| blob | 15bc52a570e66ed41fb1c75895e850de94ef5078 |
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 {
865 }
878 }
909 }
910 }
913 error:
916 }
918 /* Add an inequality to the tableau, resolving violations using
919 * restore_lexmin.
920 */
922 {
932 }
941 }
948 error:
951 }
953 /* Check if the coefficients of the parameters are all integral.
954 */
956 {
962 /* Eliminated parameter */
969 }
977 }
979 }
981 /* Check if the coefficients of the non-parameter variables are all integral.
982 */
984 {
996 }
998 }
1000 /* Check if the constant term is integral.
1001 */
1003 {
1006 }
1008 #define I_CST 1 << 0
1009 #define I_PAR 1 << 1
1010 #define I_VAR 1 << 2
1012 /* Check for first (non-parameter) variable that is non-integer and
1013 * therefore requires a cut.
1014 * For parametric tableaus, there are three parts in a row,
1015 * the constant, the coefficients of the parameters and the rest.
1016 * For each part, we check whether the coefficients in that part
1017 * are all integral and if so, set the corresponding flag in *f.
1018 * If the constant and the parameter part are integral, then the
1019 * current sample value is integral and no cut is required
1020 * (irrespective of whether the variable part is integral).
1021 */
1023 {
1042 }
1044 }
1046 /* Add a (non-parametric) cut to cut away the non-integral sample
1047 * value of the given row.
1048 *
1049 * If the row is given by
1050 *
1051 * m r = f + \sum_i a_i y_i
1052 *
1053 * then the cut is
1054 *
1055 * c = - {-f/m} + \sum_i {a_i/m} y_i >= 0
1056 *
1057 * The big parameter, if any, is ignored, since it is assumed to be big
1058 * enough to be divisible by any integer.
1059 * If the tableau is actually a parametric tableau, then this function
1060 * is only called when all coefficients of the parameters are integral.
1061 * The cut therefore has zero coefficients for the parameters.
1062 *
1063 * The current value is known to be negative, so row_sign, if it
1064 * exists, is set accordingly.
1065 *
1066 * Return the row of the cut or -1.
1067 */
1069 {
1098 }
1100 /* Given a non-parametric tableau, add cuts until an integer
1101 * sample point is obtained or until the tableau is determined
1102 * to be integer infeasible.
1103 * As long as there is any non-integer value in the sample point,
1104 * we add an appropriate cut, if possible and resolve the violated
1105 * cut constraint using restore_lexmin.
1106 * If one of the corresponding rows is equal to an integral
1107 * combination of variables/constraints plus a non-integral constant,
1108 * then there is no way to obtain an integer point an we return
1109 * a tableau that is marked empty.
1110 */
1112 {
1130 }
1132 error:
1135 }
1138 {
1145 }
1147 /* Check whether all the currently active samples also satisfy the inequality
1148 * "ineq" (treated as an equality if eq is set).
1149 * Remove those samples that do not.
1150 */
1152 {
1154 isl_int v;
1174 }
1178 error:
1181 }
1183 /* Check whether the sample value of the tableau is finite,
1184 * i.e., either the tableau does not use a big parameter, or
1185 * all values of the variables are equal to the big parameter plus
1186 * some constant. This constant is the actual sample value.
1187 */
1189 {
1202 }
1204 }
1206 /* Check if the context tableau of sol has any integer points.
1207 * Returns -1 if an error occurred.
1208 * If an integer point can be found and if moreover it is finite,
1209 * then it is added to the list of sample values.
1210 *
1211 * This function is only called when none of the currently active sample
1212 * values satisfies the most recently added constraint.
1213 */
1215 {
1246 }
1253 error:
1257 }
1259 /* First check if any of the currently active sample values satisfies
1260 * the inequality "ineq" (an equality if eq is set).
1261 * If not, continue with check_integer_feasible.
1262 */
1265 {
1267 isl_int v;
1286 }
1293 }
1295 /* For a div d = floor(f/m), add the constraints
1296 *
1297 * f - m d >= 0
1298 * -(f-(m-1)) + m d >= 0
1299 *
1300 * Note that the second constraint is the negation of
1301 *
1302 * f - m d >= m
1303 */
1305 {
1331 error:
1334 }
1336 /* Add a div specified by "div" to both the main tableau and
1337 * the context tableau. In case of the main tableau, we only
1338 * need to add an extra div. In the context tableau, we also
1339 * need to express the meaning of the div.
