Symmetric polyomino tilings, tribones, ideals, and Groebner bases
aa r X i v : . [ m a t h . C O ] J u l Symmetric polyomino tilings,tribones, ideals, and Gr¨obner bases
Manuela Muzika Dizdarevi´c
Faculty of Natural Sciencesand Mathematics, [email protected]
Rade T. ˇZivaljevi´c
Mathematical InstituteSASA, [email protected]
May 22, 2014
Abstract
We apply the theory of Gr¨obner bases to the study of signed, symmetricpolyomino tilings of planar domains. Complementing the results of Conway andLagarias we show that the triangular regions T N = T k − and T N = T k in ahexagonal lattice admit a signed tiling by three-in-line polyominoes (tribones) symmetric with respect to the 120 ◦ rotation of the triangle if and only if either N = 27 r − N = 27 r for some integer r ≥ Our general objective is to explore signed polyomino tilings which are symmetric withrespect to a group of symmetries by the methods of standard (Gr¨obner) bases of poly-nomial ideals.The tiling depicted in Figure 1, illustrating the case N = 8 of Theorem 1, shows thata triangular region in a hexagonal lattice may have a signed tiling by congruent copiesof the three-in-line tile (tribone). In the same paper Conway and Lagarias showed([5, Theorem 1.2.]) that neither this nor any other triangular region in the hexagonallattice can be tiled by tribones (if ‘negative’ tiles are not permitted).A very nice exposition of these and related results can be found in [13] and [8,Chapter 23]. Theorem 1. (Conway-Lagarias [5, Theorem 11.4])
The triangular region T N in thehexagonal lattice has a signed tiling by congruent copies by three-in-line tiles (tribones)if and only if N = 9 r or N = 9 r + 8 for some integer r ≥ . R. ˇZivaljevi´c was supported by the Grants 174017 and 174020 of the Ministry for Science andTechnological Development of Serbia. T N = T k − and T N = T k in a hexagonal lattice ∗ admit a tiling by tribones symmetric with respectto the rotation of the triangle through the angle of 120 ◦ degrees if and only if either N = 27 r − N = 27 r for some integer r ≥
0. In particular the triangle depicted inFigure 1 does not admit such a tiling.The method applied for the proof of this theorem is based on the observation that the tile homology group introduced by Conway and Lagarias in [5] (see also Reid [11, Section2]) is naturally a module over the group ring of the associated group of translations.This group ring is a quotient of a polynomial ring which allows us to reduce the tilingproblem to the ‘submodule membership problem’ and apply the theory of Gr¨obnerbases.Surprisingly enough there are very few applications of the algebraic method basedon Gr¨obner basis to problems of tilings and tessellations and the only reference we areaware of is the paper by Bodini and Nouvel [4]. The fact that the ‘tile homology group’in the sense of [11] is a module over a polynomial ring offers some obvious technicaladvantages. One of our objectives is to advertise this approach in the context of signedtilings with symmetries. These problems seem to be particularly well adopted to thealgebraic method in light of the fruitful relationship between the theory of Gr¨obnerbases and the theory of invariants of group actions [6], [12].
There are three regular lattice tilings of R , the triangular lattice L ∆ , square lattice L ✷ , and the hexagonal lattice L hex , depicted in Figure 2. If L is one of these latticetilings the associated dual lattice (point set) L ◦ is generated by all barycenters of theelementary cells of L .Let A ( L ) be the free abelian group generated by all elementary cells of the lattice L . A ‘lattice tile’ P (informally a lattice figure in L ), defined as a finite collection ∗ In the remaining case N = 3 k + 1 there is a hexagon in T N fixed by the 120 ◦ -degrees rotation. P = { c , . . . , c n } of cells in L , is associated an element P = c + . . . + c n of the group A ( L ).The problem if a given bounded lattice region (lattice figure) R admits a signed tiling with translates of prototiles R = { R , . . . , R k } is an instance of the subgroupmembership problem . Indeed, let B ( R ) be the subgroup of A ( L ) generated by alltranslates of prototiles R i and let H ( R ) = A ( L ) /B ( R ) be the associated ‘homologygroup’. Then (following [5] and [11]) such a tiling exists if and only if R ∈ B ( R ) orequivalently if the coset R + B ( R ) is the zero element in H ( R ).Let G = G ( L ) be the group of all affine transformations that keep the lattice tiling L invariant. Let Γ = Γ( L ) be its subgroup of all translations with this property. Byselecting 0 ∈ L ◦ as the zero element, L ◦ is turned into a group and there is a naturalidentification Γ = L ◦ .The group A ( L ) is a clearly a module over the group ring Z [Γ] (which is isomorphicto the ring Z [ Z ] of Laurent polynomials in two variables. This ring can be obtained(in many ways) as a quotient of the semigroup ring Z [ N d ] ∼ = Z [ x , . . . , x d ] (for some d ).This observation allows us to see the groups A ( L ) , B ( R ) and H ( R ) as modules overthe polynomial ring Z [ x , . . . , x d ] and to reduce the tiling question to the submodulemembership problem [7, Chapter 5]. In turn, in the spirit of [4], one can use the ideasand methods of Gr¨obner basis theory.Here we put some emphasis on the use of the ‘submodule membership problem’as a natural extension of the ‘ideal membership problem’, originally proposed andused by Bodini and Nouvel in [4]. This appears to be a more natural and conceptualapproach to the general tiling problems since the module A ( L ) is no longer requiredto be monogenic (cyclic over Z [ P ]) which allows us a greater freedom in choosingthe semigroup ring Z [ P ]. This property will be indispensable in the study of tilings symmetric with respect to a group of symmetries which is the main goal of this paper. The reader may find the following example, depicted in Figure 3, as a good illustrationof the main problem studied in our paper.The (3 ×
3) checkerboard C × (Figure 3) is supposed to be paved by translates oftwo types of prototiles . Each of the cells (elementary squares) is labelled (coordinatized)by a pair ( i, j ) ∈ N of integers and each tile (polyomino) is formally a union of a finite3umber of elementary cells. In the example depicted in Figure 3 there are two typesof prototiles, T = { (0 , , (1 , , (0 , } and T = { (1 , , , , (0 , } .Figure 3: A signed tiling of a 3 × ×
