On Keller's conjecture in dimension seven

A cube tiling of $\mathbb{R}^d$ is a family of pairwise disjoint cubes $[0,1)^d+T=\{[0,1)^d+t:t\in T\}$ such that $\bigcup_{t\in T}([0,1)^d+t)=\mathbb{R}^d$. Two cubes $[0,1)^d+t$, $[0,1)^d+s$ are called a twin pair if $|t_j-s_j|=1$ for some $j\in [d]=\{1,\ldots, d\}$ and $t_i=s_i$ for every $i\in [d]\setminus \{j\}$. In $1930$, Keller conjectured that in every cube tiling of $\mathbb{R}^d$ there is a twin pair. Keller's conjecture is true for dimensions $d\leq 6$ and false for all dimensions $d\geq 8$. For $d=7$ the conjecture is still open. Let $x\in \mathbb{R}^d$, $i\in [d]$, and let $L(T,x,i)$ be the set of all $i$th coordinates $t_i$ of vectors $t\in T$ such that $([0,1)^d+t)\cap ([0,1]^d+x)\neq \emptyset$ and $t_i\leq x_i$. It is known that if $|L(T,x,i)|\leq 2$ for some $x\in \mathbb{R}^7$ and every $i\in [7]$ or $|L(T,x,i)|\geq 6$ for some $x\in \mathbb{R}^7$ and $i\in [7]$, then Keller's conjecture is true for $d=7$. In the present paper we show that it is also true for $d=7$ if $|L(T,x,i)|=5$ for some $x\in \mathbb{R}^7$ and $i\in [7]$. Thus, if there is a counterexample to Keller's conjecture in dimension seven, then $|L(T,x,i)|\in \{3,4\}$ for some $x\in \mathbb{R}^7$ and $i\in [7]$.


Introduction
This paper is a continuation of the paper [10] and therefore the following introduction is limited to essential information. For a more detailed historical sketch on Keller's cube tiling conjecture and related problems on cube tilings we refer the reader to [10].
A cube tiling of R d is a family of pairwise disjoint cubes [0, 1) d + T = {[0, 1) d + t : t ∈ T } such that t∈T ([0, 1) d + t) = R d . Two cubes [0, 1) d + t, [0, 1) d + s are called a twin pair if |t j − s j | = 1 for some j ∈ [d] = {1, . . . , d} and t i = s i for every i ∈ [d] \ {j}. In 1907, Minkowski [16] conjectured that in every lattice cube tiling of R d , i.e. when T is a lattice in R d , there is a twin pair, and in 1930, Keller [8] generalized this conjecture to arbitrary cube tiling of R d . Minkowski's conjecture was confirmed by Hajós [7] in 1941. In 1940, Perron [17] proved that Keller's conjecture is true for all dimensions d ≤ 6. In 1992, Lagarias and Shor [13], using ideas from Corrádi's and Szabó's papers [18], [3], constructed a cube tiling of R 10 which does not contain a twin pair and thereby refuted Keller's cube tiling conjecture. In 2002, Mackey [15] gave a counterexample to Keller's conjecture in dimension eight, which also shows that this conjecture is false in dimension nine. In 2010, Debroni et al. [4] computed that Keller's conjecture is true for all cube tilings [0, 1) 7 + T of R 7 with T ⊂ (1/2)Z 7 or equivalently, T ⊂ a + Z 7 ∪ b + Z 7 , where fixed a, b ∈ [0, 1) 7 are such that a i = b i for every i ∈ [7]. Observe now that the condition |L(T, x, i)| ≤ 2 for some x ∈ R 7 and every i ∈ [7] means that T 1 ⊂ a + Z 7 ∪ b + Z 7 , where T 1 ⊂ T consists of all t for which ([0, 1) 7 + t) ∩ ([0, 1] 7 + x) = ∅. Thus, it is easy to show that the result of Debroni et al. proves also that the conjecture is true for cube tilings of R 7 with |L(T, x, i)| ≤ 2 for some x ∈ R 7 and every i ∈ [7]. Indeed, if there is no twin pairs in the set {[0, 1) 7 + t : t ∈ T 1 }, then extending this family to the two-periodic tiling [0, 1) 7 + T of R 7 , where T = T 1 + 2Z 7 , we obtain a cube tiling with T ⊂ a + Z 7 ∪ b + Z 7 without twin pairs, which contradicts the result of Debroni et al.
It follows from the above results that if there is a counterexample to Keller's conjecture in dimension seven, then |L(T, x, i)| ∈ {3, 4} for some x ∈ R 7 and i ∈ [7].
The proof of the crucial result (Theorem 5.2) that allows us to prove the above assertion on Keller's conjecture is based on computations, and these need reductions. The two longest and most arduous sections of the paper, Section 3 and 4, contain the preparation results for adequate reductions. So, the reader who wants to have a first overview of how discussed case of Keller's conjecture is proven may leave these sections and continue reading from Theorem 5. 2.
The present paper is organized as follows. In Section 2 we give basic notions concerning the systems of boxes and abstract words (this section is almost the same as Section 2 in [10]). These issues were developed in [6,11]. Since they are not widely known, we present them in detail. In Section 3 we give results on the structure of systems of boxes. In the next section we examine some special system of 12 boxes (written down as a system of abstract words). In Section 5, based on the result from the previous two sections, we first establish initial configurations for computations, and next we prove the theorem on the structure of the above mentioned system of 12 boxes (Theorem 5.2). In the final Section 6 using Theorem 5.2 we prove that Keller's cube tiling conjecture is true for tilings [0, 1) 7 + T of R 7 with |L(T (x, i))| = 5 for some x ∈ R 7 and i ∈ [7].

Basic notions
In this section we present the basic notions on dichotomous boxes and words (details can be found in [6,11]). We start with systems of boxes.
