| blob | 4bc8782475a5eef6aa803a23ebc0a346b1a0ff88 |
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 are currently two implementations 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, and
54 * isl_sol_for, which calls a user-defined function for each part of
55 * the solution.
56 */
61 };
64 {
68 }
75 };
78 {
83 }
86 {
88 }
91 {
104 error:
107 }
109 /* Add the solution identified by the tableau and the context tableau.
110 *
111 * The layout of the variables is as follows.
112 * tab->n_var is equal to the total number of variables in the input
113 * map (including divs that were copied from the context)
114 * + the number of extra divs constructed
115 * Of these, the first tab->n_param and the last tab->n_div variables
116 * correspond to the variables in the context, i.e.,
117 * tab->n_param + tab->n_div = context_tab->n_var
118 * tab->n_param is equal to the number of parameters and input
119 * dimensions in the input map
120 * tab->n_div is equal to the number of divs in the context
121 *
122 * If there is no solution, then the basic set corresponding to the
123 * context tableau is added to the set "empty".
124 *
125 * Otherwise, a basic map is constructed with the same parameters
126 * and divs as the context, the dimensions of the context as input
127 * dimensions and a number of output dimensions that is equal to
128 * the number of output dimensions in the input map.
129 * The divs in the input map (if any) that do not correspond to any
130 * div in the context do not appear in the solution.
131 * The algorithm will make sure that they have an integer value,
132 * but these values themselves are of no interest.
133 *
134 * The constraints and divs of the context are simply copied
135 * fron context_tab->bset.
136 * To extract the value of the output variables, it should be noted
137 * that we always use a big parameter M and so the variable stored
138 * in the tableau is not an output variable x itself, but
139 * x' = M + x (in case of minimization)
140 * or
141 * x' = M - x (in case of maximization)
142 * If x' appears in a column, then its optimal value is zero,
143 * which means that the optimal value of x is an unbounded number
144 * (-M for minimization and M for maximization).
145 * We currently assume that the output dimensions in the original map
146 * are bounded, so this cannot occur.
147 * Similarly, when x' appears in a row, then the coefficient of M in that
148 * row is necessarily 1.
149 * If the row represents
150 * d x' = c + d M + e(y)
151 * then, in case of minimization, an equality
152 * c + e(y) - d x' = 0
153 * is added, and in case of maximization,
154 * c + e(y) + d x' = 0
155 */
158 {
199 }
208 }
218 }
225 /* no unbounded */
230 else
235 /* no unbounded */
249 }
257 }
261 else
264 }
265 }
274 error:
278 }
282 {
284 }
289 {
300 error:
303 }
307 {
318 error:
321 }
324 /* Store the "parametric constant" of row "row" of tableau "tab" in "line",
325 * i.e., the constant term and the coefficients of all variables that
326 * appear in the context tableau.
327 * Note that the coefficient of the big parameter M is NOT copied.
328 * The context tableau may not have a big parameter and even when it
329 * does, it is a different big parameter.
330 */
332 {
343 }
344 }
352 }
353 }
354 }
356 /* Check if rows "row1" and "row2" have identical "parametric constants",
357 * as explained above.
358 * In this case, we also insist that the coefficients of the big parameter
359 * be the same as the values of the constants will only be the same
360 * if these coefficients are also the same.
361 */
363 {
385 }
387 }
389 /* Return an inequality that expresses that the "parametric constant"
390 * should be non-negative.
391 * This function is only called when the coefficient of the big parameter
392 * is equal to zero.
393 */
395 {
407 }
409 /* Return a integer division for use in a parametric cut based on the given row.
410 * In particular, let the parametric constant of the row be
411 *
412 * \sum_i a_i y_i
413 *
414 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
415 * The div returned is equal to
416 *
417 * floor(\sum_i {-a_i} y_i) = floor((\sum_i (-a_i mod d) y_i)/d)
418 */
420 {
434 }
436 /* Return a integer division for use in transferring an integrality constraint
437 * to the context.
438 * In particular, let the parametric constant of the row be
439 *
440 * \sum_i a_i y_i
441 *
442 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
443 * The the returned div is equal to
444 *
445 * floor(\sum_i {a_i} y_i) = floor((\sum_i (a_i mod d) y_i)/d)
446 */
448 {
461 }
463 /* Construct and return an inequality that expresses an upper bound
464 * on the given div.