3) chessboard C × is covered by 3 translated copies of prototile T and2 translated copies of prototile T . In general the translated prototiles are notrequired to be subsets of C × .(2) Each (translated) prototile is associated a weight (sign) and the total weight ofeach of the cells ( i, j ) ∈ C × is equal to 1. (The condition that the total weightof cells outside C × is zero is added if prototiles are not necessarily subsets of C × .)(3) The tiling is symmetric with respect to the main diagonal of C × in the sensethat if a translated prototile T + v (where T ∈ { T , T } and v ∈ N ) appears inthe decomposition (tiling) with a weight w ∈ Z then the diagonally symmetricprototile T ′ + v ′ is also present with the same weight.A basic observation is that each polyomino P ⊂ N can be associated a polynomial f P = P { x i y j | ( i, j ) ∈ P } , for example f T = 1 + x + y and f T = x + y + xy . The de-composition depicted in Figure 3 naturally corresponds to the following decompositionof polynomials in the ring Z [ x, y ] (or in the ring Z [ σ , σ ] of symmetric polynomials): f T + [ xf T + yf T ] − f T + xyf T = (1 + x + x )(1 + y + y ) = f C × . (1)Our immediate objective is to use the theory of Gr¨obner bases to generate such identi-ties. More generally we want to develop and study procedures and algorithms for thesystematic analysis (existence and other properties) of decompositions similar to (1).4 .2 Basic facts about polyomino tilings Informally a polyomino pattern P (or polyomino for short) is a (not necessarily con-nected) finite region consisting of cells in one of the three regular lattice tilings of theplane (Figure 2). It is sometimes more convenient to describe a polyomino as a collec-tion P = { c , . . . , c k } of elementary cells in the associated lattice L . Some authors usethe generic name polyforms for all three types of polyominoes so the L (cid:3) -polyforms arepolyominoes in the usual sense [9], L hex -polyforms are referred to as polyhexes etc.We frequently use a slightly more general (algebraic) definition of a polyomino asa multiset , subset of L , with multiple and possibly with negative elements. We willtacitly make a distinction between the geometric and algebraic definition by reservingthe term ‘weighted polyomino’ for the algebraic version. However most of the time theterm ‘polyomino’ is used interchangeably for both kinds of polyomino patterns. Definition 2.
A (weighted) polyomino P is a finite weighted subset of L (a multiset)which contains each elementary cell c ∈ L with some (positive or negative) multiplicity w c ∈ Z . In other words P = P w c c is an element of the free abelian group A ( L ) generated by all cells of the lattice tiling L . We have already seen in Section 2.1 an example of the correspondence between ageometric image (Figure 3) and an algebraic expression (equation (1)), based on thecorrespondence ( i, j ) ↔ x i y j between the cell labelled by ( i, j ) ∈ N and the associatedmonomial x i y j .More generally let S ∼ = N d be a semigroup which acts on the lattice L by transla-tions, which means that there exists a homomorphism ρ : S → Γ from S to the groupΓ = Γ( L ) of all translations that keep the lattice tiling L invariant.The group A ( L ) is naturally a module over the semigroup ring Z [ S ] ∼ = Z [ x , . . . , x d ].For example if S = Γ then Z [ S ] ∼ = Z [Γ] and A ( L ) is a Z [Γ]-module where Z [Γ] ∼ = Z [ x, x − ; y, y − ] is the ring of Laurent polynomials.Let R = { P , . . . , P k } be a collection of basic tiles (prototiles). Define B ( R ) as thesubgroup of A ( L ) generated by all translates of the prototiles P i , or equivalently as a Z [ S ]-submodule of A ( L ) generated by R .The following tautological proposition links the idea of the tile homology groupof Conway and Lagarias [5] and Reid [11] with the ‘submodule membership problem’typical for applications of Gr¨obner bases (as proposed by Bodini and Nouvel [4]). Proposition 3.
A polyomino P has a signed tiling by translates of prototiles R = { P , . . . , P k } if and only if P ∈ B ( R ) where B ( R ) is the Z [ S ] -submodule of A ( L ) generated by R . The associated class [ P ] in the tile homology module H ( R ) := A ( L ) /B ( R ) (2) is a ‘quantitative measure’ of how far is P from admitting a tiling by R . L .The following proposition serves as an illustration of the simplest case where we restrictour attention to the first quadrant of the L (cid:3) -lattice tiling. In this case S = N and Z [ S ] = Z [ x, y ]. As in Section 2.1, each polyomino P ⊂ N is associated a polynomial f P = P { x i y j | ( i, j ) ∈ P } . Proposition 4.
A polyomino pattern P ⊂ N admits a signed tiling by the first quad-rant translates of polyomino patterns P , . . . , P k if and only if, f P = h f P + . . . + h k f P k for some polynomials h , . . . , h k with arbitrary integer coefficients or equivalently if, f P ∈ h f P , . . . , f P k i . The group G = G ( L ) was introduced in Section 2 as the group of all affine transfor-mations that keep the lattice tiling L invariant. The abelian group A ( L ) is a moduleover the group ring Z [ G ]. Since Γ ⊂ G is a normal subgroup, we observe that G actson A ( L ) preserving its Z [Γ]-module structure as well, provided G acts on the ‘scalars’from Z [Γ] by conjugation.Let Q ⊂ G be a (finite) subgroup of G . Assume that the set R of prototiles isinvariant with respect to the group Q . Then Q acts on the submodule B ( R ) and thetile homology module (2). Again, one shouldn’t forget that the action of Q on scalarsfrom Z [Γ] may be non-trivial. Define B ( R ) Q = Hom Q ( Z , B ( R )) as the subgroup(submodule) of B ( R ) of elements which are invariant under the action of Q .If we restrict our attention to the subring Z [Γ] Q ⊂ Z [Γ] of Q -invariant elementsthen the action of Q on scalars from Z [Γ] Q is trivial and the Z [Γ] Q -module A ( L ) is a Q -module in the usual sense.An element of the group A ( L ) Q is referred to as an equivariant signed polyomino .The fundamental problem is to decide when a given polyomino P ∈ A ( L ) Q admits a Q -symmetric signed tiling by translates of a Q -invariant family of prototiles R . Thefollowing criterion is an equivariant analogue of Proposition 3. Proposition 5.
