15.083J Optimization over Integers. Lectures 13-14: Algebraic geometry and integer optimization

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1 J Optimization over Integers Lectures 13-14: Algebraic geometry and integer optimization Typos and errors corrected. Section 7.4 completely rewritten.

2 geometry optimization optimization 7.1. Background from algebraic 7.. Applications to binary 7.3. Grobner bases for integer 7.4. Applications of real geometry algebraic 7.5. Generating functions for integer points polyhedra in Summary 7.6. Exercises Notes and sources Chapter 7 Algebraic geometry and integer optimization Contents

3 4 Chap. 7 Algebraic geometry and integer optimization In this chapter, we i n troduce several approach e s t o i n teger optimization that have their roots in algebraic geometry. While the methods we discuss in this chapter have n o t y et resulted in practical algorithms, they are based on deep and elegant ideas from a very active eld of mathematics that could potentially lead to ecient algorithms in the future. To m a k e t h e c hapter self-contained we discuss the fundamental concepts from algebraic geometry we will use throughout the chapter. Applying some classical ideas from algebraic geometry we derive a F arkas lemma for integer optimization, and algorithms for both binary and integer optimization. We further introduce some ideas from real algebraic geometry and the problem of moments and derive a convergent (and, in the case of binary optimization, nite) sequence of semidenite relaxations for integer nonlinear optimization. Finally, w e i n troduce a powerful idea of decomposition of cones that leads to a polynomial time algorithm for counting the numbe r of integer points in a polyhedron in a xed dimension. 7.1 Background from algebraic geometry The basic algebraic objects in this chapter are polynomials o ver complex or real numbers. Denition 7.1 (Polynomials) (a) A monomial i n v ariables x = ( x 1 : : : x n) 0 is a product of the form 1 x = x n 1 x n n where = ( 1 : : : n) 0 Z +. (b) A polynomial is an expression of the form X f (x) = f x S n with S Z +, S nite, and f k (k = R or k = C ) for all S. The set of all polynomials in x 1 : : : x n with coecients in k = R or k = C is denoted by k[x 1 : : : x n]. Throughout the chapter we will use k to mean either R or C. W e will indicate explicitly when specic results only hold for k = C. While in one dimension the monomials are naturally ordered by t h e i r degree, i.e., x m > x m;1 > > x > 1, in many dimensions one can dene

4 Sec. 7.1 Background from algebraic geometry 5 several orders. Denition 7. A monomial order on k[x 1 : : : x n] is any relation > on Z n +, or equivalently, a n y relation on the set of monomials x, Z n +, satisfying: (a) > is a total order on Z n +. (b) If > and Zn +, then + > +. (c) Any nonempty subset of Z n + has a smallest element under >. Important examples of monomial orders include: (a) The lexicographic o r d e r. For every Z n +, we sa y > (or x vector dierence ; Z > lex x lex ) if, in the n, the leftmost nonzero entry is positive. For example = (1 0) 0 > (1 1 3) 0 =, since ; = (0 1 ; 3) 0 lex. (b) The graded lex order. For every n, w e say > grlex (or x > grlex x ) if, Z + Xn X n jj = i > i = jj or jj = jj and > lex : In other words, grlex orders by total degree rst and breaks ties using the lex order. For example, = (1 3) 0 > grlex (3 0) 0 = since jj = 6 > 5 = jj: (c) The graded reverse lex order. For every Z n w e say > grevlex (or x > grevlex x +, ) if, Xn X n jj = i > i = jj or jj = jj and in ; Z n the right-most nonzero entry is negative. For example, = (1 5 ) 0 > grevlex ( 3 3) 0 =, since jj = jj = 8 and ; = (;1 ;1) 0 : (d) The vector induced order. We consider a vector c Zn, and for every Zn + +, w e say > c (or x > c x ) if, c 0 > c 0 or c 0 = c 0 and > lex : Note that the requirement c 0 is important, since otherwise, Part (c) in Denition 7. is violated. Example 7.1 Under dierent monomial orders the polynomial f = 6 x 1 x 3 7x 3 ; 5x x 1x 3 is written as follows: x 3 +

5 6 Chap. 7 Algebraic geometry and integer optimization (a) Under the lex order (b) Under the grlex order (c) Under the grevlex order 3 1 f = ;5x + 4 x 1x x 1 x x x 3: f = 4 x x + 6 x 1 x x 3 ; 5x + 7 x : 1 f = 6 x 1 x x x x 3 ; 5x x 3: (d) Under the vector induced order with c = (3 ) 0 3 f = 4 x 1x 3 ; 5x x 1 x x 3 + 7x 3: 3 Note that the eld of complex numbers k = C is algebraically closed, since every nonconstant polynomial has a root in C. The eld k = R, h o wever, is not algebraically closed, since there do not exist any real roots of the equation x + 1 = 0. We n e x t i n troduce the concept of ideals and ane varieties that play a central role in algebraic geometry. Denition 7.3 (Ideals and Ane Varieties) (a) An ideal is a subset I k[x 1 : : : x n] satisfying: (i) 0 I. (ii) If f g I, then f + g I. (iii) If f I and h k[x 1 : : : x n], then hf I. (b) Given an ideal I, w e dene an ane variety V (I) V (I) = fx k n j f (x) = 0 for all f Ig: A k ey property related to ideals is that every ideal is nitely generated. Theorem 7.1 (The Hilbert Basis Theorem) Every ideal I k[x 1 : : ::x n] is nitely generated, i.e., there exist g 1 : : : g t I such that ( Xt I = h i g i h 1 : : : h t k[x 1 : : : x n] : We write I = hg 1 : : : g ti. )

6 Sec. 7.1 Background from algebraic geometry 7 Theorem 7.1 implies that if I = hg 1 : : : g ti, th en V (I) = fx k n j g i (x) = 0 for all i = 1 : : : t g: For polynomials in one variable, it turns out that any ideal in k[x] can be generated by a single polynomial, i.e., I = hf i for some polynomial f k[x]. Then V (I) is the set of roots of f. Since every nonconstant polynomial in C [x] has a root, the only way th a t V (I) = w ould be for f to be a nonzero constant. In this case, 1=f C, a n d th us 1 = (1=f ) f I, which implies that g 1 = g I for all g C [x]. This shows that I = C [x] is the only ideal of C [x] that represents the empty v ariety. It turns out that the same property holds in multiple variables, that is, in any polynomial ring, algebraic closure of k is enough to guarantee that the only ideal which represents the empty v ariety is the entire polynomial ring itself. This is the Weak Nullstellensatz, which is the basis of one of the most important mathematical results of the late nineteenth century, Hilbert's Nullstellensatz. Theorem 7. (The Weak Nullstellensatz) Let I C [x 1 : : : x n] be an ideal satisfying V (I) =. Then I = C [x 1 : : : x n]. We use I(V ) to denote the ideal that contains all polynomials t h a t vanish on a given variety V. In Exercise 7. we s h o w that for a variety V, it holds that if f m I(V ), for some integer m 1, then f I(V ). Thus, an ideal consisting of all polynomials which v anish on a variety V has the property that if some power of a polynomial belongs to the ideal, then the polynomial itself must belong to the ideal. This leads to the denition. Denition 7.4 An ideal I is radical if f m I for some integer m 1 implies that f I. The previous discussion implies that I(V ) is a radical ideal. Another po werful theorem is as follows. Theorem 7.3 (Hilbert's Nullstellensatz) The polynomials f f 1 : : : f s C [x 1 : : : x n] satisfy f I(V (f 1 : : : f s)) if and only if there exists an integer m 1 such that f m h f 1 : : : f si. Theorem 7.3 says that the only way that an arbitrary ideal I can fail to be the ideal of all polynomials v anishing on V (I) is for I to contain powers

