shatha assaad salman al-najjar - nahrain university

Ministry of Higher Education And Scientific Research Al-Nahrain University College of Science

On

The Volume And Integral Points Of A polyhedron In nR

A Thesis Submitted to the Department of Mathematics and Computer Applications/College of Science/ Al- Nahrain University In Partial Fulfillment of the Requirements For the Degree of

Doctor of Philosophy in Mathematics

By

Shatha Assaad Salman Al-Najjar

(B.Sc., 1991) (M.Sc., 1995)

Supervised by

Dr. Adil G. Naoum and Dr.Ahlam J. Khaleel

July 2005

ا د اح اا ل

nℜ

رت وت طت"#" ا! ا #&

ط(54 "ن 12ء" ا/. - ج /ا(م*(/ا()بت 7 2راه ر ل7 ا

8& -" ا(ر>;ى أ (ن

(١ ٩ ٩ ١ / ا" ا2/ج/2(رس =(م5 ) (١ ٩ ٩ ٥ / ا" ا2/ج/"ج =(م )

5> اف

أA ج8 @(8 .د ن /و و B8د د.=

ي ٢٠٠٥تز H)١٤٢٦ ا

ا(&Mوزارة ا( ا وا /. -" اج

ا(مک

Abstract

Computing the volume and integral points of a polyhedron in

nℜ is a very important subject in different areas of mathematics.

There are two representations for the polyhedron, namely the H-representation and the V-representation. For each representation we give a different method of finding the volume and number of integral points.

Moreover, the Ehrhart polynomial of a bounded polyhedron is

discussed with some methods for finding it. One of these methods is modified and we prove two theorems for computing the coefficients of the Ehrhart polynomial.

Also, a modified method for counting the number of integral

points of a bounded polyhedron is given, and it makes matrix operations on the matrix that represents the bounded polyhedron, and studies the effect of these operations on these numbers.

All of the used methods are demonstrated with different

examples.

Introduction A wide variety of pure and applied mathematics involve the problem of counting the number of integral points inside a region in space. Applications range from the very pure: number theory, toric Hilbert functions, Kostant's partition function in representation theory, Ehrhart polynomial in combinatorics to the very applied: cryptography, integer programming, statistical contingency, mass spectroscope analysis. Perhaps the most basic case is when the region is a convex bounded polyhedron. Convex polyhedra, i.e., the intersections of a finite number of half spaces of the space dℜ , are important objects in various areas of mathematics and other disciplines as seen before. In particular, the compact ones among them (polytopes), which, can equivalently be defined as the convex hulls of finitely many points in

dℜ , have been studied since ancient times, for example, platonic solids, diamonds, the great pyramids in Egypt etc., [27]. polytopes appear as building blocks of more complicated structures, e.g. in combinatorial, topology, numerical mathematics and computed aided designs. Even in physics polytopes are relevant e.g., in crystallography or string theory, [31]. Probably the most important reason of the tremendous growth of interest in the theory of convex polyhedra in the second half of 20'th century was the fact that linear programming i.e., optimizing a linear function over the solutions of a system of linear inequalities became a wide spread tool to solve practical problems in industry and military. Dantzig's simplex algorithm, developed in the 40's of the last century, showed that geometric and combinatorial knowledge of polyhedra (as the domains of linear programming problems), is quite helpful for finding and analyzing solution procedures for linear programming problems, [31]. Since the interest in the theory of convex polyhedra to a large extent comes from algorithmic problems, it is not surprising that many algorithmic questions on polyhedra rose in the past, but also inherently, convex polyhedra (in particular: polytopes) give rise to algorithmic questions, because they can be treated as finite objects by definition; this makes it possible to investigate the smaller ones among them by

computer programs like the polymake - system written by Gawilow and Jowing, [24].

Once chosen to exploit this possibility one immediately finds oneself confronted with many algorithmic challenges.

Also, the notion of the volume of a polytope is basic and intuitive; its computation has raised a lot of problems. In this thesis we attempt to answer some fundamental and practical question on volume computation of higher dimensional convex polytopes given by their vertices and / or facets. In particular, we study through extensive computational experiment typical behavior of the exact methods, including Delaunay and boundary triangulation, the triangulation scheme described by Cohen and Hickey and the methods presented by Lawrence, [13].

This thesis consists of three chapters. In chapter one we try to give a short introduction, provide a

sketch of what bounded polyhedron looks like and how they behave with many examples. Also we recall some methods for finding the number of integral points inside a convex polytope, [13], [25] and [4].

In chapter two we present some methods for computing the

coefficients of Ehrhart polynomial that depend on the concepts of Dedekind sum and residue theorem in complex analysis. Also, a method for counting these coefficients is introduced. The polytope that we take are with V-representations, [5] and [60]. We give a method for computing the coefficients of the Ehrhart polynomial, 3−dc ,

4−dc until 9−dc also we give a formula for the differentiation of the given method.

In chapter three, a method for finding the volume of H-

representation of a polytope using Laрlace transform is presented and some basic concepts and remarks about the Birkhoff polytope and their volumes are discussed with their Ehrhart polynomials, [33], [11], [7], [8] and [9]. We make a change on the matrix, which represents the polytopes and finds a general formula for the number of integral points; also we make a change of matrix operation and study the effect of this change on the number of integral points of the polytopes. To the best of our knowledge, this result seems to be new.

List of symbols

ℜ the set of all real numbers. Ζ the set of all integers. dℜ the vector space of d-dimension. dm×ℜ an dm× real matrix. dΖ the standard integer lattice. d

+ℜ d-space of vectors with positive components. Vol(P) volume of P. ext(P) extreme points of P. x greatest integer ≤x.

x least integer ≥ x.

νo(P) Voronoi cell of p. nb(S,v) nearest neighbor set of v in S. conv(nb(S,v)) Delaunay cell of v. dP ΖI number of integral points of a polyhedron.

2Ζ∂ IP number of integral points on the boundary of the

polyhedron. ),( pxδ delta function of x and p.

x the Euclidean norm of a vector x.

rank(A) rank of the matrix A. dim(P) dimension of the polytope P.

),...,,( 10 dvvv∆ simplex in dℜ with vertices d

dvvv ℜ∈,...,, 10 . ),...,det( 001 vvvv d −− determinant of dd × matrix whose columns

are 001 ,..., vvvv d −− .

ipP U+

= signed union of ip .

∏=

n

i

ix1

multiplication symbol of n

xxx ,...,,21

.

L(P,t) the Ehrhart polynomial of a polytope P. S(a,b) the Dedekind sum of a and b. Res(f(z), z=a) residue of f(z) about z = a.

),(1 nmS Stirling number of the first kind. ),(2 nmS Stirling number of the second kind.

Ζ∈

Ζ∉−−=xif

xifxxx0

2

1))((

oP interior of a polytope P.

nB Birkhoff polytope. ( )tHn Ehrhart polynomial of the Birkhoff polytope.

int |Z| = r interior of the circle |Z| = r.

Contents Abstract List of symbols Introduction Chapter One Preliminaries. 1. Representation of polyhedron. 2 2. Some methods for volume computation of a polytope. 5 1.2.1 Triangulation methods 6 1.2.2 Signed decomposition method 11

3. Methods for computing integral points. 13

Chapter Two Computing the Volume and Integral Points of V-representation of a Polytope Using

Ehrhart Polynomials. 1. Basic concepts about Ehrhart polynomials. 18 2. Counting integral points using Dedekind sums. 20 3. Counting integral points using residue theorem. 25 4. The Ehrhart coefficients. 29 5. Computing 3−dc and 4−dc in the Ehrtart polynomial. 37 6. General formula for the differentiation of h and J,...,J,J,I d21 .52

Chapter Three The Ehrhart Polynomial of H-representation of a Polytope.

1. Computing the volume of a convex polytope using Laplace 59 transform. 3.1.1 The direct method. 61

3.1.2 The indirect method. 65 2. Some basic concepts about Birkhoff polytope. 68 3. The Ehrhart polynomial of a Birkhoff polytope. 70 4. The effect of matrix operations on the number of integral points. 77

References 92

Chapter One Preliminaries

Introduction

Convex bounded polyhedrons are fundamental geometric objects that have been investigated since antiquity. The beauty of their theory is nowadays complemented by their importance for many other mathematical subjects, ranging from the integration theory, algebraic topology, and algebraic geometry (toric varieties) to the linear and

combinatorial optimization.

In this chapter we try to give a short introduction and provide a sketch of what bounded polyhedron looks like and how they behave

with many examples.

A convex polyhedron is an intersection of a finite number of half spaces of the space dℜ , and a convex polytope is a bounded convex polyhedron. Every convex polyhedron has two natural representations, a half space representation (H-representation) and a vertex representation (V-representation). In recent years various techniques of geometric computations associated with convex polyhedron have been

discovered, see [13], [31], [16] and [17].

Also, we recall some methods for finding the volume of a convex polytope and other methods for finding the number of integral points

inside a convex polytope.

This chapter consists of three sections:

In section one, some basic definitions with some useful remarks about representation of the polyhedron are presented.

In section two, some methods for computing the volume of convex polytopes are given with some illustrative examples. These are classified into two groups: triangulation methods and signed

decomposition methods.

The triangulation methods include boundary triangulation, Delaunay triangulation and Cohen & Hickey's triangulation, [13]. The

signed decomposition methods include Lawerence's method, [25]. In section three, some methods for finding the number of integral

points of a convex polytope are discussed; these methods are demonstrated with some examples.

1.1 Representation of a polyhedron

The volume of a convex bounded polyhedron is not easy to compute and the basic methods for exact computation of this volume can be classified according to whether a half space representation or a vertex representation of it, [33]. Therefore, in this section some basic definitions on a convex bounded polyhedron and its representations are given.

We start this section by the following definitions:

Definition (1.1.1), [35, p.85]:

Let bAX ≤ where dmA ×ℜ∈ is a given real matrix, and mb ℜ∈ is a

known real vector. A set : bAXXP d ≤ℜ∈= is said to be a

polyhedron.

A polyhedron P is bounded if there exists 1

+ℜ∈ω such that ω≤X for

every PX ∈ , [35, p.86].

Definition (1.1.2), [35, p.85]: Every bounded polyhedron is said to be a polytope.

Definition (1.1.3), [35, p.84]:

Let ,...,, 21 kxxxS = where d

ix ℜ∈ , ki ≤≤1 , then S is said to be

affinely independent if the unique solution of 01

=∑=

i

k

i

i xa and

01

=∑=

k

i

ia is 0=ia , for i = 1, …, k.

Recall that a polyhedron P is of dimension k, denoted by

dim(P)=k, if the maximum number of affinely independent points in P is k+1. In this case a polyhedron (polytope) is said to be k-polyhedron (k-polytope). On the other hand, a polyhedron is of a full dimensional if dim (P) = d, [35, P.86].

Remark (1.1.1):

If the polyhedron P which is defined by, : bAXXP d ≤ℜ∈= is not full dimensional, then at least one of the inequalities ii bXa ≤ ,

i=1,2,…,k, is satisfied as equality by all points of P, where ia is the i-th row of the matrix A and ib are the values of the vector b, [35, p.86].

Proposition (1.1.1), [35, p.84]:

Let : bAXXP d ≤ℜ∈= then the following statements are equivalent:

(a) φ≠≤ℜ∈ : bAXX d .

(b) rank(A)=rank(A|b), where ,, mdm bA ℜ∈ℜ∈ × A|b is the augmented matrix of the system AX=b and rank(A|b) is the maximum number of linearly independent

rows ( columns ) of A|b.

Now, if P takes the form : bAXXP d ≤ℜ∈= , the pair A|b is said to be a half space representation or simply H-representation of P,

where ,, mdm bA ℜ∈ℜ∈ × [13].

Proposition (1.1.2), [35, p.87]: Let : bAXXP d ≤ℜ∈= be a polytope, then:

dim(P)+rank( ** bA )=d,

where ** bA denotes the corresponding rows of A|b, which represent the

equality sets of the representation A|b of P, that is, : ∗∗ =ℜ∈= bXAXP d , [35,p.86].

:Definition (1.1.4), [4]

Let : bAXXP d ≤ℜ∈= be a polyhedron. If the entries of A and b have integer values then this polyhedron is said to be rational

polyhedron.

Recall that for a given convex set S, a point SX ∈ is said to be vertex (or sometimes extreme point) if it does not lie on a line segment joining two other points of this set. In this case the line joining any two

vertices is said to be an edge [41, p.98].

It can be easily seen that any polyhedron is a convex set in dℜ .

Definition (1.1.5), [38]: A lattice polytope in dℜ (sometimes called integral polytope) is a polytope whose vertices are lattice points (integral points), that is,

points in dΖ . If the lattice polytope is of dimension d then this polytope is said to be a d-dimensional lattice polytope

Definition (1.1.6), [41, p.96 ]:

Given ∑=

=d

i

ii bxa1

, where ia and b are known real constants for

di ≤≤1 . The set of points d

iixX 1 == , which satisfies the above equation, is said to be a hyperplane.

Moreover, the set of points d

iixX 1 == is called a half- space if it

satisfies the inequality ∑=

≥d

iii bxa

1

, [36, p.413].

Definition (1.1.7), [10]:

Let P be a polyhedron in dℜ . For dc ℜ∈ and ℜ∈b , the

inequality ∑=

≤d

i

ii byc1

is called valid for P if it is satisfied by all points

in P, where d

iicc 1 == . The faces of P are the sets of the form

∑=

= ==d

i

ii

d

ii bycyYP1

1 :I for some valid inequality ∑=

≤d

i

ii byc1

.

Recall that a face F is said to be proper if PF ≠≠φ . On the

other hand the faces of dimension 0 and 1 are called vertices and edges respectively. However the faces of highest dimension are termed facets.


A polytope in dℜ is said to be simple if there are exactly d edges through each vertex, and it is called simplicial if each facet contains

exactly d vertices.

It is known that a simplex in dℜ is a d-dimensional polyhedron, which has exactly d+1 vertices, [23, p.37].

Definition (1.1.9), [35, p.83 ]:

Given a non empty set dS ℜ⊆ , a point dX ℜ∈ is a convex combination of points of S if there exists a finite set of points t

iix 1 = in S

and t

+ℜ∈λ with ∑=

=t

i

i

1

1λ and ∑=

=t

i

ii xX1

λ .

It is known that the convex hull of S, denoted by conv(S) is the set of all points that are convex combinations of all points in S.

Now, if ,...,, 10 nvvvV = is a finite set of points in dℜ , the convex hull of V denoted by conv(V) is said to be convex polytope. In this case, V is called vertex representation or simply V-representation

of P, [13].

1.2 Some methods for the volume computation of a polytope

As mentioned before, computing the volume of a polytope is very important in many real life applications, so in this section we give some methods for finding it. There is a comparative study of various volume computation algorithms for polytopes in [13]. However there is no single algorithm that works well for many different types of them, [22].

For simple polytopes, triangulation-based algorithms are more

efficient and for simplicial polytopes sign-decomposition based algorithms are better, [13].

In this section, some methods for volume computation are given with different examples.

We start this section by the following remark. Remark (1.2.1):

All known algorithms for exact volume computation decompose a given polytope into simplices, and thus they all rely on the volume formula of a simplex which is given by the following proposition, [13]:

Proposition (1.2.1), [13]:

For a polytope represented by its vertices d

dvvv ℜ∈,...,, 10 , the

volume of it is given by

Vol( ),...,,( 10 dvvv∆ ) = ),...,det(!

1001 vvvv

d d −−

Where ),...,,( 10 dvvv∆ denotes the simplex in dℜ with vertices d

dvvv ℜ∈,...,, 10 and ),...,( 001 vvvv d −− is dd × matrix whose columns

are 001 ,..., vvvv d −− .

Next, there are two types of methods for exact volume

computation of the simple polytopes, which are discussed below:

I.Triangulation methods:

In these methods one has a simple polytope P in dℜ . P is

triangulated into simplices ),...,2,1( sii =∆ Us

iiP

1=

∆= . The volume of P

is simply the sum of the volumes of the simplices.

∑=

∆=s

i

iVolPVol1

)()( (1.1)

The following: boundary triangulation, Delaunay triangulation and Cohen & Hickey's combinatorial triangulation by dimensional recursion named, as triangulations method, [13].

An important difference between these methods is that the former two methods need only a V–representation while the last method requires both V- and H-representations, [13]. Before giving the signed decomposition methods, we need the following definition. Definition (1.2.1):

Let dℜ∈Ρ be a polytope, a signed union of P means, a

collection of polytopes d

k ℜ⊆ΡΡΡ ,...,, 21 such that Uk

ii

1=

Ρ=Ρ , and ji ΡΡ I

is a proper face of iΡ and jΡ , for ji ≠ .In this case we write U+

Ρ= iP ,

[30]. II. Signed decomposition methods:

Instead of triangulating a polytope P, one can decompose P into signed simplices whose signed union is exactly P. More specifically, P is represented as a signed union of simplices i∆ , i =1, 2, …, s. This

means,

Us

iii

1=

∆=Ρ σ (1.2)

Where iσ is either +1 or –1. The volume of P is, [13].

∑=

∆=s

i

iiVolPVol1

)()( σ

1.2.1 Triangulation methods

In this subsection we discuss briefly some of the known triangulation methods that compute the volume of the polytope.

(i) Boundary triangulation, [13]:

In boundary triangulation, one computes the convex hull of the perturbed points, interpreting the result in terms of the original vertices leads to a triangulation of the boundary, which by linking with a fixed interior point yields, a triangulation of P. For the convex hull computation the reverse search algorithm is chosen, [3], where only the V-representation of a polytope is required. To illustrate this method, consider the polytope which is represented by a set of vertices named a, b, c , d as given in figure (1). Using an interior point e where the boundary of a polytope is easily triangulated or already triangulated as in the case of simplicial polytopes. By linking a point e with the vertices a, b, c and d yields four triangles then, volumes of these triangles are found, summing all of these volumes the volume of this polytope is obtained.

(a) (b)

Figure (1): (a) represents a polytope P.

(b) represents a partition of the given polytope P by using the boundary triangulation method.

(ii) Delaunay triangulation, [13]:

Before we discuss this method, some basic definitions concerning the Delaunay triangulation are needed. Definition (1.2.2), [22]:

Given a set S of n distinct points in dℜ , Voronoi diagram is the partition of dℜ into n polyhedron regions (denoted by S∈ρρνο ),( ).

Each region )(ρνο is called Voronoi cell of ρ , which is defined as set

of points in dℜ that are closer to ρ than other points in S, or more

precisely ,)( ρρρνο −∈∀−≤−ℜ∈= •

• SqqXXX d .

Definition (1.2.3), [22]: Let S be a set of n points in dℜ . For each point dℜ∈ν , the nearest neighbor set denoted by )),(( νSnb of ν in S is the set of points

νρ −∈ S , which are closest to ν in Euclidean distance.


Let S be a set of n points in dℜ . A point dℜ∈ν is said to be a Voronoi vertex of S if ),( νSnb is maximal over all nearest neighbor

sets. Definition (1.2.5), [22]: Let S be a set of n points in dℜ . The convex hull of the nearest neighbor set of Voronoi vertex ν denoted by )),(( νSnbconv is said to

be a Delaunay cell of ν .

The Delaunay triangulation of S is a partition of the convex hull conv(S) into the Delaunay cells of Voronoi vertices together with their faces, [22].

Now we discuss the method of Delaunay triangulation method

that requires only the V–representation of the polytope.

The geometric idea behind a Delaunay triangulation of a d–polytope is to ' lift ' it on a paraboloid in dimension d+1. The following construction is very important to compute the Voronoi diagram, [22].

Let S be a set of n points in dℜ . For each point dS ℜ⊆∈ρ ,

consider the hyperplane tangent to the paraboloid 22

11 ... dd xxx ++=+ in 1+ℜd at ρ :

This hyperplane is represented by h (ρ ) as:

02 1

11

2 =+− +==∑∑ d

d

j

jj

d

j

j xxρρ

where ),...,2,1( djj = ρ are the coordinates of ρ , for each point ρ ,

the equality in the above equation is replaced by the inequalities )(≥ ,

which yields a system of n inequalities that is denoted by 0≥− AXb .

The polyhedron P in 1+ℜd of all solutions X to the system of inequality is a lifting of the Voronoi diagram to one higher dimensional space. [13], shows that the underlying convex hull algorithm uses the ' beneath – beyond ' method. Example (1.2.1): Consider the set of vertices:

).4,4(),4,0(),0,4(),2,1(),1,2(),0,0( 654321 ======= ρρρρρρS

Here the volume of the polytope given by these vertices is to be determined. To do so, Delaunay triangulation is used to compute the volume of this polytope.

First, write down the system of linear inequalities in three variables as explained before. That is for each 6,...,2,1, =∈ jSjρ ,

apply the inequalities:

02 3

2

1

2

1

2 ≥+−∑∑==

xxj

jj

j

j ρρ

we get a system of six inequalities

03 ≥x

0245 321 ≥+−− xxx

0425 321 ≥+−− xxx

0816 31 ≥+− xx

0816 32 ≥+− xx

08832 321 ≥+−− xxx .

The set of solutions 3ℜ∈X of the above inequalities represents a polyhedron P. By applying the cdd+ program [22], the Delaunay cells,

),,( 531 ρρρ , ),,( 321 ρρρ , ),,( 421 ρρρ , ),,( 632 ρρρ , ),,( 642 ρρρ and ),,( 653 ρρρ a

re obtained. The cell ),,( 531 ρρρ means the triangle which is represented

by three vertices 31, ρρ and 5ρ , and similarly for the other cells.

Therefore six triangles are obtained, summing the volumes of these triangles yields the volume of the polyhedron P is equal to 16.

Figure (2): represents the polyhedron P with the set of vertices ,...,, 621 ρρρ and Delaunay cells.

(iii) Triangulation by Cohen & Hickey, [13]: This recursive scheme triangulates a d- polytope P by choosing any vertex Pv∈ as an apex and connecting it with the (d-1)–dimensional simplices resulting from a triangulation of all facets of P not containing v . To be precise, denote by kθ , dk ≤≤0 , the k- dimensional faces of P, and let η be a ' map ' which associates to each

face one of its vertices. Then the pyramids with apex )( dθη and bases

among the facets 1−dθ with 1)( −∉ dd θθη form a dissection of the

polytope.

Applying the scheme recursively to all 1−dθ results in a set of

decreasing chains of faces dd θθθθ ⊂⊂⊂⊂ −110 ... such that 1)( −∉ kk θθη for dk ≤≤1 . Then the set of corresponding simplices

))(),...,(),(( 10 dθηθηθη∆ is a triangulation of P.

To implement this recursive method, an extensive use of the

double description as V-representation and H-representation is made by representing all faces as sets of vertices.

Note that in the case of Cohen & Hickey compared to a boundary triangulation all simplices in the facets containing the apex v are eliminated and therefore the number of simplices is usually reduced. Example (1.2.2):

Consider the polytope which is represented by set of vertices ,,,, 43210 ρρρρρ as illustrated in figure (3), let η be the 'map' which

assigns to each face of the polytope its vertex with the lowest number, so 0)( ρη =P , all facets which do not contain the vertex 0ρ are

examined, that is, II, III and IV. The scheme of the Cohen and Hickey is applied to facet II with 1)( ρη =II . II is intersected with all facets not

containing the vertex 1ρ , these are III, IV and V. The intersections with

IV and V are empty, so this recursion is unsuccessful. The intersection with III yields the vertex 2ρ , and the fixed vertices 10, ρρ , 2ρ forms a

first simplex. The other simplices obtained from III and IV is also marked in the figure (3). Therefore we have three triangles. Summing the volumes of these triangles yields the volume of the polytope.

Figure (3): represents the partition of the polytope by Cohen & Hickey's triangulation method.

1.2.2 Signed decomposition method In this subsection Lawrence's volume formula, which is one of the signed decomposition methods, is discussed. (i) Lawrence's volume formula, [13]: Assume the polytope P is simple and choose a vector dC ℜ∈ and

ℜ∈q such that the function ℜ→ℜdf : which is defined by

f(X)= qXCT + is not constant along any edge that connected the

vertices of the polytope P and TC is the transpose of C . Let V be the set

of vertices defining the polytope P. For each vertex Vv∈ , let vA be the

dd × – matrix composed by the rows of A which are binding at v. Then

by using [13], vA is invertible and [ ] CAT

v

v 1−=γ . The assumption

imposed on C assures that none of the entries of vγ is zero. It is shown

that

∑∏∈

=

+=Vv

d

i

v

iv

dT

Ad

qvCPVol

1

det!

)()(

γ

To illustrate this method, consider the following example.

Example (1.2.4) : Consider the polytope P which is described by the following

constraints 01 ≤− x

02 ≤− x

21 ≤x

22 ≤x

321 ≤+ xx

then

−−

=

11

10

01

10

01

A ,

=

2

1

x

xX and

=

3

2

2

0

0

b

It is easy to check that the polytope of this example is simple, therefore the Lawerence volume formula can be applied. Define a function f by 21)( xxXf −= where )1,1( −=TC and q = 0. Note that

f (X) is non – constant on each edge of the polytope in the figure (4), for example, on edge (1) which connect

21 and vv , 02 =x and 1x varies from

0 to 2. Therefore f (X) = 1

x which varies from 0 to 2 which means that

it is nonconstant on edge (1) and similarly on each edge of P. According to figure (4), it is seen that the set of vertices, which represents the polytope, is )2,0(),2,1(),1,2(),0,2(),0,0( 54321 ===== vvvvv .

