Ond-diagonal colorings

On &Diagonal Colorings

Daniel P. Sanders and Yue Zhao DEPARTMENT OF MATHEMATICS,

THE OHIO STATE UNlVERSlV COLUMBUS, OH

e-mail: [email protected]. edu e-mail: [email protected]. edu

ABSTRACT

A coloring of a graph embedded on a surface is d-diagonal if any pair of vertices that are in the same face after the deletion of a t most d edges of the graph must be colored differently. Hornak and Jendrol introduced d-diagonal colorings as a generalization of cyclic colorings and diagonal colorings. This paper proves a conjecture of Hornak and Jendrol that plane quadrangulations have d-diagonal colorings with at most 1 + 2 . colors. A similar result is proven for plane triangulations. Each of these results extends to the projective plane. Also, a lower bound for the d-diagonal chromatic number is given. 0 1996 John Wiley & Sons, Inc.

1. INTRODUCTION

A cyclic coloring of an embedded graph is a coloring of the vertices of the graph, such that if two vertices are incident with a common face, they receive different colors. A diagonal coloring of a triangulation T is a coloring of the vertices of T, such that, given two vertices x, y, if there is an edge a such that x and y are incident with a common face of T - a, they receive different colors. Hornak and Jendrol [8, 91 (in the dual) introduced an interesting generalization of these two colorings. Let two vertices 2, y of an embedded graph G be d-diagonally adjacent if there is a set S of edges such that IS\ 5 d, and such that x and y are incident with a common face of G - S. Let a d-diagonal coloring of a graph G be a coloring of the vertices of G, such that each pair of d-diagonally adjacent vertices receive different colors. Thus a cyclic coloring is a O-diagonal coloring, and a diagonal coloring is a l-diagonal coloring. Let a k-coloring of a graph G be a coloring of the vertices of G using at most k colors. Given a set A of embedded graphs, the d- diagonal chromatic number of A is the minimum number k, such that every graph in A has a d-diagonal k-coloring.

Journal of Graph Theory Vol. 22, No. 2, 155-1 66 (1 996) 0 1996 John Wiley & Sons, Inc. CCC 0364-9024/96/020155-I 2

156 JOURNAL OF GRAPH THEORY

Any bound on the d-diagonal chromatic number of a class of graphs must clearly depend on the maximum face size of these graphs. Given an embedded graph G, let A(G) (or A if the graph is clear from context) be the maximum face size of G. Ore and Plummer [lo] showed that plane graphs have 0-diagonal chromatic number at most 2A. A lower bound for this class of graphs is [:A]. Borodin [4] improved the upper bound to 2A - 3 for A 2 8. Results for small A are known. Appel and Haken [ 11 proved the Four Color Theorem, which is the case A = 3. Borodin [2, 61 showed the best possible result for A = 4, that this class of graphs has 0-diagonal chromatic number six. He [4] showed that plane graphs with A = 5, = 6, = 7, can be 0-diagonally colored with nine, eleven, and twelve colors, respectively. Also, Plummer and Toft [ 1 I] showed that 3-connected plane graphs have 0-diagonal chromatic number at most A + 9.

Bouchet, Fouquet, Jolivet, and Riviere [7] showed that plane and projective plane graphs wtih A = 3 have 1-diagonal chromatic number at most 12. They also showed that graphs embeddable on a surface of Euler characteristic x 5 0 with A = 3 have 1-diagonal chromatic number at most 6 + (49 - 24x)l/’. They showed a torus triangulation requiring 13 colors in any 1-diagonal coloring, showing that their result on the torus is best possible. They also gave a plane triangulation requiring 9 colors. Borodin (31 improved the upper bound for the plane to 11, and in [5] improved the x 5 0 case to (13 + (169 - 48~)~’’) . Sanders and Zhao [12] improved the upper bound for the plane to 10, and gave a lower bound for the projective plane of 11. One result of this paper is for the case of A = 3. Section 4 shows that plane and projective plane triangulations can be d-diagonally colored with 2 + 19 . ZdP2 colors.

Hornak and Jendrol[8] showed that the 1-diagonal chromatic number of plane quadrangulations is between 7 and 21, the lower bound being implicit. They also conjectured that the d-diagonal chromatic number of plane quadrangulations is at most 1 + 2 .3d+1. Sections 2 and 3 prove this conjecture (also for projective plane quadrangulations). A result for the plane is that the 1-diagonal chromatic number of plane quadrangulations is between 11 and 19.

