| blob | 5886fd7ec2896b94bd08640782a13658a2d514e7 |
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 }
273 error:
277 }
281 {
283 }
286 /* Store the "parametric constant" of row "row" of tableau "tab" in "line",
287 * i.e., the constant term and the coefficients of all variables that
288 * appear in the context tableau.
289 * Note that the coefficient of the big parameter M is NOT copied.
290 * The context tableau may not have a big parameter and even when it
291 * does, it is a different big parameter.
292 */
294 {
305 }
306 }
314 }
315 }
316 }
318 /* Check if rows "row1" and "row2" have identical "parametric constants",
319 * as explained above.
320 * In this case, we also insist that the coefficients of the big parameter
321 * be the same as the values of the constants will only be the same
322 * if these coefficients are also the same.
323 */
325 {
347 }
349 }
351 /* Return an inequality that expresses that the "parametric constant"
352 * should be non-negative.
353 * This function is only called when the coefficient of the big parameter
354 * is equal to zero.
355 */
357 {
369 }
371 /* Return a integer division for use in a parametric cut based on the given row.
372 * In particular, let the parametric constant of the row be
373 *
374 * \sum_i a_i y_i
375 *
376 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
377 * The div returned is equal to
378 *
379 * floor(\sum_i {-a_i} y_i) = floor((\sum_i (-a_i mod d) y_i)/d)
380 */
382 {
396 }
398 /* Return a integer division for use in transferring an integrality constraint
399 * to the context.
400 * In particular, let the parametric constant of the row be
401 *
402 * \sum_i a_i y_i
403 *
404 * where y_0 = 1, but none of the y_i corresponds to the big parameter M.
405 * The the returned div is equal to
406 *
407 * floor(\sum_i {a_i} y_i) = floor((\sum_i (a_i mod d) y_i)/d)
408 */
410 {
423 }
425 /* Construct and return an inequality that expresses an upper bound
426 * on the given div.
427 * In particular, if the div is given by
428 *
429 * d = floor(e/m)
430 *
431 * then the inequality expresses
432 *
433 * m d <= e
434 */
436 {
451 }
453 /* Given a row in the tableau and a div that was created
454 * using get_row_split_div and that been constrained to equality, i.e.,
455 *
456 * d = floor(\sum_i {a_i} y_i) = \sum_i {a_i} y_i
457 *
458 * replace the expression "\sum_i {a_i} y_i" in the row by d,
459 * i.e., we subtract "\sum_i {a_i} y_i" and add 1 d.
460 * The coefficients of the non-parameters in the tableau have been
461 * verified to be integral. We can therefore simply replace coefficient b
462 * by floor(b). For the coefficients of the parameters we have
463 * floor(a_i) = a_i - {a_i}, while for the other coefficients, we have
464 * floor(b) = b.
465 */
467 {
483 error:
486 }
488 /* Check if the (parametric) constant of the given row is obviously
489 * negative, meaning that we don't need to consult the context tableau.
490 * If there is a big parameter and its coefficient is non-zero,
491 * then this coefficient determines the outcome.
492 * Otherwise, we check whether the constant is negative and
493 * all non-zero coefficients of parameters are negative and
494 * belong to non-negative parameters.
495 */
497 {
507 }
512 /* Eliminated parameter */
522 }
533 }
535 }
537 /* Check if the (parametric) constant of the given row is obviously
538 * non-negative, meaning that we don't need to consult the context tableau.
539 * If there is a big parameter and its coefficient is non-zero,
540 * then this coefficient determines the outcome.
541 * Otherwise, we check whether the constant is non-negative and
542 * all non-zero coefficients of parameters are positive and
543 * belong to non-negative parameters.
544 */
546 {
556 }
561 /* Eliminated parameter */
571 }
582 }
584 }
586 /* Given a row r and two columns, return the column that would
587 * lead to the lexicographically smallest increment in the sample
588 * solution when leaving the basis in favor of the row.
