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Coloring of the dth power of the face-centered cubic grid
Nicolas Gastineau, Olivier Togni
To cite this version:
Nicolas Gastineau, Olivier Togni. Coloring of the dth power of the face-centered cubic grid. Dis- cussiones Mathematicae Graph Theory, University of Zielona Góra, In press, �10.7151/dmgt.2257�.
�hal-01819684�
Coloring of the d
thpower of the face-centered cubic grid
Nicolas Gastineau1 and Olivier Togni1
1LE2I FRE2005, CNRS, Arts et Métiers, Université Bourgogne Franche-Comté, F-21000 Dijon, France
June 20, 2018
Abstract
The face-centered cubic grid is a three dimensional 12-regular infinite grid. This graph represents an optimal way to pack spheres in the three-dimensional space. In this grid, the vertices represent the spheres and the edges represent the contact between spheres. We give lower and upper bounds on the chromatic number of thedthpower of the face-centered cubic grid. In particular, in the case d= 2we prove that the chromatic number of this grid is 13.
We also determine sharper bounds for d= 3and for subgraphs of of the face-centered cubic grid.
1 Introduction
For a graphG, we denote byV(G)thevertex set ofGand byE(G)⊆V(G)×V(G)theedge set ofG. We denote bydG(u, v), the usual distance between the verticesuand vin a graphG. The diameter of a graphG, denoted by diam(G)is the maximum distance between the verticesuand v in G. By G[A], we denote the graph induced by the set of vertices A⊆V(G), i.e., the graph with vertex setAand edge set{uv∈E(G)| u∈A, v∈A}.
Ak-coloring of a graphGis a mapcfromV(G)to{0,1, . . . , k−1}which satisfiesc(u)6=c(v) for everyuv∈E(G). The chromatic number χ(G)ofG is the smallest integerk such that there exists ak-coloring ofG. Thedth powerGd of a graphGis the graph obtained fromGby adding an edge between every two vertices satisfyingdG(u, v)≤d.
We recall that the survey from Kramer and Kramer [6] regroup together a good quantity of known results about the coloring of thedthpower of graphs. Note that a coloring of thedth power of graph is also called ad-distance coloring.
Our goal in this paper is to determine the chromatic number of powers of a grid called the face-centered cubic grid. The results of this paper are summarized in Table 1. These results are in the continuity of previous works [4, 5, 8] about the coloring of the dth power of the square, triangular and hexagonal grids.
The face-centered cubic grid, denoted by F, is the graph with vertex set{(i, j, k)| i∈Z, j ∈ Z, k∈Z2} ∪ {(i+ 0.5, j+ 0.5, k)|i∈Z, j∈Z, k ∈Z1}, whereZ2 is the set of the even integers andZ1=Z\Z2(the set of odd integers), and edge set{(i, j, k)(i′, j′, k′)|(|i−i′|= 1∧j=j′∧k= k′)∨(i=i′∧ |j−j′|= 1∧k=k′)∨(|i−i′|= 1/2∧ |j−j′|= 1/2∧ |k−k′|= 1)}. A subgraph of this grid is illustrated in Figure 1, showing the neighborhood of a vertex.
The layer kof the grid F is the subset of vertices{(i, j, k)| i∈Z, j ∈Z}, ifk is even or the subset of vertices{(i+ 0.5, j+ 0.5, k)|i∈Z, j∈Z}, ifk is odd. Note that the graph induced by the vertices of layerkis isomorphic to a square grid, see Figure 1.
We denote by Fk, the subgraph induced by the vertices of layerk. The graphFk
1,k2 is the subgraph ofF induced by the vertices from the layeri, fork1≤i≤k2. It can be easily remarked thatF0,k
2−k1 is isomorphic toFk
1,k2 for every two integersk1 andk2.
k 1 2 3 4 5 odd even
Lower bound onχ(Fd) 4 13 29 55 92 125d3+54d2+1912d+34 125d3+54d2+116d+ 1 Upper bound onχ(Fd) 4 13 30 65 108 (d+ 1)⌈(d+ 1)2/2⌉
Value ofχ(Fd
0,1) 4 9 16 25 36 (d+ 1)2
Lower bound onχ(F0,2d ) 4 13 24 34 52 3⌈(d+ 1)2/2⌉ −2 Upper bound onχ(Fd
0,2) 4 13 24 36 54 3⌈(d+ 1)2/2⌉
Table 1: Our results about the chromatic number of thedth power of the face-centered cubic grid.