7 8 Chap. 7 Algebraic geometry and integer optimization f m of polynomials which are not in I. This leads to the denition. Denition 7.5 Let I k[x 1 : : : x n] be an ideal. The radical of I, denoted by p I, is the set ff j f m I for some integer m 1g: It is a simple observation that an ideal is radical if and only if I = p I. It turns out (Exercise 7.3) that if I is an ideal in k[x 1 : : : x n], then p I is an ideal in k[x 1 : : : x n] containing I. F urthermore, p I is a radical ideal. Theorem 7. leads to a Farkas type lemma for a system of polynomial equations over C [x 1 : : : x n]: Theorem 7.4 Let g 1 : : : g t C [x 1 : : : x n] be polynomials. Then, the system g i (x) = 0 i = 1 : : : t (7.1) C has no solution in n C if and only if there exist polynomials h 1 : : : h t [x 1 : : : x n] such that X t h i g i = 1 : (7.) Proof. We let I = hg 1 : : : g ti and V (I) = fx C n j g i (x) = 0 for all i = 1 : : : t g : If Eq. (7.1) C does not have a solution, then V (I) =. Theorem 7. implies that I = [x 1 : : : x n]. Since the number 1 is a m em ber of k[x 1 : : : x n] and C since I is generated Pby g 1 : : : g t, there exist polynomials h 1 : : : h t t [x 1 : : : x n] such that h i g i = 1 i.e., Eq. (7.) has a solution. Conversely, suppose that Eq. (7.) has a solution. We assume, for the purpose Pof deriving a contradiction, that Eq. (7.1) has a solution x 0. Then, 1 = t h i (x 0 )g i (x 0 ) = 0 a c o n tradiction. Theorem 7.4 asserts that a certicate of the infeasibility of the system of polynomial equations (7.1) is the set of polynomials h 1 : : : h t. In this sense, Theorem 7.4 can be considered an extension of the Farkas lemma to a system of polynomial equations. In order to compute the polynomials h 1 : : : h t, w e need to be able to address the question of deciding whether a g i v en polynomial f belongs to the ideal I = hg 1 : : : g ti. Specically, Theorem 7.4 asks whether the polynomial 1 is in I. In order to develop

8 Sec. 7.1 Background from algebraic geometry 9 methods to address this question we need some additional denitions. Denition P 7.6 Given a monomial order > and a polynomial f (x) = S f x we dene (a) The multidegree of f is multideg(f ) = max S where the maximum is taken over the given order > : (b) The leading coecient o f f is (c) The leading monomial o f f (d) The leading term of f LC(f ) = f multideg(f ) : LM(f ) = x multideg(f ) : LT(f ) = LC( f ) LM(f ): To illustrate for f = ;5x x 1 x x 1 x x x 3 and the lex order, 3 multideg(f ) = (3 0 0) 0 LC(f ) = ;5 LM(f ) = x 3 1 LT(f ) = ;5x 1 : Given a monomial order, we can dene the division of a polynomial f (x) b y a polynomial g(x) as illustrated in Example 7.. A natural idea to decide if f I = hg 1 : : : g ti is then to start dividing f by g 1 : : : g t to obtain f = h 1 g h t g t + r and to say that f I if the remainder r = 0 : Note that in multivariate polynomials the remainder r depends on the order in which w e perform the division. While r = 0 implies that f I, r 6 = 0 does not necessarily imply that f = I as the next example illustrates. Example 7. Let g 1 = xy + 1, g = y ; 1 and consider the lex order. We w ant to determine whether f = xy ; x h g 1 g i. If w e divide f by g 1 rst and then by g we obtain xy ; x = y(xy + 1) + 0( y ; 1) + (;x ; y): If, however, we divide f by g rst and then by g 1 we obtain xy ; x = x(y ; 1) + 0(xy + 1) + 0 which leads to the conclusion that f h g 1 g i.

9 10 Chap. 7 Algebraic geometry and integer optimization This observation motivates the following denition. Denition 7.7 Let G = fg 1 : : : g tg be a set of polynomials and I = hg 1 : : : g ti. G is called a Grobner basis for the ideal I if f I if and only if the remainder of the division of f by the polynomials in G is always zero, independent of the order in which w e perform the division. Given an ordered collection G of polynomials and a polynomial f, w e can dene the remainder R(f G ) of the division of f by the polynomials in G in the order specied. If G, h o wever, is a Grobner basis the remainder R(f G ) is independent of the order we perform the divisions. Thus, given a Grobner basis G of an ideal I we can decide whether f I by successive divisions. It is not a priori clear that a Grobner basis exists for an arbitrary ideal. The following algorithm, however, constructs such a basis. Algorithm 7.1 The Buchberger algorithm Input: A set F = ff 1 : : : f sg of polynomials i n k[x 1 : : : x n] and a monomial order. Output: A Grobner basis G = fg 1 : : : g tg for I = hf 1 : : : f si. Algorithm: 1. (Initialization) Let G := (f 1 : : : f s) be the ordered tuple obtained from F.. For each p q G (a) Let = m ultideg(p), = m ultideg(q), i = m ax( i i), and = ( 1 : : : n) 0. (b) Compute the polynomial x x S(p q) = p ; q: LT(p) LT(q) (c) Compute the remainder r of the division of S(p q) with the ordered set G, denoted as R(S(p q) G ) using the given monomial order. (d) If r 6= 0, then G := (G r) goto Step If r = 0 for all p q G, then goto Step Output G. We remark that dierent monomial orders lead to dierent G r obner bases, since the monomial order aects the outcome of the division of polynomials.

