Minimum vertex cover with minimum overlap of one side

Question

In a bipartite graph $(X+Y,E)$, it is possible to find a minimum vertex cover in polynomial time, by Konig's theorem.

Suppose the minimum size is $s$. Is there an efficient algorithm for finding a vertex cover $S$ of size $s$, such that $S\cap X$ has a minimum cardinality?

What I could do so far is to formulate this problem as an ILP where each vertex $u\in X$ has a variable $x_u$ and each vertex $v\in Y$ has a variable $y_v$. Its LP relaxation is:

\begin{align} \text{minimize} && \sum_{u\in X} x_u && \\ \text{subject to} && x_u + y_v \geq 1 && \forall (u,v)\in E \\ && \sum_{u\in X} x_u + \sum_{v\in Y}y_v = s \\ && 1\geq x_{u}\geq 0, 1\geq y_v \geq 0 && \forall u\in X, v\in Y \end{align}

But I do not know how to check whether this ILP has an integer solution.

Is there a better approach to this problem?

Answer 1

Dulmage–Mendelsohn decomposition is a powerful tool to capture the structure of the bipartite graph, and it can be found by computing any maximum matching.

Let $G = (V = (A, B), E)$ be a bipartite graph, then $G$ can be partitioned into three parts $(C, H, R)$ and we let :

1. $C \subseteq V$ is an independent set.
2. $N(C) = H$.
3. $G[R]$ has a perfect marching.
4. $H$ is the intersection of all the minimum vertex covers.
5. Any maximum-matching in $G$ saturates all vertices in $R \cup U$.

According to the Dulmage–Mendelsohn decomposition, $H$ must be in the solution, so $S = H \cup (A \cap R)$ is a vertex cover with a minimum overlap of $B$ such that $S \cap B = H \cap B \subseteq H$.

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We can also solve this problem by using the maximum flow algorithm. Since the Vertex Cover in bipartite problem can be reduced to the minimum cut problem, this problem can be reduced to the minimum cut in the edge-weighted graphs.

Specially, we add a source node $s$ connecting every vertex in $A$ and add a sink node $t$ connecting every vertex in $B$. Suppose that $n = |V|$, the weight of every edge adjacent to the $s$ is $n$, and the weight of every edge adjacent to the $t$ is $n-1$. Besides the weight of the edge in the original graph is large enough such as $n^{2}$. Note that if the size of the minimum vertex cover in $G$ is $k$, we know that the minimum $s$-$t$ cut is in $((k-1)n, kn]$.

Answer 2

Suppose that the minimum VCP size is s.

If $|Y|=s$ then the optimal solution (without solving any problem) is $Y=1$ and $X=0$.

Else, we can solve $min$ $W_uX_u+W_vY_v$ s.t. {$x_u+y_v\ge 1$, $0\le x_u,y_v \le 1$}; i.e. the original VCP relaxation model, where $W_u=1$ and $W_v=1-\epsilon$.

Based on Nemhauser and Trotter Theorem (1975), all extreme optimal solutions have values $V_0=0$, $V_{0.5}=0.5$ or $V_1=1$ for their variables and always we have an optimal solution which is equal to the union of $V_1$ and a subset of $V_{0.5}$.

Therefore, in the optimal solution we have $z^*\approx s$ and between all optimal solutions of original problem with size $s$, the priority of vertex selection of the model is from $Y$.

Then, in optimal solution of the proposed model, select $V_1$ and then solve the problem on $V_{0.5}$. Note that, if $V_0=V_1=\emptyset$ and $V_{0.5}=V$, you can produce an adjacent optimal LP relaxation solution where $V_{0.5}\ne V$.