/*
 * Copyright 2006 Google Inc.
 *
 * Licensed 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.
 */

package com.google.common.geometry;

import com.google.common.base.Preconditions;
import com.google.common.collect.Lists;
import com.google.common.collect.Sets;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public abstract strictfp class S2EdgeIndex {
  /**
   * Thicken the edge in all directions by roughly 1% of the edge length when
   * thickenEdge is true.
   */
  private static final double THICKENING = 0.01;

  /**
   * Threshold for small angles, that help lenientCrossing to determine whether
   * two edges are likely to intersect.
   */
  private static final double MAX_DET_ERROR = 1e-14;

  /**
   * The cell containing each edge, as given in the parallel array
   * <code>edges</code>.
   */
  private long[] cells;

  /**
   * The edge contained by each cell, as given in the parallel array
   * <code>cells</code>.
   */
  private int[] edges;

  /**
   * No cell strictly below this level appears in mapping. Initially leaf level,
   * that's the minimum level at which we will ever look for test edges.
   */
  private int minimumS2LevelUsed;

  /**
   * Has the index been computed already?
   */
  private boolean indexComputed;

  /**
   * Number of queries so far
   */
  private int queryCount;

  /**
   * Empties the index in case it already contained something.
   */
  public void reset() {
    minimumS2LevelUsed = S2CellId.MAX_LEVEL;
    indexComputed = false;
    queryCount = 0;
    cells = null;
    edges = null;
  }

  /**
   * Compares [cell1, edge1] to [cell2, edge2], by cell first and edge second.
   *
   * @return -1 if [cell1, edge1] is less than [cell2, edge2], 1 if [cell1,
   *         edge1] is greater than [cell2, edge2], 0 otherwise.
   */
  private static final int compare(long cell1, int edge1, long cell2, int edge2) {
    if (cell1 < cell2) {
      return -1;
    } else if (cell1 > cell2) {
      return 1;
    } else if (edge1 < edge2) {
      return -1;
    } else if (edge1 > edge2) {
      return 1;
    } else {
      return 0;
    }
  }

  /** Computes the index (if it has not been previously done). */
  public final void computeIndex() {
    if (indexComputed) {
      return;
    }
    List<Long> cellList = Lists.newArrayList();
    List<Integer> edgeList = Lists.newArrayList();
    for (int i = 0; i < getNumEdges(); ++i) {
      S2Point from = edgeFrom(i);
      S2Point to = edgeTo(i);
      ArrayList<S2CellId> cover = Lists.newArrayList();
      int level = getCovering(from, to, true, cover);
      minimumS2LevelUsed = Math.min(minimumS2LevelUsed, level);
      for (S2CellId cellId : cover) {
        cellList.add(cellId.id());
        edgeList.add(i);
      }
    }
    cells = new long[cellList.size()];
    edges = new int[edgeList.size()];
    for (int i = 0; i < cells.length; i++) {
      cells[i] = cellList.get(i);
      edges[i] = edgeList.get(i);
    }
    sortIndex();
    indexComputed = true;
  }

  /** Sorts the parallel <code>cells</code> and <code>edges</code> arrays. */
  private void sortIndex() {
    // create an array of indices and sort based on the values in the parallel
    // arrays at each index
    Integer[] indices = new Integer[cells.length];
    for (int i = 0; i < indices.length; i++) {
      indices[i] = i;
    }
    Arrays.sort(indices, new Comparator<Integer>() {
      @Override
      public int compare(Integer index1, Integer index2) {
        return S2EdgeIndex.compare(cells[index1], edges[index1], cells[index2], edges[index2]);
      }
    });
    // copy the cells and edges in the order given by the sorted list of indices
    long[] newCells = new long[cells.length];
    int[] newEdges = new int[edges.length];
    for (int i = 0; i < indices.length; i++) {
      newCells[i] = cells[indices[i]];
      newEdges[i] = edges[indices[i]];
    }
    // replace the cells and edges with the sorted arrays
    cells = newCells;
    edges = newEdges;
  }

  public final boolean isIndexComputed() {
    return indexComputed;
  }

  /**
   * Tell the index that we just received a new request for candidates. Useful
   * to compute when to switch to quad tree.
   */
  protected final void incrementQueryCount() {
    ++queryCount;
  }

