Question

Here is an excise for graph.

Given an undirected graph G with n vertices and m edges, and an integer k, give an O(m + n) algorithm that finds the maximum induced subgraph H of G such that each vertex in H has degree ≥ k, or prove that no such graph exists.
An induced subgraph F = (U, R) of a graph G = (V, E) is a subset of U of the vertices V of G, and all edges R of G such that both vertices of each edge are in U.

My initial idea is like this:

First, this excise actually asks that we have all vertices S whose degrees are bigger than or equal to k, then we remove vertices in S who don’t have any edge connected to others. Then the refined S is H, in which all vertices have degree >= k and the edges between them is R.

In addition, it asks O(m+n), so I think I need to a BFS or DFS. Then I get stuck.

In BFS, I can know the degree of a vertex. But once I get the degree of v (a vertex), I don’t know other connected vertices except for its parent. But if the parent doesn’t have degree >= k, I can’t eliminate v as it may still be connected with others.

Any hints?

---

Edit:

According to the answer of @Michael J. Barber, I implemented it and update the code here:

Can anyone have a look at the key method of the codes public Graph kCore(Graph g, int k)? Do I do it right? Is it O(m+n)?

class EdgeNode {
   EdgeNode next;
   int y;
}

public class Graph {
   public EdgeNode[] edges;
   public int numVertices;

   public boolean directed;

   public Graph(int _numVertices, boolean _directed) {
      numVertices = _numVertices;
      directed = _directed;
      edges = new EdgeNode[numVertices];
   }

   public void insertEdge(int x, int y) {
      insertEdge(x, y, directed);
   }

   public void insertEdge(int x, int y, boolean _directed) {
      EdgeNode edge = new EdgeNode();
      edge.y = y;
      edge.next = edges[x];
      edges[x] = edge;
      if (!_directed)
          insertEdge(y, x, true);
   }

   public Graph kCore(Graph g, int k) {
      int[] degree = new int[g.numVertices];
      boolean[] deleted = new boolean[g.numVertices];
      int numDeleted = 0;
      updateAllDegree(g, degree);// get all degree info for every vertex

      for (int i = 0;i < g.numVertices;i++) {
          **if (!deleted[i] && degree[i] < k) {
           deleteVertex(p.y, deleted, g);
          }**
      }

      //Construct the kCore subgraph
      Graph h = new Graph(g.numVertices - numDeleted, false);
      for (int i = 0;i < g.numVertices;i++) {
         if (!deleted[i]) {
           EdgeNode p = g[i];
           while(p!=null) {
              if (!deleted[p.y])
                 h.insertEdge(i, p.y, true); // I just insert the good edge as directed, because i->p.y is inserted and later p.y->i will be inserted too in this loop.
                 p = p.next;
              }
           }
         }
      }

      return h;
  }

  **private void deleteVertex(int i, boolean[] deleted, Graph g) {
     deleted[i] = true;
     EdgeNode p = g[i];
     while(p!=null) {
        if (!deleted[p.y] && degree[p.y] < k) 
            deleteVertex(p.y, deleted, g);
        p = p.next;
     }
  }**

  private void updateAllDegree(Graph g, int[] degree) {
     for(int i = 0;i < g.numVertices;i++) {
         EdgeNode p = g[i];
         while(p!=null) {
            degree[i] += 1;
            p = p.next;
         }
     }
  }

}

Answer 1

A maximal induced subgraph where the vertices have minimum degree k is called a k-core. You can find the k-cores just by repeatedly removing any vertices with degree less than k.

In practice, you first evaluate the degrees of all the vertices, which is O(m). You then go through the vertices looking for vertices with degree less than k. When you find such a vertex, cut it from the graph and update the degrees of the neighbors, also deleting any neighbors whose degrees drop below k. You need to look at each vertex at least once (so doable in O(n)) and update degrees at most once for each edge (so doable in O(m)), giving a total asymptotic bound of O(m+n).

The remaining connected components are the k-cores. Find the biggest one by evaluating their sizes.