Hornak and Jendrol [9] also have results for the general case on the plane. For 8 5 A 5 11, they showed that the d-diagonal chromatic number is at most 1 + (A + 7) (A - l)d. For A 2 12, they gave an upper bound of 1 + (2A - 4)(A - l)d. Section 5 gives a lower bound for the general case.

2. REDUCIBILITY

Let a k-vertex be a vertex of degree k. Let k ( d ) := 1 + 2-3d+1. For the context of Sections 2 and 3, let a minimal graph be a minimal, with respect to the number of vertices, plane or projective plane quadrangulation that has d-diagonal chromatic number greater than k ( d ) , for d 2 1. Also, let a configuration be reducible, if it appears in no minimal graph.

Lemma 1. A 2-vertex is reducible.

Proof. Let G be a minimal graph with 2-vertex 2. From minimality, G - 2 is d- diagonally k(d)-colorable. This yields a partial d-diagonal coloring of G. Now 5 can

I be colored to give a d-diagonal k(d)-coloring of G.

Lemma 2. A face containing more than one 3-vertex is reducible.

ON D-DIAGONAL COLORINGS 157

a

FIGURE 1. The configurations for Lemma 2.

Proof. Let G be a minimal graph with a face containing two 3-vertices v, w. Assume ww E E(G). Let the vertices be labeled as in Figure la. From minimality,

G - w - w + ab is d-diagonally k(d)-colorable. This yields a partial d-diagonal coloring of G. Since each of v, w has at most k(d) - 1 d-diagonal neighbors, they can be colored to give a d-diagonal k(d)-coloring of G.

Assume vw $! E(G). Let the vertices be labeled as in Figure lb. From minimality, G - v + aw is d-diagonally k(d)-colorable. This yields a partial d-diagonal coloring of G. Since v has at most k(d) - 1 d-diagonal neighbors, it can be colored to give a d-diagonal Ic(d)-coloring of G. I

Lemma 3. A 4-vertex adjacent to more than one 3-vertex is reducible.

Proof. Let G be a minimal graph with a 4-vertex 2 adjacent to two 3-vertices v, w. By Lemma 2, and w are not incident with any face of G. Thus the vertices may be labeled as in Figure 2. From minimality, G - v - w - 5 + ab + cd is d-diagonally k(d)-colorable. This yields a partial d-diagonal coloring of G. The vertices 2, v, w may be colored in that order such that each is d-diagonally adjacent to at most k ( d ) - 1 colored vertices when it

I is colored, yielding a d-diagonal k(d)-coloring of G.

Lemma 4. Figure 3 is reducible.

Proof. Let G be a minimal graph with 3-vertex 2) and 4-vertices w, 2, y, z , configured as in Figure 3. From minimality, H := G - v + ay is d-diagonally Ic(d)-colorable. Remove the colors that were assigned to 2 and z in H . This yields a partial d-diagonal coloring of G. The vertices z , 2, v may be colored in that order such that each is d-diagonally adjacent to at most k(d) - 1 colored vertices when it is colored, yielding a d-diagonal k(d)-coloring of G. I

a C

FIGURE 2. The configuration for Lemma 3


W

a L X Y

V Z

FIGURE 3. The configuration for Lemma 4.

Lemma 5. to z, is reducible.

Figure 4, such that there is a 3-vertex distinct from w 1-diagonally adjacent

Prooj Let G be a minimal graph with 3-vertices w, w and 4-vertices z, y, z, configured as in Figure 4, such that w is different from w and is 1-diagonally adjacent to 5. From minimality, H := G - 2, - z + az + bz is d-diagonally k(d)-colorable. Remove the colors that were assigned to w, y, and z in H. This yields a partial d-diagonal coloring of G. The vertices y, z , 5 , w, w may be colored in that order such that each is d-diagonally adjacent to at most k ( d ) - 1 colored vertices when it is colored, yielding a d-diagonal k(d)-coloring

Lemma 6. of G. I

Figure 5 is reducible.