589 * Pivoting with column c will increment the sample value by a non-negative
590 * constant times a_{V,c}/a_{r,c}, with a_{V,c} the elements of column c
591 * corresponding to the non-parametric variables.
592 * If variable v appears in a column c_v, the a_{v,c} = 1 iff c = c_v,
593 * with all other entries in this virtual row equal to zero.
594 * If variable v appears in a row, then a_{v,c} is the element in column c
595 * of that row.
596 *
597 * Let v be the first variable with a_{v,c1}/a_{r,c1} != a_{v,c2}/a_{r,c2}.
598 * Then if a_{v,c1}/a_{r,c1} < a_{v,c2}/a_{r,c2}, i.e.,
599 * a_{v,c2} a_{r,c1} - a_{v,c1} a_{r,c2} > 0, c1 results in the minimal
600 * increment. Otherwise, it's c2.
601 */
604 {
620 }
641 }
643 }
645 /* Given a row in the tableau, find and return the column that would
646 * result in the lexicographically smallest, but positive, increment
647 * in the sample point.
648 * If there is no such column, then return tab->n_col.
649 * If anything goes wrong, return -1.
650 */
652 {
656 isl_int tmp;
673 else
676 }
680 error:
683 }
685 /* Return the first known violated constraint, i.e., a non-negative
686 * contraint that currently has an either obviously negative value
687 * or a previously determined to be negative value.
688 *
689 * If any constraint has a negative coefficient for the big parameter,
690 * if any, then we return one of these first.
691 */
693 {
702 }
715 }
717 }
719 /* Resolve all known or obviously violated constraints through pivoting.
720 * In particular, as long as we can find any violated constraint, we
721 * look for a pivoting column that would result in the lexicographicallly
722 * smallest increment in the sample point. If there is no such column
723 * then the tableau is infeasible.
724 */
726 {
740 }
742 error:
745 }
747 /* Given a row that represents an equality, look for an appropriate
748 * pivoting column.
749 * In particular, if there are any non-zero coefficients among
750 * the non-parameter variables, then we take the last of these
751 * variables. Eliminating this variable in terms of the other
752 * variables and/or parameters does not influence the property
753 * that all column in the initial tableau are lexicographically
754 * positive. The row corresponding to the eliminated variable
755 * will only have non-zero entries below the diagonal of the
756 * initial tableau. That is, we transform
757 *
758 * I I
759 * 1 into a
760 * I I
761 *
762 * If there is no such non-parameter variable, then we are dealing with
763 * pure parameter equality and we pick any parameter with coefficient 1 or -1
764 * for elimination. This will ensure that the eliminated parameter
765 * always has an integer value whenever all the other parameters are integral.
766 * If there is no such parameter then we return -1.
767 */
769 {
782 }
788 }
790 }
792 /* Add an equality that is known to be valid to the tableau.
793 * We first check if we can eliminate a variable or a parameter.
794 * If not, we add the equality as two inequalities.
795 * In this case, the equality was a pure parameter equality and there
796 * is no need to resolve any constraint violations.
797 */
799 {
826 }
829 error:
832 }
834 /* Check if the given row is a pure constant.
835 */
837 {
842 }
844 /* Add an equality that may or may not be valid to the tableau.
845 * If the resulting row is a pure constant, then it must be zero.
846 * Otherwise, the resulting tableau is empty.
847 *
848 * If the row is not a pure constant, then we add two inequalities,
849 * each time checking that they can be satisfied.
850 * In the end we try to use one of the two constraints to eliminate
851 * a column.
852 */
854 {
876 }
907 }
908 }
919 }
922 error:
925 }
927 /* Add an inequality to the tableau, resolving violations using
928 * restore_lexmin.
929 */
931 {
941 }
950 }
957 error:
960 }
962 /* Check if the coefficients of the parameters are all integral.
963 */
965 {
971 /* Eliminated parameter */
978 }
986 }
988 }
990 /* Check if the coefficients of the non-parameter variables are all integral.
991 */
993 {
1005 }
1007 }
1009 /* Check if the constant term is integral.
