从节点集创建组

1. 浏览一下集合列表。每套
   1. 浏览小组列表。每组G
      1. 如果S可以是G的成员(即,如果G的集合是S的子集),则将S添加到G。
      2. 如果S不能成为G的成员,但S和G集合的交集包含多个节点,则为该交集创建一个新组并将其添加到列表中。
   2. 给S一个自己的组并将其添加到列表中。
2. 删除只有一个成员集的任何组。

[1,2,5,6] [a]
[1,5] [a,b]
[1,4,5] [b]

读了c之后,它是

[1,2,5,6] [a]
[1,5] [a,b,c]
[1,4,5] [b]
[1,2,5] [a,c]

如果速度是个问题的话,还有稍微更有效的算法。

/*
Pseudocode algorithm for creating groups data from a set dataset, further explained in the project documentation. This is based on 
http://stackoverflow.com/questions/1644387/create-groups-from-sets-of-nodes

I am assuming 
- Group is a structure (class) the objects of which contain two lists: a list of sets and a list of nodes (group.nodes). Its constructor accepts a list of nodes and a reference to a Set object
- Set is a list structure (class), the objects (set)  of which contain the nodes of the list in set.nodes
- groups and sets are both list structures that can contain arbitrary objects which can be iterated with foreach(). 
- you can get the objects two lists have in common as a new list with intersection()
- you can count the number of objects in a list with length()
*/

//Create groups, going through the original sets
foreach(sets as set){
    if(groups.nodes.length==0){
        groups.addGroup(new Group(set.nodes, set));
    }
    else{
        foreach (groups as group){
                if(group.nodes.length() == intersection(group.nodes,set.nodes).length()){
                    // the group is a subset of the set, so just add the set as a member the group
                    group.addset(set);
                    if (group.nodes.length() < set.nodes.length()){
                    // if the set has more nodes than the group that already exists, 
                    // create a new group for the nodes of the set, with set as a member of that group
                    groups.addGroup(new Group(set.nodes, set));
                    }
                }

                // If group is not a subset of set, and the intersection of the nodes of the group 
                // and the nodes of the set
                // is greater than one (they have more than one person in common), create a new group with 
                // those nodes they have in common, with set as a member of that group
                else if(group.nodes.length() > intersection(group.nodes,set.nodes).length() 
                    && intersection(group.nodes,set.nodes).length()>1){
                    groups.addGroup(new Group(intersection(group.nodes,set.nodes), set);
                }
        }
    }

}

// Cleanup time!
foreach(groups as group){
    //delete any group with only one member set (for it is not really a group then)
    if (group.sets.length<2){
        groups.remove(group);
    }
    // combine any groups that have the same set of nodes. Is this really needed? 
    foreach(groups2 as group2){
        //if the size of the intersection of the groups is the same size as either of the 
        //groups, then the groups have the same nodes.
        if (intersection(group.nodes,group2.nodes).length == group.nodes.length){
            foreach(group2.sets as set2){
                if(!group.hasset(set)){
                    group.addset(set2);
                }
            }
            groups.remove(group2);
        }
        }

}