Every Planar Graph with the Distance of 5−-Cycles at Least 3 from Each Other Is DP-3-Colorable

DP-coloring was introduced by Dvořák and Postle [J. Comb. Theory Ser. B 2018, 129, 38–54]. In this paper, we prove that every planar graph in which the 5−-cycles are at distance of at least 3 from each other is DP-3-colorable, which improves the result of Montassier et al. [Inform. Process. Lett. 2008, 107, 3–4] and Yin and Yu [Discret. Math. 2019, 342, 2333–2341].


Introduction
All graphs in this paper are finite, undirected, and simple. A planar graph is a graph that can be embedded into the plane. A plane graph is a particular embedding of a planar graph into the plane. We set a plane graph G = (V, E, F), where V, E, and F are sets of vertices, edges, and faces of G, respectively. Two faces are intersecting if they have a common vertex, and are adjacent if they have a common edge. A vertex is incident to a face if it is on the face. A vertex is adjacent to a face if it is not incident to the face but adjacent to a vertex on the face. For a face f ∈ F, if the vertices on f in a cyclic order are v 1 , v 2 , · · · , v k , then we write f = [v 1 , v 2 , · · · , v k ]. The degree d(x) of x ∈ V is the number of edges incident with x. The degree d(x) of x ∈ F is the number of vertices incident with x. Let a k-vertex (k + -vertex, k − -vertex) be a vertex of degree k (at least k, at most k), and a k-face (k + -face, k − -face) be a face of degree k (at least k, at most k). The same notation will be applied to cycles. A (l 1 , l 2 , · · · , l k )-face is a k-face f = [v 1 v 2 · · · v k ] with d(v i ) = l i , respectively. A (l 1 , l 2 )-edge is an edge e = v 1 v 2 with d(v i ) = l i . Let C be a cycle of a plane graph G, |C| is the length of the cycle C. A triangle is a 3-cycle. An edge or a vertex of G is triangular if it is on a triangle. A chord in a cycle C is triangular if it splits the cycle C into a triangle and a cycle of length d(C) − 1. We use Int(C) and Ext(C) to denote the sets of vertices located inside and outside of C, respectively, and put Int(C) = G − Ext(C), Ext(C) = G − Int(C). The cycle C is called a separating cycle if Int(C) = ∅ and Ext(C) = ∅. A set of independent edges of G is called a matching. Identifying vertices means merging the vertices into a single vertex.
The distance between two vertices u and v in G, denoted by d G (u, v), is the length (number of edges) of the shortest path between u and v in G. The distance between two cycles C and C of G, denoted by d(C, C ), is defined as follows: A proper k-coloring of G is a function f : V(G) → [k] := {1, 2, · · · , k} such that for every edge uv ∈ E(G), f (u) = f (v). The smallest k such that G has a k-coloring is called the chromatic number of G and is denoted by χ(G). A list assignment of a graph G is a mapping L that assigns to each vertex v ∈ V(G) a list L(v) of colors. An L-coloring of G is a function λ : V → v∈V L(v) and λ(v) ∈ L(v) for every v ∈ V such that λ(u) = λ(v) for every edge uv ∈ E. A graph G is k-choosable if G is L-colorable for every assignment L with |L(v)| ≥ k. The smallest k such that G is k-choosable is called the choice number of G and is denoted by χ l (G).
It is well known that the problem of deciding whether a planar graph is 3-colorable is NP-complete. This provides motivation to look for sufficient conditions for planar graphs to be 3-colorable. Grötzsch [1] showed that every planar graph without triangles is 3-colorable. In 1976, Steinberg [2] conjectured that every planar graph without 4-cycle and 5-cycle is 3-colorable. Havel [3] proposed to make d ∆ large enough, where d ∆ is the smallest distance between triangles. Dvorák, Kral, and Thomas [4] showed that d ∆ ≥ 10 100 suffices. Borodin and Glebov [5] showed that planar graphs without 5-cycles and d ∆ ≥ 2 are 3-colorable.
For ordinary coloring, the identification of vertices is involved in the reduction configurations. In list coloring, since different vertices may have different lists, it is not possible for one to use the identification of vertices. To overcome this difficulty, Dvořák and Postle [12] introduced DP-coloring (under the name correspondence coloring) as a generalization of list-coloring. Definition 1. Let G be a simple graph, and L be a list assignment of V(G). For each vertex v ∈ V(G), let L v = {v} × L(v). For each edge uv in G, let M uv be a partial matching between the sets L u and L v and let M = {M uv : uv ∈ E(G)}, called the matching assignment. The matching assignment is called a k-matching assignment if L(v) = [k] for each v ∈ V(G).

