AN ALTERNATIVE ALGORITHM FOR COUNTING LATTICE POINTS IN A CONVEX POLYTOPE. 1. Introduction. Consider the (not necessarily compact) polyhedron

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1 AN ALTERNATIVE ALGORITHM FOR COUNTING LATTICE POINTS IN A CONVEX POLYTOPE JEAN B. LASSERRE AND EDUARDO S. ZERON Abstract. We provide an alternative algorithm for counting lattice points in the convex polytope {x R n Ax = b x 0}. It is based on an exact (tractable) formula for the case A Z m (m+) that we repeatedly use for the general case A Z m n.. Introduction Consider the (not necessarily compact) polyhedron (.) Ω(y) = {x R n Ax = y; x k 0} with y Z m and A Z m n for n > m; and let f : Z m R be the function (.2) y f(y) := e c x x Ω(y) N n where the vector c R n is chosen small enough (even negative) to ensure that f(y) is well defined even when Ω(y) is not compact. The notation c stands for the transpose of the vector c so that: c x = c x + + c n x n. If Ω(y) is compact then f(y) provides us with the exact number of points in the set Ω(y) N n by either choosing c := 0 or taking lim c 0 f(y) (or even rounding up f(y) to the nearest integer for c sufficiently close to zero). In recent works Barvinok [3] Barvinok and Pommersheim [4] Brion and Vergne [8] Pukhlikov and Khovanskii [] have provided nice exact formulas for f(y). For instance with y fixed Barvinok [3] considers f(y) as the generating function (evaluated at z := e c C n ) of the indicator function x I Ω(y) N n(x) for the set Ω(y) N n ; and provides a decomposition into a sum of simpler generating functions associated with supporting cones (those decomposed into unimodular cones as well). De Loera et al [0] have implemented Barvinok s counting algorithm in the software LattE which runs in time polynomial in the problem size when the dimension is fixed. Let us also mention the software developed by Verdoolaege [6] which extends the Part of this work was financially supported by the French-Mexican research cooperation program CNRS-CONACYT and CINVESTAV in México.

2 LattE software to handle parametric polytopes. In a dual approach Brion and Vergne [8] consider the generating function F : C m C of f that is (.3) z F (z) := y Z m f(y)z y for which they provide a generalized residue formula to next obtain f(y) in closed form. As a result of both approaches f(y) is finally expressed as a weighted sum over the vertices of Ω(y). Similarly Beck [5] and Beck Diaz and Robins [6] have provided a complete analysis based on residue techniques for the case of a tetrahedron (m = ); and mentioned the possibility of evaluating f(b) for general polytopes by means of residues as well. Despite of its theoretical interest Brion and Vergne s formula may not be directly tractable because it contains many products with complex coefficients (roots of unity) which makes the formula difficult to evaluate numerically. However in some cases this formula can be exploited to yield an efficient algorithm as e.g. in Baldoni-Silva De Loera and Vergne [2] for flow polytopes; in Beck and Pixton [7] for transportation polytopes; and more generally when the matrix A is totally unimodular as in the work of Cochet [9]. Finally in Lasserre and Zeron [2] we have provided two algorithms based on Cauchy residue techniques to invert the generating function F in (.3) and an alternative algebraic technique based on partial fraction expansions of the generating function (using the Hilbert NullstellenSatz). technique is to avoid computing residues. A nice feature of the latter Contribution. The goal of this paper as a sequel to [2] is to provide a recursive algorithm to compute f(y) in the spirit of the algebraic technique briefly outlined in [2 7]; but now in a more constructive and explicit way. Like in Brion and Vergne we use the generating function F in (.3) and we provide a decomposition into simpler rational fractions whose inversion is easily obtained. To avoid handling complex roots of unity we do not use residues explicitly but we build up the required decomposition in a recursive manner. Properly speaking we inductively calculate constants Q σβ and a fixed positive integer M all of them completely independent of y (and of the magnitude c as well) such that the counting function f is given by 2

3 the finite sum : f(y) = A σ β Z m β M { e c σx if x := A Q σβ σ [y β] N m 0 otherwise; where the first finite sum is computed over all invertible [m m]-square sub-matrices A σ of A. This formula is presented in Theorem 2.6 after the necessary notation is introduced in Section 2. Crucial in our algorithm is an explicit decomposition in a closed form (and so an explicit formula for f(y)) for the case n = m + ; that we next repeatedly use for the general case n > m+. Our closed form expression for the case n = m + is immediately computable and tractable for it does not contain complex coefficients such as the roots of unity in Brion and Vergne s formula. The computational complexity is O[(m+) n m Λ] where the coefficient Λ depends only on the entries of A and the vector c but not on the magnitude of y (cf. (4.2)). Actually Λ depends on the ratio between the entries of the vector c but not on the magnitude of c. However Λ is exponential in the input size of A. Thus the formulas presented in section 3 give us a simple procedure for calculating f(y) in the case n = m +. Moreover the recursive algorithm presented in section 4 is particularly attractive for calculating f(y) in all cases where n m is relatively small no matter the magnitude of y. However this algorithm becomes less efficient whenever n = m + k with relatively large values of k. Analyzing the algorithm presented in 4 against the algorithm (via integration) that we introduced in [2] we can conclude that they are complementary in the sense that the algorithm presented in [2] is attractive when m is small whereas the algorithm presented in this paper is attractive when n m is small no matter how large m and n could be. The paper is organized as follows: In 2 we present our main result an exact expression of f(y) provided that its generating function F (z) is decomposed into a sum of some rational fractions. In 3 we obtain this explicit decomposition for the case n = m + as well as the corresponding expression for f(y). In 4 we present a recursive algorithm whose output is the required decomposition for the general case n m +. 3