Figure 1: Neighborhood in the face-centered cubic gridF
The face-centered cubic grid is a grid studied in the field of programmable matter [2, 3, 7].
Programmable matter can be seen as modular robots (called modules or particles) able to fix to adjacent modules and send (receive) messages to (from) other modules fixed to the entity. Thus, the different modules form a geometric shape which is a subgraph of a grid. This grid, in the three dimensional case, is almost every time the face-centered cubic grid [7]. Note that the face-centered cubic grid is also called the cannonball grid [9].
It is easily seen that χ(F) = 4 since F contains K4 as subgraph and a 4-coloring can be obtained by using colors 1 and 2 on even layers and colors 3 and 4 on odd layers. Note that multicolorings ofF were studied by Šparl et al. [9].
This paper is organized as follows. In Section 2, we determine a formulae for the distance between vertices in the face-centered cubic grid. This formulae is used through the remaining sections of this paper. In Section 3, we determine general lower and upper bounds onχ(Fd). The lower bound corresponds to the size of a largest complete graph inFd. In Section 4, we determine the exact value ofχ(F3)and prove that29≤χ(F4)≤30. Section 5 is dedicated to the study of χ(Fd
0,1)andχ(Fd
0,2).
2 Distance between two vertices in the graph F
Letp+ be the function such thatp+(k) =kifk≥0 andp+(k) = 0otherwise.
We begin by presenting a formulae that gives the distance between any two vertices of F. Note that this formulae restricted to the vertices of a same layer ofF corresponds to the distance in the square grid. This proposition will be used through the remaining sections of this paper to calculate the distances inF.
Proposition 2.1. The distance between two vertices (i, j, k) and (i′, j′, k′)of F is given by the following formulae :
dF((i, j, k),(i′, j′, k′)) =p+
|i−i′| −|k−k′| 2
+p+
|j−j′| −|k−k′| 2
+|k−k′|.
Proof. Note that the formulae can reformulated as follows:
dF((i, j, k),(i′, j′, k′)) =
|i−i′|+|j−j′| if|i−i′|>|k−k′|/2∧ |j−j′|>|k−k′|/2;
|i−i′|+|k−k′|/2 if|i−i′|>|k−k′|/2∧ |j−j′| ≤ |k−k′|/2;
|j−j′|+|k−k′|/2 if|i−i′| ≤ |k−k′|/2∧ |j−j′|>|k−k′|/2;
|k−k′| if|i−i′| ≤ |k−k′|/2∧ |j−j′| ≤ |k−k′|/2.
In this proof, our goal is to show thatdF((i, j, k),(i′, j′, k′)) =d, if and only if p+(|i−i′| −
|k−k′|/2) +p+(|j−j′| − |k−k′|/2) +|k−k′| =d. We proceed by induction on d. For d= 0, we easily obtain that dF((i, j, k),(i′, j′, k′)) = 0 if and only ifp+(|i−i′| − |k−k′|/2) +p+(|j− j′| − |k−k′|/2) +|k−k′|= 0. Now suppose that the vertices (i′, j′, k′)at distanced′ of(i, j, k) satisfyp+(|i−i′| − |k−k′|/2) +p+(|j−j′| − |k−k′|/2) +|k−k′′|=d′, ford′ ≤d. Without loss of generality, supposei′ = 0, j′ = 0andk′ = 0. Also, without loss of generality, we can suppose thati≥0,j ≥0andk≥0(by symmetry the other cases are proven in the same way). Thus, the following is true:
dF((0,0,0),(i, j, k))≤d⇔
i+j≤d ifi > k/2∧j > k/2;
i+k/2≤d ifi > k/2∧j ≤k/2;
j+k/2≤d ifi≤k/2∧j > k/2;
k≤d ifi≤k/2∧j ≤k/2.
Now, we are going to prove thatdF((0,0,0),(i, j, k)) =d+ 1 if and only ifp+(i−k/2) +p+(j− k/2) +k=d+ 1. We cut the proof in four cases: first we consider the vertices such thati > k/2 andj > k/2, second we consider the vertices such thati > k/2∧j ≤k/2, third we consider the vertices such that i ≤k/2∧j > k/2 and finally we consider the vertices which satisfyi ≤k/2 j≤k/2. However, before doing this case analysis proof, we claim the following:
Claim 1. For every integersi,j,kandℓ >0, the following hold:
i) i+j≥ℓ, impliesdF((0,0,0),(i, j, k))≥ℓ;
ii) i+k/2≥ℓimpliesdF((0,0,0),(i, j, k)≥ℓ.