1340 * Return the index of the div or -1 if anything went wrong.
1341 */
1344 {
1368 }
1392 error:
1396 }
1399 {
1409 }
1411 }
1413 /* Return the index of a div that corresponds to "div".
1414 * We first check if we already have such a div and if not, we create one.
1415 */
1418 {
1426 }
1428 /* Add a parametric cut to cut away the non-integral sample value
1429 * of the give row.
1430 * Let a_i be the coefficients of the constant term and the parameters
1431 * and let b_i be the coefficients of the variables or constraints
1432 * in basis of the tableau.
1433 * Let q be the div q = floor(\sum_i {-a_i} y_i).
1434 *
1435 * The cut is expressed as
1436 *
1437 * c = \sum_i -{-a_i} y_i + \sum_i {b_i} x_i + q >= 0
1438 *
1439 * If q did not already exist in the context tableau, then it is added first.
1440 * If q is in a column of the main tableau then the "+ q" can be accomplished
1441 * by setting the corresponding entry to the denominator of the constraint.
1442 * If q happens to be in a row of the main tableau, then the corresponding
1443 * row needs to be added instead (taking care of the denominators).
1444 * Note that this is very unlikely, but perhaps not entirely impossible.
1445 *
1446 * The current value of the cut is known to be negative (or at least
1447 * non-positive), so row_sign is set accordingly.
1448 *
1449 * Return the row of the cut or -1.
1450 */
1453 {
1497 }
1506 }
1514 }
1516 isl_int gcd;
1530 }
1540 error:
1544 }
1546 /* Construct a tableau for bmap that can be used for computing
1547 * the lexicographic minimum (or maximum) of bmap.
1548 * If not NULL, then dom is the domain where the minimum
1549 * should be computed. In this case, we set up a parametric
1550 * tableau with row signs (initialized to "unknown").
1551 * If M is set, then the tableau will use a big parameter.
1552 * If max is set, then a maximum should be computed instead of a minimum.
1553 * This means that for each variable x, the tableau will contain the variable
1554 * x' = M - x, rather than x' = M + x. This in turn means that the coefficient
1555 * of the variables in all constraints are negated prior to adding them
1556 * to the tableau.
1557 */
1560 {
1577 }
1584 }
1597 }
1610 }
1612 error:
1615 }
1618 {
1634 error:
1637 }
1639 /* Construct an isl_sol_map structure for accumulating the solution.
1640 * If track_empty is set, then we also keep track of the parts
1641 * of the context where there is no solution.
1642 * If max is set, then we are solving a maximization, rather than
1643 * a minimization problem, which means that the variables in the
1644 * tableau have value "M - x" rather than "M + x".
1645 */
1648 {
1661 ISL_MAP_DISJOINT);
1677 }
1681 error:
1685 }
1687 /* For each variable in the context tableau, check if the variable can
1688 * only attain non-negative values. If so, mark the parameter as non-negative
1689 * in the main tableau. This allows for a more direct identification of some
1690 * cases of violated constraints.
1691 */
1694 {
1728 n++;
1729 }
1733 }
1741 }
1745 error:
1749 }
1751 /* Check whether all coefficients of (non-parameter) variables
1752 * are non-positive, meaning that no pivots can be performed on the row.
1753 */
1755 {
1767 }
1770 }
1772 /* Check whether the inequality represented by vec is strict over the integers,
1773 * i.e., there are no integer values satisfying the constraint with
1774 * equality. This happens if the gcd of the coefficients is not a divisor
1775 * of the constant term. If so, scale the constraint down by the gcd
1776 * of the coefficients.
1777 */
1779 {
1780 isl_int gcd;
1789 }
1793 }
1795 /* Determine the sign of the given row of the main tableau.
