A Note on the Linear Cycle Cover Conjecture of Gyárfás and Sárközy

A linear cycle in a 3-uniform hypergraph H is a cyclic sequence of hyperedges such that any two consecutive hyperedges intersect in exactly one element and non-consecutive hyperedges are disjoint. Let α(H) denote the size of a largest independent set of H. We show that the vertex set of every 3-uniform hypergraph H can be covered by at most α(H) edge-disjoint linear cycles (where we accept a vertex and a hyperedge as a linear cycle), proving a weaker version of a conjecture of Gyárfás and Sárközy. Mathematics Subject Classifications: 05C35, 05C69


Introduction
A well-known theorem of Pósa [3] states that the vertex set of every graph G can be partitioned into at most α(G) cycles where α(G) denotes the independence number of G (where a vertex or an edge is accepted as a cycle).Definition 1.A (linear cycle) linear path is a (cyclic) sequence of hyperedges such that two consecutive hyperedges intersect in exactly one element and two non-consecutive hyperedges are disjoint.
An independent set of a hypergraph H is a set of vertices that contain no hyperedges of H. Let α(H) denote the size of a largest independent set of H and we call it the the electronic journal of combinatorics 25(2) (2018), #P2.29 independence number of H. Gyárfás and Sárközy [2] conjectured that the following extension of Pósa's theorem holds: One can partition every k-uniform hypergraph H into at most α(H) linear cycles (here, as in Pósa's theorem, vertices and subsets of hyperedges are accepted as linear cycles).In [2] Gyárfás and Sárközy prove a weaker version of their conjecture for weak cycles (where only cyclically consecutive hyperedges intersect, but their intersection size is not restricted) instead of linear cycles.Recently, Gyárfás, Győri and Simonovits [1] showed that this conjecture is true for k = 3 if we assume there are no linear cycles in H.
In this note, we show their conjecture is true for k = 3 provided we allow the linear cycles to be edge-disjoint, instead of being vertex-disjoint.
Theorem 2. If H is a 3-uniform hypergraph, then its vertex set can be covered by at most α(H) edge-disjoint linear cycles (where we accept a single vertex or a hyperedge as a linear cycle).
Our proof uses induction on α(H).However, perhaps surprisingly, in order to make induction work, our main idea is to allow the hypergraph H to contain hyperedges of size 2 (in addition to hyperedges of size 3).First we will delete some vertices, and add certain hyperedges of size 2 into the remaining hypergraph so as to ensure the independence number of the remaining hypergraph is smaller than that of H. Then applying induction we will find edge-disjoint linear cycles (which may contain these added hyperedges) covering the remaining hypergraph.It will turn out that the added hyperedges behave nicely, allowing us to construct edge-disjoint linear cycles in H covering all of its vertices.The detailed proof is given in the next section.

Proof of Theorem 2
We call a hypergraph mixed if it can contain hyperedges of both sizes 2 and 3. A linear cycle in a mixed hypergraph is still defined according to Definition 1.We will in fact prove our theorem for mixed hypergraphs (which is clearly a bigger class of hypergraphs than 3-uniform hypergraphs).More precisely, we will prove the following stronger theorem.Theorem 3. If H is a mixed hypergraph, then its vertex set V (H) can be covered by at most α(H) edge-disjoint linear cycles (where we accept a single vertex or a hyperedge as a linear cycle).
Proof.We prove the theorem by induction on α(H).If |V (H)| = 1 or 2, then the statement is trivial.If |V (H)| ≥ 3 and α(H) = 1, then H contains all possible edges of size 2 and there is a Hamiltonian cycle consisting only of edges of size 2, which is of course a linear cycle covering V (H).
Let α(H) > 1.If E(H) = ∅, then α(H) = V (H) and the statement of our theorem holds trivially since we accept each vertex as a linear cycle.If E(H) = ∅, then let P be a longest linear path in H consisting of hyperedges h 0 , h 1 , . . ., h l (l ≥ 0).If h i is of size 3, then let h i = v i v i+1 u i+1 and if it is of size 2, then let h i = v i v i+1 .A linear subpath of P starting at v 0 (i.e., a path consisting of hyperedges h 0 , h 1 , . . ., h j for some j ≤ l) is called an initial segment of P .Let C be a linear cycle in H which contains the longest initial segment of P .If there is no linear cycle containing h 0 , then we simply let C = h 0 .

