Coloring Problems on Bipartite Graphs of Small Diameter

We investigate a number of coloring problems restricted to bipartite graphs with bounded diameter. We prove that the $k$-List Coloring, List $k$-Coloring, and $k$-Precoloring Extension problems are NP-complete on bipartite graphs with diameter at most $d$, for every $k\ge 4$ and every $d\ge 3$, and for $k=3$ and $d\ge 4$, and that List $k$-Coloring is polynomial when $d=2$ (i.e., on complete bipartite graphs) for every $k \geq 3$. Since $k$-List Coloring was already known to be NP-complete on complete bipartite graphs, and polynomial for $k=2$ on general graphs, the only remaining open problems are List $3$-Coloring and $3$-Precoloring Extension when $d=3$. We also prove that the Surjective $C_6$-Homomorphism problem is NP-complete on bipartite graphs with diameter at most $4$, answering a question posed by Bodirsky, K\'ara, and Martin [Discret. Appl. Math. 2012]. As a byproduct, we get that deciding whether $V(G)$ can be partitioned into 3 subsets each inducing a complete bipartite graph is NP-complete. An attempt to prove this result was presented by Fleischner, Mujuni, Paulusma, and Szeider [Theor. Comput. Sci. 2009], but we realized that there was an apparently non-fixable flaw in their proof. Finally, we prove that the $3$-Fall Coloring problem is NP-complete on bipartite graphs with diameter at most $4$, and give a polynomial reduction from $3$-Fall Coloring on bipartite graphs with diameter $3$ to $3$-Precoloring Extension on bipartite graphs with diameter $3$. The latter result implies that if $3$-Fall Coloring is NP-complete on these graphs, then the complexity gaps mentioned above for List $k$-Coloring and $k$-Precoloring Extension would be closed. This would also answer a question posed by Kratochv\'il, Tuza, and Voigt [Proc. of WG 2002].