1796 * The result is one of
1797 * isl_tab_row_pos: always non-negative; no pivot needed
1798 * isl_tab_row_neg: always non-positive; pivot
1799 * isl_tab_row_any: can be both positive and negative; split
1800 *
1801 * We first handle some simple cases
1802 * - the row sign may be known already
1803 * - the row may be obviously non-negative
1804 * - the parametric constant may be equal to that of another row
1805 * for which we know the sign. This sign will be either "pos" or
1806 * "any". If it had been "neg" then we would have pivoted before.
1807 *
1808 * If none of these cases hold, we check the value of the row for each
1809 * of the currently active samples. Based on the signs of these values
1810 * we make an initial determination of the sign of the row.
1811 *
1812 * all zero -> unk(nown)
1813 * all non-negative -> pos
1814 * all non-positive -> neg
1815 * both negative and positive -> all
1816 *
1817 * If we end up with "all", we are done.
1818 * Otherwise, we perform a check for positive and/or negative
1819 * values as follows.
1820 *
1821 * samples neg unk pos
1822 * <0 ? Y N Y N
1823 * pos any pos
1824 * >0 ? Y N Y N
1825 * any neg any neg
1826 *
1827 * There is no special sign for "zero", because we can usually treat zero
1828 * as either non-negative or non-positive, whatever works out best.
1829 * However, if the row is "critical", meaning that pivoting is impossible
1830 * then we don't want to limp zero with the non-positive case, because
1831 * then we we would lose the solution for those values of the parameters
1832 * where the value of the row is zero. Instead, we treat 0 as non-negative
1833 * ensuring a split if the row can attain both zero and negative values.
1834 * The same happens when the original constraint was one that could not
1835 * be satisfied with equality by any integer values of the parameters.
1836 * In this case, we normalize the constraint, but then a value of zero
1837 * for the normalized constraint is actually a positive value for the
1838 * original constraint, so again we need to treat zero as non-negative.
1839 * In both these cases, we have the following decision tree instead:
1840 *
1841 * all non-negative -> pos
1842 * all negative -> neg
1843 * both negative and non-negative -> all
1844 *
1845 * samples neg pos
1846 * <0 ? Y N
1847 * any pos
1848 * >=0 ? Y N
1849 * any neg
1850 */
1852 {
1861 isl_int tmp;
1873 }
1896 }
1902 }
1905 }
1914 }
1917 /* test for negative values */
1930 else
1939 }
1940 }
1943 /* test for positive values */
1958 }
1962 error:
1965 }
1969 /* Find solutions for values of the parameters that satisfy the given
1970 * inequality.
1971 *
1972 * We currently take a snapshot of the context tableau that is reset
1973 * when we return from this function, while we make a copy of the main
1974 * tableau, leaving the original main tableau untouched.
1975 * These are fairly arbitrary choices. Making a copy also of the context
1976 * tableau would obviate the need to undo any changes made to it later,
1977 * while taking a snapshot of the main tableau could reduce memory usage.
1978 * If we were to switch to taking a snapshot of the main tableau,
1979 * we would have to keep in mind that we need to save the row signs
1980 * and that we need to do this before saving the current basis
1981 * such that the basis has been restore before we restore the row signs.
1982 */
1985 {
2004 error:
2008 }
2010 /* Record the absence of solutions for those values of the parameters
2011 * that do not satisfy the given inequality with equality.
2012 */
2015 {
2041 error:
2044 }
2046 /* Given a main tableau where more than one row requires a split,
2047 * determine and return the "best" row to split on.
2048 *
2049 * Given two rows in the main tableau, if the inequality corresponding
2050 * to the first row is redundant with respect to that of the second row
2051 * in the current tableau, then it is better to split on the second row,
2052 * since in the positive part, both row will be positive.
2053 * (In the negative part a pivot will have to be performed and just about
2054 * anything can happen to the sign of the other row.)
2055 *
2056 * As a simple heuristic, we therefore select the row that makes the most
2057 * of the other rows redundant.
2058 *
2059 * Perhaps it would also be useful to look at the number of constraints
2060 * that conflict with any given constraint.
2061 */
2063 {
2113 r++;
2116 }
2120 }
2123 }
2129 }
2131 /* Compute the lexicographic minimum of the set represented by the main
2132 * tableau "tab" within the context "sol->context_tab".
