org.jgrapht.alg.NaiveLcaFinder Maven / Gradle / Ivy
/*
* (C) Copyright 2013-2018, by Leo Crawford, Alexandru Valeanu and Contributors.
*
* JGraphT : a free Java graph-theory library
*
* This program and the accompanying materials are dual-licensed under
* either
*
* (a) the terms of the GNU Lesser General Public License version 2.1
* as published by the Free Software Foundation, or (at your option) any
* later version.
*
* or (per the licensee's choosing)
*
* (b) the terms of the Eclipse Public License v1.0 as published by
* the Eclipse Foundation.
*/
package org.jgrapht.alg;
import org.jgrapht.*;
import java.util.*;
/**
* Find the Lowest Common Ancestor of a directed graph.
*
*
* Find the LCA, defined as Let $G = (V, E)$ be a DAG, and let $x, y \in V$ . Let $G x,y$ be the
* subgraph of $G$ induced by the set of all common ancestors of $x$ and $y$. Define SLCA (x, y) to
* be the set of out-degree 0 nodes (leafs) in $G x,y$. The lowest common ancestors of $x$ and $y$
* are the elements of SLCA (x, y). This naive algorithm simply starts at $a$ and $b$, recursing
* upwards to the root(s) of the DAG. Wherever the recursion paths cross we have found our LCA.
* from http://www.cs.sunysb.edu/~bender/pub/JALG05-daglca.pdf. The algorithm:
*
*
* 1. Start at each of nodes you wish to find the lca for (a and b)
* 2. Create sets aSet containing a, and bSet containing b
* 3. If either set intersects with the union of the other sets previous values (i.e. the set of notes visited) then
* that intersection is LCA. if there are multiple intersections then the earliest one added is the LCA.
* 4. Repeat from step 3, with aSet now the parents of everything in aSet, and bSet the parents of everything in bSet
* 5. If there are no more parents to descend to then there is no LCA
*
*
* The rationale for this working is that in each iteration of the loop we are considering all the
* ancestors of a that have a path of length n back to a, where n is the depth of the recursion. The
* same is true of b.
*
*
* We start by checking if a == b.
* if not we look to see if there is any intersection between parents(a) and (parents(b) union b)
* (and the same with a and b swapped)
* if not we look to see if there is any intersection between parents(parents(a)) and
* (parents(parents(b)) union parents(b) union b) (and the same with a and b swapped)
* continues
*
*
* This means at the end of recursion n, we know if there is an LCA that has a path of <=n to a
* and b. Of course we may have to wait longer if the path to a is of length n, but the path to
* b>n. at the first loop we have a path of 0 length from the nodes we are considering as LCA to
* their respective children which we wish to find the LCA for.
*
* @param the graph vertex type
* @param the graph edge type
*
*/
public class NaiveLcaFinder
{
private Graph graph;
/**
* Create a new instance of the naive LCA finder.
*
* @param graph the input graph
*/
public NaiveLcaFinder(Graph graph)
{
this.graph = GraphTests.requireDirected(graph, "Graph must be directed");
}
/**
* Return the first found LCA of a and b
*
* @param a the first element to find LCA for
* @param b the other element to find the LCA for
*
* @return the first found LCA of a and b, or null if there is no LCA.
*/
public V findLca(V a, V b)
{
return findLca(
Collections.singleton(a), Collections.singleton(b), new LinkedHashSet<>(),
new LinkedHashSet<>());
}
/**
* Return all the LCAs of a and b.
*
* @param a the first element to find LCA for
* @param b the other element to find the LCA for
*
* @return the set of all LCAs of a and b, or empty set if there is no LCA.
*/
@SuppressWarnings("unchecked")
public Set findLcas(V a, V b)
{
Set[] visitedSets = new Set[2];
// set of nodes visited from a
visitedSets[0] = new LinkedHashSet<>();
// set of nodes visited from b
visitedSets[1] = new LinkedHashSet<>();
doubleBfs(a, b, visitedSets);
// all common ancestors of both a and b
Set intersection;
// optimization trick: save the intersection using the smaller set
if (visitedSets[0].size() < visitedSets[1].size()) {
visitedSets[0].retainAll(visitedSets[1]);
intersection = visitedSets[0];
} else {
visitedSets[1].retainAll(visitedSets[0]);
intersection = visitedSets[1];
}
/*
* Find the set of all non-leaves by iterating through the set of common ancestors. When we
* encounter a node which is still part of the SLCA(a, b) we remove its parent(s).