465 * In particular, if the div is given by
466 *
467 * d = floor(e/m)
468 *
469 * then the inequality expresses
470 *
471 * m d <= e
472 */
474 {
489 }
491 /* Given a row in the tableau and a div that was created
492 * using get_row_split_div and that been constrained to equality, i.e.,
493 *
494 * d = floor(\sum_i {a_i} y_i) = \sum_i {a_i} y_i
495 *
496 * replace the expression "\sum_i {a_i} y_i" in the row by d,
497 * i.e., we subtract "\sum_i {a_i} y_i" and add 1 d.
498 * The coefficients of the non-parameters in the tableau have been
499 * verified to be integral. We can therefore simply replace coefficient b
500 * by floor(b). For the coefficients of the parameters we have
501 * floor(a_i) = a_i - {a_i}, while for the other coefficients, we have
502 * floor(b) = b.
503 */
505 {
521 error:
524 }
526 /* Check if the (parametric) constant of the given row is obviously
527 * negative, meaning that we don't need to consult the context tableau.
528 * If there is a big parameter and its coefficient is non-zero,
529 * then this coefficient determines the outcome.
530 * Otherwise, we check whether the constant is negative and
531 * all non-zero coefficients of parameters are negative and
532 * belong to non-negative parameters.
533 */
535 {
545 }
550 /* Eliminated parameter */
560 }
571 }
573 }
575 /* Check if the (parametric) constant of the given row is obviously
576 * non-negative, meaning that we don't need to consult the context tableau.
577 * If there is a big parameter and its coefficient is non-zero,
578 * then this coefficient determines the outcome.
579 * Otherwise, we check whether the constant is non-negative and
580 * all non-zero coefficients of parameters are positive and
581 * belong to non-negative parameters.
582 */
584 {
594 }
599 /* Eliminated parameter */
609 }
620 }
622 }
624 /* Given a row r and two columns, return the column that would
625 * lead to the lexicographically smallest increment in the sample
626 * solution when leaving the basis in favor of the row.
627 * Pivoting with column c will increment the sample value by a non-negative
628 * constant times a_{V,c}/a_{r,c}, with a_{V,c} the elements of column c
629 * corresponding to the non-parametric variables.
630 * If variable v appears in a column c_v, the a_{v,c} = 1 iff c = c_v,
631 * with all other entries in this virtual row equal to zero.
632 * If variable v appears in a row, then a_{v,c} is the element in column c
633 * of that row.
634 *
635 * Let v be the first variable with a_{v,c1}/a_{r,c1} != a_{v,c2}/a_{r,c2}.
636 * Then if a_{v,c1}/a_{r,c1} < a_{v,c2}/a_{r,c2}, i.e.,
637 * a_{v,c2} a_{r,c1} - a_{v,c1} a_{r,c2} > 0, c1 results in the minimal
638 * increment. Otherwise, it's c2.
639 */
642 {
658 }
679 }
681 }
683 /* Given a row in the tableau, find and return the column that would
684 * result in the lexicographically smallest, but positive, increment
685 * in the sample point.
686 * If there is no such column, then return tab->n_col.
687 * If anything goes wrong, return -1.
688 */
690 {
694 isl_int tmp;
711 else
714 }
718 error:
721 }
723 /* Return the first known violated constraint, i.e., a non-negative
724 * contraint that currently has an either obviously negative value
725 * or a previously determined to be negative value.
726 *
727 * If any constraint has a negative coefficient for the big parameter,
728 * if any, then we return one of these first.
729 */
731 {
740 }
753 }
755 }
757 /* Resolve all known or obviously violated constraints through pivoting.
758 * In particular, as long as we can find any violated constraint, we
759 * look for a pivoting column that would result in the lexicographicallly
760 * smallest increment in the sample point. If there is no such column
761 * then the tableau is infeasible.
762 */
764 {
778 }
780 error:
783 }
785 /* Given a row that represents an equality, look for an appropriate
786 * pivoting column.
787 * In particular, if there are any non-zero coefficients among
788 * the non-parameter variables, then we take the last of these
789 * variables. Eliminating this variable in terms of the other
790 * variables and/or parameters does not influence the property
791 * that all column in the initial tableau are lexicographically
792 * positive. The row corresponding to the eliminated variable
793 * will only have non-zero entries below the diagonal of the
794 * initial tableau. That is, we transform
795 *
796 * I I
797 * 1 into a
798 * I I
799 *
800 * If there is no such non-parameter variable, then we are dealing with
801 * pure parameter equality and we pick any parameter with coefficient 1 or -1
802 * for elimination. This will ensure that the eliminated parameter
803 * always has an integer value whenever all the other parameters are integral.