4 2. Main result 2.. Notation and definitions. The notation R Q and Z stand for the usual sets of real rational and integer numbers respectively; moreover the set of natural numbers {0 2...} is denoted by Z + or N. The notation c and A stand for the respective transposes of the vector c R n and the matrix A Z m n. Moreover the k-th column of the matrix A Z m n is denoted by A k := (A k... A mk ). When y = 0 Ω(0) in (.) is a convex cone with dual cone (2.) Ω(0) := {b R n b x 0 for every x Ω(0)}. We may now define the following open set (2.2) Γ := {c R n c > b for some b Ω(0) }. Notice that Γ and Ω(0) are both equal to R n whenever Ω(0) is the singleton {0} which is the case if Ω(y) is compact. On the other hand we will suppose from now on that the matrix A Z m n has maximal rank (see the comment before 2.2). Definition 2.. Let p N satisfy m p n and let η = {η η 2... η p } N be an ordered set with cardinality η = p and η < η 2 <... < η p n. Then (i) η is said to be a basis of order p if the [m p] sub-matrix A η := [ A η A η2 A ηp ] has maximal rank that is rank(a η ) = m. (ii) For m p n let (2.3) J p := {η {... n} η is a basis of order p} be the set of bases of order p. Notice that J n = {{ 2... n}} because A has maximal rank. Lemma 2.2. Let η be any subset of { 2... n}. (i) If η = m then η J m if and only if A η is invertible. (ii) If η = q with m < q n then η J q if and only if there exists a basis σ J m such that σ η. 4

5 Proof. (i) is immediate because A η is a square matrix; and A η is invertible if and only if A η has maximal rank. Next (ii) also follows from the fact that A η has maximal rank if and only if A η contains a square invertible sub-matrix. Lemma 2.2 automatically implies J m because the matrix A must contain at least one square invertible sub-matrix (we are supposing that A has maximal rank). Besides J p for m < p n because J m. Finally given a basis η J p for m p n and three vectors z C m c R n and w Z m we introduce the following notation (2.4) z w := z w zw 2 2 zm wm c η := (c η c η2... c ηp ) w := max{ w w 2... w m }. Definition 2.3. The vector c R n is said to be regular if for every basis σ J m+ there exist a non-zero vector v(σ) Z m+ such that : (2.5) A σ v(σ) = 0 and c σv(σ) 0. Notice that c 0 whenever c is regular. Moreover there are infinitely many vectors v Z m+ such that A σ v = 0 because rank(a σ ) = m < n. Thus the vector c R n is regular if and only if c j c πa π A j 0 π J m j π; which is the regularity condition used in Brion and Vergne [8] except we do not require c j 0 for all j =... n. As already mentioned we will suppose that the matrix A Z m n in (.) (.2) has maximal rank. That is the m rows of A v(j) = (A j... A jn ) j =... m are linearly independent. For suppose that A has not maximal rank. Then we can find 0 β Z m such that 0 = β v() + + β m v(m) and β 0. Assume that β 0. The equation y = Ax has a solution x N n if and only if x is a solution of the system of equations y j = v(j)x for 2 j m and m m y = v()x = β j v(j)x/β = y j β j /β. j=2 5 j=2

6 So if y m j=2 y jβ j /β then f(y) = 0; otherwise we can eliminate the equation y = v()x from y = Ax (because it does not depend on the free variable x) and use instead the trivial relationship β y()+ +β m y(m) = Generating function. An appropriate tool for computing the exact value of f(y) is the generating function F : C m C (2.6) z F (z) := y Z m f(y)z y with z y defined in (2.4). This generating function was already considered in Brion and Vergne [8] with λ := (ln z... ln z m ). Proposition 2.4. Let f and F be like in (.2) and (2.6) respectively and let c Γ. Then : (2.7) F (z) = on the domain (2.8) n k= ( e c A k z k z A 2k 2 z A mk m ) ( z... z m ) D with D := {ρ R m ρ > 0; e c kρ A k < k =... n}. Proof. Apply the definition (2.6) of F to obtain : F (z) = z y e c x = e c x z Ax. y Z m x N n Ax=y x N n On the other hand e c x z Ax = n k= ( ) e c k z A k z A xk mk m. The domain D in (2.8) is not empty because c Γ. Indeed a variant of Farkas Lemma (see Corollary 7.e in Schrijver [3 p. 89]) states that the system A u b has a solution if and only if b x 0 for every vector x 0 with Ax = 0. Whence the system A u b will have a solution whenever b is in the dual cone Ω(0). Moreover recalling the definition (2.2) of Γ we can deduce that A u < c has indeed a solution ŭ R m because c Γ. Thus we also have that (eŭ eŭ2... eŭm ) A k < e c k for every k n and so ρ := (eŭ... eŭm ) D. 6

7 Thus the condition e c k z A k... z A mk m < holds whenever k n and ( z... z m ) D so n ( ) F (z) = e c k z A k z A xk mk m = k= x k =0 n k= ( e c A k z k z A mk m ) Inverting the generating function. We will compute the exact value of f(y) by first determining an appropriate expansion of the generating function in the form (2.9) F (z) = σ J m Q σ (z) k σ ( ec k z A k) where the coefficients Q σ : C m C are rational functions with a finite Laurent series (2.0) z Q σ (z) = β Z m β M Q σβ z β. In (2.0) the strictly positive integer M is fixed and each Q σβ is a real number which can be computed with algebraic operations when the numbers {e c j } n j= are given. An important observation is that the integer M does not depend on the right-hand-side y. It only depends on A and c but not on the magnitude of c. Remark 2.5. The decomposition (2.9) is not unique (at all) and there are several ways to obtain such a decomposition. For instance Brion and Vergne [8 2.3 p. 85] provide an explicit decomposition of F (z) into elementary rational fractions of the form (2.) F (z) = σ J m g G(σ) j σ ( γj (g)(e c j z A j) /q) k σ δ k(g) where G(σ) is a certain set of cardinality q and the coefficients {γ j (g) δ k (g)} involve certain roots of unity. The fact that c is regular ensures that (2.) is well-defined. Thus in principle we could obtain (2.9) from (2.) but this would require a highly nontrivial analysis and manipulation of the coefficients {γ j (g) δ k (g)}. In the sequel we provide an alternative algebraic approach that avoids manipulating these complex coefficients. 7