Now, consider )0,0(1 =v , this vertex satisfies the first two constraints,

and this implies that

−−

=10

011v

A , hence 1det1

=vA and [ ]

= −

2

11

1

1

c

cAT

v

vγ

−=

−

−−

=1

1

1

1

10

011vγ

and for )0,2(2 =v , this vertex satisfies the second and third

constraints, that is,

−=

10

012vA , 1det

2=

vA And [ ]

=

−= −

1

1

1

11

2

2 T

v

v Aγ

And similarly for the other vertices we get

−=

1

23vγ ,

−=

1

24vγ And

−−

=1

15vγ

Then Lawrence's volume formula is applied to get the volume of

P.

∑∏∈

=

=Vv

i

v

iv

T

A

vCPVol 2

1

2

det !2

)()(

γ

( ) ( )

)1)(1)(1(!2

2

)1)(2)(1(!2

1

)1)(2)(1(!2

1

)1)(1)(1(!2

2

)1)(1)(1(!2

0 22222

−−−+

−−+

−++

−=

2

132414120 =+−+−++= )/()/( .

Figure (4): represents a polytope with the vertices

54321 ,,, vandvvvv

1.3 Methods for computing integral points The main objective of this section is to recall some methods for finding integral points of a polyhedron. In this work, we use the symbol

dP ΖI to denote the number of integral points in the polyhedron P,

where dΖ is the integer lattice and P is a rational polyhedron. These methods are: Method (1), [4]:

For d = 2, 2ℜ⊂P and P is an integral polyhedron. The famous formula, [42, p.240] states that

12

)(2

2 +Ζ∂

+=ΖI

IP

PareaP

That is, the number of integral points in an integral polyhedron is

equal to the area of the polyhedron plus half the number of integral points on the boundary of the polyhedron plus one. This formula is useful because it is much more efficient than the direct enumeration of integral points in a polyhedron. The area of P is computed by triangulating the polyhedron. Furthermore, the boundary ∂P is a union of finitely many straight-line intervals, and counting integral points in intervals is easy. Method (2), [4]:

Let dP ℜ⊂ be a polytope, then one can write the number of integral points in P as

dP ΖI ∑Ζ∈

=dX

PX ),(δ

where

∉∈

=PXif

PXifPX

0

1),(δ

Before we give the next method, we need the following definition.

Definition (1.3.1), [2, p.61]: The Dedekind sum of two relatively prime positive integers a and

b denoted by ),( baS can be defined as follows,

∑=

=b

i b

ai

b

ibaS

1

),(

where

Ζ∈

Ζ∉−−=xif

xifxxx0

2

1))((

and x is the greatest integer x≤ .

Remarks (1.3.1):

Dedekind sums appear in various branches of mathematics: the number theory, algebraic geometry and topology. These include the quadratic reciprocity law, random number generators [32], and lattice point problems [19]. More details about Dedekind sums are given in chapter two

Now, we are in a position that we can explain the following

method.

Method (3), [4]: Let 3ℜ⊂∆ be the tetrahedron with vertices (0, 0, 0), (a, 0, 0),

(0, b, 0), (0, 0, c) where a, b and c are pairwise coprime positive integers, then the number of integral points in ∆ can be expressed as:

),(),(

),()1

(12

1

463

cabSbacS

abcSabcc

ab

a

bc

b

accbabcacababcP

−

−−++++++++++=ΖI

This formula is useful because it reduces counting the number of integral points to a computation of Dedekind sums, which can be done efficiently. Method (4), [4]:

Let dP ℜ⊂ be an integral polytope, for positive integer t, let

tP=tX : X∈P denote the dilated polytope P. By [42, p.238] there is a polynomial p(t) called the Ehrhart polynomial of P )(tptP d =ΖI

where p(t) = 0

1

1 ... atata d

d

d

d +++ −−

furthermore, 10 =a and da is the volume of the polytope P.

To illustrate these methods, consider the following examples. Example (1.3.1):

Let us consider three points in two dimensions such that )0,0()0,1(),1,0( 321 === vandvv . Then the convex hull of 321, vandvv

is a triangle in two dimensions.

We compute the number of integral points by these methods. From method (1), one can have

12

)(2

2 +Ζ∂

+=ΖI

IP

PareaP

the area of triangle, is area(P)=0.5(1)(1)=0.5 and 32 =Ζ∂ IP which

represents the number of integral points on the boundary of the triangle. Then the number of integral points of the triangle is .31)3(5.05.02 =++=ΖIP

In method (2), we have 2ΖIP ∑

Ζ∈

=2

),(X

PXδ

1),()1,0( 11 =∈= PvthenPv δ

1),()0,1( 22 =∈= PvthenPv δ

1),()0,0( 33 =∈= PvthenPv δ

therefore 31112 =++=ΖIP , which is the number of integral points

for the triangle. Form method (4), one can have =Ζ2

IP 112 ++ aa

where 2a is the volume of the polytope and 1a is the half number of

integral points on the boundary of the polytope. In this case,

2a = 0.5(1)(1)=0.5

and 2a = 3/2

therefore 315.15.02 =++=ΖIP .

Example (1.3.1):

Let us consider the tetrahedron 3ℜ⊂∆ with vertices (0,0,0), (3,0,0), (0,5,0), (0,0,7).

Form method (3), the number of integral points in ∆ can be

expressed as,

),(),(),(

)1

(12

1

463

cabSbacSabcSabcc

ab

a

bc

b

accbabcacababcP

−−−

++++++++++=ΖI

here a=3, b=5 and c=7. Therefore

)7,15()5,21(

)3,35()105

1

7

15

3

35

5

21(

12

1

4

753352115

6

1053

SS

SP

−−

−++++++++++=ΖI

After simple computation one can get, S(35,3)=-0.05555, S(21,5)=0.2, S(15,7)= 0.35714, Thus .4035714.02.005555.0501877.15.215.173 =−−+++=ΖIP

ا

ل ا طو ب د ح ب د اھ nRا#!"! ا %&' ت ا") .(ع . -ا +وع ا+*

21"3 ح21"0ن د ا -* 2 إ ء 5و04 اV "&"1 -و2H 3"1-وھ

.ود ا ط ا#!"! ، ط+ق '&% ! ب د ا!ود

ى ھه ا+ق طرت ا . ت > <'ام >; ا+ق2 ب د ود ا*+ھ ر B+ھ"5 ! ب 0ت دة ا!ود ا*+ھ رت .وا<

C& "< ت ا!"& طرت ط+* ! ب د ا ط ا#!"! و2 أ-+اء ا

1"3 د ا .ودرا< EF2"+ھ &C د ھه ا ط ح#% ا "G ا+ق ا- %&' &1F< H!)و H'>ا.

Acknowledgements

My deepest thanks for Allah, for his generous by providing me

the strength to accomplish this work.

I would like to express my deep appreciation and gratitude to

my supervisors Dr. Adil G. Naoum and Dr. Ahlam J. Kaleel for their

help and advice during the preparation of this thesis.

Also, I would like to express my thanks to all members and

friends in the Department of Mathematics and Computer

Applications.

And all members in the central library of the University of

Technology especially Thawara and Hatham.

Shatha Assaad Al-Najjar

2005

Appendix

Before we give the proof of the relationship between the Dedekind sum and the Dedekind cotangent sum, we need the following definitions and lemma.

Definition 1 By means of the discrete Fourier series of Saw tooth function (saw

tooth function ((x)) is the first Bernoulli function ( ) ( ) ZxxBxB ∈= if 0, 11 ), and

Ρπ−Ρ

=∑

Ρπ

Ρ=

Ρikn

k

ekin 21

1

cot2

Definition 2 For a,b,c,m,n ∈N

( )

= ∑a

kcB

a

kbBcbaS n

bkmnm

mod, ,, is called Dedekind Bernoulli sum.

Definition 3 Let Naaa d ∈,,, 21 K be relatively prime to Na ∈0 define the

higher-dimensional Dedekind sum as

( ) ( )

π

π

π−= ∑−

= 00

21

1 0

1

0

10 cotcotcot1

,...,,02

a

ka

a

ka

a

ka

aaaaS d

a

kd

d

L .

Note that this sum is zero if d is odd, since the cotangent is an odd function. Lemma For m≥2

( )( ) Ρ

πΡ

=

−∑

Ρπ

Ρ+

Ρ−=

Ρkni

k

m

m

mm

m eki

mBn

B2

1

1cot2

These discrete Fourier expansions can be used to rewrite the Dedekind Bernoulli sums in terms of the Dedekind cotangent sums. Corollary If a, b, c ∈N are pairwise relatively rime and m, n ≥2 are integers with the same parity then

( )

( )( )( ) ( )

a

BBcotcot

2

1-mn

,,

1-nm

nm1-n1-a

1k

1-m

1

2

mod,

+=

−++

−

+

π

π=

=

∑

∑

a

kb

a

kc

a

a

kcB

a

kbBcbaS

nmnm

nm

nak

m

def

nm

We note that the parity assumption on m and n is no restriction; since the sums vanish if m+n is odd it is worth mentioning that a close relative of these sums namely

( )( )( )

π

π=

∑∑

=+

−−

= a

kb

a

k

aa

kbB

a

kB

mm

nma

knm

1-m1-a

1k1

21

1

cotcot2

1-m

Proof By lemma we get

( )

( )( )

( )( )

( ) ( )( )

( )( )

( )( ) ( ) ( )

π

=

π

=

+

π

=

π

=

π

=+

π

=

π

=

∑∑∑

∑ ∑∑

∑∑ ∑

∑

π

π

+

π

+

π

+

π

+=

π

+

π

+=

=

a

kcl21-a

1l

1-na

kbj21-a

1j

1-ma

kb21-a

1l

1-mn

mod

a

kcl21-a

1l

1-nma

kcl21-a

1l

1-nmnm

a

kll21-a

1

1-nn

mod

a

kbj21-a

1j

1-mm

mod,

ecotecot2a

iecot

2a

i

a-

mB

ecot2a

i

a-

nBecot

2a

i

a-

nB

a-

BB

ecot2a

in

a-

Becot

2a

im

a-

B

,,

a

l

a

jmn

a

j

a

l

a

l

a

l

a

j

a

kcB

a

kbBcbaS

nmj

m

n

ak

n

m

n

mnm

l

n

nak

m

m

nak

m

def

nm

( ) ( ) ( )lcjb21-a

1l

1-n1-a

1lj,

1-m

mod

ecotcot2a

i

+π

==

+

∑∑∑

π

π

= a

ki

ak

nm

a

l

a

jmn

( ) ( )( )

( ) ( )

( )∑

∑ ∑∑ ∑

+

π

=

π

=

−

+

π−

+

π

=

aknm

nm

iakcl

akm

m

nia

kbj

ak

m

a

BB

ea

l

a

Be

a

j

mod

2

mod

1-a

1l

1-n2

mod

1-a

1lj,

1-m

cot2a

in cot

2a

im

We use the fact that m+n=even. Note: a,b,c are relatively prime

( )

=∑π

else 0n

a if a

mod

2

ak

aink

e

if a|(jb+lc) then ajbcl mod1−= therefore

( ) ( )( )

( )( ) ( ) ( )( )

( )( )

( )( ) ( )

( ) 1

11-n

1-a

1j

1-m2

11-n

1-a

1j

1-m2

,

cotcot2

1

cotcot2

1,,

−+

−

=+

+

+

−

=+

+

−+

π−

π−=

−−+

π−

π−=

∑

∑

nmnm

nm

nm

nmnm

nm

nm

nm

a

BB

a

jbc

a

j

amn

a

BBa

a

jbc

a

j

amncbaS

Committee Certification We, the examining committee, certify that we read this thesis and have examined the student Shatha Assaad Al-Najjar in its contents and that, in our opinion it is adequate as a thesis for the degree of Ph-D of Science in Mathematics. Chairman Member Signature: Signature: Name: Dr. Nadar J. Mansour Name: Dr. Mohamad S. Kerdar Professor Professor Date: / 10 / 2005 Date: / 10 / 2005 Member Member Signature: Signature: Name: Dr. Akram M. Al-bood Name:Dr. Norii F. Athab Assist professor Assist professor Date: / 10 / 2005 Date: / 10 / 2005

Member (supervisor) Member Member(supervisor)

Signature: Signature: Signature: Name: Dr. Adil G. Naoum Name: Dr. Sahab K. Al-Saadi Name: Dr. Ahlam J. Kaleel

Professor Assist professor Assist professor Date: / 10 / 2005 Date: / 10 / 2005 Date: / 10 / 2005

I hear be certify upon the decision of the examining committee. Signature:

Name: Dr. LAITH-ABDUL AZIZ AL-ANI Dean of the College

Date: /10 /2005

Supervisors Certification

We certify that this thesis was prepared under our supervision at the University of Al-Nahrian, Colloge of Science as a partial fulfillment of the Requirements of Doctor of Philosophy of Science in Mathematics. Signature: Signature: Name: Dr. Adil G. Naoum Name: Dr. Ahlam J. Kaleel Professor Assist professor Date: / / 2005 Date: / / 2005 In view of the available recommendations, I forward this thesis for debate by examining committee. Signature: Name: Dr. Akram M. Al-Abood Assist professor

Date: / /2005

Contents Abstract List of symbols Introduction Chapter One Preliminaries.

1. Representation of polyhedron. 2 2. Some methods for volume computation of a polytope. 5

1.2.1 Triangulation methods 6 1.2.2 Signed decomposition method 11

3. Methods for computing integral points. 13

Chapter Two Computing the Volume and Integral Points of V-representation of a Polytope Using

Ehrhart Polynomials. 1. Basic concepts about Ehrhart polynomials. 18 2. Counting integral points using Dedekind sums. 20 3. Counting integral points using residue theorem. 25 4. The Ehrhart coefficients. 29 5. Computing 3−dc and 4−dc in the Ehrtart polynomial. 37 6. General formula for the differentiation of h and J,...,J,J,I d21 . 52

Chapter Three The Ehrhart Polynomial of H-representation of a Polytope.

1. Computing the volume of a convex polytope using Laplace 59 transform. 3.1.1 The direct method. 61

3.1.2 The indirect method. 65 2. Some basic concepts about Birkhoff polytope. 68 3. The Ehrhart polynomial of a Birkhoff polytope. 70 4. The effect of matrix operations on the number of integral points. 77

References 92

List of symbols

ℜ the set of all real numbers. Ζ the set of all integers. dℜ the vector space of d-vectors. dm×ℜ an dm × real matrix. dΖ the standard integer lattice. d

+ℜ d-space of vectors with positive components. Vol(P) volume of P. ext(P) extreme points of P. x greatest integer ≤x.

x least integer ≥ x.

νo(P) Voronoi cell of p. nb(S,v) nearest neighbor set of v in S. conv(nb(S,v)) Delaunay cell of v. dP ΖI number of integral points of a polyhedron.

2Ζ∂ IP number of integral points on the boundary of the

polyhedron. ),( pxδ delta function of x and p.

x the Euclidean norm of a vector x.

rank(A) rank of the matrix A. dim(P) dimension of the polytope P.

),...,,( 10 dvvv∆ simplex in dℜ with vertices d

dvvv ℜ∈,...,, 10 .

),...,det( 001 vvvv d −− determinant of dd × matrix whose columns are 001 ,..., vvvv d −− .

ipP U+

= signed union of ip .

∏=

n

i

ix1

multiplication symbol of n

xxx ,...,,21

.

L(P,t) the Ehrhart polynomial of a polytope P. S(a,b) the Dedekind sum of a and b. Res(f(z), z=a) residue of f(z) about z = a. ),(1 nmS Stirling number of the first kind. ),(2 nmS Stirling number of the second kind.

Ζ∈

Ζ∉−−=xif

xifxxx0

2

1))((

oP interior of a polytope P.

nB Birkhoff polytope. ( )tH n Ehrhart polynomial of the Birkhoff polytope.

int |Z| = r interior of the circle |Z| = r.

Dedication

To those they covered me with their love and kindness

My mother & My father

My brothers, sisters and My aunt

To my Dear husband Mahmood who supported me by all

means

To my children’s Rukia, Nayarh and Ali

With love.

Introduction A wide variety of pure and applied mathematics involve the problem of counting the number of integral points inside a region in space. Applications range from the very pure: number theory, toric Hilbert functions, Kostant's partition function in representation theory, Ehrhart polynomial in combinatorics to the very applied: cryptography, integer programming, statistical contingency, mass spectroscope analysis. Perhaps the most basic case is when the region is a convex bounded polyhedron. Convex polyhedra, i.e., the intersections of a finite number of half spaces of the space dℜ , are important objects in various areas of mathematics and other disciplines as seen before. In particular, the compact ones among them (polytopes), which, can equivalently be defined as the convex hulls of finitely many points in dℜ , have been studied since ancient times, for example, platonic solids, diamonds, the great pyramids in Egypt etc., [27]. polytopes appear as building blocks of more complicated structures, e.g. in combinatorial, topology, numerical mathematics and computed aided designs. Even in physics polytopes are relevant e.g., in crystallography or string theory, [31]. Probably the most important reason of the tremendous growth of interest in the theory of convex polyhedra in the second half of 20'th century was the fact that linear programming i.e., optimizing a linear function over the solutions of a system of linear inequalities became a wide spread tool to solve practical problems in industry and military. Dantzig's simplex algorithm, developed in the 40's of the last century, showed that geometric and combinatorial knowledge of polyhedra (as the domains of linear programming problems), is quite helpful for finding and analyzing solution procedures for linear programming problems, [31]. Since the interest in the theory of convex polyhedra to a large extent comes from algorithmic problems, it is not surprising that many algorithmic questions on polyhedra rose in the past, but also inherently, convex polyhedra (in particular: polytopes) give rise to algorithmic questions, because they can be treated as finite objects by definition; this makes it possible to investigate the smaller ones among them by computer programs like the polymake - system written by Gawilow and Jowing, [24].

Once chosen to exploit this possibility one immediately finds oneself confronted with many algorithmic challenges.

Also, the notion of the volume of a polytope is basic and intuitive; its computation has raised a lot of problems. In this thesis we attempt to answer some fundamental and practical question on volume computation of higher dimensional convex polytopes given by their vertices and / or facets. In particular, we study through extensive computational experiment typical behavior of the exact methods, including Delaunay and boundary triangulation, the triangulation scheme described by Cohen and Hickey and the methods presented by Lawrence, [13].

This thesis consists of three chapters. In chapter one we try to give a short introduction, provide a sketch of

what bounded polyhedron looks like and how they behave with many examples. Also we recall some methods for finding the number of integral points inside a convex polytope, [13], [25] and [4].

In chapter two we present some methods for computing the

coefficients of Ehrhart polynomial that depend on the concepts of Dedekind sum and residue theorem in complex analysis. Also, a method for counting these coefficients is introduced. The polytope that we take are with V-representations, [5] and [60]. We give a method for computing the coefficients of the Ehrhart polynomial, 3−dc , 4−dc until 9−dc also we give a formula for the differentiation of the given method.

In chapter three, a method for finding the volume of H-representation

of a polytope using Laрlace transform is presented and some basic concepts and remarks about the Birkhoff polytope and their volumes are discussed with their Ehrhart polynomials, [33], [11], [7], [8] and [9]. We make a change on the matrix, which represents the polytopes and finds a general formula for the number of integral points; also we make a change of matrix operation and study the effect of this change on the number of integral points of the polytopes. To the best of our knowledge, this result seems to be new.

٩٢

References

[1] J. Abate, G. L. Choudhury and W. Whitt, Numerical inversion of multidimensional Laplace transforms by the Laguerre method, http://citeseer.ist.psu.edu/abate96numerical.html,(1996), 1-24.

[2] T. M. Apostol, Modular functions and Dirichlet series in number

theory, Springer–Verlag Inc. 1976. [3] D. Avis and D. Bremner, How good are convex hull algorithms,

http://citeseer.ist.psu.edu/avis95how.html , (1995), 1-24. [4] A. Barvinok and J. E. Pommersheim, An algorithmic theory of lattice

points in polyhedra, new perspectives in geometric combinatorics, MSRI publications, 38, (1999), 91-147.

[5] M. Beck, The Reciprocity law for Dedekind sums via the constant

Ehrhart coefficient, AMM, 106, (5), (1999), 459-462. [6] M. Beck, Counting lattice points by means of the residue theorem,

RamanuJan J. 4, (3), (2000), 299-310. [7] M. Beck and D. Pixton, The Ehrhart polynomial of the Birkhoff

polytope,www.math.binghamton.edu/dennis/papers/birkhoff.html- 3k,(2002),1-14

[8] M. Beck, J. A. DE. Loera, M. Develin, J. Peifle and R. P. Stanley,

Coefficients and roots of Ehrhart polynomials, conference on integer points in polyhedra (13-17) July in Snowbird, (2003), 1-24.

[9] M. Beck and D. Pixton, The Volume of the 10th Birkhoff polytope,

www.math.binghamton.edu/dennis/papers/birkhoff.html-3k,(2003) [10] A. Bemporad, K. Fukuda and F. D. Torrisi, Convexity recognition of

the union of polyhedra, http://citeseer.ist.psu.edu/300246.html,(2000),1-16. [11] I. Bengtesson, A. Ericsson, M.Kuś, W. Tadej and K. Życzkowski,

Birkhoff `s polytope and unistochastic matrices N=3 and N=4, arXiv: math.CO/0402325 v2 24 Feb 2004, (2004), 1-30.

٩٣

[12] M. Brion and M. Vergne, Lattice points in simple polytopes, J. Amer. Math. Soc., 10, (2), (1997), 371-392.

[13] B. Büeler and A. Enge and K. Fukuda, Exact volume computation

for polytopes: A practical study, http://citeseer.ist.psu.edu/article/bueler98exact.html, (1998), 1-17. [14] C. S. Chan, D. P. Robbins and D. S. Yuen, on the volume of

a certain polytope, experiment Math.9, (1), (2000), 91-99. [15] C. S. Chan and D. P. Robbins, on the volume of the polytope of

doubly stochastic matrices, arXiv:math.CO/9806076 v1, 13 Jun 1998, (2002), 1-17.

[16] J. A. De Loera, R. Hemmecke, J. Tauzer and R.Yoshida, effective

lattice points counting in rational convex polytopes, J. Symbolic computation, (2003), 1-33.

[17] J. A. De Loera, D. Haws, R. Hemmecke, A user´s Guide for LattE

v1.1,http://www.math.ucdavis.edu/~Latte/manual/v1.1/pdf.,(2003), 1-22.

[18] R. Diaz and S. Robins, The Ehrhart polynomial of a lattice n-

simplex, Electronic research announcements of AMS. 2, (1), (1996), 1-6.

[19] R. Diaz, S. Robins, The Ehrhart polynomial of a lattice polytope,

Annals of Math. 145, (1997), 503-518. [20] P. Franklin, Functions of complex variables, Prentice – Hall, INC.

Englewood Cliffs, N. J., 1958. [21] B. A. Fuchs and B. V. Shabat, Functions of a complex variable and

some of their applications, Pergamon press LTD, 1964. [22] K. Fukuda, frequently asked questions in polyhedral computation,

http://www.ifor.math.ethz.ch/staff/fukuda/polyfaq/polfaq.html,(2004), 1-30.

[23] S. I. Gass, Linear programming, McGraw-Hill, Inc., 1985.

٩٤

[24] E. Gawrilow and M. Joswig, polymake: a Freme work for analyzing convex polytopes,

http://citeseer.ist.psu.edu/gawrilow99polymake.html, (1999), 1-31. [25] H. Greenberg, Polyhedral computation,

http://gcrg.ucsd.edu/classes/polyhcomp.pdf,(2003), 1-28 [26] P. Gritzmann and V. Klee, on the complexity of some basic

problems computational convexity, cmath. phys. sci-440, kluwer Acad. publ. Dordrecht, (1994), 373-466.

[27] J. Gubeladze, Course in information, Math 890 Discrete Geometry

Fall(2003),math.sfsu.edu/gubeladze/fall2003/discrete.pdf-63k., (2003).

[28] B. Grünbaum and G.C. Shephard, Pick´s theorem, the Amer. Math.

Monthly, 100, (2), (1993), 150-161. [29] R. Hans, Topics in analytic number theory, Springer - Verlag Berlin

Heidelberg New York, 1973. [30] M. Henk, J. R. Gebert and G.M. Ziegler, basic properties of convex

polytopes, http://citeseer.ist.psu.edu/henk97basic.html,(2004), 1-28. [31] V. Kaibel and M.E. Pfetsch, some algorithmic problems in polytope

theory, www.zib.de/pfetsch/publications/polycomplex.pdf, (2002), 1-25.

[32] D. Knuth, The art of computer programming, Vol. 2. Addison –

Wesley, Reading, Mass., 1981. [33] J. B. Lasserre and E. S. Zeron, A Laplace transform algorithm for the

volume of a convex polytope, arXiv: math.NA/0106168 v1 Jun 2001, (2001), 1-13.

[34] P. Mcmullen and G. C. Shephard, Convex polytopes and upper

bound conjecture, Cambridge University Press, 1971. [35] G. L. Nemhauser and L. A. Wolsey, Integer and combinatorial

optimization, John Wiley & Sons, Inc., 1988. [36] R.G. Parker and R. L. Radin, discrete optimization, Academic press,

Inc. 1988.

٩٥

[37] I. Pak, four questions on Birkhoff polytope, Annals of

combinatorics, 4, (2000), 83-90. [38] O. Pikhurko, lattice points in lattice polytopes,

http://citeseer.ist.psu.edu/oleg2000lattice.html,2, (2000), 1-14. [39] J. Pommersheim, Toric varieties, lattice points, and Dedekind sums,

Math. Ann. 295, (1993), 1-24. [40] G. Polya and R. E. Tarjan and D.R. Woods, Notes on introductory

combinatorics, Birkhaüser Bostan, Inc., 1983. [41] S. S. Rao, Optimization theory and applications, John Wiley & Sons,

1984. [42] R. P. Stanley, Enumerative combinatorics , Wadsworth & Brooks /

Cole Advanced Books & software, California, 1986. [43] R. P. Stanley, linear homogeneous Diophantine equations and magic

labeling of graphs, Duke Math. J.40, (1973), 607-632.