10 Sec. 7.1 Background from algebraic geometry 11 Example 7.3 We w ould like to nd the Grobner basis for I = hf 1 f i with 3 ; xy and f = x y ; y + x and the grlex order. We compute S(f 1 f ) f 1 = x as follows: multideg(f 1 ) = (3 0) 0 multideg(f ) = ( 1) 0, and thus = (3 1) 0. Thus, 3 3 x y 3 x y S(f 1 f ) = (x ; xy) ; (x y ; y + x) = ;x : x 3 x y Moreover, R(S(f 1 f ) ff 1 f g) = f 3 = ;x, and G = ( f 1 f f 3 ). We continue and obtain 3 S(f 1 f 3 ) = ( x R(S(f 1 f 3 ) ff 1 f f 3g) = f 4 = ;xy: ; xy) ; (;x)(;x ) = ;xy We add f 4 to G which now becom es G = ( f 1 f f 3 f 4 ). We continue and nd 3 S(f 1 f 4 ) = y(x Moreover, ; xy) ; (;1=)x (;xy) = ;xy S(f f 3 ) = ( x y ; y + x) ; (;y)(;x R(S(f f 3 ) G ) = f 5 = ;y + x: = yf 4 R (S(f 1 f 4 ) G ) = 0 : ) = ;y We add f 5 to G, and obtain G = ( f 1 : : : f 5 ), and we check that Thus, a Grobner basis is R(S(f i f j ) G ) = 0 for all i j = 1 : : : 5: 3 fx ; xy x y ; y + x ;x ;xy ;y + xg: + x The following theorem shows that Grobner bases are correctly computed by Algorithm 7.1. Theorem 7.5 Algorithm 7.1 terminates in nite time. It computes a Grobner basis for the ideal I = hf 1 : : : f si with respect to the given monomial order. It turns out that Grobner bases computed by Algorithm 7.1 are often bigger than necessary. W e can eliminate some of its elements by using the observation that if p G such that LT(p) h LT(G n f pg)i, th en G n f pg is also a Grobner basis. Example 7.4 Continuing Example 7.3, we observe t h a t 3 LT(f 1 ) = x = ;x LT(f 3 ) and thus we can delete f 1 from the Grobner basis. Also, and we can eliminate f. LT(f ) = x y = ;(1=)x LT(f 4 )

11 1 Chap. 7 Algebraic geometry and integer optimization We next dene the notion of a minimal Grobner basis. Denition 7.8 A minimal Grobner basis for a polynomial ideal I is a Grobner basis G for I such that (a) LC(p) = 1 for all p G. (b) For all p G, LT(p) = h LT(G n f pg)i: Example 7.5 Continuing Example 7.4, we conclude that fx x y y ; (1=)xg is a minimal Grobner basis. A given ideal, however, might h a ve m a n y minimal Grobner bases. In the example, it is easy to check t h a t G = fx + xy xy y ; (1=)xg for R being any constant is also a minimal Grobner basis. This leads to the denition. Denition 7.9 A reduced Grobner basis for a polynomial ideal I is a Grobner basis G for I such that (a) LC(p) = 1 for all p G. (b) For all p G, no monomial o f p lies in hlt(g n f pg)i: In Example 7.5, the only reduced Grobner basis is G 0 corresponding to = 0. In fact, we h a ve the following theorem. Theorem 7.6 For a given monomial order, a polynomial ideal has a unique reduced Grobner basis. Many computer algebra systems have implemented Algorithm 7.1 to compute the reduced Grobner basis for a given monomial order. However, Algorithm 7.1 has two diculties: First, it nds a lot of elements in the basis that might be deleted later as they do not belong to the reduced Grobner basis. Furthermore, because of the divisions with potentially large numbers, its numerical stability is problematic. Nevertheless, Algorithm 7.1 leads to a constructive proof of Theorem 7.1 and forms the algorithmic foundation for several applications of algebraic geometry including those in integer optimization we outline in Sections 7. and 7.3. We will also utilize the notion of an elimination ideal. Denition 7.10 Given I = hf 1 : : : f si k[x 1 : : : x n], the lth elimination ideal I l is dened as I l = I \ k[x l+1 : : : x n].

12 Sec. 7. Applications to binary optimization 13 A connection of elimination ideals and Grobner bases with the lex order is given by: Theorem 7.7 Let I k[x 1 : : : x n] be an ideal and let G be a Grobner basis of I with respect to lex order where x 1 > x > > x n. Then, for every 0 l n ; 1, the set G l = G \ k[x l+1 : : : x n] is a Grobner basis of the lth elimination ideal I l. We also have: Theorem 7.8 Let I = hq 1 : : : q si C [x 1 : : : x n] and let I 1 be the rst elimination ideal of I. F or each 1 i s, w e can write q i in the form N i 1 qi = h i (x : : : x n)x + terms in which x 1 has degree smaller than N i where N i 0 and h i C [x : : : x n] is nonzero. Suppose we h a ve a partial solution (a : : : a n) V (I 1 ). If (a : : : a n) = V (h 1 : : : h s), then there exists a 1 C such that (a 1 : : : a n) V (I). 7. Applications to binary optimization In this section, we d e v elop a Farkas type lemma and propose an algorithm for binary optimization problems using the results of Section 7.1. Let A Z mn and b Z m and consider the system Ax = b x f 0 1g n : (7.3) C We write x i f 0 1g as x i ; x i = 0 and dene k =, w h ich is algebraically closed. We l e t f i = a0 i x ; b i, i = 1 : : : m, g j = x j ; x j and N = f1 : : : n g: We let J = hf 1 : : : f m g 1 : : : g ni V (J) = fx C n j Ax = b x i ; x i = 0 i N g

13 m js 14 Chap. 7 Algebraic geometry and integer optimization and apply Theorem 7.4. Theorem 7.9 System (7.3) is infeasible if and only if there exist vectors u S Q m, S N such that u0 b = ;1 0 1 X X (7.4) u A j ; ba + u 0 A j = 0 8 S N S 6= : S js Snfjg Proof. Applying Theorem 7. to the ideal J we obtain that problem (7.3) is infeasible if and only if there exist polynomials h 1 : : : h m and r 1 : : : r n such that X m h i (x)(a i X n 0 x ; b i ) + r i (x)(x i ; x i ) = 1 : Since we a r e i n terested in x f 0 1g n it is enough to consider polynomials h 1 : : : h m of the form X Y h i (x) = u S i x j SN with u S i C. Since the matrix A and the vector b are real, it is enough to consider u S i Q by considering the real part of u S i. W e th us obtain that for all x f 0 1g n Xm X Y 1 = u S i (a i 0 x ; b i ) x j js js SN 0 1 Xm X X X Y A = u S a ij x j ; b i + a ij x j SN js j S js = 0 1 Xm X X Y = u S a ij ; b i A x j (x = x i i S) i SN js js X X X Y + u S i a ij x r SN j S = rs[fjg X Xm X XX m Y u S a ij ; b i A + u Snfjg i a ij A x j : SN js js js x j

14 1 Sec. 7. Applications to binary optimization Equating the coecients of the monomials X m u i (;b i ) = Q js x j, w e obtain Xm X us a ij ; b i A + u Snfjg i a ij A = 0 8S N S 6= : js js which leads to Eq. (7.4). We next proceed to use Algorithm 7.1 to nd a solution to system (7.3) or to detect that no feasible solution exists. 15 Algorithm 7. Input: A Z mn b Z m. Output: A feasible solution (a 1 : : : a n) to system (7.3) or a certicate that system (7.3) is infeasible. Algorithm: 1. Find a Grobner basis G of the ideal J using lex order x 1 > x > > x n. If G = f1g, then problem (7.3) has no feasible solutions.. If G 6= f1g: Consider for 1 l n ; 1, the sets G l = G \ C [x l+1 : : : x n]. Starting from index n ; 1, a n d w orking sequentially: (a) Find a n in V (G n;1 ). (b) Extend a n to (a n;1 a n) such that (a n;1 a n) V (G n; ). Continuing in this fashion, nd a such that (a : : : a n) V (G 1 ) and a 1 such that (a 1 : : : a n) V (G). (c) Output (a 1 : : : a n). We next show that Algorithm 7. correctly solves system (7.3). Theorem 7.10 Algorithm 7. either nds a feasible solution for system (7.3), or provides a certicate of infeasibility whenever the Grobner basis G = f1g. Proof. Consider the elimination ideals J 1 : : : J n;1 where J k = J \C [x k+1 : : : x n]. If we nd the Grobner basis G of the ideal J using Algorithm 7.1 with the lex order, then Theorem 7.7 implies that G k := G \ C [x k+1 : : :x n] is a

