红包系统

也不像它看起来那么简单

高并发
瞬时大流量
单SKU队列
财务审计

微博的红包服务架构

http://www.infoq.com/cn/articles/2016-hongbao-weibo

红包生成

维护公平感
制造惊喜感
预算控制

也不像它看起来那么简单

100块钱发100个红包

公平感

每个红包应当多少钱？

E(X) = { \sum X \over n }

E(X) = { \sum X \over n }

100块钱发100个红包

惊喜感

你希望拿多少钱？

D(X) \to N ( N \ggg 1)

D(X) \to N ( N \ggg 1)

发100个红包平均1元的红包

预算控制

预算需要是100元整

D(\sum X) = 0

D(\sum X) = 0

E(\sum X) = E(\sum X)

E(\sum X) = E(\sum X)

需求弄明白了

怎么实现呢？

function generateValue(max) {
 return Math.random() * max;
}

分析一下

function generateValue(max) {
 return Math.random() * max;
}

max = 2 \cdot E(X)

max = 2 \cdot E(X)

D(X) = { max \over 12}

D(X) = { max \over 12}

= { E(X) \over 6 }

= { E(X) \over 6 }

\therefore E(\sum X) = \sum E(X)

\therefore E(\sum X) = \sum E(X)

\therefore D(\sum X) = \sum D(X)

\therefore D(\sum X) = \sum D(X)

= { N \cdot E(X) \over 6 }

= { N \cdot E(X) \over 6 }

E(X) = { max \over 2 }

E(X) = { max \over 2 }

问题来了

E(X) = { max \over 2 }

E(X) = { max \over 2 }

D(X) = { max \over 12}

D(X) = { max \over 12}

E(\sum X) = { max \cdot N \over 2 }

E(\sum X) = { max \cdot N \over 2 }

D(\sum X) = { N \cdot max \over 12 }

D(\sum X) = { N \cdot max \over 12 }

D(\sum X) = 0

D(\sum X) = 0

E(\sum X) = \sum E(X)

E(\sum X) = \sum E(X)

E(X) = { \sum X \over n }

E(X) = { \sum X \over n }

D(X) \to N ( N \ggg 1)

D(X) \to N ( N \ggg 1)

需求

结论

Experiment

const maxValue = 100
const totalTimes = 1000
const distribution = {}
const recordSum = (value) => {
  const count = distribution[value] || 0
  distribution[value] = count + 1
}
const random = () => Math.random() * maxValue >> 0
const doOnce = () => compose(recordSum, sum, times)(random, totalTimes)

times(doOnce, 10000)

toPairs(distribution).forEach((arr) => console.log(`${arr[0]}, ${arr[1]}`))

http://goo.gl/0qmj6R

E(X) = { max \over 2 } = 50

E(X) = { max \over 2 } = 50

D(X) = { max \over 12} \approx 8.33

D(X) = { max \over 12} \approx 8.33

E(\sum X) = { max \cdot N \over 2 } = 50000

E(\sum X) = { max \cdot N \over 2 } = 50000

D(\sum X) = { N \cdot max \over 12 } \approx 8333.33

D(\sum X) = { N \cdot max \over 12 } \approx 8333.33

Result

E(\sum X) = { max \cdot N \over 2 } = 50000

E(\sum X) = { max \cdot N \over 2 } = 50000

D(\sum X) = { N \cdot max \over 12 } \approx 8333.33

D(\sum X) = { N \cdot max \over 12 } \approx 8333.33

微信怎么做的？

public static double getRandomMoney(LeftMoneyPackage _leftMoneyPackage) {
    // remainSize 剩余的红包数量
    // remainMoney 剩余的钱
    if (_leftMoneyPackage.remainSize == 1) {
        _leftMoneyPackage.remainSize;
        return (double) Math.round(_leftMoneyPackage.remainMoney * 100) / 100;
    }
    
    Random r = new Random();
    double min = 0.01;    
    double max = _leftMoneyPackage.remainMoney / _leftMoneyPackage.remainSize * 2;
    double money = r.nextDouble() * max;
    money = money <= min ? min : money;
    money = Math.floor(money * 100) / 100;
    _leftMoneyPackage.remainSize--;
    _leftMoneyPackage.remainMoney -= money;
    return money;
}

https://www.zhihu.com/question/22625187