Introduction

Convex bounded polyhedrons are fundamental geometric objects that have been investigated since antiquity. The beauty of their theory is nowadays complemented by their importance for many other mathematical subjects, ranging from the integration theory, algebraic topology, and algebraic geometry (toric varieties) to the linear and combinatorial optimization. In this chapter we try to give a short introduction and provide a sketch of what bounded polyhedron looks like and how they behave with many examples. A convex polyhedron is an intersection of a finite number of half spaces of the space dℜ , and a convex polytope is a bounded convex polyhedron. Every convex polyhedron has two natural representations, a half space representation (H-representation) and a vertex representation (V-representation). In recent years various techniques of geometric computations associated with convex polyhedron have been discovered, see [13], [31], [16] and [17]. Also, we recall some methods for finding the volume of a convex polytope and other methods for finding the number of integral points inside a convex polytope. This chapter consists of three sections:

In section one, some basic definitions with some useful remarks about representation of the polyhedron are presented.

In section two, some methods for computing the volume of convex polytopes are given with some illustrative examples. These are classified into two groups: triangulation methods and signed decomposition methods.

The triangulation methods include boundary triangulation, Delaunay triangulation and Cohen & Hickey's triangulation, [13]. The signed decomposition methods include Lawerence's method, [25].


٢

In section three, some methods for finding the number of integral points of a convex polytope are discussed; these methods are demonstrated with some examples. 1.1 Representation of a polyhedron

The volume of a convex bounded polyhedron is not easy to compute and the basic methods for exact computation of this volume can be classified according to whether a half space representation or a vertex representation of it, [33]. Therefore, in this section some basic definitions on a convex bounded polyhedron and its representations are given.

We start this section by the following definitions:

Definition (1.1.1), [35, p.85]:

Let bAX ≤ where dmA ×ℜ∈ is a given real matrix, and mb ℜ∈ is a known real vector. A set : bAXXP d ≤ℜ∈= is said to be a polyhedron.

A polyhedron P is bounded if there exists 1

+ℜ∈ω such that

ω≤X for every PX ∈ , [35, p.86].

Definition (1.1.2), [35, p.85]:

Every bounded polyhedron is said to be a polytope. Definition (1.1.3), [35, p.84]:

Let ,...,, 21 kxxxS = where d

ix ℜ∈ , ki ≤≤1 , then S is said to be

affinely independent if the unique solution of 01

=∑=

i

k

i

i xa and

01

=∑=

k

i

ia is 0=ia , for i = 1, …, k.

Recall that a polyhedron P is of dimension k, denoted by dim(P)=k,

if the maximum number of affinely independent points in P is k+1. In this case a polyhedron (polytope) is said to be k-polyhedron (k-polytope). On the other hand, a polyhedron is of a full dimensional if dim (P) = d, [35, P.86]. Remark (1.1.1): If the polyhedron P which is defined by, : bAXXP d ≤ℜ∈= is not full dimensional, then at least one of the inequalities ii bXa ≤ ,


٣

i=1,2,…,k, is satisfied as equality by all points of P, where ia is the i-th row of the matrix A and ib are the values of the vector b, [35, p.86].

Proposition (1.1.1), [35, p.84]:

Let : bAXXP d ≤ℜ∈= then the following statements are equivalent: (a) φ≠≤ℜ∈ : bAXX d .

(b) rank(A)=rank(A|b), where ,, mdm bA ℜ∈ℜ∈ × A|b is the augmented matrix of the system AX=b and rank(A|b) is the maximum number of linearly independent rows ( columns ) of A|b.

Now, if P takes the form : bAXXP d ≤ℜ∈= , the pair A|b is said to be a half space representation or simply H-representation of P, where ,, mdm bA ℜ∈ℜ∈ × [13]. Proposition (1.1.2), [35, p.87]:

Let : bAXXP d ≤ℜ∈= be a polytope, then:

dim(P)+rank( ** bA )=d,

where ** bA denotes the corresponding rows of A|b, which represent the

equality sets of the representation A|b of P, that is, : ∗∗ =ℜ∈= bXAXP d , [35,p.86].

Definition (1.1.4), [4]: Let : bAXXP d ≤ℜ∈= be a polyhedron. If the entries of A and b have integer values then this polyhedron is said to be rational polyhedron.

Recall that for a given convex set S, a point SX ∈ is said to be vertex (or sometimes extreme point) if it does not lie on a line segment joining two other points of this set. In this case the line joining any two vertices is said to be an edge [41, p.98].

It can be easily seen that any polyhedron is a convex set in dℜ .


A lattice polytope in dℜ (sometimes called integral polytope) is a polytope whose vertices are lattice points (integral points), that is, points


٤

in dΖ . If the lattice polytope is of dimension d then this polytope is said to be a d-dimensional lattice polytope

Definition (1.1.6), [41, p.96 ]:

Given ∑=

=d

i

ii bxa1

, where ia and b are known real constants for

di ≤≤1 . The set of points d

iixX 1 == , which satisfies the above equation, is said to be a hyperplane.

Moreover, the set of points d

iixX 1 == is called a half- space if it

satisfies the inequality ∑=

≥d

iii bxa

1

, [36, p.413].

Definition (1.1.7), [10]: Let P be a polyhedron in dℜ . For dc ℜ∈ and ℜ∈b , the inequality

∑=

≤d

i

ii byc1

is called valid for P if it is satisfied by all points in P, where

d

iicc 1 == . The faces of P are the sets of the form

∑=

= ==d

i

ii

d

ii bycyYP1

1 :I for some valid inequality ∑=

≤d

i

ii byc1

.

Recall that a face F is said to be proper if PF ≠≠φ . On the other hand the faces of dimension 0 and 1 are called vertices and edges respectively. However the faces of highest dimension are termed facets. Definition (1.1.8), [12]: A polytope in dℜ is said to be simple if there are exactly d edges through each vertex, and it is called simplicial if each facet contains exactly d vertices.

It is known that a simplex in dℜ is a d-dimensional polyhedron, which has exactly d+1 vertices, [23, p.37]. Definition (1.1.9), [35, p.83 ]: Given a non empty set dS ℜ⊆ , a point dX ℜ∈ is a convex combination of points of S if there exists a finite set of points t

iix 1 = in S

and t

+ℜ∈λ with ∑=

=t

i

i

1

1λ and ∑=

=t

i

ii xX1

λ .


٥

It is known that the convex hull of S, denoted by conv(S) is the set of all points that are convex combinations of all points in S.

Now, if ,...,, 10 nvvvV = is a finite set of points in dℜ , the convex hull of V denoted by conv(V) is said to be convex polytope. In this case, V is called vertex representation or simply V-representation of P, [13].

1.2 Some methods for the volume computation of a polytope

As mentioned before, computing the volume of a polytope is very important in many real life applications, so in this section we give some methods for finding it. There is a comparative study of various volume computation algorithms for polytopes in [13]. However there is no single algorithm that works well for many different types of them, [22].

For simple polytopes, triangulation-based algorithms are more

efficient and for simplicial polytopes sign-decomposition based algorithms are better, [13].

In this section, some methods for volume computation are given with different examples.

We start this section by the following remark. Remark (1.2.1):

All known algorithms for exact volume computation decompose a given polytope into simplices, and thus they all rely on the volume formula of a simplex which is given by the following proposition, [13]:

Proposition (1.2.1), [13]:

For a polytope represented by its vertices d

dvvv ℜ∈,...,, 10 , the volume of it is given by

Vol( ),...,,( 10 dvvv∆ ) = ),...,det(!

1001 vvvv

d d −−

Where ),...,,( 10 dvvv∆ denotes the simplex in dℜ with vertices d

dvvv ℜ∈,...,, 10 and ),...,( 001 vvvv d −− is dd × matrix whose columns are 001 ,..., vvvv d −− .

Next, there are two types of methods for exact volume computation of the simple polytopes, which are discussed below:


٦

I.Triangulation methods: In these methods one has a simple polytope P in dℜ . P is

triangulated into simplices ),...,2,1( sii =∆ Us

iiP

1=

∆= . The volume of P is

simply the sum of the volumes of the simplices.

∑=

∆=s

i

iVolPVol1

)()( (1.1)

The following: boundary triangulation, Delaunay triangulation and Cohen & Hickey's combinatorial triangulation by dimensional recursion named, as triangulations method, [13].

An important difference between these methods is that the former two methods need only a V–representation while the last method requires both V- and H-representations, [13]. Before giving the signed decomposition methods, we need the following definition. Definition (1.2.1):

Let dℜ∈Ρ be a polytope, a signed union of P means, a collection

of polytopes d

k ℜ⊆ΡΡΡ ,...,, 21 such that Uk

ii

1=

Ρ=Ρ , and ji ΡΡ I is a proper

face of iΡ and jΡ , for ji ≠ .In this case we write U+

Ρ= iP , [30].

II. Signed decomposition methods:

Instead of triangulating a polytope P, one can decompose P into signed simplices whose signed union is exactly P. More specifically, P is represented as a signed union of simplices i∆ , i =1, 2, …, s. This means,

Us

iii

1=

∆=Ρ σ (1.2)

Where iσ is either +1 or –1. The volume of P is, [13].

∑=

∆=s

i

iiVolPVol1

)()( σ

1.2.1 Triangulation methods

In this subsection we discuss briefly some of the known triangulation methods that compute the volume of the polytope.


٧

(i) Boundary triangulation, [13]: In boundary triangulation, one computes the convex hull of the

perturbed points, interpreting the result in terms of the original vertices leads to a triangulation of the boundary, which by linking with a fixed interior point yields, a triangulation of P. For the convex hull computation the reverse search algorithm is chosen, [3], where only the V-representation of a polytope is required. To illustrate this method, consider the polytope which is represented by a set of vertices named a, b, c , d as given in figure (1). Using an interior point e where the boundary of a polytope is easily triangulated or already triangulated as in the case of simplicial polytopes. By linking a point e with the vertices a, b, c and d yields four triangles then, volumes of these triangles are found, summing all of these volumes the volume of this polytope is obtained.

(a) (b)

Figure (1): (a) represents a polytope P.

(b) represents a partition of the given polytope P by using the boundary triangulation method.

(ii) Delaunay triangulation, [13]:

Before we discuss this method, some basic definitions concerning the Delaunay triangulation are needed. Definition (1.2.2), [22]:

Given a set S of n distinct points in dℜ , Voronoi diagram is the partition of dℜ into n polyhedron regions (denoted by S∈ρρνο ),( ). Each region )(ρνο is called Voronoi cell of ρ , which is defined as set of

points in dℜ that are closer to ρ than other points in S, or more precisely

,)( ρρρνο −∈∀−≤−ℜ∈= •• SqqXXX d .


٨

Definition (1.2.3), [22]: Let S be a set of n points in dℜ . For each point dℜ∈ν , the nearest neighbor set denoted by )),(( νSnb of ν in S is the set of points

νρ −∈ S , which are closest to ν in Euclidean distance. Definition (1.2.4), [22]:

Let S be a set of n points in dℜ . A point dℜ∈ν is said to be a Voronoi vertex of S if ),( νSnb is maximal over all nearest neighbor sets. Definition (1.2.5), [22]: Let S be a set of n points in dℜ . The convex hull of the nearest neighbor set of Voronoi vertex ν denoted by )),(( νSnbconv is said to be a Delaunay cell of ν .

The Delaunay triangulation of S is a partition of the convex hull conv(S) into the Delaunay cells of Voronoi vertices together with their faces, [22].

Now we discuss the method of Delaunay triangulation method that

requires only the V–representation of the polytope.

The geometric idea behind a Delaunay triangulation of a d–polytope is to ' lift ' it on a paraboloid in dimension d+1. The following construction is very important to compute the Voronoi diagram, [22].

Let S be a set of n points in dℜ . For each point dS ℜ⊆∈ρ ,

consider the hyperplane tangent to the paraboloid 22

11 ... dd xxx ++=+ in 1+ℜd at ρ :

This hyperplane is represented by h (ρ ) as:

02 1

11

2 =+− +==∑∑ d

d

j

jj

d

j

j xxρρ

where ),...,2,1( djj = ρ are the coordinates of ρ , for each point ρ , the

equality in the above equation is replaced by the inequalities )(≥ , which yields a system of n inequalities that is denoted by 0≥− AXb . The polyhedron P in 1+ℜd of all solutions X to the system of inequality is a lifting of the Voronoi diagram to one higher dimensional space. [13], shows that the underlying convex hull algorithm uses the ' beneath – beyond ' method.


٩

Example (1.2.1): Consider the set of vertices:

).4,4(),4,0(),0,4(),2,1(),1,2(),0,0( 654321 ======= ρρρρρρS Here the volume of the polytope given by these vertices is to be determined. To do so, Delaunay triangulation is used to compute the volume of this polytope.

First, write down the system of linear inequalities in three variables as explained before. That is for each 6,...,2,1, =∈ jSjρ , apply the

inequalities:

02 3

2

1

2

1

2 ≥+−∑∑==

xxj

jj

j

j ρρ

we get a system of six inequalities

03 ≥x 0245 321 ≥+−− xxx 0425 321 ≥+−− xxx

0816 31 ≥+− xx 0816 32 ≥+− xx

08832 321 ≥+−− xxx .

The set of solutions 3ℜ∈X of the above inequalities represents a polyhedron P. By applying the cdd+ program [22], the Delaunay cells,

),,( 531 ρρρ , ),,( 321 ρρρ , ),,( 421 ρρρ , ),,( 632 ρρρ , ),,( 642 ρρρ and ),,( 653 ρρρ are obtained. The cell ),,( 531 ρρρ means the triangle which is represented by three vertices 31, ρρ and 5ρ , and similarly for the other cells. Therefore six triangles are obtained, summing the volumes of these triangles yields the volume of the polyhedron P is equal to 16.


١٠

Figure (2): represents the polyhedron P with the set of vertices ,...,, 621 ρρρ and Delaunay cells.

(iii) Triangulation by Cohen & Hickey, [13]: This recursive scheme triangulates a d- polytope P by choosing any vertex Pv ∈ as an apex and connecting it with the (d-1)–dimensional simplices resulting from a triangulation of all facets of P not containing v . To be precise, denote by kθ , dk ≤≤0 , the k- dimensional faces of P, and let η be a ' map ' which associates to each face one of its vertices.

Then the pyramids with apex )( dθη and bases among the facets 1−dθ with 1)( −∉ dd θθη form a dissection of the polytope.

Applying the scheme recursively to all 1−dθ results in a set of

decreasing chains of faces dd θθθθ ⊂⊂⊂⊂ −110 ... such that 1)( −∉ kk θθη for dk ≤≤1 . Then the set of corresponding simplices

))(),...,(),(( 10 dθηθηθη∆ is a triangulation of P.

To implement this recursive method, an extensive use of the double description as V-representation and H-representation is made by representing all faces as sets of vertices.

Note that in the case of Cohen & Hickey compared to a boundary triangulation all simplices in the facets containing the apex v are eliminated and therefore the number of simplices is usually reduced. Example (1.2.2):

Consider the polytope which is represented by set of vertices ,,,, 43210 ρρρρρ as illustrated in figure (3), let η be the 'map' which

assigns to each face of the polytope its vertex with the lowest number, so

0)( ρη =P , all facets which do not contain the vertex 0ρ are examined, that is, II, III and IV. The scheme of the Cohen and Hickey is applied to facet II with 1)( ρη =II . II is intersected with all facets not containing the vertex 1ρ , these are III, IV and V. The intersections with IV and V are empty, so this recursion is unsuccessful. The intersection with III yields the vertex 2ρ , and the fixed vertices 10, ρρ , 2ρ forms a first simplex. The other simplices obtained from III and IV is also marked in the figure (3). Therefore we have three triangles. Summing the volumes of these triangles yields the volume of the polytope.


١١

Figure (3): represents the partition of the polytope by Cohen & Hickey's triangulation method.

1.2.2 Signed decomposition method In this subsection Lawrence's volume formula, which is one of the signed decomposition methods, is discussed. (i) Lawrence's volume formula, [13]: Assume the polytope P is simple and choose a vector dC ℜ∈ and

ℜ∈q such that the function ℜ→ℜdf : which is defined by

f(X)= qXCT + is not constant along any edge that connected the vertices of the polytope P and TC is the transpose of C . Let V be the set of vertices defining the polytope P. For each vertex Vv ∈ , let vA be the

dd × – matrix composed by the rows of A which are binding at v. Then by using [13], vA is invertible and [ ] CAT

v

v 1−=γ . The assumption imposed

on C assures that none of the entries of vγ is zero. It is shown that

∑∏∈

=

+=Vv

d

i

v

iv

dT

Ad

qvCPVol

1

det!

)()(

γ

To illustrate this method, consider the following example.

Example (1.2.4) : Consider the polytope P which is described by the following

constraints 01 ≤− x 02 ≤− x 21 ≤x 22 ≤x 321 ≤+ xx then


١٢

−−

=

11

10

01

10

01

A ,

=

2

1

x

xX and

=

3

2

2

0

0

b

It is easy to check that the polytope of this example is simple, therefore the Lawerence volume formula can be applied. Define a function f by 21)( xxXf −= where )1,1( −=TC and q = 0. Note that f (X) is non – constant on each edge of the polytope in the figure (4), for example, on edge (1) which connect

21 and vv , 02 =x and 1x varies from

0 to 2. Therefore f (X) = 1

x which varies from 0 to 2 which means that it is nonconstant on edge (1) and similarly on each edge of P. According to figure (4), it is seen that the set of vertices, which represents the polytope, is )2,0(),2,1(),1,2(),0,2(),0,0( 54321 ===== vvvvv .

Now, consider )0,0(1 =v , this vertex satisfies the first two constraints, and this implies that

−−

=10

011v

A , hence 1det1

=vA and [ ]

= −

2

11

1

1

c

cAT

v

vγ

−=

−

−−

=1

1

1

1

10

011vγ

and for )0,2(2 =v , this vertex satisfies the second and third constraints, that is,

−=

10

012vA , 1det

2=

vA And [ ]

=

−= −

1

1

1

11

2

2 T

v

v Aγ

And similarly for the other vertices we get

−=

1

23vγ ,

−=

1

24vγ And

−−

=1

15vγ

Then Lawrence's volume formula is applied to get the volume of P.


١٣

∑∏∈

=

=Vv

i

v

iv

T

A

vCPVol 2

1

2

det !2

)()(

γ

( ) ( )

)1)(1)(1(!2

2

)1)(2)(1(!2

1

)1)(2)(1(!2

1

)1)(1)(1(!2

2

)1)(1)(1(!2

0 22222

−−−+

−−+

−++

−=

2

132414120 =+−+−++= )/()/( .

Figure (4): represents a polytope with the vertices 54321 ,,, vandvvvv 1.3 Methods for computing integral points The main objective of this section is to recall some methods for finding integral points of a polyhedron. In this work, we use the symbol

dP ΖI to denote the number of integral points in the polyhedron P,

where dΖ is the integer lattice and P is a rational polyhedron. These methods are: Method (1), [4]:

For d = 2, 2ℜ⊂P and P is an integral polyhedron. The famous formula, [42, p.240] states that

12

)(2

2 +Ζ∂

+=ΖI

IP

PareaP

That is, the number of integral points in an integral polyhedron is

equal to the area of the polyhedron plus half the number of integral points on the boundary of the polyhedron plus one. This formula is useful because it is much more efficient than the direct enumeration of integral points in a polyhedron. The area of P is computed by triangulating the polyhedron. Furthermore, the boundary ∂P is a union of finitely many straight-line intervals, and counting integral points in intervals is easy.


١٤

Method (2), [4]:

Let dP ℜ⊂ be a polytope, then one can write the number of integral points in P as dP ΖI ∑

Ζ∈

=dX

PX ),(δ

where

∉∈

=PXif

PXifPX

0

1),(δ

Before we give the next method, we need the following definition.

Definition (1.3.1), [2, p.61]: The Dedekind sum of two relatively prime positive integers a and b

denoted by ),( baS can be defined as follows,

∑=

=b

i b

ai

b

ibaS

1

),(

where

Ζ∈

Ζ∉−−=xif

xifxxx0

2

1))((

and x is the greatest integer x≤ . Remarks (1.3.1):

Dedekind sums appear in various branches of mathematics: the number theory, algebraic geometry and topology. These include the quadratic reciprocity law, random number generators [32], and lattice point problems [19]. More details about Dedekind sums are given in chapter two

Now, we are in a position that we can explain the following

method.

Method (3), [4]: Let 3ℜ⊂∆ be the tetrahedron with vertices (0, 0, 0), (a, 0, 0),

(0, b, 0), (0, 0, c) where a, b and c are pairwise coprime positive integers, then the number of integral points in ∆ can be expressed as:


١٥

),(),(

),()1

(12

1

463

cabSbacS

abcSabcc

ab

a

bc

b

accbabcacababcP

−

−−++++++++++=ΖI

This formula is useful because it reduces counting the number of

integral points to a computation of Dedekind sums, which can be done efficiently. Method (4), [4]:

Let dP ℜ⊂ be an integral polytope, for positive integer t, let tP=tX : X∈P denote the dilated polytope P. By [42, p.238] there is a polynomial p(t) called the Ehrhart polynomial of P )(tptP d =ΖI

where p(t) = 0

1

1 ... atata d

d

d

d +++ −−

furthermore, 10 =a and da is the volume of the polytope P. To illustrate these methods, consider the following examples. Example (1.3.1):

Let us consider three points in two dimensions such that )0,0()0,1(),1,0( 321 === vandvv . Then the convex hull of 321, vandvv

is a triangle in two dimensions.

We compute the number of integral points by these methods. From method (1), one can have

12

)(2

2 +Ζ∂

+=ΖI

IP

PareaP

the area of triangle, is area(P)=0.5(1)(1)=0.5 and 32 =Ζ∂ IP which

represents the number of integral points on the boundary of the triangle. Then the number of integral points of the triangle is .31)3(5.05.02 =++=ΖIP

In method (2), we have 2ΖIP ∑

Ζ∈

=2

),(X

PXδ


١٦

1),()1,0( 11 =∈= PvthenPv δ

1),()0,1( 22 =∈= PvthenPv δ

1),()0,0( 33 =∈= PvthenPv δ therefore 31112 =++=ΖIP , which is the number of integral points for

the triangle. Form method (4), one can have =Ζ2

IP 112 ++ aa

where 2a is the volume of the polytope and 1a is the half number of integral points on the boundary of the polytope. In this case,

2a = 0.5(1)(1)=0.5 and 2a = 3/2

therefore 315.15.02 =++=ΖIP .

Example (1.3.1):

Let us consider the tetrahedron 3ℜ⊂∆ with vertices (0,0,0), (3,0,0), (0,5,0), (0,0,7).

Form method (3), the number of integral points in ∆ can be

expressed as,

),(),(),(

)1

(12

1

463

cabSbacSabcSabcc

ab

a

bc

b

accbabcacababcP

−−−

++++++++++=ΖI

here a=3, b=5 and c=7. Therefore

)7,15()5,21(

)3,35()105

1

7

15

3

35

5

21(

12

1

4

753352115

6

1053

SS

SP

−−

−++++++++++=ΖI

After simple computation one can get, S(35,3)=-0.05555, S(21,5)=0.2, S(15,7)= 0.35714, Thus .4035714.02.005555.0501877.15.215.173 =−−+++=ΖIP

Chapter Two Computing the Volume and Integral Points of V-

Representation of a Polytope Using Ehrhart Polynomial

Introduction As was shown in chapter one, the Ehrhart polynomial of a convex lattice polytope counts the number of integral points in an integral dilate of the polytope. E. Ehrhart proved that the function which counts the number of lattice points that lie inside the dilated polytope tP is a polynomial in t and it is denoted by L(P,t), which is the cardinal of

)( dtP ΖI where dΖ is the integer lattice in dℜ , [4]. Many of Ehrhart's valuable results are unknown by mathematician and computer science community at this time, since many of these results have been published in local reports and in French language, [4].

In this chapter we present some methods for computing the coefficients of Ehrhart polynomial that depend on the concepts of Dedekind sum and residue theorem in complex analysis. Also, a method for computing the coefficients of the Ehrhart polynomial is introduced with general formula that counts the derivatives in the introduced method. For our knowledge this method seems to be new. The polytopes that we take are with V – representations.

This chapter consists of six sections. In section one, some basic

definitions and remarks concerning the Ehrhart polynomial are given. Section two gives a method for finding the Ehrhart polynomial of the polytope using the formula of the Dedekind sum. Another method for finding the Ehrhart polynomial using residue theorem in complex analysis is presented in section three. In section four the Ehrhart coefficients are computed and in section five, we give a method for

computing 3−dc , 4−dc until 9−dc of the Ehrhart polynomial, the general formula for the differentiation is given in section six.

Chapter Two Computing the Volume and Integral points of V-Representation of a Polytope using Ehrhart polynomial

١٨

2.1 Basic concepts about Ehrhart polynomials As seen before, the Ehrhart polynomial is very important in many fields of mathematics. Therefore some methods for finding the coefficients of this polynomial are to be listed. To do so, some basic definitions, theorems and remarks are given in this section. We start this section by the following definitions Definition (2.1.1), [19]: Let dℜ⊂Ρ be a lattice d-polytope. For +Ζ∈t , the set tP=tX : X ∈P is said to be the dialeted polytope. The definition of the Ehrhart polynomial is given below Definition (2.1.2), [42, p.235 ]:

Let dℜ⊂Ρ be a lattice d-polytope. Define a map Ν→Ν:L by

L(P,t) = card(tP dΖI ), where 'card' means the cardinality of (tPdΖI ) and N is the set of natural numbers. It is seen that L(P,t) can be represented

as: L(P,t) =1+∑=

d

i

iitc

1

, this polynomial is said to be the Ehrhart polynomial

of a lattice d-polytope P.

