| blob | f284d25f669165b2c1ec120e3a4800f9e34b7786 |
5 /*
6 * The implementation of parametric integer linear programming in this file
7 * was inspired by the paper "Parametric Integer Programming" and the
8 * report "Solving systems of affine (in)equalities" by Paul Feautrier
9 * (and others).
10 *
11 * The strategy used for obtaining a feasible solution is different
12 * from the one used in isl_tab.c. In particular, in isl_tab.c,
13 * upon finding a constraint that is not yet satisfied, we pivot
14 * in a row that increases the constant term of row holding the
15 * constraint, making sure the sample solution remains feasible
16 * for all the constraints it already satisfied.
17 * Here, we always pivot in the row holding the constraint,
18 * choosing a column that induces the lexicographically smallest
19 * increment to the sample solution.
20 *
21 * By starting out from a sample value that is lexicographically
22 * smaller than any integer point in the problem space, the first
23 * feasible integer sample point we find will also be the lexicographically
24 * smallest. If all variables can be assumed to be non-negative,
25 * then the initial sample value may be chosen equal to zero.
26 * However, we will not make this assumption. Instead, we apply
27 * the "big parameter" trick. Any variable x is then not directly
28 * used in the tableau, but instead it its represented by another
29 * variable x' = M + x, where M is an arbitrarily large (positive)
30 * value. x' is therefore always non-negative, whatever the value of x.
31 * Taking as initial smaple value x' = 0 corresponds to x = -M,
32 * which is always smaller than any possible value of x.
33 *
34 * We use the big parameter trick both in the main tableau and
35 * the context tableau, each of course having its own big parameter.
36 * Before doing any real work, we check if all the parameters
37 * happen to be non-negative. If so, we drop the column corresponding
38 * to M from the initial context tableau.
39 */
41 /* isl_sol is an interface for constructing a solution to
42 * a parametric integer linear programming problem.
43 * Every time the algorithm reaches a state where a solution
44 * can be read off from the tableau (including cases where the tableau
45 * is empty), the function "add" is called on the isl_sol passed
46 * to find_solutions_main.
47 *
48 * The context tableau is owned by isl_sol and is updated incrementally.
49 *
50 * There is currently only one implementation of this interface,
51 * isl_sol_map, which simply collects the solutions in an isl_map
52 * and (optionally) the parts of the context where there is no solution
53 * in an isl_set.
54 */
59 };
62 {
66 }
73 };
76 {
81 }
84 {
86 }
89 {
102 error:
105 }
107 /* Add the solution identified by the tableau and the context tableau.
108 *
109 * The layout of the variables is as follows.
110 * tab->n_var is equal to the total number of variables in the input
111 * map (including divs that were copied from the context)
112 * + the number of extra divs constructed
113 * Of these, the first tab->n_param and the last tab->n_div variables
114 * correspond to the variables in the context, i.e.,
115 tab->n_param + tab->n_div = context_tab->n_var
116 * tab->n_param is equal to the number of parameters and input
117 * dimensions in the input map
118 * tab->n_div is equal to the number of divs in the context
119 *
120 * If there is no solution, then the basic set corresponding to the
121 * context tableau is added to the set "empty".
122 *
123 * Otherwise, a basic map is constructed with the same parameters
124 * and divs as the context, the dimensions of the context as input
125 * dimensions and a number of output dimensions that is equal to
126 * the number of output dimensions in the input map.
127 * The divs in the input map (if any) that do not correspond to any
128 * div in the context do not appear in the solution.
129 * The algorithm will make sure that they have an integer value,
130 * but these values themselves are of no interest.
131 *
132 * The constraints and divs of the context are simply copied
133 * fron context_tab->bset.
134 * To extract the value of the output variables, it should be noted
135 * that we always use a big parameter M and so the variable stored
136 * in the tableau is not an output variable x itself, but
137 * x' = M + x (in case of minimization)
138 * or
139 * x' = M - x (in case of maximization)
140 * If x' appears in a column, then its optimal value is zero,
141 * which means that the optimal value of x is an unbounded number
142 * (-M for minimization and M for maximization).