8 If F satisfies (2.9) then we get the following result. Theorem 2.6. Let A Z m n be of maximal rank f be as in (.2) with c Γ and assume that the generating function F in (2.6) satisfies (2.9) (2.0). Then : (2.2) f(y) = with (2.3) E σ (y β) = σ J m where c σ R m was defined in (2.4). β Z m β M Q σβ E σ (y β) { e c σx if x := A σ [y β] N m 0 otherwise; Proof. Recall that z A k = z A k z A mk m according to (2.4). On the other hand in view of (2.8) the inequality e c k z A k < holds for every k n; and so the following expansion holds as well for each σ J m : e c = e c kx k z A kx k = e c σx z Aσx. k z A k k σ x k N x N m k σ Next suppose that a decomposition (2.9) (2.0) exists. Then the following relationship is easy to establish. F (z) = (2.4) Q σ (z) e c x N m σ J m = σ J m β Z m β M σ x z Aσx Q σβ e c σ x z β+aσx. x N m Notice that both equations in (2.6) and (2.4) are equal. Hence if we want to obtain the exact value of f(y) from (2.4) we only have to sum up all terms with exponent β + A σ x equal to y. That is recalling that A σ is invertible for every σ J m (see Lemma 2.2) f(y) = { e c σ Q σβ x if x := A σ [y β] N m ; 0 otherwise; σ J m which is exactly(2.2). β Z m β M Remark 2.7. Observe that function f(y) in Theorem 2.6 can be rewritten as a weighted sum of e c x at some integral points x N n namely (2.5) f(y) = Q σβ e c x(σβ) β σ J m 8

9 where the second finite sum is calculated over all β Z m such that β M and A σ [y β] N m. Moreover each vector x(σ β) N n is an integral point. Indeed given x := A σ (y β) inside N m like in (2.3) we define the integral vector x(σ β) N n by setting the entries: { xk if j = σ [ x(σ β)] j = k for some k m 0 if j σ; for j =... n. Clearly we have that e c σ x = e c x(σβ) from which equation (2.5) follows. In addition these integral points x(σ β) N n have at most m nontrivial coordinates and their convex hull defines an integral polyhedron (that is a polyhedron with integral vertices). In view of Theorem 2.6 f(y) is easily obtained once the rational functions Q σ (z) in the decomposition (2.9) are available. As already pointed out the decomposition (2.9) (2.0) is not unique and the purpose of the next section ( 3) is to provide : - a simple decomposition (2.9) for which the expression of the coefficients Q σ are easily calculated in the case n = m + ; whereas in 4 we present : - a recursive algorithm to provide the Q σ in the general case n > m The case n = m + In this section we completely solve the case n = m+ that is we provide an explicit expression of f(y). We first need some essential intermediate algebraic calculations in order to deduce the decomposition (2.9) (2.0) of F (z) when n = m Some auxiliary rational functions. Let sgn : R Z be the sign function defined by t sgn(t) := if t > 0 if t < 0 0 otherwise. Besides adopt the convention that any sum with negative superindex : r=0 ( ) is identically equal to zero. Now given a fixed integer n > 0 for every k =... n we are going to construct auxiliary functions P k : Z n C n C such that each w P k (v w) 9

10 is a rational function of the variable w C n. Given a vector v Z n we define : (3.) P (v w) := v r=0 P 2 (v w) := (w v w sgn(v ) r. v2 ) P 3 (v w) := (w v wv 2 r=0 v3 2 ) w sgn(v 2) r 2 r=0 w sgn(v 3) r 3. := P n (v w) :=. n j= w v j j v n r=0 w sgn(vn) r n. Obviously we have that P k (v w) = 0 whenever v k = 0. Moreover we claim that : Lemma 3.. Let v Z n and w C n. The functions P k defined in (3.) satisfy (3.2) n ( ) w sgn(v k) k P k (v w) = w v. k= Proof. First notice that ( w sgn(v ) ) ( P (v w) = w sgn(v ) ) v r=0 w sgn(v ) r = w v. Previous equalities are obvious when v = 0. We have similar formulas for 2 k n ( w sgn(v k) k ) P k (v w) = ( w v k k ) k j= w v j j = k Therefore adding together all the terms in equation (3.2) yields n ( ) n w sgn(v k) k P k (v w) = w v j j. k= j= j= w v j j k j= w v j j. 0

11 3.2. Solving the case n = m +. We now use the algebraic expansions of 3. to calculate the function f(y) in (.2) where Ω(y) is given in (.) and A Z m (m+) is a maximal rank matrix. Theorem 3.2. Let n = m + be fixed A Z m n a maximal rank matrix and let c Γ be regular. Let v Z n be a non-zero vector such that Av = 0 and c v 0 (cf. Definition 2.3). Define the vector (3.3) w := (e c z A e c 2 z A 2... e cn z An ). Then : (i) The generating function F (z) in (2.6) has the expansion (3.4) F (z) = n Q k (z) j k ( ec jz A j) = k= σ J m Q σ (z) j σ ( ec jz A j) where the rational functions z Q k (z) are defined by : P k (v w)/( e c v ) if v k > 0 (3.5) Q k (z) := w k P k(v w)/( e c v ) if v k < 0 0 otherwise; for k n. Each function P k in (3.5) is defined as in (3.). Notice that the first sum in equation (3.4) is done only over the indexes k for which v k 0 because Q k (z) = 0 whenever v k = 0. (ii) Given y Z m the function f(y) in (.2) is directly obtained by applying Theorem 2.6. Proof. (i) Since c is regular let v Z n be a vector such that Av = 0 and c v 0 see (2.5) in Definition 2.3. Let w C n be the vectors defined in (3.3). We can easily deduce that (3.6) w v = n ( e c j z A ) v j j = e c v z Av = e c v. j= Next let z Q k (z) be the rational function defined in (3.5). Then from Lemma 3. (3.7) n ( e c k z A ) k Q k (z) = k= n ( ) w sgn(v k) Pk (v w) k e c v k= = wv e c v =.