A non-empty set K ⊆ X = X 1 × · · · × X d is called a box if K = K 1 × · · · × K d and K i ⊆ X i for each i ∈ [d]. By Box(X) we denote the set of all boxes in X. The set X will be called a d-box. The box K is said to be proper if K i = X i for each i ∈ [d]. Two boxes K and G in X are called dichotomous if there is i ∈ [d] such that K i = X i \G i . A suit is any collection of pairwise dichotomous boxes. A suit is proper if it consists of proper boxes. A non-empty set F ⊆ X is said to be a polybox if there is a suit F for F , i.e. if F = F . A polybox F is rigid if it has exactly one suit. (Figure 2 presents the suit for a rigid polybox. The polyboxes F 3,A and F 3,A ′ in Figure 3 are not rigid). The important property of proper suits is that, for every proper suits F and G for a polybox F , we have |F | = |G |. Thus, we can define a box number |F | 0 = the number of boxes in any proper suit for F (In Figure 3 we have | F 3,A | 0 = 3). A proper suit for a d-box X is called a minimal partition of X (Figure 3). Every minimal partition of a d-box has 2 d boxes.
Two boxes K, G ⊂ X are said to be a twin pair if K j = X j \ G j for some j ∈ [d] and K i = G i for every i ∈ [d] \ {j}. Observe that, the suit for a rigid polybox can not contain a twin pair.
Every two cubes [0, 1) d + t and [0, 1) d + p in an arbitrary cube tiling [0, 1) d + T of R d satisfy Keller's condition ( [8]): there is i ∈ [d] such that t i − p i ∈ Z \ {0}, where t i and p i are ith coordinates of the vectors t and p. For any cube [0, 1] d + x, where x = (x 1 , . .., x d ) ∈ R d , the family F x = {([0, 1) d + t) ∩ ([0, 1] d + x) = ∅ : t ∈ T } is a partition of the cube [0, 1] d + x, in which, because of Keller's condition, every two boxes K, G ∈ F x are dichotomous, i.e. there is i ∈ [d] such that K i and G i are disjoint and K i ∪ G i = [0, 1] + x i . Moreover, since cubes in cube tilings are half-open, every box K ∈ F x is proper, and consequently the family F x is a minimal partition of [0, 1] d + x. The structure of the partition F x reflects the local structure of the cube tiling [0, 1) d + T . Obviously, a cube tiling [0, 1) d + T contains a twin pair if and only if the partition F x contains a twin pair for some x ∈ R d (see Figure 1).
Below we sketch our approach to the problem of the existence of twin pairs in a cube tiling [0, 1) 7 + T of R 7 with |L(T, x, i)| = 5. To do this we describe the structure of a minimal partition. A graph-theoretic description of this structure can be found in [2] (see also [14]) . Let Figure 3). Let F be a minimal partition and let A ⊂ X i be a set such that there is Since the boxes in F are pairwise dichotomous, the set ( The boxes in F are proper, and hence |F i,A n ∪ F i,(A n ) ′ | ≥ 2 for every i ∈ [d] and n ∈ [k]. Thus k ≤ 2 d−1 and consequently |L(T, x, i)| ≤ 2 d−1 for every cube tiling [0, 1) d + T , x ∈ R d and i ∈ [d], as |L(T, x, i)| is the number of all i-cylinders in the partition F x .
If K is a box in X, G is a family of boxes, x ∈ X and i ∈ [d], then , two 3-cylinders and its suits.
Let now [0, 1) 7 + T be a cube tiling of R 7 and let F x be as defined above. If |L(T, x, i)| = 5 for some i ∈ [7], then Assume that there are no twin pairs in the tiling [0, 1) 7 + T . Then F x does not contain a twin pair. It follows from [10, Theorem 3.1] (see Theorem 3.1 in Section 3) that |F i,A k | ≥ 12 for every k ∈ [5]. Thus, there is at least one k ∈ [5] such that |F i,A k | = 12 because |F x | = 128 and |F i,A k | = |F i,(A k ) ′ | for every k ∈ [5]. The main effort in the paper will be rely on describing precisely the structure of all twin pairs free suits Knowing this structure, we will be able to prove that Keller's conjecture is true for a cube tiling [0, 1) Similarly like in [10] instead of suits we will consider systems of abstract words. We collect below basic notions concerning words (details can be found in [11]).
A set S of any objects will be called an alphabet, and the elements of S will be called letters. A permutation s → s ′ of the alphabet S such that s ′′ = (s ′ ) ′ = s and s ′ = s is said to be a complementation. We add an extra element * to the set S and the set S ∪ { * } is denoted by * S. We set * ′ = * . Each sequence of letters s 1 · · · s d from the set * S is called a word. The set of all words of length d is denoted by ( * S) d , and by S d we denote the set of all words s 1 · · · s d such that About such defined f we will say that it preserves dichotomies.
We will exploit some abstract but very useful realization of polybox codes. This sort of realization was invented in [1], where it was the crucial tool in proving the main theorem of that paper.
Let S be an alphabet with a complementation, and let The equicomplementary realization of the code V is the family If S is finite, s 1 , . . . , s n ∈ S and s i ∈ {s j , s ′ j } for every i = j, then In the paper we will assume that S is finite. The value of the realization  1 Let w, u, v ∈ S d , and let D be a simple partition of the d-boxw. If boxesw ∩ȗ andw ∩v belong to D, then there is a simple partition code C ⊂ S d such that u, v ∈ C. In particular, ifw ∩ȗ andw ∩v form a twin pair, then u and v are a twin pair.
Observe that, rigid polybox codes can not contain a twin pair.
Let g : S d × S d → Z be defined by the formula  Let s * = * · · · * ∈ ( * S) d and letḡ(·, s * ) : ( * S) d → Z be defined as follows: The proofs of the last two results can be found in [10].

Polybox codes with a few words
To show that Keller's conjecture is true in dimension seven for a cube tiling [0, 1) 7 + T for which |L(T, x, i)| ≥ 6, it was sufficient to prove the following theorem ([10]): , are disjoint and equivalent polybox codes without twin pairs, then |V | ≥ 12.
To show that the conjecture is true in dimension seven for a cube tiling [0, 1) 7 + T with |L(T, x, i)| = 5, we have to know precisely the structure of all twin pairs free disjoint and equivalent polybox codes V and W , with 12 words each, in dimensions four, five and six. To find this structure we need to know the structure of some polybox codes having a few words. We start with partition codes.
