1 Introduction
Graphs considered in this paper are undirected, simple and finite (unless otherwise noted). Given a graph with vertex set and edge set , for convenience, we often identify a triangle in with its edge set. A subset of is called a triangle cover if it intersects each triangle of . Let denote the minimum cardinality of a triangle cover of , referred to as the triangle covering number of . A set of pairwise edgedisjoint triangles in is called a triangle packing of . Let denote the maximum cardinality of a triangle packing of , referred to as the triangle packing number of . It is clear that holds for every graph . Our research is motivated by the following conjecture raised by Tuza [1] in 1981.
Conjecture 1.1 (Tuza’s Conjecture [1]).
holds for every graph .
To the best of our knowledge, the conjecture is still unsolved in general. If it is true, then the upper bound 2 is sharp as shown by and – the complete graphs of orders and .
Related work.
The only known universal upper bound smaller than 3 was given by Haxell [2], who shown that holds for all graphs . Haxell’s proof [2] implies a polynomialtime algorithm for finding a triangle cover of cardinality at most 66/23 times that of a maximal triangle packing. Other partial results on Tuza’s conjecture concern with special classes of graphs.
Tuza [3] proved his conjecture holds for planar graphs, free chordal graphs and graphs with vertices and at least edges. The proof for planar graphs [3] gives an elegant polynomialtime algorithm for finding a triangle cover in planar graphs with cardinality at most twice that of a maximal triangle packing. The validity of Tuza’s conjecture on the class of planar graphs was later generalized by Krivelevich [4] to the class of graphs without subdivision. Haxell and Kohayakawa [5] showed that for tripartite graphs , where . Haxell, Kostochka and Thomasse [6] proved that every free planar graph satisfies .
Regarding the tightness of the conjectured upper bound 2, Tuza [3] noticed that infinitely many graphs attain the conjectured upper bound . Cui, Haxell and Ma [7] characterized planar graphs satisfying ; these graphs are edgedisjoint unions of ’s plus possibly some vertices and edges that are not in triangles. Baron and Kahn [8] proved that Tuza’s conjecture is asymptotically tight for dense graphs.
Fractional and weighted variants of Conjecture 1.1 were studied in literature. Krivelevich [4] proved two fractional versions of the conjecture: and , where and are the values of an optimal fractional triangle cover and an optimal fractional triangle packing of , respectively. The result was generalized by Chapuy et al. [9] to the weighted version, which amounts to packing and covering triangles in multigraphs (obtained from by adding multiple edges). The authors [9] showed that and ; the arguments imply an LPbased 2approximation algorithm for finding a minimum weighted triangle cover.
Our contributions.
Along a different line, we establish new sufficient conditions for validity of Tuza’s conjecture by comparing the triangle packing number, the number of triangles and the number of edges. Given a graph , we use is a triangle in to denote the set consisting of the (edge sets of) triangles in . Without loss of generality, we focus on the graphs in which every edge is contained in some triangle. These graphs are called irreducible.
Theorem 1.2.
Let be an irreducible graph. Then a triangle cover of with cardinality at most can be found in polynomial time, which implies , if one of the following conditions is satisfied: (i) , (ii) , (iii) .
The primary idea behind the theorem is simple: any one of conditions (i) – (iii) allows us to remove at most edges from to make the resulting graph satisfy ; the removed edges and the edges in a minimum triangle cover of form a triangle cover of with size at most . The idea is realized by establishing new results on linear 3uniform hypergraphs (see Section 2); the most important one states that such a hypergraphs could be made acyclic by removing a number of vertices that is no more than a third of the number of its edges. A key observation here is that hypergraph is linear and 3uniform.
To show the qualities of conditions (i) – (iii) in Theorem 1.2, we obtain the following result which complements to the constants , and 2 in these conditions with , and , respectively.
Theorem 1.3.
Tuza’s conjecture holds for every graph if there exists some real such that Tuza’s conjecture holds for every irreducible graph satisfying one of the following properties: (i’) , (ii’) , (iii’) .
We also investigate Tuza’s conjecture on classical ErdősRényi random graph , and prove that provided and .
It is worthwhile pointing out that strengthening Theorem 1.2, our arguments actually establish stronger results for linear 3uniform hypergraphs (see Theorem 4.1).
The rest of paper is organized as follows. Section 2 proves theoretical and algorithmic results on linear 3uniform hypergraphs concerning feedback sets, which are main technical tools for establishing new sufficient conditions for Tuza’s conjecture in Section 3. Section 4 concludes the paper with extensions and future research directions.
2 Hypergraphs
This section develops hypergraph tools for studying Tuza’s conjecture. The theoretical and algorithmic results are of interest in their own right.