Prooj Let G be a minimal graph with 3-vertices w,w and 4-vertices z,y, configured as in Figure 5. From minimality, G - w - w - II: - y + ab + cd + ce is d-diagonally k (d ) - colorable. This yields a partial d-diagonal coloring of G. The vertices y , ~ , w,w may be colored in that order such that each is d-diagonally adjacent to at most k ( d ) - 1 colored vertices when it is colored, yielding a d-diagonal k(d)-coloring of G. m



V X Y

a C W

b d

e

FIGURE 5. The configuration for Lemma 6.

Lemma 7. Figure 6 is reducible.

Proof. Let G be a minimal graph with 3-vertices u, v, w, 4-vertices 2, y, and 5-vertex z , configured as in Figure 6. From minimality, H := G - v + ay is d-diagonally k(d)- colorable. Remove the colors that were assigned to u, w, 2, y, and z in H . This yields a partial d-diagonal coloring of G. The vertices z , y, 2, w, v, u may be colored in that order such that each is d-diagonally adjacent to at most k(d) - 1 colored vertices when it is

I colored, yielding a d-diagonal k(d)-coloring of G.

Lemma 8. Figure 7, such that either m or n is a 3-vertex, is reducible.

Proof. Let G be a minimal graph with 3-vertices u, v, w, 4-vertices 2, y, and 5-vertex z , configured as in Figure 7, such that either w = m, or w = n. From minimality, H := G-u- y+az+ bx is d-diagonally k(d)-colorable. Remove the colors that were assigned to v, w, 2, and z in H . This yields a partial d-diagonal coloring of G. The vertices z , 2, y, u, v, w may be colored in that order such that each is d-diagonally adjacent to at most k(d) - 1 colored

I vertices when it is colored, yielding a d-diagonal k(d)-coloring of G.



a


3. DISCHARGING

The main theorem of this paper will be proven in this section by means of the Discharging Method, the technique used to solve the Four Color Theorem [l]. From Euler’s formula, and the fact that each face is of size four, the equality below follows for plane quadrangulations (the same equation with the 8 replaced by a 4 is true for projective plane quadrangulations):

(4- deg(x)) = 8. x E V ( G )

This value 4- deg(z) is the initial charge of the vertex z. Note that the sum of the initial charges is eight, a positive number. Thus there is some term of the sum that is positive, or that there is a vertex with initial positive charge, thus a vertex of degree at most three. Although by Lemma 1 a vertex of degree two does not present a problem, a vertex of degree three cannot be eliminated. This is circumvented by discharging the initial positive charges, or locally redistributing this charge into vertices with initial negative charge. This will define the charge of a vertex, and again the sum of the charges is positive. Thus some vertex has positive charge. Each way that that can happen will yield a structure proved reducible in Section 2.

The discharging will be done by some rules. Given a vertex x of a plane graph G, let a vertex y be a partner of x if there is a face of G incident with both x and y, and yet y is not a neighbor of x. If y is either a neighbor or a partner of x, then y is in the wheel of x. The positive charge will be moved from a 3-vertex 21 to the vertices in its wheel of degree at least five. Each vertex in the wheel of w will receive at most f from w.

Rule I Rule 2 Rule 3

FIGURE 8. The first three discharging rules


Rules 1-3 are represented in Figure 8. A triangle indicates a vertex of degree three, a square, degree four, and a circle, degree at least five. No symbol indicates a vertex of any degree. For each rule, a charge of f is sent from the triangle to the circle.

be a vertex of degree three, and w be in the wheel of 'u of degree at least five that did not receive by Rules 1-3. Note that there are two neighbors of w that are in the wheel of v. Let k be the number of those that are 4-vertices. Send a charge of (2 + k)/12 from v to w.

Thus each neighbor of degree at least five in the wheel of a 3-vertex v receives either from v.

Theorem 1. The d-diagonal chromatic number of plane or projective plane quadrangulations is at most I + 2 . 3d+'.

Suppose that there is a plane or projective plane quadrangulation that requires k ( d ) + 1 colors in a d-diagonal coloring. Let G be a minimal graph. By Lemma 1, the minimum degree of G is three. Let G be discharged according to Rules 1 4 above. Let charge (z) be the charge assigned to each vertex by these rules.