1010 */
1012 {
1015 }
1017 #define I_CST 1 << 0
1018 #define I_PAR 1 << 1
1019 #define I_VAR 1 << 2
1021 /* Check for first (non-parameter) variable that is non-integer and
1022 * therefore requires a cut.
1023 * For parametric tableaus, there are three parts in a row,
1024 * the constant, the coefficients of the parameters and the rest.
1025 * For each part, we check whether the coefficients in that part
1026 * are all integral and if so, set the corresponding flag in *f.
1027 * If the constant and the parameter part are integral, then the
1028 * current sample value is integral and no cut is required
1029 * (irrespective of whether the variable part is integral).
1030 */
1032 {
1051 }
1053 }
1055 /* Add a (non-parametric) cut to cut away the non-integral sample
1056 * value of the given row.
1057 *
1058 * If the row is given by
1059 *
1060 * m r = f + \sum_i a_i y_i
1061 *
1062 * then the cut is
1063 *
1064 * c = - {-f/m} + \sum_i {a_i/m} y_i >= 0
1065 *
1066 * The big parameter, if any, is ignored, since it is assumed to be big
1067 * enough to be divisible by any integer.
1068 * If the tableau is actually a parametric tableau, then this function
1069 * is only called when all coefficients of the parameters are integral.
1070 * The cut therefore has zero coefficients for the parameters.
1071 *
1072 * The current value is known to be negative, so row_sign, if it
1073 * exists, is set accordingly.
1074 *
1075 * Return the row of the cut or -1.
1076 */
1078 {
1107 }
1109 /* Given a non-parametric tableau, add cuts until an integer
1110 * sample point is obtained or until the tableau is determined
1111 * to be integer infeasible.
1112 * As long as there is any non-integer value in the sample point,
1113 * we add an appropriate cut, if possible and resolve the violated
1114 * cut constraint using restore_lexmin.
1115 * If one of the corresponding rows is equal to an integral
1116 * combination of variables/constraints plus a non-integral constant,
1117 * then there is no way to obtain an integer point an we return
1118 * a tableau that is marked empty.
1119 */
1121 {
1139 }
1141 error:
1144 }
1146 /* Check whether all the currently active samples also satisfy the inequality
1147 * "ineq" (treated as an equality if eq is set).
1148 * Remove those samples that do not.
1149 */
1151 {
1153 isl_int v;
1173 }
1177 error:
1180 }
1182 /* Check whether the sample value of the tableau is finite,
1183 * i.e., either the tableau does not use a big parameter, or
1184 * all values of the variables are equal to the big parameter plus
1185 * some constant. This constant is the actual sample value.
1186 */
1188 {
1201 }
1203 }
1205 /* Check if the context tableau of sol has any integer points.
1206 * Returns -1 if an error occurred.
1207 * If an integer point can be found and if moreover it is finite,
1208 * then it is added to the list of sample values.
1209 *
1210 * This function is only called when none of the currently active sample
1211 * values satisfies the most recently added constraint.
1212 */
1214 {
1236 }
1243 error:
1247 }
1249 /* First check if any of the currently active sample values satisfies
1250 * the inequality "ineq" (an equality if eq is set).
1251 * If not, continue with check_integer_feasible.
1252 */
1255 {
1257 isl_int v;
1276 }
1283 }
1285 /* For a div d = floor(f/m), add the constraints
1286 *
1287 * f - m d >= 0
1288 * -(f-(m-1)) + m d >= 0
1289 *
1290 * Note that the second constraint is the negation of
1291 *
1292 * f - m d >= m
1293 */
1295 {
1321 error:
1324 }
1326 /* Add a div specified by "div" to both the main tableau and
1327 * the context tableau. In case of the main tableau, we only
1328 * need to add an extra div. In the context tableau, we also
1329 * need to express the meaning of the div.
1330 * Return the index of the div or -1 if anything went wrong.
1331 */
1334 {
1358 }
1382 error:
1386 }
1389 {
1399 }
1401 }
1403 /* Return the index of a div that corresponds to "div".