Introduction
Graph coloring problems are among the most fundamental and studied problems in graph theory, due to its practical and theoretical importance. A proper coloring of a graph G is a function f : V (G) → N such that f (u) = f (v) for every uv ∈ E(G), and the k-Coloring problem asks whether a given graph G admits a proper coloring using at most k colors. It is well-known that the k-Coloring problem is NP-complete for every fixed k ≥ 3 [16]. In this paper, we study two of the most general coloring problems: list coloring and graph homomorphism.
In the k-List Coloring problem, we are given a graph G together with a function L which assigns to each u ∈ V (G) a subset of allowed colors with |L(u)| ≤ k for every u ∈ V (G). This is called a list assignment of G. The question is whether G admits a proper coloring f such that f (u) ∈ L(u) for every u ∈ V (G); if the answer is "yes", we say that G is L-colorable. Observe that this generalizes the k-Coloring problem: it suffices to consider L(u) = {1, . . . , k} for every u ∈ V (G). Thus k-List Coloring is NP-complete for every fixed k ≥ 3 [13,16]. Another natural coloring problem that can be modeled as a list coloring problem is the k-Precoloring Extension problem, where some vertices have fixed colors; these precolored vertices therefore have lists of size 1, while the remaining ones all have list equal to {1, . . . , k}.
A long-standing question about coloring problems is whether one can decide in polynomial time if a graph with diameter 2 can be properly colored using at most 3 colors (see e.g. [15,23]), and only recently the answer for the 3-Coloring problem on graphs with diameter at most 3 has been settled negatively by Mertzios and Spirakis [22]. Interestingly enough, even though bipartite graphs play a special role in the investigation of list colorings [10], up to our knowledge no results about the complexity of the problem on bipartite graphs with small diameter exist. We prove that 3-Precoloring Extension is NP-complete even on bipartite graphs with diameter 4. We mention that Kratochvíl [17] proved that the problem is NP-complete on planar bipartite graphs, answering a question posed by Hujter and Tuza [14].
It is well-known that a k-coloring can also be seen as a homomorphism to K k (the complete graph on k vertices). Given graphs G and H, a homomorphism from G to H is a function f : V (G) → V (H) that respects edges, i.e., such that f (u)f (v) ∈ E(H) whenever uv ∈ E(G). When H is fixed, the H-Homomorphism problem consists in deciding whether G has a homomorphism to H, while the List H-Homomorphism is defined in a similar way as the k-List Coloring problem, i.e., each vertex of G can only be mapped to a subset of the vertices of H given in a list. It is known that H-Homomorphism is polynomial if H is bipartite, and NP-complete otherwise [11]. A dichotomy is also known for the List H-Homomorphism problem: Feder and Hell [4] proved that if H is a reflexive graph (a graph is reflexive if every vertex of H has a loop), then the problem is polynomial if H is a bipartite interval graph and NP-complete otherwise. Furthermore, Feder et al. [5] proved that if H has no loops, then the problem is polynomial when H is bipartite and H, the complement graph of H, is a circular-arc graph, and NP-complete otherwise.
We say that a homomorphism f from a graph G to a graph H is surjective if every vertex v ∈ V (H) is the image of some vertex u ∈ V (G). If f is edge-surjective, we say that f is an H-compaction of G. If H is a subgraph of G and f (v) = v for every v ∈ V (H), we say that f is a retraction of G to H. We denote the related problems by Surjective H-Homomorphism, H-Compaction, and Retract(G, H), respectively, where in the first two problems H is considered to be a fixed graph. Even though the NP-completeness of the Surjective H-Homomorphism problem when H is not bipartite trivially follows from [11], we are still far from a dichotomy for this problem. Martin and Paulusma [21] proved that it is NP-complete when H is the reflexive cycle on 4 vertices, and Golovach et al. [8] proved that the problem is polynomial when H is a path, and NP-complete for many other cases (e.g. linear forests and trees of pathwidth at most 2). Also, Golovach et al. [7] recently proved that the problem is NP-complete when H has exactly 2 vertices u and v such that {uu, vv} ⊆ E(H), provided uv / ∈ E(H). We remark that a proof by Fleischner et al. [6] implies that Retract(G, H) is NPcomplete on bipartite graphs when H is isomorphic to C 6 . We give a stronger version of this result that is used to prove some of our hardness results. Namely, we show that Retract(G, H) is NP-complete even if V (H) dominates one of the parts of G and each vertex of H is within distance 2 of every vertex in the other part of G. Then, we use this result to show that 3-Precoloring Extension, C 6 -Compaction, and Surjective C 6 -Homomorphism are NP-complete even when restricted to bipartite graphs with diameter 4. The complexity of the latter problem was asked by Bodirsky et al. [1]. We also refer the reader to [1] for a nice survey on surjective homomorphisms and related problems.
Given a graph G, a biclique of G is a non-empty complete bipartite subgraph of G. In the k-Biclique problem, the task is to decide whether V (G) can be partitioned into k bicliques. When k is given as input, the problem is NP-complete even when restricted to bipartite graphs [12]. An attempt to show that 3-Biclique is NP-complete on bipartite graphs was presented by Fleischner et al. [6]. Unfortunately, there is a mistake in their proof as we show in this article. Nevertheless, applying our result for Surjective C 6 -Homomorphism, we get that the 3-Biclique problem is indeed NP-complete on bipartite graphs, even when the bipartite complement of G has diameter 4. It is worth mentioning that 2-Biclique is NP-complete on general graphs [21] (with a very technical reduction that uses an abstract algebraic meta-theorem) but polynomial on bipartite graphs [6]. Hence, the NP-completeness of 3-Biclique on bipartite graphs is best possible.
Given a proper k-coloring f of G, a vertex v is called a b-vertex if the neighborhood of v contains one vertex of each color (distinct from that of v), and f is a k-fall-coloring of G if every vertex of G is a b-vertex. In the k-Fall Coloring problem, we ask whether an input graph G admits a k-fall-coloring. We show that if 3-Fall Coloring is NP-complete on bipartite graphs with diameter 3, then we get a complete dichotomy for the list coloring problems on bipartite graphs with diameter constraints. Also, this would answer a question posed by Kratochvíl et al. [18]. Although we do not know if 3-Fall Coloring is NP-complete on bipartite graphs with diameter 3, we show that it is NP-complete on bipartite graphs with diameter 4, strengthening a result of Laskar and Lyle [19].
Organization. In Section 2 we present the formal notation and definitions. In Section 3 we give an almost complete classification for the investigated list coloring problems in terms of the number of colors and the diameter of the input graph G, taking into accout the results presented in the current article. In Section 4 we show our hardness result for Retract(G, H) when H is isomorphic to C 6 and use it to show that 3-Precoloring Extension is NPcomplete on bipartite graphs with diameter 4. In Section 5 we show that C 6 -Compaction and Surjective C 6 -Homomorphism are NP-complete on bipartite graphs with diameter 4.
In Section 6, we point out the flaw in the hardness proof for 3-Biclique on bipartite graphs presented in [6], and show how to obtain the result using the results presented in Section 5. Finally, in Section 7 we provide a reduction from 3-Fall Coloring to 3-Precoloring Extension that preserves the diameter of the input graph, and prove that 3-Fall Coloring is NP-complete when G is bipartite with diameter 4.

Definitions and notation
All graphs that we consider are simple, that is, they do not have loops nor multiple edges. We can also assume the graphs are connected as otherwise it suffices to solve the problems on the connected components. For basic definitions on graph theory, we refer the reader to [25].
The diameter of a graph G is the maximum length of a shortest path in G. We denote by B the class of bipartite digraphs and, for a fixed positive integer d, by B d the class of bipartite graphs with diameter at most d, and by D d the class of graphs with diameter at most d. Given a bipartite graph G with vertex bipartition (X, Y ), the bipartite complement of G is the bipartite graph with vertex bipartition (X, Y ) containing exactly the non-edges of G, i.e., the bipartite graph G B = (X ∪ Y, E ) such that uv ∈ E if and only if u ∈ X, v ∈ Y , and uv / ∈ E(G). For an integer ≥ 1, we denote by [ ] the set {1, . . . , }. Given problems P and P , we write P P to denote that there exists a polynomial reduction from P to P (hence, P is at least as hard as P ). Also, given a graph class G and a problem P , we denote P restricted to G by P | G .
Given sets A and B, a function f : A → B, and an element b ∈ B, the set of elements of A whose image is b is denoted by f −1 (b). Also, given X ⊆ A, we denote by f (X) the set containing the image of X, i.e., f (X) = {b ∈ B | there exists x ∈ X s.t. f (x) = b}. If f (X) = {b}, we abuse notation and write f (X) = b.
Given a graph G and a positive integer k, we say that a function f : , and f is a k-fall-coloring of G if every vertex of G is a b-vertex. A list assignment of G is a function L : V (G) → 2 [N] that assigns to each vertex a finite subset of positive integers. A partial k-coloring of G is a function p : V → [k] where V ⊆ V (G) and p is a proper coloring of the subgraph of G induced by V . If V is not given, we denote it by dom(p). Given a partial k-coloring p of G, we say that f is a k-extension of p if f is a proper k-coloring of G such that f (u) = p(u) for every u ∈ dom(p).
A biclique of a graph G is a subgraph H which is a complete bipartite graph that contains at least one edge.
The coloring problems investigated in this article are formally defined below, adopting the notation from other papers; see e.g. [23]. In each of the problems defined below, we consider k to be a fixed integer with k ≥ 1.