Remark (2.1.1): Let 2ℜ⊂Ρ be a lattice 2-polytope, the Ehrhart polynomial of P is

given by

L (P,t) = 12

12 +Β+Α tt

where A is the area of the polytope and B is the number of lattice points on the boundary of P, [28]. Theorem (2.1.1), [8]: Let dℜ⊂Ρ be a lattice d-polytope, with the Ehrhart polynomial

L(P,t) =∑=

d

i

i

itc0

. Then dc is the volume of P, while the constant term is

one, which is equal to the Euler characteristic of P.

The other coefficients of L(P,t) are not easily accessible. In fact, a method of computing these coefficients was unknown until quite recently, [4], [12] and [19].


١٩

Before we give the next theorems, we need the following definitions:

Definition (2.1.3), [35, p.86]: Let P be a polytope which is defined by : bAXXP d ≤ℜ∈= ,

X∈Ρ is said to be an interior point of Ρ if ii bXa < for i=1,2,…,m. , where

ia is the i-th row of the matrix A, ib is the i-th row of the vector b and m<d. Definition (2.1.4), [42, p.18]: Let k and j be two given positive integer and c(k,j) is the number of permutations ),...,( 1 kcc=π with exactly j cycles. Then the Stirling numbers of the first kind of k and j, denoted by S1(k,j), are defined as

),()1(),(1 jkcjkS jk−−= , and c(k,j) can be found from the recurrence

relation c(k,j)=(k-1)c(k-1,j)+c(k-1,j), k,j ≥1,

Theorem(2.1.2),(The reciprocity theorem for Ehrhart polynomial), [42 ,p.238], [19]:

The function L(P,t) satisfies the reciprocity law:

L (P,-t) = (-1) Pdim L(P o ,t) where Po is the interior of P and t is an integer. Remark (2.1.2): From theorem (2.1.2) the study of the polytope Po is essentially equivalent to the study of the polytope P, where Po is the interior of P.

A linear relation satisfied by the coefficients of all Ehrhart polynomials is established by [8] which is a continuation of the pioneering work of [Stanley,1980,1991; Betke&Mcmullen, 1985; Hibi,1995] in [8] who established several relations of linear inequalities for the coefficients and are given in the following theorems. Theorem (2.1.3), [8]: Let dℜ⊂Ρ be a lattice d-polytope with the Ehrhart polynomial

L(P,t)= ∑=

+d

i

i

itc1

1 then, )!1(

)1,()1(),()1( 11

1 −+−+−≤ −−−

d

rdScrdSc rd

d

rd

r for

r=1,2,…, d-1, where S1(k, j) denotes the Stirling numbers of the first kind of k and j.


٢٠

Theorem (2.1.4), [8]: Let dℜ⊂Ρ be a lattice d-polytope with the Ehrhart polynomial

L(P,t)= ∑=

+d

i

i

itc1

1 . Then the following inequalities are valid:

!

1

dcd ≥

)!1(2

11 −

+≥− d

dcd

∑=

− ≥−d

ii

id c0

0)1( .

2.2 Counting integral points using Dedekind sums In this section we describe the relation between the Dedekind sum and the Ehrhart polynomial of a polytope and discussed a theorem that counts the number of integral points in a polytope. This section starts by some basic definitions and remarks that are useful to discuss the method for counting the integral points of a polytope.

Recall that the Dedekind sum of two relatively prime positive integers a and b, denoted by S(a,b), is defined as

∑=

=b

i b

ai

b

ibaS

1

))))((((),(

where

Ζ∈

Ζ∉−−=xif

xifxxx0

2

1))((

and x is the greatest integer x≤ . Remark(2.2.1): The discrete Fourier expansions can be used to rewrite the Dedekind sum in terms of the Dedekind cotangent sum, that is, for two relatively prime positive integers a and b:

∑=−

=

1

1)cot()cot(

4

1),(

b

k b

k

b

ka

bbaS

ππ

where S(a,b) is the Dedekind sum of a and b, [2, p. 72].


٢١

Theorem (2.2.1), [5]: Let P denote the simplex in dℜ with the vertices (0,0,…,0), ( 1a ,0,…,0),(0, 2a ,0,…,0),…,(0,…,0, da ), where Ν∈daa ,...,1 are relatively

prime. Denote the corresponding Ehrhart polynomial by ∑=

=d

j

j

jtctPL0

),( .

Then mc is the coefficient of )1( +− ms in the Laurent series at s = 0 of

++

++

++

++∑

=−

+

))(coth(1

))(coth(1...))(coth(1))(coth(12! 1 21

1

irs

irsa

irsa

irsam dr

md

m

ρπ

πππρ

π ρ

where daa ...1=ρ . Definition (2.2.2) [34, p. 29]: Let K be a closed bounded convex set in dℜ . Then a hyperplane H of dℜ is said to support K if φ≠KH I , and K is contained in one of the closed half - spaces determined by H. Theorem (2.2.2), (Dedekind's reciprocity law), [5]: For two relatively prime positive integers a and b

+++−=+a

b

abb

aabSbaS

1

12

1

4

1),(),( (2.1)

Proof: Let P denote a simplex in 2ℜ with the vertices (0,0), (a,0), (0,b), where a, b∈N are relatively prime. From theorem (2.2.1), 0c is the

coefficient of 1−s in the Laurent series at s = 0 of

++

++

++∑=

))(coth(1))(coth(1))(coth(14 1

irsab

irsb

irsaab

ab

r

ππππ

(2.2) The Laurent series of each factor depends on r: that is for any Ν∈c , we

write the series of ))(coth1( irsc

++ π such that c divides r or not.

To illustrate this, we consider these two cases: First case: if c divides r, then there exists an integer m such that r = mc therefore,


٢٢

( )

+=

++=

++c

smi

c

sirs

c

πππcoth1coth1coth1

it is known that, Laurent series of coth(z) about z = 0 is:

π<+++−+= zz

z

n

Bzz

zz

n

n ...,)2(

)!2(...

45

1

3

11)coth(

2

23

where the sBn ' are Bernoulli numbers that are defined by the following steps:

Step (1): Expand 1

1

−Ze as

12

0

122

11

1

1 +∞

=+∑+−=

−i

i

iZzA

ze

where ,...30240

1,

720

1,

12

1531 =−== AAA

Step (2): Define, the n-th Bernoulli number nB by, [29, p.6]

11 =B , )!2(

)1( 21

12 n

BA nn

n

−− −= and 012 =+nB for n=1,2,…

By replacing z by c

sπ in coth(z) we get

1),(3

)coth( 3 <Ο++=c

sss

cs

c

c

s ππ

π

Let )(3

1)coth(1 31 ssc

sc

c

sSc Ο+++=+= − π

ππ .

Second case: assume that c does not divide r.

Let ))(coth1()( irsc

sf ++= π

then )coth(1)0(c

rif

π+=

also

+π

+ππ=′ )(coth)(csc)( irsc

irsc

hc

sf then

).coth()(csc)0(c

ir

c

irh

cf

πππ=′

Therefore, the Maclaurin series of f(s) takes the form

=cR )()coth(1 sc

ri Ο++ π .

Therefore, from the above one can get


٢٣

If c|r then )(3

1 31 ssc

sc

Sc Ο+++= − ππ

and

If c ∤ r then )()coth(1 sc

riRc Ο++= π

Introduce the notation

χ =c

r cif0

r cif1

so ))(coth1( irsc

++ π can be written as

)1(RS)irs(c

coth1 cccc χχπ −+=

++ and (2.2) becomes

( )∑=

−+−+−+ab

rababababbbbbaaaa RSRSRS

ab 1

))1())(1())(1((4

χχχχχχπ

Expand this into eight terms, which are

(

))1)(1)(1()1)(1( )1(

)1( )1()1)(1(

)1()1(4 1

babaabbabaabababab

aaabbbaaababbbbababbaa

ab

rbbababaaababbbaaababbbaa

RRRRRS

RRSRSSRRS

RSSRSSSSSab

χχχχχχχχχχχχχχχ

χχχχχχχχχπ

−−−+−−+−−+−+−−

+−+−+∑=

and consider each term according to the number of cS factors 1. Terms with one cS factor are

)()1()1()1( abaabbaababbaabbaababbbaa RRSRRSRRS χχχχχχχχχ −=+−−=−− (2.3) and similarly

)()1()1( abbababababaabb RRSRRS χχχχχ −=−− (2.4)

The term with abS is zero (note that ababbaba χχχχχ == and

abbaχχχ = ).

To compute the contribution of (2.3) we need the support

of aba χχ − in 1,2, …, ab which is 1bk1ka −≤≤•• ;

thus its contribution to (2.2) is,


٢٤

))coth(1))(coth(1(4

1

1∑

−

=

++b

r ab

ika

b

ikaa

ab

πππ

π

))cot(1))(cot(1(4

1 1

1∑

−

=

−−=b

r b

ki

b

kai

b

ππ

−

−+

−= ∑−

= b

kcot

b

kacoti

b

kcot

b

kacot1

b4

1 1b

1r

ππππ

The imaginary part in the preceding sum is zero, because the original generating function is real. The obtained result is

),(4

1

4

1baS

b−− .

Similarly (2.4) gives a contribution of ),(4

1

4

1abS

a−− .

2. There are no terms with two cS factors,

0)1()1( =−=− abababbaababbbaa RSSRSS χχχχχ and

0)1()1( =−=− bababbaababbbaa SRSSRS χχχχχ 3. Finally, the term ababbaababbbaa SSSSSS χχχχ = has a support ab, and gives a contribution of

)333

(4 πππ

πππ

πππ

πππ

π abba

a

abb

a

aba

ab

ba

ab+++++

).111

(4

1)

1(

12

1 +++++=aba

b

b

a

ab

Adding all contributions, with the fact that 0c is the coefficient of 1−s in the Laurent series at s = 0 of (2.2) and 10 =c , [42, p.235] we get

),(),()1

(12

1

4

31 abSbaS

a

b

b

a

ab−−+++=

Then


٢٥

)1

(12

1

4

1),(),(

a

b

abb

aabSbaS +++−=+

which is the Dedekind reciprocity law. 2.3 Counting integral points using the residue theorem This section is concerned with a method given in [6] to count the integral points of a given polytope by means of the residue theorem.

To do so, let dd ℜ⊂Ζ be a d–dimensional integer lattice, so the

polytope oP is defined as

><ℜ∈= ∑=

01:),...,(1

1 k

d

k k

kd

d xanda

xxxPo , with

vertices )a,0,...,0(),...,0,...,0,a,0(),0,...,0,a(),0,...,0,0( d21 , where d21 a,...,a,a are positive integers.

Recall that for Ν∈t , L(P,t) is the number of integral points in the

dilated polytope tP. Let us use the notation: d21 a...aaA = , dk21k a...a...aaA = (where ka

means the factor ka is omitted). Then L(P,t) can be written as,

≥≤∈= ∑=

0mallandta

m:)m...,m,m(card)t,P(L k

d

1k k

kdd21 Ζ .

Thus

ta...aa

)a...aa(m...)a...aa(m)a...aa(m

d21

1d21dd312d321 ≤+++ −

In other words,

tA

AmAmAm dd ≤+++ ...2211

Therefore,

AtAmd

kkk ≤∑

=1

Hence, there exists a non negative real number m such that

0m,m,AtmAm k

d

1kkk ≥=+∑

=


٢٦

Therefore, by using [6] one can get,

=+∈= ∑=

≥

+ tAmAm:)m,m,...,m(card)t,P(Ld

1k0m,km

kk

1d

d1 Ζ

L(P,t) can be interpreted as the Taylor coefficient of tAΖ for the function

.1

1

1

1...

1

1

1

1

...)1...)(1...)...(1...)(1(

...

21

2211

2

22

1

11

2222

0000

Ζ−Ζ−Ζ−Ζ−=

+Ζ+Ζ++Ζ+Ζ++Ζ+Ζ++Ζ+Ζ+=

Ζ⋅ΖΖ⋅Ζ ∑∑∑∑

====

d

dd

d

dd

AAA

AAAAAA

m

m

m

Am

m

Am

m

Am

Equivalently

=ΖΖ−Ζ−Ζ−Ζ−

Ζ=−−

0,)1)(1)...(1)(1(

Re),(21

1

dAAA

tA

stPL

=Ζ

Ζ−Ζ−Ζ−Ζ−Ζ+Ζ−Ζ

=−−

0,)1)(1)...(1)(1(

11Re

21

1

dAAA

tA

s

+

Ζ−Ζ−Ζ−Ζ−Ζ−Ζ

=−−

)1)(1)...(1)(1(

1Re

21

1

dAAA

tA

s

=Ζ

Ζ−Ζ−Ζ−Ζ−Ζ 0,

)1)(1)...(1)(1(

1

21 dAAA

+

=Ζ

ΖΖ−Ζ−Ζ−Ζ−−Ζ=

−

0,)1)(1)...(1)(1(

1Re

21 dAAA

tA

s

=Ζ

ΖΖ−Ζ−Ζ−Ζ−0,

)1)(1)...(1)(1(

1Re

21 dAAAs

then

),( tPL 10,)1)(1)...(1)(1(

1Re

21+

=ΖΖΖ−Ζ−Ζ−Ζ−

−Ζ=−

dAAA

tA

s (2.5)


٢٧

Let ΖΖ−Ζ−Ζ−Ζ−

−Ζ=Ζ−

− )1)(1)...(1)(1(

1)(

21 dAAA

tA

tf

Expression (2.5) counts the number of lattice points in tP. Therefore we need to compute the residue of )(Ζ−tf at Z=0 and use the residue theorem for the sphere ∞UC . In this notation,

1)0),((Re),( +=ΖΖ= −tfstPL (2.6)

Let djk1 ,1 :1\C jkaa

A

≤<≤=∈= ΖΖΩ The roots of )(Ζ−tf of unity in Ω, 0 and 1 are the only poles of )(Ζ−tf Now, by [21, p. 273] one can get

=ΖΖ

=∞=ΖΖ −− 0),1

(Re)),((Re tt fsfs . Therefore,

)1)(1)...(1(

)1()

1(

1

21 1...

−Ζ−Ζ−ΖΖ−Ζ=

++++

− d

d

AA

AAAtA

t zf

then, 0)0,1

(Re =

=ΖΖ−tfs .

The following lemma is needed for proving the next theorem.

Lemma (2.3.1), [20, p.204]:

The sum of the residues of a rational function at all the poles in the finite plane, together with the residue at infinity, is zero.

This theorem appears in [6] without a proof. Here, we prove it for

the sake of completeness.

Theorem (2.3.1), [6]: Let P be a polytope defined as

>≤ℜ∈= ∑=

01:),...,(1

1 k

d

k k

kd

d xanda

xxxP , (2.7)

with vertices )0,...,0,a(),0,...,0,0( 1 , ),0,...,0(),...,0,...,0,,0( 2 daa ,where


٢٨

daa ,...,1 are positive integers, and )(Ζ−tf and Ω are defined as

ΖΖ−Ζ−Ζ−Ζ−−Ζ=Ζ

−

− )1)(1)...(1)(1(1

)(21 dAAA

tA

tf ,

and

djk1 ,1 :1\C jkaa

A

≤<≤=∈= ΖΖΩ , then

)),((Re)1),((Re1),( ∑Ω∈

−− =ΖΖ−=ΖΖ−=λ

λtt fsfstPL .

Proof: Using lemma (2.3.1), the poles of )(Ζ−tf are at 1, 0 and the roots

of unity therefore,

0)),((Re

)),((Re)1),((Re)0),((Re

=∞=ΖΖ+

=ΖΖ+=ΖΖ+=ΖΖ

−

Ω∈−−− ∑

t

ttt

fs

fsfsfsλ

λ

So )),((Re)1),((Re)0),((Re ∑

Ω∈−−− =ΖΖ−=ΖΖ−==ΖΖ

λ

λttt fsfsfs

After substations the results in (2.6), we get

)),((Re)1),((Re1),( ∑Ω∈

−− =ΖΖ−=ΖΖ−=λ

λtt fsfstPL .

Theorem(2.3.2), [6]:

Let P be a lattice d-polytope given by expression (2.7). Then, the function L(P,t) is a polynomial in t. Proof: Let Ω∈λ be a B-th root of unity, where B is the product of some of the sak ' . Express tA−Ζ in terms of its power series about λ=Ζ . The

coefficients of this power series involve various derivatives of tA−Ζ , evaluated at λ=Ζ . Introduce a change of variable:

( )ωωΖ logB/1expB1

== . A suitable branch of the logarithm is chosen such that ( ) λ=)1log(B/1exp . The terms depending on t in the power

series of tA−Ζ consist therefore the derivatives of the function BtA−

Ζ , evaluated at z=1. For this, the coefficients of the power series of tA−Ζ are polynomials in t. The fact that L(P,t) is simply the sum of all these residues, L(P,t) is a polynomial in t.


٢٩

2.4 The Ehrhart coefficients In this section, some details for deriving formula of Ehrhart coefficients are given. For each coefficient of the Ehrhart polynomial

0

1

1 ...),( ctctctPL d

d

d

d +++= −− .

A formula for finding these coefficients can be derived with a small modification of )(Ζtf .

Consider the function,


−

)1)(1)...(1)(1(

)1()(

21 dAAA

ktA

kg

ΖΖ−Ζ−Ζ−Ζ−

−Ζ

=Ζ

−−

=∑

)1)(1)...(1)(1(

)1(

)(21

)(

0

dAAA

jjktAk

j

k

j

k

g

If 0)1(0

=−

−∑

=

jk

j j

k is inserted in the numerator of the above equation, we

get


−

−−Ζ

=Ζ∑∑

=

−−

=

)1)(1)...(1)(1(

)1()1(

)(21

0

)(

0

dAAA

k

j

jjjktAk

j

k

j

k

j

k

g

∑−

=

−−


−Ζ−

=Ζ1

0

)(

)1)(1)...(1)(1(

)1()1(

)(21

k

jAAA

jktAj

k d

j

k

g

∑−

=−− Ζ−

=

1

0

)( )()1(k

j

jkt

j fj

k

Recall that, 1)0 ),(f(sRe)t,P(L t +== − ΖΖ , using this relation, we obtain,

)0,)()1((Re)0),((Re1

0

)( =ΖΖ−

==ΖΖ ∑

−

=−−

k

j

jkt

j

k fj

ksgs


٣٠

)0,)((Re)1(1

0

)( =ΖΖ−

=∑−

=−−

k

j

jkt

j fsj

k

∑−

=

−−−

=

1

0

)1))(,(()1(k

j

j tjkPLj

k

∑ ∑−

=

−

=

−

−−−

=1

0

1

0

)1())(,(()1(k

j

jk

j

j

j

ktjkPL

j

k

∑−

=

−+−−

=Ζ1

0

)1())(,(()1()(k

j

kj

k tjkPLj

kg .

The following lemma is needed to derive the formula of the

coefficients of the Ehrhart polynomial. But, before that we give the definition of the Stirling number of the second kind and its properties.

Definition(2.4.1), [42, p. 33]: The Stirling number of the second kind ),(2 kmS is defined as, the number of partitions of an m –set into k – blocks.

The following properties of ),(2 kmS are known, [42, p. 33-35].

0),(2 =kmS if k > m (2.8) 1)1,(2 =mS (2.9) 1),(2 =mmS (2.10)

∑=

−−

=k

j

mjk jj

k

kkmS

0

2 )1(!

1),( (2.11)

)1,1(),1(),( 222 −−+−= kdSkdkSkmS Lemma(2.4.1), [6]:

Suppose that 0

1

1 ...),( ctctctPL d

d

d

d +++= −− , then for dk ≤≤1

∑=

==d

km

mm2k tc)k,m(S!k)0),(g(sRe ΖΖ

where ),(2 kmS denotes the Stirling number of the second kind of m and k

and 0c =1. Proof:

Suppose that

∑ ∑−

= =

=−−

1

0 0,))(,()1(

k

j

d

m

mmk

j tbtjkPLj

k (2.12)


٣١

So that for m > 0,

∑ ∑−

= =

−−

=−−

=

1

0 0

, )1()()1(k

j

k

j

mjk

m

m

m

j

mk jj

kcjkc

j

kb

using the identity (2.11) of the Stirling number so ),(! 2, kmSkcb mmk = for m > 0.

and by formula (2.8), we conclude that 0, =mkb for km <≤1 . The

constant term in formula (2.12) is

∑−

=

−−=−

=1

0

000, )1()1(k

j

kj

k ccj

kb

Therefore,

∑ ∑−

= =

=−−

1

0 02 ),(!))(,()1(

k

j

d

m

mm

j tkmSkctjkPLj

k

∑=

−−=d

km

km

m tkmSck )1(),(! 2

Then,

∑ ∑−

= =

=−+−−

1

02 ),(!)1())(,()1(

k

j

d

km

mm

kj tkmScktjkPLj

k

Therefore,

∑=

==ΖΖd

km

m

mk tckmSkgs ),(!)0),((Re 2.

The following theorem appears in [6] without proof. Here we prove

it for the sake of completeness. Theorem (2.4.1), [6]:

Let P be a lattice d-polytope given by expression (2.7), with the Ehrhart polynomial 0

1

1 ...),( ctctctPL d

d

d

d +++= −− , then for dk ≤≤1

))),((Re)1),(((Re!

1),(2 λ

λ∑∑

Ω∈=

=ΖΖ+=ΖΖ−=k

kk

d

km

m

m gsgsk

tckmS


٣٢

where dj...jj1 ,1 :1\C 1k21

a...a

A

k1kj1j ≤<<<≤=∈= +

+ΖΖΩ Proof:

Since )0,)()1((Re)0),((Re1

0)( =ΖΖ−

==ΖΖ ∑

−

=−−

k

jjkt

j

k fj

ksgs

and from

∑Ω∈

−−− =ΖΖ−=ΖΖ−==ΖΖλ

λ)),((Re)1),((Re)0),((Re ttt fsfsfs

therefore,

∑Ω∈

−−−−−− =ΖΖ−=ΖΖ−==ΖΖλ

λ)),((Re)1),((Re)0),((Re )()()( jktjktjkt fsfsfs

Now

)]),((Re)1),(([Re)1(

)0,)((Re)1(

)()(

1

0

1

0)(

λ=ΖΖ+=ΖΖ−

−=

=ΖΖ−

∑∑

∑

Ω∈λ−−−−

−

=

−

=−−

jktjkt

jk

j

k

jjkt

j

fsfsj

k

fsj

k

∑

Ω∈λ

λ=ΖΖ−=ΖΖ−==ΖΖ )),((Re)1),((Re)0),((Re kkk gsgsgs

By using lemma(2.4.1) we get

))),((Re)1),(((Re!

1),(2 λ

λ∑∑

Ω∈=

=ΖΖ+=ΖΖ−=k

kk

d

km

m

m gsgsk

tckmS .

The following corollary appears in [6] without proof. Here we prove

it for the sake of completeness.

Corollary (2.4.1), [6]: For m > 0, mc is the coefficient of mt in

))),((Re)1),(((Re!

1 λ=ΖΖ+=ΖΖ−∑

Ω∈λ m

mm gsgsm


٣٣

Proof:

By using the fact that 1),(2 =mmS in theorem (2.4.1), we get the result of corollary (2.4.1).

Now, we give the following proposition with its proof, which we

need for proving the next theorem. Proposition (2.4.1), [20, p. 265]:

Let f (z) be analytic inside and on a simple closed curve C except at a pole a of order m inside C. The residue of f (z) at a is given by

)()()!1(

1lim

1

1

1 zfazdz

d

ma m

m

m

az−

−=

−

−

→− (2.13)

Proof:

If f (z) has a pole a of order m, then the Laurent series of f (z) at a is

...)()(...)()(

)( 2

2101

1

1 +−+−++−

++−

+−

= −−

+−− azaazaaaz

a

az

a

az

azf

m

m

m

m

Then multiplying both sides by maz )( − , we have

...)()(...)()()( 0

1

11 +−+−++−+=− −−+−−

mm

mm

m azaazaazaazfaz (2.14) From Taylor's theorem one can see that the coefficient of 1)( −− maz in the expansion (2.14) is

)()()!1(

1lim

1

1

1 zfazdz

d

ma m

m

m

az−

−=

−

−

→− .

Remark(2.4.1):

We know that from the definitions of the Ehrhart polynomial, the leading coefficient is the volume of the polytope and the constant term is one; these are termed as the trivial coefficient of the Ehrhart polynomial, the other coefficients are nontrivial.


٣٤

Theorem (2.4.2) [6]:

Let dℜ⊂Ρ be a lattice d-polytope, with vertices )0,...,0,0( , ),...,0,0(),...,0,...,0,( 1 daa where d21 a...,a,a are pairwise relatively prime

integers. The first nontrivial Ehrhart coefficients 2−dc , 3d ≥ is given by,

)),(...),(()!2(

1112 dddd aASaASC

dc −−−

−=−

where ),( baS denotes the Dedekind sum and

)...1

(12

1)...(

4

1

1

1,12,1

d

dddd a

A

a

A

AAAdC +++++++= − ,

d21 a...aaA = , dk21k a...a...aaA = (where ka means the factor ka is omitted),

and kjA , denotes dkj aaaa ...ˆ...ˆ..1 .