12 Multiplying together the generating function (2.7) with the left side of (3.7) yields the expansion (3.8) F (z) = n k= which gives us the first equality in (3.4). Q k (z) j k ( ec jz A j) ; (ii) Since c Γ the function F (z) is indeed the generating function of f(y). Next consider the ordered sets (3.9) σ(k) = { 2 k k + n} for k =... n. In order to apply Theorem 2.6 we only need to prove that each square sub-matrix A σ(k) is indeed invertible for every k =... n with v k 0. Recall that Q k (z) = 0 whenever v k = 0 and that σ(k) is an element of J m precisely when A σ(k) is invertible. We know that A R m n has maximal rank so A has m linearly independent columns. With no loss of generality we may assume that the first m columns A k are linearly independent for k =... m. Thus the square matrix A σ(n) = [A A 2 A m ] is invertible recall that n = m + ; and so σ(n) = {... m} defined in (3.9) is an element of J m. On the other hand since A σ(n) is invertible and the vector v Z n satisfies 0 = Av = A σ(n) (v v 2... v m ) + A n v n with v 0 we automatically have that the n-entry v n 0; and so the n-column of A is equal to A n = m v j j= v n A j. Whence for every k m with v k 0 the square matrix A σ(k) = [A... A k A k+... A m A n ] is clearly invertible because the column A k of A σ(n) has been substituted with the linear combination A n = m v j j= v n A j whose coefficient v k /v n is different from zero. That is the set σ(k) in (3.9) is an element of J m for every k m with v k 0. Therefore the expansion (3.8) can be re-written F (z) = Q k (z) v k 0 j σ(k) ( ec j z A j) = Q σ (z) j σ ( ec j z A j) 2 σ J m

13 with { Qk (z) if σ = σ(k) for some index k with v Q σ (z) k 0 0 otherwise. And so a closed form of f(y) is obtained by applying Theorem 2.6. Remark 3.3. In the case where n = m+ and Ω(y) is compact a naive way to evaluate f(y) is as follows. Suppose that B := [A... A m ] is invertible. One may then calculate ρ := max{x m+ Ax = y x 0}. Thus the evaluation of f(y) reduces to summing up x ec x over all vectors x = (ˆx x m+ ) N m+ such that x m+ [0 ρ] N and ˆx := B [y A m+ x m+ ]. This may work very well for reasonable values of ρ; but clearly ρ depends on the magnitude of y. On the other hand the computational complexity of the algorithm presented in 3 does not depend on y. Indeed the bound M in (2.2) of Theorem 2.6 does not depend at all on y. Moreover the algorithm also applies to the case where Ω(y) is not compact. To illustrate the difference consider the following trivial example where n = 2 m = A = [ ] and c = [0 a] with a 0. The generating function F (z) in (2.6) and (2.7) is the rational function F (z) = ( z)( e a z). Setting v = ( ) and w = (z e a z) one obtains = ( z) Q (z) + ( e a z) Q 2 (z) = ( z) z P (v w) e a + ( e a z) P 2(v w) e a = ( z) z e a + ( ea z) z e a an illustration of the Hilbert Nullstellensatz applied to the two polynomials z ( z) and z ( e a z) which have no common zero in C. And so the generating function F (z) gets expanded to (3.0) F (z) = z ( e a )( e a z) + z ( e a )( z). Finally using Theorem 2.6 we obtain f(y) in closed form by (3.) f(y) = ea(y+) e a + e0(y+) e a = = e(y+)a e a. Looking back at (2.0) we may see that M = (and obviously does not depend on y) and so the evaluation of f(y) via (2.2) in Theorem 2.6 (as 3

14 described in (3.) is done in 2 elementary steps no matter the magnitude of y. On the other hand the naive procedure would require y elementary steps. Remark 3.4. We have already mentioned that the expansion of the generating function F (z) is not unique. In the trivial example of Remark 3.3 we may also expand F (z) as the following sum of linear fractions F (z) = e a (e a )( e a z) (e a )( z) which is not the same as the expansion in (3.0). However applying Theorem 2.6 again yields the same formula (3.) for f(y) An algorithm for the case n = m +. Let R[z z ] be the set of polynomials with real coefficients and entries in z t and zt for t m; so negative exponents are allowed. That is a rational function Q is an element of R[z z ] if and only if Q has a finite Laurent series like in 2.0. Considering Theorems 2.6 and 3.2 the algorithm for the case n = m + can be written as follows : Procedure Solve(A c y). Input: A Z m n full rank and n = m + ; c R n regular; y Z m. Output: {Q k } n k= R[z z ] in (3.4) and f(y) in (2.2). Step 0: Compute a vector v Z n such that Av = 0 and c v 0. Step : Compute polynomials {w P k (v w)} from (3.). Step 2: Compute polynomials {z Q k (z)} from (3.5); and let L k := {Q kβ 0} β N m be the list of nonzero coefficients of Q k (z) for all nonzero Q k (z). Set f(y) := 0. Set k :=. Step 3: While k n do: If v k = 0 go to Label{skip}. If v k 0 let A σ Z m m be the nonsingular matrix obtained from A by deleting k-column A k ; and let c σ R m be the vector obtained from c by deleting k-entry c k. While L k do: Pick Q kβ L k and solve A σ x = y β. Set L k := L k \ {Q kβ }. If x N m then f(y) = f(y) + e c σx Q kβ. Label{skip}. Set k := k +. 4

