A graph is said to be a König graph if the size of its maximum matching is equal to the size of its minimum vertex cover. The König Edge Deletion problem asks if in a given graph there exists a set of at most k edges whose deletion results in a König graph. While the vertex version of the problem (König vertex deletion) has been shown to be fixed-parameter tractable more than a decade ago, the fixed-parameter-tractability of the König Edge Deletion problem has been open since then, and has been conjectured to be W[1]-hard in several papers. In this paper, we settle the conjecture by proving it W[1]-hard. We prove that a variant of this problem, where we are given a graph G and a maximum matching M and we want a k-sized König edge deletion set that is disjoint from M, is fixed-parameter-tractable.2. Let U be the set of unsaturated vertices of M, then S ∩ U = ∅. From the above observation it follows that for every minimum vertex cover S and maximum matching M of a König graph G, M lies across the edges (S, V(G) \ S) and M saturates S. Moreover, if there exists a maximum matching M and minimum vertex cover S such that M saturates S and M lies across (S, V(G) \ S), then |M| = |S| and hence G must be König. So we have the following characterization of König graphs. Lemma 2.1. [7, 26] A graph G is König if and only if for every minimum vertex cover S of G, there exists a matching across (S, V(G) \ S) that saturates S.As for any vertex cover S and a matching M of G, |S| ≥ |M|, we also have the following equivalent characterization using the fact that if there is a matching M and a vertex cover S such that |M| = |S|, then M must be a maximum matching and S must be a minimum vertex cover. E) is König if and only if for some vertex cover S of G, there exists a matching across (S, V(G) \ S) that saturates S.In passing, we observe the following interesting property of König graphs that we couldn't find in literature. Though we don't use it in the rest of our paper, we feel that this can be of independent interest. Lemma 2.3. A graph G is König if and only if vc f (G) = vc(G), that is, its vertex cover number is equal to its fractional vertex cover number.Proof. If G is König, then vc f (G) = vc(G) as observed in Subsection 2.2. To prove the converse, find the optimum solution to the vertex cover linear program (that is, the fractional vertex cover) of G that satisfies the properties in Theorem 1.By definition, S 0 induces an independent set. As vc f (G) = vc(G), by the second property of the theorem, S 1/2 = ∅ as there exists an optimum solution to the vertex cover linear program which is integral. Hence (S 1 , S 0 ) is a partition of the vertex set and S 1 is a vertex cover. By the third property of the theorem, there exists a matching across (S 1 , S 0 ) that saturates S 1 . Hence the lemma follows from Lemma 2.2. 4