content: package org.apache.lucene.search;
  
  /**
   * Licensed to the Apache Software Foundation (ASF) under one or more
   * contributor license agreements.  See the NOTICE file distributed with
   * this work for additional information regarding copyright ownership.
   * The ASF licenses this file to You under the Apache License, Version 2.0
   * (the "License"); you may not use this file except in compliance with
   * the License.  You may obtain a copy of the License at
   *
   *     http://www.apache.org/licenses/LICENSE-2.0
   *
   * Unless required by applicable law or agreed to in writing, software
   * distributed under the License is distributed on an "AS IS" BASIS,
   * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
   * See the License for the specific language governing permissions and
   * limitations under the License.
   */
  
  import java.io.IOException;
  import java.util.ArrayList;
  
  final class SloppyPhraseScorer extends PhraseScorer {
      private int slop;
    private boolean checkedRepeats; // flag to only check in first candidate doc in case there are no repeats
    private boolean hasRepeats; // flag indicating that there are repeats (already checked in first candidate doc)
    private PhraseQueue pq; // for advancing min position
    private PhrasePositions[] nrPps; // non repeating pps ordered by their query offset
  
      SloppyPhraseScorer(Weight weight, PhraseQuery.PostingsAndFreq[] postings, Similarity similarity,
                         int slop, byte[] norms) {
          super(weight, postings, similarity, norms);
          this.slop = slop;
      }
  
      /**
       * Score a candidate doc for all slop-valid position-combinations (matches) 
       * encountered while traversing/hopping the PhrasePositions.
       * <br> The score contribution of a match depends on the distance: 
       * <br> - highest score for distance=0 (exact match).
       * <br> - score gets lower as distance gets higher.
       * <br>Example: for query "a b"~2, a document "x a b a y" can be scored twice: 
       * once for "a b" (distance=0), and once for "b a" (distance=2).
       * <br>Possibly not all valid combinations are encountered, because for efficiency  
       * we always propagate the least PhrasePosition. This allows to base on 
       * PriorityQueue and move forward faster. 
       * As result, for example, document "a b c b a"
       * would score differently for queries "a b c"~4 and "c b a"~4, although 
       * they really are equivalent. 
       * Similarly, for doc "a b c b a f g", query "c b"~2 
       * would get same score as "g f"~2, although "c b"~2 could be matched twice.
       * We may want to fix this in the future (currently not, for performance reasons).
       */
      @Override
    protected float phraseFreq() throws IOException {
      int end = initPhrasePositions();
      //printPositions(System.err, "INIT DONE:");
      if (end==Integer.MIN_VALUE) {
        return 0.0f;
      }
      
      float freq = 0.0f;
      PhrasePositions pp = pq.pop();
      int matchLength = end - pp.position;
      int next = pq.size()>0 ? pq.top().position : pp.position;
      //printQueue(System.err, pp, "Bef Loop: next="+next+" mlen="+end+"-"+pp.position+"="+matchLength);
      while (pp.nextPosition() && (end=advanceRepeats(pp, end)) != Integer.MIN_VALUE)  {
        if (pp.position > next) {
          //printQueue(System.err, pp, "A: >next="+next+" matchLength="+matchLength);
          if (matchLength <= slop) {
            freq += getSimilarity().sloppyFreq(matchLength); // score match
          }      
          pq.add(pp);
          pp = pq.pop();
          next = pq.size()>0 ? pq.top().position : pp.position;
          matchLength = end - pp.position;
          //printQueue(System.err, pp, "B: >next="+next+" matchLength="+matchLength);
        } else {
          int matchLength2 = end - pp.position;
          //printQueue(System.err, pp, "C: mlen2<mlen: next="+next+" matchLength="+matchLength+" matchLength2="+matchLength2);
          if (matchLength2 < matchLength) {
            matchLength = matchLength2;
          }
        }
      }
      if (matchLength <= slop) {
        freq += getSimilarity().sloppyFreq(matchLength); // score match
      }    
      return freq;
    }
  
    /**
     * Advance repeating pps of an input (non-repeating) pp.
     * Return a modified 'end' in case pp or its repeats exceeds original 'end'.
     * "Dirty" trick: when there are repeats, modifies pp's position to that of 
     * least repeater of pp (needed when due to holes repeaters' positions are "back").
     */
    private int advanceRepeats(PhrasePositions pp, int end) throws IOException {
      int repeatsEnd = end;
      if (pp.position > repeatsEnd) {
        repeatsEnd = pp.position;
      }
      if (!hasRepeats) {
        return repeatsEnd;
      }
      int tpPos = tpPos(pp);
      for (PhrasePositions pp2=pp.nextRepeating; pp2!=null; pp2=pp2.nextRepeating) {
        while (tpPos(pp2) <= tpPos) {
          if (!pp2.nextPosition()) {
            return Integer.MIN_VALUE;
          }
        }
        tpPos = tpPos(pp2);
        if (pp2.position > repeatsEnd) {
          repeatsEnd = pp2.position;
        }
        // "dirty" trick: with holes, given a pp, its repeating pp2 might have smaller position.
        // so in order to have the right "start" in matchLength computation we fake pp.position.
        // this relies on pp.nextPosition() not using pp.position.
        if (pp2.position < pp.position) { 
          pp.position = pp2.position;     
        }
      }
      return repeatsEnd;
    }
  