804 * If there is no such parameter then we return -1.
805 */
807 {
820 }
826 }
828 }
830 /* Add an equality that is known to be valid to the tableau.
831 * We first check if we can eliminate a variable or a parameter.
832 * If not, we add the equality as two inequalities.
833 * In this case, the equality was a pure parameter equality and there
834 * is no need to resolve any constraint violations.
835 */
837 {
864 }
867 error:
870 }
872 /* Check if the given row is a pure constant.
873 */
875 {
880 }
882 /* Add an equality that may or may not be valid to the tableau.
883 * If the resulting row is a pure constant, then it must be zero.
884 * Otherwise, the resulting tableau is empty.
885 *
886 * If the row is not a pure constant, then we add two inequalities,
887 * each time checking that they can be satisfied.
888 * In the end we try to use one of the two constraints to eliminate
889 * a column.
890 */
892 {
903 }
916 }
947 }
948 }
951 error:
954 }
956 /* Add an inequality to the tableau, resolving violations using
957 * restore_lexmin.
958 */
960 {
970 }
979 }
986 error:
989 }
991 /* Check if the coefficients of the parameters are all integral.
992 */
994 {
1000 /* Eliminated parameter */
1007 }
1015 }
1017 }
1019 /* Check if the coefficients of the non-parameter variables are all integral.
1020 */
1022 {
1034 }
1036 }
1038 /* Check if the constant term is integral.
1039 */
1041 {
1044 }
1046 #define I_CST 1 << 0
1047 #define I_PAR 1 << 1
1048 #define I_VAR 1 << 2
1050 /* Check for first (non-parameter) variable that is non-integer and
1051 * therefore requires a cut.
1052 * For parametric tableaus, there are three parts in a row,
1053 * the constant, the coefficients of the parameters and the rest.
1054 * For each part, we check whether the coefficients in that part
1055 * are all integral and if so, set the corresponding flag in *f.
1056 * If the constant and the parameter part are integral, then the
1057 * current sample value is integral and no cut is required
1058 * (irrespective of whether the variable part is integral).
1059 */
1061 {
1080 }
1082 }
1084 /* Add a (non-parametric) cut to cut away the non-integral sample
1085 * value of the given row.
1086 *
1087 * If the row is given by
1088 *
1089 * m r = f + \sum_i a_i y_i
1090 *
1091 * then the cut is
1092 *
1093 * c = - {-f/m} + \sum_i {a_i/m} y_i >= 0
1094 *
1095 * The big parameter, if any, is ignored, since it is assumed to be big
1096 * enough to be divisible by any integer.
1097 * If the tableau is actually a parametric tableau, then this function
1098 * is only called when all coefficients of the parameters are integral.
1099 * The cut therefore has zero coefficients for the parameters.
1100 *
1101 * The current value is known to be negative, so row_sign, if it
1102 * exists, is set accordingly.
1103 *
1104 * Return the row of the cut or -1.
1105 */
1107 {
1136 }
1138 /* Given a non-parametric tableau, add cuts until an integer
1139 * sample point is obtained or until the tableau is determined
1140 * to be integer infeasible.
1141 * As long as there is any non-integer value in the sample point,
1142 * we add an appropriate cut, if possible and resolve the violated
1143 * cut constraint using restore_lexmin.
1144 * If one of the corresponding rows is equal to an integral
1145 * combination of variables/constraints plus a non-integral constant,
1146 * then there is no way to obtain an integer point an we return
1147 * a tableau that is marked empty.
1148 */
1150 {
1168 }
1170 error:
1173 }
1176 {
1183 }
1185 /* Check whether all the currently active samples also satisfy the inequality
1186 * "ineq" (treated as an equality if eq is set).
1187 * Remove those samples that do not.
1188 */
1190 {
1192 isl_int v;
1212 }
1216 error:
1219 }
1221 /* Check whether the sample value of the tableau is finite,
1222 * i.e., either the tableau does not use a big parameter, or
1223 * all values of the variables are equal to the big parameter plus
1224 * some constant. This constant is the actual sample value.