Note that Claim 1.i) is obtained by observing that every two adjacent vertices (i1, j1, k1) and (i2, j2, k2) satisfy |i1−i2|+|j1−j2| ≤ 1. Note that every path beginning by (0,0,0) going to (i, j, k)should contain at leastkedges between vertices of different layers. Also, every two adjacent vertices(i1, j1, k1)and(i2, j2, k2), with|k1−k2|= 1, are such that|i1−i2|= 1/2. Consequently, i+k/2≥ℓimpliesdF((0,0,0),(i, j, k)≥i−k/2 +k≥ℓand Claim 1.ii) follows.
Case 1: i > k/2 and j > k/2. First, ifi+j≤d, then, by induction hypothesis, we havedF((0,0,0),(i, j, k))≤ d. Moreover, by Claim 1.i), we obtain thati+j≥d+ 2impliesdF((0,0,0),(i, j, k))≥d+ 2.
Consequently,dF((0,0,0),(i, j, k)) =d+ 1 impliesi+j=d+ 1.
Second, suppose that i+j=d+ 1. Note that the vertex(k/2, k/2, k)is at distancekfrom (0,0,0). By using the distance in a square grid (the vertices of layer k induce a square grid in F), the vertex (k/2, k/2, k)is at distance at most i−k/2 +j−k/2 = d+ 1−k from (i, j, k). Thus,dF((0,0,0),(i, j, k))≤d+ 1. Moreover, by Claim 1.i), we obtain that i+j≥d+ 1impliesdF((0,0,0),(i, j, k))≥d+ 1. Finally, we have thati+j=d+ 1implies dF((0,0,0),(i, j, k)) =d+ 1.
Case 2 : i > k/2 and j≤k/2. First, ifi+k/2≤d, then, by induction hypothesisdF((0,0,0),(i, j, k))≤ d. Moreover, by Claim 1.ii), we obtain that i+k/2≥d+ 2impliesdF((0,0,0),(i, j, k))≥ d+ 2. Consequently,dF((0,0,0),(i, j, k)) =d+ 1impliesi+k/2 =d+ 1.
Second, suppose thati+k/2 =d+ 1. We can easily notice thatdF((0,0,0),(k/2, j, k))≤k.
By using the distance in a square grid (the vertices of layerkinduce a square grid inF), we havedF((k/2, j, k),(i, j, k)≤i−k/2. Thus, dF((0,0,0),(i, j, k))≤k+i−k/2 =i+k/2.
Moreover, by Claim 1.ii), we have thati+k/2≥d+ 1impliesdF((0,0,0),(i, j, k))≥d+ 1.
Thus, we obtain thatdF((0,0,0),(i, j, k)) =d+ 1.
Case 3 : i≤k/2 and j > k/2. The proof is the same than in Case 2 by interchanging the role ofiand j.
Case 4 : i≤k/2 and j≤k/2. First, ifk≤d, then, by induction hypothesis,dF((0,0,0)(i, j, k))≤ d. Moreover, it can be easily noticed that k > d+ 1 implies dF((0,0,0)(i, j, k))> d+ 1.
Consequently, we obtain that dF((0,0,0),(i, j, k)) =d+ 1 impliesk=d+ 1.
Second, suppose that k=d+ 1. Note that the vertices(i, j, d+ 1) are at distance at most d+1from(0,0,0). Consequently, we have thatk=d+1impliesdF((0,0,0),(i, j, k)) =d+1.
3 Chromatic number of Fd
In this section, we determine general lower and upper bounds onχ(Fd). Note that there is a gap between the lower and upper bounds. For the majority of the classical grids (square and triangular grids), the value of the chromatic number ofGdcorresponds to the size of the largest clique inGd. However, that it is not the case for the face-centered cubic grid (see Section 4) and that explains this gap between our lower and upper bounds.
