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1//===- Barvinok.cpp - Barvinok's Algorithm ----------------------*- C++ -*-===//2//3// Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions.4// See https://llvm.org/LICENSE.txt for license information.5// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception6//7//===----------------------------------------------------------------------===//8 9#include "mlir/Analysis/Presburger/Barvinok.h"10#include "mlir/Analysis/Presburger/Utils.h"11#include "llvm/ADT/Sequence.h"12#include <algorithm>13 14using namespace mlir;15using namespace presburger;16using namespace mlir::presburger::detail;17 18/// Assuming that the input cone is pointed at the origin,19/// converts it to its dual in V-representation.20/// Essentially we just remove the all-zeroes constant column.21ConeV mlir::presburger::detail::getDual(ConeH cone) {22  unsigned numIneq = cone.getNumInequalities();23  unsigned numVar = cone.getNumCols() - 1;24  ConeV dual(numIneq, numVar, 0, 0);25  // Assuming that an inequality of the form26  // a1*x1 + ... + an*xn + b ≥ 027  // is represented as a row [a1, ..., an, b]28  // and that b = 0.29 30  for (auto i : llvm::seq<int>(0, numIneq)) {31    assert(cone.atIneq(i, numVar) == 0 &&32           "H-representation of cone is not centred at the origin!");33    for (unsigned j = 0; j < numVar; ++j) {34      dual.at(i, j) = cone.atIneq(i, j);35    }36  }37 38  // Now dual is of the form [ [a1, ..., an] , ... ]39  // which is the V-representation of the dual.40  return dual;41}42 43/// Converts a cone in V-representation to the H-representation44/// of its dual, pointed at the origin (not at the original vertex).45/// Essentially adds a column consisting only of zeroes to the end.46ConeH mlir::presburger::detail::getDual(ConeV cone) {47  unsigned rows = cone.getNumRows();48  unsigned columns = cone.getNumColumns();49  ConeH dual = defineHRep(columns);50  // Add a new column (for constants) at the end.51  // This will be initialized to zero.52  cone.insertColumn(columns);53 54  for (unsigned i = 0; i < rows; ++i)55    dual.addInequality(cone.getRow(i));56 57  // Now dual is of the form [ [a1, ..., an, 0] , ... ]58  // which is the H-representation of the dual.59  return dual;60}61 62/// Find the index of a cone in V-representation.63DynamicAPInt mlir::presburger::detail::getIndex(const ConeV &cone) {64  if (cone.getNumRows() > cone.getNumColumns())65    return DynamicAPInt(0);66 67  return cone.determinant();68}69 70/// Compute the generating function for a unimodular cone.71/// This consists of a single term of the form72/// sign * x^num / prod_j (1 - x^den_j)73///74/// sign is either +1 or -1.75/// den_j is defined as the set of generators of the cone.76/// num is computed by expressing the vertex as a weighted77/// sum of the generators, and then taking the floor of the78/// coefficients.79GeneratingFunction80mlir::presburger::detail::computeUnimodularConeGeneratingFunction(81    ParamPoint vertex, int sign, const ConeH &cone) {82  // Consider a cone with H-representation [0  -1].83  //                                       [-1 -2]84  // Let the vertex be given by the matrix [ 2  2   0], with 2 params.85  //                                       [-1 -1/2 1]86 87  // `cone` must be unimodular.88  assert(abs(getIndex(getDual(cone))) == 1 && "input cone is not unimodular!");89 90  unsigned numVar = cone.getNumVars();91  unsigned numIneq = cone.getNumInequalities();92 93  // Thus its ray matrix, U, is the inverse of the94  // transpose of its inequality matrix, `cone`.95  // The last column of the inequality matrix is null,96  // so we remove it to obtain a square matrix.97  FracMatrix transp = FracMatrix(cone.getInequalities()).transpose();98  transp.removeRow(numVar);99 100  FracMatrix generators(numVar, numIneq);101  transp.determinant(/*inverse=*/&generators); // This is the U-matrix.102  // Thus the generators are given by U = [2  -1].103  //                                      [-1  0]104 105  // The powers in the denominator of the generating106  // function are given by the generators of the cone,107  // i.e., the rows of the matrix U.108  std::vector<Point> denominator(numIneq);109  ArrayRef<Fraction> row;110  for (auto i : llvm::seq<int>(0, numVar)) {111    row = generators.getRow(i);112    