143 * We currently assume that the output dimensions in the original map
144 * are bounded, so this cannot occur.
145 * Similarly, when x' appears in a row, then the coefficient of M in that
146 * row is necessarily 1.
147 * If the row represents
148 * d x' = c + d M + e(y)
149 * then, in case of minimization, an equality
150 * c + e(y) - d x' = 0
151 * is added, and in case of maximization,
152 * c + e(y) + d x' = 0
153 */
156 {
197 }
206 }
216 }
223 /* no unbounded */
228 else
233 /* no unbounded */
247 }
255 }
259 else
262 }
263 }
272 error:
276 }
280 {
282 }
287 {
298 error:
301 }
305 {
316 error:
319 }
322 /* Store the "parametric constant" of row "row" of tableau "tab" in "line",
323 * i.e., the constant term and the coefficients of all variables that
324 * appear in the context tableau.
325 * Note that the coefficient of the big parameter M is NOT copied.
326 * The context tableau may not have a big parameter and even when it
327 * does, it is a different big parameter.
328 */
330 {
341 }
342 }
350 }
351 }
352 }
354 /* Check if rows "row1" and "row2" have identical "parametric constants",
355 * as explained above.
356 * In this case, we also insist that the coefficients of the big parameter
357 * be the same as the values of the constants will only be the same
358 * if these coefficients are also the same.
359 */
361 {
383 }
385 }
387 /* Return an inequality that expresses that the "parametric constant"
388 * should be non-negative.
389 * This function is only called when the coefficient of the big parameter
390 * is equal to zero.
391 */
393 {
405 }
407 /* Return a integer division for use in a parametric cut based on the given row.
408 * In particular, let the parametric constant of the row be
409 *
410 * \sum_i a_i y_i
411 *
412 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
413 * The div returned is equal to
414 *
415 * floor(\sum_i {-a_i} y_i) = floor((\sum_i (-a_i mod d) y_i)/d)
416 */
418 {
432 }
434 /* Return a integer division for use in transferring an integrality constraint
435 * to the context.
436 * In particular, let the parametric constant of the row be
437 *
438 * \sum_i a_i y_i
439 *
440 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
441 * The the returned div is equal to
442 *
443 * floor(\sum_i {a_i} y_i) = floor((\sum_i (a_i mod d) y_i)/d)
444 */
446 {
459 }
461 /* Construct and return an inequality that expresses an upper bound
462 * on the given div.
463 * In particular, if the div is given by
464 *
465 * d = floor(e/m)
466 *
467 * then the inequality expresses
468 *
469 * m d <= e
470 */
472 {
487 }
489 /* Given a row in the tableau and a div that was created
490 * using get_row_split_div and that been constrained to equality, i.e.,
491 *
492 * d = floor(\sum_i {a_i} y_i) = \sum_i {a_i} y_i
493 *
494 * replace the expression "\sum_i {a_i} y_i" in the row by d,
495 * i.e., we subtract "\sum_i {a_i} y_i" and add 1 d.
496 * The coefficients of the non-parameters in the tableau have been
497 * verified to be integral. We can therefore simply replace coefficient b
498 * by floor(b). For the coefficients of the parameters we have
499 * floor(a_i) = a_i - {a_i}, while for the other coefficients, we have
500 * floor(b) = b.
501 */
503 {
520 error:
523 }
525 /* Check if the (parametric) constant of the given row is obviously
526 * negative, meaning that we don't need to consult the context tableau.
527 * If there is a big parameter and its coefficient is non-zero,
528 * then this coefficient determines the outcome.
529 * Otherwise, we check whether the constant is negative and
530 * all non-zero coefficients of parameters are negative and
531 * belong to non-negative parameters.
532 */
534 {
544 }
549 /* Eliminated parameter */
559 }
570 }
572 }
574 /* Check if the (parametric) constant of the given row is obviously
575 * non-negative, meaning that we don't need to consult the context tableau.