If v ∈ ( * S) d , and σ is a permutation of the set [d], then v σ = v σ(1) · · · v σ(d) . For every i ∈ [d] let h i : * S → * S be a bijection such that h i ( * ) = * and (h i (l ′ )) ′ = h i (l) for every l ∈ S. We say that polybox codes V, W ⊂ ( * S) d are isomorphic if there are σ and h 1 , . . . , h d If |V | = 4 and V contains only one twin pair, then If |V | = 5 and V does not contain a twin pair, then If |V | = 6 and V does not contain a twin pair, then The above partition codes are given up to an isomorphism.
There are two solutions of the equation Since the words are pairwise dichotomous, it can be easily checked that in both cases there are Thus, we have to determine all partitions of a 3-dimensional box into four pairwise dichotomous boxes with only one twin pair. It is easy to see that the first solution corresponds to partitions with more than one twin pair (examples of such partitions are presented in Figure 4b We consider the two halves of the 4box X  If |V | = |W | = 2 and d ≥ 2, then The above polybox codes are given up to an isomorphism.
We can assume thatȗ ∩w and, by (2.1), we can choose x ∈ȗ \w and y ∈v \w such that (x) i = (y) i and (x) i ∈ (w) i . The words in W are pairwise dichotomous, and thus there is a word in W , say p, such that x, y ∈p. Note that p i = w ′ i and consequently, ( Clearly, z ∈w ∪p, and thus, z ∈q. Then, by (2.1), Then z 1 ∈ȓ, and since p and r are dichotomous, we have r i = p ′ i = w i . Now it can be easily seen that (w) i ∪ (ȓ) i = (ȗ) i , which implies that w and r form a twin pair, a contradiction. This completes the proof that w i = * .
We now show that exactly one box from the set E(W ) has nonempty intersection with both boxesȗ andv. Assume on the contrary that there are exactly two boxes in E(W ), sayw andp, having nonempty intersections withȗ andv simultaneously. Then, as we have just shown, w i = p i = * and then (w) i and (p) i are a twin pair, and consequently, w and p are a twin pair, which is a contradiction. Therefore, Therefore, (q) i = (ȓ) i and consequently, q and r are a twin pair, a contradiction.
, two of the three boxes (w) i , (p) i , (q) i are a twin pair, and therefore there is a twin pair among the words w, p, q, which is impossible.
Similarly (3.2) and the proof of Lemma 3.2 (the case |V | = 4), there is a twin pair in the set {w, p, q, r}, which is a contradiction.
In the first case the d-boxȗ is partitioned into four pairwise dichotomous boxes, and thus the structure of this partition is given by (3.2) which contains exactly one twin pair. Hence, the partition {ȗ ∩w,p,q,ȓ} contains one twin pair, and W does not contain a twin pair. Therefore, the boxȗ ∩w must be one of the twins. We may assume thatp is the second one. Thus, there are i 1 , i 2 , i 3 ∈ [d], i 1 < i 2 < i 3 and the letters l 1 , l 2 , l 3 ∈ S such that (we assume without loss of generality that We consider the first case as in the rest of them we obtain isomorphic forms. Let In the second case there are two possibilities: Since now the d-boxȗ is divided into three pairwise dichotomous boxes, the structure of the partition {ȗ∩w,p,q} is given by (3.1). Clearly, as above, we may assume that the boxesȗ ∩w andp are the only twin pair in this partition. Thus, there are Thus, we have to exclude the case (p) B c = l ′ 1 l 2 l 3 and (q) B c = * l ′ 2 l 3 , for otherwise p and r are a twin pair (if we take k = i 2 , then (r) B c = l 1 l ′ 2 l ′ 3 and the case (p) B c = l 1 l ′ 2 l 3 and (q) B c = * l 2 l 3 has to be excluded). Sincev =v ∩w ∪ȓ, Therefore there is exactly one j ∈ B c such that w j = r ′ j and w j = * . Assume without loss of generality Permuting the letters at the third and the fourth position in every word in V and W we get the form as it is in the lemma.
Similarly like above, the two darkest boxes in Figure e are one box which is a realization of the word l ′ 1 l ′ 2 l 3 * .
The sketches of the proofs of the rest of the cases of Lemma 3. 3.
Let |V | = 2, |W | = 3, and let V = {v, u} and W = {w, p, q}. In the same way as above we show that there is exactly one word in W , say w, and there Ifȗ =ȗ ∩w ∪p ∪q andv =v ∩w, then the structure the partition {ȗ ∩w,p,q} of the d-boxȗ is given by (3.1), and (v) i = (w) i . This case is illustrated in Figure 6b.
Ifȗ =ȗ ∩w ∪p andv =v ∩w ∪q, then the boxesȗ ∩w,p are a twin pair andv ∩w,q are a twin pair. Note that (ȗ) i ∩ (q) i = ∅, for otherwise p and q are a twin pair, which is impossible. This case is illustrated in Figure 6c.
Since in this case realizations are 3-dimensional boxes, we will establish only the valuesḡ(v i , s * ) and g(w i , s * ) for i = 1, 2, 3. Recall thatḡ(v i , s * ) = 2 k if and only if the words v i contains k stars and, by Lemma then v 2 and v 3 form a twin pair. This is easy to see thatḡ(w 1 , s * ) = 4, g(w 2 , s * ) = 2,ḡ(w 3 , s * ) = 1. This case is illustrated in Figure 6e.
It can be easily verified that the caseḡ(v 1 , s * ) = 4,ḡ(v 2 , s * ) = 1 and Similarly, this is not hard to find the forms of V and W in the case when Finally, this is easy to check that the casesḡ We now collect the above results in the forms in which they will be used later in the paper.