denominator[i] = Point(row);113  }114 115  // The vertex is v \in Z^{d x (n+1)}116  // We need to find affine functions of parameters λ_i(p)117  // such that v = Σ λ_i(p)*u_i,118  // where u_i are the rows of U (generators)119  // The λ_i are given by the columns of Λ = v^T U^{-1}, and120  // we have transp = U^{-1}.121  // Then the exponent in the numerator will be122  // Σ -floor(-λ_i(p))*u_i.123  // Thus we store the (exponent of the) numerator as the affine function -Λ,124  // since the generators u_i are already stored as the exponent of the125  // denominator. Note that the outer -1 will have to be accounted for, as it is126  // not stored. See end for an example.127 128  unsigned numColumns = vertex.getNumColumns();129  unsigned numRows = vertex.getNumRows();130  ParamPoint numerator(numColumns, numRows);131  SmallVector<Fraction> ithCol(numRows);132  for (auto i : llvm::seq<int>(0, numColumns)) {133    for (auto j : llvm::seq<int>(0, numRows))134      ithCol[j] = vertex(j, i);135    numerator.setRow(i, transp.preMultiplyWithRow(ithCol));136    numerator.negateRow(i);137  }138  // Therefore Λ will be given by [ 1    0 ] and the negation of this will be139  //                              [ 1/2 -1 ]140  //                              [ -1  -2 ]141  // stored as the numerator.142  // Algebraically, the numerator exponent is143  // [ -2 ⌊ - N - M/2 + 1 ⌋ + 1 ⌊ 0 + M + 2 ⌋ ] -> first  COLUMN of U is [2, -1]144  // [  1 ⌊ - N - M/2 + 1 ⌋ + 0 ⌊ 0 + M + 2 ⌋ ] -> second COLUMN of U is [-1, 0]145 146  return GeneratingFunction(numColumns - 1, SmallVector<int>(1, sign),147                            std::vector({numerator}),148                            std::vector({denominator}));149}150 151/// We use Gaussian elimination to find the solution to a set of d equations152/// of the form153/// a_1 x_1 + ... + a_d x_d + b_1 m_1 + ... + b_p m_p + c = 0154/// where x_i are variables,155/// m_i are parameters and156/// a_i, b_i, c are rational coefficients.157///158/// The solution expresses each x_i as an affine function of the m_i, and is159/// therefore represented as a matrix of size d x (p+1).160/// If there is no solution, we return null.161std::optional<ParamPoint>162mlir::presburger::detail::solveParametricEquations(FracMatrix equations) {163  // equations is a d x (d + p + 1) matrix.164  // Each row represents an equation.165  unsigned d = equations.getNumRows();166  unsigned numCols = equations.getNumColumns();167 168  // If the determinant is zero, there is no unique solution.169  // Thus we return null.170  if (FracMatrix(equations.getSubMatrix(/*fromRow=*/0, /*toRow=*/d - 1,171                                        /*fromColumn=*/0,172                                        /*toColumn=*/d - 1))173          .determinant() == 0)174    return std::nullopt;175 176  // Perform row operations to make each column all zeros except for the177  // diagonal element, which is made to be one.178  for (unsigned i = 0; i < d; ++i) {179    // First ensure that the diagonal element is nonzero, by swapping180    // it with a row that is non-zero at column i.181    if (equations(i, i) != 0)182      continue;183    for (unsigned j = i + 1; j < d; ++j) {184      if (equations(j, i) == 0)185        continue;186      equations.swapRows(j, i);187      break;188    }189 190    Fraction diagElement = equations(i, i);191 192    // Apply row operations to make all elements except the diagonal to zero.193    for (unsigned j = 0; j < d; ++j) {194      if (i == j)195        continue;196      if (equations(j, i) == 0)197        continue;198      // Apply row operations to make element (j, i) zero by subtracting the199      // ith row, appropriately scaled.200      Fraction currentElement = equations(j, i);201      equations.addToRow(/*sourceRow=*/i, /*targetRow=*/j,202                         /*scale=*/-currentElement / diagElement);203    }204  }205 206  // Rescale diagonal elements to 1.207  for (unsigned i = 0; i < d; ++i)208    equations.scaleRow(i, 1 / equations(i, i));209 210  // Now we have reduced the equations to the form211  // x_i + b_1' m_1 + ... + b_p' m_p + c' = 0212  // i.e. each variable appears exactly once in the system, and has coefficient213  // one.214  //215  // Thus we have216  // x_i = - b_1' m_1 - ... - b_p' m_p - c217  // and so we return the negation of the last p + 1 columns