如何使用Java模擬退火算法優(yōu)化Hash函數(shù)
一、背景
現(xiàn)有個處理股票行情消息的系統(tǒng)，其架構(gòu)如下：
由于數(shù)據(jù)量巨大，系統(tǒng)中啟動了 15 個線程來消費行情消息。消息分配的策略較為簡單：對 symbol 的 hashCode 取模，將消息分配給其中一個線程進(jìn)行處理。 經(jīng)過驗證，每個線程分配到的 symbol 數(shù)量較為均勻，于是系統(tǒng)愉快地上線了。
運(yùn)行一段時間后，突然收到了系統(tǒng)的告警，但此時并非消息峰值時間段。經(jīng)過排查后，發(fā)現(xiàn)問題出現(xiàn)在 hash 函數(shù)上：
雖然每個線程被分配到的 symbol 數(shù)量較為均衡，但是部分熱門 symbol 的報價消息量會更多，如果熱門 symbol 集中到特定線程上，就會造成線程負(fù)載不均衡，使得系統(tǒng)整體的吞吐量大打折扣。
為提高系統(tǒng)的吞吐量，有必要消息分發(fā)邏輯進(jìn)行一些改造，避免出現(xiàn)熱點線程。為此，系統(tǒng)需要記錄下某天內(nèi)每個 symbol 的消息量，然后在第二天使用這些數(shù)據(jù)，對分發(fā)邏輯進(jìn)行調(diào)整。具體的改造的方案可以分為兩種：
- 放棄使用 hash 函數(shù)
- 對 hash 函數(shù)進(jìn)行優(yōu)化
二、放棄 hash 函數(shù)
問題可以抽象為：
將 5000 個非負(fù)整數(shù)分配至 15 個桶(bucket)中，并盡可能保證每個桶中的元素之和接近（每個桶中的元素個數(shù)無限制）。
每個整數(shù)元素可能的放置方法有 15 種，這個問題總共可能的解有 155000種，暴力求解的可能性微乎其微。作為工程問題，最優(yōu)解不是必要的，可以退而求其次尋找一個可接受的次優(yōu)解：
根據(jù)所有 symbol 的消息總數(shù)計算一個期望的分布均值(expectation)。將每個 symbol 的消息數(shù)按照 symbol 的順序進(jìn)行排列，最后將這組數(shù)組劃分為 15 個區(qū)間，并且盡可能使得每個區(qū)間元素之和與 expection 接近。使用一個有序查找表記錄每個區(qū)間的首個 symbol，后續(xù)就可以按照這個表對數(shù)據(jù)進(jìn)行劃分。
public class FindBestDistribution {
static final int NUM_OF_SYMBOLS = 5000;
static final int NUM_OF_BUCKETS = 15;
public static void main(String[] args) {
// 生成樣本
IntStream ints = ThreadLocalRandom.current().ints(0, 1000);
PrimitiveIterator.OfInt iterator = ints.iterator();
Map<String,Integer> symbolAndCount = new TreeMap<>();
for (int i=0; i<NUM_OF_SYMBOLS; i++) {
symbolAndCount.put(Integer.toHexString(i).toUpperCase(), iterator.next());
}
// 按照 symbol 劃分每個桶的數(shù)量
TreeMap<String, Integer> distribution = findBestDistribution(symbolAndCount);
// 測試效果
int[] buckets = new int[NUM_OF_BUCKETS];
for (Map.Entry<String, Integer> entry : symbolAndCount.entrySet()) {
Map.Entry<String, Integer> floor = distribution.floorEntry(entry.getKey());
int bucketIndex = floor == null ? 0 : floor.getValue();
buckets[bucketIndex] += entry.getValue();
}
System.out.printf("buckets: %s\n", Arrays.toString(buckets));
}
public static TreeMap<String, Integer> findBestDistribution(Map<String,Integer> symbolAndCount) {
// 每個桶均勻分布的情況（最優(yōu)情況）
int avg = symbolAndCount.values().stream().mapToInt(Integer::intValue).sum() / NUM_OF_BUCKETS;
// 嘗試將 symbol 放入不同的桶
int bucketIdx = 0;
int[] buckets = new int[NUM_OF_BUCKETS];
String[] bulkheads = new String[NUM_OF_BUCKETS-1];
for (Map.Entry<String, Integer> entry : symbolAndCount.entrySet()) {
// 如果首個 symbol 數(shù)據(jù)量過大，則分配給其一個獨立的桶
int count = entry.getValue();
if (count / 2 > avg && bucketIdx == 0 && buckets[0] == 0) {
buckets[bucketIdx] += count;
continue;
}
// 評估將 symbol 放入桶后的效果
// 1. 如果桶中的數(shù)量更接近期望，則將其放入當(dāng)前桶中
// 2. 如果桶中的數(shù)量更遠(yuǎn)離期望，則將其放入下個桶中
double before = Math.abs(buckets[bucketIdx] - avg);
double after = Math.abs(buckets[bucketIdx] + count - avg);
if (after > before && bucketIdx < buckets.length - 1) {
bulkheads[bucketIdx++] = entry.getKey();
}
buckets[bucketIdx] += count;
}
System.out.printf("expectation: %d\n", avg);
System.out.printf("bulkheads: %s\n", Arrays.toString(bulkheads));
TreeMap<String,Integer> distribution = new TreeMap<>();
