Find which vertices to delete from graph to get smallest largest component

Given a graph $G = (V, E)$, find $k$ vertices $\{v^_1,\dots,v^_k\}$, which removal would result in a graph with smallest largest component.

I assume for large $n = |V|$ and large $k$ the problem is difficult (NP-hard), but I am interested in small values of $k$ ($k \in \{1, 2, 3, 4\}$).

For $k = 1$, I think it is possible to find best vertex $\{v^*_1\}$ to remove by performing single depth-first-search of the graph (i.e., checking articulation points).

For $k = 2$, it would be possible to find best vertices $\{v^_1, v^_2\}$ by performing $n$ depth-first searches (each of them for graph $G_i = G / \{v_i\}$). A similar approach could be applied in the case $k > 2$.

I wonder if there is any better solution than that.

(Related: counting the minimum number of vertices without necessarily enumerating them)

The problem you are describing is known as Component Order Connectivity in the field of vulnerability measures of graphs. The decision version of the problem is as follows:


Component Order Connectivity:


Input: Graph $G = (V,E)$, integers $k$ and $\ell$


Question: Does there exist a set of vertices $X \subseteq V$ of size at most $k$ such that the size of the largest component of $G - X$ is at most $\ell$?

The problem is obviously NP-complete since it generalizes vertex cover; the case when $\ell=1$ is vertex cover. Hence the problem cannot be fixed parameter tractable when parameterized by $\ell$ (unless $FPT = W[1]$). The problem is also known to be $W[1]$-hard when parameterized by $k$. Hence, we have to resort to algorithms with an exponential running time in $k+\ell$.

Very interesting question. For input $G,k,\ell$, a brute force approach would be:

branching(G,k,l):
    Find a connected set of vertices D of G of size l+1
    if no such D exists:
            return True // no component of size > l
    for v in D:
        if branching(G-v,k-1,l):
            return True
    return False

The algorithm runs in time $(\ell + 1)^k \cdot n^2$.

Observe that any yes instance $G,k,\ell$ of the problem has treewidth, and indeed pathwidth at most $k+\ell$. This can be observed by seeing that taking a deletion set $X$ of size at most $k$ yields a graph $G-X$ where every connected component has size at most $\ell$. Hence, a valid path decomposition is simply to construct one bag for each of the components in $G-X$ and then add all of $X$ to every bag. It follows that any yes instance has $|E(G)| \leq n(k+\ell)$.

A related problem has been studied in the past under the name Graph Integrity, or Vertex Integrity to distinguish the vertex deletion version and the edge deletion version:


Vertex Integrity:


Input: Graph $G = (V,E)$, integer $p$


Question: Does there exist a set of vertices $X \subseteq V$ such that $|X| + \max_{D \in cc(G-X)}|D| \leq p$?

That is, the sum of the deletion set and the size of the maximal component should be minimized. This problem is also NP-hard. See, e.g.,

• Clark, L.H., Entringer, R.C., Fellows, M.R.: Computational complexity of integrity. J. Combin. Math. Combin. Comput 2, 179–191 (1987)

• Fellows, M., Stueckle, S.: The immersion order, forbidden subgraphs and the complexity of network integrity. J. Combin. Math. Combin. Comput 6, 23–32 (1989)

StackExchange Computer Science Q#12789, answer score: 10