of the matrix.218  //219  // We copy these columns and return them.220  ParamPoint vertex =221      equations.getSubMatrix(/*fromRow=*/0, /*toRow=*/d - 1,222                             /*fromColumn=*/d, /*toColumn=*/numCols - 1);223  vertex.negateMatrix();224  return vertex;225}226 227/// This is an implementation of the Clauss-Loechner algorithm for chamber228/// decomposition.229///230/// We maintain a list of pairwise disjoint chambers and the generating231/// functions corresponding to each one. We iterate over the list of regions,232/// each time adding the current region's generating function to the chambers233/// where it is active and separating the chambers where it is not.234///235/// Given the region each generating function is active in, for each subset of236/// generating functions the region that (the sum of) precisely this subset is237/// in, is the intersection of the regions that these are active in,238/// intersected with the complements of the remaining regions.239std::vector<std::pair<PresburgerSet, GeneratingFunction>>240mlir::presburger::detail::computeChamberDecomposition(241    unsigned numSymbols, ArrayRef<std::pair<PresburgerSet, GeneratingFunction>>242                             regionsAndGeneratingFunctions) {243  assert(!regionsAndGeneratingFunctions.empty() &&244         "there must be at least one chamber!");245  // We maintain a list of regions and their associated generating function246  // initialized with the universe and the empty generating function.247  std::vector<std::pair<PresburgerSet, GeneratingFunction>> chambers = {248      {PresburgerSet::getUniverse(PresburgerSpace::getSetSpace(numSymbols)),249       GeneratingFunction(numSymbols, {}, {}, {})}};250 251  // We iterate over the region list.252  //253  // For each activity region R_j (corresponding to the generating function254  // gf_j), we examine all the current chambers R_i.255  //256  // If R_j has a full-dimensional intersection with an existing chamber R_i,257  // then that chamber is replaced by two new ones:258  // 1. the intersection R_i \cap R_j, where the generating function is259  // gf_i + gf_j.260  // 2. the difference R_i - R_j, where the generating function is gf_i.261  //262  // At each step, we define a new chamber list after considering gf_j,263  // replacing and appending chambers as discussed above.264  //265  // The loop has the invariant that the union over all the chambers gives the266  // universe at every step.267  for (const auto &[region, generatingFunction] :268       regionsAndGeneratingFunctions) {269    std::vector<std::pair<PresburgerSet, GeneratingFunction>> newChambers;270 271    for (const auto &[currentRegion, currentGeneratingFunction] : chambers) {272      PresburgerSet intersection = currentRegion.intersect(region);273 274      // If the intersection is not full-dimensional, we do not modify275      // the chamber list.276      if (!intersection.isFullDim()) {277        newChambers.emplace_back(currentRegion, currentGeneratingFunction);278        continue;279      }280 281      // If it is, we add the intersection and the difference as chambers.282      newChambers.emplace_back(intersection,283                               currentGeneratingFunction + generatingFunction);284      newChambers.emplace_back(currentRegion.subtract(region),285                               currentGeneratingFunction);286    }287    chambers = std::move(newChambers);288  }289 290  return chambers;291}292 293/// For a polytope expressed as a set of n inequalities, compute the generating294/// function corresponding to the lattice points included in the polytope. This295/// algorithm has three main steps:296/// 1. Enumerate the vertices, by iterating over subsets of inequalities and297///    checking for satisfiability. For each d-subset of inequalities (where d298///    is the number of variables), we solve to obtain the vertex in terms of299///    the parameters, and then check for the region in parameter space where300///    this vertex satisfies the remaining (n - d) inequalities.301/// 2. For each vertex, identify the tangent cone and compute the generating302///    function corresponding to it. The generating function depends on the303///    parametric expression of the vertex and the (non-parametric) generators304///    of the tangent cone.305/// 