Gröbner Bases and Tilingsfernique/isfahan/hashemi...w’al-muqabala” gives us the word algebra and can be considered as the ﬁrst book to be written on algebra. Amir Hashemi Gro¨bner

IntroductionGrobner bases

Computation of Grobner BasesApplications

Grobner Bases over Integers

Grobner Bases and Tilings

Amir Hashemi

Isfahan University of Technology & IPM

CIMPA-IRAN school on Tilings and Tessellations, Isfahan, Iran, 2015

Amir Hashemi Grobner Bases and Tilings




Abu Abdallah Muhammed Ibn Mussa

Al−Khwarizmi

Al-Khwarizmi, Persian mathematician,

was born in Khwarizm (Kheva) around

the year 770, in a village south of the

Oxus River in a region that is now named

the Uzbekistan (Persian empire, the former

Iran). He died around the year 840. He

is known for having introduced the concept

of mathematical algorithm. The word

algorithm derives from his name. His algebra treatise “Hisab al-jabr

w’al-muqabala” gives us the word algebra and can be considered as the

first book to be written on algebra.





Outline of talk

1 Introduction

2 Grobner basesMonomial OrderingsGrobner Bases

3 Computation of Grobner BasesHistory of Grobner BasesMathematical Softwares

4 ApplicationsElimination TheorySolving Polynomial Systems

5 Grobner Bases over IntegersApplication to Tilings





Notations

✄ K; a field e.g. K = R,Q, . . .

✄ x1, . . . , xn; a sequence of variables

✄ A polynomial is a sum of products of numbers andvariables, e.g.

f = x1x2 + 12x1 − x32

✄ R = K[x1, . . . , xn]; set of all polynomials

✄ f1, . . . , fk ∈ R and F = {f1, . . . , fk}

✄ I = 〈F 〉 = {p1f1 + · · ·+ pkfk | pi ∈ R}





Notations (cont.)

The variety of a system: Consider F = {f1, . . . , fk}

✄

f1 = 0...

fk = 0.

✄ V(f1, . . . , fk) = V(F ) : The solutions of the system

✄ E.g.

x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1

✄ V(f1, f2, f3) = {(1,√2, 1), . . .} where

f1 = x2 + y2 + z2 − 4, · · ·





What is Algebraic Geometry?

Studying the solutions of a polynomial systemi.e. computing V(F ) using algebraic tools:

✄ Algebra✄ Geometry✄ Computer Science (Computational Algebraic Geometry)





What is Algebraic Geometry?

Studying the solutions of a polynomial systemi.e. computing V(F ) using algebraic tools:

✄ Algebra✄ Geometry✄ Computer Science (Computational Algebraic Geometry)

Example: I = 〈x2 − y2 − z, z − 2〉 ⊂ R[x, y, z]





Variables versus Unknowns

If we consider the equation x2 − x+ 1 = 0; i.e. we look forits zeros then x is called an unknown







If we consider the polynomial x2 − x+ 1 then x is called avariable








So, the polynomial x2 − x+ 1 is never 0.









For example if we consider the polynomial ax2 − bx+ c = 0 inx then a = b = c = 0.









For example if we consider the polynomial ax2 − bx+ c = 0 inx then a = b = c = 0.

Equation ❀ UnknownPolynomial ❀ V ariable





Univariate Polynomial Ring

Let K be a field and K[x] the ring of polynomials in x







If f1, . . . , fk ∈ K[x] then 〈f1, . . . , fk〉 = 〈gcd(f1, . . . , fk)〉







If f1, . . . , fk ∈ K[x] then 〈f1, . . . , fk〉 = 〈gcd(f1, . . . , fk)〉K[x] is a PID, e.g. 〈x− 1, x2 − 1〉 = 〈x− 1〉







If f1, . . . , fk ∈ K[x] then 〈f1, . . . , fk〉 = 〈gcd(f1, . . . , fk)〉K[x] is a PID, e.g. 〈x− 1, x2 − 1〉 = 〈x− 1〉Thus, gcd computations (using Euclid algorithm) can solvemany problems in K[x]








Suppose thatf := x3 − 6x2 + 11x− 6

g := x3 − 10x2 + 29x− 20









g := x3 − 10x2 + 29x− 20

Since gcd(f, g) = x− 1 then V(f, g) = V(〈f, g〉) = V(x− 1)









g := x3 − 10x2 + 29x− 20


The only solution of the system f = g = 0 is x = 1









g := x3 − 10x2 + 29x− 20


The only solution of the system f = g = 0 is x = 1

Membership Problem: x2 − 1 ∈ 〈f, g〉 because x− 1 | x2 − 1.





Two Questions

☞ R = K[x1, . . . , xn]; a multivariate polynomial ring☞ F ⊂ R; a finite set of polynomials☞ I ⊂ R; an ideal

Membership problem: h ∈ I?

Solving polynomial systems: V(F )?

In practice, the answer to these questions is not easy!





Two Questions

☞ R = K[x1, . . . , xn]; a multivariate polynomial ring☞ F ⊂ R; a finite set of polynomials☞ I ⊂ R; an ideal

Membership problem: h ∈ I?

Solving polynomial systems: V(F )?

In practice, the answer to these questions is not easy!

Grobner bases can answer them!





