示例#1
0
  /**
   * Answer a list of the named hierarchy roots of a given {@link OntModel}. This will be similar to
   * the results of {@link OntModel#listHierarchyRootClasses()}, with the added constraint that
   * every member of the returned iterator will be a named class, not an anonymous class expression.
   * The named root classes are calculated from the root classes, by recursively replacing every
   * anonymous class with its direct sub-classes. Thus it can be seen that the values in the list
   * consists of the shallowest fringe of named classes in the hierarchy.
   *
   * @param m An ontology model
   * @return A list of classes whose members are the named root classes of the class hierarchy in
   *     <code>m</code>
   */
  public static List<OntClass> namedHierarchyRoots(OntModel m) {
    List<OntClass> nhr = new ArrayList<OntClass>(); // named roots
    List<OntClass> ahr = new ArrayList<OntClass>(); // anon roots

    // do the initial partition of the root classes
    partitionByNamed(m.listHierarchyRootClasses(), nhr, ahr);

    // now push the fringe down until we have only named classes
    while (!ahr.isEmpty()) {
      OntClass c = ahr.remove(0);
      partitionByNamed(c.listSubClasses(true), nhr, ahr);
    }

    return nhr;
  }
示例#2
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  /**
   * Answer the shortest path from the <code>start</code> resource to the <code>end</code> RDF node,
   * such that every step on the path is accepted by the given filter. A path is a {@link List} of
   * RDF {@link Statement}s. The subject of the first statement in the list is <code>start</code>,
   * and the object of the last statement in the list is <code>end</code>.
   *
   * <p>The <code>onPath</code> argument is a {@link Filter}, which accepts a statement and returns
   * true if the statement should be considered to be on the path. To search for an unconstrained
   * path, pass {@link Filter#any} as an argument. To search for a path whose predicates match a
   * fixed restricted set of property names, pass an instance of {@link PredicatesFilter}.
   *
   * <p>If there is more than one path of minimal length from <code>start</code> to <code>end</code>
   * , this method returns an arbitrary one. The algorithm is blind breadth-first search, with loop
   * detection.
   *
   * @param m The model in which we are seeking a path
   * @param start The starting resource
   * @param end The end, or goal, node
   * @param onPath A filter which determines whether a given statement can be considered part of the
   *     path
   * @return A path, consisting of a list of statements whose first subject is <code>start</code>,
   *     and whose last object is <code>end</code>, or null if no such path exists.
   */
  public static Path findShortestPath(
      Model m, Resource start, RDFNode end, Filter<Statement> onPath) {
    List<Path> bfs = new LinkedList<Path>();
    Set<Resource> seen = new HashSet<Resource>();

    // initialise the paths
    for (Iterator<Statement> i = m.listStatements(start, null, (RDFNode) null).filterKeep(onPath);
        i.hasNext(); ) {
      bfs.add(new Path().append(i.next()));
    }

    // search
    Path solution = null;
    while (solution == null && !bfs.isEmpty()) {
      Path candidate = bfs.remove(0);

      if (candidate.hasTerminus(end)) {
        solution = candidate;
      } else {
        Resource terminus = candidate.getTerminalResource();
        if (terminus != null) {
          seen.add(terminus);

          // breadth-first expansion
          for (Iterator<Statement> i = terminus.listProperties().filterKeep(onPath);
              i.hasNext(); ) {
            Statement link = i.next();

            // no looping allowed, so we skip this link if it takes us to a node we've seen
            if (!seen.contains(link.getObject())) {
              bfs.add(candidate.append(link));
            }
          }
        }
      }
    }

    return solution;
  }