3. [Clauss-Loechner decomposition] Identify the regions in parameter space306///    (chambers) where each vertex is active, and accordingly compute the307///    GF of the polytope in each chamber.308///309/// Verdoolaege, Sven, et al. "Counting integer points in parametric310/// polytopes using Barvinok's rational functions." Algorithmica 48 (2007):311/// 37-66.312std::vector<std::pair<PresburgerSet, GeneratingFunction>>313mlir::presburger::detail::computePolytopeGeneratingFunction(314    const PolyhedronH &poly) {315  unsigned numVars = poly.getNumRangeVars();316  unsigned numSymbols = poly.getNumSymbolVars();317  unsigned numIneqs = poly.getNumInequalities();318 319  // We store a list of the computed vertices.320  std::vector<ParamPoint> vertices;321  // For each vertex, we store the corresponding active region and the322  // generating functions of the tangent cone, in order.323  std::vector<std::pair<PresburgerSet, GeneratingFunction>>324      regionsAndGeneratingFunctions;325 326  // We iterate over all subsets of inequalities with cardinality numVars,327  // using permutations of numVars 1's and (numIneqs - numVars) 0's.328  //329  // For a given permutation, we consider a subset which contains330  // the i'th inequality if the i'th bit in the bitset is 1.331  //332  // We start with the permutation that takes the last numVars inequalities.333  SmallVector<int> indicator(numIneqs);334  for (unsigned i = numIneqs - numVars; i < numIneqs; ++i)335    indicator[i] = 1;336 337  do {338    // Collect the inequalities corresponding to the bits which are set339    // and the remaining ones.340    auto [subset, remainder] = poly.getInequalities().splitByBitset(indicator);341    // All other inequalities are stored in a2 and b2c2.342    //343    // These are column-wise splits of the inequalities;344    // a2 stores the coefficients of the variables, and345    // b2c2 stores the coefficients of the parameters and the constant term.346    FracMatrix a2(numIneqs - numVars, numVars);347    FracMatrix b2c2(numIneqs - numVars, numSymbols + 1);348    a2 = FracMatrix(349        remainder.getSubMatrix(0, numIneqs - numVars - 1, 0, numVars - 1));350    b2c2 = FracMatrix(remainder.getSubMatrix(0, numIneqs - numVars - 1, numVars,351                                             numVars + numSymbols));352 353    // Find the vertex, if any, corresponding to the current subset of354    // inequalities.355    std::optional<ParamPoint> vertex =356        solveParametricEquations(FracMatrix(subset)); // d x (p+1)357 358    if (!vertex)359      continue;360    if (llvm::is_contained(vertices, vertex))361      continue;362    // If this subset corresponds to a vertex that has not been considered,363    // store it.364    vertices.emplace_back(*vertex);365 366    // If a vertex is formed by the intersection of more than d facets, we367    // assume that any d-subset of these facets can be solved to obtain its368    // expression. This assumption is valid because, if the vertex has two369    // distinct parametric expressions, then a nontrivial equality among the370    // parameters holds, which is a contradiction as we know the parameter371    // space to be full-dimensional.372 373    // Let the current vertex be [X | y], where374    // X represents the coefficients of the parameters and375    // y represents the constant term.376    //377    // The region (in parameter space) where this vertex is active is given378    // by substituting the vertex into the *remaining* inequalities of the379    // polytope (those which were not collected into `subset`), i.e., into the380    // inequalities [A2 | B2 | c2].381    //382    // Thus, the coefficients of the parameters after substitution become383    // (A2 • X + B2)384    // and the constant terms become385    // (A2 • y + c2).386    //387    // The region is therefore given by388    // (A2 • X + B2) p + (A2 • y + c2) ≥ 0389    //390    // This is equivalent to A2 • [X | y] + [B2 | c2].391    //392    // Thus we premultiply [X | y] with each row of A2393    // and add each row of [B2 | c2].394    FracMatrix activeRegion(numIneqs - numVars, numSymbols + 1);395    for (unsigned i = 0; i < numIneqs - numVars; i++) {396      activeRegion.setRow(i, vertex->preMultiplyWithRow(a2.getRow(i)));397      activeRegion.addToRow(i, b2c2.getRow(i), 1);398    }399 400    // We convert the representation of the active region to an integers-only401    // form so as to store it as a PresburgerSet.402    IntegerPolyhedron activeRegionRel(403        PresburgerSpace::getRelationSpace(0, numSymbols, 0, 0), activeRegion);404 405    // Now, we compute the generating function at this vertex.406    // We collect the inequalities corresponding to each vertex to compute407    // the tangent cone at that vertex.408    //409    // We only need the coefficients of the variables (NOT the parameters)410    // as the generating function only depends on these.411    // We translate the cones to be pointed at the origin by making the412    // constant terms zero.413    ConeH tangentCone = defineHRep(numVars);414    for (unsigned j = 0, e = subset.getNumRows(); j < e; ++j) {415      SmallVector<DynamicAPInt> ineq(numVars + 1);416      for (unsigned k = 0; k < numVars; ++k)417        ineq[k] = subset(j, k);418      tangentCone.addInequality(ineq);419    }420    // We assume that the tangent cone is unimodular, so there is no need421    // to decompose it.422    //423    // In the general case, the unimodular decomposition may have several424    // cones.425    GeneratingFunction vertexGf(numSymbols, {}, {}, {});426    SmallVector<std::pair<int, ConeH>, 4> unimodCones = {{1, tangentCone}};427    for (const std::pair<int, ConeH> &signedCone : unimodCones) {428      auto [sign, cone] = signedCone;429      vertexGf = vertexGf +430                 computeUnimodularConeGeneratingFunction(*vertex, sign, cone);431    }432    // We store the vertex we computed with the generating function of its433    // tangent cone.434    regionsAndGeneratingFunctions.emplace_back(PresburgerSet(activeRegionRel),435                                               vertexGf);436  } while (std::next_permutation(indicator.begin(), indicator.end()));437 438  // Now, we use Clauss-Loechner decomposition to identify regions in parameter439  // space where each vertex is active. These regions (chambers) have the440  // property that no two of them have a full-dimensional intersection, i.e.,441  // they may share "facets" or "edges", but their intersection can only have442  // up to numVars - 1 dimensions.443  //444  // In each chamber, we sum up the generating functions of the active vertices445  // to find the generating function of the polytope.446  return computeChamberDecomposition(numSymbols, regionsAndGeneratingFunctions);447}448 449/// We use an iterative procedure to find a vector not orthogonal450/// to a given set, ignoring the null vectors.451/// Let the inputs be {x_1, ..., x_k}, all vectors of length n.452///453/// In the following,454/// vs[:i] means the elements of vs up to and including the i'th one,455/// <vs, us> means the dot product of vs and us,456/// vs ++ [v] means the vector vs with the new element v appended to it.457///458/// We proceed iteratively; for steps d = 0, ... n-1, we construct a vector459/// which is not orthogonal to any of {x_1[:d], ..., x_n[:d]}, ignoring460/// the null vectors.461/// At step d = 0, we let vs = [1]. Clearly this is not orthogonal to462/// any vector in the set {x_1[0], ..., x_n[0]}, except the null ones,463/// which we ignore.464/// At step d > 0 , we need a number v465/// s.t. <x_i[:d], vs++[v]> != 0 for all i.466/// => <x_i[:d-1], vs> + x_i[d]*v != 0467/// => v != - <x_i[:d-1], vs> / x_i[d]468/// We compute this value for all x_i, and then469/// set v to be the maximum element of this set plus one. Thus470/// v is outside the set as desired, and we append it to vs471/// to obtain the result of the d'th step.472Point mlir::presburger::detail::getNonOrthogonalVector(473    ArrayRef<Point> vectors) {474  unsigned dim = vectors[0].size();475  assert(llvm::all_of(476             vectors,477             [&dim](const Point &vector) { return vector.size() == dim; }) &&478         "all vectors need to be the same size!");479 480  SmallVector<Fraction> newPoint = {Fraction(1, 1)};481  Fraction maxDisallowedValue = -Fraction(1, 0),482           disallowedValue = Fraction(0, 1);483 484  for (unsigned d = 1; d < dim; ++d) {485    // Compute the disallowed values  - <x_i[:d-1], vs> / x_i[d] for each i.486    maxDisallowedValue = -Fraction(1, 0);487    for (const Point &vector : vectors) {488      if (vector[d] == 0)489        