精确去重和Roaring BitMap （咆哮位图）

基本概念

Roaring BitMap 以下简称 RBM,中文翻译为咆哮位图,它本质上是定义了一个很大的 bit 数组，每个元素对应到 bit 数组的其中一位，一个Integer是32-bit, 一共有Integer.MAX_VALUE = 2 ^ 32个值，32-bit的unsigned integer的集合(共2 ^ 32 = 42 9496 7296个)
这个数足够覆盖一款产品的user数或item数(item 泛指是新闻，商品等)
由定义可知，其去重是针对int 型数据进行操作，对于非Integer类型的数据(比如String类型)可以通过数据字典映射成Integer,比如数据库的ID

基本位图实现存在问题

• 即使只存一个数(最大的),存储空间也是512MB
  对于原始的Bitmap来说，这就需要2 ^ 32长度的bit数组
  通过计算可以发现(2 ^ 32 / 8 bytes = 512MB), 一个普通的Bitmap需要耗费512MB的存储空间
  不管业务值的基数有多大，这个存储空间的消耗都是恒定不变的，这显然是不能接受的。
  redis 的基本的位图实现就存在这个问题
  REDIS BITMAP 测试

---

localhost:0>flushdb
localhost:0>info
# Memory
used_memory:690288
used_memory_human:674.11K
used_memory_rss:652376
used_memory_rss_human:637.09K #表示该进程所占物理内存的大小
used_memory_peak:667584400
used_memory_peak_human:637.09K #是过去Redis内存使用的峰值
localhost:0> setbit a 4294967295 1 #将最大位置1，此时基数为1
# Memory
used_memory:541755832
used_memory_human:516.66M #set 一个值，直接占512m内存
used_memory_rss:541717944
used_memory_rss_human:516.62M
used_memory_peak:667584400
used_memory_peak_human:636.66M

附redis raoring-bitmap 实现 https://github.com/aviggiano/redis-roaring

Roaring Bitmap实现

• RBM实现源理
  将32位无符号整数按照高16位分桶，即最多可能有2¹⁶=65536个桶，论文内称为container。存储数据时，按照数据的高16位找到container（找不到就会新建一个），再将低16位放入container中。也就是说，一个RBM就是很多container的集合。
  RoaringBitmap 构造源码

 RoaringArray highLowContainer = null;

  /**
   * Create an empty bitmap
   */
  public RoaringBitmap() {
    highLowContainer = new RoaringArray();
  }

RoaringArray

static final int INITIAL_CAPACITY = 4;
  short[] keys = null;//高16位数组 有序数组，方便二分查找

  Container[] values = null;//低16位容器数组

  int size = 0;

  protected RoaringArray() {
    this(INITIAL_CAPACITY);
  }

• 低16位容器数组存储示意图

  
  低16位容器数组存储示意图
• 前1000个62的倍数
• 高16位为1，低16位为0到99 0x00010000-0x00010063
• 高16位为2, 低16位的所有偶数

Container 实现

Container只需要处理低16位的数据。

• ArrayContainer

public final class ArrayContainer extends Container implements Cloneable {
  private static final int DEFAULT_INIT_SIZE = 4;
  private static final int ARRAY_LAZY_LOWERBOUND = 1024;
  static final int DEFAULT_MAX_SIZE = 4096;// containers with DEFAULT_MAX_SZE or less integers
                                           // should be ArrayContainers
  private static final long serialVersionUID = 1L;
  protected int cardinality = 0;
  short[] content;
  /**
   * Create an array container with default capacity
   */
  public ArrayContainer() {
    this(DEFAULT_INIT_SIZE);
  }

short[] content，将16位value直接存储。
short[] content始终保持有序且不重，方便使用二分查找。
根据源码可以看出，常量DEFAULT_MAX_SIZE值为4096，当容量超过这个值的时候会将当前Container替换为BitmapContainer。

/**
   * running time is in O(n) time if insert is not in order.
   */
  @Override
  public Container add(final short x) {
    //基数为0或当前值大于容器中的最大值（因为有序，最后一个即最大值）
    if (cardinality == 0 || (cardinality > 0
            && toIntUnsigned(x) > toIntUnsigned(content[cardinality - 1]))) {
//基数>=4096则转为BitmapContainer
      if (cardinality >= DEFAULT_MAX_SIZE) {
        return toBitmapContainer().add(x);
      }
//如果容器空间不路，则扩容[见下方扩容源代码]
      if (cardinality >= this.content.length) {
        increaseCapacity();
      }
      content[cardinality++] = x;
    } else {
//如果当前值在容器范围内，则找到该值对应的位置[见下二分查找源码]
      int loc = Util.unsignedBinarySearch(content, 0, cardinality, x);
//未找到
      if (loc < 0) {
        // Transform the ArrayContainer to a BitmapContainer
        // when cardinality = DEFAULT_MAX_SIZE
//虽然在容器范围内也可能会涉及到容器升级和扩容
        if (cardinality >= DEFAULT_MAX_SIZE) {
          return toBitmapContainer().add(x);
        }
        if (cardinality >= this.content.length) {
          increaseCapacity();
        }
        // insertion : shift the elements > x by one position to
        // the right
        // and put x in it's appropriate place
        System.arraycopy(content, -loc - 1, content, -loc, cardinality + loc + 1);
        content[-loc - 1] = x;
        ++cardinality;
      }
    }
    return this;
  }