Let z be a 3-vertex of G (with initial charge 1). By Lemma 2, every vertex in the wheel of z has degree at least four. Consider the face F of G - z of size six. By Lemma 4, F has no four consecutive vertices of degree four in G. If F has three consecutive vertices of degree four in G, then z either sends out f to two neighbors by Rule 3 or it sends out f to two partners by Rule 2, and charge (z) 5 0. If F has two pairs of consecutive vertices of degree four in G, then z sends out f to a neighbor by Rule 3 and another f to a partner by Rule 1. If the previous cases do not occur, and F has a pair of consecutive vertices of degree four in G, then z sends out f to a neighbor by Rule 3, and a total charge of at least f by Rule 4, and charge (z) 5 0. If F has no two consecutive vertices of degree four in G, then z sends out a total charge of 1 by Rule 4, and charge (z) = 0.

Note that a 4-vertex has initial charge 0, and receives no charge. Let v be a 5-vertex of G (with initial charge -1). Let the vertices of the face of G - v

of size ten in a planar cyclic ordering be v l , . . . , v10, such that v1 is a neighbor of v. Let wl l be the neighbor of v1 that is a partner of 212. Let 2112 be the neighbor of 213 that is a partner of v2.

Rule 4 is as follows: Let

1 1 1 2 , 3 , 4, or

Proof.

Case 1. Assume v receives f from v2 by Rule 2.

Let 2113 be the vertex neighbor to each of v2,011,v12. Without loss of generality, by Rule 2, deg(v2) = 3, deg(v3) = 4, deg(v12) = 4, and deg(v13) = 4. By Lemma 5, each of vl,v4, v5, 216, v7, vg, vlo has degree at least four. From the Discharging Rules, sends in at most f, and charge (v) 5 0.

Case 2. Assume v receives f from 212 by Rule 1.

Thus by Rule 1, deg(v1) = 4, deg(v2) = 3, deg(v3) = 4, deg(vl1) = 4, and deg(vl2) = 4. By Lemma 3, deg(v4) 2 4 and deg(vlo) 2 4. If deg(v5) = 3, then by Lemma 2, deg(v6) 2 4 and deg(v7) 2 4, and by Lemma 8, deg(v8) 2 4 and deg(vg) 2 4, and v

has only two 3-vertices in its wheel, and charge (v) 5 0. Thus, assume deg(v5) 2 4 and symmetrically, deg(vg) 2 4. If deg(v7) 5 4, then by Lemmas 2 and 3, at most one of 216,v7,V8 is a 3-vertex. Since each 3-vertex sends in at most f to v, in this case, charge (v) 5 0. Thus assume deg(v7) 2 5, and thus Rule 1 cannot apply to either v6 or 2)s. By Case 1, Rule 2 does not apply to either v6 or 2)8. Thus each of v6, v8 sends in at most $ by Rule 4, and charge (w) I 0.


Assume v receives i from v1. Without loss of generality, by Rule 3, deg(vl) = 3, deg(v2) = 4, and deg(v3) = 4. By Lemma 2, deg(vg) 2 4 and deg(v1o) 2 4. By Lemma 6, deg(v4) 2 4. If deg(v5) = 3, then by Lemma 2, deg(v6) 2 4 and deg(v7) 2 4, and by Lemma 7, deg(v8) 2 4. Thus v has only two 3-vertices in its wheel, and charge (v) 5 0. Thus assume deg(v5) 2 4. If deg(v7) 5 4, then by Lemmas 2 and 3, charge (v) 5 0, similar to before. Thus assume deg(v7) 2 5. From Cases 1 and 2, neither v6 nor w8 can send in i. Thus each of v6, v8 sends in at most i by Rule 4, and charge (w) 5 0.

Thus each vertex sends in at most By Lemma 2, v has at most two neighbors of degree three. If v has two neighbors of

degree three, by Lemma 2, v has at most three 3-vertices in its wheel, and charge (v) 5 0. If v has one neighbor of degree three, by Lemma 2, v has at most three 3-vertices as partners. If it has less than three, clearly charge (v) 5 0. Otherwise, by Lemma 3, two of its partners send in at most f and one sends in i; again charge (w) 5 0. Finally, consider if v has no neighbors of degree three. If v has 3 partners of degree three, each sends in at most 5 . If v has 4 partners of degree three, each sends in at most i, by Lemma 3. If has 5 partners of degree three, each sends in i, by Lemma 3. In each case, charge v) 5 0.