for (int i=0; i<bulkheads.length; i++) {
distribution.put(bulkheads[i], i+1);
}
return distribution;
}
}
該方法存在的問題：
- 分配策略并不是最優(yōu)解，且無法對其分片效果進(jìn)行直觀的評估。
- 當(dāng)區(qū)間數(shù)量較多時，查找表本身可能成為一個潛在的性能瓶頸。
- 可能的組合受到 key 的順序限制，極大地限制了可能的解空間。
三、優(yōu)化 hash 函數(shù)
換個角度來看，造成分布不均勻的原因不是數(shù)據(jù)，而是 hash 函數(shù)本身。
項目中使用的 hash 函數(shù)是 JDK String 中的原生實現(xiàn)。經(jīng)過查閱資料，發(fā)現(xiàn)該實現(xiàn)其實是 BKDRHash 的 seed = 31 的特殊情況。這樣意味著：通過調(diào)整 seed 的值，可以改變 hash 函數(shù)的特性并使其適配特定的數(shù)據(jù)分布。
int BKDRHash(char[] value, int seed) {
int hash = 0;
for (int i = 0; i < value.length; i++) {
hash = hash * seed + value[i];
}
return hash & 0x7fffffff;
}
那么問題來了，應(yīng)該如何評估某個 seed 的分布的優(yōu)劣？
3.1、評價函數(shù)
一種可行的方法是計算每個 seed 對應(yīng)的 bucket 分布的標(biāo)準(zhǔn)差，標(biāo)準(zhǔn)差越小則分布越均勻，則該 seed 越優(yōu)。
然而這一做法只考慮了每個 bucket 與均值之間的誤差，無法量化不同 bucket 之間的誤差。為了能夠直觀的量化 bucket 之間分布差異的情況，考慮使用下面的評估函數(shù)：
ouble calculateDivergence(long[] bucket, long expectation) {
long divergence = 0;
for (int i=0; i<bucket.length; i++) {
final long a = bucket[i];
final long b = (a - expectation) * (a - expectation);
for (int j=i+1; j<bucket.length; j++) {
long c = (a - bucket[j]) * (a - bucket[j]);
divergence += Math.max(b, c);
}
}
return divergence; // the less the better
}
該數(shù)值越小，則證明 seed 對應(yīng)的分布越均勻，其對應(yīng)的 hash 函數(shù)越優(yōu)。
3.2、訓(xùn)練策略
seed 是一個 32bit 的無符號整數(shù)，其取值范圍為 0 ～ 232-1。在 5000 個 symbol 的情況下，單線程嘗試遍歷所有 seed 的時間約為 25 小時。
通常情況下 symbol 的數(shù)量會超過 5000，因此實際的搜索時間會大于這個值。此外，受限于計算資源限制，無法進(jìn)行大規(guī)模的并行搜索，因此窮舉法的耗時是不可接受的。
幸好本例并不要求最優(yōu)解，可以引入啟發(fā)式搜索算法，加快訓(xùn)練速度。由于本人在這方面并不熟悉，為了降低編程難度，最終選擇了模擬退火(simulated annealing)算法。它模擬固體退火過程的熱平衡問題與隨機(jī)搜索尋優(yōu)問題的相似性來達(dá)到尋找全局最優(yōu)或近似全局最優(yōu)的目的。
相較于最簡單的爬山法，模擬退火算法通以一定的概率接受較差的解，從而擴(kuò)大搜索范圍，保證解近似最優(yōu)。
/**
* Basic framework of simulated annealing algorithm
* @param <X> the solution of given problem
*/
public abstract class SimulatedAnnealing<X> {
protected final int numberOfIterations; // stopping condition for simulations
protected final double coolingRate; // the percentage by which we reduce the temperature of the system
protected final double initialTemperature; // the starting energy of the system
protected final double minimumTemperature; // optional stopping condition
protected final long simulationTime; // optional stopping condition
protected final int detectionInterval; // optional stopping condition
protected SimulatedAnnealing(int numberOfIterations, double coolingRate) {
this(numberOfIterations, coolingRate, 10000000, 1, 0, 0);
}
protected SimulatedAnnealing(int numberOfIterations, double coolingRate, double initialTemperature, double minimumTemperature, long simulationTime, int detectionInterval) {
this.numberOfIterations = numberOfIterations;
this.coolingRate = coolingRate;
this.initialTemperature = initialTemperature;
this.minimumTemperature = minimumTemperature;
this.simulationTime = simulationTime;
this.detectionInterval = detectionInterval;
}
protected abstract double score(X currentSolution);
protected abstract X neighbourSolution(X currentSolution);