continue;490      disallowedValue =491          -dotProduct(ArrayRef(vector).slice(0, d), newPoint) / vector[d];492 493      // Find the biggest such value494      maxDisallowedValue = std::max(maxDisallowedValue, disallowedValue);495    }496    newPoint.emplace_back(maxDisallowedValue + 1);497  }498  return newPoint;499}500 501/// We use the following recursive formula to find the coefficient of502/// s^power in the rational function given by P(s)/Q(s).503///504/// Let P[i] denote the coefficient of s^i in the polynomial P(s).505/// (P/Q)[r] =506/// if (r == 0) then507///   P[0]/Q[0]508/// else509///   (P[r] - {Σ_{i=1}^r (P/Q)[r-i] * Q[i])}/(Q[0])510/// We therefore recursively call `getCoefficientInRationalFunction` on511/// all i \in [0, power).512///513/// https://math.ucdavis.edu/~deloera/researchsummary/514/// barvinokalgorithm-latte1.pdf, p. 1285515QuasiPolynomial mlir::presburger::detail::getCoefficientInRationalFunction(516    unsigned power, ArrayRef<QuasiPolynomial> num, ArrayRef<Fraction> den) {517  assert(!den.empty() && "division by empty denominator in rational function!");518 519  unsigned numParam = num[0].getNumInputs();520  // We use the `isEqual` method of PresburgerSpace, which QuasiPolynomial521  // inherits from.522  assert(llvm::all_of(num,523                      [&num](const QuasiPolynomial &qp) {524                        return num[0].isEqual(qp);525                      }) &&526         "the quasipolynomials should all belong to the same space!");527 528  std::vector<QuasiPolynomial> coefficients;529  coefficients.reserve(power + 1);530 531  coefficients.emplace_back(num[0] / den[0]);532  for (unsigned i = 1; i <= power; ++i) {533    // If the power is not there in the numerator, the coefficient is zero.534    coefficients.emplace_back(i < num.size() ? num[i]535                                             : QuasiPolynomial(numParam, 0));536 537    // After den.size(), the coefficients are zero, so we stop538    // subtracting at that point (if it is less than i).539    unsigned limit = std::min<unsigned long>(i, den.size() - 1);540    for (unsigned j = 1; j <= limit; ++j)541      coefficients[i] = coefficients[i] -542                        coefficients[i - j] * QuasiPolynomial(numParam, den[j]);543 544    coefficients[i] = coefficients[i] / den[0];545  }546  return coefficients[power].simplify();547}548 549/// Substitute x_i = t^μ_i in one term of a generating function, returning550/// a quasipolynomial which represents the exponent of the numerator551/// of the result, and a vector which represents the exponents of the552/// denominator of the result.553/// If the returned value is {num, dens}, it represents the function554/// t^num / \prod_j (1 - t^dens[j]).555/// v represents the affine functions whose floors are multiplied by the556/// generators, and ds represents the list of generators.557static std::pair<QuasiPolynomial, std::vector<Fraction>>558substituteMuInTerm(unsigned numParams, const ParamPoint &v,559                   const std::vector<Point> &ds, const Point &mu) {560  unsigned numDims = mu.size();561#ifndef NDEBUG562  for (const Point &d : ds)563    assert(d.size() == numDims &&564           "μ has to have the same number of dimensions as the generators!");565#endif566 567  // First, the exponent in the numerator becomes568  // - (μ • u_1) * (floor(first col of v))569  // - (μ • u_2) * (floor(second col of v)) - ...570  // - (μ • u_d) * (floor(d'th col of v))571  // So we store the negation of the dot products.572 573  // We have d terms, each of whose coefficient is the negative dot product.574  SmallVector<Fraction> coefficients;575  coefficients.reserve(numDims);576  for (const Point &d : ds)577    coefficients.emplace_back(-dotProduct(mu, d));578 579  // Then, the affine function is a single floor expression, given by the580  // corresponding column of v.581  ParamPoint vTranspose = v.transpose();582  std::vector<std::vector<SmallVector<Fraction>>> affine;583  affine.reserve(numDims);584  for (unsigned j = 0; j < numDims; ++j)585    affine.push_back({SmallVector<Fraction>{vTranspose.getRow(j)}});586 587  QuasiPolynomial num(numParams, coefficients, affine);588  num = num.simplify();589 590  std::vector<Fraction> dens;591  dens.reserve(ds.size());592  // Similarly, each term in the denominator has exponent593  // given