org.jgrapht.alg

Class NaiveLcaFinder<V,E>

java.lang.Object

org.jgrapht.alg.NaiveLcaFinder<V,E>

public class NaiveLcaFinder<V,E>
extends Object

Constructor Summary

Constructors

Constructor and Description
NaiveLcaFinder(DirectedGraph<V,E> graph) Find the Lowest Common Ancestor of a directed graph.

Method Summary

Methods

Modifier and Type	Method and Description
V	findLca(V a, V b) Return the first found LCA of a and b
Set<V>	findLcas(V a, V b) Return all the LCA of a and b.

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

NaiveLcaFinder

public NaiveLcaFinder(DirectedGraph<V,E> graph)

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 ∈ 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.

Method Detail

findLca

public V findLca(V a,
        V b)

Return the first found LCA of a and b

Parameters:

a - the first element to find LCA for

b - the other element to find the LCA for

Returns:

the first found LCA of a and b, or null if there is no LCA.

findLcas

public Set<V> findLcas(V a,
              V b)

Return all the LCA of a and b. Currently not implemented

Parameters:

a - the first element to find LCA for

b - the other element to find the LCA for

Returns:

the set of all LCA of a and b, or empty set if there is no LCA.