15 4. The general case n > m + We now consider the case n > m + and obtain the decomposition (2.9) that permits to compute f(y) by invoking Theorem 2.6. The idea is to use the results of 3 recursively and we exhibit a decomposition (2.9) in the general case n > m + by induction. The following result is proved with the same arguments as in the proof of Theorem 3.2. Proposition 4.. Let A Z m n be a maximal rank matrix and c Γ be regular. Suppose that the generating function F in (2.6) (2.7) has the expansion (4.) F (z) = π J p Q π (z) j π ( ec jz A j) for some integer p with m < p n and for some rational functions z Q π (z) explicitly known and with a finite Laurent s series expansion (2.0). Then F also has the expansion (4.2) F (z) = π J p Q π (z) j π ( ec jz A j) where the rational functions z Q π (z) are constructed explicitly and have a finite Laurent s series expansion (2.0). Proof. Let π J p be any given basis with m < p n and such that Q π (z) 0 in (4.). We are going to build up simple rational functions z R π η (z) where η J p such that the expansion (4.3) j π ( ec jz A j) = η J p Rπ η (z) j η ( ec jz A j) holds. Invoking Lemma 2.2 there exists a basis σ J m such that σ π. Pick any index g π \ σ so the basis σ := σ {g} is indeed an element of J m+ because of Lemma 2.2 again. Next since c is regular pick a vector v Z m+ such that A σ v = 0 and c σv 0 like in (2.5). The statements below follow from the same arguments as in the proof of Theorem 3.2(i) so we briefly outline the proof. Define the vector (4.4) w := (e c z A e c 2 z A 2 e cn z An ) C n ; 5

16 so that with same notation as in (2.4) w σ C m+. Like in (3.6) we may deduce that (w σ ) v = e c σ v. Moreover define the rational functions P k (v w σ )/( e c σv ) if v k > 0 (4.5) Rk π (z) := [w σ ] k P k(v w σ )/( e c σ v ) if v k < 0 0 otherwise; where the functions P k are defined as in (3.) for k m +. Thus (4.6) = m+ k= m+ ( [w σ ] k ) Rk π (z) = k= like in (3.7). From (4.6) one easily obtains m+ (4.7) j π ( ec jz A j) = Rk π (z) k= ( e c σk z A σk ) R π k (z) j π j σ k e c jz A. j Notice that the sum in (4.7) is done only over the integers k for which v k 0 because R π k (z) = 0 whenever v k = 0. Next we use the same arguments as in the proof of Theorem 3.2(ii). With no loss of generality suppose that the ordered sets σ σ π are given by : (4.8) π := { 2... p} σ := σ {p} and p σ. Notice that σ k σ for k m and σ m+ = p. Besides consider the ordered sets (4.9) η(k) = {j π j σ k } for k =... m +. We next show that each sub-matrix A η(k) has maximal rank for every k =... m + with v k 0. Notice that η(k) = p because π = p; whence the set η(k) is indeed an element of J p precisely when A η(k) has maximal rank. Now we have that σ η(m + ) for the index σ m+ = p is contained in π\ σ. Therefore since σ J m Lemma 2.2 implies that η(m+) in (4.9) is an element of J p the square sub-matrix A σ is invertible and A η(m+) has maximal rank. On the other hand the vector v Z m+ satisfies 0 = A σ v = A σ (v v 2... v m ) + A p v m+ with v 0 so we may conclude that the last entry v m+ 0. We can now express the p-column of A as the finite sum A p = m v j j= v m+ A σj. Whence for every k m with v k 0 the matrix A η(k) = [A A σk A σk + A p ] 6

17 has maximal rank because the column A σk of A σ(m+) has been substituted with the linear combination A p = m v j j= v m+ A σj whose coefficient v k /v m+ is different from zero. Thus the set η(k) defined in (4.9) is an element of J p for every k m with v k 0. Therefore (4.7) can be re-written j π ( ec jz A j) = v k 0 = which is the desired identity (4.3) with η J p Rk π (z) j η(k) ( ec j z A j) Rπ η (z) j η ( ec j z A j) R π η (z) { R π k (z) if η = η(k) for some index k with v k 0 0 otherwise. On the other hand it is easy to see that all rational functions R π k and Rη π have finite Laurent series (2.0) because each Rk π is defined in terms of P k in (4.5) and each rational function P k in (3.) also has a finite Laurent series. Finally (4.2) follows easily. Compounding (4.) and (4.3) together yields (4.0) F (z) = η J p π J p Rπ η (z) Q π (z) k η ( ec k z A k) so that the decomposition (4.2) holds by setting Q η identically equal to the finite sum π J p R π η Q π for every η J p. Notice that the sum in (4.) runs over the bases of order p whereas the sum in (4.2) runs over the bases of order p. Hence repeated applications of Proposition 4. yields a decomposition of the generating function F into a sum over the bases of order m which is the decomposition described in (2.9) (2.0). Namely Corollary 4.2. Let A Z m n be a maximal rank matrix and let c Γ be regular. Let f be as in (.2) and F be its generating function (2.6) (2.7). Then : (i) F (z) has the expansion (4.) F (z) = σ J m Q σ (z) k σ ( ec k z A k) 7

18 for some rational functions z Q σ (z) which can be built up explicitly and with finite Laurent series (2.0). (ii) For every y Z m the function f(y) is obtained from Theorem 2.6. Proof. The point (i) is proved by induction. Notice that (2.7) can be rewritten F (z) = π J n k π ( ec k z A k) because J n = {{ 2... n}} and A has maximal rank (see (2.3)). Thus from Proposition 4. (4.2) holds for p = n as well. And more generally repeated applications of Proposition 4. show that (4.2) holds for all m p < n. However (4.) is precisely (4.2) with p = m. On the other hand (ii) follows because as c Γ F (z) is the generating function of f(y) and has the decomposition (4.) required to apply Theorem An algorithm for the general case n > m +. Let R[z z ] be the set of polynomials with real coefficients and entries in z t and zt so negative exponents are allowed. In view of Theorems 2.4 and 2.6 the recursive algorithm for the general case reads as follows: Procedure Solve2(A c y). Input: A Z m n full rank and n > m + ; c R n regular; y Z m ; Output: f(y) as in (2.2). Step 0 initialization : Calculate the set of bases J m so that σ J m if and only if the square matrix A σ = [A σ A σm ] is invertible. Set p := n and z Q π (z) with π := { 2... n}. Q π (z) R[z z ] and : F (z) = Q π (z) π J p j π ( ec jz A j) Step : While p m + do: with For every π J p with Q π (z) 0 do: Pick a basis σ J m such that σ π. J p = {{ 2... n}}. Pick a point g π \ σ; and let σ := σ {g} J m+. 8 Thus