by the dot product of μ with u_i.594  for (const Point &d : ds) {595    // This term in the denominator is596    // (1 - t^dens.back())597    dens.emplace_back(dotProduct(d, mu));598  }599 600  return {num, dens};601}602 603/// Normalize all denominator exponents `dens` to their absolute values604/// by multiplying and dividing by the inverses, in a function of the form605/// sign * t^num / prod_j (1 - t^dens[j]).606/// Here, sign = ± 1,607/// num is a QuasiPolynomial, and608/// each dens[j] is a Fraction.609static void normalizeDenominatorExponents(int &sign, QuasiPolynomial &num,610                                          std::vector<Fraction> &dens) {611  // We track the number of exponents that are negative in the612  // denominator, and convert them to their absolute values.613  unsigned numNegExps = 0;614  Fraction sumNegExps(0, 1);615  for (const auto &den : dens) {616    if (den < 0) {617      numNegExps += 1;618      sumNegExps += den;619    }620  }621 622  // If we have (1 - t^-c) in the denominator, for positive c,623  // multiply and divide by t^c.624  // We convert all negative-exponent terms at once; therefore625  // we multiply and divide by t^sumNegExps.626  // Then we get627  // -(1 - t^c) in the denominator,628  // increase the numerator by c, and629  // flip the sign of the function.630  if (numNegExps % 2 == 1)631    sign = -sign;632  num = num - QuasiPolynomial(num.getNumInputs(), sumNegExps);633}634 635/// Compute the binomial coefficients nCi for 0 ≤ i ≤ r,636/// where n is a QuasiPolynomial.637static std::vector<QuasiPolynomial>638getBinomialCoefficients(const QuasiPolynomial &n, unsigned r) {639  unsigned numParams = n.getNumInputs();640  std::vector<QuasiPolynomial> coefficients;641  coefficients.reserve(r + 1);642  coefficients.emplace_back(numParams, 1);643  for (unsigned j = 1; j <= r; ++j)644    // We use the recursive formula for binomial coefficients here and below.645    coefficients.emplace_back(646        (coefficients[j - 1] * (n - QuasiPolynomial(numParams, j - 1)) /647         Fraction(j, 1))648            .simplify());649  return coefficients;650}651 652/// Compute the binomial coefficients nCi for 0 ≤ i ≤ r,653/// where n is a QuasiPolynomial.654static std::vector<Fraction> getBinomialCoefficients(const Fraction &n,655                                                     const Fraction &r) {656  std::vector<Fraction> coefficients;657  coefficients.reserve((int64_t)floor(r));658  coefficients.emplace_back(1);659  for (unsigned j = 1; j <= r; ++j)660    coefficients.emplace_back(coefficients[j - 1] * (n - (j - 1)) / (j));661  return coefficients;662}663 664/// We have a generating function of the form665/// f_p(x) = \sum_i sign_i * (x^n_i(p)) / (\prod_j (1 - x^d_{ij})666///667/// where sign_i is ±1,668/// n_i \in Q^p -> Q^d is the sum of the vectors d_{ij}, weighted by the669/// floors of d affine functions on p parameters.670/// d_{ij} \in Q^d are vectors.671///672/// We need to find the number of terms of the form x^t in the expansion of673/// this function.674/// However, direct substitution (x = (1, ..., 1)) causes the denominator675/// to become zero.676///677/// We therefore use the following procedure instead:678/// 1. Substitute x_i = (s+1)^μ_i for some vector μ. This makes the generating679/// function a function of a scalar s.680/// 2. Write each term in this function as P(s)/Q(s), where P and Q are681/// polynomials. P has coefficients as quasipolynomials in d parameters, while682/// Q has coefficients as scalars.683/// 3. Find the constant term in the expansion of each term P(s)/Q(s). This is684/// equivalent to substituting s = 0.685///686/// Verdoolaege, Sven, et al. "Counting integer points in parametric687/// polytopes using Barvinok's rational functions." Algorithmica 48 (2007):688/// 37-66.689QuasiPolynomial690mlir::presburger::detail::computeNumTerms(const GeneratingFunction &gf) {691  // Step (1) We need to find a μ such that we can substitute x_i =692  // (s+1)^μ_i. After this substitution, the exponent of (s+1) in the693  // denominator is (μ_i • d_{ij}) in each term. Clearly, this cannot become694  // zero. Hence we find a vector μ that is not orthogonal to any of the695  // d_{ij} and substitute x accordingly.696  std::vector<Point> allDenominators;697  for (ArrayRef<Point> den : gf.getDenominators())698    