From 13c4d60d66018a828361293c9f71e5cda1d16c11 Mon Sep 17 00:00:00 2001 From: simon Date: Wed, 24 Aug 2005 17:32:39 +0000 Subject: [PATCH] Brand new difficulty level in Solo. The other day Gareth and I independently discovered an advanced reasoning technique in Map, and then it occurred to me that since Solo can also be considered as a graph-colouring game the same technique ought to be applicable. And it is; so here's a new difficulty level, `Extreme', which sits just above Advanced. Grids graded `Extreme' by new-Solo will of course fall into old-Solo's `Unreasonable' category (since they're not soluble using the old set of non-recursive methods). A brief and unscientific experiment suggests that about one in six Unreasonable grids generated by old-Solo are classified Extreme by the new solver; so the remaining Unreasonable mode (now containing a subset of the grids it used to) hasn't actually become much harder. git-svn-id: svn://svn.tartarus.org/sgt/puzzles@6209 cda61777-01e9-0310-a592-d414129be87e --- solo.c | 234 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 227 insertions(+), 7 deletions(-) diff --git a/solo.c b/solo.c index c6d39eb..ac98afd 100644 --- a/solo.c +++ b/solo.c @@ -116,8 +116,8 @@ typedef unsigned char digit; enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4, SYMM_REF4D, SYMM_REF8 }; -enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, - DIFF_SET, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE }; +enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_NEIGHBOUR, + DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE }; enum { COL_BACKGROUND, @@ -177,6 +177,7 @@ static int game_fetch_preset(int i, char **name, game_params **params) { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } }, { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } }, { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } }, + { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_NEIGHBOUR } }, { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } }, #ifndef SLOW_SYSTEM { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } }, @@ -236,6 +237,8 @@ static void decode_params(game_params *ret, char const *string) string++, ret->diff = DIFF_INTERSECT; else if (*string == 'a') /* advanced */ string++, ret->diff = DIFF_SET; + else if (*string == 'e') /* extreme */ + string++, ret->diff = DIFF_NEIGHBOUR; else if (*string == 'u') /* unreasonable */ string++, ret->diff = DIFF_RECURSIVE; } else @@ -264,6 +267,7 @@ static char *encode_params(game_params *params, int full) case DIFF_SIMPLE: strcat(str, "db"); break; case DIFF_INTERSECT: strcat(str, "di"); break; case DIFF_SET: strcat(str, "da"); break; + case DIFF_NEIGHBOUR: strcat(str, "de"); break; case DIFF_RECURSIVE: strcat(str, "du"); break; } } @@ -298,7 +302,7 @@ static config_item *game_configure(game_params *params) ret[3].name = "Difficulty"; ret[3].type = C_CHOICES; - ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Unreasonable"; + ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable"; ret[3].ival = params->diff; ret[4].name = NULL; @@ -335,9 +339,7 @@ static char *validate_params(game_params *params, int full) /* ---------------------------------------------------------------------- * Solver. * - * This solver is used for several purposes: - * + to generate filled grids as the basis for new puzzles (by - * supplying no clue squares at all) + * This solver is used for two purposes: * + to check solubility of a grid as we gradually remove numbers * from it * + to solve an externally generated puzzle when the user selects @@ -389,6 +391,29 @@ static char *validate_params(game_params *params, int full) * the numbers' possible positions (or the spaces' possible * contents). * + * - Mutual neighbour elimination: find two squares A,B and a + * number N in the possible set of A, such that putting N in A + * would rule out enough possibilities from the mutual + * neighbours