(a) If the polybox code V does not contain a twin pair, then it contains at least five words. The code V has exactly five words and does not contain a twin pair if and only if it is of the form where u ∈ S d and u ∈ V , the code V does not contain twin pairs and the words w, u are dichotomous but they do not form a twin pair, If |P | = 1, |Q| = 5 and V does not contain a twin pair, then there is a (2,4), (3, 3)} and V does not contain a twin pair, then the structure of (P ) i and (Q) i is such as in Lemma 3.3, but in all those polybox codes we put w j instead of * , if the star appears at the j-th position, and l k ∈ {w i k , w ′ i k } for k ∈ {1, 2, 3, 4}. (d) Let P and Q be such as in (c). If |P | = 1 and 1 ≤ |Q| ≤ 4, then there is a twin pair in V .
Proof of (a). For |V | = 5 it can be found in [10], and the case |V | = 6 is proven in the same manner.
Suppose that |W | = 5 and |U| = 5. Again by (a), there is a set Assume without loss of generality that i = i 1 . Since the structure of U is such as predicted in (a) and and hence w and u are a twin pair, a contradiction.
If W = U, then u = w, which contradicts the assumption. Suppose now that |W | = 6 and assume on the contrary that |V | = 6. Then V = W . By (3.4) in Lemma 3.2 and (a), Note that the structure of (W \ {v 1 }) i 4 is such as in (a). In particular, (v 1 ) i 4 is one and only word which is covered by Consequently, w and u are a twin pair, a contradiction.
Proof of (c). The setw ∩ E(V i,l ) ∪w ∩ E(V i,l ′ ) is an i-cylinder in the d-boxw because {w ∩v : v ∈ V } is a suit forw (Figure 7).
Since V does not contain twin pairs, the set of boxes {w ∩v : v ∈ V } is a partition of the d-boxw into pairwise dichotomous boxes which, by Lemma 2.1, does not contain twin pairs (Figure 7).
Since the set {w ∩v We prove only the case |P | = 1, |Q| = 5. The rest of the cases is proven in the very similar way (compare Example 3.5). Let Thus, (w ∩v) i ⊆ (w ∩ȗ) i for every v ∈ Q, and then Ew j ∩ Ev j ⊆ Ew j ∩ Eu j for every j ∈ [d] \ {i}. It follows that, by (2.1), if w j = u j , then v j = u j . Moreover, by Lemma 2.1, the boxes of the partition {w ∩v : v ∈ Q} do not form twin pairs. Therefore, a code of the partition {(w ∩v) i : v ∈ Q} of the box (w ∩ȗ) i is given by (3.3). Since for every j ∈ A = {i 1 < i 2 < i 3 } (A is such as in (a)) there is l ∈ S such that v j = l and q j = l ′ for some v, q ∈ Q and Ew j ∩ Eu j = Ew j ∩ Ev j ∪ Ew j ∩ Eq j , it must be, by (2.1), w j = u j for every j ∈ A. Thus, (P ) (3.2) and the proof of Lemma 3.2 (the case |V | = 4), this partition contains a twin pair. In the same manner as in (c) we show that there is a twin pair in Q. Figure 7 the five boxes on the left are a realization of the polybox code V = {aaa, a ′ a ′ a ′ , baa ′ , a ′ ba, aa ′ b}, and the box in the middle is a realization of the word w = bbb. Since w ⊑ V , we havew ⊂ E(V ). Thus, the 3-boxw is divided into pairwise dichotomous boxesw ∩v for v ∈ V , and the set ( 3 : v ∈ Q} is divided twice into pairwise dichotomous boxes without twin pairs. Since |Q| = |P | = 2, we apply Lemma 3.3 for the case |V | = |W | = 2 to get the structure of (Q) 3 and (P ) 3   4

Example 3.5 In
The structure of equivalent polyboxes codes with 12 words: necessary conditions In this section we determine necessary conditions which have to be fulfilled by disjoint and equivalent twin pairs free polyboxes codes V and W having 12 words each. This conditions will serve us to establish the initial configurations of words for the computations. Similarly like in [10] we define a graph on a polybox code V . A pair of In [10], we proved the following two lemmas.
and if n + m ≤ 2d − 1, then for some i ∈ [d] and l ∈ S.
By d(G) we denote the average degree of a graph G. For fixed x ∈ ES and i ∈ [d] let where {x} stands at the ith position. If V ⊂ ( * S) d is a polybox code, then the slice π i x ∩ E(V ) is a "flat" polybox in (ES) d (boxes which are contained in this polybox have the factor {x} at the ith position). Therefore we define a polybox (π i The polybox (π i x ∩ E(V )) i does not depend on a particular choice of a polybox code, because if W is an equivalent polybox code to V , then E(V ) = E(W ), and hence (π i x ∩ E(V )) i = (π i x ∩ E(W )) i . We will slice a polybox E(V ) by the set π i x for various x ∈ ES. In particular, we will pay attention whether the polybox code {(v) i : v ∈ V, π i x ∩ v = ∅} is rigid or it contains a twin pair.
In [10] we showed that any polybox code without twin pairs having at most seven words is rigid. Now we need a slightly better rigidity result: does not contain a twin pair and |V | ≤ 9, then it is rigid. Proof. We will show that if W is an equivalent polybox code to V and V ∩ W = ∅, then |V | > 9. We proceed by induction on d. In [10, Corollary 3.5] we showed that for d ≤ 3 every polybox code V ⊂ S d without twin pairs is rigid. Thus, the lemma is true for d ≤ 3. Let d ≥ 4.
We may assume that for every i ∈ [d] there is at least one letter l ∈ S such that V i,l = ∅ and V i, and E(V i,s ) = E(W i,s ). By the inductive hypothesis, V i,l = W i,l and V i,s = W i,s , and hence V ∩ W = ∅, a contradiction.
and let x ∈ Ea and y ∈ Ea ′ . It follows from the inductive hypothesis that the polybox code {(v) i : v ∈ V i,l } is rigid for l ∈ {a, a ′ }, and therefore . Taking x ∈ Ea ∩ Eb ′ and y ∈ Ea ′ ∩ Eb ′ , in the same manner as above, we show that Thus, in the rest of the proof we assume that V i,l = ∅ for every i ∈ [d] and l ∈ S.
