K Nearest Neighbor问题的解决——KD-TREE Implementation

命题一：  
已知的1000个整数的数组，给定一个整数，要求查证是否在数组中出现？ 

命题二：  
已知1000个整数的数组，给定一个整数，要求查找数组中与之最接近的数字？ 

命题三：  
已知1000个Point（包含X与Y坐标）结构的数组，给定一个Point，要求查找数组中与之最接近（比如：欧氏距离最短）的点。 

命题四：  
已知1,000,000个向量，每个向量为128维；给定一个向量，要求查找数组中与之最接近的K个向量 

• 对于命题一，如果不考虑桶式、哈希等方式，常用的方法应该是排序后，使用折半查找。
• 对于命题二，与命题一类似，比较折半查找得出的结果，以及附近的各一个元素，即可。整个过程相当于是把这个包含1000个数组的数据结构做成一颗二叉树，最后只需比较叶子节点与其父节点即可。
• 对于命题三、四其中命题三和四就是所谓的Nearest Neighbor问题。一种近似解决的方法就是KD-TREE

高维向量的KNN检索问题，在图像等多媒体内容搜索中是相当关键的。关于高维向量的讨论，网上资料比较少；在此，我将一些心得分享给大家。 
与二叉树相比，KD-TREE也采用类似的划分方式，只不过树中的各节点均是高维向量，因此划分的方式，采用随机或指定的方式选取一个维度，在该指定维度上进行划分；整体的思想就是采用多个超平面对数据集空间进行两两切分，这一点，有点类似于数据挖掘中的决策树。 

一个运用KD-TREE分割二维平面的DEMO如下： 

 

KD-Tree build的代码如下： 

 private ClusterKDTree(Clusterable[] points, int height, boolean randomSplit){
     if ( points.length == 1 ){
         cluster = points[0];
     }
     else {
         splitIndex = chooseSplitDimension//选取切分维度
             (points[0].getLocation().length,height,randomSplit);
         splitValue = chooseSplit(points,splitIndex);//选取切分值

         Vector<Clusterable> left = new Vector<Clusterable>();
         Vector<Clusterable> right = new Vector<Clusterable>();
         for ( int i = 0; i < points.length; i++ ){
             double val = points[i].getLocation()[splitIndex];
             if ( val == splitValue && cluster == null ){
                 cluster = points[i];
             }
             else if ( val >= splitValue ){
                 right.add(points[i]);
             } else {
                 left.add(points[i]);
             }
         }

         if ( right.size() > 0 ){
             this.right = new ClusterKDTree(right.toArray(new
             Clusterable[right.size()]),
             randomSplit ? splitIndex : height+1, randomSplit);
         }
         if ( left.size() > 0 ){
             this.left = new ClusterKDTree(left.toArray(new
             Clusterable[left.size()]),randomSplit ? splitIndex : height+1,
             randomSplit);
         }
     }
 }

 private int chooseSplitDimension(int dimensionality,int height,boolean random){
     if ( !random ) return height % dimensionality;
     int rand = r.nextInt(dimensionality);
     while ( rand == height ){
         rand = r.nextInt(dimensionality);
     }
     return rand;
 }

 private double chooseSplit(Clusterable points[],int splitIdx){
     double[] values = new double[points.length];
     for ( int i = 0; i < points.length; i++ ){
     values[i] = points[i].getLocation()[splitIdx];
     }
     Arrays.sort(values);
     return values[values.length/2];//选取中间值以保持树的平衡
 }

构建完一颗KD-TREE之后，如何使用它来做KNN检索呢？我用下面的图来表示（20s的GIF动画）： 



使用KD-TREE，经过一次二分查找可以获得Query的KNN（最近邻）贪心解，代码如下： 

 private Clusterable restrictedNearestNeighbor(Clusterable point, SizedPriorityQueue<ClusterKDTree> values){
     if ( splitIndex == -1 ) {
         return cluster; //已近到叶子节点
     }

     double val = point.getLocation()[splitIndex];
     Clusterable closest = null;
     if ( val >= splitValue && right != null || left == null ){
         //沿右边路径遍历，并将左边子树放进队列
         if ( left != null ){
             double dist = val - splitValue;
             values.add(left,dist);
         }
         closest = right.restrictedNearestNeighbor(point,values);
     }
     else if ( val < splitValue && left != null || right == null ) {
         //沿左边路径遍历，并将右边子树放进队列
         if ( right != null ){
             double dist = splitValue - val;
             values.add(right,dist);
         }
         closest = left.restrictedNearestNeighbor(point,values);
     }
     //current distance of the 'ideal' node
     double currMinDistance = ClusterUtils.getEuclideanDistance(closest,point);
     //check to see if the current node we've backtracked to is closer
     double currClusterDistance = ClusterUtils.getEuclideanDistance(cluster,point);
     if ( closest == null || currMinDistance > currClusterDistance ){
         closest = cluster;
         currMinDistance = currClusterDistance;
     }
     return closest;
 }

事实上，仅仅一次的遍历会有不小的误差，因此采用了一个优先级队列来存放每次决定遍历走向时，另一方向的节点。SizedPriorityQueue代码的实现，可参考我的另一篇文章： 
http://grunt1223.iteye.com/blog/909739  

一种减少误差的方法（BBF：Best Bin First）是回溯一定数量的节点： 

 public Clusterable restrictedNearestNeighbor(Clusterable point, int numMaxBinsChecked){
     SizedPriorityQueue<ClusterKDTree> bins = new SizedPriorityQueue<ClusterKDTree>(50,true);
     Clusterable closest = restrictedNearestNeighbor(point,bins);
     double closestDist = ClusterUtils.getEuclideanDistance(point,closest);
     //System.out.println("retrieved point: " + closest + ", dist: " + closestDist);
     int count = 0;
     while ( count < numMaxBinsChecked && bins.size() > 0 ){
         ClusterKDTree nextBin = bins.pop();
     //System.out.println("Popping of next bin: " + nextBin);
     Clusterable possibleClosest = nextBin.restrictedNearestNeighbor(point,bins);
         double dist = ClusterUtils.getEuclideanDistance(point,possibleClosest);
         if ( dist < closestDist ){
         closest = possibleClosest;
         closestDist = dist;
     }
     count++;
     }
     return closest;
 }

可以用如下代码进行测试： 

 public static void main(String args[]){
     Clusterable clusters[] = new Clusterable[10];
     clusters[0] = new Point(0,0);
     clusters[1] = new Point(1,2);
     clusters[2] = new Point(2,3);
     clusters[3] = new Point(1,5);
     clusters[4] = new Point(2,5);
     clusters[5] = new Point(1,1);
     clusters[6] = new Point(3,3);
     clusters[7] = new Point(0,2);
     clusters[8] = new Point(4,4);
     clusters[9] = new Point(5,5);
     ClusterKDTree tree = new ClusterKDTree(clusters,true);
     //tree.print();
     Clusterable c = tree.restrictedNearestNeighbor(new Point(4,4),1000);
     System.out.println(c);
 }