GraphGists

反叛军财务系统

引言

在这里，我将通过一个故事来介绍一个日常问题的解决方案。有多少次，你与其他反叛军一起进行超空间旅行，或者与几位绝地武士执行任务，或者只是与其他人一起做某事时，需要分摊费用？

在这些情况下，经常会出现某人欠钱给另一人，而后者又欠钱给前者的情况。

在这里，我将讲述共和国如何使用Neo4j优雅地解决了这个问题的故事。

故事

很久很久以前，在一个遥远的星系里，有四个人一起旅行，为了拯救共和国。

他们很勇敢，但超空间的通行费和燃料可不是免费的；他们不富裕，也没有一个中心化的财务系统，而是由他们的积蓄构成的分布式系统。

幸运的是，除了原力之外，他们还有一个Neo4j数据库来处理这个分布式财务问题。

所以有勇敢的绝地女孩蕾伊，悔改的芬恩，老派的汉和毛茸茸的外星人丘巴卡。

他们有一个任务：从Neo4j星球获取管理共和国财务的技术，并找到正确使用它的方法。

第一部分很容易，他们带着一个U盘到达了那个星球，获得了数据库，但他们必须找到正确使用它的方法来解决奥加纳的问题。他们决定用他们的返程旅途进行一个小实验。

出发前，他们将自己作为节点添加到数据库中，名字作为属性，并标记为Friends。

CREATE (f1:Friend {name:'Rey'})
CREATE (f2:Friend {name:'Finn'})
CREATE (f3:Friend {name:'Han'})
CREATE (f4:Friend {name:'Chewbacca'})
RETURN f1, f2, f3, f4

然后他们需要加满燃料，于是汉为千年隼号加了1000共和国信用点（从现在起我们称之为£）的燃料，这样其他人每人欠他250£（1000 / 4 = 250）。

他们决定将250£表示为Expense节点，船员与汉之间的关系是一个Credit节点，其中包含欠债权人的总金额——在本例中是250£。每个Credit通过:CREDIT关系链接到债权人，并通过:DEBIT关系链接到债务人。在Cypher中，以汉为债权人，蕾伊为债务人的表示方式是

MATCH (creditor:Friend {name:'Han'}), (debitor:Friend {name:'Rey'})
MERGE (creditor)-[:CREDIT]->(c:Credit {Creditor: creditor.name, Debitor: debitor.name})-[:DEBIT]->(debitor)
MERGE (c)<-[:EXPENSE]-(exp:Expense {Creditor: creditor.name, Debitor: debitor.name, Amount: 250, Description: 'Fuel Millenium Falcon'})
WITH c
MATCH (c)<-[:EXPENSE]-(exp)
WITH c, SUM(exp.Amount) AS total
SET c.Amount = total;
MATCH (creditor:Friend {name:'Han'}), (debitor:Friend)
WHERE debitor.name IN ['Finn', 'Chewbacca']
MERGE (creditor)-[:CREDIT]->(c:Credit {Creditor: creditor.name, Debitor: debitor.name})-[:DEBIT]->(debitor)
MERGE (c)<-[:EXPENSE]-(exp:Expense {Creditor: creditor.name, Debitor: debitor.name, Amount: 250, Description: 'Fuel Millenium Falcon'})
WITH c
MATCH (c)<-[:EXPENSE]-(exp)
WITH c, SUM(exp.Amount) AS total
SET c.Amount = total;

对于整个团队，则变成

MATCH (n)
RETURN n

加完油后，他们开始了旅程，过了一会儿到了午餐时间，于是他们在服务区停下，丘巴卡在那里买了些夸瑞恩食物，花了80£。

同样地，他们将这笔费用添加到他们的Neo4j数据库中，这样他们每人欠丘巴卡20£

这时有一个有趣的发现：丘巴卡欠汉250£，而汉欠丘巴卡20£。这意味着他们的信用是相互关联的。

MATCH p=(c:Friend {name:'Chewbacca'})-[:CREDIT|DEBIT*2]-(h:Friend {name:'Han'})
RETURN p

他们想描述这种关联，所以他们将同一个人两个信用之间的关系定义为:CHAIN，这表示他的进账和出账信用之间存在资金流。

在这种情况下，汉的借记和信用之间存在一个:CHAIN关系，丘巴卡也是一样

MATCH (credit:Credit)<-[:CREDIT]-(creditor:Friend {name:'Han'})<-[:DEBIT]-(debit :Credit)
MERGE (credit)<-[chain:CHAIN]-(debit);

MATCH (credit:Credit)<-[:CREDIT]-(creditor:Friend {name:'Chewbacca'})<-[:DEBIT]-(debit :Credit)
MERGE (credit)<-[chain:CHAIN]-(debit);