Definition 2.
A cover of G is a graph G L,M (simply write G) satisfying the following two conditions: (1) the vertex set of G is the disjoint union of L v for all v ∈ V(G); (2) the edge set of G is the matching assignment M.
Note that the induced subgraph G[L v ] is an independent set for each vertex v ∈ V(G). Definition 3. Let G be a simple graph, and G be a cover of G. An M-coloring of G is an independent set I in G such that |I L v | = 1 for each vertex v ∈ V(G). The graph G is DP-k-colorable if, for each k-list assignment L and each matching assignment M over L, it has an M-coloring. The minimum k such that G is DP-k-colorable is the DP-chromatic number of G, denoted by χ DP (G).
For each e ∈ E(G L,M ), M is obtained form M by replacing the vertex (v, c 1 ) by (v, c 2 ). Thus, G L ,M is obtained from G L,M by replacing the vertex (v, c 1 ) by (v, c 2 ). Then, G can be M -colorable when G is M-colorable by changing the color of v to c 2 when φ(v) = c 1 , and vice versa. We say that M is obtained from M by renaming at the vertex v.
An edge uv ∈ E(G) is straight in a k-matching assignment M if every (u, c 1 )(v, c 2 ) ∈ E(M u,v ) satisfies c 1 = c 2 . One can construct a cover of any graph G based on a list assignment for G, thus showing that list coloring is a special case of DP-coloring and, in particular, χ DP (G) ≥ χ l (G) for all graphs G. DP-coloring is quite different from list coloring-for example, Bernshteyn [13] showed that the DP-chromatic number of every graph G with average degree d is Ω(d/ log d), while Alon [14] proved that χ l (G) = Ω(log d) and the bound is sharp. DP-coloring is a generalization of list-coloring. Dvořák and Postle [12] proved that every planar graph G without cycles of length from 4 to 8 is 3-choosable. They also noted that χ DP (G) ≤ 3 if G is a planar graph without {3, 4}-cycles. Liu and Li [15] proved that every planar graph G without adjacent cycles of length at most 8 is 3-choosable.
In this paper, we improve the results in Theorems 1 and 2. A 9-cycle C is bad if it is in a subgraph of G isomorphic to the graphs in Figure 1, and is the outer cycle of the subgraph. A 9-cycle is good if it is not bad. Theorem 3. Let G be a planar graph in which the 5 − -cycles are at distance of at least 3 from each other. Let C 0 be a 8 − -cycle or a good 9-cycle in G. Then, each DP-3-coloring of C 0 can be extended to G. Corollary 1. Every planar graph in which the 5 − -cycles are at a distance of at least 3 from each other is DP-3-colorable (thus also 3-choosable).
Proof. Let G be a planar graph. Either G is 4 − -cycles free or it is not 4 − -cycles free. In the first case, as proved in Reference [12], G is DP-3-colorable. Thus, we only have to consider the case when G contains a 4 − -cycle. As proved in Reference [12], the 4 − -cycle can be precolored. Then, by Theorem 3, G is DP-3-colorable extended from the coloring of the 4 − -cycle when the 5 − -cycles are at distance of at least 3 from each other.

Proof of Theorem 3
We will prove Theorem 3 by reductio ad absurdum. Let's start by a temporary assumption that the theorem is wrong. Then, there has to be a non-empty set of counterexamples to this theorem. Assume that G is a minimal (least number of vertices) counterexample to Theorem 3. Let C 0 be a 8 − -cycle or a good 9-cycle in G.

Proof.
Let v be a 2 − -vertex in V(G − C 0 ). By the minimality of G, each DP-3-coloring of C 0 can be extended to G − v. Then, the coloring of G − v can be extended to G by selecting a color φ(v) for v such that, for each neighbor u of v, ((u, φ(u) (v, φ(v)) ∈ E(M uv ), a contradiction.