576 * If there is a big parameter and its coefficient is non-zero,
577 * then this coefficient determines the outcome.
578 * Otherwise, we check whether the constant is non-negative and
579 * all non-zero coefficients of parameters are positive and
580 * belong to non-negative parameters.
581 */
583 {
593 }
598 /* Eliminated parameter */
608 }
619 }
621 }
623 /* Given a row r and two columns, return the column that would
624 * lead to the lexicographically smallest increment in the sample
625 * solution when leaving the basis in favor of the row.
626 * Pivoting with column c will increment the sample value by a non-negative
627 * constant times a_{V,c}/a_{r,c}, with a_{V,c} the elements of column c
628 * corresponding to the non-parametric variables.
629 * If variable v appears in a column c_v, the a_{v,c} = 1 iff c = c_v,
630 * with all other entries in this virtual row equal to zero.
631 * If variable v appears in a row, then a_{v,c} is the element in column c
632 * of that row.
633 *
634 * Let v be the first variable with a_{v,c1}/a_{r,c1} != a_{v,c2}/a_{r,c2}.
635 * Then if a_{v,c1}/a_{r,c1} < a_{v,c2}/a_{r,c2}, i.e.,
636 * a_{v,c2} a_{r,c1} - a_{v,c1} a_{r,c2} > 0, c1 results in the minimal
637 * increment. Otherwise, it's c2.
638 */
641 {
657 }
678 }
680 }
682 /* Given a row in the tableau, find and return the column that would
683 * result in the lexicographically smallest, but positive, increment
684 * in the sample point.
685 * If there is no such column, then return tab->n_col.
686 * If anything goes wrong, return -1.
687 */
689 {
693 isl_int tmp;
710 else
713 }
717 error:
720 }
722 /* Return the first known violated constraint, i.e., a non-negative
723 * contraint that currently has an either obviously negative value
724 * or a previously determined to be negative value.
725 *
726 * If any constraint has a negative coefficient for the big parameter,
727 * if any, then we return one of these first.
728 */
730 {
739 }
752 }
754 }
756 /* Resolve all known or obviously violated constraints through pivoting.
757 * In particular, as long as we can find any violated constraint, we
758 * look for a pivoting column that would result in the lexicographicallly
759 * smallest increment in the sample point. If there is no such column
760 * then the tableau is infeasible.
761 */
763 {
777 }
779 error:
782 }
784 /* Given a row that represents an equality, look for an appropriate
785 * pivoting column.
786 * In particular, if there are any non-zero coefficients among
787 * the non-parameter variables, then we take the last of these
788 * variables. Eliminating this variable in terms of the other
789 * variables and/or parameters does not influence the property
790 * that all column in the initial tableau are lexicographically
791 * positive. The row corresponding to the eliminated variable
792 * will only have non-zero entries below the diagonal of the
793 * initial tableau. That is, we transform
794 *
795 * I I
796 * 1 into a
797 * I I
798 *
799 * If there is no such non-parameter variable, then we are dealing with
800 * pure parameter equality and we pick any parameter with coefficient 1 or -1
801 * for elimination. This will ensure that the eliminated parameter
802 * always has an integer value whenever all the other parameters are integral.
803 * If there is no such parameter then we return -1.
804 */
806 {
819 }
825 }
827 }
829 /* Add an equality that is known to be valid to the tableau.
830 * We first check if we can eliminate a variable or a parameter.
831 * If not, we add the equality as two inequalities.
832 * In this case, the equality was a pure parameter equality and there
833 * is no need to resolve any constraint violations.
834 */
836 {
863 }
866 error:
869 }
871 /* Check if the given row is a pure constant.
872 */
874 {
879 }
881 /* Add an equality that may or may not be valid to the tableau.
882 * If the resulting row is a pure constant, then it must be zero.
883 * Otherwise, the resulting tableau is empty.