*/
Set nonLeaves = new LinkedHashSet<>();
for (V node : intersection) {
for (E edge : graph.incomingEdgesOf(node)) {
if (graph.getEdgeTarget(edge).equals(node)) {
V source = graph.getEdgeSource(edge);
if (intersection.contains(source))
nonLeaves.add(source);
}
}
}
// perform the actual removal of non-leaves
intersection.removeAll(nonLeaves);
return intersection;
}
/**
* Perform a simultaneous bottom-up bfs from a and b, saving all visited nodes. Once a a node
* has been visited from both a and b, it is no longer expanded in our search (we know that its
* ancestors won't be part of the SLCA(x, y) set).
*/
@SuppressWarnings("unchecked")
private void doubleBfs(V a, V b, Set[] visitedSets)
{
Queue[] queues = new Queue[2];
queues[0] = new ArrayDeque<>();
queues[1] = new ArrayDeque<>();
queues[0].add(a);
queues[1].add(b);
visitedSets[0].add(a);
visitedSets[1].add(b);
for (int ind = 0; !queues[0].isEmpty() || !queues[1].isEmpty(); ind ^= 1) {
if (!queues[ind].isEmpty()) {
V node = queues[ind].poll();
if (!visitedSets[0].contains(node) || !visitedSets[1].contains(node))
for (E edge : graph.incomingEdgesOf(node)) {
if (graph.getEdgeTarget(edge).equals(node)) {
V source = graph.getEdgeSource(edge);
if (!visitedSets[ind].contains(source)) {
queues[ind].add(source);
visitedSets[ind].add(source);
}
}
}
}
}
}
/**
* Recurse through the descendants of aSet and bSet looking for the LCA of a and b, which are
* members of sets aSeenSet and bSeenSet respectively, along with all elements on the paths from
* every member of aSet and bSet
*/
private V findLca(
Set aSet, Set bSet, LinkedHashSet aSeenSet, LinkedHashSet bSeenSet)
{
while (true) {
// if there is no LCA...
if ((aSet.size() == 0) && (bSet.size() == 0)) {
return null;
}
// does aSet intersect with bSeenSet
if (!Collections.disjoint(aSet, bSeenSet)) {
return overlappingMember(aSet, bSeenSet);
}
// does bSet intersect with aSeenSet
if (!Collections.disjoint(bSet, aSeenSet)) {
return overlappingMember(bSet, aSeenSet);
}
if (!Collections.disjoint(aSet, bSet)) {
return overlappingMember(aSet, bSet);
}
aSeenSet.addAll(aSet);
bSeenSet.addAll(bSet);
aSet = allParents(aSet);
// no point doing the same again (and it can stop us getting stuck in
// an infinite loop)
aSet.removeAll(aSeenSet);
bSet = allParents(bSet);
bSet.removeAll(bSeenSet);
}
}
/**
* Find the immediate parents of every item in the given set, and return a set containing all
* those parents
*
* @param vertexSet the set of vertex to find parents of
*
* @return a set of every parent of every vertex passed in
*/
private Set allParents(Set vertexSet)
{
HashSet result = new HashSet<>();
for (V e : vertexSet) {
for (E edge : graph.incomingEdgesOf(e)) {
if (graph.getEdgeTarget(edge).equals(e)) {
result.add(graph.getEdgeSource(edge));
}
}
}
return result;
}
/**
* Return a single vertex that is both in $x$ and $y$. If there is more than one then select the
* first element from the iterator returned from $y$, after all the elements of $x$ have been
* removed. this allows an orderedSet to be passed in to give predictable results.
*
* @param x set containing vertex
* @param y set containing vertex, which may be ordered to give predictable results
*
* @return the first element of $y$ that is also in $x$, or null if no such element
*/
private V overlappingMember(Set x, Set y)
{
y.retainAll(x);
return y.iterator().next();
}
}
// End NaiveLcaFinder.java