15 16 Chap. 7 Algebraic geometry and integer optimization Grobner basis of J k, further implying that V (G k ) = V (J k ). Observe th a t one of the following two cases holds: (a) The Grobner basis G = f1g. Then, the ideal J = C [x 1 : : : x n] indicating that V (J ) is empty, since we are working in an algebraically closed eld. Therefore, system (7.3) is infeasible. (b) The Grobner basis G 6= f1g. In this case, we notice that G n;1 has elements x n ; 1 or x n or x n ; x n belonging to it. This follows from the observation that J n;1 is a polynomial ideal in one variable, namely a subset in C [x n ], and therefore needs only one generator. Moreover, we know t h a t xn ; x n is in the ideal, and thus, given that the only possible solutions in the corresponding variety are x n = 0 or x n = 1, it follows that the only possible elements of G n;1 are x n ; 1 or x n or x n ; x n. W e i n terpret this to mean that points 1 or 0 or both 1 and 0 are partial solutions, respectively. Thus, we n d a n V (J n;1 ). Subsequently, w e obtain a n;1 by extending the partial solution a n to (a n;1 a n) V (J n; ). That is, we take a partial solution in V (J k ) and extend it to a partial solution in V (J k;1 ) and so on. By Theorem 7.8 this is always possible as the appropriate leading coecients (starting from f i and g j ) in our setting are constants. Continuing in this way and applying Theorem 7.8 we can obtain a solution (a 1 : : : a n) in V (J ). It is important t o o b s e r v e that using dierent a n from V (J n;1 ), we can nd all solutions to system (7.3). Moreover, it is im portant to use the lex monomial order. The results do not hold for other orders. Example 7.6 Consider the problem x 1 + x + 3 x x x x 6 = 15 x j f 0 1g 8 j: The ideal we consider is J = hx 1 + x + 3 x x x x 6 ; 15 x 1 ; x 1 x ; x : : : x 6 ; x 6 i: The reduced Grobner basis of J with lex order is G = fx 6 ; x 6 x 5 + x 6 ; 1 x 4 + x 6 ; 1 x 3 + x 6 ; 1 x + x 6 ; 1 x 1 + x 6 ; 1g: Therefore, we h a ve G 5 = fx 6 ; x 6 g, indicating that a 6 = 1 and a 6 = 0 are both partial solutions. Starting from a 6 = 1, we get a 1 = a = : : : = a 5 = 0. Starting from a 6 = 0, w e get a 1 = : : : = a 5 = 1. Therefore, the only two feasible solutions are ( ) 0 and ( ) 0. We next outline some structural properties of the reduced Grobner basis obtained from Algorithm 7. using the lex order x 1 > x > > x n. If the solution to system (7.3) is unique, then all the reduced Grobner basis elements are of the form x i ; a i with a i f 0 1g. Example 7.7 Consider the system x 1 + x + x 3 = x 1 + x = x + x 3 = 1

16 Sec. 7. Applications to binary optimization 17 x 1 x x 3 f 0 1g having a unique solution x 1 = x = 1, x 3 = 0. Algorithm 7. yields this solution through the reduced Grobner basis G = fx 3 x ; 1 x 1 ; 1g. As we h a ve seen in the proof of Theorem 7.10, if the reduced Grobner basis G 6= f1g, then the polynomials in G can be partitioned in n sets S n : : : S 1 as follows: (a) Set S n contains only one polynomial, which is either x n, x n ; 1 or x n ; x n. (b) Set S n;1 contains polynomials in x n and x n;1. (c) Set S i, i = n ; : : : 1 c o n tains polynomials in x n x n;1 : : : x i. This property follows from the fact that the underlying term order is an elimination order. This \diagonalized structure" of the reduced Grobner basis { a generalization of Gaussian elimination for the nn linear system of equations { is useful in enumerating the feasible binary solutions. Starting with S n, w e a ssig n v alues for x n = 0 (if x n S n ), or x n = 1 (if x n ;1 S n ), or x n = 0 1 (if x n ; x n S n ). Having chosen x n, w e backsolve for x n;1 using the polynomials in S n;1, and proceed to assign values to the other variables recursively. Example 7.8 Consider the system x 1 + x + 3 x x x 5 = 6 x 1 : : : x 6 f 0 1g: The reduced Grobner basis for the lex order is G = fx 5 ; x 5 x 4 x 5 x 4 ; x 4 x 3 + x 4 + x 5 ; 1 x + x 5 ; 1 x 1 + x 4 + x 5 ; 1g: The sets are as follows: S 5 = fx 5 ; x 5 g S 4 ; x 4 g, S 3 = fx 3 + x = fx 4 x 5 x x5 ; 1g S = fx + x 5 ; 1g, S 1 = fx 1 + x 4 + x 5 ; 1g: Algorithm 7. enumerates all binary solutions: ( ) 0 ( ) 0 ( ) 0 : We next use Grobner bases to solve the optimization problem minimize c 0 x subject to Ax = b (7.5) x j = x j j = 1 : : : n : C P We w ork in n [x 1 : : : x n y ] and we let h = y ; j=1 c j x j f i = a0 i x ; b i, i = 1 : : : m and g j = x j ; x j, j = 1 : : : n. We consider V = V (f 1 : : : f m g 1 : : : g n h ). Following the same approach as that used for the feasibility problem, with lex order x 1 > x > : : : > x n > y, w e notice that the Grobner basis G of J = hf 1 : : : f m g 1 : : : g n h i is either f1g (indicating infeasibility) or we will have G intersected with k[y] w hich

17 j=1 j=1 j=1 18 Chap. 7 Algebraic geometry and integer optimization leads to a polynomial in y. This polynomial in y has the interpretation that every root of the polynomial is an achievable cost, i.e., there exists a feasible solution x that has cost equal to the root. Therefore, we can nd the minimum root, and work backwards to get the associated x j values. Example 7.9 Consider the problem minimize x 1 + x + 3 x x 4 subject to x 1 + x + x 3 + x 4 = 3 x j = x j j = 1 : : : 4: The reduced Grobner basis with lex order of J = hx 1 + x + 3 x x 4 ; y x 1 + x + 4 x 3 + x 4 ; 6 x 1 ; x 1 x ; x x 3 ; x 3 x 4 ; x 4 i is G = f10 ; 74y + 15 y 3 ; y ; 0 + x y ; y ; x 3 + y ; x 3 y x 3 ; x 3 ; 3 ; x ; x 3 ; x y ; y 3 ; x 1 ; x 3 ; 11y + y g which suggests that all feasible y's are y = 4 5 6, i.e., the roots of the rst polynomial of the Grobner basis. If, however, we h a ve additional information that 4 y 5, we can add the polynomial (y ; 4)(y ; 5) to J and rerun algorithm 7.. The reduced Grobner basis with lex order is then: G = f0 ; 9y + y ; 1 + x 3 ; 4 ; x + y 5 ; x 1 ; yg: 7.3 Grobner bases for integer optimization We consider the integer optimization problem in standard form n minimize c 0 x subject to Ax = b (7.6) x Z n + where c Z +, b Z m and A Z mn. Note that for ease of exposition we + + consider rst the case that all data in the problem are nonnegative i n tegers. We i n troduce an indeterminate z i, i = 1 : : : m for the ith constraint a0 i x = b i, and exponentiate to obtain z a i1 x 1 + : : : + a in x n i = z b i i i = 1 : : :m: Multiplying these equalities we obtain m n! n m n m xj m a ij xj a x j a ij zi = z ij i i i Y Y ; Y Y; Y Y Y = z = z b i : (7.7)

