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這篇文章主要講解了“Java模擬退火算法優(yōu)化Hash函數(shù)的方法”,文中的講解內(nèi)容簡單清晰,易于學習與理解,下面請大家跟著小編的思路慢慢深入,一起來研究和學習“Java模擬退火算法優(yōu)化Hash函數(shù)的方法”吧!
一、背景
二、放棄 hash 函數(shù)
三、優(yōu)化 hash 函數(shù)
3.1、評價函數(shù)
3.2、訓練策略
3.3、ForkJoin 框架
3.4、效果
現(xiàn)有個處理股票行情消息的系統(tǒng),其架構如下:
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)進行優(yōu)化
問題可以抽象為:
將 5000 個非負整數(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,則將其放入當前桶中 // 2. 如果桶中的數(shù)量更遠離期望,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行直觀的評估。
當區(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。
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,應該如何評估某個 seed 的分布的優(yōu)劣?
一種可行的方法是計算每個 seed 對應的 bucket 分布的標準差,標準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 對應的分布越均勻,其對應的 hash 函數(shù)越優(yōu)。
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搜索尋優(yōu)問題的相似性來達到尋找全局最優(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); } }
為了達到更好的搜索效果,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)域的搜索結果。
使用 JDK 提供的 ForkJoinPool 與 RecursiveTask 能很好地完成以上任務。
@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; } }
將改造后的代碼部署到測試環(huán)境后,某日訓練日志:
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]
當晚使用 BKDRHash(seed=31)
對新的交易日數(shù)據(jù)的進行分片:
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]
當晚使用 BKDRHash(seed=362333)
對新的交易日數(shù)據(jù)的進行分片:
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,基本達到當初設想的優(yōu)化效果。
感謝各位的閱讀,以上就是“Java模擬退火算法優(yōu)化Hash函數(shù)的方法”的內(nèi)容了,經(jīng)過本文的學習后,相信大家對Java模擬退火算法優(yōu)化Hash函數(shù)的方法這一問題有了更深刻的體會,具體使用情況還需要大家實踐驗證。這里是億速云,小編將為大家推送更多相關知識點的文章,歡迎關注!
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