    /**
     * Initialize PhrasePositions in place.
     * There is a one time initialization for this scorer (taking place at the first doc that matches all terms):
     * <ul>
     *  <li>Detect groups of repeating pps: those with same tpPos (tpPos==position in the doc) but different offsets in query.
     *  <li>For each such group:
     *  <ul>
     *   <li>form an inner linked list of the repeating ones.
     *   <li>propagate all group members but first so that they land on different tpPos().
     *  </ul>
     *  <li>Mark whether there are repetitions at all, so that scoring queries with no repetitions has no overhead due to this computation.
     *  <li>Insert to pq only non repeating PPs, or PPs that are the first in a repeating group.
     * </ul>
     * Examples:
     * <ol>
     *  <li>no repetitions: <b>"ho my"~2</b>
     *  <li>repetitions: <b>"ho my my"~2</b>
     *  <li>repetitions: <b>"my ho my"~2</b>
     * </ol>
     * @return end (max position), or Integer.MIN_VALUE if any term ran out (i.e. done) 
     */
    private int initPhrasePositions() throws IOException {
      int end = Integer.MIN_VALUE;
      
      // no repeats at all (most common case is also the simplest one)
      if (checkedRepeats && !hasRepeats) {
        // build queue from list
        pq.clear();
        for (PhrasePositions pp=min,prev=null; prev!=max; pp=(prev=pp).next) {  // iterate cyclic list: done once handled max
          pp.firstPosition();
          if (pp.position > end) {
            end = pp.position;
          }
          pq.add(pp);         // build pq from list
        }
        return end;
      }
      
      //printPositions(System.err, "Init: 1: Bef position");
      
      // position the pp's
      for (PhrasePositions pp=min,prev=null; prev!=max; pp=(prev=pp).next) {  // iterate cyclic list: done once handled max  
        pp.firstPosition();
      }
      
      //printPositions(System.err, "Init: 2: Aft position");
      
      // one time initialization for this scorer (done only for the first candidate doc)
      if (!checkedRepeats) {
        checkedRepeats = true;
        ArrayList<PhrasePositions> ppsA = new ArrayList<PhrasePositions>();
        PhrasePositions dummyPP = new PhrasePositions(null, -1, -1);
        // check for repeats
        for (PhrasePositions pp=min,prev=null; prev!=max; pp=(prev=pp).next) {  // iterate cyclic list: done once handled max
          if (pp.nextRepeating != null) {
            continue; // a repetition of an earlier pp
          }
          ppsA.add(pp);
          int tpPos = tpPos(pp);
          for (PhrasePositions prevB=pp, pp2=pp.next; pp2!= min; pp2=pp2.next) {
            if (
                pp2.nextRepeating != null  // already detected as a repetition of an earlier pp
                || pp.offset == pp2.offset // not a repetition: the two PPs are originally in same offset in the query! 
                || tpPos(pp2) != tpPos) {  // not a repetition
              continue; 
            }
            // a repetition
            hasRepeats = true;
            prevB.nextRepeating = pp2;  // add pp2 to the repeats linked list
            pp2.nextRepeating = dummyPP; // allows not to handle the last pp in a sub-list
            prevB = pp2;
          }
        }
        if (hasRepeats) {
          // clean dummy markers
          for (PhrasePositions pp=min,prev=null; prev!=max; pp=(prev=pp).next) {  // iterate cyclic list: done once handled max
            if (pp.nextRepeating == dummyPP) {
              pp.nextRepeating = null;
            }
          }
        }
        nrPps = ppsA.toArray(new PhrasePositions[0]);
        pq = new PhraseQueue(nrPps.length);
      }
      
      //printPositions(System.err, "Init: 3: Aft check-repeats");
      
      // with repeats must advance some repeating pp's so they all start with differing tp's
      if (hasRepeats) {
        for (PhrasePositions pp: nrPps) {
          if ((end=advanceRepeats(pp, end)) == Integer.MIN_VALUE) {
            return Integer.MIN_VALUE; // ran out of a term -- done (no valid matches in current doc)
          }
        }
      }
      
      //printPositions(System.err, "Init: 4: Aft advance-repeats");
      
      // build queue from non repeating pps 
      pq.clear();
      for (PhrasePositions pp: nrPps) {
        if (pp.position > end) {
          end = pp.position;
        }
        pq.add(pp);
      }
      
      return end;
    }
    
    /** Actual position in doc of a PhrasePosition, relies on that position = tpPos - offset) */
    private final int tpPos(PhrasePositions pp) {
      return pp.position + pp.offset;
    }
    
  //  private void printPositions(PrintStream ps, String title) {
  //    ps.println();
  //    ps.println("---- "+title);
  //    int k = 0;
  //    if (nrPps!=null) {
  //      for (PhrasePositions pp: nrPps) {
  //        ps.println("  " + k++ + "  " + pp);
  //      }
  //    } else {
  //      for (PhrasePositions pp=min; 0==k || pp!=min; pp = pp.next) {  
  //        ps.println("  " + k++ + "  " + pp);
  //      }
  //    }
  //  }
  
  //  private void printQueue(PrintStream ps, PhrasePositions ext, String title) {
  //    ps.println();
  //    ps.println("---- "+title);
  //    ps.println("EXT: "+ext);
  //    PhrasePositions[] t = new PhrasePositions[pq.size()];
  //    if (pq.size()>0) {
  //      t[0] = pq.pop();
  //      ps.println("  " + 0 + "  " + t[0]);
  //      for (int i=1; i<t.length; i++) {
  //        t[i] = pq.pop();
  //        assert t[i-1].position <= t[i].position : "PQ is out of order: "+(i-1)+"::"+t[i-1]+" "+i+"::"+t[i];
  //        ps.println("  " + i + "  " + t[i]);
  //      }
  //      // add them back
  //      for (int i=t.length-1; i>=0; i--) {
  //        pq.add(t[i]);
  //      }
  //    }
  //  }
  }
  