Let us suppose that there are i ∈ [d] and two letters in S, say a and b, such that the polybox code (V i,a ∪ V i,b ) i does not contain a twin pair, i.e. there are no i-siblings in V i,a ∪ V i,b . This means, by the inductive hypothesis, that the polybox code Therefore, we assume that for every l, s ∈ S, l ∈ {s, s ′ }, the set (V i,l ∪ V i,s ) i contains a twin pair, i.e. there is an i-siblings in V i,l ∪ V i,s . We It can be easily seen that there are i, j ∈ [d], i = j, and l, s ∈ {a, b} such that |V i,l ∪ V i,l ′ | = 7 and |V j,s ∪ V j,s ′ | = 7. We can assume without loss of generality that l = s = a, as i = j. Let where N(u 0 ) and N(v 0 ) denote the set of all neighbors of vertices u 0 and v 0 , respectively. We have Note that u = v, for otherwise u ∈ N(u 0 ) ∪ N(v 0 ). Assume without loss of generality that u i = b and v i = a. Since p i , p j ∈ {a, a ′ } for every p ∈ (N(u 0 ) ∪ N(v 0 )) \ {u, v}, the only vertices from the set N(u 0 ) ∪ N(v 0 ) which can be adjacent to the vertex w are u and v. This means that, there is no i-siblings q, t ∈ V such that q i = b ′ and t i = a ′ , which is a contradiction.
Let now d = 4 and assume without loss of generality that u 0 = aaaa and v 0 = ba ′ aa. By just considered case, we assume that for every i, j ∈ [d], i = j and l, s ∈ {a, b} we have |V i,l ∪ V i,l ′ | ≤ 6 or |V j,s ∪ V j,s ′ | ≤ 6. Thus, it suffices to consider three cases: n 2 = n 3 = n 4 = 2; n 2 = 3, n 3 = 2, n 4 = 1, and n 2 = n 3 = 2, n 4 = 1, where and i ∈ {2, 3, 4}, then v j ∈ {a, a ′ } for every j ∈ {2, 3, 4} \ {i}.) Therefore, in these two cases, the maximal number of edges with ends in N(u 0 ) ∪ N(v 0 ) is achieved if the vertices v 1 , . . . , v 6 are arranged as presented in Figure 8a and 8b for the first and the second case, respectively (recall that the graph G does not contain triangles). Since d(w) ≤ 4, where {w} = V \ (N(u 0 ) ∪ N(v 0 )), we have |E | < 16, which contradicts the assumption on the number of edges in G.
In the third case we assume that v 1 Observe that, now the vertices v 2 , . . . , v 6 can be joined with v 1 , but similarly as above if v n , v m are adjacent, where n, m ∈ {2, 3, 4, 5, 6}, then Since d(w) ≤ 4, the number of edges with ends in the set N(u 0 ) ∪ N(v 0 ) has to be at least 12 because |E | ≥ 16. This can be done only if v 1 , v 3 , v 5 ∈ N(u 0 ) and v 2 , v 4 , v 6 ∈ N(v 0 ) or v 1 , v 2 , v 4 ∈ N(v 0 ) and v 3 , v 5 , v 6 ∈ N(u 0 ). We will consider the first case (the second case is considered in the same way.) To obtain 12 edges with ends in the set N(u 0 ) ∪ N(v 0 ), which is the maximal number of edges with ends in this set, the vertices v 1 , . . . , v 6 have to be arranged as pictured in Figure 9. Since d(v 1 ) = 4 and the graph G does not contain triangles, it must be d(w) < 4, and then |E | < 16, a contradiction.  In the next two lemmas we give forbidden distributions of words in the considered codes V and W .
In the similar way as in the first part of the proof of Lemma 4.3 we show that  3. It can be easily checked, using (2.1), that for every l i , s i ∈ {a, a ′ }, Observe now that, again by (2.1), the point y can be chosen such that y i ∈ Eb \ Ea, and then y ∈ E(W i,b ) \ E(V i,a ). Thus, y ∈ E(W ) and y ∈ E(V ), a contradiction.
Before we consider the case when u and v are an i-siblings note that Then, by Statement 3.4 (b) and (a), respectively, |W i,a | ≥ 7 and |W i,a ′ | ≥ 5, and thus |W i,a | = 7 and |W i,a ′ | = 5, as |V | = 12.
. Thus, the structures of V i,a and V i,a ′ are such as predicted in Statement 3.4 Let now u and v be an i-siblings. Then (v) i and (u) i are a twin pair. We can assume without loss of generality that (   Since must be covered by a box from the set E(W i,s ), and thus W i,s = ∅. Consequently, we may assume that |W i,l | = 5.
Suppose now that for every r ∈ S, r ∈ {l, l ′ } we have W i,r = ∅ or W i,r ′ = ∅. Then, by (2.1), there is x ∈ El such that E(W ) ∩ π i x = E(W i,l ) ∩ π i x . By Lemma 4.3, the polybox code {(w) i : w ∈ W i,l } is rigid, and therefore (v) i = (w) i for some w ∈ W i,l and v ∈ V i,l because Hence, there is r ∈ S, r ∈ {l, l ′ } such that the sets W i,r and W i,r ′ are non-empty. Clearly, r = s and then |W i,s | = |W i,s ′ | = 1 because |W i,l | ≥ 5, |W i,l ′ | ≥ 5 and |V | = 12. It follows from Statement 3.4 (d) that W i,s ⊑ V i,s and W i,s ′ ⊑ V i,s ′ , and thus, by Statement 3.4 (a), |V i,s | ≥ 5 and |V i,s ′ | ≥ 5. Since , |V i,s | = |V i,s ′ |, we have |V i,s | + |V i,s ′ | ≥ 11, and consequently |V | > 12, a contradiction.