Let R be a Q -invariant (finite) set of prototiles. A Q -invariant poly-omino P ∈ A ( L ) Q has an equivariant, signed tiling by translates of prototiles R if andonly if P ∈ B ( R ) Q where B ( R ) is the Z [ S ] Q -submodule of A ( L ) generated by R . The setting of Proposition 5 is exactly the same as before (Proposition 3), howeverthe emphasis is now on the Z [Γ] Q -module structure on A ( L ) Q and B ( R ) Q . In order toapply this criterion one is supposed to determine the ring of invariants Z [ S ] Q and thestructure of the module B ( R ) Q . Both goals can be achieved with the aid of the theoryof Gr¨obner bases, see [6, Section 7] for necessary tools.6 Hexagonal polyomino with symmetries
Let G hex be the group of symmetries of the hexagonal tiling L hex of the plane depictedin Figure 4. Our objective is to study L hex -tiling problems which are symmetric withrespect to some (finite) subgroup of G hex . Our initial focus is on subgroups whichact without fixed points (invariant hexagons) so let S be the group of all elementsin G hex which keep the vertex o fixed, and let Z be its subgroup generated by the120 ◦ -rotation.The group G hex has a free abelian subgroup D = Γ( L hex ) ∼ = Z of rank 2 which isgenerated by three translations (vectors) t x , t y , t y satisfying the condition t x + t y + t z = 0.The associated group ring P = Z [ D ] is isomorphic to the ring Z [ x, y ; x − , y − ] of Lau-rent polynomials in two variables. For our purposes a more convenient representationis P = Z [ x, y, z ] / h xyz − i (Figure 4) where variables x, y, z correspond to vectors t x = −→ bc , t y = −→ ca, t z = −→ ab .Figure 4: The hexagonal tiling group A hex as a module over P = Z [ x, y, z ] / h xyz − i .Let A hex be the (infinite dimensional) free abelian group generated by all elementaryhexagonal cells of the lattice L hex . The group A hex is a finitely generated module overthe ring P , indeed it is generated by the three neighboring cells a, b, c with commonvertex O , depicted in Figure 4.For added clarity from here on the lattice L hex is represented by its dual lattice L ◦ hex of barycenters of all hexagons (the black dots in Figure 5). Consequently the ‘lattice’ L ◦ hex is a geometric object (a periodic set of points).The lattice (discrete subgroup of Z ) D , generated by vectors t x , t y , t z is in thisfigure represented by white dots. The fact that A hex is a module over P is simply areformulation of the fact that the lattice of white dots acts on the set of black dots.The lattice D is sometimes (Section 3.4) referred to as the xyz -lattice. The lattice(discrete group) E generated by black dots is referred to as the abc -lattice since it is7igure 5: The lattice of black dots as a module over the lattice of white dots.generated by vectors t a , t b , t c , where t a + t b + t c = 0 (Figure 5). The group ring of E is Q = Z [ a, b, c ] / h abc − i .Note that E has three types of points (Figure 5 on the right) which reflects the factthat the ‘white dot lattice’ D is a sublattice of E of index 3. S and Z on Z [ x, y, z ] Here we collect some basic facts about the symmetric group S and the cyclic group Z actions on Z [ x, y, z ] induced by the permutations of variables x, y, z . As usualfor a given G -module M the associated submodule of G -invariant elements is M G .Elementary symmetric polynomials are σ = x + y + z, σ = xy + yz + zx, σ = xyz .The S -invariant polynomials in Z [ x, y, z ] which form a Z -basis are σ p = x p y p z p (where p ≥ x p y p z q ) = x p y p z q + y p z p x q + z p x p y q (for p = q ), and for p = q = r = p , H ( x p y q z r ) = x p y q z r + y p z q x r + z p x q y r + y p x q z r + x p z q y r + z p y q x r . (3)Basic Z -invariant polynomials in Z [ x, y, z ] are, x p y p z p (where p ≥
0) and ∆( x p y q z r ) = x p y q z r + y p z q x r + z p x q y r , (4)where ( p, q, r ) = ( p, p, p ). There is an involution I on the set Z [ x, y, z ] Z of Z -invariantpolynomials defined by I ( p ( x, y, z )) = p ( y, x, z ). The map α : Z [ x, y, z ] S → Z [ x, y, z ] Z is a monomorphism and the image Im( α ) is the fixed point set of the involution I .More explicitly, α ( x p y p z p ) = x p y p z p , α (∆( x p y p z q )) = ∆( x p y p z q ) and α ( H ( x p y q z r )) = ∆( x p y q z r ) + I (∆( x p y q z r )) . (5)From here we deduce the following proposition.8 roposition 6. There is a commutative diagram / / h xyz − i S α ′ (cid:15) (cid:15) / / Z [ x, y, z ] S α (cid:15) (cid:15) / / Z [ σ , σ ] / / α ′′ (cid:15) (cid:15) / / h xyz − i Z / / Z [ x, y, z ] Z / / ( Z [ x, y, z ] / h xyz − i ) Z / / where h xyz − i ⊂ Z [ x, y, z ] is the principal ideal generated by xyz − , with the splithorizontal exact sequences and injective vertical homomorphisms α, α ′ and α ′′ . Proof:
Since Z [ x, y, z ] S = Z [ σ , σ , σ ] and h xyz − i S = Z [ σ , σ , σ ]( σ − S -invariant and Z -invariant polynomials in Z [ x, y, z ] ((3) and (4)) allows to describe in a similar fashion invariant polynomialsin the ideal (submodule) h xyz − i = h σ − i . For example the basic S -invariantpolynomials in h σ − i S are, σ p ( σ − , ∆( x p y p z q )( σ − , H ( x p y q z r )( σ − . (7)Similarly, the basic Z -invariant polynomials in h σ − i Z areFigure 6: 3 d -representation of invariant polynomials (Proposition 6) σ p ( σ −
1) and ∆( x p y p z q )( σ − . (8)There is an exact sequence of Z [ Z ]-modules,0 → h xyz − i −→ Z [ x, y, z ] −→ Z [ x, y, z ] / h xyz − i → h xyz − i is as a Z -submodule of Z [ x, y, z ] freely generated by binomials x p y q z r ( xyz −
1) = x p +1 y q +1 z r +1 − x p y q z r . This binomial is Z -invariant if and only if9 = q = r . From here we easily deduce the structure of h xyz − i as a Z [ Z ]-module, inparticular we observe that there is a decomposition h xyz − i ∼ = T ⊕ F of Z [ Z ]-moduleswhere T is a trivial and F a free Z [ Z ]-module.It follows that H ( Z ; h xyz − i ) ∼ = 0 and from the long exact sequence of cohomol-ogy we obtain the exactness of the second row of (6),0 → h σ − i Z −→ Z [ x, y, z ] Z −→ ( Z [ x, y, z ] / h xyz − i ) Z → . (10)In particular, ( Z [ x, y, z ] / h xyz − i ) Z ∼ = Z [ x, y, z ] Z / h xyz − i Z . (11)The injectivity of α and α ′ follows from (5). In order to establish the injectivity of α ′′ we observe that (in light of (4) and (8)) Z [ x, y, z ] Z / h xyz − i Z is isomorphic to thesubmodule of Z [ x, y, z ] Z generated by 1 = x y z , ∆( x p ) = x p + y p + z p and ∆( x p y q ) = x p y q + y p z q + z p x q (where ( p, q ) = (0 , Z [ σ , σ ] ∼ = Z [ x, y, z ] S / h xyz − i S is generated by 1 = x y z , ∆( x p ) , ∆( x p y p )and H ( x p y q ). For example one of these H -polynomials (or hexagons) is depicted inFigure 6. From here we observe that α ′′ is injective since it satisfies analogues offormulas (5). (cid:3) P Z of Z -invariant polynomials We begin the analysis of Z -invariant hexagonal tilings by describing the structure ofthe ring P Z = ( Z [ x, y, z ] / h xyz − i ) Z of Z -invariant polynomials.The ring P is as a Z -module freely generated by monomials x p y q z r which are notdivisible by xyz , that is monomials x p y q z r which satisfy at least one of conditions p = 0 , q = 0 , r = 0. Each of these monomials is associated a white dot in the XY Z -coordinate system (Figure 7).The ring P Z of Z -invariant polynomials is as a free Z -module generated by poly-nomials (‘triangles’) ∆( x p y q ) = x p y q + y p z q + z p x q , where either p > q >
0, andthe constant monomial 1 = x y z . Note that ∆( x p y q ) is the Z -symmetrized versionof x p y q ; we also write ∆(1) = 3 = 1 + 1 + 1. Moreover, x p y q is the leading monomialof ∆( x p y q ) in a monomial order such that x > y > z . Consequently, ∆( y k ) = ∆( z k ) isalmost always recorded as ∆( x k ). Lemma 7.
The following identities hold in the ring P = Z [ x, y, z ] / h xyz − i . ∆( xy )∆( x p − y q − ) = ∆( x p y q ) + ∆( x p − y q − ) + ∆( x p − y q − ) ( p ≥ , q ≥
2) (12)∆( x )∆( x p − y q ) = ∆( x p y q ) + ∆( x p − y q +1 ) + ∆( x p − y q − ) for p ≥ q ≥ x )∆( x p y q − ) = ∆( x p y q ) + ∆( x p +1 y q − ) + ∆( x p − y q − ) for p ≥ q ≥ x )∆( x p − ) = ∆( x p ) + ∆( x p − y ) + ∆( xy p − ) for p ≥ xy )∆( x p − ) = ∆( x p y ) + ∆( x p − ) + ∆( xy p ) for p ≥ . (16)10igure 7: Polynomial ∆( x y ) = x y + y z + z y as one of basic ∆-polynomials whichgenerate (over Z ) all Z -invariant polynomials in P .By the injectivity of the map α ′′ (Proposition 6) the ring Z [ σ , σ ] can be seen asa subring of the ring P = Z [ x, y, z ] / h xyz − i , so both P and P Z are modules over Z [ σ , σ ]. Theorem 8.
