找到目标的目标即朋友的朋友有力指导图

Question

I am using D3.JS libraries force directed graph,

I am using the following code to find target nodes

graph.links.forEach(function(d) {
          linkedByIndex[d.source.index + "," + d.target.index] = 1;
          linkedByIndex[d.target.index + "," + d.source.index] = 1;
                  });
    });


    function neighboring(a, b) {
      return a.index == b.index || linkedByIndex[a.index + "," + b.index];
    }

How can I find and highlight target's target ? Can anybody help me with this ?

EDIT:

Solved the problem in the following way:

First created the adjacent matrix:

var adjMat=null;
var adjMatSq=null;


adjMat=new Array(graph.nodes.length);
        for(var i = 0;i < graph.nodes.length; ++i)
        {
            adjMat[i]=new Array(graph.nodes.length);            
            for(var j = 0; j < graph.nodes.length; ++j)
            {
                adjMat[i][j] = 0;
            }
        }

Then assigned the values to the adjacent matrix:

graph.links.forEach(function(d) {
          adjMat[d.source.index][d.target.index] = adjMat[d.target.index][d.source.index] = 1;
          adjMat[d.source.index][d.source.index] = 1; 
        });

        adjMatSq=matrixMultiply(adjMat, adjMat);
    });

Then I found the square of matrix so that I'll be able to get the second degree nodes:

function matrixMultiply(m1,m2)
        {
            var result = [];
          for(var j = 0; j < m2.length; j++) {
            result[j] = [];
            for(var k = 0; k < m1[0].length; k++) {
                var sum = 0;
              for(var i = 0; i < m1.length; i++) {
                    sum += m1[i][k] * m2[j][i];
                }
                result[j].push(sum);
            }
        }
        return result;
        }

Defined a function to find the second degree nodes:

    function areAtSecondDegree(a,c)
    {
        return adjMatSq[a.index][c.index] == 1;
    }

My Code Plnkr

Answer 1

The principle is as follows. Given a specific node, determine its immediate neighbours. To get second degree neighbours, take this list of neighbours you have just determined, and for each get the list of neighbours again. The union of all those lists is the list of first and second degree neighbours.

In code this could look like this:

var connected = {};
  var friends = graph.nodes.filter(function(o) { return areFriends(d, o); });
  friends.forEach(function(o) {
      connected[o.name] = 1;
      // second pass to get second-degree neighbours
      graph.nodes.forEach(function(p) {
        if(areFriends(o, p)) {
          connected[p.name] = 1;
        }
      });
  });

If you want to have arbitrary n degree neighbours, you would need to change this to accommodate the fact that you don't know how many iterations of this to run.

Complete demo here .