  /**
   * If the index hasn't been computed yet, looks at how much work has gone into
   * iterating using the brute force method, and how much more work is planned
   * as defined by 'cost'. If it were to have been cheaper to use a quad tree
   * from the beginning, then compute it now. This guarantees that we will never
   * use more than twice the time we would have used had we known in advance
   * exactly how many edges we would have wanted to test. It is the theoretical
   * best.
   *
   *  The value 'n' is the number of iterators we expect to request from this
   * edge index.
   *
   *  If we have m data edges and n query edges, then the brute force cost is m
   * * n * testCost where testCost is taken to be the cost of
   * EdgeCrosser.robustCrossing, measured to be about 30ns at the time of this
   * writing.
   *
   *  If we compute the index, the cost becomes: m * costInsert + n *
   * costFind(m)
   *
   *  - costInsert can be expected to be reasonably stable, and was measured at
   * 1200ns with the BM_QuadEdgeInsertionCost benchmark.
   *
   *  - costFind depends on the length of the edge . For m=1000 edges, we got
   * timings ranging from 1ms (edge the length of the polygon) to 40ms. The
   * latter is for very long query edges, and needs to be optimized. We will
   * assume for the rest of the discussion that costFind is roughly 3ms.
   *
   *  When doing one additional query, the differential cost is m * testCost -
   * costFind(m) With the numbers above, it is better to use the quad tree (if
   * we have it) if m >= 100.
   *
   *  If m = 100, 30 queries will give m*n*testCost = m * costInsert = 100ms,
   * while the marginal cost to find is 3ms. Thus, this is a reasonable thing to
   * do.
   */
  public final void predictAdditionalCalls(int n) {
    if (indexComputed) {
      return;
    }
    if (getNumEdges() > 100 && (queryCount + n) > 30) {
      computeIndex();
    }
  }

  /**
   * Overwrite these functions to give access to the underlying data. The
   * function getNumEdges() returns the number of edges in the index, while
   * edgeFrom(index) and edgeTo(index) return the "from" and "to" endpoints of
   * the edge at the given index.
   */
  protected abstract int getNumEdges();

  protected abstract S2Point edgeFrom(int index);

  protected abstract S2Point edgeTo(int index);

  /**
   * Appends to "candidateCrossings" all edge references which may cross the
   * given edge. This is done by covering the edge and then finding all
   * references of edges whose coverings overlap this covering. Parent cells are
   * checked level by level. Child cells are checked all at once by taking
   * advantage of the natural ordering of S2CellIds.
   */
  protected void findCandidateCrossings(S2Point a, S2Point b, List<Integer> candidateCrossings) {
    Preconditions.checkState(indexComputed);
    ArrayList<S2CellId> cover = Lists.newArrayList();
    getCovering(a, b, false, cover);

    // Edge references are inserted into the map once for each covering cell, so
    // absorb duplicates here
    Set<Integer> uniqueSet = new HashSet<Integer>();
    getEdgesInParentCells(cover, uniqueSet);

    // TODO(user): An important optimization for long query
    // edges (Contains queries): keep a bounding cap and clip the query
    // edge to the cap before starting the descent.
    getEdgesInChildrenCells(a, b, cover, uniqueSet);

    candidateCrossings.clear();
    candidateCrossings.addAll(uniqueSet);
  }

  /**
   * Returns the smallest cell containing all four points, or
   * {@link S2CellId#sentinel()} if they are not all on the same face. The
   * points don't need to be normalized.
   */
  private static S2CellId containingCell(S2Point pa, S2Point pb, S2Point pc, S2Point pd) {
    S2CellId a = S2CellId.fromPoint(pa);
    S2CellId b = S2CellId.fromPoint(pb);
    S2CellId c = S2CellId.fromPoint(pc);
    S2CellId d = S2CellId.fromPoint(pd);

    if (a.face() != b.face() || a.face() != c.face() || a.face() != d.face()) {
      return S2CellId.sentinel();
    }

    while (!a.equals(b) || !a.equals(c) || !a.equals(d)) {
      a = a.parent();
      b = b.parent();
      c = c.parent();
      d = d.parent();
    }
    return a;
  }