1225 */
1227 {
1240 }
1242 }
1244 /* Check if the context tableau of sol has any integer points.
1245 * Returns -1 if an error occurred.
1246 * If an integer point can be found and if moreover it is finite,
1247 * then it is added to the list of sample values.
1248 *
1249 * This function is only called when none of the currently active sample
1250 * values satisfies the most recently added constraint.
1251 */
1253 {
1284 }
1291 error:
1295 }
1297 /* First check if any of the currently active sample values satisfies
1298 * the inequality "ineq" (an equality if eq is set).
1299 * If not, continue with check_integer_feasible.
1300 */
1303 {
1305 isl_int v;
1324 }
1331 }
1333 /* For a div d = floor(f/m), add the constraints
1334 *
1335 * f - m d >= 0
1336 * -(f-(m-1)) + m d >= 0
1337 *
1338 * Note that the second constraint is the negation of
1339 *
1340 * f - m d >= m
1341 */
1343 {
1369 error:
1372 }
1374 /* Add a div specified by "div" to both the main tableau and
1375 * the context tableau. In case of the main tableau, we only
1376 * need to add an extra div. In the context tableau, we also
1377 * need to express the meaning of the div.
1378 * Return the index of the div or -1 if anything went wrong.
1379 */
1382 {
1406 }
1430 error:
1434 }
1437 {
1447 }
1449 }
1451 /* Return the index of a div that corresponds to "div".
1452 * We first check if we already have such a div and if not, we create one.
1453 */
1456 {
1464 }
1466 /* Add a parametric cut to cut away the non-integral sample value
1467 * of the give row.
1468 * Let a_i be the coefficients of the constant term and the parameters
1469 * and let b_i be the coefficients of the variables or constraints
1470 * in basis of the tableau.
1471 * Let q be the div q = floor(\sum_i {-a_i} y_i).
1472 *
1473 * The cut is expressed as
1474 *
1475 * c = \sum_i -{-a_i} y_i + \sum_i {b_i} x_i + q >= 0
1476 *
1477 * If q did not already exist in the context tableau, then it is added first.
1478 * If q is in a column of the main tableau then the "+ q" can be accomplished
1479 * by setting the corresponding entry to the denominator of the constraint.
1480 * If q happens to be in a row of the main tableau, then the corresponding
1481 * row needs to be added instead (taking care of the denominators).
1482 * Note that this is very unlikely, but perhaps not entirely impossible.
1483 *
1484 * The current value of the cut is known to be negative (or at least
1485 * non-positive), so row_sign is set accordingly.
1486 *
1487 * Return the row of the cut or -1.
1488 */
1491 {
1535 }
1544 }
1552 }
1554 isl_int gcd;
1568 }
1578 error:
1582 }
1584 /* Construct a tableau for bmap that can be used for computing
1585 * the lexicographic minimum (or maximum) of bmap.
1586 * If not NULL, then dom is the domain where the minimum
1587 * should be computed. In this case, we set up a parametric
1588 * tableau with row signs (initialized to "unknown").
1589 * If M is set, then the tableau will use a big parameter.
1590 * If max is set, then a maximum should be computed instead of a minimum.
1591 * This means that for each variable x, the tableau will contain the variable
1592 * x' = M - x, rather than x' = M + x. This in turn means that the coefficient
1593 * of the variables in all constraints are negated prior to adding them
1594 * to the tableau.
1595 */
1598 {
1615 }
1622 }
1635 }
1648 }
1650 error:
1653 }
1656 {
1672 error:
1675 }
1677 /* Construct an isl_sol_map structure for accumulating the solution.
1678 * If track_empty is set, then we also keep track of the parts
1679 * of the context where there is no solution.
1680 * If max is set, then we are solving a maximization, rather than
1681 * a minimization problem, which means that the variables in the
1682 * tableau have value "M - x" rather than "M + x".
1683 */
1686 {
1699 ISL_MAP_DISJOINT);
1715 }
1719 error:
1723 }
1725 /* For each variable in the context tableau, check if the variable can
1726 * only attain non-negative values. If so, mark the parameter as non-negative
1727 * in the main tableau. This allows for a more direct identification of some
1728 * cases of violated constraints.
1729 */
1732 {
1766 n++;
1767 }
1771 }
1779 }
1783 error:
1787 }
1789 /* Check whether all coefficients of (non-parameter) variables
1790 * are non-positive, meaning that no pivots can be performed on the row.