Proof.
First of all, we show that C 0 is not a separating cycle. Otherwise, if C 0 is a separating cycle, we may extend the coloring of C 0 to both Int(C 0 ) and Ext(C 0 ), respectively, and then combine them to get a coloring of G, a contradiction.
Let C = C 0 be a separating 8 − -cycle or separating good 9-cycle in G. By the minimality of G, the coloring of C 0 can be extended to Ext(C). Now that C is colored, thus the coloring of C can be extended to Int(C) by the minimality of G again. Combining the inside and outside of C, we have a coloring of G extended from the coloring of C 0 , a contradiction.
Lemma 3. C 0 is the boundary of the out face of the embedding of G.
Proof. C 0 is not a separating cycle by Lemma 2. Thus, either Int(C 0 ) or Ext(C 0 ) is empty. Without loss of generality, we assume that Int(C 0 ) is empty, and we can redraw the graph to make Ext(C 0 ) empty instead. Lemma 4. If a 9 − -cycle C in G has an internal chord e, then |C| ∈ {7, 8, 9} and either e is triangular, or |C| = 8 and e splits C into a 4-cycle and a 6-cycle, or |C| = 9 and e splits C into a 4-cycle and a 7-cycle, or |C| = 9 and e splits C into a 5-cycle and a 6-cycle.
Proof. Due to fact that the cycles of lengths 3, 4, and 5 in G are at a distance of at least 3 from each other, C cannot have a chord if |C| ≤ 6 and can have only a triangular one when |C| = 7. If |C| = 8 and e is not triangular, then e splits C into a 4-cycle and a 6-cycle. If |C| = 9 and e is not triangular, then e splits C into a 4-cycle and a 7-cycle, or e splits C into a 5-cycle and a 6-cycle.
By Lemmas 3 and 4, if a bad 9-cycle C (one type in Figure 1) is a subgraph in G, then C must be induced.
Proof. If C 0 contains a chord e, then e is one of the types described in Lemma 4. By Lemma 2, G has no separating 8 − -cycles. Thus, G contains no other vertices and the coloring on C 0 is also a coloring of G, a contradiction.
The following lemma from [21] provides a powerful tool to prove the reducibility. Lemma 6. Let k ≥ 3 and H be a subgraph of G. If the vertices of H can be ordered as v 1 , v 2 , · · · , v l such that the following hold Lemma 7. Let f be an internal 6-face in G. If f is a (3, 3, 3, 3, 3, 3)-face, then f cannot be adjacent to an internal face f 1 with 5 or less vertices such that all vertices on f 1 are vertices with degree 3. 4, 5}), so that v 1 v 2 is the common edge of f and f 1 , and all vertices on f 1 are 3-vertices. Order the vertices on f and f 1 as v 1 , w 4 , w 3 , w 2 , w 1 , v 2 , v 3 , · · · , v i (i ∈ {3, 4, 5}). Let H be the set of vertices in the list. Since all vertices in H are from the internal faces f and f 1 , no vertex in C 0 is going to be removed by such subtraction.
Because G is a minimal counterexample, by Lemma 6, every DP-3-coloring of G − H can be extended to G, a contradiction.
Let f be a (3, 3, 3, 3, 3, 3)-face adjacent to a 3-face f . We call the vertex v on f but not on f the roof of f , and f the base of v.
Lemma 8. Let f be an internal 6-face in G and f 1 be an internal face with 5 or less vertices which is adjacent to f . If f 1 has one 4-vertex, while the other vertices incident with f 1 are vertices with degree 3, then each of the following holds: (a) f cannot be adjacent to another face with five or less vertices; Proof.
(a) follows from the condition on the distance of 5 − -faces. 4)-edge of f 1 and f , and all other vertices incident with f and f 1 are 3-vertices. Let . Let H be the set of vertices in the list. By Lemma 6, every DP-3-coloring of G − H can be extended to G, a contradiction.
. Let H be the set of vertices in the list. By Lemma 6, every DP-3-coloring of G − H can be extended to G, a contradiction. If j > 2, then order the vertices on f , f 1 and . Let H be the set of vertices in the list. By Lemma 6, every DP-3-coloring of G − H can be extended to G, a contradiction.
] be an internal 6-face that is adjacent to an internal face with five or less vertices ). If all of the vertices on f 1 are vertices with degree 3, Proof. We assume that d(v 3 ) = d(v 6 ) = 3, and we use u to denote the neighbor of w 1 that is not on f 1 . First, we may rename the lists of vertices in ) by identifying v 4 and u. We claim that no new cycles of length from 3 to 5 multiple edges or loop are created. Otherwise, there is a path of length 1, 2, 3, 4 or 5 from u to v 4 in G, which together with w 1 , v 2 , v 3 , v 4 forms a cycle C, 5 ≤ d(C) ≤ 9. Because f 1 is a 5 − -face, C cannot be a 5-cycle.