The ring P Z = ( Z [ x, y, z ] / h xyz − i ) Z of Z -invariant polynomials isisomorphic ( as a module over P σ = Z [ σ , σ ]) to the free module P σ · ⊕ P σ · θ of ranktwo where θ = ∆( x y ) ( Figure 7 ) . Moreover, Θ = Θ( σ , σ , θ ) := θ − ( σ σ − θ + ( σ + σ − σ σ + 9) = 0 (17) so there is an isomorphism of rings, ( Z [ x, y, z ] / h xyz − i ) Z ∼ = Z [ σ , σ , θ ] / h Θ i (18) where h Θ i is the principal ideal in Z [ σ , σ , θ ] generated by Θ . Proof:
By definition ∆( x ) = σ , ∆( xy ) = σ , θ = ∆( x y ) and θ ′ = ∆( xy ) . Let usshow that the ring P Z of Z -invariant polynomials in P is generated as a Z [ σ , σ ]-module by 1 and θ . • If p ≥ q ≥ x p y q )can be expressed in terms of lexicographically smaller ∆-polynomials, multipliedby elements of Z [ σ , σ ]. • If p ≥ q = 1 then by Lemma 7 (equation (13)) the polynomial ∆( x p y ) canbe also reduced to lexicographically smaller ∆-polynomials. • If p ≥ x p ) is by Lemma 7 (equation (15)) reducible to lexicographicallysmaller ∆-polynomials.It follows that all ∆-polynomials can be expressed in terms of θ, θ ′ , σ , σ and 1. Since θ + θ ′ = σ σ − P Z = Z [ σ , σ ] · Z [ σ , σ ] · θ. (19)11he sign “+” in formula (19) can be replaced by “ ⊕ ”. Indeed, if P + Qθ = 0 for some P, Q ∈ Z [ σ , σ ] then (by interchanging variables x and y ) we have P + Qθ ′ = 0 whichis possible only if P = Q = 0.By (19) θ = P + Qθ for some P, Q ∈ Z [ σ , σ ] which by direct calculation leads toequation (17) and the isomorphism (18). (cid:3) abc -lattice E and the xyz -lattice D The ‘white dot’ lattice or the xyz -lattice D (Section 3.1) is generated by vectors (trans-lations) t x , t y , t z (Figures 4 and 5). It is convenient to introduce the abc -lattice E asthe lattice generated by vectors t a , t b , t c (Figures 8 and 9).The lattice D is a sublattice of E of index 3. The set of black dots (Figures 5, 8and 9) is clearly one of the cosets of the quotient lattice E/D ∼ = Z .The fact that E/D ∼ = Z explains why there are three types of dots in these images.In order to avoid clutter we will in subsequent sections continue to draw only blackand white dots, however one should keep in mind the whole of the background lattice E and the presence of ‘invisible’ dots (circled asterisks in Figures 8 and 9).Figure 8: The lattice E has three types of dots, black, white, and asterisks in circles.Let ι : E → Z / Z be the homomorphism which sends the generators a, b, c to 1 ∈ Z .Then D = E = ι − (0) is the set of white dots and E = ι − (1) is the set of black dots. abc -ring Q and the xyz -ring P Let Q = Z [ a, b, c ] / h abc − i be the semigroup ring of the lattice E and let Q Z =( Z [ a, b, c ] / h abc − i ) Z be the associated ring of Z -invariant polynomials. All structureresults that apply to the ring P = Z [ x, y, z ] / h xyz − i apply to the ring Q as well. Inparticular there is an isomorphism (Theorem 8),( Z [ a, b, c ] / h abc − i ) Z ∼ = Z [ s , s , t ] / h Θ i (20)where Θ = Θ( s , s , t ) is the polynomial described in equation (17) and s = a + b + c s = ab + bc + ca t = ∆( a b ) = a b + b c + c a. (21)12igure 9: Three types of dots correspond to the cosets of the sublattice D ⊂ E .The homomorphism ι : E → Z / Z introduced in Section 3.4 allows us to define a Z -grading in the rings Q and Q Z . Indeed, the monomial m = a p b q c r ∈ Q is gradedby its ‘degree mod 3’ i.e. the mod 3 class of deg( m ) = p + q + r .The xyz -ring P , as the group ring of the lattice D of white dots, corresponds to theelements in the abc -ring Q graded by 0. Indeed, this follows from the fact (Figure 9)that x = ac , y = ba , z = cb .Moreover, the P -submodule of Q of elements graded by 1 (monomials with thedegree congruent to 1 mod 3) is precisely the submodule generated by the monomialsassociated with the black dots.Recall that the P -module generated by black dots is precisely the module A hex from Section 3.1. This observation allows us to reduce the “submodule membershipproblem” in the P -module A hex to the corresponding “ideal membership problem” inthe ring Q . We are primarily interested in Z -invariant polynomials so in the followingsection we show how the similar “submodule membership problem” in the P Z -module A Z hex can be reduced to the corresponding “ideal membership problem” in the ring Q Z (Proposition 9). Q Z and A Z hex as a P Z -module The free abelian group A hex , generated by all elementary 2-cells of the hexagonal lattice L hex (or equivalently all 0-dimensional cells of its dual lattice L ◦ hex ), is a module overthe ring P = Z [ x, y, z ] / h xyz − i .∆-polynomials already appeared in the description of the ring P Z of Z -invariantpolynomials in P (Figure 7). Following the idea of the ‘Newton polygon construction’,a polynomial ∆( x p y q ) = x p y q + y p z q + z p y q is visualized as a triangle with vertices inthe ‘white dot’-lattice, invariant with respect to the action of group Z .Similarly the ∆-polynomials ∆( a p b q ) = a p b q + b p c q + c p a q are, together with 1 = a b c , Z -generators of the ring Q Z . An immediate consequence is that P Z is a subring13f Q Z .Finally, the ‘black dot’ ∆-polynomials ∆( x p y q a ) = x p y q a + y p z q b + z p x q c (Figure 10)form a Z -basis of the group A Z hex of Z -invariant elements of A hex .The ring Q Z inherits the Z -gradation from the ring Q . Moreover P Z is preciselythe subset of all elements graded by 0 ∈ Z while A Z hex is generated by ‘black dottriangles’ which are precisely the triangles graded by 1 ∈ Z . This characterization isa basis of the following fundamental proposition. Proposition 9.