  /**
   * Returns the smallest cell containing both points, or Sentinel if they are
   * not all on the same face. The points don't need to be normalized.
   */
  private static S2CellId containingCell(S2Point pa, S2Point pb) {
    S2CellId a = S2CellId.fromPoint(pa);
    S2CellId b = S2CellId.fromPoint(pb);

    if (a.face() != b.face()) {
      return S2CellId.sentinel();
    }

    while (!a.equals(b)) {
      a = a.parent();
      b = b.parent();
    }
    return a;
  }

  /**
   * Computes a cell covering of an edge. Clears edgeCovering and returns the
   * level of the s2 cells used in the covering (only one level is ever used for
   * each call).
   *
   *  If thickenEdge is true, the edge is thickened and extended by 1% of its
   * length.
   *
   *  It is guaranteed that no child of a covering cell will fully contain the
   * covered edge.
   */
  private int getCovering(
      S2Point a, S2Point b, boolean thickenEdge, ArrayList<S2CellId> edgeCovering) {
    edgeCovering.clear();

    // Selects the ideal s2 level at which to cover the edge, this will be the
    // level whose S2 cells have a width roughly commensurate to the length of
    // the edge. We multiply the edge length by 2*THICKENING to guarantee the
    // thickening is honored (it's not a big deal if we honor it when we don't
    // request it) when doing the covering-by-cap trick.
    double edgeLength = a.angle(b);
    int idealLevel = S2Projections.MIN_WIDTH.getMaxLevel(edgeLength * (1 + 2 * THICKENING));

    S2CellId containingCellId;
    if (!thickenEdge) {
      containingCellId = containingCell(a, b);
    } else {
      if (idealLevel == S2CellId.MAX_LEVEL) {
        // If the edge is tiny, instabilities are more likely, so we
        // want to limit the number of operations.
        // We pretend we are in a cell much larger so as to trigger the
        // 'needs covering' case, so we won't try to thicken the edge.
        containingCellId = (new S2CellId(0xFFF0)).parent(3);
      } else {
        S2Point pq = S2Point.mul(S2Point.minus(b, a), THICKENING);
        S2Point ortho =
            S2Point.mul(S2Point.normalize(S2Point.crossProd(pq, a)), edgeLength * THICKENING);
        S2Point p = S2Point.minus(a, pq);
        S2Point q = S2Point.add(b, pq);
        // If p and q were antipodal, the edge wouldn't be lengthened,
        // and it could even flip! This is not a problem because
        // idealLevel != 0 here. The farther p and q can be is roughly
        // a quarter Earth away from each other, so we remain
        // Theta(THICKENING).
        containingCellId =
            containingCell(S2Point.minus(p, ortho), S2Point.add(p, ortho), S2Point.minus(q, ortho),
                S2Point.add(q, ortho));
      }
    }

    // Best case: edge is fully contained in a cell that's not too big.
    if (!containingCellId.equals(S2CellId.sentinel())
        && containingCellId.level() >= idealLevel - 2) {
      edgeCovering.add(containingCellId);
      return containingCellId.level();
    }

    if (idealLevel == 0) {
      // Edge is very long, maybe even longer than a face width, so the
      // trick below doesn't work. For now, we will add the whole S2 sphere.
      // TODO(user): Do something a tad smarter (and beware of the
      // antipodal case).
      for (S2CellId cellid = S2CellId.begin(0); !cellid.equals(S2CellId.end(0));
          cellid = cellid.next()) {
        edgeCovering.add(cellid);
      }
      return 0;
    }
    // TODO(user): Check trick below works even when vertex is at
    // interface
    // between three faces.

    // Use trick as in S2PolygonBuilder.PointIndex.findNearbyPoint:
    // Cover the edge by a cap centered at the edge midpoint, then cover
    // the cap by four big-enough cells around the cell vertex closest to the
    // cap center.
    S2Point middle = S2Point.normalize(S2Point.div(S2Point.add(a, b), 2));
    int actualLevel = Math.min(idealLevel, S2CellId.MAX_LEVEL - 1);
    S2CellId.fromPoint(middle).getVertexNeighbors(actualLevel, edgeCovering);
    return actualLevel;
  }