1404 * We first check if we already have such a div and if not, we create one.
1405 */
1408 {
1416 }
1418 /* Add a parametric cut to cut away the non-integral sample value
1419 * of the give row.
1420 * Let a_i be the coefficients of the constant term and the parameters
1421 * and let b_i be the coefficients of the variables or constraints
1422 * in basis of the tableau.
1423 * Let q be the div q = floor(\sum_i {-a_i} y_i).
1424 *
1425 * The cut is expressed as
1426 *
1427 * c = \sum_i -{-a_i} y_i + \sum_i {b_i} x_i + q >= 0
1428 *
1429 * If q did not already exist in the context tableau, then it is added first.
1430 * If q is in a column of the main tableau then the "+ q" can be accomplished
1431 * by setting the corresponding entry to the denominator of the constraint.
1432 * If q happens to be in a row of the main tableau, then the corresponding
1433 * row needs to be added instead (taking care of the denominators).
1434 * Note that this is very unlikely, but perhaps not entirely impossible.
1435 *
1436 * The current value of the cut is known to be negative (or at least
1437 * non-positive), so row_sign is set accordingly.
1438 *
1439 * Return the row of the cut or -1.
1440 */
1443 {
1487 }
1496 }
1504 }
1506 isl_int gcd;
1520 }
1530 error:
1534 }
1536 /* Construct a tableau for bmap that can be used for computing
1537 * the lexicographic minimum (or maximum) of bmap.
1538 * If not NULL, then dom is the domain where the minimum
1539 * should be computed. In this case, we set up a parametric
1540 * tableau with row signs (initialized to "unknown").
1541 * If M is set, then the tableau will use a big parameter.
1542 * If max is set, then a maximum should be computed instead of a minimum.
1543 * This means that for each variable x, the tableau will contain the variable
1544 * x' = M - x, rather than x' = M + x. This in turn means that the coefficient
1545 * of the variables in all constraints are negated prior to adding them
1546 * to the tableau.
1547 */
1550 {
1567 }
1574 }
1587 }
1600 }
1602 error:
1605 }
1608 {
1620 error:
1623 }
1625 /* Construct an isl_sol_map structure for accumulating the solution.
1626 * If track_empty is set, then we also keep track of the parts
1627 * of the context where there is no solution.
1628 * If max is set, then we are solving a maximization, rather than
1629 * a minimization problem, which means that the variables in the
1630 * tableau have value "M - x" rather than "M + x".
1631 */
1634 {
1647 ISL_MAP_DISJOINT);
1663 }
1667 error:
1671 }
1673 /* For each variable in the context tableau, check if the variable can
1674 * only attain non-negative values. If so, mark the parameter as non-negative
1675 * in the main tableau. This allows for a more direct identification of some
1676 * cases of violated constraints.
1677 */
1680 {
1714 n++;
1715 }
1719 }
1727 }
1731 error:
1735 }
1737 /* Check whether all coefficients of (non-parameter) variables
1738 * are non-positive, meaning that no pivots can be performed on the row.
1739 */
1741 {
1753 }
1756 }
1758 /* Check whether the inequality represented by vec is strict over the integers,
1759 * i.e., there are no integer values satisfying the constraint with
1760 * equality. This happens if the gcd of the coefficients is not a divisor
1761 * of the constant term. If so, scale the constraint down by the gcd
1762 * of the coefficients.
1763 */
1765 {
1766 isl_int gcd;
1775 }
1779 }
1781 /* Determine the sign of the given row of the main tableau.
1782 * The result is one of
1783 * isl_tab_row_pos: always non-negative; no pivot needed
1784 * isl_tab_row_neg: always non-positive; pivot
1785 * isl_tab_row_any: can be both positive and negative; split
1786 *
1787 * We first handle some simple cases
1788 * - the row sign may be known already
1789 * - the row may be obviously non-negative
1790 * - the parametric constant may be equal to that of another row
1791 * for which we know the sign. This sign will be either "pos" or
1792 * "any". If it had been "neg" then we would have pivoted before.