Let be a hypergraph with vertex set and edge set . For convenience, we use to denote the number of edges in . If hypergraph satisfies and , we call a subhypergraph of , and write . For each , the degree is the number of edges in that contain . We say is an isolated vertex of if . Let be a positive integer, hypergraph is called regular if for each , and uniform if for each . Hypergraph is linear if for any pair of distinct edges .
A vertexedge alternating sequence of is called a path (of length ) between and if are distinct, are distinct, and for each . We consider each vertex of as a path of length 0. Hypergraph is said to be connected if there is a path between any pair of distinct vertices in . A maximal connected subhypergraph of is called a component of . Obviously, is connected if and only if it has only one component.
A vertexedge alternating sequence , where , is called a cycle (of length ) if are distinct, are distinct, and for each , where . We consider the cycle as a subhypergraph of with vertex set and edge set . For any (resp. ), we write for the subhypergraph of obtained from by deleting all vertices in and all edges incident with some vertices in (resp. deleting all edges in and keeping vertices). If is a singleton set , we write instead of . For any , the hypergraph is often written as , and as if .
A vertex (resp. edge) subset of is called a feedback vertex set or FVS (resp. feedback edge set or FES) of if it intersects the vertex (resp. edge) set of every cycle of . A vertex subset of is called a transversal of if it intersects every edge of . Let , and denote, respectively, the minimum cardinalities of a FVS, a FES, and a transversal of . A matching of is an nonempty set of pairwise disjoint edges of . Let denote the maximum cardinality of a matching of . It is easy to see that , and . Our discussion will frequently use the trivial observation that if no cycle of contains any element of some subset of , then and have the same set of FVS’s, and . The following theorem is one of main contributions of this paper.
Theorem 2.1.
Let be a linear uniform hypergraph. Then .
Proof.
Suppose that the theorem failed. We take a counterexample with such that is as small as possible. Obviously . Without loss of generality, we can assume that has no isolated vertices. Since is linear, any cycle in is of length at least 3.
If there exists which does not belong to any cycle of , then . The minimality of implies , giving , a contradiction. So we have

Every edge in is contained in some cycle of .
If there exists with , then , where the first inequality is due to the minimality of . Given a minimum FVS of , it is clear that is a FVS of with size , a contradiction to . So we have

for all .
Suppose that there exists with . Let be the unique edge that contains . Recall from (1) that is contained in a cycle , where . By (2), we have for all . In particular implies , and in turn enforces . Let be a minimum FVS of . It follows from (2) that
and in , edge intersects at most one other edge, and therefore is not contained by any cycle. Thus is a FVS of , and hence a FVS of , implying that is a FVS of . We deduce that . Therefore shows a contradiction to the minimality of . Hence the vertices of all have degree at least 2, which together with (2) gives

is regular.
Let be a shortest cycle in , where . For each , suppose that , where .
Because is a shortest cycle, for each pair of distinct indices , we have if and only if and are not adjacent in , i.e., . This fact along with the linearity of says that are distinct. By (3), each is contained in a unique edge , . We distinguish among three cases depending on the values of . In each case, we construct a proper subhypergraph of with and which shows a contradiction to the minimality of .
Case 1. :
Let be a minimum FVS of . Setting and , it follows from (3) that
and in , each edge in intersects exactly one other edge, and therefore is not contained by any cycle. Thus is also acyclic, so is , saying that is a FVS of . We deduce that . Therefore shows a contradiction.
Case 2. :
Consider the case where or . Relabeling the vertices and edges if necessary, we may assume without loss of generality that . Let be a minimum FVS of . Set , if and , otherwise. In any case we have and
Note from (3) that in , each edge in can intersect at most one other edge, and therefore is not contained by any cycle. Thus is also acyclic, so is . Thus is a FVS of , and . This gives , a contradiction.
Consider the case where and . As are distinct and , we have . Observe that is a cycle in of length 4. The minimality of enforces . Therefore consist of 6 distinct edges. Let be a minimum FVS of . It follows from (3) that
In , both and intersect only one other edge, which is , and any cycle through must contain or . It follows that none of is contained by a cycle of . Thus is acyclic, so is , saying that is a FVS of . Hence . In turn shows a contradiction.
Case 3. :
Let be a minimum FVS of . Setting and , we have and
In , each edge in intersects at most one other edge, and therefore is not contained by any cycle. Thus is acyclic, so is . Hence is a FVS of , yielding and a contradiction .
The combination of the above three cases complete the proof. ∎
We remark that the upper bound in Theorem 2.1 is best possible. See Figure 1 for illustrations of five 3uniform linear hypergraphs attaining the upper bound. It is easy to prove that the maximum degree of every extremal hypergraph (those with ) is at most three. It would be interesting to characterize all extremal hypergraphs for Theorem 2.1.