Proof:

Consider ΖΖ−Ζ−Ζ−Ζ−

−Ζ=Ζ−−

− )1)(1)...(1)(1(

)1()(

21

2

2 dAAA

dtA

dg

since d21 a,...,a,a are pairwise relatively prime, ( )Zgd 2− has simple poles

at the d21 a,...,a,a - th roots of unite. Let λλ ≠= 11a . Then,

λ=Ζ

Ζ−Ζ−Ζ−−Ζ

λλ−λ−=λ=ΖΖ

−−

− ,)1)...(1)(1(

)1(Re

)1)(1(

1)),((Re

321

2

2 dAAA

dtA

Ad sgs

A change of variable

== ωωΖ logB

1expB

1 is made, where a suitable

branch of logarithm such that λ=

)1log(

a

1exp

1

, thus

=−−

−−−

==ΖΖ−−

− 1,)1)...(1(

)1(Re

)1)(1(

1)),((Re

21

2

1

2 ωωω

ωλλλλ

λdBB

dtB

Ad sa

gs

where dkkd aaaBaaB ...ˆ...,,..., 22 == . We change ω by Z. Claim

22

1,)1)...(1(

)1(Re

2

−−−

−=

=ΖΖ−Ζ−

−Ζ d

BB

dtB

tsd

to prove this, first note that the Taylor series of 2)1( −− −Ζ dtB about Z = 1 is


٣٥

))1(()1()()1( 2222 −−−−− −ΖΟ+−Ζ−=−Ζ ddddtB tB Now for Ν∈m

m

1

1

1lim1,

1

1sRe

m1m

−=−

−=

=− → Ζ

ΖΖΖ Ζ

Putting all of this together, we obtain

))...((

)(1,

)1)...(1(

)1(Re

2

22

2

d

d

BB

dtB

BB

tBs

d −−−=

=ΖΖ−Ζ−

−Ζ −−−

2d

2d

d

2d

2

2d

d

2d

22d

ta ...a

a ...at −−−

−−−

−=−=

as desired. Therefore

)1)(1()),((Re

1

1

2

2 λλλ

−−−==ΖΖ

−

− A

d

d a

tgs

Adding up all the residues at the tha −1 roots of unity ≠ 1, we get

∑≠=

− =ΖΖλλ

λ1

21

)),((Rea

dgs ∑−

=

−

−−−=

1

11

2 1

1 )1)(1(

1a

kkkA

d

a

t

ξξ

where ξ is a primitive tha −1 roots of unity. This finite sum is practically a Dedekind sum:

∑∑−

=

−

=

−++

−++=

−−

1a

1kk

k

kA

kA

1

1a

1kkkA

1

1

1

11

1 1

11

1

11

a4

1

)1)(1(

1

a

1

ξξ

ξξ

ξξ

After simple computations one can get,

∑−

= −−

1

11

1

1 )1)(1(

11 a

kkkAa ξξ

)1(4

11

1

−= aa

−− ∑−

=

1

1 11

1

1

1

cotcot4

1 a

k a

k

a

kA

a

ππ

∑−

=

+

1a

1k 11

1

1

1

a

kcot

a

kAcot

a4

i

ππ

the imaginary terms disappear and the cotangent sum can be rewrite in terms of the Dedekind sum , so we get


٣٦

∑−

= −−

1

11

1

1 )1)(1(

11 a

kkkAa ξξ

),(4

1

4

111

1

aASa

−−=

and therfore

∑≠=

−−

−−−==

λλ

λΖΖ1

11

1

2d

2d1a

)a,A(Sa4

1

4

1t)),(g(sRe .

Similar expressions for the residues at the other roots of unity are obtained, so that corollary (2.4.1) give us for 3≥d

−−−

+++−−

−=− )a,A(S...)a,A(S

a

1...

a

1

a

1

4

1C

4

d

)!2d(

1c dd11

d21

2d

(2.15) where C is the coefficient of 2−dt of )1),((Re 2 =ΖΖ−dgs , Next, )0),((Re)1),((Re 22 =Ζ==ΖΖ Ζ

−− egesgs d

Z

d

=Ζ−−−−

−=ΖΖΖΖ

−Ζ−

0,)1)(1)...(1)(1(

)1(Re

21

2

eeee

es

dAAA

dtA

.

Since ))(()()1( 122 −−−Ζ− ΖΟ+Ζ−=− dddtA ttAe and

)(12

1

2

1

1

1 31 ΖΟ+Ζ−+Ζ−=−

−Ζe

,

the coefficient of 2−dt of )1),((Re 2 =ΖΖ−dgs turns out to be

+−−=

+−

d

dd

AAAC

...

)1(

12

1)(

1

12 +− +

d

d

AA

A

...

)1(

2

11

+

−+−

+

11

1

...

)1(...

d

d

d

AA

A

4

1

++− −

......

)1(

2

1

d

d

AA+−

−

−

11

1

...

)1(

d

d

AA+− −

d

d

AA ...

)1(

3

1

−+−

−

−−

21

1

42

1

...

)1(...

...

)1(

d

d

d

d

AAAAA

+++−

+++++−= −

d

ddd

d a

A

a

A

AAA

aa...

1

12

1...

1...

1

4

1

1

1,12,1

1

By substituting this into equation (2.15) we get the result.


٣٧

2.5 Computing 3−dc and 4−dc in The Ehrhart Polynomial

As seen before, the leading coefficient of the Ehrhart polynomial represents the volume of the polytope, the second coefficient represents half of the surface area of the polytope and the constant term is one, while the other coefficients are unknown.

In this section we find the non trivial coefficients 3−dc and 4−dc for

the d–polytope with d≥4 and d≥5 respectively, where P is represented by a list of vertices 0,...,0,0( ), ),0,...,0,( 1a )0,...,0,,0( 2a )a,0,...,0,0(,..., d ,

such that daa ,...,1 are pairwise relatively prime positive integers. By corollary (2.4.1), if we define )(3 Zgd − as


−−

− )1)(1()1)(1(

)1()(

21

3

3 dAAA

dtA

dgL

where d21 aaaA ⋅⋅⋅= , dk21k aaaaA ⋅⋅⋅⋅⋅⋅= and ka means that the factor

ka is omitted, then the poles of the function )(3 Zgd − are at Z = 0, 1 and

the roots of unity.

We find the residues of the function )(3 Zgd − at these poles.

Since daa ,...,1 are pairwise relatively prime therefore )(3 Zgd − has simple

poles at daa ,...,1 -th roots of unity. Let λλ ≠=11a and since,

daaA ⋅⋅⋅= 1 , daaaA ⋅⋅⋅= 321 ,…, 121 −⋅⋅⋅= dd aaaA , therefore

ΖΖ−Ζ−⋅⋅⋅Ζ−Ζ−

−Ζ=Ζ−⋅⋅⋅⋅⋅⋅⋅⋅⋅

−⋅⋅⋅−

− )1)(1()1)(1(

)1()(

121312

1 3)(

3 ddd

d

aaaaaaaa

daat

dg .

Now at λ=Z ,

01 2 ≠− ⋅⋅⋅ daaλ and 01 ≠− λ . Therefore

A change of variables

== ωωΖ log

a

1exp

1

a1

1 is made, where

a suitable branch of logarithm such that λ=

)1log(

a

1exp

1

, thus


٣٨

=−−

−

−−==ΖΖ

−−

−

1,)1)...(1(

)1(Re

)1)(1(

1)),((Re

2

1

3

1

3

ωωω

ω

λλλλ

λ

dBB

dtB

Ad

s

ags

where dk32kd32 a...a...aaB,a,...,a,aB == . since )0),((Re)1),((Re =Ζ==ΖΖ ΖΖ efesfs , then

=Ζ−Ζ−

−Ζ −−

1,)1)...(1(

)1(Re

2

3

ZsdBB

dtB

=Ζ−−

−= Ζ

−Ζ−Ζ

0,)1)...(1(

)1(Re

2

3

dBZB

dtB

ee

ees

Let α = tB, then

)0,)1)...(1(

)1((Re

2

3

=Ζ−−

−ΖΖ

−Ζ−Ζ

dBB

dtB

ee

ees = )0,

)1)...(1(

)1((Re

2

3

=Ζ−−

−ΖΖ

−Ζ−Ζ

dBB

d

ee

ees

α

.

By writing the Maclaurin series for exponential function one can get,

=Ζ

−−−−

−−−−

−++−+−

−

0,...

!2)(

11...!2)(

11

)1(...!3

)(

!2

)(1

Re22

22

332

ZBZB

ZBZB

ZZZe

sd

d

d

Z

L

ααα

after simple computations the above residue can be written as,

=

+++

++

−+−

−⋅⋅⋅−−

−

−++−

−

0,...

!3)ZB(

!2)ZB(

1...!3)ZB(

!2)ZB(

1

...!3

)Z(

!2

)Z(1

z)B()B(

e)(sRe

2

dd

2

22

3d2

1d3d

d2

Z3d

Ζ

ααα

L

Let

3

33

22

...!4!3!2

1−

+−+−=d

ZZZIααα

1

3322

222

2 ...!4!3!2

1−

++++= ZB

ZB

ZB

J

1

3332

233

3 ...!4!3!2

1−

++++= ZB

ZB

ZB

J


٣٩

M 1

33

22

...!4!3!2

1−

++++= ZB

ZB

ZB

J dddd

then

=−−

− −−

0,)e1)...(e1(

)1e(esRe

d2 BB

3d

ΖΖΖ

ΖαΖ

=

−−−=

−

0),()()(

)(Re 22

2

3

ZJIJZBB

es d

d

Zd

LL

α

for the function )()()(

)(22

2

3

d

d

Zd

JIJZBB

eL

L −−− −α we have a pole of order two at

zero.

Let d

Z JJIJeZ L32)( =φ , and ))...((

)(

2

3

d

d

BB −−=

−αγ

After simple computations on γ , we get B

t d 3−

=γ . By the formula for

finding the residues given by (2.13), if we consider

2

)()(

Z

ZZf

γφ= , then !1

)0()0),((Re

γφ′==ΖΖfs ,where

d

Z

d

Z JJIJeJJJIeZZ ′++′+=′ LL 3232 ...)()( φφ . Let

d

Z JJJIeK L321′= , d

Z JJJIeK L322′= ,…, d

Z

d JJIJeK ′= L32 therefore

dKKKZZ ++++=′ ...)()( 21φφ at Z = 0, we compute )0(φ′ , after simple computations we get

!2

)3()...(

!2

11)0( 32

−−+++−=′ dBBB d

αφ

therefore,

−−+++−==−

!2

)3()...(

!2

11)0),((Re 32

3 dBBB

B

tZZfs d

d α .

Let

+++−= )...(!2

11

132 dBBB

BD

Therefore,


٤٠

3

1

3 )1)(1()),((Re

1

−− −−

==ΖΖ d

Ad ta

Dgs

λλλ .

all the 1a -th roots of unity ≠ 1 are added up to get

∑∑≠=

−

≠=− −−

==ΖΖλλλλ λλ

λ11

3

1

31

11 )1)(1(

1)),((Re

aa

A

d

d a

Dtgs .

Let ξ be a primitive tha −1 roots of unity, therefore

∑≠=

−

−−λλ λλ11

3

11 )1)(1(

1

a

A

d

a

Dt∑

−

=

−

−−=

1

11

3 1

1 )1)(1(

1a

kkkA

d

a

Dt

ξξ

then

∑−

= −−

1

11

1

1 )1)(1(

11a

kkkAa ξξ )1(2

11

)1(2

111 1

11

1

1

11

k

kka

kkA

kAkA

a ξξξ

ξξξ

−++−⋅

−++−= ∑

−

=

−++⋅

−++= ∑

−

=k

ka

kkA

kA

a ξξ

ξξ

1

11

1

11

4

1 1

11

1

1

1

∑−

=

−+⋅

−++

−++

−++=

1

11

1

1

1

1

1

1

1

1

1

1

1

1

11

4

1 a

kk

k

kA

kA

kA

kA

k

k

a ξξ

ξξ

ξξ

ξξ

−+⋅

−++

−++

−++= ∑ ∑ ∑

−

=

−

=

−

=

1

1

1

1

1

11

1 1 1

1

1

1

1

1

1

1

1

1

1

1

11

4

1 a

k

a

k

a

kkA

kA

k

k

kA

kA

k

k

a ξξ

ξξ

ξξ

ξξ

−+⋅

−++

−++

−++−= ∑∑

−

=

−

=

1

1

1

11

1

1

1

1

11

1

1

1

1

1

1

1

1

1

1

4

1)1(

4

1 a

kkA

kA

k

ka

kkA

kA

k

k

aa

a ξξ

ξξ

ξξ

ξξ

Now, since 1

1

1 a=ξ , then by using the formula for finding the roots in

the complex plane, i

a

k

k er 1

2 π

= , k = 0,1,…, 11 −a . We obtain

∑−

=

−++

−+1

1

1

1

1

1

1

1

1a

kkA

kA

k

k

ξξ

ξξ =∑

−

=

−

++−

+1

1)

2(

)2

(

)2

(

)2

(1

1

1

1

1

1

1

1

1

1

1a

ki

a

Ak

ia

Ak

ia

k

ia

k

e

e

e

eπ

π

π

π

.

and


٤١

∑−

=

−+⋅

−+1

1

1

1

1

11

11a

kkA

kA

k

k

ξξ

ξξ

∑−

=

−

+⋅

−

+=1

1)

2(

)2

(

)2

(

)2

(1

1

1

1

1

1

1

1

1

1

1a

ki

a

Ak

ia

Ak

ia

k

ia

k

e

e

e

eπ

π

π

π

But

−+−==

iz

iz

e

ei

z

zz

2

2

1

1

)sin(

)cos()cot(

hence =

−

++−

+∑

−

=

1

1)

2(

)2

(

)2

(

)2

(1

1

1

1

1

1

1

1

1

1

1a

ki

a

Ak

ia

Ak

ia

k

ia

k

e

e

e

eπ

π

π

π

∑−

=

+−1

1 1

1

1

1

cotcot1a

k a

kA

a

k

i

ππ

and

∑−

=

−+⋅

−+1

1

1

1

1

11

11a

kkA

kA

k

k

ξξ

ξξ

∑−

=

π⋅π−=1

1 1

1

1

1

cotcota

k a

kA

a

k

∑−

=

⋅−=

1

1 1

1

1

1

cotcota

k a

kA

a

k ππ

therefore

−+⋅

−++

−++

−++− ∑∑

−

=

−

=

1

1

1

11

1

1

1

1

11

1

1

1

1

1

1

1

1

1

1

4

1)1(

4

1 a

kkA

kA

k

ka

kkA

kA

k

k

aa

a ξξ

ξξ

ξξ

ξξ

)1(4

11

1

−= aa ∑

−

=

++

1

1 1

1

11

1

cotcot4

a

k a

kA

a

k

a

i ππ∑

−

=

⋅−

1

1 1

1

11

1

cotcot4

1 a

k a

kA

a

k

a

ππ

The imaginary terms disappear, and then the above equation can be written as

4

1

14

1

a− ∑

−

=

⋅−

1

1 1

1

11

1

cotcot4

1 a

k a

kA

a

k

a

ππ4

1=14

1a

− ),(4

411

1

1 aASa

a−

where ),( 11 aAS is the Dedekind sum of 11 and aA . Hence

)),(4

1

4

1()),((Re 11

1

3

1

31

aASa

Dtgs d

da

−−==ΖΖ −

≠=−∑

λλ

λ .

Similar expressions are obtained for the residues at the other roots of unity.

Now we find the residue at )(3 Zgd − at Z = 1, we have


٤٢

)0),((Re)1),((Re 33 =Ζ==ΖΖ Ζ−

Ζ− egesgs dd

then

==ΖΖ− )1),((Re 3dgs

=Ζ−−−−

−ΖΖΖΖ

−Ζ−Ζ

0,)1)(1)...(1)(1(

)1(Re

21

3

eeeee

ees

ZAAA

dtA

d

.

By writing the Maclaurin series for exponential function we get,

=Ζ

−−−−⋅

−−−−

−−−−

−++

α−

α+α−

−

0,

...!2

11...!2

)(11...

!2

)(11

)1(...!3

)(

!2

)(1

Re222

22

332

ZZ

ZAZA

ZAZA

ZZZ

sd

d

d

L

where tA=α , then the above residue becomes

=Ζ

+++

+++

++

−

α+

α−

⋅⋅⋅α

−

−

0,

...!3!2

1...!3

)(

!2

)(1...

!3

)(

!2

)(1

...!3

)(

!2

)(1

)()(

)(Re

222

11

32

4

1

3

ZZZAZAZAZA

ZZ

ZAAs

dd

d

d

d

L

the function for which we want to find the residue has a pole of order four at zero. Let

++⋅

++

+++

++−=

−

...!3!2

1...!3

)(

!2

)(1...

!3

)(

!2

)(1

...!3

)(

!2

)(1

)(222

11

32

ZZZAZAZAZA

ZZ

Zdd

d

L

αα

φ ,

d

d

AA ...1

3−

= αγ and 4

)()(

z

zzf

γφ=

By the formula for finding the residue given by (2.13), we get

!3

)0()0),(f(sRe

)3( γφΖΖ == .

Let 3

33

22

...!4!3!2

1−

+−+−=d

ZZZIααα ,

1

32 ...Z!4

1Z

!3

1Z

!2

11h

−

++++=


٤٣

1

33

122

111 ...Z

!4

AZ

!3

AZ

!2

A1J

−

++++= ,1

33

222

222 ...Z

!4

AZ

!3

AZ

!2

A1J

−

++++= ,…,

and 1

33

d22

ddd ...z

!4

Az

!3

Az

!2

A1J

−

++++= .

Then hJJIJZ dL21)( =φ and

hJJIJhJJIJhJJJIhJJJIZ dddd′+′++′+′=′ LLLL 21212121 ...)(φ

let

hJJJIK dL211′= , hJJJIK dL212

′= ,…, hJJIJK dd′=+ L211 and

hJJIJK dd′=+ L212

hence

2121 ...)( ++ ++++=′dd KKKKZφ and 2121 ...)( ++ ′′+′′++′′+′′=′′′

dd KKKKZφ Now,

3

33

22

...!4

)(

!3

)(

!21

−

+−+−=d

ZtA

ZtA

ZtA

I

therefore

( )

++−⋅

+−+−−=′

−

...!3

)(2

!2...

!4

)(

!3

)(

!21)3(

24

33

22

ZtAtA

ZtA

ZtA

ZtA

dZId

Differentiating I ′ to get I ′′ and I ′′′ , then put Z = 0 in the obtained expression to get

1)0( =I

−−=′!2

)3()0(tA

dI

+−−−=′′!3

)(2)

!2)(4()3()0(

22 tAtA

ddI


٤٤

−+

−

−+−−

+

−−−−=′′′

!4

)(!3

!3

)(2

2)4(

!3

)(2))(4(

!2)6)(5()3()0(

322

3

tAtAtAd

tAtAd

tAdddI

For

1

33

122

111 ...

!4!3!21

−

++++= ZA

ZA

ZA

J

( )

+++

++++−=′−

...!4

3

!3

2

!2...

!4!3!21 2

3

1

2

11

2

33

122

111 Z

AZ

AAZ

AZ

AZ

AZJ

Differentiate J ′ to get J ′′ and J ′′′ , then put Z = 0 in the obtained expressions to get

1)0(1 =J , !2

)0( 11

AJ −=′ ,

!3)0(

2

11

AJ =′′ and 0)0(1 =′′′J .

In a similar way, we get the other differentiation of

dJJJ ,...,, 32 and h, then

=

=

⋅⋅⋅

−

0),()()(

)(Re

41

3

ZZzAA

tAs

d

d

φ )0(!3)()(

)( )3(3

1

3

φ−−

⋅⋅⋅

d

d

d t

AA

A .

Let !3

)0(

)()(

)( )3(

1

3 φ⋅⋅⋅⋅

=−

d

d

AA

AC

So by corollary (2.4.1) we get for d≥4, 3−dc , which is the coefficient of

3−dt of

))),((Re)1),(((Re)!3(

1

3

33 λλ∑

−Ω∈−− =ΖΖ+=ΖΖ

−−

d

dd gsgsd

So

−

−−−

++−

−−=− CaASaAS

aa

dD

dc dd

d

d ),(...),(1

...1

4

1

4)!3(

111

1

3

Thus we have proved the following:


٤٥

Theorem (2.5.1): Let P denote the polytope in )4( ≥ℜ dd with vertices

)a,...,0,0(),...,0,...,0,a(),0,...,0,0( d1 where daa ,...,1 are pairwise relatively prime positive integers. Then 3−dc is given by

−

−−−

++−

−−=− CaASaAS

aa

dD

dc dd

d

d ),(...),(1

...1

4

1

4)!3(

111

1

3

where S(a,b) is the Dedekind sum of a and b,

+++−= )...(!2

11

132 dBBB

BD ,

!3

)0(

)()(

)( )3(

1

3 φ⋅

⋅⋅⋅=

−

d

d

AA

AC ,

+++⋅

+++

+++

++−=

−

...!3

Z!2

Z1...

!3)ZA(

!2)ZA(

1...!3)ZA(

!2)ZA(

1

...!3)tBZ(

!2)tBZ(

1)Z(

22

dd

2

11

3d2

L

φ .

d21 a...aaA = , dk21k a...a...aaA = , ka means the factor ka is omitted,

dk32kd32 a...a...aaB and a...aaB == . In a similar way, we get 4−dc for the d–polytope P (d≥5), where P is

represented by a list of vertices ),...,0,0(),...,0,...,0,(),0,...,0( 1 daa , daa ,...,1 are pairwise relatively prime positive integers. By corollary (2.4.1), if we define )(4 Zgd − as


−−

− )1)(1)...(1)(1(

)1()(

21

4

4dAAA

dtA

dg

where daaA ⋅⋅⋅= 1 , ddk aaaA ⋅⋅⋅⋅⋅⋅= ˆ

1 and ka means that the factor ka

is omitted. The poles of the function ( )Zg d 4− are at Z = 0, 1 and the roots of unity. Now we compute the residues at these poles.

Since daa ,...,1 are pairwise relatively prime therefore ( )Zgd 4−

has simple poles at daa ,...,1 -th roots of unity.

Let λλ ≠= 11a and since, daaA ⋅⋅⋅= 1 , daaaA ⋅⋅⋅= 321

Therefore


٤٦

ΖΖ−⋅⋅⋅Ζ−Ζ−−Ζ=Ζ ⋅⋅⋅⋅⋅⋅

−⋅⋅⋅−

− )1()1)(1(

)1()(

322

1 4)(

4 dd

d

aaaaa

daat

dg .

Now at λ=z

01 1 ≠− Aλ and 01 ≠− λ . Therefore

=−−−

−−−

==−−

− λΖΖΖΖ

Ζλλλ

λΖΖ ,)1)...(1)(1(

)1(sRe

)1)(1(

1)),(g(sRe

d321 AAA

4dtA

A4d

We make a change of variables

== ωωΖ log

a

1exp

1

a1

1 , where a suitable

branch of logarithm such that λ=

)1log(

a1

exp1

, thus,

=

−−−

−−==

−−

− 1,)1)...(1(

)1(sRe

a)1)(1(

1)),(g(sRe

d21 BB

4dtB

1

A4d ωωω

ωλλλλ

λΖΖ

where dk32kd32 a...a...aaB,a...aaB == .

By the same steps followed before, we obtain with α =tB

)0,)1)...(1(

)1((Re

2

4

=Ζ−−

−ΖΖ

−−Ζ

dBB

dZ

ee

ees

α

=Ζ

+++

++

−

α+

α−

−⋅⋅⋅−α−=

−

−++−

−

0,

...!3

)(

!2

)(1...

!3

)(

!2

)(1

...!3

)(

!2

)(1

)()(

)(Re

2222

42

142

4

ZBZBZBZB

ZZ

ZBB

es

dd

d

dd

d

Zd

L

let

4

33

22

...!4!3!2

1−

+−+−=d

ZZZIααα

13322

222

2 ...)!4!3!2

1( −++++= ZB

ZB

ZB

J

M

133

22

...)!4!3!2

1( −++++= ZB

ZB

ZB

J dddd


٤٧

then

=Ζ

−−−

ΖΖ

−Ζα−Ζ

0,)1()1(

)1(Re

2

4

dBB

d

ee

ees

L

=

−−α−=

−

0),()()(

)(Re 23

2

4

ZJIJZBB

es d

d

Zd

LL

So for the function in the above residue we have a pole of order three at zero.

Let d

Z JJIJeZ L32)( =φ and ))...((

)(

2

4

d

d

BB −−=

−αγ

after simple commutations on γ we get 2

4

B

t d −

−=γ

By the formula for finding the residues given by (2.13), if we consider

3

)()(

Z

ZZf

γφ= , then !2

)0()0),((Re

γφ ′′==ΖΖfs

Let )0(!2

12

φ ′′=B

D

Therefore,

4

1

4 )1)(1()),((Re

1

−− −−

−==ΖΖ d

Ad ta

Dgs

λλλ .

all the 1a -th roots of unity ≠ 1 are added up to get

∑∑≠=

−

≠=− −−

−==ΖΖλλλλ λλ

λ11

4

1

41

11 )1)(1(

1)),((Re

aa

A

d

d a

Dtgs .

With the same procedure that we used to get 3−d

C , we obtain

−−−== −

≠=−∑ )a,A(S

a41

41

Dt)),(g(sRe 11

1

4d

1

4d1a λλ

λΖΖ .

Similar expressions are obtained for the residues at the other roots of unity.