3.1 Lower bound on the chromatic number of Fd
The following Proposition is a consequence of a result of Bjornholm [1] which has proven that the set Aℓ ={(i, j, k)∈V(F)| dF((0,0,0)(i, j, k))≤ℓ}is such that |Aℓ|= (2ℓ+ 1)(5ℓ2+ 5ℓ+ 3)/3 (On-line encyclopedia of integer sequences number: A005902). Since diam(G[Aℓ]) = 2ℓ and by settingd=ℓ/2, we obtain the following result.
Proposition 3.1 ([1]). If dis even, thenχ(Fd)≥125d3+54d2+116d+ 1.
LetD0={(0,0,0),(1,0,0),(0.5,0.5,1),(0.5,−0.5,1)}. Note thatD0induces a complete graph of four vertices in the gridF. LetDℓ={u∈F| minv∈D0(dF(u, v)) =ℓ}.
In the two following Lemmas, we calculate the size of a largest clique in Fd when dis odd.
From this value, we will infer a lower bound onχ(Fd)(as in Proposition 3.1) for an oddd.
Lemma 3.2. For any ℓ≥1,|Dℓ|= 10ℓ2+ 10ℓ+ 4.
Proof. Note that we have|D0|= 4and thatD1={(−0.5,0.5,−1),(−0.5,−0.5,−1),(0.5,0.5,−1), (0.5,−0.5,−1),(1.5,0.5,−1),(1.5,−0.5,−1)} ∪ {(0,1,0), (0,−1,0),(1,1,0),(1,−1,0),(−1,0,0), (2,0,0)} ∪ {(−0.5,0.5,1), (1.5,0.5,1), (−0.5,−0.5,1), (1.5,−0.5,1), (0.5,1.5,1), (0.5,−1.5,1)} ∪ {(0,−1,2),(1,−1,2),(0,0,2),(1,0,2), (0,1,2), (1,1,2)}. By induction on ℓ, we will prove that there are(ℓ+ 1)(ℓ+ 2) vertices in layers−ℓandℓ+ 1in Dℓ and that there are2 + 4ℓvertices in layerj in Dℓ, for−ℓ < j < ℓ+ 1.
Ifℓ= 1, we can easily remark that there are6 vertices of D1 in each layeri,i∈ {−1,0,1,2}.
Thus,|D1|= 24. Now, suppose by induction that there areℓ(ℓ+ 1)vertices in layers−(ℓ−1)and ℓin Dℓ−1 and that there are2 + 4(ℓ−1)vertices in the layeriinDℓ−1, for−(ℓ−1)< i < ℓ.
It can be easily verified that there are(ℓ+ 2)(ℓ+ 1)vertices in layers−ℓandℓ+ 1inDℓ. Now, since we have supposed that there are2 + 4(ℓ−1)vertices in layeriinDℓ−1, for−(ℓ−1)< i < ℓ, we obtain that there are2 + 4ℓvertices in layeriinDℓ, for−(ℓ−1)< i < ℓ(since there are four more vertices in each layer inDℓ). Also, it can be easily verified that there are2 + 4ℓvertices in layers−(ℓ−1)andℓ inDℓ and the property follows.
By calculation, we obtain:
|Dℓ|= 2(ℓ+ 1)(ℓ+ 2) + 2ℓ(2 + 4ℓ) = 2ℓ2+ 6ℓ+ 4 + 8ℓ2+ 4ℓ= 10ℓ2+ 10ℓ+ 4.
For an integerℓ≥0, letBℓ=∪0≤i≤ℓDi.
Lemma 3.3. For any integerℓ≥0, diam(F[Bℓ]) = 2ℓ+ 1.
Proof. Note thatF[B0]is an induced complete graph of four vertices and thus diam(F[B0]) = 1.
By construction, we have diam(F[Bℓ+1]) = diam(F[Bℓ]) + 2. Thus, by induction, we have diam(F[Bℓ]) = 2ℓ+ 1.
As in Proposition 3.1, we infer the following lower bound onχ(Fd), for an oddd.
Proposition 3.4. If dis odd, then
χ(Fd)≥ 5 12d3+5
4d2+19 12d+3
4.
Proof. Since, by Lemma 3.3, diam(F[Bℓ]) = 2ℓ+ 1, we have χ(F2ℓ+1)≥ |Bℓ|. By Lemma 3.2,
|Dℓ|= 10ℓ2+ 10ℓ+ 4and by calculation:
|Bℓ|=Pℓ
i=0|Di|=Pℓ
i=0(10i2+ 10i+ 4) = 10Pℓ
i=0i2+ 10Pℓ
i=0i+ 4ℓ
= 10(16ℓ(2ℓ2+ 3ℓ+ 1)) + 10(12ℓ(ℓ+ 1)) + 4(ℓ+ 1) =103ℓ3+ 10ℓ2+323ℓ+ 4.