Let now |V i,l | = 1 and 2 ≤ |V i,l ′ | ≤ 4. By Statement 3.4 (d), V i,l ⊑ W i,l and V i,l ′ ⊑ W i,l ′ , and by Statement 3.4 (a) and (b), respectively, |W i,l | ≥ 5 and |W i,l ′ | ≥ 7. Thus, |W i,l | = 5 and |W i,l ′ | = 7 because |V | = 12. This means that W i,s ∪ W i,s ′ = ∅ for every s ∈ S \ {l, l ′ }, and therefore the set An important role in determining the structure of the polybox codes V and W will play the following lemma. If V i,s = ∅ or V i,s ′ = ∅ for every s ∈ S, then in the similar way as at the beginning of the proof of Lemma 4.3 we show that V i,l and W i,l are equivalent and similarly, V i,p and W i,p are equivalent. By Theorem 3.1, then the polybox codes V i,l and W i,l are equivalent and similarly, V i,l ′ and W i,l ′ are equivalent. Then, by Theorem 3.1, If W i,s = ∅, where s ∈ {l, l ′ }, then in the same manner as in the second part of the proof of the previous lemma we show that the set E(W i,s ∪W i,s ′ ) is an i-cylinder, which gives |W | ≥ |W i,s | + |W i,s ′ | ≥ 24, and thus |V | > 12.
Suppose now on the contrary that |V | = 12. Furthermore, assume that Suppose that V i,s = ∅ for at least one s ∈ {l, l ′ }. By (2.1), we can choose x ∈ El ∩ Es ′ and y ∈ El ′ ∩ Es ′ such that If V i,l is not rigid, then, by Lemma 4.3, |V i,l | ≥ 10. Since |V | = 12, we have |V i,l | = 10, |V i,l ′ | = 1 and |V i,s | = 1. Then V i,s ⊑ W i,s , and consequently, by Statement 3.4 (a), |W i,s | ≥ 5. Therefore, for every x ∈ El ∩ Es ′ and y ∈ El ′ ∩ Es we have Now it is easy to see (compare [10, Lemma 3.6]) that |V | ≥ 15, a contradiction.
If the codes V i,l and V i,l ′ are rigid, then W i,l ∪ W i,l ′ = ∅, for otherwise taking w ∈ W i,l we get, by (4.3) and the rigidity of is an i-cylinder (compare the proof of the previous lemma), and since V i,l and V i,l are rigid, the set V i,l ∪ V i,l ′ consists of twin pairs, which is a contradiction.
Thus we may assume that the sets V i,l , V i,l ′ , V i,s and V i,s ′ are non-empty.
(We still assume that |V | = 12.) It follows from the above that Let x ∈ Ea ∩ Eb be such that the set {(v) i : v ∈ V, π i x ∩v = ∅} does not contain a twin pair.
If |(π i x ∩ E(V )) i | 0 ≤ 9, then, by Lemma 4.3, the polybox code We now once again indicate a forbidden distribution of words in V and W .
is impossible. Proof. By Lemma 4.6, we may assume that for every i ∈ [d] and l, s ∈ {a, a ′ , b, b ′ }, l ∈ {s, s ′ }, there are i-siblings in the set V i,l ∪ V i,s .
Suppose on the contrary that V has the distribution |V i,l | = |V i,l ′ | = 3 for every i ∈ [d] and l ∈ {a, b}. Let G = (V, E ) be a graph of siblings in V . Note that, it follows from the assumption on i-siblings in V that |E | ≥ 4d.
Let u 0 , v 0 ∈ V be such that . We can assume without loss generality, can be joined only with u or v. This means that there is no i-siblings q, t ∈ V such that q i = b ′ and t i = a ′ , a contradiction.
In the next lemma we show that to find the structure of the polyboxes V and W we may assume that they are written down in the alphabet S = {a, a ′ , b, b ′ }.  Proof. It follows from Theorem 3.1 that |V | ≥ 12. Suppose on the contrary that |V | = 12. In the same way as in the first part of the proof of Lemma 4.6 we show that for every i ∈ [d] there are at least two letters l, s ∈ S, l ∈ {s, s ′ }, such that then, in the same way as in the proof of Lemma 4.5, we show that the set E(V i,l ∪ V i,l ′ ) is an i-cylinder, and consequently |V | > |V i,l |∪|V i,l ′ | ≥ 24, a contradiction. Thus, W i,l ∪W i,l ′ = ∅ and W i,s ∪ W i,s ′ = ∅ Suppose that V i,r = ∅ and V i,r ′ = ∅ for some r ∈ S \ {l, l ′ , s, s ′ }. For every w ∈ W such thatw ∩ E(V i,r ) = ∅ we have w i = r, and hence V i,r ⊑ W i,r from where, by Statement 3.4 (a), we obtain |W i,r | ≥ 5. Then Since |W i,r | ≥ 5 and W i,s ∪ W i,s ′ = ∅, by Lemma 4.5, |W | > 12, a contradiction.
Let d = 4. Since V can be extended to a partition code U ⊂ S 4 , i.e.
But then |U i,p ∪ U i,p ′ | ≥ 6 for p ∈ {l, s, r}, and thus |U| > 16 which is a contradiction.
Let now d ≥ 5. For every p ∈ {l, s, r} there is w ∈ W such thatw ∩ E(V i,p ) = ∅ andw ∩ E(V i,p ′ ) = ∅, for otherwise V i,p ⊑ W i,p and V i,p ′ ⊑ W i,p ′ for some p, and thus, by Statement 3.4 (b), |W i,p | ≥ 7 and |W i,p ′ | ≥ 7. Then |W | > 12, a contradiction. By Statement 3.4 (d), for every p ∈ {l, s, r} there are i 1 (p), where l k (r) ∈ {s k (r), s k (r) ′ } for k = 1, 2. We consider the first case (the second case is considered in the same manner). In the first case we get a contradiction to (4.5). In the second case, by Lemma 4.5, |V | > 12, which is also a contradiction. To Then in the same way as in the second part of Lemma 4.5 we show that the set E(W i,c ∪ W i,c ′ ) is an i-cylinder and consequently |W i,c | + |W i,c ′ | ≥ 24, a contradiction.
At the end of this section we show that the computations will be made mainly for d = 4, 5 and only in one case for d = 6.
In particular, for every i ∈ [6] there are at least 4 edges with the colour i, and therefore |E | ≥ 24.