Now we find the residue of )(4 Zgd − at Z = 1 by using

)0),((Re)1),((Re 44 =Ζ==ΖΖ Ζ−− egesgs d

Z

d then

==ΖΖ− )1),((Re 4dgs

=Ζ−−−−

−ΖΖΖΖΖ

−Ζ−Ζ

0,)1)(1)...(1)(1(

)1(Re

21

4

eeeee

ees

dAAA

dtA

.

By the Maclaurin series of the exponential function with tA=α , the above residue becomes


٤٨

=Ζ

+++⋅

+++

+++

−+−

⋅⋅⋅−

−

−

0,

...!3!2

1...!3

)(

!2

)(1...

!3

)(

!2

)(1

...!3

)(

!2

)(1

)()(

)(Re

22211

42

51

4

ZZZAZAZAZA

ZZ

ZAAs

dd

d

d

d

L

ααα

for the function in the above residue, we have a pole of order five at zero.

Let d

d

AA ...1

4−

−= αγ and 5

)()(

Z

ZZf

γφ=

where =)(Zφ

+++⋅

+++

+++

++−

−

...!3

Z

!2

Z1...

!3

)ZA(

!2

)ZA(1...

!3

)ZA(

!2

)ZA(1

...!3

)Z(

!2

)Z(1

22dd

2

11

4d2

L

αα

by the formula for finding the residues given by (2.14), we get

!4

)0()0),((Re

)4( γφ ⋅==ΖΖfs

Let 4d

33

22

...Z!4

Z!3

Z!2

1I−

+−+−= ααα , 1

32 ...Z!4

1Z

!3

1Z

!2

11h

−

++++=

1

33

122

111 ...Z

!4

AZ

!3

AZ

!2

A1J

−

++++= ,…,1

33d2

2dd

d ...Z!4

AZ

!3

AZ

!2

A1J

−

++++=

then hJJIJZ dL21)( =φ Now for

4d

33

22

...Z!4

)tA(Z

!3

)tA(Z

!2

tA1I

−

+−+−=

The derivative is

( )

++−⋅

+−+−−=′−

...!3

)(2

!2...

!4

)(

!3

)(

!21)4(

24

33

22

ZtAtA

ZtA

ZtA

ZtA

dZId

Differentiating I ′ to getI ′′ , I ′′′ and )4(I , then put Z = 0 in the obtained expression to get

1)0( =I


٤٩

−−=′!2

)4()0(tA

dI

+

−−−=′′!3

)(2

!2)5()4()0(

2tAtAddI

−+

−−+

−−+

−−−−=′′′

!4

)tA(!3

!3

)tA(2

2

tA)5d(

!3

)tA(2)tA)(5d(

!2

tA)6d)(5d()4d()0(I

322

3

)]!4

)tA(!3)(

2

tA)(5d()

!3

)tA(2)(5d()

!3

)tA(2()

!2

tA)((6d)(5d(

)!4

)tA(!3)(

2

tA)(2)(5d()

!3

)tA(2)(2)(5d()

!3

)tA(2)(

2

tA2)(6d)(5d(

)!3

)tA(2)(

2

)tA(3)(6d)(5d()

!2

tA)(7d)(6d)(5d)[(4d()0(I

32

222

32

22

2

2

24)4(

−−−+−+−−−

+−−−+−+−−−

+−−+−−−−−=

For

1

33

122

111 ...Z

!4

AZ

!3

AZ

!2

A1J

−

++++=

( )

+++⋅

++++−=′−

...!4

3

!3

2

!2...

!4!3!21 2

3

1

2

11

2

33

122

111 Z

AZ

AAZ

AZ

AZ

AZJ

Differentiating J ′ to get J ′′ , J ′′′ and )4(1J , then put Z = 0 in the

obtained expression to get

1)0(1 =J , 2

11

AJ −=′ ,

!3)0(

2

11

AJ =′′ , 0)0(1 =′′′J and

30)0(

41)4(

1

AJ −=

In a similar way, we get the other differentiation of

dJJJ ,...,, 32 and h, then

=

=

⋅⋅⋅

−

0),()()(

)(Re

5

1

4

ZZZAA

tAs

d

d

φ )0(!4)()(

)( )4(4

1

4

φ−−

⋅⋅⋅− d

d

d t

AA

A

Let !4

)0(

)()(

)( )4(

1

4 φ⋅⋅⋅⋅

−=−

d

d

AA

AC

So by corollary (2.4.1) we get for d≥5, 4−dc , which is the coefficient of 4−dt of


٥٠

))),((Re)1),(((Re)!4(

1

4

44 λλ∑

−Ω∈−− =ΖΖ+=ΖΖ

−−

d

dd gsgsd

So

−

−−−

++−

−−=− C)a,A(S...)a,A(S

a

1...

a

1

a

1

4

1

4

dD

)!4d(

1c dd11

d21

4d.

Thus we have proved the following: Theorem (2.5.2): Let P denote the polytope in )5( ≥ℜ dd with vertices

)a,...,0,0(),...,0,...,0,a(),0,...,0,0( d1 where daa ,...,1 are pairwise

relatively prime positive integers. Then 4−dc is given by

−

−−−

++−

−−=− C)a,A(S...)a,A(S

a

1...

a

1

a

1

4

1

4

dD

)!4d(

1c dd11

d21

4d

where S(a,b) denotes the Dedekind sum of a and b,

02

2

2

!2

))((

== Z

Z

B

ZedZ

d

Dφ

, !4

)0(

)()(

)( )4(

1

4 φ⋅⋅⋅⋅

−=−

d

d

AA

AC ,

...)!2

1(...)!2

)(1...)...(

!2

)(1(

...)!3

)(

!2

)(1(

)(1

42

++⋅++++

++−=

−

ZZAZA

tBZtBZ

Zd

d

φ .

daaA ...1= , dkk aaaA ...ˆ...1= , ka means the factor ka is omitted and

dk2kd32 a...a...aB and a...aaB == .

With the same procedure we get the coefficients 965 ,...,, −−− ddd ccc under certain conditions. From the above methods, we note that )(zφ with their definitions are needed to differentiate more than once, so we obtain a general form for this differentiation given in sec. (2.6). Example (2.5.1): Let P be a 4-polytope with vertices (0,0,0,0), (7,0,0,0), (0,2,0,0), (0,0,3,0) and (0,0,0,5). It is easy to check that this polytope satisfies the conditions of theorem (2.5.1).


٥١

In this case the Ehrhart polynomial of the polytope is given by

.),( 01

22

33

44 ctctctctctPL ++++=

by theorem (2.5.1) 1c can be computed as

]C)5,A(S

)3,A(S)2,A(S)7,A(S5

1

3

1

2

1

7

1

4

1

4

4(D

)!34(

1c

4

3211

−−

−−−

+++−−−=

where ( )( )432 BBB5.01B

1D ++−= ,

+++⋅

+++

+++

++−=

−

...!3

Z

!2

Z1...

!3

)ZA(

!2

)ZA(1...

!3

)ZA(

!2

)ZA(1

...!3

)tBZ(

!2

)tBZ(1

)Z(22

dd

2

11

3d2

L

φ

After simple computations one can get

42,70,105,30,210 4321 ===== AAAAA

30

1D , 6B,10B,15B,30B 432 ===== .

!3

)0(

900

1 )3(φ⋅=C where

( ) ( )( )

++++++

++++

+−=...

!21

1...31...51...2

151...151

...151)(

ZZZZZ

tZZφ

let

...151 +−= tZI , ( ) 1

1 ...Z151J −++= , 1

2 ...Z2

151J

−

++= , ( ) 1

3 ...Z51J −++= ,

1

4 ...)31( −++= ZJ and 1

...Z!2

11h

−

++=

then hJJJIJZ 4321)( =φ , we find )(Zφ ′ , )(Zφ ′′ and )()3( Zφ then put Z=0 in

)()3( Zφ to get the value of C which is equal to

C= )0(!3

1

900

1 )3(φ


٥٢

Also we need to compute S(30,7), S(105,2), S(70,3) and S(42,5),

which are equal to 0.071428, 0, 0.055555 and 0 respectively. By substituting 1c in the above formula one can get:

) ( )]52482.296)5,42(S

)3,70(S)2,105(S)7,30(S5

1

3

1

2

1

7

1

4

1

4

4

60

29

)!34(

1c1

−−

−−−

+++−

−−−=

39864.296−= .

From, )),(...),(()!2(

1112 dddd aASaASC

dc −−−

−=−

, where

)...1

(12

1)...(

4

1

1

1,12,1

d

dddd a

A

a

A

AAAdC +++++++= −

one can get the value of 2c , which is equal to,

)]5,()3,()2,()7,([)!24(

1432142 ASASASASCc −−−−

−=

where

( )

++++++++=

4

4

3

3

2

2

1

14,33,22,14

1

12

14

4

1

a

A

a

A

a

A

a

A

AAAAC

15432,1 == aaA , 35413,2 == aaA , and 14214,3 == aaA , therefore

( ) 125000.1212698.037698.24!2

12 =−=c .

2.6 General formula for the differentiation of h,J,...,J,J,I d21

In this section, we get a general form for the differentiation of the terms hJJJI d and ,...,,, 21 that appears throughout the process of finding the coefficients of the Ehrhart polynomial, we begin by considering

[ ] ( ) ,...2,1 32 == jhJJJIeI d

jZjL

where [ ]jI means that only I in the expression hJJIJe d

ZL32 is

differentiated j times.


٥٣

Let d

d

J

J

J

JE

′++

′+= ...1

2

21 , then

[ ] IEII ′+=′′1

2 ,

[ ] [ ] IEIIEII ′′+′′++=′′′1

21

3 )( ,

[ ] [ ] [ ] [ ] IEIIEIEIIEII ′+′′+′++′′′++= 1

2

1

22

1

3

1

4)4( )2()2( ,

[ ] [ ] [ ] [ ] [ ]

[ ] [ ],)3(

3)(33)3(23

11

2

1

211

31

321

)4(41

5)5(

IEIEIIE

IEEIIEIEIIEII

+′′′′+′′+′′+

′+′′′+′++++=

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ] [ ],)4(

)64(46123

)46(46)4(

241

)4(1

21

3

1

2

11

2

1

2

1

3

11

22

1

)4(4

1

24

1

33

1

42

1

)5(5

1

6)6(

IEIEIIE

IIEIEEIEEIEEIE

IIEIEIEIEIIEII

+′+′′+′′′+

′′′+′′+′′+′+′+′++′++++++=

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ],10

151551020

103030)5(

)105()1010()10(

51010)5(

211

2211

321

211

21

21

311

21

31

31

21

411

)5(1

2)4(1

3

1

)4(4

1

)5(5

1

25

1

34

1

43

1

52

1

)6(6

1

7)7(

IEE

IEEIEIEEIEEIEE

IEEIEEIEEIEIIE

IIEIIEIIE

IEIEIEIEIIEII

′′′+

′+′+′′′+′′+′′+

′+′+′+′+′′++

′′′+′′′++′′++′+++++++=

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ],)(1560156045

904561530

2060601560

9060)6()156(

)2015()1520()615(

15205)6(

231

2111

211

311

221

21

32

11

22

1

42

1

2)4(

11

2

1

2

1

3

11

21

31

31

21

411

21

41

31

31

4

1

2

1

5

11

)6(

1

2)5(

1

3)4(

1

)4(4

1

)5(5

1

)6(6

1

361

351

531

621

)7(71

8)8(

IEIEEEIEEIEEIEE

IEEIEIEIEEIEEIEE

IEEIEEIEEIEEIEE

IEEIEEIEIIEIIE

IIEIIEIIE

IEIEIEIEIIEII

′+′′′+′′′′+′′′+′+

′+′′+′++′′′+′′′+

′′+′′+′′+′+′+

′+′+′+′′++′′′+++′′′++′′++′+++++++=

In order to differentiate dJJJ ,,, 32 K we need to find a general

formula for these differentiations so we work on these elements and find a general formula. To illustrate this, consider for example,


٥٤

1...)

!4!3!21(

2

2133

222

222 −

=++++= −Zwe

ZwZ

wZ

wZ

wJ

then ZwJJe Zw

2222 =− . By assuming the implicit differentiation for both

sides of the above equation, we get

222 )()( 2 wJdZ

dJe

dZ

d Zw =−

and the second derivative of the above equation is

0)()( 22

2

22

2

2 =− JdZ

dJe

dZ

d Zw

when we differentiate 22 Je Zw d-times we get a shape like a binomial

formula dba )( + = dddd bbadd

bdaa ++−++ −− ...!2

)1( 221 .

Therefore,

0)( )(

2222 =−+ mmZw JwJe

where )(

2mJ is the m-th derivative of 2J , since 2w is constant therefore mw2

means 2w raised to the power m. For example,

let ZweJh 2

2= , then

)( 222222222 JwJeJeweJh ZwZwZw +′=+′=′ ,

),2( 2

22222222

22222

22222 JwJwJeeJwJeweJweJh ZwzBzBZwZw +′+′′=′++′+′′=′′

).33( 2

3

22

2

22222 JwJwJwJeh Zw +′+′′+′′′=′′′

And so on. Therefore

]1[222 JJJIeJ d

Z =′=′ L ,

22

]2[

22 JEJJ ′+=′′ where d

d

J

J

I

IE

′++

′+= ...12 .

JEJEJEJJ ′′+′++=′′′ 22

2

2

]2[

22

]3[

22 2 .

22222

]2[

222

3

2

]2[

2

2

2

]3[

22

]4[

2

)4(

2 3333 JEEJEJEJEJEJEJJ ′′+′′′+′+′+++= .


٥٥

and similarly for highest derivative. By arranging them together we obtain

d

Z JJIeJ L22′=′ ,

,22

]2[22 JEJJ ′+=′′

JEJJEJJ ′′+′′++=′′′ 22

]2[

22

]3[

22 )( ,

222

]2[

22

]2[

2

2

22

]3[

22

]4[

2

)4(

2 )2()2( JEJJEJEJJEJJ ′′′+′′+′++′′′++= ,

,3)3(

)33()3()3(]2[

222222

]2[

22

2

]3[

22

]2[

2

3

2

]3[

2

2

2

)4(

2

]4[

22

]5[

2

)5(

2

JEEJEJJE

JJEJEJEJJEJJ

′+′′′′+′′+′′+

′′′+′+++++=

( ) ,)(34 612

)4()64()46(

46)4(

]2[

2

2

2

]2[

222

]2[

2

2

22

]3[

2222

4

2

2]2[

22)3(

2]3[

22)4(

2]4[

22

]2[2

42

]3[2

32

]4[2

22

)5(2

]5[22

]6[2

)6(2

JEJEEJEEJEEJE

JJEJJEJJE

JEJEJEJJEJJ

′+′′+′+′+′+′′+′′′++′′++′

++++++=

,10)(15)(155

10201030

30)5()105(

)1010()510(

51010)5(

]2[

222

]3[

22

2

2

]3[

2

2

2

]2[

222

]2[22

22

]3[222

]2[2

322

]3[2

222

]4[2222

)5(22

]2[2

)4(22

]3[22

)4(2

]4[22

)5(2

]5[22

]2[2

52

]3[2

42

]4[2

32

]5[2

22

)6(2

]6[22

]7[2

)7(2

JEEJEEJEJEE

JEEJEEJEEJEE

JEEJEJJEJJE

JJEJJEJE

JEJEJEJJEJJ

′′′+′+′+′′′+

′′+′′+′+′+

′+′+′′++′′′+′′′++′′++′++

+++++=

,)(10601560

)(45)(90)(15)(456

1530206060

15609060

)6()136()2015(

)1520()615(6

152015)6(

]2[2

22

]2[2222

]2[222

]3[222

]2[

2

2

2

2

2

]3[

2

2

22

]2[

2

3

2

]4[

2

2

2

]2[

2

)4(

22

]2[

22

2

2

]3[

222

]2[

22

3

2

]3[

22

2

2

]4[

222

]2[22

42

]2[22

32

]4[2

222

]5[2222

)6(2

2

]2[

2

)5(

2

)3(

2

]3[

2

)4(

2

)4(

2

]4[

22

)5(2

]5[22

)6(2

]6[22

]2[2

62

]3[2

52

]4[2

42

]5[2

32

]6[2

22

)7(2

]7[22

]8[2

)8(2

JEJEEEJEEJEE

JEEJEEJEJEJEE

JEEJEEJEEJEEJEE

JEEJEEJEEJEEJE

JJEJJEJJE

JJEJJEJEJE

JEJEJEJJEJJ

′′+′′′+′′′′+′′′+′+′+′+′+

+′′′+′′′+′′+′′+′′+′+′′+′+′+′

+′′+++++′′′++′′++′++

++++++=


٥٦

( ) ( )( ) ( ) ( )( )

( )

JEEEJEEE

JEEEJEEJEEJEJEE

JEEJEEJEEJEJEE

JEEJEEJEJEEJEE

JEEJEEJEEJEEJEE

JEEJEEJEEJEE JEE

JEEJEEJEEJEJJE

JJEJJEJJE

JJEJJEJEJE7

JEJEJEJEJJEJJ

.105210

42035)(10)(7021

105210)(105)(105)(105

)(315)(315)(105721

423510510535

14021014021105

2102101057

21735213535

2135721

21353521)7(

]2[222

]2[222

22

]3[

2222

]2[

222

]2[

2

2

22

]3[

2

2

2

]2[

2

)4(

22

]3[222

]4[222

]2[2

322

]3[2

32

]2[2

22

32

]3[

2

2

2

2

2

]4[

2

2

22

]5[

2

2

2

]2[

2

)5(

22

]2[

2

)4(

2

2

2

]3[

2

)4(

22

]2[

22

3

2

]3[

22

2

2

]4[

222

]2[

22

4

2

]3[22

32

]4[22

22

]5[222

]2[22

52

]3[22

42

]4[

22

3

2

]5[

2

2

22

]6[

2222

)7(

22

]2[

2

)6(

2

32

]3[2

)5(2

)4(2

]4[2

)4(2

)5(2

]5[22

)6(

2

]6[

22

)7(

2

]7[

22

]2[

2

7

2

]3[

2

6

2

]4[2

52

]5[2

42

]6[2

32

]7[2

22

)8(2

]8[22

]9[2

)9(2

′′′′+′′′+′′′+′′′′′+′′+′′+′

+′′′′+′′′+′+′+′+′+′+′++

++′′′+′′′+′′′+′′+′′+′′+′′+′+′

+′+′+′+′+′′+++++++′′′

++′′++′+++++++++=

Since our work is for finding the coefficients of the Ehrhart polynomial until 9−dc , so the derivatives that we are needed are until 9-th derivative.

By similar procedure we get the derivatives of dJJJ ,,, 43 K and h

that are used in the definition of φ(z) in the preceding sections. When we arrange the obtained results we get a triangle like a Polya triangle [40, p.20] where the contents of the triangle are the coefficients of

,...E,E 3

2

2

2 in the expression ,..., )5(

2

)4(

2 JJ

[ ]

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ]

[ ]

[ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ]

[ ] [ ] [ ]2

2

3

2

4

2

2

2

5

2

6

2

7

2

8

2

9

2

3

2

4

2

5

2

6

2

7

2

2

2

3

2

4

2

5

2

2

2

3

2

)10(2

8

2

)9(

2

62

72

)8(2

4

2

5

2

6

2

)7(

2

2

2

3

2

4

2

5

2

)6(

2

2

2

3

2

4

2

)5(

2

2

2

3

2

)4(

2

2

2

182856705628

1721353521

16152015

151010

146

13

1

JJJ

J

JJJJE

JJJJJ

JJJJ

JJ

J

EJ

JEJ

JJEJ

JJJEJ

JJEJ

JEJ

E

Also, the first terms of the coefficients of K,, 22 EE ′′′ in the

expression of ,..., )5(2

)4(2 JJ are


٥٧

[ ]

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ]

[ ]

[ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]22

32

42

52

62

72

22

32

42

52

22

32

)7(2

)9(2

62

)6(2

)8(2

42

52

)5(2

)7(2

22

32

42

)4(2

)6(2

22

322

)5(2

222

)4(2

2

1721353521

16152015

151010

146

13

1

JJJJJJ

JJJJ

JJ

EJ

JEJ

JJEJ

JJJEJ

JJEJ

JEJ

E

′′′′′

′

The second terms of the coefficients of ,..., 22 EE ′′′ in the expression

of ,..., )5(2

)4(2 JJ are

222)4(

2)5(

2)6(

2)7(

2

222

)4(

2

)5(

2

222

2

)8(2

)6(

2

)4(2

2

)9(2

)7(

2

)8(

2

)5(2

)6(2

)7(2

2

)4(

2

)5(

2

)6(

2

222

)4(

2

)5(

2

222)4(

2

22

17213535217

161520156

1510105

1464

133

11

JJJJJJJ

JJJJJ

JJJ

J

E

J

J

J

J

EJ

JEJ

JJEJ

JJJEJ

JJEJ

EE

′′′′′′′′′′′′

′′′′′′′′′′′′

′′′′′′′′′′′′

′′′

The coefficients of ,...EE ,EE 2

2222

′′ in the expression of ,..., 22 JJ ′′′′′ are arranged as follows.

[ ]

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ]

[ ]

[ ] [ ]

[ ] [ ] [ ] [ ]22

32

42

52

22

32

62

62

)9(2

42

522

52

)8(2

22

32

422

42

)7(2

22

322

32

)6(2

222

22

)5(2

22

)4(2

2

2

21105210210105

15609060

103030

612

3

0

00

JJJJ

JJ

JEEJ

JJEEJ

JJJEEJ

JJEEJ

JEEJ

EEJ

JJ

′′

′′

′′

′′′′′

The diagonal of the above results is the second column of the

preceding Polya triangle, and the first column for the above results is obtained as follows:


٥٨

By multiplying the diagonal by 4,5,6… we get the line under the diagonal, which are: (3)(4)=12, (6)(5)=30, (10)(6)=60, (15)(7)=105, The general formula of the differentiation is given by

( ) [ ] ( ) [ ] ( )( ) WJJmEJJ mmmm ++−+= −− 1

2

1

2222 2 where 1<m≤8 and W can be obtained from the given tables as follow when m=3 then

( ) [ ] [ ] ( )( )2

2

2

22

3

2

3

2 JJEJJ ++= when m=4 then

( ) [ ] [ ] ( )( ) WJJEJJ +++= 3

2

3

22

4

2

4

2 2 from the tables, W can be found as follows

[ ] [ ] ( )( ) 22

2

2

2

22

2

2

2

2 2 JEJJEJEW ′′′++′+= .

Chapter Three The Ehrhart Polynomial of H-representation

Of a Polytope Introduction As seen before the computation of the volume of a polytope is very important in many applications. For this importance many researches concerning the volume of the polytopes and in particular with the Birkhoff polytopes are found. In this chapter a method for computing the coefficients of the Ehrhart polynomial of the Birkhoff polytope is discussed. Make a change on the matrix, which represents the polytope and find a formula for the number of integral points. These changes of the matrix are matrix operations on the rows (columns) of the matrix. We try to lessen the effect of their changes on the number of integral points. We further discuss a method for finding the volume of a polytope using Laplace transform.

Chapter three consists of four sections. In section one, a method for finding the volume of H-representation of polytope using Laplace transforms. In section two some basic concepts and remarks about the Birkhoff polytopes and their volumes are given. In section three, the Ehrhart polynomial of Birkhoff polytope is presented and in section four, further properties about the number of integral points are obtained under certain conditions, also we give the relation between the number of integral points of the original polytopes and the changing polytopes. 3.1 Computing the volume of a polytope using Laplace transform, [33]

This section is devoted to the computation of the volume of a polytope with H-representation using Laplace transform.

The idea of the method is to consider the volume of

bAXXP d ≤ℜ∈=••

+ , dnA ×ℜ∈ and ny ℜ∈ , as a function g(b) where

ℜ→ℜn:g which provides a simple expression of its Laplace transform

CCG n →: , and the inverse Laplace transform to G, which, in the next section, can efficiently be done by repeated applications of Cauchy residue theorem for the evaluation of one–dimensional complex integrals.

Chapter Three The Ehrhart polynomial of H-Representation of a polytope

٦٠

Let ny ℜ∈ and dnA ×ℜ∈ such that the convex polyhedron,

yAXX)y(P d ≤ℜ∈=••

+ (3.1)

is compact, that is, P(y) is a convex polytope. The symbol +ℜ stands for the semi closed interval ℜ⊂∞),0[ .

Now consider the function ℜ→ℜng : defined by

))(()()(

yPVoldxygyP

== ∫ (3.2)

and let CCG n →: be its n-dimensional Laplace transform, that is, [1] ∫

ℜ

><−=n

dyygeG y )()( ,λλ (3.3)

where >⋅⋅< , is the Euclidean product defined on nℜ . Theorem (3.1.1), [33]:

Let )( yAXXyP d ≤ℜ∈=••

+ be a polytope, and let g and G be

defined by

))(()()(

yPVoldxygyP

== ∫ , ∫ℜ

><−=n

dyygeG y )()( ,λλ

respectively. Assume that X=0 is the only solution of the system

00 ≤≥•• AXX . Then:

>

>=

∏ ∏= =

0)Re(

0)Re( with,

)(

1)(

1 1

λλλλ

λT

n

i

d

jj

T

iAA

G (3.4)

Moreover

∫∫∞+

∞−

><∞+

∞−=

ic

ic

y,ic

icn

n

n

1

1

d)(Ge...)i2(

1)y(g

λλ

πλ (3.5)

where the real constants nccc ,...,, 21 are chosen such that c>0 satisfies

0cA T > . Proof:

Apply the definition of G given by (3.3) to obtain

∫ ∫ℜ ≤≥

><−

=

n

dydxe)(GyAx,0x

y,λλ


٦١

∫ ∫+ℜ ≥

><−

=

d

dxdyeAxy

y,λ

0)Re( with ,dxe1

d

T x,An

1ii

>= ∫∏ +ℜ

><−

=

λλ

λ

then after simple computations, one can get

>

>=

∏ ∏= =

0)Re(

0)ARe( with ,

)A(

1)(G

Tn

1i

d

1jj

T

i

λλλλ

λ

and (3.5) is obtained by a direct application of the inverse Laplace transform, [1]. It remains to show that, the domain

0)Re(,0)Re( >> λλ TA is non empty. This flows from the fact that a special version of Farka’s lemma due to Carver, [35, p.30], which states

that 00 >>•• uAu T has a solution nu ℜ∈ if and only if (x, y) = 0 is the

only solution of the system 0,00 ≥≥=+•• yxyAx . In other words,

x = 0 is the only solution of 00 ≤≥•• Axx , which is indeed the condition

given by this theorem.

