I'm interested in the complexity status of the following problem.

Input: a graph $G=(V,E)$ and two natural numbers $k$ and $d$.

Output: Yes, if there exists a subset $E' \subseteq E$ of cardinality at most $k$ such that the maximum degree of $G-E' = (V,E \setminus E')$ is at most $d$. No, otherwise.


1 Answer 1


Equivalently, your problem may be stated as follows: You are looking for a subgraph, in which every vertex degree lies in the interval $[0,d]$ and that contains as many edges as possible.

This problem is well-known to be solvable in polynomial time.
For instance, it is discussed in Chapter 10 (and in particular in Exercise 10.1.4) of the book

M.D. Plummer and L. Lovász:
Matching Theory
Volume 29 of Annals of Discrete Mathematics, Elsevier, 1986.

  • $\begingroup$ Thanks, Gamow! It is exactly what I was looking for. $\endgroup$
    – Victor
    Mar 30, 2019 at 14:11
  • $\begingroup$ Gamow, isn't Exercise 10.1.4 about (inclusion-wise) maximal subgraph $H$? How do we derive the desired result about a subgraph that contains as many edges as possible? $\endgroup$
    – Victor
    Mar 30, 2019 at 19:07
  • $\begingroup$ @Victor: No, no, no. Just study the chapter, and you will see that it discusses the max-cardinality version. (Finding an inclusion-wise maximal subgraph would be trivial, and far below the level of Lovasz and Plummer.) $\endgroup$
    – Gamow
    Mar 30, 2019 at 20:11
  • $\begingroup$ Gamow, indeed, finding an inclusion-wise maximal subgraph is trivial. I was just confused by the fact that the terms "maximal" and "minimal" are defined in the inclusion-wise sense in the book. $\endgroup$
    – Victor
    Mar 30, 2019 at 21:23

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