The biclique covering number of grids

We determine the exact value of the biclique covering number for all grid graphs.

(see [6,3]). In this paper, we determine the biclique covering number for all grids. Let G p,q be the p × q grid. Recall that G p,q has vertex set Since G p,q G q,p , our main result determines the biclique covering number of all grid graphs. This settles an open problem raised by Denis Cornaz at the 9 th Cargèse Workshop on Combinatorial Optimization in 2018. Note that for p even and q p(p − 1), Theorem 1 implies that bc(G p,q ) = pq /2 − 1.

Main result
Since we are only interested in biclique covers of minimum size, we may assume that biclique covers only consist of maximal bicliques (under edge inclusion). For brevity, we call such biclique covers simply covers. For G p,q , this implies that covers consist of elements which are isomorphic to K 1,3 , K 1,4 , or the 4-cycle. We first establish the following upper bound.
Proof. Observe that G p,q is a bipartite graph with bipartition (X, Y ) where |X| = pq /2 and |Y | = pq /2 . Therefore taking the set of stars centered at vertices in X gives a cover of size pq /2 . See Figure 1.
We now establish the following lower bounds.  Lemma 3. For all integers 1 p q, We define a special subset S(G p,q ) of edges of G p,q inductively as follows. If p ∈ {1, 2}, we let S(G p,q ) be the set of all horizontal edges of G p,q . If p 3, let Out be the outer cycle of G p,q and define S(G p,q ) = E(Out) ∪ S(G p,q − V (Out)). See Figure 2a and Figure 2b. An easy induction gives  On the other hand, every biclique of G p,q contains at most 2 edges of S(G p,q ). Therefore, bc(G p,q ) | S(Gp,q)| /2 , which completes the proof.
Next, we give values of p and q for which bc(G p,q ) = pq /2 − 1.
Proof. By Lemma 3, it suffices to construct a cover C of size pq /2 − 1. Since q − 1 = k(p − 1) + 2 for some integers 0 < k, we can decompose G p,q as k copies of G p,p and copies of G p,3 such that each of the copies of G p,3 is between two copies of G p,p . Since p is even, G p,p has two covers of size p 2 /2 − 1, whose set of 4-cycles are the 4-cycles of the two diagonals of G p,p , respectively. See Figure 3 for the case p = 6.
The cover C is constructed as follows. For every two consecutive copies of G p,p , we use one cover from Figure 3 on one copy and the other cover from Figure 3 on the other copy. Note that this results in p /2 − 1 bicliques that are in both copies. For every copy of G p,3 between two copies of G p,p we again use the two covers of size p 2 /2 − 1 on the two copies of G p,p , and then we use p /2 + 1 bicliques in G p,3 to cover the remaining edges, see Figure 4. The total construction when p = 6 and q = 25 is given in Figure 5. Observe that: Figure 3. Two covers of G6,6 of size 17.  We now establish that we may choose a cover with special properties, before completing the proof of the main theorem. Recall that Out is the outer cycle of G p,q . Let C be a cover of G p,q . A boundary element of C is an element of C containing at least one edge of Out. A boundary 4-cycle is a boundary element that is a 4-cycle and a boundary star is a boundary element that is a star.
Lemma 5. For every cover of G p,q , there exists a cover C of the same size with the following properties: (i) all boundary stars of C are pairwise edge-disjoint; (ii) no edge is contained in both a boundary 4-cycle and a boundary star of C.
Proof. If two boundary stars meet in an edge, we replace them with a star and a 4-cycle which cover a superset of edges, as in Figure 6. If a boundary 4-cycle and a boundary star are not edge-disjoint, then we replace them with two 4-cycles which cover a superset of edges, as in Figure 7.  By repeatedly performing these two replacement rules, we eventually obtain a cover C satisfying (i) and (ii).
We now complete the proof of Theorem 1 by establishing the following converse to Lemma 4. Lemma 6. Let 1 p q, with p even. If G p,q has a cover of size pq /2 − 1, then q − 1 = k(p − 1) + 2 for some integers 0 < k.
Proof. The lemma clearly holds if p = 2, so we may assume p 4. Let C be a cover of G p,q of size pq /2 − 1 satisfying properties (i) and (ii) of Lemma 5. We begin by defining some objects required for the proof. Let Out be the outer cycle of G p,q . The corners of G p,q are the vertices (1, 1), (1, p), (q, 1), and (q, p). Let H be the subgraph of G p,q induced by the edges of the boundary 4-cycles of C. A fence is a connected component of H. The size of a fence is the number of boundary 4-cycles it contains. A link is a connected component of Out \ E(H) containing at least two vertices and no corners. See Figure 8 for an illustration of fences and links.
Suppose L is a link contained on the topmost path of G p,q . Let L left be the path starting from the left end of L by first proceeding down twice and then alternating between proceeding right and down. We define L right by interchanging left and right in the definition of L left . The staircase generated by L is the subgraph of G p,q contained in the region bounded by L, L left , L right , and (possibly) a subpath of the bottommost path of G p,q , see Figure 9a and Figure 9b. Staircases for links contained in the leftmost, rightmost, and bottommost paths of G p,q are defined by rotating G p,q so that the link is on the topmost path, applying the above definition, and then rotating back.