Multivariate Polynomial Ring

F ⊂ K[x1, . . . , xn]






F ⊂ K[x1, . . . , xn]

V(F ) doesn’t depend on the generators of 〈F 〉






F ⊂ K[x1, . . . , xn]

V(F ) doesn’t depend on the generators of 〈F 〉f := 2x2 + 3y2 − 11, g := x2 − y2 − 3






F ⊂ K[x1, . . . , xn]


Since 〈f, g〉 = 〈x2 − 4, y2 − 1〉 then V (f, g) = {(±2,±1)}






F ⊂ K[x1, . . . , xn]


Since 〈f, g〉 = 〈x2 − 4, y2 − 1〉 then V (f, g) = {(±2,±1)}So we look for a more convenient generating set for 〈F 〉






F ⊂ K[x1, . . . , xn]


Since 〈f, g〉 = 〈x2 − 4, y2 − 1〉 then V (f, g) = {(±2,±1)}So we look for a more convenient generating set for 〈F 〉To bring it into computer, we need |F | < ∞






F ⊂ K[x1, . . . , xn]


Since 〈f, g〉 = 〈x2 − 4, y2 − 1〉 then V (f, g) = {(±2,±1)}So we look for a more convenient generating set for 〈F 〉To bring it into computer, we need |F | < ∞By Hilbert Basis Theorem there exist f1, . . . , fm ∈ F so that

〈F 〉 = 〈f1, . . . , fm〉





Monomial OrderingsGrobner Bases

Polynomial Ring

☞ K a field

✄ We denote the monomial xα1

1 · · · xαn

n by Xα withα = (α1, . . . , αn)

✄ {monomials in R} ↔ Zn≥0

✄ If Xα is a monomial and a ∈ K, then aXα is a term

✄ A polynomial is a finite sum of terms.

☞ R = K[x1, . . . , xn]; the ring of all polynomials.






Definition

A monomial ordering is a total ordering ≺ on the set of monomialsXα = xα1

1 · · · xαn

n ,α = (α1, . . . , αn) ∈ Zn≥0 such that,

Xα ≺ Xβ ⇒ Xα+γ ≺ Xβ+γ and

1 ≺ Xα for all α

Convention

For a monomial Xα = xα1

1 · · · xαn

n , We call |α| := α1 + · · ·+ αn,the total degree of Xα






Some important monomial orderings

Lex(Lexicographic) Ordering

Xα ≺lex Xβ if leftmost nonzero of α− β is < 0

Degree Reverse Lex Ordering

Xα ≺drl Xβ if

|α| < |β|or

|α| = |β| and rightmost nonzero of α− β is > 0

Examplex ≺ x3 x2y3 ≺lex(x,y) x

3y2 x3y ≺drl(x,y) xy5






Notations

☞ R = K[x1, . . . , xn], f ∈ R

☞ ≺ a monomial ordering on R

☞ I ⊂ R an ideal

LM(f): The greatest monomial (with respect to ≺) in f5x3y2 + 4x2y3 + xy + 1

LC(f): The coefficient of LM(f) in f5x3y2 + 4x2y3 + xy + 1

LT(f): LC(f)LM(f)5x3y2+4x2y3 + xy + 1

LT(I): 〈LT(f) | f ∈ I〉






Definition

✄ I ⊂ K[x1, . . . , xn]

✄ ≺ A monomial ordering

✄ A finite set G ⊂ I is a Grobner Basis for I w.r.t. ≺, if

LT(I) = 〈LT(g) | g ∈ G〉

Existence of Grobner bases

Each ideal has a Grobner basis

Example

I = 〈xy − x, x2 − y〉, y ≺lex xLT(I) = 〈x2, xy, y2〉A Grobner basis is: {xy − x, x2 − y, y2 − y}.






Unicity

Definition

✄ G ⊂ I is a reduced Grobner basis w.r.t. ≺ if

G is a Grobner basis for I w.r.t. ≺∀g ∈ G, LC(g) = 1∀f, g ∈ G, LT(f) ∤ any term of g

Theorem

Every ideal I has a unique reduced Grobner basis w.r.t. ≺





History of Grobner BasesMathematical Softwares

Division Algorithm in K[x1, . . . , xn]

Theorem

Fix a monomial ordering ≺ and let F := (f1, . . . , fk) be an ordereds−tuple of polynomials in K[x1, . . . , xn] Then, everyf ∈ K[x1, . . . , xn] can be written as

f = q1f1 + · · ·+ qkfk + r

where qi, r ∈ K[x1, . . . , xn] and either r = 0 or no term of r isdivisible by any of LT(f1), . . . ,LT(fk). We call r, the remainderon division of f by F .






Algorithm 1 Division Algorithm

Require: f, f1, . . . , fk and ≺Ensure: q1, . . . , qk, rq1 := 0; · · · ; qk := 0;p := f ;while ∃fi s.t. LT(fi) divides a term m in p do

qi := qi +m

LT(fi)

p := p− ( mLT(fi)

)fiend whilereturn q1, . . . , qk, p






Example

Divide f = xy2 + 1 by f1 = xy + 1, f2 = y + 1 and y ≺lex x






Example

Divide f = xy2 + 1 by f1 = xy + 1, f2 = y + 1 and y ≺lex xf → (xy2 + 1)− y(xy + 1) = 1− y → (1− y) + (y + 1) = 2






Example

Divide f = xy2 + 1 by f1 = xy + 1, f2 = y + 1 and y ≺lex xf → (xy2 + 1)− y(xy + 1) = 1− y → (1− y) + (y + 1) = 2So we can write f = y(xy + 1) + (−1)(y + 1) + 2






Example







Example


Example

✄ f = xy2+x2y+y2, f1 = y2−1, f2 = xy−1, y ≺lex x

✄ f = (x+ 1)f1 + xf2 + 2x+ 1

✄ f = (x+ y)f2 + f1 + x+ y + 1

⇒ The remainder is not unique.