Let us denote the subhypergraph of H induced on
We will show that α(H ) < α(H) and any linear cycle cover of H can be extended to a linear cycle cover of H by adding C and extending the red edges by v 0 .
The following claim shows that the independence number of H is smaller than the independence number of H.This fact will later allow us to apply induction.Claim 4. If I is an independent set in H , then I ∪ v 0 is an independent set in H.
Proof.Suppose by contradiction that h ⊆ (I ∪ v 0 ) for some h ∈ E(H).Then, clearly v 0 ∈ h because otherwise I is not an independent set in H . Now let us consider different cases depending on the size of h ∩ (V (P ) \ V (C)).If |h ∩ (V (P ) \ V (C))| = 0 then, by adding h to P , we can produce a longer path than P , a contradiction.If |h ∩ (V (P ) \ V (C))| = 1, let h∩(V (P )\V (C)) = {x}.Then the linear subpath of P between v 0 and x together with h forms a linear cycle which contains a larger initial segment of P than C, a contradiction.
Let us take smallest i and j such that x ∈ h i and y ∈ h j (i.e., if x ∈ h i ∩ h i+1 then let us take h i ).If i = j, say i < j without loss of generality, then the linear subpath of P between v 0 and x together with h forms a linear cycle with longer initial segment of P than C, a contradiction.Therefore, i = j but in this case, {x, y} is a red edge and so at most one of them can be contained in I, contradicting the assumption that h = v 0 xy ⊆ (I ∪ v 0 ).Hence, I ∪ v 0 is an independent set in H, as desired.
The following claim will allow us to construct linear cycles in H from red edges.Claim 5.The set of hyperedges of every linear cycle in H contains at most one red edge.
Proof.Suppose by contradiction that there is a linear cycle C in H containing at least two hyperedges which are red edges.Then there is a linear subpath P of C consisting of hyperedges h 0 , h 1 , . . ., h m such that h 0 := v s u s and h m := v t u t (where s > t) are red edges but h k is not a red edge for any 1 ≤ k ≤ m − 1.Let us first take the smallest i such that V (P ) ∩ h i = ∅ and then the smallest j such that h j ∩ h i = ∅.It is easy to see that |V (P ) ∩ h i | ≤ 2 (since i was smallest).If h j ∩ h i = 1, then the linear cycle consisting of hyperedges h 1 , . . ., h j and h i , h i−1 , . . ., h 0 and v 0 v s u s contains a larger initial segment of P than C (as then notice that h j+1 ∩ h i = 1.Now the linear cycle consisting of the hyperedges h m−1 , h m−2 , . . ., h j+1 and h i , h i−1 , . . ., h 0 and v 0 v t u t contains a larger initial segment of P than C, a contradiction.By Claim 4, α(H ) ≤ α(H) − 1.So by induction hypothesis, V (H ) can be covered by at most α(H)−1 edge-disjoint linear cycles (where we accept a single vertex or a hyperedge as a linear cycle).Now let us replace each red edge {x, y} with the hyperedge xyv 0 of H. Claim 5 ensures that in each of these linear cycles, at most one of the hyperedges is a red edge.Therefore, it is easy to see that after the above replacement, linear cycles of H remain as linear cycles in H and they cover V (H ) = V (H) \ V (C).Now the linear cycle C, together with these linear cycles give us at most α(H) − 1 + 1 = α(H) edge-disjoint linear cycles covering V (H), completing the proof.