884 *
885 * If the row is not a pure constant, then we add two inequalities,
886 * each time checking that they can be satisfied.
887 * In the end we try to use one of the two constraints to eliminate
888 * a column.
889 */
891 {
903 }
916 }
947 }
948 }
951 error:
954 }
956 /* Add an inequality to the tableau, resolving violations using
957 * restore_lexmin.
958 */
960 {
971 }
980 }
987 error:
990 }
992 /* Check if the coefficients of the parameters are all integral.
993 */
995 {
1001 /* Eliminated parameter */
1008 }
1016 }
1018 }
1020 /* Check if the coefficients of the non-parameter variables are all integral.
1021 */
1023 {
1035 }
1037 }
1039 /* Check if the constant term is integral.
1040 */
1042 {
1045 }
1047 #define I_CST 1 << 0
1048 #define I_PAR 1 << 1
1049 #define I_VAR 1 << 2
1051 /* Check for first (non-parameter) variable that is non-integer and
1052 * therefore requires a cut.
1053 * For parametric tableaus, there are three parts in a row,
1054 * the constant, the coefficients of the parameters and the rest.
1055 * For each part, we check whether the coefficients in that part
1056 * are all integral and if so, set the corresponding flag in *f.
1057 * If the constant and the parameter part are integral, then the
1058 * current sample value is integral and no cut is required
1059 * (irrespective of whether the variable part is integral).
1060 */
1062 {
1081 }
1083 }
1085 /* Add a (non-parametric) cut to cut away the non-integral sample
1086 * value of the given row.
1087 *
1088 * If the row is given by
1089 *
1090 * m r = f + \sum_i a_i y_i
1091 *
1092 * then the cut is
1093 *
1094 * c = - {-f/m} + \sum_i {a_i/m} y_i >= 0
1095 *
1096 * The big parameter, if any, is ignored, since it is assumed to be big
1097 * enough to be divisible by any integer.
1098 * If the tableau is actually a parametric tableau, then this function
1099 * is only called when all coefficients of the parameters are integral.
1100 * The cut therefore has zero coefficients for the parameters.
1101 *
1102 * The current value is known to be negative, so row_sign, if it
1103 * exists, is set accordingly.
1104 *
1105 * Return the row of the cut or -1.
1106 */
1108 {
1137 }
1139 /* Given a non-parametric tableau, add cuts until an integer
1140 * sample point is obtained or until the tableau is determined
1141 * to be integer infeasible.
1142 * As long as there is any non-integer value in the sample point,
1143 * we add an appropriate cut, if possible and resolve the violated
1144 * cut constraint using restore_lexmin.
1145 * If one of the corresponding rows is equal to an integral
1146 * combination of variables/constraints plus a non-integral constant,
1147 * then there is no way to obtain an integer point an we return
1148 * a tableau that is marked empty.
1149 */
1151 {
1169 }
1171 error:
1174 }
1177 {
1184 }
1186 /* Check whether all the currently active samples also satisfy the inequality
1187 * "ineq" (treated as an equality if eq is set).
1188 * Remove those samples that do not.
1189 */
1191 {
1193 isl_int v;
1213 }
1217 }
1219 /* Check whether the sample value of the tableau is finite,
1220 * i.e., either the tableau does not use a big parameter, or
1221 * all values of the variables are equal to the big parameter plus
1222 * some constant. This constant is the actual sample value.
1223 */
1225 {
1238 }
1240 }
1242 /* Check if the context tableau of sol has any integer points.
1243 * Returns -1 if an error occurred.
1244 * If an integer point can be found and if moreover it is finite,
1245 * then it is added to the list of sample values.
1246 *
1247 * This function is only called when none of the currently active sample
1248 * values satisfies the most recently added constraint.
1249 */
1251 {
1282 }
1289 error:
1293 }
1295 /* First check if any of the currently active sample values satisfies
1296 * the inequality "ineq" (an equality if eq is set).
1297 * If not, continue with check_integer_feasible.