1791 */
1793 {
1805 }
1808 }
1810 /* Check whether the inequality represented by vec is strict over the integers,
1811 * i.e., there are no integer values satisfying the constraint with
1812 * equality. This happens if the gcd of the coefficients is not a divisor
1813 * of the constant term. If so, scale the constraint down by the gcd
1814 * of the coefficients.
1815 */
1817 {
1818 isl_int gcd;
1827 }
1831 }
1833 /* Determine the sign of the given row of the main tableau.
1834 * The result is one of
1835 * isl_tab_row_pos: always non-negative; no pivot needed
1836 * isl_tab_row_neg: always non-positive; pivot
1837 * isl_tab_row_any: can be both positive and negative; split
1838 *
1839 * We first handle some simple cases
1840 * - the row sign may be known already
1841 * - the row may be obviously non-negative
1842 * - the parametric constant may be equal to that of another row
1843 * for which we know the sign. This sign will be either "pos" or
1844 * "any". If it had been "neg" then we would have pivoted before.
1845 *
1846 * If none of these cases hold, we check the value of the row for each
1847 * of the currently active samples. Based on the signs of these values
1848 * we make an initial determination of the sign of the row.
1849 *
1850 * all zero -> unk(nown)
1851 * all non-negative -> pos
1852 * all non-positive -> neg
1853 * both negative and positive -> all
1854 *
1855 * If we end up with "all", we are done.
1856 * Otherwise, we perform a check for positive and/or negative
1857 * values as follows.
1858 *
1859 * samples neg unk pos
1860 * <0 ? Y N Y N
1861 * pos any pos
1862 * >0 ? Y N Y N
1863 * any neg any neg
1864 *
1865 * There is no special sign for "zero", because we can usually treat zero
1866 * as either non-negative or non-positive, whatever works out best.
1867 * However, if the row is "critical", meaning that pivoting is impossible
1868 * then we don't want to limp zero with the non-positive case, because
1869 * then we we would lose the solution for those values of the parameters
1870 * where the value of the row is zero. Instead, we treat 0 as non-negative
1871 * ensuring a split if the row can attain both zero and negative values.
1872 * The same happens when the original constraint was one that could not
1873 * be satisfied with equality by any integer values of the parameters.
1874 * In this case, we normalize the constraint, but then a value of zero
1875 * for the normalized constraint is actually a positive value for the
1876 * original constraint, so again we need to treat zero as non-negative.
1877 * In both these cases, we have the following decision tree instead:
1878 *
1879 * all non-negative -> pos
1880 * all negative -> neg
1881 * both negative and non-negative -> all
1882 *
1883 * samples neg pos
1884 * <0 ? Y N
1885 * any pos
1886 * >=0 ? Y N
1887 * any neg
1888 */
1890 {
1899 isl_int tmp;
1911 }
1934 }
1940 }
1943 }
1952 }
1955 /* test for negative values */
1968 else
1977 }
1978 }
1981 /* test for positive values */
1996 }
2000 error:
2003 }
2007 /* Find solutions for values of the parameters that satisfy the given
2008 * inequality.
2009 *
2010 * We currently take a snapshot of the context tableau that is reset
2011 * when we return from this function, while we make a copy of the main
2012 * tableau, leaving the original main tableau untouched.
2013 * These are fairly arbitrary choices. Making a copy also of the context
2014 * tableau would obviate the need to undo any changes made to it later,
2015 * while taking a snapshot of the main tableau could reduce memory usage.
2016 * If we were to switch to taking a snapshot of the main tableau,
2017 * we would have to keep in mind that we need to save the row signs
2018 * and that we need to do this before saving the current basis
2019 * such that the basis has been restore before we restore the row signs.
2020 */
2023 {
2042 error:
2046 }
2048 /* Record the absence of solutions for those values of the parameters
2049 * that do not satisfy the given inequality with equality.
2050 */
2053 {
2079 error:
2082 }
2084 /* Given a main tableau where more than one row requires a split,
2085 * determine and return the "best" row to split on.
2086 *
2087 * Given two rows in the main tableau, if the inequality corresponding
2088 * to the first row is redundant with respect to that of the second row
2089 * in the current tableau, then it is better to split on the second row,
2090 * since in the positive part, both row will be positive.
2091 * (In the negative part a pivot will have to be performed and just about
2092 * anything can happen to the sign of the other row.)
