我有两个有效树的图(即单根,没有循环).它们差别很小：其中一个叶子另一个叶子没有叶子.有没有办法使用Gremlin或其他可能的图形查询语言,如Cypher,以返回表示这两棵树之间差异的图形？

DOT示例：

graph A {
  a -- b -- c;
}

graph B {
  a -- b -- c;
  b -- d;
}

graph C = graph B - graph A : // <-- How do I do that?

graph C {
  b -- d;
}

最佳答案 以下junit测试是解决此问题的简单可行解决方案.

请参阅算法的注释.

有更好的方法可以做到这一点,但这只是为了表明有一个解决方案.

package com.graph.test;

import static org.junit.Assert.assertEquals;

import java.util.ArrayList;
import java.util.List;

import org.apache.tinkerpop.gremlin.structure.Direction;
import org.apache.tinkerpop.gremlin.structure.Edge;
import org.apache.tinkerpop.gremlin.structure.Graph;
import org.apache.tinkerpop.gremlin.structure.Vertex;
import org.apache.tinkerpop.gremlin.tinkergraph.structure.TinkerGraph;
import org.junit.Test;

/**
 * https://stackoverflow.com/questions/38786038/is-it-possible-to-calculate-difference-between-tree-graphs-in-gremlin
 * https://stackoverflow.com/questions/16553343/diff-for-directed-acyclic-graphs
 * @author wz
 *
 * Algorithm:
 *    Step one: Intersect all vertices and edges of a and b
 *    Step two: Remove all edges that are the same in a and b
 *    Step tree: Remove all non connected vertices
 *    
 *    Step three is not implemented here and one and two are very inefficient. This
 *    is just to show the principle and have a viable solution
 *
 */
public class TestGraphDiff {

  public static final String SORTKEY="sortKey";
  public static final String LINK="link";

  public Graph getGraph() {
    Graph g = TinkerGraph.open();
    Vertex a = g.addVertex("a");
    Vertex b = g.addVertex("b");
    Vertex c = g.addVertex("c");
    b.addEdge(LINK,a);
    c.addEdge(LINK, b);
    return g;
  }

  @Test
  public void testGraphDiff() {
    Graph A=getGraph();
    Graph B=getGraph();
    Vertex d = B.addVertex("d");
    Vertex b= B.traversal().V().hasLabel("b").next();
    d.addEdge(LINK, b);
    Graph C=diff(A,B);
    // there should be one edge left
    assertEquals(1,C.traversal().E().count().next().longValue());
    // there should be two vertices left
    assertEquals(2,C.traversal().V().count().next().longValue());
    // the left over edge should be "b-d"
    assertEquals("b-d",C.traversal().E().next().property(SORTKEY).value().toString());

  }

  /**
   * add a sort key to the given edge
   * @param e
   * @return - the sort key
   */
  private String addSortKey(Edge e) {
    String sortKey=e.inVertex().label()+"-"+e.outVertex().label();
    e.property(SORTKEY,sortKey);
    return sortKey;
  }

  /**
   * get the difference of two Graphs
   * @param a
   * @param b
   * @return
   */
  public Graph diff(Graph a, Graph b) {
    // step one: intersect both graphs
    Graph c= TinkerGraph.open();
    // copy all vertices of a 
    a.traversal().V().forEachRemaining(v->c.addVertex(v.label()));
    // copy all edges of a
    a.traversal().E().forEachRemaining(e->{
      Vertex iva = e.inVertex();
      Vertex ova = e.outVertex();
      Vertex ivc=c.traversal().V().hasLabel(iva.label()).next();
      Vertex ovc=c.traversal().V().hasLabel(ova.label()).next();
      addSortKey(e);
      Edge edge=ovc.addEdge(e.label(), ivc);   
      addSortKey(edge);
    });
    // copy all vertices of b that are not yet in c
    b.traversal().V().forEachRemaining(v->{
     if (c.traversal().V().hasLabel(v.label()).count().next().longValue()==0) {
       c.addVertex(v.label());
     }
     // copy all edges of b that are not yet in c
     b.traversal().E().forEachRemaining(e->{
       Vertex ivb = e.inVertex();
       Vertex ovb = e.outVertex();
       Vertex ivc=c.traversal().V().hasLabel(ivb.label()).next();
       Vertex ovc=c.traversal().V().hasLabel(ovb.label()).next();
       String sortKey=addSortKey(e);
       if (!c.traversal().E().has(SORTKEY,sortKey).hasNext()) {
         Edge edge = ovc.addEdge(e.label(), ivc);
         addSortKey(edge);
       }
     });
    });
    // step two: remove all edges that are "the same" in both graphs
    // inefficient version - would be much quicker with sorted lists
    // of edges ...
    c.traversal().E().forEachRemaining(e->{
      String sortKey=e.property(SORTKEY).value().toString();
      if (a.traversal().E().has(SORTKEY, sortKey).hasNext() &&
          c.traversal().E().has(SORTKEY, sortKey).hasNext()) {
        e.remove();
      }
    });
    // step three remove all "unconnected" vertices
    c.traversal().V().forEachRemaining(v->{
      if (!v.edges(Direction.BOTH).hasNext())
        v.remove();
    });
    return c;
  }

}