k-Precoloring Extension (henceforth abbreviated as k-PreExt) Input:
A graph G = (V, E), a positive integer k, and a partial k-coloring p of G. Question: Does p have a k-extension? k-Biclique Partition Input: A graph G = (V, E). Question: Does G have a biclique vertex partition with at most k parts?

Input:
A graph G = (V, E). Question: Does G admit a k-fall-coloring?
we say that f is an retraction from G to H. Finally, if f is a retraction to H and edge-surjective, we say that f is an H-compaction.
The investigated homomorphism problems are listed below. Observe that in the first two, graph H is considered to be fixed.

Input:
A graph G = (V, E). Question: Does G have a surjective homomorphism to H?

Input:
A graph G = (V, E). Question: Does G admit an H-compaction?

Input:
A graph G = (V, E) and a subgraph H ⊆ G. Question: Is there a retraction of G to H? A 3-uniform hypergraph is a hypergraph such that each hyperedge has size exactly 3. A 2-coloring of a hypergraph G is a function f : V (G) → {1, 2} such that f (e) = {1, 2} for every hyperedge e ∈ E(G) (i.e., no hyperedge is monochromatic). The problem of 2-coloring 3-uniform hypergraphs, formally defined below, is known to be NP-complete [20]. This problem will be used in the reductions of Sections 4 and 7.
We let n denote the number of vertices of the input graph of the problem under consideration.

List coloring vs. diameter
In this section, we investigate the complexity of k-PreExt, List k-Coloring, and k-List Coloring on bipartite graphs with diameter d, for a fixed integer d ≥ 2. We provide a complete picture about the hardness of these problems, leaving as open cases only 3-PreExt and List 3-Coloring for d = 3. First, notice that there is a straightforward reduction from k-PreExt to List k- . Furthermore, List k-Coloring is a particular case of k-List Coloring, since each vertex in an instance of the former problem has a list assignment of size at most k. From these two remarks, we get k-PreExt List k-Coloring k-List Coloring. (1) Since the reductions discussed above do not change the input graph, we remark that Equation (1) holds when we restrict the problems to graphs in B d . We remark that there is a polynomial-time algorithm for the 2-List Coloring problem [20,24], hence for 2-PreExt and List 2-Coloring as well. In what follows we investigate the complexity of these problems for k ≥ 3. Observe that if k-PreExt is proved to be NP-complete for some k, then the same holds for the other problems, and the next result shows that it also implies that (k + 1)-PreExt| B3 is NP-complete.

Proposition 1. Let k ≥ 1 be a fixed integer. Then k-PreExt|
Proof: Let G ∈ B with parts X and Y , p be a partial k-coloring of G, and let (G , p ) be an instance of (k + 1)-PreExt| D3 obtained from (G, p) as follows. G is obtained from G by adding two new vertices x and y, all edges from x to vertices in Y , and all edges from y to vertices in X. Let p be obtained from p by giving color k + 1 to both x and y. Now, any extension of p defines an extension of p and vice-versa, since color k + 1 can only appear in the new vertices. Let us argue about the diameter of G . Recall that we can assume that G is connected, and therefore has no isolated vertices.
Consider a pair u, v ∈ V (G ). If they are within the same part, say X, then either they are both adjacent to y, or u = x in which case (x, w, v) is a path, where w is any neighbor of v in Y − {y} (recall that G has no isolated vertices). If u ∈ X and v ∈ Y , then either {u, v} ∩ {x, y} = ∅, in which case (u, y, w, v) is a (u, v)-path for any neighbor w of v with w ∈ X − {x}; or {u, v} = {x, y} and (x, w, w , y) is a (u, v)-path for any edge ww ∈ E(G); or |{u, v} ∩ {x, y}| = 1, in which case uv ∈ E(G ). Thus, we get that G has indeed diameter at most 3. Table 1 (resp. Table 2) presents the complexity of the problems discussed in this section for k = 3 (resp. for every fixed k ≥ 4) restricted to bipartite graphs with diameter at most 2, 3, and 4. Let us explain how these tables are filled. In Section 4 we prove that 3-PreExt| B4 is NP-complete. Note that, from Equation (1) and the fact that B d ⊆ D d+1 , we get that row 4 downwards in Table 1 is filled with NP-completeness results. Also, by Proposition 1, we get that for every fixed k ≥ 4, row 3 downwards in Table 2 is filled with NP-completeness results. In Section 3.1 we prove that List k-Coloring is polynomial for every fixed k ≥ 1 when G is a complete bipartite graph (or equivalently, a bipartite graph with diameter 2), and thus Equation (1) and the reduction discussed before it tell us that the same result holds for k-PreExt since no changes in the input graph are needed to reduce the former problem to the latter. Finally, Gravier [9] proved that 3-List Coloring is NP-complete on complete bipartite graphs 1 . The latter two sentences justify the second row of both tables. Table 1 Row labeled i presents the complexity of the corresponding problems restricted to bipartite graphs with diameter at most i. Table 2 Row labeled i presents the complexity of the corresponding problems restricted to bipartite graphs with diameter at most i, for every fixed integer k ≥ 4.