Let u 0 , v 0 ∈ V be such that : v, u ∈ V and v, u are adjacent}.
We may assume without loss of generality that u 0 = aaaaaa, v 0 = ba ′ aaaa.
By the assumption on the distribution of words in V and Lemma 4.5, for every i ∈ [6] and l ∈ {a, b}.
It follows from (4.6) that in the first case there are at least two words in the set {w 1 , w 2 , w 3 , w 4 }, say these are w 1 and w 2 , which have the letters b or b ′ at at least three positions i, j, k ∈ {3, 4, 5, 6}, which means that they cannot be adjacent to vertices from the set N(u 0 )∪N(v 0 ). In the second case there are at least three such words; assume that these are w 1 , w 2 and w 3 . It is easy to verify that in the first case there are at most 8 edges with ends in the set N(u 0 ) ∪ N(v 0 ), and in the second case there are at most 12 such edges (compare the second part of the proof of Lemma 4.3). The maximal number of edges with ends in {w 1 , w 2 , w 3 , w 4 } is four as the graph G does not contain triangles. Thus, in the second case in order to obtain |E | ≥ 24 it must be d(w 4 ) ≥ 8, which is impossible since d(v) ≤ 6 for every v ∈ V . For the same reason in the first case it must be d(w 3 ) = d(w 4 ) = 6. Then for every v ∈ (N(u 0 ) ∪ N(v 0 )) \ {u 0 , v 0 } the vertices w 3 and w 4 are adjacent to v. Since G does not contain triangles, w 3 and w 4 cannot be adjacent, and hence |E | < 24, which contradicts the assumption on E .

Computations
In this section we describe the computations which lead to the determination of all possible twin pairs free equivalent and disjoint polybox codes V, W ⊂ S d having 12 words each, where S = {a, a ′ , b, b ′ } and d ∈ {4, 5, 6}. The structure of such polybox codes V, W is given in Theorem 5. 2.
The longest part of the paper was devoted to the preparations of the computations, since it seems hopeless to do this without any initial configurations of words, where by an initial configuration of words we mean a some number of words or their fragments in the constructing code V (see tables in this section).

Initial configurations of the words.
It follows from Corollary 5.1 that there is w ∈ W such thatw ∩ E(V i,b ) = ∅ andw ∩ E(V i,b ′ ) = ∅ when the polybox code V has the distributions of words 1 − 5 and 9 − 13 and there is w ∈ W such thatw ∩ E(V i,a ) = ∅ and w ∩ E(V i,a ′ ) = ∅ when V has the distributions 6 and 7.
Now we show how the initial configurations of words are established. We do it in detail for the distributions 1, 2, 4 and 9. The rest configurations are determined in the similar manner.
Let V has the distribution of words of the form 1, 2, 4 or 9 of Corollary 5.1. Then, by Statement 3.4 Figure  7). Clearly, without loss of generality we may assume that w i 1 = w i 2 = b and p = b · · · b (recall that V, W ⊂ {a, a ′ , b, b ′ } d ). Then, by Statement 3.4 (c), l 1 , l 2 ∈ {a, a ′ }, so we may assume that l 1 = l 2 = a.
Let V be such as in the case 1 of Corollary 5.1, and let i = 1, i 1 = 2, i 2 = 3 and d = 5. For the computations we take a = +1, a ′ = −1, b = +2 and b ′ = −2. We arrange the words from the set V 1,+2 ∪ V 1,−2 as the first four rows of the matrix M of the size 12 × 5. The rest eight words from V 1,+1 ∪ V 1,−1 are unknown; we know only their first letters, i.e. +1 and −1 and that |V 1,+1 | = 7 and |V 1,−1 | = 1. The matrix M has a form These four full-length words and the eight first letters of the rest of words from V , seven letters +1 and one −1, are the initial configurations of words for the computations corresponding to the case 1 of Corollary 5.1 for d = 5. Our task is to compute the missing cells of the matrix M such that the rows in the resulting matrix form the polybox code V having 12 words without twin pairs and, by Lemma 4.6, for every i ∈ [5] and every l, s ∈ {+1, −1, +2, −2}, l ∈ {s, s ′ }, the set V i,l ∪ V i,s contains an i-siblings.
In the case 1,2,4, and 9 the initial configurations of words are the following (we write M in the form of a table): where (+2) at the end of v i means that the letter +2 has to be placed at the fifth position in the words v i , i = 1, 2, 3, 4, for the case d = 5. This notation will be used also below.
As we noted above, in the cases 6 and 7 we havew ∩ E(V i,a ) = ∅ and w ∩ E(V i,a ′ ) = ∅. In this two cases we have w i = b. Thus, by Lemma 4.5, |{w ∩v = ∅ : v ∈ V i,a }| = 5 and |{w ∩v = ∅ : v ∈ V i,a ′ }| = 1. Therefore, by Statement 3.4 (c), Finally, we consider the initial configurations of words corresponding to the case 8 of Corollary 5. 1. In this case, by Lemma 4.9, besides computations for d = 4, 5 we have to make computations for d = 6. Observe first that, by Statement 3.4 (d), V i,b ⊑ W i,b and V i,b ′ ⊑ W i,b ′ , and thus, by Statement 3.4 (a), |W i,b | ≥ 5 and |W i,b ′ | ≥ 5, and since, by Lemma 4.6, |W i,a | ≥ 1 and |W i,a ′ | ≥ 1, we have |W i,b | = 5, |W i,b ′ | = 5 and |W i,a | = 1, |W i,a ′ | = 1. Clearly, W i,a ⊑ V i,a and W i,a ′ ⊑ V i,a ′ , and thus the structure of V i,a and V i,a ′ are such as in Statement 3.4 (a).
Below we present an algorithm using in the computations.

