Пример #1
0
  /**
   * Sets the enum to operate in linear fashion, as we have found a looping transition at position:
   * we set an upper bound and act like a TermRangeQuery for this portion of the term space.
   */
  private void setLinear(int position) {
    assert linear == false;

    int state = runAutomaton.getInitialState();
    int maxInterval = 0xff;
    for (int i = 0; i < position; i++) {
      state = runAutomaton.step(state, seekBytesRef.bytes[i] & 0xff);
      assert state >= 0 : "state=" + state;
    }
    for (int i = 0; i < allTransitions[state].length; i++) {
      Transition t = allTransitions[state][i];
      if (t.getMin() <= (seekBytesRef.bytes[position] & 0xff)
          && (seekBytesRef.bytes[position] & 0xff) <= t.getMax()) {
        maxInterval = t.getMax();
        break;
      }
    }
    // 0xff terms don't get the optimization... not worth the trouble.
    if (maxInterval != 0xff) maxInterval++;
    int length = position + 1; /* position + maxTransition */
    if (linearUpperBound.bytes.length < length) linearUpperBound.bytes = new byte[length];
    System.arraycopy(seekBytesRef.bytes, 0, linearUpperBound.bytes, 0, position);
    linearUpperBound.bytes[position] = (byte) maxInterval;
    linearUpperBound.length = length;

    linear = true;
  }
Пример #2
0
  /**
   * Returns the next String in lexicographic order that will not put the machine into a reject
   * state.
   *
   * <p>This method traverses the DFA from the given position in the String, starting at the given
   * state.
   *
   * <p>If this cannot satisfy the machine, returns false. This method will walk the minimal path,
   * in lexicographic order, as long as possible.
   *
   * <p>If this method returns false, then there might still be more solutions, it is necessary to
   * backtrack to find out.
   *
   * @param state current non-reject state
   * @param position useful portion of the string
   * @return true if more possible solutions exist for the DFA from this position
   */
  private boolean nextString(int state, int position) {
    /*
     * the next lexicographic character must be greater than the existing
     * character, if it exists.
     */
    int c = 0;
    if (position < seekBytesRef.length) {
      c = seekBytesRef.bytes[position] & 0xff;
      // if the next byte is 0xff and is not part of the useful portion,
      // then by definition it puts us in a reject state, and therefore this
      // path is dead. there cannot be any higher transitions. backtrack.
      if (c++ == 0xff) return false;
    }

    seekBytesRef.length = position;
    visited[state] = curGen;

    Transition transitions[] = allTransitions[state];

    // find the minimal path (lexicographic order) that is >= c

    for (int i = 0; i < transitions.length; i++) {
      Transition transition = transitions[i];
      if (transition.getMax() >= c) {
        int nextChar = Math.max(c, transition.getMin());
        // append either the next sequential char, or the minimum transition
        seekBytesRef.grow(seekBytesRef.length + 1);
        seekBytesRef.length++;
        seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) nextChar;
        state = transition.getDest().getNumber();
        /*
         * as long as is possible, continue down the minimal path in
         * lexicographic order. if a loop or accept state is encountered, stop.
         */
        while (visited[state] != curGen && !runAutomaton.isAccept(state)) {
          visited[state] = curGen;
          /*
           * Note: we work with a DFA with no transitions to dead states.
           * so the below is ok, if it is not an accept state,
           * then there MUST be at least one transition.
           */
          transition = allTransitions[state][0];
          state = transition.getDest().getNumber();

          // append the minimum transition
          seekBytesRef.grow(seekBytesRef.length + 1);
          seekBytesRef.length++;
          seekBytesRef.bytes[seekBytesRef.length - 1] = (byte) transition.getMin();

          // we found a loop, record it for faster enumeration
          if (!finite && !linear && visited[state] == curGen) {
            setLinear(seekBytesRef.length - 1);
          }
        }
        return true;
      }
    }
    return false;
  }