19 Let A σ := [A j ] j σ Z m (m+) and c σ := [c j ] j σ R m+. Apply steps 0 2 of Solve(A σ c σ y). That is : Compute v Z m+ such that A σ v = 0 and c σv 0. Compute polynomials {w σ P k (v w σ )} from (3.). Compute polynomials {z Rk π (z)} from (4.5); so that : m+ j π ( ec jz A j) = Rk π (z) e c jz A. j k= For every η J p do: Set R π η := 0. j π j σ k For every k m + with v k 0 do: Set η[k] := π \ {σ k } J p and R π η[k] (z) := Rπ k (z) R[z z ]. We finally have the identity : Q π (z) j π ( ec jz A j) = η J p Rπ η (z)q π (z) j η ( ec jz A j). For every η J p do: Set Q η (z) := π J p R π η (z)q π (z). Hence each Q η (z) R[z z ] and : F (z) = Q η (z) j η ( ec jz A j). Set p := p. η J p Step 2: We have obtained the decomposition : F (z) = Q π (z) j π ( ec jz A j) where each Q π(z) R[z z ]. π J m Since F (z) is now in the form (2.9) required to apply Theorem 2.6 we thus obtain f(y) from (2.2) in Theorem Computational complexity. First observe that the procedures Solve and Solve2 defined in 3.3 and 4. compute the coefficients of the polynomials Q σ (z) in the decomposition (2.2) of F (z). This computation involves only algebraic operations provided the matrix A Z m n and the vector c R n are given. Recall that a main step in these procedures is to calculate a vector v Z m+ such that A σ v = 0 and c σv 0. Thus for practical implementation one should directly consider working with a rational vector c Q n. Next concerning the real numbers {e c k} k used in the procedures Solve and Solve2 according to 3.3 and 4. one may easily see that the 9

20 entries e c k can be treated symbolically. Indeed we need the numerical value of each e c k only at the very final step i.e. for evaluating f(y) via (2.2) in Theorem 2.6; see the illustrative example in the next section. Therefore only at the very final stage one needs a good rational approximation of e c j in Q n to evaluate f(y). Having said these the computational complexity is essentially determined by the number of coefficients {Q σβ } in equation (2.2); or equivalently by the number of nonzero coefficients of the polynomials {Q σ (z)} in the decomposition (2.9) (2.0). Define (4.2) Λ := max σ J m+ { min{ v Aσ v = 0 c σv 0 v Z m+ } }. In the case n = m + (see 3.) each polynomial Q σ (z) has at most Λ terms. This follows directly from (3.) and (3.5). For n = m + 2 we have at most (m + ) 2 polynomials Q σ (z) in (2.9); and again each one of them has at most Λ non-zero coefficients. Therefore in the general case n > m we end up with at most (m + ) n m Λ terms in (2.2). Thus the computational complexity is equal to O[(m+) n m Λ]. As a nice feature of the algorithm notice that the computational complexity does not depends on the right-hand-side y Z m. Moreover notice that the constant Λ does not change (at all) if we multiply the vector c Q n for any real r 0 because c σv 0 if and only if rc σv 0. Hence we can also conclude that the computational complexity does not depends on the magnitude of c it only depends on the ratio between the entries of c. However as shown in the following simple example kindly suggested by an anonymous referee Λ is exponential in the input size of A. Indeed if A = [ a a 2 a + (a + ) 2 ] then necessarily every solution v Z 3 of Av = 0 is an integer multiple of the vector (a 2 + a 2a ) and so Λ = O(a 2 ). Finally the constant M > 0 in (2.0) and (2.2) depends polynomially on Λ. 20

21 5. Illustrative Example Consider the following example with n = 6 m = 3 and data A := c := (c... c 6 ) so that F (z) is equal to the rational fraction ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 3 z z 3 )( e c 4 z )( e c 5 z2 )( e c 6 z3 ). Let us apply the algorithm developed in 4.. Step 0: The set J 3 is composed by all bases {i j k} with i j k 6 but the exceptions: { 4 5} and {3 4 6}. Set z Q π (z) when π = { }. Step : p = 6. With π = { } J 6 define the vector w = ( e c z z 2 2 e c 2 z z 2 z 2 3 e c 3 z z 3 e c 4 z e c 5 z 2 e c 6 z 3 ). Notice that the element k in the base π indeed represents the k-th column of A the k-th entry of w and the k-th factor in the denominator of F (z). Now choose σ := {4 5 6} and σ := { }. Let v := ( 0 ) Z 4 solve A σ v = 0. We obviously have that w σ = (e c 3 z z 3 e c 4 z e c 5 z 5 e c 6 z 3 ) and (w σ ) v = e c 4+c 6 c 3. Therefore applying equations (3.) and (4.5) we get R π (z) = (ec 3 z z 3 ) e c 4+c 6 c 3 R2 π (z) = (ec 3 z z 3 ) e c 4+c 6 c 3 R3 π (z) = 0 R4 π (z) = e(c 4 c 3 ) z3 e c 4+c 6 c 3. Hence we can easily calculate that : = ( e c 3 z z 3 )R π (z) + ( e c 4 z )R π 2 (z) + ( e c 6 z 3 )R π 4 (z). Notice that the term ( e c 3 z z 3 )R π (z) will kill the element 3 in the base π. Moreover the terms ( )R2 π(z) and ( )Rπ 3 (z) will also kill the respective entries 4 and 6 in the base π so 2