18 Sec. 7.3 Grobner bases for integer optimization 19 We further introduce the indeterminates w j, j = 1 : : : n, and the mapping : C [w 1 : : : w n]! C [z 1 : : : z m] dened as: Y m a ij i (w j ) = z and for a polynomial g C [w 1 : : : w n], (g(w 1 : : : w n)) = g((w 1 ) : : : (w n )): The following proposition is a direct manipulation of the denitions and is left as an exercise. Proposition 7.1 A v ector x Z n + is feasible in problem (7.6) if and only x1 x if maps the monomial w x = w n Q 1 w n to the monomial z b = m z bi i : Example 7.10 We consider the integer feasibility problem 4x x + x 3 = 37 x x + x 4 = 0 x 1 x x 3 x 4 Z + The mapping : C [w 1 w w 3 w 4 ]! C [z 1 z ] is given by (w 1 ) = z 1 z (w ) = z 1 z (w 3 ) = z 1 (w 4 ) = z : The set of feasible solutions are all the integer points (x 1 x x 3 x 4 ) 0 such that x x (w x1 w w w x4 ) = z 1 z : Q m a ij i Let f j = (w j ) = z and consider the ideal I = hf 1 ; w 1 : : : f n ; w n i C [z 1 : : : z m w 1 : : : w n]: We further consider monomial orders in C [z 1 : : : z m w 1 : : : w n] with the property t h a t a n y monomial containing one of the z i 's is greater than any monomial c o n taining only w j 's. For example, the monomial order z 1 > : : : > z m > w n > : : : > w 1 is one such order. We consider a Grobner basis G for the ideal I and the monomial order above. Let f C [z 1 : : : z m ] and let g = R(f G ) be the remainder of the division of f by the polynomials in

19 0 Chap. 7 Algebraic geometry and integer optimization G. W e next characterize R(f G ) and the structure of G. Proposition 7. Let I = hf 1 ; w 1 : : : f n ; w n i. Let G be a Grobner basis G for the ideal I and the monomial order z 1 > : : : > z m > w n > : : : > w 1. (a) All elements of G are dierences of two monomials. (b) If f is a monomial i n C [z 1 : : : z m ], then R(f G ) is also a monomial. Proof. (a) We initialize Algorithm 7.1 with polynomials f i ; w i, i = 1 : : : n that are all dierences of two monomials. In the computation of an S-polynomial in Algorithm 7.1 to compute G, w e are subtracting two polynomials each of which is a dierence of two monomials. Since the leading terms cancel, the resulting S-polynomial is again a dierence of two monomials. Arguing inductively, the result follows. (b) In the remainder calculation, we rst subtract from f a m ultiple of an element o f G, w h i c h from part (a) is the dierence of two monomials in a way that the leading terms cancel. Given that f is a monomial, then the remainder will be a monomial. C o n tinuing with the divisions, we conclude that R(f G ) is a monomial. Example 7.11 The ideal I in Example 7.10 is I = hz 1 z ; w 1 z 1 z ; w z 1 ; w 3 z ; w i: For the monomial order z 1 > z > w 4 > w 3 > w > w 1 the Grobner basis is 4 3 G = fz 1 ; w 3 z ; w 3 w 4 w 3 ; w 1 w 4w 3 w ; w w w 4 w 3 w 1 ; w w 4 w 1 ; w 3 w w ; w 1 g i.e., all the elements are dierences of two monomials. The remainder is R(z z G ) = w 1 w w 3 : This monomial corresponds to the feasible solution x 1 = 4 x = 4 x 3 = 1 x 4 = 0. The previous example demonstrates how w e nd a feasible solution to problem (7.6). In order to obtain an optimal solution we need to use a monomial order that incorporates information about the cost vector c. Specically, for c Z n +, w e consider a monomial order in which as before any monomial containing one of the z i 's is greater than any monomial containing only w j 's and in addition the monomials w x, w y are ordered according to the vector c induced order, so that if c 0 x > c 0 y, then w x > w y. If c 0 x = c 0 y,

20 Sec. 7.3 Grobner bases for integer optimization 1 we break ties arbitrarily. W e consider the following algorithm. Algorithm 7.3 Input: A Z mn b Z c Z Output: An optimal solution to problem (7.6). m Algorithm: Q a 1. Let f ij j = m z i and I = hf 1 ; w 1 : : : f n ; w n i:. Compute a Grobner basis of I using the c induced order. Q 3. Compute the remainder m g = R bi z i G. n x x 1 4. C If g [w 1 : : : w n], i.e., g = w x = w 1 n 1 w n C, then (x : : : x n) 0 is an optimal solution to problem (7.6) If g 6 [w 1 : : : w n] problem (7.6) is infeasible. We next show that Algorithm 7.3 is correct. Theorem 7.11 Algorithm 7.3 correctly solves problem (7.6). Proof. Suppose that w x = R(z b G ), i.e., (w x ) = z b. We assume for the purpose of deriving a contradiction that there exists a vector y such that (w y ) = z b and c 0 y < c 0 x. Let h = w x ; w y. W e have that (h) = zb ;z b = 0, which implies (see Exercise 7.9) that h I. Hence R(h G) = 0. However, since we use the c induced monomial order, the leading term of h must be w x. Since w x = R(z b G ), the remainder R(h G) cannot be zero, leading to a contradiction. Note that if g 6 C [w 1 : : : w n], g = R(z b G ) is a polynomial in both z w. Since the variables w are lower in the order than z, if there were a feasible solution, the variables z would have been eliminated. Therefore, if g 6 C [w 1 : : : w n], then there is no feasible solution. Example 7.1 In Example 7.10 we consider the objective function minimize 3x 3 + x 4 : 8 Applying Algorithm 7.3 we nd the monomial w 1 w w 4, leading to the optimal solution x 1 = 8 x = 1 x 3 = 0 x 4 = 1 : We nally discuss how to modify Algorithm 7.3 to allow for A b c having negative e n tries. Note that since we can write problem (7.6) as minimize y subject to c 0 x ; y = 0 Ax = b x Z n +