llvm::append_range(allDenominators, den);699  Point mu = getNonOrthogonalVector(allDenominators);700 701  unsigned numParams = gf.getNumParams();702  const std::vector<std::vector<Point>> &ds = gf.getDenominators();703  QuasiPolynomial totalTerm(numParams, 0);704  for (unsigned i = 0, e = ds.size(); i < e; ++i) {705    int sign = gf.getSigns()[i];706 707    // Compute the new exponents of (s+1) for the numerator and the708    // denominator after substituting μ.709    auto [numExp, dens] =710        substituteMuInTerm(numParams, gf.getNumerators()[i], ds[i], mu);711    // Now the numerator is (s+1)^numExp712    // and the denominator is \prod_j (1 - (s+1)^dens[j]).713 714    // Step (2) We need to express the terms in the function as quotients of715    // polynomials. Each term is now of the form716    // sign_i * (s+1)^numExp / (\prod_j (1 - (s+1)^dens[j]))717    // For the i'th term, we first normalize the denominator to have only718    // positive exponents. We convert all the dens[j] to their719    // absolute values and change the sign and exponent in the numerator.720    normalizeDenominatorExponents(sign, numExp, dens);721 722    // Then, using the formula for geometric series, we replace each (1 -723    // (s+1)^(dens[j])) with724    // (-s)(\sum_{0 ≤ k < dens[j]} (s+1)^k).725    for (auto &j : dens)726      j = abs(j) - 1;727    // Note that at this point, the semantics of `dens[j]` changes to mean728    // a term (\sum_{0 ≤ k ≤ dens[j]} (s+1)^k). The denominator is, as before,729    // a product of these terms.730 731    // Since the -s are taken out, the sign changes if there is an odd number732    // of such terms.733    unsigned r = dens.size();734    if (dens.size() % 2 == 1)735      sign = -sign;736 737    // Thus the term overall now has the form738    // sign'_i * (s+1)^numExp /739    // (s^r * \prod_j (\sum_{0 ≤ k < dens[j]} (s+1)^k)).740    // This means that741    // the numerator is a polynomial in s, with coefficients as742    // quasipolynomials (given by binomial coefficients), and the denominator743    // is a polynomial in s, with integral coefficients (given by taking the744    // convolution over all j).745 746    // Step (3) We need to find the constant term in the expansion of each747    // term. Since each term has s^r as a factor in the denominator, we avoid748    // substituting s = 0 directly; instead, we find the coefficient of s^r in749    // sign'_i * (s+1)^numExp / (\prod_j (\sum_k (s+1)^k)),750    // Letting P(s) = (s+1)^numExp and Q(s) = \prod_j (...),751    // we need to find the coefficient of s^r in P(s)/Q(s),752    // for which we use the `getCoefficientInRationalFunction()` function.753 754    // First, we compute the coefficients of P(s), which are binomial755    // coefficients.756    // We only need the first r+1 of these, as higher-order terms do not757    // contribute to the coefficient of s^r.758    std::vector<QuasiPolynomial> numeratorCoefficients =759        getBinomialCoefficients(numExp, r);760 761    // Then we compute the coefficients of each individual term in Q(s),762    // which are (dens[i]+1) C (k+1) for 0 ≤ k ≤ dens[i].763    std::vector<std::vector<Fraction>> eachTermDenCoefficients;764    std::vector<Fraction> singleTermDenCoefficients;765    eachTermDenCoefficients.reserve(r);766    for (const Fraction &den : dens) {767      singleTermDenCoefficients = getBinomialCoefficients(den + 1, den + 1);768      eachTermDenCoefficients.emplace_back(769          ArrayRef<Fraction>(singleTermDenCoefficients).drop_front());770    }771 772    // Now we find the coefficients in Q(s) itself773    // by taking the convolution of the coefficients774    // of all the terms.775    std::vector<Fraction> denominatorCoefficients;776    denominatorCoefficients = eachTermDenCoefficients[0];777    for (unsigned j = 1, e = eachTermDenCoefficients.size(); j < e; ++j)778      denominatorCoefficients = multiplyPolynomials(denominatorCoefficients,779                                                    eachTermDenCoefficients[j]);780 781    totalTerm =782        totalTerm + getCoefficientInRationalFunction(r, numeratorCoefficients,783                                                     denominatorCoefficients) *784                        QuasiPolynomial(numParams, sign);785  }786 787  return totalTerm.simplify();788}789