2133 * On entry the sample value of the main tableau is lexicographically
2134 * less than or equal to this lexicographic minimum.
2135 * Pivots are performed until a feasible point is found, which is then
2136 * necessarily equal to the minimum, or until the tableau is found to
2137 * be infeasible. Some pivots may need to be performed for only some
2138 * feasible values of the context tableau. If so, the context tableau
2139 * is split into a part where the pivot is needed and a part where it is not.
2140 *
2141 * Whenever we enter the main loop, the main tableau is such that no
2142 * "obvious" pivots need to be performed on it, where "obvious" means
2143 * that the given row can be seen to be negative without looking at
2144 * the context tableau. In particular, for non-parametric problems,
2145 * no pivots need to be performed on the main tableau.
2146 * The caller of find_solutions is responsible for making this property
2147 * hold prior to the first iteration of the loop, while restore_lexmin
2148 * is called before every other iteration.
2149 *
2150 * Inside the main loop, we first examine the signs of the rows of
2151 * the main tableau within the context of the context tableau.
2152 * If we find a row that is always non-positive for all values of
2153 * the parameters satisfying the context tableau and negative for at
2154 * least one value of the parameters, we perform the appropriate pivot
2155 * and start over. An exception is the case where no pivot can be
2156 * performed on the row. In this case, we require that the sign of
2157 * the row is negative for all values of the parameters (rather than just
2158 * non-positive). This special case is handled inside row_sign, which
2159 * will say that the row can have any sign if it determines that it can
2160 * attain both negative and zero values.
2161 *
2162 * If we can't find a row that always requires a pivot, but we can find
2163 * one or more rows that require a pivot for some values of the parameters
2164 * (i.e., the row can attain both positive and negative signs), then we split
2165 * the context tableau into two parts, one where we force the sign to be
2166 * non-negative and one where we force is to be negative.
2167 * The non-negative part is handled by a recursive call (through find_in_pos).
2168 * Upon returning from this call, we continue with the negative part and
2169 * perform the required pivot.
2170 *
2171 * If no such rows can be found, all rows are non-negative and we have
2172 * found a (rational) feasible point. If we only wanted a rational point
2173 * then we are done.
2174 * Otherwise, we check if all values of the sample point of the tableau
2175 * are integral for the variables. If so, we have found the minimal
2176 * integral point and we are done.
2177 * If the sample point is not integral, then we need to make a distinction
2178 * based on whether the constant term is non-integral or the coefficients
2179 * of the parameters. Furthermore, in order to decide how to handle
2180 * the non-integrality, we also need to know whether the coefficients
2181 * of the other columns in the tableau are integral. This leads
2182 * to the following table. The first two rows do not correspond
2183 * to a non-integral sample point and are only mentioned for completeness.
2184 *
2185 * constant parameters other
2186 *
2187 * int int int |
2188 * int int rat | -> no problem
2189 *
2190 * rat int int -> fail
2191 *
2192 * rat int rat -> cut
2193 *
2194 * int rat rat |
2195 * rat rat rat | -> parametric cut
2196 *
2197 * int rat int |
2198 * rat rat int | -> split context
2199 *
2200 * If the parametric constant is completely integral, then there is nothing
2201 * to be done. If the constant term is non-integral, but all the other
2202 * coefficient are integral, then there is nothing that can be done
2203 * and the tableau has no integral solution.
2204 * If, on the other hand, one or more of the other columns have rational
2205 * coeffcients, but the parameter coefficients are all integral, then
2206 * we can perform a regular (non-parametric) cut.
2207 * Finally, if there is any parameter coefficient that is non-integral,
2208 * then we need to involve the context tableau. There are two cases here.
2209 * If at least one other column has a rational coefficient, then we
2210 * can perform a parametric cut in the main tableau by adding a new
2211 * integer division in the context tableau.
2212 * If all other columns have integral coefficients, then we need to
2213 * enforce that the rational combination of parameters (c + \sum a_i y_i)/m
2214 * is always integral. We do this by introducing an integer division
2215 * q = floor((c + \sum a_i y_i)/m) and stipulating that its argument should
2216 * always be integral in the context tableau, i.e., m q = c + \sum a_i y_i.