public X simulateAnnealing(X currentSolution) {
final long startTime = System.currentTimeMillis();
// Initialize searching
X bestSolution = currentSolution;
double bestScore = score(bestSolution);
double currentScore = bestScore;
double t = initialTemperature;
for (int i = 0; i < numberOfIterations; i++) {
if (currentScore < bestScore) {
// If the new solution is better, accept it unconditionally
bestScore = currentScore;
bestSolution = currentSolution;
} else {
// If the new solution is worse, calculate an acceptance probability for the worse solution
// At high temperatures, the system is more likely to accept the solutions that are worse
boolean rejectWorse = Math.exp((bestScore - currentScore) / t) < Math.random();
if (rejectWorse || currentScore == bestScore) {
currentSolution = neighbourSolution(currentSolution);
currentScore = score(currentSolution);
}
}
// Stop searching when the temperature is too low
if ((t *= coolingRate) < minimumTemperature) {
break;
}
// Stop searching when simulation time runs out
if (simulationTime > 0 && (i+1) % detectionInterval == 0) {
if (System.currentTimeMillis() - startTime > simulationTime)
break;
}
}
return bestSolution;
}
}
/**
* Search best hash seed for given key distribution and number of buckets with simulated annealing algorithm
*/
@Data
public class SimulatedAnnealingHashing extends SimulatedAnnealing<HashingSolution> {
private static final int DISTRIBUTION_BATCH = 100;
static final int SEARCH_BATCH = 200;
private final int[] hashCodes = new int[SEARCH_BATCH];
private final long[][] buckets = new long[SEARCH_BATCH][];
@Data
public class HashingSolution {
private final int begin, range; // the begin and range for searching
private int bestSeed; // the best seed found in this search
private long bestScore; // the score corresponding to bestSeed
private long calculateDivergence(long[] bucket) {
long divergence = 0;
for (int i=0; i<bucket.length; i++) {
final long a = bucket[i];
final long b = (a - expectation) * (a - expectation);
for (int j=i+1; j<bucket.length; j++) {
long c = (a - bucket[j]) * (a - bucket[j]);
divergence += Math.max(b, c);
}
}
return divergence; // the less the better
}
private HashingSolution solve() {
if (range != hashCodes.length) {
throw new IllegalStateException();
}
for (int i=0; i<range; i++) {
Arrays.fill(buckets[i], hashCodes[i] = 0);
}
for (KeyDistribution[] bucket : distributions) {
for (KeyDistribution distribution : bucket) {
Hashing.BKDRHash(distribution.getKey(), begin, hashCodes);
for (int k = 0; k< hashCodes.length; k++) {
int n = hashCodes[k] % buckets[k].length;
buckets[k][n] += distribution.getCount();
}
}
}
int best = -1;
long bestScore = Integer.MAX_VALUE;
for (int i = 0; i< buckets.length; i++) {
long score = calculateDivergence(buckets[i]);
if (i == 0 || score < bestScore) {
bestScore = score;
best = i;
}
}
if (best < 0) {
throw new IllegalStateException();
}
this.bestScore = bestScore;
this.bestSeed = begin + best;
return this;
}
@Override
public String toString() {
return String.format("(seed:%d, score:%d)", bestSeed, bestScore);
}
}