Example


Example

✄ f = xy2+x2y+y2, f1 = y2−1, f2 = xy−1, y ≺lex x

✄ f = (x+ 1)f1 + xf2 + 2x+ 1

✄ f = (x+ y)f2 + f1 + x+ y + 1

⇒ The remainder is not unique.

Theorem

{g1, . . . , gk} is a Grobner basis iff for each f ∈ R the remainder off by G is unique.






Buchberger’s Criterion

Definition

S-polynomial

Spoly(f, g) =xγ

LT(f)f − xγ

LT(g)g

xγ = lcm(LM(f),LM(g))

Spoly(x3y2+xy3, xyz−z3) = z(x3y2+xy3)−x2y(xyz−z3) = zxy3+x2yz3

Buchberger’s Criterion

✄ G is a Grobner basis for 〈G〉✄ ∀gi, gj ∈ G, remainder((Spoly(gi, gj), G) = 0






Algorithm 2 Buchberger’s Algorithm

Require: F := (f1, . . . , fs) and ≺Ensure: A Grobner basis for the ideal 〈f1, . . . , fs〉 w.r.t. ≺

G := FB := {{f, g}|f, g ∈ F}while B 6= ∅ do

Select and remove a pair {f, g} from BLet r be the remainder of Spoly(f, g) by Fif r 6= 0 then

B := B ∪ {{h, r} | h ∈ G}G := G ∪ {r}

end ifend whilereturn G






Example

I = 〈f1, f2〉 = 〈xy − x, x2 − y〉 y ≺lex x

G := {f1, f2}

Spoly(f1, f2) = xf1 − yf2 = y2 − x2f2−−−−→ y2 − y = f3

G := {f1, f2, f3}

Spoly(fi, fj)G−−−−→ 0

G := {f1, f2, f3} is a Grobner basis for I






Buchberger, 65 :

Developing the theory of Grobner basesBuchberger criteria

Lazard, 83 :

Using linear algebra

Gebauer, Moller, 88 :

Installing Buchberger criteria

Faugere, 93, 94, 02 :

FGLMF4 algorithm (intensive linear algebra)F5 algorithm

Gao, Volny, 10 :

G2V (with Guan)GVW (with Wang)






Magma

University of Sydney, Australia

Maple

University of Waterloo, Canada

Singular

University of Kaiserslautern, Germany

Macaulay 2

University of cornell, USA

Cocoa

University of Geneva, Italy





Elimination TheorySolving Polynomial Systems

Ideal Membership

Theorem

f ∈ I iff f ❀G 0 where G is a GB of I

Example

✄ I = 〈xy − x, x2 − y〉✄ Is y2 + y ∈ I?






Ideal Membership

Theorem

f ∈ I iff f ❀G 0 where G is a GB of I

Example

✄ I = 〈xy − x, x2 − y〉✄ Is y2 + y ∈ I?

✄ y ≺lex x

✄ The reduced Grobner basis of I isG = {xy − x, x2 − y, y2 − y}

⇒ y2 + y ❀G 2y 6= 0, and thus y2 + y /∈ I.






Ideal Membership (cont.)

Theorem (Weak Hilbert’s Nullstellensatz)

V(F ) = ∅ ⇔ 1 ∈ 〈F 〉 ⇔ 1 ∈ G

Example

✄ Let F be the set{x2+3y+z−1, x−3y2−z2, x−y, y2−zxy−x, x2−y}

✄ I = 〈F 〉✄ z ≺lex y ≺lex x

✄ The reduced Grobner basis of I is = {1}⇒ V(F ) = ∅






Elimination

Theorem

✄ I ⊂ K[x1, . . . , xn]; an ideal

✄ G ⊂ I; a GB of I for xn ≺lex · · · ≺lex x1

⇒ G ∩K[xi, . . . , xn] generates I ∩K[xi, . . . , xn].

Example

✄ I = 〈x2 + y2 + z2 − 4, x2 + 2y2 − 5, xz − 1〉✄ Grobner basis of I w.r.t. z ≺lex x ≺lex y

{1− 3z2 + 2z4,−1− z2 + y2,−3z + 2z3 + x}

⇒ I ∩K[z, x] = 〈1− 3z2 + 2z4,−3z + 2z3 + x〉.






Solving Polynomial Systems

x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1







x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1

✄ I = 〈x2 + y2 + z2 − 4, x2 + 2y2 − 5, xz − 1〉







x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1


{1− 3z2 + 2z4,−1− z2 + y2,−3z + 2z3 + x}







x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1


{1− 3z2 + 2z4,−1− z2 + y2,−3z + 2z3 + x}✄ The zero set corresponding to G is equal to the that

of the initial system.







x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1




3z2 − 2z4 = 1−z2 + y2 = 1

−3z + 2z3 + x = 0







x2 + y2 + z2 = 4x2 + 2y2 = 5

xz = 1




3z2 − 2z4 = 1−z2 + y2 = 1

−3z + 2z3 + x = 0

(1,±√2, 1), (−1,±

√2,−1), (

√2,±

√6�2, 1�

√2), (−

√2,±

√6�2,−1�

√2)






(1,±√2, 1), (−1,±

√2,−1), (

√2,±

√6�2, 1�

√2), (−

√2,±

√6�2,−1�

√2)






Implicitation

Theorem

Consider a rational parametrization:

x1 = f1(t1,...,tm)g1(t1,...,tm)

...

xn = fn(t1,...,tm)gn(t1,...,tm)

✄ I = 〈x1g1 − f1, . . . , xngn − fn, 1− yg1 · · · gn〉✄ Note that I ⊂ K[x1, . . . , xn, t1, . . . , tm, y]

⇒ I ∩K[x1, . . . , xn] provides an implicitation.