of A and B that there would be no possibilities + * left for B. Thereby rule out N in A. + * + The simplest case of this is if B has two possibilities + * (wlog {1,2}), and there are two mutual neighbours of A and + * B which have possibilities {1,3} and {2,3}. Thus, if A + * were to be 3, then those neighbours would contain 1 and 2, + * and hence there would be nothing left which could go in B. + * + There can be more complex cases of it too: if A and B are + * in the same column of large blocks, then they can have + * more than two mutual neighbours, some of which can also be + * neighbours of one another. Suppose, for example, that B + * has possibilities {1,2,3}; there's one square P in the + * same column as B and the same block as A, with + * possibilities {1,4}; and there are _two_ squares Q,R in + * the same column as A and the same block as B with + * possibilities {2,3,4}. Then if A contained 4, P would + * contain 1, and Q and R would have to contain 2 and 3 in + * _some_ order; therefore, once again, B would have no + * remaining possibilities. + * * - Recursion. If all else fails, we pick one of the currently * most constrained empty squares and take a random guess at its * contents, then continue solving on that basis and see if we @@ -627,6 +652,7 @@ static int solver_intersect(struct solver_usage *usage, struct solver_scratch { unsigned char *grid, *rowidx, *colidx, *set; + int *mne; }; static int solver_set(struct solver_usage *usage, @@ -825,6 +851,158 @@ static int solver_set(struct solver_usage *usage, return 0; } +/* + * Try to find a number in the possible set of (x1,y1) which can be + * ruled out because it would leave no possibilities for (x2,y2). + */ +static int solver_mne(struct solver_usage *usage, + struct solver_scratch *scratch, + int x1, int y1, int x2, int y2) +{ + int c = usage->c, r = usage->r, cr = c*r; + int *nb[2]; + unsigned char *set = scratch->set; + unsigned char *numbers = scratch->rowidx; + unsigned char *numbersleft = scratch->colidx; + int nnb, count; + int i, j, n, nbi; + + nb[0] = scratch->mne; + nb[1] = scratch->mne + cr; + + /* + * First, work out the mutual neighbour squares of the two. We + * can assert that they're not actually in the same block, + * which leaves two possibilities: they're in different block + * rows _and_ different block columns (thus their mutual + * neighbours are precisely the other two corners of the + * rectangle), or they're in the same row (WLOG) and different + * columns, in which case their mutual neighbours are the + * column of each block aligned with the other square. + * + * We divide the mutual neighbours into two separate subsets + * nb[0] and nb[1]; squares in the same subset are not only + * adjacent to both our key squares, but are also always + * adjacent to one another. + */ + if (x1 / r != x2 / r && y1 % r != y2 % r) { + /* Corners of the rectangle. */ + nnb = 1; + nb[0][0] = cubepos(x2, y1, 1); + nb[1][0] = cubepos(x1, y2, 1); + } else if (x1 / r != x2 / r) { + /* Same row of blocks; different blocks within that row. */ + int x1b = x1 - (x1 % r); + int x2b = x2 - (x2 % r); + + nnb = r; + for (i = 0; i < r; i++) { + nb[0][i] = cubepos(x2b+i, y1, 1); + nb[1][i] = cubepos(x1b+i, y2, 1); + } + } else { + /* Same column of blocks; different blocks within that column. */ + int y1b = y1 % r; + int y2b = y2 % r; + + assert(y1 % r != y2 % r); + + nnb = c; + for (i = 0; i < c; i++) { + nb[0][i] = cubepos(x2, y1b+i*r, 1); + nb[1][i] = cubepos(x1, y2b+i*r, 1); + } + } + + /* + * Right. Now loop over each possible number. + */ + for (n = 1; n <= cr; n++) { + if (!cube(x1, y1, n)) + continue; + for (j = 0; j < cr; j++) + numbersleft[j] = cube(x2, y2, j+1); + + /* + * Go over every possible subset of each neighbour list, + * and see if its union of possible numbers minus n has the + * same size as the subset. If so, add the numbers in that + * subset to the set of things