22 F (z) = + + = R π(z) ( e c z z2 2)( ec 2 z z 2 z3 2)( ec 4 z )( e c 5 z2 )( e c 6 z3 ) R2 π(z) ( e c z z2 2)( ec 2 z z 2 z3 2)( ec 3 z z 3 )( e c 5 z2 )( e c 6 z3 ) R3 π(z) ( e c z z2 2)( ec 2 z z 2 z3 2)( ec 3 z z 3 )( e c 4 z )( e c 5 z2 ) 3 j= Q η5 [j](z) k η 5 [j] ( ec k z A k) with η 5 [j] J 5 j = 2 3. One has η 5 [] = { } η 5 [2] = { } η 5 [3] = { } and Q η5 [j](z) = Rj π (z) for j = 2 3. Step : p = 5. - Analyzing η 5 [] = { } J 5 choose σ = {4 5 6} and σ := { 4 5 6}. Let v := ( 2 0) Z 4 solve A σ v = 0. We have that w σ = (e c z z 2 2 ec 4 z e c 5 z 2 e c 6 z 3 ) so we get R η 5[] (z) = (ec z z 2 2 ) e c +c 4 +2c 5 R η 5[] 2 (z) = R η 5[] (e c z z 2 2 ) e c +c 4 +2c 5 3 (z) = (ec 4 c z2 2 )( + ec 5 z 2 ) e c +c 4 +2c 5 R η 5[] 4 (z) = 0. One may easily verify that = ( e c z z 2 2)R η 5[] (z) + ( e c 4 z )R η 5[] 2 (z) + ( e c 5 z 2 )R η 5[] 3 (z). Notice that the terms associated to R η 5[] (z) R η 5[] 2 (z) and R η 5[] 3 (z) kill the respective entries 4 and 5 in the base η 5 []. - Analyzing η 5 [2] = { } J 5 choose σ = {3 5 6} and σ := { }. Let v := ( ) Z 4 solve A σ v = 0. We have that w σ = (e c 2 z z 2 z 2 3 ec 3 z z 3 e c 5 z 2 e c 6 z 3 ) so we get 22

23 R η 5[2] (z) = (ec 2 z z 2 z 2 3 ) e c 2+c 3 +c 5 +c 6 R η 5[2] 2 (z) = R η 5[2] 3 (z) = R η 5[2] 4 (z) = We may easily verify that (e c 2 z z 2 z3 2) e c 2+c 3 +c 5 +c 6 e c 3 c 2 z2 z 3 e c 2+c 3 +c 5 +c 6 e c 5+c 3 c 2 z3 e c 2+c 3 +c 5 +c 6. = ( e c 2 z z 2 z3 2)Rη 5[2] (z) + ( e c 3 z z 3 )R η 5[2] 2 (z)+ +( e c 5 z 2 )R η 5[2] 3 (z) + ( e c 6 z 3 )R η 5[2] 4 (z). Notice that the terms associated to R η 5[2] (z) R η 5[2] 2 (z) R η 5[2] 3 (z) and R η 5[2] 4 (z) kill the respective entries and 6 in the base η 5 [2]. - Analyzing η 5 [3] = { } J 5 choose σ = {3 4 5} and σ := { }. Let v := ( 2 ) Z 4 solve A σ v = 0. We have that w σ = (e c 2 z z 2 z 2 3 ec 3 z z 3 e c 4 z e c 5 z 2 ) so we get R η 5[3] (z) = (e c 2 z z 2 z 2 3 ) e c 2+2c 3 c 4 +c 5 R η 5[3] 2 (z) = (ec 2 z z 2 z 2 3 ) ( + e c 3 z z 3 ) e c 2+2c 3 c 4 +c 5 3 (z) = (ec 4 z ) (e 2c 3 c 2 z z2 ) e c 2+2c 3 c 4 +c 5 R η 5[3] R η 5[3] 4 (z) = We may easily verify that e 2c 3 c 2 c 4 z 2 e c 2+2c 3 c 4 +c 5. = ( e c 2 z z 2 z3 2)Rη 5[3] (z) + ( e c 3 z z 3 )R η 5[3] 2 (z)+ +( e c 4 z )R η 5[3] 3 (z) + ( e c 5 z 2 )R η 5[3] 4 (z). Notice that the terms associated to R η 5[3] (z) R η 5[3] 2 (z) R η 5[3] 3 (z) and R η 5[3] 4 (z) kill the respective entries and 5 in the base η 5 [3]. Therefore we have the following expansion of F (z). 23

24 F (z) = = Q η5 [](z)r η 5[] (z) ( e c 2 z z 2 z 2 3 )( ec 4 z )( e c 5 z2 )( e c 6 z3 ) Q η5 [](z)r η 5[] 2 (z) + Q η5 [2](z)R η 5[2] 2 (z) ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 5 z2 )( e c 6 z3 ) Q η5 [](z)r η 5[] 3 (z) ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 4 z )( e c 6 z3 ) Q η5 [2](z)R η 5[2] (z) ( e c z z 2 2 )( ec 3 z z 3 )( e c 5 z2 )( e c 6 z3 ) Q η5 [2](z)R η 5[2] 3 (z) ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 3 z z 3 )( e c 6 z3 ) Q η5 [2](z)R η 5[2] 4 (z) + Q η5 [3](z)R η 5[3] 3 (z) ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 3 z z 3 )( e c 5 z2 ) Q η5 [3](z)R η 5[3] (z) ( e c z z 2 2 )( ec 3 z z 3 )( e c 4 z )( e c 5 z2 ) Q η5 [3](z)R η 5[3] 2 (z) ( e c z z 2 2 )( ec 2 z z 2 z 2 3 )( ec 4 z )( e c 5 z2 ) Q η5 [3](z)R η 5[3] 4 (z) ( e c z z2 2)( ec 2 z z 2 z3 2)( ec 3 z z 3 )( e c 4 z ) 9 j= Q η4 [j](z) k η 4 [j] ( ec k z A k) with η 4 [j] J 4 j = Step : p = 4 = m+. At this step we obtain the required decomposition (4.) that is we express F (z) as the sum (5.) F (z) = j Q η3 [j](z) k η 3 [j] ( ec k z A k) with η 3 [j] J 3 = J m j. The above sum contains only 6 terms (not detailed here) out of the potentially ( 6 3) = 20 terms. For illustration we only provide the term Q η3 [j](z) relative to the basis η 3 [j] = {2 5 6} J 3. - With η 4 [] = { } J 4 choose σ := {4 5 6} and σ := { }. Let v := ( 2) Z 4 solve A σ v = 0. We have that w σ = (e c 2 z z 2 z 2 3 ec 4 z e c 5 z 2 e c 6 z 3 ) so we get 24