Implicitation (cont.)

Example

Consider the folium of Descartes:

{

x1 = 3t1+t3

x2 = 3t2

1+t3

✄ I = 〈x1(1 + t3)− 3t, x2(1 + t3)− 3t2, 1− y(1 + 3t2)〉✄ G := A GB of I w.r.t. x1 ≺lex x2 ≺lex t ≺lex y

⇒ Since x31 + x32 − 3x1x2 ∈ G then

x31 + x32 = 3x1x2

is an implicitation of the curve.






Computing Minimal Polynomial

Definition

Let M be a square matrix over the field K. The monic polynomialm ∈ K[x] where

〈m〉 = 〈f ∈ K[x] | f(M) = 0〉

is called minimal polynomial of M

Example

Minimal polynomial of

2 1 0 00 2 0 00 0 1 10 0 −2 4

is x3 − 7x2 + 16x− 12






Algorithm for Computing Minimal Polynomial

Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0

I := 〈7a2 + a1 + a0, 10a2 + 2a1, 15a2 + 3a1, 22a2 + 4a1 + a0〉







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0

I := 〈7a2 + a1 + a0, 10a2 + 2a1, 15a2 + 3a1, 22a2 + 4a1 + a0〉G := {2a1 − 5a0, 2a2 + a0} GB w.r.t. a0 ≺lex a1 ≺lex a2







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0


m := Remainder(m,G) = −12a0x

2 + 52a0x+ a0







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0



2 + 52a0x+ a0

a0 6= 0, so m := m

− 1

2a0

= x2 − 5x− 2







Input M :=

(

1 23 4

)

,M2 =

(

7 1015 22

)

m := a2x2 + a1x+ a0

a2M2 + a1M + a0I2×2 =

(

7a2 + a1 + a0 10a2 + 2a115a2 + 3a1 22a2 + 4a1 + a0

)

= 0



2 + 52a0x+ a0

a0 6= 0, so m := m

− 1

2a0

= x2 − 5x− 2

⇒ Minimal polynomial of M is x2 − 5x− 2






Algorithm

Algorithm 3 MinPoly

Require: Mn×n

Ensure: Minimal polynomial m of M in xI := 〈∑n

k=0 akMk[i, j] | i, j = 1, . . . , n〉

G := A Grobner basis for I w.r.t. a0 ≺lex . . . ≺lex and :=| G |m := Remainder(

∑d

i=0 aixi, G|ad+1=···=an=0)

m := mLC(m)

Return(m)






A Surprising Example

Timing

M : a random 90× 90 matrix

time (sec) memory (Gb)

MinPoly 19m39s 8.3Maple 7h26m33s 74.6






Definition A simple graph G is proper k−colorable if one canassign k colors to the vertices of G such that theadjacent vertices receive different colors

Three questions:

Is a simple graph k−colorable?If yes, how many solutions?If yes, how we can find all the solutions?

Example 3−colorable?






Assign xi to the vertex i

Correspond a 3-th root of unity to a color

I := 〈x31 − 1, . . . , x35 − 1〉I := I + 〈x2i + xixj + x2j | (i, j) is an edge〉

Theorem

The elements of V(I) provide the different possible colorings and|V(I)| = the number of colorings

Algorithm for Graph Coloring

✄ Compute a Grobner basis for I

✄ if 1 ∈ G ⇒ G is not k−colorable

✄ else, compute V(I) using G






The corresponsing polynomial system:

F := [x31 − 1, . . . , x24 + x4x5 + x25];

with(Groebner); G:=Basis(F,plex(x1, . . . , x5));

G = [x35 − 1, x24 + x4x5 + x25, x5 + x3 + x4, x2 − x4, x1 + x4 + x5]

Theorem

The set of zeros of F is equal to the one of G.

x5 ∈ {1, ω, ω2}, we let x5 = 1

by second equation x4 6= x5, so x4 ∈ {ω, ω2}, we set x4 = ω

x3 = ω2, x2 = ω, x1 = ω2.






V(I) =

(ω ω2 ω ω2 1 )(ω2 ω ω2 ω 1 )(ω 1 ω 1 ω2)(1 ω 1 ω ω2)(ω2 1 ω2 1 ω )(1 ω2 1 ω2 ω )






Jacobian Conjecture

Theorem (Inverse function theorem)

✄ Consider F : Kn 7→ Kn withF (a1, . . . , an) = (f1(a1, . . . , an), . . . , fn(a1, . . . , an))

✄ Suppose that J(F ) = det

∂f1∂x1

· · · ∂f1∂xn

......

...∂fn∂x1

· · · ∂fn∂xn

⇒ If J(F )(p) 6= 0 then F has a local inverse at p ∈ Kn.






Jacobian Conjecture

Theorem (Inverse function theorem)

✄ Consider F : Kn 7→ Kn withF (a1, . . . , an) = (f1(a1, . . . , an), . . . , fn(a1, . . . , an))

✄ Suppose that J(F ) = det

∂f1∂x1

· · · ∂f1∂xn

......

...∂fn∂x1

· · · ∂fn∂xn

⇒ If J(F )(p) 6= 0 then F has a local inverse at p ∈ Kn.