List k-Coloring is polynomial on complete bipartite graphs
In this section we prove that List k-Coloring can be solved in linear time on complete bipartite graphs, for every fixed k ≥ 1. As mentioned before, this implies that the same holds for k-PreExt by Equation (1) and the fact that the reduction does not modify the input graph.
Given a family F of subsets of [k], we say that We present a linear-time algorithm for the following auxiliary problem which, to the best of our knowledge, has not been studied before in the literature. Afterwards, we provide a reduction from the List k-Coloring problem on compete bipartite graphs to it. In what follows, S denotes [k] \ S. For the necessity, let f be a proper coloring of G respecting L. Let S be the set containing all colors used on vertices in A. That is, define S = {f (v) | v ∈ A}. By construction, S is a hitting set for A. Since G is a complete bipartite graph, we have that f (u) = f (v) for every pair of vertices u, v with u ∈ A and v ∈ B and thus S is a hitting set for B, and the necessity follows since no color used in B can occur in A.
For the sufficiency, let S be a hitting set of A such that S is a hitting set B. We generate a proper coloring of G as follows: for each vertex a ∈ A, choose for f (a) any color in the set S ∩ L(a), and for each b ∈ B, choose for f (b) any color in the set S ∩ L(b). Since S is a hitting set of A and S is a hitting set of B, f (v) is well defined for every v ∈ V (G). Furthermore, no color used on a vertex in A is used on a vertex in B since S ∩ S = ∅.

Lemma 3.
There is an algorithm running in time O(2 k · k 2 · n) for the k-Complementary Hitting-sets problem.
Proof: Given a subset S ⊆ [k], applying brute force one can verify whether S and S are hitting-sets of A and B, respectively, in time (|S|· |A|+|S|· |B|)·M , where M is the maximum size of a subset in A ∪ B. This is O(k 2 · n), where n = |A| + |B|. Therefore, by checking every possible subset of [k], we get that the k-Complementary Hitting-sets problem can be solved in time O(2 k · k 2 · n).
From Lemmas 2 and 3 we get the following theorem.
In particular, it can be solved in linear time for every fixed integer k ≥ 1.
In fact, Theorem 4 states that List k-Coloring| B2 is fixed-parameter tractable parameterized by k, using terminology from parameterized complexity (cf. for instance [2]). Also, by considering the disjoint union of instances of List k-Coloring| B2 it is easy to obtain a so-called and-cross-composition, hence refuting the existence of polynomial kernels for List k-Coloring| B2 parameterized by k under standard complexity assumptions; see [2] for the missing definitions.

4
Retraction to C 6 In this section we prove that Retract(G, H) is NP-complete when H is isomorphic to a C 6 , even under very strict constraints that imply that G ∈ B 4 . These constraints are useful in the next section, where we prove NP-completeness of C 6 -Compaction and Surjective C 6 -Homomorphism on B 4 using a reduction from Retract(G, H). Let G be a graph. We say that set X ⊆ V (G) dominates a set Y ⊆ V (G) if every y ∈ Y is in X or has a neighbor in X. For u, v ∈ V (G), we denote by dist(u, v) the distance between u and v in G. The proof of the following theorem consists of an appropriate modification of a reduction of Kratochvíl [17].
Proof: We reduce from the 3-Uniform 2-Col problem. For this, consider a 3-uniform hypergraph H = (V, E), and let G be the bipartite graph with bipartition (V, E) such that ue ∈ E(G) if and only if u ∈ e. Add vertices p V 1 , p V 2 , p V 3 and p E 1 , p E 2 , p E 3 to parts V and E, respectively, and make the subgraph induced by these vertices be the cycle . Add an edge between each v ∈ V and p E 3 . This ensures that any retraction f from G to C is such that f (V ) = {p V 1 , p V 2 }. Now, for each hyperedge e ∈ E, we replace some of the edges incident to e with an edge gadget defined as follows. For easier reference, consider V = {v 1 , . . . , v n } and E = {e 1 , . . . , e m }. Let e j ∈ E, and let i 1 , i 2 , i 3 be the indices of the vertices within e j . Remove edges v i1 e j and v i2 e j from G, and replace them with the gadget of Figure 1. For better visibility, sometimes we make more than one copy of some of the vertices of C in the figure (for instance, vertex p E 3 is represented 3 different times, but all occurrences correspond to the same vertex). Dotted and solid vertices are used to present the bipartition of G. The purpose of this gadget is to ensure that hyperedge e j cannot be monochromatic, as discussed below.
We need to prove that G is a bipartite graph, that C has the desired properties, and that f has a retraction to C if and only if H has an appropriate 2-coloring. We first prove the latter. So, let first f be a retraction to C; we prove that f restricted to V is a 2-coloring, using colors p V 1 , p V 2 , such that no hyperedge is monochromatic. Suppose otherwise and let We prove that f must be like in Figure 2, a contradiction since in this case e j has neighbors in f −1 (p E i ) for every i ∈ [3], and therefore cannot be mapped to C. Labels of vertices that do not have their image forced are left blank. Recall that C can also be seen as the complete bipartite graph minus the perfect matching But now we get that v 1,j is adjacent to p E 1 and to f −1 (p E 2 ), and therefore must be in p V 3 . This implies that d 1,j is adjacent to p V 2 and f −1 (p V 3 ), and hence is colored with p E 1 . Finally, we get that v 1,j is adjacent to p E 2 and f −1 (p E 1 ), and must be colored with p V 3 . The analysis for the right hand side of the gadget is similar. For the case where we have the situation depicted in Figure 3, and it follows similarly. Therefore, no edge is monochromatic, as we wanted to prove.
Conversely, suppose now that f is a 2-coloring of H with no monochromatic hyperedge, and let f be obtained from f by mapping to p V i all the vertices colored with i, for i ∈ {1, 2}. Let e j = {v i1 , v i2 , v i3 }. We show how to map the vertices within the edge gadget related to e j . The possibilities are the following: Map the vertices as in Figure 2, except for v i3 , and note that we can map e j to p E 1 , and that the blank vertices can be mapped to C; Map the vertices as in Figure 2, except for v i3 , and note that e j can be mapped to p V 2 , and that the blank vertices can be mapped to C; map the left-hand side as in Figure 2, and the right-hand side as in Figure 3. Note that e j can be mapped to p ∈ {p E 1 , p E 2 } \ {f (v i3 )}, and that the blank vertices can be mapped to C; : map the left-hand side as in Figure 3, and the right-hand side as in Figure 2. Note that e j can be mapped to p E 3 , and that the blank vertices can Figure 2 Coloring of edge gadget when f ({vi 1 , vi 2 , vi 3 }) = p V 1 . Vertices vi 1 , vi 2 and vi 3 are emphasized. The image of a vertex is put inside of the node, and its label appears next to it, with the exception of ej.
be mapped to C.
Finally, we need to prove that G is bipartite, and that C has the desired properties. To see that G is bipartite, just observe that the dotted and solid vertices in Figure 1 form a bipartition of G; let (X, Y ) be such partition, where V ⊆ X and E ⊆ Y (i.e., X contains the dotted vertices, and Y the solid ones). More formally, one can see that As for the properties of C, observe first that every x ∈ X is adjacent to p E i , for some i ∈ {1, 2, 3}, i.e., It remains to prove that every p ∈ Y C is at distance at most 2 from every y ∈ Y . Given e j ∈ E, denote by Y j the subset of vertices in Y contained in a gadget related to e j ; hence Y = Y C ∪ j∈[m] Y j and it suffices to prove that, given some j ∈ [m], every p ∈ Y C is at distance at most 2 from every vertex in Y j . For p E 1 , it follows from the fact that