Let Y be a 6-vertex of G (with initial charge - 2 ) . By Lemma 2, v has at most three neighbors of degree three. If v has three neighbors of degree three, by Lemma 2, v has no partners of degree three, and charge (v) 5 -;. If v has two neighbors of degree three, by Lemma 2, v has at most two partners of degree three, and charge (v) 5 0. If v has one neighbor of degree three, there is only a question if w has exactly four partners of degree three, but in this case, by Lemma 3, at least two of them send in i, and charge (w) 5 -i. If v has no neighbors of degree three, there is only a question if v has at least five partners of degree three, but in this case, by Lemma 3, at least three of them send in i, and charge

Let v be a 7-vertex of G (with initial charge -3). There is only a question if v has at least seven 3-vertices in its wheel. By Lemma 2, the only way that this can happen is if v has no neighbors of degree three, and seven partners of degree three. But in this case, all seven send in i, and charge (v) = - (1 1/6).

Let v be a vertex of G of degree k 2 8. By Lemma 2, v has at most k 3-vertices in its wheel. Thus charge (w) 5 (4 - k ) + (k/2) 2 0.

Thus no vertex has positive charge. This contradicts Euler’s formula, for, as mentioned

to v.

(v) L 0.

above, CzEV(G) charge 2 4. I

4. TRIANGULATIONS

Appel and Haken [l] showed the 0-diagonal chromatic number of plane triangulations is four. The 1-diagonal chromatic number of plane triangulations is either nine or ten (see [12]). This section gives an upper bound for the d-diagonal number for d 2 2.

Let j ( d ) := 2+ 19.2d-2. For this section, a graph is minimal if it is a plane or projective plane triangulation with d-diagonal chromatic number greater than j ( d ) , for d 2 2 . Let a configuration be reducible if it appears in no minimal graph.

Lemma 9.

Proof. Let G be a minimal graph with vertex z of degree at most four. Let H be G - z, with an edge added to triangulate if deg(z) = 4, or a multiple edge deleted if deg(z) = 2 , or a loop and a multiple edge deleted if deg(z) = 1. From minimality, H has a d-diagonal

A vertex of degree at most four is reducible.


j(d)-coloring. This yields a partial d-diagonal coloring of G. Now z can be colored to I

Let a j , k-edge be an edge incident with a j-vertex and a k-vertex. Let an i, j , k-face be a triangle incident with an i-vertex, a j-vertex, and a k-vertex. Given a graph G and an edge or triangle a of G, let G . (Y be the graph with cy contracted to a vertex.

give a d-diagonal j(d)-coloring of G.

Lemma 10. A 5, 5-edge is reducible.

Proof. Let G be a minimal graph with 5, 5-edge zy. From minimality, G . zy has a d-diagonal j(d)-coloring. Let 2 be the vertex representing the edge zy in G.zy. Removing the color from z yields a partial d-diagonal coloring of G, since d 2 2. But now z, y can each be colored, since there are less than j ( d ) vertices d-diagonally adjacent to each. This

I gives a d-diagonal j(d)-coloring of G.

Lemma 11. A 5, 6, 6-face is reducible.

Proof. Let G be a minimal graph with 5, 6, 6-face syz (with z a 5-vertex). From minimality, G . zyz has a d-diagonal j(d)-coloring. Let w be the vertex representing the triangle zyz in G . zyz. Removing the color from w yields a partial d-diagonal coloring of G. Color z the color that w received. The vertices y, z may be colored in that order such that each is d-diagonally adjacent to at most j ( d ) - 1 colored vertices when it is colored.

I

A different equality applies to triangulations than what applied to quadrangulations. The following equality follows from Euler’s formula for plane triangulations (the equality with the 12 replaced by a 6 is true for projected plane triangulations):

This yields a d-diagonal j(d)-coloring of G.

(6 - deg(z)) = 12. x E V ( G )

For this section, let the initial charge of a vertex z be 6 - degfz). Let a graph be discharged, if its vertices first receive the initial charges, and then each 5-vertex sends a charge of 5 to each of its neighbors of degree at least seven. This simple discharge rule will yield the result below.

Theorem 2. is at most 2 + 19. 2d-2.

The d-diagonal chromatic number of plane or projective plane triangulations

Proof. Suppose that there is a plane or projective plane triangulation that requires j ( d ) + 1 colors in a d-diagonal coloring. Let G be a minimal graph. By Lemma 9, the minimum degree of G is five. Let G be discharged.