•
If v 1 and v 6 are in Int(C), see Figure 2a, then C is not a bad 9-cycle. Otherwise, since v 1 and v 6 are in Int(C), C must be the type (b) in Figure 1. Thus, v 1 and v 6 must be in a 3-cycle which is adjacent to f , a contradiction to Lemma 8(a). Thus, C is a separating {6, 7, 8}-cycle or separating good 9-cycle, a contradiction to Lemma 2.

•
If v 1 and v 6 are not in Int(C), see Figure 2b. Due to d(v 3 ) = 3 and f being a 6-cycle, by Lemma 2 and 4, v 3 is incident with an edge e in Int(C). Either the other vertex incident to e is in the cycle C, or it is not. If it is in the cycle C, then edge e is its chord, 7 ≤ d(C) ≤ 9 and e must be incident with a 5 − -cycle C by Lemma 4. The distance between f 1 and C is at most 1, a contradiction. If the other vertex incident to e is not in cycle C, then C is a separating cycle. If C is a bad 9-cycle, recall that e is incident v 3 and in Int(C), the v 3 must be incident with or adjacent to a 3-cycle. Thus, the distance between f 1 and the 3-cycle is at most 2, a contradiction. Thus, C must be a separating {6, 7, 8}-cycle or separating good 9-cycle, a contradiction to Lemma 2. Now, we claim that there is no new chord in C 0 of G . Otherwise, u is on C 0 and v 4 is adjacent to a vertex v 4 which is on C 0 . Then, there is a path between u and v 4 on C 0 with length at most four, which forms a {5, 6, 7, 8, 9}-cycle with w 1 , v 2 , v 3 , v 4 in G. Similar to the proof process above, this does not occur.
Because f 1 is a 5 − -cycle, the distance from v 4 to other 5 − -cycles is at least one and the distance from u to other 5 − -cycles is at least two. Thus, the 5 − -cycles are at a distance of at least 3 from each other in G . Since C 0 is not a bad 9-cycle in G and every bad 9-cycles in G is induced, C 0 is not a bad 9-cycle in G . Because C 0 still is the boundary of the out face of the embedding of G and no new chord in C 0 is formed in G and G is a minimal counterexample, the DP-3-coloring of C 0 can be extended to G . Now, we color v 4 and u with the color of the identified vertex and keep the colors of all vertices in G . Now, we color v 3 first, and then color w 1 with the color of v 3 . We can do this because the edges in {uw 1 , w 1 v 2 , v 1 v 2 , v 2 v 3 , v 3 v 4 } are straight and the color of v 3 is different from the color of v 4 and u.
be an internal face with five or less vertices such that all vertices of f are vertices with degree 3, so that Proof. Assume that d(w 2 ) ≤ 4. Since there is no separating 6-cycle and every 6-cycle has no chords by Lemma 2 and Lemma 4, the 6-cycle w 1 u 1 u 2 w 2 w 2 w 1 and w 2 v 1 v 2 w 3 w 3 w 2 are both 6-faces. If d(w 1 ) = 3, then, by Lemma 9, d(w 2 ) = 4. Let w 2 be the fourth neighbor of w 2 . Since the cycles of length 3, 4, and 5 in G are at a distance of at least 3 from each other, w 2 is not one vertex in {w 1 , u 1 , u 2 , w 2 , w 2 , · · · , w i , w 1 , w 3 , v 2 , v 1 }. We may rename the list of vertices in {w 2 , w 2 , w 1 , w 3 , w 3 } so that the edges {w 2 w 2 , w 2 w 2 , w 1 w 2 , w 2 w 3 , w 3 w 3 } are straight.
Consider the graph G obtained from G − {w 1 , u 1 , u 2 , w 2 , w 2 , · · · , w i , w 1 } (i ∈ {3, 4, 5}) by identifying w 3 and w 2 . Since the cycles of length 3, 4, and 5 in G are a distance of at least 3 from each other, the distance between w 3 and w 2 in G is at least 2 and the distance between v 1 and w 2 in G is at least 4. We claim that no new cycles of length from 2 to 5 are created. Otherwise, there is a path of length 2 to 5 from w 3 and w 2 in G, which together with w 2 , w 2 , w 3 forms a {6, 7, 8, 9}-cycle C. If u 1 is in Int(C), then C is not a bad 9-cycle. Otherwise, since u 1 is in Int(C), u 1 , u 2 , w 1 and w 1 are in Int(C). Then, C is not isomorphic to a bad 9-cycle in Figure 1. Thus, C is a separating {6, 7, 8}-cycle or separating good 9-cycle, a contradiction to Lemma 2. Let v 1 be in Int(C) (Since the distance between v 1 and w 2 in G is at least 4, v 1 is not on C). Because the cycles of length 3,4, and 5 in G are at a distance of at least 3 from each other, v 1 cannot be on triangles. Then, C is not isomorphic to a bad 9-cycle see Figure 3b. Thus, C is a separating {6, 7, 8}-cycle or separating good 9-cycle, a contradiction to Lemma 2. Now, we claim that there is no new chord in C 0 of G . Otherwise, w 2 is on C 0 and w 3 is adjacent to a vertex w 3 which is on C 0 ; then, there is a path between w 2 and w 3 on C 0 with length at most four, which forms a {5, 6, 7, 8, 9}-cycle with w 3 , w 2 , w 2 , w 3 in G. Similar to the proof process above, this does not occur.
Because f is a 5 − -cycle, the distance from w 3 to other 5 − -cycles is at least two and the distance from w 2 to other 5 − -cycles is at least one. Thus, the 5 − -cycles are at a distance of at least 3 from each other in G . Since C 0 is not a bad 9-cycle in G and every bad 9-cycles in G is induced, C 0 is not a bad 9-cycle in G . Because C 0 still is the boundary of the out face of the embedding of G and no new chord in C 0 is formed in G and G is a minimal counterexample, the DP-3-coloring of C 0 can be extended to G . Now, color w 2 and w 3 with the color of the identified vertex and keep the colors of all vertices in G . Now, we color w 2 first, and then color w 3 with the color of w 2 . We can do this because the edges in {w 2 w 2 , w 2 w 2 , w 2 w 3 , w 3 w 3 } are straight and the color of w 2 is different from the color of w 2 and w 3 . If d( f 1 ) = 3, then we color u 2 , u 1 , w 1 , w 1 , w 2 in the order. If d( f 1 ) = 4 or 5, then we color u 2 , u 1 , w 1 , w 4 , (w 5 ), w 1 , w 2 in the order. Then, G has been colored, a contradiction.
We are now ready to present a discharging procedure that will complete the proof of the Theorem 3. Let each vertex v ∈ V(G) have an initial charge of µ(v) = 2d(v) − 6, and each face f = f 0 in our fixed plane drawing of G have an initial charge of µ( f ) = d( f ) − 6. Let µ( f 0 ) = d( f 0 ) + 6. By Euler's Formula, ∑ x∈V F µ(x) = 0.
Let µ * (x) be the charge of x ∈ V F after the discharge procedure. To lead to a contradiction, we shall prove that µ For shortness, let F k = { f : f be a k-face and V( f ) C 0 = ∅}. The discharging rules: (R1): If v is an internal 4 + -vertex.

