Yes, getOtherNode. Someone's sleepy.
Sean
On Sep 30, 2013, at 5:14 AM, Beth Kar wrote:
> Thank you Sean, I tried your approach and so far it works fine for me.
> However, currently I only work on a relatively small network of agents (15).
> However, since there is no method getOtherEdge(from) I guess you meant
> getOtherNode(from), right?
>
> So, the code I used is:
>
> public Edge getEdge(Network network, Object from, Object to){
> Bag b = network.getEdgesOut(from);
>
> for (int i=0; i<b.size(); i++){
>
> Edge e = (Edge)(b.get(i));
>
> if(e.getOtherNode(from).equals(to)){
> return e;
> }
>
> }
>
> return null;
> }
>
> and then:
>
> if(getedge(AgentA, AgentB) == null){
> network.addEdge(AgentA, AgentB, null);
> }
>
> Beth.
>
>
> On Fri, 27 Sep 2013 22:09:31 -0400, Sean Luke <[log in to unmask]> wrote:
>
>> The implementations discussed so far have some problems. So some notes on
> people's implementations. First Tony's:
>>
>>> public boolean areNeighbors(Network network, Object agentA, Object agentB) {
>>> Bag edges = network.getEdges(agentA, null);
>>> Iterator<Edge> iter = edges.iterator();
>>> while (iter.hasNext()) {
>>> Edge e = iter.next();
>>> if (e.getOtherNode(agentA)==agentB)
>>> return true;
>>> }
>>> return false;
>>> }
>>
>> The problem with this approach is threefold. First, it searches the entire
> network looking for an edge between two nodes -- this is O(E), which is
> *extremely* slow, particularly when Network has an O(1) lookup for this!
> Second, if agentA isn't on either end of an edge, then getOtherNode(agentA)
> will return a node -- and in theory that node could be agentB. So using it
> like this will have bugs. Third, using == is probably not preferable to
> using .equals(...) since Network uses hash lookups (which use .equals(...)
> internally).
>>
>> Beth's approach:
>>
>>> Bag edges = mynetwork.getEdges(agentA, null);
>>> int length = edges.size();
>>> boolean exists = false;
>>>
>>> for (int node=0; node<len; node++) {
>>> Edge e = (Edge)(edges.get(node));
>>> if(e.getOtherNode(agentA) == agentB) exists = true;
>>> else continue;
>>> }
>>>
>>> if (!exists) mynetwork.addEdge(agentA, agentB, null);
>>
>> This has the same three issues.
>>
>> I suggest instead doing this. Let the two objects be "from" and "to"
> instead of agentA and agentB. If your graph is undirected, it doesn't
> matter which is from and which is to.
>>
>> public Edge getEdge(Network network, Object from, Object to)
>> {
>> Bag b = network.getEdgesOut(from); // will never be null
>> for(int i = 0; i < b.size(); i++)
>> {
>> Edge e = (Edge)(b.get(i));
>> if (e.getOtherEdge(from).equals(to))
>> return e;
>> }
>> return null;
>> }
>>
>> If an edge exists, this method returns it, else it returns null. This
> should be a gazillion times faster for large networks.
>>
>> I've not tested this to see if it works, but if it doesn't, it should be a
> simple bugfix. If we can get some verification, I'll add a similar method
> to Network proper.
>>
>> Sean, embarrassed that this wasn't here in the first place...
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