1793 *
1794 * If none of these cases hold, we check the value of the row for each
1795 * of the currently active samples. Based on the signs of these values
1796 * we make an initial determination of the sign of the row.
1797 *
1798 * all zero -> unk(nown)
1799 * all non-negative -> pos
1800 * all non-positive -> neg
1801 * both negative and positive -> all
1802 *
1803 * If we end up with "all", we are done.
1804 * Otherwise, we perform a check for positive and/or negative
1805 * values as follows.
1806 *
1807 * samples neg unk pos
1808 * <0 ? Y N Y N
1809 * pos any pos
1810 * >0 ? Y N Y N
1811 * any neg any neg
1812 *
1813 * There is no special sign for "zero", because we can usually treat zero
1814 * as either non-negative or non-positive, whatever works out best.
1815 * However, if the row is "critical", meaning that pivoting is impossible
1816 * then we don't want to limp zero with the non-positive case, because
1817 * then we we would lose the solution for those values of the parameters
1818 * where the value of the row is zero. Instead, we treat 0 as non-negative
1819 * ensuring a split if the row can attain both zero and negative values.
1820 * The same happens when the original constraint was one that could not
1821 * be satisfied with equality by any integer values of the parameters.
1822 * In this case, we normalize the constraint, but then a value of zero
1823 * for the normalized constraint is actually a positive value for the
1824 * original constraint, so again we need to treat zero as non-negative.
1825 * In both these cases, we have the following decision tree instead:
1826 *
1827 * all non-negative -> pos
1828 * all negative -> neg
1829 * both negative and non-negative -> all
1830 *
1831 * samples neg pos
1832 * <0 ? Y N
1833 * any pos
1834 * >=0 ? Y N
1835 * any neg
1836 */
1838 {
1847 isl_int tmp;
1859 }
1882 }
1888 }
1891 }
1900 }
1903 /* test for negative values */
1916 else
1925 }
1926 }
1929 /* test for positive values */
1944 }
1948 error:
1951 }
1955 /* Find solutions for values of the parameters that satisfy the given
1956 * inequality.
1957 *
1958 * We currently take a snapshot of the context tableau that is reset
1959 * when we return from this function, while we make a copy of the main
1960 * tableau, leaving the original main tableau untouched.
1961 * These are fairly arbitrary choices. Making a copy also of the context
1962 * tableau would obviate the need to undo any changes made to it later,
1963 * while taking a snapshot of the main tableau could reduce memory usage.
1964 * If we were to switch to taking a snapshot of the main tableau,
1965 * we would have to keep in mind that we need to save the row signs
1966 * and that we need to do this before saving the current basis
1967 * such that the basis has been restore before we restore the row signs.
1968 */
1971 {
1991 error:
1995 }
1997 /* Record the absence of solutions for those values of the parameters
1998 * that do not satisfy the given inequality with equality.
1999 */
2002 {
2029 error:
2032 }
2034 /* Given a main tableau where more than one row requires a split,
2035 * determine and return the "best" row to split on.
2036 *
2037 * Given two rows in the main tableau, if the inequality corresponding
2038 * to the first row is redundant with respect to that of the second row
2039 * in the current tableau, then it is better to split on the second row,
2040 * since in the positive part, both row will be positive.
2041 * (In the negative part a pivot will have to be performed and just about
2042 * anything can happen to the sign of the other row.)
2043 *
2044 * As a simple heuristic, we therefore select the row that makes the most
2045 * of the other rows redundant.
2046 *
2047 * Perhaps it would also be useful to look at the number of constraints
2048 * that conflict with any given constraint.
2049 */
2051 {
2101 r++;
2104 }
2108 }
2111 }
2117 }
2119 /* Compute the lexicographic minimum of the set represented by the main
2120 * tableau "tab" within the context "sol->context_tab".
2121 * On entry the sample value of the main tableau is lexicographically
2122 * less than or equal to this lexicographic minimum.