（这里是模型清晰图片的链接，遗憾的是我无法链接Dropbox图片 :( ）

神奇之处就在这里，因为他们发现可以请求Neo4j查找信用之间是否存在抵消并进行结算

MATCH (creditor:Friend {name:'Chewbacca'})-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0
FOREACH (credit in ns| SET credit.Amount = credit.Amount - Compensation)
RETURN Chain_length, People_involved, Compensation;

在这种情况下，丘巴卡和汉之间沿着长度为2的链条发生了20£的抵消。查询MATCH从丘巴卡开始，因为他最近的信用更新可能产生了一条链条。

到目前为止都很简单，但让我们继续故事，看看这个工具的强大之处。

他们恢复旅程后，遭遇了一场流星雨，护盾受损了。

于是他们去取了一个机械零件来替换损坏的零件，蕾伊支付了1200£（每人300£）。

这使得网络更加复杂，因为他们得到了更多的抵消

MATCH (creditor:Friend)-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0 AND
ALL(n in People_involved where
          1=size([m in People_involved where m=n]))
RETURN Chain_length, People_involved, Compensation
ORDER BY Compensation DESC, Chain_length DESC

现在棘手的部分来了：我们有不同的链条，补偿金额不同，但有一些共同的信用节点。这使得我们无法像处理丘巴卡和汉那样一次性抵消所有事项。我们必须谨慎地逐条链进行补偿。最好的解决方案是补偿金额最多且链条最长的那一条（当补偿金额相同时，最好优先满足人数最多的链条）。

MATCH (creditor:Friend)-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0 AND
ALL(n in People_involved where
          1=size([m in People_involved where m=n]))

WITH Chain_length, People_involved, Compensation, ns
ORDER BY Compensation DESC, Chain_length DESC
LIMIT 1
FOREACH (credit in ns| SET credit.Amount = credit.Amount - Compensation)
RETURN Chain_length, People_involved, Compensation

继续故事，他们拿到护盾零件后，去了机械师那里，芬恩支付了修理费600£（每人150£），他们将这笔费用添加到系统中。

与此同时，丘巴卡和汉决定去酒馆喝一杯班萨奶酒和一杯绝地啤酒，丘巴卡支付了50£（每人25£）。

数据库再次提供了帮助，找到了一些抵消来平账。

MATCH (creditor:Friend)-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0 AND
ALL(n in People_involved where
          1=size([m in People_involved where m=n]))

WITH Chain_length, People_involved, Compensation, ns
ORDER BY Compensation DESC, Chain_length DESC
LIMIT 1
FOREACH (credit in ns| SET credit.Amount = credit.Amount - Compensation)
RETURN Chain_length, People_involved, Compensation

并且多次运行查询后，他们达到了所有款项都被抵消的地步。

MATCH (creditor:Friend)-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0 AND
ALL(n in People_involved where
          1=size[m in People_involved where m=n]))

WITH Chain_length, People_involved, Compensation, ns
ORDER BY Compensation DESC, Chain_length DESC
LIMIT 1
FOREACH (credit in ns| SET credit.Amount = credit.Amount - Compensation)
RETURN Chain_length, People_involved, Compensation
MATCH (creditor:Friend)-[:CREDIT]->(credit:Credit),
p = (credit)-[:CHAIN*1..]->(credit)
WITH TAIL(NODES(p)) AS ns, credit

WITH ns,
	SIZE(ns) AS Chain_length,
    [x IN ns | x.Creditor] AS People_involved,
    REDUCE(comp = credit.Amount, n in ns | CASE WHEN comp < n.Amount THEN comp ELSE n.Amount END) AS Compensation
WHERE Compensation > 0 AND
ALL(n in People_involved where
          1=size([m in People_involved where m=n]))

WITH Chain_length, People_involved, Compensation, ns
ORDER BY Compensation DESC, Chain_length DESC
LIMIT 1
FOREACH (credit in ns| SET credit.Amount = credit.Amount - Compensation)
RETURN Chain_length, People_involved, Compensation

他们知道这项技术是新革命的开端。这确实可以极大地简化反叛军的整个财务系统，并节省大量金钱和时间。

他们飞回自己的星球，向奥加纳将军展示了他们的发现。

故事的其余部分广为人知，但直到此刻，没有人解释过在这个如此分散的共和国里，贫穷的反叛军是如何管理他们的资金的。

未来发展

在我的工作中，我介绍了一种抵消链条的方法，这种方法有点机械化，但我相信有更好的方法。首先，描述一种查询来抵消所有不相连的子网络将非常有趣，此外，如果能找到一种一次性抵消所有事项的方法就太好了。

结论

这是一个有趣的故事，用来解释图数据库真正可以发挥最大作用的一个用例。它可以描述这样一个内在相互连接的网络，并对数据执行非常有描述性的查询。我的工作只是解决寻找共享信用子网络问题的一种方法，并且可以作为开发分布式经济系统的一个想法。我真的要感谢Neo4j，因为它描述连接数据的方法开启了新的思维方式，并使人们能够更好地洞察信息的真实价值。

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