private final KeyDistribution[][] distributions; // key and its count（2-dimensional array for better performance）
private final long expectation; // the expectation count of each bucket
private final int searchOutset;
private int searchMin, searchMax;
/**
* SimulatedAnnealingHashing Prototype
* @param keyAndCounts keys for hashing and count for each key
* @param numOfBuckets number of buckets
*/
public SimulatedAnnealingHashing(Map<String, Integer> keyAndCounts, int numOfBuckets) {
super(100000000, .9999);
distributions = buildDistribution(keyAndCounts);
long sum = 0;
for (KeyDistribution[] batch : distributions) {
for (KeyDistribution distribution : batch) {
sum += distribution.getCount();
}
}
this.expectation = sum / numOfBuckets;
this.searchOutset = 0;
for (int i = 0; i< buckets.length; i++) {
buckets[i] = new long[numOfBuckets];
}
}
/**
* SimulatedAnnealingHashing Derivative
* @param prototype prototype simulation
* @param searchOutset the outset for searching
* @param simulationTime the expect time consuming for simulation
*/
private SimulatedAnnealingHashing(SimulatedAnnealingHashing prototype, int searchOutset, long simulationTime) {
super(prototype.numberOfIterations, prototype.coolingRate, prototype.initialTemperature, prototype.minimumTemperature,
simulationTime, 10000);
distributions = prototype.distributions;
expectation = prototype.expectation;
for (int i = 0; i< buckets.length; i++) {
buckets[i] = new long[prototype.buckets[i].length];
}
this.searchOutset = searchOutset;
this.searchMax = searchMin = searchOutset;
}
@Override
public String toString() {
return String.format("expectation: %d, outset:%d, search(min:%d, max:%d)", expectation, searchOutset, searchMin, searchMax);
}
private KeyDistribution[][] buildDistribution(Map<String, Integer> symbolCounts) {
int bucketNum = symbolCounts.size() / DISTRIBUTION_BATCH + Integer.signum(symbolCounts.size() % DISTRIBUTION_BATCH);
KeyDistribution[][] distributions = new KeyDistribution[bucketNum][];
int bucketIndex = 0;
List<KeyDistribution> batch = new ArrayList<>(DISTRIBUTION_BATCH);
for (Map.Entry<String, Integer> entry : symbolCounts.entrySet()) {
batch.add(new KeyDistribution(entry.getKey().toCharArray(), entry.getValue()));
if (batch.size() == DISTRIBUTION_BATCH) {
distributions[bucketIndex++] = batch.toArray(new KeyDistribution[0]);
batch.clear();
}
}
if (batch.size() > 0) {
distributions[bucketIndex] = batch.toArray(new KeyDistribution[0]);
batch.clear();
}
return distributions;
}
@Override
protected double score(HashingSolution currentSolution) {
return currentSolution.solve().bestScore;
}
@Override
protected HashingSolution neighbourSolution(HashingSolution currentSolution) {
// The default range of neighbourhood is [-100, 100]
int rand = ThreadLocalRandom.current().nextInt(-100, 101);
int next = currentSolution.begin + rand;
searchMin = Math.min(next, searchMin);
searchMax = Math.max(next, searchMax);
return new HashingSolution(next, currentSolution.range);
}
public HashingSolution solve() {
searchMin = searchMax = searchOutset;