  /**
   * Filters a list of entries down to the inclusive range defined by the given
   * cells, in <code>O(log N)</code> time.
   *
   * @param cell1 One side of the inclusive query range.
   * @param cell2 The other side of the inclusive query range.
   * @return An array of length 2, containing the start/end indices.
   */
  private int[] getEdges(long cell1, long cell2) {
    // ensure cell1 <= cell2
    if (cell1 > cell2) {
      long temp = cell1;
      cell1 = cell2;
      cell2 = temp;
    }
    // The binary search returns -N-1 to indicate an insertion point at index N,
    // if an exact match cannot be found. Since the edge indices queried for are
    // not valid edge indices, we will always get -N-1, so we immediately
    // convert to N.
    return new int[]{
        -1 - binarySearch(cell1, Integer.MIN_VALUE),
        -1 - binarySearch(cell2, Integer.MAX_VALUE)};
  }

  private int binarySearch(long cell, int edge) {
    int low = 0;
    int high = cells.length - 1;
    while (low <= high) {
      int mid = (low + high) >>> 1;
      int cmp = compare(cells[mid], edges[mid], cell, edge);
      if (cmp < 0) {
        low = mid + 1;
      } else if (cmp > 0) {
        high = mid - 1;
      } else {
        return mid;
      }
    }
    return -(low + 1);
  }

  /**
   * Adds to candidateCrossings all the edges present in any ancestor of any
   * cell of cover, down to minimumS2LevelUsed. The cell->edge map is in the
   * variable mapping.
   */
  private void getEdgesInParentCells(List<S2CellId> cover, Set<Integer> candidateCrossings) {
    // Find all parent cells of covering cells.
    Set<S2CellId> parentCells = Sets.newHashSet();
    for (S2CellId coverCell : cover) {
      for (int parentLevel = coverCell.level() - 1; parentLevel >= minimumS2LevelUsed;
          --parentLevel) {
        if (!parentCells.add(coverCell.parent(parentLevel))) {
          break; // cell is already in => parents are too.
        }
      }
    }

    // Put parent cell edge references into result.
    for (S2CellId parentCell : parentCells) {
      int[] bounds = getEdges(parentCell.id(), parentCell.id());
      for (int i = bounds[0]; i < bounds[1]; i++) {
        candidateCrossings.add(edges[i]);
      }
    }
  }

  /**
   * Returns true if ab possibly crosses cd, by clipping tiny angles to zero.
   */
  private static boolean lenientCrossing(S2Point a, S2Point b, S2Point c, S2Point d) {
    // assert (S2.isUnitLength(a));
    // assert (S2.isUnitLength(b));
    // assert (S2.isUnitLength(c));

    double acb = S2Point.crossProd(a, c).dotProd(b);
    double bda = S2Point.crossProd(b, d).dotProd(a);
    if (Math.abs(acb) < MAX_DET_ERROR || Math.abs(bda) < MAX_DET_ERROR) {
      return true;
    }
    if (acb * bda < 0) {
      return false;
    }
    double cbd = S2Point.crossProd(c, b).dotProd(d);
    double dac = S2Point.crossProd(c, a).dotProd(c);
    if (Math.abs(cbd) < MAX_DET_ERROR || Math.abs(dac) < MAX_DET_ERROR) {
      return true;
    }
    return (acb * cbd >= 0) && (acb * dac >= 0);
  }

  /**
   * Returns true if the edge and the cell (including boundary) intersect.
   */
  private static boolean edgeIntersectsCellBoundary(S2Point a, S2Point b, S2Cell cell) {
    S2Point[] vertices = new S2Point[4];
    for (int i = 0; i < 4; ++i) {
      vertices[i] = cell.getVertex(i);
    }
    for (int i = 0; i < 4; ++i) {
      S2Point fromPoint = vertices[i];
      S2Point toPoint = vertices[(i + 1) % 4];
      if (lenientCrossing(a, b, fromPoint, toPoint)) {
        return true;
      }
    }
    return false;
  }

