Caffeine应用w-TinyLFU算法实现数据频次的记录，主要解决两个问题：
- 尽量降低数据的占用空间：将每个key的最大命中次数上限设为15，用4个bit表示。每个long为64位，可分为16个4bit，这样每个long型整数可以保存16个。
- 尽量保持数据相对"新鲜"：达到一定条件时，就将所有key的命中频次减半（降频）。
做法：
-
用一个Long型一维数组保存所有数据，每个数组元素包括16个分段（下面将每个数组元素称为一个slot，每个4bit分段称为一个counter）
-
使用hash算法将每个key的频次映射到某个Counter中。并借鉴BloomFilter思想，用多种hash算法降低hash重复带来的误差。
-
每个key计算一个hash值，再用hash值计算出4种index，表示在哪些slot中；用hash值计算出在16个counter中的位置，每个slot中的counter下标相差1位，counter下标从右到左为0-16。
-
每次命中后将4个counter值加1，最多15。
-
由于每个元素的hash值都可能跟其他key相同，所以每次频次加1时，可能将其他key的频次加1。所以取这4个key的最小值最为该key的频次
image.png
/**
* A probabilistic multiset for estimating the popularity of an element within a time window. The
* maximum frequency of an element is limited to 15 (4-bits) and an aging process periodically
* halves the popularity of all elements.
*
* @author ben.manes@gmail.com (Ben Manes)
*/
final class FrequencySketch<E> {
/*
* This class maintains a 4-bit CountMinSketch [1] with periodic aging to provide the popularity
* history for the TinyLfu admission policy [2]. The time and space efficiency of the sketch
* allows it to cheaply estimate the frequency of an entry in a stream of cache access events.
*
* The counter matrix is represented as a single dimensional array holding 16 counters per slot. A
* fixed depth of four balances the accuracy and cost, resulting in a width of four times the
* length of the array. To retain an accurate estimation the array's length equals the maximum
* number of entries in the cache, increased to the closest power-of-two to exploit more efficient
* bit masking. This configuration results in a confidence of 93.75% and error bound of e / width.
*
* The frequency of all entries is aged periodically using a sampling window based on the maximum
* number of entries in the cache. This is referred to as the reset operation by TinyLfu and keeps
* the sketch fresh by dividing all counters by two and subtracting based on the number of odd
* counters found. The O(n) cost of aging is amortized, ideal for hardware prefetching, and uses
* inexpensive bit manipulations per array location.
*
* [1] An Improved Data Stream Summary: The Count-Min Sketch and its Applications
* http://dimacs.rutgers.edu/~graham/pubs/papers/cm-full.pdf
* [2] TinyLFU: A Highly Efficient Cache Admission Policy
* https://dl.acm.org/citation.cfm?id=3149371
*/
//hash种子，用来计算4个不同的hash值
static final long[] SEED = { // A mixture of seeds from FNV-1a, CityHash, and Murmur3
0xc3a5c85c97cb3127L, 0xb492b66fbe98f273L, 0x9ae16a3b2f90404fL, 0xcbf29ce484222325L};
static final long RESET_MASK = 0x7777777777777777L; //二进制 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111 0111
static final long ONE_MASK = 0x1111111111111111L; //二进制 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001
//size累加到sampleSize时，执行减半操作
int sampleSize;
//hash&tableMask相当于取模，得到slot的index
int tableMask;
//记录frequency的一维数组
long[] table;
//所有counter之和
int size;
/**
* Creates a lazily initialized frequency sketch, requiring {@link #ensureCapacity} be called
* when the maximum size of the cache has been determined.
*/
@SuppressWarnings("NullAway.Init")
public FrequencySketch() {}
/**
* Initializes and increases the capacity of this <tt>FrequencySketch</tt> instance, if necessary,
* to ensure that it can accurately estimate the popularity of elements given the maximum size of
* the cache. This operation forgets all previous counts when resizing.
* 初始化table，若maximumSize=0，table长度为1（无界缓存，所有数据都被记录下来，也就无需记录每个数据的频次了）
* 否则table长度为大于等于maximumSize的最小的2的整数倍。maximumSize大于0时，sampleSize=10倍数组长度。
* 数组长度最大为Integer.MAX_VALUE的二分之一，也就是2147483647/2=1073741823，大概10.7亿
* tableMask=数组长度减1
*
* 比如：
* maximumSize=0，table长度为1，sampleSize=Integer.MAX_VALUE
* maximumSize=10，table长度为16，sampleSize=160
* maximumSize=100，table长度为128，sampleSize=1280
*
* @param maximumSize the maximum size of the cache
*/
public void ensureCapacity(@NonNegative long maximumSize) {
requireArgument(maximumSize >= 0);
int maximum = (int) Math.min(maximumSize, Integer.MAX_VALUE >>> 1);
if ((table != null) && (table.length >= maximum)) {
return;
}
table = new long[(maximum == 0) ? 1 : Caffeine.ceilingPowerOfTwo(maximum)];
tableMask = Math.max(0, table.length - 1);
sampleSize = (maximumSize == 0) ? 10 : (10 * maximum);
if (sampleSize <= 0) {
sampleSize = Integer.MAX_VALUE;
}
size = 0;
}
/**
* Returns if the sketch has not yet been initialized, requiring that {@link #ensureCapacity} is
* called before it begins to track frequencies.
*/
public boolean isNotInitialized() {
return (table == null);
}
/**
* Returns the estimated number of occurrences of an element, up to the maximum (15).