21 Chap. 7 Algebraic geometry and integer optimization we can assume without loss of generality t h a t c Z n +. W e next consider the case that A b contain negative e n tries. We i n troduce a new variable t such that tz 1 z m ; 1 = 0 (note t = 1 =(z 1 z m )). Then each of the terms Ym Y m a ij i zi z b i can be rewritten as Y Ym m t k j a b z ij i i t k z i with k j = ; minf0 a 1j : : : a mj g 0 a ij = a ij +k j 0, k = ; minf0 b 1 : : : b mg 0 b i = b i + k 0: Thus, Eq. (7.7) becomes! n m x Y Y j Y m a ij i i j=1 t k j z = t k z b i modulo the relation tz 1 z m ; 1 = 0. T h us, we consider the mapping that maps w j to the remainder Y m a ij i (wj ) = t kj z mod htz 1 z m ; 1i and extend this mapping to a general polynomial g C [w 1 : : : w n] as before. Then, the analog of Proposition 7.1 is: Proposition 7.3 A v ector x Z n + is feasible in problem (7.6) having A b with general entries if and only if (w x ) ; t k z b is divisible by tz 1 z m ; 1. Let f j = t kj Q m a ij i z and consider the ideal J = htz 1 z m ; 1 f 1 ; w 1 : : : f n ; w n i C [z 1 : : : z m t w 1 : : : w n]: We further consider monomial orders in C [z 1 : : : z m t w 1 : : : w n] with the property that any monomial c o n taining one of the z i 's or t is greater than any monomial containing only w j 's. We consider a Grobner basis G for the ideal J for the lex order above. Let f C [z 1 : : : z m t ] and let g = R(f G ) the remainder of f on G. F rom Proposition 7.(b), we obtain that if f is a monomial in C [z 1 : : : z m t ], then R(f G ) is also a monomial. This

22 Sec. 7.3 Grobner bases for integer optimization 3 property together with Proposition 7.3 leads to the following algorithm. Algorithm 7.4 Input: A Z mn b Z m c Z n +. Output: An optimal solution to problem (7.6). a ij Algorithm: Q 1. Let f j = t kj m z i and J = htz 1 z m ; 1 f 1 ; w 1 : : : f n ; w n i:. Compute a Grobner basis of J using the c induced order. 3. Compute the remainder g = R t k Q m z i b i G. 4. C If g [w (x 1 : : : w n], i.e., g = w x, th en : : : x n) 0 1 is an optimal solution to problem (7.6) if g 6 C [w 1 : : : w n], problem (7.6) is infeasible. We illustrate Algorithm 7.4 on the following example. Example 7.13 We consider the problem minimize x 1 ; x subject to 3x 1 ; x = 1 x 1 + x = 7 x1 x Z +: which w e rewrite to have nonnegative cost vector as: minimize y subject to x 1 ; x ; y = 0 3x 1 ; x = 1 x 1 + x = 7 x1 x Z +: The ideal J in this case is 3 3 J = htz 1 z z 3 ; 1 z 1 z z 3 ; w 1 t z 1 z 3 ; w t ; w 3 i: Let G be a Grobner basis for the term order induced by ( 0 0 1) 0. W e com pute 7 3 R(z z 3 G ) = w 1 w w 3, which leads to the optimal solution x1 = 3 x = 1 y =.

23 4 Chap. 7 Algebraic geometry and integer optimization 7.4 Applications of real algebraic geometry In this section, we use results from algebraic geometry over the reals to provide a convergent (and in the case of binary optimization, nite) sequence of semidenite relaxations for the general polynomial optimization problem Z = m inim ize f(x) subject to g i (x) 0 i = 1 : : : m (7.8) x R n where f g i R[x] are polynomials dened as: X f(x) = f x Z+ X n g i (x) = g i x Z+ n where there are only nitely many nonzero f and g i. Let K = fx R n j g i (x) 0 i = 1 : : : mg: Note that problem (7.8) can model binary optimization, by taking f(x) = 0 c 0 x and taking as the polynomials g i (x), a i x ; b i, x j ; x j and ;x j + x j (to model x j ; x j = 0 ). Problem (7.8) can also model bounded integer optimization, as well as bounded mixed integer nonlinear optimization. Problem (7.8) can be written as: maximize subject to f(x) ; 0 8 x K: (7.9) This leads us to consider conditions for polynomials to be nonnegative. We rst consider nonnegative polynomials in one dimension. Theorem 7.1 A polynomial f R[x] of degree d is nonnegative i f and only if it Pcan be written as a sum of squares of other polynomials, i.e., f(x) = k i(x)], w ith h i R[x], i = 1 : : : k. [h Proof. P Clearly, i f f(x) = k [h i(x)], then f(x) 0. P Conversely, suppose that the polynomial f(x) = d f i x i i=0 of degree d is nonnegative for all x. Then, the real roots of f(x) should have e v en multiplicity, otherwise f(x) w ould alter its sign in a neighborhood of a root. Let i, i = 1 : : : r be its real roots with corresponding multiplicity

24 Sec. 7.4 Applications of real algebraic geometry 5 m i. Its complex roots can be arranged in conjugate pairs, a j +ib j, a j ;ib j, j = 1 : : : h. Then, Yr Y h ; f (x) = f d (x ; i ) mi (x ; a j ) + b : j j=1 Note that the leading coecient f d needs to be positive. By expanding the terms in the products, we see that f (x) can be written as a sum of squares of h polynomials. Clearly, i n m ultiple dimensions if a polynomial can be written as a sum of squares of other polynomials, then it is nonnegative. However, it is possible for a polynomial in higher dimensions to be nonnegative without being a sum of squares. In Exercise 7.10 we show that the polynomial f (x y z) = x 4 y + x y 4 + z 6 ; 3x y z is nonnegative without being a sum of squares of polynomials. The next theorem generalizes Theorem 7.1 in multiple dimensions and accomodates constraints. Let M = f1 : : : m g. W e dene the set 8 9 < X Y = : P (g 1 : : : g m) = g R[x] g(x) = q J (x) g i (x) J M ij with q J (x) are polynomials that have a sum of squares representation. Theorem 7.13 (Positivstellensatz) Let K = fx R n j g i (x) 0 i = 1 : : : m g and g R[x]. (a) g 0 on K if and only if there exist q h P (g 1 : : : g m) and k Z + such that gq = g k + h: (b) g > 0 on K if and only if there exist q h P (g 1 : : : g m) such that gq = 1 + h: (c) g = 0 on K if and only if there exist h P (g 1 : : : g m) and k Z + such that g k + h = 0 : If the set K is compact we can obtain a more precise certicate of positivity o n K. Theorem 7.14 If K is compact, then the polynomial g > 0 on K if and only if g P (g 1 : : : g m), t h a t i s X Y g(x) = q J (x) g i (x) (7.10) J M ij with q J (x) are polynomials that have a sum of squares representation.