  /**
   * Appends to candidateCrossings the edges that are fully contained in an S2
   * covering of edge. The covering of edge used is initially cover, but is
   * refined to eliminate quickly subcells that contain many edges but do not
   * intersect with edge.
   */
  private void getEdgesInChildrenCells(S2Point a, S2Point b, List<S2CellId> cover,
      Set<Integer> candidateCrossings) {
    // Put all edge references of (covering cells + descendant cells) into
    // result.
    // This relies on the natural ordering of S2CellIds.
    S2Cell[] children = null;
    while (!cover.isEmpty()) {
      S2CellId cell = cover.remove(cover.size() - 1);
      int[] bounds = getEdges(cell.rangeMin().id(), cell.rangeMax().id());
      if (bounds[1] - bounds[0] <= 16) {
        for (int i = bounds[0]; i < bounds[1]; i++) {
          candidateCrossings.add(edges[i]);
        }
      } else {
        // Add cells at this level
        bounds = getEdges(cell.id(), cell.id());
        for (int i = bounds[0]; i < bounds[1]; i++) {
          candidateCrossings.add(edges[i]);
        }
        // Recurse on the children -- hopefully some will be empty.
        if (children == null) {
          children = new S2Cell[4];
          for (int i = 0; i < 4; ++i) {
            children[i] = new S2Cell();
          }
        }
        new S2Cell(cell).subdivide(children);
        for (S2Cell child : children) {
          // TODO(user): Do the check for the four cells at once,
          // as it is enough to check the four edges between the cells. At
          // this time, we are checking 16 edges, 4 times too many.
          //
          // Note that given the guarantee of AppendCovering, it is enough
          // to check that the edge intersect with the cell boundary as it
          // cannot be fully contained in a cell.
          if (edgeIntersectsCellBoundary(a, b, child)) {
            cover.add(child.id());
          }
        }
      }
    }
  }

  /*
   * An iterator on data edges that may cross a query edge (a,b). Create the
   * iterator, call getCandidates(), then hasNext()/next() repeatedly.
   *
   * The current edge in the iteration has index index(), goes between from()
   * and to().
   */
  public static class DataEdgeIterator {
    /**
     * The structure containing the data edges.
     */
    private final S2EdgeIndex edgeIndex;

    /**
     * Tells whether getCandidates() obtained the candidates through brute force
     * iteration or using the quad tree structure.
     */
    private boolean isBruteForce;

    /**
     * Index of the current edge and of the edge before the last next() call.
     */
    private int currentIndex;

    /**
     * Cache of edgeIndex.getNumEdges() so that hasNext() doesn't make an extra
     * call
     */
    private int numEdges;

    /**
     * All the candidates obtained by getCandidates() when we are using a
     * quad-tree (i.e. isBruteForce = false).
     */
    ArrayList<Integer> candidates;

    /**
     * Index within array above. We have: currentIndex =
     * candidates.get(currentIndexInCandidates).
     */
    private int currentIndexInCandidates;

    public DataEdgeIterator(S2EdgeIndex edgeIndex) {
      this.edgeIndex = edgeIndex;
      candidates = Lists.newArrayList();
    }

    /**
     * Initializes the iterator to iterate over a set of candidates that may
     * cross the edge (a,b).
     */
    public void getCandidates(S2Point a, S2Point b) {
      edgeIndex.predictAdditionalCalls(1);
      isBruteForce = !edgeIndex.isIndexComputed();
      if (isBruteForce) {
        edgeIndex.incrementQueryCount();
        currentIndex = 0;
        numEdges = edgeIndex.getNumEdges();
      } else {
        candidates.clear();
        edgeIndex.findCandidateCrossings(a, b, candidates);
        currentIndexInCandidates = 0;
        if (!candidates.isEmpty()) {
          currentIndex = candidates.get(0);
        }
      }
    }

    /**
     * Index of the current edge in the iteration.
     */
    public int index() {
      Preconditions.checkState(hasNext());
      return currentIndex;
    }

    /**
     * False if there are no more candidates; true otherwise.
     */
    public boolean hasNext() {
      if (isBruteForce) {
        return (currentIndex < numEdges);
      } else {
        return currentIndexInCandidates < candidates.size();
      }
    }

    /**
     * Iterate to the next available candidate.
     */
    public void next() {
      Preconditions.checkState(hasNext());
      if (isBruteForce) {
        ++currentIndex;
      } else {
        ++currentIndexInCandidates;
        if (currentIndexInCandidates < candidates.size()) {
          currentIndex = candidates.get(currentIndexInCandidates);
        }
      }
    }
  }
}