Example

✄ We consider F (x, y) = (excos(y), exsin(y)) then

✄ J(F ) = e2x and therefore

⇒ F has a local inverse at each point (no global).






Jacobian Conjecture (cont.)

Jacobian conjecture

✄ If f1, . . . , fn ∈ C[x1, . . . , xn] and J(F ) ∈ C \ {0}⇒ F has a global inverse.

History

✄ It was first posed in 1939 by Ott-Heinrich Keller,

✄ Wang in 1980 proved it for polynomials of degree 2,

✄ It is still open for degree ≥ 3.







Theorem

✄ Let I = 〈y1 − f1, . . . , yn − fn〉.✄ G = Basis(I, plex(x1, . . . , xn, y1, . . . , yn)).

⇒ Then F has a global inverse iffG = {x1 − g1, . . . , xn − gn} with gi ∈ K[y1, . . . , yn].







Theorem



Example

✄ F (x1, x2) = (x1 + (x1 + x2)3, x2 − (x1 + x2)

3)

✄ G = {x1 − y1 + (y1 + y2)3, x2 − y1 − (y1 + y2)

3}.







Theorem



Example

✄ F (x1, x2) = (x1 + (x1 + x2)3, x2 − (x1 + x2)

3)

✄ G = {x1 − y1 + (y1 + y2)3, x2 − y1 − (y1 + y2)

3}.⇒ Thus F has a global local inverse (J(F ) = 1),







Theorem



Example

✄ F (x1, x2) = (x1 + (x1 + x2)3, x2 − (x1 + x2)

3)

✄ G = {x1 − y1 + (y1 + y2)3, x2 − y1 − (y1 + y2)

3}.⇒ Thus F has a global local inverse (J(F ) = 1),

⇒ F−1(y1, y2) = (y1 − (y1 + y2)3, y1 + (y1 + y2)

3).






Lattice

Definition: Orderly Lattice

A poset (L,≤) is called a Lattice, if

∀a, b ∈ L, sup{a, b}, inf{a, b} ∈ L






Lattice

Definition: Orderly Lattice

A poset (L,≤) is called a Lattice, if

∀a, b ∈ L, sup{a, b}, inf{a, b} ∈ L

Definition: Algebraic Lattice

L : a set;

∨,∧ : two binary operations on L;

(L,∨,∧) is a Lattice, if for all a, b, c ∈ L :

Commutative a ∨ b = b ∨ a, a ∧ b = b ∧ aAssociative a∨ (b∨ c) = (a∨ b)∨ c, a∧ (b∧ c) = (a∧ b)∧ c

Absorptance a ∨ (a ∧ b) = a, a ∧ (a ∨ b) = a






(L,≤) = (L,∨,∧)

Orderly ⇔ Algebraic

Orderly ⇒ Algebraic Let: a ∨ b = sup{a, b}, a ∧ b = inf{a, b}Algebraic ⇒ Orderly Define: a ≤ b ⇔ a ∨ b = b (or a ∧ b = a)






(L,≤) = (L,∨,∧)

Orderly ⇔ Algebraic

Orderly ⇒ Algebraic Let: a ∨ b = sup{a, b}, a ∧ b = inf{a, b}Algebraic ⇒ Orderly Define: a ≤ b ⇔ a ∨ b = b (or a ∧ b = a)

Example

Let A be a set;(P (A),⊆) = (P (A),∪,∩)






Distributive Lattice

Definition

A lattice L is distributive, if for each a, b, c ∈ L

a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and

a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)






Distributive Lattice

Definition

A lattice L is distributive, if for each a, b, c ∈ L

a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and

a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)

Theorem

L is distributive if and only if none of its sublattices is isomorphicto M3 or N5.






Problem

Lattice Distributivity Problem

✄ How can we determine whether a finite lattice is distributive?






✄ L: a finite lattice

✄ IL := 〈zazb − za∧bza∨b | a, b ∈ L〉 ⊂ K[za | a ∈ L]






✄ L: a finite lattice

✄ IL := 〈zazb − za∧bza∨b | a, b ∈ L〉 ⊂ K[za | a ∈ L]

Theorem: (a)⇒ (b): Hibi, 1987, (b)⇒ (a): Qureshi, 2012

The following are equivalent:

(a) L is distributive,

(b) GL := {zazb − za∧bza∨b | a, b ∈ L are incomparable} is aGrobner basis for IL w.r.t. each rank reverse lexicographicordering.






Example:






Example:

GM3= {zbzc − zaze, zbzd − zaze, zczd − zaze} is not a Grobner

basis w.r.t. ze ≺ zb ≺ zc ≺ zd ≺ za, because:






Example:



LM(Spoly(zbzc − zaze, zbzd − zaze)) = zdzaze /∈ LM(GM3)






Example:




Result

M3 is not distributive.






Example:




Result

M3 is not distributive.