2123 * Pivots are performed until a feasible point is found, which is then
2124 * necessarily equal to the minimum, or until the tableau is found to
2125 * be infeasible. Some pivots may need to be performed for only some
2126 * feasible values of the context tableau. If so, the context tableau
2127 * is split into a part where the pivot is needed and a part where it is not.
2128 *
2129 * Whenever we enter the main loop, the main tableau is such that no
2130 * "obvious" pivots need to be performed on it, where "obvious" means
2131 * that the given row can be seen to be negative without looking at
2132 * the context tableau. In particular, for non-parametric problems,
2133 * no pivots need to be performed on the main tableau.
2134 * The caller of find_solutions is responsible for making this property
2135 * hold prior to the first iteration of the loop, while restore_lexmin
2136 * is called before every other iteration.
2137 *
2138 * Inside the main loop, we first examine the signs of the rows of
2139 * the main tableau within the context of the context tableau.
2140 * If we find a row that is always non-positive for all values of
2141 * the parameters satisfying the context tableau and negative for at
2142 * least one value of the parameters, we perform the appropriate pivot
2143 * and start over. An exception is the case where no pivot can be
2144 * performed on the row. In this case, we require that the sign of
2145 * the row is negative for all values of the parameters (rather than just
2146 * non-positive). This special case is handled inside row_sign, which
2147 * will say that the row can have any sign if it determines that it can
2148 * attain both negative and zero values.
2149 *
2150 * If we can't find a row that always requires a pivot, but we can find
2151 * one or more rows that require a pivot for some values of the parameters
2152 * (i.e., the row can attain both positive and negative signs), then we split
2153 * the context tableau into two parts, one where we force the sign to be
2154 * non-negative and one where we force is to be negative.
2155 * The non-negative part is handled by a recursive call (through find_in_pos).
2156 * Upon returning from this call, we continue with the negative part and
2157 * perform the required pivot.
2158 *
2159 * If no such rows can be found, all rows are non-negative and we have
2160 * found a (rational) feasible point. If we only wanted a rational point
2161 * then we are done.
2162 * Otherwise, we check if all values of the sample point of the tableau
2163 * are integral for the variables. If so, we have found the minimal
2164 * integral point and we are done.
2165 * If the sample point is not integral, then we need to make a distinction
2166 * based on whether the constant term is non-integral or the coefficients
2167 * of the parameters. Furthermore, in order to decide how to handle
2168 * the non-integrality, we also need to know whether the coefficients
2169 * of the other columns in the tableau are integral. This leads
2170 * to the following table. The first two rows do not correspond
2171 * to a non-integral sample point and are only mentioned for completeness.
2172 *
2173 * constant parameters other
2174 *
2175 * int int int |
2176 * int int rat | -> no problem
2177 *
2178 * rat int int -> fail
2179 *
2180 * rat int rat -> cut
2181 *
2182 * int rat rat |
2183 * rat rat rat | -> parametric cut
2184 *
2185 * int rat int |
2186 * rat rat int | -> split context
2187 *
2188 * If the parametric constant is completely integral, then there is nothing
2189 * to be done. If the constant term is non-integral, but all the other
2190 * coefficient are integral, then there is nothing that can be done
2191 * and the tableau has no integral solution.
2192 * If, on the other hand, one or more of the other columns have rational
2193 * coeffcients, but the parameter coefficients are all integral, then
2194 * we can perform a regular (non-parametric) cut.
2195 * Finally, if there is any parameter coefficient that is non-integral,
2196 * then we need to involve the context tableau. There are two cases here.
2197 * If at least one other column has a rational coefficient, then we
2198 * can perform a parametric cut in the main tableau by adding a new
2199 * integer division in the context tableau.
2200 * If all other columns have integral coefficients, then we need to
2201 * enforce that the rational combination of parameters (c + \sum a_i y_i)/m
2202 * is always integral. We do this by introducing an integer division
2203 * q = floor((c + \sum a_i y_i)/m) and stipulating that its argument should
2204 * always be integral in the context tableau, i.e., m q = c + \sum a_i y_i.