1298 */
1301 {
1303 isl_int v;
1322 }
1329 }
1331 /* For a div d = floor(f/m), add the constraints
1332 *
1333 * f - m d >= 0
1334 * -(f-(m-1)) + m d >= 0
1335 *
1336 * Note that the second constraint is the negation of
1337 *
1338 * f - m d >= m
1339 */
1341 {
1368 error:
1371 }
1373 /* Add a div specified by "div" to both the main tableau and
1374 * the context tableau. In case of the main tableau, we only
1375 * need to add an extra div. In the context tableau, we also
1376 * need to express the meaning of the div.
1377 * Return the index of the div or -1 if anything went wrong.
1378 */
1381 {
1405 }
1429 error:
1433 }
1436 {
1446 }
1448 }
1450 /* Return the index of a div that corresponds to "div".
1451 * We first check if we already have such a div and if not, we create one.
1452 */
1455 {
1463 }
1465 /* Add a parametric cut to cut away the non-integral sample value
1466 * of the give row.
1467 * Let a_i be the coefficients of the constant term and the parameters
1468 * and let b_i be the coefficients of the variables or constraints
1469 * in basis of the tableau.
1470 * Let q be the div q = floor(\sum_i {-a_i} y_i).
1471 *
1472 * The cut is expressed as
1473 *
1474 * c = \sum_i -{-a_i} y_i + \sum_i {b_i} x_i + q >= 0
1475 *
1476 * If q did not already exist in the context tableau, then it is added first.
1477 * If q is in a column of the main tableau then the "+ q" can be accomplished
1478 * by setting the corresponding entry to the denominator of the constraint.
1479 * If q happens to be in a row of the main tableau, then the corresponding
1480 * row needs to be added instead (taking care of the denominators).
1481 * Note that this is very unlikely, but perhaps not entirely impossible.
1482 *
1483 * The current value of the cut is known to be negative (or at least
1484 * non-positive), so row_sign is set accordingly.
1485 *
1486 * Return the row of the cut or -1.
1487 */
1490 {
1534 }
1543 }
1551 }
1553 isl_int gcd;
1567 }
1577 error:
1581 }
1583 /* Construct a tableau for bmap that can be used for computing
1584 * the lexicographic minimum (or maximum) of bmap.
1585 * If not NULL, then dom is the domain where the minimum
1586 * should be computed. In this case, we set up a parametric
1587 * tableau with row signs (initialized to "unknown").
1588 * If M is set, then the tableau will use a big parameter.
1589 * If max is set, then a maximum should be computed instead of a minimum.
1590 * This means that for each variable x, the tableau will contain the variable
1591 * x' = M - x, rather than x' = M + x. This in turn means that the coefficient
1592 * of the variables in all constraints are negated prior to adding them
1593 * to the tableau.
1594 */
1597 {
1614 }
1621 }
1634 }
1647 }
1649 error:
1652 }
1655 {
1671 error:
1674 }
1676 /* Construct an isl_sol_map structure for accumulating the solution.
1677 * If track_empty is set, then we also keep track of the parts
1678 * of the context where there is no solution.
1679 * If max is set, then we are solving a maximization, rather than
1680 * a minimization problem, which means that the variables in the
1681 * tableau have value "M - x" rather than "M + x".
1682 */
1685 {
1698 ISL_MAP_DISJOINT);
1714 }
1718 error:
1722 }
1724 /* For each variable in the context tableau, check if the variable can
1725 * only attain non-negative values. If so, mark the parameter as non-negative
1726 * in the main tableau. This allows for a more direct identification of some
1727 * cases of violated constraints.
1728 */
1731 {
1765 n++;
1766 }
1770 }
1778 }
1782 error:
1786 }
1788 /* Check whether all coefficients of (non-parameter) variables
1789 * are non-positive, meaning that no pivots can be performed on the row.