Note

b = b ∧ (c ∨ d) 6= (b ∧ c) ∨ (b ∧ d) = e






Automatic Geometry Theorem Proving

Example

Pappus theorem: P,Q and R are collinear






Coordinate of points:D := (0, 0) E := (u1, 0) F := (u2, 0)A := (u3, u4) B := (u5, u6) C := (u7, x1)P := (x2, x3) Q := (x4, x5) R := (x6, x7)






Since A,B,C are colinear, we have u5−u3

u6−u4= u7−u3

x1−u4, and so

from the collinearity of points we obtain:







u6−u4= u7−u3

x1−u4, and so


Hypothesis polynomials

h1 := x1u3 + u6u7 − u6u3 − x1u5 − u4u7 + u4u5h2 := u4u1 + x3u3 − x3u1 − u4x2h3 := u5x3 − u6x2h4 := u4u2 + x5u3 − x5u2 − u4x4h5 := u7x5 − x1x4h6 := u6u2 + x7u5 − x7u2 − u6x6h7 := x1u1 + x7u7 − x7u1 − x1x6







u6−u4= u7−u3

x1−u4, and so


Hypothesis polynomials

h1 := x1u3 + u6u7 − u6u3 − x1u5 − u4u7 + u4u5h2 := u4u1 + x3u3 − x3u1 − u4x2h3 := u5x3 − u6x2h4 := u4u2 + x5u3 − x5u2 − u4x4h5 := u7x5 − x1x4h6 := u6u2 + x7u5 − x7u2 − u6x6h7 := x1u1 + x7u7 − x7u1 − x1x6

Conclusion polynomialg := x7x2 + x5x6 − x5x2 − x7x4 − x3x6 + x3x4






✄ I := 〈h1, . . . , hr〉 ⊂ R[x1, . . . , xn, u1, . . . , um]






✄ I := 〈h1, . . . , hr〉 ⊂ R[x1, . . . , xn, u1, . . . , um]






✄ I := 〈h1, . . . , hr〉 ⊂ R[x1, . . . , xn, u1, . . . , um]

Theorem

Conclusion is true iff the reduced Grobner basis of

〈h1, . . . , hr, 1− yg〉 ⊂ R(u1, . . . , um)[x1, . . . , xn, y]

equals to {1}






✄ I := 〈h1, . . . , hr〉 ⊂ R[x1, . . . , xn, u1, . . . , um]

Theorem

Conclusion is true iff the reduced Grobner basis of

〈h1, . . . , hr, 1− yg〉 ⊂ R(u1, . . . , um)[x1, . . . , xn, y]

equals to {1}

The reduced Grobner basis of

〈h1, . . . , h7, 1− yg〉 ⊂ R(u1, . . . , u7)[x1, . . . , x7, y]

is {1}, so the Pappus theorem is proved.






Frobenius Problem

a1 ≤ a2 ≤ · · · ≤ ar natural numbers

gcd(a1, . . . , ar) = 1

N(a1, . . . , ar) := max{n ∈ N | n 6= λ1a1 + · · · + λrar}Frobenius problem : Find N(a1, . . . , ar)

Example

N(5, 7) = 23

N(11, 13, 17) = 53

N(238, 569, 791) = 18205






✄ I := 〈x1 − ta1 , . . . , xr − tar〉 ⊂ Q[x1, . . . , xr, t]






✄ I := 〈x1 − ta1 , . . . , xr − tar〉 ⊂ Q[x1, . . . , xr, t]

✄ G:= A Grobner basis for I w.r.t. xi ≺lex t






✄ I := 〈x1 − ta1 , . . . , xr − tar〉 ⊂ Q[x1, . . . , xr, t]







✄ I := 〈x1 − ta1 , . . . , xr − tar〉 ⊂ Q[x1, . . . , xr, t]


Theorem

N = λ1a1 + · · · + λrar iff remainder(tN , G) = xλ1

1 · · · xλr

r






Example

McNugget numbers N(6, 9, 20) = 43






Example


J := 〈x1 − t6, x2 − t9, x3 − t20〉 :






Example


J := 〈x1 − t6, x2 − t9, x3 − t20〉 :G:=Basis(J, plex(t, x1, x2, x3));






Example



NormalForm(t41, G, plex(t, x1, x2, x3));






Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20






Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20







Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20


tx1x42






Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20


tx1x42







Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20


tx1x42


x1x22x3 ⇒ 44 = 1× 6 + 2× 9 + 1× 20

· · ·






Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20


tx1x42


x1x22x3 ⇒ 44 = 1× 6 + 2× 9 + 1× 20

· · ·NormalForm(t50, G, plex(t, x1, x2, x3));






Example




x21x2x3 ⇒ 41 = 2× 6 + 1× 9 + 1× 20


tx1x42


x1x22x3 ⇒ 44 = 1× 6 + 2× 9 + 1× 20

· · ·NormalForm(t50, G, plex(t, x1, x2, x3));

x21x22x3 ⇒ 50 = 2× 6 + 2× 9 + 1× 20






N Remainder λi,s

53 x42t −54 x1x

22x3 [1, 2, 1]

55 x51 [5, 0, 0]56 x32x3 [0, 3, 1]57 x41x2 [4, 1, 0]58 x1x2x

23 [1, 1, 2]

59 x31x22 [3, 2, 0]

60 x22x23 [0, 2, 2]

61 x21x32 [2, 3, 0]

62 x1x33 [1, 0, 3]

63 x1x42 [1, 4, 0]

64 x2x33 [0, 1, 3]

⇒ N(11, 13, 17) = 53






Solving Linear Diophantine Systems [Conti-Traverso, 1991]

Consider aij , bi ∈ N and the following Diophantine systems:

a11σ1 + · · · + a1nσn = b1...

am1σ1 + · · · + amnσn = bm.

I := 〈y1 − xa111 · · · xam1

m , . . . , yn − xa1n1 · · · xamn

m 〉








a11σ1 + · · · + a1nσn = b1...

am1σ1 + · · · + amnσn = bm.

I := 〈y1 − xa111 · · · xam1

m , . . . , yn − xa1n1 · · · xamn

m 〉G = Grobner basis of I w.r.t yi ≺lex xj








a11σ1 + · · · + a1nσn = b1...

am1σ1 + · · · + amnσn = bm.