25 6 Chap. 7 Algebraic geometry and integer optimization Note that the number of terms in Eq. (7.10) is exponential. The next theorem establishes a certicate of positivity o n K, under a mild assumption on K. Theorem 7.15 Suppose that the set K is compact and there exists a polynomial h(x) of the form X m h(x) = h 0 (x) + h i (x)g i (x) such that fx R n j h(x) 0g is compact and h i (x), i = 0 1 : : : m are polynomials that have a sum of squares representation. Then, the polynomial g > 0 on K if and only if there exist p i R[x], i = 0 1 : : : m that are sums of squares such t h a t X m g(x) = p 0 (x) + p i (x)g i (x): (7.11) In contrast to Eq. (7.10), the number of terms in Eq. (7.11) is linear. While the assumption of Theorem 7.15 may seem restrictive, it is satised in several cases: (a) For binary optimization problems, that is, when K includes the in- equalities x j x j and x j x j for all j = 1 : : : n. (b) If all the g j 's are linear, i.e., K is a polyhedron. (c) If there is one polynomial g k such that the level set fx R n j g k (x) 0g is compact. More generally, one way to ensure that the assumption of Theorem 7.15 holds is to add to K the extra quadratic constraint g m+1 (x) = a ; kxk 0 for some a suciently large. It is also important to emphasize that we do not assume that K is convex (it may e v en be disconnected). From Theorem 7.15 it is natural to investigate when a polynomial is a sum of squares. We consider the vector v d (x) = ( x ) jjd = (1 x d 1 : : : x n x x 1x : : : x n;1x n x n : : : x d : : : x n) ; ; n+d of all the Pmonomials x of degree less than or equal to d, which has dimenn sion s = d = : i=0 i d Proposition 7.4 The polynomial g(x) of degree d has a sum of squares decomposition if and only if there exists a positive semidenite matrix Q for which g(x) = v d (x) 0 Qv d (x).

26 Sec. 7.4 Applications of real algebraic geometry 7 Proof. Suppose there exists an ss matrix Q 0 for which g(x) = v d (x) 0 Qv d (x). Then Q = HH 0 for some matrix s k H, and thus, X k g(x) = v d (x) 0 HH 0 v d (x) = (H 0 v d (x)) i : Since (H 0 v d (x)) i is a polynomial, then g(x) is expressed as a sum of squares of the polynomials (H 0 v d (x)) i. Conversely, suppose that g(x) has a sum of squares decomposition P` g(x) = h i (x) : Let h i be t h e v ector of coecients of the polynomial h i (x), i.e., h i (x) = h i 0 v d (x): Thus, X ` g(x) = v d (x) 0 h i h0 i v d (x) = v d (x) 0 Qv d (x) P` with Q = h i h 0 0, and the proposition follows. i Theorem 7.15 and Proposition 7.4 jointly imply that one can use semidenite optimization to provide a sequence of semidenite relaxations for problem (7.9). Assuming that the set K satises the assumption of Theorem 7.15, then f (x) ; 0, for all x K if and only if X m f (x) ; = p 0 (x) + p i (x)g i (x) (7.1) where p i (x), i = 0 1 : : : m have a sum of squares representation. Theorem 7.15 does not specify the degree of the polynomials p i (x). Thus, we select a bound d on the degree of the polynomials p i (x), and we apply Proposition 7.4 to each of the polynomials p i (x), that is, p i (x) is a sum of squares if and only if p i (x) = v d (x) 0 Q i v d (x) with Q i 0, i = 0 1 : : : m. Substituting to Eq. (7.1), we obtain that Q i, i = 0 1 : : : m satisfy linear equations that we denote as L( Q 0 Q 1 : : : Q m ) = 0 : Thus, we can nd a lower bound to problem (7.8) by solving the semidenite optimization problem Z d = m a x s:t: L( Q 0 Q 1 : : : Q m ) = 0 (7.13) Q i 0 i = 0 1 : : : m : Problem (7.13) involves semidenite optimization over m +1 s s matrices. From the construction Z d Z. It turns out that as d increases, Z d converges to Z. Moreover, for binary optimization, there exists a nite d for which Z d = Z. Problem (7.13) provides a systematic way to nd convergent semidefinite relaxations to problem (7.8). While the approach is both general

27 8 Chap. 7 Algebraic geometry and integer optimization (it applies to very general nonconvex problems including nonlinear mixed integer optimization problems) and insightful from a theoretical point o f view, it is only practical for values of d = 1, as large scale semidenite optimization problems cannot be solved in practice. In many situations, however, Z 1 or Z provide strong bounds. Let us consider an example. Example 7.14 Consider the problem We attempt to write minimize f (x 1 x ) = x 1 + x 1 x ; x 1 x + 5 x : f (x 1 x ) = x 1 + x 1 x ; x 1 x + 5 x x 1 q 11 q 1 q 13 x 1 B C B CB C x q 1 q q 3 A@ x A 0 1 x 1 x q 13 q 3 q 33 x 1 x = q 11 x 1 + q x + ( q 13 + q 1 )x 1 x + q 13 x 1 x + q 3 x 1 x : In order to have a n id e n tity, w e obtain q 11 = q = 5 q 33 + q 1 = ;1 q 13 = q 3 = 0 : Using semidenite optimization, we nd a particular solution such that Q 0 is given by It follows that ;3 1 0 B C 1 B C Q ;3 5 0 A = HH 0 H = ;3 1 A : f (x 1 x ) = (x 1 ; 3x + x 1 x ) 1 + (x + 3 x 1 x ) and thus the optimal solution value is = 0 and the optimal solution is x 1 = x = Generating functions for integer points in polyhedra In this section, we i n troduce generating functions for integer points in polyhedra for the purpose of counting the numbe r o f i n teger points in a bounded rational polyhedron P R n. The problem naturally arises in several branches of mathematics, specically combinatorics, representation theory and number theory, while it plays a central role in statistics, specically in the computation of the exact conditional distribution for contingency tables.

28 Sec. 7.5 Generating functions for integer points in polyhedra 9 Given a rational polyhedron P R n (not necessarily bounded) we introduce the generating function: X f(p x) = x : P \Zn Note that if P is bounded f(p e) is exactly the numbe r o f i n teger points in P. Let v be a v ertex of P. T he supporting cone of P at v is: K(P v) = v+fu R n j there exists > 0 : v+u P for all 0 g: Our point of departure is the following theorem: Theorem 7.16 Let P be a rational polyhedron and let V be the set of its vertices. Then, X f(p x) = f(k(p v) x): (7.14) vv Proof outline Let P (y) = fy R n j Ax = yg be a polyhedron in standard form with + m A Z mn and y Z. W e i n troduce the generating function X (z x) = f(p (y) x)z ;y Since we obtain that = yzm X X yz m A=y Zn + X Z + = x z ;A : n x z ;y Y n ; ;a1k ;a mk x z ;A k = x k z 1 z ;ak z m k=1 Yn X 1 ; ;a1k ;a mk k (z x) = x k z 1 z ;ak z m k=1 k =0 n Y 1 = 1 ;ak ;a ; x mk k z ;a1k z z m k=1 1 under a1k ak a the condition that jz z z mk 1 m j > jx k j: Given that we know the generating function (z x) in closed form, we can use Cauchy's residue theorem to nd the coecients f(p (y) x) of the generating function (z x).