1790 */
1792 {
1804 }
1807 }
1809 /* Check whether the inequality represented by vec is strict over the integers,
1810 * i.e., there are no integer values satisfying the constraint with
1811 * equality. This happens if the gcd of the coefficients is not a divisor
1812 * of the constant term. If so, scale the constraint down by the gcd
1813 * of the coefficients.
1814 */
1816 {
1817 isl_int gcd;
1826 }
1830 }
1832 /* Determine the sign of the given row of the main tableau.
1833 * The result is one of
1834 * isl_tab_row_pos: always non-negative; no pivot needed
1835 * isl_tab_row_neg: always non-positive; pivot
1836 * isl_tab_row_any: can be both positive and negative; split
1837 *
1838 * We first handle some simple cases
1839 * - the row sign may be known already
1840 * - the row may be obviously non-negative
1841 * - the parametric constant may be equal to that of another row
1842 * for which we know the sign. This sign will be either "pos" or
1843 * "any". If it had been "neg" then we would have pivoted before.
1844 *
1845 * If none of these cases hold, we check the value of the row for each
1846 * of the currently active samples. Based on the signs of these values
1847 * we make an initial determination of the sign of the row.
1848 *
1849 * all zero -> unk(nown)
1850 * all non-negative -> pos
1851 * all non-positive -> neg
1852 * both negative and positive -> all
1853 *
1854 * If we end up with "all", we are done.
1855 * Otherwise, we perform a check for positive and/or negative
1856 * values as follows.
1857 *
1858 * samples neg unk pos
1859 * <0 ? Y N Y N
1860 * pos any pos
1861 * >0 ? Y N Y N
1862 * any neg any neg
1863 *
1864 * There is no special sign for "zero", because we can usually treat zero
1865 * as either non-negative or non-positive, whatever works out best.
1866 * However, if the row is "critical", meaning that pivoting is impossible
1867 * then we don't want to limp zero with the non-positive case, because
1868 * then we we would lose the solution for those values of the parameters
1869 * where the value of the row is zero. Instead, we treat 0 as non-negative
1870 * ensuring a split if the row can attain both zero and negative values.
1871 * The same happens when the original constraint was one that could not
1872 * be satisfied with equality by any integer values of the parameters.
1873 * In this case, we normalize the constraint, but then a value of zero
1874 * for the normalized constraint is actually a positive value for the
1875 * original constraint, so again we need to treat zero as non-negative.
1876 * In both these cases, we have the following decision tree instead:
1877 *
1878 * all non-negative -> pos
1879 * all negative -> neg
1880 * both negative and non-negative -> all
1881 *
1882 * samples neg pos
1883 * <0 ? Y N
1884 * any pos
1885 * >=0 ? Y N
1886 * any neg
1887 */
1889 {
1900 isl_int tmp;
1912 }
1935 }
1941 }
1944 }
1953 }
1956 /* test for negative values */
1969 else
1978 }
1979 }
1982 /* test for positive values */
1997 }
2001 error:
2004 }
2008 /* Find solutions for values of the parameters that satisfy the given
2009 * inequality.
2010 *
2011 * We currently take a snapshot of the context tableau that is reset
2012 * when we return from this function, while we make a copy of the main
2013 * tableau, leaving the original main tableau untouched.
2014 * These are fairly arbitrary choices. Making a copy also of the context
2015 * tableau would obviate the need to undo any changes made to it later,
2016 * while taking a snapshot of the main tableau could reduce memory usage.
2017 * If we were to switch to taking a snapshot of the main tableau,
2018 * we would have to keep in mind that we need to save the row signs
2019 * and that we need to do this before saving the current basis
2020 * such that the basis has been restore before we restore the row signs.
2021 */
2024 {
2043 error:
2047 }
2049 /* Record the absence of solutions for those values of the parameters
2050 * that do not satisfy the given inequality with equality.
2051 */
2054 {
2080 error:
2083 }
2085 /* Given a main tableau where more than one row requires a split,
2086 * determine and return the "best" row to split on.