which would be ruled out + * from (x2,y2) if n were placed at (x1,y1). + */ + memset(set, 0, nnb); + count = 0; + while (1) { + /* + * Binary increment: change the rightmost 0 to a 1, and + * change all 1s to the right of it to 0s. + */ + i = nnb; + while (i > 0 && set[i-1]) + set[--i] = 0, count--; + if (i > 0) + set[--i] = 1, count++; + else + break; /* done */ + + /* + * Examine this subset of each neighbour set. + */ + for (nbi = 0; nbi < 2; nbi++) { + int *nbs = nb[nbi]; + + memset(numbers, 0, cr); + + for (i = 0; i < nnb; i++) + if (set[i]) + for (j = 0; j < cr; j++) + if (j != n-1 && usage->cube[nbs[i] + j]) + numbers[j] = 1; + + for (i = j = 0; j < cr; j++) + i += numbers[j]; + + if (i == count) { + /* + * Got one. This subset of nbs, in the absence + * of n, would definitely contain all the + * numbers listed in `numbers'. Rule them out + * of `numbersleft'. + */ + for (j = 0; j < cr; j++) + if (numbers[j]) + numbersleft[j] = 0; + } + } + } + + /* + * If we've got nothing left in `numbersleft', we have a + * successful mutual neighbour elimination. + */ + for (j = 0; j < cr; j++) + if (numbersleft[j]) + break; + + if (j == cr) { +#ifdef STANDALONE_SOLVER + if (solver_show_working) { + printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n", + solver_recurse_depth*4, "", + 1+x1, 1+YUNTRANS(y1), 1+x2, 1+YUNTRANS(y2)); + printf("%*s ruling out %d at (%d,%d)\n", + solver_recurse_depth*4, "", + n, 1+x1, 1+YUNTRANS(y1)); + } +#endif + cube(x1, y1, n) = FALSE; + return +1; + } + } + + return 0; /* nothing found */ +} + static struct solver_scratch *solver_new_scratch(struct solver_usage *usage) { struct solver_scratch *scratch = snew(struct solver_scratch); @@ -833,11 +1011,13 @@ static struct solver_scratch *solver_new_scratch(struct solver_usage *usage) scratch->rowidx = snewn(cr, unsigned char); scratch->colidx = snewn(cr, unsigned char); scratch->set = snewn(cr, unsigned char); + scratch->mne = snewn(2*cr, int); return scratch; } static void solver_free_scratch(struct solver_scratch *scratch) { + sfree(scratch->mne); sfree(scratch->set); sfree(scratch->colidx); sfree(scratch->rowidx); @@ -850,7 +1030,7 @@ static int solver(int c, int r, digit *grid, int maxdiff) struct solver_usage *usage; struct solver_scratch *scratch; int cr = c*r; - int x, y, n, ret; + int x, y, x2, y2, n, ret; int diff = DIFF_BLOCK; /* @@ -1107,6 +1287,45 @@ static int solver(int c, int r, digit *grid, int maxdiff) } /* + * Mutual neighbour elimination. + */ + for (y = 0; y+1 < cr; y++) { + for (x = 0; x+1 < cr; x++) { + for (y2 = y+1; y2 < cr; y2++) { + for (x2 = x+1; x2 < cr; x2++) { + /* + * Can't do mutual neighbour elimination + * between elements of the same actual + * block. + */ + if (x/r == x2/r && y%r == y2%r) + continue; + + /* + * Otherwise, try (x,y) vs (x2,y2) in both + * directions, and likewise (x2,y) vs + * (x,y2). + */ + if (!usage->grid[YUNTRANS(y)*cr+x] && + !usage->grid[YUNTRANS(y2)*cr+x2] && + (solver_mne(usage, scratch, x, y, x2, y2) || + solver_mne(usage, scratch, x2, y2, x, y))) { + diff = max(diff, DIFF_NEIGHBOUR); + goto cont; + } + if (!usage->grid[YUNTRANS(y)*cr+x2] && + !usage->grid[YUNTRANS(y2)*cr+x] && + (solver_mne(usage, scratch, x2, y, x, y2) || + solver_mne(usage, scratch, x, y2, x2, y))) { + diff = max(diff, DIFF_NEIGHBOUR); + goto cont; + } + } + } + } + } + + /* * If we reach here, we have made no deductions in this * iteration, so the algorithm terminates. */ @@ -2731,6 +2950,7 @@ int main(int argc, char **argv) ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)": ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)": ret==DIFF_SET ? "Advanced (set elimination required)": + ret==DIFF_NEIGHBOUR ? "Extreme (mutual neighbour elimination required)": ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)": ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)": ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)": -- 2.11.0