Let K ⊂ A Z hex be a P Z -submodule of A Z hex . Let I K ⊂ Q Z be the idealin Q Z generated by K . Suppose that p ∈ A Z hex . Then, p ∈ K ⇐⇒ p ∈ I K . (22) In other words the ‘submodule membership problem’ is reduced to the ‘ideal membershipproblem’ in the ring Q Z . Proof:
The implication p ∈ K ⇒ p ∈ I K is clear. The opposite implication is equallyeasy. Indeed, if p = α p + . . . + α k p k for some elements p i ∈ K and homogeneous (inthe sense of the Z -gradation) elements α i ∈ Q Z then we can assume that all α i ∈ P Z (the other terms cancel out). (cid:3) Figure 10: ∆-polynomials in the ‘black dot lattice’ generate the group A Z hex .14 .7 Submodule of A Z hex generated by tribones A three-in-line hexagonal polyomino or a tribone is a translate of one of the followingthree types (Figure 10), T x = x − + 1 + x = ab + 1 + ac T y = y − + 1 + y = bc + 1 + ba T z = z − + 1 + z = ca + 1 + cb (23)If A = x p y q a is a ‘black dot’ in the angle XOY then the three basic tribones centeredat the point A are T x ( A ) = x p y q aT x , T y ( A ) = x p y q aT y , T z ( A ) = x p y q aT z . (24)For example, T x ( a ) = ( x − + 1 + x ) a T y ( a ) = ( y − + 1 + y ) a T z ( a ) = ( z − + 1 + z ) a (25)The Z -symmetric triplets of tribones, associated to tribones (24), are∆( T x ( A )) , ∆( T y ( A )) , ∆( T z ( A )) (26)where for example,∆( T y ( A )) = ∆( x p y q aT y ) = ∆( x p y q − a ) + ∆( x p y q a ) + ∆( x p y q +1 a ) . (27) Theorem 10.
The submodule K trib ⊂ A Z hex of Z -invariant polyominoes (polyhexes)which admit a signed, symmetric tiling by tribones is generated, as a module over P Z ,by the Z -symmetric triplets of tribones, ∆( T x ( a )) , ∆( T y ( a )) , ∆( T z ( a )) , ∆( T x ( ax )) , ∆( T y ( ax )) , ∆( T z ( ax )) . (28) Proof:
Let A = M a = x p y q a ∈ A hex be a ‘black dot’ in the angle ∠ XOY where M = x p y q ∈ P is the corresponding ‘white dot’ (Figure 11). The three tribonescentered at A are M aT x , M aT y , M aT z . We want to show that each of the associated Z -symmetric triplets ∆( M aT x ) , ∆( M aT y ) , ∆( M aT z ) is in the P Z -module generatedby the six elements listed in the equation (28).Figure 11 depicts the case where M = x y and the chosen tribone is M aT y = x y ( y − + 1 + y ) a . Using this particular example we describe a general reductionprocedure which allows us to express the triplets ∆( M aT x ) , ∆( M aT y ) , ∆( M aT z ) bytriplets strictly closer to the origin O .Let ∆( x ) = x + y + z = σ and ∆( xy ) = xy + yz + zx = σ be the two basic ‘whitedot’ triangles (∆-polynomials) depicted at the center of Figure 11. By translating thesetriangles we surround the point A by six triangles forming a regular hexagon.Let us assume that the origin O is not contained in the interior of this hexagon. Itfollows that one of the six triangles (the shaded triangle ABC in Figure 11) has the15igure 11: ∆-polynomials and the structure of the P Z -module of Z -invariant tiles.property that the segment OA intersects the side BC opposite to A . (Observe thatthe origin O can be on the segment BC and this happens precisely if A ∈ { ya, xya } .)An immediate consequence is that the lengths of all segments OB, OC, OX arestrictly smaller than the length of OA . Indeed, by construction ∠ ACO ≥ ◦ ≥ ∠ CAO , and at least one of these inequalities is strict. In our case the triangle
ABC isa translate of ∆( xy ) (otherwise we would use ∆( x )). Since ∆( xy ) X = A + B + C weobserve that∆( xy )( y − + 1 + y ) X = ( y − + 1 + y ) A + ( y − + 1 + y ) B + ( y − + 1 + y ) C (29)Let ∆ Y = ∆(( y − + 1 + y ) Y ). By symmetrizing the equality (29) and adding (that isby applying the ∆-operator on both sides of (29)) we finally obtain,∆( xy )∆ X = ∆ A + ∆ B + ∆ C . (30)Summarizing, we see that each Z -symmetric triplet of tribones ∆ A can be expressedin terms of triplets closer to the origin, provided the hexagon associated to A does notcontain the origin O in its interior.The only remaining possibilities are A ′ = a and A ′′ = xa which accounts for the six Z -symmetric triplets of tribones listed in the theorem. (cid:3) .8 The ideal I K trib In this section we express the generating polynomials for the ideal I K trib ⊂ Q Z (listedin (28)) in terms of variables s , s , t which appear in the description of the ambientring Q Z (equation (20) in Section 3.5)). Proposition 11. ∆( T x ( a )) = − s + 2 s ∆( T y ( a )) = 3 s − s + s t ∆( T z ( a )) = s s − s − s t ∆( T x ( ax )) = − s s + 2 s − s t + s t ∆( T y ( ax )) = − s + s s − s ∆( T z ( ax )) = 3 s − s s − s + s s + s t − s t (31) Proof:
The proof is by direct calculations which follow the algorithm described in theproof of Theorem 8. For example∆( T x ( a )) = ∆( a b + a + a c ) = 2∆( a b ) + ∆( a ) = 2 s − s + s = 2 s − s . Similarly since∆( a b ) = ∆( a )∆( a b ) − ∆( a b ) − ∆( a ) = s t − ( s − s ) − s = s t − s + s we deduce that,∆( T y ( a )) = ∆( c + a + a b ) = 2 s + ∆( a b ) = s t − s + 3 s . Let I ⊂ Z [ x, y ] be the ideal, I = h x + x , y + y , xy + ( xy ) i . (32)If p − q ∈ I we say that p and q are congruent mod I and write p ≡ I q . In thissection we collect some elementary congruences mod I which are needed for subsequentcalculations. Lemma 12. If m − n is divisible by then, x m ≡ I x n y m ≡ I y n ( xy ) m ≡ I ( xy ) n . (33) Lemma 13. L k := 1 + x + . . . x k − ≡ I if k ≡ mod if k ≡ mod
31 + x if k ≡ mod . (34)17 emma 14. (cid:3) k := (1 + x + . . . x k − )(1 + y + . . . y k − ) ≡ I if k ≡ mod if k ≡ mod x y if k ≡ mod . (35)Figure 12: Pictorial proof of the relation ∆ n − ≡ I n ∆ .The Newton polygon of the polynomial (cid:3) k is the square with vertices { (0 , , (0 , k − , ( k − , , ( k − , k − } which has precisely k integer points on each of its sides.Similarly the polynomial∆ k = L k + xyL k − + ( xy ) L k − + . . . + ( xy ) k − L (36)collects all monomials associated to the integer points in the triangle with the vertices { (0 , , ( k − , , ( k − , k − } (∆ is depicted in Figure 12). Lemma 15. ∆ n − ≡ I n ∆ ≡ I n (1 + x + xy )∆ n ≡ I ∆ n − ≡ I n ∆ ∆ n +1 ≡ I n ∆ + 1 (37) ∆ -polynomial of a triangular region We use the calculations from Section 4.1 to determine the ∆-polynomial of a triangularregion T N depicted in Figure 13, where N = 3 k − T N areblack dots which are associated the monomials, ax k − y k − by k − z k − cz k − x k − . It is sufficient to determine the ≡ I class of the polynomial A k described as the sum of allmonomials in T N which belong to the cone xOy . By inspection we see that A k = B k + C k where B k corresponds to the parallelogram with vertices a, ax k − , ay k − , a ( xy ) k − and18igure 13: Decomposition of T N into blocks. C k is the polynomial associated to the triangle with vertices axy k , ax k − y k , ax k − y k − .Since B k = a · (cid:3) k and C k = axy k ∆ k − we can use Lemmas 12, 14 and 15 to evaluate A k . Proposition 16. A k = dax ∆ if k = 3 da + daxy ∆ if k = 3 d + 1 ax y + axy ( d ∆ + 1) if k = 3 d + 2 . (38) Proof: B k = a · (cid:3) k ≡ I k = 3 da if k = 3 d + 1 ax y if k = 3 d + 2 . (39)Since, axy k ≡ I ax if k = 3 daxy if k = 3 d + 1 axy if k = 3 d + 2 , (40) C k = axy k ∆ k − ≡ I dax ∆ if k = 3 ddaxy ∆ if k = 3 d + 1 axy ( d ∆ + 1) if k = 3 d + 2 . (41)Since A k = B k + C k the result follows by adding equations (39) and (41). (cid:3) Proposition 17.
The ∆ -polynomial of A k is equal to ∆( A k ) = P + dQ where, ( k = 3 d ) P = 0 Q = 3 s − s s + s s k = 3 d + 1) P = s Q = 9 s − s s + s s + 4 s t − s s t + s t ( k = 3 d + 2) P = 11 s + s − s s + 5 s + s s − s s + 4 s t − s s t + s t + s t Q = 24 s + s − s s + s s − s s + 4 s s + 8 s t − s t − s s t + 3 s t . Proof:
It follows from Proposition 16 that, A k = d ( a c + a c + a c ) if k = 3 da + d ( a c + a c + a c ) if k = 3 d + 1( a c + a ) + d ( a + a c + a c ) if k = 3 d + 2 . (42)The rest of the proof is by direct calculation, by hand or preferably by a computeralgebra system. (cid:3) K trib ⊂ A Z hex We want to test if the polynomials ∆( A k ) described in Proposition 17 belong (fordifferent values of k ) to the submodule K trib described in Theorem 10 (Section 3.7).In light of Proposition 9 this question is reduced to the ‘ideal membership problem’for the associated ideal I K trib in the ring Q Z ∼ = Z [ s , s , t ] / h Θ i and in turn to the ‘idealmembership problem’ for the ideal J K trib := I K trib ∪ { Θ } ⊂ Z [ s , s , t ] . (43)Here Θ is the polynomial defined in Theorem 8 (equation 18) and again, in the contextof the abc -ring Q Z , in Section 3.5.In this section we determine the Groebner basis for the ideal J K trib . Proposition 18.
The Groebner basis G = G trib of the ideal J K trib ⊂ Z [ s , s , t ] withrespect to the lexicographic order of variables s , s , t is given by the following list ofpolynomials:
27 + 9 t + 3 t −
27 + t s + 3 s t + s t s s t s s + s s + s ts s s s s + s + 3 t + t (44)As the first application of Proposition 18 we calculate the remainders P G and Q G ofpolynomials P and Q introduced in Proposition 17 on division by the Groebner basis G = G trib . Proposition 19.
Let P i and Q i be the polynomials such that to ∆( A d + i ) = P i + dQ i (Proposition 17). Then the remainders of these polynomials on division by the Groebnerbasis G = G trib are the following: P G = 0 Q G = − s P G = s Q G = − s P G = − s Q G = − s (45)20 Main results
Theorem 20.