Corollary 6. 3-PreExt| D4 is NP-complete, even if every vertex in one of the parts is adjacent to some precolored vertex.
Proof: Given a bipartite graph G and a C 6 , C in G with the properties stated in Theorem 5, it suffices to precolor p E i , p V i with i for each i ∈ [3]. One can see that G has a 3-extension for this precoloring if and only if G has a retraction to C. Also, the fact that {p E 1 , p E 2 , p E 3 } dominates X gives us the property claimed in the statement. It remains to verify that diam(G) ≤ 4. For this, first we argue that dist(x, y) ≤ 3 for every x ∈ X and y ∈ Y ; indeed, either xy ∈ E(G) or (x, p E i , w, y) is a path between x and y, where p E i is any vertex in N (x) ∩ C and (p E i , w, y) is the (p E i , y)-path of length 2 guaranteed in the statement of Theorem 5. Now, for x, x ∈ X, let y ∈ N (x) (it exists by construction); we know from the Figure 3 Coloring of edge gadget when f ({vi 1 , vi 2 , vi 3 }) = p V 2 . Vertices vi 1 , vi 2 and vi 3 are emphasized. The image of a vertex is put inside of the node, and its label appears next to it, with the exception of ej.
previous sentence that dist(y, x ) ≤ 3 and thus it follows that dist(x, x ) ≤ 4. The argument for y, y ∈ Y is symmetric.

Surjective C 6 -homomorphism
In this section we first prove that C 6 -Compaction is NP-complete on B 4 , and as a byproduct we get that Surjective C 6 -homomorphism is NP-complete (cf. Corollary 9). Notice that in the H-Compaction problem, H is not necessarily a fixed subgraph of G. However, we show that, under some assumptions, an H-compaction for G coincides with an H -retraction for some choice of H ⊆ G with H isomorphic to H.
In what follows, we present a reduction from the Retract(G, H) problem, where G ∈ B 4 and H ⊆ G isomorphic to C 6 , to the C 6 -Compaction problem. For this, we write H as (h 1 , . . . , h 6 ), and let X, Y be the parts of G, with {h 1 , h 3 , h 5 } ⊆ X. We first introduce the gadget in Figure 4 related to a vertex u ∈ X \ V (H). The cycle (h 1 , . . . , h 6 ) represents a circular permutation of (h 1 , . . . , h 6 ) such that h 1 ∈ {h 1 , h 3 , h 5 }; we call this the (h 1 , h 4 )-gadget (this is because h 1 h 4 is the diagonal related to this gadget). We obtain the input graph G of Surjective C 6 -homomorphism from G by adding an (h 1 , h 4 )-gadget related to each u ∈ X \ V (H), for each possible pair (h 1 , h 4 ) such that h 1 ∈ {h 1 , h 3 , h 5 }, namely for (h 1 , h 4 ), (h 3 , h 6 ), and (h 5 , h 2 ) (see Figure 6). Note that V (G ) \ V (G) consists of precisely the vertices {a, b, c, d, e, g} introduced for each choice of u ∈ X \ V (H) and each choice of h 1 , for a total of 6(|X| − 3) new vertices. We use the circular permutation in the following lemma to avoid making analogous arguments for each type of gadget separately. G ∈ B 4 , and H ⊆ G be isomorphic to a C 6 ; write H = (h 1 , · · · , h 6 ). Also, let G be obtained from G by adding the gadget represented in Figure 4 for every u ∈ X, and every (h 1 , h 4 6 (1, . . . , 6). Suppose that (h 1 , · · · , h 6 ) is a circular permutation of H such that

