[Treebase-devel] Determining implicit root

[Hilmar L. <hl...@dr...>]

Rutger - 

I'm curious and am probably missing something and therefore don't fully follow why the below works. Would you mind explaining? By edges pointing into a node, do you mean incoming (from ancestor) or outgoing (to descendant) edges? It sounds like outgoing (or otherwise there wouldn't be two, right), but why does NeXML then require a node with one outgoing edge at the root of the tree, artificial or not? Or do you mean outgoing plus incoming?

So I'm obviously missing something, but can you point me to what it is?

-hilmar

Sent with a tap.

On May 19, 2011, at 9:00 AM, rv...@us... wrote:

> Revision: 873
>          http://treebase.svn.sourceforge.net/treebase/?rev=873&view=rev
> Author:   rvos
> Date:     2011-05-19 13:00:28 +0000 (Thu, 19 May 2011)
> 
> Log Message:
> -----------
> Added additional loop to identify the root node based on the implicit topology (instead of by looking for the @root attribute).
> 
> Modified Paths:
> --------------
>    trunk/treebase-web/src/main/webapp/scripts/jsphylosvg-min.js
> 
> Modified: trunk/treebase-web/src/main/webapp/scripts/jsphylosvg-min.js
> ===================================================================
> --- trunk/treebase-web/src/main/webapp/scripts/jsphylosvg-min.js    2011-05-19 12:57:49 UTC (rev 872)
> +++ trunk/treebase-web/src/main/webapp/scripts/jsphylosvg-min.js    2011-05-19 13:00:28 UTC (rev 873)
> @@ -761,6 +761,27 @@
>            }
>        }
>        
> +        // RAV: the root attribute is used to indicate whether an implicitly rooted
> +        // topology should, from a biological p.o.v., be considered as such. However,
> +        // valid NeXML tree structures are always rooted, even if they don't have
> +        // the root attribute, in the sense that there's always going to be one node
> +        // that doesn't have two edges pointing into it. It seems to me that we can
> +        // still render these trees, we just have to find which node is the one that
> +        // doesn't have two edges pointing in. This loop does that.
> +        if (!root){
> +            for(i = 0; i < nexNodes.length; i++) {
> +                var targetCount = 0;
> +                for(j = 0; j < nexEdges.length; j++) {
> +                    if(nexEdges[j].target == nexNodes[i].id) {
> +                        targetCount++;
> +                    }
> +                }
> +                if ( targetCount < 2 ) {
> +                    root = nexNodes[i];                    
> +                }
> +            }
> +        }
> +        
>        if(root){
>            root = recursiveParse(root);
>        
> 
> 
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