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如何使用Java模擬退火算法優(yōu)化Hash函數(shù)

 更新時(shí)間:2021年06月16日 14:57:43   作者:buttercup  
為了解決局部最優(yōu)解問(wèn)題,1983年,Kirkpatrick等提出了模擬退火算法(SA)能有效的解決局部最優(yōu)解問(wèn)題。模擬退火算法包含兩個(gè)部分即Metropolis算法和退火過(guò)程。Metropolis算法就是如何在局部最優(yōu)解的情況下讓其跳出來(lái),是退火的基礎(chǔ)

一、背景

現(xiàn)有個(gè)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(dòng)了 15 個(gè)線程來(lái)消費(fèi)行情消息。消息分配的策略較為簡(jiǎn)單:對(duì) symbol 的 hashCode 取模,將消息分配給其中一個(gè)線程進(jìn)行處理。 經(jīng)過(guò)驗(yàn)證,每個(gè)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(shí)間后,突然收到了系統(tǒng)的告警,但此時(shí)并非消息峰值時(shí)間段。經(jīng)過(guò)排查后,發(fā)現(xiàn)問(wèn)題出現(xiàn)在 hash 函數(shù)上:

雖然每個(gè)線程被分配到的 symbol 數(shù)量較為均衡,但是部分熱門(mén) symbol 的報(bào)價(jià)消息量會(huì)更多,如果熱門(mén) symbol 集中到特定線程上,就會(huì)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(diǎn)線程。為此,系統(tǒng)需要記錄下某天內(nèi)每個(gè) symbol 的消息量,然后在第二天使用這些數(shù)據(jù),對(duì)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(duì) hash 函數(shù)進(jìn)行優(yōu)化

二、放棄 hash 函數(shù)

問(wèn)題可以抽象為:

將 5000 個(gè)非負(fù)整數(shù)分配至 15 個(gè)桶(bucket)中,并盡可能保證每個(gè)桶中的元素之和接近(每個(gè)桶中的元素個(gè)數(shù)無(wú)限制)。

每個(gè)整數(shù)元素可能的放置方法有 15 種,這個(gè)問(wèn)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(wèn)題,最優(yōu)解不是必要的,可以退而求其次尋找一個(gè)可接受的次優(yōu)解:

根據(jù)所有 symbol 的消息總數(shù)計(jì)算一個(gè)期望的分布均值(expectation)。將每個(gè) 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(gè)區(qū)間,并且盡可能使得每個(gè)區(qū)間元素之和與 expection 接近。使用一個(gè)有序查找表記錄每個(gè)區(qū)間的首個(gè) symbol,后續(xù)就可以按照這個(gè)表對(duì)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(gè)桶的數(shù)量
        TreeMap<String, Integer> distribution = findBestDistribution(symbolAndCount);

        // 測(cè)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(gè)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(gè) symbol 數(shù)據(jù)量過(guò)大,則分配給其一個(gè)獨(dú)立的桶
            int count = entry.getValue();
            if (count / 2 > avg && bucketIdx == 0 && buckets[0] == 0) {
                buckets[bucketIdx] += count;
                continue;
            }

            // 評(píng)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(gè)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(wèn)題:

  • 分配策略并不是最優(yōu)解,且無(wú)法對(duì)其分片效果進(jìn)行直觀的評(píng)估。
  • 當(dāng)區(qū)間數(shù)量較多時(shí),查找表本身可能成為一個(gè)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(gè)角度來(lái)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(xiàng)目中使用的 hash 函數(shù)是 JDK String 中的原生實(shí)現(xiàn)。經(jīng)過(guò)查閱資料,發(fā)現(xiàn)該實(shí)現(xiàn)其實(shí)是 BKDRHash 的 seed = 31 的特殊情況。這樣意味著:通過(guò)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(wèn)題來(lái)了,應(yīng)該如何評(píng)估某個(gè) seed 的分布的優(yōu)劣?

3.1、評(píng)價(jià)函數(shù)

一種可行的方法是計(jì)算每個(gè) seed 對(duì)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(gè) bucket 與均值之間的誤差,無(wú)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(píng)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(duì)應(yīng)的分布越均勻,其對(duì)應(yīng)的 hash 函數(shù)越優(yōu)。

3.2、訓(xùn)練策略

seed 是一個(gè) 32bit 的無(wú)符號(hào)整數(shù),其取值范圍為 0 ~ 232-1。在 5000 個(gè) symbol 的情況下,單線程嘗試遍歷所有 seed 的時(shí)間約為 25 小時(shí)。

通常情況下 symbol 的數(shù)量會(huì)超過(guò) 5000,因此實(shí)際的搜索時(shí)間會(huì)大于這個(gè)值。此外,受限于計(jì)算資源限制,無(wú)法進(jìn)行大規(guī)模的并行搜索,因此窮舉法的耗時(shí)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(guò)程的熱平衡問(wèn)題與隨機(jī)搜索尋優(yōu)問(wèn)題的相似性來(lái)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(jiǎn)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(gè)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(cè)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(duì)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(duì)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(duì)比日志發(fā)現(xiàn) hash 經(jīng)過(guò)優(yōu)化后,分區(qū)的均勻程度有了顯著的上升,并且熱點(diǎn)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(qǐng)關(guān)注腳本之家其它相關(guān)文章!

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