*
* 获取元素的频次，由于每个元素都有4个hash算法，在4个位置记录了4个频次，取其中最小的频次作为该元素的频次
* 获取元素的counter下标和slot下标，再取对应的4bit的数据
*
* @param e the element to count occurrences of
* @return the estimated number of occurrences of the element; possibly zero but never negative
*/
@NonNegative
public int frequency(@NonNull E e) {
if (isNotInitialized()) {
return 0;
}
int hash = spread(e.hashCode());
int start = (hash & 3) << 2; //start为counter下标，这个算法start只可能为0 4 8 12的其中一种
int frequency = Integer.MAX_VALUE;
for (int i = 0; i < 4; i++) {
int index = indexOf(hash, i); //index为每个slot下标
//table[index]为当前slot的long整数。
//假设start为4，二进制为0100，table[index]的二进制为0100 1010... 1101 0011 1000 0011 1001
//(start + i) << 2 左移两位0001 0000，也就是十进制16，其实就是从右数，下标为4的counter的二进制的右边第一位，这个counter就是1101
//(table[index] >>> 16也就是0000 0000 0000 0000 0100 1010... 1101
//oxfL也就是15，二进制为0000 0000...0000 1111，跟"0000 0000 0000 0000 0100 1010... 1101"相与，就是只保留后4位，1101，也就是这个counter的值=该元素在该位置的频次
int count = (int) ((table[index] >>> ((start + i) << 2)) & 0xfL);
frequency = Math.min(frequency, count);
}
return frequency;
}
/**
* Increments the popularity of the element if it does not exceed the maximum (15). The popularity
* of all elements will be periodically down sampled when the observed events exceeds a threshold.
* This process provides a frequency aging to allow expired long term entries to fade away.
*
* 增加元素的频次
* 获取4个slot位置，获取counter位置，并为每个counter加1
*
* 假设slot1的counter下标=c。slot1的counter下标等于c+1，slot2的counter下标等于c+2，slot3的counter下标等于c+3
* 这么做是因为计算counter的算法得到的只能为0,4,8,12这4种情况，为了使用所有16个counter，所以这么处理（也可能是理论原因所以这么计算counter）
* @param e the element to add
*/
public void increment(@NonNull E e) {
if (isNotInitialized()) {
return;
}
//hash后再打乱一次，使hashcode更加均匀
int hash = spread(e.hashCode());
//使用hash值二进制后两位计算counter下标
int start = (hash & 3) << 2;
//用不同seed获取4个slot下标
// Loop unrolling improves throughput by 5m ops/s
int index0 = indexOf(hash, 0);
int index1 = indexOf(hash, 1);
int index2 = indexOf(hash, 2);
int index3 = indexOf(hash, 3);
//尝试在每个counter上加1，最多15
boolean added = incrementAt(index0, start);
added |= incrementAt(index1, start + 1);
added |= incrementAt(index2, start + 2);
added |= incrementAt(index3, start + 3);
if (added && (++size == sampleSize)) {
reset();
}
}
/**
* Increments the specified counter by 1 if it is not already at the maximum value (15).
* 这里跟frequency方法类似，目的是找到counter所在的4位bit，然后加1，最多=15
* offset为counter的4位bit的最低位在64位bit中的下标。
* 64bit分为16个counter。从右往左数，j为counter下标，offset为64bit的下标
*
* @param i the table index (16 counters)
* @param j the counter to increment
* @return if incremented
*/
boolean incrementAt(int i, int j) {
int offset = j << 2;
//mask为掩码，counter的那4个bit，在mask中相同位置的4bit为1111，其他位置为0。
long mask = (0xfL << offset);
if ((table[i] & mask) != mask) {//(table[i] & mask) != mask，表示那4bit不全为1，也就是不等于15
//1L也就是0001，左移offset位，也就是将这4位移到跟counter相同的位置，然后相加。
//比如table[index]的二进制为0100 1010...1101 0011 1000 0011 1001，offset=12，counter的4bit为1101，十进制=13
//1L << offset的结果为 0000 0000...0001 0000 0000 0000 0000。相加后1101变成1110，十进制=14
//虽然counter只加了1，但table[i]加了"1 0000 0000 0000 0000"，十进制507376
table[i] += (1L << offset);
return true;
}
return false;
}
/** Reduces every counter by half of its original value. */
void reset() {
int count = 0;
for (int i = 0; i < table.length; i++) {
count += Long.bitCount(table[i] & ONE_MASK);//16个counter中频次为奇数的个数
//table[i] >>> 1，整体右移1位，其中每4个bit也右移1位，相当于除2。但每个counter的高位是上一个bit的低位，可能为1
//& RESET_MASK，抹去新counter的最高位，保留低三位。最终实现每个counter除2
// 1100 1001 0001 0010
// 右移1位 0110 0100 1000 1001
// 相与后 0110 0100 0000 0001
table[i] = (table[i] >>> 1) & RESET_MASK;
}
//新size=老size/2-奇数数据/4。没明白为什么要除以4
size = (size >>> 1) - (count >>> 2);
}
/**
* Returns the table index for the counter at the specified depth.
*
* @param item the element's hash
* @param i the counter depth
* @return the table index
*/
int indexOf(int item, int i) {
long hash = (item + SEED[i]) * SEED[i];
hash += (hash >>> 32);
return ((int) hash) & tableMask;
}
/**
* Applies a supplemental hash function to a given hashCode, which defends against poor quality
* hash functions.
*/
int spread(int x) {
x = ((x >>> 16) ^ x) * 0x45d9f3b;
x = ((x >>> 16) ^ x) * 0x45d9f3b;
return (x >>> 16) ^ x;
}
}
网友评论