);
+      }
+    }
+
+    if (needsSorting) {
+      // because of the predecessors, we have put the nodeinfos in the map in an incorrect order.
+      // we need to correct that before we try to sort...
+      TreeSet<NodeInfo<ID, N>> tmp = new TreeSet<>(Comparator.comparingInt(a -> a.sourcePosition));
+      tmp.addAll(nodeInfos.values());
+      nodeInfos.clear();
+      tmp.forEach(ni -> nodeInfos.put(ni.id, ni));
+
+      // now we're ready to produce the results
+      sort(nodeInfos, results);
+    } else {
+      // we don't need to sort, but we need to keep the expected behavior of removing the duplicates
+      results = new ArrayList<>(nodeInfos.values());
+    }
+
+    return results.stream().map(ni -> ni.node).collect(Collectors.toList());
+  }
+
+  private void sort(LinkedHashMap<ID, NodeInfo<ID, N>> nodes, List<NodeInfo<ID, N>> results) {
+
+    while (!nodes.isEmpty()) {
+      NodeInfo<ID, N> curr = removeFirstIndependent(nodes);
+      if (curr != null) {
+        // yay, simple. Just add the found independent node to the results.
+        results.add(curr);
+      } else {
+        // ok, there is a cycle in the graph. Let's remove all the nodes in the first cycle we find
+        // from our predecessors map, add them to the result in their original order and try to
+        // continue normally
+
+        // find the first cycle in the predecessors (in the original list order)
+        Iterator<NodeInfo<ID, N>> nexts = nodes.values().iterator();
+        List<NodeInfo<ID, N>> cycle;
+        do {
+          curr = nexts.next();
+          cycle = findCycle(curr, nodes);
+        } while (cycle.isEmpty() && nexts.hasNext());
+
+        // If we ever find a graph that doesn't have any independent node, yet we fail to find a
+        // cycle in it, the universe must be broken.
+        if (cycle.isEmpty()) {
+          throw new IllegalStateException(
+              String.format(
+                  "Failed to find a cycle in a graph that doesn't seem to  have any independent"
+                      + " node. This should never happen. Please file a bug. Current state of the"
+                      + " sorting is: nodes=%s, results=%s",
+                  nodes.toString(), results.toString()));
+        }
+
+        cycle.sort(Comparator.comparingInt(a -> a.sourcePosition));
+
+        for (NodeInfo<ID, N> n : cycle) {
+          removePredecessorMapping(nodes, n);
+          results.add(n);
+        }
+      }
+    }
+  }
+
+  private void removePredecessorMapping(Map<ID, NodeInfo<ID, N>> nodes, NodeInfo<ID, N> node) {
+    forgetNodeInSuccessors(node, nodes);
+    nodes.remove(node.id);
+  }
+
+  private void forgetNodeInSuccessors(NodeInfo<ID, N> node, Map<ID, NodeInfo<ID, N>> nodes) {
+    if (node.successors != null) {
+      for (ID succ : node.successors) {
+        NodeInfo<ID, N> succNode = nodes.get(succ);
+        if (succNode != null) {
+          succNode.predecessors.remove(node.id);
+        }
+      }
+    }
+  }
+
+  private List<NodeInfo<ID, N>> findCycle(NodeInfo<ID, N> node, Map<ID, NodeInfo<ID, N>> nodes) {
+    // bail out quickly if there are no preds - should be fairly common occurrence hopefully
+    Set<ID> preds = node.predecessors;
+    if (preds == null || preds.isEmpty()) {
+      return emptyList();
+    }
+
+    List<NodeInfo<ID, N>> ret = new ArrayList<>();
+    List<ID> todo = new ArrayList<>(preds);
+
+    Set<ID> visited = new HashSet<>();
+
+    while (!todo.isEmpty()) {
+      ID n = todo.remove(0);
+
+      if (visited.contains(n)) {
+        continue;
+      }
+
+      visited.add(n);
+
+      NodeInfo<ID, N> predNode = nodes.get(n);
+      if (predNode != null) {
+        todo.addAll(predNode.predecessors);
+        ret.add(predNode);
+
+        if (predNode.equals(node)) {
+          // we found the cycle to our original node
+          return ret;
+        }
+      }
+    }
+
+    return emptyList();
+  }
+
+  private NodeInfo<ID, N> removeFirstIndependent(Map<ID, NodeInfo<ID, N>> nodes) {
+    Iterator<Entry<ID, NodeInfo<ID, N>>> it = nodes.entrySet().iterator();
+
+    while (it.hasNext()) {
+      Entry<ID, NodeInfo<ID, N>> e = it.next();
+      if (e.getValue().predecessors.isEmpty()) {
+        it.remove();
+        NodeInfo<ID, N> ret = e.getValue();
+        forgetNodeInSuccessors(ret, nodes);
+        return ret;
+      }
+    }
+
+    return null;
+  }
+
+  private static final class NodeInfo<ID, N> {
+    private ID id;
+    private int sourcePosition;
+    private Set<ID> predecessors;
+    private Set<ID> successors;
+    private N node;
+  }
+}