Search for all connected indirect graph components

Question

I have a list of objects (non-oriented edges) as shown below:

pairs = [

 pair:["a2", "a5"],
 pair:["a3", "a6"],
 pair:["a4", "a5"],
 pair:["a7", "a9"]

];

I need to find all the components (linked nodes) in separate groups. Therefore, from these pairs I need to get:

groups = [
  group1: ["a2", "a5", "a4"],
  group2: ["a3", "a6"],
  group3: ["a7", "a9"]
];

I actually read some answers here and searched for it, and I realized that this is called “finding related components in a graph”, but cannot find any code example . I use JavaScript on Node.js, but any example with other languages ​​will be very useful. Thank you

Answer 1

, .

, , . , , 2 ..

- , . , .

JavaScript. node.js :

// Breadth First Search function
// v is the source vertex
// all_pairs is the input array, which contains length 2 arrays
// visited is a dictionary for keeping track of whether a node is visited
var bfs = function(v, all_pairs, visited) {
  var q = [];
  var current_group = [];
  var i, nextVertex, pair;
  var length_all_pairs = all_pairs.length;
  q.push(v);
  while (q.length > 0) {
    v = q.shift();
    if (!visited[v]) {
      visited[v] = true;
      current_group.push(v);
      // go through the input array to find vertices that are
      // directly adjacent to the current vertex, and put them
      // onto the queue
      for (i = 0; i < length_all_pairs; i += 1) {
        pair = all_pairs[i];
        if (pair[0] === v && !visited[pair[1]]) {
          q.push(pair[1]);
        } else if (pair[1] === v && !visited[pair[0]]) {
          q.push(pair[0]);
        }
      }
    }
  }
  // return everything in the current "group"
  return current_group;
};

var pairs = [
  ["a2", "a5"],
  ["a3", "a6"],
  ["a4", "a5"],
  ["a7", "a9"]
];

var groups = [];
var i, k, length, u, v, src, current_pair;
var visited = {};

// main loop - find any unvisited vertex from the input array and
// treat it as the source, then perform a breadth first search from
// it. All vertices visited from this search belong to the same group
for (i = 0, length = pairs.length; i < length; i += 1) {
  current_pair = pairs[i];
  u = current_pair[0];
  v = current_pair[1];
  src = null;
  if (!visited[u]) {
    src = u;
  } else if (!visited[v]) {
    src = v;
  }
  if (src) {
    // there is an unvisited vertex in this pair.
    // perform a breadth first search, and push the resulting
    // group onto the list of all groups
    groups.push(bfs(src, pairs, visited));
  }
}

// show groups
console.log(groups);

UPDATE. , , . . " " , , ( , ).

// Converts an edgelist to an adjacency list representation
// In this program, we use a dictionary as an adjacency list,
// where each key is a vertex, and each value is a list of all
// vertices adjacent to that vertex
var convert_edgelist_to_adjlist = function(edgelist) {
  var adjlist = {};
  var i, len, pair, u, v;
  for (i = 0, len = edgelist.length; i < len; i += 1) {
    pair = edgelist[i];
    u = pair[0];
    v = pair[1];
    if (adjlist[u]) {
      // append vertex v to edgelist of vertex u
      adjlist[u].push(v);
    } else {
      // vertex u is not in adjlist, create new adjacency list for it
      adjlist[u] = [v];
    }
    if (adjlist[v]) {
      adjlist[v].push(u);
    } else {
      adjlist[v] = [u];
    }
  }
  return adjlist;
};