Algorithm
Input: The set M d r (n +1 , n −1 , n +2 ) Output: A polybox code V having 12 words without twin pairs such that for every i ∈ [d] and every l, s ∈ {+1, −1, +2, −2}, l / ∈ {s, s ′ }, the set V i,l ∪ V i,s contains an i-siblings 1: V := ∅ 2: W := M d r (n +1 , n −1 , n +2 ) 3: k := r 4: repeat 5: while k < 12 do 6: select a word v from the set S d +1 ∪ S d −1 ∪ S d −2 which is dichotomous to every word of W and does not form a twin pair with any word of W 7: W := W ∪ {v} 8: k := k + 1 9: end while 10: if |W 1,+1 | = n +1 , |W 1,−1 | = n −1 , |W 1,−2 | = n −2 and for every i ∈ , 5, 6}, are disjoint and equivalent polybox codes without twin pairs having twelve words and V is extensible to a partition code, then there is a set A = {i 1 < i 2 < i 3 < i 4 } ⊂ [d] such that and (V ) A = (W ) A = {(p) A } for some p ∈ S d . The representation of V and W is given up to an isomorphism. Proof. To show that V and W do not contain a twin pair observe that the partition code W 1 contains four twin pairs ls ′ ss, ls ′ s ′ s; l ′ l ′ s ′ l, l ′ l ′ sl; ssl ′ l ′ , s ′ sl ′ l ′ ; slls ′ , s ′ lls ′ , and the partition code W 2 also contains four twins ls ′ ss, l ′ s ′ ss; l ′ l ′ s ′ l, ll ′ s ′ l; ssl ′ l ′ , ssll ′ ; s ′ lls ′ , s ′ ll ′ s ′ , where the marked words form the set W 1 ∩ W 2 . Therefore, the polybox codes V = W 1 \ W 1 ∩ W 2 and W = W 2 \ W 1 ∩ W 2 are disjoint and equivalent, they do not contain twin pairs and |V | = |W | = 12.
By Corollary 5.1, the distributions of the letters in V cannot be other than the distributions 1 − 13 of this corollary. Thus, the code V has to contain at least one initial configuration from the list of the initial configurations of words given in the tables in this section.
The computations showed that the initial configurations corresponding to the cases 1 − 7 and 9 − 12 cannot be completed to a twin pair free polybox code V with twelve words such that for every i ∈ [d] and l, s ∈ {a, a ′ , b, b ′ }, l / ∈ {s, s ′ }, the set V i,l ∪ V i,s contains an i-siblings (compare Lemma 4.6). Similarly, this cannot be done in the case 8 for d = 5 and 6. The only configurations that can be completed to a polybox code V with the above properties are the case 8 for d = 4 and the configurations given in the last two tables for the case 13. The structures of all these complemented codes V (for d = 4 we have (V ) A c = V as A c = ∅) with 12 words are, up to an isomorphism, as stated in the theorem. The polybox code W is a complementation of the words W 1 ∩ W 2 = {ls ′ ss, l ′ l ′ s ′ l, ssl ′ l ′ , slls ′ } to a partition codes. It can be easily check that there is exactly one such complementation which is disjoint with V , and its structure is such as given in the theorem. (The words ls ′ ss, l ′ l ′ s ′ l, ssl ′ l ′ , slls ′ can be complemented to a partition code exactly in two ways: V and W ).
Since for d = 5 and 6 all initial configurations cannot be completed to a twin pair free polybox code with twelve words such that for every i ∈ [d] and l, s ∈ {a, a ′ , b, b ′ }, l / ∈ {s, s ′ }, the set V i,l ∪ V i,s contains an i-siblings, it follows that for the above two dimensions we have, by Lemma 4.6, (V ) A = (W ) A = {(p) A } for some p ∈ S d , and (V ) A c , (W ) A c are such as in dimension four.

Remark 5.2
The representation of V and W given in Theorem 5.2 can be found in [12]. The codes V and W were used by Lagarias and Shor [13,14] and later on by Mackey [15] to construct the counterexamples to Keller's cube tiling conjecture. In the context of this conjecture one of these codes was Let W ⊂ U 4,l ′ be the set of all w ∈ U 4,l ′ such that (u) 4 ⊑ (W ) 4 and (w) 4 ∩ (ȗ) 4 = ∅. Since U is a partition code, W = ∅. Clearly, for every i ∈ [7], i = 4, and w ∈ W we have w i = u ′ i , for otherwise (w) 4 ∩ (ȗ) 4 = ∅ which contradicts the definition of W . Every w ∈ W is dichotomous to the words v, q and p, and therefore w 1 = w 2 = s and w 3 = w 4 = l ′ for every w ∈ W . Then lll ⊑ (W ) {1,2,3,4} .
If (W ) {1,2,3,4} does not contain a twin pair, then its structure is such as in where s i ∈ {l, l ′ } for i = 1, 2, 3. But then ssl ′ l ′ s ′ 1 s 2 l ∈ U 7,l which is not true. Therefore, the code (W ) {1,2,3,4} contains a twin pair, and hence W contains a twin pair.
From the result of Debroni et al., [10,Corollary 4.2] and the above corollary we obtain the following Corollary 6.3 If there is a counterexamples to Keller's conjecture in dimension seven, then |L(T, x, i)| ∈ {3, 4} for some x ∈ R 7 and i ∈ [7].
In [10] we extended the notion of a d-dimensional Keller graph: if S is an alphabet with a complementation, then a d-dimensional Keller graph on the set S d is the graph in which two vertices u, v ∈ S d are adjacent if they are dichotomous but do not form a twin pair.
From Theorem 6.1 we obtain the following Corollary 6.4 Every clique in a 7-dimensional Keller graph on S 7 which contains at least five vertices u 1 , . . . , u 5 such that u n i ∈ {u m i , (u m i ) ′ } for some i ∈ [7] and every n, m ∈ {1, ..., 5}, n = m, has less than 2 7 elements. Proof. Assume on the contrary that there is a clique U containing vertices u 1 , . . . , u 5 and |U| = 2 7 . Thus, U is a partition code without twin pairs. Since u n i ∈ {u m i , (u m i ) ′ } for every n, m ∈ {1, ..., 5}, n = m, it follows that |U i,u m i | ≤ 12 for some m ∈ [5]. By Theorem 6.1, there is a twin pair in U, a contradiction.