数组扩容逻辑

private void increaseCapacity(boolean allowIllegalSize) {
    int newCapacity = (this.content.length == 0) ? DEFAULT_INIT_SIZE
        : this.content.length < 64 ? this.content.length * 2
            : this.content.length < 1067 ? this.content.length * 3 / 2
                : this.content.length * 5 / 4;
    // never allocate more than we will ever need
    if (newCapacity > ArrayContainer.DEFAULT_MAX_SIZE && !allowIllegalSize) {
      newCapacity = ArrayContainer.DEFAULT_MAX_SIZE;
    }
    // if we are within 1/16th of the max, go to max
    if (newCapacity > ArrayContainer.DEFAULT_MAX_SIZE - ArrayContainer.DEFAULT_MAX_SIZE / 16
        && !allowIllegalSize) {
      newCapacity = ArrayContainer.DEFAULT_MAX_SIZE;
    }
    this.content = Arrays.copyOf(this.content, newCapacity);
  }

数组的二分查找(找到则返回value的位置，未找到则返回当前值将要insert 的位置，为与找到值的位置区分，这里返回负值)

/**
   * Look for value k in array in the range [begin,end). If the value is found, return its index. If
   * not, return -(i+1) where i is the index where the value would be inserted. The array is assumed
   * to contain sorted values where shorts are interpreted as unsigned integers.
   *
   * @param array array where we search
   * @param begin first index (inclusive)
   * @param end last index (exclusive)
   * @param k value we search for
   * @return count
   */
  public static int unsignedBinarySearch(final short[] array, final int begin, final int end,
      final short k) {
    if (USE_HYBRID_BINSEARCH) {
      return hybridUnsignedBinarySearch(array, begin, end, k);
    } else {
      return branchyUnsignedBinarySearch(array, begin, end, k);
    }
  }

为什么是4096的时侯升级容器?

bitmap vs array

1. bitmap存储空间恒定为8K，最大的基数可达到8*1024*8=65536个
2. array的基数与存储空间成正比，即基数越大，占用空占越多
   通过以上两点我们取两者交相交的点为平衡点，即小于该点array更省空间，大于该点bitmap更省空间。

上图中的前两个container基数都没超过4096，所以均为ArrayContainer。

• BitmapContainer

/**
 * Simple bitset-like container.
 */
public final class BitmapContainer extends Container implements Cloneable {
  protected static final int MAX_CAPACITY = 1 << 16;

//保存低16位，所以最大值为2¹⁶

  // the parameter is for overloading and symmetry with ArrayContainer
  protected static int serializedSizeInBytes(int unusedCardinality) {
    return MAX_CAPACITY / 8;
  }
  final long[] bitmap;
  int cardinality;
  // nruns value for which RunContainer.serializedSizeInBytes ==
  // BitmapContainer.getArraySizeInBytes()
  private final int MAXRUNS = (getArraySizeInBytes() - 2) / 4;

  /**
   * Create a bitmap container with all bits set to false
*构造最大的long 值数组，每个long 64位
   */
  public BitmapContainer() {
    this.cardinality = 0;
    this.bitmap = new long[MAX_CAPACITY / 64];//1024个long
  }

这种Container使用long[]存储位图数据。我们知道，每个Container处理16位整形的数据，也就是0~65535，因此根据位图的原理，需要65536个比特来存储数据，每个比特位用1来表示有，0来表示无。每个long有64位，因此需要1024个long来提供65536个bit。
因此，每个BitmapContainer在构建时就会初始化长度为1024的long[]。这就意味着，不管一个BitmapContainer中只存储了1个数据还是存储了65536个数据，占用的空间都是同样的8kb。
上图中的第三个container基数远远大于4096，所以要用BitmapContainer存储。
bit map container add方法源码分析

@Override
  public Container add(final short i) {
    final int x = Util.toIntUnsigned(i);
//当前值/64取整找到long数组的索引
    final long previous = bitmap[x / 64];

//找到当前值所在long 的第几位,并将该位置1，1L<<x 等价于1x<<(x%64)
关于位移操作详情官方解释
https://docs.oracle.com/javase/specs/jls/se10/html/jls-15.html#jls-15.19

The operators << (left shift), >> (signed right shift), and >>> (unsigned right shift) are called the shift operators. The left-hand operand of a shift operator is the value to be shifted; the right-hand operand specifies the shift distance.
The type of the shift expression is the promoted type of the left-hand operand.