2087 *
2088 * Given two rows in the main tableau, if the inequality corresponding
2089 * to the first row is redundant with respect to that of the second row
2090 * in the current tableau, then it is better to split on the second row,
2091 * since in the positive part, both row will be positive.
2092 * (In the negative part a pivot will have to be performed and just about
2093 * anything can happen to the sign of the other row.)
2094 *
2095 * As a simple heuristic, we therefore select the row that makes the most
2096 * of the other rows redundant.
2097 *
2098 * Perhaps it would also be useful to look at the number of constraints
2099 * that conflict with any given constraint.
2100 */
2102 {
2152 r++;
2155 }
2159 }
2162 }
2168 }
2170 /* Compute the lexicographic minimum of the set represented by the main
2171 * tableau "tab" within the context "sol->context_tab".
2172 * On entry the sample value of the main tableau is lexicographically
2173 * less than or equal to this lexicographic minimum.
2174 * Pivots are performed until a feasible point is found, which is then
2175 * necessarily equal to the minimum, or until the tableau is found to
2176 * be infeasible. Some pivots may need to be performed for only some
2177 * feasible values of the context tableau. If so, the context tableau
2178 * is split into a part where the pivot is needed and a part where it is not.
2179 *
2180 * Whenever we enter the main loop, the main tableau is such that no
2181 * "obvious" pivots need to be performed on it, where "obvious" means
2182 * that the given row can be seen to be negative without looking at
2183 * the context tableau. In particular, for non-parametric problems,
2184 * no pivots need to be performed on the main tableau.
2185 * The caller of find_solutions is responsible for making this property
2186 * hold prior to the first iteration of the loop, while restore_lexmin
2187 * is called before every other iteration.
2188 *
2189 * Inside the main loop, we first examine the signs of the rows of
2190 * the main tableau within the context of the context tableau.
2191 * If we find a row that is always non-positive for all values of
2192 * the parameters satisfying the context tableau and negative for at
2193 * least one value of the parameters, we perform the appropriate pivot
2194 * and start over. An exception is the case where no pivot can be
2195 * performed on the row. In this case, we require that the sign of
2196 * the row is negative for all values of the parameters (rather than just
2197 * non-positive). This special case is handled inside row_sign, which
2198 * will say that the row can have any sign if it determines that it can
2199 * attain both negative and zero values.
2200 *
2201 * If we can't find a row that always requires a pivot, but we can find
2202 * one or more rows that require a pivot for some values of the parameters
2203 * (i.e., the row can attain both positive and negative signs), then we split
2204 * the context tableau into two parts, one where we force the sign to be
2205 * non-negative and one where we force is to be negative.
2206 * The non-negative part is handled by a recursive call (through find_in_pos).
2207 * Upon returning from this call, we continue with the negative part and
2208 * perform the required pivot.
2209 *
2210 * If no such rows can be found, all rows are non-negative and we have
2211 * found a (rational) feasible point. If we only wanted a rational point
2212 * then we are done.
2213 * Otherwise, we check if all values of the sample point of the tableau
2214 * are integral for the variables. If so, we have found the minimal
2215 * integral point and we are done.
2216 * If the sample point is not integral, then we need to make a distinction
2217 * based on whether the constant term is non-integral or the coefficients
2218 * of the parameters. Furthermore, in order to decide how to handle
2219 * the non-integrality, we also need to know whether the coefficients
2220 * of the other columns in the tableau are integral. This leads
2221 * to the following table. The first two rows do not correspond
2222 * to a non-integral sample point and are only mentioned for completeness.
2223 *
2224 * constant parameters other
2225 *
2226 * int int int |
2227 * int int rat | -> no problem
2228 *
2229 * rat int int -> fail
2230 *
2231 * rat int rat -> cut
2232 *
2233 * int rat rat |
2234 * rat rat rat | -> parametric cut
2235 *
2236 * int rat int |
2237 * rat rat int | -> split context
2238 *
2239 * If the parametric constant is completely integral, then there is nothing
2240 * to be done. If the constant term is non-integral, but all the other
2241 * coefficient are integral, then there is nothing that can be done
2242 * and the tableau has no integral solution.