The proof of Theorem 2.1 actually gives a recursive combinatorial algorithm for finding in polynomial time a FVS of size at most on a linear 3uniform hypergraph .
Note that Algorithm 1 never visits isolated vertices (it only scans along the edges of the current hypergraph). The number of iterations performed by the algorithm is upper bounded by . Since is 3uniform, the condition in any step is checkable in time. Any cycle in Step 7 or Step 9 can be found in time.^{1}^{1}1The shortest path between any pair of vertices can be find in time using breadth first search. A shortest cycle can be find by checking all possibilities. Thus Algorithm 1 runs in time.
Corollary 2.2.
Given any linear 3uniform hypergraph , Algorithm 1 finds in time a FVS of with size at most .∎
Lemma 2.3.
If is a connected linear uniform hypergraph without cycles, then .
Proof.
We prove by induction on . The base case where is trivial. Inductively, we assume that and the lemma holds for all connected acyclic linear 3uniform hypergraph of edges fewer than . Take arbitrary . Since is connected, acyclic and 3uniform, contains exactly three components , . Note that for each , hypergraph with is connected, linear, 3uniform and acyclic. By the induction hypothesis, we have for . It follows that . ∎
Given any hypergraph , we can easily find a minimal (not necessarily minimum) FES in time: Go through the edges of the trivial FES in any order, and remove the edge from the FES immediately if the edge is redundant. The redundancy test can be implemented using Depth First Search.
Lemma 2.4.
Let be a linear uniform hypergraph with components. If is a minimal FES of , then . In particular, .
Proof.
Suppose that contains exactly components , . It follows from Lemma 2.3 that for each . Thus , which means . To establish the lemma, it suffices to prove .
In case of , we have and . In case of , suppose that . Because is a minimal FES of , for each , there is a cycle in such that , and is a path in connecting two of the three vertices in . Considering being obtained from be removing sequentially, for , since , the presence of path implies that the removal of can create at most one more component. Therefore we have as desired. ∎
Given a hypergraph with vertices and edges, let be the incidence matrix. From , we may construct a bipartite graph with bipartition such that there is an edge of between and if and only if in .
Suppose that is acyclic. It is easy to see that is acyclic. Thus falls within the class of restricted totally unimodular (RTUM) matrices defined by Yannakakis [10]. As the name indicates, RTUM matrices are all totally unimodular. Hence the total unimodularity and LP duality give the wellknown result [11] that . Moreover, since is RTUM, both a minimum transversal and a maximum matching of can be found in time using Yanakakis’s combinatorial algorithm [10] based on the current best combinatorial algorithms for the matching problem and the maximum weighted independent set problem on a bipartite mulitgraph with vertices and edges, where the bipartite matching problem can be solved with the minimumcost flow algorithm in time (see Section 21.5 and Page 356 of [12]) and the maximum weighted independent set problem can be solved with maximum flow algorithm in time (See Pages 300301 of [10]).
3 Triangle packing and covering
This section establish several new sufficient conditions for Conjecture 1.1 as well as their algorithmic implications on finding minimum triangle covers. Section 3.1 deals with graphs of high triangle packing numbers. Section 3.2 investigates irreducible graphs with many edges. Section 3.3 discusses ErdősRényi graphs with high densities.
To each graph , we associate a hypergraph , referred to as triangle hypergraph of , such that the vertices and edges of are the edges and triangles of , respectively. Since is simple, it is easy to see that is uniform and linear, and . Note that , and if is irreducible, i.e., . Note that the number of nonisolated vertices of is upper bounded by .
3.1 Graphs with many edgedisjoint triangles
We investigate Tuza’s conjecture for graphs with large packing numbers, which are firstly compared with the number of triangles, and then with the number of edges.
Theorem 3.1.
If graph and real number satisfy , then a triangle cover of with size at most can be found in time, which implies .
Proof.
We consider the triangle hypergraph of which is 3uniform and linear. By Corollary 2.2, we can find in time a FVS of with . Since , it follows that . As is acyclic, Theorem 2.5 enables us to find in time a minimum transversal of such that . We observe that and is trianglefree. Hence is a triangle cover of with size
which proves the theorem. ∎
The special case of in the above theorem gives the following result providing a new sufficient condition for Tuza’s conjecture.
Corollary 3.2.
If graph satisfies , then .∎
The condition in Corollary 3.2 applies, in some sense, only to the class of large scale sparse graphs (which, e.g., does not include complete graphs on four or more vertices). The mapping from the real number in the condition to the coefficient in the conclusion of Theorem 3.1 shows the tradeoff between conditions and conclusions. As in Corollary 3.2, maps to hitting the boundary of Tuza’s conjecture. It remains to study graphs with . The next theorem (Theorem 3.3) tells us that actually we only need to take care of graphs with
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