If the promoted type of the left-hand operand isint, then only the five lowest-order bits of the right-hand operand are used as the shift distance.（int 只有低5位有效）It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

If the promoted type of the left-hand operand islong, then only the six lowest-order bits of the right-hand operand are used as the shift distance.(long 只有低6位有效)It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive.

    long newval = previous | (1L << x);
//赋新值
    bitmap[x / 64] = newval;
    if (USE_BRANCHLESS) {
      cardinality += (previous ^ newval) >>> x;
    } else if (previous != newval) {
      ++cardinality;
    }
    return this;
  }

• RunContainer
  该容器由RLE实现

/**
 * This container takes the form of runs of consecutive values (effectively, run-length encoding).
 *
 * Adding and removing content from this container might make it wasteful so regular calls to
 * "runOptimize" might be warranted.
 */
public final class RunContainer extends Container implements Cloneable {
  private static final int DEFAULT_INIT_SIZE = 4;
  private static final boolean ENABLE_GALLOPING_AND = false;
  private short[] valueslength;//主要存储结构 we interleave values and lengths, so
  // that if you have the values 11,12,13,14,15, you store that as 11,4 where 4 means that beyond 11
  // itself, there are
  // 4 contiguous values that follows.
  // Other example: e.g., 1, 10, 20,0, 31,2 would be a concise representation of 1, 2, ..., 11, 20,
  // 31, 32, 33
  int nbrruns = 0;// how many runs, this number should fit in 16 bits.

• Run-Length Encoding(RLE)
  Run-length encoding是被许多bitmap文件格式支持的数据压缩算法
  RLE工作方式是减少重复字符的物理尺寸，被编码成两个字节，第一个字节表示字符数量，第二个字节表示本身字符值。
  字符串包含4个不同字符
  例如：

AAAAAAbbbXXXXXt

使用RLE，可以形成4个2字节包

6A3b5X1t

当手动执行runOptimize 方法时会触发优化

/**
   * Use a run-length encoding where it is more space efficient
   *
   * @return whether a change was applied
   */
  public boolean runOptimize() {
    boolean answer = false;
    for (int i = 0; i < this.highLowContainer.size(); i++) {
      Container c = this.highLowContainer.getContainerAtIndex(i).runOptimize();
      if (c instanceof RunContainer) {
        answer = true;
      }
      this.highLowContainer.setContainerAtIndex(i, c);
    }
    return answer;
  }

ArrayContainer 优化逻辑

@Override
  public Container runOptimize() {
    // TODO: consider borrowing the BitmapContainer idea of early
    // abandonment
    // with ArrayContainers, when the number of runs in the arrayContainer
    // passes some threshold based on the cardinality.
    int numRuns = numberOfRuns();
    int sizeAsRunContainer = RunContainer.serializedSizeInBytes(numRuns);
//如果RunContainer 比当前的容器省空间，则升级为 RunContainer。BitMap 同理
    if (getArraySizeInBytes() > sizeAsRunContainer) {
      return new RunContainer(this, numRuns); // this could be maybe
                                              // faster if initial
                                              // container is a bitmap
    } else {
      return this;
    }
  }

numberOfRuns 计算run 的个数

@Override
  int numberOfRuns() {
    if (cardinality == 0) {
      return 0; // should never happen
    }
    int numRuns = 1;
    int oldv = toIntUnsigned(content[0]);
    for (int i = 1; i < cardinality; i++) {
      int newv = toIntUnsigned(content[i]);
      if (oldv + 1 != newv) {
        ++numRuns;
      }
      oldv = newv;
    }
    return numRuns;
  }

计算RunContainer 可能占用的容量

protected static int serializedSizeInBytes(int numberOfRuns) {
    return 2 + 2 * 2 * numberOfRuns; // each run requires 2 2-byte entries.
//值和长度成对出现，分别两个字节
  }

容器转换总结

• ArrayContainer：
  如果插入值后容量超过4096，则自动转换为BitmapContainer。因此正常使用的情况下不会出现容量超过4096的ArrayContainer。
  调用runOptimize()方法时，会比较和RunContainer的空间占用大小，选择是否转换为RunContainer。
• BitmapContainer：
  如果删除某值后容量低至4096，则会自动转换为ArrayContainer。因此正常使用的情况下不会出现容量小于4096的BitmapContainer。
  调用runOptimize()方法时，会比较和RunContainer的空间占用大小，选择是否转换为RunContainer。
• RunContainer：
  只有在调用runOptimize()方法才会发生转换，会分别和ArrayContainer、BitmapContainer比较空间占用大小，然后选择是否转换。

参考

《Better bitmap performance with Roaring bitmaps》
《Consistently faster and smaller compressed bitmaps with Roaring》
The Java® Language Specification
https://www.jianshu.com/p/818ac4e90daf
https://blog.csdn.net/luanpeng825485697/article/details/101110798