2243 * If, on the other hand, one or more of the other columns have rational
2244 * coeffcients, but the parameter coefficients are all integral, then
2245 * we can perform a regular (non-parametric) cut.
2246 * Finally, if there is any parameter coefficient that is non-integral,
2247 * then we need to involve the context tableau. There are two cases here.
2248 * If at least one other column has a rational coefficient, then we
2249 * can perform a parametric cut in the main tableau by adding a new
2250 * integer division in the context tableau.
2251 * If all other columns have integral coefficients, then we need to
2252 * enforce that the rational combination of parameters (c + \sum a_i y_i)/m
2253 * is always integral. We do this by introducing an integer division
2254 * q = floor((c + \sum a_i y_i)/m) and stipulating that its argument should
2255 * always be integral in the context tableau, i.e., m q = c + \sum a_i y_i.
2256 * Since q is expressed in the tableau as
2257 * c + \sum a_i y_i - m q >= 0
2258 * -c - \sum a_i y_i + m q + m - 1 >= 0
2259 * it is sufficient to add the inequality
2260 * -c - \sum a_i y_i + m q >= 0
2261 * In the part of the context where this inequality does not hold, the
2262 * main tableau is marked as being empty.
2263 */
2265 {
2293 n_split++;
2298 }
2316 }
2329 }
2339 }
2367 }
2368 done:
2372 error:
2376 }
2378 /* Compute the lexicographic minimum of the set represented by the main
2379 * tableau "tab" within the context "sol->context_tab".
2380 *
2381 * As a preprocessing step, we first transfer all the purely parametric
2382 * equalities from the main tableau to the context tableau, i.e.,
2383 * parameters that have been pivoted to a row.
2384 * These equalities are ignored by the main algorithm, because the
2385 * corresponding rows may not be marked as being non-negative.
2386 * In parts of the context where the added equality does not hold,
2387 * the main tableau is marked as being empty.
2388 */
2391 {
2405 else
2437 }
2440 error:
2444 }
2448 {
2450 }
2452 /* Check if integer division "div" of "dom" also occurs in "bmap".
2453 * If so, return its position within the divs.
2454 * If not, return -1.
2455 */
2458 {
2476 }
2478 }
2480 /* The correspondence between the variables in the main tableau,
2481 * the context tableau, and the input map and domain is as follows.
2482 * The first n_param and the last n_div variables of the main tableau
2483 * form the variables of the context tableau.
2484 * In the basic map, these n_param variables correspond to the
2485 * parameters and the input dimensions. In the domain, they correspond
2486 * to the parameters and the set dimensions.
2487 * The n_div variables correspond to the integer divisions in the domain.
2488 * To ensure that everything lines up, we may need to copy some of the
2489 * integer divisions of the domain to the map. These have to be placed
2490 * in the same order as those in the context and they have to be placed
2491 * after any other integer divisions that the map may have.
2492 * This function performs the required reordering.
2493 */
2496 {
2503 common++;
2510 }
2518 }
2521 }
2523 error:
2526 }
2528 /* Compute the lexicographic minimum (or maximum if "max" is set)
2529 * of "bmap" over the domain "dom" and return the result as a map.
2530 * If "empty" is not NULL, then *empty is assigned a set that
2531 * contains those parts of the domain where there is no solution.
2532 * If "bmap" is marked as rational (ISL_BASIC_MAP_RATIONAL),
2533 * then we compute the rational optimum. Otherwise, we compute
2534 * the integral optimum.
2535 *
2536 * We perform some preprocessing. As the PILP solver does not
2537 * handle implicit equalities very well, we first make sure all
2538 * the equalities are explicitly available.
2539 * We also make sure the divs in the domain are properly order,
2540 * because they will be added one by one in the given order
2541 * during the construction of the solution map.
2542 */
2546 {
2564 }
2581 }
2589 error:
2593 }