HashingSolution initialSolution = new HashingSolution(searchOutset, SEARCH_BATCH);
return simulateAnnealing(initialSolution);
}
public SimulatedAnnealingHashing derive(int searchOutset, long simulationTime) {
return new SimulatedAnnealingHashing(this, searchOutset, simulationTime);
}
}
3.3、ForkJoin 框架
為了達(dá)到更好的搜索效果，可以將整個搜索區(qū)域遞歸地劃分為兩兩相鄰的區(qū)域，然后在這些區(qū)域上執(zhí)行并發(fā)的搜索，并遞歸地合并相鄰區(qū)域的搜索結(jié)果。
使用 JDK 提供的 ForkJoinPool 與 RecursiveTask 能很好地完成以上任務(wù)。
@Data
@Slf4j
public class HashingSeedCalculator {
/**
* Recursive search task
*/
private class HashingSeedCalculatorSearchTask extends RecursiveTask<HashingSolution> {
private SimulatedAnnealingHashing simulation;
private final int level;
private final int center, range;
private HashingSeedCalculatorSearchTask() {
this.center = 0;
this.range = Integer.MAX_VALUE / SimulatedAnnealingHashing.SEARCH_BATCH;
this.level = traversalDepth;
this.simulation = hashingSimulation;
}
private HashingSeedCalculatorSearchTask(HashingSeedCalculatorSearchTask parent, int center, int range) {
this.center = center;
this.range = range;
this.level = parent.level - 1;
this.simulation = parent.simulation;
}
@Override
protected HashingSolution compute() {
if (level == 0) {
long actualCenter = center * SimulatedAnnealingHashing.SEARCH_BATCH;
log.info("Searching around center {}", actualCenter);
HashingSolution solution = simulation.derive(center, perShardRunningMills).solve();
log.info("Searching around center {} found {}", actualCenter, solution);
return solution;
} else {
int halfRange = range / 2;
int leftCenter = center - halfRange, rightCenter = center + halfRange;
ForkJoinTask<HashingSolution> leftTask = new HashingSeedCalculatorSearchTask(this, leftCenter, halfRange).fork();
ForkJoinTask<HashingSolution> rightTask = new HashingSeedCalculatorSearchTask(this, rightCenter, halfRange).fork();
HashingSolution left = leftTask.join();
HashingSolution right = rightTask.join();
return left.getBestScore() < right.getBestScore() ? left : right;
}
}
}
private final int poolParallelism;
private final int traversalDepth;
private final long perShardRunningMills;
private final SimulatedAnnealingHashing hashingSimulation;
/**
* HashingSeedCalculator
* @param numberOfShards the shard of the whole search range [Integer.MIN_VALUE, Integer.MAX_VALUE]
* @param totalRunningHours the expect total time consuming for searching
* @param symbolCounts the key and it`s distribution
* @param numOfBuckets the number of buckets
*/
public HashingSeedCalculator(int numberOfShards, int totalRunningHours, Map<String, Integer> symbolCounts, int numOfBuckets) {
int n = (int) (Math.log(numberOfShards) / Math.log(2));
if (Math.pow(2, n) != numberOfShards) {
throw new IllegalArgumentException();
}
this.traversalDepth = n;
this.poolParallelism = Math.max(ForkJoinPool.getCommonPoolParallelism() / 3 * 2, 1); // conservative estimation for parallelism
this.perShardRunningMills = TimeUnit.HOURS.toMillis(totalRunningHours * poolParallelism) / numberOfShards;
this.hashingSimulation = new SimulatedAnnealingHashing(symbolCounts, numOfBuckets);
}
@Override