Also, by calculation:
|B(d−1)/2|= 5 12d3+5
4d2+19 12d+3
4. Thus, ifdis odd, then we obtain that
χ(Fd)≥ |B(d−1)/2|= 5 12d3+5
4d2+19 12d+3
4.
0 1 2 12
13
11 12 10
2 3 1
9
4
Fk Fk−1
Fk−2 Fk+1 Fk+2
8 9 7 6
10 11 9
8 6 5
12 11 4 5
3 2 3 1
7 6 5
4 5 3 2
6 7 5 4
2 1
8 7
9 10 8
7 8 6
11 12 10
Figure 2: The colors of the vertices at distance at most 2 of a fixed vertexuof color0in the layer kof the graphF (square: u; circle: vertex at distancedof u,1≤d≤2).
3.2 Upper bound on the chromatic number of Fd
We determine the following upper bound on the chromatic number ofFd. Note that this upper bound is reached ford= 1 and is not reached ford= 2 ord= 3(see Section 4).
Theorem 3.5. For any d≥1,χ(Fd)≤(d+ 1)⌈(d+ 1)2/2⌉.
Proof. By Proposition 2.1, the distance between two vertices(i, j, k)and(i′, j′, k)from the same layer is|i−i|+|j−j′|. This distance corresponds to the distance between two vertices(i, j)and (i′, j′)in the square grid. Moreover, the distance between two vertices(i, j, k)and(i′, j′, k′)is at leastd+ 1if|k−k′| ≥d+ 1.
Therefore, by using (d+ 1)different patterns of ⌈(d+ 1)2/2⌉colors used for coloring the dth power of the square grid [4] it is possible to colorFd. We use theath pattern to color the layerb, ifb≡a (mod d+ 1). Since in each pattern we use ⌈(d+ 1)2/2⌉different colors, in total we have used(d+ 1)⌈(d+ 1)2/2⌉colors.
4 Chromatic number of the 2nd and the 3th power of F
4.1 Chromatic number of the 2nd power of F
Note that the value of χ(F2) corresponds to the lower bound of Proposition 3.1 which itself corresponds to the size of the largest clique inF2.
Theorem 4.1. χ(F2) = 13.
Proof. By Proposition 3.1, we haveχ(F2)≥40/12 + 5 + 11/3 + 1 = 13. Now, it remains to prove thatχ(F2)≤13. We set the following coloring function:
∀(i, j, k)∈V(F), c((i, j, k)) =i−2j+9k
2 (mod 13).
We claim thatc is a coloring function of the2nd power of F. Letube a vertex inF such that c(u) = 0. In Figure 2, we represent the graph induced by the vertices at distance at most2from u. It is easy to verify that no vertex of this subgraph exceptuhas color0.
In order to prove that every two vertices(i, j, k)and(i′, j′, k′)satisfyingc((i, j, k)) =c((i′, j′, k′)) are such thatdF((i, j, k),(i′, j′, k′))>2, we consider three cases: first the casek′=k, second the case|k−k′|= 1and finally the case|k−k′|= 2.
Case 1: k′ =k. The vertices(i′, j′, k)at distance at most 2 from(i, j, k)are such that|i−i′|+|j− j′| ≤2. First, if1≤ |i−i′| ≤2andj=j′, then we have(i−i′)6≡0 (mod 13). Consequently, sincec((i, j, k))−c((i′, j′, k)≡(i−i′) (mod 13), we havec((i, j, k))6=c((i′, j′, k)). Second, if 1 ≤ |j−j′| ≤2 and i =i′, then we have 2(j′−j)6≡ 0 (mod 13). Consequently, since c((i, j, k))−c((i′, j′, k)≡2(j′−j) (mod 13), we havec((i, j, k)) 6=c((i′, j′, k)). Finally, if
|i−i′|= 1and|j−j′|= 1, then we have(i−i′) + 2(j′−j)6≡0 (mod 13). Consequently, sincec((i, j, k))−c((i′, j′, k)≡(i−i′)+ 2(j′−j) (mod 13), we havec((i, j, k))6=c((i′, j′, k)).