2093 *
2094 * As a simple heuristic, we therefore select the row that makes the most
2095 * of the other rows redundant.
2096 *
2097 * Perhaps it would also be useful to look at the number of constraints
2098 * that conflict with any given constraint.
2099 */
2101 {
2151 r++;
2154 }
2158 }
2161 }
2167 }
2169 /* Compute the lexicographic minimum of the set represented by the main
2170 * tableau "tab" within the context "sol->context_tab".
2171 * On entry the sample value of the main tableau is lexicographically
2172 * less than or equal to this lexicographic minimum.
2173 * Pivots are performed until a feasible point is found, which is then
2174 * necessarily equal to the minimum, or until the tableau is found to
2175 * be infeasible. Some pivots may need to be performed for only some
2176 * feasible values of the context tableau. If so, the context tableau
2177 * is split into a part where the pivot is needed and a part where it is not.
2178 *
2179 * Whenever we enter the main loop, the main tableau is such that no
2180 * "obvious" pivots need to be performed on it, where "obvious" means
2181 * that the given row can be seen to be negative without looking at
2182 * the context tableau. In particular, for non-parametric problems,
2183 * no pivots need to be performed on the main tableau.
2184 * The caller of find_solutions is responsible for making this property
2185 * hold prior to the first iteration of the loop, while restore_lexmin
2186 * is called before every other iteration.
2187 *
2188 * Inside the main loop, we first examine the signs of the rows of
2189 * the main tableau within the context of the context tableau.
2190 * If we find a row that is always non-positive for all values of
2191 * the parameters satisfying the context tableau and negative for at
2192 * least one value of the parameters, we perform the appropriate pivot
2193 * and start over. An exception is the case where no pivot can be
2194 * performed on the row. In this case, we require that the sign of
2195 * the row is negative for all values of the parameters (rather than just
2196 * non-positive). This special case is handled inside row_sign, which
2197 * will say that the row can have any sign if it determines that it can
2198 * attain both negative and zero values.
2199 *
2200 * If we can't find a row that always requires a pivot, but we can find
2201 * one or more rows that require a pivot for some values of the parameters
2202 * (i.e., the row can attain both positive and negative signs), then we split
2203 * the context tableau into two parts, one where we force the sign to be
2204 * non-negative and one where we force is to be negative.
2205 * The non-negative part is handled by a recursive call (through find_in_pos).
2206 * Upon returning from this call, we continue with the negative part and
2207 * perform the required pivot.
2208 *
2209 * If no such rows can be found, all rows are non-negative and we have
2210 * found a (rational) feasible point. If we only wanted a rational point
2211 * then we are done.
2212 * Otherwise, we check if all values of the sample point of the tableau
2213 * are integral for the variables. If so, we have found the minimal
2214 * integral point and we are done.
2215 * If the sample point is not integral, then we need to make a distinction
2216 * based on whether the constant term is non-integral or the coefficients
2217 * of the parameters. Furthermore, in order to decide how to handle
2218 * the non-integrality, we also need to know whether the coefficients
2219 * of the other columns in the tableau are integral. This leads
2220 * to the following table. The first two rows do not correspond
2221 * to a non-integral sample point and are only mentioned for completeness.
2222 *
2223 * constant parameters other
2224 *
2225 * int int int |
2226 * int int rat | -> no problem
2227 *
2228 * rat int int -> fail
2229 *
2230 * rat int rat -> cut
2231 *
2232 * int rat rat |
2233 * rat rat rat | -> parametric cut
2234 *
2235 * int rat int |
2236 * rat rat int | -> split context
2237 *
2238 * If the parametric constant is completely integral, then there is nothing
2239 * to be done. If the constant term is non-integral, but all the other
2240 * coefficient are integral, then there is nothing that can be done
2241 * and the tableau has no integral solution.
2242 * If, on the other hand, one or more of the other columns have rational
2243 * coeffcients, but the parameter coefficients are all integral, then
2244 * we can perform a regular (non-parametric) cut.
2245 * Finally, if there is any parameter coefficient that is non-integral,
2246 * then we need to involve the context tableau. There are two cases here.
2247 * If at least one other column has a rational coefficient, then we
2248 * can perform a parametric cut in the main tableau by adding a new
2249 * integer division in the context tableau.