The length of a path is its number of edges, and the length of a staircase is the length of the link that generates it. An edge is thick if it is covered by at least two bicliques in C. A pyramid P is a staircase of length 2p − 4 containing a size-2 fence B such that the thick edge in B is the only thick edge in P, see Figure 10. We call B the tip of the pyramid.   Proof. Let L be a link of length k connecting two fences B and B (see Figure 11a) and let S be the staircase generated by L. Towards a contradiction, suppose S does not contain a thick edge. Since C satisfies properties (i) and (ii) of Lemma 5, k is even and P is covered by exactly k/2 boundary stars as in Figure 11b. Let E 1 be the set of edges in S with one end on L and not covered by a boundary element of C (blue edges in Figure 11b). Each edge e in E 1 must be covered by a K 1,4 , else e is thick, see Figure 11c. Let e 1 and e 2 be the two horizontal edges of S which intersect (V (B) ∪ V (B )) \ V (L) (blue edges in Figure 11c). Since neither e 1 nor e 2 are thick, they must be covered by 4-cycles as in Figure 11d. By repeating this argument, we either find a thick edge (red edge in Figure 11e), or we reach the bottommost path of G p,q . Let F be the set of edges f ∈ E(S) such that f is covered by a K 1,4 of C and f has exactly one endpoint on the bottommost path of G p,q . By the above argument, F is non-empty. Moreover, each edge in F is thick (see Figure 11f).
A double staircase is a pair (S, S ), where S and S are distinct staircases such that there is exactly one thick edge in S ∪ S . Suppose S has length 2a and S has length 2b. Observe that either a + b = p − 2 or a + b = p and C must cover S ∪ S as in Figures 12a and 12b, respectively.
We define the waste of C to be Let τ be the number of copies of K 1,3 that appear in C and t i be the number of edges of G p,q that are covered exactly i times by C. By double counting, we obtain: Since |C| = pq /2 − 1 and |E(G p,q )| = 2pq − p − q, (1) immediately gives the following: Note that a fence can have size at most N := 2p + 2q − 8. For i ∈ [N ], let b i be the number of fences of size i of C.  Proof. If b N = 1, then C contains 2p + 2q − 8 boundary 4-cycles. These boundary 4-cycles cover 6p + 6q − 24 edges of G p,q . Thus, 2pq − 7p − 7q + 24 edges are not covered by boundary 4-cycles. It follows that where the last inequality follows since 4 p q. However, this contradicts |C| = pq /2 − 1.
Let β be the number of edges that are covered twice by boundary 4-cycles of C, and let c ∈ [4] be the number of corners of G p,q covered by the set of fences. We now provide a lower bound on the waste of C.
Proof. The inequality follows immediately from (2). For the equality, first observe β = Next, because each boundary star covers two edges of Out, each edge of Out not covered by a fence contributes 1 /2 to τ . Since Out contains 2p + 2q − 4 edges and the set of fences cover c + Claim 10. The set of staircases of C can be enumerated as P 1 , . . . P n , S 1 , S 1 , . . . , S m , S m such that P i is a pyramid for all i ∈ [n] and (S j , S j ) is a double staircase for all j ∈ [m]. Moreover, b 2 = n, b i = 0 for all i 3, and w(C) = τ + n + m.
Proof. The number of staircases of C is at least At most b 2 fences can be the tips of pyramids, so at least b 1 + N −1 i=3 b i + c − 4 staircases are not pyramids. Each of these staircases contains a thick edge by Claim 7, and each thick edge can be in at most 2 staircases. Therefore, these staircases contribute at least (b 1 + to the waste of C which is not counted in β + τ . By Claim 9, By (3), w(C) = p + q − 4. Hence, we must have b i = 0 for i 3 and equality throughout the above argument. In particular, there are exactly b 2 pyramids. The remaining staircases each contain exactly one thick edge, and each of these thick edges is in exactly two staircases. This gives the required enumeration of the staircases of C and implies w(C) = τ + n + m.
Proof. Let L be the leftmost path of G p,q . Since q p, no link contained in L can generate a pyramid or be part of a double staircase. By Claim 10, no link is contained in L. It follows that L intersects at most one fence. Since p is even, it follows that L intersects exactly one fence B left . Since all fences are of size 1 or 2 and p is even, B left is of size 1 and L \ E(B left ) is the disjoint union of two even-length paths P a and P b . We assume that P a is above P b and that P a and P b have lengths a and b, respectively. By the same argument, the rightmost path of G p,q intersects exactly one fence B right . We first consider the case when B left is a fence that does not contain a corner. Suppose L 1 is the first link along the topmost path of G p,q and S 1 is the staircase generated by L 1 . By Claim 10, there are no thick edges outside of staircases. Therefore, B left and the boundary stars force the leftmost portion of G p,q to be covered by C as in Figure 13a. This forces S 1 to be in a double staircase (S 1 , S 1 ), where S 1 has length 2b and S 1 has length 2a (see Figures 14a and 14b), or S 1 has length 2b + 2 and S 1 has length 2a + 2 (see Figures 14c  and 14d).
Suppose L 2 is the next link along the topmost path of G p,q . Because of the previous double staircase (S 1 , S 1 ), S 2 must be in a double staircase (S 2 , S 2 ) where S 2 has length 2a and S 2 has length 2b (see Figures 14a  and 14c), or S 2 has length 2a+2 and S 2 has length 2b+2 (see Figures 14b and 14d). Repeating the argument, we obtain a sequence of double staircases (S 1 , S 1 ), . . . , (S k−1 , S k−1 ), where (S k−1 , S k−1 ) is the last double staircase. Note that it is possible that this sequence is empty, corresponding to the case that there are no staircases and k − 1 = 0.
Again, because there are no thick edges outside of staircases, the fence B right and the boundary stars force the rightmost portion of G p,q to be covered as in Figure 13b. For all i ∈ [k − 1], let i and i be the lengths of S i and S i respectively. Let be the number of times that { i , i } = {2a + 2, 2b + 2}. It follows that q − 1 = k(p − 1) + 2 , as required.
The case when B left contains a corner is treated exactly as above, resulting in the construction from Figure 5.
This completes the entire proof.