Suppose that we want to compute the volume of the convex

polytope bAXXP d ≤ℜ∈=••

+ with b>0, that is, we must evaluate g(y) at

the point by = . Without a loss of generality, we may assume that 1=iy for every i=1,…,n. The problem is then computing h(1) of the function ℜ→ℜ+:h given by,

∫ ∫∞+

∞−

∞+

∞−

><==ic

ic

ic

ic

ez

nn

n

n

n dGei

zegzh1

1

, )(...)2(

1)()( λλ

πλ (3.6)

where ...),1,1(=ie be the vector in iℜ for 1≥i , such that its element is

one, and the real vector nc ℜ∈<0 , satisfies 0cA T > .

Computing the complex integral (3.5) can be done in two ways, directly as given in sec. (3.1.1) or indirectly as given in sec. (3.1.2).


٦٢

3.1.1 The direct method, [33]: For better understanding the direct method, consider the case of

a polytope P with n=2. Let dA ×ℜ∈ 2 be such that X=0 is the only solution of

0AX0X ≤≥•• . Moreover, suppose that ][ baAT = with dba ℜ∈, and

assume that: i) 0≠jjba and jj ba ≠ for all j =1,2,…,d.

ii) kkjj baba ≠ for all j, k =1,2,…,d.

then:

>

>++=

∏=

0)Re(0)Re(

with ,)(

1)(

21

12121

λλλλλλλ

λba

baG d

jjj

Next, fix 1c and 02 >c such that 021 >+ cbca jj for every j =1,2,…,d, and

compute the integral (3.6) as follows. First evaluate the integral,

1

1211

1

1

1

1

)(2

1 λλλλπ

λ

dba

e

iI

ic

icd

jjj

z

∫∏

∞+

∞−

=

+= (3.7)

by using Cauchy residue technique. That is:

a) Close the path of integration by adding a semicircle Γ of radius R large enough.

b) Evaluate the closed integral using Cauchy's residue theorem. c) Show that the integral along Γ converges to zero when ∞→R .

Now, since we are integrating with respect to 1λ and we must

evaluate h(z) at z=1, the semicircle Γ must be added on the left side of the integration path 11)Re( c=λ because 1ze λ converges to zero when

( ) −∞→1Re λ . Therefore, we must consider only poles of ),( 1 ⋅λG whose real parts is strictly less than 1c (with 2λ being fixed). Recall that

122 /)/Re( cabcab jjjj <−=−λ for each j=1,2,…,d, and ),( 1 ⋅λG has only

poles of the first order (with2λ being fixed). Then, the evaluation of (3.7) follows, and

∑ ∏∏ =≠

λ−

=

+−λ−+

λ=

d

j

jkkjjk

dj

zab

d

jj

d babab

e

bI

jj

1 2

)/(

12

1 )/(

1 2


٦٣

Therefore,

( )2

ic

ic 2

21

z

dIe

i2

1)z(h

2

2

2

λλ

λπ

λ

∫∞+

∞−=

∑ ∫ ∏∫∏ =

∞+

∞−

≠

+

−∞+

∞−

=

+ −−=

d

1j2

ic

ic

jkjkjkjj

1d2

z)a/b1(d

j

2

ic

ic d

1jj

1d2

z

d)baab(ba

ea

i2

1d

b

e

i2

1 2

2

2jj2

2

2

λλπ

λλπ

λλ

these integrals must be evaluated according to whether yab jj )/1( − is

positive or negative. Thus recalling that Z > 0, each integral is equal to: i) its residue at the pole 22 0 c<=λ when jj ab /1− is positive, and

ii) zero if jj ab /1− is negative because there is no pole on the right

side of 22)Re( c=λ that is

−−

−= ∑ ∏∏ <≠

=

1/

1

)(

)(1

!)(

jj ab

jkjkjkjj

d

jj

d

jj

d

baabba

ba

bd

zzh (3.8)

Observe that the above formula is not symmetrical in the

parameters a, b. This is because we have chosen to integrate first with respect to 1λ ; and the set 1/: <jj abj is different from the set

1/: >jj baj , which would have been considered if we have to integrate

first with respect to 2λ .

In the latter case, we would have obtained.

−−

−= ∑ ∏∏ <≠

=

1/

1

)(

)(1

!)(

jj ba

jkjkjkjj

d

jj

d

jj

d

abbaba

ab

ad

zzh

which is (3.8) by interchanging a and b. Example(3.1.1),[33]: Let ( )1,1,1e3 = be a vector in 3ℜ and let 2

3)ze(P ℜ⊂ be the polytope, which is defined by,


٦٤

zxx2;zx2x2;zxxX)ze(P 212121

2

3 ≤−≤+−≤+ℜ∈=••

+ . To find

the volume of this polytope, write )ze(P 3 as

zAXX)ze(P 2

3 ≤ℜ∈=••

+ , where

−−=

12

2 2

1 1

A ,

choose 1 and 2,3 321 === ccc , so that 321 22 ccc −> and 231 2ccc −>

that satisfies the conditions c>0 and 0cA T > .

∫ ∫ ∫∞+

∞−

∞+

∞−

∞+

∞−

++=ic

ic

ic

ic

ic

ic

z)(

3

1

1

2

2

3

3

321 d)(Ge)i2(

1)z(h λλ

πλλλ ,

with )2)(22(

1)(

321321321 λλλλλλλλλλ

−++−=G ,

Integrate first with respect to 1λ ; that is, evaluate the residue of ( )λG at the poles 3211 22,0 λλλλ −== and 231 2λλλ −= since 0<z, 10 c< ,

132 )22Re( c<− λλ and 123 )2Re( c<− λλ . We obtain,

3

24322

2

2

3

3

)( ...)2(

1)( λλ

πddIII

izh

ic

ic

ic

ic∫ ∫∞+

∞−

∞+

∞−++= .

where,

)2)((2 232332

)(

2

32

λλλλλλ

λλ

−−=

+− zeI

)3/4)((6 232332

)3(

3

32

λλλλλλ

λλ

−−=

− zeI

)3/4)(2(3 232332

)2(

4

32

λλλλλλ

λλ

−−=

− zeI

Next, integrate 2I with respect to 3λ , we must consider the poles of 2I on the left side of 1)Re( 3 =λ , that is the pole 03 =λ , since 2)Re( 2 =λ . Thus,

we get 324

2

λ

λze−, and the next integration with respect to 2λ yields

8

2z−.

When integrate 3I with respect to 3λ , we have to consider the poles

23 λλ = and 3

4 23

λλ = , on the right side of 1)Re( 3 =λ ; and we get


٦٥

+−−8

3

2

1 3/52

32

22 λλ

λ

zz ee

Recall that the path of integration has a negative orientation, so the

negative values of residues have to be considered. The next integration

with respect to 2λ yields

−48

2512z .

Finally, when integrating ),( 324 λλI with respect to 3λ , we must consider

only the pole 03 =λ , we get 328

2

λ

λze−; the next integration with respect to 2λ

yields zero. Hence, adding up the above three partial results, yields,

48

17

48

251

8

1)(

22 z

zzh =

−+−= .

3.1.2 The indirect method, [33]: The indirect method permits avoid evaluating integrals of exponential function in (3.6) which will be illustrated in this section. We want to compute (3.6) where 0≠b and c>0 are real vectors in nℜ with 0cA T > . From (3.1) it is deduced that g (b)=0 whenever 0≤b , so that the last entry 0>nb ; and the following simple change of variables are done. Let >=< bs ,λ and >=< bcd ,1 so that,

n

n

jjj

n b

bs

−=

∑−

=

1

1

λλ and

[ ] 1

c

1

1

1

1

1

1

1

... ˆ...)2(

1)( −

∞+

∞−

∞+

∞−

∞+

∞−∫ ∫ ∫−

−

= n

ic

i

ic

ic

id

id

zs

ndddsGe

izh

n

n

λλπ

where

−= ∑

−

=−− j

n

j n

j

n

n

n

n b

b

b

sG

bsG λλλλλ

1

11111 ,,...,

1),,...,(ˆ

h(z) can be rewrite as follows:

∫∞+

∞−=

id

id

zs dssHei

zh1

1

)()2(

1)(

π, with

111

1

1

1

1

ˆ)2(

1)( −

∞+

∞−

∞+

∞−− ∫ ∫−

−

= n

ic

ic

ic

ic nd d G

isH

n

n

λλπ

LL


٦٦

where H(s) is the Laplace transform of h(z)=g(zb), and is called the associated transform of )(λG .

Let nA ×ℜ∈ 2 such that X=0 is the only solution of 0AX0X ≤≥•• .

Write ][ baAT = with dba ℜ∈, , assume that 0≠jjba for all j=1,…,d and

kjbaba kkjj ≠≠ allfor

Then,

>

>+=

∏=

,0)Re(

0)Re(h wit,

)(

1)(

12121

λλλλλλ

λtAba

G d

jjj

fix 12 λλ −= s and choose real constants 01 >c and 02 >c such that

021 >+ cbca jj for every j=1,2,…,d. Notice that 21)Re( ccs += . H(s) is

obtained by integrating ),( 11 λλ −sG along the line 11)Re( c=λ , which yields,

∫∏ +−−

= ∞+∞−

=

ic

ic d

jj

dsbbasi

sHjj

1

1 1

11

11 ))(()(

1

2

1)( λ

λλλπ

Next, determine which poles of ),( 11 λλ −sG are on the left(right)

side of the integration path )Re( 11 c=λ in order to apply the Cauchy

residue theorem. Let jjjj bajJbajJ ==>= ••

••+ , 0 and

jj bajJ <= ••− . Then the poles on the left side of )Re( 11 c=λ are

01 =λ and )/(1 jjj basb −−=λ for all +∈ Jj since 1)/()Re( cbasb jjj <−− .

Besides, the poles on the right side of )Re( 11 c=λ are s=1λ and )/(1 jjj basb −−=λ for all −∈ Jj .

Finally, notice that ),( 11 λλ −sG has only poles of the first order. Hence, computing the residues of poles on the left side of )Re( 11 c=λ yields,

+−−−

+= ∑ ∏∏∏ +∈≠∉

−

∉∈Jj

jkjkkjkjjj

jd

jj

Jjj

Jjj bsaasbbas

ba

sbssbsH

,

2

0

0

00

)(

)(11)(


٦٧

After moving terms around, we obtain

−−

−= ∑ ∏∏ >≠

=

+jj ba

jkkjkjjj

d

jj

d

jj

d abbaba

ba

bssH

)(

)(11)(

1

1

Notice that the previous equation holds even for the case 00 ≠J .

Finally, after integration with respect to s, we get

−−

−= ∑ ∏∏ >≠

=

jj ba

jkkjkjjj

d

jj

d

jj

d

abbaba

ba

bd

zzh

)(

)(1

!)(

1

Now, computing the negative value of residues of poles on the

right side of )Re( 11 c=λ yields,

−−

−= ∑ ∏∏ >≠

=

+jj ab

jkkjkjjj

d

jj

d

jj

d baabba

ab

assH

)(

)(11)(

1

1

and after integration with respect to s, one also get the following

−−

−= ∑ ∏∏ <≠

=

1/

1

)(

)(1

!)(

jj ba

jkjkjkjj

d

jj

d

jj

d

abbaba

ab

ad

zzh in particular case 00 =J

Example(3.1.2),[33]: Consider example(3.1.1). By setting )1,1,1(3 =e , let 2

3)( ℜ⊂Ρ ze be the polytope which is defined by

zxx2 ;zx2x2 ;zxxX)ze(P 212121

2

3 ≤−≤+−≤+ℜ∈= +

We can choose 2,1 321 === ccc and 123 λλλ −−= s , so that, 4)Re( 1 == ds , and

2

1

1

1

1 12

),()2(

1)( λλλ

πddsM

isH

i

i

i

i∫ ∫∞+

∞−

∞+

∞−=

with )32)(42)((

1),(

21212121 ssssM

−+−−−−=

λλλλλλλλλ


٦٨

We first integrate with respect to 1λ . Only the real parts of poles

01 =λ and 2/)3( 21 λλ −= s are less than 1. Therefore, the residue of the 0-pole yields:

2

1

1 2222 )3)(42)((

1

2

1 λλλλλπ

dsssi

i

i∫∞+

∞− −−− (3.9)

whereas the residue of ( )2/)3( 2λ−s -pole yields

2

1

1 2222 )53)()(3(

4

2

1 λλλλλπ

dsssi

i

i∫∞+

∞− −+− (3.10)

Applying again the Cauchy residue theorem to (3.9) at the

pole 02 =λ yields )2/(1 3s− . Similarly, applying the Cauchy residue theorem to (3.10) at the poles

02 =λ and s2 −=λ yields 324

29

s, finally we obtain ( )

324

17

ssH = and so

248

17)(

zzh = , which is the area of the polytope.

3.2 Some basic concepts about Birkhoff polytopes The set of doubly stochastic nn × matrices form a convex set called Birkhoff polytope. In this section, we describe the Birkhoff polytope with some methods for finding its volume.

We start this section by the following definition.

Definition (3.2.1), [7]: The n-th Birkhoff polytope nB is the set of all doubly stochastic

nn × matrices, that is, those matrices with non negative real coefficients in which every row and column sums to one. In other words, the n-th Birkhoff polytope nB is defined as


٦٩

≤≤=

≤≤=ℜ∈

=∑

∑+

kjk

jjk

n

nn2n1n

n22221

n11211

n nj1 allfor 1x

nk1 allfor 1x:

x...xx

...

...

...

xxx

x...xx

B2

nB is a convex polytope. Remarks (3.2.1):

There are different ways for computing the volume of Birkhoff polytopes. One of the recent attempts to compute )( nBvol relies on the theory of counting functions for the integer points in the polytopes. Recall that Ehrhart has proved that for a polytope nP ℜ⊂ with integral vertices, the number, )(),( ntPcardtPL Ζ= I , is a polynomial in the positive integer variable t, this counting function, has three properties, which are [7],

• The degree of ),( tPL is the dimension of P.

• The leading term of ),( tPL is the relative volume of P.

• Since ),( tPL is a polynomial, therefore it can be evaluated at nonpositive integers. These evaluations yield 1)0,( =PL

),()1(),( )dim( tPLtPL P o−=− (3.11) where oP is the interior of P.

We will denote the Ehrhart polynomial of the Birkhoff polytope

nB as ),()( tBLtH nn = , where nH is a polynomial in t of degree 2)1( −n , to do this count, note that the last row and column are fixed by the conditions that the row and column sums should equal one. The remaining 2)1( −n entries can be chosen freely; therefore the dimension

of nB is 2)1( −n , which is the degree of the Ehrhart polynomial of the Birkhoff polytope.

The first two of these polynomials are easily computed, [7] which

are, 1)( ,1)( 21 +== ttHtH


٧٠

We will compute these polynomials in the next section, the first nontrivial polynomial was computed see [7], is

2

3 2

2t

2

3t3)t(H

++

+−=

The properties of nH are found in [7]. From (3.11), it is deduced that

)()1()( 1 tHtnH n

n

n

−−=−− and

0)1(...)2()1( =+−==−=− nHHH nnn

This allows the following strategy for computing nH , and

therefore, the volume of nB : compute the first

−2

1n values of nH , use

the above symmetry and trivial values of nH , and calculate the polynomial nH by interpolation, [7], [17].

3.3 The Ehrhart polynomial of a Birkoff polytope In this section we give some theorems that can be used to count the Ehrhart polynomial of the Birkhoff polytope. One can view the Birkhoff polytope nB as given in the form of

bAX:XP d =ℜ∈= + (3.12)

where

=

1010010

0....0.00.0

...

.0..

111

1...1

00.

0.0000

0.0000

01...100

0001...1

A

We illustrate how to represent the Birkhoff polytope in the form (3.12). For example if n = 2 then from the definition of the Birkhoff polytope,


٧١

=

2221

1211

2 aa

aaB

where

1

1

1

1

2212

2111

2221

1211

=+=+=+=+

aa

aa

aa

aa

this means that

=

1

1

1

1

1010

0101

1100

0011

22

21

12

11

a

a

a

a

Let

=

=

=

1

1

1

1

,

1010

0101

1100

0011

22

21

12

11

band

a

a

a

a

XA

Then bAX:XP d =ℜ∈= + which is the H-representation of the polytope. Now, some theorems that concern this type of polytopes are given.

Theorem (3.3.1), [7]: Consider a convex rational polytope P given by (3.12) for some ( dm× )-matrix A and m-dimensional vector b. Denote the columns of A by dcc ,...,1 . Then

∫∫=

−−−−−−

= −−−=

mm

d

m

Z

ccc

tb

m

tbtb

Z

mdZ

ZZZ

ZZZ

itPL

ξξπ )1()1)(1()2(

1),(

21

21

11

11

2

1

1

L

LL

where 1,...,0 1 << mξξ are distinct real numbers and t is a positive integer. Proof: Consider the function

)1()1)(1(

)(21

21 11

2

1

1

d

m

ccc

tb

m

tbtb

ZZZ

ZZZZf

−−−=

−−−−−−

L

L


٧٢

Here, (the standard multivariate notation nw

n

wW vvV ...`1

1= ) is used. f(Z) will be integrated with respect to each variable over a circle with a small radius:

∫∫== mmZ

mm

Z

dZdZZZfξξ

11 ),...,(11

LL (3.13)

here 1,...,0 1 << mξξ are chosen such that, all of the )1(

1kcZ−

can be

expanded into power series about 0. Since the integral over one variable will give the respective residue at 0, integration can be done with respect to one variable at a time. When f is expanded into its Laurent series about 0, each term has the form 11

1 ,...,11 −−−− mm tbr

m

tbr ZZ , where mrr ,...,1 are nonnegative integers. Thus, in the integral (3.13) will give a contribution precisely. Remark(3.3.1): A special case of theorem (3.3.1) can be obtained by putting b = (1,1,…,1) nZ 2∈ , therefore the Ehrhart polynomial of the n-th Birkhoff polytope is given by

∫ ∫ −−−=

++

−−

dZZZZZZZ

ZZ

itH

nnnn

t

n

nn )1()1)(1(

)(

)2(

1)(

22211

121

2L

LL

π

Here, each integral is over a circle with radius < 1 centered at the

origin; all appearing radii should be different.

The following theorem gives a general formula of the Ehrhart polynomial of the Birkhoff polytope with b = (1,1,…,1).

Theorem (3.3.2), [7]: For any distinct 1,...,0 1 << nξξ ,

1 1

11

111 )(

)()2(

1)( dZdZ

ZZ

ZZZ

itH n

n

Z Z

n

k

kjjk

nt

kt

nnnnn

LLL∫ ∫ ∑∏= ==

≠

−+−−

−=

ξ ξπ.


٧٣

Proof: From remark (3.3.1), the last n variables of )(tH n can separate and

obtain

( ) 1111

11

21 )1()1)(1(2

1

)2(

1)( dZdZdZdZ

ZZZZZZ

Z

iZZZ

itH nn

n

n

tt

nnn LL

LL −

−−−−

∫ ∫ ∫

−−−ππ=

The radius of integration circle of the inner most integral may be chosen to be smaller than the radii of the other integration paths. Then the innermost integral is computed which is equal to the residue at zero of

)1()1(

1

11 ZZZZZ n

t −−+L

and, by residue theorem, equal to the negative of the sum of the residues at 11

1 ,, −−nZZ K . ( Note that here we use the fact that t>0 ).

The residues at these poles are computed: the one, say, at 1

1

Zcan be

calculated as

)1()1(

1

1

1

lim

)1()1(

11lim

2

1

1

1

1

11

11

1

1

ZZZZZZZ

ZZ

ZZZZZZZ

n

t

ZZ

n

t

ZZ

−−⋅

−

−=

−−⋅

−

+→

+→

L

L

)()()1()1(

1

121

1

1

11

2

1

1

1 n

nt

n

t

ZZZZ

Z

Z

Z

Z

ZZ

Z −−−=

−−⋅−=

−++

LL

Similarly we calculate the residues at the other poles, which yield the Ehrhart polynomial of the Birkhoff polytope.

In the next example we will illustrate the computation of 1H . Example (3.3.1):

From theorem (3.3.2) we get,

.122

1

2

1

2

1)( 1

1

111

1

11

1111

=⋅=

== ∫∫=

−

=

−−

ii

dZZi

dZZZi

tHzz

tt

ππ

ππ ξξ


٧٤

The next example will explain the computation of 3H and hence

the volume of 3B . Example (3.3.2), [7]:

By using theorem (3.3.2) we get

dZZZZZ

Z

ZZZZ

Z

ZZZZ

ZZZZ

itH

t

ttt

3

2313

23

3212

22

3121

21

1

32133

))((

))(())(()(

)2(

1)(

−−

+

−−+

−−=

+

++−−

∫π

We have to order the radii of the integration paths, for each variable; we choose 10 123 <<<< ξξξ we use this fact after multiplying out the cubic:

integrating, for example, the term 3

13

3

23

523

12

11

)()( ZZZZ

ZZZ ttt

−−

+−−−−

with respect to 3Z

gives 0, as this function is analytic at the 3Z - origin and 321 , ξ>ZZ .

After using this observation for all the terms stemming from the cubic, the only integrals surviving are

∫ −−

−−−−+

dZZZZZ

ZZZ

i

ttt

331

321

13

12

521

3 )()()2(

1

π

and

∫ −−−− −−+

dZZZZZZZ

ZZZ

i

tt

)()()()2(

3

32

2

31

3

21

132

31

3π

The first integral factors and yields, again by residue calculus

∫ −−

−−−−+

dZZZZZ

ZZZ

i

ttt

3

31

3

21

13

12

521

3 )()()2(

1

π

.2

2

)2)(1(2

1

2

1

)()2(

1

2

1

23

152

1

1

2

31

152

13

+=

−−−−−π

=

−π=

∫

∫

−−+

−−+

t

dZZttZi

dZdZZZ

ZZ

i

tt

tt


٧٥

for the second integral, it is most efficient to integrate with respect to 2Z first.

∫ −−−− −−+

dZZZZZZZ

ZZZ

i

tt

)()()()2(

3

32

2

31

3

21

132

31

3π ∫ −−=

−+

135

31

33

12 )()2(

3dZdZ

ZZ

ZZ

i

tt

π

∫

+−=−−−−−−−−= −−+

4

33)3)(2)(1)((

!4

1.

2

31

41

31

tdZZttttZ

itt

π

Adding up the last two lines finally gives

14

9

8

15

4

3

8

1

4

33

2

2)( 234

2

3 ++++=

+−

+= tttt

tttH

In general, the relative volume of the fundamental domain of the sub lattice of

2nΖ in the affine space spanned by nB is 1−nn , [7]. Therefore, to obtain the volume of 3B , the leading term of 3H has to be multiplied by the relative volume of the fundamental domain of the sub lattice of 9Ζ in the affine space spanned by 3B which is equal to 9;

hence 8

9)B(vol 3 =

Now for n = 4, the number of integrals that has to evaluate to

compute 4H is only slightly higher. By theorem (3.3.2)

.)(

.)()2(

1)(

4

4

1

31

432144

11 22 33 44

dZZZ

ZZZZZ

itH

k

kjjk

t

kt

Z Z Z Z

−= ∑∏∫ ∫ ∫ ∫

=≠

+−−

= = = =ξ ξ ξ ξπ

Again we have a choice of ordering the radii; 10 1234 <<<<< ξξξξ are used. After multiplying out the quadric, five integrals have to be calculated; their evaluation (again straight forward by means of the residue theorem) is as follows:

∫ −−−

−−−−−−+

dZZZZZZZ

ZZZZ

i

tttt

4

41

4

31

4

21

14

13

12

1131

4 )()()()2(

1

π

3

1

3

4

1

1113

14 3

3

)(.