2205 * Since q is expressed in the tableau as
2206 * c + \sum a_i y_i - m q >= 0
2207 * -c - \sum a_i y_i + m q + m - 1 >= 0
2208 * it is sufficient to add the inequality
2209 * -c - \sum a_i y_i + m q >= 0
2210 * In the part of the context where this inequality does not hold, the
2211 * main tableau is marked as being empty.
2212 */
2214 {
2242 n_split++;
2247 }
2265 }
2278 }
2288 }
2316 }
2317 done:
2321 error:
2325 }
2327 /* Compute the lexicographic minimum of the set represented by the main
2328 * tableau "tab" within the context "sol->context_tab".
2329 *
2330 * As a preprocessing step, we first transfer all the purely parametric
2331 * equalities from the main tableau to the context tableau, i.e.,
2332 * parameters that have been pivoted to a row.
2333 * These equalities are ignored by the main algorithm, because the
2334 * corresponding rows may not be marked as being non-negative.
2335 * In parts of the context where the added equality does not hold,
2336 * the main tableau is marked as being empty.
2337 */
2340 {
2354 else
2386 }
2389 error:
2393 }
2397 {
2399 }
2401 /* Check if integer division "div" of "dom" also occurs in "bmap".
2402 * If so, return its position within the divs.
2403 * If not, return -1.
2404 */
2407 {
2425 }
2427 }
2429 /* The correspondence between the variables in the main tableau,
2430 * the context tableau, and the input map and domain is as follows.
2431 * The first n_param and the last n_div variables of the main tableau
2432 * form the variables of the context tableau.
2433 * In the basic map, these n_param variables correspond to the
2434 * parameters and the input dimensions. In the domain, they correspond
2435 * to the parameters and the set dimensions.
2436 * The n_div variables correspond to the integer divisions in the domain.
2437 * To ensure that everything lines up, we may need to copy some of the
2438 * integer divisions of the domain to the map. These have to be placed
2439 * in the same order as those in the context and they have to be placed
2440 * after any other integer divisions that the map may have.
2441 * This function performs the required reordering.
2442 */
2445 {
2452 common++;
2459 }
2467 }
2470 }
2472 error:
2475 }
2477 /* Compute the lexicographic minimum (or maximum if "max" is set)
2478 * of "bmap" over the domain "dom" and return the result as a map.
2479 * If "empty" is not NULL, then *empty is assigned a set that
2480 * contains those parts of the domain where there is no solution.
2481 * If "bmap" is marked as rational (ISL_BASIC_MAP_RATIONAL),
2482 * then we compute the rational optimum. Otherwise, we compute
2483 * the integral optimum.
2484 *
2485 * We perform some preprocessing. As the PILP solver does not
2486 * handle implicit equalities very well, we first make sure all
2487 * the equalities are explicitly available.
2488 * We also make sure the divs in the domain are properly order,
2489 * because they will be added one by one in the given order
2490 * during the construction of the solution map.
2491 */
2495 {
2513 }
2530 }
2538 error:
2542 }
2550 };
2553 {
2556 }
2559 {
2561 }
2563 /* Add the solution identified by the tableau and the context tableau.
2564 *
2565 * See documentation of sol_map_add for more details.
2566 *
2567 * Instead of constructing a basic map, this function calls a user
2568 * defined function with the current context as a basic set and
2569 * an affine matrix reprenting the relation between the input and output.
2570 * The number of rows in this matrix is equal to one plus the number
2571 * of output variables. The number of columns is equal to one plus
2572 * the total dimension of the context, i.e., the number of parameters,
2573 * input variables and divs. Since some of the columns in the matrix
2574 * may refer to the divs, the basic set is not simplified.
2575 * (Simplification may reorder or remove divs.)
2576 */
2579 {
2612 /* no unbounded */
2625 }
2633 }
2640 }
2650 error:
2654 }
2658 {
2660 }
2666 {
2695 error:
2699 }
2703 {
2705 }
2711 {
2732 }
2737 error:
2741 }
2747 {
2749 }
2755 {
2757 }