25 R η 4[] (z) = (ec 2 z z 2 z 2 3 ) e 2c 6+c 5 +c 4 c 2 R η 4[] 2 (z) = R η 4[] 3 (z) = R η 4[] (e c 2 z z 2 z3 2) e 2c 6+c 5 +c 4 c 2 e c 4 c 2 (z 2 z3 2) e 2c 6+c 5 +c 4 c 2 4 (z) = (ec 4+c 5 c 2 z3 2 )( + ec 6 z 3 ) e 2c 6+c 5 +c 4 c 2. = ( e c 2 z z 2 z3 2)Rη 4[] (z) + ( e c 4 z )R η 4[] 2 (z)+ +( e c 5 z 2 )R η 4[] 3 (z) + ( e c 6 z 3 )R η 4[] 4 (z). Notice that the term associated to R η 4[] 2 kills the entry 4 in the base η 4 [] = { } so we are getting the desired base η 3 [] = {2 5 6}. - With η 4 [2] = { 2 5 6} J 4 choose σ := {2 5 6} and σ := { 2 5 6}. Let v := ( 2) Z 4 solve A σ v = 0. We have that w σ = (e c z z 2 2 ec 2 z z 2 z 2 3 ec 5 z 2 e c 6 z 3 ) so we get R η 4[2] (z) = R η 4[2] 2 (z) = R η 4[2] 3 (z) = (e c z z2 2) e c 2+c 5 c 2c 6 (e c z z2 2) e c 2+c 5 c 2c 6 e c 2 c z2 z2 3 e c 2+c 5 c 2c 6 R η 4[2] 4 (z) = (ec 6 z 3 ) (e c 2 c +c 5 z 2 3 )( + (ec 6 z 3 ) ) e c 2+c 5 c 2c 6. = ( e c z z2 2)Rη 4[2] (z) + ( e c 2 z z 2 z3 2)Rη 4[] 2 (z)+ +( e c 5 z 2 )R η 4[] 3 (z) + ( e c 6 z 3 )R η 4[] 4 (z). Notice that the term associated to R η 4[2] kills the entry in the base η 4 [2] = { 2 5 6} so we are getting the desired base η 3 [] = {2 5 6}. Therefore working on the base η 3 [] we obtain the numerator 25

26 Q η3 [](z) = Q η4 []R η 4[] 2 + Q η4 [2]R η 4[2] [ ] = Q η5 [](z)r η 5[] (z) + R η 4[] 2 + [ Q η5 [](z)r η 5[] 2 (z) + Q η5 [2](z)R η 5[2] 2 (z) ] R η 4[2]. Step 2: The value of f(y) is obtained from (2.2) in Theorem 2.6 using the expression (5.) of F (z). References [] W. Baldoni-Silva M. Vergne. Residues formulae for volumes and Ehrhart polynomials of convex polytopes arxiv:math.co/ v 200. [2] W. Baldoni-Silva J.A. De Loera M. Vergne. Counting integer flows in networks Found. Comp. Math. to appear. [3] A.I. Barvinok. Computing the volume counting integral points and exponentials sums Discr. Comp. Geom. 0 (993) [4] A.I. Barvinok J.E. Pommersheim. An algorithmic theory of lattice points in polyhedral in: New Perspectives in Algebraic Combinatorics MSRI Publications 38 (999) [5] M. Beck. Counting Lattice Points by means of the Residue Theorem. Ramanujan Journal 4 (2000) [6] M. Beck R. Diaz S. Robins. The Frobenius problem rational polytopes and Fourier-Dedekind sums J. Numb. Theor. 96 (2002) 2. [7] M. Beck D. Pixton. The Ehrhart polynomial of the Birkhoff polytope Discr. Comp. Math. 30 (2003) [8] M. Brion M. Vergne. Residue formulae vector partition functions and lattice points in rational polytopes J. Amer. Math. Soc. 0 (997) [9] C. Cochet. Réduction des graphes de Goretsky-Kottwitz-MacPherson; nombres de Kostka et coefficients de Littlewodd-Richardson Thèse de Doctorat: Mathématiques Université Paris 7 Paris Décembre [0] J.A. De Loera R. Hemmecke J. Tauzer and R. Yoshida. Effective lattice point counting in rational convex polytopes J. of Symb. Comp. to appear. [] A.V. Pukhlikov A.G. Khovanskii. A Riemann-Roch theorem for integrals and sums of quasipolynomials over virtual polytopes St. Petersburg Math. J. 4 (993) [2] J.B. Lasserre E.S. Zeron. On counting integral points in a convex rational polytope Math. Oper. Res. 28 (2003) [3] A. Schrijver. Theory of Linear and Integer Programming John Wiley & Sons Chichester 986. [4] A. Szenes M. Vergne. Residue formulae for vector partitions and Euler-MacLaurin sums. Adv. in Appl. Math. 30 (2003) [5] A. Szenes. Residue theorem for rational trigonometric sums and Verlinde s formula. Duke Math. J. 8 (2003) [6] S. Verdoolaege K. Beyls M. Bruynooghe R. Seghir V. Loechner. Analytical Computation of Ehrhart Polynomials and its Applications for Embedded Systems Technical report # 376 Computer Science Department KUL University Leuwen Belgium. 26

27 LAAS-CNRS 7 Avenue du Colonel Roche 3077 Toulouse Cédex 4 France. address: lasserre@laas.fr Depto. Matemáticas CIVESTAV-IPN Apdo. Postal 4740 Mexico D.F México. address: eszeron@math.cinvestav.mx 27