Let z be a 5-vertex of G (with initial charge 1). By Lemma 10, z has no neighbors of degree five. By Lemma 11,z has no two consecutive neighbors of degree six. Thus z has at least three neighbors of degree at least seven, and sends out at least 1.

Note that a 6-vertex has initial charge 0 and receives no charge. Let y be a vertex of G of degree k 2 7. By Lemma 10, y has no two consecutive

neighbors of degree five. Thus y has at most Lk/2J neighbors of degree five, and y receives at most [k/2J/3 from its neighbors. Thus charge (y) 5 6 - k + [k/2]/3 5 0.


Thus no vertex has positive charge. This contradicts Euler's formula, since CzEV(G) charge (z) 2 6. I

Note that the authors can replace the constant 19 in Theorem 2 by 73/4, but the proof is not as nice.

5. LOWER BOUNDS

As noted in the introduction, in [9] Hornak and Jendrol prove an upper bound for the d-diagonal chromatic number of plane graphs (when A 2 8). They failed to give a lower bound. This section mentions a simple lower bound for the general case, and a construction yielding better results for low d and A.

Theorem 3. For even d, the d-diagonal chromatic number of 3-connected plane graphs is at least A(A - l)d/2.

Let integer A and even integer d be given. Let Go be a cycle on A vertices, embedded in the plane. For each k 2 1, let Gk be the graph obtained from Gk- by adding, for each edge ay incident to the infinite face of G k - 1 , a path v1 . . . va-2 embedded in the infinite face together with edges zvl and yvn-2. For each k 2 0, let H k be the graph obtained from Gk by adding a vertex z in the infinite face, and adding edges from z to every vertex incident with the infinite face. It can be checked that each Hk is 3-connected, and has 1 + A(A - l ) k vertices. Also note that, given two vertices v, w of f f k such that neither is z, then u and w are (2k)-diagonally adjacent. Setting k := d/2 gives the result.

Proof.

Theorem 4. at least 1 + 3(A - 2)(A - l)(d-1)/2.

For odd d, the d-diagonal chromatic number of 3-connected plane graphs is

Proof. Let integer A and odd integer d be given. Let Go be the graph with vertex set {'Q, . . . ,V3~-6} and edge Set (viv2,. . . , v3A-703A-6, ~ ~ A - 6 ~ 1 , ~ 0 ~ A - 2 , ~ 0 ~ 2 A - 4 , ~ 0 ~ 3 A - 6 } ,

FIGURE 9. A graph requiring eleven colors.


embedded in the plane, with wo not incident with the infinite face. Let Gk and Hk be constructed as in the proof of Theorem 2. Again, Hk is 3-connected. This time Hk has 2 + 3(A - 2)(A - l)k vertices. Here, given any two vertices w, w of Hk, neither equal to z , then u and w, are (2k + 1)-diagonally adjacent. Setting k := (d - 1)/2 gives the result.

The following construction improves the lower bound for d 5 3 and A = 4.

Theorem 5. quadrangulations is at least [( [ g d + 31 (d + 3))/2] + 1.

For 1 5 d 5 3, the d-diagonal chromatic number of 3-connected plane

Proof. Let integer d satisfying 1 5 d 5 3 be given. Let G d be the graph with vertex set {w,,~ : z E (1,. . . , [(5/3)d+3]},y E (1,. . . ,d+3}}U{a,b} and edge set { ~ ~ , ~ v ~ , ~ + l : z E {1 , . . . , [ (5 /3)d+31},~ E {l,. . .d+2)}U{w,,yu,-i, ,+i : z E (2, ...,I( 5 / 3 ) d + 3 1 ) , ~ E {1,...,d+2))U{vi,ywr(5/3)d+31,y+l : Y E {1,.. . ,d+2))U{avz,i : z E { I , . . . , [(5/3)d+ 31)) U {WZ,d+3b : z E (1,. . . , [(5/3)d + 31)). For an example, G1 is shown in Figure 9, where a vertex (a or b) is meant to be at infinity. Note that G d has ([(5/3)d+3])(d+3)+2 vertices, no three of which are pairwise d-diagonally nonadjacent. The result follows.

A corollary of Theorems 1 and 5 is that the 1-diagonal chromatic number of plane quadrangulations is between 11 and 19.

ACKNOWLEDGMENTS

The research of DPS was supported by the Office of Naval Research, Grant Number N00014-92-J- 1965.

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Received April 10, 1995

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