We prove that if
To see that f (h 1 ) = 1, note that the images of the gadget related to u must be as depicted in Figure 5 (fixed values appear inside the vertex, while implied values appear between parenthesis next to the vertex). Since g is adjacent to u ∈ f −1 (1) and h 5 ∈ f −1 (5), we get that f (g) = 6. Similarly, we get f (c) = 2 since it is adjacent to u ∈ f −1 (1) and h 3 ∈ f −1 (3). Note that this implies that f (d) = 3 and f (e) = 5, which in turn implies that f (a) = 6 and f (b) = 2. Because a, b ∈ N (h 1 ), we get f (h 1 ) = 1 as we wanted to prove. Recall that Theorem 5 states that the restriction of Retract(G, H) used in the following lemma is NP-complete. for every h ∈ X H and x ∈ X. If G is obtained as above, then G is a bipartite graph with diam(G ) ≤ 4, and Retract(G, H) is "yes" if and only if C 6 -Compaction(G ) is "yes".
Proof: Write H as before and suppose X H = {h 1 , h 3 , h 5 }. Denote by A, B, C the sets of vertices containing the vertices labeled with a, b, c in the gadgets in Figure 6, respectively. To see that G is bipartite, observe the coloring in grey and black in Figure 6, with X being gray. Now, denote by (X , Y ) the bipartition of G such that X ⊆ X , and observe that X = X ∪ A, and Y = Y ∪ B ∪ C. We prove that the same property holding for X H in G also holds in G . The fact that diam(G ) ≤ 4 then follows by the same argument given in Corollary 6. We discuss each property separately: 1. X H dominates Y : we know that X H dominates Y by assumption. Also, one can verify in Figure 6 that every vertex in B ∪ C is adjacent to some vertex in {h 1 , h 3 , h 5 }; 2. dist(x, h) ≤ 2 for every x ∈ X and h ∈ X H : we know that this holds when x ∈ X by assumption. Consider a vertex x ∈ A. It suffices to show that this holds when x is within an (h 1 , h 4 )-gadget, since the other cases are symmetric. So let u be such that x is within the (h 1 , h 4 )-gadget related to u. If x = a u 1,1,4 , then (x, b u 1,1,4 , h 1 ), (x, c u 1,1,4 , h 3 ), and (x, h 4 , h 5 )) are paths of length 2 between x and each h ∈ {h 1 , h 3 , h 5 }, as we wanted to show. An analogous argument holds if x = a u 2,1,4 . Now, we prove the second part of the theorem. Let f be an H-retraction of G. Since H is isomorphic to C 6 , it suffices to extend f to G . Also, because the gadgets are symmetric, we Conversely, suppose that G has a C 6 -compaction f , and write the target C 6 as H = (1, . . . , 6). We want to prove that f (h i ) = i, for every i ∈ [6], so that f also gives an H-retraction of G. Note that if, at some point, we get that f (h i ) = i either for every odd i, or for every even i, then we are done.
First, we prove that |f (X H )| > 1. So suppose without loss of generality that f (X H ) = 1, and note that in this case f (H) ⊆ {1, 2, 6}. Because G is bipartite and f is vertex-surjective, we know that f −1 (4) = ∅, and since f −1 (4) ⊆ Y we get a contradiction to property (1) of the set X H , since Y would not be dominated by X H . We then may assume that |f (X H )| = 2, since the proof is finished when it is equal to 3. This means that two vertices among {h 1 , h 3 , h 5 } get distinct images. By relabeling H if necessary, we can assume that the possible cases are the following: • f (h 3 ) = 3 and f (h 5 ) = 5: this implies that f (h 4 ) = 4. We know by Lemma 7 that f −1 (1)∩X = ∅ as otherwise f (h 1 ) = 1 and the lemma follows. Also, note that {h 2 , h 4 , h 6 } dominates A. Hence, since ∅ = f −1 (1) ⊆ A, we must get that either f (h 2 ) = 2 or f (h 6 ) = 6. In fact, if we get that f (h 2 ) = 2, then by relabeling (1, . . . , 6) to (3,4,5,6,1,2) and doing the same with H, we get the situation f (h i ) = i for each i ∈ {3, 4, 5, 6}. So suppose this is the case and note that we can assume that f (h 1 ) = 5 and f (h 2 ) = 4, since if f (h 1 ) = 1 and f (h 2 ) = 2 the proof is complete. We want to prove that f −1 (1) = ∅, thus getting a contradiction. Recall that f −1 (1) ⊆ A, and let a ∈ A be such that f (a) = 1. Also let u ∈ X be such that a is within some diagonal gadget related to u.
Proof: Let G be a bipartite graph with diameter 4. The first statement follows by Theorem 5 and Lemma 8. We argue that G has a C 6 -compaction if and only G has a surjective C 6homomorphism. Clearly a C 6 -compaction is a surjective C 6 -homomorphism, so it remains to show that given a surjective C 6 -homomorphism f of G, it follows that f is a C 6 -compaction as well. Suppose otherwise, and denote the target C 6 by (1, . . . , 6). Without loss of generality, suppose that edge 12 has no pre-image, i.e., there is no uv ∈ E(G) such that f (u) = 1 and f (v) = 2. This means that every path between u ∈ f −1 (1) and v ∈ f −1 (2) is mapped to (1,6,5,4,3,2), a contradiction since in this case we get dist(u, v) ≥ 5, contradicting the hypothesis that diam(G) ≤ 4.

