From: Simon Tatham Date: Tue, 30 Aug 2005 17:44:18 +0000 (+0000) Subject: Implemented a couple more reasoning modes for Extreme difficulty X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=commitdiff_plain;h=e7a02ae3330d9d7f6d81d35985efee47e26f57be;p=sgt-puzzles.git Implemented a couple more reasoning modes for Extreme difficulty level: positional set elimination (which is so obvious I really should have thought of it myself, though it's tricky to spot) and forcing chains (which are a type of one-level proof by contradiction, findable through a simple breadth-first search without requiring recursion, but so ludicrously powerful that they are able to solve _two thirds_ of grids that the pre-Extreme Solo generated and rated as Unreasonable). Of course this makes Unreasonable mode harder still... [originally from svn r6239] --- diff --git a/solo.c b/solo.c index 1d4425a..229a058 100644 --- a/solo.c +++ b/solo.c @@ -116,7 +116,7 @@ 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_NEIGHBOUR, +enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE }; enum { @@ -177,7 +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 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME } }, { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } }, #ifndef SLOW_SYSTEM { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } }, @@ -238,7 +238,7 @@ static void decode_params(game_params *ret, char const *string) else if (*string == 'a') /* advanced */ string++, ret->diff = DIFF_SET; else if (*string == 'e') /* extreme */ - string++, ret->diff = DIFF_NEIGHBOUR; + string++, ret->diff = DIFF_EXTREME; else if (*string == 'u') /* unreasonable */ string++, ret->diff = DIFF_RECURSIVE; } else @@ -267,7 +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_EXTREME: strcat(str, "de"); break; case DIFF_RECURSIVE: strcat(str, "du"); break; } } @@ -652,7 +652,10 @@ static int solver_intersect(struct solver_usage *usage, struct solver_scratch { unsigned char *grid, *rowidx, *colidx, *set; - int *mne; + int *neighbours, *bfsqueue; +#ifdef STANDALONE_SOLVER + int *bfsprev; +#endif }; static int solver_set(struct solver_usage *usage, @@ -867,8 +870,8 @@ static int solver_mne(struct solver_usage *usage, int nnb, count; int i, j, n, nbi; - nb[0] = scratch->mne; - nb[1] = scratch->mne + cr; + nb[0] = scratch->neighbours; + nb[1] = scratch->neighbours + cr; /* * First, work out the mutual neighbour squares of the two. We @@ -1003,6 +1006,201 @@ static int solver_mne(struct solver_usage *usage, return 0; /* nothing found */ } +/* + * Look for forcing chains. A forcing chain is a path of + * pairwise-exclusive squares (i.e. each pair of adjacent squares + * in the path are in the same row, column or block) with the + * following properties: + * + * (a) Each square on the path has precisely two possible numbers. + * + * (b) Each pair of squares which are adjacent on the path share + * at least one possible number in common. + * + * (c) Each square in the middle of the path shares _both_ of its + * numbers with at least one of its neighbours (not the same + * one with both neighbours). + * + * These together imply that at least one of the possible number + * choices at one end of the path forces _all_ the rest of the + * numbers along the path. In order to make real use of this, we + * need further properties: + * + * (c) Ruling out some number N from the square at one end + * of the path forces the square at the other end to + * take number N. + * + * (d) The two end squares are both in line with some third + * square. + * + * (e) That third square currently has N as a possibility. + * + * If we can find all of that lot, we can deduce that at least one + * of the two ends of the forcing chain has number N, and that + * therefore the mutually adjacent third square does not. + * + * To find forcing chains, we're going to start a bfs at each + * suitable square, once for each of its two possible numbers. + */ +static int solver_forcing(struct solver_usage *usage, + struct solver_scratch *scratch) +{ + int c = usage->c, r = usage->r, cr = c*r; + int *bfsqueue = scratch->bfsqueue; +#ifdef STANDALONE_SOLVER + int *bfsprev = scratch->bfsprev; +#endif + unsigned char *number = scratch->grid; + int *neighbours = scratch->neighbours; + int x, y; + + for (y = 0; y < cr; y++) + for (x = 0; x < cr; x++) { + int count, t, n; + + /* + * If this square doesn't have exactly two candidate + * numbers, don't try it. + * + * In this loop we also sum the candidate numbers, + * which is a nasty hack to allow us to quickly find + * `the other one' (since we will shortly know there + * are exactly two). + */ + for (count = t = 0, n = 1; n <= cr; n++) + if (cube(x, y, n)) + count++, t += n; + if (count != 2) + continue; + + /* + * Now attempt a bfs for each candidate. + */ + for (n = 1; n <= cr; n++) + if (cube(x, y, n)) { + int orign, currn, head, tail; + + /* + * Begin a bfs. + */ + orign = n; + + memset(number, cr+1, cr*cr); + head = tail = 0; + bfsqueue[tail++] = y*cr+x; +#ifdef STANDALONE_SOLVER + bfsprev[y*cr+x] = -1; +#endif + number[y*cr+x] = t - n; + + while (head < tail) { + int xx, yy, nneighbours, xt, yt, xblk, i; + + xx = bfsqueue[head++]; + yy = xx / cr; + xx %= cr; + + currn = number[yy*cr+xx]; + + /* + * Find neighbours of yy,xx. + */ + nneighbours = 0; + for (yt = 0; yt < cr; yt++) + neighbours[nneighbours++] = yt*cr+xx; + for (xt = 0; xt < cr; xt++) + neighbours[nneighbours++] = yy*cr+xt; + xblk = xx - (xx % r); + for (yt = yy % r; yt < cr; yt += r) + for (xt = xblk; xt < xblk+r; xt++) + neighbours[nneighbours++] = yt*cr+xt; + + /* + * Try visiting each of those neighbours. + */ + for (i = 0; i < nneighbours; i++) { + int cc, tt, nn; + + xt = neighbours[i] % cr; + yt = neighbours[i] / cr; + + /* + * We need this square to not be + * already visited, and to include + * currn as a possible number. + */ + if (number[yt*cr+xt] <= cr) + continue; + if (!cube(xt, yt, currn)) + continue; + + /* + * Don't visit _this_ square a second + * time! + */ + if (xt == xx && yt == yy) + continue; + + /* + * To continue with the bfs, we need + * this square to have exactly two + * possible numbers. + */ + for (cc = tt = 0, nn = 1; nn <= cr; nn++) + if (cube(xt, yt, nn)) + cc++, tt += nn; + if (cc == 2) { + bfsqueue[tail++] = yt*cr+xt; +#ifdef STANDALONE_SOLVER + bfsprev[yt*cr+xt] = yy*cr+xx; +#endif + number[yt*cr+xt] = tt - currn; + } + + /* + * One other possibility is that this + * might be the square in which we can + * make a real deduction: if it's + * adjacent to x,y, and currn is equal + * to the original number we ruled out. + */ + if (currn == orign && + (xt == x || yt == y || + (xt / r == x / r && yt % r == y % r))) { +#ifdef STANDALONE_SOLVER + if (solver_show_working) { + char *sep = ""; + int xl, yl; + printf("%*sforcing chain, %d at ends of ", + solver_recurse_depth*4, "", orign); + xl = xx; + yl = yy; + while (1) { + printf("%s(%d,%d)", sep, 1+xl, + 1+YUNTRANS(yl)); + xl = bfsprev[yl*cr+xl]; + if (xl < 0) + break; + yl = xl / cr; + xl %= cr; + sep = "-"; + } + printf("\n%*s ruling out %d at (%d,%d)\n", + solver_recurse_depth*4, "", + orign, 1+xt, 1+YUNTRANS(yt)); + } +#endif + cube(xt, yt, orign) = FALSE; + return 1; + } + } + } + } + } + + return 0; +} + static struct solver_scratch *solver_new_scratch(struct solver_usage *usage) { struct solver_scratch *scratch = snew(struct solver_scratch); @@ -1011,13 +1209,21 @@ 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); + scratch->neighbours = snewn(3*cr, int); + scratch->bfsqueue = snewn(cr*cr, int); +#ifdef STANDALONE_SOLVER + scratch->bfsprev = snewn(cr*cr, int); +#endif return scratch; } static void solver_free_scratch(struct solver_scratch *scratch) { - sfree(scratch->mne); +#ifdef STANDALONE_SOLVER + sfree(scratch->bfsprev); +#endif + sfree(scratch->bfsqueue); + sfree(scratch->neighbours); sfree(scratch->set); sfree(scratch->colidx); sfree(scratch->rowidx); @@ -1286,6 +1492,24 @@ static int solver(int c, int r, digit *grid, int maxdiff) } } + /* + * Row-vs-column set elimination on a single number. + */ + for (n = 1; n <= cr; n++) { + ret = solver_set(usage, scratch, cubepos(0,0,n), cr*cr, cr +#ifdef STANDALONE_SOLVER + , "positional set elimination, number %d", n +#endif + ); + if (ret < 0) { + diff = DIFF_IMPOSSIBLE; + goto got_result; + } else if (ret > 0) { + diff = max(diff, DIFF_EXTREME); + goto cont; + } + } + /* * Mutual neighbour elimination. */ @@ -1310,14 +1534,14 @@ static int solver(int c, int r, digit *grid, int maxdiff) !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); + diff = max(diff, DIFF_EXTREME); 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); + diff = max(diff, DIFF_EXTREME); goto cont; } } @@ -1325,6 +1549,14 @@ static int solver(int c, int r, digit *grid, int maxdiff) } } + /* + * Forcing chains. + */ + if (solver_forcing(usage, scratch)) { + diff = max(diff, DIFF_EXTREME); + goto cont; + } + /* * If we reach here, we have made no deductions in this * iteration, so the algorithm terminates. @@ -2950,7 +3182,7 @@ 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_EXTREME ? "Extreme (complex non-recursive techniques required)": ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)": ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)": ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":