1
If v is an internal 4-vertex and incident with a 3-face f , then v gives 3 2 to its incident 3-face, 1 2 to its base when v has a base and 1 2 to its incident (3, 3, 3, 3, 3, 4 + )-face that is adjacent to f .

2
If v is an internal 5 + -vertex and incident with a 3-face f , then v gives 3 2 to its incident 3-face, 1 2 to its base when v has a base and 1 2 to its incident 6-face that is adjacent to f . 3 If v is incident with a 4-face, then v gives 2 to its incident 4-face. 4 If v is incident with a 5-face, then v gives 1 to its incident 5-face. 5 If v is an internal 4-vertex. If v is adjacent to a 5 − -face f such that all vertices on f are 3-vertices, then v gives 1 to its adjacent 5 − -face f and v gives 1 2 to its incident 6-faces if any that are not adjacent to f . 6 If v is an internal 5 + -vertex. If v is adjacent to a 5 − -face f such that all vertices on f are 3-vertices, then v gives 2 to its adjacent 5 − -face f and 1 2 to its incident 6-faces if any that are not adjacent to f . 7 If v is an internal 4 + -vertex. If v is not on a 5 − -face nor adjacent to a 5 − -face f such that all vertices on f are 3-vertices, then v gives 1 2 to its incident 6-faces. Each non-internal 6-face or 7 + -face other than f 0 gives 1 to each of its adjacent 5 − -face, if any, and gives the rest to the outer face f 0 . (R3): The outer face f 0 get µ(v) from each v ∈ C 0 , gives 3 to each face in F 3 , 2 to each face in F 4 , 1 to each face in F 5 , and 1 to each face in F 6 adjacent to an internal face with 5 or less vertices.
Finally, assume that v is not on a 5 − -face or adjacent to a 5 − -face f that all vertices on f are 3-vertices, then, by (R1) 7 , its final charge is Because of this, the cycle of lengths 3, 4, and 5 are at a distance of at least 3 from each other. If v is on a 5 − -face f . If d( f ) = 3, then, by (R1) 2 , v gives 3 2 to the 3-face, and 1 2 to its base or incident (3, 3, 3, 3, 3, 4 + )-faces that is adjacent to f . Thus, If v is adjacent to a 5 − -face f that all vertices on f are 3-vertices. Thus, by (R1) 6 , v gives at most 2 to the 5 − -face, and 1 2 to each other incident 6-faces that are not adjacent to the 5 − -face. Hence, Finally, assume that v is not on a 5 − -face or adjacent to a 5 − -face f that all vertices on f are 3-vertices, then, by (R1) 7 , its final charge is µ