3-Biclique Partition on bipartite graphs with diameter 4
In this section we first show how Corollary 9 implies the NP-completeness of the 3-Biclique Partition problem, and then we present the flaw in the proof in [6]. Recall that G B denotes the bipartite complement of G.
Corollary 10. Let G be a bipartite graph such that G B has diameter at most 4. Then, deciding whether G has a 3-biclique partition is NP-complete.
Proof: Denote the parts of G by X and Y . It suffices to notice that V 1 , V 2 , V 3 is a 3-biclique partition of G if and only if the function f defined as follows is a surjective C 6 -homomorphism The theorem thus follows from Corollary 9.
Fleischner et al. [6] presented a hardness proof for the k-Biclique Partition problem for a general k ≥ 3, but since their reduction for values of k larger than 3 can be obtained using the trick presented in Proposition 1, for the sake of simplicity we consider k = 3. Still for the sake of simplicity, we work on the bipartite complement of their construction. Their reduction is from List k-Coloring. So, consider G bipartite with parts X, Y , and a list assignment L such that L(u) ⊆ {1, 2, 3} for every u ∈ V (G). Let G be obtained from G by adding a cycle C = (x 1 , y 2 , x 3 , y 1 , x 2 , y 3 , x 1 ) (which alternatively can be seen as the complete bipartite graph on these vertices minus the perfect matching {x i y i | i ∈ [3]}). Then, for every u ∈ X, add an edge from u to y i if and only if i / ∈ L(u). One can see that G has an L-coloring if and only if G has a retraction to C, if and only if G B has a retraction to its subgraph induced by {x i , y i | i ∈ [3]} (which is a perfect matching; denote this subgraph by C B ). Clearly, if G B has a retraction to C B , then G B has a 3-biclique partition. However, in the reverse implication, it is not necessarily true that a 3-biclique partition V 1 , V 2 , V 3 will map edge x i y i inside of V i for every i ∈ [3], as the authors claim. For example, when G is simply an edge uv, and Nevertheless, their proof does imply that Retract(G, H) is NP-complete when G is bipartite and H is isomorphic to C 6 . We provided another proof of this fact in Theorem 5 because we needed stronger constraints on G and H in order to prove Corollaries 9 and 10.

k-Fall Coloring
In this section we investigate the complexity of k-Fall Coloring on bipartite graphs with diameter at most d for every pair k, d. As in the case of list coloring problems, again the only case that we leave open is when k = 3 and d = 3. We conjecture that this case is also NP-complete, and we prove in Proposition 14 that, if so, then the cases left open in Table 1 will also be NP-complete. We start with the following technical lemma.
Lemma 11. Let G be a bipartite graph with vertex bipartition (X, Y ) and f be a k-fall- Proof: Towards a contradiction, suppose that 1 / ∈ f (X), which implies that every v ∈ X has some neighbor in f −1 (1) ⊆ Y ; however, there can be no u ∈ Y with f (u) = 1, otherwise 1 / ∈ f (N [u]), so we have f −1 (1) = Y . But in this case, since k ≥ 3 and G is bipartite, |f (N [v])| ≤ 2 for every v ∈ X, contradicting the hypothesis that f is a k-fall-coloring.
The following is analogous to Proposition 1.