2217 * Since q is expressed in the tableau as
2218 * c + \sum a_i y_i - m q >= 0
2219 * -c - \sum a_i y_i + m q + m - 1 >= 0
2220 * it is sufficient to add the inequality
2221 * -c - \sum a_i y_i + m q >= 0
2222 * In the part of the context where this inequality does not hold, the
2223 * main tableau is marked as being empty.
2224 */
2226 {
2254 n_split++;
2259 }
2277 }
2290 }
2300 }
2328 }
2329 done:
2333 error:
2337 }
2339 /* Compute the lexicographic minimum of the set represented by the main
2340 * tableau "tab" within the context "sol->context_tab".
2341 *
2342 * As a preprocessing step, we first transfer all the purely parametric
2343 * equalities from the main tableau to the context tableau, i.e.,
2344 * parameters that have been pivoted to a row.
2345 * These equalities are ignored by the main algorithm, because the
2346 * corresponding rows may not be marked as being non-negative.
2347 * In parts of the context where the added equality does not hold,
2348 * the main tableau is marked as being empty.
2349 */
2352 {
2366 else
2398 }
2401 error:
2405 }
2409 {
2411 }
2413 /* Check if integer division "div" of "dom" also occurs in "bmap".
2414 * If so, return its position within the divs.
2415 * If not, return -1.
2416 */
2419 {
2437 }
2439 }
2441 /* The correspondence between the variables in the main tableau,
2442 * the context tableau, and the input map and domain is as follows.
2443 * The first n_param and the last n_div variables of the main tableau
2444 * form the variables of the context tableau.
2445 * In the basic map, these n_param variables correspond to the
2446 * parameters and the input dimensions. In the domain, they correspond
2447 * to the parameters and the set dimensions.
2448 * The n_div variables correspond to the integer divisions in the domain.
2449 * To ensure that everything lines up, we may need to copy some of the
2450 * integer divisions of the domain to the map. These have to be placed
2451 * in the same order as those in the context and they have to be placed
2452 * after any other integer divisions that the map may have.
2453 * This function performs the required reordering.
2454 */
2457 {
2464 common++;
2471 }
2479 }
2482 }
2484 error:
2487 }
2489 /* Compute the lexicographic minimum (or maximum if "max" is set)
2490 * of "bmap" over the domain "dom" and return the result as a map.
2491 * If "empty" is not NULL, then *empty is assigned a set that
2492 * contains those parts of the domain where there is no solution.
2493 * If "bmap" is marked as rational (ISL_BASIC_MAP_RATIONAL),
2494 * then we compute the rational optimum. Otherwise, we compute
2495 * the integral optimum.
2496 *
2497 * We perform some preprocessing. As the PILP solver does not
2498 * handle implicit equalities very well, we first make sure all
2499 * the equalities are explicitly available.
2500 * We also make sure the divs in the domain are properly order,
2501 * because they will be added one by one in the given order
2502 * during the construction of the solution map.
2503 */
2507 {
2525 }
2542 }
2550 error:
2554 }
2562 };
2565 {
2568 }
2571 {
2573 }
2575 /* Add the solution identified by the tableau and the context tableau.
2576 *
2577 * See documentation of sol_map_add for more details.
2578 *
2579 * Instead of constructing a basic map, this function calls a user
2580 * defined function with the current context as a basic set and
2581 * an affine matrix reprenting the relation between the input and output.
2582 * The number of rows in this matrix is equal to one plus the number
2583 * of output variables. The number of columns is equal to one plus
2584 * the total dimension of the context, i.e., the number of parameters,
2585 * input variables and divs. Since some of the columns in the matrix
2586 * may refer to the divs, the basic set is not simplified.
2587 * (Simplification may reorder or remove divs.)
2588 */
2591 {
2624 /* no unbounded */
2637 }
2645 }
2652 }
2662 error:
2666 }
2670 {
2672 }
2678 {
2707 error:
2711 }
2715 {
2717 }
2723 {
2744 }
2749 error:
2753 }
2759 {
2761 }
2767 {
2769 }