// Breadth First Search using adjacency list
var bfs = function(v, adjlist, visited) {
  var q = [];
  var current_group = [];
  var i, len, adjV, nextVertex;
  q.push(v);
  visited[v] = true;
  while (q.length > 0) {
    v = q.shift();
    current_group.push(v);
    // Go through adjacency list of vertex v, and push any unvisited
    // vertex onto the queue.
    // This is more efficient than our earlier approach of going
    // through an edge list.
    adjV = adjlist[v];
    for (i = 0, len = adjV.length; i < len; i += 1) {
      nextVertex = adjV[i];
      if (!visited[nextVertex]) {
        q.push(nextVertex);
        visited[nextVertex] = true;
      }
    }
  }
  return current_group;
};

var pairs = [
  ["a2", "a5"],
  ["a3", "a6"],
  ["a4", "a5"],
  ["a7", "a9"]
];

var groups = [];
var visited = {};
var v;

// this should look like:
// {
//   "a2": ["a5"],
//   "a3": ["a6"],
//   "a4": ["a5"],
//   "a5": ["a2", "a4"],
//   "a6": ["a3"],
//   "a7": ["a9"],
//   "a9": ["a7"]
// }
var adjlist = convert_edgelist_to_adjlist(pairs);

for (v in adjlist) {
  if (adjlist.hasOwnProperty(v) && !visited[v]) {
    groups.push(bfs(v, adjlist, visited));
  }
}

console.log(groups);

Answer 2

, , Disjoint Set Forest.

, .

• , . root(v). . , root(v) . , root(v) .
• , - . , -1. , parent[v] = -1 v, , .
• , , . . . . rank. 1s
• , . , .

:

parent[v] = -1 for v in Vertices
rank[v] = 1 for v in Vertices
root (v):
   processed = []
   while parent[v] != -1
      processed << v
      v = parent[v]
   for vertex : processed
      parent = v // optimisation: here we move the assoc. trees to be directly connected the root
   return v

join (v1, v2):
  if rank[v1] < rank[v2]:
     parent[v1] = v2
  if rank[v1] > rank[v2]:
     parent[v2] = v1
  parent[v2] = v1
  rank[v1]++

merge_trees (v1, v2)
  root1 = root(v1)
  root2 = root(v2)
  if root1 == root2:
    // already in same tree nothing else to be done here
    return true
  else
    // join trees
    join (v1, v2)
    return false

main:
   numberTrees = size(Vertives)
   for edge: edges
      if merge_trees(edge.begin, edge.end):
         numberTrees--
   print numberTrees // this is the number you are interested in.

, . , , , . join . , , .

Answer 3

# , , ""

// It will return an object like Vertices{node: EdgesTo{node,node}, node:...}
function toGraph(arr) {
    var graph = {}; // this will hold the node "IDs"
    for (var i = 0; i < arr.length; i++) {
        // "create node" if the it not added in the graph yet
        graph[arr[i][0]] = graph[arr[i][0]] || {};
        graph[arr[i][1]] = graph[arr[i][1]] || {};
        // add bidirectional "edges" to the "vertices"
        // Yes, we set the value to null, but what important is to add the key.
        graph[arr[i][0]][arr[i][1]] = null;
        graph[arr[i][1]][arr[i][0]] = null;
    }
    return graph;
}

(DFS, BFS)

DFS:

// to be called after getting the result from toGraph(arr)
function getSubGraphs(graph) {
    var subGraphs = []; // array of connected vertices
    var visited = {};
    for (var i in graph) { // for every node...
        var subGraph = dfs(graph, i, visited); // ... we call dfs
        if (subGraph != null) // if vertex is not added yet in another graph
        subGraphs.push(subGraph);
    }
    return subGraphs;
}

// it will return an array of all connected nodes in a subgraph
function dfs(graph, node, visited) {
    if (visited[node]) return null; // node is already visited, get out of here.
    var subGraph = [];
    visited[node] = true;
    subGraph.push(node);
    for (var i in graph[node]) {
        var result = dfs(graph, i, visited);
        if (result == null) continue;
        subGraph = subGraph.concat(result);
    }
    return subGraph;
}

And you end up calling it like that getSubGraphs(toGraph(myArray));and doing whatever you need.

Fiddle Here