)2(

1

+=

−= ∫∫

−−+ t

dZdZZZ

ZZ

i

tt

π,


٧٦

∫ −−−−−− −−−−+

dZZZZZZZZZZZ

ZZZZ

i

ttt

))(()()()()2(

4

4232

3

41

3

31

4

21

14

13

22

821

4π

21

2

2

3

1

1

4

21

22

821

4 )()()()2(

4dZdZdZ

ZZZZ

Z

ZZ

ZZ

i

tt

−−⋅

−−= ∫∫

−−+

π

dZZZ

Z

ZZ

Z

ZZ

Zt

ZZ

Zt

ZZ

ZZ

i

tt

tttt

−−

−

+

−++

−

+⋅

−=

−−−−

−−−−+−+

∫

3

21

12

3

31

11

2

21

21

21

31

7

21

12

821

2

)()(2

)()1(2

)(2

22

)()2(

4

π

+−

++

+++

+

+=

9

824

9

78

8

6)1(8

7

5

2

28

tttt

tt,

∫ −−−−−

−−−−+

dZZZZZZZZZZZ

ZZZZ

i

ttt

))(()()()()2(

4

4332

3

41

4

31

3

21

14

23

12

821

4π

∫ −−−=

−−−−+

dZZZZZZZ

ZZZ

i

ttt

)()()()2(

4

42

7

41

3

21

14

12

821

3π

dZZZ

Z

ZZ

Zt

ZZ

Zt

ZZ

ZZ

i

ttttt

−+

−++

−

+⋅

−=

−−−−−−−−+

∫ 2

41

11

2

41

21

41

31

7

41

14

821

2 )()()1(

)(2

2

)()2(

4

π

++

++

+++

+

+=

9

7

9

7

8

6)1(

7

5

2

24

tttt

tt,

∫ −−−−−

−−−−++

dZZZZZZZZZZZ

ZZZZ

i

tttt

2

42

2

32

2

41

2

31

4

21

14

13

52

51

4 )()()()()()2(

6

π

21

2

2

21

1

4

21

52

51

4 ))(()()2(

6dZdZdZ

ZZZZ

Z

ZZ

ZZ

i

ttt

−−⋅

−= ∫∫

−−++

π


٧٧

dZZZ

Z

ZZ

Zt

ZZ

Zt

ZZ

ZZ

i

ttttt

))(

4)(

)1(4)(

)1(()()2(

66

21

22

2

5

21

32

2

4

21

42

22

4

21

5

2

5

1

2 ∫ −+

−+−

−+⋅

−=

−−−−−−++

π

++

++−

++=

9

524

8

5)1(24

7

5)1(6 2

ttt

tt ,

∫ −−−−−−−=

−−+

dZZZZZZZZZZZZZ

ZZZZ

i

tt

))(()()()()()2(

12

4342

2

32

2

41

3

31

3

21

14

23

22

51

4π

∫ −−−−=

−−+

dZZZZZZZ

ZZZ

i

tt

3

42

5

41

3

21

14

22

51

3 )()()()2(

12

π

dZZZ

Z

ZZ

Z

ZZZZ

ZZ

i

tt

))(

6)(

6)(

1(

)()2(

125

41

2

4

4

21

4

3

41

5

41

1

4

5

1

2 ∫ −+

−+

−⋅

−−=

+−+

π

+−

+−

+−=

9

572

8

572

7

512

ttt

After simple computations one can get

118

65

63

379

5670

3511710

43

540

1109

3

2

135

19

630

11

11340

11)(

23

456789

4

++++

+++++=

ttt

tttttttH

and hence

2835

176

11340

114)( 3

4 =⋅=Bvol .

3.4 The effect of matrix operations on the number of integral points

Consider the polytope P that is defined as : bAXX d ≤ℜ∈ , where ddA ×ℜ∈ and db ℜ∈ . In this section we study the change of the number

of integral points of P upon performing the usual matrix operations on the matrix A, these operations include interchanging two rows (columns), the


٧٨

addition of a row (column) to another row (column), and the transpose of the matrix A.

The study is when n = 2, 3 and n = 4, the computation shows that there is no change in the number of integral points in the case of transpose and in the interchanging of rows or column under certain conditions.

Now we discuss the method for the polytope : bAXX d ≤ℜ∈ , where ddA ×ℜ∈ and db ℜ∈ . In two cases which are:

(I) The case of 22× matrix:

Suppose that

=

2221

1211

aa

aaA , ( )tb 11= . In order to find L(P, t) we use

∫∫=

−−−−

=−−

=mm

d

m

Z

cc

tb

m

tb

Z

mdZ

ZZ

ZZ

itPL

ξξπ )1)...(1(

......

)2(

1),(

1

1

11

11

1 where dcc ,...,1 are the

column vectors of A, here 1,...,01

<< ξξm

are distinct real numbers and

t a positive integer.

We use the standard multivariate notation nw

n

wW vvV L`1

1= .

∫ ∫= =

−−−−

−−=

11 22

21

21

)1)(1()2(

1),(

1

2

1

12

ξ ξπ Z Zcc

tbtb

dZZZ

ZZ

itPL , here 10 12 <<< ξξ .

Let 1,1 2211 <=<= ξξ ZZ and ( ) ( )2212221111 , aacaac == .

By assuming t = 1, 121 == bb we get

∫ ∫= =

−−

−−=

11 22

22122111 12

2121

22

21

2 )1)(1()2(

1)1,(

ξ ξπ Z Zaaaa

dZdZZZZZ

ZZ

iPL

Now let,

)1)(1(

1)(

22122111

2121

2

2

2

1

aaaa ZZZZZZZf

−−=

the poles of f(z) at 2ZZ = are

1. A pole at zero:

Since 22int0 ξ=∈ Z , therefore it is a pole of order two.

2. 01 2111

21 =− aa ZZ then 11

21

1

2

1a

a

ZZ = , 21

11

1

1

2 )1

( a

aZZ = , this is a pole if 22 ξ<Z

therefore,


٧٩

21

11

21

11

11

2

11

a

a

a

a

ZZ

ξ== , 2

1

1 ξξ

> ( because 10 12 <<< ξξ )

then 21

11

1

1

2 )1

( a

aZZ = ∉ )int( 22 ξ=Z .

3. 01 2212

21 =− aa ZZ then 12

22

1

2

1a

a

ZZ = , 22

12

1

1

2 )1

( a

aZZ =

therefore, 22

12

1

2

1

a

aZ

ξ= , but 2

1

1 ξξ

> ( because 10 12 <<< ξξ ) then

22

12

1

1

2 )1

( a

aZZ = ∉ )int( 22 ξ=Z .

Therefore we have pole of order two at zero only, and its residue has to be computed as follows,

for )1)(1(

1)(

22122111

2121

2

2

2

1

aaaa ZZZZZZZf

−−= , we determine the residues on

the circle 22 ξ=Z with radius 2ξ and center zero, if we consider

2

2

)()(

Z

ZZf

Φ= , where

=Φ )(Z)1)(1(

122122111

2121

2

1

aaaa ZZZZZ −−

1Z can be regarded as a constant, then

[]2

2

1

2221

2

1

2

1

221

2

221

)1)(()1(

)1).(()1(1

)(

222221

222121

−−−

−−−

−−−+

−−−−=Φ′

aaa

aaa

ZZaZ

ZZaZZ

Z

ββα

βαα

[]1

2

2

2

1

222

1

2

2

2

1

2212

1

)1()1(

)1()1(1

)(

212222

222121

−−−

−−−

−⋅−+

−⋅−=Φ′

aaa

aaa

ZZZa

ZZZaZ

Z

αββ

βαα

where 11

1

aZ=α and 12

1

aZ=β therefore

1. If 121 >a and 122 >a then 0)0( =Φ′ therefore L(P,1) = 0 .

Now if 121 =a and 122 >a then


٨٠

[ ]1

2

2

2

1

222

1

2

2

221

2 )1()1()1()1(1

)( 2222 −−−−− −⋅−+−⋅−=Φ′ ZZZaZZZ

Z aa αβββαα

2

111)0( −=Φ′ aZ

then

∫=

−=11

11

1

2

12

1)1,(

ξπ Z

a dZZi

PL

2. If 211 >a or 111 <a , then L(P,1) = 0.

3. If 111 =a then ∫=

=11

1

1

1

2

1)1,(

ξπ Z

dZZi

PL =1

If 121 >a and 122 =a then

[ ]1

2

2

221

2 )1()1(1

)( 21 −− −⋅−=Φ′ aZZZ

Z αββ

[ ] 2

12

1

121

)0( −==Φ′ aZZ

β

∫=

−=11

12

1

2

12

1)1,(

ξπ Z

a dZZi

PL

4. If 212 >a or 112 <a then L(P,1) = 0

5. If 112 =a then 11

2

1)1,(

11

1

1

== ∫=ξπ Z

dZZi

PL

If 12221 == aa , then [ ]βα +=Φ′2

1

1

1)(

ZZ

2

1

2

111211)( −− +=Φ′ aa ZZZ then ∫

=

−− +=21

1211

1

2

1

2

1 )(2

1),(

ξπ Z

aa dZZZi

tPL

6. If 211 >a or 111 <a , and 212 >a or 112 =a , then L(P,1) = 0

7. If 111 =a and 112 =a then L(P,1) = 2

This means we have three cases for n = 2, which are


٨١

Case 1: L(P,1) = 0

If

(1) 11, 2221

2221

1211 >>

= aanda

aa

aaA

(2) )12(1,1 111122

22

1211 <>>

= oraaanda

a

aaA

(3) )1 2( 1,1 121221

21

1211 <>>

= aoraanda

a

aaA

(4) )1,(2,,11 12111112

1211 <>

= aaoraa

aaA

Case 2: L(P,1) = 1

If

(1) 1,1

122

22

12 >

= a

a

aA

(2) 1,1

121

21

11 >

= a

a

aA

(3) 1 2,11

11212

12 <>

= aora

aA

(4) 1 2,11

11111

11 <>

= aora

aA

Case 3: L(P,1) = 2

If

=

11

11A

Now we look at the change in L(P,1) when a matrix operation is performed on the matrix A.

That is,

1. We first look at the operation of taking the adjoin, the relation between the number of integral points for : 2 bAXXP =ℜ∈= and

: 2 bXAXP T =ℜ∈= are equal and,


٨٢

a. L(P,1) = 0 if there exists a row nR of the matrix A with index n such that 1>nja , where nja are the elements of the row nR , and

there exists a column nC of the matrix A with index n such that 1>ina , where ina are the elements of the column njiCn ,...,1,, = .

b. L(P,1) = 1 if there exists a column jC of the matrix A with

element ija such that 1=ija and 21 >ka , jk ≠ and there exists

a row jR of the matrix A with index j such that njiaij ,...,1,,1 == .

c. L(P,1) = 2 if there exists two columns 1j

C2

, jC of the matrix A

such that 11

=ija and 12

=ija , 211 ,,2 jkjka k ≠≠≥ and there exists

two rows 21

, ii RR of the matrix A such that 1121

== jiji aanda and

for 21 jjC and

2121 ,:,,2,21

iimjjkaR mkii ≠≠> .

2. Interchanging of rows or columns:

The number of integral points for the original polytope and the polytope obtained by interchanging of rows or columns of the matrix A are equal and,

a. L(P,1) = 0 if there exists a row jR of the matrix A with index j

such that 2>jia , where jia are the elements of the row jR , and

column iC of the matrix A, nji ,...,1, = .

b. L(P,1) = 1 if there exists a column jC of the matrix A with

element ija such that 1=ija and 2>ika ,

njijk ,...,1,, =≠∀ .

c. L(P,1) = 2 if there exists two columns of the matrix A such that its elements are equal to one.

3. Sum of two rows or columns:

The relation between the numbers of integral points for the original polytope and the polytope obtained by adding two rows or columns of the matrix A are given as follows:

a. L(P,1) = 0, the conditions are the same as L(P, 1) = 0 for interchanging of rows or columns.

b. L(P, 1) = 1, in this case the number of integral points of the original polytopes and the changing polytopes may not be equal


٨٣

under any conditions. For example:

=

51

21A , where L(P, 1) =1,

but when we add two columns the result is

=

56

23B and

L(P, 1) = 0.

c. L(P, 1) = 2, also in this case, the numbers of integral points of the original polytopes and the changing polytopes may not be equal.

c. Multiplying by a constant c > 0:

The relation between the numbers of integral points for the original polytope and the polytope obtained by multiplying a row or column by a constant c > 0 are are given as follows:

a. L(P,1)=0, the conditions are the same as L(P,1)=0 for interchanging of rows or columns.

b. L(P, 1) = 1, in this case, the number of integral points of the original polytopes and the changing polytopes may not be equal

any where for example:

=

51

21A , where L(P, 1) = 1, but when

we multiply by a constant c > 0 the result is

=

53

23B and

L(P, 1) = 0.

c. L(P, 1) = 2, also in this case, the number of integral points of the original polytopes and the changing polytopes may not be equal under any condition.

(II) The case of 33× matrix:

Now if

=

333231

232221

131211

aaa

aaa

aaa

A

This means that the polytope : 3 bAXXP =ℜ∈= 3, Ζ∈b . By theorem (3.3.1) with t = 1 and 1321 === bbb , we get

∫ ∫ ∫= = =

−−−

−−−=

11 22 33

321 )1)(1)(1()2(

1)1,(

23

22

21

3ξ ξ ξπ Z Z Z

cccdZ

ZZZ

ZZZ

iPL

where

( )3121111 aaac = , ( )3222122 aaac = , ( )3323133 aaac = ,1,,0 123 << ξξξ .


٨٤

Then

)1()1()1(

)2(

1)1,(

332313322212312111

11 22 33

321321321

1232

32

22

1

3

aaaaaaaaa

Z Z Z

ZZZZZZZZZ

dZdZdZZZZ

iPL

−⋅−⋅−

=

−−−

= = =∫ ∫ ∫

ξ ξ ξπ

the integrand is

)1()1()1( 332313322212312111

321321321

23

22

21

aaaaaaaaa ZZZZZZZZZ

ZZZ

−⋅−⋅−

−−−

now we find the poles for the above function:

1. A pole of order one at zero,

Since 33int0 ξ=∈ Z .

2. 01 312111

321 =− aaa ZZZ then 1312111

321 =aaa ZZZ

31

2111

1

21

3

1 a

aa ZZZ

=

31

1

2111

21

3

1a

aa ZZ

Z =

=

31

21

31

11

11

1

a

a

a

a

ξξ

11

11

>ξξ

)int( 333 ξ=∉ ZZ .

Similarly for the other points.

By assuming

=Φ )(Z)1)(1)(1(

1333231

333

aaa ZZZ γβα −−−

where 231322122111

212121 and , aaaaaa ZZZZZZ === γβα

by generalized Cauchy integral formula at Z = 0, we get

1

3

1

3

2

3

1

333

1

3

1

3

2

3

1

332

133

1

3312

3

)1()1()1(

)1()1()1)()(1(

)1)(1)(()1()(

32313333

33313232

33323131

−−−−

−−−−

−−−

−−−+−−−−−+

−−−−−=Φ′

aaaa

aaaa

aaaa

ZZZZa

ZZZZa

ZZZaZZ

βαγγγαββ

γβαα


٨٥

regarding 1Z and 2Z as constants.

Now

If 11,1 333231 >>> aandaa then 0)0( =Φ′ , therefore

L(P, 1) = 0 with 11,1 333231 >>> aandaa .

Ι) If 1,1 333231 >= aaanda

1

3

1

3

2

3

1

333

1

3

1

3

23

1

3321

31

32

331

)1()1( )1()1()1(

)1()1()1()1()(3233333331

32323332

−−−−−−

−−−−−

−−−+−−

−+−−−=′aaaaa

aaaa

ZZZZaZZ

ZZaZZZaZ

βαγγγαββγβααΦ

2111

2131)0( aa ZZa ==Φ′ α

∫ ∫= =

=11 22

2111

122

2

2

1

212)2(

1)1,(

ξ ξπ Z Z

aa

dZdZZZ

ZZ

iPL

∫ ∫= =

−−=11 22

2111

12

2

2

2

12)2(

1

ξ ξπ Z Z

aa dZdZZZi

i) If 12 2121 <> aora , then L(P,1) = 0.

ii) If 121 =a then 1)0( =Φ′ . And

1

2

1

11

11

)2(

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

then

a) If 12 1111 <> aora then L(P, 1) = 0.

b) If 111 =a then L(P, 1) = 1.

Π) If 1,1 333132 >= aaanda then 2212

2132)( aa ZZaZ ==Φ′ β and

∫ ∫= =

=11 22

2212

122

2

2

1

212)2(

1)1,(

ξ ξπ Z Z

aa

dZdZZZ

ZZ

iPL

∫∫=

−

=

−=22

22

11

12

12

2

2

2

12)2(

1

ξξπ Z

a

Z

a dZdZZZi

i) If 12 2222 <> aora then 0)0( =Φ′ , this mean that L(P, 1) = 0.

ii) If 122 =a then 1)0( =Φ′ and 1

2

1

11

12

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If 12 1112 <> aora then L(P, 1) = 0.


٨٦

b) If 112 =a then 1)0( =Φ′ and hence L(P, 1) = 1.

III) If 1,1 323133 >= aaanda then

2313

2133)( aa ZZaZ ==Φ′ γ and hence

∫ ∫= =

−−=11 22

2313

12

2

2

2

12)2(

1)1,(

ξ ξπ Z Z

aa dZdZZZi

PL

∫∫=

−

=

−=22

23

11

13

12

2

2

2

12)2(

1

ξξπ Z

a

Z

a dZdZZZi

i) If 12 2323 <> aora then 0)0( =Φ′ , this means that L(P, 1) = 0.

ii) If 123 =a then 1)0( =Φ′ and 1

2

1

11

13

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

Then,

a) If 12 1313 <> oraa then L(P, 1) = 0.

I1) If 11,1 333231 >== aandaa

1

3

1

3

2

3

1

333

1

3

1

3

2

3

1

3

1

3

2

3

)1()1()1(

)1()1()1(

)1()1()1()(

3333

3132

33

−−−−

−−−

−−−

−−−+−−−+

−−−=Φ′

ZZZZa

ZZZ

ZZZZ

aa

aa

a

βαγγγαββ

γβαα

then 22122111

2121)0( aaaa ZZZZ +=+=Φ′ βα

hence ∫ ∫= =

+=11 22

22122111

122

2

2

1

21212)2(

1)1,(

ξ ξπ Z Z

aaaa

dZdZZZ

ZZZZ

iPL

∫ ∫= =

−−−− +=11 22

22122111

12

2

2

2

1

2

2

2

12)(

)2(

1

ξ ξπ Z Z

aaaa dZdZZZZZi

i) If ( 12 2121 <> aora ) and ( 12 2222 <> aora ) then L(P, 1) = 0.

ii) If ( 121 212221 <>= aoraanda ) then 1

21211)( −−=Φ′ ZZZ a ,

hence 1

2

1

11

11

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If 12 1111 <> oraa then 0)0( =Φ′ , this means that L(P, 1) = 0.

b) If 111 =a then L(P, 1) = 1.

iii) If )12(1 212122 <>= aoraanda then


٨٧

1

2

2

112)( −−=Φ′ ZZZ a , hence 1

2

1

11

12

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If 12 1212 <> aora then 0)0( =Φ′ , this means that L(P, 1) = 0.

b) If 112 =a then L(P, 1) = 1. iv) If 12212 == aa , then

1

2

1

2

1

11

1211 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ

a) If )12( 1111 <> aora and )12( 1212 <> aora , then L(P, 1) = 0.

b) If 111 =a and )12( 1212 <> aora then L(P, 1) = 1.

c) If 112 =a and )12( 1111 <> aora then L(P, 1) = 1.

d) If 111 =a and 112 =a then L(P, 1) = 2. II1) If 11 323331 >== aandaa , then

23132111

2121)0( aaaa ZZZZ +=+=Φ′ γα , then

∫ ∫= =

+=11 22

23132111

122

2

2

1

21212)2(

1)1,(

ξ ξπ Z Z

aaaa

dZdZZZ

ZZZZ

iPL

∫ ∫= =

−−−− +=11 22

23132111

12

2

2

2

1

2

2

2

12)(

)2(

1

ξ ξπ Z Z

aaaa dZdZZZZZi

i) If )12( 2121 <> aora and )12( 2323 <> aora , then L(P, 1) = 0.

ii) If 121 =a and )12( 2323 <> aora , then

1

2

2

111)( −−=Φ′ ZZZ a , hence 1

2

1

11

11

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If )12( 1111 <> aora then 0)0( =Φ′ and L(P, 1) = 0.

b) If 111 =a then L(P, 1) = 1.

iii) If 123 =a and )12( 2121 <> aora , then

1

2

2

113)( −−=Φ′ ZZZ a , hence 1

2

1

11

13

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If )12( 1313 <> aora then L(P, 1) = 0.

b) If 113 =a then L(P, 1) = 1.


٨٨

iv) If 11 2321 == aanda then

1

2

1

2

1

11

1311 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ

a) If )12( 1111 <> aora and )12( 1313 <> aora , then L(P, 1) = 0.

b) If 111 =a and )12( 1313 <> aora , then L(P, 1) = 1.

c) If 11 1311 == aanda then L(P, 1) = 2.

III1) If 11 313332 >== aandaa then

23132212

2121)( aaaa ZZZZZ +=+=Φ′ γβ , hence

∫ ∫= =

+=11 22

23132212

122

2

2

1

21212)2(

1)1,(

ξ ξπ Z Z

aaaa

dZdZZZ

ZZZZ

iPL

∫ ∫= =

−−−− +=11 22

23132212

12

2

2

2

1

2

2

2

12)(

)2(

1

ξ ξπ Z Z

aaaa dZdZZZZZi

i) If )12( 2222 <> aora and )12( 2323 <> aora , then L(P, 1) = 0.

ii) If 122

=a and )12( 2323 <> aora , then

1

2

2

112)( −−=Φ′ ZZZ a , hence 1

2

1

11

12

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If 12 1212 <> aora then 0)0( =Φ′ , this means that L(P, 1) = 0.

b) If 112 =a then L(P, 1) = 1.

iii) If 123 =a and )12( 2222 <> aora , then

1

2

2

113)( −−=Φ′ ZZZ a , hence 1

2

1

11

13

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If 12 1313 <> aora then L(P, 1) = 0.

b) If 113 =a then L(P, 1) = 1. iv) If 12322 == aa then

1

2

1

2

1

11

1312 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ

a) If )12( 1212 <> aora and )12( 1313 <≥ aora , then L(P, 1) = 0.

b) If 112 =a and )12( 1313 <> aora , then L(P, 1) = 1.


٨٩

IV1) If 1333231 === aaa then

231322122111

212121)( aaaaaa ZZZZZZZ ++=++=Φ′ γβα

∫ ∫= =

++=11 22

231322122111

122

2

2

1

2121212)2(

1)1,(

ξ ξπ Z Z

aaaaaa

dZdZZZ

ZZZZZZ

iPL

∫ ∫= =

−−−−−− ++=11 22

231322122111

12

2

2

2

1

2

2

2

1

2

2

2

12)(

)2(

1

ξ ξπ Z Z

aaaaaa dZdZZZZZZZi

i) If )12( 2121 <> aora , )12( 2222 <> aora and

)12( 2323 <> aora then L(P, 1) = 0.

ii) If 121 =a , )12( 2222 <> aora and )12( 2323 <> aora then

1

2

1

11

12

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If )12( 1212 <> aora , then L(P, 1) = 0.

b)If 112 =a then L(P, 1) = 1.

iv) If 123 =a , )12( 2121 <> aora and )12( 2222 <> aora then

1

2

1

11

13

2

1)1,( dZZ

iPL

Z

a

∫=

−=ξπ

a) If )12( 1313 <> aora then L(P, 1) = 0.

b) If 113 =a then L(P, 1) = 1.

v) If 12221 == aa and )12( 2323 <> aora then

1

2

1

2

1

11

1211 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ

a) If )12( 1111 <> aora and )12( 1212 <> aora then L(P, 1) = 0.



d) If 11211 == aa then L(P, 1) = 2. vi) If 12322 == aa and )12( 2121 <> aora then

1

2

1

2

1

11

1312 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ



٩٠



d) If 11312 == aa then L(P, 1) = 2. vii) If 12321 == aa and )12( 2222 <> aora then

1

2

1

2

1

11

1311 )(2

1)1,( dZZZ

iPL

Z

aa

∫=

−− +=ξπ




d) If 11311 == aa then L(P, 1) = 2. viii) If 1232221 === aaa then

1

2

1

2

1

2

1

11

131211 )(2

1)1,( dZZZZ

iPL

Z

aaa

∫=

−−− ++=ξπ

a) If )1aor2a(),1aor2a( 12121111 <><> and )12( 1313 <≥ aora then

L(P, 1) = 0.


c) If 112 =a , )1aor2a( and )1a or 2a( 11111313 <><> then L(P,1)=1

d) If 113 =a , )12( 1111 <> aora and )12( 1212 <> aora then L(P, 1)=1

e) If 11211 == aa and )12( 1313 <> aora then L(P, 1) = 2.

f) If 11311 == aa and )12( 1212 <> aora then L(P, 1) = 2.

g) If 11312 == aa and )12( 1111 <> oraa then L(P, 1) = 2.

h) If 1131211 === aaa then L(P, 1) = 3.

The obtained results in the case of n=3 are related to the number of integral points for the original polytope and the changing polytope are the same as the case when n=2.


٩١

The general case is:

If

=

nnnn

n

aaa

aaa

A

..

..

..

..

..

21

11211

where njia ji ,...,1,, =

are positive integers and t=1, )1...1(=b . For the polytope

: bAXXP d ≤ℜ∈= .

Our obtained results are, The number of integral points is:

1. L(P,1)=0

This will hold if nR or any row of the matrix A has an element niain ,...,1,2 => , where nR is the n-th row of the matrix A, and the

number of integral points does not change when:

a) Interchanging any two columns (rows).

b) Summing two columns (rows).

c) the matrix A is transposed, the condition on A is 1≥ina and njianj ,...,1,,1 =≥ .

2. L(P,1)=1 When nR or any row of the matrix A has an element

niain ,...,1,2 => , where nR is the n-th row of the matrix A, and the

number of integral points does not change when:

a) Interchanging any two columns (rows).

But the number of integral points is change if:

b) Multiplying by a constant c>0.

c) Summing two columns (rows).

d) The matrix A is transposed, the conditions on A that are needed to get the same number of integral points are:

i) The elements of the first row are one.

ii) There exist two rows and columns and where their elements are one.

shatha assaad salman al-najjar - nahrain university

Documents