public String toString() {
int numberOfShards = (int) Math.pow(2, traversalDepth);
int totalRunningHours = (int) TimeUnit.MILLISECONDS.toHours(perShardRunningMills * numberOfShards) / poolParallelism;
return "HashingSeedCalculator(" +
"numberOfShards: " + numberOfShards +
", perShardRunningMinutes: " + TimeUnit.MILLISECONDS.toMinutes(perShardRunningMills) +
", totalRunningHours: " + totalRunningHours +
", poolParallelism: " + poolParallelism +
", traversalDepth: " + traversalDepth + ")";
}
public synchronized HashingSolution searchBestSeed() {
long now = System.currentTimeMillis();
log.info("SearchBestSeed start");
ForkJoinTask<HashingSolution> root = new HashingSeedCalculatorSearchTask().fork();
HashingSolution initSolution = hashingSimulation.derive(0, perShardRunningMills).solve();
HashingSolution bestSolution = root.join();
log.info("Found init solution {}", initSolution);
log.info("Found best solution {}", bestSolution);
if (initSolution.getBestScore() < bestSolution.getBestScore()) {
bestSolution = initSolution;
}
long cost = System.currentTimeMillis() - now;
log.info("SearchBestSeed finish (cost:{}ms)", cost);
return bestSolution;
}
}
3.4、效果
將改造后的代碼部署到測試環(huán)境后，某日訓(xùn)練日志：
12:49:15.227 85172866 INFO hash.HashingSeedCalculator - Found init solution (seed:15231, score:930685828341164)
12:49:15.227 85172866 INFO hash.HashingSeedCalculator - Found best solution (seed:362333, score:793386389726926)
12:49:15.227 85172866 INFO hash.HashingSeedCalculator - SearchBestSeed finish (cost:10154898ms)
12:49:15.227 85172866 INFO hash.TrainingService -
Training result: (seed:362333, score:793386389726926)
Buckets: 15
Expectation: 44045697
Result of Hashing.HashCode(seed=362333): 21327108 [42512742, 40479608, 43915771, 47211553, 45354264, 43209190, 43196570, 44725786, 41999747, 46450288, 46079231, 45116615, 44004021, 43896194, 42533877]
Result of Hashing.HashCode(seed=31): 66929172 [39723630, 48721463, 43365391, 46301448, 43931616, 44678194, 39064877, 45922454, 43171141, 40715060, 33964547, 49709090, 58869949, 34964729, 47581868]
當(dāng)晚使用 BKDRHash(seed=31) 對新的交易日數(shù)據(jù)的進(jìn)行分片：
04:00:59.001 partition messages per minute [45171, 68641, 62001, 80016, 55977, 61916, 55102, 49322, 55982, 57081, 51100, 70437, 135992, 37823, 58552] , messages total [39654953, 48666261, 43310578, 46146841, 43834832, 44577454, 38990331, 45871075, 43106710, 40600708, 33781629, 49752592, 58584246, 34928991, 47545369]
當(dāng)晚使用 BKDRHash(seed=362333) 對新的交易日數(shù)據(jù)的進(jìn)行分片：
04:00:59.001 partition messages per minute [62424, 82048, 64184, 47000, 57206, 69439, 64430, 60096, 46986, 58182, 54557, 41523, 64310, 72402, 100326] , messages total [44985772, 48329212, 39995385, 43675702, 45216341, 45524616, 41335804, 44917938, 44605376, 44054821, 43371892, 42068637, 44000817, 42617562, 44652695]
對比日志發(fā)現(xiàn) hash 經(jīng)過優(yōu)化后，分區(qū)的均勻程度有了顯著的上升，并且熱點分片也被消除了，基本達(dá)到當(dāng)初設(shè)想的優(yōu)化效果。
以上就是如何使用Java模擬退火算法優(yōu)化Hash函數(shù)的詳細(xì)內(nèi)容，更多關(guān)于Java 模擬退火算法優(yōu)化Hash的資料請關(guān)注腳本之家其它相關(guān)文章！
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