Lemma 12.
Every face other than f 0 in G has a nonnegative final charge.
Let d( f ) = 5. If f contains some vertices of C 0 , then f gets 1 from f 0 by (R3), so µ * ( f ) = 0. Let f be an internal face. If f contains a 4 + -vertex, then, by (R1) 4 f gets 1 from each of the incident or non-internal 6-faces, then it sent 1 to f by (R2) and µ * ( f ) ≥ d( f ) − 6 + 1 = 0. Thus, we may assume that it is adjacent to five internal 6-faces. By Lemma 9, f is adjacent to at least three internal 4 + -vertices, so, by (R1) 5 and 6 , f gets 1 + 1 from the three Let d( f ) = 6. If f contains vertices of C 0 or f is not adjacent to an internal 3-face, then, by (R2) and (R3), µ * ( f ) = 0. Now, we assume that f is an internal 6-face that is adjacent to an internal 3-face  from f when f contains a 4 + -vertex that is not adjacent to the 3-face, in which case, by (R1) 7 f gets 1 2 from the 4 If f is an internal (4, 3, 3, 3, 3, 3)-face, then f gets 1 2 from v 2 by (R1) 1 , or else f contains another 4 + -vertex, then, by (R1) 7 f gets 1 2 from the 4 + -vertex. Thus, If d( f ) ≥ 7. Since the cycles of lengths 3, 4, and 5 in G are a distance of at least 3 from each other, f is adjacent to at most We call a bad 6-face f in F 6 Special if f is adjacent to one internal 5 − -face.

Lemma 13.
The final charge of f 0 is positive.
• If 3 = 1, then d( f 0 ) = 9 and x = 0. Since C 0 is not a bad 9-cycle, C 0 is adjacent to a 7 + -face f and f is adjacent to the 3-face. Thus, by (R2), f gives at least 1 to f 0 , that is, x ≥ 1, a contradiction.
Proof of Theorem 3. By Lemmas 11-13, ∑ x∈V F µ * (x) > 0, a contradiction to ∑ x∈V F µ(x) = 0. Thus, the counterexample can't exist, which, in turn, would show that the theorem is true for all cases.

Conclusions
The coloring theory of graphs plays a very important role in combinatorial optimization, computer theory, allocation of wireless communication channels, network data transmission, and so on. For example, the efficient design of airline schedules, the design of computer coding programs, etc.
It is well known that the problem of deciding whether a planar graph is 3-colorable is NP-complete. The Three Color Problem is very much alive, replete with an assortment of established results and an abundance of open problems. In addition, 3-coloring is the first significant graph coloring problem and, on the plane, the only unqualified graph coloring problem remaining. DP-coloring is one generalization of list coloring, which is a stronger version of proper coloring. It is very difficult to determine whether a graph is DP-colorable, or even whether a planar graph is DP-3-colorable.
It is unknown if Theorem 3 is most possible in the sense that there exists a planar graph with 5 − -cycles that are a distance of at least 2 from each other is not DP-3-colorable.
Let d 1 denote the least integer k such that every planar graph with 5 − -cycles are at distance from each other of at least k is DP-3-colorable. Problem 1. What is the exact value of d 1 ?
Let d 2 denote the least integer k such that every planar graph with 5 − -cycles are at distance from each other of at least k is 3-choosable.

Problem 2.
What is the exact value of d 2 ?
It follows from Theorem 3 that d 1 ≤ 3 and d 2 ≤ 3.