2250 * If all other columns have integral coefficients, then we need to
2251 * enforce that the rational combination of parameters (c + \sum a_i y_i)/m
2252 * is always integral. We do this by introducing an integer division
2253 * q = floor((c + \sum a_i y_i)/m) and stipulating that its argument should
2254 * always be integral in the context tableau, i.e., m q = c + \sum a_i y_i.
2255 * Since q is expressed in the tableau as
2256 * c + \sum a_i y_i - m q >= 0
2257 * -c - \sum a_i y_i + m q + m - 1 >= 0
2258 * it is sufficient to add the inequality
2259 * -c - \sum a_i y_i + m q >= 0
2260 * In the part of the context where this inequality does not hold, the
2261 * main tableau is marked as being empty.
2262 */
2264 {
2292 n_split++;
2297 }
2315 }
2328 }
2338 }
2366 }
2367 done:
2371 error:
2375 }
2377 /* Compute the lexicographic minimum of the set represented by the main
2378 * tableau "tab" within the context "sol->context_tab".
2379 *
2380 * As a preprocessing step, we first transfer all the purely parametric
2381 * equalities from the main tableau to the context tableau, i.e.,
2382 * parameters that have been pivoted to a row.
2383 * These equalities are ignored by the main algorithm, because the
2384 * corresponding rows may not be marked as being non-negative.
2385 * In parts of the context where the added equality does not hold,
2386 * the main tableau is marked as being empty.
2387 */
2390 {
2404 else
2436 }
2439 error:
2443 }
2447 {
2449 }
2451 /* Check if integer division "div" of "dom" also occurs in "bmap".
2452 * If so, return its position within the divs.
2453 * If not, return -1.
2454 */
2457 {
2475 }
2477 }
2479 /* The correspondence between the variables in the main tableau,
2480 * the context tableau, and the input map and domain is as follows.
2481 * The first n_param and the last n_div variables of the main tableau
2482 * form the variables of the context tableau.
2483 * In the basic map, these n_param variables correspond to the
2484 * parameters and the input dimensions. In the domain, they correspond
2485 * to the parameters and the set dimensions.
2486 * The n_div variables correspond to the integer divisions in the domain.
2487 * To ensure that everything lines up, we may need to copy some of the
2488 * integer divisions of the domain to the map. These have to be placed
2489 * in the same order as those in the context and they have to be placed
2490 * after any other integer divisions that the map may have.
2491 * This function performs the required reordering.
2492 */
2495 {
2502 common++;
2509 }
2517 }
2520 }
2522 error:
2525 }
2527 /* Compute the lexicographic minimum (or maximum if "max" is set)
2528 * of "bmap" over the domain "dom" and return the result as a map.
2529 * If "empty" is not NULL, then *empty is assigned a set that
2530 * contains those parts of the domain where there is no solution.
2531 * If "bmap" is marked as rational (ISL_BASIC_MAP_RATIONAL),
2532 * then we compute the rational optimum. Otherwise, we compute
2533 * the integral optimum.
2534 *
2535 * We perform some preprocessing. As the PILP solver does not
2536 * handle implicit equalities very well, we first make sure all
2537 * the equalities are explicitly available.
2538 * We also make sure the divs in the domain are properly order,
2539 * because they will be added one by one in the given order
2540 * during the construction of the solution map.
2541 */
2545 {
2563 }
2580 }
2588 error:
2592 }
2600 };
2603 {
2606 }
2609 {
2611 }
2613 /* Add the solution identified by the tableau and the context tableau.
2614 *
2615 * See documentation of sol_map_add for more details.
2616 *
2617 * Instead of constructing a basic map, this function calls a user
2618 * defined function with the current context as a basic set and
2619 * an affine matrix reprenting the relation between the input and output.
2620 * The number of rows in this matrix is equal to one plus the number
2621 * of output variables. The number of columns is equal to one plus
2622 * the total dimension of the context, i.e., the number of parameters,
2623 * input variables and divs. Since some of the columns in the matrix
2624 * may refer to the divs, the basic set is not simplified.
2625 * (Simplification may reorder or remove divs.)
2626 */
2629 {
2662 /* no unbounded */
2675 }
2683 }
2690 }
2700 error:
2704 }
2708 {
2710 }
2716 {
2745 error:
2749 }
2753 {
2755 }
2761 {
2782 }
2787 error:
2791 }
2797 {
2799 }
2805 {
2807 }