Let T N = T k − be the Z -symmetric triangular region in the hexagonallattice depicted in Figure 13 where N is the number of black dots (hexagons) on theedge of the triangle. Then T N admits a Z -symmetric, signed tiling by three-in-linepolyominoes (tribones) if and only if k = 9 r for some integer r . The first such triangleis T , in particular the triangle T shown in Figure 1 does not have a Z -symmetric,signed tiling by tribones. Proof:
By Proposition 17 the polynomial ∆( A k ), equal to the sum of all monomialscovered by the triangular region T N = T k − , can be expressed (as a polynomial invariables s , s , t ) as the sum ∆( A k ) = P + dQ . More explicitly (taking into accountthe different cases of Proposition 17) we write ∆( A d + i ) = P i + dQ i where i ∈ { , , } .By Proposition 19 remainders of these polynomials on division by the Groebnerbasis G = G trib are displayed in the table (45).The leading terms of the Groebner basis (44) are the following,3 t t s t s s t s s s ts s s s s (46)By inspection we see that the polynomial P i + dQ i can be reduced to zero if andonly i = 0 and d = 3 r for some integer r , or in other words if and only if k = 3 d = 9 r ( N = 27 r − (cid:3) N = 3 k The triangular region T N in the case N = 3 k + 1 has a fixed point (black dot) withrespect the Z action so we focus on the remaining case N = 3 k .In this case the convex hull of the set of all black dots in the intersection of T k with theangle xOy is the trapeze with vertices associated the monomials a, ax k , ay k − , ax k − y k − .We denote the sum of all these monomials by A k . This trapeze is divided into a rhom-bus, with vertices at a, ax k − , ay k − , a ( xy ) k − and an equilateral triangle with verticesat ax k , ax k y k − , ax k − y k − . The sums of monomials in these two regions are respec-tively B k and C k , so by definition A k = B k + C k .The mod I class of the polynomial B k = a · (cid:3) k is as in the case N = 3 k − C k = ax k ∇ k where ∇ k is the sum of all monomialsin the triangle with vertices at 1 , y k − , ( xy ) k − . This triangle is obtained from ∆ k byinterchanging variables x and y so the following lemma is an immediate consequenceof Lemma 15. Lemma 21. ∇ n − ≡ I n ∇ ≡ I n (1 + x + xy ) ∇ n ≡ I ∇ n − ≡ I n ∇ ∇ n +1 ≡ I n ∇ + 1 (47)21imilarly, ax k ≡ I a if k = 3 dax if k = 3 d + 1 ax if k = 3 d − , (48) C k = ax k ∇ k ≡ I da ∇ if k = 3 dax ( d ∇ + 1) if k = 3 d + 1 dax ∇ if k = 3 d − . (49)and finally, Proposition 22. A k = da ∇ = d ( a + a b + a c ) k = 3 da + ax ( d ∇ + 1) = ( a + a c ) + d ( a c + a c + a c ) k = 3 d + 1 ax y + dax ∆ = a c + d ( a c + a c + a c ) k = 3 d − . (50)The associated ∆-polynomials are given by the following proposition. Proposition 23.
The ∆ -polynomial of A k is equal to ∆( A k ) = P + dQ where, ( k = 3 d ) P = 0 Q = s s − s ( k = 3 d + 1) P = − s + s Q = − s s − s + s s − s t ( k = 3 d − P = 7 s − s s + 3 s + s s − s s + 4 s t − s s t + s t + s t Q = 2 s s + 4 s − s s + s . Proposition 24.
Let P i and Q i be the polynomials such that to ∆( A d + i ) = P i + dQ i (Proposition 23). Then the remainders of these polynomials on division by the Groebnerbasis G = G trib are the following: P G = 0 Q G = s P G = − s + s Q G = s P − G = s Q − G = s (51)The analysis similar to the proof of Theorem 20 leads to the following result. Theorem 25.
Let T N = T k be the Z -symmetric triangular region in the hexagonallattice which has k hexagons on one of its sides. Then T N admits a Z -symmetric,signed tiling by three-in-line polyominoes (tribones) if and only if k = 9 r for someinteger r . The first such triangle is T . .2 Examples illustrating Theorems 20 and 25 Here we describe explicit Z -invariant tilings predicted by Theorems 20 and 25.We begin with the case N = 3 k − r −
1. Figure 13 describes a decompositionof the triangular region T N into triangular and rhombic blocks. The side length of thetriangle with the vertices axy k , ax k − y k , ax k − y k − is k − r − a, ax k − , ay k − , a ( xy ) k − is k = 9 r so it can be paved by tribones. This signedtiling is extended to the whole region T N by rotations through the angle of 120 ◦ and240 ◦ .The case N = 3 k = 27 r is treated similarly. The intersection of T k with the angle xOy is the trapeze with vertices at a, ax k , ay k − , ax k − y k − (Section 5.1). This trapezeis divided into a rhombus, with vertices at a, ax k − , ay k − , a ( xy ) k − and an equilateraltriangle with vertices at ax k , ax k y k − , ax k − y k − . Both figures admit a (signed) tribonetiling. Indeed, the triangle has the side length k = 9 r (and the Theorem 1 applies)while the side length of the rhombus is as before k = 9 r . Our main objective was to illustrate the method of Gr¨obner bases in the case of equiv-ariant tribone tilings of the hexagonal lattice. However the method has many otheradvantages and some of them were mentioned already in the original paper of Bodiniand Nouvel [4].Perhaps the main reason why this method is so well adapted for lattice tiling prob-lem is its close connection with the already developed methods and tools used in latticegeometry.
Following [1] and [2] we define the integer-point transform of a finite subset K ⊂ N asthe polynomial f K = P { x α y β | ( α, β ) ∈ N } .Brion’s formula, see [1, Chapter 8] and [2, Section 9.3], is a versatile tool for evaluat-ing the integer-point transform of convex polytopes. It allows us to replace a polynomialwith a large number of monomials by a very short expression involving only rationalfunctions. It was tacitly used throughout the paper for an independent checking ofsome of the formulas. More systematic application of these ideas will be presented in[10]. Let P be a convex polytope with vertices in N d and let f P be its integer-point transform.The ‘discrete volume’ of Q , defined as the number of integer points inside P , can be23valuated as the remainder of f P on division by the ideal I = h x − , x − , . . . , x d − i . Let J ⊂ Z [ x , . . . , x d ] be an ideal, say the ideal associated to a set R of prototiles in N d . Let G = G J be the Gr¨obner basis of J with respect to some term order. It maybe tempting to ask (at least for some carefully chosen ideals J ) what is the geomet-ric and combinatorial significance of the remainder f GQ of the integer-point transformpolynomial f Q on division by the Gr¨obner basis G . Acknowledgements:
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