Proposition 12. Let k ≥ 3 be a fixed positive integer. Then,
Proof: Let G ∈ B with parts X and Y , and G be obtained from G by adding new vertices x, y together with all edges from x to vertices in Y and all edges from y to vertices in X. This is the same graph as the one constructed in Proposition 1, and thus we already know that G has diameter 3. Now, we prove that G has a k-fall-coloring if and only if G has a (k + 1)-fall-coloring. If f is a k-fall-coloring of G, then let f be obtained from f by coloring x and y with k + 1. Every vertex of X and Y is adjacent to the new color, so they continue to be b-vertices, and x and y are b-vertices by Lemma 11. Now, let f be a (k + 1)-fall-coloring of G , and suppose without loss of generality that f (x) = k + 1. Again, by Lemma 11 we get that each part must contain every color. Therefore, because x is complete to Y , we get that the only vertex on Y ∪ {y} that can be colored with k + 1 is y, i.e., f (y) = k + 1. In this case, one can see that f restricted to G must define a k-fall-coloring of G.
We get the following partial classification of the problem. As we already mentioned, the only open case is when k = 3 and d = 3. Proof: Observe that if G is a complete bipartite graph, i.e., a bipartite graph with diameter 2, then every coloring that uses more than 2 colors will have a non-b-vertex, hence the answer to k-Fall-Coloring is trivially "no" when k ≥ 3 and G is a complete bipartite graph (that is, d = 2). When k ≤ 2, then either G has an isolated vertex and the answer is "no", or it does not and the answer is "yes" since any 2-coloring is also a 2-fall-coloring. For k ≥ 4, it is known that 3-Fall-Coloring is NP-complete on bipartite graphs [19], which applying Proposition 12 and induction on k gives us that k-Fall-Coloring| B d is also NP-complete for every d ≥ 3. For the remaining case, we prove in Theorem 15 that 3-Fall-Coloring| B4 is NP-complete. Now, we prove that 3-Fall-Coloring is NP-complete even restricted to bipartite graphs with diameter 4. We mention that our proof is an improvement on the proof presented by Laskar and Lyle [19], where the constructed graphs have diameter 6, although the authors do not mention that in their proof.
Proof: We make a reduction from 3-Uniform 2-Col. Consider a 3-uniform hypergraph G on vertices V = {v 1 , . . . , v n } and hyperedges E = {e 1 , . . . , e m }, and let G be constructed as follows (see Figure 9 for an illustration). Add V and E to the set of vertices of G , together with a copy v i of each vertex v i ∈ V ; denote by V the set {v i | v i ∈ V }. Also, add two new vertices v, v , and make v complete to V and v complete to V . Finally, add an edge between e j and each v i ∈ e j for every e j ∈ E(G), and add the matching {v i v i | i ∈ [n]}. We prove that G is a bipartite graph with diameter 4, and that G is a yes-instance of 3-Uniform 2-Col if and only if G is a yes-instance of 3-Fall-Coloring.
First, note that (V ∪ E ∪ {v}, V ∪ {v }) is a bipartition of G . To see that G has diameter 4, first note that G − E(G) consists of a perfect matching between V and V , together with a vertex v complete to V and a vertex v complete to V . Observe that this subgraph has diameter 3, with the most distant pairs of vertices being v and v , and v i and v j with i = j. Now, consider a hyperedge e ∈ E(G). Below, we show that the distance between e and each other vertex of G is at most 4. Figure 9 Graph G related to hypergraph G = (V, E).
then (e, v j , v, v i ) is a path in G ; • d(e, v i ) ≤ 4 for every v i ∈ V : if v i ∈ e, then (e, v i , v i ) is a path in G . Otherwise, let v j ∈ e; then (e, v j , v, v i , v i ) is a path in G ; • d(e, e ) ≤ 4, for every e ∈ E(G) \ e: if there exists v i ∈ e ∩ e , then (e, v i , e ) is a path in G . Otherwise, let v i ∈ e and v j ∈ e , then (e, v i , v, v j , e ) is a path in G . Now, we prove that G is a yes-instance of 3-Uniform 2-Col if and only if G is a yesinstance of 3-Fall-Coloring. First, consider a 2-coloring f of G with no monochromatic hyperedge, and suppose that the used colors are {2, 3}. We extend f to a 3-fall-coloring f of G . For this, color every x ∈ E ∪ {v, v } with 1, and color v i with c ∈ {2, 3} \ f (v i ). One can verify that, because no hyperedge of G is monochromatic in f , the obtained coloring is a fall-coloring of G .
Finally, consider a 3-fall-coloring f of G , and suppose, without loss of generality, that f (v) = 1. This and the fact that v is a b-vertex imply that f (V ) = {2, 3}. Hence, for every e ∈ E(G), since N G (e) ⊆ V and 1 / ∈ f (V ), in order for e to be a b-vertex we must have that f (e) = 1, and that f (N G (e)) = {2, 3}. Therefore, the coloring f restricted to V is a 2-coloring of G with no monochromatic hyperedge.

3-List Coloring is NP-complete on complete bipartite graphs
In this section we provide an alternative proof the one given by Gravier [9] for the following result.
Theorem 16. Let k ≥ 3 be a fixed integer. The k-List Coloring| B2 problem is NPcomplete.
Proof: k-List Coloring is clearly in NP since, given a k-coloring f , one can check in polynomial time whether f satisfies the list constraints. We first prove that (3-List Coloring) D2 is NP-hard, again by a reduction from 3-  (1) or f −1 (2), we know that v i cannot be chosen as a color for vertices of both parts, i.e., this is a proper coloring of G . Also, the chosen colors clearly satisfy the list assignments.
Now, let f : V (G ) → {v 1 , . . . , v n } be a proper coloring of G that satisfies the list assignment. Let f be obtained from f by coloring v i with 1 if v i ∈ f (A), and with 2 otherwise. For each i ∈ {1, . . . , m}, we know by construction that f (f (a i )) = 1. Also, since f is a proper coloring and G is complete bipartite, we know that f (b i ) / ∈ f (A); hence, we get that f (f (b i )) = 2 and e i is not monochromatic.
Finally, note that (k-List Coloring) D2 ((k + 1)-List Coloring) D2 , since an instance of the former is also an instance of the latter. The theorem thus follows by induction on k, starting from k = 3.

3-b-colorings and 3-fall-colorings
In this section, for the sake of completeness, we presented a proof of Faik [3].
Theorem 17 (Faik [3]). Let G be a bipartite graph with diameter at most 3. If f is a 3-b-coloring of G, then f is a 3-fall-coloring of G.
Proof: Let (X, Y ) be the bipartition of V (G). By Lemma 11,it holds that that f (X) = f (Y ) = [3]. Note that if u, v are within the same part, then N (u) ∩ N (v) = ∅, as otherwise their distance would be at least 4. So, let u ∈ X be of color 1. Because there exists v ∈ X of color 2 and since N (u) ∩ N (v) = ∅, we get that u must have a neighbor of color 3, namely the common neighbor with v. The analogous holds when picking any v ∈ X of color 3; therefore u is a b-vertex. Clearly this argument can be applied to every u ∈ X ∪ Y just by renaming the colors and the parts.