Thus, no vertex at distance at most 2 from (i, j, k) has the same color than (i, j, k) in the layerk.
Case 2: |k′−k|= 1. The vertices(i′, j′, k+ 1)((i′, j′, k−1), respectively) at distance at most 2 from(i, j, k)are such that|i−i′|= 1/2∧|j−j′| ≤3/2or such that|i−i′| ≤3/2∧|j−j′|= 1/2.
Therefore, we have(i−i′) + 2(j′−j)6≡9/2 (mod 13)((i−i′) + 2(j′−j)6≡ −9/2 (mod 13), respectively). Thus, since c((i, j, k))−c((i′, j′, k+ 1)≡(i−i′) + 2(j′−j)−9/2 (mod 13) ( c((i, j, k))−c((i′, j′, k−1) ≡ (i−i′) + 2(j′−j) + 9/2 (mod 13), respectively), we have c((i, j, k))6=c((i′, j′, k+ 1))(c((i, j, k))6=c((i′, j′, k−1)), respectively). Therefore, no vertex at distance at most 2 from(i, j, k)has the same color than(i, j, k)in the layerk+ 1(k−1, respectively).
Case 3: |k′−k|= 2. The vertices (i, j, k+ 2) ((i′, j′, k−2), respectively) at distance at most 2 from(i, j, k)are such that|i−i′| ≤1∧ |j−j′| ≤1. Therefore, we have(i−i′) + 2(j′−j)6≡9 (mod 13)((i−i′)+2(j′−j)6≡4 (mod 13), respectively). Thus, sincec((i, j, k))−c((i′, j′, k+ 2)≡(i−i′) + 2(j′−j)−9 (mod 13)(c((i, j, k))−c((i′, j′, k−2)≡(i−i′) + 2(j′−j) + 9 (mod 13), respectively), we havec((i, j, k))6=c((i′, j′, k+ 2)) (c((i, j, k))6=c((i′, j′, k−2)), respectively). Therefore, no vertex at distance at most 2 from (i, j, k) has the same color than(i, j, k)in the layerk+ 2(k−2, respectively).
Finally, since every vertex of layerd, ford≤k−3ord≥k+ 3is at distance at least 3 of(i, j, k), we obtain that no vertex at distance at most2 from(i, j, k)has the same color than (i, j, k).
4.2 Chromatic number of the 3th power of F
Note that the value of χ(F3) is smaller than the upper bound of Proposition 3.5 (which is 32, ford= 3). In the following theorem, we denote by i(modj) or byi (modj) the smallest positive integerasuchi≡a (mod 30)
Theorem 4.2. χ(F3)≤30.
Proof. We set the following function: ∀(i, j, k)∈V(F), c((i, j, k)) =
i(mod5)+ 5j+ 15k/2 (mod 30) ifk≡0 (mod 2);
(i−1/2)(mod5)+ 5(j−1/2) + 8 + 15(k−1)/2 (mod 30) otherwise.
We claim thatcis a coloring function of the3thpower ofF. It remains to prove that every two vertices(i, j, k)and(i′, j′, k′)satisfyingc((i, j, k)) =c((i′, j′, k′))are such thatdF((i, j, k),(i′, j′, k′))>
3. This can be proven by a (tedious) case analysis similar with the one of the proof of Theorem 4.1. Instead, we only give an illustrative argument.
Letu be a vertex in F. Without loss of generality, suppose u has color 0. In Figure 3, we represent the graph induced by the vertices at distance at most3fromu. It is easy to verify that no vertex of this subgraph exceptuhas color 0.
For verticesu1,u2,u3andu4ofF, we denote byT(u1, u2, u3, u4), the vertex set{u∈F|uv∈ E(F), v∈ {u1, u2, u3, u4}}. Note that in the case{u1, u2, u3, u4}induces a complete graph inF, T(u1, u2, u3, u4)induces inF a graph isomorphic to the graphB1from Lemma 3.3. Consequently, in this case, diam(T(u1, u2, u3, u4)) = 3.
Lemma 4.3. If there exists a 28-coloring of F3, then the following is true about every vertex (i, j, k):
(i) one vertex among(i+ 3, j, k+ 2),(i+ 2.5, j+ 0.5, k+ 3)and(i+ 2.5, j−0.5, k+ 3)has the same color than (i, j, k);