29 30 Chap. 7 Algebraic geometry and integer optimization The application of complex variables theory leads to Eq. (7.14). The derivation uses complex analysis heavily and it is ommited as it is outside the scope of the book. Example 7.15 Let P = [0 1]. Clearly, f (P x ) = 1 + x: Moreover, for each o f its two v ertices, K 0 = K (P 0) = [0 1) and K 1 = K (P 1) = (;1 1]. Then, for jxj < 1, and for jxj > 1, We observe that f (K 0 x ) = x f (K 1 x ) = x X 1 k 1 = 1 ; x k=0 X 1 k 1 k=;1 = x + : 1 ; (1=x) 1 1 f (K 0 x ) + f (K 1 x ) = + x + = 1 + x 1 ; x 1 ; (1=x) verifying Theorem Note that there is no common value of x such that the two innite sums converge. Thus, the generating functions in Theorem 7.16 need to be understood symbolically as formal power series. Therefore, the need arises to compute f(k x) for rational cones K. We rst examine the case that cone K is described by linearly independent integral vectors. Proposition 7.5 Let K = cone(u 1 : : : u k ), n where u 1 : : : u k are lin- P o early independent v ectors in Z n. Let S = k i u i 0 i < 1 : Then,! X Y k 1 S\Zn f(k x) = x : (7.15) 1 ; x ui Proof. Let K \ Z n, i.e., X k = i u i i 0: Let Xk X k 1 = ( i ; b i c)u i and = b i cu i : n Therefore, = 1 +, 1 S \ Z and is a nonnegative i n teger combination of u 1 : : : u k. In fact, this representation is unique. It is

30 Sec. 7.5 Generating functions for integer points in polyhedra 31 also clear that the sum of an integer point i n S and a nonnegative i n teger combination of u 1 : : : u k is an integer point i n K. Therefore,! 0 1 X X X K\Z 1 S\Z (m1 ::: m k )Z + f(k x) = x x m1 u1 +:::+m k u k A = x : n n k The second term is a multiple geometric series which s u m s u p t o Y k 1 1 ; x ui and Eq. (7.15) follows. We next show that for certain types of cones K the computation of f(k x) is signicantly simpler. Denition 7.11 Let u 1 : : : u k Z n be linearly independent v ectors, K = cone(u 1 : : : u k ) and L k = span(u 1 : : : u k ): We say that K is a unimodular cone if u 1 : : : u n k is a basis of the lattice L k \ Z. n o It follows that if the supporting cone K = v + cone( u 1 : : : u k ) is P unimodular, then (v+s)\z n = fvg, where S = k i u i 0 i < 1 : Clearly, i s v Z n, th en v = v. T h us, from Proposition 7.5 we obtain x v f(k v) = : (7.16) (1 ; x u1 ) (1 ; x uk ) We next show h o w to decompose a given cone into unimodular cones. For simplicity of exposition and without loss of generality w e assume that the cone is fully dimensional, i.e., K = cone( u 1 : : : u n ). Let A be the matrix with columns u 1 : : : u n. W e dene index(k) = jdet(a)j. N ote that jdet(a)j is equal to the numbe r o f i n teger points in the parallelepiped np k o i u i 0 i < 1. In particular, K is a unimodular cone if and only if jdet(a)j = 1.

31 3 Chap. 7 Algebraic geometry and integer optimization Algorithm 7.5 Input: K = cone(u 1 : : : u n ) with u 1 : : : u n Zn linearly independent. P Output: A collection K i, i I of unimodular cones such t h a t K = i K i, where i f; 1 1g. ii Algorithm: 1. If K is unimodular, set U 1 = fkg U = otherwise U 1 =, U = fkg.. While U 6= a. Select a cone C = cone(u 1 : : : u n ) U and set A = [u 1 : : : u n ]. b. Apply the basis reduction algorithm?? to nd a reduced basis A for the lattice L(A ;1 ). Compute by e n umeration the shortest nonzero vector w with respect to k k 1 in the lattice L(A ;1 ) as follows: min kaxk 1 s:t: jx j j p n n(n;1)=4 j = 1 : : : n x Z n : Let z = Aw. c. Set K i = cone(u 1 : : : u i;1 z u i+1 : : : u n ) and A i = [ u 1 : : : u i;1 z u i+1 : : : u n ] for i = 1 : : : n. d. For i = 1 : : : n, set Ki = K if det(a) and det(a i ) have the same sign, Ki = ; K, otherwise. If K i is unimodular, then U 1 := U 1 [ f K i g, otherwise U := U [ f K i g. 3. Output the elements of the set U 1. Example 7.16 Let K = cone(( 7) 0 (1 0) 0 ) and 1 A ;1 = 0 1=7 A = ;=7 with det(a) = ;7. The reduced basis A of the lattice L(A ;1 ) is The vector w = (1 =7 ;=7) 0 1=7 3=7 A = : ;=7 1=7 is the shortest vector in L(A ;1 ) with respect to the norm k k 1, and thus z = Aw = (0 1) 0. Then, performing Step c of Algorithm

32 Sec. 7.5 Generating functions for integer points in polyhedra we obtain the cones K 1 = cone(( 7) 0 (0 1) 0 ) and K = cone((0 1) 0 (1 0) 0 ) with A 1 = A = : The cone K is unimodular with det(a ) = ;1 of the same sign as det(a). Then, U1 = fk g and K = 1 : The cone K 1 has determinant, and thus U = fk 1 g and K1 = ;1. We c o n tinue Algorithm 7.5 with cone K 1 and matrix A 1. We obtain that a reduced basis for L(A ;1 1 ) is Thus, w = (1 = ;1=) 0 1= 1= A 1 = : ;1= 1= and z = A 1 w = (1 3) 0. Step c generates the two cones K 3 = cone(( 7) 0 (1 3) 0 ) and K 4 = cone((1 3) 0 (0 1) 0 ). Both cones are unimodular with K3 = 1 and K4 = ;1. We t h us obtain the decomposition K = K + K 3 ; K 4 : We n e x t s h o w that Algorithm 7.5 is polynomial when the dimension is xed. Proposition 7.6 Algorithm 7.5 is correct and runs in polynomial time when the dimension n is xed. Proof. We will show rst that the indices of the cones K i decrease. Let ( ) X n B = z R n z = u i x i = Ax jx i j j det(a)j ;1=n : The set B is symmetric around the origin (if z B, th en ;z B), is convex and has volume vol(b) = njdet(a)j=jdet(a)j = n : By Theorem?? there exists a nonzero integral point z B. T h us, there exists some x = A ;1 z L (A ;1 ) with kxk 1 j det(a)j ;1=n : Since in Step b, Algorithm 7.5 nds a nonzero w L (A ;1 ) with the Psmallest kwk 1, w e have kwk 1 kxk 1 j det(a)j ;1=n : Since z = Aw = n w i u i, w e obtain index(k i ) = jdet[u 1 : : : u i;1 z u i+1 : : : u n ]j = jw i jjdet[u 1 : : : u i;1 u i u i+1 : : : u n ]j k wk 1 jdet(a)j j det(a)j (n;1)=n = index(k) (n;1)=n : For all K i U, index(k i ) decreases after each iteration. Thus, index(k i ) will become one, i.e., all the resulting cones in U will be unimodular, and thus the decomposition given by Algorithm 7.5 is correct.