I := 〈y1 − xa111 · · · xam1

m , . . . , yn − xa1n1 · · · xamn


xb11 · · · xbmm ❀G h








a11σ1 + · · · + a1nσn = b1...

am1σ1 + · · · + amnσn = bm.

I := 〈y1 − xa111 · · · xam1

m , . . . , yn − xa1n1 · · · xamn



h = yσ1

1 · · · yσn

n ⇔ (σ1, . . . , σn) is a natural solution.






Solving Linear Diophantine Systems (cont.)

Example

✄

{

3σ1 +2σ2 +σ3 + σ4 = 104σ1 +σ2 +σ3 + = 5.







Example

✄

{

3σ1 +2σ2 +σ3 + σ4 = 104σ1 +σ2 +σ3 + = 5.

✄ I := 〈y1 − x31x42, y2 − x21x2, y3 − x1x2, y4 − x1〉,







Example

✄

{

3σ1 +2σ2 +σ3 + σ4 = 104σ1 +σ2 +σ3 + = 5.

✄ I := 〈y1 − x31x42, y2 − x21x2, y3 − x1x2, y4 − x1〉,

✄ Grobner basis of I w.r.t yi ≺lex xj is:

{y2 − y3y4, y1y4 − y43 , x2y4 − y3, x2y33 − y1, x1 − y4}







Example

✄

{

3σ1 +2σ2 +σ3 + σ4 = 104σ1 +σ2 +σ3 + = 5.

✄ I := 〈y1 − x31x42, y2 − x21x2, y3 − x1x2, y4 − x1〉,

✄ Grobner basis of I w.r.t yi ≺lex xj is:

{y2 − y3y4, y1y4 − y43 , x2y4 − y3, x2y33 − y1, x1 − y4}

⇒ x101 x52 ❀G y53y54 and (0, 0, 5, 5) is a solution.






Integer Solutions

z; a new variable

I := 〈y1−xa111 · · · xam1

m , . . . , yn−xa1n1 · · · xamn

m , y1 · · · ynz − 1〉G = Grobner basis of I w.r.t z ≺lex yi ≺lex xj


Theorem (Hashemi, 2013)

The has an integer solution iff h ∈ K[y1, . . . , yn, z]. Moreover, ifthe remainder is yα1

1 · · · yαn

n zα, then (α1 − α, . . . , αn − α) is asolution of the system.






Example

12σ1 + 7σ2 + 9σ3 = 125σ2 + 8σ3 + 10σ4 = 0

15σ1 + 21σ3 + 69σ4 = 3.

I = 〈y1 − x121 x15

3 , y2 − x71x

52, y3 − x9

1x82x

213 , y4 − x10

2 x693 , y1y2y3y4z − 1〉

G = Grobner basis of I w.r.t z ≺lex yi ≺lex xj

x121 x33 ❀G z336y2931 y6563 y2484

(293, 0, 656, 248) − (336, 336, 336, 336) = (−43,−336, 320,−88)an integer solution

Since z2056y18191 y2382 y37863 y15814 − 1 ∈ I, we have the generalsolution (−237k − 43,−1818k − 336, 1730k + 320,−475k − 88)






Grobner Bases and Tilings

Amir Hashemi

Isfahan University of Technology & IPM

CIMPA-IRAN school on Tilings and Tessellations, Isfahan, Iran, 2015





Application to Tilings

Hakim Abolfath Omar ebn Ibrahim KhayyamNieshapuri (18 May 1048 -4 December 1131), born inNishapur in North Eastern Iran, was a great Persianpolymath, philosopher, mathematician, astronomer andpoet, who wrote treatises on mechanics, geography,mineralogy, music, and Islamic theology.

Khayyam was famous during his times as a mathematician. Hewrote the influential treatise on demonstration of problems ofalgebra (1070), which laid down the principles of algebra, part ofthe body of Persian Mathematics that was eventually transmittedto Europe. In particular, he derived general methods for solvingcubic equations and even some higher orders.






He wrote on the triangular array of binomial

coefficients known as Khayyam-Pascal’s triangle.

In 1077, Khayyam also wrote a book published in

English as ”On the Difficulties of Euclid’s

Definitions”. An important part of the book is

concerned with Euclid’s famous parallel postulate

and his approach made their way to Europe, and

may have contributed to the eventual development

of non-Euclidean geometry.






Notations

So far, we have considered R = K[x1, . . . , xn]

and discussed Grobner bases over fields + applications

In this part, we consider A = Z[x1, . . . , xn]

We discuss Grobner bases over integers

and review its application in tilings.






Main Problem

☞ Main obstacle in theory of Grobner bases over Z is division.Let us consider the polynomial ring R = K[x1, . . . , xn]. In thisring we have 2x | 3x. However that does not hold in

A = Z[x1, . . . , xn]






Main Problem


A = Z[x1, . . . , xn]

i.e. 2x ∤ 3x






Main Problem


A = Z[x1, . . . , xn]

i.e. 2x ∤ 3x

Further, in A we have 3x ∤ x and 2x ∤ x however x ∈ 〈2x, 3x〉






Main Problem


A = Z[x1, . . . , xn]

i.e. 2x ∤ 3x

Further, in A we have 3x ∤ x and 2x ∤ x however x ∈ 〈2x, 3x〉Thus f = x+ 1 is reducible by {f1 = 2x+ 3, f2 = 3x− 5},i.e. f ❀ f − (f2 − f1) = (x+ 1)− (3x− 5− 2x− 3) = 9.






