高效增量实现偏序集

Question

我正在用SQLAlchemy实现一个结构，这个结构有一个数学特性，叫做部分有序集合，我需要能够一次添加或删除一条边。

在我目前最好的设计中，我使用了两个邻接表，一个是分配列表（大概就是Hass图中的边），因为我需要保留哪些节点对是明确设置为有序的，另一个邻接表是第一个的传递闭包，这样我就可以高效地查询一个节点是否相对于另一个节点是有序的。现在，每次在分配邻接列表中添加或删除边时，我都会重新计算传递闭包。

它看起来像这样：

assignment = Table('assignment', metadata,
    Column('parent', Integer, ForeignKey('node.id')),
    Column('child', Integer, ForeignKey('node.id')))

closure = Table('closure', metadata,
    Column('ancestor', Integer, ForeignKey('node.id')),
    Column('descendent', Integer, ForeignKey('node.id')))

class Node(Base):
    __tablename__ = 'node'
    id = Column(Integer, primary_key=True)

    parents = relationship(Node, secondary=assignment,
        backref='children',
        primaryjoin=id == assignment.c.parent,
        secondaryjoin=id == assignment.c.child)

    ancestors = relationship(Node, secondary=closure,
        backref='descendents',
        primaryjoin=id == closure.c.ancestor,
        secondaryjoin=id == closure.c.descendent,
        viewonly=True)

    @classmethod
    def recompute_ancestry(cls.conn):
        conn.execute(closure.delete())
        adjacent_values = conn.execute(assignment.select()).fetchall()
        conn.execute(closure.insert(), floyd_warshall(adjacent_values))

其中floyd_warshall()是同名算法的实现。

这让我遇到了两个问题。第一个是，这似乎不是很高效，但我不太确定可以用什么算法来替代。

第二个问题更实际一些，就是每次分配发生时都必须明确调用Node.recompute_ancestry()，而且只能在分配被刷新到会话中并且连接正确之后。如果我想在ORM中看到这些变化，我还得再刷新一次会话。我觉得如果能把重新计算祖先的操作用ORM来表达，那会简单得多。

Answer 1

在插入数据的时候，更新闭包，并且要用ORM的方式来做：

def add_assignment(parent, child):
"""And parent-child relationship between two nodes"""
    parent.descendants += child.descendants + [child]
    child.ancestors += parent.ancestors + [parent] 
    parent.children += child

如果你需要删除一些分配，这样用纯SQL会更快：

def del_assignment(parent, child):
    parent.children.remove(child)
    head = [parent.id] + [node.id for node in parent.ancestors]
    tail = [child.id] + [node.id for node in child.descendants]
    session.flush()
    session.execute(closure.delete(), and_(
          closure.c.ancestor.in_(head), 
          closure.c.descendant.in_(tail)))
    session.expire_all()

Answer 2

好吧，我自己找到了问题的解决办法。简单来说，就是在父节点的祖先的后代和子节点的后代的祖先的交集上应用Floyd-Warshall算法，但只把结果应用到父节点的祖先和子节点的后代的并集上。我花了很多时间在这个上，最后决定把这个过程发在我的博客上，不过这里有代码。

from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

association_table = Table('edges', Base.metadata,
    Column('predecessor', Integer, 
           ForeignKey('nodes.id'), primary_key=True),
    Column('successor', Integer, 
           ForeignKey('nodes.id'), primary_key=True))

path_table = Table('paths', Base.metadata,
    Column('predecessor', Integer, 
           ForeignKey('nodes.id'), primary_key=True),
    Column('successor', Integer, 
           ForeignKey('nodes.id'), primary_key=True))

class Node(Base):
    __tablename__ = 'nodes'
    id = Column(Integer, primary_key=True)
    # extra columns

    def __repr__(self):
        return '<Node #%r>' % (self.id,)

    successors = relationship('Node', backref='predecessors',
        secondary=association_table,
        primaryjoin=id == association_table.c.predecessor,
        secondaryjoin=id == association_table.c.successor)

    before = relationship('Node', backref='after',
        secondary=path_table,
        primaryjoin=id == path_table.c.predecessor,
        secondaryjoin=id == path_table.c.successor)

    def __lt__(self, other):
        return other in self.before

    def add_successor(self, other):
        if other in self.successors:
            return
        self.successors.append(other)
        self.before.append(other)
        for descendent in other.before:
            if descendent not in self.before:
                self.before.append(descendent)
        for ancestor in self.after:
            if ancestor not in other.after:
                other.after.append(ancestor)

    def del_successor(self, other):
        if not self < other:
            # nodes are not connected, do nothing!
            return
        if not other in self.successors:
            # nodes aren't adjacent, but this *could*
            # be a warning...
            return

        self.successors.remove(other)

        # we buld up a set of nodes that will be affected by the removal
        # we just did.  
        ancestors = set(other.after)
        descendents = set(self.before)

        # we also need to build up a list of nodes that will determine
        # where the paths may be.  basically, we're looking for every 
        # node that is both before some node in the descendents and
        # ALSO after the ancestors.  Such nodes might not be comparable
        # to self or other, but may still be part of a path between
        # the nodes in ancestors and the nodes in descendents.
        ancestors_descendents = set()
        for ancestor in ancestors:
            ancestors_descendents.add(ancestor)
            for descendent in ancestor.before:
                ancestors_descendents.add(descendent)

        descendents_ancestors = set()
        for descendent in descendents:
            descendents_ancestors.add(descendent)
            for ancestor in descendent.after:
                descendents_ancestors.add(ancestor)
        search_set = ancestors_descendents & descendents_ancestors

        known_good = set() # This is the 'paths' from the 
                           # original algorithm.  

        # as before, we need to initialize it with the paths we 
        # know are good.  this is just the successor edges in
        # the search set.
        for predecessor in search_set:
            for successor in search_set:
                if successor in predecessor.successors:
                    known_good.add((predecessor, successor))

        # We now can work our way through floyd_warshall to resolve
        # all adjacencies:
        for ancestor in ancestors:
            for descendent in descendents:
                if (ancestor, descendent) in known_good:
                    # already got this one, so we don't need to look for an
                    # intermediate.  
                    continue
                for intermediate in search_set:
                    if (ancestor, intermediate) in known_good \
                            and (intermediate, descendent) in known_good:
                        known_good.add((ancestor, descendent))
                        break # don't need to look any further for an
                              # intermediate, we can move on to the next
                              # descendent.  


        # sift through the bad nodes and update the links
        for ancestor in ancestors:
            for descendent in descendents:
                if descendent in ancestor.before \
                        and (ancestor, descendent) not in known_good:
                    ancestor.before.remove(descendent)