Main Problem


A = Z[x1, . . . , xn]

i.e. 2x ∤ 3x







Main Problem


A = Z[x1, . . . , xn]

i.e. 2x ∤ 3x


Thus, LT(f) is divisible by none of LT(f1) and LT(f2)however, it is reducible by a combination of LT(f1) and LT(f2)






Example

Example

✄ A = Z[x, y] and y ≺lex x

✄ f = xy − 1, f1 = 7x+ 3, f2 = 3y − 5

✄ We note that LT(f1) ∤ LT(f) and LT(f2) ∤ LT(f)

✄ however, xy ∈ 〈7x, 3y〉Z[x,y] ❀ xy = y(7x)− 2x(3y)

✄ f − (yf1 − 2xf2) = −3y − 10x− 1.






Grobner Bases

Definition

✄ I ⊂ A

✄ ≺ a monomial ordering

✄ A finite set G ⊂ I is a Grobner Basis for I w.r.t. ≺, if

∀f ∈ I ∃g ∈ G s.t. LT(g) | LT(f)

Existence of Grobner bases

Each ideal has a Grobner basis

Membership problem

f ∈ I iff f ❀G 0






Example

☞ A = Z[x, y] and y ≺lex x

f1 = 4x+ 1, f2 = 6y + 1 and I = 〈f1, f2〉






Example


f1 = 4x+ 1, f2 = 6y + 1 and I = 〈f1, f2〉f := 3yf1 − 2xf2 = 3y − 2x and LT(f) = −2x /∈ 〈4x, 6y〉






Example








Example



{f1, f2} is not a Grobner basis for I






Grobner Bases Calculation over Integers

The process is the same as Buchberger’s algorithm over fields

The only difference is the division algorithm






Grobner Bases Calculation over Integers

The process is the same as Buchberger’s algorithm over fields

The only difference is the division algorithm

Example

✄ A = Z[x, y] and y ≺lex x

✄ f1 = 2x+ 1, f2 = 3y + 1 and I = 〈f1, f2〉Spoly(f1, f2) =

lcm(LT(f1),LT(f2))LT(f1)

f1 − lcm(LT(f1),LT(f2))LT(f2)

f2

Spoly(f1, f2) =6xy2x f1 − 6xy

3y f2 = 3yf1 − 2xf2 = 3y − 2x

✄ Spoly(f1, f2) ❀ Spoly(f1, f2) + f1 = 3y + 1 ❀f2 0.

So, {f1, f2} is a Grobner basis for I






Polynomial Model

Definition

A celle is the squarec(i, j) = {(x, y) | i ≤ x < i+ 1, j ≤ y < j + 1}To a celle c(i, j) we associate the polynomial Pc(i,j) = xiyj

y

x

(i, j)

i

j






Polynomial Model (cont.)

☞ A polyomino is a finite union of celles

Definition

T = ∪(i,j)∈Λc(i, j) then define PT =∑

(i,j)∈Λ Pc(i,j).

☞ We associate 1 + x+ xy + xy2 to the following polyomino

y

x1 x

xy

xy2






Z-Tiling

P : A polyomino

F : A finite set of polyominoes

Z-tiling problem is a finite number of translated polyominoes from Fto cover P so that the sum of signs on P at each cell is +1.






Example

We consider the following polyomio P

y

x1 x

y2

y3

y

xy3

☞ We associate the polynomial f = 1 + x+ y + · · ·+ xy3.






Example (cont.)

and we would like to cover it by the following polyomino T

y

x1 x

xy

xy2

☞ we associate the polynomial f1 = 1 + x+ xy + xy2.






Example (cont.)

☞ so we shall consider all (four) rotations of T , e.g.

y

x1

y

y2xy2

☞ we associate the polynomial f2 = 1 + y + y2 + xy2.☞ Let f2, f3, f4 be associated poly of all rotations






Example (cont.)

☞ In addition, we shall consider all rotations of the reflection of T

y

x1

y

y2

x

☞ we associate the polynomial g1 = 1 + y + y2 + x.☞ Let g2, g3, g4 be associated poly of all rotations






Example (cont.)

To check whether P can be covered by T (and its rotationsand reflection), one can verify whether

f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]






Example (cont.)


f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]

we can do this check by Grobner bases over integers






Example (cont.)


f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]


Let G be the Grobner basis of this ideal






Example (cont.)


f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]



since the remainder of f by G is zero,(indeed we have f = f1 + yf2)






Example (cont.)


f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]









Example (cont.)


f ∈ 〈f1, . . . , f4, g1, . . . , g4〉 ⊂ Z[x, y]




Thus, we can cover P by T .






Example (cont.)

Thus we have

y

xf1

f2

y

☞ f = f1 + yf2






Thanks for your attention...






References

[1] David Cox, John Little and Donal O’Shea:Ideals, varieties, and algorithms.Undergraduate Texts in Mathematics, Springer-Verlag, 2006.[2] David Cox, John Little and Donal O’Shea:Using algebraic geometry.Graduate Texts in Mathematics, Springer-Verlag, 2005.[3] Thomas Becker and Volker Weispfenning:Grobner bases, A computational approach to commutative algebra.Graduate Texts in Mathematics, Springer-Verlag, New York, 1993.[4] William W. Adams and Philippe Loustaunau:An introduction to Grobner bases.American Mathematical Society, 1994.


Documents

Gröbner Bases and Tilingsfernique/isfahan/hashemi...w’al-muqabala” gives us the word algebra and can be considered as the ﬁrst book to be written on algebra. Amir Hashemi Gro¨bner